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ajax.data
Add or modify data submitted to the server upon an Ajax request.
Description
When making an Ajax request to the server, DataTables will construct a data object internally, with the data it requires to be sent to the server for the request. What this data contains will depend upon the processing mode DataTables is operating in:
- For client-side processing no additional data is submitted to the server
- For server-side processing (
serverSide) the draw request parameters are submitted - see the server-side processing manual.
The
ajax.data option provides the ability to add additional data to the request, or to modify the data object being submitted if required.
In principle it operates in exactly the same way as jQuery's
$.ajax.data property, in that it can be given as an object with parameters and values to submit, but DataTables extends this by also providing it with the ability to be a function, to allow the data to be re-evaluated upon each Ajax request (see above).
Types
object
- Description:
As an object, the
ajax.dataoption is used to extend the data object that DataTables constructs internally to submit to the server. This provides an easy method of adding additional, static, parameters to the data to be sent to the server. For dynamically calculated values, use
ajax.dataas a function (see below).
function ajax.data( data, settings )
- Description:
As a function, the
ajax.dataoption can be used to modify the data DataTables submits to the server upon an Ajax request, by manipulating the original data object DataTables constructs internally, or by replacing it completely.
This provides the ability to submit additional information to the server upon an Ajax request, with the function being executed upon each Ajax request, allowing values to be dynamically calculated. For example, a value could be read from a text input field to act as an additional search option.
- Parameters:
- Returns:
If there is no return value from the function (i.e.
undefined) then the original data object passed into the function by DataTables will be used for the request (the function may have manipulated its values).
If an object is returned, then that object will be used as the data for the request. It will not be merged with the original data object constructed by DataTables before being sent.
If a string is returned, this string it will be used in the Ajax request body rather than individual HTTP parameters being sent. This is particularly useful for sending JSON encoded data in the request body so the server can decode it directly, rather than individual HTTP parameters being sent. See example below for how to use
JSON.stringify()to achieve this.
Examples
Add an extra parameter (
user_id in this case), of a static value to the data submitted:
$('#example').dataTable( { "ajax": { "url": "data.json", "data": { "user_id": 451 } } } );
Add data to the request by manipulating the data object (no return from the function):
$('#example').dataTable( { "ajax": { "url": "data.json", "data": function ( d ) { d.extra_search = $('#extra').val(); } } } );
Add data to the request (returning an object):
$('#example').dataTable( { "ajax": { "url": "data.json", "data": function ( d ) { return $.extend( {}, d, { "extra_search": $('#extra').val() } ); } } } );
Submit data as JSON in the request body:
$('#example').dataTable( { "ajax": { "url": "data.json", "contentType": "application/json", "type": "POST", "data": function ( d ) { return JSON.stringify( d ); } } } );
Related
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Sage 300cloud Key Features | 36,286 |
\begin{document}
\title{Numerical analysis of a finite element formulation of the P2D model for Lithium-ion cells}
\author{R. Bermejo \\
{\small
$$
Dpto. Matem\'{a}tica Aplicada a la Ingenier\'{i}a Industrial
ETSII Universidad Polit\'{e}cnica de Madrid. }\\
[0pt] {\small e-mail: [email protected] }}
\date{}
\maketitle
\begin{abstract}
The mathematical P2D model is a system of strongly coupled nonlinear
parabolic-elliptic equations that describes the electrodynamics of
lithium-ion batteries. In this paper, we present the numerical
analysis of a finite element-implicit Euler scheme for such a model.
We obtain error estimates for both the spatially semidiscrete and
the fully discrete systems of equations, and establish the existence
and uniqueness of the fully discrete solution.
\end{abstract}
\textit{Keywords:} P2D model, lithium-ion batteries, nonlinear,
parabolic, elliptic, finite elements, error estimates.
\textit{2010:MSC:} 65M60, 35M13, 35Q99.
\section{Introduction}
In this paper, we present the numerical analysis of a finite
element-implicit Euler method to calculate the numerical solution of
the so called pseudo-two-dimensional (P2D) model. proposed by J.
Newman and coworkers \cite{Doyle}. This is a mathematical model
based on the electrochemical kinetics and continuun mechanics laws,
which consists of a system of coupled nonlinear parabolic-elliptic
equations to model the physical-chemical phenomena governing the
behavior of lithium ion batteries. The P2D model is very much used
in engineering studies. A good presentation of it can be found in
\cite{NT} and \cite{Plet}. A lithium-ion battery system is composed
of a number of lithium-ion cells. A typical cell consists of three
regions, namely, a porous negative electrode (which plays the role
of anode of the cell in the discharge process) connected to the
negative terminal collector of the battery, a separator that is an
electron insulator allowing the flow of lithium ions between the
anode and the cathode, and a porous positive electrode (which plays
the role of cathode during the discharge process) connected to the
positive terminal, see Fig. 1. We must point out that in the charge
process the negative electrode plays the role of cathode and the
positive electrode is the anode. The electrodes are composite porous
structures of highly packed active lithium particles, typically
Li$_{\mathrm{x}}$C$_{\mathrm{6}}$ in the negative electrode and
metal oxide, such as
Li$_{\mathrm{1-x}}$Mn$_{\mathrm{2}}O_{\mathrm{4}}$, in the positive
electrode, plus a binder and a polymer that act as conductive
agents. Furthermore, the cell is filled with the electrolyte that
occupies the holes left free by the particles and the filler
material. The electrolyte is a lithium salt dissolved in an organic
solvent. In the description of the model it is customary to consider
two phases: the electrolyte phase and the solid phase, the latter is
composed of the solid particles of the electrodes.
\begin{figure}[th]
\begin{center}
\includegraphics[height=10cm]{fig_1.eps}
\end{center}
\caption{\textit{Upper panel}: a cross-section of a cell along the
x-direction. The lithium ions travel from the anode to the cathode
during the discharge process and in the opposite direction during
the charge process. \textit{Middel panel}: the cell model as a
non-denumerable collection of solid spheres plus the separator,
there is one sphere of radius $R_{s}(x)$
at each point ${x}$ of the electrodes. \textit{Lower panel}: the domain $
D_{3}$ }
\label{figure1}
\end{figure}
The P2D model of a lithium-ion cell considers that the dynamics is only
relevant along the $x$-axis, neglecting what happens along the $y$-axis and $
z$-axis, because the ratios $\frac{L_{x}}{L_{y}}$ and $\frac{L_{x}}{L_{z}}
=O(10^{-3})$, $L_{x}$,\ $L_{y}$ and $L_{z}$ being the characteristic
length scales along the corresponding axes. The main modeling
assumptions are the following : (1) The active particles of the
electrodes are assumed to be spheres of radius $R_{s}$ which may be
different in each electrode. (2)\ Side reactions are neglected and
no gas phase is present. (3) The transport of lithium ions is due to
diffusion and migration in the electrolyte solution, and in the
solid particles the atoms of lithium move between vacancies in the
crystalline structure of the particles due to local diffusion in
concentration. By longitudinal and latitudinal symmetry
considerations, the diffusion in the active particles is only in the
radial direction. (4) The electrochemical reaction of lithium
insertion and extraction processes follows the Buttler-Volmer law.
(5)\ The effective transport coefficients are calculated by the
Bruggeman relation, i.e., $\mu ^{eff}=\mu \varepsilon ^{p}$ (p=1.5),
where $\mu $ is a generic transport coefficient and $\varepsilon $
is the component volume fraction of the material in the composite
electrodes and separator.To formulate the
equations of the model we distinguish the following domains.
\begin{equation*}
\left\{
\begin{array}{l}
D_{\mathrm{n}}=(0,L_{n}),\ D_{\mathrm{s}}=(L_{n},L_{n}+\delta
)\text{,\ }D_{\mathrm{p}}=(L_{n}+\delta ,L),\ L_{p}:=L-(L_{n}+\delta) , \\
\\
D_{1}=(0,L),\ D_{2}=D_{\mathrm{n}}\cup D_{\mathrm{p}}\text{ \
\textrm{and\
\ }}D_{3}=\cup _{x\in D_{2}}\left\{ x\right\} \times (0,R_{\mathrm{s}}(x)),
\end{array}
\right.
\end{equation*}
where $D_{\mathrm{n}}$, $D_{\mathrm{s}}$ and $D_{\mathrm{p}}$ denote
the domains of the negative electrode, the separator and the
positive electrode respectively. Notice that $D_{1}$ represents the
cell domain, $D_{2}$ is a domain that is the union of two disjoint
domains corresponding to the electrodes, and $D_{3}$ is in a certain
sense a modeling space accounting for the spherical balls of radius
$R_{\mathrm{s}}(x)$ that represent at each
$x\in D_{2}$ the solid active particles, such that when $x\in D_{\mathrm{n}}$
, $R_{\mathrm{s}}(x)=R_{\mathrm{s}}^{-}$, and when $x\in $
$D_{\mathrm{p}}$, $R_{\mathrm{s}}(x)=R_{\mathrm{s}}^{+}$. The
variables of the model are the following:\ for the electrolyte
phase, the molar concentration of lithium ions $u(x,t)$, and the
electric potential $\phi _{1}(x,t)$, $x\in D_{1}$; for the solid
phase, the molar concentration of lithium $v(x;r,t)$, $x\in D_{2}$
and $r\in \left( 0;R_{\mathrm{s}}(x)\right) $, and the electric
potential $\phi _{2}(x,t)$, $x\in D_{2}$. Another important variable
is the
so called molar flux of lithium ions exiting the solid particles, $
J(x,u,v,\phi _{1},\phi _{2},U)/F$, $F$ being the Faraday constant.
The mathematical expression of $J$ is given by the Buttler-Volmer
law, see (\ref{Ja}).
Many numerical models to integrate the P2D model have been proposed.
The first one is the Dualfoil model developed by J. Newman and his
collaborators \cite{dual}, this is a model that uses second order
finite differences for space discretization of the differential
operators combined with the first order backward Euler time stepping
scheme; the Dualfoil model is distributed as free software, which is
being updated through time. Later on, authors such as \cite{ML} and
\cite{SW}, just to cite a few, have developed their own codes by
using second order finite volume for space discretizations combined
with the first order in time implicit Euler scheme for time
discretization. Other authors make the numerical simulations with
COMSOL multi-physics package that uses finite elements for space
discretizations of the equations, the resulting system of nonlinear
differential equations is integrated by different time stepping
schemes, in particular, conventional DAE solvers, such as DASK
\cite{NOR}. New numerical models have recently been proposed to
improve the computational efficiency, to this respect, we mention
the operator splitting technique of \cite{Farkas}, the orthogonal
collocation method for space discretization combined with the first
order implicit Euler scheme for time discretization of \cite{KZ},
and the implicit-explicit Runge-Kutta-Chebyshev finite element
method of \cite{BG}. Despite the activity in the development of
numerical methods no rigorous numerical analysis of such methods has
been published so far; so, to the best of our knowledge, this is the
first paper presenting the analysis of a numerical method developed
to integrate the P2D model.
The layout of the paper is the following. In Section 2 we introduce
the governing equations of the P2D model together with the
functional framework needed for the numerical analysis. Section 3 is
devoted to the semidiscrete space discretization of the model in a
finite element framework. The error analysis of the semi-discrete
solution is performed in Section 4. Since this analysis is long,
then we have split the section into three subsections in order to
make more palatable its presentation. Subsection 4.1 is a collection
of auxiliary results; subsections 4.2 and 4.3 deal with the error
estimates for the potentials and the concentrations, respectively.
The fully discrete model and its error analysis is presented in
Section 5, which is also split into subsections. Since the fully
discrete model is a nonlinear system of elliptic and fully discrete
parabolic equations at each time instant $t_{n}$, then we have also
studied the existence and uniqueness of the solution by applying
Minty-Browder theorem \cite{Zeid} for the elliptic equations, and
Brower\'{}s fixed point theorem for the parabolic equations.
\section{The governing equations of the isothermal P2D model}
We consider the governing equations of the isothermal P2D model for
the variables $u(x,t)$, $v(x;r,t)$, $\phi_{1}(x,t)$ and
$\phi_{2}(x,t)$ presented in Chapters 3 and 4 of \cite{Plet}.
However, to facilitate both the formulation of the numerical method
to integrate these equations and its numerical analysis, it is
convenient to make the changes of variable introduced in \cite{Kro}
and \cite {WXZ}. Thus, in order to make homogeneous the Neumann type
boundary conditions for the potential $\phi_{2}$ one considers the
function $H(x,t)$ given by the expression
\begin{equation*}
H(x,t)=\left\{
\begin{array}{l}
-\displaystyle\frac{(x-L_{n})^{2}I(t)}{2\sigma L_{n}A},\ \ x\in D_{\mathrm{n}
}, \\
\\
\displaystyle\frac{(x-(L_{n}+\delta ))^{2}I(t)}{2\sigma L_{p}A},\ \ x\in D_{
\mathrm{p}},
\end{array}
\right.
\end{equation*}
where $I(t)$ denotes the applied current, $A$ is the area of the
plate and $\sigma $ is a positive coefficient defined below, and
replace $\phi _{2}(x,t)$ by $\phi _{2}(x,t)+H(x,t)$; likewise, we
replace the potential $\phi _{1}(x,t)$ by $\phi _{1}(x,t)+\alpha \ln
u(x,t)$, with $\alpha =\alpha(u) =\displaystyle\frac{2RT\kappa(u)
}{F}(t_{+}^{0}-1)$, where $\kappa(u)>0$ denotes the effective
electrolyte phase ionic conductivity; $t_{+}^{0}>0$ is the so called
transfer number, which is assumed to be constant; $R$ is the
universal gas constant and $T$ denotes the absolute temperature
inside the cell, which is assumed to be constant in the isothermal
model; this latter change of variable for $\phi_{1}(x,t)$ simplifies
the expression of the equation for the potential of the electrolyte
phase written in Chapter 4 of \cite{Plet}, making it more manageable
from a computational viewpoint. Another important variable, as we
mentioned above, is the reaction current density $J$. The reaction
rate is coupled to phase
potentials by the Buttler-Volmer kinetic expression.
\begin{equation} \label{Ja}
J=J(x,u,v_{s},T,\eta )=\left\{
\begin{array}{l}
a_{s}i_{0}\displaystyle\left( \exp \frac{\alpha _{a}F}{RT}\left( \eta -\frac{
R_{SEI}}{a_{s}}J\right) \right. \\
\\
-\displaystyle\left. \exp \frac{-\alpha _{c}F}{RT}\left( \eta -\frac{R_{SEI}
}{a_{s}}J\right) \right) \ \ \text{if \ }x\in D_{n}
\cup D_{p} , \\
\\
0\ \ \text{if \ }x\in D_{s}.
\end{array}
\right.
\end{equation}
In this expression, $v_{s}=v(x;R_{s}(x),t)$ denotes the lithium
concentration on the surface of the active particles; $a_{s}=a_{s}(x)=\displaystyle
\frac{3\varepsilon _{s}(x)}{R_{s}(x)}$ is the active area per
electrode unit volume; $\varepsilon _{s}(x)$ denotes the volume
fraction of the active material, $\varepsilon _{s}(x)=\varepsilon
_{s}^{-}>0$ for $x \in D_{n}$ and $\varepsilon _{s}(x)=\varepsilon
_{s}^{+}>0$ for $x \in D_{p}$; $\alpha _{a}\in (0,1)$ and $\alpha
_{c}\in (0,1)$ are anodic and cathodic transfer coefficients for an
electron reaction; $R_{SEI}$ represents the solid interface
resistance, usually, $R_{SEI}=0$ in the engineering literature
unless the model also considers aging phenomena of the battery, so
in this paper we take $R_{SEI}=0$.
\begin{equation*}
\eta =\eta (x,\phi _{1},\phi _{2},U)=\left\{
\begin{array}{l}
\phi _{2}(x,t)-\phi _{1}(x,t)-U(x,v_{s})\text{ if\ }x\in
D_{n} \cup D_{p}, \\
\\
0\text{ \textrm{if} }x\in D_{s},
\end{array}
\right.
\end{equation*}
where $U$ stands for the equilibrium potential at the solid
electrolyte interface, which is assumed to be known. $i_{0}$ is the
exchange current density, i.e.,
\begin{equation} \label{i0}
i_{0}=i_{0}(u,v_{s})=ku^{\alpha _{a}}(v_{\max }-v_{s})^{\alpha
_{a}}v_{s}^{\alpha _{c}}\text{ if }x\in D_{n} \cup L_{n},
\end{equation}
here, $v_{\max }$ is the maximum concentration of lithium in the
solid phase, which may have different values in the positive and
negative electrodes, so
\begin{equation*}
v_{\max }=v_{\mathrm{\max }}(x)=\left\{
\begin{array}{l}
v_{\mathrm{\max }}^{-}\text{ \textrm{if} }x\in D_{n}, \\
v_{\mathrm{\max }}^{+}\text{ \textrm{if} }x\in D_{p},
\end{array}
\right.
\end{equation*}
the coefficient $k$ represents the kinetic rate constant,
\begin{equation*}
k=k(x)=\left\{
\begin{array}{l}
k^{-}\text{ \textrm{if} }x\in D_{n}, \\
k^{+}\text{ \textrm{if} }x\in D_{p}.
\end{array}
\right.
\end{equation*}
Considering the above mentioned changes of variable and taking the
transfer coefficients $\alpha _{a}$ and $\alpha _{c}$ equal to
$0.5$, as many engineering papers do, the expression for the
reaction current that we use in the paper is
\begin{equation}
J=J(x,u,v_{s},\phi _{1},\phi _{2},\overline{U})=\left\{
\begin{array}{l}
a_{2}(x)i_{0}\sinh \left( \beta \eta \right) ,\ \forall x\in D_{2},
\\
\\
0\ \mathrm{for\ }x\notin D_{2},
\end{array}
\right. \label{F}
\end{equation}
where $\beta = \displaystyle\frac{F}{2RT}$; $a_{2}(x)=3\varepsilon
_{s}(x)/R_{s}(x)$; and
\begin{equation}
\eta =\phi _{2}-\phi _{1}-\alpha \ln u-\overline{U}, \label{F1}
\end{equation}
$\overline{U}=\overline{U}(x,t,v_{s})=U(v_{s})-H(x,t)$. Noting that
the boundaries $\partial D_{1}$ and $\partial D_{2}$ of the
domains $D_{1}$ and $D_{2}$ are $\partial D_{1}:=\left\{ 0,L\right\} $ and $
\partial D_{2}:=\left\{ 0,L_{n},L_{n}+\delta ,L\right\} $, we formulate the
equations of the model as follows.
\bigskip
\textit{Concentration }$u(x,t)$\textit{\ in the electrolyte phase}.
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\frac{\partial u}{\partial
t}-\displaystyle\frac{\partial}{\partial x}(k_{1}\frac{
\partial u}{\partial x})=a_{1}(x)J\text{ \ \textrm{in} \ }D_{1}\times
(0,T_{\text{\textrm{end}}}), \\
\\
\displaystyle\frac{\partial u}{\partial x}\mid _{\partial D_{1}\times (0,T_{
\text{\textrm{end}}})}=0,\ u(x,0)=u^{0}(x)\ \text{\textrm{in} }D_{1}.
\end{array}
\right. \label{rf1}
\end{equation}
\textit{Concentration }$v(x;\,r,t)$\textit{\ in the solid phase}. For almost
every $x\in D_{2}$,
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\frac{\partial v}{\partial t}-\frac{k_{2}}{r^{2}}\frac{
\partial }{\partial r}\left( r^{2}\frac{\partial v}{\partial r}\right) v=0
\text{\ \ \textrm{in}\ \ }D_{3}\times (0,T_{\text{\textrm{end}}}), \\
\\
\displaystyle\frac{\partial v}{\partial r}\mid _{r=0}=0,\ k_{\mathrm{2}}
\displaystyle\frac{\partial v}{\partial r}\mid _{r=R_{\mathrm{s}}(x)}=
\displaystyle\frac{-J}{a_{2}(x)F},\ v(x;r,0)=v^{0}(x;r)\ \
\text{\textrm{in}
\ }D_{3}.
\end{array}
\right. \label{rf2}
\end{equation}
\textit{Electrolyte potential} $\phi _{\mathrm{1}}(x,t)$.
\begin{equation}
\left\{
\begin{array}{l}
-\displaystyle\frac{\partial }{\partial x}(\kappa (u)\frac{\partial \phi _{
\mathrm{1}}}{\partial x})=J\text{ \ \textrm{in} \ }D_{1}\times (0,T_{\text{
end}}), \\
\\
\displaystyle\frac{\partial \phi _{\mathrm{1}}}{\partial x}\mid _{\partial
D_{1}\times (0,T_{\text{end}})}=0, \\
\\
\int_{D_{1}}\phi _{\mathrm{1}}(x,t)dx=0.
\end{array}
\right. \label{rf3}
\end{equation}
\textit{Solid phase potential} $\phi _{\mathrm{2}}(x,t)$.
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\frac{\partial }{\partial x}\left( \sigma \frac{\partial \phi _{
\mathrm{2}}}{\partial x}\right) =J+g\text{ \ \textrm{in} \ }D_{2}\times
(0,T_{\text{\textrm{end}}}), \\
\\
\sigma \displaystyle\frac{\partial \phi _{\mathrm{2}}}{\partial x}\mid
_{\partial D_{2}\times (0,T_{\text{\textrm{end}}})}=0, \\
\\
g(x,t)=\left\{
\begin{array}{r}
\displaystyle\frac{-I(t)}{L_{n}A},\ x\in D_{\textrm{n}}, \\
\\
\displaystyle\frac{I(t)}{L_{p}A},\ x\in D_{\textrm{p}},
\end{array}
\right.
\end{array}
\right. \label{rf4}
\end{equation}
where $a_{1}(x)=\displaystyle\frac{1-t_{+}^{0}}{3\varepsilon
_s(x)F}$. In these equations, $k_{1}(x)>0$ and $k_{2}(x)>0$
represent effective diffusion coefficients in the electrolyte and
solid phases respectively, and $\sigma (x) $ denotes the effective
electric conductivity in the solid phase. The functions $a_{1}(x),\
a_{2}(x)\ $, $\sigma (x)$ and $k_{2}(x)$ are considered to be
piecewise positive constant functions in the sense that they have
different constant values in the negative electrode, separator and
positive electrode.
We also have to consider that for $t\in (0,T_{\text{\textrm{end}}})$, $
J(x,u,v_{\mathrm{s}},\phi _{\mathrm{1}},\phi _{\mathrm{2}})$
satisfies the
algebraic conditions
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\int_{D_{1}}Jdx=\int_{D_{2}}Jdx=0, \\
\\
\displaystyle\int_{D_{\textrm{n}}}Jdx=I(t),\ \int_{D_{\textrm{p}}}Jdx=-I(t).
\end{array}
\right. \label{rf5}
\end{equation}
Notice that the first row of algebraic conditions follow directly
from (\ref{rf3}) and the definition of $J$, whereas the second row
conditions translates the boundary conditions of the solid phase
potential. It is worth remarking the conservative properties enjoyed
by both $u(x,t)$
and $v(x;r,t)$; namely, for all $t\in \left[ 0,T_{\mathrm{end}}\right] $
\begin{equation*}
\int_{D_{1}}u(x,t)dx=\int_{D_{1}}u(x,0)dx,
\end{equation*}
and
\begin{equation*}
\int_{D_{2}}\int_{0}^{R_{s}(x)}v(x;r,t)r^{2}drdx=\int_{D_{2}}
\int_{0}^{R_{s}(x)}v(x;r,0)r^{2}drdx.
\end{equation*}
These relations are readily obtained by integrating (\ref{rf1}) and (\ref
{rf2}) and using the corresponding boundary conditions. Moreover, it
can be shown \cite{Kro} that for $t\geq 0$ and $x\in D_{1}$,
$u(x,t)>0$, similarly, for $(x,r)\in D_{3}$, $0<v(x;r,t)<v_{\max }$.
Let $D$ denote a generic open bounded domain in $\mathbb{R}$;
hereafter, the closure of a domain $D$ is denoted $\overline{D}$.
The functional spaces that we use in this paper are the following.
The Sobolev
spaces $H^{m}(D)$, $m$ being a nonnegative integer, when $m=0$, $
H^{0}(D):=L^{2}(D)$; the Lebesgue spaces $L^{p}(D),\ 1\leq p\leq \infty $;
the spaces of measurable radial functions \cite{Diaz}
\begin{equation*}
H_{r}^{q}(0,R):=\left\{ v:(0,R)\rightarrow \mathbb{R}\text{:\textrm{\ }}
\left\Vert v\right\Vert
_{H_{r}^{q}(0,R)}^{2}=\sum_{j=0}^{q}\int_{0}^{R}\left( \frac{d^{j}v}{dr^{j}}
\right) ^{2}r^{2}dr<\infty \right\} ,
\end{equation*}
$q$ being a nonnegative integer, when $q=0$ we set $
H_{r}^{0}(0,R):=L_{r}^{2}(0,R)$; also, for $p$ being a nonnegative
integer,
the normed spaces of measurable functions
\begin{equation*}
H^{p}(D_{2};H_{r}^{q}(0,R_{s}(\cdot ))):=\left\{ v:D_{2}\rightarrow
H_{r}^{q}(0,R_{s}(\cdot )):\left\Vert v\right\Vert
_{H^{p}(D_{2};H_{r}^{q}(0,R_{s}(\cdot )))}<\infty \right\} ,
\end{equation*}
where $\left\Vert v\right\Vert _{H^{p}(D_{2};H_{r}^{q}(0,R_{s}(\cdot
)))}^{2}=$ $\displaystyle\sum_{j=0}^{p}\int_{D_{2}}\left\Vert \frac{
\partial ^{j}v(x;\cdot )}{\partial x^{j}}\right\Vert
_{H_{r}^{q}(0,R_{s}(x))}^{2}dx$; and the spaces
\begin{equation*}
H_{\ \ r}^{p,q}(D_{2}\times (0,R_{s}(\cdot
)))=H^{p}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))\cap
L^{2}(D_{2},H_{r}^{q}(0,R_{s}(\cdot )))
\end{equation*}
with norm
\begin{equation*}
\left\Vert u\right\Vert _{H_{\ \ r}^{p,q}(D_{2}\times (0,R_{s}(\cdot
)))}^{2}:=\left\Vert u\right\Vert
_{H^{p}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\left\Vert
u\right\Vert _{L^{2}(D_{2},H_{r}^{q}(0,R_{s}(\cdot )))}^{2}.
\end{equation*}
Notice that $H_{\ \ r}^{0,0}(D_{2}\times (0,R_{s}(\cdot
)))=L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot ))).$
Since the variables of the model depend on time, then we also introduce the
normed spaces $L^{p}(0,t;X)$, where $1\leq p\leq \infty $, and $(X,\
\left\Vert \cdot \right\Vert _{X})$ being a real Banach space.
\begin{equation*}
L^{p}(0,t;X):=\left\{ v:(0,t)\rightarrow X\text{ \textrm{strongly measurable
such that\ \ }}\left\Vert v\right\Vert _{L^{p}(0,t;X)}<\infty \right\} ,
\end{equation*}
with $\left\Vert v\right\Vert _{L^{p}(0,t;X)}=\displaystyle\left(
\int_{0}^{t}\left\Vert v(\tau )\right\Vert _{X}^{p}d\tau \right)
^{1/p}$ when $1\leq p<\infty $, and for $p=\infty $, $\left\Vert
v\right\Vert _{L^{\infty }(0,t;X)}=ess\ \sup_{0<\tau <t}\left\Vert
v(\tau )\right\Vert _{X}$. Other spaces used in the paper are
$W(D_{1}):=\left\{ v\in
H^{1}(D_{1}):\displaystyle\int_{D_{1}}vdx=0\right\} $, which is a
closed subspace of $H^{1}(D_{1})$ where the potential $\phi
_{1}(x,t)$ is calculated, and the space of $q$ times continuously
differentiable functions defined on $D$, $C^{q}(D)$, when $q=0$,
$C^{0}(D):=C(D)$.
Next, we introduce the following regularity assumptions on the data and the
molar flux $J$ \cite{Diaz}.
\textbf{A1)}
\begin{equation*}
u^{0}\in H^{1}(D),\ u^{0}>0,\ v^{0}\in C(\overline{D}_{3}),\ 0<v^{0}<v_{\max
},\ I(t)\in C_{part}([0,T^{\ast }]),0<T_{end}\leq T^{\ast }<\infty ,
\end{equation*}
where $C_{part}$ denotes the set of piecewise continuous functions, i.e.,
\begin{equation*}
C_{\mathrm{part}}([a,b])=\left\{ g:[a,b]\rightarrow \mathbb{R}:\exists
a=t_{0}<t_{1}<\cdots t_{N}=b\text{ such that }g\in C\left(
[t_{i-1},t_{i}]\right) \right\} .
\end{equation*}
\textbf{A2)} For $0<a<c<+\infty $, $k_{0}$ and $\sigma _{0}$
positive constants,
\begin{equation*}
k_{1}\in L^{\infty }(D_{1}),\ k_{1}\geq k_{0}>0,\ k_{2}\in \lbrack
a,c],\ \kappa \in C^{2}\left( (0,+\infty )\right) ,\ \sigma \in
L^{\infty }(D_{2}),\ \kappa \geq \kappa _{0}>0,\ \sigma \geq \sigma
_{0}>0.
\end{equation*}
\ \ \ \ \ Moreover, $\underline{k_{1}}:=\inf_{x\in
{D_{1}}}{k_{1}(x)}$ and $\underline{k_{2}}:=\min_{x\in
{D_{2}}}{k_{2}(x)}$.
\bigskip
\textbf{A3)} For all $(x,u,v_{s},\eta )\in D_{2}\times (0,+\infty )\times
(0,v_{s,\mathrm{\max }})\times \mathbb{R}$,
\begin{equation*}
J\in C^{2}(D_{2}\times (0,+\infty )\times (0,v_{s,\mathrm{\max }})\times
\mathbb{R}),\ \displaystyle\frac{\partial J}{\partial \eta }>0.
\end{equation*}
A weak formulation to (\ref{rf1})-(\ref{rf4}) is the following. Find
\begin{equation*}
\left\{
\begin{array}{l}
u\in {L}^{2}\left( 0,T_{\mathrm{end}};\ H^{1}(D_{1}\right) ),\ \displaystyle
\frac{du}{dt}\in L^{2}(0,T_{\mathrm{end}};\ H^{1}(D_{1}))^{\ast }, \\
\\
v{\in L}^{2}\left( 0,T_{\mathrm{end}};L^{2}(D_{2,}\
H_{r}^{1}(0,R_{s}(\cdot
))\right) ),\ \displaystyle\frac{dv}{dt}\in {L}^{2}\left( 0,T_{\mathrm{end}
};L^{2}(D_{2,}\ H_{r}^{1}(0,R_{s}(\cdot ))\right) )^{\ast }, \\
\\
\phi _{\mathrm{1}}\in L^{2}(0,T_{\mathrm{end}};\ W(D_{1}))\text{
\textrm{and}
}\phi _{\mathrm{2}}\in L^{2}(0,T_{\mathrm{end}};\ H^{1}(D_{2})),
\end{array}
\right.
\end{equation*}
such that
\begin{equation}
\int_{D_{1}}\frac{\partial {u}}{\partial {t}}wdx+\int_{D_{1}}k_{1}\frac{
\partial u}{\partial x}\frac{dw}{dx}dx=\int_{D_{1}}a_{1}Jwdx\ \forall w\in
H^{1}(D_{1}); \label{w1}
\end{equation}
for a.e.$x\in D_{2}$ and for all $w\in H_{r}^{1}(0,R_{s}(x))$
radially symmetric
\begin{equation}
\int_{0}^{R_{s}(x)}\frac{\partial {v}}{\partial {t}}wr^{2}dr+
\int_{0}^{R_{s}(x)}k_{2}\frac{\partial v}{\partial r}\frac{\partial w}{
\partial r}r^{2}dr=\frac{-R_{\mathrm{s}}^{2}(x)Jw(R_{s}(x))}{a_{2}(x)F}; \label{w2}
\end{equation}
\begin{equation}
\int_{D_{1}}\kappa (u)\frac{\partial \phi _{1}}{\partial x}\frac{dw}{dx}
dx=\int_{D_{1}}Jwdx\text{ \ }\forall w\in H^{1}(D_{1}); \label{w3}
\end{equation}
and
\begin{equation}
\int_{D_{2}}\sigma \frac{\partial \phi _{2}}{\partial x}\frac{dw}{dx}
dx=-\int_{D_{2}}\left( J+g\right) wdx,\forall w\in H^{1}(D_{2}),
\label{w4}
\end{equation}
where $L^{2}(0,T_{\mathrm{end}};\ H^{1}(D_{1}))^{\ast }$ and
${L}^{2}\left( 0,T_{\mathrm{end}};L^{2}(D_{2,}\
H_{r}^{1}(0,R_{s}(\cdot ))\right) )^{\ast }$ denote the respective
dual spaces of\ $L^{2}(0,T_{\mathrm{end}};\ H^{1}(D_{1}))$ and
${L}^{2}\left( 0,T_{\mathrm{end}};L^{2}(D_{2,}\
H_{r}^{1}(0,R_{s}(\cdot ))\right) )$.
\begin{remark}
\label{remark1} Following the arguments of \cite{Diaz}, where its
non-isothermal P2D model includes an additional time dependent non
linear ordinary equation for the bulk temperature $T(t)$, one can
formulate an alternative definition
of the weak solution to (\ref{rf1}
)-(\ref{rf4}) based on its Definition 2.7 and prove, under the
assumptions \textbf{A1}-\textbf{A3} and for a
partition $t_{0}<t_{1}<\cdots <t_{N}$ of $[0,T_{
\mathrm{end}}]$, $T_{\mathrm{end}}$ being small enough, that there
is a unique
weak solution $\left( u,v,\phi _{1},\phi _{2}\right) $ in each interval $
[t_{n},t_{n+1}]$, such that $\left( u,v,\phi _{1},\phi _{2}\right)
\in C([t_{n},t_{n+1}];K_{Z})$, where $K_{Z}:=H^{1}(D_{1})\times
L^{2}(D_{2,}\ H_{r}^{1}(0,R_{s}(\cdot )))\times W(D_{1})\times
H^{1}(D_{2})$. $\left( u(t_{n}),v(t_{n})\right)$ being the initial
condition in such an interval. Also, Kr$\ddot{\text{o}}$ner
\cite{Kro} proves a local existence and uniqueness theorem for the
weak solution of the isothermal P2D model under less general
assumptions than in \cite{Diaz}.
\end{remark}
\section{The semidiscrete finite element formulation of the isothermal P2D
model}
We use $H^{1}$-conforming linear finite elements ($P_{1}-$finite
elements) for the space approximation of the variables $u(x,t)$,
$\phi _{\mathrm{1}}(x,t)$ and $\phi _{\mathrm{2}}(x,t)$; however,
$v(x;r,t)$ is approximated by nonconforming $P_{0}-$finite elements
in the $x-$coordinate and $H^{1}$-conforming $P_{1}-$finite elements
in the $r- $coordinate. The family of meshes $D_{1h}$ constructed on
the domain $D_{1}$ includes the points $x=0$, $x=L_{n}$,
$x=L_{n}+\delta $ and $x=L$ as mesh points; since these points are
also boundary points of $D_{2}$, then they are also considered as
mesh points in the family of meshes $D_{2h}$. Figure 2 illustrates
the families of meshes that we are going to describe next. Noting
that $D_{2}\subset D_{1}$, we choose the family of meshes $D_{2h}$
as a subset of $D_{1h}$. Let $NE_{1}$ and $
NE_{2}$ be the number of elements of $D_{1h}$ and $D_{2h}$
respectively, and let $M_{1}$ and $M_{2}$ be the number of mesh
points of such meshes, then, for $i=1,2$, we have that
\begin{equation*}
D_{ih}=\left\{ e_{m}\right\} _{m=1}^{NE_{i}}\ \text{\textrm{and}}\ \overline{
D}_{i}=\cup _{m=1}^{NE_{i}}e_{m},
\end{equation*}
where the $m$th element, $e_{m}:=\left\{ x:x_{1}^{m}\leq x\leq x_{2}^{m}\right\} $, and $
h_{m}:=x_{2}^{m}-x_{1}^{m}$ is the length of the element $e_{m}$;
the points $\ x_{1}^{m}$ $x_{2}^{m}$ are denoted element nodes. We
set $h=\max_{m}h_{m}$, and $ \gamma =h^{-1}\min_{m}h_{m}$. The
parameter $\gamma $ is a measure of the uniformity of the meshes.
The collection of all the element nodes defines
the set of nodes, $\left\{ x_{l}\right\} _{l=1}^{M_{i}}$, of the mesh $
D_{ih} $. To construct the family of meshes $\widehat{D}_{3h\Delta
r}$ on $D_{3}$, we recall that $D_{3}=\cup _{x\in D_{2}}\left\{
x\right\} \times (0,R_{s}(x))$, where $R_{s}(x)$ is the radius of
the solid spherical particle associated with the point $\{x\}$.
Thus, for each mesh point $\left\{ x_{l}\right\} \in D_{2h}$ we
define the radial vertical domain $D_{r}^{(l)}:=\left\{ r\in
\mathbb{R}
:0<r<R_{s}(x_{l})\right\} $, which represents the spherical particle at $
x_{l}$, and let
\begin{equation*}
D_{\Delta r}^{(l)}=\left\{ e_{k}^{(l)}\right\} _{k=1}^{NE^{(l)}}\text{
\textrm{such that }}\overline{D}_{r}^{(l)}=\cup _{k=1}^{NE^{(l)}}e_{k}^{(l)},
\end{equation*}
where $NE^{(l)}$ denotes the number of elements in the interval $
[0,R_{s}(x_{l})]$ and $\Delta r_{k}^{(l)}$ is the width of the
element $e_{k}^{(l)}:=\{r:r_{1}^{(l)k}\leq r\leq r_{2}^{(l)k}\}$, we
set $\Delta r=\max_{l}(\max_{k}\Delta r_{k}^{(l)})$ and $\gamma
_{r}=\Delta r^{-1}\min_{l}(\min_{k}\Delta r_{k}^{(l)}).$ The set of
mesh points in each mesh $D_{\Delta r}^{(l)}$ is denoted $\left\{
r_{j}^{(l)}\right\} _{j=1}^{M^{(l)}}$.
Furthermore, let $\left\{ \widehat{e}_{l}\right\} _{l=1}^{M_{2}}$ be
the collection of nonconforming elements of the mesh
$\widehat{D}_{2h}$ which are associated with the nodes $\left\{
x_{l}\right\} $, they are defined as
follows: if $\left\{ x_{l}\right\} $ is not a boundary point, then
\begin{equation*}
\widehat{e}_{l}:=\left\{ x\in D_{2}:x_{l}-\frac{h_{l-1}}{2}\leq x<x_{l}+
\frac{h_{l}}{2}\right\} ;
\end{equation*}
on the contrary, if $\left\{ x_{l}\right\} $ is a left boundary point, then
\begin{equation*}
\widehat{e}_{l}:=\left\{ x\in \overline{D}_{2}:x_{l}\leq x<x_{l}+\frac{
h_{l}}{2}\right\} ,
\end{equation*}
and if $\left\{ x_{l}\right\} $ is a right boundary point, then
\begin{equation*}
\widehat{e}_{l}:=\left\{ x\in
\overline{D}_{2}:x_{l}-\frac{h_{l-1}}{2}\leq x\leq x_{l}\right\} .
\end{equation*}
We define the meshes $\widehat{D}_{3h\Delta r}$ as
\begin{equation*}
\widehat{D}_{3h\Delta r}:=\left\{ \widehat{e}_{l}\times
\overline{D}_{\Delta r}^{(l)}\right\} _{l=1}^{M_{2}}\
\text{\textrm{such that }}\overline{D}_{3}:=\cup
_{l=1}^{M_{2}}\widehat{e}_{l}\times \overline{D}_{\Delta r}^{(l)}.
\end{equation*}
\begin{figure}[th!]
\begin{center}
\includegraphics[height=10cm]{fig_2.eps}
\end{center}
\caption{\textit{Panel (a)}: the mesh for the domain $D_{1}$, which
includes the negative electrode (anode in the figure) $D_{a}$, the
separator $D_{s}$ and the positive electrode (cathode in the figure)
$D_{c}$. \textit{
Panel (b)}: the mesh for the domain $D_{2}=D_{a}\cup D_{c}$.
\textit{Panel (c)}: the mesh of nonconforming elements for the domain $D_{2}$. \textit{
Panel (d)}: the mesh for the domain $D_{3}$.} \label{figure2}
\end{figure}
The families of conforming linear finite element spaces associated with
these meshes are the following. For $i=1,2,$
\begin{equation*}
V_{h}^{(1)}(\overline{D}_{i}):=\left\{ v_{h}\in C(\overline{D}
_{i}):\forall e_{m}\in D_{ih},\ v_{h}\mid _{e_{m}}\in P_{1}(e_{m})\right\} ,
\end{equation*}
where $P_{1}(e_{m})$ denotes the set of linear polynomials defined on $e_{m}$
.\ Let $\left\{ \psi _{l}(x)\right\} _{l=1}^{M_{i}}$ be the set of nodal
basis functions for the space $V_{h}^{(1)}(\overline{D}_{i})$, then any
function $v_{h}\in V_{h}^{(1)}(\overline{D}_{i})$ can be written as
\begin{equation*}
v_{h}(x)=\sum_{l=1}^{M_{i}}V_{l}\psi _{l}(x),\ \mathrm{where\ }
V_{l}=v_{h}(x_{l})\text{.}
\end{equation*}
Note that $V_{h}^{(1)}(\overline{D}_{i})\subset H^{1}(D_{i})$. The
nonconforming finite element space associated with the mesh
$\widehat{D}_{2h}$ is defined as
\begin{equation*}
V_{h}^{(0)}(\overline{D}_{2}):=\left\{ v_{h}\in L^{2}(D_{2}):\forall
\widehat{e}_{l}\in \widehat{D}_{2h},\ v_{h}\mid
_{\widehat{e}_{l}}\in P_{0}(\widehat{e}_{l})\right\} ,
\end{equation*}
where $P_{0}(\widehat{e}_{l})$ is the set of polynomials of degree
zero defined on $\widehat{e}_{l}$. Let $\left\{ \chi _{l}(x)\right\}
_{l=1}^{M_{2}}$ be the set of nodal basis functions for $V_{h}^{(0)}(
\overline{D}_{2})$,
\begin{equation*}
\chi _{l}(x)=\left\{
\begin{array}{r}
1\text{ \ \textrm{if} \ }x\in \widehat{e}_{l}, \\
\\
0\ \ \text{\textrm{otherwise}},
\end{array}
\right.
\end{equation*}
then any function $v_{h}(x)\in V_{h}^{(0)}(\overline{D}_{2})$ is expressed as
\begin{equation*}
v_{h}(x)=\sum_{l=1}^{M_{2}}V_{l}\chi _{l}(x),\ \mathrm{where\ }
V_{l}=v_{h}(x_{l}).
\end{equation*}
It is worth remarking that for $1\leq p<\infty$, the $L^{p}$-norm of $
v_{h}(x)\in V_{h}^{(0)}(\overline{D}_{2})$ is given as $\left\Vert
v_{h}\right\Vert _{L^{p}(D_{2})}^{p}=\sum_{l=1}^{M_{2}}\widehat{h}
_{l}V_{l}^{p} $, where $\widehat{h}_{l}$ denotes the length of the
element $\widehat{e}_{l}$, and the $L^{2}$-inner product of
$v_{h},w_{h} \in V_{h}^{(0)}(\overline{D}_{2})$,
$\displaystyle\int_{D_{2}}v_{h}w_{h}dx=\sum_{l=1}^{M_{2}}
\widehat{h}_{l}V_{l}W_{l}$. Next, we introduce the finite element
space $V_{\Delta
r}^{(1)}(\overline{D}_{r}^{(l)})$. For $1\leq l\leq M_{2}$,
\begin{equation*}
V_{\Delta r}^{(1)}(\overline{D}_{r}^{(l)}):=\left\{ v_{\Delta r}\in C(
\overline{D}_{r}^{(l)}):\forall e_{k}^{(l)}\in D_{\Delta r}^{(l)},\
v_{\Delta r}^{(l)}(r)\mid _{e_{k}^{(l)}}\in P_{1}(e_{k}^{(l)})\right\} .
\end{equation*}
So, if $\left\{ \alpha _{j}^{(l)}(r)\right\} _{j=1}^{M^{(l)}}$
denotes the set of nodal basis of $V_{\Delta
r}(\overline{D}_{r}^{(l)})$, any function $v_{\Delta r}^{(l)}\in
V_{\Delta r}^{(1)}(\overline{D}_{r}^{(l)})\subset
H_{r}^{1}(0,R_{s}(x_{l}))$ can be written as
\begin{equation*}
v_{\Delta r}^{(l)}(r)=\sum_{j=1}^{M^{(l)}}V_{j}^{(l)}\alpha
_{j}^{(l)}(r),\ \mathrm{where\
}V_{j}^{(l)}=v_{h}^{(l)}(r_{j})\text{.}
\end{equation*}
Regarding the meshes $\widehat{D}_{3h\Delta r}$, we define the finite element space $
V_{h\Delta r}(\overline{D}_{3})$ as follows. For $1\leq l\leq M_{2}$ and $
1\leq k\leq NE^{(l)}$
\begin{equation*}
V_{h\Delta r}(\overline{D}_{3}):=\displaystyle\left\{ {v}_{h\Delta
r}{\in H}
_{\ \ r}^{0,1}{(D_{2}\times (0,R_{s})(\cdot )):v_{h\Delta r}(x;r)\mid _{
\widehat{e}_{l}\times e_{k}^{(l)}}\in P_{0}(\widehat{e}_{l})\otimes
P_{1}(e_{k}^{(l)})}\right\},
\end{equation*}
noting that when $x\in \widehat {e}_{l}$, $R_{s}(x)=R_{s}(x_{l})$.
Hence, any function $v_{h\Delta r}(x;r)\in V_{h\Delta
r}(\overline{D}_{3})$
is of the form
\begin{equation*}
v_{h\Delta r}(x;r)=\sum_{l=1}^{M_{2}}\sum_{j=1}^{M^{(l)}}V_{lj}\chi
_{l}(x)\alpha _{j}^{(l)}(r),\ \mathrm{where\ }V_{lj}=v_{h\Delta
r}(~x_{l},r_{j}),
\end{equation*}
or equivalently, using the notation $v_{\Delta r}^{(l)}(r)$ to
denote $v_{h\Delta
r}(x_{l};r)$, we can write
\begin{equation} \label{vhr}
v_{h\Delta r}(x;r)=\sum_{l=1}^{M_{2}}v_{\Delta r}^{(l)}(r)\chi _{l}(x).
\end{equation}.
The function $v_{sh}(x):=v_{h\Delta r}(x;R_{s}(x))$ is given by the
expression
\begin{equation*}
v_{sh}(x;R_{s}(x))=\sum_{l=1}^{M_{2}}V_{lM^{(l)}}\chi _{l}(x),
\end{equation*}
so that $v_{sh}(x)\in V_{h}^{(0)}(\overline{D}_{2})$. We calculate
$\phi _{1h}(x,t)$, which is the approximation to $\phi_{1}(x,t)$, in
the finite
dimensional space
\begin{equation*}
W_{h}(\overline{D}_{1}):=\left\{ v_{h}\in V_{h}^{(1)}(\overline{D}
_{1}):\int_{D_{1}}v_{h}dx=0\right\} ,\ W_{h}(\overline{D}_{1})\subset
W(D_{1}).
\end{equation*}
Thus, the finite element formulation is as follows. For all $t\in (0,T_{
\mathrm{end}})$, the semi-discrete approximation $(u_{h}(t),v_{h\Delta
r}(t),\phi _{1h}(t),\phi _{2h}(t))\in V_{h}^{(1)}(\overline{D}_{1})\times
V_{h\Delta r}(\overline{D}_{3})\times W_{h}(\overline{D}_{1})\times
V_{h}^{(1)}(\overline{D}_{2}),$ $(u_{h}^{0},v_{h}^{0})\in V_{h}^{(1)}(
\overline{D}_{1})\times V_{h\Delta r}(\overline{D}_{3})$, is solution to the
following system of equations.
\begin{equation}
\int_{D_{1}}\frac{\partial u{_{h}}}{\partial {t}}w_{h}dx+\int_{D_{1}}k_{1}
\frac{\partial u_{h}}{\partial x}\frac{dw_{h}}{dx}dx=
\int_{D_{1}}a_{1}J_{h}w_{h}dx\ \forall w_{h}\in V_{h}^{(1)}(\overline{D}
_{1}). \label{num1}
\end{equation}
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\displaystyle\frac{\partial v{
_{h\Delta r}}}{\partial {t}}w_{h\Delta r}r^{2}drdx+\displaystyle
\int_{D_{2}}\int_{0}^{R_{s}(x)}k_{2}\displaystyle\frac{\partial v{_{h\Delta
r}}}{\partial {r}}\frac{\partial w_{h\Delta r}}{\partial r}r^{2}drdx \\
\\
=-\displaystyle\int_{D_{2}}\frac{R_{s}^{2}(x)J_{h}w_{h\Delta
r}(x,R_{s}(x))}{a_{2}(x)F}dx\ \ \forall w_{h\Delta r}\in V_{h\Delta
r}(\overline{D}_{3}).
\end{array}
\right. \label{num2}
\end{equation}
\begin{equation}
\int_{D_{1}}\kappa (u_{h})\frac{\partial \phi _{1h}}{\partial x}\frac{dw_{h}
}{dx}dx=\int_{D_{1}}J_{h}w_{h}dx\ \ \forall w_{h}\in V_{h}^{(1)}(\overline{D}
_{1}). \label{num3}
\end{equation}
\bigskip
\begin{equation}
\int_{D_{2}}\sigma \frac{\partial \phi _{2h}}{\partial x}\frac{dw_{h}}{dx}
dx=-\int_{D_{2}}\left( J_{h}+g\right) w_{h}dx\ \ \forall w_{h}\in
V_{h}^{(1)}(\overline{D}_{2}). \label{num4}
\end{equation}
\bigskip
\begin{equation}
\int_{D_{1}}J_{h}dx=\int_{D_{2}}J_{h}dx=0, \text{ with
}\int_{D_{\mathrm{n}}}J_{h}dx=I(t)=- \int_{D_{\textrm{p}}}J_{h}dx.
\label{num5}
\end{equation}
In this system,
\begin{equation}
\begin{array}{l}
J_{h}:=J(x,u_{_{h}},v_{sh},\eta _{h})=a_{2}(x)i_{0h}\sinh \left(
\beta \eta _{h}\right), \
i_{0h}:=i_{0}(u_{h},v_{sh}) \\
\\
\eta _{h}:=\phi _{1h}-\phi _{2h}-\alpha_{h} \ln
u_{h}-\overline{U}_{h}(v_{sh}),\ \alpha_{h} =\alpha(u_{h}).
\end{array}
\label{num6}
\end{equation}
Note that $J_{h}$ also depends on $t$ through $u_{h}$, $v_{sh}$ and $\eta
_{h}$.
\begin{remark}
\label{remark2} We must note, see (\ref{vhr}), that $v_{h\Delta r}$
and $w_{h\Delta r}$ are elementwise constant functions in the $x$
direction, and $J_{h}$ is a piecewise continuous function in $x$ for
which it makes sense to consider the approximation,
$I_{h}^{(0)}J_{h}\in V_{h}^{(0)}(\overline{D}_{2})$, see in
Subsection 4.1 the definition of the interpolant $I_{h}^{(0)}$.
Then, approximating $J_{h}$ by $I_{h}^{(0)}J_{h}$ one readily shows,
by performing the integral on $D_{2}$, that (\ref{num2}) can be
recast as follows: for all mesh-point $\left\{ x_{l}\right\} \in
D_{2h}$, calculate $v_{\Delta
r}^{(l)}(r,t)$ $\in V_{\Delta r}^{(1)}(\overline{D}_{r}^{(l)})$ such that
\begin{equation} \label{num2_1}
\begin{array}{r}
\displaystyle\int_{0}^{R_{s}(x_{l})}\frac{\partial v_{\Delta r}^{(l)}(r,t)}{
\partial t}w_{\Delta r}^{(l)}(r)r^{2}dr+\int_{0}^{R_{s}(x_{l})}\displaystyle
k_{2}\frac{\partial v_{\Delta r}^{(l)}(r,t)}{\partial
r}\frac{\partial
w_{\Delta r}^{(l)}(r)}{\partial r}r^{2}dr \\
\\
=-R_{s}^{2}(x_{l})(a_{2}(x_{l})F)^{-1}J_{h}\mid _{x_{l}}w_{\Delta
r}^{(l)}(R_{s}(x_{l}))\ \ \forall w_{\Delta r}^{(l)}\in V_{\Delta r}^{(1)}(
\overline{D}_{r}^{(l)}).
\end{array}
\end{equation}
Once $v_{\Delta r}^{(l)}(r,t)$ is known, one calculates $v_{h\Delta
r}$ by the expression (\ref{vhr}).
\end{remark}
Notice that this equation is the finite element approximation of (\ref{w2})
for $w\in H_{r}^{1}(0,R_{s}(x_{l}))$, see \cite{BG}.
Based on Remark \ref{remark1} and since $H^{1}(D_{i})\hookrightarrow
C(\overline{D}_{i})$, we introduce the following spaces which are
used
in the error analysis and in the application of the fixed point theorems in Section 4.
\begin{equation*}
\begin{array}{l}
S_{P}:=\left\{ u\in C_{\mathrm{part}}(\overline{D}_{1}\times \lbrack 0,T_{
\mathrm{end}}]):\displaystyle\frac{1}{P}\leq u\leq P\right\} , \\
\\
S_{Q}:=\left\{ w\in C_{\mathrm{part}}(\overline{D}_{2}\times \lbrack 0,T_{
\mathrm{end}}]):\displaystyle\frac{1}{Q}\leq w\leq 1-\frac{1}{Q}\right\} , \\
\\
S_{Q}^{\ast }:=\left\{ w\in C_{\mathrm{part}}^{\ast
}(\overline{D}_{2}\times
\lbrack 0,T_{\mathrm{end}}]):\frac{1}{Q}\leq w\leq 1-\frac{1}{Q}\right\} ,
\end{array}
\end{equation*}
where $P$ and $Q$ are constants sufficiently large; for $i=1,2$, $C_{\mathrm{part}
}(\overline{D}_{i}\times [0,T_{\mathrm{end}}])$ denotes the set of
piecewise
continuous functions in time and continuous in space and $C_{\mathrm{part}
}^{\ast }(\overline{D}_{2}\times [0,T_{\mathrm{end}}])$ denotes the
set of piecewise continuous functions in both time and space.
$S_{P}$ is the candidate pool for the concentration $u$ and its
approximate $u_{h}$, whereas $S_{Q}$ and $S_{Q}^{\ast}$ play the
same role for the concentrations $\frac{v_{s}}{v_{\max }}$ and
$\frac{v_{sh}}{v_{\max }}$ respectively. So, we make the following
assumption.
\textbf{A4)\ }There exist constants $P,\ Q$ and $K$ sufficiently
large such that that for almost
every $t\in \lbrack 0,T_{\mathrm{end}}]$ the following bounds hold:
\begin{equation}
\frac{1}{P}\leq u(t),\ u_{h}(t)\leq P,\text{ \ }\ \frac{1}{Q}\leq
\frac{v_{s}(t)}{v_{\max }},\ \frac{v_{sh}(t)}{v_{\max }}\leq \left(
1-\frac{1}{Q}\right) , \label{estK1}
\end{equation}
and for $i=1,2$,
\begin{equation}
\left\Vert \frac{\partial \phi _{1}(t)}{\partial x}\right\Vert _{L^{\infty
}(D_{1})},\ \left\Vert \phi _{i}(t)\right\Vert _{H^{1}(D_{i})},\ \left\Vert
\phi _{i}(t)\right\Vert _{L^{\infty }(D_{i})},\ \left\Vert \phi
_{ih}(t)\right\Vert _{L^{\infty }(D_{i})}\leq K. \label{estK2}
\end{equation}
\section{Error analysis for the semidiscrete problem}
We present in this section the error analysis for the semidiscrete
potentials and concentrations. Since the development of such an
analysis is long, we have split its presentation in a sequence of
three subsections. In the first one, we introduce some auxiliary
results needed for the error analysis. The second subsection deals
with the error estimate for the potentials. Observing that the P2D
model is a nonlinear coupled system of equations, then the error for
the potentials depends on the error estimates for the
concentrations, the analysis of which is carried out in the last
subsection.
\subsection{Auxiliary results}
It is well known \cite{Ci} that for the finite element spaces $V_{h}^{(p)}(\overline{D}
_{i})$ ($i=1,2; p=0,1$) the following approximation property holds. For $
1\leq s\leq p+1$,
\begin{equation}
\inf_{v_{h}\in V_{h}^{(p)}(D_{i})}\left\{ \left\Vert
v-v_{h}\right\Vert _{L^{2}(D_{i})}+h\left\Vert \frac{d^{p}\left(
v-v_{h}\right) }{dx^{p}}\right\Vert _{L^{2}(D_{i})}\right\} \leq
Ch^{s}\left\vert v\right\vert _{H^{s}(D_{i})}, \label{pre1}
\end{equation}
where it should be understood that for $p=0$,
$\displaystyle\frac{d^{p}\left( v-v_{h}\right) }{dx^{p}}=v-v_{h}$.
For symmetric radial functions defined in the interval $[0,R]$, let
$V_{\Delta r}^{(1)}[0,R]$ be a linear finite element space where we
approximate such functions, one can prove, following the approach
used to
prove Lemmas 1 and 2 in \cite{ET}, that when $w\in H_{r}^{2}(0,R)$,
\begin{equation}
\inf_{w_{\Delta r}\in V_{\Delta r}^{(1)}[0,R]}\left\{ \left\Vert
w-w_{\Delta r}\right\Vert _{L_{r}^{2}(0,R)}+\Delta r\left\Vert
\frac{d\left( w-w_{\Delta r}\right) }{dr}\right\Vert
_{L_{r}^{2}(0,R)}\right\} \leq C\Delta r^{2}\left\vert w\right\vert
_{H_{r}^{2}(0,R)}. \label{pre2}
\end{equation}
We consider the interpolants $I_{h}^{(1)}:C(\overline{D}_{i})\rightarrow $ $
V_{h}^{(1)}(\overline{D}_{i})$, $I_{h}^{(0)}:C(\overline{D}
_{2})\rightarrow V_{h}^{(0)}(\overline{D}_{2}),$ and the elliptic projection
$P_{1}:H^{1}(D_{1})\rightarrow V_{h}^{(1)}(D_{1})$ such that for $u\in
H^{1}(D_{1})$
\begin{equation}
\int_{D_{1}}\left(
k_{1}\frac{d(P_{1}u-u)}{dx}\frac{du_{h}}{dx}+\lambda
(P_{1}u-u)u_{h}\right) dx=0\ \ \forall u_{h}\in V_{h}^{(1)}(D_{1}),
\label{pre3}
\end{equation}
where $\lambda >0$ is a constant; the error analysis for elliptic
problems suggests that a good choice is $\lambda
=\underline{k_{1}}$. By virtue of (\ref{pre1}) it follows that\
there exists a constant $C$ independent of $h$ such that
\begin{equation}
\left\Vert v-I_{h}^{(0)}v\right\Vert _{L^{2}(D_{2})}\leq Ch\left\Vert
v\right\Vert _{H^{1}(D_{2})}; \label{pre4}
\end{equation}
for $1\leq m\leq 2,\ 0\leq l\leq 1,$
\begin{equation}
\left\Vert v-I_{h}^{(1)}v\right\Vert _{H^{l}(D_{i})}\leq
Ch^{m-l}\left\Vert v\right\Vert _{H^{m}(D_{i})}; \label{pre4.1}
\end{equation}
and from the well known error analysis for elliptic problems
\cite{Ci}
\begin{equation}
\left\Vert u-P_{1}u\right\Vert _{H^{l}(D_{1})}\leq
Ch^{m-l}\left\Vert u\right\Vert _{H^{m}(D_{1})}. \label{pre5}
\end{equation}
Likewise, for symmetric radial functions $v\in H_{r}^{q}(0,R)$, we define
the elliptic projector $P_{1}^{r}:H_{r}^{1}(0,R)\rightarrow V_{\Delta
r}^{(1)}[0,R]$ as the solution of the problem
\begin{equation}
\int_{0}^{R}\left( k_{2}\frac{d(P_{1}^{r}w-w)}{dr}\frac{dw_{\Delta r}}{dr}
+\lambda (P_{1}^{r}w-w)w_{\Delta r}\right) r^{2}dr=0\ \ \forall
w_{\Delta r}\in V_{\Delta r}^{(1)}[0,R], \label{pre6}
\end{equation}
with $\lambda >0$; as before, a good choice now is
$\lambda=\underline{k_{2}}$. By virtue of (\ref{pre2}) it follows
that there exists a constant $C$ independent of $\Delta r$ such that
\begin{equation}
\left\Vert w-P_{1}^{r}w\right\Vert _{L_{r}^{2}(0,R)}+\Delta r\left\Vert
\frac{d\left( w-P_{1}^{r}w\right) }{dr}\right\Vert _{L_{r}^{2}(0,R)}\leq
C\Delta r^{2}\left\vert w\right\vert _{H_{r}^{2}(0,R)}. \label{pre7}
\end{equation}
$P_{1}^{r}$ can be extended to functions of $x$ and $r$ in an
$L^{2}$-sense. Thus, for $(x,r)\in D_{2}\times (0,R(\cdot ))$ we
define the extended projection $P_{1}^{r}:H_{\ \
r}^{1,1}(D_{2}\times (0,R(\cdot )))\rightarrow L^{2}(D_{2})\otimes
V_{\Delta
r}^{(1)}[0,R(\cdot )]$ as
\begin{equation}
\int_{D_{2}}\int_{0}^{R(x)}\left( k_{2}\frac{\partial (P_{1}^{r}w-w)}{\partial r
}\frac{dw_{\Delta r}}{dr}+\lambda (P_{1}^{r}w-w)w_{\Delta r}\right)
r^{2}drdx=0\ \ \forall w_{\Delta r}\in L^{2}(D_{2})\otimes V_{\Delta
r}^{(1)}[0,R(\cdot )]. \label{pre8}
\end{equation}
Assuming that $w\in H_{\ \ r}^{1,1}(D_{2}\times (0,R(\cdot ))$ is such that
for a.e. $x\in D_{2}$, $w(x,\cdot )\in H_{r}^{2}(0,R(x))$, then by virtue of (\ref
{pre7})
\begin{equation*}
\left\Vert \left( w-P_{1}^{r}w\right) (x,\cdot )\right\Vert
_{L_{r}^{2}(0,R(x))}+\Delta r\left\Vert \frac{\left( \partial \left(
w-P_{1}^{r}w\right) \right) (x,\cdot )}{\partial r}\right\Vert
_{L_{r}^{2}(0,R(x))}\leq C\Delta r^{2}\left\vert w(x,\cdot )\right\vert
_{H_{r}^{2}(0,R(x))}.
\end{equation*}
Noting that $H_{\ \ r}^{0,q}(D_{2}\times (0,R(\cdot
)))=L^{2}(D_{2},L_{r}^{2}(0,R(\cdot )))\cap L^{2}(D_{2},H_{r}^{q}(0,R(\cdot )))$
, then it readily follows that for $q=0,1$
\begin{equation}
\left\Vert w-P_{1}^{r}w\right\Vert _{H_{\ \ r}^{0,q}(D_{2}\times
(0,R(\cdot )))}\leq C\Delta r^{2-q}\left\Vert w\right\Vert _{H_{\ \
r}^{0,2}(D_{2}\times (0,R(\cdot )))}. \label{pre10}
\end{equation}
We shall also consider the $x$-Lagrange interpolant for functions
that depend on $x$ and $r$, $I_{0}^{x}:H_{\ \ r}^{1,1}(D_{2}\times
(0,R(\cdot )))\rightarrow V_{h}^{(0)}(\overline{D}_{2})\otimes
H_{r}^{1}(0,R(\cdot ))$. Thus,
for $v(x;r)\in H_{\ \ r}^{1,1}(D_{2}\times (0,R(\cdot )))$
\begin{equation} \label{pre4.1}
I_{0}^{x}v(x;r)=\sum_{l=1}^{M_{2}}v(x_{l};r)\chi _{l}(x) \
\textrm{with} \ r\in (0,R(x_{l})).
\end{equation}
Since $I_{0}^{x}$ can be viewed as an extended Lagrange interpolant $
I_{h}^{(0)}:C(\overline{D}_{2})\rightarrow
V_{h}^{(0)}(\overline{D}_{2})$
, then based on (\ref{pre4}) one can show that
\begin{equation}
\left\Vert v-I_{0}^{x}v\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot
)))}\leq Ch\left\Vert v\right\Vert
_{H^{1}(D_{2};L_{r}^{2}(0,R_{s}(\cdot )))}. \label{uv3}
\end{equation}
\begin{lemma}
\label{lem1} Let $I=(a,b),\ 0\leq a<b$, be a bounded interval with
$\overline{I}=[a,b]$, and let $f\in H^{1}(I)$. There exists an
arbitrarily small number $\epsilon $ and a positive constant
$C(\epsilon )$
such that
\begin{equation}
\left\Vert f\right\Vert _{L^{\infty }(I)}\leq \epsilon \left\Vert \frac{df}{
dx}\right\Vert _{L^{2}(I)}+C(\epsilon )\left\Vert f\right\Vert _{L^{2}(I)}.
\label{pre11}
\end{equation}
\end{lemma}
\begin{proof}
Since $H^{1}(I)\hookrightarrow C(\overline{I})$, then $f\in C(
\overline{I})$ and so does $f^{2}$, so for any $y,\ z\in I$, $y<z$,
we have
that
\begin{equation*}
\begin{array}{r}
f^{2}(z)-f^{2}(y)=\displaystyle\int_{y}^{z}\frac{df^{2}}{dx}dx=2\displaystyle
\int_{y}^{z}f\frac{df}{dx}dx \\
\\
\leq \epsilon ^{2}\displaystyle\int_{I}\left\vert \frac{df}{dx}\right\vert
^{2}dx+\epsilon ^{-2}\int_{I}f^{2}dx\text{.}
\end{array}
\end{equation*}
Since exists $x^{\ast }\in I$ such that $f^{2}(x^{\ast })=\min_{x\in
I}f^{2}(x)$, then letting $y=x^{\ast }$ it follows that
\begin{equation*}
f^{2}(y)\leq \frac{1}{\left\vert I\right\vert }\int_{I}f^{2}dx
\end{equation*}
Substituting this estimate the result follows.
\end{proof}
The next result is a rewording of Lemma 2.4 of \cite{SE}. Let
$F^{1}(R)$ be the closure of $C^{\infty }-$ functions with respect
to the $H_{r}^{1}(0,R)$-norm and with the property that their with
first derivative vanishes at $r=0$.
\begin{lemma}
\label{lem2} If $v\in F^{1}(R),$ then for all $0<a<R,$
1)\ $v\in H^{1}(a,R).$
2)\ There exists an arbitrarily small number $\epsilon $ and a positive
(possibly large) constant $C(\epsilon )$, both depending on $a$, such that
\begin{equation}
\left\Vert v\right\Vert _{L^{\infty }(a,R)}\leq \epsilon \left\Vert \frac{dv
}{dr}\right\Vert _{L_{r}^{2}(0,R)}+C(\epsilon )\left\Vert v\right\Vert
_{L_{r}^{2}(0,R)}. \label{pre12}
\end{equation}
\end{lemma}
\begin{proof}
To prove 1) we note that for any $v\in F^{1}(R)$,
\begin{equation}
\begin{array}{l}
\left\Vert v\right\Vert _{H_{r}^{1}(0,R)}^{2}=\displaystyle
\int_{0}^{R}v^{2}r^{2}dr+\displaystyle\int_{0}^{R}\left\vert \frac{dv}{dr}
\right\vert ^{2}r^{2}dr \\
\\
\geq \displaystyle\int_{a}^{R}v^{2}r^{2}dr+\displaystyle\int_{a}^{R}\left
\vert \frac{dv}{dr}\right\vert ^{2}r^{2}dr \\
\\
\geq a^{2}\displaystyle\int_{a}^{R}v^{2}dr+\displaystyle\int_{a}^{R}\left
\vert \frac{dv}{dr}\right\vert ^{2}dr=a^{2}\left\Vert v\right\Vert
_{H^{1}(a,R)}^{2}.
\end{array}
\label{pre13}
\end{equation}
So, any sequence $\{v_{n}\}$ that converges with respect to the $
H_{r}^{1}(0,R)-$norm also converges with respect to the $H^{1}(a,R)-$norm.
As for the point 2), we notice that from (\ref{pre11}) and (\ref{pre13}) it
readily follows (\ref{pre12}).
\end{proof}
\begin{lemma} \label{lem3}
For each $(x,t)\in D_{2}\times [0,T_{\mathrm{end}}]$ we
have the following estimates.
\begin{equation}
\begin{array}{l}
\left\vert i_{0}-i_{0h}\right\vert \leq C\left\vert
u-u_{h}\right\vert
+C\left\vert v_{s}-v_{hs}\right\vert , \\
\\
\left\vert \ln u-\ln u_{h}\right\vert \leq C\left\vert u-u_{h}\right\vert ,
\\
\\
\left\vert \overline{U}(v_{s})-\overline{U}(v_{sh})\right\vert \leq
C\left\vert v_{s}-v_{sh}\right\vert .
\end{array}
\label{pre10.1}
\end{equation}
\end{lemma}
\begin{proof}
Noting that the functions $x\rightarrow \sqrt{x},\ x\rightarrow
\sqrt{1-x},\ x\rightarrow \ln x$ and $x\rightarrow
\overline{U}(x)$ are smooth bounded and Lipschitz functions in any bounded interval $
[a,b],\ 0<a<b$, and that the composition and multiplication of
bounded Lipschitz functions results in a Lipschitz function, then
the estimates follow. The constant $C$ in (\ref{pre10.1}) depends on
the constants $P$ and $Q$ of (\ref{estK1}).
\end{proof}
\begin{lemma}
\label{lem4}Let us consider $J$ and its approximate $J_{h}$, then
for a.e. $t\in [0,T_{end}]$ there exists a
positive constant $C$ such that
\begin{equation}
\begin{array}{r}
\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}\leq C\left\{ \left\Vert
\phi _{2}(t)-\phi _{2h}(t)\right\Vert _{L^{2}(D_{2})}^{2}+\left\Vert \phi
_{1}(t)-\phi _{1h}(t)\right\Vert _{L^{2}(D_{1})}^{2}\right. \\
\\
\left. +\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}^{2}\right\} .
\end{array}
\label{pre14}
\end{equation}
\end{lemma}
\begin{proof}
Recalling the expressions for $J$, see (\ref{i0})-(\ref{F}), and
$J_{h}$, see (\ref{num6}), using the assumption \textbf{A4} and the
bounds (\ref{estK1}) and (\ref{estK2}),
we have that for all $(x,t)\in D_{2}\times \lbrack 0,T_{
\mathrm{end}}]$
\begin{equation*}
\begin{array}{l}
\left\vert J-J_{h}\right\vert =\left\vert a_{2}i_{0}\sinh (\beta \eta
)-a_{2}i_{0h}\sinh (\beta \eta _{h})\right\vert \\
\\
\leq \left\vert a_{2}i_{0}\left( \sinh (\beta \eta )-\sinh (\beta \eta
_{h})\right) \right\vert +\left\vert a_{2}\left( i_{0}-i_{0h}\right) \sinh
(\beta \eta _{h})\right\vert \\
\\
\leq C\left\vert \sinh (\beta \eta )-\sinh (\beta \eta
_{h})\right\vert +\left\Vert a_{2}\sinh (\beta \eta _{h})\right\Vert
_{L^{\infty }(D_{2}\times [0,T_{end}])}\left\vert
i_{0}-i_{0h}\right\vert \\
\\
\leq C\left\vert \sinh (\beta \eta )-\sinh (\beta \eta
_{h})\right\vert +C\left\vert i_{0}-i_{0h}\right\vert.
\end{array}
\end{equation*}
where due to the bounds (\ref{estK1}) and (\ref{estK2}) the
constants $C=C(P,Q,K)$. Now, by virtue of the mean value theorem
there exists $z\in (\eta ,\eta _{h})$
such that
\begin{equation*}
\left\vert \sinh (\beta \eta )-\sinh (\beta \eta _{h})\right\vert
\leq \beta \left\vert \cosh (z)\right\vert \left\vert \eta -\eta
_{h}\right\vert \leq C\left\vert \eta -\eta _{h}\right\vert,
\end{equation*}
and resorting again to (\ref{estK1}) and (\ref{estK2}) it follows
that
\begin{equation*}
\left\Vert z\right\Vert _{L^{\infty }(D_{2}\times \lbrack 0,T_{\mathrm{end}
}])}\leq \left\Vert \eta \right\Vert _{L^{\infty }(D_{2}\times \lbrack 0,T_{
\mathrm{end}}])}+\left\Vert \eta _{h}\right\Vert _{L^{\infty }(D_{2}\times
\lbrack 0,T_{\mathrm{end}}])}\leq C
\end{equation*}
Hence, applying Lemma \ref{lem3} yields
\begin{equation*}
\left\vert J-J_{h}\right\vert \leq C\left( \left\vert \phi _{2}-\phi
_{2h}\right\vert +\left\vert \phi _{1}-\phi _{1h}\right\vert +\left\vert
u-u_{h}\right\vert +\left\vert v_{s}-v_{sh}\right\vert \right)
\end{equation*}
From this estimate it follows (\ref{pre14}).
\end{proof}
\subsection{Error estimates for the potentials}
To estimate the error for the potentials $\phi _{1}$ and $\phi _{2}$
is convenient to introduce the spaces $V:=H^{1}(D_{1})\times
H^{1}(D_{2}):=\left\{ w=\left( w_{1},w_{2}\right) :w_{1}\in
H^{1}(D_{1}),\
w_{2}\in H^{1}(D_{2})\right\} $ and $V_{h}\subset V$, where $
V_{h}:=V_{h}^{(1)}(\overline{D}_{1})\times V_{h}^{(1)}(\overline{D}_{2})$. $
V $ is a Hilbert space with norm
\begin{equation*}
\left\Vert w\right\Vert _{V}=\left( \left\Vert w_{1}\right\Vert
_{H^{1}(D_{1})}^{2}+\left\Vert w_{2}\right\Vert _{H^{1}(D_{2})}^{2}\right)
^{1/2},
\end{equation*}
and seminorm
\begin{equation*}
\left\vert v\right\vert _{V}=\left( \left\vert w_{1}\right\vert
_{H^{1}(D_{1})}^{2}+\left\vert w_{2}\right\vert _{H^{1}(D_{2})}^{2}\right)
^{1/2}.
\end{equation*}
Considering the bilinear forms $a_{1}:H^{1}(D_{1})\times
H^{1}(D_{1})\rightarrow \mathbb{R}$ and $a_{2}:H^{1}(D_{2})\times
H^{1}(D_{2})\rightarrow \mathbb{R}$,
\begin{equation*}
\left\{
\begin{array}{l}
a_{1}(\phi _{1},\psi _{1})=\displaystyle\int_{D_{1}}\kappa
(u)\frac{d\phi _{1}}{dx}\frac{d\psi _{1}}{dx}dx,\ \ \\
\\
a_{2}(\phi _{2},\psi _{2})=\displaystyle\int_{D_{2}}\sigma \frac{d\phi _{2}}{dx}\frac{d\psi _{2}}{dx}dx,
\end{array}
\right.
\end{equation*}
we can define the bilinear form $a:V\times V\rightarrow \mathbb{R}$
as follows. Let $\Phi $ and $\Psi $ $\in V$, $\Phi =(\phi _{1},\phi
_{2})$ and
$\Psi =(\psi _{1},\psi _{2})$, then
\begin{equation*}
a(\Phi ,\Psi )=a_{1}(\phi _{1},\psi _{1})+a_{2}(\phi _{2},\psi _{2}).
\end{equation*}
Furthermore, concerning the right hand side terms of (\ref{w3}) and (\ref{w4}
), we introduce the operator $B:V\rightarrow V^{\ast }$, $V^{\ast }$ being
the dual for $V$, as
\begin{equation*}
\left\langle B(\Phi ),\Psi \right\rangle =\int_{D_{2}}J\left( \psi _{2}-\psi
_{1}\right) dx.
\end{equation*}
Hence, we can recast the equations (\ref{w3}) and (\ref{w4}) as follows.
Find $\Phi \in L^{2}(0,T_{end}; W(D_{1})\times H^{1}(D_{2}))$ such that
\begin{equation}
a(\Phi ,\Psi )+\left\langle B(\Phi ),\Psi \right\rangle =-\int_{D_{2}}g\psi
_{2}dx\ \ \forall \Psi \in V. \label{pre15}
\end{equation}
Likewise, the finite element solutions $\phi _{1h}$ and $\phi _{2h}$
that satisfy (\ref{num3}) and (\ref{num4}) respectively, can be
formulated as follows. For all $t\in [0,T_{end}]$, find $\Phi _{h}\in W_{h}(\overline{D}_{1})\times V_{h}^{(1)}(
\overline{D}_{2})$ such that
\begin{equation}
a_{h}(\Phi _{h},\Psi _{h})+\left\langle B_{h}(\Phi _{h}),\Psi
_{h}\right\rangle =-\int_{D_{2}}g\psi _{2h}dx\ \ \forall \Psi _{h}\in V_{h},
\label{eep1}
\end{equation}
where
\begin{equation}
a_{h}(\Phi _{h},\Psi _{h})=a_{1h}(\phi _{1h},\psi _{1h})+a_{2}(\phi
_{2h},\psi _{2h}), \label{eep1_1}
\end{equation}
with
\begin{equation*}
a_{1h}(\phi _{1h},\psi _{1h})=\int_{D_{1}}\kappa (u_{h})\frac{d\phi
_{1h}}{dx}\frac{d\psi _{1h}}{dx}dx,
\end{equation*}
and
\begin{equation}
\left\langle B_{h}(\Phi _{h}),\Psi _{h}\right\rangle
=\int_{D_{2}}J_{h}\left( \psi _{2h}-\psi _{1h}\right) dx. \label{eep1_2}
\end{equation}
\begin{remark} \label{solpots}
The existence and uniqueness of $\Phi$ is proven in \cite{Diaz}
under the same kind of assumptions as \textbf{A1}-\textbf{A4},
whereas in \cite{WXZ} and \cite{Kro} the existence is proven
applying Schauder Fixed Point Theorem \cite{GT} and the uniqueness
using the fact that $\phi_{1}(x) \in W(D_{1})$.
\end{remark}
As for the bilinear form $a$ and the operator $B$, we have the
following result.
\begin{lemma}
\label{lem5} Assuming that \textbf{A1-A4} hold, we have that: (i)\
the
bilinear form $a$ is continuous, (ii)\ the operator $B$ is monotone, i.e.,
\begin{equation}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Phi -\widehat{\Phi
}\right\rangle \geq 0, \label{pre16}
\end{equation}
bounded and continuous in the sense that for all $\Psi \in V$ there exists a
constant $C$ such that
\begin{equation}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Psi \right\rangle \leq
C\left( \left\Vert \Phi -\widehat{\Phi }\right\Vert _{V}\right)
\left( \left\Vert \psi _{1}\right\Vert _{L^{2}(D_{1})}+\left\Vert
\psi _{2}\right\Vert _{L^{2}(D_{2})}\right) . \label{pre17}
\end{equation}
\end{lemma}
\begin{proof}
It is easy to prove the continuity of the bilinear form $a$ if one takes
into account the regularity assumption\ \textbf{A2}. To prove (\ref{pre16})
we note that
\begin{equation*}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Phi -\widehat{\Phi }\right\rangle
=\int_{D_{2}}a_{2}i_{0}\left( \sinh (\beta \eta )-\sinh (\beta \widehat{\eta
})\right) \left( \left( \phi _{2}-\widehat{\phi }_{2}\right) -\left( \phi
_{1}-\widehat{\phi }_{1}\right) \right) dx,
\end{equation*}
where $\eta =\phi _{2}-\phi _{1}-\alpha \ln u-\overline{U}$ and $\widehat{
\eta }=\widehat{\phi }_{2}-\widehat{\phi }_{1}-\alpha \ln
u-\overline{U}$. Since for all $x\in D_{2}$ $a_{2}i_{0}>0$, then by
virtue of \textbf{A4} we can choose a constant $C(P,Q)$ such that
for all $x\in D_{2}$ $C(P,Q)\leq a_{2}i_{0}$, and by the mean value
theorem $\sinh (\beta \eta )-\sinh (\beta \widehat{\eta })\geq \beta
(\eta -\widehat{\eta })$, then one readily
obtains
\begin{equation}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Phi -\widehat{\Phi }\right\rangle
\geq C\int_{D_{2}}\left( \left( \phi _{2}-\phi _{1}\right) -\left( \widehat{
\phi }_{2}-\widehat{\phi }_{1}\right) \right) ^{2}dx\geq 0.
\label{pre17.1_1}
\end{equation}
To prove that $B$ is bounded we notice that for all $\Phi \in V$
\begin{equation*}
\left\langle B(\Phi ),\Phi \right\rangle \leq \int_{D_{2}}\left\vert
J\right\vert \times \left\vert \phi _{2}-\phi _{1}\right\vert
dx=\int_{D_{2}}\left\vert a_{2}i_{0}(\sinh (\beta \eta )\right\vert \times
\left\vert \phi _{2}-\phi _{1}\right\vert dx,
\end{equation*}
but $\left\vert J\right\vert $ is bounded by virtue of \textbf{A4},
then using the Cauchy-Schwarz inequality it readily follows that
there exists a
bounded positive constant $C$ such that
\begin{equation*}
\left\langle B(\Phi ),\Phi \right\rangle \leq C\left\Vert \Phi \right\Vert
_{V},
\end{equation*}
so $B$ is bounded. To prove that $B$ is continuous, we again notice that
\begin{equation*}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Psi \right\rangle \leq
\int_{D_{2}}\left\vert a_{2}i_{0}(\sinh (\beta \eta )-\sinh (\beta \widehat{
\eta }))\right\vert \times \left\vert \psi _{2}-\psi _{1}\right\vert dx,
\end{equation*}
so, arguing as in the proof of Lemma \ref{lem4} we have that there exists a
positive constant $C$ such that
\begin{equation*}
\left\vert a_{2}i_{0}(\sinh (\beta \eta )-\sinh (\beta \widehat{\eta }
))\right\vert \leq C\left\vert \eta -\widehat{\eta }\right\vert =C\left\vert
\left( \phi _{2}-\widehat{\phi }_{2}\right) -\left( \phi _{1}-\widehat{\phi }
_{1}\right) \right\vert .
\end{equation*}
Substituting this estimate in the above inequality and making use of the
Cauchy-Schwarz inequality it follows that
\begin{equation*}
\begin{array}{c}
\left\langle B(\Phi )-B(\widehat{\Phi }),\Psi \right\rangle \leq C\left(
\left\Vert \phi _{2}-\widehat{\phi }_{2}\right\Vert
_{L^{2}(D_{2})}+\left\Vert \phi _{1}-\widehat{\phi }_{1}\right\Vert
_{L^{2}(D_{1})}\right) \left( \left\Vert \psi _{2}\right\Vert
_{L^{2}(D_{2})}+\left\Vert \psi _{1}\right\Vert _{L^{2}(D1)}\right) \\
\\
\leq C\left\Vert \Phi -\widehat{\Phi }\right\Vert _{V}\left( \left\Vert \psi
_{2}\right\Vert _{L^{2}(D_{2})}+\left\Vert \psi _{1}\right\Vert
_{L^{2}(D1)}\right) .
\end{array}
\end{equation*}
\end{proof}
\begin{corollary}
\label{cor1} The discrete bilinear form $a_{h}$ defined in (\ref{eep1_1}) is
continuous. The discrete operator $B_{h}$ defined in (\ref{eep1_2}) is
monotone, bounded and continuous.
\end{corollary}
\begin{theorem}
\label{Theorem 1} For a.e. $t\in \lbrack 0,T_{\mathrm{end}}]$, let
the solution of (\ref{pre15}), $\Phi (t)=\left( \phi _{1}(t),\phi
_{2}(t)\right)$, be in $H^{2}(D_{1})\times H^{2}(D_{2})$. There
exists a constant $C$
independent of $h$ such that
\begin{equation}
\begin{array}{c}
\left\Vert \Phi (t)-\Phi _{h}(t)\right\Vert _{V}^{2}\leq C\left(
h^{2}(\left\Vert \phi _{1}(t)\right\Vert _{H^{2}(D_{1})}^{2}+\left\Vert \phi
_{2}(t)\right\Vert _{H^{2}(D_{2})}^{2})\right. \\
\\
\left. +\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}^{2}\right) .
\end{array}
\label{eep}
\end{equation}
\end{theorem}
\begin{proof}
Setting $\Psi =\Psi _{h}$ in (\ref{pre15}) and subtracting (\ref{eep1})
yields
\begin{equation*}
a(\Phi ,\Psi _{h})-a_{h}(\Phi _{h},\Psi _{h})+\left\langle B(\Phi )-B\left(
\Phi _{h}\right) ,\Psi _{h}\right\rangle =-\left\langle B(\Phi
_{h})-B_{h}\left( \Phi _{h}\right) ,\Psi _{h}\right\rangle .
\end{equation*}
Noting that
\begin{equation*}
a(\Phi ,\Psi _{h})-a_{h}(\Phi _{h},\Psi _{h})=a_{h}(\Phi -\Phi
_{h},\Psi _{h})+\int_{D_{1}}\left( \kappa (u)-\kappa (u_{h})\right)
\frac{\partial \phi _{1}}{\partial x}\frac{d\psi _{1h}}{dx}dx,
\end{equation*}
it follows that
\begin{equation}
\begin{array}{r}
a_{h}(\Phi -\Phi _{h},\Psi _{h})+\left\langle B(\Phi )-B\left( \Phi
_{h}\right) ,\Psi _{h}\right\rangle =\left\langle B_{h}(\Phi
_{h})-B\left( \Phi
_{h}\right) ,\Psi _{h}\right\rangle \\
\\
-\displaystyle\int_{D_{1}}\left( \kappa (u)-\kappa (u_{h})\right) \frac{\partial \phi _{1}}{\partial x}\frac{d\psi _{1h}}{dx}dx.
\end{array}
\label{eep2}
\end{equation}
To estimate the terms of this expression we choose $\Psi
_{h}=I_{h}^{(1)}\Phi -\Phi _{h}=$ $\left( I_{h}^{(1)}\phi _{1}-\phi
_{1h},I_{h}^{(1)}\phi _{2}-\phi _{2h}\right) $, $I_{h}^{(1)}$ being
the Lagrange interpolant on
$V_{h}=V_{h}^{(1)}(\overline{D}_{1})\times
V_{h}^{(1)}(\overline{D}_{2})$, this means that $ I_{h}^{(1)}\phi
_{1}\in V_{h}^{(1)}(\overline{D}_{1})$ and $I_{h}^{(1)}\phi _{2}\in
V_{h}^{(1)}(\overline{D}_{2})$. For convenience, we shall split the
expression for $\Psi _{h}$ as
\begin{equation*}
\Psi _{h}=\left( \Phi -\Phi _{h}\right) +\left( I_{h}^{(1)}\Phi
-\Phi \right) .
\end{equation*}
Replacing this expression for $\Psi _{h}$ in (\ref{eep2}) we have that
\begin{equation}
\left\{
\begin{array}{l}
a_{h}(\Phi -\Phi _{h},\Phi -\Phi _{h})+\left\langle B(\Phi )-B\left(
\Phi
_{h}\right) ,\Phi -\Phi _{h}\right\rangle \\
\\
=a_{h}(\Phi -\Phi _{h},\Phi -I_{h}^{(1)}\Phi )+\left\langle B(\Phi
)-B\left( \Phi
_{h}\right) ,\Phi -I_{h}^{(1)}\Phi \right\rangle \\
\\
+\left\langle B_{h}(\Phi _{h})-B\left( \Phi _{h}\right) ,\left( \Phi -\Phi
_{h}\right) +\left( I_{h}^{(1)}\Phi -\Phi \right) \right\rangle \\
\\
+\displaystyle\int_{D_{1}}\left( \kappa (u)-\kappa (u_{h})\right)
\frac{\partial \phi _{1}}{\partial x}\frac{\partial \left( \left(
\phi _{1}-\phi
_{1h}\right) +\left( I_{h}^{(1)}\phi _{1}-\phi _{1h}\right) \right) }{\partial x}
dx.
\end{array}
\right. \label{eep3}
\end{equation}
We bound the terms of (\ref{eep3}). We start by showing that there
exists a positive constant $\alpha$ such that the term on the left
hand side satisfies
\begin{equation}
a_{h}(\Phi -\Phi _{h},\Phi -\Phi _{h})+\left\langle B(\Phi )-B\left(
\Phi _{h}\right) ,\Phi -\Phi _{h}\right\rangle \geq \alpha
\left\Vert \Phi -\Phi _{h}\right\Vert _{V}^{2}. \label{eep4}
\end{equation}
To do so we note that by virtue of (\ref{pre17.1_1})
\begin{equation}
\begin{array}{l}
a_{h}(\Phi -\Phi _{h},\Phi -\Phi _{h})+\left\langle B(\Phi )-B\left(
\Phi
_{h}\right) ,\Phi -\Phi _{h}\right\rangle \\
\\
\geq a_{1h}(\phi _{1}-\phi _{1h},\phi _{1}-\phi _{1h})+a_{2}(\phi
_{2}-\phi
_{2h},\phi _{2}-\phi _{2h}) \\
\\
+C\displaystyle\int_{D_{2}}\left( \left( \phi _{2}-\phi _{1}\right)
-\left( \phi
_{2h}-\phi _{1h}\right) \right) ^{2}dx.
\end{array}
\label{eep3.1}
\end{equation}
Since $\phi _{1}-\phi _{1h}\in W(D_{1})$, we can use \textbf{A2 }and
Poincar\'{e}-Wirtinger inequality to bound $a_{1}(\phi _{1}-\phi
_{1h},\phi _{1}-\phi _{1h})$ from below as
\begin{equation*}
a_{1h}(\phi _{1}-\phi _{1h},\phi _{1}-\phi _{1h})\geq
c_{1}\left\Vert \phi _{1}-\phi _{1h}\right\Vert _{H^{1}(D_{1})}^{2},
\end{equation*}
where the constant $c_{1}=\kappa _{0}\left( 1+C_{P}\right) ^{-1}$,
$C_{P}$
being the constant of the Poincar\'{e}-Wirtinger inequality; using again \textbf{
A2}, we bound\textbf{\ }the term $a_{2}(\phi _{2}-\phi _{2h},\phi _{2}-\phi
_{2h})$ as
\begin{equation*}
a_{2}(\phi _{2}-\phi _{2h},\phi _{2}-\phi _{2h})\geq \sigma _{0}\left\vert
\phi _{2}-\phi _{2h}\right\vert _{H^{1}(D_{2})}^{2}.
\end{equation*}
Applying Young inequality we find that there exists a constant
$\gamma \in \left( 0,1\right) $ such that
\begin{equation*}
\begin{array}{l}
C\displaystyle\int_{D_{2}}\left( \left( \phi _{2}-\phi _{1}\right)
-\left( \phi
_{2h}-\phi _{1h}\right) \right) ^{2}dx. \\
\\
\geq C\left[ (1-\frac{4}{\gamma })\left\Vert \phi _{1}-\phi
_{1h}\right\Vert _{L^{2}(D_{2})}^{2}+(1-\gamma )\left\Vert \phi
_{2}-\phi _{2h}\right\Vert
_{L^{2}(D_{2})}^{2}\right] \\
\\
\geq C\left[ (1-\frac{4}{\gamma })\left\Vert \phi _{1}-\phi
_{1h}\right\Vert _{L^{2}(D_{1})}^{2}+(1-\gamma )\left\Vert \phi
_{2}-\phi _{2h}\right\Vert
_{L^{2}(D_{2})}^{2}\right].
\end{array}
\end{equation*}
Now, we can choose the constants $C$ and $\gamma $ such that $
c_{2}=c_{1}+C(1-\frac{4}{\gamma })>0$, and substitute these bounds in (\ref
{eep3.1}) to obtain the inequality (\ref{eep4}), where $\alpha =\min
(c_{1},c_{2},\sigma _{0},(1-\gamma )C)$. Next, we bound the terms on
the right hand side. By continuity of the bilinear form and
Young inequality, we find that there exists a small positive number $
\epsilon _{1}$ and a constant $C(\epsilon _{1})$ such that
\begin{equation}
\begin{array}{r}
a_{h}(\Phi -\Phi _{h},\Phi -I_{h}^{(1)}\Phi )\leq C\left\Vert \Phi
-\Phi
_{h}\right\Vert _{V}\left\Vert \Phi -I_{h}^{(1)}\Phi \right\Vert _{V} \\
\\
\leq \epsilon _{1}\left\Vert \Phi -\Phi _{h}\right\Vert _{V}^{2}+C(\epsilon
_{1})\left\Vert \Phi -I_{h}^{(1)}\Phi \right\Vert _{V}^{2}.
\end{array}
\label{eep5}
\end{equation}
To bound $\left\langle B(\Phi )-B\left( \Phi _{h}\right) ,\Phi
-I_{h}^{(1)}\Phi \right\rangle $ we note that
\begin{equation*}
\left\langle B(\Phi )-B\left( \Phi _{h}\right) ,\Phi
-I_{h}^{(1)}\Phi \right\rangle \leq \int_{D_{2}}\left\vert
a_{2}i_{0}\right\vert \left\vert \int_{\beta \eta _{h}}^{\beta \eta
}\cosh \xi d\xi \right\vert \left\vert \Phi -I_{h}^{(1)}\Phi
\right\vert dx.
\end{equation*}
By virtue of (\ref{estK1}) and (\ref{estK2}) and the mean value theorem for
the integral
\begin{equation*}
\begin{array}{l}
\left\langle B(\Phi )-B\left( \Phi _{h}\right) ,\Phi
-I_{h}^{(1)}\Phi \right\rangle \leq
C\displaystyle\int_{D_{2}}\left\vert \eta -\eta
_{h}\right\vert \left\vert \Phi -I_{h}^{(1)}\Phi \right\vert dx \\
\\
\leq C\displaystyle\int_{D_{2}}\left( \left\vert \phi _{2}-\phi
_{2h}\right\vert +\left\vert \phi _{1}-\phi _{1h}\right\vert +\left\vert
u-u_{h}\right\vert +\left\vert v_{s}-v_{sh}\right\vert \right) \left\vert
\Phi -I_{h}^{(1)}\Phi \right\vert dx.
\end{array}
\end{equation*}
Applying Young inequality yields
\begin{equation}
\begin{array}{c}
\left\langle B(\Phi )-B\left( \Phi _{h}\right) ,\Phi
-I_{h}^{(1)}\Phi \right\rangle \leq \epsilon _{2}\left( \left\Vert
\Phi -\Phi _{h}\right\Vert
_{V}^{2}+\left\Vert u-u_{h}\right\Vert _{L^{2}(D_{1})}^{2}\right. \\
\\
\left. +\left\Vert v_{s}-v_{sh}\right\Vert _{L^{2}(D_{2})}^{2}\right)
+C(\epsilon _{2})\left\Vert \Phi -I_{h}^{(1)}\Phi \right\Vert _{V}^{2},
\end{array}
\label{eep6}
\end{equation}
where $\epsilon _{2}$ is a small positive number and $C(\epsilon _{2})$ is a
constant. Next, noting that by virtue of (\ref{estK1}) and (\ref{estK2}) $
\left\vert \sinh (\beta \eta _{h})\right\vert $ is bounded in $D_{2}$, then
\begin{equation*}
\left\langle B_{h}(\Phi _{h})-B\left( \Phi _{h}\right) ,\left( \Phi
-\Phi _{h}\right) +\left( I_{h}^{(1)}\Phi -\Phi \right)
\right\rangle \leq C\int_{D_{2}}a_{2}\left\vert
i_{0}-i_{0h}\right\vert \left\vert \left( \Phi -\Phi _{h}\right)
+\left( I_{h}^{(1)}\Phi -\Phi \right) \right\vert dx.
\end{equation*}
Again, using Lemma \ref{lem3} and Young inequality we obtain that there
exist a small number $\epsilon _{3}$ and a constant $C(\epsilon _{3})$ such
that
\begin{equation}
\begin{array}{l}
\left\langle B_{h}(\Phi _{h})-B\left( \Phi _{h}\right) ,\left( \Phi
-\Phi _{h}\right) +\left( I_{h}^{(1)}\Phi -\Phi \right)
\right\rangle \leq \epsilon
_{3}\left\Vert \Phi -\Phi _{h}\right\Vert _{V}^{2} \\
\\
+\epsilon _{3}\left( \left\Vert u-u_{h}\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert v_{s}-v_{sh}\right\Vert
_{L^{2}(D_{2})}^{2}\right) +C(\epsilon _{3})\left\Vert \Phi
-I_{h}^{(1)}\Phi
\right\Vert _{V}^{2}.
\end{array}
\label{eep7}
\end{equation}
To bound the last term on the right hand side of (\ref{eep3}) we note that $
\left\Vert \frac{\partial \phi _{1}}{\partial x}\right\Vert
_{L^{\infty }(D_{1})}$ is bounded and by virtue of assumption
\textbf{A4}, and by the mean value theorem, $\left\vert \left(
\kappa (u)-\kappa (u_{h})\right) \right\vert \leq C\left\vert
u-u_{h}\right\vert $, then\textbf{\ }it follows
that
\begin{equation}
\begin{array}{l}
\displaystyle\int_{D_{1}}\left( \kappa (u)-\kappa (u_{h})\right)
\frac{\partial \phi _{1}}{\partial x}\frac{\partial \left( \left(
\phi _{1}-\phi _{1h}\right) +\left( I_{h}^{(1)}\phi _{1}-\phi
_{1h}\right) \right)
}{\partial x}dx \\
\\
\leq C\displaystyle\int_{D_{1}}\left\vert u-u_{h}\right\vert
\left\vert \frac{\partial \left( \left( \phi _{1}-\phi _{1h}\right)
+\left( I_{h}^{(1)}\phi
_{1}-\phi _{1h}\right) \right) }{\partial x}\right\vert dx \\
\\
\leq \epsilon _{4}\left\Vert \Phi -\Phi _{h}\right\Vert _{V}^{2}+C(\epsilon
_{4})\left( \left\Vert u-u_{h}\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
\Phi -I_{h}^{(1)}\Phi \right\Vert _{V}^{2}\right) .
\end{array}
\label{eep8}
\end{equation}
Letting$\ \epsilon _{1}+\cdots +\epsilon _{4}=\alpha /2$ and noting that,
see (\ref{pre4.1}),
\begin{equation*}
\begin{array}{c}
\left\Vert \Phi -I_{h}^{(1)}\Phi \right\Vert _{V}^{2}=\left\Vert
\phi _{1}-I_{h}^{(1)}\phi _{1}\right\Vert
_{H^{1}(D_{1})}^{2}+\left\Vert \phi
_{2}-I_{h}^{(1)}\phi _{2}\right\Vert _{H^{1}(D_{2})}^{2} \\
\\
\leq Ch^{2}\left( \left\Vert \phi _{1}\right\Vert
_{H^{2}(D_{1})}^{2}+\left\Vert \phi _{2}\right\Vert
_{H^{2}(D_{2})}^{2}\right) .
\end{array}
\end{equation*}
the estimate (\ref{eep}) follows from (\ref{eep3})-(\ref{eep8}).
\end{proof}
\subsection{Error estimates for the concentrations $u(x,t)$ and $v(x;r,t)$}
{We wish to estimate $u(x,t)-u_{h}(x,t)$ and $v(x;r,t)-v_{h\Delta
r}(x;r,t)$ in the $L^{2}-$norm assuming that both $u(x,t)$ and
$v(x;r,t)$ are as
regular as required. Following the standard approach, we decompose $
u-u_{h}$ as
\begin{equation} \label{uv1}
u-u_{h}=(u-P_{1}u)+(P_{1}u-u_{h})\equiv \rho _{u}+\theta _{u},
\end{equation}
where $P_{1}$ is the elliptic projector defined in (\ref{pre3}), and
note that $\theta _{u}(x,t)\in V_{h}^{(1)}(\overline{D}_{1})$. To
carry out a decomposition of this kind for $v(x;r,t)-v_{h\Delta
r}(x;r,t)$, at first we can try using the extended elliptic
projector $P_{1}^{r}:H_{\ \ r}^{1,1}(D_{2}\times (0,R_{s}(\cdot
)))\rightarrow L^{2}(D_{2})\otimes V_{\Delta r}^{(1)}[0,R_{s}(\cdot
)]$ defined in (\ref{pre8}) and assume that for all $t$,
$v(x;r,t)\in H_{\ \ r}^{1,1}(D_{2}\times (0,R_{s}(\cdot )))$, then
we find that for a. e. $x\in D_{2}$
\begin{equation*}
P_{1}^{r}v(x;r,t)=\sum_{j=1}^{M}P_{1}^{r}v(x;r_{j},t)\alpha _{j}(r),
\end{equation*}
here, $M$ denotes the number of mesh points in $[0,R_{s}(\cdot )]$,
$\left\{ \alpha _{j}(r)\right\} _{j=1}^{M}$ the nodal basis of the
linear finite element space $V_{\Delta r}^{(1)}[0,R_{s}(\cdot
)]\subset H_{r}^{1}(0,R_{s}(\cdot ))$ and the function
$P_{1}^{r}v(x;r_{j},t)\in
L^{2}(D_{2})$; so, in general, $P_{1}^{r}v(x;r,t)$ is not in $V_{h\Delta r}(\overline{D}
_{3})$ and, consequently, it does not make sense to use $P_{1}^{r}v(x;r,t)-v_{h
\Delta r}(x;r,t)$ for such type of decomposition; however, recalling
the interpolant $I_{0}^{x}:H_{\ \ r}^{1,1}(D_{2}\times (0,R(\cdot
)))\rightarrow V_{h}^{(0)}(\overline{D}_{2})\otimes
H_{r}^{1}(0,R(\cdot ))$ defined in (\ref{pre4.1}) and further
assuming that $P_{1}^{r}v(x;r_{j},t)\in H^{1}(D_{2})$, then it
follows that $P_{1}^{r}v(x;r,t)\in H_{\ \
r}^{1,1}(D_{2}\times (0,R(\cdot )))$ and, therefore, we can define $
I_{0}^{x}P_{1}^{r}v(x;r,t)$ as
\begin{equation*}
I_{0}^{x}P_{1}^{r}v(x;r,t)=\sum_{l=1}^{M_{2}}
\sum_{j=1}^{M^{(l)}}P_{1}^{r}v(x_{l};r_{j},t)\alpha
_{j}^{(l)}(r)\chi _{l}(x);
\end{equation*}
this expression implies that $I_{0}^{x}P_{1}^{r} v(x;r,t)\in
V_{h\Delta r}(\overline{D}_{3})$, so it makes sense to set
\begin{equation*}
\theta _{v}(x;r,t)=I_{0}^{x}P_{1}^{r}v(x;r,t)-v_{h\Delta r}(x;r,t).
\end{equation*}
Now, using again the extended $P_{1}^{r}$ elliptic projector we
define
\begin{equation*}
\rho _{v}(x;r,t)=v(x;r,t)-P_{1}^{r}v(x;r,t)\in H_{\ \
r}^{1,1}(D_{2}\times (0,R(\cdot ))),
\end{equation*}
and consequently
\begin{equation*}
I_{0}^{x}\rho
_{v}(x;r,t)=I_{0}^{x}v(x;r,t)-I_{0}^{x}P_{1}^{r}v(x;r,t).
\end{equation*}
Then, from all these considerations we can write that
\begin{equation} \label{uv2}
v(x;r,t)-v_{h\Delta r}(x;r,t)=(v-I_{0}^{x}v)(x;r,t)+I_{0}^{x}\rho
_{v}(x;r,t)+\theta _{v}(x;r,t).
\end{equation}
From (\ref{uv1}) and (\ref{uv2}) it follows that
\begin{equation*}
\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}\leq \left\Vert \rho
_{u}(t)\right\Vert _{L^{2}(D_{1})}+\left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})},
\end{equation*}
and
\begin{equation*}
\begin{array}{r}
\left\Vert v(t)-v_{h\Delta r}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot )))}\leq \left\Vert
v(t)-I_{0}^{x}v(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot )))} \\
\\
+\left\Vert I_{0}^{x}\rho _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot )))}+\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot )))}.
\end{array}
\end{equation*}
The estimates for $\rho _{u}$ and $\rho _{v}$ are given in (\ref{pre5}) and (
\ref{pre10}) respectively, i.e.,
\begin{equation} \label{uv2_1}
\left\{
\begin{array}{l}
\left\Vert \rho _{u}(t)\right\Vert _{L^{2}(D_{1})}\leq Ch^{2}\left\Vert
u(t)\right\Vert _{H^{2}(D_{1})}, \\
\\
\left\Vert \rho _{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R(\cdot
)))}\leq C\Delta r^{2}\left\Vert v(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R(\cdot )))},
\end{array}
\right.
\end{equation}
then, it remains to calculate the estimates for $\theta _{u}$ and
$\theta _{v} $; but before going into the details of such
calculations, we present
new estimates for $\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}$ and $
\left\Vert v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}^{2}$, which
depend on $\theta _{u}$ and $\theta _{v}$ respectively, and will be
useful for the subsequent part of the analysis.
\begin{lemma}
\label{lem6} Assuming that the regularity assumptions required in the
estimates hold, there exist an arbitrarily small positive number $\epsilon $
and constants $C$ and $C(\epsilon )$ independent of $h$ and $\Delta r$ such
that
\begin{equation}
\begin{array}{r}
\left\Vert v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}^{2}\leq C
h^{2}\left\vert v_{s}(t)\right\vert
_{H^{1}(D_{2})}^{2}+C(\epsilon)\Delta r^{2}\left\Vert
v(t)\right\Vert _{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot
)))}^{2} \\
\\
+C(\epsilon )\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\Vert \frac{\partial \theta _{v}(t)}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2},
\end{array}
\label{uv6.0}
\end{equation}
and
\begin{equation}
\begin{array}{c}
\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}\leq Ch^{2}\left(
\left\Vert \phi _{1}(t)\right\Vert _{H^{2}(D_{1})}^{2}+\left\Vert \phi
_{2}(t)\right\Vert _{H^{2}(D_{2})}^{2}+h^{2}\left\Vert u(t)\right\Vert
_{H^{2}(D_{1})}^{2}\right) \\
\\
+C\left\Vert v_{s}(t)-v_{sh}(t)\right\Vert
_{L^{2}(D_{2})}^{2}+C\left\Vert
\theta _{u}(t)\right\Vert _{L^{2}(D_{1})}^{2}.
\end{array}
\label{uv6}
\end{equation}
\end{lemma}
\begin{proof}
To calculate the estimate (\ref{uv6.0}) we set $v_{s}-v_{sh}=\left(
v_{s}-I_{h}^{0}v_{s}\right) +\left( I_{h}^{0}v_{s}-v_{sh}\right) $, so using
(\ref{pre4}) it follows that
\begin{equation*}
\left\Vert v_{s}(t)-v_{sh}(t)\right\Vert
_{L^{2}(D_{2})}^{2}=Ch^{2}\left\vert v_{s}(t)\right\vert
_{H^{1}(D_{2})}^{2}+2\left\Vert I_{h}^{0}v_{s}(t)-v_{sh}(t)\right\Vert
_{L^{2}(D_{2})}^{2}.
\end{equation*}
Since $I_{h}^{0}v_{s}-v_{sh}=\left( I_{h}^{0}v-v_{h\Delta r}\right)
(x;R_{s}(x),t)=\left( I_{0}^{x}\rho _{v}+\theta _{v}\right)
(x;R_{s}(x),t)$ (recalling the definition of $\theta_{v}$) then by
virtue of the definition of the $L^{2}$-norm for functions of
$V_{h}^{0}(\overline{D}_{2})$
presented in Section 3, we have that
\begin{equation*}
\left\Vert I_{h}^{0}v_{s}(t)-v_{sh}(t)\right\Vert
_{L^{2}(D_{2})}^{2}\leq 2\sum_{l=1}^{M_{2}}\widehat{h}_{l}\left(
\rho _{v}^{2}(x_{l};R_{s}(x_{l}),t)+\theta
_{v}^{2}(x_{l};R_{s}(x_{l}),t)\right) .
\end{equation*}
To estimate $\rho _{v}^{2}(x_{l};R_{s}(x_{l}),t)$ and $\theta
_{v}^{2}(x_{l};R_{s}(x_{l}),t)$ we make use of Lemma \ref{lem2}
noting that there exists a real number $a$, $0<a<R_{s}(x_{l})$, such
that $\rho _{v}^{2}(x_{l};R_{s}(x_{l}),t)\leq \left\Vert \rho
_{v}(x_{l};r,t)\right\Vert _{L^{\infty }(a,R(x_{l}))}^{2}$ and
$\theta _{v}^{2}(x_{l};R_{s}(x_{l}),t)\leq \left\Vert \theta
_{v}(x_{l};r,t)\right\Vert _{L^{\infty }(a,R(x_{l}))}^{2}$, thus by
virtue of (\ref{pre12}) it follows that there are a real number
$\epsilon$ and positive constant $C(\epsilon)>\epsilon$ such that
\begin{equation*}
\begin{array}{r}
\displaystyle\sum_{l=1}^{M_{2}}\widehat{h}_{l}\rho
_{v}^{2}(x_{l};R_{s}(x_{l}),t)\leq \displaystyle\epsilon \sum_{l=1}^{M_{2}}
\widehat{h}_{l}\int_{0}^{R_{s}(x_{l})}r^{2}\left\vert \frac{\partial
\rho
_{v}(x_{l};r,t)}{\partial r}\right\vert ^{2}dr \\
\\
+\displaystyle C(\epsilon )\sum_{l=1}^{M_{2}}\widehat{h}
_{l}\int_{0}^{R_{s}(x_{l})}r^{2}\left\vert \rho _{v}(x_{l};r,t)\right\vert
^{2}dr \\
\\
\leq C(\epsilon )\left\Vert I_{0}^{x}\rho _{v}(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{1}(0,R_{s}(\cdot )))}^{2} \\
\\
\leq C(\epsilon )\left\Vert \rho _{v}(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{1}(0,R_{s}(x)))} \\
\\
\leq C(\epsilon )\Delta r^{2}\left\Vert v(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2},
\end{array}
\end{equation*}
because by approximation theory $\left\Vert I_{0}^{x}\rho
_{v}(t)\right\Vert _{L^{2}(D_{2},H_{r}^{1}(0,R_{s}(x)))}\leq
C\left\Vert \rho _{v}(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{1}(0,R_{s}(x)))}$, this latter term
being estimated according to (\ref{pre10}). Similarly,
\begin{equation}
\begin{array}{r}
\displaystyle\sum_{l=1}^{M_{2}}\widehat{h}_{l}\theta
_{v}^{2}(x_{l};R_{s}(x_{l}),t)\leq \epsilon \displaystyle\sum_{l=1}^{M_{2}}
\widehat{h}_{l}\int_{0}^{R_{s}(x_{l})}r^{2}\left\vert \frac{\partial
\theta _{v}(x_{l};r,t)}{\partial r}\right\vert ^{2}dr \\
\\
+C(\epsilon )\displaystyle\sum_{l=1}^{M_{2}}\widetilde{h}_{l}
\int_{0}^{R_{s}(x_{l})}r^{2}\left\vert \theta _{v}(x_{l};r,t)\right\vert
^{2}dr \\
\\
\leq \epsilon \left\vert \theta _{v}(t)\right\vert
_{L^{2}(D_{2},H_{r}^{1}(0,R_{s}(\cdot )))}^{2}+C(\epsilon )\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\end{array}
\label{theta}
\end{equation}
So, putting these bounds together the result (\ref{uv6.0}) follows. To
calculate the estimate (\ref{uv6}) we notice that by virtue of (\ref{pre14})
and Theorem \ref{Theorem 1}
\begin{equation*}
\begin{array}{r}
\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}\leq C\left(
h^{2}\left(\left\Vert \phi _{1}(t)\right\Vert
_{H^{2}(D_{1})}^{2}+\left\Vert \phi
_{2}(t)\right\Vert _{H^{2}(D_{2})}^{2}\right)\right. \\
\\
\left. +\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}^{2}\right) .
\end{array}
\end{equation*}
Since $u-u_{h}=\rho _{u}+\theta _{u}$, then taking into account (\ref{pre5})
\begin{equation*}
\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}\leq Ch^{4}\left\Vert
u(t)\right\Vert _{H^{2}(D_{1})}^{2}+2\left\Vert \theta _{u}\right\Vert
_{L^{2}(D_{1})}^{2},
\end{equation*}
so the result (\ref{uv6}) follows.
\end{proof}
Next, we calculate an estimate for $\theta _{v}$. To this end, we
obtain, based on equation (\ref{w2}), the integral equation for
$I_{0}^{x}v(x;r,t)$ that will be used for that purpose. Thus, for
each one of the mesh points $\{x_{l}\}_{l=1}^{M_{2}}$ of $D_{2h}$
the equation (\ref{w2}) reads
\begin{equation}
\int_{0}^{R_{s}(x_{l})}\frac{\partial {v}^{(l)}}{\partial {t}}
w^{(l)}r^{2}dr+\int_{0}^{R_{s}(x_{l})}k_{2}\frac{\partial v^{(l)}}{\partial r
}\frac{\partial w^{(l)}}{\partial r}r^{2}dr=-\left(
R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}Jw^{(l)}(R_{s}(x))\right)|
_{x=x_{l}}, \label{I0v1}
\end{equation}
where $v^{(l)}:=v(x_{l};r,t)\in H_{r}^{1}(0,R_{s}(x_{l})$ and
$w^{(l)}:=w(x_{l};r)\in H_{r}^{1}(0,R_{s}(x_{l}))$. Using the nodal
basis functions $\left\{ \chi _{l}(x)\right\} _{l=1}^{M_{2}} $ of
the finite element space $V_{h}^{(0)}(\overline{D}_{2})$ we can write
\begin{equation*}
I_{0}^{x}{v(x;r,t)=}\sum_{l}^{M_{2}}v^{(l)}(r,t)\chi _{l}(x)\mathrm{\ and\ \
}w(x;r)=\sum_{l=1}^{M_{2}}w^{(l)}(r)\chi _{l}(x).
\end{equation*}
Now, noting that
\begin{equation*}
\int_{D_{2}}\int_{0}^{R_{s}(x)}\left( I_{0}^{x}{v}\right)
wr^{2}drdx=\sum_{l=1}^{M_{2}}\int_{D_{2}}\chi _{l}^{2}(x)\left(
\int_{0}^{R_{s}(x)}v^{(l)}(r,t)w^{(l)}(r)r^{2}dr\right) dx,
\end{equation*}
and for $x\in \widehat{e}_{l}$, $R_{s}(x)=R_{s}(x_{l})$, then it follows that
\begin{equation*}
\sum_{l=1}^{M_{2}}\int_{D_{2}}\chi _{l}^{2}(x)\left(
\int_{0}^{R_{s}(x)}v^{(l)}(r,t)w^{(l)}(r)r^{2}dr\right) dx=\sum_{l=1}^{M_{2}}
\widehat{h}_{l}\int_{0}^{R_{s}(x_{l})}v^{(l)}(r,t)w^{(l)}(r)r^{2}dr.
\end{equation*}
Hence, (\ref{I0v1}) becomes
\begin{equation}
\int_{D_{2}}\int_{0}^{R_{s}(x)}\left( \frac{\partial I_{0}^{x}{v}}{\partial {
t}}w+k_{2}\frac{\partial I_{0}^{x}v}{\partial r}\frac{\partial w}{\partial r}
\right)
r^{2}drdx=-\int_{D_{2}}I_{0}^{x}(R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}Jw(x;R_{s}(x)))dx.
\label{uv4}
\end{equation}
We proceed to formulate the equation for $\theta _{v}$. From (\ref{uv2}) it
follows that $v_{h\Delta r}=I_{0}^{x}v-(I_{0}^{x}\rho _{v}+\theta _{v})$,
then replacing this expression for $v_{h\Delta r}$ in (\ref{num2}) and using
(\ref{pre8}) and (\ref{uv4}) it follows that for all $w_{h\Delta r}\in
V_{h\Delta r}(\overline{D}_{3})$,
\begin{equation}
\begin{array}{r}
\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\left( \frac{\partial \theta _{v}
}{\partial {t}}w_{h\Delta r}+k_{2}\frac{\partial \theta _{v}}{\partial r}
\frac{\partial w_{h\Delta r}}{\partial r}\right) r^{2}drdx=\lambda
\int_{D_{2}}\int_{0}^{R_{s}(x)}I_{0}^{x}\rho _{v}w_{h\Delta r}r^{2}drdx \\
\\
-\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\frac{\partial I_{0}^{x}\rho
_{v}}{\partial {t}}w_{h\Delta r}r^{2}drdx \\
\\
-\displaystyle\int_{D_{2}}I_{h}^{0}(R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}(J(x)-J_{h}(x))w_{sh}(x))dx
\\
\\
+\displaystyle\int_{D_{2}}\left(
R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}J_{h}(x)-I_{h}^{0}(R_{s}^{2}(x)a_{2}^{-1}F^{-1}(x)J_{h}(x)\right) )w_{sh}(x)dx,
\end{array}
\label{uv5}
\end{equation}
where we have made use of the following properties of the interpolant $
I_{0}^{x}$: (i) for $w_{h\Delta r}(x,r)\in V_{h\Delta r}(\overline{D}_{3})$,
$w_{h\Delta r}(x,r)=I_{0}^{x}w_{h\Delta r}(x,r)$, and (ii) when $r=R_{s}(x)$
, we can define the function $w_{sh}(x)=w_{h\Delta r}(x,R_{s}(x))$ such that
$w_{sh}(x)=I_{0}^{x}w_{h\Delta r}(x,R_{s}(x))=I_{h}^{0}w_{sh}(x)$.\ Setting $
w_{h\Delta r}=\theta _{v}$ yields
\begin{equation}
\begin{array}{l}
\displaystyle\frac{1}{2}\frac{d}{dt}\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\displaystyle \underline{k_{2}}\left\Vert \frac{\partial
\theta _{v}(t)}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2} \\
\\
\leq \lambda \left\Vert I_{0}^{x}\rho _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
+\displaystyle\left\Vert \frac{\partial I_{0}^{x}\rho _{v}(t)}{\partial t}
\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
+C\left\Vert I_{h}^{(0)}\left( (J-J_{h})\theta _{vs}(t)\right)
\right\Vert _{L^{1}(D_{2})}+C\left\Vert
(J_{h}-I_{(0)}^{0}J_{h})\theta _{vs}(t)\right\Vert
_{L^{1}(D_{2})}\equiv \sum_{i=1}^{4}R_{i},
\end{array}
\label{uv7}
\end{equation}
where,\ $\theta _{vs}(t)=\theta _{v}(x;R_{s}(x),t)$ is the value of $\theta
_{v}$ on the surface of the sphere of radius $R_{x}(x)$ associated with the
point $\{x\}$ of $D_{2}$.
\begin{lemma}
\label{lem7} There exists a constant $C$ independent of $h$ and
$\Delta r$, but depending on $\underline{k_{1}}$ and
$\underline{k_{2}}$, such that
\begin{equation}
\begin{array}{c}
\displaystyle\frac{d}{dt}\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\displaystyle
\underline{k_{2}}\left\Vert \frac{\partial \theta _{v}(t)}{\partial
r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2} \\
\\
\leq Ch^{2}\left( \left\Vert \phi _{1}(t)\right\Vert
_{H^{2}(D_{1})}^{2}+\left\Vert \phi _{2}(t)\right\Vert
_{H^{2}(D_{2})}^{2}\right) \\
\\
+Ch^{2}\left( \left\vert v_{s}(t)\right\vert
_{H^{1}(D_{2})}^{2}+h^{2}\left\Vert u(t)\right\Vert _{H^{2}(D_{1})}^{2}+
\displaystyle\left\Vert \frac{\partial J}{\partial x}\right\Vert
_{L^{2}(D_{2})}^{2}\right) \\
\\
+C\Delta r^{2}\left( \left\Vert v(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\Delta r^{2}\displaystyle
\left\Vert \frac{\partial v(t)}{\partial t}\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) \\
\\
+C\left(\left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\right).
\end{array}
\label{uv7.1}
\end{equation}
\end{lemma}
\begin{proof}
We bound the terms $R_{1},\ldots ,R_{4}$ on the right hand side of (\ref{uv7}
). Noting that $\left\Vert I_{0}^{x}\rho _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\leq C\left\Vert \rho
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}$, then by
virtue of Young inequality and (\ref{pre10}) it follows that
\begin{equation}
R_{1}\leq C\Delta r^{4}\left\Vert v(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}+C\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\label{uv8}
\end{equation}
To bound $R_{2}$ we notice that
\begin{equation*}
\begin{array}{r}
\displaystyle\left\Vert \frac{\partial I_{0}^{x}\rho _{v}(t)}{\partial t}
\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\leq C\displaystyle
\left\Vert \frac{\partial \rho _{v}(t)}{\partial t}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
= C\displaystyle\left\Vert (I-P_{1}^{r})\frac{\partial v(t)}{\partial t}
\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
\leq C\Delta r^{2}\displaystyle\left\Vert \frac{\partial v(t)}{\partial t}
\right\Vert _{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}.
\end{array}
\end{equation*}
Hence, by Young inequality it follows that
\begin{equation}
R_{2}\leq C\Delta r^{4}\left\Vert \frac{\partial v(t)}{\partial t}
\right\Vert _{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}+C\left\Vert
\theta _{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\label{uv9}
\end{equation}
We bound the term $R_{3}$. Thus, we have that
\begin{equation*}
\begin{array}{l}
C\left\Vert I_{h}^{0}\left( (J-J_{h})\theta _{vs}(t)\right)
\right\Vert
_{L^{1}(D_{2})}=C\displaystyle\sum_{l=1}^{M_{2}}\widehat{h}_{l}\left\vert
\left( J(x_{l})-J_{h}(x_{l}\right) )\theta
_{v}(x_{l};R_{s}(x_{l}),t)\right\vert \\
\\
\leq
\displaystyle\frac{C^{2}}{2}\sum_{l=1}^{M_{2}}\widehat{h}_{l}\left(
J(x_{l})-J_{h}(x_{l}\right) )^{2}+\displaystyle\frac{1}{2}\sum_{l=1}^{M_{2}}
\widehat{h}_{l}\theta _{v}^{2}(x_{l};R_{s}(x_{l}),t) \\
\\
\leq \displaystyle\frac{C^{2}}{2}\left\Vert
I_{h}^{0}(J-J_{h})\right\Vert
_{L^{2}(D_{2})}^{2}+\displaystyle\frac{1}{2}\sum_{l=1}^{M_{2}}\widehat{h}
_{l}\theta _{v}^{2}(x_{l};R_{s}(x_{l}),t).
\end{array}
\end{equation*}
Estimating the last term on the right hand side of this inequality as we did
before in the proof of Lemma \ref{lem6}, see (\ref{theta}), and noting that $
\left\Vert I_{h}^{0}(J-J_{h})\right\Vert _{L^{2}(D_{2})}\leq C\left\Vert
J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}$, it readily follows\textbf{\ }that
\begin{equation}
R_{3}\leq C\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}+C(\epsilon
)\left\Vert \theta _{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\epsilon \left\Vert \frac{\partial \theta _{v}(t)}{\partial r}
\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}. \label{uv10}
\end{equation}
To bound $R_{4}$ we notice that $
J_{h}-I_{h}^{(0)}J_{h}=(J_{h}-J)+(J-I_{h}^{(0)}J)+I_{h}^{(0)}(J-J_{h})$,
so by the triangle inequality it follows that
\begin{equation*}
\begin{array}{l}
C\left\Vert (J_{h}-I_{h}^{(0)}J_{h})\theta _{vs}(t)\right\Vert
_{L^{1}(D_{2})}\leq C\left\Vert (J_{h}-J)\theta _{vs}(t)\right\Vert
_{L^{1}(D_{2})} \\
\\
+C\left\Vert (J-I_{h}^{(0)}J)\theta _{vs}(t)\right\Vert
_{L^{1}(D_{2})}+C\left\Vert I_{h}^{(0)}(J-J_{h})\theta
_{vs}(t)\right\Vert
_{L^{1}(D_{2})}.
\end{array}
\end{equation*}
Noting that $\left\Vert ab\right\Vert _{L^{1}(D_{2})}\leq \frac{\varepsilon
}{2}\left\Vert a\right\Vert _{L^{2}(D_{2})}^{2}+\frac{1}{2\varepsilon }
\left\Vert b\right\Vert _{L^{2}(D_{2})}^{2}$ and applying the same argument
as we have just done to bound $R_{3}$, we obtain that
\begin{equation*}
\begin{array}{r}
\left\Vert (J_{h}-I_{h}^{0}J_{h})\theta _{vs}(t)\right\Vert
_{L^{1}(D_{2})}\leq C\left( \left\Vert J-J_{h}\right\Vert
_{L^{2}(D_{2})}^{2}+\left\Vert J-I_{h}^{0}J\right\Vert
_{L^{2}(D_{2})}^{2}\right) \\
\\
+\sum_{l=1}^{M_{2}}\widehat{h}_{l}\theta
_{v}^{2}(x_{l};R_{s}(x_{l}),t).
\end{array}
\end{equation*}
We bound the last term of this inequality as we have done for
$R_{3},$ and by virtue of assumption \textbf{A3 }set $\left\Vert
J-I_{h}^{(0)}J\right\Vert
_{L^{2}(D_{2})}^{2}\leq Ch^{2}\left\Vert \frac{\partial J}{\partial x}
\right\Vert _{L^{2}(D_{2})}^{2}$. Hence,
\begin{equation}
\begin{array}{r}
R_{4}\leq C(\epsilon )\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\vert \frac{\partial \theta _{v}(t)}{\partial r}\right\vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2} \\
\\
+C\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{2})}^{2}+Ch^{2}\displaystyle
\left\Vert \frac{\partial J}{\partial x}\right\Vert _{L^{2}(D_{2})}^{2}.
\end{array}
\label{uv10_1}
\end{equation}
Letting $\epsilon=\underline{k_{2}}/8$ in (\ref{uv6}), (\ref{uv10})
and (\ref{uv10_1}),
and replacing (\ref{uv8})-(\ref{uv10_1}) in (\ref{uv7}), the result (\ref
{uv7.1}) follows.
\end{proof}
Next, we proceed to calculate an estimate for $\theta _{u}$. Thus,
subtracting (\ref{num1}) from (\ref{w1}) it readily follows that
\begin{equation*}
\int_{D_{1}}\frac{\partial \theta _{u}}{\partial {t}}w_{h}dx+
\int_{D_{1}}k_{1}\frac{\partial \theta _{u}}{\partial x}\frac{dw_{h}}{dx}
dx=\lambda \int_{D_{1}}\rho _{u}w_{h}dx-\int_{D_{1}}\frac{\partial \rho _{u}
}{\partial {t}}w_{h}dx+\int_{D_{1}}a_{1}\left( J-J_{h}\right) w_{h}dx.
\end{equation*}
Setting $w_{h}=\theta _{u}$ in this equation yields
\begin{equation*}
\begin{array}{r}
\displaystyle\frac{1}{2}\frac{d}{dt}\left\Vert \theta
_{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\underline{k_{1}}\displaystyle\left\Vert
\frac{\partial \theta _{u}(t)}{\partial x}\right\Vert
_{L^{2}(D_{1})}^{2}\leq C\displaystyle\left(
\left\Vert \rho _{u}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert \frac{
\partial \rho _{u}(t)}{\partial t}\right\Vert _{L^{2}(D_{1})}^{2}\right) \\
\\
+C\left\Vert J-J_{h}\right\Vert _{L^{2}(D_{1})}^{2}+C\left\Vert \theta
_{u}(t)\right\Vert _{L^{2}(D_{1})}^{2}.
\end{array}
\end{equation*}
By virtue of Lemma \ref{lem6} we have the following result.
\begin{lemma}
\label{lem8} There exists a constant $C$ independent of $h$ and
$\Delta r$, but depending on $\underline{k_{1}}$ and
$\underline{k_{2}}$, such that
\begin{equation}
\begin{array}{l}
\displaystyle\frac{d}{dt}\left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\displaystyle \underline{k_{1}}\left\Vert
\frac{\partial \theta
_{u}(t)}{\partial x}\right\Vert _{L^{2}(D_{1})}^{2} \\
\\
\leq C\left\{ h^{2}(\left\Vert \phi _{1}(t)\right\Vert
_{H^{2}(D_{1})}^{2}+\left\Vert \phi _{2}(t)\right\Vert
_{H^{2}(D_{2})}^{2}+\left\vert v_{s}(t)\right\vert
_{H^{1}(D_{2})}^{2})\right. \\
\\
\left. +h^{4}\left( \left\Vert u(t)\right\Vert
_{H^{2}(D_{1})}^{2}+\displaystyle\left\Vert \frac{\partial
u(t)}{\partial t}\right\Vert _{H^{2}(D_{1})}^{2}\right) +\Delta
r^{2}\left\Vert v(t)\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right\} \\
\\
+C\left( \left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\right)+\displaystyle
\frac{\underline{k_{2}}}{4} \left\Vert \frac{\partial {\theta
_{v}(t)}}{\partial r} \right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}.
\end{array}
\label{uv11}
\end{equation}
\end{lemma}
We are now in a position to establish the main result of this subsection.
\begin{theorem}
\label{Theorem 2} Let $(u,v,\phi _{1},\phi _{2})$ and $(u_{h},v_{h\Delta
r},\phi _{1h},\phi _{2h})$ be the solutions to (\ref{w1})-(\ref{w4}) and (
\ref{num1})-(\ref{num4}) respectively, with
\begin{equation*}
\left\Vert u(0)-u_{h}(0)\right\Vert _{L^{2}(D_{1})}\leq Ch^{2}\ \mathrm{and\
}\left\Vert v(0)-v_{h\Delta r}(0)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\leq C(h+\Delta r^{2}).
\end{equation*}
Furthermore, for $0\leq t\leq T_{\mathrm{end}},$ the following regularity
assumptions hold:
R1)$\ u$ and $\displaystyle \frac{\partial u}{\partial t}\in L^{2}(0,T_{
\mathrm{end}};H^{2}(D_{1})),$
\bigskip
R2) $v$ and $\displaystyle \frac{\partial v(t)}{\partial t}\in L^{2}(0,T_{
\mathrm{end}};L^{2}(D_{2};H_{r}^{2}(0,R_{s}(\cdot )))),$ and $v_{s}\in
L^{2}(0,T_{\mathrm{end}};H^{1}(D_{2})),$
\bigskip
R3) $\phi _{1}\in L^{2}(0,T_{\mathrm{end}};H^{2}(D_{1}))$, $\phi _{2}\in
L^{2}(0,T_{\mathrm{end}};H^{2}(D_{2})),$ and $J\in L^{2}(0,T_{\mathrm{end}
};H^{1}(D_{2}));$
\bigskip
then there is a constant $C(t,\underline{k_{1}},\underline{k_{2}})$ such that
\begin{equation}
\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v(t)-v_{h\Delta r}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\int_{0}^{t}\left\Vert \Phi (\tau )-\Phi _{h}(\tau )\right\Vert
_{V}^{2}d\tau \leq C(h^{2}+\Delta r^{2}). \label{uv12}
\end{equation}
\end{theorem}
\begin{proof}
Bounding $\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}$ as $
2\left( \left\Vert \rho _{u}(t)\right\Vert ^{2}+\left\Vert \theta
_{u}(t)\right\Vert ^{2}\right) $ and using the estimate of Lemma \ref{lem6}
for $\left\Vert v_{s}(t)-v_{sh}(t)\right\Vert _{L^{2}(D_{2})}$, Theorem \ref
{Theorem 1} yields
\begin{equation*}
\begin{array}{l}
\left\Vert \Phi (t)-\Phi _{h}(t)\right\Vert _{V}^{2}\leq Ch^{2}\left\vert
v_{s}(t)\right\vert _{H^{1}(D_{2})}^{2}+Ch^{4}\left\Vert u(t)\right\Vert
_{H^{2}(D_{1})}^{2}+C\left\Vert \theta _{u}\right\Vert _{L^{2}(D_{1})}^{2}
\\
\\
+C(\epsilon )\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\Vert \frac{\partial \theta _{v}(t)}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\end{array}
\end{equation*}
Considering this estimate together with those of Lemmas \ref{lem7} and \ref
{lem8} it readily follows that
\begin{equation*}
\begin{array}{r}
\displaystyle\frac{d}{dt}\left( \left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) +\left\Vert \Phi
(t)-\Phi _{h}(t)\right\Vert _{V}^{2}\leq C(h^{2}+\Delta r^{2}) \\
\\
+C\left( \left\Vert \theta _{u}(t)\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) .
\end{array}
\end{equation*}
Then, Gronwall inequality yields
\begin{equation}
\left\Vert \theta _{u}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
\theta _{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\int_{0}^{t}\left\Vert \Phi (\tau )-\Phi _{h}(\tau
)\right\Vert _{V}^{2}d\tau \leq
C(t,\underline{k_{1}},\underline{k_{2}})(h^{2}+\Delta r^{2}).
\label{uv13}
\end{equation}
The terms $\left\Vert \theta _{u}(0)\right\Vert _{L^{2}(D_{1})}^{2}$ and $
\left\Vert \theta _{v}(0)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}$ are considered to be
zero. From (\ref{uv1}), (\ref{uv2}), (\ref{uv2_1}) and (\ref{uv3}) it follows that
\begin{equation*}
\begin{array}{r}
\left\Vert u(t)-u_{h}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v(t)-v_{h\Delta r}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}\leq 2\left( \left\Vert \rho _{u}(t)\right\Vert
_{L^{2}(D_{1})}+\left\Vert \theta _{u}(t)\right\Vert _{L^{2}(D_{1})}\right)
\\
\\
+C\left( h^{2}\left\Vert
v(t)\right\Vert_{H^{1}(D_{2};L_{r}^{2}(0,R_{s}(\cdot)))} +\left\Vert
\rho _{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
))}^{2}+\left\Vert \theta
_{v}(t)\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot ))}^{2}\right) .
\end{array}
\end{equation*}
So, by combining this inequality with (\ref{uv13}) we obtain (\ref{uv12}).
\end{proof}
\section{Fully discrete model}
We consider now the fully discrete model based on the time stepping backward
Euler scheme. This scheme has been used to discretize in time the equations
of the P2D model either with finite differences \cite{dual}, finite volumes
\cite{SW}, \cite{ML}, or finite elements. For convenience, hereafter we
shall use the notation $a^{n}:=a(x,t_{n})$, where $n$ is a nonnegative
integer and $t_{n}=n\Delta t$, $\Delta t$ being a uniform time step. The
formulation of the fully discrete model is as follows. Assuming that at time
$t_{n-1}$,\ $n=1,2,\ldots ,N$, the solution $(u_{h}^{n-1},v_{h\Delta
r}^{n-1},\phi _{1h}^{n-1},\phi _{1h}^{n-1})\in V_{h}^{(1)}(\overline{D}
_{1})\times V_{h\Delta r}(\overline{D}_{3})\times W_{h}(\overline{D}
_{1})\times V_{h}^{(1)}(\overline{D}_{2})$ is known, calculate $
(u_{h}^{n},v_{h\Delta r}^{n},\phi _{1h}^{n},\phi _{1h}^{n})\in V_{h}^{(1)}(
\overline{D}_{1})\times V_{h\Delta r}(\overline{D}_{3})\times W_{h}(
\overline{D}_{1})\times V_{h}^{(1)}(\overline{D}_{2})$ as solution of the
system
\begin{equation}
\int_{D_{1}}\widetilde{\partial}
_{t}u^{n}{_{h}}w_{h}dx+\int_{D_{1}}k_{1}\frac{du_{h}^{n}}{dx}\frac{dw_{h}}{dx}dx=\int_{D_{1}}a_{1}J_{h}^{n}w_{h}dx\
\forall w_{h}\in V_{h}^{(1)}(\overline{D}_{1}). \label{fdm1}
\end{equation}
\begin{equation}
\left\{
\begin{array}{l}
\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\displaystyle\widetilde{\partial}
_{t}v^{n}_{h\Delta r}w_{h\Delta r}r^{2}drdx+\displaystyle\int_{D_{2}}
\int_{0}^{R_{s}(x)}k_{2}\displaystyle\frac{\partial v^{n}{_{h\Delta r}}}{
\partial {r}}\frac{\partial w_{h\Delta r}}{\partial r}r^{2}drdx \\
\\
=-\displaystyle\int_{D_{2}}\frac{R_{s}^{2}(x)J_{h}^{n}w_{h\Delta
r}(x,R_{s}(x))}{a_{2}(x)F}dx\ \ \forall w_{h\Delta r}\in V_{h\Delta r}(
\overline{D}_{3}).
\end{array}
\right. \label{fdm2}
\end{equation}
\begin{equation}
\int_{D_{1}}\kappa (u_{h}^{n})\frac{d\phi _{1h}^{n}}{dx}
\frac{dw_{h}}{dx}dx=\int_{D_{1}}J_{h}^{n}w_{h}dx\ \ \forall w_{h}\in
V_{h}^{(1)}(\overline{D}_{1}). \label{fdm3}
\end{equation}
\bigskip
\begin{equation}
\int_{D_{2}}\sigma \frac{d\phi _{2h}^{n}}{dx}\frac{dw_{h}}{dx
}dx=-\int_{D_{2}}\left( J_{h}^{n}+g\right) w_{h}dx\ \ \forall w_{h}\in
V_{h}^{(1)}(\overline{D}_{2}). \label{fdm4}
\end{equation}
\bigskip
\begin{equation}
\int_{D_{2}}J_{h}^{n}dx=0\text{ with }\int_{D_{\mathrm{a}
}}J_{h}^{n}dx=I(t_{n})=-\int_{D_{\text{\textrm{c}}}}J_{h}^{n}dx,
\label{fdm5}
\end{equation}
where
\begin{equation}
\begin{array}{l}
\widetilde{\partial}
_{t}u_{h}^{n}=\displaystyle\frac{u_{h}^{n}-u_{h}^{n-1}}{\Delta t},
\text{\textrm{\ \ }}\widetilde{\partial}
_{t}v_{h\Delta r}^{n}=\displaystyle\frac{
v_{h\Delta r}^{n}-v_{h\Delta r}^{n-1}}{\Delta t}, \\
\\
J_{h}^{n}=J(x,u_{_{h}}^{n},v_{sh}^{n},\eta _{h}^{n})=a_{2}(x)i_{0h}^{n}\sinh
\left( \beta \eta _{h}^{n}\right),\ i_{0h}^{n}=i_{0}(u_{h}^{n},v_{sh}^{n}) , \\
\\
\eta _{h}^{n}=\phi _{1h}^{n}-\phi _{2h}^{n}-\alpha^{n}_{h} \ln u_{h}^{n}-\overline{U}
_{h}(v_{sh}^{n}),\ \alpha^{n}_{h}=\alpha(u_{h}^{n}).
\end{array}
\label{fdm6}
\end{equation}
\subsection{On the existence and uniqueness of the solution of the fully
discrete model}
To prove that the system (\ref{fdm1})-(\ref{fdm4}) has a unique
solution, we first show that assuming $(u_{h}^{n},v_{h\Delta
r}^{n},v_{sh}^{n})\in V_{h}^{(1)}(\overline{D}_{1})\times V_{h\Delta
r}(\overline{D}_{3})\times V_{h}^{(0)}(\overline{D}_{2})$ and the
assumptions \textbf{A1-A4} hold, the system
(\ref{fdm3})-(\ref{fdm4}) has a unique solution $(\phi
_{1h}^{n},\phi _{2h}^{n})\in W_{h}(\overline{D}_{1})\times
V_{h}^{(1)}(\overline{D}_{2})$; then, returning to the system
(\ref{fdm1})-(\ref{fdm2}) and applying a well-known \ consequence of
Brower\'{}s fixed point theorem, which is
presented as Corollary 1.1 in \cite{GR}, we prove that there exists $
(u_{h}^{n},v_{h\Delta r}^{n})\in V_{h}^{(1)}(\overline{D}_{1})\times
V_{h\Delta r}(\overline{D}_{3})$.
\begin{lemma}
\label{lem9} Assuming that for all $n$, $(u_{h}^{n},v_{h\Delta
r}^{n},v_{sh}^{n})\in V_{h}^{(1)}(\overline{D}_{1})\times V_{h\Delta r}(
\overline{D}_{3})\times V_{h}^{(0)}(\overline{D}_{2})$, and the
assumptions \textbf{A1-A4} hold, then the system
(\ref{fdm3})-(\ref{fdm4}) has a unique solution $(\phi
_{1h}^{n},\phi _{2h}^{n})\in W_{h}(\overline{D}_{1})\times
V_{h}^{(1)}(\overline{D}_{2})$.
\end{lemma}
\begin{proof}
Looking at (\ref{fdm3})-(\ref{fdm4}) and in order to apply
Minty-Browder theorem to prove
the existence of a solution, we define\ the functions
\begin{equation*}
\widehat{\eta }_{h}^{n}:=-\alpha_{h}^{n} \ln u_{h}^{n}-\overline{U}
_{h}(v_{sh}^{n})\ \mathrm{and\ }\widehat{J}
_{h}^{n}:=J_{h}^{n}(x,u_{h}^{n},v_{sh}^{n},\widehat{\eta }_{h}^{n}),
\end{equation*}
it is worth noticing that $\widehat{\eta }_{h}^{n}$ is equal to
$\eta _{h}^{n}$ when the potentials $\phi _{1h}^{n}$ and $\phi
_{1h}^{n}$ are zero. Now, going back to Section 4.2 and using
$\widehat{J}_{h}^{n}$, we
define the operators $\widehat{B}_{h}:V_{h}\rightarrow V_{h}^{\ast }$ and $
A_{h}:V_{h}\rightarrow V_{h}^{\ast }$ as follows: for all
$n=1,2,..,N$,
\begin{equation*}
\left\langle \widehat{B}_{h}(\Phi _{h}^{n}),\Psi _{h}\right\rangle
=\int_{D_{2}}(J_{h}^{n}-\widehat{J}_{h}^{n})(\psi _{2h}-\psi
_{1h})dx\ \ \forall \Psi _{h}\in V_{h}.
\end{equation*}
and
\begin{equation*}
\left\langle A_{h}(\Phi _{h}^{n}),\Psi _{h}\right\rangle =a_{h}(\Phi
_{h}^{n},\Psi _{h})+\left\langle \widehat{B}_{h}(\Phi _{h}^{n}),\Psi
_{h}\right\rangle .
\end{equation*}
Notice that when $\Phi _{h}^{n}=(0,0)$, $\left\langle
\widehat{B}_{h}(\Phi _{h}^{n}),\Psi _{h}\right\rangle =0$ because
$J_{h}^{n}=\widehat{J}_{h}^{n}$. Now, we can recast
(\ref{fdm3})-(\ref{fdm4}) as
follows. Find $\Phi _{h}^{n}:=(\phi _{1}^{n},\phi _{1}^{n})\in W_{h}(
\overline{D}_{1})\times V_{h}^{(1)}(\overline{D}_{2})$ such that
\begin{equation}
\left\langle A_{h}(\Phi _{h}^{n}),\Psi _{h}\right\rangle
=-\int_{D_{2}}g\psi _{2h}dx-\int_{D_{2}}\widehat{J}_{h}^{n}(\psi
_{2h}-\psi _{1h})dx\ \ \forall \Psi _{h}\in V_{h}. \label{opA}
\end{equation}
We can prove, using the same arguments as in Lemma \ref{lem5}, that the operator $
\widehat{B}_{h}$ is monotone, bounded and continuous satisfying an
inequality as (\ref{pre17}); since the bilinear form $a_{h}$ is
continuous and semi-definite positive, then it follows that the
operator $A_{h}$ is monotone, bounded and continuous satisfying an
inequality as (\ref{pre17}). In order to prove that (\ref{opA}) has
a solution it remains to show that $A_{h}$ is coercive, i.e.,
$\forall \Phi _{h}^{n}\in W_{h}(\overline{D}_{1})\times
V_{h}^{(1)}(\overline{D}_{2})$,
there exists a positive constant $\alpha $ such that
\begin{equation*}
\left\langle A_{h}(\Phi _{h}^{n}),\Phi _{h}^{n}\right\rangle \geq
\alpha \left\Vert \Phi _{h}^{n}\right\Vert _{V}^{2}.
\end{equation*}
This can be easily done by considering the following facts: 1)\
$A_{h}$ is monotone; 2) it is easy to check, using the same
arguments as in Theorem \ref{Theorem 1} to prove (\ref{eep4}), that
$\forall \Phi _{h}^{n}$, $\overline{\Phi }_{h}^{n}$ $\in
W_{h}(\overline{D}_{1})\times V_{h}^{(1)}(\overline{D}_{2})$
\begin{equation*}
a_{h}(\Phi _{h}^{n}-\overline{\Phi }_{h}^{n},\Phi _{h}^{n}-\overline{\Phi }
_{h}^{n})+\left\langle \widehat{B}_{h}(\Phi _{h}^{n})-\widehat{B}_{h}(
\overline{\Phi }_{h}^{n}),\Phi _{h}^{n}-\overline{\Phi }_{h}^{n}\right
\rangle \geq \alpha \left\Vert \Phi _{h}^{n}-\overline{\Phi }
_{h}^{n}\right\Vert _{V}^{2},
\end{equation*}
then taking$\ \overline{\Phi }_{h}^{n}=(0,0)$ it follows the
coerciveness of $A_{h}$. Hence, the Minty-Browder theorem
\cite{Zeid} guaranties the existence of a solution $\Phi _{h}^{n}$
of (\ref{opA}). To prove the uniqueness of this
solution we follow the argument put forward in \cite{WXZ} to prove the uniqueness of the exact solution,
and assume that there two solutions $
\Phi _{h}^{n}:=(\phi _{1h}^{n},\phi _{2h}^{n})$ and $\overline{\Phi }
_{h}^{n}:=\left( \overline{\phi }_{1h}^{n},\overline{\phi
}_{2h}^{n}\right) $ of (\ref{fdm3})-(\ref{fdm4}), then setting,
$z_{1h}=\phi _{1h}^{n}-\overline{\phi }_{1h}^{n}$ and $z_{2h}=\phi
_{2h}^{n}-\overline{\phi }_{2h}^{n}$,\ from (\ref{fdm3})\ it
follows that
\begin{equation*}
\int_{D_{1}}\kappa (u_{h}^{n})\frac{dz_{1h}}{dx}\frac{dw_{h}}{dx}
dx=\int_{D_{2}}\left( J_{h}^{n}-\overline{J}_{h}^{n}\right) w_{h}dx\
\ \forall w_{h}\in V_{h}^{\left( 1\right) }(\overline{D}_{1})
\end{equation*}
and from (\ref{fdm4})
\begin{equation*}
\int_{D_{1}}\sigma
\frac{dz_{2h}}{dx}\frac{dv_{h}}{dx}dx=-\int_{D_{2}}\left(
J_{h}^{n}-\overline{J}_{h}^{n}\right) v_{h}dx\ \ \forall v_{h}\in
V_{h}^{\left( 1\right) }(\overline{D}_{2}),
\end{equation*}
where\ $J_{h}^{n}=J_{h}^{n}(x,u_{h}^{n},v_{sh}^{n},\eta _{h}^{n})$ and $
\overline{J}_{h}^{n}=J_{h}^{n}(x,u_{h}^{n},v_{sh}^{n},\overline{\eta }
_{h}^{n})$, with $\overline{\eta }_{h}^{n}=\overline{\phi }_{2h}^{n}-
\overline{\phi }_{1h}^{n}-\alpha_{h}^{n} \ln u_{h}^{n}-\overline{U}
_{h}(v_{sh}^{n}) $. Setting $w_{h}=z_{1h}$ and $v_{h}=z_{2h}$ and
applying the mean value
theorem one readily obtains that
\begin{equation*}
\int_{D_{1}}\kappa (u_{h}^{n})\left( \frac{dz_{1h}}{dx}\right)
^{2}dx+\int_{D_{1}}\sigma \left( \frac{dz_{2h}}{dx}\right)
^{2}dx+\int_{D_{2}}\frac{\partial J_{h}^{n}\left( \xi \right)
}{\partial \eta^{n} }(z_{2h}-z_{1h})^{2}=0,
\end{equation*}
here $\frac{\partial J_{h}^{n}\left( \xi \right) }{\partial \eta^{n}
}>0$ according to assumption \textbf{A3}. The first term of this
expression
implies that for all $n$,
\begin{equation*}
z_{1h}=\phi _{1h}^{n}-\overline{\phi }_{1h}^{n}=K_{1},
\end{equation*}
but the constant $K_{1}=0$ because $\phi _{1h}^{n}$ and $\overline{\phi }
_{1h}^{n}$ are in $W_{h}(\overline{D}_{1})$, so $\phi _{1h}^{n}=\overline{
\phi }_{1h}^{n}$ Similarly, from the second and third terms it follows that $
z_{2h}=0$, and consequently $\phi _{2h}^{n}=\overline{\phi
}_{2h}^{n}$. Hence, we have just proved that for all $n$ there is a
unique solution $(\phi _{1h}^{n},\phi _{2h}^{n})$.
\end{proof}
\begin{lemma}
\label{lem10} Let $(\phi _{1h}^{n},\phi _{2h}^{n})\in W_{h}(\overline{D}
_{1})\times V_{h}^{(1)}(\overline{D}_{2})$ be the solution to (\ref{fdm3})-(
\ref{fdm4}). There exists a unique solution $\left( u_{h}^{n},v_{h\Delta
r}^{n}\right) \in V_{h}^{(1)}(\overline{D}_{1})\times V_{h\Delta r}(
\overline{D}_{3})$ to the system (\ref{fdm1})-(\ref{fdm2}).
\end{lemma}
\begin{proof}
We start proving the existence of $u_{h}^{n}\in V_{h}^{(1)}(\overline{D}
_{1}) $ as solution of (\ref{fdm1}). To this end, we write (\ref{fdm1}) as $
F_{h}(u_{h}^{n})=0$, where $F_{h}:V_{h}^{(1)}(\overline{D}_{1})\rightarrow
V_{h}^{(1)}(\overline{D}_{1})$ is a continuous mapping defined by the
relation
\begin{equation*}
\begin{array}{r}
\displaystyle\int_{D_{1}}F_{h}(\chi _{h})w_{h}dx=\displaystyle
\int_{D_{1}}\left( \chi _{h}-u_{h}^{n-1}\right) w_{h}dx+\Delta
t\int_{D_{1}}k_{1}\frac{d\chi _{h}^{n}}{dx}\frac{dw_{h}}{dx}
dx \\
\\
-\displaystyle\Delta t\int_{D_{1}}a_{1}J_{h}^{n}(\chi _{h})w_{h}dx=0\ \
\forall w_{h}\in V_{h}^{(1)}(\overline{D}_{1}),
\end{array}
\end{equation*}
here,\ $J_{h}^{n}(\chi _{h})=J(x,\chi _{h},v_{sh},\phi _{1h}^{n},\phi _{2h}^{n},
\overline{U}_{h}(v_{sh})),$ with $v_{sh}$ being picked up from
$S_{Q}^{\ast } $ because we assume that $v_{h\Delta r}^{n}$ belongs
to this space;
moreover, we also assume that $\chi _{h}$ is in $S_{P}$. According to Brower
\'{}s fixed point theorem, the equation $F_{h}(\chi _{h})=0$ has a solution$
\ \chi _{h}\in B_{q}:=\left\{ v_{h}\in V_{h}^{(1)}(\overline{D}
_{1}):\left\Vert v_{h}\right\Vert _{L^{2}(D_{1})}\leq q\right\} $,
if $\int_{D_{1}}F_{h}(\chi _{h})\chi_{h}dx>0$ for $\left\Vert \chi
_{h}\right\Vert _{L^{2}(D_{1})}=q$. On account of the assumptions
$v_{sh}\in S_{Q}^{\ast }$ and $\chi _{h}\in S_{P}$, it follows that
there exists a constant $C_{1}=C_{1}(P,Q,K)$ such that\
$\int_{D_{1}}a_{1}J_{h}^{n}(\chi _{h})\chi_{h}dx\leq C_{1}\left\Vert
\chi _{h}\right\Vert _{L^{2}(D_{1})}$.
Hence,
\begin{equation*}
\begin{array}{r}
\displaystyle\int_{D_{1}}F_{h}(\chi _{h})\chi _{h}dx\geq \left\Vert
\chi _{h}\right\Vert _{L^{2}(D_{1})}^{2}-\left\Vert
u_{h}^{n-1}\right\Vert
_{L^{2}(D_{1})}^{2}+\Delta tk_{0}\left\Vert \frac{d\chi _{h}^{n}}{
dx}\right\Vert _{L^{2}(D_{1})}^{2}-\Delta tC_{1}\left\Vert \chi
_{h}\right\Vert _{L^{2}(D_{1})} \\
\\
\geq \left\Vert \chi _{h}\right\Vert _{L^{2}(D_{1})}^{2}-\left\Vert
u_{h}^{n-1}\right\Vert _{L^{2}(D_{1})}^{2}-\Delta tC_{1}\left( 1+\left\Vert
\chi _{h}\right\Vert _{L^{2}(D_{1})}\right) \left\Vert \chi _{h}\right\Vert
_{L^{2}(D_{1})}.
\end{array}
\end{equation*}
Then, taking $\Delta t\leq \Delta t_{0}<1/C_{1}$, $\int_{D_{1}}F_{h}(\chi
_{h})\chi _{h}dx$ is positive for $\left\Vert \chi _{h}\right\Vert
_{L^{2}(D_{1})}$ sufficiently large. This shows the existence of the
solution $u_{h}^{n}\in V_{h}^{(1)}(\overline{D}_{1})$. Next, we prove the
uniqueness. To this end, we consider that there exist $X$ and $Y$ $\in
V_{h}^{(1)}(\overline{D}_{1})$ satisfying (\ref{fdm1}), so
\begin{equation*}
\int_{D_{1}}\left( X-Y\right) w_{h}dx+\Delta t\int_{D_{1}}k_{1}\frac{
d(X-Y)}{dx}\frac{dw_{h}}{dx}dx=\Delta t\int_{D_{1}}a_{1}\left(
J_{h}^{n}(X)-J_{h}^{n}(Y)\right) w_{h}dx\ \ \forall w_{h}\in
V_{h}^{(1)}(\overline{D}_{1}).
\end{equation*}
Setting $w_{h}=X-Y$ and invoking the arguments of Lemmas \ref{lem3} and \ref
{lem4} yields
\begin{equation*}
\left\Vert X-Y\right\Vert _{L^{2}(D_{1})}^{2}+k_{1}\Delta t\left\Vert \frac{
d(X-Y)}{dx}\right\Vert _{L^{2}(D_{1})}^{2}\leq \Delta
tC_{2}\left\Vert X-Y\right\Vert _{L^{2}(D_{1})}^{2},
\end{equation*}
where the constant $C_{2}=C_{2}(P,Q,K)$. Thus, taking $\Delta t\leq \Delta
t_{0}<1/C_{2}$ it follows that $X=Y$. It remains to prove the existence and
uniqueness of $v_{h\Delta r}^{n}$, but the arguments to be used for such a
proof are the same as for $u_{h}^{n}$, so we omit them.
\end{proof}
\subsection{Error estimates for the fully discrete solution}
As in Section 4.3, we write for $t=t_{n}$
\begin{equation}
\left\{
\begin{array}{l}
u^{n}-u_{h}^{n}=\rho _{u}^{n}+\theta _{u}^{n}, \\
\\
v^{n}-v_{h\Delta r}^{n}=v^{n}-I_{0}^{x}v^{n}+I_{0}^{x}\rho _{v}^{n}+\theta
_{v}^{n}.
\end{array}
\right. \label{fdm7}
\end{equation}
\begin{theorem}
\label{Theorem 3} Let $(u_{h}^{n},v_{h\Delta r}^{n},\phi _{1}^{n},\phi
_{2}^{n})$ be the solution to (\ref{fdm1})-(\ref{fdm4}). Then, under proper
regularity assumptions there exists a constant $C$ such that for $\Delta t$
small
\begin{equation}
\left\Vert u^{n}-u_{h}^{n}\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert
v^{n}-v_{h\Delta r}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\Delta t\sum_{j=1}^{t_{n}}\left\Vert \Phi ^{j}-\Phi
_{h}^{j}\right\Vert _{V}^{2}d\tau \leq C(h^{2}+\Delta r^{2}+\Delta t^{2}).
\label{fdm8}
\end{equation}
The constant $C$ is of the form $C(\Gamma )\exp
(C(\underline{k_{1}},\underline{k_{2}})t_{n}$, $C(\Gamma )$ being
another constant that depends on the exact solution
$(u,v,\phi_{1},\phi_{2})$, see (\ref{cgamma}) below.
\end{theorem}
\begin{proof}
Since $\rho _{u}^{n}$, $v^{n}-I_{0}^{x}v^{n}+I_{0}^{x}\rho _{v}^{n}$ and $
\left\Vert \Phi ^{j}-\Phi _{h}^{j}\right\Vert _{V}^{2}$ are
estimated as in Section 4.3, we shall address our attention to the
estimations for $\theta _{u}^{n}$ and $\theta _{v}^{n}$. We start
with the calculation for $\theta
_{v}^{n}$. For this purpose, we recast (\ref{uv4}) for $t=t_{n}$ as
\begin{equation}
\begin{array}{c}
\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\left(\widetilde{\partial}_{t}I_{0}^{x}{v}
^{n}w+k_{2}\frac{\partial I_{0}^{x}v^{n}}{\partial r}\frac{\partial w}{
\partial r}\right) r^{2}drdx \\
\\
=-\displaystyle
\int_{D_{2}}I_{0}^{x}(R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}J^{n}w(x;R_{s}(x)))dx \\
\\
+\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\left(\widetilde{\partial}_{t}I_{0}^{x}{v}
^{n}w-\frac{\partial I_{0}^{x}v^{n}}{\partial t}w\right) r^{2}drdx.
\end{array}
\label{eef2}
\end{equation}
Setting, as we did in Section 4.3, $v_{h\Delta
r}^{n}=I_{0}^{x}v^{n}-\left(
I_{0}^{x}\rho _{v}^{n}+\theta _{v}^{n}\right) $ in (\ref{fdm2}) and using (
\ref{pre8}) and (\ref{eef2} ) yields for $t=t_{n}$
\begin{equation*}
\begin{array}{r}
\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\left(\widetilde{\partial}_{t}\theta
_{v}^{n}w_{h\Delta r}+k_{2}\frac{\partial \theta _{v}^{n}}{\partial
r}\frac{
\partial w_{h\Delta r}}{\partial r}\right) r^{2}drdx=\lambda
\int_{D_{2}}\int_{0}^{R_{s}(x)}I_{0}^{x}\rho _{v}^{n}w_{h\Delta r}r^{2}drdx
\\
\\
-\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\widetilde{\partial}_{t}I_{0}^{x}\rho
_{v}^{n}w_{h\Delta r}r^{2}drdx \\
\\
-\displaystyle\int_{D_{2}}I_{h}^{0}(R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}(J^{n}(x)-J_{h}^{n}(x))w_{hs}(x))dx
\\
\\
+\displaystyle\int_{D_{2}}\left(
R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}J_{h}^{n}(x)-I_{h}^{0}(R_{s}^{2}(x)a_{2}^{-1}(x)F^{-1}J_{h}^{n}(x))\right)
w_{hs}(x)dx,
\\
\\
-\displaystyle\int_{D_{2}}\int_{0}^{R_{s}(x)}\left(\widetilde{\partial}_{t}I_{0}^{x}v^{n}-\frac{\partial
I_{0}^{x}v^{n}}{\partial t}\right)
r^{2}drdx
\end{array}
\end{equation*}
Letting $w_{h\Delta r}=\theta _{v}^{n}$, $w_{hs}=\theta _{vs}^{n}$,
and noting that for $a$ and $b$ real numbers,
$2(a-b)b=a^{2}-b^{2}\textcolor{blue}{-}(a-b)^{2}$,
it follows that
\begin{equation}
\begin{array}{l}
\displaystyle\frac{1}{2}\widetilde{\partial}_{t}\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\displaystyle \underline{k_{2}}\left\Vert \frac{\partial
\theta _{v}^{n}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2} \\
\\
\leq \lambda \left\Vert I_{0}^{x}\rho _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
+\displaystyle\left\Vert \widetilde{\partial}_{t}I_{0}^{x}\rho
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))} \\
\\
+C\left\Vert I_{h}^{0}\left( (J^{n}-J_{h}^{n})\theta _{vs}^{n}\right)
\right\Vert _{L^{1}(D_{2})}+C\left\Vert (J_{h}^{n}-I_{h}^{0}J_{h}^{n})\theta
_{vs}^{n}\right\Vert _{L^{1}(D_{2})} \\
\\
+\left\Vert \widetilde{\partial}_{t}I_{0}^{x}v^{n}-\frac{\partial I_{0}^{x}v^{n}}{
\partial {t}}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\equiv \sum_{i=1}^{5}R_{i}^{n}.
\end{array}
\label{eef4}
\end{equation}
We bound the right hand side of this inequality applying the same arguments
as in (\ref{uv7}). Thus, we have that
\begin{equation*}
R_{1}^{n}\leq C\Delta r^{4}\left\Vert v^{n}\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}+C\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}
\end{equation*}
\begin{equation*}
R_{2}^{n}\leq \frac{C\Delta r^{4}}{\Delta t}\int_{t_{n-1}}^{t_{n}}\left\Vert
\frac{\partial v}{\partial t}\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}dt+C\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\end{equation*}
\begin{equation*}
R_{3}^{n}\leq C\left\Vert J^{n}-J_{h}^{n}\right\Vert
_{L^{2}(D_{2})}^{2}+C(\epsilon )\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \left\Vert \frac{
\partial \theta _{v}^{n}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\end{equation*}
\begin{equation*}
\begin{array}{r}
R_{4}^{n}\leq C\left\Vert J^{n}-J_{h}^{n}\right\Vert
_{L^{2}(D_{2})}^{2}+C(\epsilon )\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\vert \frac{\partial \theta _{v}^{n}}{\partial r}\right\vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2} \\
\\
+Ch^{2}\displaystyle\left\Vert \frac{\partial J^{n}}{\partial x}\right\Vert
_{L^{2}(D_{2})}^{2}.
\end{array}
\end{equation*}
In both $R_{3}^{n}$ and $R_{4}^{n}$ the term $C\left\Vert
J^{n}-J_{h}^{n}\right\Vert _{L^{2}(D_{2})}^{2}$ is bounded by Lemma \ref
{lem6} for $t=t_{n}$; thus, using the notation $\Gamma =(u,v,v_{s},\phi
_{1},\phi _{2})$, we can set that
\begin{equation}
\begin{array}{l}
C\left\Vert J^{n}-J_{h}^{n}\right\Vert _{L^{2}(D_{2})}^{2}\leq
C(\Gamma )(h^{2}+\Delta r^{2})+C\left\Vert \theta
_{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2} \\
\\
+C(\epsilon )\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\Vert \frac{\partial \theta _{v}^{n}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2},
\end{array}
\label{eef5}
\end{equation}
where the constant $C(\Gamma )$ is given as
\begin{equation} \label{cgamma}
\begin{array}{c}
C(\Gamma )=C\max \left( \left\Vert \phi _{1}\right\Vert _{L^{\infty }(0,T_{
\mathrm{end}};H^{2}(D_{1}))},\left\Vert \phi _{2}\right\Vert
_{L^{\infty }(0,T_{\mathrm{end}};H^{2}(D_{2}))},\left\Vert
v_{s}\right\Vert _{L^{\infty
}(0,T_{\mathrm{end}};H^{1}(D_{2}))},\right. \\
\\
\left. \left\Vert u\right\Vert _{L^{\infty }(0,T_{\mathrm{end}
};H^{2}(D_{1}))},\left\Vert v\right\Vert _{L^{\infty }(0,T_{\mathrm{end}
};L^{2}(D_{2};H_{r}^{2}(0,R_{s}(\cdot))))}\right) .
\end{array}
\end{equation}
Hence, we can write
\begin{equation*}
\begin{array}{r}
R_{3}^{n}+R_{4}^{n}\leq C(\Gamma )(h^{2}+\Delta r^{2})+Ch^{2}\displaystyle
\left\Vert \frac{\partial J^{n}}{\partial x}\right\Vert
_{L^{2}(D_{2})}^{2}+C\left\Vert \theta _{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2} \\
\\
+C(\epsilon )\left\Vert \theta _{v}(t)\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\epsilon \displaystyle
\left\Vert \frac{\partial \theta _{v}(t)}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}.
\end{array}
\end{equation*}
To estimate the term $R_{5}^{n}$, we notice that by approximation
theory
\begin{equation*}
\left\Vert \widetilde{\partial}_{t}I_{0}^{x}v^{n}-\frac{\partial I_{0}^{x}v^{n}}{
\partial {t}}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\leq
C\left\Vert \widetilde{\partial}_{t}v^{n}-\frac{\partial v^{n}}{\partial {t}}
\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}
\end{equation*}
and
\begin{equation*}
\widetilde{\partial}_{t}v^{n}-\frac{\partial v^{n}}{\partial {t}}=\frac{-1}{\Delta t}
\int_{t_{n-1}}^{t_{n}}(t-t_{n-1})\frac{\partial ^{2}v}{\partial
t^{2}}dt,
\end{equation*}
so,
\begin{equation*}
\left\Vert \widetilde{\partial}_{t}I_{0}^{x}v^{n}-\frac{\partial I_{0}^{x}v^{n}}{
\partial {t}}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}\leq
C\left( \Delta t\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial ^{2}v}{
\partial {t}^{2}}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}dt\right) ^{1/2}.
\end{equation*}
Applying Young inequality yields
\begin{equation*}
R_{5}^{n}\leq C\Delta t\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial ^{2}v
}{\partial {t}^{2}}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}dt+C\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}
\end{equation*}
Collecting these bounds in (\ref{eef4}) and letting $\epsilon
=\underline{k_{2}}/2$
yields
\begin{equation}
\begin{array}{r}
\displaystyle\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\displaystyle\Delta
t\underline{k_{2}}\left\Vert \frac{\partial \theta
_{v}^{n}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}\leq
\displaystyle\left\Vert \theta _{v}^{n-1}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+F_{v}^{n} \\
\\
+C(\underline{k_{2}})\Delta t\left( \left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\left\Vert \theta
_{u}^{n}\right\Vert _{L^{2}(D_{1})}^{2}\right) ,
\end{array}
\label{eef6.0}
\end{equation}
where
\begin{equation}
\begin{array}{c}
F_{v}^{n}=\Delta t\left( C(\Gamma )(h^{2}+\Delta r^{2})+Ch^{2}\displaystyle
\left\Vert \frac{\partial J^{n}}{\partial x}\right\Vert
_{L^{2}(D_{2})}^{2}+C\Delta r^{4}\left\Vert v^{n}\right\Vert
_{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) \\
\\
+C\Delta r^{4}\displaystyle\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial v
}{\partial t}\right\Vert _{L^{2}(D_{2},H_{r}^{2}(0,R_{s}(\cdot
)))}^{2}dt+C\Delta t^{2}\displaystyle\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{
\partial ^{2}v}{\partial {t}^{2}}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}dt.
\end{array}
\label{eef6.1}
\end{equation}
To calculate an estimate for $\theta _{u}^{n}$, we observe that
subtracting (\ref{fdm1}) from (\ref{w1}) and setting $\theta
_{u}^{n}=e_{u}^{n}-\rho
_{u}^{n}$ it follows that
\begin{equation*}
\begin{array}{r}
\displaystyle\int_{D_{1}}\widetilde{\partial}_{t}\theta _{u}^{n}w_{h}dx+\displaystyle\int_{D_{1}}k_{1}
\frac{d\theta _{u}^{n}}{dx}\frac{dw_{h}}{dx}dx=\lambda
\int_{D_{1}}\rho _{u}^{n}w_{h}dx-\int_{D_{1}}\partial \rho
_{u}^{n}w_{h}dx
\\
\\
+\displaystyle\int_{D_{1}}a_{1}\left( J^{n}-J_{h}^{n}\right) w_{h}dx+
\displaystyle\int_{D_{1}}\left( \widetilde{\partial}_{t}u^{n}-\frac{\partial u^{n}}{
\partial t}\right) w_{h}dx.
\end{array}
\end{equation*}
Letting $w_{h}=\theta _{u}^{n}$ yields
\begin{equation*}
\begin{array}{r}
\displaystyle\frac{1}{2}\widetilde{\partial}_{t}\left\Vert \theta
_{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2}+\underline{k_{1}}\displaystyle\left\Vert
\frac{d\theta
_{u}^{n}}{dx}\right\Vert _{L^{2}(D_{1})}^{2}\leq C\displaystyle
\left( \left\Vert \rho _{u}^{n}\right\Vert _{L^{2}(D_{1})}^{2}+\frac{1}{
\Delta t}\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial \rho _{u}}{\partial
t}\right\Vert _{L^{2}(D_{1})}^{2}dt\right) \\
\\
+C\Delta t\displaystyle\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial ^{2}u
}{\partial ^{2}t}\right\Vert _{L^{2}(D_{1})}^{2}dt+C\left\Vert
J^{n}-J_{h}^{n}\right\Vert _{L^{2}(D_{1})}^{2}+C\left\Vert \theta
_{u}^{n}\right\Vert _{L^{2}(D_{1})}^{2}.
\end{array}
\end{equation*}
Then by virtue of (\ref{pre5}) and Lemma \ref{lem6} it follows that
\begin{equation} \label{eef6.01}
\begin{array}{l}
\left\Vert \theta _{u}^{n}(t)\right\Vert _{L^{2}(D_{1})}^{2}+\Delta
t\underline{k_{1}}
\displaystyle\left\Vert \frac{d\theta _{u}^{n}}{dx}
\right\Vert _{L^{2}(D_{1})}^{2}\leq \left\Vert \theta
_{u}^{n-1}(t)\right\Vert _{L^{2}(D_{1})}^{2}+F_{u}^{n} \\
\\
+C(\underline{k_{1}},\underline{k_{2}})\Delta t\left( \left\Vert
\theta _{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\left\Vert \theta _{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2}\right)+\displaystyle\frac{\underline{k_{2}}}{2}\left\Vert
\frac{\partial {\theta_{v}^{n}}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}
\end{array}
\end{equation}
where
\begin{equation}
F_{u}^{n}=\Delta tC(\Gamma )(h^{2}+\Delta r^{2})+Ch^{4}\displaystyle
\int_{t_{n-1}}^{t_{n}}\left\Vert \frac{\partial u}{\partial t}\right\Vert
_{H^{2}(D_{1})}^{2}dt+C\Delta t^{2}\displaystyle\int_{t_{n-1}}^{t_{n}}\left
\Vert \frac{\partial ^{2}u}{\partial ^{2}t}\right\Vert _{L^{2}(D_{1})}^{2}dt
\label{eef7}
\end{equation}
It remains to estimate $\left\Vert \Phi ^{n}-\Phi _{h}^{n}\right\Vert _{V}$.
Returning to the proof of Theorem \ref{Theorem 2} we have that for $t=t_{n}$
\begin{equation}
\left\Vert \Phi ^{n}-\Phi _{h}^{n}\right\Vert _{V}^{2}\leq C(\Gamma
)h^{2}+C(\epsilon )\left( \left\Vert \theta _{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2}+\left\Vert \theta _{v}^{n}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) +\epsilon \displaystyle
\left\Vert \frac{\partial \theta _{v}^{n}}{\partial r}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}. \label{eef8}
\end{equation}
Thus, setting $\epsilon=\underline{k_{2}}/2$ in (\ref{eef8}) and
adding (\ref{eef6.0}), (\ref{eef6.01}) and (\ref{eef8}) we obtain
that
\begin{equation*}
\begin{array}{l}
\left( 1-C(\underline{k_{1}},\underline{k_{2}})\Delta t\right)
\left( \left\Vert \theta _{u}^{n}\right\Vert
_{L^{2}(D_{1})}^{2}+\displaystyle\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}\right) +\Delta t\left\Vert \Phi ^{n}-\Phi
_{h}^{n}\right\Vert _{V}^{2}\leq
F_{v}^{n}+F_{u}^{n} \\
\\
+\left\Vert \theta _{u}^{n-1}\right\Vert _{L^{2}(D_{1})}^{2}+\displaystyle
\left\Vert \theta _{v}^{n-1}\right\Vert
_{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}
\end{array}
\end{equation*}
For $\Delta t$ small
\begin{equation*}
\begin{array}{r}
\left\Vert \theta _{u}^{n}\right\Vert _{L^{2}(D_{1})}^{2}+\displaystyle
\left\Vert \theta _{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}+\Delta t\left\Vert \Phi ^{n}-\Phi _{h}^{n}\right\Vert _{V}^{2}\leq
C\left( F_{v}^{n}+F_{u}^{n}\right) \\
\\
\left( 1+C(\underline{k_{1}},\underline{k_{2}})\Delta t\right)
\left( \left\Vert \theta _{u}^{n-1}\right\Vert
_{L^{2}(D_{1})}^{2}+\displaystyle\left\Vert \theta
_{v}^{n-1}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}\right) .
\end{array}
\end{equation*}
Hence, by repeated application and taking $\left\Vert \theta
_{u}^{0}\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert \theta
_{v}^{0}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot
)))}^{2}=0$, it results that
\begin{equation*}
\left\Vert \theta _{u}^{n}\right\Vert _{L^{2}(D_{1})}^{2}+\left\Vert \theta
_{v}^{n}\right\Vert _{L^{2}(D_{2},L_{r}^{2}(0,R_{s}(\cdot )))}^{2}+\Delta
t\sum_{j=1}^{n}\left\Vert \Phi ^{j}-\Phi _{h}^{j}\right\Vert _{V}^{2}\leq C
\displaystyle\sum_{j=1}^{n}\frac{F_{v}^{j}+F_{u}^{j}}{\left(
1+C(\underline{k_{1}},\underline{k_{2}}))\Delta t\right) ^{j-n}}.
\end{equation*}
Noting that $\left( 1+C(k_{1},k_{2})\Delta t\right) \leq
e^{C(k_{1},k_{2})\Delta t}$, then we can write
\begin{equation*}
\begin{array}{c}
\displaystyle\sum_{j=1}^{n}\frac{F_{v}^{j}+F_{u}^{j}}{\left(
1+C(\underline{k_{1}},\underline{k_{2}})\Delta t\right) ^{j-n}}\leq
e^{C(\underline{k_{1}},\underline{k_{2}})t_{n}}\sum_{j=1}^{n}F_{v}^{j}+F_{u}^{j}\ \ \left( \mathrm{
by\ (\ref{eef6.1})\ and\ (\ref{eef7})}\right) \\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq C(\Gamma
)e^{C(\underline{k_{1}},\underline{k_{2}})t_{n}}\left( h^{2}+\Delta r^{2}+\Delta t^{2}\right) .
\end{array}
\end{equation*}
This completes the proof
\end{proof}
\section{Acknowledgements}
This research has been partially funded by grant PGC-2018-097565-B100 of
Ministerio de Ciencia, Innovaci\'{o}n y Universidades of Spain. | 95,569 |
\begin{document}
\baselineskip=18pt
\begin{abstract}
Let $G$ be a connected reductive group over an algebraic closure $\bar{\mathbb{F}}_q$ of a finite field ${\mathbb{F}_q}$. In this paper it is proved that the infinite dimensional Steinberg module of $kG$ defined by N. Xi in 2014 is irreducible when $k$ is a field of positive characteristic and char$k\ne $char$\mathbb{F}_q$. For certain special linear groups, we show that the Steinberg modules of the groups are not quasi-finite with respect to some natural quasi-finite sequences of the groups.
\end{abstract}
\maketitle
\def\Cal{\mathcal}
\def\bold{\mathbf}
\def\ca{\mathcal A}
\def\cdz{\mathcal D_0}
\def\cd{\mathcal D}
\def\cdo{\mathcal D_1}
\def\bold{\mathbf}
\def\l{\lambda}
\def\le{\leq}
N. Xi studied some induced representations of infinite reductive groups with Frobenius maps (see [X]). In particular, he defined Steinberg modules for any reductive groups by extending Steinberg's construction of Steinberg modules for finite reductive groups. These Steinberg modules are infinite dimensional when the reductive groups are infinite.
Let $G$ be a connected reductive group over the algebraic closure $\bar{\mathbb{F}}_q$ of a finite field ${\mathbb{F}_q}$ and $k$ a field. Xi proved that the Steinberg module of the group algebra $kG$ of $G$ over $k$ is irreducible if $k$ is the field of complex numbers or $k=\bar{\mathbb{F}}_q$ ( In fact, his proof works when char k=0 or char $\mathbb{F}_q$ ) . In this paper we prove that if $k$ has positive characteristic and char$k\ne $ char $\mathbb{F}_q$, then the Steinberg module of $kG$ remains irreducible (see Theorem 2.2).
The reductive group $G$ is quasi-finite in the sense of [X, 1.8]. For quasi-finite groups Xi introduced the concept of quasi-finite irreducible module and raised the question whether an irreducible $kG$-module is always quasi-finite. For certain special linear groups, we show that the Steinberg modules of the groups are not quasi-finite with respect to some natural quasi-finite sequences of the groups (see Proposition 3.2).
\section{Preliminaries}
\def\ind{{\text {Ind}}}
\def\Hom{{\text {Hom}}}
\def\res{{\text {Res}}}
\def\bbf{\mathbb{F}_q}
\def\bbfa{\mathbb{F}_{q^a}}
\def\bbbf{\bar{\mathbb{F}}_q}
\def\bbc{\mathbb{C}}
In this section we recall some basic facts for reductive groups defined over a finite field, for details we refer to [C].
\subsection{} Let $G$ be a connected reductive group over an algebraically closure $\bbbf$
of a finite field $\mathbb{F}_q$ of $q$ elements, where $q$ is a power of a prime $p$. Assume that $G$ is defined over $\mathbb{F}_q$. Then
$G$ has a Borel subgroup $B$ defined over $\bbf$ and $B$ contains
a maximal torus $T$ defined over $\bbf$. The unipotent radical $U$ of $B$ is defined over $\bbf$. For any power $q^a$ of $q$,
we denote by $G_{q^a}$ the $\mathbb{F}_{q^a}$-points of $G$ and shall identify $G$ with its $\bar{\mathbb{F}}_q$-points. Then we have $G=\bigcup_{a=1}^{\infty}G_{q^a}$.
Similarly we define $B_{q^a}$, $T_{q^a}$ and $U_{q^a}$.
\subsection{} Let $N=N_G(T)$ be the normalizer of $T$ in $G$. Then $B$ and $N$ form a $BN$-pair of $G$.
Let $R\subset\text{Hom}(T,\bbbf^*)$ be the root system of $G$ and $R^+$ the set of positive roots determined by $B$. For $\alpha\in R^+$, let $U_\alpha$ be the corresponding root subgroup of $U$.
For any simple root $\alpha$ in $R$, let $s_\alpha$ be the corresponding simple reflection in the Weyl group $W=N/T$. For $w\in W$, $U$ has two subgroups $U_w$ and $U'_w$ such that $U=U'_wU_w$ and $wU'_xw^{-1}\subseteq U$. If $w=s_\alpha$ for some simple root $\alpha$, then $U_w=U_\alpha$ and we simply write $U'_\alpha$ for $U'_w$, which equals $\prod_{\beta\in R^+-\{\alpha\}}U_\beta$. In general,
let $w=s_{\alpha_i}\cdots s_{\alpha_2}s_{\alpha_1}$ be a reduced expression of $w$. Set $\beta_{j}=s_{\alpha_1}s_{\alpha_2}\cdots s_{\alpha_{j-1}}(\alpha_j)$ for $j=1,...,i$. Then
\def\st{\stackrel}
\def\sc{\scriptstyle}
\medskip
\noindent (a) $U_w=U_{\beta_i}\cdots U_{\beta_2}U_{\beta_1}$ and $U'_w=\displaystyle{\prod_{\st{\sc \beta\in R^+}
{w(\beta)\in R^+}}U_\beta}$.
\medskip
\noindent (b) If $\alpha$ and $\beta$ are positive roots and $w(\alpha)=\beta$, then $n_wU_\alpha n_w^{-1}=U_\beta$, where $n_w$ is a representative of $w$ in $N$.
\medskip
Now assume that $w_0=s_{\alpha_r}\cdots s_{\alpha_2}s_{\alpha_1}$ is a reduced expression of the longest element of $W$. Set $\beta_{j}=s_{\alpha_1}s_{\alpha_2}\cdots s_{\alpha_{j-1}}(\alpha_j)$ for $j=1,...,r$. Then
\medskip
\noindent (c) For any $1\le i\le j\le r$, $U_{\beta_j}\cdots U_{\beta_{i+1}}U_{\beta_i}$ is a subgroup of $U$ and
$U_{\beta_j}\cdots U_{\beta_{i+1}}U_{\beta_i}=U_{\beta_i}U_{\beta_j}\cdots U_{\beta_{i+1}}.$
\medskip
\def\va{\varepsilon_\alpha}
The roots subgroups $U_\alpha,\ \alpha\in R^+$, are also defined over $\bbf$. For each positive root, we fix an isomorphism $\varepsilon_\alpha:\bbbf\to U_\alpha$ such that $t\va(c)t^{-1}=\va(\alpha(t)c)$. Set $U_{\alpha,q^a}=\va(\bbfa)$.
\section{Infinite dimensional Steinberg modules}
In this section the main result (Theorem 2.2) of this paper is proved, which says that certain infinite dimensional Steinberg modules are irreducible.
\subsection{} Let $k$ be a field. For any one dimensional representation $\theta$ of $T$ over $k$, let $k_\theta$ be the corresponding $kT$-module.
We define the $k G$-module $M(\theta)=k G\otimes_{k B}k_\theta$. When $\theta$ is trivial representation of $T$ over $k$, we write $M(tr)$ for $M(\theta)$ and let $1_{tr}$ be a nonzero element in $k_\theta$. We shall also write $x1_{tr}$ instead of $x\otimes 1_{tr}$ for $x\in kG$.
For $w\in W=N/T$, the element $w1_{tr}$ is defined to be $n_w1_{tr}$, where $n_w$ is a representative in $N$ of $w$. This is well defined since $T$ acts on $k_\theta$ trivially.
Let $\eta=\sum_{w\in W} (-1)^{l(w)}w1_{tr}\in M(tr),$ where $l:W\to \mathbb{N}$ is the length function of $W$. Then $kU\eta$ is a submodule of $M(tr)$ and is called a Steinberg module of $G$, denoted by St, see [X, Prop. 2.3]. Xi proved that St is irreducible if $k$ is the field of complex numbers or $k=\bbbf$ (see [X, Theorem 3.2]). His argument in fact works for proving that St is irreducible whenever char$k=0$ or char$k=$char$\bbf$. The main result of this paper is the following.
\subsection{Theorem.} Assume that $k$ is a field of positive characteristic and char$k\ne$ char$\mathbb{F}_q$. Then the Steinberg module St is irreducible.
Combining Xi's result we have the following result.
\subsection{Corollary.} The Steinberg module St of $kG$ is irreducible for any field $k$.
\subsection{} We need some preparation to prove the theorem.
Let $s_{\alpha_r}s_{\alpha_{r-1}}\cdots s_{\alpha_1}$ be a reduced expression of the longest element $w_0$ of $W$. Set $\beta_i=s_{\alpha_1}\cdots s_{\alpha_{i-1}}(\alpha_i)$. Then $R^+$ consists of $\beta_r,\beta_{r-1},...,\beta_1$, and $U=U_{\beta_r}U_{\beta_{r-1}}\cdots U_{\beta_1}$. Let $n_i$ be a representative in $N=N_G(T)$ of $s_{\alpha_i}$ and set $n=n_rn_{r-1}\cdots n_1$. Note that the elements $z\eta,\ z\in U$, form a basis of St.
\subsection{Lemma.} Let $u\in U_{q^a}$. If $u$ is not the neutral element $e$ of $U$, then the sum of all coefficients of $nu\eta$ in terms the basis $z\eta,\ z\in U$, is 0.
Proof. Let $\alpha_i$ and $\beta_i$ be as in subsection 2.4. Then $u=u_{r}u_{r-1}\cdots u_1,$ $ u_m\in U_{\beta_m}$. Assume that $u_1=u_2=\cdots u_{i-1}=e$ but $u_i\ne e$, where $e$ is the neutral element of $G$. We use induction on $i$ to prove the lemma. Note that $n_i\eta=-\eta$ for $i=1,2,...,r$.
Assume that $i=r$. Then $nu\eta=(-1)^{r-1}n_ru'_r\eta$, where $u'_r=n_{r-1}\cdots n_1u_rn_1^{-1}\cdots n_{r-1}^{-1}\in U_{\alpha_r}$. According to the proof of [S, Lemma 1] (see also proof of [X, Proposition 2.3]), there exists $x_r\in U_{\alpha_r}$ such that $n_ru'_r\eta=(x_r-1)\eta$. So the lemma is true in this case.
Now assume that the lemma is true for $r,r-1,...,i+1$, we show that it is also true for $i$. In this case we have $nu=(-1)^{i-1}n_r\cdots n_i u'_{r}u'_{r-1}\cdots u'_i\eta$, where $u'_j=n_{i-1}\cdots n_1u_jn_1^{-1}\cdots n_{i-1}^{-1}\in n_{i-1}\cdots n_1U_{\beta_j}n_1^{-1}\cdots n_{i-1}^{-1}= U_{\gamma_j}$, where $\gamma_j=s_{\alpha_i}\cdots s_{\alpha_{j-1}}(\alpha_j)$, $j=i,i+1,...,r$. Then $u'_i\ne e$. Note that $\gamma_i=\alpha_i$. According to the proof of [S, Lemma 1], there exists $x_i\in U_{\gamma_i}$ such that $n_iu'_i\eta=(x_i-1)\eta$.
If $u'_r\cdots u'_{i+1}=e$, we are done. Now assume that $u'=u'_r\cdots u'_{i+1}\ne e$. Since both $M_i=U_{\gamma_r}U_{\gamma_{r-1}}\cdots U_{\gamma_{i}}$ and $M_{i+1}=U_{\gamma_r}U_{\gamma_{r-1}}\cdots U_{\gamma_{i+1}}$ are subgroups of $U$ and $M_{i+1}x_i= x_iM_{i+1}$,we see that $u'x_i=x_iu''$ for some $u''\in M_{i+1}$ and $u''\ne e$. Thus
$$(-1)^{i-1}nu\eta=n_r\cdots n_{i+1}u'(x_i-1)\eta=n_r\cdots n_{i+1}x_iu''\eta-n_r\cdots n_{i+1}u'\eta.$$
By induction hypotheses, we know that the sum of the coefficients of $n_r\cdots n_{i+1}u''\eta$ and the sum of the coefficients of $n_r\cdots n_{i+1}u'\eta$ are 0. Since $n_r\cdots n_{i+1}x_in_{i+1}^{-1}\cdots n_r^{-1}\in U$, we see that the sum of the coefficients of $nu\eta=(-1)^{i-1}n_r\cdots n_{i+1}u'(x_i-1)\eta$ is 0.
The lemma is proved.
\subsection{Lemma.} Let $V$ be a nonzero submodule of St. Then there exists an integer $a$ such that $\sum_{x\in U_{q^a}}x\eta$ is in $V$.
Proof. Let $v$ be a nonzero element in $V$. Then $v\in kU_{q^a}\eta$ for some integer $a$. Let $v=\sum_{y\in U_{q^a}}a_yy\eta.$ We may assume that $a_e\ne 0$. Otherwise choose $y\in U_{q^a}$ such that $a_y$ is nonzero and replace $v$ by $y^{-1}v$.
By Lemma 2.5 we see that the sum $A$ of all the coefficients of $nv$ in terms of the basis $z\eta,$ $z\in U$, is $(-1)^{l(w_0)}a_e\ne 0$. Thus $\sum_{\in U_{q^a}}xnv=A\sum_{x\in U_{q^a}} x\eta$. The lemma is proved.
\subsection{} Now we can prove the theorem. We show that St=$kGv$ for any nonzero element $v$ in St. Let $V=kGv$. Let $\alpha_i$ and $\beta_i$ be as in subsection 2.4. For any positive integer $b$, set $X_{i,q^b}=U_{\beta_r,q^b}U_{\beta_{r-1},q^b}\cdots U_{\beta_i,q^b}$. Then $X_{i,q^b}$ is a subgroup of $U$ and $X_{i,q^b}=X_{i+1,q^b}U_{\beta_i,q^b}$. Clearly $X_{i,q^b}$ is a subgroup $X_{i,q^{b'}}$ if $\mathbb{F}_{q^b}$ is a subfield of $\mathbb{F}_{q^{b'}}.$
We use induction on $i$ to show that there exists positive integer $b_i$ such that the element $\sum_{x\in X_{i,q^{b_i}}}x\eta$ is in $V$. For $i=1$, this is true by Lemma 2.6. Now assume that $\sum_{x\in X_{i,q^{b_i}}}x\eta$ is in $V$, we show that $\sum_{x\in X_{i+1,q^{b_{i+1}}}}x\eta$ is in $V$ for some $b_{i+1}$.
Let $c_1,...,c_{q^{b_i}+1}$ be a complete set of representatives of all cosets of $\mathbb{F}^*_{q^{b_i}}$ in $\mathbb{F}^*_{q^{2b_i}}$. Choose $t_1,...,t_{q^{b_i}+1}\in T$ such that $\beta_i(t_j)=c_j$ for $j=1,...,q^{b_i}+1$. Note that $t^{-1}\eta=\eta$ for any $t\in T$. Thus
$$\sum_{j=1}^{q^{b_i}+1}t_j\sum_{x\in U_{\beta_i,q^{b_i}}}x\eta=q^{b_i}\eta+\sum_{x\in U_{\beta_i,q^{2b_{i}}}}x\eta.$$
Since $X_{i,q^{b_i}}=X_{i+1,q^{b_i}}U_{\beta_i,q^{b_i}}$ and $\sum_{x\in X_{i,q^{b_i}}}x\eta$ is in $V$, we see
\begin{equation*}
\begin{split}
\xi=& \sum_{j=1}^{q^{b_i}+1}t_j\sum_{x\in X_{i,q^{b_i}}}x\eta\\
=&\sum_{j=1}^{q^{b_i}+1}t_j\sum_{y\in X_{i+1,q^{b_i}}}y\sum_{x\in U_{\beta_i,q^{b_i}}}x\eta\\
=&\sum_{y\in X_{i+1,q^{b_i}}}\sum_{j=1}^{q^{b_i}+1}t_jyt_j^{-1}(t_j\sum_{x\in U_{\beta_i,q^{b_i}}}x\eta)\in V.
\end{split}
\end{equation*}
Choose $b_{i+1}$ such that all $\alpha_m(t_j)$ ($r\ge m\ge i$) are contained in $\mathbb{F}_{q^{b_{i+1}}}$. Then $\mathbb{F}_{q^{b_{i+1}}}$ contains $\mathbb{F}_{q^{2b_{i}}}$. Thus $t_jyt_j^{-1}$ is in $X_{i+1,q^{b_i+1}}$ for any $y\in X_{i+1,q^{b_i}}$. Let $Z\in kG$ be the sum of all elements in $X_{i+1,q^{b_i+1}}$. Then we have
\noindent (1) $\displaystyle {Z\xi=q^{(r-i)b_i}Z\sum_{j=1}^{q^{b_i}+1}t_j\sum_{x\in U_{\beta_i,q^{b_i}}}x\eta=q^{(r-i)b_i}Z(q^{b_i}\eta+\sum_{x\in U_{\beta_i,q^{2b_{i}}}}x\eta)\in V.}$
Since $\sum_{x\in X_{i,q^{b_i}}}x\eta$ is in $V$, we have $\displaystyle \sum_{x\in X_{i,q^{2b_{i}}}}x\eta\in V.$ Thus
\noindent(2) $\displaystyle Z\sum_{x\in X_{i,q^{2b_{i}}}}x\eta=q^{2(r-i)b_i}Z\sum_{x\in U_{\beta_i,q^{2b_i}}}x\eta\in V.$
Since $q\ne 0$ in $k$, combining (1) and (2) we see that $Z\eta\in V$, i.e., $\sum_{x\in X_{i+1,q^{b_{i+1}}}}x\eta$ is in $V$.
Note that $X_{r,q^{b_r}}=U_{\beta_r,q^{b_r}}$. Now we have
$$\sum_{x\in U_{r,q^{b_r}}}x\eta\in V\quad\text{and}\quad\sum_{x\in U_{r,q^{2b_r}}}x\eta\in V.$$
The above arguments show that
$$q^{b_r}\eta+\sum_{x\in U_{r,q^{2b_r}}}x\eta\in V.$$
Therefore $q^{b_r}\eta$ is in $V$. So $V$ contains $kG\eta$=St, hence $V=$St. The theorem is proved.
\subsection{Remark.} Let St${}_a=kG_{q^a}\eta$. Then St${}_a$ is the Steinberg module of $kG_{q^a}$, which is not irreducible in general. As an example, say, $G=SL_2(\bbbf)$ and $q$ is odd, char$k=2$. Since $q^a+1$ is always divisible by 2=char$k$, St${}_a$ is not irreducible for any positive number $a$ (see [S, Theorems 3]). However, by Theorem 2.2, St is irreducible $kG$-module.
\section{Non-quasi-finite irreducibility of certain Steinberg modules}
In this section we show that for certain special linear groups the Steinberg modules of the groups are not quasi-finite with respect to some natural quasi-finite sequences of the groups, see Proposition 3.2.
\subsection{} By definition, a group
$G$ is quasi-finite if $G$ has a sequence $G_1,\ G_2,\ ,...,\ G_n,\
... $ of finite subgroups
such that $G$ is the union of all $G_i$ and for any positive integers $i,j$ there exists integer $r$
such that $G_i$ and $G_j$ are contained in $G_r$. The sequence $G_1,G_2,G_3,...$ is called a quasi-finite sequence of $G$. An irreducible module (or representation) $M$ of
$G$ is {\it quasi-finite} (with respect to the quasi-finite sequence
$G_1,G_2,G_3,...)$ if it has a sequence of subspaces $M_1$, $ M_2$,
$ M_3, $ ... of $M$ such that (1) each $M_i$ is an irreducible
$G_i$-submodule of $M$, (2) if $G_i$ is a subgroup of $G_j$, then
$M_i$ is a subspace of $M_j$, (3) $M$ is the union of all $M_i$. The
sequence $M_1$, $ M_2$, $ M_3, $ ... will be called a quasi-finite
sequence of $M$. See [X, 1.8]
The following question was raised in [4, 1.8]: is every irreducible G-module quasi-finite (with respect to a certain quasi-finite sequence
of G).
The main result of this section is the following result.
\subsection{Proposition.} Let $G=SL_n(\bbbf)$ and $k$ a field of positive characteristic. Assume that char$k$ divides $(1+q^{a})(1+q^{a}+q^{2a})\cdots(1+q^{a}+\cdots+q^{(n-1)a})$ for all positive integers $a$. If a quasi-finite sequence of $G$ is a subsequence of $SL_n(\bbf),$ $SL_n(\mathbb{F}_{q^2}),$ $ SL_n(\mathbb{F}_{q^3}),$ $ SL_n(\mathbb{F}_{q^4}),...$, then the Steinberg module St of $kG$ is not quasi-finite with respect to the quasi-finite sequence.
Proof. Let $G_1,\ G_2,\ ,...,\ G_n,\
... $ be a quasi-finite sequence of $G$. Assume that the quasi-finite sequence is a subsequence of $SL_n(\bbf),$ $SL_n(\mathbb{F}_{q^2}),$ $ SL_n(\mathbb{F}_{q^3}),$ $ SL_n(\mathbb{F}_{q^4}),...$. If St is quasi-finite with respect to this quasi-finite sequence, then there exists a sequence of subspaces $M_1$, $ M_2$,
$ M_3, $ ... of St such that (1) each $M_i$ is an irreducible
$G_i$-submodule of $M$, (2) if $G_i$ is a subgroup of $G_j$, then
$M_i$ is a subspace of $M_j$, (3) $M$ is the union of all $M_i$.
Choose a nonzero element $v\in M_1$. By the proof of Theorem 2.2, there exists $x\in kG$ such that $xv=\eta$. Since $G$ is the union of all $H_a=SL_n(\mathbb{F}_{q^a})$, there exists positive integer $i$ such that $x\in kH_i$. Since $G$ is the union of all $G_a$, there exists $j$ such that $H_i$ is included in $G_j$. Note that $G_j=H_{j'}$ for some positive integer $j'$. Choose integer $d$ such that $G_d$ includes both $G_1$ and $G_j$. Then $M_d$ includes $M_1$ and $M_j$. Moreover, $x\in kG_d$, so that $xv=\eta$ is in $M_d$ and $M_d$ includes $kG_d\eta$. Since $G_d=H_{d'}$ for some $d'$ and $kH_{d'}\eta$ is not irreducible, $M_d$ is not irreducible. This contradicts the assumption that $M_d$ is irreducible $G_d$-module. The proposition is proved.
\subsection{Remark} Assume that $n$ is power of a prime $p'$ and $k$ has characteristic $p'$. If $n,q$ are coprime, then $n$ divides $(1+q^{a})(1+q^{a}+q^{2a})\cdots(1+q^{a}+\cdots+q^{(n-1)a})$ for all positive integers $a$. To see this, let $A_m= 1+q^{a}+\cdots+q^{(m-1)a}$, $m=2,...,n$. Since $n,q$ are coprime, if $A_m\equiv 1$(mod $n$), then $m>2$ and $A_{m-1}$ is divisible by $n$. If $A_m\not\equiv 1$(mod $n$) for $m=2,...,n$, then either some $A_m$ is divisible by $n$ or $A_m\equiv A_l$(mod $n$) for some $n\ge m>l\ge 2$. Since $n,q$ are coprime, we have $m\ge l+2$. Then $A_{m-l}$ is divisible by $n$. By Proposition 2.2, in this case, the Steinberg module St of $kSL_n(\bbbf)$ is not quasi-finite for a quasi-finite sequence of $G$ whenever it is a subsequence of $SL_n(\bbf),$ $SL_n(\mathbb{F}_{q^2}),$ $ SL_n(\mathbb{F}_{q^3}),$ $ SL_n(\mathbb{F}_{q^4}),...$.
However, it is not clear that whether St is quasi-finite with other quasi-finite sequences of $SL_n(\bbbf)$.
For other reductive groups, one discusses similarly.
\subsection{} Assume that $G$ is quasi-finite and has sequence of normal subgroups $\{1\}=G_0\subset G_1\subset \cdots\subset
G_n=G$ such that all
$G_i/G_{i-1}$ are abelian. Xi asks whether any irreducible $\mathbb{C}G$-module is isomorphic to the induced module of a one dimensional
module of a subgroup of $G$ (see [X, 1.12]). The question has a negative answer ever for finite groups, for instance, the two-dimensional irreducible complex representation of $SL_2(\mathbb F_3)$ is an counterexample. Perhaps for the question the condition of all
$G_i/G_{i-1}$ being abelian should be further strengthened to all
$G_i/G_{i-1}$ being cyclic.
\bigskip
{\bf Acknowledgement.} I am very grateful to Professor Nanhua Xi for guidance and great helps in writing the paper. The work was done during my visit to the Academy of Mathematics and Systems Science, Chinese Academy of Sciences as a visiting student for the academic year 2014-2015. I thank the Academy of Mathematics and Systems Science for hospitality.
\bigskip | 207,466 |
TITLE: Subgroups of $\mathrm{GL}(n,\mathbb{Z})$ which are not finitely generated
QUESTION [11 upvotes]: The group $\mathrm{GL}(n,\mathbb{Z})$ is finitely generated: take for example diagonal matrices, permutations and one elementary matrix (upper triangular). Are there some simple / nice examples of non-finitely generated subgroups? If possible, for $n$ not too big.
REPLY [2 votes]: I want to use the same idea as Lee Mosher, based around the well-known free subgroup, but give a more explicit example (that is, the generators have a nicer form). In the free group $F(a, b)$ the set $\{a^kba^k:k\in\mathbb{Z}\}$ freely generates a free group on countably many generators (for a proof, see here). So, the matrices of the following form generate a non-finitely generated group of $GL(2,\mathbb{Z})$.
$$
\left(
\begin{array}{cc}
1&0\\
2&1
\end{array}
\right)^k
\left(
\begin{array}{cc}
1&2\\
0&1
\end{array}
\right)
\left(
\begin{array}{cc}
1&0\\
2&1
\end{array}
\right)^k
$$
These matrices are easily seen to have the following form.
$$
\left(
\begin{array}{cc}
4k+1&2\\
4k(2k+1)&4k+1
\end{array}
\right)
$$
As an aside, $\operatorname{GL}(n, \mathbb{Z})$ also contains finitely generated groups which are not finitely presentable. I cannot think of an obvious proof of this fact, and indeed my reasoning requires some serious results! The first result is about small cancellation groups. Small cancellation groups are groups with presentations where the relators do not interact very much with one another (any cancellation between the relators is "small"). It is a recent result, using some rather complicated machinery, that such groups are linear over $\mathbb{Z}$, so given a small cancellation group $G$ there exists some $n$ such that $G$ embeds into $\operatorname{GL}(n, \mathbb{Z})$. It is then a result of Rips that there exist small cancellation groups which contain finitely generated subgroups which are not finitely presentable: see Rips' paper Subgroups of small cancellation groups. | 115,799 |
TITLE: Why are there some constraints on linearity of expectation?
QUESTION [2 upvotes]: I'm quoting from here:
Expectations are linear, for example, $ E[αf(x) +βg(x)] =αE[f(x)]
+βE[g(x)],$ when α and β are fixed (not random and not depending on x).
My question about contraints on expectation property. It is clear to me that $ α $ and $ β$ should not depend on $ x$, however I haven't understand why they can not be a random variable which are drawn from a distribution that is independent from distribution of $ x$. Is there a special propert of integration that doesn't allow the random variables to taken out of integration without changing the remaining part of it?
REPLY [2 votes]: Let's start by the core idea. The linearity of the expectation can be formulated in a more general way as:
Let $X_1, X_2,\ldots,X_n$ be random variables and $a_1, a_2,\ldots,a_n$ are scalars. Then $E[a_1X_1+a_2X_2+\cdots+a_nX_n] = a_1E[X_1]+a_2E[X_2]+\cdots+a_nE[X_n]$.
From this definition of linearity we have two main points:
The $X_i$'s doesn't have to be independent (one constraint less!)
The $a_i$'s are scalars, not random variables.
Linearity of the expectation comes from the linearity of the integration (summation) and you would see that this in turn is related to the important concept of a linear combination, where the coefficients of the combination are scalars.
Now, suppose that we insist in using independent random variables instead of scalars as coefficients of the linear combination. Let's explore what are actually the properties that we would use with your definition.
Let me denote $f(X)$ as $X_1$ and $g(X)$ as $X_2$.
Let $A$ and $B$ be random variables independent of $X$ and $Y = AX_1+BX_2$. Let $W = AX_1$ and $Z=BX_2$. We have then $Y = W + Z$. How do we compute $E[Y]$? Let's see:
\begin{align}
E[Y] &= \iint (w+z)f_{W,Z}(w,z)dwdz\\
&= \iint wf_{W,Z}(w,z)dwdz + \iint zf_{W,Z}(w,z)dwdz\\
&= \int w \left(\int f_{W,Z}(w,z)dz\right) dw + \int z \left(\int f_{W,Z}(w,z)dw\right) dz\\
&= \int w f_{W}(w) dw + \int z f_{Z}(z) dz\\
&= E[W] + E[Z],
\end{align}
where in the second inequality we applied the linearity property of the integral, in the fourth one we compute the marginal PDF of one variable from the integration of the joint PDF over the other random variable and in the last one we use the definition of expected value.
Now, how do we compute $E[W]$?
\begin{align}
E[W] = E[AX_1] &= \iint \alpha x_1f_{A,X_1}(\alpha,x_1)d\alpha dx_1\\
&= \int \alpha f_{A}(\alpha) d\alpha \int x_1f_{X_1}(x_1) dx_1 = E[A]E[X_1]
\end{align}
but here we do not use the linearity of the integral! What we have used in the second line to factor out is the independence of the random variables and a corollary of the Fubini's theorem!
So, to get the result
$$E[Y] = E[A]E[f(X)]+E[B]E[g(X)],$$
what we are actually applying is two properties of the expected value, the linearity property and the property of the correlation (expected value of the product) of independent random variables. | 25,585 |
\begin{document}
\begin{abstract}
We study nodal del Pezzo 3-folds of degree $1$
(also known as double Veronese cones)
with~$28$ singularities, which is the maximal possible number
of singularities for such varieties. We show that they are in one-to-one
correspondence with smooth plane quartics and use this correspondence to study their automorphism groups.
As an application, we find all $G$-birationally rigid varieties of this kind, and
construct an infinite number of non-conjugate embeddings of the group $\mathfrak{S}_4$ into the space Cremona group.
\end{abstract}
\address{\emph{Hamid Ahmadinezhad}
\newline
\textnormal{Department of Mathematical Sciences, Loughborough University
\newline
Loughborough LE11 3TU, UK
\newline
\texttt{[email protected]}}}
\address{ \emph{Ivan Cheltsov}\newline \textnormal{School of Mathematics, The
University of Edinburgh
\newline \medskip Edinburgh EH9 3JZ, UK
\newline
National Research University Higher School of Economics,
Laboratory of Algebraic Geometry,
\newline
6 Usacheva street, Moscow, 117312, Russia
\newline
\texttt{[email protected]}}}
\address{ \emph{Jihun Park}\newline \textnormal{Center for Geometry and Physics, Institute for Basic Science
\newline \medskip 77 Cheongam-ro, Nam-gu, Pohang, Gyeongbuk, 37673, Korea \newline
Department of
Mathematics, POSTECH
\newline
77 Cheongam-ro, Nam-gu, Pohang, Gyeongbuk, 37673, Korea \newline
\texttt{[email protected]}}}
\address{\emph{Constantin Shramov}\newline \textnormal{Steklov Mathematical Institute of Russian Academy of Sciences
\newline
\medskip 8 Gubkina street, Moscow, 119991, Russian
\newline
National Research University Higher School of Economics, Laboratory of Algebraic Geometry,
\newline
6 Usacheva street, Moscow, 117312, Russia
\newline \texttt{[email protected] }}}
\maketitle
\tableofcontents
\section{Introduction}
\label{section:intro}
It is a common experience that varieties with the maximal possible
number of singularities in a given class have fascinating geometric properties.
Classical examples of such varieties include the Burkhardt quartic (see~\cite{JSBV},
\cite{CheltsovPrzyjalkowskiShramov-BarBur}), the Segre cubic, and many others.
The Segre cubic, which is the unique three-dimensional cubic hypersurface with $10$ isolated
singular points, can be thought of as a \emph{del Pezzo 3-fold}. Recall that a del Pezzo 3-fold is a
Fano 3-fold with at most terminal
Gorenstein singularities
whose canonical class is divisible by $2$ in the Picard group; in particular, the singularities of such 3-folds
are always isolated.
Del Pezzo 3-folds are classified according to their \emph{degrees}, which are defined as the cubes of the half-anticanonical classes.
Such varieties with a maximal possible number of singularities were
described in~\cite[Theorem~7.1]{Prokhorov}.
The smaller degrees such del Pezzo 3-folds have, the more intriguing their geometry is.
Del Pezzo 3-folds of degree $4$ can have at most six singularities,
and there exists a unique such variety with exactly six singular points.
It has been studied in \cite{Avilov-IntTwoQuadrics}.
There are many works devoted to the geometry of the Segre cubic (see~\mbox{\cite[\S2]{Dolgachev-Segre}}, \cite{Finkelnberg},
\cite[\S5]{CheltsovShramov-A6}, \cite{Avilov-Segre}, and \cite{Avilov-RealSegre}).
Del Pezzo 3-folds of degree~$2$ with~$16$ singular points, which is the maximal possible
number of singularities in this case, are
quartic double solids branched along Kummer quartic surfaces.
Thus their properties are closely related to those of
the corresponding abelian surfaces. In particular, such del Pezzo 3-folds are in one-to-one correspondence with smooth curves of genus~$2$
(see for instance~\mbox{\cite[Propositions~1.2(i) and~1.3]{Keum}}).
The current paper studies the geometry
of del Pezzo 3-folds of degree~$1$ with $28$ singular points,
which is the maximal possible number of singularities for such varieties
(see~\mbox{\cite[Theorem~1.7]{Prokhorov}} and~\mbox{\cite[Remark~1.8]{Prokhorov}}).
Each del Pezzo 3-fold of degree~$1$ is a hypersurface of degree six in the weighted projective space~\mbox{$\mathbb{P}(1,1,1,2,3)$} (see \cite[Theorem~3.2.5(i)]{Isk-Prokh}). This implies that it is a double cover of the cone over the Veronese surface in~$\mathbb{P}^5$.
For this reason, del Pezzo 3-folds of degree $1$ are often called \emph{double Veronese cones}.
Let $V$ be a double Veronese cone with $28$ singular points.
Then $-K_V\sim 2H$, where $H$ is an ample Cartier divisor on $V$ such that $H^3=1$.
The existence of the following diagram was proven in \cite{Prokhorov}.
\begin{equation}
\label{equation:Prokhorov-diagram}
\xymatrix{
\widehat{\mathbb{P}^3}\ar@{->}[d]_{\pi}\ar@{->}[rr]^{\phi}&&V\ar@{-->}[d]^{\kappa}\\%
\mathbb{P}^3\ar@{-->}[rr]&&\mathbb{P}^2}
\end{equation}
In this diagram, $\phi$ is a small resolution of all singular points of the 3-fold $V$,
the morphism $\pi$ is the blow up of $\mathbb{P}^3$ at seven distinct points $P_1,\ldots,P_7$ that satisfy certain explicit
generality conditions, the rational map $\kappa$ is given by the linear system $|H|$,
and the rational map from $\mathbb{P}^3$ to $\mathbb{P}^2$ is given by the linear system (net) of quadrics passing through $P_1,\ldots,P_7$.
Furthermore, Prokhorov implicitly verified
in~\cite{Prokhorov}
that each singular point of $V$ is an isolated ordinary double point (a node).
Note that $V$ is not $\mathbb{Q}$-factorial
since $\mathrm{Cl}(V)\cong\mathbb{Z}^8$ and $\mathrm{Pic}(V)\cong\mathbb{Z}$.
Nevertheless, the class group~$\mathrm{Cl}(V)$ can be naturally equipped with the intersection form:
\begin{equation}
\label{equation:E7}
\big(D_1,D_2\big)=D_1\cdot D_2\cdot H
\end{equation}
for any two Weil divisors $D_1$ and $D_2$ on the 3-fold $V$. Furthermore,
Prokhorov proved in~\mbox{\cite[Theorem~1.7]{Prokhorov}} (cf.~\mbox{\cite[Remark~1.8]{Prokhorov}}) that
the orthogonal complement to the canonical divisor~$K_V$ in~$\mathrm{Cl}(V)$ is the lattice~$\mathbf{E}_7$ (cf.~\mbox{\cite[p.~107]{DolgachevOrtland}}).
The same lattice appears in the Picard group of the double cover of $\mathbb{P}^2$ branched along a smooth plane quartic curve.
This suggests a possible relation between double Veronese cones with $28$ nodes and smooth plane quartic curves.
As far as we know, such a relation was first pointed out in \cite[\S\,IX.6]{DolgachevOrtland}. A goal in the present
paper is to make this
relation precise and detailed.
The net of quadrics in \eqref{equation:Prokhorov-diagram}
has exactly eight distinct base points,
including the points~\mbox{$P_1,\ldots,P_7$}; such a collection of eight points is
called a \emph{regular Cayley octad} (see Definition~\ref{definition:Aronhold-heptad} for details).
Moreover, the Hessian curve of this net is a smooth plane quartic curve, which we denote by $C$.
One can show that the isomorphism class of the curve $C$ is independent of the choice of the commutative diagram~\eqref{equation:Prokhorov-diagram}.
In fact, the curve $C$ is projectively dual
to the discriminant curve of the rational elliptic fibration~$\kappa$
(see~\S\ref{section:1-1} for details).
The first main result of the present paper is
\begin{theorem}\
\label{theorem:one-to-one}
Assigning the curve $C$ to the 3-fold $V$ gives
a one-to-one correspondence between the isomorphism classes of smooth plane quartic curves and the isomorphism classes of $28$-nodal double Veronese cones.
\end{theorem}
Two alternative proofs of Theorem~\ref{theorem:one-to-one} will be presented in this paper.
The first proof starts with a plane quartic curve $C$ and produces an explicit equation that defines the corresponding 3-fold $V$ in the weighted projective space $\mathbb{P}(1,1,1,2,3)$ (see equation~\eqref{eq:V1} and~\S\ref{section:equations} for details). Another proof uses machinery from the theory of nets of quadrics to recover a
commutative diagram~\eqref{equation:Prokhorov-diagram}
and check that the resulting 3-fold does not depend on the choices we have to make on the way.
Observe
that for a given $28$-nodal double Veronese cone
the commutative diagram~\eqref{equation:Prokhorov-diagram} is not unique.
However, we can refine Theorem~\ref{theorem:one-to-one} to characterize all such diagrams
in terms of the corresponding smooth plane quartic curve.
Essentially, the diagram~\eqref{equation:Prokhorov-diagram} depends only on the choice of seven distinct points $P_1,\ldots,P_7$ in $\mathbb{P}^3$ up to projective transformations. We will show that there are exactly~$288$ diagrams~\eqref{equation:Prokhorov-diagram} up to projective transformations of $\mathbb{P}^3$ for a given general $28$-nodal double Veronese cone $V$,
and will also give a way to compute the number of such diagrams for an arbitrary~$V$ (see Corollaries~\ref{corollary:Aronhold-sets} and~\ref{corollary:288-diagram}). Here, the number $288=36\cdot 8$
equals
the number of even theta characteristics of $C$ times the number of choices for a point in the corresponding regular Cayley octad.
In other words, it is the number of Aronhold systems
on $C$. We refer the reader to~\S\ref{section:theta} for details.
Note that there are other classical constructions that associate certain 3-folds to smooth plane quartics.
We recall some of them in \S\ref{section:question}.
It would be interesting to find the connections between these varieties and
$28$-nodal double Veronese cones.
The second result of the present paper is the following.
\begin{theorem}
\label{theorem:Aut}
One has
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathrm{Aut}(C),
$$
where the generator of the cyclic group $\mumu_2$ of order $2$
acts on $V$ by the Galois involution of the double cover of the Veronese cone.
\end{theorem}
We point out that Theorem~\ref{theorem:Aut} can be obtained from \cite[\S\,IX.6]{DolgachevOrtland} and Torelli theorem (see for instance~\cite{Weil}).
However, we provide a more elementary proof. Another advantage of our proof is its explicit nature, enabling an easier access to details of the group actions on the $3$-folds, which will be useful later on when we study their birational properties.
Note that the commutative diagram \eqref{equation:Prokhorov-diagram} is not necessarily $\mathrm{Aut}(V)$-equivariant.
However, the rational map $\kappa$ in \eqref{equation:Prokhorov-diagram} always is.
This gives a natural action of the group $\mathrm{Aut}(V)$ on the plane $\mathbb{P}^2$.
This action agrees with the direct product structure in Theorem~\ref{theorem:Aut}
(see the proof of Theorem~\ref{theorem:Aut} in~\S\ref{section:1-1}).
The automorphism groups of smooth plane quartic curves are completely classified (see~\mbox{\cite[Theorem~6.5.2]{Dolgachev}}, \cite{Henn}, or \cite{KuribayashiKomiya}).
This gives us the full list of the automorphism groups of $28$-nodal double Veronese cones.
\begin{example}
\label{example:V1-S4}
Let $C$ be the plane quartic curve given by
\begin{equation}\label{eq:S4-quartic-general}
x^4+y^4+z^4+\lambda(y^2z^2+x^2z^2+x^2y^2)=0,
\end{equation}
where $\lambda$ is a constant different from $-2$, $2$, and $-1$ (in these three cases, the curves are singular).
The corresponding $28$-nodal double Veronese cone $V$ is the hypersurface of degree $6$ in~\mbox{$\mathbb{P}(1,1,1,2,3)$}
that is given by
$$
w^2=v^3-g_4(s,t,u)v+g_6(s,t,u),
$$
where $s$, $t$, $u$, $v$ and $w$ are weighted homogeneous coordinates on $\mathbb{P}(1,1,1,2,3)$ whose weights
are equal to $1$, $1$, $1$, $2$ and $3$, respectively. From \S\ref{section:equations} and Appendix~\ref{section:appendix} we obtain
$$
g_4(s,t,u)=\frac{(\lambda^2 + 12)}{3}(s^4 +t^4+u^4)+ \frac{2(\lambda^2+6\lambda)}{3}(t^2u^2+s^2u^2+s^2t^2)
$$
and
\[\begin{split}
g_6(s,t,u)&=\frac{2( - \lambda^3 + 12\lambda^2 + 12\lambda)}{9}(t^4u^2+t^2u^4+s^4u^2+s^2u^4+s^4t^2 + s^2t^4) +\\
&+\frac{2(-\lambda^3 + 36\lambda)}{27}(s^6+t^6+u^6)+\frac{4(8\lambda^3 - 9\lambda^2 + 36 )}{9}s^2t^2u^2.
\end{split}\]
Let $\mathfrak{G}$ be the subgroup in $\mathrm{Aut}(V)$ that is generated by
\[\begin{split}
&[s:t:u:v:w]\mapsto [-s:-t:u:v:w],\\
&[s:t:u:v:w]\mapsto [u:s:t:v:w],\\
&[s:t:u:v:w]\mapsto [t:s:u:v:-w].\\
\end{split}\]
Then $\mathfrak{G}\cong\mathfrak{S}_4$. Let $\tau\in\mathrm{Aut}(V)$ be the Galois involution of the double cover of the Veronese cone.
If~$\lambda\ne 0$ and $\lambda^2+3\lambda+18\ne 0$, then $\tau$ and
$\mathfrak{G}$ generate the group
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathfrak{S}_4
$$
(see \cite[Theorem~6.5.2]{Dolgachev}, \cite{Henn}, or \cite{KuribayashiKomiya}).
If~\mbox{$\lambda=0$}, then $C$ is the Fermat quartic, so that
$$
\mathrm{Aut}(V)\cong\mumu_2\times(\mumu_4^2\rtimes\mathfrak{S}_3),
$$
where $\mumu_n$ denotes the cyclic group of order $n$.
Finally, if $\lambda^2+3\lambda+18=0$, then $C$ is isomorphic to the Klein quartic, so that
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathrm{PSL}_2(\mathbb{F}_7).
$$
\end{example}
It follows from \eqref{equation:Prokhorov-diagram} that $28$-nodal double Veronese cones are rational.
Thus, their automorphism groups are embedded into the space Cremona group, that is the group of birational selfmaps of $\mathbb{P}^3$.
For a given subgroup $G\subset\mathrm{Aut}(V)$,
to study its conjugacy class in the space Cremona group,
one should understand $G$-equivariant birational geometry of the 3-fold $V$.
From this point of view, the most interesting groups $G$ are such that
$\mathrm{rk}\,\mathrm{Cl}(V)^G=1$, and $V$ is $G$-birationally (super-)rigid (see \cite[Definition~3.1.1]{CheltsovShramov-V5}).
In this paper, we prove
\begin{theorem}
\label{theorem:rigid}
Let $V$ be a $28$-nodal double Veronese cone and $G$ be a subgroup of $\mathrm{Aut}(V)$.
Then the following four conditions are equivalent:
\begin{enumerate}
\item $\mathrm{rk}\,\mathrm{Cl}(V)^G=1$ and $V$ is $G$-birationally rigid;
\item $\mathrm{rk}\,\mathrm{Cl}(V)^G=1$ and $V$ is $G$-birationally super-rigid;
\item $\mathrm{rk}\,\mathrm{Cl}(V)^G=1$ and $G$ does not fix any point in $\mathbb{P}^2$;
\item $V$ is a double Veronese cone from Example~\ref{example:V1-S4} and $G$ contains the subgroup $\mathfrak{G}$.
\end{enumerate}
\end{theorem}
It is known that the plane Cremona group,
that is the group of birational selfmaps of~$\mathbb{P}^2$, contains
an infinite number of non-conjugate subgroups isomorphic to
the symmetric group~$\mathfrak{S}_4$
(see~\mbox{\cite[\S8]{DolgachevIskovskikh}},
and cf.~\mbox{\cite[Theorem~1.5]{Lucas}}).
Theorem~\ref{theorem:rigid} implies the following result.
\begin{corollary}
\label{corollary:S4}
The space Cremona group contains an infinite number of non-conjugate subgroups isomorphic to $\mathfrak{S}_4$.
\end{corollary}
It is known from \cite{CheltsovShramov-PSL} and \cite[Theorem~2.2]{Ahmadinezhad} that
the space Cremona group contains at least four non-conjugate subgroups isomorphic to $\mathrm{PSL}_2(\mathbb{F}_7)$.
Applying Theorem~\ref{theorem:rigid} to the case of the Klein quartic curve, we obtain
\begin{corollary}
\label{corollary:Klein}
The space Cremona group contains at least five non-conjugate subgroups isomorphic to $\mathrm{PSL}_2(\mathbb{F}_7)$.
\end{corollary}
Let $V$ be the 3-fold from Example~\ref{example:V1-S4}.
Then $\mathrm{Aut}(V)$ contains a subgroup $\mathfrak{G}^\prime\cong\mathfrak{S}_4$ that is not conjugate to the subgroup $\mathfrak{G}$.
The subgroup $\mathfrak{G}^\prime$ is generated by
\[\begin{split}
&[s:t:u:v:w]\mapsto [-s:-t:u:v:w],\\
&[s:t:u:v:w]\mapsto [u:s:t:v:w],\\
&[s:t:u:v:w]\mapsto [t:s:u:v:w].\\
\end{split}\]
In \S\ref{section:S4-Veronese}, we show that there exists a $\mathfrak{G}^\prime$-equivariant birational map $V\dasharrow X$,
where $X$ is a del Pezzo 3-fold of degree $2$ with $16$ singular points such that
$\mathrm{rk}\,\mathrm{Cl}(X)^{\mathfrak{G}^\prime}=1$.
It would be interesting to determine whether $X$ is $\mathfrak{G}^\prime$-birationally rigid or not (cf. \cite{Avilov-15points}).
We have established a relation between $28$-nodal double Veronese cones and smooth plane quartics.
Despite this establishment, the connection between the $\mathbf{E}_7$ latices appearing in their class groups and the Picard groups
of the double covers of $\mathbb{P}^2$ branched in the corresponding smooth quartic curves does not seem to be straightforward.
We discuss this phenomenon in \S\ref{section:question}.
\medskip
The plan of the paper is as follows.
In~\S\ref{section:preliminaries}
we collect several auxiliary results
that will be used later in our proofs.
In~\S\ref{section:nets-of-quadrics} we recall some assertions concerning nets of quadrics in~$\mathbb{P}^3$. The base points of the nets of quadrics described therein will be utilized in order to construct $28$-nodal double Veronese cones in~\S\ref{section:construction}. This construction is originally due to~\cite{Prokhorov}.
In~\S\ref{section:equations} we construct a $28$-nodal double Veronese cone starting from a smooth plane quartic curve.
We also provide an explicit equation~\eqref{eq:V1} for such a 3-fold in terms of covariants
of the quartic. In~\S\ref{section:1-1} we put the previous results together to establish a one-to-one correspondence between
$28$-nodal double Veronese cones and smooth plane quartics, which proves Theorem~\ref{theorem:one-to-one}.
Also, in~\S\ref{section:1-1} we
study the automorphism groups of $28$-nodal double Veronese cones and prove Theorem~\ref{theorem:Aut}.
In~\S\ref{section:theta} we
give an alternative proof of Theorem~\ref{theorem:one-to-one}.
Also, in~\S\ref{section:theta} we prove Corollary~\ref{corollary:Aronhold-sets}
which relates commutative diagrams~\eqref{equation:Prokhorov-diagram} with even theta characteristics on the smooth plane quartic.
In~\S\ref{section:S4-Veronese}
we investigate the $\mathfrak{S}_4$-equivariant birational geometry of
double Veronese cones introduced in Example~\ref{example:V1-S4},
and, in particular, describe the action of the group~$\mathfrak{S}_4$ on their Weil divisor class groups.
In~\S\ref{section:rigidity}
we study $G$-birationally rigid nodal double Veronese cones and
prove Theorem~\ref{theorem:rigid}.
In~\S\ref{section:question} we discuss some open questions
concerning $28$-nodal double Veronese cones.
Finally, in Appendix~\ref{section:appendix} we collect
some formulae allowing to construct explicit equations
of $28$-nodal double Veronese cones from the equations of plane quartics.
\medskip
\textbf{Acknowledgements.}
The authors are grateful to Fabrizio Catanese, Igor Dolgachev,
Alexander Kuznetsov, Marco Pacini, and Yuri Prokhorov
for inspiring discussions.
Part of this work was done during their stay at the Erwin Schr\"odinger International Institute for Mathematics and Physics in Vienna in August 2018.
They are grateful to the institute for excellent working conditions.
Jihun Park would also like to thank Adrien Dubouloz for his hospitality at Institut de Math\'ematiques de Bourgogne in August 2019, where he worked on finalizing the paper.
Ivan Cheltsov was supported by the Royal Society
grant No.~\mbox{IES\textbackslash R1\textbackslash 180205}
and the
HSE University Basic Research Program,
Russian Academic Excellence Project~\mbox{5-100}.
Jihun Park was supported by IBS-R003-D1, Institute for Basic Science in Korea.
Constantin Shramov was supported by
the HSE University Basic Research Program,
Russian Academic Excellence Project~\mbox{5-100},
by the Young Russian Mathematics award, and by the Foundation for the
Advancement of Theoretical Physics and Mathematics ``BASIS''.
\section{Preliminaries}
\label{section:preliminaries}
In this section we collect several auxiliary results
that will be used later in our proofs.
The following fact is well known to experts. It was pointed out to us by
Alexander Kuznetsov. This will be utilized in \S\ref{section:construction}.
\begin{lemma}[{cf.~\cite[Remark~4.2.7]{CheltsovShramov-V5}}]
\label{lemma:normal-bundle-blow-up}
Let $Y$ be a smooth variety, $C$ be a smooth curve on~$Y$, and~$P$ be a point on $C$.
Let $\pi\colon\widehat{Y}\to Y$ be the blow up of $Y$ at $P$, and let $\widehat{C}$ be the proper
transform of $C$ on $\widehat{Y}$.
Then
$$
\mathcal{N}_{\widehat{C}/\widehat{Y}}\cong \mathcal{N}_{C/Y}(-P).
$$
\end{lemma}
\begin{proof}
The differential of the morphism $\pi$ provides the
morphisms~\mbox{$d\pi\colon T_{\widehat{C}/\widehat{Y}}\to \pi^*T_{C/Y}$}
and~\mbox{$d\pi\colon \mathcal{N}_{\widehat{C}/\widehat{Y}}\to \pi^*\mathcal{N}_{C/Y}$}.
They can be included into the following commutative diagram:
$$
\xymatrix{
0\ar@{->}[r]& T_{\widehat{C}}\ar@{->}[r]\ar@{->}[d]_{\cong}&
T_{\widehat{C}/\widehat{Y}}\ar@{->}[r]\ar@{->}[d]^{d\pi}& \mathcal{N}_{\widehat{C}/\widehat{Y}}\ar@{->}[r]\ar@{->}[d]^{d\pi} &0\phantom{.}\\
0\ar@{->}[r]& \pi^*T_C\ar@{->}[r]& \pi^*T_{C/Y}\ar@{->}[r]& \pi^*\mathcal{N}_{C/Y}\ar@{->}[r] &0.
}
$$
Let $E$ be the exceptional divisor of $\pi$.
Note that the fiber of $\mathcal{N}_{\widehat{C}/\widehat{Y}}$ over the point~\mbox{$\widehat{P}=\widehat{C}\cap E$}
is identified with the tangent space to $E$ at
$\widehat{P}$. Since the differential $d\pi$ vanishes
on the latter space, we conclude that the
morphism~\mbox{$d\pi\colon \mathcal{N}_{\widehat{C}/\widehat{Y}}\to \pi^*\mathcal{N}_{C/Y}$}
factors through the morphism
$$
\theta\colon \mathcal{N}_{\widehat{C}/\widehat{Y}}\to \pi^*\mathcal{N}_{C/Y}(-k\widehat{P})
$$
for some
positive integer $k$.
Comparing the degrees of these vector bundles, we conclude that~\mbox{$k=1$}. Now
$\theta$ is a morphism of vector bundles of the same rank and degree, and it is an isomorphism
over a general point of $\widehat{C}$. This implies that $\theta$ is an isomorphism.
\end{proof}
\begin{corollary}
\label{corollary:normal-bundle-1}
Let $L$ be a line in $\mathbb{P}^3$, and let
$P_1, P_2$ be two points on $L$. Let $\pi\colon \widehat{\mathbb{P}}^3\to\mathbb{P}^3$
be the blow up of $\mathbb{P}^3$ at $P_1$ and $P_2$, and denote by $\widehat{L}$ the proper transform
of $L$ on $\widehat{\mathbb{P}}^3$. Then
$$
\mathcal{N}_{\widehat{L}/\widehat{\mathbb{P}}^3}\cong\mathcal{O}_{\mathbb{P}^1}(-1)\oplus\mathcal{O}_{\mathbb{P}^1}(-1).
$$
\end{corollary}
\begin{corollary}\label{corollary:normal-bundle-2}
Let $C$ be a twisted cubic curve in $\mathbb{P}^3$, and let
$P_1, \ldots, P_6$ be six distinct points on~$C$. Let $\pi\colon \widehat{\mathbb{P}}^3\to\mathbb{P}^3$
be the blow up at the points $P_1,\ldots,P_6$, and denote by $\widehat{C}$ the proper transform
of $C$ on $\widehat{\mathbb{P}}^3$. Then
$$
\mathcal{N}_{\widehat{C}/\widehat{\mathbb{P}}^3}\cong\mathcal{O}_{\mathbb{P}^1}(-1)\oplus\mathcal{O}_{\mathbb{P}^1}(-1).
$$
\end{corollary}
\begin{proof}
Note that
$$
\mathcal{N}_{C/\mathbb{P}^3}\cong\mathcal{O}_{\mathbb{P}^1}(5)\oplus\mathcal{O}_{\mathbb{P}^1}(5),
$$
(see~\cite[Proposition~6]{EisenbudVanDeVen}).
It remains to apply Lemma~\ref{lemma:normal-bundle-blow-up}.
\end{proof}
In \S\ref{section:rigidity}
we will need the following auxiliary result that
is a special equivariant version of a result of Kawakita (see \cite[Theorem~1.1]{Kawakita}).
\begin{lemma}
\label{lemma:mult-2}
Let $Y$ be a 3-fold, and let
$P$ be its smooth point. Let $G$
be a
subgroup of~\mbox{$\mathrm{Aut}(Y)$}
that fixes the point~$P$.
Let $\mathcal{M}_Y$ be a $G$-invariant mobile linear system on $Y$.
Suppose that~$P$ is a center of non-canonical singularities of the
log pair $(Y,\lambda\mathcal{M}_Y)$ for some rational number $\lambda$.
If the Zariski tangent space $T_{P,Y}$ is an irreducible
representation of the group~$G$,
then
$$
\mathrm{mult}_{P}\big(\mathcal{M}_Y\big)>\frac{2}{\lambda}.
$$
\end{lemma}
\begin{proof}
Suppose that $\mathrm{mult}_{P}(\mathcal{M}_Y)\leqslant\frac{2}{\lambda}$.
Let us show that $T_{P,Y}$ is a reducible representation of the group~$G$.
Let $\sigma\colon \hat{Y}\to Y$ be the blow up at the point $P$, let $E$ be the $\sigma$-exceptional surface,
and let~$\mathcal{M}_{\hat{Y}}$ be the proper transform of the linear system $\mathcal{M}_Y$ via $\sigma$.
Then
$$
K_{\hat{Y}}+\lambda\mathcal{M}_{\hat{Y}}+\Big(\lambda\mathrm{mult}_{P}(\mathcal{M}_Y)-2\Big)E\sim_{\mathbb{Q}}\sigma^*\big(K_Y+\lambda\mathcal{M}_Y\big),
$$
so that $E$ contains a non-canonical center of the log pair $(\hat{Y},\lambda\mathcal{M}_{\hat{Y}})$.
Then $E$ contains a non-log canonical center of the log pair $(\hat{Y},E+\lambda\mathcal{M}_{\hat{Y}})$.
It then follows from the inversion of adjunction (for instance, see \cite[Theorem~5.50]{KollarMori98})
that $(E,\lambda\mathcal{M}_{\hat{Y}}\vert_{E})$ is not log canonical.
Let $\mu$ be the positive rational number such that $(E,\mu\mathcal{M}_{\hat{Y}}\vert_{E})$ is strictly log canonical.
Then~\mbox{$\mu<\lambda$}, so that $\mu\mathrm{mult}_{P}(\mathcal{M}_Y)<2$.
On the other hand, we have
\begin{equation}
\label{equation:P2-equivalence}
\mu\mathcal{M}_{\hat{Y}}\vert_{E}\sim_{\mathbb{Q}}\mu\mathrm{mult}_{P}\big(\mathcal{M}_Y\big)L,
\end{equation}
where $L$ is a line in $E\cong\mathbb{P}^2$.
Let $Z$ be a log canonical center of the pair $(E,\mu\mathcal{M}_{\hat{Y}}\vert_{E})$.
If $Z$ is a curve, then the multiplicity of the restriction $\mu\mathcal{M}_{\hat{Y}}\vert_{E}$ at this curve is at least $1$ by \cite[Exercise~6.18]{CoKoSm03},
so~that \eqref{equation:P2-equivalence} implies that~$Z$ is a line.
The same arguments imply that other curves in $E$ cannot
be log canonical centers of the
pair~\mbox{$(E,\mu\mathcal{M}_{\hat{Y}}\vert_{E})$}
simultaneously with $Z$ in this case,
so that $Z$ is $G$-invariant.
This implies that $T_{P,Y}$ is a reducible representation of the group $G$.
To complete the proof, we may assume that the locus of log canonical singularities
of the log pair $(E,\mu\mathcal{M}_{\hat{Y}}\vert_{E})$ consists of finitely many points.
On the other hand, it follows from \eqref{equation:P2-equivalence} that the divisor $-(K_E+\mu\mathcal{M}_{\hat{Y}}\vert_{E})$ is ample.
Thus, using Koll\'ar--Shokurov connectedness principle (for example, see \cite[Corollary~5.49]{KollarMori98}),
we see that this locus is connected. Then it consists of one point.
This again means that $T_{P,Y}$ is a reducible representation of the group $G$.
\end{proof}
The following two lemmas will also be used in~\S\ref{section:rigidity}.
\begin{lemma}\label{lemma:4-general-points}
Let $G$ be a finite group acting on $\mathbb{P}^2$ without fixed points.
If a $G$-orbit on $\mathbb{P}^2$ has at least four points, then it
contains four points such
that no three of them are collinear.
\end{lemma}
\begin{proof}
Let $\Sigma$ be a $G$-orbit with at least four points. The $G$-action has no invariant line, otherwise it would have a fixed point outside the invariant line.
Thus $\Sigma$ contains
at least three non-collinear points, say, $P_1$, $P_2$, and $P_3$.
For each $i=1,2,3$ denote by $L_i$ the line determined by the two points~${\{P_1, P_2, P_3\}\setminus\{P_i\}}$.
If $\Sigma$ contains a point outside $L_1\cup L_2\cup L_3$, then
we are done. Suppose that this is not a case.
Let~$R_1$ be a point of $\Sigma$ different from $P_1$, $P_2$, and $P_3$.
We may assume that $R_1\in L_1$. Since the line $L_1$ is not $G$-invariant, there is a line $L_1'$ different from $L_1$ and containing three points of $\Sigma$. We are then able to choose four distinct points of $\Sigma\setminus \left(L_1\cap L_1'\right)$ that lie on $L_1$ or $L_1'$. Such four points satisfy the desired property.
\end{proof}
\begin{lemma}\label{lemma:must-contain-S4}
Let $G$ be a finite group acting on $\mathbb{P}^2$ without fixed points.
Suppose that there is a $G$-invariant smooth
quartic curve $C$ in $\mathbb{P}^2$. Then
\begin{itemize}
\item the curve $C$ is given
by equation~\eqref{eq:S4-quartic-general};
\item the group
$G$ contains a subgroup isomorphic to $\mathfrak{S}_4$;
\item the projective plane $\mathbb{P}^2$ can be identified
with a projectivization of an irreducible three-dimensional
representation of $\mathfrak{S}_4$.
\end{itemize}
\end{lemma}
\begin{proof}
The list of all groups preserving plane quartics
together with the equations
of the corresponding quartics can be found
in~\mbox{\cite[Theorem~6.5.2]{Dolgachev}}, \cite{Henn}, or \cite{KuribayashiKomiya}. Thus, it remains to check
which of them have fixed points on $\mathbb{P}^2$.
\end{proof}
\section{From Aronhold heptads to smooth plane quartics}
\label{section:nets-of-quadrics}
In this section we present some auxiliary assertions concerning nets of quadrics in $\mathbb{P}^3$.
The theory of nets of quadrics is a classical subject. We refer the reader to \cite{Beauville-Prym},
\cite{Edge1}, \cite{Edge2}, \cite{Edge3}, \cite{Edge4}, \cite{Edge5}, \cite{Wall1}, \cite{Wall2}, \cite{DolgachevOrtland},
\cite[\S6.3.2]{Dolgachev}, \cite{Tyurin}, \cite{Reid}, and references therein for many aspects of this theory.
For seven distinct points $P_1,\ldots, P_7$ in $\mathbb{P}^3$ denote by $\mathcal{L}(P_1, \ldots, P_7)$ the linear system of the quadrics passing through the points $P_1,\ldots, P_7$ in $\mathbb{P}^3$.
Before we proceed, let us employ a simple assertion that tells us the dimension of $\mathcal{L}(P_1, \ldots, P_7)$.
\begin{lemma}\label{lemma:CB}
Let $P_1,\ldots, P_r$ be points in a zero-dimensional complete intersection of three quadric surfaces in $\mathbb{P}^3$.
\begin{itemize}
\item If $r\leqslant 7$, then the $r$ points $P_1,\ldots, P_r$ impose independent conditions on quadric surfaces.
\item If $r=8$, then they do not impose independent conditions on quadric surfaces.
\end{itemize}
\end{lemma}
\begin{proof}
The first assertion immediately follows from \cite[Conjecture~CB11]{EGH}.
The conjecture is partially
proven in \cite[2.2]{EGH}, which allows us to apply it to our case.
The second assertion is obvious since the eight points form a complete intersection of three quadric surfaces.
\end{proof}
The following assertion is a key observation that connects 28-nodal double Veronese cones and nets of quadrics in~$\mathbb{P}^3$.
\begin{lemma}[{cf. \cite[Lemma~IX.5]{DolgachevOrtland}}]
\label{lemma:Aronhold-condition}
For seven distinct points $P_1,\ldots,P_7$ in $\mathbb{P}^3$,
the two conditions
\begin{itemize}
\item[(A)] every element of $\mathcal{L}(P_1, \ldots, P_7)$ is
irreducible;
\item[(B)] the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ consists
of eight distinct points,
\end{itemize}
are satisfied if and only if
the following three conditions hold:
\begin{itemize}
\item[(A${}^\prime$)] no four points of $P_1,\ldots,P_7$ are coplanar
(and in particular no three are collinear);
\item[(B${}^\prime$)] all points $P_1,\ldots,P_7$
are not contained in a single twisted cubic;
\item[(C${}^\prime$)] for each $i$, the twisted cubic passing through the points of $\{P_1,\ldots,P_7\}\setminus \{P_i\}$ and the line passing through the point $P_i$ and one point in $\{P_1,\ldots,P_7\}\setminus \{P_i\}$ meet neither twice nor tangentially.
\end{itemize}
\end{lemma}
\begin{proof}
First, suppose that conditions (A) and (B) hold.
If a plane $\Pi_1$ contains four points in $\{P_1,\ldots, P_7 \}$, then we have another plane $\Pi_2$ that contains the remaining three points. The divisor $\Pi_1+\Pi_2$ is a reducible member in $\mathcal{L}(P_1, \ldots, P_7)$, which
is a contradiction to condition~(A).
Therefore, condition~(A${}^\prime$) holds.
If the seven points $P_1,\ldots,P_7$ are contained in a twisted cubic, then every quadric surface passing through $P_1,\ldots,P_7$
must contain this twisted cubic. Thus,
the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ is one-dimensional,
which contradicts condition~(B).
Therefore, condition~(B${}^\prime$) must hold.
Suppose that for some $i, j$, say $i=1$, $j=2$, the twisted cubic curve $C$ passing through the six points $P_2,\ldots, P_7$ and the line $L$ passing through $P_1$ and $P_2$ meet at $P_2$ and at some point~$P$.
If they meet tangentially at $P_2$, then we put $P=P_2$.
Since the linear system of quadric surfaces containing~$C$ is two-dimensional, there are two distinct quadric surfaces $Q_1$ and $Q_2$ that contain the curve $C$ and the point $P_1$. Note that $Q_1$ and $Q_2$ belong to the linear system~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}.
The intersection of~$Q_1$ and $Q_2$ is a curve of degree $4$ that contains the curve $C$.
Since $L$ passes through three points~${P_1,\, P_2}$, and $P$ of $Q_i$, $i=1,2$, it must be contained in~$Q_i$, $i=1,2$.
Therefore, the intersection of $Q_1$ and~$Q_2$ consists of $C$ and~$L$.
This implies that for an element~$Q$ in $\mathcal{L}(P_1, \ldots, P_7)$,
the intersection of the quadrics $Q_1$, $Q_2$ and $Q$
either is a curve or consists of~$7$ points (i.e., $P_1+2P_2+P_3+\ldots+P_7$),
which gives a contradiction to condition~(B).
Therefore, condition~(C${}^\prime$) holds.
Now assume that conditions (A${}^\prime$), (B${}^\prime$),
and (C${}^\prime$) hold.
Suppose that there is a reducible member in $\mathcal{L}(P_1, \ldots, P_7)$. Then it consists of two planes containing the seven points $P_1,\ldots, P_7$.
Then one of the planes must contain four of the seven points,
which is a contradiction to condition~(A${}^\prime$).
Therefore, condition~(A) must be satisfied.
Suppose that condition (B) does not hold, i.e.,
the support of the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ either consists of only seven points or contains a curve.
Note that $\dim \mathcal{L}(P_1, \ldots, P_7)\geqslant 2$.
Let~$C$ be the twisted cubic curve passing through the six points $P_2,\ldots, P_7$. Then there are two distinct quadric surfaces $Q_1$ and $Q_2$ that contain the curve $C$ and the point $P_1$. They are elements
of~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}.
The intersection of $Q_1$ and $Q_2$ consists of the twisted cubic curve~$C$ and a line~$L$. Since $C$ does not contain the point $P_1$
by condition (B${}^\prime$), the line $L$ must contain~$P_1$. Note that $L$ intersects $C$ twice or tangentially.
If the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ is zero-dimensional,
then we choose an element $Q_3$ of~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}
that contains neither $C$ nor $L$.
The intersection of the quadrics $Q_1$, $Q_2$ and~$Q_3$
then consists of the seven points $P_1, \ldots, P_7$ set-theoretically. Otherwise it would consists of eight points including $P_1, \ldots, P_7$, and every element in $\mathcal{L}(P_1, \ldots, P_7)$ would pass through the eight intersection points by Lemma~\ref{lemma:CB}.
Since the three quadric surfaces meet at eight points counted with multiplicities, the line $L$ and $C$ must meet at one of the six points $P_2,\ldots, P_7$; otherwise~$Q_3$ would contain either $C$ or $L$. This is a contradiction
to condition (C${}^\prime$) since $L$ meets~$C$ twice or tangentially.
If the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ contains a curve, then for an arbitrary member $Q$
in~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} the intersection of the quadrics $Q_1$, $Q_2$ and $Q$ contains either $C$ or $L$. Since the dimension of the linear system of quadrics containing both $C$ and $P_1$ is $1$, the base locus
of~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} cannot contain~$C$.
Choose two general quadric surfaces that contain both the point~$P_7$ and the twisted cubic curve passing through $P_1, \ldots, P_6$. They are elements in~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}. Therefore, their intersection consists of the twisted cubic curve passing through the points $P_1, \ldots, P_6$, and the line $L$. Therefore, $L$ must meet $C$ at $P_7$. This contradicts condition (C${}^\prime$)
since $L$ and $C$ meet twice or tangentially.
Therefore, condition~(B) holds.
\end{proof}
\begin{definition}\label{definition:Aronhold-heptad}
A set of seven distinct points $P_1,\ldots, P_7$
in $\mathbb{P}^3$ is called an \emph{Aronhold heptad} if every element of $\mathcal{L}(P_1, \ldots, P_7)$ is irreducible and
the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ consists of eight distinct points.
The set of such eight distinct points is called a \emph{regular Cayley octad}. To be precise, the seven points $P_1,\ldots, P_7$ are called an Aronhold heptad of the regular Cayley octad.
\end{definition}
For an Aronhold heptad $P_1,\ldots, P_7$ in $\mathbb{P}^3$, Lemma~\ref{lemma:CB} immediately implies that there exists a unique regular Cayley octad that contains $P_1,\ldots, P_7$. It also shows that
the linear system~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} is a net.
Conversely, seven points of a regular Cayley octad
always form an Aronhold heptad.
\begin{lemma}\label{lemma:pencil-base-locus-ij}
Let $P_1,\ldots,P_8$ be a regular Cayley octad in $\mathbb{P}^3$.
For all $1\leqslant i<j\leqslant 8$, denote by~$L_{ij}$
the line passing through the points $P_i$ and $P_j$, and
denote by $T_{ij}$ the twisted cubic passing through the six points of
$\{P_1,\ldots,P_8\}\setminus\{P_i,P_j\}$. Let $\mathcal{L}_{ij}$ be the linear subsystem
of~\mbox{$\mathcal{L}(P_1,\ldots,P_7)$} that consists of the quadrics passing through $T_{ij}$.
Then $\mathcal{L}_{ij}$ is a pencil, and the base locus
of~$\mathcal{L}_{ij}$ is the union $T_{ij}\cup L_{ij}$.
\end{lemma}
\begin{proof}
The first assertion follows from Lemma~\ref{lemma:CB}.
To prove the second assertion, note that the base locus of $\mathcal{L}_{ij}$ is a union of
$T_{ij}$ and some line. On the other hand, the base locus must contain all the points
$P_1,\ldots,P_8$. Since the twisted cubic $T_{ij}$
passes neither through $P_i$ nor
through $P_j$, the line in the base locus
of~$\mathcal{L}_{ij}$ must pass through both of these
points, i.e., this line must be~$L_{ij}$.
\end{proof}
Suppose that we are given seven points $P_1,\ldots, P_7$ in $\mathbb{P}^3$ such that $\dim \mathcal{L}(P_1, \ldots, P_7)=2$. Then there are three linearly independent quadric homogeneous polynomials $F_0$, $F_1$, $F_2$ over $\mathbb{P}^3$ that generate the linear system $\mathcal{L}(P_1, \ldots, P_7)$, i.e., an element
in~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}
is defined by the quadric homogeneous equation
\begin{equation}
\label{equation:matrix}xF_0+yF_1+zF_2=0
\end{equation}
for some $[x:y:z]\in \mathbb{P}^2$. We can express~\eqref{equation:matrix} as a $4\times 4$ symmetric matrix $M(P_1, \ldots, P_7)$ with entries of linear forms in $x, y,z$. Then $\det \left(M(P_1, \ldots, P_7)\right)=0$ defines a plane quartic curve $H(P_1, \ldots, P_7)$ in~$\mathbb{P}^2$.
This plane quartic curve is called the \emph{Hessian quartic}
of the net~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}.
\begin{remark}\label{remark:determinantal-hypersurface}
The hypersurface $\mathfrak{D}$ in $\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)\cong\mathbb{P}^9$
defined by the determinant polynomial is singular exactly at non-zero $4\times 4$ symmetric matrices of coranks at least~2.
The three linearly independent quartic homogeneous polynomials $F_0$, $F_1$, $F_2$ determine a plane $\Pi( F_0, F_1, F_2)$
in~\mbox{$\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)$}.
Therefore, the plane quartic curve $H(P_1, \ldots, P_7)$ is singular if and only if the plane $\Pi( F_0, F_1, F_2)$ either passes through a point corresponding to a non-zero $4\times 4$ symmetric matrix of corank at least~2, or tangentially intersects the hypersurface $\mathfrak{D}$ at a point corresponding to a $4\times 4$ symmetric matrix of corank~$1$.
\end{remark}
\begin{lemma}\label{lemma:smooth-quartic}
Seven distinct points $P_1,\ldots, P_7$ in $\mathbb{P}^3$
form an Aronhold heptad
if and only if the linear system $\mathcal{L}(P_1, \ldots, P_7)$ is a net and
the plane quartic curve $H(P_1, \ldots, P_7)$ is smooth.
Moreover, every smooth plane quartic curve can be obtained in this way.
\end{lemma}
\begin{proof}
Assume that the plane quartic curve $H(P_1, \ldots, P_7)$ is singular.
Since the net~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}
does not contain any reducible member, it follows from Remark~\ref{remark:determinantal-hypersurface} that
the plane $\Pi(F_0, F_1, F_2)$ tangentially intersects the hypersurface $\mathfrak{D}$ at a point corresponding to a~\mbox{$4\times 4$} symmetric matrix of corank $1$
in~\mbox{$\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)$}.
We may assume that $F_0$ corresponds to such a $4\times 4$ symmetric
matrix of corank~$1$.
Denote by $A_0, A_1, A_2$ the $4\times 4$ symmetric matrices corresponding to $F_0$, $F_1$, $F_2$, respectively.
We may also assume that the matrix $A_0$ is a diagonal matrix with the last diagonal entry $0$. Since the lines $xA_0+yA_1$ and $xA_0+zA_2$ in $\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)$ tangentially intersect the surface $\mathfrak{D}$ at $A_0$, both the entry of $A_1$ at the 4th row and the 4th column and the entry of~$A_2$ at the 4th row and the 4th column are zero. This means that the point $[0:0:0:1]$ is a base point of~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$}.
However, $F_0$ is singular at $[0:0:0:1]$, and hence the base locus of~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} cannot consist of eight distinct points. This means that the points $P_1,\ldots, P_7$ do not
form an Aronhold heptad.
Now assume that the curve $H(P_1, \ldots, P_7)$ is smooth.
Suppose that the net $\mathcal{L}(P_1, \ldots, P_7)$ contains a reducible member. Then the plane~\mbox{$\Pi( F_0, F_1, F_2)$} passes through a point corresponding to a non-zero $4\times 4$ symmetric matrix of corank at least~2, and hence $H(P_1, \ldots, P_7)$
is singular by Remark~\ref{remark:determinantal-hypersurface}, which gives a contradiction.
Suppose that the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ is not zero-dimensional. It then must be one-dimensional. Furthermore, each irreducible curve $B$ of the base locus must be of degree at most~$3$ since $\mathcal{L}(P_1, \ldots, P_7)$ is a net.
If $B$ is a line, then there is a point $P_i$ outside $B$. Let $\Pi$ be the plane determined by $B$ and $P_i$. Each quadric from
$\mathcal{L}(P_1, \ldots, P_7)$ cuts out in $\Pi$ the line $B$ and a line passing though $P_i$. Such lines form at most a pencil. Therefore, there is a member in $\mathcal{L}(P_1, \ldots, P_7)$ containing $\Pi$, which leads to a contradiction.
If $B$ is a conic, two members in $\mathcal{L}(P_1, \ldots, P_7)$ intersect along $B$ and another conic $B'$. Then either~$B$ or~$B'$ must contain
four points of $P_1, \ldots, P_7$.
This again implies that there is a reducible member
in~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} and leads to a contradiction.
If $B$ is a cubic curve, two members in $\mathcal{L}(P_1, \ldots, P_7)$
intersect along $B$ and a line. Since a line cannot
contain more than two points of $P_1, \ldots, P_7$,
the curve $B$ must contain at least~$5$ points
of~\mbox{$P_1, \ldots, P_7$}. If $B$ is singular, then $B$ is planar, and hence $\mathcal{L}(P_1, \ldots, P_7)$ contains a reducible member. Therefore,~$\mathcal{L}(P_1, \ldots, P_7)$ must be the net determined by the twisted cubic~$B$. In such a case, we can directly show that $H(P_1, \ldots, P_7)$ is singular.
Consequently, the base locus of $\mathcal{L}(P_1, \ldots, P_7)$ is zero-dimensional. Suppose that the base locus of~$\mathcal{L}(P_1, \ldots, P_7)$ does not consist of eight distinct points. Then there is a point $P_i$, say~$P_1$,
with the following property: three general elements in $\mathcal{L}(P_1, \ldots, P_7)$
meet at $P_1$ so that their local intersection index at~$P_1$
is at least $2$.
This implies that there is an element in~\mbox{$\mathcal{L}(P_1, \ldots, P_7)$} singular at $P_1$,
and hence~\mbox{$H(P_1, \ldots, P_7)$} cannot be smooth.
Indeed, we may assume that the singular quadric is defined
by a~\mbox{$4\times 4$} diagonal matrix $A_0$
with the $4$th diagonal entry equal to zero.
The singular point~$P_1$ is located
at $[0:0:0:1]$. Then the net $\mathcal{L}(P_1, \ldots, P_7)$ can be generated by the net of~\mbox{$4\times 4$} symmetric matrices
\[xA_0+yA_1+zA_2,\]
where $A_1$ and $A_2$ are $4\times 4$ symmetric matrices each of which has $0$ for the entry at the 4th row and the 4th column.
Then
$$
\det \left(xA_0+yA_1+zA_2\right)=0
$$
defines a quartic curve singular at~$[1:0:0]$. The obtained contradiction
shows that the base locus of~$\mathcal{L}(P_1, \ldots, P_7)$ consists of eight distinct points,
so that $P_1,\ldots, P_7$ form an Aronhold heptad.
A given smooth quartic curve can be defined by the determinant of a $4\times 4$ symmetric linear matrix.
This fact together with the first assertion of the lemma implies the second assertion.
\end{proof}
\begin{remark}
The representation of a smooth plane quartic as the determinant
of a symmetric linear matrix that we used in the proof of Lemma~\ref{lemma:smooth-quartic}
goes back to~1855 (see~\cite{Hesse}) and~1902 (see~\cite{Dixon}).
For a contemporary proof, see \cite[Proposition~4.2]{Beauville-determinantal}.
\end{remark}
The following well-known fact concerning intersections of quadrics in $\mathbb{P}^3$ will be necessary for the present paper. It should be remarked here that they have been extensively researched in \cite{Reid} and \cite{Tyurin} in much wider settings.
\begin{lemma}
\label{lemma:Q1Q2}
Let $Q_1$ and $Q_2$ be two distinct
quadrics in $\mathbb{P}^3$ and $E$ be the intersection of $Q_1$ and $Q_2$. Then the following are equivalent.
\begin{itemize}
\item the intersection $E$ is smooth and of codimension $2$ in $\mathbb{P}^3$.
\item The line $\mathcal{L}$ determined by $Q_1$ and $Q_2$ in $\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)\cong\mathbb{P}^9$ intersects the determinant hypersurface $\mathfrak{D}$ (see Remark~\ref{remark:determinantal-hypersurface}) at four distinct points.
\item After a suitable linear coordinate change, $E$ can be defined by the equations
\[
x_0^2+x_1^2+x_2^2+x_3^2=\lambda_0 x_0^2+\lambda_1x_1^2+\lambda_2x_2^2+\lambda_3x_3^2=0
\]
in $\mathbb{P}^3$, where $\lambda_i$'s are four distinct constants.
\end{itemize}
If these equivalent conditions hold, then
the intersection $E$ is isomorphic to the double cover of the line $\mathcal{L}$ in $\mathbb{P}\left(\mathrm{Sym}(4, \mathbb{C})\right)$ branched exactly at the four intersection points $\mathcal{L}\cap\mathfrak{D}$.
\end{lemma}
\begin{proof}
The equivalence immediately follows from \cite[Proposition~2.1]{Reid}.
As a matter of fact, the last statement is also instantly implied by the result in \cite[Theorem~4.8]{Reid} that deals with much more general situation.
However, in $\mathbb{P}^3$, the conditions for \cite[Theorem~4.8]{Reid} turn into a tangible state, so that an elementary and short proof could be presented as follows.
Suppose that $E$ is defined in $\mathbb{P}^3$ by the equations
\[
x_0^2+x_1^2+x_2^2+x_3^2=\lambda_0 x_0^2+\lambda_1x_1^2+\lambda_2x_2^2+\lambda_3x_3^2=0
\]
for some four distinct constants $\lambda_0, \ldots, \lambda_3$. To complete the proof, it is enough to show that $E$ is isomorphic to a double cover of $\mathbb{P}^1$ branched at four distinct points whose cross-ratio is
\begin{equation}\label{eq:cross-ratio-elliptic}
\frac{(\lambda_1-\lambda_3)(\lambda_2-\lambda_0)}{(\lambda_1-\lambda_0)(\lambda_2-\lambda_3)}
\end{equation}
modulo permutation.
Note that $\lambda_0\neq\lambda_i$ for $1\leqslant i\leqslant 3$ because $E$ is smooth. Rewrite the equations of~$E$ as
\[x_0^2+x_1^2+x_2^2+x_3^2=(\lambda_0-\lambda_1)x_1^2+(\lambda_0-\lambda_2)x_2^2+(\lambda_0-\lambda_3)x_3^2=0.\]
We first consider the projection of $E$ to the plane $\Pi$ defined by $x_0=0$ centered at the point~\mbox{$[1:0:0:0]$}. The image is the conic $R$ defined by
$$
(\lambda_0-\lambda_1)x_1^2+(\lambda_0-\lambda_2)x_2^2+(\lambda_0-\lambda_3)x_3^2=0
$$
on~$\Pi$. Furthermore, $E$ is a double cover of $R$. Let $[0:\alpha:\beta:\gamma]$ be a point on this conic. The equation
\[\mu^2+\nu^2(\alpha^2+\beta^2+\gamma^2)=0\]
in $[\mu:\nu]\in\mathbb{P}^1$ determines the points
$[\mu:\nu\alpha:\nu\beta:\nu\gamma]$ of $E$ over the
point~\mbox{$[0:\alpha:\beta:\gamma]$}. Therefore, the double cover is
branched at~\mbox{$[0:\alpha:\beta:\gamma]$} if and only if
$$
\alpha^2+\beta^2+\gamma^2=0.
$$
Consequently, the four branch points on $R$ are
the intersection points of the conics
\[
\aligned
&x_1^2+x_2^2+x_3^2=0,\\
&\lambda_1x_1^2+\lambda_2x_2^2+\lambda_3x_3^2=0
\endaligned
\]
on $\Pi$. These four points are $\left[0:\sqrt{\lambda_3-\lambda_2}:\pm\sqrt{\lambda_1-\lambda_3}:\pm\sqrt{\lambda_2-\lambda_1}\right]$.
We now consider the projection of $R$ centered at the point
$$
[0:\sqrt{\lambda_0-\lambda_2}: \sqrt{\lambda_1-\lambda_0}:0]
$$
to the line $L$ defined by $x_1=0$ on $\Pi$. This projection defines an isomorphism of $R$ onto $L$. Therefore,~$E$ is the double cover of $L$ branched at the points
\[
\left[0:0:\sqrt{\lambda_3-\lambda_2}\sqrt{\lambda_1-\lambda_0}\pm\sqrt{\lambda_0-\lambda_2}\sqrt{\lambda_1-\lambda_3}: \pm\sqrt{\lambda_0-\lambda_2}\sqrt{\lambda_2-\lambda_1}\right].
\]
It is easy to check that the cross-ratio of the above four points
\[\frac{\pm\sqrt{\lambda_3-\lambda_2}\sqrt{\lambda_1-\lambda_0}\pm\sqrt{\lambda_0-\lambda_2}\sqrt{\lambda_1-\lambda_3}}{\sqrt{\lambda_0-\lambda_2}\sqrt{\lambda_2-\lambda_1}}\]
is equal to the desired number~\eqref{eq:cross-ratio-elliptic}
modulo permutation.
This completes the proof.
\end{proof}
For an Aronhold heptad $P_1,\ldots,P_7$, the
projective plane $\check{\mathbb{P}}^2$ projectively dual to the net of quadrics~\mbox{$\mathcal{L}(P_1,\ldots,P_7)\cong\mathbb{P}^2$}
can be identified with the base of the elliptic fibration $\varkappa$
that is obtained from $\mathbb{P}^3$ by blowing up the points of the corresponding regular Cayley octad.
It is well-known that the projectively dual curve
of the Hessian quartic curve can be interpreted in terms of singular fibers of~$\varkappa$.
Namely, we have the following
\begin{lemma}\label{lemma:Hessian-dual}
Let $P_1,\ldots,P_7$ be an Aronhold heptad, and let $\check{H}\subset\check{\mathbb{P}}^2$ be the curve
parameterizing singular fibers of~$\varkappa$.
Then $\check{H}$ is projectively dual to $H(P_1,\ldots,P_7)$.
\end{lemma}
\begin{proof}
Choose a fiber $\widetilde{E}$ of the elliptic fibration $\varkappa$.
It is the proper transform
of an intersection curve~$E$ of two quadrics in
$\mathcal{L}(P_1,\ldots,P_7)$. Furthermore,
the fiber $\widetilde{E}$ is smooth if and only if the intersection curve $E$ is smooth because it
cannot be singular at the points of the regular Cayley octad.
On the other hand, Lemma~\ref{lemma:Q1Q2} implies that the intersection curve $E$ is smooth if and only if
the corresponding pencil of quadrics
contains exactly four singular quadrics.
This verifies that
the projectively dual curve of~$\check{H}$ coincides with
the Hessian curve of the net $\mathcal{L}(P_1,\ldots,P_7)$.
\end{proof}
\section{From Aronhold heptads to double Veronese cones}
\label{section:construction}
In this section we review the birational construction of $28$-nodal double Veronese cones
due to~\cite{Prokhorov}.
The following result is mainly a part of \cite[Theorem~7.1]{Prokhorov}. We provide its proof for the reader's convenience and for clarification.
\begin{proposition}
\label{proposition:Prokhorov-construction}
Let $P_1,\ldots,P_7$ be seven points in $\mathbb{P}^3$ that form an Aronhold heptad.
Let~\mbox{$\pi\colon\widehat{\mathbb{P}}^3\to\mathbb{P}^3$} be the blow up of $\mathbb{P}^3$ at the points $P_1,\ldots,P_7$,
and let~\mbox{$\phi\colon \widehat{\mathbb{P}}^3\dasharrow V$}
be the map given by the linear system $|-2K_{\widehat{\mathbb{P}}^3}|$.
Then
\begin{itemize}
\item the map $\phi$ is a birational morphism;
\item the exceptional locus of $\phi$ is a disjoint union of the proper transforms of the lines
passing through pairs of the points $P_i$ and the twisted cubics passing through six-tuples
of the points $P_i$;
\item the variety $V$ is a $28$-nodal double Veronese cone.
\end{itemize}
\end{proposition}
\begin{proof}
Note that by Lemma~\ref{lemma:Aronhold-condition} there are $21$ lines passing through pairs of the points $P_i$,
and $7$ twisted cubic curves passing through six-tuples
of the points $P_i$. Furthermore, due to Lemma~\ref{lemma:Aronhold-condition} their proper transforms on $\widehat{\mathbb{P}}^3$ are disjoint. It is easy to see that these proper transforms intersect the anticanonical class of~$\widehat{\mathbb{P}}^3$ trivially.
We claim that any other irreducible curve intersects the anticanonical class neither trivially nor negatively. To see this, let $L$ be an irreducible curve on $\mathbb{P}^3$ whose proper transform on $\widehat{\mathbb{P}}^3$ non-positively intersects the anticanonical class of $\widehat{\mathbb{P}}^3$. It means
\begin{equation}\label{eq:sum-mult-2deg}
\sum_{i=1}^{7}\mathrm{mult}_{P_i}L\geqslant 2\deg (L).
\end{equation}
Choose a point $P$ on $L$ other than the base point of $\mathcal{L}(P_1,\ldots,P_7)$. Then the quadric surfaces in~\mbox{$\mathcal{L}(P_1,\ldots,P_7)$} passing through the extra point $P$ form a pencil. It follows from~\eqref{eq:sum-mult-2deg} that every quadric surface in this pencil contains the curve~$L$. If $\deg L=4$, then we see from~\eqref{eq:sum-mult-2deg} that~$L$ must be singular
at some of the points $P_i$. This means that the intersection of three quadrics froms~ $\mathcal{L}(P_1,\ldots,P_7)$
is not reduced at that point, i.e., the base locus of $\mathcal{L}(P_1,\ldots,P_7)$ consists of less than $8$ points, which
is a contradiction. If $\deg L=2$, then $L$ is a conic, and by~\eqref{eq:sum-mult-2deg}
it contains four of the points $P_i$. This is impossible by Lemma~\ref{lemma:Aronhold-condition}. Therefore,
one has either $\deg L=3$ or $\deg L=1$.
If $L$ is a singular cubic curve, then it is planar. However,~\eqref{eq:sum-mult-2deg} implies that
it contains at least four of the points $P_i$. This is impossible by Lemma~\ref{lemma:Aronhold-condition}.
Therefore, $L$ is either a twisted cubic or a line.
In both cases we see from~\eqref{eq:sum-mult-2deg} that $L$ is one of the above $28$ curves.
Let $\tilde{\phi}$ be the morphism defined by the linear system $|-nK_{\widehat{\mathbb{P}}^3}|$ for large enough $n$.
Now Corollaries~\ref{corollary:normal-bundle-1}~and~\ref{corollary:normal-bundle-2} imply that $\tilde{\phi}$ is a flopping contraction, which implies that $\tilde{\phi}$ satisfies the properties of the first and the second assertions. They also verify that the $28$ proper transforms are contracted to~$28$ nodes on~$V$.
Since $\widehat{\mathbb{P}}^3$ is a smooth weak Fano 3-fold and $\tilde{\phi}$ contracts all the curves that trivially intersect the anticanonical class of $\widehat{\mathbb{P}}^3$, the 3-fold $V$ is a Fano 3-fold with exactly $28$ singular points.
Let $Q$ be a quadric surface passing through the Aronhold heptad, then
\[
-K_{\widehat{\mathbb{P}}^3}\sim 2\left(\pi^*(Q)-(F_1+\ldots +F_7)\right),
\]
where $F_i$'s are the exceptional surfaces of $\pi$. This shows that $-K_V$ is divisible by $2$.
Since
$$
\left(-K_V\right)^3=\left(-K_{\widehat{\mathbb{P}}^3}\right)^3=8,
$$
the variety $V$ is a del Pezzo 3-fold of degree $1$.
Now it is easy to check that the divisor $-2K_V$ is very ample, which implies that the
morphism $\tilde{\phi}$ coincides with the map $\phi$ defined by the linear system~$|-2K_{\widehat{\mathbb{P}}^3}|$.
This completes the proof.
\end{proof}
The following observation was explained to us by Yuri Prokhorov.
\begin{example}
Let $C$ be a twisted cubic in $\mathbb{P}^3$. Choose six distinct points $P_1,\ldots,P_6$ on $C$.
Let~$P$ be a point on $C$ different from $P_1,\ldots,P_6$, and let $L$
be the line passing through $P_6$ and~$P$. Choose a point $P_7\in L\setminus\{P_6, P\}$.
Then such seven points cannot be an Aronhold heptad since they violate the conditions in Lemma~\ref{lemma:Aronhold-condition}.
Furthermore, if we define the 3-folds $\widehat{\mathbb{P}}^3$ and~$V$
and the map $\phi$ as in Proposition~\ref{proposition:Prokhorov-construction},
then $\phi$ contracts (in particular) the proper transforms of~$C$ and~$L$ on~$\widehat{\mathbb{P}}^3$.
These curves are not disjoint, which implies that the 3-fold $V$ is not nodal.
The same holds if $L$ is chosen to be a tangent line at the point $P_6$ to~$C$.
\end{example}
The following result is implicitly contained (but not clearly stated)
in \cite{Prokhorov}.
It is implied by the results of~\cite{Prokhorov}
together with a simple additional observation.
\begin{theorem}\label{theorem:28-can-be-constructed}
Let $V$ be a double Veronese cone with $28$ singular points. Then $V$ can be constructed from
some Aronhold heptad as in Proposition~\ref{proposition:Prokhorov-construction}.
\end{theorem}
\begin{proof}
It first follows from \cite[Theorem~1.7]{Prokhorov} and~\cite[Remark~1.8]{Prokhorov}
that the rank of the divisor class group of $V$ is $8$.
Then \cite[Theorem~7.1(i)]{Prokhorov} implies that $V$ can be obtained in the same way as in Proposition~\ref{proposition:Prokhorov-construction} with seven points of $\mathbb{P}^3$ such that
no four of them are coplanar.
It remains to verify that the seven points form a Aronhold heptad.
However, \cite[Theorem~7.1(ii)]{Prokhorov} shows that having~$28$ singular points is equivalent to ($\mathrm{A}'$), ($\mathrm{B}'$), and ($\mathrm{C}'$) in Lemma~\ref{lemma:Aronhold-condition}. Therefore, the seven points in the construction must form an Aronhold heptad by Lemma~\ref{lemma:Aronhold-condition}.
\end{proof}
Proposition~\ref{proposition:Prokhorov-construction} and Theorem~\ref{theorem:28-can-be-constructed}
immediately imply the following
\begin{corollary}\label{corollary:28-node}
If a double Veronese cone has $28$ singular points, then they are all nodes.
\end{corollary}
We conclude this section with another result that is a part of \cite[Theorem~7.1]{Prokhorov}. Its proof
is similar to that of Proposition~\ref{proposition:Prokhorov-construction}.
\begin{proposition}
\label{proposition:Prokhorov-construction-2}
Let $P_1,\ldots,P_6$ be six points in $\mathbb{P}^3$ such that no four of them are coplanar.
Let~\mbox{$\pi\colon\widehat{\mathbb{P}}^3\to\mathbb{P}^3$} be the blow up of $\mathbb{P}^3$ at the points $P_1,\ldots,P_6$,
and let~\mbox{$\phi\colon \widehat{\mathbb{P}}^3\dasharrow W$}
be the map given by the linear system $|-K_{\widehat{\mathbb{P}}^3}|$.
Then
\begin{itemize}
\item the map $\phi$ is a birational morphism;
\item the exceptional locus of $\phi$ is a disjoint union of the proper transforms of the lines
passing through pairs of the points $P_i$ and the twisted cubic passing through all the points~$P_i$;
\item the variety $W$ is a $16$-nodal double cover of $\mathbb{P}^3$ branched over a quartic surface.
\end{itemize}
\end{proposition}
\section{From smooth plane quartics to
double Veronese cones}
\label{section:equations}
In this section we construct a $28$-nodal double Veronese cone starting from a smooth plane quartic curve.
We also provide an explicit equation~\eqref{eq:V1} for such a 3-fold in terms of covariants
of the plane quartic curve.
Let $\check{\mathbb{P}}^2$ be the dual of the projective plane $\mathbb{P}^2$.
We use a homogeneous coordinate system~\mbox{$[x:y:z]$} for $\mathbb{P}^2$ and $[s:t:u]$ for~$\check{\mathbb{P}}^2$.
We start with a smooth quartic curve $C$ in the projective plane $\mathbb{P}^2$ given by an equation
\begin{equation}\label{equation:quartic}
H(x,y,z)=\sum_{i+j+k=4}a_{ijk}x^iy^jz^k=0.
\end{equation}
We regard $[s:t:u]$ as a general point in $\check{\mathbb{P}}^2$. Then the corresponding line on $\mathbb{P}^2$ is a general line~$L_{s,t,u}$ given by
\[sx+ty+uz=0.\]
The line $L_{s,t,u}$ hits the quartic $C$ at four distinct points $x_1$, $x_2$, $x_3$, $x_4$ lying on $L_{s,t,u}\setminus\{z=0\}$. We may regard these four points as points on the affine line, so that we could define their cross-ratio as follows:
\[\lambda(x_1, x_2, x_3, x_4)=\frac{(x_1-x_3)(x_4-x_2)}{(x_1-x_2)(x_4-x_3)}.\]
This has six different values according to the order of the four points. However, the following $j$-function is invariant with respect to the reordering $x_1$, $x_2$, $x_3$, $x_4$.
\begin{equation}\label{equation:j-function}
\begin{split}
j(x_1, x_2, x_3, x_4)&=256\frac{\left(1-\lambda(x_1, x_2, x_3, x_4)\left(1-\lambda(x_1, x_2, x_3, x_4)\right)\right)^3}
{\lambda(x_1, x_2, x_3, x_4)^2\left(1-\lambda(x_1, x_2, x_3, x_4)\right)^2}\\%
&=2^8\frac{\left((x_1-x_2)^2(x_4-x_3)^2-(x_1-x_3)(x_4-x_2)(x_4-x_1)(x_3-x_2)\right)^3}
{(x_1-x_2)^2(x_1-x_3)^2(x_1-x_4)^2(x_2-x_3)^2(x_2-x_4)^2(x_3-x_4)^2}.
\end{split}
\end{equation}
By plugging $z=-\frac{sx+ty}{u}$ into \eqref{equation:quartic}, we obtain
\begin{equation}\label{equation:quartic2}
\begin{split}
u^4H\left(x,y,-\frac{sx+ty}{u}\right)&=\sum_{i+j+k=4}a_{ijk}(-u)^{4-k}\left(sx+ty\right)^kx^iy^j\\
&=\sum_{r=0}^4b_{4-r}x^ry^{4-r},
\end{split}
\end{equation}
where
\[b_r=\sum_{j=0}^{r}\sum_{i+k=4-j}(-1)^ka_{ijk}\binom{k}{k+j-r}s^{k+j-r}t^{r-j}u^{4-k}.\]
Then we have the following identities for elementary symmetric functions of $x_1$, $x_2$, $x_3$, $x_4$:
\[\aligned
& x_1+x_2+x_3+x_4=-\frac{b_1}{b_0};\\
& x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=\frac{b_2}{b_0};\\
& x_2x_3x_4+x_1x_2x_4+x_1x_2x_4+x_1x_2x_3=-\frac{b_3}{b_0};\\
& x_1x_2x_3x_4=\frac{b_4}{b_0}.
\endaligned\]
Since the denominator and the numerator of the $j$-function in \eqref{equation:j-function} are symmetric polynomials in~$x_1, x_2, x_3, x_4$,
the $j$-function in \eqref{equation:j-function}
may be regarded as a rational function in~$b_0, b_1, b_2, b_3, b_4$.
Indeed, one has
\[
\begin{split}
j(b_0, b_1, b_2, b_3, b_4)&=1728\frac{4h_2(b_0, b_1, b_2, b_3, b_4)^3}{4h_2(b_0, b_1, b_2, b_3, b_4)^3-27h_{3}(b_0, b_1, b_2, b_3, b_4)^2},\\
\end{split}
\]
where
\[h_2(b_0, b_1, b_2, b_3, b_4)=\frac{1}{3}\Big(-3b_1b_3 +12 b_0b_4+b_2^2\Big);\]
\[h_{3}(b_0, b_1, b_2, b_3, b_4)=\frac{1}{27}\Big(72b_0b_2b_4 - 27b_0b_3^2 - 27b_1^2b_4 + 9b_1b_2b_3 - 2b_2^3\Big)\]
(see Appendix~\ref{section:appendix} for a more detailed
computation).
Regarding $h_2 (b_0, b_1, b_2, b_3, b_4)$ and $h_3 (b_0, b_1, b_2, b_3, b_4)$ as polynomials in $s,t,u$, we see from Appendix~\ref{section:appendix}
that
\[h_2 (b_0, b_1, b_2, b_3, b_4)=u^4g_4(s,t,u),\]
\[h_3 (b_0, b_1, b_2, b_3, b_4)=u^6g_6(s,t,u),\]
where $g_4(s,t,u)$ and $g_6(s,t,u)$ are homogeneous polynomials of degrees $4$ and $6$, respectively, in $s, t, u$.
Consequently, the rational function $j(b_0, b_1, b_2, b_3, b_4)$ may be regarded as a rational function $j_C$ in~$s, t, u$, so that
it is a rational function on $\check{\mathbb{P}}^2$. More precisely, one has
\begin{equation}\label{equation:j-function2}
j_C(s,t,u)=1728\frac{4g_4(s,t,u)^3}{4g_4(s,t,u)^3-27g_{6}(s,t,u)^2}.
\end{equation}
\begin{remark}\label{remark:harmonic}
The equation $g_4(s,t,u)=0$ of degree $4$ describes the points of $\check{\mathbb{P}}^2$
corresponding to lines that intersect~$C$ by equianharmonic quadruples of points (see for instance
\cite[\S2.3.4]{Dolgachev} for terminology). In other words, these lines with the quadruples of points define elliptic curves of $j$-invariant $0$. Elliptic curves of $j$-invariant $0$ are isomorphic to the Fermat plane cubic curve.
Similarly, the equation $g_6(s,t,u)=0$ of degree $6$ describes the points of $\check{\mathbb{P}}^2$
corresponding to lines that intersect~$C$ by harmonic quadruples of points. In this case, lines with the quadruples of points define elliptic curves of $j$-invariant $1728$. Note that neither the cross-ratio nor the $j$-invariant is defined for the lines corresponding to the $24$ intersection points (counted with multiplicities) of $g_4(s,t,u)=0$ and~$g_6(s,t,u)=0$.
\end{remark}
\begin{lemma}\label{lemma:j-not-constant}
The $j$-function $j_C(s,t,u)$ is not constant. In particular, $g_4(s,t,u)$ cannot be a zero polynomial.
\end{lemma}
\begin{proof}
Among the lines on $\mathbb{P}^2$, there are both tangent lines to $C$ and non-tangent lines to $C$.
This implies that the $j$-function, which is defined by the cross-ratios of the intersection points, is not constant.
\end{proof}
We can use the polynomials $g_4(s,t,u)$ and $g_6(s,t,u)$ to write down the equation
of the projectively dual curve of the plane quartic~$C$.
\begin{proposition}
\label{proposition:discriminant}
The equation
\begin{equation}\label{equation:discriminant}
4g_4(s,t,u)^3-27g_6(s,t,u)^2=0
\end{equation}
of degree $12$ defines the projectively dual curve $\check{C}$
of the quartic curve $C$ in $\check{\mathbb{P}}^2$.
\end{proposition}
\begin{proof}
The curve $\check{C}$ has degree $12$.
By construction the rational function $j_C(s,t,u)$ is well-defined at least outside the projectively dual curve $\check{C}$. On the other hand,
we know from Lemma~\ref{lemma:j-not-constant} that~\mbox{$g_4(s,t,u)$} is not a zero polynomial.
Thus it follows from \eqref{equation:j-function2} that $j_C(s,t,u)$ has poles along the curves defined by some of
the factors of~\mbox{$4g_4(s,t,u)^3-27g_6(s,t,u)^2$}. These poles must be contained in the projectively dual curve~$\check{C}$. Since $\check{C}$ is an irreducible curve of degree $12$,
it must be defined by the homogeneous equation~\eqref{equation:discriminant}
of degree $12$.
\end{proof}
Recall that a
bitangent line to the quartic curve $C$, which is not a tangent line at a hyper-inflection point of $C$, yields an ordinary double point on the projectively dual curve $\check{C}$. Meanwhile, the tangent line to $C$ at an ordinary inflection point of $C$ produces an ordinary cusp in $\check{C}$ and the tangent line to~$C$ at a hyper-inflection point of $C$ generates a triple point of $\check{C}$ that is analytically isomorphic to a singularity defined by equation~\mbox{$s^3=t^4$}.
These are all the possible types of singularities of~$\check{C}$.
Let $C_4$ be the quartic curve defined by $g_4(s,t,u)=0$, and let $C_6$ be the sextic curve defined by~$g_6(s,t,u)=0$ in $\check{\mathbb{P}}^2$.
\begin{lemma}\label{lemma:C4-meets-C6}
Let $P$ be an intersection point of the curves $C_4$ and $C_6$.
Then one of the following possibilities occur.
\begin{itemize}
\item The curves $C_4$ and $C_6$ are smooth at $P$, and meet
transversally at $P$; in this case
$P$ is an ordinary cusp of the curve $\check{C}$.
\item The curve $C_6$ has a double point at $P$
and the local intersection index of $C_4$ and $C_6$ at $P$ is~$2$;
in this case $P$
is a triple point of $\check{C}$ analytically isomorphic to~\mbox{$s^3=t^4$}.
\end{itemize}
\end{lemma}
\begin{proof}
The point $P$ is a singular point of $\check{C}$. The defining equation \eqref{equation:discriminant} of $\check{C}$ shows that $C_4$ is smooth at $P$ and that
$P$ cannot be an ordinary double point of $\check{C}$.
Thus $\check{C}$ has either an ordinary cusp
or a triple point analytically
isomorphic to $s^3=t^4$ at $P$. Note that $P$ is a triple point of $\check{C}$ if and only if~$C_6$ is singular at $P$. Furthermore, in such a case, $P$ must be a double point of $C_6$.
The defining equation~\eqref{equation:discriminant} tells that
$P$ is an ordinary cusp of $\check{C}$ if $C_4$ and $C_6$ are transversal at $P$.
On the other hand, if $C_6$ is smooth and tangent to $C_4$ at $P$, then $\check{C}$ has a double point worse than a simple cusp at~$P$, in the sense that its log canonical threshold is smaller than~$5/6$.
We now suppose that $P$ is a double point of $C_6$. Then, \eqref{equation:discriminant} shows that $\check{C}$ has a triple point at~$P$.
Therefore, to complete the proof, it is enough to show that the singularity of $\check{C}$ at $P$ cannot be a triple point analytically
isomorphic to $s^3=t^4$ if the tangent line to $C_4$ at $P$ is one of the Zariski tangent lines of $C_6$ at $P$. This can be easily checked by blowing up. Indeed, if the tangent line to $C_4$ at $P$ is one of the Zariski tangent lines of $C_6$ at $P$, then the proper transform of $\check{C}$ via the blow up of $\check{\mathbb{P}}^2$ at~$P$ is still singular at some
point over~$P$, which is not the case for the singularity~\mbox{$s^3=t^4$}.
\end{proof}
We now consider the hypersurface $V$ of degree $6$ in $\mathbb{P}(1,1,1,2,3)$ given by an equation
\begin{equation}\label{eq:V1}
-w^2+v^3-g_4(s,t,u)v+g_6(s,t,u)=0,
\end{equation}
where $\mathrm{wt}(w)=3, \mathrm{wt}(v)=2$.
We denote by $F(s,t,u,v,w)$ the left hand side of~\eqref{eq:V1},
and set
$$
G(s,t,u)=4g_4(s,t,u)^3-27g_6(s,t,u)^2.
$$
Then $G(s,t,u)=0$ defines the projectively dual curve $\check{C}$ of the smooth quartic curve $C$ in $\check{\mathbb{P}}^2$
by Proposition~\ref{proposition:discriminant}.
Denote by $V_{v}^{sing}$ the set of singular points of $V$ in the section by $v=0$ and by $V_{o}^{sing}$ the set of singular points of $V$ outside the section by $v=0$. On the other hand, denote by $\check{C}_v^{sing}$ the set of singular points of~$\check{C}$ at which~$C_6$ is singular, and denote by~$\check{C}_o^{sing}$ the set of singular points of~$\check{C}$ outside~$C_6$.
\begin{lemma}\label{lemma:correspondence}
Define the maps as follows:
\[\begin{split}\check{C}_o^{sing} &\longrightarrow V_{o}^{sing},\\
[a_0:a_1:a_2]&\mapsto \left[a_0:a_1:a_2:\frac{3g_6(a_0,a_1,a_2)}{2g_4(a_0,a_1,a_2)} :0\right];\\
&\\
\check{C}_v^{sing} &\longrightarrow V_{v}^{sing},\\
[a_0:a_1:a_2]&\mapsto [a_0:a_1:a_2:0:0].\\
\end{split}\]
Then both are one-to-one correspondences.
\end{lemma}
\begin{proof}
Let $[a_0:a_1:a_2]$ be a point in the set $\check{C}_o^{sing}$.
Observe that
$g_6(a_0,a_1,a_2)\ne 0$,
and hence~\mbox{$g_4(a_0,a_1,a_2)\ne 0$}.
Since $[a_0:a_1:a_2]\in\check{C}_o^{sing}$, we have
$$
\frac{\partial{F}}{\partial v}\left(a_0,a_1,a_2,\frac{3g_6(a_0,a_1,a_2)}{2g_4(a_0,a_1,a_2)},0\right)=\frac{27g_6(a_0,a_1,a_2)^2-4g_4(a_0,a_1,a_2)^3}{4g_4(a_0,a_1,a_2)^2}=0.
$$
Also, since
$$
12g_4(a_0,a_1,a_2)^2\frac{\partial g_4}{\partial s}(a_0,a_1,a_2)-54g_6(a_0,a_1,a_2)\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)=0,
$$
we have
\[\begin{split}
\frac{\partial{F}}{\partial s}\left(a_0,a_1,a_2,\frac{3g_6(a_0,a_1,a_2)}{2g_4(a_0,a_1,a_2)},0\right)&=-\frac{3g_6(a_0,a_1,a_2)}{2g_4(a_0,a_1,a_2)}\frac{\partial g_4}{\partial s}(a_0,a_1,a_2)+\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)\\
&= -\frac{27g_6(a_0,a_1,a_2)^2}{4g_4(a_0,a_1,a_2)^3}\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)+\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)=0.
\end{split}
\]
Similarly, we see
that $\frac{\partial{F}}{\partial t}$ and $\frac{\partial{F}}{\partial u}$
vanish at the point $\left[a_0:a_1:a_2:\frac{3g_6(a_0,a_1,a_2)}{2g_4(a_0,a_1,a_2)}:0\right]$, so that this point belongs to $V_{o}^{sing}$.
Conversely, suppose that $[a_0:a_1:a_2: v_0 :w_0]$ belongs to $V_{o}^{sing}$. Then $w_0=0$ and~\mbox{$v_0\ne 0$}.
Furthermore, taking the partial derivative of
\eqref{eq:V1} with respect to $v$, we obtain
$$
g_4(a_0,a_1,a_2)=3v_0^2.
$$
Using \eqref{eq:V1} again, we get
$$
g_6(a_0,a_1,a_2)=-v_0^3+v_0g_4(a_0,a_1,a_2)=2v_0^3.
$$
Then
\[\begin{split}\frac{\partial{G}}{\partial s}(a_0,a_1,a_2)&=12g_4(a_0,a_1,a_2)^2\frac{\partial g_4}{\partial s}(a_0,a_1,a_2)-54g_6(a_0,a_1,a_2)\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)\\
&= 108v_0^4\frac{\partial g_4}{\partial s}(a_0,a_1,a_2)-108v_0^3\frac{\partial g_6}{\partial s}(a_0,a_1,a_2)=-108v_0^3 \frac{\partial F}{\partial s} (a_0,a_1,a_2,v_0,w_0)=0.
\end{split}\]
Similarly we obtain
\[\frac{\partial{G}}{\partial t}(a_0,a_1,a_2)=\frac{\partial{G}}{\partial u}(a_0,a_1,a_2)=0.\]
This implies that $[a_0:a_1:a_2]$ belongs to $\check{C}_{o}^{sing}$.
Suppose that $[a_0:a_1:a_2]$ is in $\check{C}_v^{sing}$. Then
\[\frac{\partial{g_6}}{\partial s}(a_0,a_1,a_2)= \frac{\partial{g_6}}{\partial t}(a_0,a_1,a_2)=\frac{\partial{g_6}}{\partial u}(a_0,a_1,a_2)=0.\]
Moreover, we have
$g_6(a_0,a_1,a_2)=0$, and thus also
$g_4(a_0,a_1,a_2)=0$.
These imply
\[\frac{\partial{F}}{\partial w}\left(a_0,a_1,a_2,0,0\right)=\frac{\partial{F}}{\partial v}\left(a_0,a_1,a_2,0,0\right)=0,\]
\[\frac{\partial{F}}{\partial s}\left(a_0,a_1,a_2,0,0\right)=\frac{\partial{F}}{\partial t}\left(a_0,a_1,a_2,0,0\right)=\frac{\partial{F}}{\partial u}\left(a_0,a_1,a_2,0,0\right)=0.\]
Therefore, $[a_0:a_1:a_2:0:0]$ lies in $V_{v}^{sing}$.
Conversely, suppose that $[a_0:a_1:a_2:0:0]$ belongs to $V_v^{sing}$. Then
\[0=\frac{\partial{F}}{\partial v}\left(a_0,a_1,a_2,0,0\right)=-g_4(a_0,a_1,a_2). \]
In particular, this gives $g_6(a_0,a_1,a_2)=0$. We have
\[0=\frac{\partial{F}}{\partial s}\left(a_0,a_1,a_2,0,0\right) =\frac{\partial{g_6}}{\partial s}(a_0,a_1,a_2),\]
\[0=\frac{\partial{F}}{\partial t}\left(a_0,a_1,a_2,0,0\right) =\frac{\partial{g_6}}{\partial t}(a_0,a_1,a_2),\]
\[0=\frac{\partial{F}}{\partial u}\left(a_0,a_1,a_2,0,0\right) =\frac{\partial{g_6}}{\partial u}(a_0,a_1,a_2).\]
Therefore, we conclude that the point $[a_0:a_1:a_2]$ belongs to $\check{C}_v^{sing}$.
\end{proof}
\begin{theorem}\label{theorem:28-node}
The del Pezzo 3-fold $V$ has exactly $28$ singular points.
\end{theorem}
\begin{proof}
Let $\delta_o$ be the number of the ordinary bitangent lines,
i.e., lines tangent to $C$ at two distinct points, of the smooth
quartic curve $C$, and let
$\delta_s$ be the number of the hyper-inflection points of the smooth quartic curve $C$. The tangent line to $C$ at a hyper-inflection point is a bitangent
line to $C$ that is tangent to $C$ at a single point with multiplicity $4$. Let $\iota$ be the number of ordinary inflection points of $C$. Then we can derive
\[
\delta_o+\delta_s=28,\quad
\iota+2\delta_s=24
\]
from the classical Pl\"ucker formulae (for instance, see \cite[\S2.4]{GH}).
In particular, the projectively dual curve~$\check{C}$ has exactly
\[\delta_o+\delta_s+\iota=52-2\delta_s\]
singular points.
The ordinary bitangent lines define $\delta_o$
ordinary double points on $\check{C}$. Such singular points cannot
lie on $C_6$. If so, then they would also lie on $C_4$ by Proposition~\ref{proposition:discriminant}, and would
not be ordinary double points of $\check{C}$ by Lemma~\ref{lemma:C4-meets-C6}.
Meanwhile, by Lemma~\ref{lemma:C4-meets-C6}
the curves $C_4$ and $C_6$ meet \begin{itemize}
\item either transversally
\item or in such a way that $C_6$ has a double point at $P$, and
the local intersection index of $C_4$ and~$C_6$ at~$P$ is~$2$.
\end{itemize}
Furthermore,
intersection points of the former type yield ordinary cusps, and ones of the latter type produce triple points analytically isomorphic to $s^3=t^4$.
The number of intersection points of the former type is $\iota$ and the number of intersection points of the latter type is $\delta_s$.
We may then conclude that
\[ \#\left| \check{C}_o^{sing}\right |=\delta_o, \quad \#\left| \check{C}_v^{sing}\right |=\delta_s.\]
Since $\delta_o+\delta_s=28$, Lemma~\ref{lemma:correspondence} immediately implies the statement.
\end{proof}
The following assertion can be deduced from Theorem~\ref{theorem:28-node} and Corollary~\ref{corollary:28-node}.
However, we present a computational proof here to show how the singularities of $\check{C}$ and the singularities of~$V$ interact with each other.
\begin{proposition}\label{proposition:node}
The singular points of the del Pezzo 3-fold $V$ are all ordinary double points.
\end{proposition}
\begin{proof}
Let $P$ be a point in $V_{v}^{sing}$. After a suitable coordinate changes, we may assume
that
$$
P=[0:0:1:0:0].
$$
By Lemmas~\ref{lemma:C4-meets-C6} and~\ref{lemma:correspondence}, the curve
$C_4$ is smooth at~$[0:0:1]$, the curve $C_6$ has a double point at~$[0:0:1]$,
and the local intersection
index of $C_4$ and $C_6$ at~\mbox{$[0:0:1]$} equals to~$2$.
We may also assume that the
tangent line to $C_4$ at~$[0:0:1]$ is defined by $s=0$
and the Zariski tangent cone
of $C_6$ at $[0:0:1]$ is defined by $t(t+\alpha s)=0$, where $\alpha$ is a (possibly zero) constant.
Put
\[
\begin{array}{ll}
f(s,t,v,w)=F(s,t,1,v,w), & g(s,t)=G(s,t,1),\\
\tilde{g}_4(s,t,v,w)=g_4(s,t,1),& \tilde{g}_6(s,t,v,w)=g_6(s,t,1).
\end{array}
\]
We regard $s,t,v,w$ as local coordinates around the point $P$ that corresponds to the origin in~$\mathbb{C}^4$, which
we denote by $p$.
We then consider the Hessian of $f$ at the point $p$:
$$
\mathrm{Hess}(f)(p)=2\left(\frac{\partial \tilde{g}_4}{\partial s}(p)\right)^2\frac{\partial^2 \tilde{g}_6}{\partial t^2}(p)\ne 0.
$$
This implies that the point $P$ is an ordinary double point of $V$.
We now consider a point $R$ in $V_{o}^{sing}$. By appropriate coordinate changes, we may assume that
$$
R=[0:0:1:v_0:0],
$$
where $v_0$ is a non-zero constant. In the affine chart defined by $u\ne 0 $ in~\mbox{$\mathbb{P}(1,1,1,2,3)$}, the point~$R$ corresponds to $q=(0,0,v_0,0)$. In the affine chart defined by $u\ne0 $ in $\check{\mathbb{P}}^2$, we denote by $o$ the origin.
Since $q$ is a singular point of the hypersurface defined by $f=0$, we conclude that~$o$
is a singular point of the curve defined by $g=0$ by Lemma~\ref{lemma:correspondence}. Furthermore, since $q$ is from~$V_{o}^{sing}$, the origin $o$ is an ordinary double point of $g=0$.
From this one can obtain the following relations:
\begin{equation*}
\begin{array}{ll}
v_0=\frac{3\tilde{g}_6(q)}{2\tilde{g}_{4}(q)}, & 27\left(\tilde{g}_6(q)\right)^3=4\left(\tilde{g}_{4}(q)\right)^3,\\
3\tilde{g}_6(q)\frac{\partial \tilde{g}_4}{\partial s}(q)=2\tilde{g}_4(q)\frac{\partial \tilde{g}_6}{\partial s}(q), &
3\tilde{g}_6(q)\frac{\partial \tilde{g}_4}{\partial t}(q)=2\tilde{g}_4(q)\frac{\partial \tilde{g}_6}{\partial t}(q),\\
9\tilde{g}_6(q)\frac{\partial \tilde{g}_6}{\partial s}(q)=2\tilde{g}_4(q)^2\frac{\partial \tilde{g}_4}{\partial s}(q), &
9\tilde{g}_6(q)\frac{\partial \tilde{g}_6}{\partial t}(q)=2\tilde{g}_4(q)^2\frac{\partial \tilde{g}_4}{\partial t}(q).\\
\end{array}
\end{equation*}
Using them,
we are able to derive
\[
48\left(\tilde{g}_{4}(q)\right)^4\mathrm{Hess}(f)(q)=-\tilde{g}_{6}(q)\mathrm{Hess}(g)(o).
\]
(Since the computation for this is messy and tedious, it is omitted.)
The point $[0:0:1]$ is an ordinary double point of $\check{C}$, so that $\mathrm{Hess}(g)(o)\ne 0$. This completes the proof.
\end{proof}
Below, we list several particular examples of $28$-nodal double Veronese cones
constructed from smooth plane quartics with interesting automorphism groups (cf.~\cite[Table~6]{DolgachevIskovskikh}).
In each case the 3-fold is given by equation~\eqref{eq:V1}.
\begin{example}
\label{example:V1-Klein}
We consider the Klein quartic for $C$. It is given by the equation
\[x^3y+y^3z+z^3x=0.\]
Its automorphism group is isomorphic to $\mathrm{PSL}_2(\mathbb{F}_7)$.
We have
$g_4(s,t,u)=s^3t+t^3u+u^3s$ and
$$
g_6(s,t,u)=\frac{1}{8}\left(3s^5u - 15s^2t^2u^2 + 3st^5 + 3tu^5\right).
$$
\end{example}
\begin{example}
\label{example:Fermat}
Let $C$ be the Fermat plane quartic curve defined by
$$
x^4+y^4+z^4=0.
$$
Its automorphism group is isomorphic to $\mumu_4^2\rtimes\mathfrak{S}_3$.
We have $g_4(s,t,u)=4(s^4+t^4+u^4)$ and~\mbox{$g_6(s,t,u)=16s^2t^2u^2$}.
\end{example}
For explicit equations of smooth plane quartics with an action of
the symmetric group $\mathfrak{S}_4$ and the corresponding
$28$-nodal double Veronese cones we refer the reader to
Example~\ref{example:V1-S4}.
\begin{example}
\label{example:Aut-trivial}
Suppose that the smooth quartic curve $C$ is given by
$$
x^4+y^4+z^4+x^3y+2x^3z=0.
$$
Then $C$ has no non-trivial automorphisms. We have
$g_4(s,t,u)=4(s^4 - st^3 - 2su^3 + t^4 + u^4)$ and
$$
g_6(s,t,u)=-16s^3t^2u - 8s^3tu^2 + 16s^2t^2u^2 - 4t^6 + 4t^5u - t^4u^2 -4t^2u^4 + 4tu^5 - u^6.
$$
\end{example}
\section{One-to-one correspondence}
\label{section:1-1}
In the present section we put the previous results together to establish a one-to-one
correspondence between $28$-nodal double Veronese cones and smooth plane quartics, which is asserted in
Theorem~\ref{theorem:one-to-one}. We also study automorphism groups of
$28$-nodal double Veronese cones and prove Theorem~\ref{theorem:Aut}.
In this section, for the projective plane we keep the same two notations as in~\S\ref{section:equations}, i.e., $\mathbb{P}^2$ with a homogeneous coordinate system $[x:y:z]$ and its projectively dual plane $\check{\mathbb{P}}^2$ with~$[s:t:u]$.
\begin{remark}\label{remark:kappa-elliptic}
For an arbitrary del Pezzo 3-fold of degree $1$,
let $\kappa\colon V\dasharrow \check{\mathbb{P}}^2$ be the rational map given by
the half-anticanonical linear system on $V$. The half-anticanonical linear system has a unique
base point. The 3-fold $V$ is smooth at the base point,
so that a general fiber of $\kappa$
is an elliptic curve by Bertini theorem and adjunction formula.
\end{remark}
In the remaining part of this section we use
$\kappa$ to denote the rational half-anticanonical elliptic fibration on a double
Veronese cone~$V$.
Now we construct a smooth plane quartic from a $28$-nodal double Veronese cone.
\begin{lemma}\label{lemma:V-to-quartic}
Let $V$ be a $28$-nodal double Veronese cone.
Denote by $\check{C}$ the
discriminant curve of the rational elliptic fibration $\kappa$, and let $C$ be the projectively dual curve of~$\check{C}$.
Then $C$ is a smooth plane quartic.
Furthermore, for any construction of $V$ from an Aronhold heptad as in Proposition~\ref{proposition:Prokhorov-construction},
the curve $C$ is isomorphic to the Hessian curve of the corresponding net of quadrics.
\end{lemma}
\begin{proof}
It follows from
Theorem~\ref{theorem:28-can-be-constructed}
that $V$ can be constructed out of some Aronhold heptad in~$\mathbb{P}^3$
as in Proposition~\ref{proposition:Prokhorov-construction}.
Choose one construction like this, and let $\mathcal{L}$ be the net of quadrics defined by the Aronhold heptad.
The rational elliptic fibration~$\kappa$ is given by the proper transforms of the members of the net $\mathcal{L}$. Using Lemma~\ref{lemma:Hessian-dual}, we see that the projectively dual curve of~$\check{C}$
is the Hessian curve of the net $\mathcal{L}$, which is a smooth quartic curve by Lemma~\ref{lemma:smooth-quartic}.
\end{proof}
Recall that to any (rational) fibration whose general fiber is an elliptic curve we can associate the $j$-function
of its fibers, which is a rational function on the base of the fibration.
\begin{remark}\label{remark:V-j}
Let $V$ be a del Pezzo 3-fold of degree $1$.
The 3-fold
$V$ can be defined in the weighted projective space
$\mathbb{P}(1,1,1,2,3)$ with weighted homogeneous coordinates $s,t,u,v$, and $w$ by an equation of the form
\begin{equation*}
w^2=v^3-h_4(s,t,u)v+h_6(s,t,u),
\end{equation*}
where $h_4$ and $h_6$ are homogeneous polynomials of degrees $4$ and $6$, respectively.
Since the singularities of $V$ are isolated, the polynomial
$4h_4(s,t,u)^3-27h_{6}(s,t,u)^2$ is not a zero polynomial.
The rational elliptic fibration $\kappa$
is given by the projection to the coordinates
$s$, $t$, and $u$.
The corresponding $j$-function can be written as
$$
j_V(s,t,u)=1728\frac{4h_4(s,t,u)^3}{4h_4(s,t,u)^3-27h_{6}(s,t,u)^2}
$$
(see for instance \cite[\S3.1.1]{Dolgachev}).
In particular, the discriminant curve of $\kappa$ is given in $\check{\mathbb{P}}^2$ by the equation
$$
4h_4(s,t,u)^3-27h_6(s,t,u)^2=0.
$$
\end{remark}
Recall from \S\ref{section:equations} that any (smooth) quartic curve $C$ in the projective plane $\mathbb{P}^2$
defines a $j$-function $j_C$
on the dual projective plane $\check{\mathbb{P}}^2$ that computes the $j$-invariant of the quartuple of intersection points
of $C$ with the corresponding line. It is given by the formula~\eqref{equation:j-function2}.
\begin{lemma}\label{lemma:jV-vs-jC}
Let $V$ be a $28$-nodal double Veronese cone, and let $C$ be the smooth plane quartic curve
corresponding to $V$ via Lemma~\ref{lemma:V-to-quartic}.
Then $j_V(s,t,u)=j_C(s,t,u)$ for all $s,t,u$.
\end{lemma}
\begin{proof}
By Theorem~\ref{theorem:28-can-be-constructed}, the 3-fold $V$ is constructed out of some net
$\mathcal{L}$ of quadrics in $\mathbb{P}^3$ as in Proposition~\ref{proposition:Prokhorov-construction}.
Let $P$ be a general point of $\check{\mathbb{P}}^2$. The point $P$ determines a one-dimensional linear
subsystem~$\mathcal{L}_P$ of the net $\mathcal{L}$.
The fiber $E_P$ of the half-anticanonical
elliptic fibration $\kappa$ over $P$
is the proper transform of the base locus of the pencil $\mathcal{L}_P$. The base locus is given by the intersection of two general quadrics $Q_1$ and $Q_2$ in $\mathcal{L}_P$. The pencil $\mathcal{L}_P$ corresponds to the line~$L$ in $\mathbb{P}^2$ dual to the point $P$.
The singular quadrics in $\mathcal{L}_P$ correspond
to the four intersection points of the line $L$ and the Hessian curve~$C$
of the net $\mathcal{L}$. Lemma~\ref{lemma:Q1Q2} implies that the curve $E_P$ is an elliptic curve isomorphic to the double cover of $L$ branched at the four intersection points of $L$ and $C$.
This means that $j_V(P)=j_C(P)$.
\end{proof}
We will need the following assertion on
$28$-nodal double Veronese cones and their half-anticanonical elliptic fibrations.
\begin{lemma}
\label{lemma:V-from-elliptic-fibration}
Let $V$ be a $28$-nodal double Veronese cone.
Then $V$ is uniquely
defined by the $j$-function~$j_V$ of the
fibers of the rational elliptic fibration $\kappa$.
\end{lemma}
\begin{proof}
Suppose that there exists another $28$-nodal double Veronese cone $V'$
such that~\mbox{$j_V=j_{V'}$}.
The 3-folds $V$ and $V'$ can be defined in the weighted projective space
$\mathbb{P}(1,1,1,2,3)$ with weighted homogeneous coordinates $s,t,u,v$, and $w$ by equations
\begin{equation}\label{eq:V-first}
w^2=v^3-h_4(s,t,u)v+h_6(s,t,u)
\end{equation}
and
\begin{equation}\label{eq:V-second}
w^2=v^3-k_4(s,t,u)v+k_6(s,t,u),
\end{equation}
respectively, where $h_d$ and $k_d$ are homogeneous polynomials
of degree $d$. By Remark~\ref{remark:V-j} the $j$-functions can be written as
$$
j_V(s,t,u)=1728\frac{4h_4(s,t,u)^3}{4h_4(s,t,u)^3-27h_{6}(s,t,u)^2}
$$
and
$$
j_{V'}(s,t,u)=1728\frac{4k_4(s,t,u)^3}{4k_4(s,t,u)^3-27k_{6}(s,t,u)^2}.
$$
Recall that neither $h_4$ nor $k_4$ is a zero polynomial by Lemmas~\ref{lemma:jV-vs-jC} and~\ref{lemma:j-not-constant}.
Since
$$
j_V(s,t,u)=j_{V'}(s,t,u)
$$
for all $s,t$, $u$, we can easily conclude that
$$
k_4(s,t,u)=ah_4(s,t,u),\quad k_6(s,t,u)=bh_6(s,t,u),
$$
for some non-zero constants $a$ and $b$ with $a^3=b^2$.
Thus equation~\eqref{eq:V-second}
takes the form
$$
w^2=v^3-ah_4(s,t,u)v+bh_6(s,t,u).
$$
Put $c=\frac{b}{a}$. Then $c^2=a$ and $c^3=b$.
The automorphism of $\mathbb{P}(1,1,1,2,3)$ defined by
$$
\left[s:t:u:v:w\right]\mapsto \left[s:t:u:\frac{v}{c}:\frac{w}{c\sqrt{c}}\right]
$$
brings it to the form~\eqref{eq:V-first},
which means that $V$ and $V'$ are isomorphic.
\end{proof}
Now we are able to prove Theorem~\ref{theorem:one-to-one}.
\begin{proof}[First proof of Theorem~\ref{theorem:one-to-one}]
By Lemma~\ref{lemma:V-to-quartic}, a $28$-nodal double Veronese cone uniquely determines a smooth plane quartic curve that is the projectively dual curve of the discriminant curve $\check{C}$ of the half-anticanonical rational elliptic fibration~$\kappa$.
Now let $C$ be a smooth plane quartic curve. We then construct a $28$-nodal double Veronese cone~$V$ from $C$ as in~\S\ref{section:equations} (see equation~\eqref{eq:V1}).
Using Remark~\ref{remark:V-j} and Proposition~\ref{proposition:discriminant}, we see that the projectively dual curve of the discriminant curve of the rational elliptic fibration~$\kappa$ is isomorphic to~$C$.
It remains to show that if $V'$ is a $28$-nodal double Veronese cone such that
the projectively dual curve of the discriminant curve of its half-anticanonical rational elliptic fibration is isomorphic to $C$,
then $V'$ is isomorphic to $V$. By Lemma~\ref{lemma:jV-vs-jC} the $j$-functions of $V$ and $V'$ depend only on~$C$,
so that
$$
j_V(s,t,u)=j_C(s,t,u)=j_{V'}(s,t,u)
$$
for all $s$, $t$, $u$.
According to Lemma~\ref{lemma:V-from-elliptic-fibration} this means that
$V'$ is isomorphic to~$V$.
\end{proof}
We now move our attention to automorphism groups of $28$-nodal double Veronese cones.
Recall that if $V$ is such a 3-fold, we denote by $\tau$ the Galois involution
of the double cover~\mbox{$V\to \mathbb{P}(1,1,1,2)$} given by the linear system $|2H|=|-K_V|$.
The automorphism $\tau$ is contained in the center of the group~\mbox{$\Aut(V)$}.
It is easy to see that it preserves the fibers of the rational
elliptic fibration~$\kappa$.
\begin{proof}[Proof of Theorem~\ref{theorem:Aut}]
Since the map $\kappa\colon V\dasharrow \check{\mathbb{P}}^2$
is $\Aut(V)$-equivariant, we have a group homomorphism
$$
\check{\Xi}\colon \Aut(V)\to \Aut(\check{\mathbb{P}}^2).
$$
Its kernel $\mathcal{K}\subset\Aut(V)$ acts on every fiber of
$\kappa$ preserving the unique base point $O$ of $|H|$. Let~$E$
be a general fiber of $\kappa$.
Then $E$ is a general elliptic curve by Lemma~\ref{lemma:j-not-constant}, so that
the stabilizer of the point~$O$ in $\Aut(E)$
is isomorphic to $\mumu_2$.
Therefore, $\mathcal{K}$ is a subgroup of $\mumu_2$. On the other hand, $\mathcal{K}$
obviously contains the Galois involution
$\tau$ of the double cover given by the anticanonical linear system of $V$.
Thus we conclude that $\mathcal{K}\cong\mumu_2$.
Consider the action of the group $\check{\Xi}(\Aut(V))$ on the projective plane $\check{\mathbb{P}}^2$
which is the target of the map $\kappa$. By construction the action of $\check{\Xi}(\Aut(V))$
preserves the discriminant curve~$\check{C}$ of the rational elliptic fibration $\kappa$,
and hence its action on the projectively dual plane of the projective plane $\check{\mathbb{P}}^2$, which is $\mathbb{P}^2$, preserves the curve~$C$ projectively dual to $\check{C}$.
Therefore, we obtain a group homomorphism
\begin{equation*}
\Xi\colon\Aut(V)\to\Aut(C)
\end{equation*}
whose kernel $\mathcal{K}$ is isomorphic to $\mumu_2$.
Note that at this moment we claim neither that $\Xi$ is surjective nor that $\mathcal{K}$ splits as a direct factor in $\Aut(V)$.
However, we know that $\mathcal{K}$ is contained in the center
of the group $\Aut(V)$.
Now let $G$ be the automorphism group of the smooth plane quartic $C$.
Then $G$ is a finite group, and there is a natural action of
$G$ on the plane $\check{\mathbb{P}}^2$ which is projectively dual
to the plane~$\mathbb{P}^2$ where the curve~$C$ sits.
By Remark~\ref{remark:harmonic} the latter action preserves the curves given
by equations~\mbox{$g_4(s,t,u)=0$} and~\mbox{$g_6(s,t,u)=0$}. This means that there exist
group characters ~\mbox{$\chi_4, \chi_6\colon G\to\mathbb{C}^*$} such that for any~\mbox{$\gamma\in G$} one has
\[
g_4\big(\gamma(s,t,u)\big)=\chi_4(\gamma)\cdot g_4(s,t,u), \quad
g_6\big(\gamma(s,t,u)\big)=\chi_6(\gamma)\cdot g_6(s,t,u).
\]
On the other hand, the elements of $G$ preserve the equivalence class of the cross-ratio
of a quadruple of intersection points of a line in $\mathbb{P}^2$ and the curve $C$; in other words, it preserves
the $j$-function~\mbox{$j_C(s,t,u)$}. By~\eqref{equation:j-function2} this gives
$$
\chi_4(\gamma)^3=\chi_6(\gamma)^2
$$
for every $\gamma\in G$.
Set
$$
\chi_2(\gamma)=\frac{\chi_6(\gamma)}{\chi_4(\gamma)}, \quad \gamma\in G.
$$
Then $\chi_2$ is a character of $G$ such that $\chi_2(\gamma)^2=\chi_4(\gamma)$ and $\chi_2(\gamma)^3=\chi_6(\gamma)$.
Since the image~\mbox{$\chi_2(G)\subset\mathbb{C}^*$} is a cyclic group, there is a well-defined
character
\[
\chi\colon G\to \mathbb{C}^*
\]
such that $\chi(\gamma)^2=\chi_2(\gamma)$ for all $\gamma\in G$.
We have
\[
\chi(\gamma)^4=\chi_4(\gamma), \quad \chi(\gamma)^6=\chi_6(\gamma).
\]
Given an element $\gamma\in G$, we write
$$
\gamma(s,t,u)=[s':t':u']
$$
and define an automorphism of $\mathbb{P}(1,1,1,2,3)$ by
$$
[s:t:u:v:w]\mapsto
[s':t':u':\chi(\gamma)^2v:\chi(\gamma)^3w].
$$
Since the 3-fold $V$ is given by \eqref{eq:V1} in $\mathbb{P}(1,1,1,2,3)$, $G$ preserves the hypersurface
$V$. This provides a group homomorphism
$$
\Xi'\colon \Aut(C)\to\Aut(V).
$$
It is easy to see that $\Xi\circ\Xi'=\mathrm{id}_{\Aut(C)}$.
This implies that $\Aut(V)\cong\mathcal{K}\rtimes\Aut(C)$.
Since $\mathcal{K}\cong\mumu_2$ is contained in the center
of $\Aut(V)$, we conclude that $\Aut(V)\cong\mathcal{K}\times\Aut(C)$.
\end{proof}
\begin{corollary}\label{corollary:Aut-V-vs-S}
Let~$V$ be a $28$-nodal double Veronese cone,
let $C$ be the smooth plane quartic curve corresponding to $V$
by Theorem~\ref{theorem:one-to-one},
and let~$S$ be the del Pezzo surface of degree~$2$ constructed as the double
cover of $\mathbb{P}^2$ branched along $C$. Then
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathrm{Aut}(C)\cong \mathrm{Aut}(S).
$$
\end{corollary}
\section{Theta characteristics}
\label{section:theta}
As we have seen in \S\ref{section:construction}, Aronhold heptads, which are parts of regular Cayley octads, play a role in constructing $28$-nodal double Veronese cones.
On the other hand, neither an Aronhold heptad nor a regular Cayley octad is uniquely associated to a
$28$-nodal double Veronese cone. In what follows, correspondences between regular Cayley octads, smooth plane quartic curves, and $28$-nodal double Veronese cones
will be investigated in more detail, and the necessary additional data that is required to
have natural one-to-one correspondences will be clarified.
So far we used Proposition~\ref{proposition:Prokhorov-construction} to construct $28$-nodal
double Veronese cones starting from Aronhold heptads. However, it turns out that the construction
essentially depends only on the choice of the regular Cayley octad.
This can be easily understood through Geiser type involutions.
\begin{lemma}\label{lemma:V-i-j}
Let $P_1,\ldots,P_8$ be a regular Cayley octad in $\mathbb{P}^3$.
Then for all $i=1,\ldots, 8$ the $28$-nodal double Veronese cones
defined by the Aronhold heptads $\{P_1,\ldots,P_8\}\setminus\{P_i\}$ are isomorphic to each other.
\end{lemma}
\begin{proof}
For each $i=1,\ldots, 8$, let $\widehat{\mathbb{P}}^3_i$ be the blow up of $\mathbb{P}^3$ at the points $P_1,\ldots, P_8$ except the point~$P_i$. Let $\widehat{\mathbb{P}}^3_{ij}$ be the weak Fano 3-fold obtained by blowing up $\mathbb{P}^3$ at the six points of~\mbox{$\{P_1,\ldots, P_8\}\setminus\{P_i,P_j\}$} and let $W_{ij}$ be its anticanonical model.
For $1\leqslant k\ne l\leqslant 8$, denote by $L_{kl}$
the line passing through the points $P_k$ and $P_l$, and
denote by~$T_{kl}$ the twisted cubic passing through the six points of
$\{P_1,\ldots,P_8\}\setminus\{P_k,P_l\}$. Let $\widehat{L}_{kl}$ and~$\widehat{T}_{kl}$ be the proper transforms of these curves
on $\widehat{\mathbb{P}}^3_{ij}$.
Proposition~\ref{proposition:Prokhorov-construction-2} implies that
that $W_{ij}$ is the double cover of
$\mathbb{P}^3$ branched over a quartic surface and that the map $\phi_{ij}: \widehat{\mathbb{P}}^3_{ij}\to W_{ij} $ given by the anticanonical linear system is a birational morphism that contracts exactly the $15$ curves $\widehat{L}_{kl}$ for $k<l$ with $\{k,l\}\cap\{i,j\}=\varnothing$, and the curve
$\widehat{T}_{ij}$. Denote by~$\bar{L}_{kl}$ and $\bar{T}_{kl}$ the images on $W_{ij}$
of the curves $\widehat{L}_{kl}$ and $\widehat{T}_{kl}$ not contracted by~$\phi_{ij}$, respectively.
Note that the points $\bar{P}_i$ and $\bar{P}_j$ on $W_{ij}$ corresponding to $P_i$ and $P_j$ are conjugate to each other
with respect to the Galois involution $\varepsilon$ of the double cover.
The double cover $\zeta\colon W_{ij}\to\mathbb{P}^3$ is given by the half-anticanonical linear system,
so that the pull-backs of planes under $\zeta$
should be the proper transforms of the quadrics in the original $\mathbb{P}^3$ passing through
the six points of $\{P_1,\ldots,P_8\}\setminus\{P_i,P_j\}$.
The images of the curves $\zeta(\bar{T}_{ki})$ and~$\zeta(\bar{T}_{kj})$, where $k\not\in\{i,j\}$, are
lines in $\mathbb{P}^3$ (note however that the images $\zeta(\bar{T}_{kl})$ for~\mbox{$\{k,l\}\cap\{i,j\}=\varnothing$}
are conics).
Therefore, by Lemma~\ref{lemma:pencil-base-locus-ij} for $k\not\in\{i,j\}$ one has
$$
\aligned
&\varepsilon(\bar{T}_{ki})=\bar{L}_{ki}, \quad \varepsilon(\bar{L}_{ki})=\bar{T}_{ki},\\
&\varepsilon(\bar{T}_{kj})=\bar{L}_{kj}, \quad \varepsilon(\bar{L}_{kj})=\bar{T}_{kj}.
\endaligned
$$
The involution $\varepsilon\colon W_{ij}\to W_{ij}$ gives rise to a birational
map
$$
\hat{\varepsilon}\colon \widehat{\mathbb{P}}^3_i\dasharrow\widehat{\mathbb{P}}^3_j.
$$
Since $\varepsilon(\bar{P}_i)=\bar{P}_j$, we conclude that $\hat{\varepsilon}$ is actually
an isomorphism.
Denote by $\widehat{L}_{kl}^i$ and $\widehat{T}_{kl}^i$ (respectively, $\widehat{L}_{kl}^j$ and $\widehat{T}_{kl}^j$)
the proper transforms of the curves~$L_{kl}$ and~$T_{kl}$
on~$\widehat{\mathbb{P}}^3_{i}$ (respectively, $\widehat{\mathbb{P}}^3_{j}$). Then for $k\not\in\{i,j\}$ one has
$$
\aligned
&\hat{\varepsilon}(\widehat{T}_{ki}^i)=\widehat{L}_{ki}^j, \quad \hat{\varepsilon}(\widehat{L}_{ki}^i)=\widehat{T}_{ki}^j,\\
&\hat{\varepsilon}(\widehat{T}_{kj}^i)=\widehat{L}_{kj}^j, \quad \hat{\varepsilon}(\widehat{L}_{kj}^i)=\widehat{T}_{kj}^j.
\endaligned
$$
Also, we observe that for $\{k,l\}\cap\{i,j\}=\varnothing$
$$
\hat{\varepsilon}(\widehat{L}_{kl}^i)=\widehat{L}_{kl}^j, \quad \hat{\varepsilon}(\widehat{T}_{ij}^i)=\widehat{T}_{ij}^j.
$$
Therefore, the pluri-anticanonical
maps $\phi_i$ and $\phi_j$ of $\widehat{\mathbb{P}}^3_{i}$ and $\widehat{\mathbb{P}}^3_{i}$ both factor through $W_{ij}$,
and the pluri-anticanonical
models of $\widehat{\mathbb{P}}^3_{i}$ and $\widehat{\mathbb{P}}^3_{i}$,
that are obtained by the construction described in Proposition~\ref{proposition:Prokhorov-construction},
are both isomorphic to one and the same double Veronese cone $V$.
$$
\xymatrix{
\widehat{\mathbb{P}}^3_i\ar@{->}[rrd]\ar@{->}[rrddd]_{\phi_i}\ar@{->}[rrrr]^{\hat{\varepsilon}} &&&& \widehat{\mathbb{P}}^3_j\ar@{->}[lld]\ar@{->}[llddd]^{\phi_j} \\
&& \widehat{\mathbb{P}}^3_{ij}\ar@{->}[d]^{\phi_{ij}} && \\
&&W_{ij}\ar@{->}[d]&&\\
&&V&& \\
}
$$
This completes the proof of the statement.
\end{proof}
\begin{remark}
A two-dimensional analog of Lemma~\ref{lemma:V-i-j} is the following simple observation. Let~\mbox{$R_1,\ldots,R_9$}
be points in $\mathbb{P}^2$ in general position such that they are the intersection of two cubic curves.
Then the del Pezzo surfaces of degree $1$ obtained by blowing up the eight points of~$\{R_1,\ldots,R_9\}\setminus\{R_i\}$ are isomorphic to each other for all~\mbox{$1\leqslant i\leqslant 9$}.
\end{remark}
Let $\mathcal{N}$ be the set that consists of
isomorphism classes of regular Cayley octads modulo
projective transformations.
Let $\mathcal{T}$ be the set
that consists of
the pairs $(C,\theta)$, where $C$ is a smooth plane quartic
considered up to isomorphism, and $\theta$ is an even theta characteristic
on $C$.
A given regular Cayley octad defines a net $\mathcal{L}$ of quadrics in~$\mathbb{P}^3$. The net $\mathcal{L}$ yields its Hessian quartic curve~\mbox{$H(\mathcal{L})$} in~$\mathbb{P}^2$,
which is smooth by Lemma~\ref{lemma:smooth-quartic}.
Meanwhile, the singular points of quadrics in the net~$\mathcal{L}$ sweep out a smooth curve of degree $6$ in~$\mathbb{P}^3$ (see \cite[Lemme~6.8]{Beauville-Prym} and \cite[Proposition~6.10]{Beauville-Prym}), which is called the \emph{Steinerian curve} of the net and is denoted by $S(\mathcal{L})$. There is an even theta characteristic $\theta(\mathcal{L})$ such that the linear system~\mbox{$|K_{H(\mathcal{L})}+\theta(\mathcal{L})|$}
defines an isomorphism of $H(\mathcal{L})$ with $S(\mathcal{L})$ ( see \cite[Proposition~6.10]{Beauville-Prym}
and \cite[Lemme~6.12]{Beauville-Prym}). With such Hessian quartic curves and their even theta characteristics, we define the map
\[
\Theta\colon\mathcal{N}\to \mathcal {T}
\]
by assigning $\Theta(\mathcal{L})=(H(\mathcal{L}), \theta(\mathcal{L}))$.
\begin{theorem}\label{theorem:bijection-Theta}
The map $\Theta$ is bijective.
\end{theorem}
\begin{proof}
See \cite[Proposition~6.23]{Beauville-Prym}.
\end{proof}
Suppose that the points $P_1,\ldots,P_7$ form an Aronhold heptad, and denote by $P_8$ the eighth base point of the net of quadrics $\mathcal{L}$ defined by the Aronhold heptad. Choose a pair of points $P_i$, $P_j$ in the regular Cayley octad $P_1,\ldots,P_8$. This pair yields a pencil in the net $\mathcal{L}$ that consists of the quadrics containing the line passing through $P_i$ and $P_j$. The pencil contains exactly one or two singular quadrics
(see \cite[Lemme~6.6(i)]{Beauville-Prym}). Furthermore, the pencil defines
a line in the projective plane where the Hessian quartic
curve $H(\mathcal{L})$ sits, and this
line is bitangent to the Hessian quartic curve (see \cite[Theorem~6.3.5]{Dolgachev}).
Such a bitangent line defines an odd theta characteristic of $H(\mathcal{L})$.
It will be denoted by $\theta_{ij}(\mathcal{L})$. {It is obvious that~\mbox{$\theta_{ij}(\mathcal{L})=\theta_{kl}(\mathcal{L})$} if and only if
$\{i,j\}=\{k,l\}$. In particular, every odd theta characteristic on~$H(\mathcal{L})$ can be represented by~\mbox{$\theta_{ij}(\mathcal{L})$} for some $i$ and $j$.
For each choice of four distinct indices $i,j,k,l$ in $\{1,\ldots,8\}$, set
\begin{equation}\label{equation:even}
\theta_{i,jkl}=\theta_{ij}(\mathcal{L})+\theta_{ik}(\mathcal{L})+\theta_{il}(\mathcal{L})-K_{H(\mathcal{L})}.
\end{equation}
Before we proceed, notice that
\begin{equation}\label{eq:3-relation}
\theta_{ij}(\mathcal{L})+\theta_{ik}(\mathcal{L})+\theta_{jk}(\mathcal{L})=K_{H(\mathcal{L})}+\theta(\mathcal{L})
\end{equation}
for each choice of three distinct indices $i,j,k$ in $\{1,\ldots,8\}$ since the left hand side is the section of the Steinerian curve $S(\mathcal{L})$ by the plane containing the points $P_i$, $P_j$, and $P_k$ in $\mathbb{P}^3$.
\begin{lemma}
\label{lemma:even-from-7-odd}
Let $r$ be a fixed index in $\{1,\ldots,8\}$.
\begin{itemize}
\item The theta characteristic $\theta_{r, ijk}$ is even for three distinct indices $i,j,k$ in $\{1,\ldots,8\}\setminus\{r\}$.
\item Every even theta characteristic on $H(\mathcal{L})$ except $\theta(\mathcal{L})$ can be represented as $\theta_{r, ijk}$ for
some choice of three distinct indices $i$, $j$, $k$ in $\{1,\ldots,8\}\setminus\{r\}$.
\item One has
\begin{equation}\label{eq:even-from-7-odd}
\theta(\mathcal{L})=-3K_{H(\mathcal{L})}+\sum\limits_{i\neq r}\theta_{ri}(\mathcal{L}).
\end{equation}
\end{itemize}
\end{lemma}
\begin{proof}
For the first statement, see \cite[Theorem~6.3.6]{Dolgachev} or~\cite[Proposition~IX.4]{DolgachevOrtland}.
Since there are exactly $35$ even theta characteristics excluding $\theta(\mathcal{L})$, in order to prove the second statement, it is enough to verify that $\theta_{r, i_1j_1k_1}\ne \theta_{r, i_2j_2k_2}$ if $\{i_1,j_1,k_1\}\ne \{ i_2,j_2,k_2\}$.
Suppose that~$\theta_{r, i_1j_1k_1}= \theta_{r, i_2j_2k_2}$. For convenience, assume that $r=8$. Then
\[\theta_{8i_1}(\mathcal{L})+\theta_{8j_1}(\mathcal{L})+\theta_{8k_1}(\mathcal{L})+\theta_{8i_2}(\mathcal{L})+\theta_{8j_2}(\mathcal{L})+\theta_{8k_2}(\mathcal{L})-3K_{H(\mathcal{L})}
=\theta_{r, i_1j_1k_1}+\theta_{r, i_2j_2k_2}-K_{H(\mathcal{L})}=0.\]
If $\{i_1,j_1,k_1\}\cap \{ i_2,j_2,k_2\}=\varnothing$, then the above equality and~\eqref{eq:3-relation} imply
\[\theta_{i_1i_2}(\mathcal{L})+\theta_{j_1j_2}(\mathcal{L})+\theta_{k_1k_2}(\mathcal{L})=3\theta(\mathcal{L})=K_{H(\mathcal{L})}+\theta(\mathcal{L}).\]
This means that the six points $P_{i_1}, P_{i_2}, P_{j_1}, P_{j_2}, P_{k_1}, P_{k_2}$ lie on a single plane, which is impossible
by Lemma~\ref{lemma:Aronhold-condition}.
If $\{i_1,j_1,k_1\}\cap \{ i_2,j_2,k_2\}\ne\varnothing$, then we may assume that $i_1=i_2$. Then
\[\theta_{8j_1}(\mathcal{L})+\theta_{8k_1}(\mathcal{L})=\theta_{8j_2}(\mathcal{L})+\theta_{8k_2}(\mathcal{L}).\]
This together with~\eqref{eq:3-relation} implies
\[\theta_{j_1k_1}(\mathcal{L})=\theta_{j_2k_2}(\mathcal{L}),\]
and hence $\{i_1,j_1,k_1\}=\{ i_2,j_2,k_2\}$.
Now let us prove the third statement. Suppose that the right hand side of~\eqref{eq:even-from-7-odd} is an odd theta characteristic~$\theta_{mn}(\mathcal{L})$. Then $m, n\ne r$; otherwise $\theta_{r, ijk}$ would yield the same even theta characteristic for two different choices of three distinct indices $i$, $j$, $k$,
which is not the case by the second statement. For convenience, we may assume that $r=8$, $m=2$, and $n=1$. Then our assumption reads
$$
\theta_{21}(\mathcal{L})=-3K_{H(\mathcal{L})}+\sum\limits_{i=1}^7\theta_{8i}(\mathcal{L}).
$$
By~\eqref{eq:3-relation} this yields
\begin{equation}\label{eq:128}
-K_{H(\mathcal{L})}+\sum\limits_{i=3}^7\theta_{8i}(\mathcal{L})=\theta_{21}(\mathcal{L})+\theta_{81}(\mathcal{L})+\theta_{82}(\mathcal{L})=K_{H(\mathcal{L})}+\theta(\mathcal{L}).
\end{equation}
Note that
$$
K_{H(\mathcal{L})}+\theta(\mathcal{L})=\theta_{43}(\mathcal{L})+\theta_{83}(\mathcal{L})+\theta_{84}(\mathcal{L})
$$
by~\eqref{eq:3-relation}. Using this, we
deduce from~\eqref{eq:128} that
\[-K_{H(\mathcal{L})}+\sum\limits_{i=5}^7\theta_{8i}(\mathcal{L})=\theta_{43}(\mathcal{L}).\]
This contradicts the first statement. Therefore, the right hand side of~\eqref{eq:even-from-7-odd} is an even theta characteristic.
We now suppose that the right hand side of~\eqref{eq:even-from-7-odd} is not $\theta(\mathcal{L})$. Then, due to the second statement, we may assume that $r=8$ and
$$
-3K_{H(\mathcal{L})}+\sum\limits_{i=1}^7\theta_{8i}(\mathcal{L})= \theta_{81}(\mathcal{L})+\theta_{82}(\mathcal{L})+\theta_{83}(\mathcal{L})-K_{H(\mathcal{L})}.
$$
However, this implies an absurd identity $\theta_{8,456}=\theta_{87}(\mathcal{L})$. Consequently, the right hand side of~\eqref{eq:even-from-7-odd} must be $\theta(\mathcal{L})$.
\end{proof}
Let $\mathcal{L}_{\theta_{i,jkl}}$ be the net of quadrics
corresponding to the plane quartic $H(\mathcal{L})$ and
the even theta characteristic $\theta_{i,jkl}$
via the bijection from Theorem~\ref{theorem:bijection-Theta}, and let~$S_{\theta_{i,jkl}}$ be its Steinerian curve.
It turns out that these curves (and also regular Cayley octads and nets of quadrics)
are related by standard Cremona transformations of~$\mathbb{P}^3$.
\begin{proposition}\label{proposition:Cremona-Cayley-octads}
Up to projective transformation, the curve $S_{\theta_{i,jkl}}$
is obtained from $S(\mathcal{L})$ by the standard Cremona transformation centered at the points $P_i$, $P_j$, $P_k$, $P_l$,
and also by the standard Cremona transformation centered at the complementary set of points $P_m$, $P_n$, $P_s$, $P_t$.
\end{proposition}
\begin{proof}
See \cite[Proposition~IX.4]{DolgachevOrtland}.
\end{proof}
\begin{remark}\label{remark:other-35}
By Lemma~\ref{lemma:even-from-7-odd},
the $35$ even theta characteristics on $H(\mathcal{L})$ other than $\theta(\mathcal{L})$ are all of the form~\eqref{equation:even} for a fixed index $i$ (one can take for instance $i=8$). This means that
the even theta characteristics obtained as in Proposition~\ref{proposition:Cremona-Cayley-octads}
by the $35$ Cremona transformations
associated to choices of four
points out of the seven points $P_1,\ldots,P_7$ (alternatively, by the~$35$ Cremona transformations
associated to the point $P_8$ and choices of three points out of~$P_1,\ldots,P_7$) are pairwise different.
\end{remark}
Using Proposition~\ref{proposition:Cremona-Cayley-octads}, we deduce the following.
\begin{corollary}\label{corollary:Cremona-Cayley-octads}
Up to projective transformation, the net of quadrics
$\mathcal{L}_{\theta_{i,jkl}}$ consists of the proper transforms of the quadrics from the net
$\mathcal{L}$ with respect to the standard Cremona transformation~$\varsigma$ centered at the points $P_i$, $P_j$, $P_k$, $P_l$
(or with respect to the standard Cremona transformation centered at the complementary set of points $P_m$, $P_n$, $P_s$, $P_t$). Furthermore, the corresponding regular Cayley octad $P_1',\ldots,P_8'$ can be constructed as follows:
the points $P_i', P_j', P_k', P_l'$ are the images of the divisors contracted by $\varsigma$, and $P_r'=\varsigma(P_r)$ for $r=m, n, s, t$.
\end{corollary}
\begin{proof}
After a suitable projective transformation
we may assume that
$$
P_i=[1:0:0:0],\quad
P_j=[0:1:0:0],\quad
P_k=[0:0:1:0],\quad
P_l=[0:0:0:1].
$$
Then the standard Cremona transformation $\varsigma$ is the selfmap of $\mathbb{P}^3$ defined by $$
\varsigma(x_0,x_1,x_2,x_3)=\left[\frac{1}{x_0}:\frac{1}{x_1}:\frac{1}{x_2}:\frac{1}{x_3}\right].
$$
Every quadric surface in $\mathcal{L}$ is defined by a quadric homogeneous polynomial of the form
\[\sum_{\alpha\ne \beta}a_{\alpha\beta}x_\alpha x_\beta=0,\]
where $a_{\alpha\beta}=a_{\beta\alpha}.$
Its proper transform by $\varsigma$ is defined by the quadric homogeneous polynomial
\[\sum_{\{\alpha, \beta, \gamma, \delta\}=\{0,1,2,3\}}a_{\alpha\beta}x_\gamma x_\delta=0.\]
The four coordinate points and the four points $\varsigma(P_m)$, $\varsigma(P_n)$, $\varsigma(P_s)$, $\varsigma(P_t)$ are the eight intersection points of three quadrics. Lemma~\ref{lemma:CB} then implies that the net $\mathcal{L}'$ of quadric surfaces passing though these eight points is exactly the net of the proper transforms of the quadric surfaces in $\mathcal{L}$.
Observe that the symmetric matrix corresponding to the former quadric surface and the one corresponding to the latter have the same determinant. It then follows from Lemma~\ref{lemma:smooth-quartic} that the four coordinate points and the four points $\varsigma(P_m)$, $\varsigma(P_n)$, $\varsigma(P_s)$, $\varsigma(P_t)$ form a regular Cayley octad.
It also immediately follows from our observation that the Steinerian curve
$S(\mathcal{L}')$ is the proper transform of the Steinerian curve
$S(\mathcal{L})$ with respect to $\varsigma$. Hence by
Proposition~\ref{proposition:Cremona-Cayley-octads} we have
$$
S(\mathcal{L}')=S_{\theta_{i,jkl}}
$$
up to projective transformation. By Theorem~\ref{theorem:bijection-Theta}
this implies that
$$
\mathcal{L}'=\mathcal{L}_{\theta_{i,jkl}},
$$
and the assertion follows.
\end{proof}
For a given smooth plane quartic curve $C$ equipped with an even theta characteristic $\theta$, its associated net $\mathcal{L}$ of quadrics determines a regular Cayley octad $P_1,\ldots, P_8$ in $\mathbb{P}^3$,
which in turn allows us to construct a unique $28$-nodal double Veronese cone $V_{C,\theta}$ (see Lemma~\ref{lemma:V-i-j}).
Let $\theta'$
be another even theta characteristic on $C$, and let $V_{C,\theta'}$ be the corresponding
$28$-nodal double Veronese cone.
\begin{lemma}\label{lemma:V-theta-thetaprime}
The 3-folds $V_{C,\theta}$ and $V_{C,\theta'}$
are isomorphic.
\end{lemma}
\begin{proof}
By Remark~\ref{remark:other-35}, there are indices $1\leqslant i<j<k\leqslant 7$ such that
\[
\theta'=\theta_{8i}(\mathcal{L})+\theta_{8j}(\mathcal{L})+\theta_{8k}(\mathcal{L})-K_{C}.
\]
For convenience, let us say $i=5, j=6, k=7$.
Corollary~\ref{corollary:Cremona-Cayley-octads} tells us that the net of quadrics~$\mathcal{L}'$ associated
to~\mbox{$(C, \theta')$} via Theorem~\ref{theorem:bijection-Theta} can be obtained
by applying the standard Cremona transformation~$\varsigma$
centered at the points $P_1$, $P_2$, $P_3$, $P_4$ to the original net $\mathcal{L}$, up to projective equivalence.
Furthermore, the net~$\mathcal{L}'$ is defined by the regular Cayley octad $P_1',\ldots,P_8'$, where
$P_1',\ldots,P_4'$ are the images of the divisors contracted by $\varsigma$, and set
$P_i'=\varsigma(P_i)$ for $i=5,\ldots,8$. The 3-folds~\mbox{$V_{C,\theta}$} and~\mbox{$V_{C,\theta'}$} are constructed from the Aronhold heptads
$P_1,\ldots,P_7$ and $P_1',\ldots,P_7'$, respectively.
Let $\widehat{\mathbb{P}}^3$ be the blow up of $\mathbb{P}^3$ at~\mbox{$P_1,\ldots, P_7$}. Denote by $F_i$ the exceptional divisor on $\widehat{\mathbb{P}}^3$ over the point $P_i$. Let $L_{ij}$ be the line in~$\mathbb{P}^3$ passing through the points~$P_i$ and~$P_j$ with $i< j$, and let~$\widehat{L}_{ij}$ be its proper transform on $\widehat{\mathbb{P}}^3$.
Flopping the six curves~$\widehat{L}_{ij}$ with~\mbox{$1\leqslant i<j\leqslant 4$}, we obtain a birational map~$\xi$
from~$\widehat{\mathbb{P}}^3$ to another weak Fano 3-fold $\widehat{\mathbb{P}}^{3+}$.
Thus, we obtain the following diagram:
$$
\xymatrix{
\widehat{\mathbb{P}}^3\ar@{-->}[rr]^{\xi}\ar@{->}[dr]^{\nu}&&\ \widehat{\mathbb{P}}^{3+}\ar@{->}[dl]_{\nu^+}\\%
&\bar{V}&
}
$$
Here $\nu$ and $\nu^+$ are contractions of the flopping curves of $\xi$ and $\xi^{-1}$, respectively.
Keeping in mind the decomposition of the standard Cremona transformation into blow ups and blow downs, we see
that $\widehat{\mathbb{P}}^{3+}$ can be represented as the blow up
$\pi^+\colon\widehat{\mathbb{P}}^{3+}\to\mathbb{P}^3$ at the points~\mbox{$P_1',\ldots,P_7'$}. More precisely,
let $\Pi_i$, $1\leqslant i\leqslant 4$, be the plane
passing through the three points among $P_1,\ldots,P_4$ except $P_i$.
The proper transforms of the four planes $\Pi_i$ in $\widehat{\mathbb{P}}^{3+}$ are four disjoint surfaces $\Pi_i^+$
isomorphic to $\mathbb{P}^2$. Let $F_i^+$, $5\leqslant i\leqslant 7$,
be the proper transforms in $\widehat{\mathbb{P}}^{3+}$ of the surfaces $F_i$.
Then the surfaces~$\Pi_1^+,\,\Pi_2^+,\,\Pi_3^+,\,\Pi_4^+,\,F_5^+,\,F_6^+$, and~$F_7^+$ are contracted to the seven points $P_1',\ldots, P_7'$ on $\mathbb{P}^3$, respectively.
By Proposition~\ref{proposition:Prokhorov-construction}, the 3-folds
$V_{C,\theta}$ and $V_{C,\theta'}$ are the images of the pluri-anticanonical morphisms~$\phi$ and~$\phi^+$ of
$\widehat{\mathbb{P}}^{3}$ and $\widehat{\mathbb{P}}^{3+}$, respectively.
Since $\xi$ is a flop, both $\phi$ and $\phi^+$ factor through~$\bar{V}$,
so that
both~$V_{C,\theta}$ and~$V_{C,\theta'}$ are obtained by contracting one and the same curves on~$\bar{V}$.
Thus,~$V_{C,\theta}$ and $V_{C,\theta'}$ are isomorphic to one and the same double Veronese cone $V$.
$$
\xymatrix{
\widehat{\mathbb{P}}^3\ar@{->}[ddd]_{\pi}\ar@{->}[ddrr]_{\phi}\ar@{-->}[rrrr]^{\xi}\ar@{->}[drr]^{\nu}&&&&\ \widehat{\mathbb{P}}^{3+}\ar@{->}[ddd]^{\pi^{+}}\ar@{->}[ddll]^{\phi^{+}}\ar@{->}[dll]_{\nu^+}\\%
&&\bar{V}\ar@{->}[d]&&\\
&&V&&\\
\mathbb{P}^3\ar@{-->}[rrrr]^{\varsigma}&&&&\mathbb{P}^3}
$$
This completes the proof of the lemma.
\end{proof}
Lemmas~\ref{lemma:V-i-j} and~\ref{lemma:V-theta-thetaprime} give the following
\begin{corollary}\label{corollary:all-V-same}
Let $C$ be a smooth plane quartic curve.
Then for any choice of an even theta characteristic $\theta$ on $C$, and for
any choice of an Aronhold heptad in the regular Cayley octad corresponding to~$\theta$,
the $28$-nodal double Veronese cones constructed from these Aronhold heptads as in
Proposition~\ref{proposition:Prokhorov-construction} are isomorphic to each other.
\end{corollary}
Corollary~\ref{corollary:all-V-same} allows us to put together the following proof.
\begin{proof}[Second proof of Theorem~\ref{theorem:one-to-one}]
Given a $28$-nodal double Veronese cone, we associate to it a smooth plane quartic curve as in Lemma~\ref{lemma:V-to-quartic}.
Let $C$ be a smooth plane quartic curve. Choose an even theta characteristic on $C$. By
Theorem~\ref{theorem:bijection-Theta}, this provides us a regular Cayley octad.
Choose an Aronhold heptad from this regular Cayley octad, and construct a $28$-nodal double Veronese cone
as in Proposition~\ref{proposition:Prokhorov-construction}.
By Corollary~\ref{corollary:all-V-same} the result does not depend on the choice of the even theta characteristic and the
Aronhold heptad, and by Theorem~\ref{theorem:28-can-be-constructed} every $28$-nodal double Veronese cone can be obtained
in this way.
Using Lemma~\ref{lemma:V-to-quartic} once again, we see that the above two constructions are mutually
inverse.
\end{proof}
Recall from \cite[\S\,IX.2]{DolgachevOrtland} that a (unordered) set of seven distinct odd theta characteristics~\mbox{$\theta_1,\ldots,\theta_7$} on a smooth plane quartic curve~$C$ is called an \emph{Aronhold system} if they satisfy the condition that~${\theta_i+\theta_j+\theta_k-K_C}$ is an even theta characteristic for each choice
of three distinct indices $i,j,k$.
This is equivalent to the requirement that the six points of any three of~\mbox{$\theta_1,\ldots,\theta_7$} are not contained in a conic because
\[\left|2K_C-\theta_i-\theta_j-\theta_k\right|=\left|\theta_i+\theta_j+\theta_k-K_C\right|.\]
A given smooth quartic has exactly~$288$ Aronhold systems (see \cite[Proposition~IX.3]{DolgachevOrtland}).
It follows from Lemma~\ref{lemma:even-from-7-odd} that if $P_1,\ldots,P_8$ is a regular Cayley octad, and
$\mathcal{L}$ is the corresponding net of quadrics in $\mathbb{P}^3$, then for every fixed index
$r\in\{1,\ldots,8\}$ the seven odd theta characteristics~\mbox{$\theta_{ri}(\mathcal{L})$, $i\neq r$,} form
an Aronhold system. It turns out that all Aronhold systems arise in this way.
\begin{lemma}[{cf. \cite[Proposition~6.3.11]{Dolgachev}}]
\label{lemma:all-288}
Let $C$ be a smooth plane quartic, and let $\Xi$ be an Aronhold system on $C$. Then
there exist a unique even theta characteristic $\theta$ on $C$ and a unique point~$P_8$
in the regular Cayley octad
$P_1,\ldots,P_8$ corresponding to $(C,\theta)$ via Theorem~\ref{theorem:bijection-Theta},
such that
\begin{equation}\label{eq:Aronhold-from-Aronhold}
\Xi=\{\theta_{8i}(\mathcal{L})\mid 1\leqslant i\leqslant 7\},
\end{equation}
where $\mathcal{L}$ is the net of quadrics defined by the regular Cayley octad
$P_1,\ldots,P_8$.
\end{lemma}
\begin{proof}
For a given even theta characteristic $\theta$ on $C$, we obtain a regular Cayley octad~\mbox{$P_1,\ldots,P_8$} in~$\mathbb{P}^3$ from Theorem~\ref{theorem:bijection-Theta}. This gives us eight distinct Aronhold systems $\Xi_1,\ldots,\Xi_8$ on $C$ as in~\eqref{eq:Aronhold-from-Aronhold}.
By Lemma~\ref{lemma:even-from-7-odd} one has
$$
\theta=-3K_C+ \sum_{\vartheta\in\Xi_r }\vartheta
$$
for every $r\in\{1,\ldots,8\}$. Hence none of the Aronhold systems $\Xi_r$ can coincide with an Aronhold system
constructed in the same way starting from any other even theta characteristic $\theta'\neq \theta$
on~$C$. Since there are exactly $288$ Aronhold systems and exactly $36$ even theta characteristics on~$C$,
the above construction exhausts all the Aronhold systems on $C$, and the assertion follows.
\end{proof}
Lemmas~\ref{lemma:even-from-7-odd} and~\ref{lemma:all-288} imply the following.
\begin{corollary}
Let $C$ be a smooth plane quartic, and let $\Xi$ be an Aronhold system on $C$. Then
\begin{equation}\label{equation:even-theta-from-Aronhold}
\theta=-3K_C+ \sum_{\vartheta\in\Xi }\vartheta
\end{equation}
is an even theta characteristic on $C$. Moreover,
for every even theta characteristic $\theta$ on $C$ there exist exactly eight
Aronhold systems such that $\theta$ is obtained from them as in~\eqref{equation:even-theta-from-Aronhold},
and each of these eight Aronhold systems is in turn obtained from $\theta$ as in Lemma~\ref{lemma:all-288}.
\end{corollary}
To proceed we need to define a certain equivalence relation on the set of Aronhold systems.
Let~$\mathcal{T}_a$ be the set that consists of the pairs $(C, \Xi)$, where $C$ is a smooth quartic curve considered up to isomorphism, and $\Xi$ is an Aronhold system on $C$. Two members $(C_1, \Xi_1)$ and~\mbox{$(C_2, \Xi_2)$} in~$\mathcal{T}_a$ are considered to be equivalent if
\begin{itemize}
\item $C_1=C_2$;
\item one has
\[\sum_{\vartheta\in\Xi_1}\vartheta = \sum_{\vartheta\in\Xi_2}\vartheta;\]
\item there is an automorphism $\sigma\colon C_1\to C_2$ such that $\sigma^*(\Xi_2)=\Xi_1$.
\end{itemize}
In other words, the equivalence comes from the action of stabilizers of the
even theta characteristic attached to $(C,\Xi)$ by~\eqref{equation:even-theta-from-Aronhold}
in the automorphism group of $C$.
Let $\mathcal{T}_A$ be the set of the equivalence classes of members in $\mathcal{T}_a$.
Let $\mathcal{A}$ be the set of Aronhold heptads in $\mathbb{P}^3$ up to projective transformations.
Define a map
\[\bigtriangleup\colon \mathcal{A}\to \mathcal{T}_A\]
by assigning to a given Aronhold heptad
the Hessian curve of the corresponding net of quadrics together with the Aronhold system constructed as in~\eqref{eq:Aronhold-from-Aronhold}.
\begin{theorem}\label{theorem:diagram-one-to-one}
The map $\bigtriangleup$ is bijective.
\end{theorem}
\begin{proof}
It is enough to show that for a given smooth plane quartic curve $C$ the map $\Delta$ induces one-to-one correspondence between members of $\mathcal{A}$ with the Hessian curve $C$ and members of $\mathcal{T}_A$ with the first component $C$.
It immediately follows from Lemma~\ref{lemma:all-288} that the map $\bigtriangleup$ is surjective.
Suppose that $\bigtriangleup (X)=\bigtriangleup(Y) $ for Aronhold heptads $X$ and $Y$. First of all, the Hessian curves of the nets of quadrics determined by $X$ and $Y$, respectively, are one and the same smooth quartic curve~$C$. Let $\Xi_X$ and $\Xi_Y$ be the Aronhold systems determined by $X$ and $Y$, respectively.
Then
$$
-3K_C+ \sum_{\vartheta\in\Xi_X }\vartheta=-3K_C+ \sum_{\vartheta\in\Xi_Y }\vartheta
$$
is one and the same even theta characteristic $\theta$ on $C$.
Furthermore, since $\bigtriangleup (X)=\bigtriangleup(Y) $, there is an automorphism $\sigma$ of $C$ such that $\Xi_X=\sigma^*(\Xi_Y)$. Then the automorphism $\sigma$ preserves the linear system~$|K_C+\theta |$ that is the linear system of hyperplane sections of the Steinerian curve of the net of quadrics $\mathcal{L}$ in $\mathbb{P}^3$
corresponding to~$(C, \theta)$ via Theorem~\ref{theorem:bijection-Theta}.
This means that the automorphism $\sigma$ considered as an automorphism of the Steinerian curve
is induced from a projective transformation of~$\mathbb{P}^3$ preserving the regular Cayley octad defined by $\mathcal{L}$. Thus, we conclude that $X=Y$ in~$\mathcal{A}$.
\end{proof}
\begin{corollary}\label{corollary:Aronhold-sets}
Let $V$ be a $28$-nodal double Veronese cone, and let $C$ be the smooth plane quartic curve
corresponding to $V$ via Theorem~\ref{theorem:one-to-one}. Then
there exists a natural one-to-one correspondence
between the set of
the diagrams~\eqref{equation:Prokhorov-diagram}
considered up to projective transformations of $\mathbb{P}^3$ and the set of members in $ \mathcal{T}_A$ with the first component in the pair isomorphic to $C$.
\end{corollary}
\begin{proof}
This immediately follows from Theorem~\ref{theorem:diagram-one-to-one}, because by Corollary~\ref{corollary:all-V-same} there is an obvious one-to-one correspondence
between the set of
the diagrams~\eqref{equation:Prokhorov-diagram}
considered up to projective transformations of~$\mathbb{P}^3$ and the set of members in $ \mathcal{A}$ with the Hessian curve $C$.
\end{proof}
\begin{corollary}\label{corollary:288-diagram}
For a general smooth plane quartic curve $C$, there are exactly $288$ diagrams~\eqref{equation:Prokhorov-diagram} associated to $C$,
up to projective transformations of $\mathbb{P}^3$.
\end{corollary}
If the smooth plane quartic curve $C$ is not general, the number of diagrams~\eqref{equation:Prokhorov-diagram} associated to $C$
up to projective transformations of $\mathbb{P}^3$
may be smaller than~$288$. Note however that this number is always at least~$36$.
\begin{example}
Let $C$ be the Klein quartic curve given in $\mathbb{P}^2$ by the equation
\[x^3y+y^3z+z^3x=0.\]
Then $\Aut(C)\cong\mathrm{PSL}_2(\mathbb{F}_7)$.
There exists a unique $\Aut(C)$-invariant even theta characteristic~$\theta_0$ on~$C$ (see \cite[Example~2]{Burns} or \cite[Example~2.8]{Dolgachev97}).
The remaining $35$ even theta characteristics split into three $\Aut(C)$-orbits of
lengths $7$, $7$, and $21$, respectively (see \cite[(8.3)]{DolgachevKanev}).
Furthermore, according to \cite[Theorem~10.1]{Jeurissen}
the set of $288$ Aronhold systems on $C$ splits into four $\Aut(C)$-orbits of lengths~$8,\,56,\,56$, and $168$,
respectively. This means that the stabilizer of every even theta characteristic~$\theta$ in~$\Aut(C)$ acts transitively
on the set of eight Aronhold systems corresponding to $\theta$ via~\eqref{equation:even-theta-from-Aronhold}.
Now let $V$ be the $28$-nodal double Veronese cone corresponding to $C$ via Theorem~\ref{theorem:one-to-one}.
Corollary~\ref{corollary:Aronhold-sets} tells us that $V$ admits exactly~$36$
diagrams~\eqref{equation:Prokhorov-diagram}
up to projective transformations of~$\mathbb{P}^3$.
One of them corresponds to the $\Aut(C)$-invariant theta characteristic $\theta_0$.
The group~\mbox{$\Aut(C)$} acts on the corresponding three-dimensional projective space
preserving the regular Cayley octad.
Note however that since the regular Cayley octad
in $\mathbb{P}^3$ is a single $\Aut(C)$-orbit (see for instance \cite[Lemma~3.2]{CheltsovShramov-PSL}),
this diagram is not equivariant with respect to the whole
group~\mbox{$\Aut(C)$}. It is equivariant with respect to the
subgroup~\mbox{$\mumu_7\rtimes\mumu_3\subset\Aut(C)$}.
\end{example}
\medskip
There is another natural
way to recover the smooth plane quartic $C$ corresponding to the $28$-nodal double Veronese cone $V$.
Let $P_1,\ldots, P_7$ be seven points in $\mathbb{P}^3$ that form an Aronhold heptad. Let $P_8$ be the eighth base point of the net $\mathcal{L}(P_1,\ldots, P_7)$. Let $\rho\colon\mathbb{P}^3\dasharrow \mathbb{P}^2$ be the linear projection centered at~$P_8$.
\begin{lemma}\label{lemma:Gale}
No three of the seven points $\rho(P_1),\ldots , \rho(P_7)$ are collinear (so that in particular no two of them coincide),
and no six points among them are contained in a conic.
\end{lemma}
\begin{proof}
If three points, say $\rho(P_1),\rho(P_2)$, and $\rho(P_3)$, lie on a line,
then the four points $P_1,P_2, P_3$, and~$P_8$ lie on a plane.
This implies that there is an element in~\mbox{$\mathcal{L}(P_1,\ldots, P_7)$} that contains the plane.
Otherwise the restriction of the net $\mathcal{L}(P_1,\ldots, P_7)$ to the plane would define a net of conics passing through the four points
$P_1,P_2, P_3$, and $P_8$.
We now suppose that six points, say $\rho (P_1),\ldots, \rho (P_6)$, lie on a conic. Then there is a quadric surface
in $\mathbb{P}^3$ passing through the six points $P_1,\ldots, P_6$ together with $P_8$. It follows from Lemma~\ref{lemma:CB}
that the quadric surface should pass through the point $P_7$.
If the base point $P_8$ of the net $\mathcal{L}(P_1,\ldots, P_7)$
is a singular point of a member of the net,
then the base locus of the net~\mbox{$\mathcal{L}(P_1,\ldots, P_7)$}
could not consists of eight distinct points.
\end{proof}
\begin{remark}\label{remark:Gale-recover}
The heptad $\rho(P_1),\ldots , \rho(P_7)$ is the \emph{Gale transform}
of the heptad~\mbox{$P_1,\ldots,P_7$} (see~\cite{EP00}, and in particular the examples following \cite[Corollary~3.2]{EP00}).
Thus, the points~\mbox{$\rho(P_1),\ldots , \rho(P_7)$} uniquely define the Aronhold heptad~\mbox{$P_1,\ldots,P_7$} and
the regular Cayley octad~\mbox{$P_1,\ldots,P_8$} up to projective transformations of~$\mathbb{P}^3$.
\end{remark}
Now let $\alpha\colon S\to \mathbb{P}^2$ be the blow up
of $\mathbb{P}^2$ at the seven points $\rho(P_1),\ldots , \rho(P_7)$.
By Lemma~\ref{lemma:Gale} the surface $S$ is a smooth del Pezzo surface of degree $2$, so that its anticanonical linear system
yields the double cover $\varphi\colon S\to\mathbb{P}^2$ branched along some smooth plane quartic curve $C$.
The quartic curve $C$ coincides with the Hessian curve of the net
of quadrics $\mathcal{L}(P_1,\ldots, P_7)$
(see~\mbox{\cite[Proposition~6.3.11]{Dolgachev}}
or~\mbox{\cite[Proposition~IX.2]{DolgachevOrtland}}).
The above maps are diagramed as follows:
\begin{equation}
\label{equation:non-commutative}
\xymatrix{
V\ar@{-->}[d]_{\kappa}&\widehat{\mathbb{P}}^3\ar@{->}[d]^{\pi}\ar@{->}[l]_{\phi}&&\ S\ar@{->}[dl]_{\alpha}\ar@{->}[d]^{\varphi}\\%
\mathbb{P}^2&\mathbb{P}^3\ar@{-->}[r]_{\rho}&\mathbb{P}^2&\mathbb{P}^2}
\end{equation}
Here $\phi$ is a small resolution of all singular points of the 3-fold $V$,
the morphism $\pi$ is the blow up of~$\mathbb{P}^3$ at the seven distinct points $P_1,\ldots,P_7$, and the rational map $\kappa$ is given by the half-anticanonical linear system of $V$.
\begin{remark}\label{remark:Aronhold}
The image of the exceptional curve of $\alpha$ over the point $\rho(P_i)$ by the double cover morphism $\varphi$ is the bitangent line of $C$ corresponding to the odd theta characteristic~$\theta_{8i}(\mathcal{L})$
(see~\mbox{\cite[Proposition~IX.2]{DolgachevOrtland}}).
These seven odd theta characteristic form an Aronhold system.
Moreover, this Aronhold system is the same
as the one given by~\eqref{eq:Aronhold-from-Aronhold}.
\end{remark}
\begin{remark}
Following the diagram \eqref{equation:non-commutative}, one may easily figure out the last statement in Lemma~\ref{lemma:Q1Q2}. Indeed, a fiber of the rational map $\kappa$ is given by the base locus of a pencil contained in the net~$\mathcal{L}(P_1,\ldots, P_7)$. In case of a smooth fiber, the fiber is the proper transform of a smooth intersection $E$ of two quadrics in the net. The intersection $E$ is a curve of degree $4$ passing through eight points $P_1,\ldots, P_8$ which are the base points of the net. Thus the projection $\rho$ sends~$E$ to a cubic curve passing through the seven points $\rho(P_1),\ldots, \rho(P_7)$. The proper transform of this cubic curve on the del Pezzo surface $S$ is a smooth member of the anticanonical linear system of $S$. Since the double cover $\varphi$ is given by $|-K_S|$, the latter proper transform is the pull-back of a line on $\mathbb{P}^2$ by $\varphi$. It is the double cover of the line branched at the distinct four points at which the line and the quartic curve $C$ intersect. The point of the dual projective plane corresponding to this line is exactly the point over which the original fiber lies.
\end{remark}
\section{$\mathfrak{S}_4$-symmetric $28$-nodal double Veronese cones}
\label{section:S4-Veronese}
In this section
we study the $\mathfrak{S}_4$-equivariant birational geometry of
double Veronese cones introduced in Example~\ref{example:V1-S4}.
Let $C$ and $V$ be the plane quartic curve and the corresponding $28$-nodal double Veronese cone in Example~\ref{example:V1-S4}.
We consider the automorphisms of $\mathbb{P}(1,1,1,2,3)$
\[\begin{split}
&\tau\colon [s:t:u:v:w]\mapsto [s:t:u:v:-w],\\
&\sigma_1\colon [s:t:u:v:w]\mapsto [-s:-t:u:v:w],\\
&\sigma_2\colon [s:t:u:v:w]\mapsto [u:s:t:v:w],\\
&\sigma_3\colon [s:t:u:v:w]\mapsto [t:s:u:v:-w].
\end{split}
\]
Since they keep $V$ invariant, they may be regarded as automorphisms of $V$.
Then the involution~$\tau$ is the Galois involution of the double cover of the Veronese cone.
Moreover, by Theorem~\ref{theorem:Aut}, we have
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathrm{Aut}(C),
$$
where the subgroup $\mumu_2$ is generated by $\tau$.
Let $\mathfrak{G}$ be the subgroup in $\mathrm{Aut}(V)$ that is generated by $\sigma_1$, $\sigma_2$, and $\sigma_3$,
and let $\mathfrak{G}^\prime$ be the subgroup in $\mathrm{Aut}(V)$ generated by $\sigma_1$, $\sigma_2$, and $\tau\circ\sigma_3$.
Then $\mathfrak{G}\cong\mathfrak{G}^\prime\cong\mathfrak{S}_4$, and both of them are projected isomorphically to the same subgroup in $\mathrm{Aut}(C)$.
Meanwhile, the subgroups $\mathfrak{G}$ and $\mathfrak{G}^\prime$ are not conjugate in $\mathrm{Aut}(V)$.
To facilitate the computations, introduce an auxiliary variable
\begin{equation}\label{eq:bar-v-vs-v}
\bar{v}=v-\mu(s^2+t^2+u^2),
\end{equation}
where $\mu=\frac{2\lambda}{3}$.
The defining equation of the 3-fold $V$ may then be rewritten as
\begin{multline}\label{eq:V-bar-v}
w^2=\bar{v}\Big(\bar{v}^2+3\mu\bar{v}(s^2+t^2+u^2)+3\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u)\Big)+\\
+\mu^3(s^2+t^2+u^2)^3-\mu(s^2+t^2+u^2)g_4(s,t,u)+g_6(s,t,u).
\end{multline}
One has
$$
\mu^3(s^2+t^2+u^2)^3-\mu(s^2+t^2+u^2)g_4(s,t,u)+g_6(s,t,u)=4(\lambda-2)^2(\lambda+1)s^2t^2u^2.
$$
Set $\gamma=2(\lambda-2)\sqrt{\lambda+1}$. We see from~\eqref{eq:V-bar-v}
that the 3-fold $V$ contains the surfaces $\Pi_{\pm}$ given by
equations
$$
\bar{v}=w\mp\gamma stu=0,
$$
that is, by equations
\begin{equation}\label{eq:Pi-pm-eqs}
\left\{\aligned
&v=\mu(s^2+t^2+u^2),\\
&w=\pm\gamma stu.
\endaligned
\right.
\end{equation}
Note also that both $\Pi_+$ and $\Pi_-$ are $\mathfrak{G}^\prime$-invariant,
and both of them are not $\mathbb{Q}$-Cartier divisors.
In particular, we see that $\mathrm{rk}\,\mathrm{Cl}(V)^{\mathfrak{G}^\prime}\ne 1$.
Let us introduce a new coordinate
$$
r=\frac{w-\gamma stu}{\bar{v}}.
$$
On the 3-fold $V$, one has
$$
r=\frac{\bar{v}^2+3\mu\bar{v}(s^2+t^2+u^2)+3\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u)}{w+\gamma stu}.
$$
This gives us a birational map of $V$ to the complete intersection in $\mathbb{P}(1,1,1,1,2,3)$ given by
\begin{equation}\label{eq:w-exclusion}
\left\{\aligned
&r\bar{v}=w-\gamma stu,\\
&r(w+\gamma stu)=\bar{v}^2+3\mu\bar{v}(s^2+t^2+u^2)+3\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u).\\
\endaligned
\right.
\end{equation}
Now excluding the variable $w$ using the first equation
in~\eqref{eq:w-exclusion} and expressing $\bar{v}$ in terms of
$s$, $t$, $u$, and~$v$ by~\eqref{eq:bar-v-vs-v},
we obtain a quartic hypersurface in~\mbox{$\mathbb{P}(1,1,1,1,2)$}
with homogeneous coordinates $s$, $t$, $u$, $r$, and $v$
that is given by
$$
r(r\bar{v}+2\gamma stu)=\bar{v}^2+3\mu\bar{v}(s^2+t^2+u^2)+3\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u)
$$
or equivalently by
\begin{equation}\label{eq:double-cover-from-V1}
v^2+v\Big(\mu(s^2+t^2+u^2)-r^2\Big)+\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u)+\mu r^2(s^2+t^2+u^2)-2\gamma rstu=0.
\end{equation}
We denote this 3-fold by $W$.
We have constructed a birational map $V\dasharrow W$ that fits into the following $\mathfrak{G}^\prime$-equivariant diagram
\begin{equation}
\label{equation:unprojection}
\xymatrix{
\widehat{V}\ar@{->}[rr]\ar@{->}[d]&&V\ar@{-->}[lld]\\%
W&&}
\end{equation}
Here the horizontal arrow is the blow up of the
surface $\Pi_+$, and the vertical arrow
is the blow down of the proper transform of the surface $\Pi_-$ to the point
$$
[s:t:u:r:v]=[0:0:0:1:0],
$$
which is a smooth point of the 3-fold $W$.
The 3-fold $W$ is the double cover of $\mathbb{P}^3$ branched over a quartic surface
$$
\Big(\mu(s^2+t^2+u^2)-r^2\Big)^2-4\Big(\mu^2(s^2+t^2+u^2)^2-g_4(s,t,u)+\mu(s^2+t^2+u^2)r^2-2\gamma rstu\Big)=0,
$$
where we consider $s$, $t$, $u$ and $r$ as homogeneous coordinates on $\mathbb{P}^3$.
This quartic surface has exactly~$16$ nodes, which means that it is a Kummer surface, so that $W$ would also have exactly~$16$ nodes.
Let $S$ be the surface in $W$ that is cut out by $r=0$.
Then $S$ is $\mathfrak{G}^\prime$-invariant.
It is the double cover of $\mathbb{P}^2$ branched over the quartic curve
$$
4g_4(s,t,u)-3\mu^2(s^2+t^2+u^2)^2=0,
$$
where we consider $s$, $t$ and $u$ as homogeneous coordinates on $\mathbb{P}^2$.
Keeping in mind that $\mu=\frac{2\lambda}{3}$, we see that this curve is given by
\begin{equation}\label{eq:branch-curve-section}
s^4+t^4+u^4+\lambda(t^2u^2+s^2u^2+s^2t^2)=0.
\end{equation}
Therefore, it is isomorphic to our original curve $C$. In particular, the surface $S$ is a smooth del Pezzo surface.
\begin{lemma}
\label{lemma:S4-reps}
Let $\mathbb{U}=\mathrm{Pic}(S)\otimes\mathbb{C}$.
One has an isomorphism of $\mathfrak{G}^{\prime}$-representations
\begin{equation}
\label{eq:W3W3prime}
\mathbb{U}\cong\mathbb{I}\oplus\mathbb{I}^\prime\oplus\mathbb{W}_3\oplus\mathbb{W}_3^\prime,
\end{equation}
where $\mathbb{I}$ is the trivial representation, $\mathbb{I}^\prime$ is the sign representation,
and $\mathbb{W}_3$ and $\mathbb{W}_3^\prime$ are non-isomorphic irreducible three-dimensional representations.
\end{lemma}
\begin{proof}
Recall from~\eqref{eq:double-cover-from-V1} that the surface $S$ is given in $\mathbb{P}(1,1,1,2)$
with weighted homogeneous coordinates $s,t,u$, and $v$ by equation
\begin{equation*}
v^2+\frac{2\lambda}{3}v(s^2+t^2+u^2)+\frac{4\lambda^2}{9}(s^2+t^2+u^2)^2-g_4(s,t,u)=0,
\end{equation*}
where
\[g_4(s,t,u)=\frac{\lambda^2+12}{3}(s^4+t^4+u^4)+\frac{2(\lambda^2+6\lambda)}{3}(t^2u^2+s^2u^2+t^2s^2).\]
As before, the parameter $\lambda$ varies in $\mathbb{C}\setminus\{\pm 2, -1\}$.
Let $\mathrm{v}$ be a new variable defined as
\[
\mathrm{v}=\frac{v}{2}+\frac{\lambda}{6}(s^2+t^2+u^2).
\]
One can verify that, after this change of coordinates, $S$ is given by the following equation in~\mbox{$\mathbb{P}(1,1,1,2)$} with weighted homogeneous coordinates $s,t,u$, and $\mathrm{v}$, cf.~\eqref{eq:branch-curve-section}:
\begin{equation*}
\mathrm{v}^2=s^4+t^4+u^4+\lambda(t^2u^2+s^2u^2+t^2s^2).
\end{equation*}
Note that $\mathfrak{G}^{\prime}$ is the subgroup in $\Aut(S)$ generated by the transformations
\[\begin{split}
&[s:t:u:\mathrm{v}]\mapsto [-s:-t:u:\mathrm{v}],\\
&[s:t:u:\mathrm{v}]\mapsto [u:s:t:\mathrm{v}],\\
&[s:t:u:\mathrm{v}]\mapsto [t:s:u:\mathrm{v}].
\end{split}\]
Let $\mathfrak{G}^{\prime\prime}$
be the subgroup in $\Aut(S)$ generated by the transformations
\[\begin{split}
&[s:t:u:\mathrm{v}]\mapsto [-s:-t:u:\mathrm{v}],\\
&[s:t:u:\mathrm{v}]\mapsto [u:s:t:\mathrm{v}],\\
&[s:t:u:\mathrm{v}]\mapsto [t:s:u:-\mathrm{v}].
\end{split}\]
Then $\mathfrak{G}^{\prime\prime}\cong\mathfrak{G}^{\prime}$, but these subgroups are not conjugate in $\Aut(S)$.
They intersect by the even elements, and every odd element of $\mathfrak{G}^{\prime}$ differs
from the corresponding element of $\mathfrak{G}^{\prime\prime}$ by the Galois involution of the anticanonical double cover $S\to\mathbb{P}^2$.
This involution acts as a multiplication by~$-1$ on the orthogonal complement
to $K_S$ with respect to the intersection form in~\mbox{$\mathrm{Pic}(S)\otimes\mathbb{C}$}.
Thus, to prove~\eqref{eq:W3W3prime}, it is enough to show that
$$
\mathbb{U}\cong\mathbb{I}^{\oplus 2}\oplus\mathbb{W}_3\oplus\mathbb{W}_3^\prime
$$
as a representation of the group $\mathfrak{G}^{\prime\prime}$.
For ease of notation, take
$\alpha,\beta\in\mathbb{C}$ so that
\[
\alpha^2=\frac{-\lambda+2\sqrt{-\lambda-1}}{\lambda+2}
\quad\text{and}\quad
\beta=\frac{\lambda}{2}(1+\alpha^2).
\]
Consider the following twelve $(-1)$-curves on the surface $S$:
\[\begin{array}{ll}
L_1 : s+\alpha t=\mathrm{v}-\beta t^2-u^2=0; & L_2 : s-\alpha t=\mathrm{v}-\beta t^2-u^2=0;\\
L_3 : t+\alpha s=\mathrm{v}+\beta s^2+u^2=0; & L_4 : t-\alpha s=\mathrm{v}+\beta s^2+u^2=0; \\
L_5 : t+\alpha u=\mathrm{v}-\beta u^2-s^2=0; & L_6 : t-\alpha u=\mathrm{v}-\beta u^2-s^2=0; \\
L_{7} : s+\alpha u=\mathrm{v}+\beta u^2+t^2=0;& L_{8}: s-\alpha u=\mathrm{v}+\beta u^2+t^2=0.\\
L_9 : u+\alpha t=\mathrm{v}+\beta t^2+s^2=0; & L_{10} : u-\alpha t=\mathrm{v}+\beta t^2+s^2=0; \\
L_{11} : u+\alpha s=\mathrm{v}-\beta s^2-t^2=0;& L_{12} : u-\alpha s=\mathrm{v}-\beta s^2-t^2=0.
\end{array}\]
They form a single $\mathfrak{G}^{\prime\prime}$-orbit.
Moreover, these $12$ curves split into a disjoint union of the following pairs of intersecting lines:
$$
L_1\cup L_2, \quad L_3\cup L_4, \quad L_5\cup L_6, \quad L_7\cup L_{8}, \quad L_9\cup L_{10}, \quad L_{11}\cup L_{12}.
$$
Hence, there exists a $\mathfrak{G}^{\prime\prime}$-equivariant conic bundle
$\zeta\colon S\to\mathbb{P}^1$ such that these pairs are exactly its singular fibers.
This implies that $\mathrm{Pic}(S)^{\mathfrak{G}^{\prime\prime}}$ is generated by $K_S$ and a fiber of this conic bundle.
In particular, the surface $S$ is not $\mathfrak{G}^{\prime\prime}$-minimal.
With \cite[Table~7]{DolgachevIskovskikh}, we conclude that $\mathfrak{G}^{\prime\prime}$
is the \emph{odd lift} of $\mathfrak{S}_4$ to $\mathrm{Aut}(S)$
(see \cite[\S6.6]{DolgachevIskovskikh} for the terminology).
Therefore, the subgroup $\mathfrak{G}^\prime$ is the \emph{even lift} of $\mathfrak{S}_4$ to $\mathrm{Aut}(S)$.
Hence, by \cite[Table~7]{DolgachevIskovskikh}, the surface~$S$ is $\mathfrak{G}^{\prime}$-minimal,
so that $\mathbb{U}$ contains a unique trivial $\mathfrak{G}^{\prime}$-representation (generated by $K_S$).
This in turn implies that $\mathbb{U}$ does not contain
summands isomorphic to the sign representation of the group~$\mathfrak{G}^{\prime\prime}$.
As $\mathfrak{G}^{\prime\prime}$-representation,
$\mathbb{U}$ splits as
\begin{equation}\label{eq:U-preliminary-splitting}
\mathbb{U}\cong\mathbb{I}^{\oplus 2}\oplus\overline{\mathbb{U}},
\end{equation}
where $\overline{\mathbb{U}}$ is some six-dimensional representation of $\mathfrak{G}^{\prime\prime}$
that does not contain one-dimensional subrepresentations.
We conclude that $\overline{\mathbb{U}}$ splits into a sum of two- and three-dimensional irreducible~$\mathfrak{G}^{\prime\prime}$-representations.
If there is a two-dimensional summand in $\overline{\mathbb{U}}$, then it must be a sum of three two-dimensional summands.
In this case the action of $\mathfrak{G}^{\prime\prime}$
on $\mathrm{Pic}(S)$ would be not faithful, which contradicts \cite[Corollary~8.2.40]{Dolgachev}.
Therefore, we see that $\overline{\mathbb{U}}$ is a sum of two three-dimensional irreducible representations of $\mathfrak{G}^{\prime\prime}$. Note that in the splitting~\eqref{eq:U-preliminary-splitting}
the summand~$\mathbb{I}^{\oplus 2}$ is generated by $K_S$ and a fiber of the conic bundle $\zeta$,
while $L_1$, $L_3$, $L_5$, $L_7$, $L_9$, and $L_{11}$ form a basis in~$\overline{\mathbb{U}}$.
Let $\sigma$ be the element of the group $\mathfrak{G}^{\prime\prime}$ that acts by
$$
[s:t:u:\mathrm{v}]\mapsto[t:s:u:-\mathrm{v}].
$$
Then
$$
\sigma(L_1)=L_3, \quad \sigma(L_3)=L_1, \quad \sigma(L_5)=L_{7}, \quad \sigma(L_7)=L_5, \quad \sigma(L_9)=L_{11},
\quad \sigma(L_{11})=L_9.
$$
Thus, the trace of $\sigma$ in $\overline{\mathbb{U}}$ equals $0$.
The element $\sigma$ corresponds to a transposition in $\mathfrak{G}^{\prime\prime}\cong\mathfrak{S}_4$.
This means that $\overline{\mathbb{U}}\cong\mathbb{W}_3\oplus\mathbb{W}_3^\prime$ as $\mathfrak{G}^{\prime\prime}$-representation.
Therefore, the isomorphism~\eqref{eq:W3W3prime} of $\mathfrak{G}^{\prime}$-representations holds.
\end{proof}
The commutative diagram~\eqref{equation:unprojection} gives a surjective $\mathfrak{G}^{\prime}$-module homomorphism of
$\mathrm{Cl}(V)$ onto~\mbox{$\mathrm{Cl}(W)$} with one-dimensional kernel generated by the surface $\Pi_-$.
On the other hand, restricting divisors in~$\mathrm{Cl}(W)$ to $S$, we obtain a $\mathfrak{G}^{\prime}$-module
homomorphism $$
\mathrm{Cl}(W)\to\mathrm{Pic}(S),
$$
which is injective (cf. \cite[Corollary~3.9.3]{Prokhorov}).
Note that $\mathrm{Cl}(W)$ is of rank $7$, while $\mathrm{Pic}(S)$ is of rank~$8$.
As $\mathfrak{G}^{\prime}$-representations,
we have
\begin{equation*}
\mathrm{Cl}(V)\otimes\mathbb{C}\cong\mathbb{I}^{\oplus 2}\oplus\mathbb{V},
\end{equation*}
where $\mathbb{I}$ is a trivial $\mathfrak{G}^{\prime}$-representation, and $\mathbb{V}$ is some six-dimensional $\mathfrak{G}^{\prime}$-representation.
Note that the summand~\mbox{$\mathbb{I}^{\oplus 2}$} is generated by $K_V$ and $\Pi_-$.
We see that $\mathbb{V}$ is a subrepresentation of~\mbox{$\mathrm{Pic}(S)\otimes\mathbb{C}$},
so that~$\mathbb{V}$ is a sum of two irreducible
three-dimensional $\mathfrak{G}^{\prime}$-representations~$\mathbb{W}_3$ and~$\mathbb{W}_3^\prime$ by Lemma~\ref{lemma:S4-reps}.
As a by-product, we obtain
\begin{corollary}
\label{corollary:class-group-V-G-prime}
There is an isomorphism of $\mathfrak{G}^{\prime}$-representations
\begin{equation*}
\Cl(V)\otimes\mathbb{C}\cong
\mathbb{I}^{\oplus 2}\oplus\mathbb{W}_3\oplus\mathbb{W}_3^\prime.
\end{equation*}
In particular,
$\mathrm{rk}\,\mathrm{Cl}(V)^{\mathfrak{G}^\prime}=2$
and $\mathrm{rk}\,\mathrm{Cl}(W)^{\mathfrak{G}^\prime}=1$.
\end{corollary}
Recall that the subgroups $\mathfrak{G}$ and $\mathfrak{G}^{\prime}$ are both isomorphic to $\mathfrak{S}_4$.
They intersect by the even elements, and every odd element of $\mathfrak{G}$ differs
from the corresponding element of $\mathfrak{G}^{\prime}$ by the Galois involution $\tau$. This involution acts as a multiplication by $-1$ on the orthogonal complement, with respect to
the intersection form \eqref{equation:E7}, to $K_V$ in $\mathrm{Cl}(V)\otimes\mathbb{C}$.
Thus, Corollary~\ref{corollary:class-group-V-G-prime}
implies that~\mbox{$\mathrm{Cl}(V)\otimes\mathbb{C}$} contains
a unique trivial subrepresentation of the group $\mathfrak{G}$.
\begin{corollary}
\label{corollary:class-group-V-G}
One has $\mathrm{rk}\,\mathrm{Cl}(V)^{\mathfrak{G}}=1$.
\end{corollary}
\section{Birational rigidity}
\label{section:rigidity}
In this section we study $G$-birationally
rigidity of nodal double Veronese cones.
Let $V$ be a nodal del Pezzo 3-fold of degree $1$
(not necessarily with 28 nodes).
Then $V$ can be given in $\mathbb{P}(1,1,1,2,3)$ with weighted homogeneous coordinates $s$, $t$, $u$, $v$, and $w$
(of weights $1$, $1$, $1$, $2$, and~$3$, respectively)
by equation
$$
w^2=v^3+vh_4(s,t,u)+h_6(s,t,u)
$$
for some homogeneous polynomials $h_4(s,t,u)$ and $h_6(s,t,u)$ of degree $4$ and $6$, respectively.
Then
$$
-K_V\sim 2H,
$$
where $H$ is an ample Cartier divisor on $V$ such that $H^3=1$.
The linear system $|H|$ has one base point, which we denote by $O$.
Recall that the 3-fold $V$ is smooth at the point~$O$.
The linear system $|2H|$ is free from base points and gives a double cover $V\to\mathbb{P}(1,1,1,2)$.
As before, denote by $\tau$ the Galois involution of this double cover.
Let $\kappa\colon V\dasharrow\mathbb{P}^2$ be the projection given by
$$
[s:t:u:v:w]\mapsto[s:t:u].
$$
Then $\kappa$ is a rational map given by the linear system $|H|$.
In particular, $\kappa$ is $\mathrm{Aut}(V)$-equivariant. Furthermore,
there exists an $\mathrm{Aut}(V)$-equivariant commutative diagram
$$
\xymatrix{
&\widetilde{V}\ar@{->}[ld]_{\psi}\ar@{->}[rd]^{\tilde{\kappa}}&\\%
V\ar@{-->}[rr]_{\kappa}&&\mathbb{P}^2}
$$
where $\psi$ is the blow up at the point $O$,
and $\tilde{\kappa}$ is a morphism whose general fiber is an elliptic curve.
Note that every fiber of the morphism $\tilde{\kappa}$ is irreducible.
Moreover, we have an exact sequence of groups
$$
\xymatrix{1\ar@{->}[r]&\Gamma\ar@{->}[r]&\mathrm{Aut}(V)\ar@{->}[r]&\mathrm{Aut}(\mathbb{P}^2),}
$$
where $\Gamma$ is a finite subgroup in $\mathrm{Aut}(V)$ that contains $\tau$.
If $h_4(x,y,z)$ is not a zero polynomial, then~$\Gamma$ is generated by $\tau$.
If $h_4(x,y,z)$ is a zero polynomial, then $\Gamma$ is a cyclic group of order $6$
that is generated by $\tau$ and a map given by
$$
[s:t:u:v:w]\mapsto[s:t:u:\epsilon_3 v:w],
$$
where $\epsilon_3$ is a primitive cube root
of unity (cf. \cite[\S6.7]{DolgachevIskovskikh}).
Let $G$ be a finite subgroup of $\mathrm{Aut}(V)$.
Denote its image in $\mathrm{Aut}(\mathbb{P}^2)$ by $\overline{G}$.
Suppose that
$$
\mathrm{rk}\,\mathrm{Cl}\big(V\big)^G=1,
$$
so that $V$ is a $G$-Mori fibre space (see \cite[Definition~1.1.5]{CheltsovShramov-V5}).
If $\mathbb{P}^2$ contains a $\overline{G}$-fixed point,
then there exists a $G$-equivariant commutative diagram as follows:
$$
\xymatrix{
&\check{V}\ar@{->}[ld]_{\alpha}\ar@{->}[rrrdd]^{\upsilon}&\\%
V\ar@{-->}[d]_{\kappa}&&&&\\
\mathbb{P}^2\ar@{-->}[rrrr]&&&&\mathbb{P}^1}
$$
Here $\mathbb{P}^2\dasharrow\mathbb{P}^1$ is the projection from the $\overline{G}$-fixed point,
$\alpha$ is the maximal extraction whose center is the fiber of $\kappa$ over the $\overline{G}$-fixed point,
and $\upsilon$ is a fibration into del Pezzo surfaces of degree $1$.
Thus, in this case, the 3-fold $V$ cannot be $G$-birationally rigid.
If $V$ is smooth and $\mathbb{P}^2$ contains no $\overline{G}$-fixed points,
then $V$ is $G$-birationally super-rigid.
This follows from \cite{Grinenko2003} and \cite{Grinenko2004} (cf. \cite[Remark~1.19]{CheltsovShramov2009}).
It is conjectured that the following two conditions are equivalent
for a nodal double Veronese cone $V$ with an action of
a finite group $G$ such that~\mbox{$\mathrm{rk}\,\mathrm{Cl}\big(V\big)^G=1$}:
\begin{itemize}
\item $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points,
\item the 3-fold $V$ is $G$-birationally rigid.
\end{itemize}
At present, its proof is out of reach.
Instead, we give a very simple proof of a weaker result,
which implies Theorem~\ref{theorem:rigid}.
To state the result, note that a surface $\Pi$ in the 3-fold $V$ is said to be a \emph{plane} if~$\Pi\cong\mathbb{P}^2$ and $H^2\cdot\Pi=1$.
For instance, by \cite[Theorem~7.2]{Prokhorov}
every $28$-nodal double Veronese cone contains exactly $126$ planes.
\begin{theorem}
\label{theorem:V1-G-rigid}
Let $V$ be a nodal
double Veronese cone with an action of
a finite group $G$ such that $\mathrm{rk}\,\mathrm{Cl}\big(V\big)^G=1$.
Suppose that the following three conditions are satisfied:
\begin{enumerate}
\item $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points;
\item the singular locus $\mathrm{Sing}(V)$ does not contain $G$-orbits of length $3$;
\item for every $G$-irreducible curve $D$ in $V$ such that $H\cdot D$ is either $2$ or $3$, and $\kappa(D)$ is a curve in $\mathbb{P}^2$ of degree $H\cdot D$,
there is a plane in $V$ that contains~$D$.
\end{enumerate}
Then $V$ is $G$-birationally super-rigid.
\end{theorem}
\begin{corollary}
\label{corollary:V1-G-rigid}
Let $V$ and $G$ be as in Theorem~\ref{theorem:V1-G-rigid}. Suppose that $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points.
If $\mathbb{P}^2$ does not contain $\overline{G}$-invariant curves of degree $2$ or $3$,
then $V$ is $G$-birationally super-rigid.
\end{corollary}
\begin{remark}
\label{remark:Z-3-singular-points}
We cannot drop condition (2) in Theorem~\ref{theorem:V1-G-rigid}.
Suppose that $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points,
but $\mathrm{Sing}(V)$ contains a $G$-orbit of length $3$.
Let $\beta\colon \breve{V}\to V$ be the blow up at this $G$-orbit.
Then it follows from
\cite{JahnkeRadloff} and
\cite[Theorem~1.5]{CheltsovShramovPrzyjalkowski}
that there exists a crepant $G$-birational morphism~$\nu\colon\breve{V}\to X$
such that $X$ is a double cover of $\mathbb{P}^3$ branched
in an irreducible sextic surface.
Let $\iota$ be the involution in $\mathrm{Aut}(X)$ of this double cover,
and put
$$
\rho=\beta\circ\nu^{-1}\circ\iota\circ\nu\circ\beta^{-1}.
$$
Then $\rho\in\mathrm{Bir}^{G}(V)$.
Moreover, if $\nu$ is small, then $\rho$ is not biregular.
\end{remark}
Before proving Theorem~\ref{theorem:V1-G-rigid}, let us use it to prove Theorem~\ref{theorem:rigid}.
\begin{proof}[Proof of Theorem~\ref{theorem:rigid}]
Suppose that the $28$-nodal double Veronese cone $V$ is $G$-birationally rigid.
Then the group $\overline{G}$ does not have
fixed points on $\mathbb{P}^2$, and thus also has no
fixed points on its projectively dual plane. Let $C$ be the smooth plane quartic curve
that is constructed as
the projectively dual curve of the discriminant curve $\check{C}$ of the half-anticanonical rational elliptic fibration
$\kappa\colon V\dasharrow \mathbb{P}^2$ (see Theorem~\ref{theorem:one-to-one} or Lemma~\ref{lemma:V-to-quartic}).
Then the group $\overline{G}$ acts faithfully on~$C$.
We see from Lemma~\ref{lemma:must-contain-S4}
that $C$ can be given
by equation~\eqref{eq:S4-quartic-general}, the group~$\overline{G}$ must contain a subgroup isomorphic to~$\mathfrak{S}_4$,
and~$\mathbb{P}^2$ (as well as its projective dual) can be identified
with a projectivization of an irreducible three-dimensional
representation of $\mathfrak{S}_4$.
Using once again Theorem~\ref{theorem:one-to-one}
and formula~\eqref{eq:V1}, we conclude that $V$ is a double Veronese cone from
Example~\ref{example:V1-S4}.
Furthermore, by Theorem~\ref{theorem:Aut} the group $G$ contains a subgroup isomorphic to~$\mathfrak{S}_4$.
Now it follows from Corollary~\ref{corollary:class-group-V-G-prime} that $G$ contains a subgroup
conjugate to~$\mathfrak{G}$.
Now we suppose that $V$ is a double Veronese cone
from Example~\ref{example:V1-S4}, and $G=\mathfrak{G}$.
Then $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points,
and $\mathrm{rk}\,\mathrm{Cl}(V)^G=1$ by Corollary~\ref{corollary:class-group-V-G}.
To complete the proof, we have to show that $V$ is $G$-birationally super-rigid.
Suppose that this is not a case.
Let us seek for a contradiction.
Note that $\overline{G}\cong\mathfrak{S}_4$, and $\mathbb{P}^2$ contains a unique $\overline{G}$-orbit of length $3$.
This orbit consists of the points~$[0:0:1]$, $[0:1:0]$ and $[1:0:0]$.
Taking partial derivatives of~\eqref{eq:V1} and using the expressions for $g_4$ and $g_6$
provided in Example~\ref{example:V1-S4},
we see that~\mbox{$\mathrm{Sing}(V)$}
does not contain points that are mapped by $\kappa$ to $[0:0:1]$, $[0:1:0]$ or $[1:0:0]$.
In particular, we see that the singular locus $\mathrm{Sing}(V)$ does not contain $\overline{G}$-orbits of length $3$.
Using Theorem~\ref{theorem:V1-G-rigid}, we conclude that $V$ contains a $G$-irreducible curve $D$ on $V$ such
that~\mbox{$H\cdot D$} is either $2$ or $3$,
the curve $D$ is not contained in any plane in $V$,
and $\kappa(D)$ is a curve in $\mathbb{P}^2$ of degree~$H\cdot D$.
Observe that there exist a
unique $\overline{G}$-invariant conic
and a unique $\overline{G}$-invariant
cubic curve in $\mathbb{P}^2$.
The former curve is given by $s^2+t^2+u^2=0$,
and the latter curve is given by $stu=0$.
Suppose that $H\cdot D=3$.
Then $\kappa(D)$ is given by $stu=0$, so that $D$ consists of three irreducible components that are mapped isomorphically by $\kappa$
to the lines $s=0$, $t=0$ and $u=0$ in $\mathbb{P}^2$.
Let $D_s$ be the irreducible component of the curve $D$ that
is mapped by $\kappa$ to the line~\mbox{$s=0$}. Then the stabilizer $G_s$ of $D_s$ in $G\cong \mathfrak{S}_4$
is isomorphic to the dihedral group of order $8$. However,
its action on $D_s$ is not faithful, but the action of its
quotient isomorphic to $\mumu_2\times\mumu_2$ is faithful. Therefore,
a general $G_s$-orbit in $D_s$ has length $4$. Consider the pencil
$\mathcal{R}$ of $G$-invariant
surfaces in~$V$ generated by the surfaces
$v=0$ and $s^2+t^2+u^2=0$. Since a surface from $\mathcal{R}$
has intersection number $2$ with $D_s$,
we conclude that there is a surface $R$ in $\mathcal{R}$
that contains $D_s$. Since~$R$ is $G$-invariant, it contains
the whole curve $D$.
The surface $R$ is given by equation
$$
v=\mu(s^2+t^2+u^2)
$$
for some $\mu\in\mathbb{C}$. Thus, the curve $D$ is contained in the subset in $V$ that is cut out by
$$
\left\{\aligned
&v=\mu(s^2+t^2+u^2),\\
&stu=0.\\
\endaligned
\right.
$$
If $\mu=\frac{2\lambda}{3}$, then
the equation $v=\mu(s^2+t^2+u^2)$ cuts out two planes in $V$, which are given by
$$
\left\{\aligned
&v=\mu(s^2+t^2+u^2),\\
&w=\gamma stu,\\
\endaligned
\right.
$$
where $\gamma=\pm 2(\lambda-2)\sqrt{\lambda+1}$ (see~\eqref{eq:Pi-pm-eqs}).
Thus, in this case, the curve $D$ is contained in a plane, which is impossible by assumption.
Hence, we conclude that $\mu\ne\frac{2\lambda}{3}$.
We see that the irreducible component $D_s$
is contained in the subset in $V$ that is cut out by
$$
\left\{\aligned
&v=\mu(t^2+u^2),\\
&s=0.\\
\endaligned
\right.
$$
If this subset were irreducible and reduced, then it would coincide with $D_s$, which would contradict~${H\cdot D_s=1}$.
Hence, the above subset is either reducible or non-reduced.
Algebraically, this simply means that the polynomial
$$
P(t,u)=\mu^3(t^2+u^2)^3-\mu g_4(0,t,u)(t^2+u^2)+g_6(0,t,u)
$$
must be a complete square (this includes the possibility
for the above polynomial to be zero, which corresponds to the
non-reduced case).
The polynomial $P(t,u)$ simplifies as
\[\left(\mu^3-\mu\frac{\lambda^2+12}{3}-\frac{2\lambda(\lambda-6)(\lambda+6)}{27}\right)(t^2+u^2)^3
-4(\lambda-2)\left(\mu-\frac{2}{3}\lambda\right)(t^2+u^2)t^2u^2.\]
This shows that the polynomial $P(t,u)$ cannot be a zero polynomial because $\lambda\ne 2$ and $\mu\ne \frac{2\lambda}{3}$.
Suppose that~$P(t,u)$ is a complete square. Then it must be of the form
\[
P(t,u)=(a_3t^3+a_2t^2u+a_1tu^2+a_0u^3)^2,
\]
where $a_i$'s are constants depending on $\lambda$ and $\mu$. We can see directly from the simplified~$P(t,u)$ above
that $P(t,u)$ is invariant under switching variables $t$ and $u$ and that it contains $t$ and $u$ with only even exponents. We therefore obtain
$$
a_0^2=a_3^2, \quad a_1^2+2a_0a_2=a_2^2+2a_1a_3, \quad a_0a_1=a_2a_3=0, \quad a_0a_3+a_1a_2=0.
$$
However, these yield $a_0=a_1=a_2=a_3=0$, which is impossible. Consequently, one has~\mbox{$H\cdot D\ne 3$}.
We see that $H\cdot D=2$.
Then $\kappa(D)$ is given by equation $s^2+t^2+u^2=0$ in $\mathbb{P}^2$.
In particular, one has~\mbox{$D\cong \mathbb{P}^1$}.
The action of $G\cong\mathfrak{S}_4$ on $D$ is faithful,
so that any $G$-orbit in $D$ has length at least~$6$. Consider the pencil
$\mathcal{R}$ as above.
Since a surface from $\mathcal{R}$
has intersection number $4$ with~$D_s$,
we conclude that every surface in $\mathcal{R}$
contains $D$.
In particular, $D$ is contained in a surface given by equation
$$
v=\frac{2\lambda}{3}(s^2+t^2+u^2).
$$
As above, we conclude that $D$ is contained in a plane,
which gives a contradiction.
This completes the proof of Theorem~\ref{theorem:rigid}.
\end{proof}
Now we are equipped to prove Theorem~\ref{theorem:V1-G-rigid}.
Recall that $V$ is a nodal
double Veronese cone with an action of
a finite group $G$ such that $\mathrm{rk}\,\mathrm{Cl}\big(V\big)^G=1$,
and assumptions (1)--(3) of Theorem~\ref{theorem:V1-G-rigid} hold.
Namely, we
suppose that $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points;
that the singular locus $\mathrm{Sing}(V)$
does not contain $G$-orbits of length $3$;
that for any $G$-irreducible curve $D$ in $V$ such that~\mbox{$H\cdot D$} is either $2$ or $3$,
and $\kappa(D)$ is a curve in $\mathbb{P}^2$ of degree $H\cdot D$,
there is a plane in $V$ containing the curve $D$.
\begin{remark}\label{remark:overline-G-orbits}
Since there are no $\overline{G}$-fixed points on
$\mathbb{P}^2$, there are also no $\overline{G}$-invariant lines.
This implies that there are no $\overline{G}$-orbits
contained in a line. In other words, every
$\overline{G}$-orbit contains three non-collinear points,
and in particular no $\overline{G}$-orbits
of length $2$.
Furthermore, by Lemma~\ref{lemma:4-general-points}
every
$\overline{G}$-orbit contains four points such that
no three of them are collinear.
\end{remark}
To~prove Theorem~\ref{theorem:V1-G-rigid}, we have to show that
the 3-fold~$V$ is $G$-birationally super-rigid.
Suppose that it is not.
Then there exists a $G$-invariant mobile linear system $\mathcal{M}$ on the 3-fold~$V$ such that
$$
\mathcal{M}\sim nH,
$$
and the log pair $(V,\frac{2}{n}\mathcal{M})$ is not canonical
(see for instance \cite[Theorem~3.3.1]{CheltsovShramov-V5}).
Let $Z$ be a $G$-center of non-canonical
singularities of the log pair $(V,\frac{2}{n}\mathcal{M})$.
Recall that $O$ denotes the (unique) base point of the linear
system $|H|$ on $V$, and $O$ is a smooth point of $V$.
\begin{lemma}
\label{lemma:Z-not-O}
The $G$-center $Z$ cannot be the point $O$.
\end{lemma}
\begin{proof}
Let $M$ be a general surface in $\mathcal{M}$,
and let $E$ be a general fiber of the map $\kappa$.
Then~\mbox{$E\not\subset M$}, so that
$$
n=M\cdot E\geqslant\mathrm{mult}_{O}\big(E\big)\cdot\mathrm{mult}_{O}\big(M\big)\geqslant\mathrm{mult}_{O}\big(M\big).
$$
On the other hand, the Zariski tangent space $T_{O,V}$ is an irreducible representation of the group~$G$,
because $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points.
Thus, using Lemma~\ref{lemma:mult-2}, we conclude that $O$
is not a center of non-canonical singularities
of the log pair $(V,\frac{2}{n}\mathcal{M})$.
\end{proof}
\begin{lemma}
\label{lemma:Z-not-smooth-point}
If $Z$ is $0$-dimensional, then $Z\subset\mathrm{Sing}(V)$.
\end{lemma}
\begin{proof}
Suppose that the $G$-orbit $Z$ consists of smooth points of the 3-fold $V$.
Let $M_1$ and~$M_2$ be general surfaces in the linear system $\mathcal{M}$.
Then
\begin{equation}
\label{equation:Z-not-smooth-point}
\mathrm{mult}_{P}\big(M_1\cdot M_2\big)>n^2
\end{equation}
for every $P\in Z$ by \cite[Corollary~3.4]{Corti2000}.
Let us show that this leads to a contradiction.
By Lemma~\ref{lemma:Z-not-O}, we have $Z\ne O$, so that $\kappa(Z)$ is a
well-defined
$\overline{G}$-orbit.
Moreover, $\kappa(Z)$ consists of at least $3$ points,
and contains three non-collinear points
by Remark~\ref{remark:overline-G-orbits}.
Denote them by $P_1$, $P_2$ and~$P_3$.
Let $E_1$, $E_2$ and $E_3$ be fibers of the map $\kappa$ over the points $P_1$, $P_2$ and $P_3$, respectively.
Then
$$
M_1\cdot M_2=m\big(E_1+E_2+E_3\big)+\Delta,
$$
where $m$ is a non-negative integer, and $\Delta$ is an effective one-cycle whose support does not contain the curves $E_1$, $E_2$ and $E_3$.
Then $m\leqslant\frac{n^2}{3}$, since
$$
n^2=H\cdot M_1\cdot M_2=H\cdot\Big(m\big(E_1+E_2+E_3\big)+\Delta\Big)=3m+H\cdot\Delta\geqslant 3m.
$$
Let $O_1$, $O_2$ and $O_3$ be points in $Z$ that are mapped to $P_1$, $P_2$ and $P_3$, respectively.
Let~\mbox{$\delta=\mathrm{mult}_{O_i}\big(E_i\big)$}. Then either $\delta=1$ or $\delta=2$.
Moreover, it follows from \eqref{equation:Z-not-smooth-point} that
$$
\mathrm{mult}_{O_i}\big(\Delta\big)>n^2-m\delta
$$
for each point $O_i$.
Let $\mathcal{B}$ be a linear subsystem in $|2H|$ consisting of all
surfaces that contain the curves $E_1,\,E_2$ and~$E_3$.
Then $\mathcal{B}$ does not have other base curves,
since $P_1$, $P_2$ and $P_3$ are not collinear.
Thus, for a general surface $B$ in $\mathcal{B}$, we have
$$
2n^2-6m=B\cdot\Delta\geqslant\sum_{i=1}^{3}\mathrm{mult}_{O_i}\big(\Delta\big)>3\Big(n^2-m\delta\Big).
$$
This implies
$$
-4m\geqslant m\delta-6m>n^2,
$$
which is a contradiction.
\end{proof}
\begin{lemma}
\label{lemma:Z-not-singular-point}
The center $Z$ is $1$-dimensional.
\end{lemma}
\begin{proof}
Suppose that $\dim Z=0$, so that $Z$ is a $G$-orbit. Then $Z\subset\mathrm{Sing}(V)$ by Lemma~\ref{lemma:Z-not-smooth-point}.
Let~$r$ be the length of the $\overline{G}$-orbit $\kappa(Z)$.
Note that $r\geqslant 3$ by Remark~\ref{remark:overline-G-orbits}.
Denote the points of $\kappa(Z)$ by $P_1,\ldots,P_r$.
Let $E_i$ be the fiber of the map $\kappa$ over the point $P_i$.
Then $E_i$ is an irreducible curve because~\mbox{$E_i\cdot H=1$}.
Since the arithmetic genus of $E_i$ equals $1$,
we see that $E_i$ has at most one
singular point, and thus
contains at most one
singular point of the 3-fold $V$.
Hence, since~\mbox{$E_i\cap Z\ne\varnothing$}, each curve~$E_i$ contains exactly one singular point of $V$,
which must be a singular point of the curve~$E_i$.
In particular, we see that
the length of the $G$-orbit $Z$ also equals~$r$.
Let $O_i$ be the singular point of the curve $E_i$.
Then $Z$ consists of the points $O_1,\ldots,O_r$.
By assumption, we have $r\geqslant 4$.
The $\overline{G}$-orbit $\kappa(Z)$ contains four points in $\mathbb{P}^2$ such that no three of
them are collinear by Remark~\ref{remark:overline-G-orbits}.
Without loss of generality, we may assume that these points are $P_1$, $P_2$, $P_3$, and~$P_4$.
Let $\mathcal{B}$ be a linear subsystem in $|2H|$ consisting of all
surfaces that contain the curves $E_1$, $E_2$, $E_3$ and $E_4$.
Then $\mathcal{B}$ does not contain other base curves.
Let~$\beta\colon \overline{V}\to V$ be the blow up of the points $O_1$, $O_2$, $O_3$ and $O_4$,
and let $F_1$, $F_2$, $F_3$, $F_4$ be the $\beta$-exceptional surfaces that are mapped to the points $O_1$, $O_2$, $O_3$, $O_4$, respectively.
Denote by~$\overline{\mathcal{B}}$ the proper transforms on $\overline{V}$ of the linear system $\mathcal{B}$.
Then
$$
-K_{\overline{V}}\sim\overline{\mathcal{B}}\sim\beta^*\big(2H\big)-\big(F_1+F_2+F_3+F_4\big),
$$
which implies that the divisor $-K_{\overline{V}}$ is nef, because it intersects the proper transform of each curve $E_i$ trivially.
Denote by $\overline{\mathcal{M}}$ the proper transform of the linear system $\mathcal{M}$ on the 3-fold $\overline{V}$.
Then
$$
\overline{\mathcal{M}}\sim_{\mathbb{Q}}\beta^*\big(nH\big)-m\big(F_1+F_2+F_3+F_4\big)
$$
for some non-negative rational number $m$.
Let $\overline{M}_1$ and $\overline{M}_2$ be general surfaces in $\overline{\mathcal{M}}$.
Then
$$
0\leqslant\overline{B}\cdot\overline{M}_1\cdot\overline{M}_2=2n^2-8m^2,
$$
which gives $m\leqslant\frac{n}{2}$. This is impossible by \cite[Theorem~3.10]{Corti2000}.
\end{proof}
Thus, Lemma~\ref{lemma:Z-not-singular-point} concludes that $Z$ is a $G$-irreducible curve.
Denote by $Z_1,\ldots,Z_r$ the irreducible components of $Z$.
Thus, if $r=1$, then $Z=Z_1$ is an irreducible curve.
In this case, $\kappa(Z)$ is neither a point nor a line, because $\mathbb{P}^2$ does not contain $\overline{G}$-fixed points.
In particular, we see that $H\cdot Z\ne 1$.
Since $Z$ is a $G$-center of non-canonical singularities of the log pair $(V,\frac{2}{n}\mathcal{M})$, we have
$$
\mathrm{mult}_{Z_i}\big(\mathcal{M}\big)>\frac{n}{2}.
$$
Let $M_1$ and $M_2$ be general surfaces in $\mathcal{M}$.
Then
$$
n^2=H\cdot M_1\cdot M_2\geqslant \sum_{i=1}^{r}\mathrm{mult}_{Z_i}\big(M_1\cdot M_2\big)>\frac{n^2}{4}rH\cdot Z_i,
$$
so that we have the following four possibilities:
\begin{itemize}
\item[($\mathrm{A}$)] $r=1$, $H\cdot Z=2$, and $\kappa(Z)$ is a smooth conic in $\mathbb{P}^2$;
\item[($\mathrm{B}$)] $r=1$, $H\cdot Z=3$, and $\kappa(Z)$ is a smooth cubic in $\mathbb{P}^2$;
\item[($\mathrm{C}$)] $r=3$, $H\cdot Z=3$, and $\kappa(Z)$ is a union of three lines;
\item[($\mathrm{D}$)] $r=3$, $H\cdot Z=3$,
and $\kappa(Z)$ is a $\overline{G}$-orbit of length $3$.
\end{itemize}
We claim that the case ($\mathrm{D}$) is impossible.
Indeed, if $\kappa(Z)$ is a $\overline{G}$-orbit of length $3$,
then the linear system~$|H|$ contains a surface $H_{12}$ passing through
$Z_1$ and $Z_2$. Let $M$ be a general element of
the linear system $\mathcal{M}$. Then
$$
n=H\cdot M\cdot H_{12}\geqslant \mathrm{mult}_{Z_1}\big(M\big)+
\mathrm{mult}_{Z_2}\big(M\big)>n,
$$
which is absurd.
Thus, we see that $H\cdot Z$ is either $2$ or $3$ and that $\kappa(Z)$ is a curve of degree $H\cdot Z$.
Therefore, by assumption, the 3-fold $V$ has a plane $\Pi$ that contains the curve $Z$.
Let $L$ be a general line in this plane, so that $H\cdot L=1$,
the intersection $L\cap Z$ consists of exactly $H\cdot Z$ points,
and $L$ is not contained in the base locus of the linear system $\mathcal{M}$.
Then, for a general surface $M\in\mathcal{M}$, we have
$$
n=L\cdot M\geqslant\mathrm{mult}_Z\big(M\big)\cdot\big| L\cap Z\big|\geqslant2\mathrm{mult}_Z\big(M\big)\geqslant2\mathrm{mult}_{Z}\big(\mathcal{M}\big)>n,
$$
which is a contradiction. This completes the proof of Theorem~\ref{theorem:V1-G-rigid}.
\section{Questions and problems}
\label{section:question}
In this section we discuss several open questions concerning
$28$-nodal double Veronese cones.
\medskip
\textbf{Del Pezzo surfaces of degree $2$.}
Let~$V$ be a $28$-nodal double Veronese cone,
let $C$ be the plane quartic curve corresponding to $V$
by Theorem~\ref{theorem:one-to-one},
and let~$S$ be the del Pezzo surface of degree $2$ constructed as the double
cover of $\mathbb{P}^2$ branched along $C$.
We know from Corollary~\ref{corollary:Aut-V-vs-S}
that
$$
\mathrm{Aut}(V)\cong\mumu_2\times\mathrm{Aut}(C)\cong\Aut(S),
$$
although there is no obvious choice for a natural isomorphism
between $\Aut(V)$ and $\Aut(S)$.
Also, one has
$$
\Cl(V)\cong\mathbb{Z}^8\cong\Pic(S).
$$
Recall from \cite[Corollary~7.1.4]{Prokhorov}
that the group $\Aut(V)$ acts faithfully on $\Cl(V)$,
while by~\mbox{\cite[Corollary~8.2.40]{Dolgachev}} the group
$\Aut(S)$ acts faithfully on~\mbox{$\Pic(S)$}.
It would be interesting to know
if there exists a natural identification of the corresponding
representations. More precisely, we ask the following:
\begin{question}
\label{question:V1-C4}
Does there exist an isomorphism $\mathrm{Aut}(V)\cong\mathrm{Aut}(S)$
under which $\mathrm{Cl}(V)\otimes\mathbb{C}$
is isomorphic to $\mathrm{Pic}(S)\otimes\mathbb{C}$ as representations
of this group?
\end{question}
Recall that both $\Cl(V)$ and $\Pic(S)$ have a naturally
defined intersection form.
The canonical class~$K_V$ is invariant with respect to the action of
the group $\Aut(V)$ on $\Cl(V)$, and the canonical class $K_S$ is
invariant with respect to the action of $\Aut(S)$ on $\Pic(S)$.
Moreover, in some cases these are the only trivial subrepresentations
in $\mathrm{Cl}(V)\otimes\mathbb{C}$ and
$\mathrm{Pic}(S)\otimes\mathbb{C}$; for instance, this is the case
when $\Aut(C)\cong\mathrm{PSL}_2(\mathbb{F}_7)$.
Therefore, if there exists a natural identification
of automorphism groups as required in Question~\ref{question:V1-C4},
then for any $28$-nodal Veronese double cone $V$ and the corresponding
del Pezzo surface $S$ there must exist an identification
of the seven-dimensional representations
of the group~\mbox{$\Aut(V)\cong\Aut(S)$} in the orthogonal
complement~\mbox{$\mathrm{Cl}(V)^{\perp K_V}\otimes\mathbb{C}$} to~$K_V$ in
$\mathrm{Cl}(V)\otimes\mathbb{C}$ and in
the orthogonal complement $\mathrm{Pic}(S)^{\perp K_S}\otimes\mathbb{C}$
to $K_S$ in~\mbox{$\mathrm{Pic}(S)\otimes\mathbb{C}$}.
Recall that both lattices $\mathrm{Cl}(V)^{\perp K_V}$ and
$\mathrm{Pic}(S)^{\perp K_S}$ are isomorphic to the lattice
$\mathbf{E}_7$ (see~\mbox{\cite[Theorem~1.7]{Prokhorov}}
and \cite[Remark~1.8]{Prokhorov} for the former, and \cite[\S8.2.6]{DolgachevIskovskikh}
for the latter).
We point out that despite this fact in general
one cannot hope for an equivariant
identification of the lattices~\mbox{$\mathrm{Cl}(V)^{\perp K_V}$} and
$\mathrm{Pic}(S)^{\perp K_S}$ that takes into account the intersection
forms, even if the answer to Question~\ref{question:V1-C4}
is positive.
Indeed, let $C$ be a smooth plane quartic curve with an action of the symmetric group
$\mathfrak{S}_4$ as in Example~\ref{example:V1-S4}.
Let us use the notation of~\S\ref{section:S4-Veronese}.
It follows from Corollary~\ref{corollary:class-group-V-G-prime}
and Lemma~\ref{lemma:S4-reps} that
each of the $\mathfrak{G}^\prime$
representations $\mathrm{Cl}(V)\otimes\mathbb{C}$
and $\mathrm{Pic}(S)\otimes\mathbb{C}$ contains a unique
two-dimensional subrepresentation.
This implies that the lattice $\Cl(V)$ contains a unique
$\mathfrak{G}^\prime$-invariant primitive sublattice $\Lambda_V$ of rank~$2$,
and the lattice $\Pic(S)$ contains a unique
$\mathfrak{G}^\prime$-invariant primitive sublattice~$\Lambda_S$ of rank~$2$.
However, the sublattice $\Lambda_V$ is generated by the half-anticanonical
divisor $H$ and the plane $\Pi_{-}$, so that the intersection
form on $\Lambda_V$ is given by the matrix
$$
\left(
\begin{array}{cc}
H^2 & H\cdot\Pi_{-}\\
H\cdot\Pi_{-} & \Pi_{-}^2
\end{array}
\right)=
\left(
\begin{array}{cc}
1 & 1\\
1 & -1
\end{array}
\right).
$$
On the other hand, it follows from the proof of
Lemma~\ref{lemma:S4-reps} that
the sublattice $\Lambda_S$ is generated by~$-K_S$
and a fiber $F$ of a conic bundle, so that the intersection
form on $\Lambda_S$ is given by the matrix
$$
\left(
\begin{array}{cc}
K_S^2 & -K_S\cdot F\\
-K_S\cdot F & F^2
\end{array}
\right)=
\left(
\begin{array}{cc}
2 & 2\\
2 & 0
\end{array}
\right).
$$
Thus, we see that the lattices $\Lambda_V$
and $\Lambda_S$ are not isomorphic.
One situation when we can easily establish the isomorphism required
in Question~\ref{question:V1-C4} is as follows.
Let $G$ be a subgroup in $\mathrm{Aut}(\mathbb{P}^3)$ such that there exists
a $G$-invariant Aronhold heptad~\mbox{$P_1,\ldots,P_7$}.
Then the diagram~\eqref{equation:non-commutative}
is $G$-equivariant, which implies that
$\mathrm{Cl}(V)\otimes\mathbb{C}$ is isomorphic
to~\mbox{$\mathrm{Pic}(S)\otimes\mathbb{C}$}
as $G$-representations.
\medskip
\textbf{Elliptic fibration.}
The following construction was pointed out to us by Alexander
Kuznetsov.
Let~$C$ be a smooth plane quartic, and let $S$ be the corresponding
del Pezzo surface of degree~$2$ with the anticanonical morphism
$\varphi\colon S\to\mathbb{P}^2$. Let $\check{\mathbb{P}}^2$ be the projectively dual plane of $\mathbb{P}^2$.
Let $p_1$ and $p_2$ be the projections
of the flag variety
$$
\mathcal{F}=\mathrm{Fl}(1,2,3)\subset\mathbb{P}^2\times\check{\mathbb{P}}^2
$$
to the first and the second factor of $\mathbb{P}^2\times\check{\mathbb{P}}^2$,
respectively.
Consider the fiber product
$$
\xymatrix{
& Y\ar@{->}[ld]\ar@{->}[rd]^{\gamma} & & \\
S\ar@{->}[rd]_{\varphi} & &
\mathcal{F}\ar@{->}[ld]^{p_1}\ar@{->}[rd]_{p_2} & \\
& \mathbb{P}^2 & & \check{\mathbb{P}}^2
}
$$
Note that there is a natural action of the group
$\Aut(S)\cong\mumu_2\times\Aut(C)$ on $Y$.
The morphism
$$
\kappa_Y=p_2\circ\gamma\colon Y\to\check{\mathbb{P}}^2
$$
is an elliptic fibration
whose discriminant curve is projectively dual to~$C$.
Now let $V$ be the $28$-nodal double Veronese cone constructed from the plane quartic~$C$ by Theorem~\ref{theorem:one-to-one}.
Then the half-anticanonical map $\kappa\colon V\dasharrow\check{\mathbb{P}}^2$ is
a rational elliptic fibration. Let~$\tilde{V}$ be the blow up of
$V$ at the unique base point of $\kappa$, and let $\tilde{\kappa}\colon \tilde{V}\to\check{\mathbb{P}}^2$
be the corresponding regular elliptic fibration. Then the discriminant curve of $\tilde{\kappa}$
is projectively dual to~$C$.
Moreover, it follows from Lemma~\ref{lemma:Q1Q2} that over every point in $\check{\mathbb{P}}^2$ the fibers
of the elliptic fibrations $\tilde{\kappa}$
and~$\kappa_Y$ are isomorphic to each other (cf. Lemma~\ref{lemma:V-from-elliptic-fibration}).
Furthermore, it follows from Lemma~\ref{lemma:Q1Q2} that (at least over the complement
to the discriminant curve) the elliptic fibration $\kappa_Y$ is the relative $\mathrm{Pic}^2$ of the elliptic
fibration~$\tilde{\kappa}$.
Indeed, let $E$ be a smooth fiber of $\tilde{\kappa}$, and let $P_1$ and $P_2$ be two points on $E$.
Recall that $E$ is a proper transform of the base locus~$E_{\mathcal{Q}}$ of some pencil $\mathcal{Q}$ of quadrics in $\mathbb{P}^3$.
The two points $P_1, P_2\in E_{\mathcal{Q}}$ define a line~$L$ passing through them.
Thus they define a unique quadric $Q$ in $\mathcal{Q}$ containing the line $L$, and also
the family of lines on $\mathcal{Q}$ which contains $L$. Note that the fiber of
the elliptic fibration~$\kappa_Y$ over the point $\tilde{\kappa}(E)$ can be identified with the double cover
of $\mathcal{Q}\cong\mathbb{P}^1$ branched in the locus of singular quadrics of $\mathcal{Q}$
or, in other words, with the Hilbert scheme of lines
contained in the quadrics of $\mathcal{Q}$. Therefore, the pair of points~$(P_1,P_2)$
defines a point on the fiber of $\kappa_Y$ over~$\tilde{\kappa}(E)$. This point
depends only on the class of the pair~$(P_1,P_2)$ in $\mathrm{Sym}^2(E)$, since every other pair from
the same class determines the same quadric~$Q$ and a line from the same family of lines on $Q$ where $L$ lies.
We conclude that the fiber of~$\kappa_Y$ over $\tilde{\kappa}(E)$ is identified with~$\mathrm{Pic}^2(E)$.
It remains to notice that this construction can be performed in the family, and thus the assertion
about the fibrations follows.
One of the consequences of the above construction is as follows.
Since the elliptic fibration~$\tilde{\kappa}$ is obtained by blowing up
$V$ at a single point, it comes with a distinguished section (which corresponds to the
eighth base point of the net of quadrics in the construction of $V$ from an Aronhold heptad).
Therefore, there exist a natural birational map between $V$ and~$Y$.
\begin{question}
Is there a more explicit geometric relation between the 3-fold $Y$ and the
$28$-nodal double Veronese cone $V$?
\end{question}
\medskip
\textbf{Intersections of quadrics and cubics.}
The following construction was also explained to us by Alexander
Kuznetsov. Let $P_1,\ldots,P_8$ be a regular Cayley octad, and let $\mathcal{L}$ be the corresponding net of quadrics in $\mathbb{P}^3$. Then the lines contained in the quadrics of $\mathcal{L}$ are parameterized by a three-dimensional variety
$Z$ (cf.~\cite[\S1.2]{Reid}). Then $Z$ can be regarded as a subvariety of the Grassmannian~$\mathrm{Gr}(2,4)$.
Let~$\mathcal{U}^\vee$ be the universal quotient bundle on~$\mathrm{Gr}(2,4)$ (or, which is the same, the dual
of the tautological subbundle).
Furthermore, $Z$ can be described as the degeneracy locus
of the morphism of vector bundles
$$
\mathcal{O}_{\mathrm{Gr}(2,4)}\to \mathrm{S}^2\mathcal{U}^\vee
$$
corresponding to~$\mathcal{L}$. Therefore, the 3-fold $Z$ is an intersection of a quadric
and a cubic in~$\mathbb{P}^5$. Note that $Z$ has $28$ singular points corresponding to the lines
passing through pairs of points $P_i$ and $P_j$. In particular, its intermediate Jacobian is trivial.
\begin{question}
Is there a natural geometric relation between the 3-fold $Z$ and the
$28$-nodal double Veronese cone $V$ constructed from the net of quadrics~$\mathcal{L}$?
Is the 3-fold $Z$ rational?
\end{question}
\begin{question}
Does the 3-fold $Z$ depend on the choice of the net of quadrics~$\mathcal{L}$, or only on the
corresponding plane quartic curve?
\end{question}
\medskip
\textbf{Other 3-folds constructed from plane quartics.}
There are other interesting classes of 3-folds related to
plane quartic curves. For instance, a construction due to S.\,Mukai assigns
to a general plane quartic
a smooth Fano 3-fold of genus $12$ (and anticanonical degree $22$),
see~\cite{Mukai92}, \cite{Mukai04}, and~\cite{RanestadSchreyer} for details.
\begin{question}
Is there a natural geometric relation between the
$28$-nodal double Veronese cone~$V$ and the Fano 3-fold $V_{22}$
of genus 12 constructed from the same smooth plane quartic~$C$?
\end{question}
We point out that both of the above varieties are rational, so
there always exists a birational map between them. However,
one cannot choose such a map to be $\Aut(C)$-equivariant. Indeed,
if~\mbox{$\Aut(C)\cong\mathrm{PSL}_2(\mathbb{F}_7)$},
then both $V$ and $V_{22}$
are $\mathrm{PSL}_2(\mathbb{F}_7)$-birationally super-rigid
by Theorem~\ref{theorem:rigid} and~\cite[Theorem~1.10]{CheltsovShramov-PSL}.
\medskip
\textbf{Degenerations.}
The following question was asked by Vyacheslav Shokurov and Yuri Prokhorov.
\begin{question}
Can one extend the one-to-one correspondence between smooth plane quartics and
$28$-nodal double Veronese cones given by Theorem~\ref{theorem:one-to-one}
to the (mildly) singular case
(like, double Veronese cones with $c\mathbf{A}_1$-singularities, or
double Veronese cones with $26$ nodes and one nice $c\mathbf{A}_2$-singularity,
or plane quartics with a single node)?
Furthermore, can one include in this one-to-one correspondence
the plane quartic that is a union of a cuspidal cubic and its tangent line in the cusp
(in other words, the quartic that gives rise to a del Pezzo surface of degree $2$ with an $\mathbf{E}_7$-singularity)?
\end{question}
\bigskip
\appendix
\section{Computer-aided calculations}
\label{section:appendix}
This appendix explains, from a computational point of view,
how~\eqref{equation:j-function2} is obtained.
In particular, a way to extract explicit formulae
of the covariants $g_4(s,t,u)$ and $g_6(s,t,u)$ from a given
plane quartic~\eqref{equation:quartic} is described,
which enables us to produce concrete examples of $28$-nodal double
Veronese cones from smooth quartic curves as in Example~\ref{example:V1-S4}.
We point out that there is a method to write down
the polynomial $g_4(s,t,u)$ using a smaller amount of
explicit computations. It is
based on the expression for the cubic invariant in the coefficients of the
plane quartic given in~\cite[\S293]{Salmon} (see
the end of~\cite[\S1]{Ottaviani}).
Since the denominator and the numerator of the $j$-function
in \eqref{equation:j-function} are symmetric polynomials
in~\mbox{$x_1, x_2, x_3, x_4$}, the $j$-function
in~\eqref{equation:j-function} may be regarded as a rational function
in $b_0, b_1, b_2, b_3, b_4$ of~\eqref{equation:quartic2}.
Using the simple MAGMA (\cite{Magma}) function
\begin{minipage}[m]{.95\linewidth}
\vspace{0.5cm}
\hrule width 15cm
\vspace{0.1cm}
\noindent Computing the numerator (\texttt{N}) and the denominator (\texttt{D}) in terms of symmetric functions
\vspace{0.1cm}
\hrule width 15cm
\begin{alltt}
Q<x1,x2,x3,x4> := PolynomialRing(RationalField(), 4);
Num:=(x1-x2)^2*(x4-x3)^2-(x1-x3)*(x4-x2)*(x4-x1)*(x3-x2);
Den:=(x1-x2)^2*(x1-x3)^2*(x1-x4)^2*(x2-x3)^2*(x2-x4)^2*(x3-x4)^2;
Q<b1, b2, b3, b4> := PolynomialRing(RationalField(), 4);
I, N := IsSymmetric(Num, Q);
I, D := IsSymmetric(Den, Q);
D;
N;
\end{alltt}
\hrule width 15cm
\end{minipage}
\vspace{0.5cm}
\noindent we are able to obtain the denominator and the numerator of the $j$-function (respectively \texttt{D} and \texttt{N}), which after homogenization with respect to $b_0$ result in
\[\aligned & (x_1-x_2)^2(x_4-x_3)^2-(x_1-x_3)(x_4-x_2)(x_4-x_1)(x_3-x_2)=\frac{1}{b_0^2}\Big(-3b_1b_3 +12 b_0b_4+b_2^2\Big);\\
& (x_1-x_2)^2(x_1-x_3)^2(x_1-x_4)^2(x_2-x_3)^2(x_2-x_4)^2(x_3-x_4)^2 =
\frac{1}{b_0^6}\Big(-27b_1^4b_4^2 + 18b_1^3b_2b_3b_4 - \\ &- 4b_1^3b_3^3 - 4b_1^2b_2^3b_4 + b_1^2b_2^2b_3^2
+ 144b_0b_1^2b_2b_4^2 - 6b_0b_1^2b_3^2b_4 - 80b_0b_1b_2^2b_3b_4 + 18b_0b_1b_2b_3^3 -\\&-
192b_0^2b_1b_3b_4^2 +16b_0b_2^4b_4 - 4b_0b_2^3b_3^2 - 128b_0^2b_2^2b_4^2 + 144b_0^2b_2b_3^2b_4 -
27b_0^2b_3^4 + 256b_0^3b_4^3\Big).\\ \endaligned
\]
Therefore, regarded as a rational function in $b_0, b_1, b_2, b_3, b_4$, the $j$-function in \eqref{equation:j-function} can be expressed as follows:
\[\begin{split}j(b_0, b_1, b_2, b_3, b_4)&=\frac{2^8(-3b_1b_3 +12 b_0b_4+b_2^2)^3}{\left\{
\aligned
&-27b_1^4b_4^2 + 18b_1^3b_2b_3b_4 - 4b_1^3b_3^3 - 4b_1^2b_2^3b_4 + b_1^2b_2^2b_3^2
+ 144b_0b_1^2b_2b_4^2 -\\
&- 6b_0b_1^2b_3^2b_4 - 80b_0b_1b_2^2b_3b_4 + 18b_0b_1b_2b_3^3 -
192b_0^2b_1b_3b_4^2 +\\&+ 16b_0b_2^4b_4 - 4b_0b_2^3b_3^2 - 128b_0^2b_2^2b_4^2 + 144b_0^2b_2b_3^2b_4 -
27b_0^2b_3^4 + 256b_0^3b_4^3\\
\endaligned\right\}}\\
&=1728\frac{4h_2(b_0, b_1, b_2, b_3, b_4)^3}{4h_2(b_0, b_1, b_2, b_3, b_4)^3-27h_{3}(b_0, b_1, b_2, b_3, b_4)^2},\\
\end{split}\]
where
\[h_2(b_0, b_1, b_2, b_3, b_4)=\frac{1}{3}\Big(-3b_1b_3 +12 b_0b_4+b_2^2\Big);\]
\[ h_{3}(b_0, b_1, b_2, b_3, b_4)=\frac{1}{27}\Big(72b_0b_2b_4 - 27b_0b_3^2 - 27b_1^2b_4 + 9b_1b_2b_3 - 2b_2^3\Big).\]
Since $b_0,\ldots, b_4$ are homogeneous polynomials of
degree $4$ in $s$, $t$, and~$u$,
we can obtain expressions for~\mbox{$h_2(b_0, b_1, b_2, b_3, b_4)$} and
$h_{3}(b_0, b_1, b_2, b_3, b_4)$ as homogeneous polynomials of degrees $8$ and $12$, respectively, in $s, t, u$.
Starting with a quartic equation
\[
\sum_{i+j+k=4} a_{ijk}x^iy^jz^k=0,
\]
where $a_{ijk}$ are independent variables,
we can calculate $b_0,\ldots, b_4$, using the identities
given after \eqref{equation:quartic2}, as in the code below.
\begin{minipage}[m]{.95\linewidth}
\vspace{0.5cm}
\hrule width 11.7cm
\vspace{0.1cm}
\noindent Calculating $h_2$ and $h_3$ in terms of $s,t,u$ with general coefficients $a_{ijk}$
\vspace{0.1cm}
\hrule width 11.7cm
\begin{alltt}
P<a400,a310,a301,a220,a202,a211,a130,a103,a121,a112,a040,a031,a022,a013,a004,
s,t,u>:=PolynomialRing(Rationals(),18);
b0:=a004*s^4 - a103*s^3*u + a202*s^2*u^2 - a301*s*u^3 + a400*u^4;
b1:=4*a004*s^3*t - a013*s^3*u - 3*a103*s^2*t*u + a112*s^2*u^2 + 2*a202*s*t*u^2
-a211*s*u^3 - a301*t*u^3 + a310*u^4;
b2:=6*a004*s^2*t^2 - 3*a013*s^2*t*u + a022*s^2*u^2 - 3*a103*s*t^2*u
+ 2*a112*s*t*u^2 - a121*s*u^3 + a202*t^2*u^2 - a211*t*u^3 + a220*u^4;
b3:=4*a004*s*t^3 - 3*a013*s*t^2*u + 2*a022*s*t*u^2 - a031*s*u^3 - a103*t^3*u
+ a112*t^2*u^2 - a121*t*u^3 + a130*u^4;
b4:=a004*t^4 - a013*t^3*u + a022*t^2*u^2 - a031*t*u^3 + a040*u^4;
h2:=-3*b1*b3 + b2^2 + 12*b4*b0;
h3:=8/3*b0*b2*b4-b0*b3^2-b1^2*b4+1/3*b1*b2*b3-2/27*b2^3;
\end{alltt}
\vspace{0.1cm}
\hrule width 11.7cm
\end{minipage}
\vspace{0.5cm}
\noindent It can be checked, for example using the command
\texttt{Factorisation(h2)},
that $h_2$ and $h_3$ are divisible by $u^4$ and~$u^6$, respectively. Therefore, we are able to put
\[h_2 (b_0, b_1, b_2, b_3, b_4)=u^4g_4(s,t,u)\quad\text{and}\quad
h_3 (b_0, b_1, b_2, b_3, b_4)=u^6g_6(s,t,u),\]
where $g_4(s,t,u)$ and $g_6(s,t,u)$ are
homogeneous polynomials of degrees $4$ and $6$, respectively,
in~\mbox{$s, t, u$}.
Consequently, the rational function $j(b_0, b_1, b_2, b_3, b_4)$ can be presented as a rational function $j_C$
in~$\check{\mathbb{P}}^2$ with variables \mbox{$s, t, u$} via
\[j_C(s,t,u)= 1728\frac{4g_4(s,t,u)^3}{4g_4(s,t,u)^3-27g_{6}(s,t,u)^2}\cdot\] | 99,813 |
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TITLE: Finding a closed form expression for this sum
QUESTION [3 upvotes]: For non-negative $n$, find
$$
\sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k}.
$$
I can't figure this out. Any ideas?
REPLY [3 votes]: The sum is a convolution of the sequence $a_n = \binom{2n}{n}$ with itself. This suggests generating functions. Multiplying generating functions for two sequences results in a generating function for the convolution of the two sequences:
If $f(x) = \sum_{n \geq 0} a_n x^n$ and $g(x) = \sum_{n \geq 0} b_n x^n$, then $$f(x)g(x) = \sum_{n \geq 0} \left(\sum_{k=0}^n a_k b_{n-k}\right) x^n$$
The generating function $f$ for the sequence $\binom{2n}{n}$ is given by $$f(x) = \frac{1}{\sqrt{1 - 4x}}$$
Since $[f(x)]^2 = \frac{1}{1-4x}$, which is the generating function for $4^n$, the given sum is $4^n$.
There are approaches that don't involve generating functions, but the generating function calculation is very quick.
REPLY [3 votes]: The sum has the form $d_n = \sum_{k=0}^n c_k c_{n-k}$, where $c_k = \binom{2k}{k}$. The sum is known as the Cauchy product. The generating function for $d_n$ is the second power of the generating function of $c_n$.
$$
\sum_{n=0}^\infty d_n x^n = \left( \sum_{k=0}^\infty c_k x^k \right)^2 = \left( \frac{1}{\sqrt{1-4x}} \right)^2 = \frac{1}{1-4x} = \sum_{k=0}^\infty 2^{2k} x^k
$$
Thus
$$
\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} = 2^{2n}
$$ | 99,451 |
TITLE: How to construct a matrix given the null basis of A?
QUESTION [2 upvotes]: Construct a $4\times4$ matrix $A$ such that $\{(1,2,3,4),(1,1,2,2)\}$ is a basis of $N(A)$.
So I know that $A$ will have two pivot columns and two free columns, but beyond this I'm not sure how to approach/solve.
REPLY [3 votes]: You could for instance say that you have a matrix
\begin{equation}
A
=
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{bmatrix}
\end{equation}
such that
\begin{equation}
AX = 0
\end{equation}
where
\begin{equation}
X =
\begin{bmatrix}
1 & 1 \\
2 & 1 \\
3 & 2\\
4 & 2
\end{bmatrix}
\end{equation}
Hence we have to find $a_{ij}$'s such that
\begin{equation}
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44} \\
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
2 & 1 \\
3 & 2\\
4 & 2
\end{bmatrix}
=
0
\end{equation}
i.e. we have the following system to solve:
\begin{align}
a_{11} + 2a_{12} + 3a_{13} + 4a_{14} & = 0 \\
a_{21} + 2a_{22} + 3a_{23} + 4a_{24} & = 0 \\
a_{31} + 2a_{32} + 3a_{33} + 4a_{34} & = 0 \\
a_{41} + 2a_{42} + 3a_{43} + 4a_{44} & = 0 \\
a_{11} + a_{12} + 2a_{13} + 2a_{14} & = 0 \\
a_{21} + a_{22} + 2a_{23} + 2a_{24} & = 0 \\
a_{31} + a_{32} + 2a_{33} + 2a_{34} & = 0 \\
a_{41} + a_{42} + 2a_{43} + 2a_{44} & = 0
\end{align}
This is exhaustive as it is a system of 8 equations in 16 unknowns. Therefore, we will have infinitely many solutions. \\
For example, you could construct a projector matrix $P_X$ that spans the columns of $X$, i.e.
\begin{equation}
P_X = X(X^TX)^{-1}X^T
=
\begin{bmatrix}
0.5455 & -0.0909 & 0.4545 & -0.1818\\
-0.0909 & 0.1818 & 0.0909 & 0.3636\\
0.4545 & 0.0909 & 0.5455 & 0.1818\\
-0.1818 & 0.3636 & 0.1818 & 0.7273
\end{bmatrix}
\end{equation}
And then you can say that my matrix $A$ spans the null space of $P_X$, i.e.
\begin{equation}
A = I - P_X
=
\begin{bmatrix}
0.4545 & 0.0909 & -0.4545 & 0.1818\\
0.0909 & 0.8182 & -0.0909 & -0.3636\\
-0.4545 & -0.0909 & 0.4545& -0.1818\\
0.1818 & -0.3636 & -0.1818 & 0.2727
\end{bmatrix}
\end{equation}
Now check,
\begin{equation}
AX =(I - P_X)X = X - P_XX = X - X(X^TX)^{-1}X^TX = X-X = 0
\end{equation}
and voila there you have a matrix with nullspace being the columns of $X$. | 152,936 |
Writing this entry actually hurts me, more than I can say, but it needs to be written. I had a drink with an old friend who works in a sphere very close to mine – interactive TV advertising – and we really had a difference of opinion that has thown me sideways. Our conversation, after the usual gossip and catching up, veered to the advertising industry, which I normally keep as far from as practically possible. Our chat turned towards the future and this is where we diverged dramatically (as we always do). He insisted that the media buying and owning staus quo would remain in play indefintely whatever happened technologically in the near future – I have a more optimistic view that we might finally be liberated from the strangle of heavyweight advertising interests.
It got nasty, he cited the fact that News International could demand a war when required, buy respected movie ‘review’ opinion when needed for any advertisers need. I throw back the destruction of flakey hollywood prospective blockbusters after a first night SMS frenzy from a disappointed crowd.
The battle between old monolithic “do what we tell yo” media and new distributed “I’ll ask my friends, thanks” media has shaken me. I am shocked with the fierceness of his fight, the belief in the power of advertising. Am I following the wrong path here? Is this a battle that dispite the best minds of our generation, we just get bought out by Murdoch and eat the corporate dollar? Is there a point? | 191,737 |
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Fully revised and updated, this new edition now includes coverage of such topics as 3D TV, social networks, high-efficiency video compression and conferencing, wireless and mobile networks, and their attendant technologies. Read more...
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TITLE: Law of cosines for $n$ dimensional vectors
QUESTION [0 upvotes]: The law of cosine is
$$\|b-a\|^{2}=\|b\|^{2}+\|a\|^{2}-2\|b\|\|a\| \cos \theta $$
Is there a way to prove this when vectors $a, b$ have more than 3 components?
For 3 or less components one can prove it using geometry but what about higher dimensions?
REPLY [0 votes]: In any dimension, the norm is related to the scalar product by :
$$\| a\|^2 = a\cdot a$$
and the angle $\theta$ between two vectors $a,b$ is defined by :
$$a\cdot b = \|a\|\|b\|\cos \theta$$
Therefore, :
\begin{align}
\|a- b\|^2 &= (a-b)\cdot (a-b) \\
&= a\cdot a - a \cdot b - b\cdot a + b\cdot b \\
&= \|a\|^2 + \|b\|^2 - 2 a\cdot b\\
&= \|a\|^2 + \|b\|^2 - 2 \| a\|\|b\|\cos \theta
\end{align} | 80,173 |
( Fic awaits, my good man! Tally-ho!Collapse )
( Fic awaits, my good man! Tally-ho!Collapse )
When I was very young,
a "rough-and-tumbler," as you called me
kicking up dust, invigorated with the spark of the eyes
and the strong fists and the fearless bravado of youth,
I couldn't have known.
But then, one day, when I realized I was taller
than I once was, and the dust had settled around me,
and the lion in my heart had taught himself to behave,
I understood.
When the dust had cleared, I saw you standing in its wake
windblown, tangled, and fine, just fine.
You smiled at me.
I had never realized how you eyes shone before.
It wasn't a sudden whirlwind of love,
like you used to read about in your stories.
It blanketed out like steady rain over a garden.
You knelt on the ground to tie my shoe.
I brushed your hair when you were sick.
A drip here, a drop there, and one day we looked up
And realized we were drenched in love.
Now, I understand.
So please, allow me to love you as you love me.
Shield me from my shadows
I will fight monsters while you lie in sleep.
Selfish pride has left me in its wake
Help me stand when my legs are frail
Teach me to walk again
Lie awake with me when I cannot sleep.
I stand before you, no longer a prideful man
You taught me to need again.
I will sing with you in the mornings
I will stand with you on the front lines
I will eat lunch with you in the quiet house
And I will be brave with you when the bombs explode.
Feeling the cold, moist dirt underneath the soles of his feet
Chasing about rare little patches of sunshine.
Fine gold flecks of dust to be found beneath.
Speaking with spiders and dormice,
addressing the great old council of trees
That stand so tall and regally, kind and willowed
Bestowing crowns of flowers and stems upon his head.
He burrows into the forest floor, sometimes
Under the curling roots of some kind old watchful tree
Under the leaves, into the sweet, rich soil
And he sleeps like a field mouse, in the deep
spirit of the forest, watching the stars blink and glimmer against the watery indigo sky.
In the summer nights, the whole forest sleeps
beneath shadows and scattered light.
The size of a spiderling in a glorious, overgrown world.
I had a baby doll, the sort with a torso made out of fabric but the head, arms, and legs made out of... whatever it is dolls are usually made of. She had a little white and yellow onesie that you could take on and off so you could switch out outfits. Anyway, I got her for my first birthday from my aunt and uncle, and I loved her. I named her Baby Eyelashes... still don't quite know what led me to choose that name, other than the fact that she had eyelashes (and not even actual ones, just little itty bitty ones painted on her face). But Baby Eyelashes she was, and I took her all over the place. She went through it all. Ketchup stains from the McDonald's Playplace. Being confiscated by airport security and run through the little luggage car-wash to check her for deadly weapons before Papa left for his flight while I watched with a mixture of confusion and fear. Getting accidentally thrown against walls, dropped on the floor, slept on and occasionally forgotten about. She's a good doll. Currently, she's with my other stuffed animals in my room.
So, over the past few days, me, K, and Mama all made s'mores and watched North & South, and it was lovely. The plot, the characters, the cinematography and setting and music—it was beautiful, it was all beautiful. I love this breed of film, the sort that puts you in another place and weaves you in with the characters, all their joys and sorrows.
Watch it. Just watch it.
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep."
My heart dreams at Christmastime.
The sentry of nutcrackers on my mantel comes to life: tiny, laconic, but otherwise faithful Peabody; standoffish but good-hearted Meriwether; Ambrose, tall, young, new to the house and quite unsure of himself, but all good intentions and bright intellect; Old General, from the times of my mother's childhood, Andy Williams and Christmas parties, stout and sure and of good sense, as in his youth; and then Nutcracker. Closest to my heart. I received him in the December of my eighth year. Taller and younger than Old General, shorter and older than Ambrose, he is brave and kind-hearted and will give anything for the good of his friends. He is wonderful, and he is mine.
In my mind, I am an explorer of the Endless Snow, walking among hills and across rippling gray lakes in snowshoes and parkas and woolen scarves, blowing clouds of breath up into the air, waking snow-swans and sables and great cetaceans from the deep.
Maria and the Nutcracker
She stopped when she saw him, pushed the hair out of her face and grinned narrowly at him, drenched in her glory. How could a girl so stone-hard make him so weak?
We made it, her smile said, written across her pearly teeth and in the corners of her eyes. It was something so victorious, a silent girl standing in the rain. It was her rain.
Suddenly, she was running to him, and her hand was pulling his—her hands were so cold—and she led him out into the rain, smiling. Both her hands were on his, and she squeezed them.
"Thank you," she said, "thank you so much," as they stood there, their hearts beating and their clothes ruining and their souls rising.
"No, thank you," Tobias said. "It was all you, Marlena."
They looked up into the rain, the blinding gray-white sky. And at once, Marlena began to laugh, softly at first but slowly building, until it was so delighted and pure and happy that all he could do was stand and listen and not dare to breathe. He never wanted to forget the sound, laughter amidst rain. It was a beautiful thing.
So many stories in my mind, shy glances and tender hearts, birds and spiders and tiny toy soldiers falling in love with humans. "One day," he says, with the wood of his hands together, click click click, with his broken leg and his soft, soft eyes, "one day, Peabody will be a human too." And wood would turn to flesh, and a small, small soldier would turn tall and strong, and then Eleanor would never turn away again.
A sparrow falls in love with a young man in the city, a porch-sitting boy with eyes in every shade of blue, with a soft glance at her as she sings on the telephone wire. She holds his hand after gray feathers turn to blond hair, sweet black eyes turn hazel. "I'm terribly sorry," she says, turning her head side to side as birds do, and her voice is a bird's soft song, lovely. "You're the only human I know, I'm afraid, and I'm so very lost—"
"It's alright," he says, his bare feet matching hers. His voice feels hard and bristly against hers, so delicate and feathery, but he can't know, he can never know how disarmingly gentle his voice is. "Do you like tea?" he asks, as he slides back the door, and the warm deep smell of tea leaves and peaches and humans meet them at the threshold. Everything smells like him, and she breathes in so deeply, barely able to take it in.
"Yes," she says, "yes, tea would be wonderful."
"Some say they touch the sky,." | 363,366 |
Shirley Chisholm Early Childhood Center
The. Art, music and drama round out the day. All children receive healthy and nutritious meals each day – breakfast, lunch and snack
Our programs support English, dual-language learners, and children with IEPs. Families are engaged to become active partners in supporting their child’s language development. To apply, visit My Schools NYC.
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Brownsville, Brooklyn is a vibrant, evolving, and unique neighborhood; and SCO is proud of our partnership with the community for over three decades.
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- One dog walks
- Group walks
- Doggy daycare at my property
- Dog sitting at owners property
- Dog boarding at my property
- Care for other animals (e.g. cat visits)
Our aim is to ensure that your animal has the most enjoyable experience with us as possible, feeling comfortable on the walk, in transit and be able to express themselves in a controlled environment.
The Dog Walking service we offer is tailored to suit all breeds and ages, as we are aware that no one dog is the same. Therefore, the walker will always be conscious and considerate, ensuring to reflect this in their practice.
We understand that everyone’s situation is different, whether you are at work, called away from home, hospitalized, whatever the issue, we are here to support you with the exercise and all other day to day commitments of having an animal and will care for it as if it were one of our own.
We offer a free initial consultation, this is time for us to meet you and your pet’s in your home, discuss and arrange our services, answer any queries and for us to take notes on your requirements.
My name is Laura Murray and I live in Christchurch with my family. For the majority of my working life I have been employed in the Animal Sector. I studied and gained an Advanced National Certificate in Animal Management at Kingston Maurward College in Dorchester in 2010. This course covered topics like, Animal Health, Science, Dog Grooming and Welfare.
Following college I worked for a local dog groomer, to brush up on my skills. I also worked as a volunteer on a Dairy farm in Hampshire for 18 months and latterly worked in paid employment on a Dairy farm in New Zealand which was an amazing personal and educational experience. This broadened my perspectives on animal care and reinforced my own ethics that animal welfare is paramount.
My career aspirations led me to establish my own Pet Service business, Pets Inc, on returning to the UK.
As a dog owner myself I understand how hard it is to cater for our animals at home with busy lives and routines, so it is important to me that you are confident that myself and PETS Inc. will be a home from home for your companion. | 283,945 |
Roden Gray | Accessories | Diagonal Stripe Small Tablet Holder Thom Browne Diagonal Stripe Small Tablet Holder $910.00 CAD $1,300.00 CAD Colour: White Free returns Secure checkout Details Fit info Designer Tablet holder in white pebble grain leather with red, white, and blue diagonal stripe on front and back. 100% Leather. Zipper closure. Tri-colour grosgrain tab. Measures approximately 11 ⅝” x 7 ⅞” / 29.5cm x 20 | 42,406 |
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TITLE: Is there a number system which eliminates singularities over rational functions?
QUESTION [6 upvotes]: I tutor some high school students in Algebra 2 and Pre-Calculus. This is a questions from one of them. Here's the context:
Often, when students are first introduced to the concepts of domain and range, they are told that the domain is "all of the allowed inputs," where an input is allowed if it (a) is a real number and (b) produces a real number as an output.
In general, they are taught that there are two situations that break these rules by accepting a real number as an input but then giving back something that is ill-defined. The two situations are zeroes in the denominators of rational functions and negative numbers under even radicals.
Later, complex numbers will be introduced, and it will be explained that they solve various problems including the ill-defined-ness of the roots of negative numbers. However, the issue of dividing by zero is never resolved.
Are there any number systems which resolve the issue of dividing by zero similar to how complex numbers resolve taking the square root of negative numbers?
While it is not necessary, preferably this number system would be a natural extension of the real or complex numbers that avoids introducing new singularities. It seems to me that the answer to this question might depend on whether or not there are any fields where the additive identity has a multiplicative inverse.
REPLY [2 votes]: If there was it would be very different from the real number system. We can show that the existence of $x$ such that $x=\frac{1}{0}$ leads to a contradicion. We know that the equation $0*1=0*2$ is true. If we multiply both sides by $x$ we obtain $\frac{0}{0}*1=\frac{0}{0}*2$. The $\frac{0}{0}$ terms can be cancelled from both sides to give $1=2$ which is a contradicition.
The "number system" I think you are looking for is the extended complex numbers which can be visualized with the Riemann Sphere. | 127,102 |
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DataControl’s Smart Tag technology makes it possible to track data usage throughout your organization. You can document data usage and destruction for compliance with Sarbanes Oxley, PCI, and HIPAA.
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"is unconstitutional and therefore invalid".
.
"I'm happy I won't be getting any more criminal records for possession," Ruaan Wilson, 29, told AFP before pausing for a puff.
"Now, we can get police to focus on real drugs and thugs," he added.
".
Activists have held marches over the years to demand that the law be changed to allow people to smoke "weed", which is called "dagga" in South Africa.."
Previously, possessing, growing or using marijuana for personal use - even in small quantities - exposed users to fines of up to hundreds of dollars as well as jail time. Penalties for selling it were far higher.
South Africa' Legalises Use of cannabis
Reviewed by letstalk naija
on
September 19, 2018
Rating: | 111,700 |
TITLE: Commutator of algebraic subgroups is connected
QUESTION [5 upvotes]: Let $G$ be an algebraic group over an
algebraically closed field. If $H$ and
$K$ are closed subgroups and one of
them is connected, then their
commutator $[H,K]$ is also connected.
Is there an easy way to see this fact?
The proof that I see in Springer's book on Linear Algebraic Groups is very long-winded. (Of course, he obtains many results on the way). My question is, is the given statement true for any topological group? If yes, does that proof apply to the algebraic groups' case?
(Remark: The definition of a topological group includes the Hausdorff property, whereas algebraic groups are not so).
REPLY [1 votes]: This fact is Corollary 16.4.1 on page 40 of the course notes from (the first semester of) Brian Conrad's course on linear algebraic groups. In the notes there is a proof of this fact is modern language, over an arbitrary field. The notes can be found here: http://math.stanford.edu/~conrad/252Page/handouts/alggroups.pdf | 179,210 |
Tag: PowerApps […]
Four Benefits of PowerApps
January 15, 2019
Businesses rely on forms to collect data from employees in the field or in the office. Following the retirement of InfoPath several years ago, […] […]
Power BI – What’s the BIG deal?
January 31, 2018
By now, most people are aware of BIG data terms that are out there in industry. Most organizations are finally gaining traction in the […] […] | 16,723 |
LIFE FOR CHOPPING WIFE Posted: Tuesday, August 1, 2017LIFE FOR CHOPPING WIFEA HIGH COURT judge yesterday issued a stern warning to men who perpetrate domestic violence against their partners when the relationship ends - you will be punished for your crime.He may kill her next timeJudge jails man for life over cutlass attack on ex-loverDaddy rapes mommyFIVE children were left traumatised after they witnessed their father raping their mother on the hood of the man's car, along a lonely track in Wallerfield early yesterday morning.Pay for slaveryPRESIDENT Anthony Carmona yesterday publicly supported a call to have European governments, whose countries benefited from slavery in the West Indies, to pay reparations to the descendants of African slaves.PM: Scars of slavery still evidentPM: Greatness possibleIn his Emancipation Day message, Rowley called on citizens to, "draw on each other's talents, skills and strength." "We cannot yield under the whip of internal and external challenges.Kamla: Freedom under threatOPPOSITION Leader Kamla Persad-Bissessar warns that personal freedom is under threat today from the criminal element, even as she lauded the historical liberation of slavesIllegal Chinese immigrants heldON THE eve of the Emancipation Day holiday, police raided a building on Charlotte Street in Port of Spain and detained eight illegal Chinese immigrants16 Venezuelans caught in TobagoTWO pirogues were intercepted and their crew of 16 Venezuelan nationals detained by members of the Coast Guard on Sunday afternoon in Tobago waters.Way cleared for Toco portCABINET has approved the construction of the port in Toco to service the Tobago sea bridge, a consultant is already on board, and tender documents for contractors to bid to build the port are expected to go out shortlyAlmost dead teen stuns doctorsCop's sister slain in GrandeAmbush after day at church harvestNurse slain outside homeMissing man found dead in carAlthough Lincoln Boodoo's lifeless body bore no marks of violence when he was found yesterday, his family remains confused as to how he died.Fear mounts for missing bank workerArsonist jailed for 15 yearsA 28-YEAR-OLD mentally ill man, who admitted setting fire to his father's home, has been sentenced to 15 years' hard labour and ordered taken to the St Ann's Psychiatric Hospital for assessment.2 couples in court for sacrilegeLakshmi Mahadeo was among four people appearing before a San Fernando magistrate charged with breaking into a Hindu temple and stealing items.
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, Perth & Kinross.
He’s now working out of his large garage and it’s taken him the best part of a year to install machinery and equipment – alongside making commissions for customers.
“At the public exhibition at the school after graduation I had someone very interested in my work and she got in touch with me a month afterwards to commission some pieces,” says Colin.
“Since then, she has been a very good repeat customer and has also recommended me to other people. Those recommendations have won me, among others, a kitchen in Haddington and Japanese tea ceremony pieces for a customer in London.”
Colin, originally from Birmingham, moved north to work in outdoor education but, over the years, found himself less and less outside and more and more behind a desk – although he did find time to volunteer with his local mountain rescue team.
However, being naturally creative, he decided to enroll on a professional course at the Chippendale school and change the course of his career.
“I love the practical precision of furniture making, and the disciplines involved in turning a design idea into a beautiful piece of furniture. But I also enjoy the creativity that goes into making a desk or cabinet into something absolutely unique,” he says.
That practical precision is demonstrated in a bespoke unit (pictured) made from flamed beech, elm, ash and oak, with bog oak inlayed lines. The doors have a curved split up the middle perfectly matching each other, with the inlayed lines flowing over and around them.
Showing a talent for marketing, Colin displayed some of his work at a local bike shop and, once again, a commission for a kitchen followed – as well as smaller commissions from local people in the area.
And underlining how one thing can lead onto another, the bike shop expanded to include a café and Colin designed and made the counter.
This attracted much interest and, you guessed it, a local artist has commissioned Colin to make a counter for his small gallery in Dundee, which will be fitted next week.
“I have thoroughly enjoyed the last year. It has been challenging, but it’s also been a pleasure to discuss ideas with customers and then designing and making pieces for them. It’s lovely to get to know clients, an important first step in designing exactly what they want,” he says.
A stand-out piece that he made at the Chippendale school was a steam-bent desk in olive ash and oak which he created “to test the limits of what can be achieved with steam bending,” said Colin.
It was a wonderful signature piece (main picture), and we’re delighted it has led to a promising start in the professional world of fine furniture design.
Recent Comments | 355,661 |
Spelling and grammar should be checked by a native English speaker to ensure the clarity and accuracy of manuscript if English is not authors’ first language. Alternatively, authors may contact Wallace Academic Editing, who has an agreement with AAQR to give a 10% discount on editing for papers previously submitted to AAQR. To receive the discount authors can mention their Manuscript ID when sending their paper for editing. Wallace Editing sites are: | 383,858 |
Added to Shopping Bag
Céline Cabas Phantom Yellow Lambskin Tote
Listed by Abbey R
Price: Sold for:
131 people saved this item
- Exterior scuffs or marks
Scuffing and marks throughout. Color transfer throughout.
Celine Cabas Phantom Yellow Tote
Handle Drop 8"
Made in Italy
Interior has one zip pocket
Care booklet included
Tags included, not attached
Can be worn as a shoulder bag
Type:
Totes
Measurements:
15"L x 16"H x 2.5"W
Fabric:
Lambskin | 197,364 |
Vitamins are essential to being alive. Our ancestors have been guzzling them since before they had spines, and they’re a key element in a number of chemical reactions that support our immune system, metabolism, and yes, our shiny, shiny hair. If we deprive ourselves of them, we’re subject to rickets, scurvy and all sorts of awfulness.
Certainly, vitamin supplements have done a lot of good. In third-world countries vitamin A deficiency can cause a whole range of eye diseases – which can be immediately repaired by a small supplementation in their diet.
Studies looking into populations that live longer and were generally healthier found they had high concentrations of vitamins A, C and E in their blood. But here’s the important bit: they got those vitamins by eating lots of fruit and veg.
This perception that we need vitamins supplements is a naive, drug-discovery approach.
Dr Trent Watson, a representative of the Dieticians Association of Australia, says just because people who eat a boatload of fruit get more vitamins, but getting the vitamins without the fruit does not mean they will be just as healthy.
“What we’ve got to be careful of is not getting correlation mixed up with causation,” he advises. “There’s a correlation between high fruit and veg (intake) and health. But then, that was extrapolated by supplement companies saying these are obviously markers within these fruit and veg that provide these benefits, so we’ll just isolate those and supplement them.”
Dr Watson is adamant there’s no substitute for eating well. “The bottom line is that you can get all your essential vitamins from the food you consume.” he says. “I think, in the wash-up, this perception that we need vitamins supplements is a naive, drug-discovery approach.”
What about those time-honoured remedies – like fighting a cold with vitamin C? “There’s no evidence to support that Vitamin C is protective against us getting a cold, or reducing the duration with which you have it,” he says.
How about conquering your hangover with a fizzy vitamin B? “As a consequence of drinking a flavoured water that happens to have B vitamins in it, then maybe you’ll reduce dehydration. But B group vitamins don’t have any effect.”
And Dr Watson is hardly alone in suggesting you might as well flush your B-group vitamins as eat it. “Don’t waste your money” read a recent editorial by the Annals of Internal Medicine.
“The message is simple. Most supplements do not prevent chronic disease or death, their use is not justified, and they should be avoided.”
The scathing editorial preceded the findings from two studies that suggested that 6000 doctors over 65 who ate a daily multivitamin for ten years showed no improvement in retaining cognitive function than those who ate a placebo. Another study of 294,718 people in 2013 found that supplements made no impact on preventing cardiovascular disease.
“What you’re seeing is the marketing machine,” says Dr Watson. “They’re almost creating this worry through marketing to the general population, saying, ‘You need this stuff because food doesn’t have the vitamins and minerals.’ That’s not the case.”
Consumer watchdog Choice agrees.
“Marketing messages, often backed up by high profile sporting celebrities, give the impression that in order to be fit and healthy, we all need multivitamins,” says Choice spokesperson, Ingrid Just. “If you have a healthy diet and you’re not a person with specific nutritional requirements, there’s a good chance you’re wasting your money.”
If you have a healthy diet and you’re not a person with specific nutritional requirements, there’s a good chance you’re wasting your money.
While it’s galling that vitamin supplements actually may be totally useless, what’s more worrying is that there’s increasing evidence that in some cases, they can actually be harmful.
“‘Some is good, more is better,’ is not the case with this situation,” explains Dr Watson. “In fact, the law of toxicology says that anything, in the right dose, is toxic.”
Recent research has come up with some troubling findings. According to a New York Times article, a 2004 study found that taking vitamins A, C, E, beta carotene and selenium to prevent intestinal cancers increased mortality. Another found that in 19 trials of nearly 136,000 people eating vitamin E supplements increased mortality. Another discovered that vitamin E in high doses can lead to an increased incidence of cardiovascular disease. Yet another found that in 19 trials of covering 136,000 people, supplemental vitamin E increased mortality. A study in the 1990s into the effects of supplementing vitamin E and beta carotene in smokers had to be stopped because it was leading to increased instances of lung cancer.
“In medicine, our first course as practitioners is first to do no harm. Now, I think what’s happening here is that that’s not the case in some of these situations,” says Dr Watson.
Needless to say, Dr Watson doesn’t take his daily multi-group.
Of course, there are a number of perfectly legitimate reasons to supplement one’s diet – if you actually have a deficiency. But, surprise, surprise, wandering into the supermarket and buying armfuls of Health™ probably won’t work.
For real advice on getting your vitamin intake right, see a dietician or your local GP. | 126,480 |
USA U-17 defeat Ireland 4-1
COLORADO SPRINGS, C.O. – The USA U-17 Women’s National Team had a strong performance against Ireland this afternoon to clinch their second win of the tour. Excellent full field passing and a new system designed to strengthen the midfield, enabled USA to defeat Ireland 4-1.
Alyssa Chillano (Phoenixville, PA) notched USA’s first goal of the game with a drive from the top of the circle. Chillano’s shot was deflected over Ireland’s goalkeepers head by their own defender. Ireland kept the game tied going into the half, with a shot sneaking by USA goalkeeper Samantha Zeiders (Hummelstown, Pa).
The second half remained scoreless for both opponents until almost the twenty minute mark. Chillano had her second goal of the match off of a penalty corner. Up 2-1 over Ireland, USA put on the pressure for the remainder of the match. Two more goals were score by USA’s Gabrielle Major (Royersford, Pa) and Tara Vittese (Cherry Hill, NJ). Vittese’s goal came in the last second of the game as time expired. USA finished the match up 4-1 over Ireland.
“Today was a proud moment to see how the USA side came together and played as a team in their first international match against Ireland,” said Char Morett, Assistant Coach. “A new system put in place to strength the midfield worked well despite a lack of opportunity to practice it. This definitely illustrates the girls knowledge of the game is increasing.”
The win against Ireland sets USA up with momentum and confidence heading into the DOMO Easter Tournament.
### | 223,770 |
Not gonna lie, we’ve kinda been obsessed with HoCo and Mars in the past month or so.
The way they effortlessly blend r&b, soul, and hip hop is the reason we fell in love with the duo – smooth vocals and solid bars, with a soulful instrumental underneath. And “Wndrlnd” is indeed a wonderful example of that.
This is the closing track of their Atlas album, which you should totally check out – “Cherry Roses” & “Sunflower” are also from there. | 32,091 |
SORTIE OF THE IMPERIAL FLEET
Only once in the two years that had passed since the hard- pressed, bitterly fought days of Coral Sea, Midway, and Guadalcanal had the Japanese Fleet ventured out in strength to offer battle. Even in the critical actions of 1942 nothing like full scale commitments had been made on either side, and while the Battle of the Philippine Sea brought out a large part of the enemy fleet the engagement had been confined to air action. None of our many landing operations in 1943 and, with the single exception of the Marianas, none in the following year had been challenged by major forces of the Japanese Navy.
The westward sweep of our Pacific offensive had by the fall of 1944 converged in two mighty thrusts aimed at the Philippine Islands, flanking them from the east and south. On September 15 simultaneous landings were made on Peleliu Island of the Palau Group and on Morotai in the Moluccas. The Peleliu landing brought the Central Pacific Forces, under the command of Admiral Chester W. Nimitz, to the Western Caroline Islands within five hundred miles east of Mindanao. The path of their advance had been westward from the Gilbert Islands through the Marshalls and Marianas. By the landing on Morotai General Douglas MacArthur advanced the frontier of his Southwest Pacific Forces, which had pushed northwest along the coast of New Guinea, to within three hundred miles southeast of Mindanao. The Philippines lay ahead as the next great objective.
Would the landing in the Philippines precipitate the long-
-7- | 385,812 |
> Jaskell Shell
Jaskell Shell is an interactive command line interpreter for Jaskell. It is handy for experimenting simple jaskell expression or java statement real quick.
In order to use Jaskell Shell. You need to download Jaskell jar file. Make sure the jfuntuil.jar, jparsec.jar and jaskell.jar are all in classpath. Then run "java jfun.jaskell.shell.Shell"
The shell will then show:
The version number may vary depending on the version you are running on.
The ">" character is a command prompt.
Now assume we want to try "javax.swing.JOptionPane.showInputDialog()", we can type in:
Hit the "Enter" key twice to evaluate this expression. (This is because shell allows you to enter expression in multiple lines, the first "enter" key may just start another line of the expression.)
A swing dialog will be popped up by now. Enter "13" in the input box, hit "Enter", the "13" will be shown as the value of the expression in the shell.
Note, we use square brackets instead of parenthesis for java method. This can be understood as a reflection call where an array of objects is passed to each method.
The above code is straight-forward, but not handy enough. Suppose we need to run "showInputDialog", "showConfirmationDialog" etc repeatedly, we certainly don't want to keep typing in "javax.swing.JOptionPane" again and again.
In order to do this, we can give "javax.swing.JOptionPane" an alias, kinda like typedef in C:
When "Enter" key is hit twice, the value of the expression is shown after "=>". In this case, the expression value is the JOptionPane class.
Now, we can use the shell command "?" to look at all the static methods supported by this "dialog":
Since "?" is a shell command, not a jaskell function, we just need to hit "Enter" once. It then shows us all the static methods of JOptionPane.
In order to run "showMessageDialog", we type in:
A "hello world" message dialog will be popped up.
To simplify the syntax even more, we can:
The value of this expression is a function called "say". This function expects one parameter.
Now we can "say" a lot of things:
blah blah blah.
The following expression starts a thread that prints "hello world":
We could use the prelude function "const" to simplify the expression so that it becomes:
As we have done already, we can simplify the println as:
In fact, println function is already defined, you can use it without defining it. We define "println" here primarily as a demo of how functions can be defined to make syntax simpler.
Using the same technique, we can define a "run" function that makes running task in thread easy:
Let's see how we can use the "run" function:
The last "say nice" would have blocked the shell if it were not running in a thread. But calling it with "run", the shell is not blocked.
There are of course many other uses. Any jaskell expression can be typed in this shell.
To see what functions are available globally:
To see what functions are available in a certain namespace:
To get the value of the previously evaluated expression:
To define more than one names in one line:
To exit shell:
To get help of shell commands: | 154,839 |
Bosphorus – Anaplus Bosphori Tracii
by Guillaume Sanson
Royal cartography: rare and spectacular
Detail
Date of first edition: 1666
Date of this map: 1666
Dimensions (not including margins): 41 x 53 cm
Condition: Excellent. Strong paper and wide margins. Soft outlining old colouring.
Condition rating: A+
Verso: blanc
Price (without VAT): €800,00 (FYI +/- $944,00 / £712,00)
We charge the following expedition costs in euro:
– Benelux: 20 euro
– Rest of Europe: 30 euro
– Rest of the World: 50 euro
In stock
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This is the rare second state of the view, with the roundel at the right including the portrait of Sultan Murad III, which is blank in state 1.
Black Sea in the Antiquity – Pontus Euxinus. Aequor Iasonis pulsatum remige Primumby Abraham OrteliusPrice (without VAT): €1 000,00 / $1 180,00 / £890,00
Where history never stops…
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Unique, rare and large frog’s view
The rise of French cartography
Starting in the fourth quarter of the 17th century Paris (and thus French cartography) will gradually start replacing Amsterdam as the central production site for mapmaking.
Nicolas Sanson (1600- 1667) is more or less regarded as the father of French cartography. He became chief cartographer under Louis XIV. One of his sons, Guillaume (1633-1703) will take over this position. The latter’s map of 1666 shows the Bosphorus Strait, with a length of 32 km between the Sea of Marmara to the south and the Black Sea in the north (Pontus Euxinus). The Bosphorus separates Europe from Asia. Bottom left in pink lies Constantinople.
The data for this map were provided by Pierre Gilles and go back to 1561.
Sanson mentions in the cartouche that he has obtained a privilege for 20 years to produce this map. | 222,016 |
Buffalo Crochet and Public Knitters Unite have UNITED! Rather than the organizers paying for two separate meetups we have combined them into one group.
There will be both Knitters and Crocheters at every meetup. If you would like to learn how to knit or crochet, come to our dedicated learning nights (the last Wednesday of every month!).
Hope to see you soon! Happy. | 210,886 |
TITLE: models versus complete theories
QUESTION [1 upvotes]: To properly ask my question, I need to set up some definitions of whose I don't know, if I use the words in the usual sense:
Let $\mathcal{L}$ be a formal language of first-order predicate logic with equality. With the word theory, I mean a collection of formulas in that language, that is
closed under the usual rules of inference, so every formula, that can be derived from the theory is already a member of it.
consistent, so the formula $\top \leftrightarrow \bot$ is no member of it.
I will call a theory $T$ in the language $\mathcal{L}$ complete iff for any sentence $A$, that can be formed in the language $\mathcal{L}$, one of the two sentences
$$ A \leftrightarrow \top \qquad \text{or} \qquad A\leftrightarrow \bot$$
is a member of $T$.
Note that a complete theory contains a formula like $\exists x : A(x)$, if and only if there is a term $t$, such that it contains the formula $A(t)$ (so the $\exists$-quanitifier has its intended "meaning").
For some theory $T$ in the language $\mathcal{L}$, a syntactic model for $T$ consists of
a language $\mathcal{L}'$, that contains the same function symbols, relation symbols and variables as the language $\mathcal{L}$, contains every constant symbol of $\mathcal{L}$ but may contain even more constant symbols.
and a complete theory $T'$ in the language $\mathcal{L}$', that contains every sentence of $T$, when interpreted as a sentence in the language $\mathcal{L}'$.
For some theory $T$ in the languahe $\mathcal{L}$, a set model for $T$ is an $\mathcal{L}$-structure, satisfying each formula of $T$, interpreted in standard semantics (just the usual definition of a model here).
I observe now, that there is an "almost-bijection" between syntactic models and set models for a theory:
Let $M$ be a set model for the theory $T$. For any member $x \in M$, add a constant symbol $\mathsf{c}_x$ to the language of $T$ and let $T'$ be the collection of all sentences in the enhanced language, that are true in $M$ (interpreted via standard semantics).
Let $(\mathcal{L}',T')$ be a syntactic model for the theory $T$ in the language $\mathcal{L}$. Define an $\mathcal{L}$-structure $M$ by building equivalence classes of $\mathcal{L}'$-terms containing no variables, where two terms $t,s$ are equivalent, if the equation $s=t$ is a member of $T'$. Interpret the function symbols in the obvious way. Now for every $n$-ary relation symbol $R$, define the coresponding set theoretical relation as
$$ \{ (t_1, \dots , t_n) \in M^n \mid ( R(t_1, \dots ,t_n) \leftrightarrow \top) \in T' \} $$
Thus, we get a set model for $T$.
These two operations are almost inverse, but the operation $\{\text{syntactic models}\} \to \{\text{set models}\} \to \{\text{syntactic models}\}$ adds a constant symbol for each equivalence class of constant terms.
It seems to me that the notion of syntactic model can fully replace set models. Also every prove of Gödel's completeness theorem, that I have seen, goes on to construct a syntactic model first and then transform it into the set model. It follows that the completeness theorem still holds, if we replace set models by syntactic models.
Now comes my question: Why do we even go that detour around set theory, when defining semantics? Why don't we only work with syntactic models?
The advantage of syntactic models is, that they don't stress the ontology more than necessary. If you believe in strings and the notion of formal derivability, it's not hard for you to belive in syntactic models as well. You don't need to belive in or even think of general power sets, functions being defined as sets of Kuratowski pairs, uncountable infinities or the notion of cardinality at all.
REPLY [5 votes]: As Eric Wofsey points out, there are some errors implicit in your question which have to be addressed before it can be answered. The issues around variables are a side issue which I'll ignore. The important error is the claim
Note that a complete theory contains a formula like $\exists x : A(x)$, if and only if there is a term $t$, such that it contains the formula $A(t)$ (so the $\exists$-quanitifier has its intended "meaning").
which is quite false. One easy way to see that this fails even for complete theories is by thinking about models (in the usual semantic sense): for any structure $\mathcal{M}$ the theory $Th(\mathcal{M}):=\{\varphi: \mathcal{M}\models\varphi\}$ is complete (this is a trivial consequence of the negation clause in the Tarski definition of truth), so the theory of any structure in a language with only relation symbols is a counterexample to your claim.
However, this does isolate a key property of theories. Say that a theory $T$ has the strong witness property if for every formula $\varphi(x_1,...,x_n)$ there are constant symbols $c_1,...,c_n$ in the language of $T$ such that $$T\vdash[\exists x_1,...,x_n\varphi(x_1,...,x_n)\implies\varphi(c_1,...,c_n)].$$ The right notion of "syntactic model of $T$" then is just consistent complete extension of $T$ with the strong witness property.
Having said that, you're quite right that we can rephrase everything in terms of syntactic models (once we've redefined them properly per the above). But this has the drawback of making the interaction between logic and the rest of math more cumbersome: mathematicians outside logic care about mathematical structures more than sets of formal sentences. That is, the language of (semantic) models isn't something logicians introduce, it's something we abstract from existing mathematical discourse (and indeed it only captures part of the picture - think of, say, topological spaces). Similarly, students coming into logic tend to already have "semantic mathematics" in the background. So it's genereally more natural.
Moreover, as you yourself observed syntactic models and semantic models are more-or-less equivalent, so we actually don't buy any more philosophical tameness by focusing only on the latter: "a language" is no less complicated than "a set," and restricting attention to countable languages is no less complicated than restricting attention to countable models in the usual sense. | 9,572 |
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TITLE: Derangement problem!
QUESTION [5 upvotes]: Is the solution of the problem, in how many ways can the digits $$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$$
be arranged so that no even digit is in its original position, is $5!D_5$.
Where $D_n$ = $n! \left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}+....+(-1)^n\dfrac{1}{n!} \right)$ and denotes derangement number.
Here we find expression of $D_n$ using inclusion exclusion principle.
$P_i$: $i^{th}$ object is at it's place.
$N(P_i)$ Number of object having property $P_i$
so we have to find $N(P_1P_2....P_n)$
REPLY [4 votes]: We count the permutations in which no even number ends up in its original position (good permutations) by dividing into cases.
There are $5$ odd numbers. Perhaps $0$ end up their original positions, or perhaps exactly $1$, or perhaps exactly $2$, and so on up to $5$.
There are $D_{10}$ good permutations in which $0$ odds end up in their original position.
For good permutations with exactly $1$ odd in its original position, the odd can be chosen in $\binom{5}{1}$ ways. Everybody else must move, giving a total of $\binom{5}{1}D_9$.
For exactly $2$ odd in their original positions, the odds can be chosen in $\binom{5}{2}$ ways. Everybody else must move, giving a total of $\binom{5}{2}D_8$.
And so on (three more cases). The total count is
$$\sum_{k=0}^5 \binom{5}{k}D_{10-k}.$$ | 198,943 |
TITLE: $\underset{(x,y) \rightarrow (0,0)}{\text{lim}} \frac{xy}{y-x^3}$
QUESTION [1 upvotes]: Evaluate
$$\underset{(x,y) \rightarrow (0,0)}{\text{lim}} \frac{xy}{y-x^3}$$
My attempt: I've tried to use polar coordinates $x=r\cos \theta, \; y = r \sin \theta$:
$$\underset{(x,y) \rightarrow (0,0)}{\text{lim}} \frac{xy}{y-x^3} =\underset{r \rightarrow 0}{\text{lim}} \frac{r^2 \sin \theta \cos \theta}{r(\sin \theta - r^2 \cos \theta)} = 0 $$
but the book says it doesn't exists. What am I doing wrong?
REPLY [4 votes]: Hint: See what happens if $(x,y)$ approaches $(0,0)$ along the curve $y=x^3+x^4$.
Remark: As to what you are doing wrong, implicitly you are assuming that $\theta$ is constant as $(x,y)$ approaches $(0,0)$, that we are approaching $(0,0)$ along a ray. | 107,542 |
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Applying same animation to different elements on hover?
dmaller replied to dmaller's topic in GSAPThat works great, thank you very much PointC! | 186,674 |
Good News. Bad News.
I’m going to start with the bad news so I can get it off my chest and move on to much happier things.
Recently, I saw on the NAPP member site that a new 20% off discount was available for any single purchase through the Adobe.com store until March 2nd. With that, on top of Adobe’s introductory pricing on CS4 ending at the end of February (which has since been extended), I decided that I couldn’t wait to upgrade anymore.
I was talking with my friend and co-worker, Eric, about it and offered to use my discount to help him upgrade to CS4 Design Premium when I placed my order to upgrade to InDesign CS4 and Photoshop CS4. So I placed the order, saving around $200 between us, and waited eagerly for it to arrive.
Two things then happened. First, I got a sinking feeling one day that I had been in the mindset of ordering the PC versions for myself that I inadvertantly ordered Eric, a Mac user, the PC version of CS4 Design Premium. Big OOPS! And then I was expecting to receive the software on Monday but FedEx decided that the correct address I put on the order was not the right one.
All got worked out, though. I called Adobe customer service and spoke with a really nice rep who helped me get the Mac version on its way to replace the one I had mistakenly ordered. To be entirely honest, it was one of the best customer service calls I’ve ever had. He even honored the pricing from the previous order. After driving over to FedEx to pick up the undelievered package and talking to a manager to work out the address issue – since the driver wasn’t back yet – the software finally arrived the next morning.
Then the fun started… I found out, the hard way, that as it turns out it is against Adobe’s upgrade policy to allow customers who own a full version of a program that is part of a bundle to upgrade to a stand-alone version of the same software. Meaning, even though I own CS3 Design Premium, which includes InDesign CS3, I can not upgrade to InDesign CS4. I would either have to upgrade to CS4 Design Premium – which I don’t need – or purchase the full version of InDesign CS4 – which isn’t going to happen either. So I have to send it back.
The up-side is that I can keep the upgrade to Photoshop CS4 since I have copies of Photoshop going back as far as version 6. It’s the more important program for me to have anyway.
I’m not all worked up about it but it really frustrates me that even though I own a full version of InDesign CS3 – albeit as part of a bundle – I am ineligible to upgrade a single product. It was not very clear on the Adobe site and, in the end, they are crediting my card so they’re losing out on the sale. It won’t affect whether or not I will buy from them again, obviously, but I appeal to Adobe to revisit their upgrade policy so it would make more sense and do a better job of serving the customers.
Thus endeth the rant.
On to the good news! Kim and I have been wanting to escape apartment living for a while now and after some searching we are elated to share that we have found a house! Not just “a” house, but an amazing house!
We’re doing the mortgage paperwork today and we hope to close in the next month and a half or so – give or take. We’re going to be doing another walkthrough on Sunday afternoon so we’ll be able to take some shots of the inside too. It’s in Johnson City, NY, which is only about 15 minutes from where we live now in Endicott, NY. It’s about 25 minutes to Kim’s school and roughly 10 minutes to work for me.
It’s cute on the outside and down-right beautiful inside. We’re so excited about it and we’re still pinching ourselves because it’s all finally happening!
Andy
Congratulations to both you and Kim.
Adobe’s pricing and upgrade structures are certainly not consumer-friendly at times. In the future, you may also want to check into education pricing available to Kim. Not always the best choice, but it can be beneficial at times.
Kevin Mullins Photography
Hi Jason – congratulations on the move. I moved “out of the city” a couple of years ago and never looked back.
As for Adobe – pha! Don’t get me going. I had a legitimate copy of CS3 Master Collection costing well over 1500 Sterling at the time. I wanted to upgrade just to Photoshop CS4 as I wanted the 3D stuff. No Chance! In the end I had to buy a new retail version of CS4 Extended. I understand they need to make a profit, and I love their software, but they really need to think again about upgrade paths for current customers. There really isn’t any serious competitor (in terms of Photoshop) to keep them on their feet in this area in my humble opinion.
Mike Palmer
Congrats on the house
Your Photo Tips
Hey Jason,
Congrats on the house! My wife and I just got out of an apartment ourselves into our very first home at the beginning of this month and it’s been great. I will say that buying an older home does come with the immediate back-breaking joy of being a homeowner but we just couldn’t see ourselves in one of those mass produced cookie cutter homes. You know, the kind where you get lost in your neighborhood because every house looks the same! The house you picked out looks really great and hope everything with the closing goes well.
One tip I have is to get multiple quotes for your insurance. We kinda got screwed on ours and it’s a little bit of a pain getting it resolved. You live and you learn, right?
Jason D. Moore
Thanks for all of your kind words, tips and well-wishes guys! It’s a very exciting/freeing/overwhelming thing to do and we can’t wait!
It is an older house – built in the late 50s – but most of the first floor has been redone in the last couple of years along with the roof.
As for Adobe, I think you’re right, Kevin. Since there really isn’t any serious competition there’s no incentive for Adobe to lower their price point. I just think it’s ridiculous to limit the choices of current customers. I know it’s not like they will lose me as a customer over it but I’ll start taking an even closer look at what I really need to buy.
InDesign would’ve been nice, and I was more than willing to grab it, but it was only worth it, to me, at the upgrade price, not so much if I have to get the full version.
They produce great, professional software and there is certainly an argument for pricing it so high, but to be so strict on how you can upgrade is nuts. | 298,332 |
\begin{document}
\title{Detection of Derivative Discontinuities\\ in Observational Data}
\author{Dimitar Ninevski\orcidID{0000-0003-0101-8686} \and
Paul O'Leary\orcidID{0000-0002-1367-8270}}
\authorrunning{D. Ninevski and P. O'Leary}
\institute{University of Leoben, A8700 Leoben, Austria
\email{[email protected]}\\
\url{http://automation.unileoben.ac.at}}
\maketitle
\begin{abstract}
This paper presents a new approach to the detection of discontinuities in the n-th derivative of observational data. This is achieved by performing two polynomial approximations at each interstitial point. The polynomials are coupled by constraining their coefficients to ensure continuity of the model up to the (n-1)-th derivative; while yielding an estimate for the discontinuity of the n-th derivative. The coefficients of the polynomials correspond directly to the derivatives of the approximations at the interstitial points through the prudent selection of a common coordinate system. The approximation residual and extrapolation errors are investigated as measures for detecting discontinuity. This is necessary since discrete observations of continuous systems are discontinuous at every point. It is proven, using matrix algebra, that positive extrema in the combined approximation-extrapolation error correspond exactly to extrema in the difference of the Taylor coefficients. This provides a relative measure for the severity of the discontinuity in the observational data. The matrix algebraic derivations are provided for all aspects of the methods presented here; this includes a solution for the covariance propagation through the computation. The performance of the method is verified with a Monte Carlo simulation using synthetic piecewise polynomial data with known discontinuities. It is also demonstrated that the discontinuities are suitable as knots for B-spline modelling of data. For completeness, the results of applying the method to sensor data acquired during the monitoring of heavy machinery are presented.
\keywords{Data analysis \and Discontinuity detection \and Free-knot splines.}
\end{abstract}
\section{Introduction}
In the recent past \emph{physics informed data science} has become a focus of research activities, e.g.,~\cite{Owhadi2015Bayesian}. It appears under different names e.g., \emph{physics informed}~\cite{Raissi2019}; \emph{hybrid learning}~\cite{Saxena2019}; \emph{physics-based}~\cite{yaman2019selfsupervised}, etc.; but with the same basic idea of embedding physical principles into the data science algorithms. The goal is to ensure that the results obtained obey the laws of physics and/or are based on physically relevant features. Discontinuities in the observations of continuous systems violate some very basic physics and for this reason their detection is of fundamental importance. Consider Newton's second law of motion,
\begin{equation}
F(t) = \dt{} \left\{ m(t) \, \dt{}y(t) \right\} =
\dot{m}(t)\,\dot{y}(t) + m(t)\,\ddot{y}(t).
\end{equation}
Any discontinuities in the observations of $m(t)$, $\dot{m}(t)$, $y(t)$, $\dot{y}(t)$ or $\ddot{y}(t)$ indicate a violation of some basic principle: be it that the observation is incorrect or something unexpected is happening in the system. Consequently, detecting discontinuities is of fundamental importance in physics based data science. A function $s(x)$ is said to be $C^n$ discontinuous, if $s \in C^{n-1}\setminus C^n$, that is if $s(x)$ has continuous derivatives up to and including order $n-1$, but the $n$-th derivative is discontinuous. Due to the discrete and finite nature of the observational data, only jump discontinuities in the $n$-th derivative are considered; asymptotic discontinuities are not considered. Furthermore, in more classical data modelling, $C^n$ jump discontinuities form the basis for the locations of knots in B-Spline models of observational data~\cite{Wahba1990}.
\subsection{State of the Art}
There are numerous approaches in the literature dealing with estimating regression functions that are smooth, except at a finite number of points. Based on the methods, these approaches can be classified into four groups: local polynomial methods, spline-based methods, kernel-based methods and wavelet methods. The approaches vary also with respect to the available a priori knowledge about the number of points of discontinuity or the derivative in which these discontinuities appear. For a good literature review of these methods, see \cite{Gijbels2005}. The method used in this paper is relevant both in terms of local polynomials as well as spline-based methods; however, the new approach requires no a priori knowledge about the data.
In the local polynomial literature, namely in \cite{Horvath2002} and \cite{Spokoiny1998}, ideas similar to the ones presented here are investigated. In these papers, local polynomial approximations from the left and the right side of the point in question are used. The major difference is that neither of these methods use constraints to ensure that the local polynomial approximations enforce continuity of the lower derivatives. Additionally, it is not clear whether only co-locative points are considered as possible change points, or interstitial points are also considered. Furthermore, both papers use different residuals to determine the existence of a change point. The latter also focuses on optimal subset selection, which is not the focus of this paper.
In \cite{Qiu1998} on the other hand, one polynomial instead of two is used, and the focus is mainly on detecting $C^0$ and $C^1$ discontinuities. Additionally, the number of change-points must be known a-priori, so only their location is approximated; the required a-priori knowledge make the method unsuitable in real sensor based system observation.
In the spline-based literature there are heuristic methods (top-down and bottom-up) as well as optimization methods. For a more detailed state of the art on splines, see \cite{Dung2017}. Most heuristic methods use a discrete geometric measure to calculate whether a point is a knot, such as: discrete curvature, kink angle, etc, and then use some (mostly arbitrary) threshold to improve the initial knot set. In the method presented here, which falls under the category of bottom-up approaches, the selection criterion is based on calculus and statistics, which allows for incorporation of the fundamental physical laws governing the system, in the model, but also ensures mathematical relevance and rigour.
\subsection{The New Approach}
This paper presents a new approach to detecting $C^n$ discontinuities in observational data. It uses constrained coupled polynomial approximation to obtain two estimates for the $n^\text{th}$ Taylor coefficients and their uncertainties, at every interstitial point. These correspond approximating the local function by polynomials, once from the left $\P{f}(x,\V{\alpha})$ and once from the right $\P{g}(x,\V{\beta})$. The constraints couple the polynomials to ensure that $\alpha_i = \beta_i \,\,\, \text{for every}\, i \in [0 \ldots n-1]$. In this manner the approximations are $C^{n-1}$ continuous at the interstitial points, while delivering an estimate for the difference in the $n^\text{th}$ Taylor coefficients. All the derivations for the coupled constrained approximations and the numerical implementations are presented. Both the approximation and extrapolation residuals are derived. It is proven that the discontinuities must lie at local positive peaks in the extrapolation error. The new approach is verified with both known synthetic data and on real sensor data obtained from observing the operation of heavy machinery.
\section{Detecting $C^n$ Discontinuities}
Discrete observations $s(x_i)$ of a continuous system $s(x)$ are, by their very nature, discontinuous at every sample. Consequently, we will require some measure for discontinuity, with uncertainty, which provides the basis for a statistical hypothesis test.
The observations are considered to be the co-locative points, denoted by $x_i$ and collectively by the vector $\V{x}$; however, we wish to estimate the discontinuity at the interstitial points, denoted by $\zeta_i$ and collectively as $\V{\zeta}$. Using interstitial points, one ensures that each data point is used for only one polynomial approximation at a time. Furthermore, in the case of sensor data, one expects the discontinuities to happen between samples. Consequently the data is segmented at the interstitial points, i.e. between the samples. This requires the use of interpolating functions and in this work we have chosen to use polynomials.
Polynomials have been chosen because of their approximating, interpolating and extrapolating properties when modelling continuous systems: The Weierstrass approximation theorem \cite{Weierstrass1885} states that if $f(x)$ is a continuous real-valued function defined on the real interval $x \in [a, b]$, then for every $\epsilon > 0$, there exists a polynomial $p(x)$ such that for all $x \in [a, b]$, the supremum norm $\|f(x) - p(x)\|_{\infty} < \epsilon$. That is \emph{any} function $f(x)$ can be approximated by a polynomial to an arbitrary accuracy $\epsilon$ given a sufficiently high degree.
The basic concept (see Figure~\ref{fig:ContinuityDef}) to detect a $C^n$ discontinuity is: to approximate the data to the left of an interstitial point by the polynomial $\P{f}(x,\V{\alpha})$ of degree $d_L$ and to the right by $\P{g}(x,\V{\beta})$ of degree $d_R$, while constraining these approximations to be $C^{n-1}$ continuous at the interstitial point. This approximation ensures that,
\begin{equation}
\label{eqn.constraints}
\P{f}^{(k-1)}(\zeta_i) = \P{g}^{(k-1)}(\zeta_i),
\hspace{2ex}
\text{for every} \hspace{1ex} k\, \in \left[1 \ldots n\right].
\end{equation}
while yielding estimates for $\P{f}^{(n)}(\zeta_i)$ and $\P{g}^{(n)}(\zeta_i)$ together with estimates for their variances $\lambda_{f(\zeta_i)}$ and $\lambda_{g(\zeta_i)}$. This corresponds exactly to estimating the Taylor coefficients of the function twice for each interstitial point, i.e., once from the left and once from the right. It they differ significantly, then the function's $n^\text{th}$ derivative is discontinuous at this point.
\begin{figure}[h]
\centering
\includegraphics{figures/ContinuityDefFull.pdf}
\caption{Schematic of a finite set of discrete observations (dotted circles) of a continuous function. The span of the observation is split into a left and right portion at the interstitial point (circle), with lengths $l_L$ and $l_R$ respectively. The left and right sides are considered to be the functions $f(x)$ and $g(x)$; modelled by the polynomials $\P{f}(x,\V{\alpha})$ and $\P{g}(x,\V{\beta})$ of degrees $d_L$ and $d_R$.}
\label{fig:ContinuityDef}
\end{figure}
The Taylor series of a function $f(x)$ around the point $a$ is defined as,
\begin{equation}
\label{eqn.Taylor}
f(x) = \sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(a\right)}{k!}\left(x-a\right)^k
\end{equation}
for each $x$ for which the infinite series on the right hand side converges. Furthermore, any function which is $n+1$ times differentiable can be written as
\begin{equation}
f(x) = \tilde{\P{f}}(x) + R(x)
\end{equation}
where $\tilde{\P{f}}(x)$ is an $n^\text{th}$ degree polynomial approximation of the function $f(x)$,
\begin{equation}
\tilde{\P{f}}(x) = \sum_{k=0}^{n}\frac{f^{\left(k\right)}\left(a\right)}{k!}\left(x-a\right)^k
\end{equation}
and $R(x)$ is the remainder term. The Lagrange form of the remainder $R(x)$ is given by
\begin{equation}
R(x) = \frac{f^{\left(n+1\right)}\left(\xi\right)}{\left(n+1\right)!}\left(x-a\right)^{n+1}
\end{equation}
where $\xi$ is a real number between $a$ and $x$.
A Taylor expansion around the origin (i.e. $a = 0$ in Equation~\ref{eqn.Taylor}) is called a Maclaurin expansion; for more details, see~\cite{Burden1989}. In the rest of this work, the $n^\text{th}$ Maclaurin coefficient for the function $f(x)$ will be denoted by
\begin{equation}
t_{f}^{(n)} \defas \frac{f^{\left(n\right)}\left(0\right)}{n!}.
\end{equation}
The coefficients of a polynomial $\P{f}(x,\V{\alpha}) = \alpha_n x^n + ... +\alpha_1 x + \alpha_0$ are closely related to the coefficients of the Maclaurin expansion of this polynomial. Namely, it's easy to prove that
\begin{equation}
\alpha_k = t_\P{f}^{(k)}, \hspace{2ex}
\text{for every} \hspace{1ex} k\, \in \left[0 \ldots n\right].
\end{equation}
A prudent selection of a common local coordinate system, setting the interstitial point as the origin, ensures that the coefficients of the left and right approximating polynomials correspond to the derivative values at this interstitial point. Namely, one gets a very clear relationship between the coefficients of the left and right polynomial approximations, $\V{\alpha}$ and $\V{\beta}$, their Maclaurin coefficients, $t_{\P{f}}^{(n)}$ and $t_{\P{g}}^{(n)}$, and the values of the derivatives at the interstitial point
\begin{equation}
\label{eqn.TaylorVSPoly}
t_{\P{f}}^{(n)} = \alpha_n = \frac{\P{f}^{\left(n\right)}\left(0\right)}{n!}
\hspace{4ex}
\text{and}
\hspace{4ex}
t_{\P{g}}^{(n)} = \beta_n = \frac{\P{g}^{\left(n\right)}\left(0\right)}{n!}.
\end{equation}
From equation~\ref{eqn.TaylorVSPoly} it is clear that performing a left and right polynomial approximation at an interstitial point is sufficient to get the derivative values at that point, as well as their uncertainties.
\section{Constrained and Coupled Polynomial Approximation}
\label{CoupledPoly}
The goal here is to obtain $\Delta t_{\P{fg}}^{\left(n\right)} \defas t_{\P{f}}^{\left(n\right)} - t_{\P{g}}^{\left(n\right)}$ via polynomial approximation. To this end two polynomial approximations are required; whereby, the interstitial point is used as the origin in the common coordinate system, see Figure~\ref{fig:ContinuityDef}. The approximations are coupled~\cite{olearyDirect2006} at the interstitial point by constraining the coefficients such that $\alpha_i = \beta_i, \, \text{for every} \, i \in [0\ldots n-1]$. This ensures that the two polynomials are $C^{n-1}$ continuous at the interstitial points. This also reduces the degrees of freedom during the approximation and with this the variance of the solution is reduced. For more details on constrained polynomial approximation see~\cite{KlopfensteinConditional1964,olearyConstrained2019}.
To remain fully general, a local polynomial approximation of degree $d_L$ is performed to the left of the interstitial point with the support length $l_L$ creating $\P{f}(x,\V{\alpha})$; similarly to the right $d_R$, $l_R$, $\P{g}(x,\V{\beta})$. The $x$ coordinates to the left, denoted as $\V{x}_L$ are used to form the left Vandermonde matrix $\M{V}_L$, similarly $\V{x}_R$ form $\M{V}_R$ to the right. This leads to the following formulation of the approximation process,
\begin{equation}
\V{y}_L = \M{V}_L \, \V{\alpha}
\hspace{5mm}
\text{and}
\hspace{5mm}
\V{y}_R = \M{V}_R \, \V{\beta}.
\end{equation}
\begin{equation}
\begin{bmatrix}
\M{V}_L & \M{0} \\
\M{0} & \M{V}_R
\end{bmatrix}
\,
\begin{bmatrix}
\V{\alpha} \\
\V{\beta}
\end{bmatrix}
=
\begin{bmatrix}
\V{y}_L \\
\V{y}_R
\end{bmatrix}
\end{equation}
A $C^{n-1}$ continuity implies $\alpha_i = \beta_i, \,\text{for every}\, i \in [0\ldots n-1]$ which can be written in matrix form as
\begin{equation}
\left[
\arraycolsep=4pt
\begin{array}{cc|cc}
\M{0}& \M{I}_{n-1}& \M{0}& - \M{I}_{n-1}
\end{array}
\right]
\,
\begin{bmatrix}
\V{\alpha} \\
\V{\beta}
\end{bmatrix}
=
\V{0}
\end{equation}
Defining
\begin{equation}\notag
\M{V} \defas
\begin{bmatrix}
\M{V}_L & \M{0} \\
\M{0} & \M{V}_R
\end{bmatrix},
\,
\V{\gamma} \defas
\begin{bmatrix}
\V{\alpha} \\
\V{\beta}
\end{bmatrix},
\,
\V{y} \defas
\begin{bmatrix}
\V{y}_L \\
\V{y}_R
\end{bmatrix}
\,
\text{and}
\,
\M{C} \defas
\left[
\arraycolsep=2pt
\begin{array}{cc|cc}
\M{0}& \M{I}_{n-1}& \M{0}& - \M{I}_{n-1}
\end{array}
\right]
\end{equation}
We obtain the task of least squares minimization with homogeneous linear constraints,
\begin{empheq}[box=\widefbox]{align}
\min_{\V{\gamma}} \hspace{5mm} & \| \V{y} - \M{V} \, \V{\gamma} \|_2^2 \notag\\
\text{Given} \hspace{5mm} & \M{C} \, \V{\gamma} = \V{0}.\label{eqn.min1}
\end{empheq}
Clearly $\V{\gamma}$ must lie in the null-space of $\M{C}$; now, given $\M{N}$, an ortho-normal vector basis set for $\nullS{\M{C}}$, we obtain,
\begin{equation}
\V{\gamma} = \M{N} \, \V{\delta}.
\end{equation}
Back-substituting into Equation~\ref{eqn.min1} yields,
\begin{equation}
\min_{\V{\delta}} \| \V{y} - \M{V} \, \M{N} \, \V{\delta} \|_2^2
\end{equation}
The least squares solution to this problem is,
\begin{equation}
\V{\delta} = \left( \M{V} \, \M{N} \right)^+ \, \V{y},
\end{equation}
and consequently,
\begin{empheq}[box=\widefbox]{align}
\V{\gamma} = \begin{bmatrix}
\V{\alpha}\\
\V{\beta}
\end{bmatrix} =
\M{N} \, \left( \M{V} \, \M{N} \right)^+ \, \V{y}
\end{empheq}
Formulating the approximation in the above manner ensures that the difference in the Taylor coefficients can be simply computed as
\begin{equation}
\Delta t_{\P{fg}}^{\left(n\right)} = t_{\P{f}}^{\left(n\right)} - t_{\P{g}}^{\left(n\right)} = \alpha_n = \beta_n.
\end{equation}
Now defining $\V{d} = [1, \, \V{0}_{d_L -1}, \, -1, \, \V{0}_{d_R -1}]^\mathrm{T}$, $\Delta t_{\P{fg}}^{\left(n\right)}$ is obtained from $\V{\gamma}$ as
\begin{equation}
\label{eqn.DeltaTaylor}
\Delta t_{\P{fg}}^{\left(n\right)} = \VT{d}\V{\gamma} = \VT{d}\M{N} \, \left( \M{V} \, \M{N} \right)^+ \, \V{y}.
\end{equation}
\subsection{Covariance Propagation}
Defining, $\M{K} = \M{N} \, \left( \M{V} \, \M{N} \right)^+$, yields,
$\V{\gamma} = \M{K} \, \V{y}$.
Then given the covariance of $\V{y}$, i.e., $\M{\Lambda}_{\V{y}}$, one gets that,
\begin{empheq}[box=\widefbox]{align}
\M{\Lambda}_{\V{\gamma}} = \M{K} \, \M{\Lambda}_{\V{y}} \, \MT{K}.
\end{empheq}
Additionally, from equation~\ref{eqn.DeltaTaylor} one could derive the covariance of the difference in the Taylor coefficients
\begin{equation}
\M{\Lambda}_{\V{\Delta}} = \V{d}\M{\Lambda}_{\V{\gamma}}\VT{d}
\end{equation}
Keep in mind that, if one uses approximating polynomials of degree $n$ to determine a discontinuity in the $n^{\text{th}}$ derivative, as done so far, $\M{\Lambda}_{\V{\Delta}}$ is just a scalar and corresponds to the variance of $\Delta t_{\P{fg}}^{\left(n\right)}$.
\section{Error Analysis}
\label{Residuals}
In this paper we consider three measures for error:
\begin{enumerate}
\item the norm of the approximation residual;
\item the combined approximation and extrapolation error;
\item the extrapolation error.
\end{enumerate}
\subsection{Approximation Error}
The residual vector has the form
\begin{equation*}
\V{r} =\V{y}-\M{V}\V{\gamma} =\begin{bmatrix}
\V{y}_L-\M{V}_L\V{\alpha}\\
\V{y}_R-\M{V}_R\V{\beta}
\end{bmatrix}.
\end{equation*}
The approximation error is calculated as
\begin{align*}
&E_a = \| \V{r}\|_2^2 = \| \V{y}_L-\M{V}_L\V{\alpha}\|_2^2 + \| \V{y}_R-\M{V}_R\V{\beta}\|_2^2\\
= &\left(\V{y}_L-\M{V}_L\V{\alpha}\right)^\mathrm{T}\left(\V{y}_L-\M{V}_L\V{\alpha}\right) + \left(\V{y}_R-\M{V}_R\V{\beta}\right)^\mathrm{T}\left(\V{y}_R-\M{V}_R\V{\beta}\right)\\
= &\VT{y}\V{y} - 2\VT{\alpha}\MT{V}_L\V{y}_L + \VT{\alpha}\MT{V}_L\M{V}_L\V{\alpha} - 2\VT{\beta}\MT{V}_R\V{y}_R + \VT{\beta}\MT{V}_R\M{V}_R\V{\beta}.
\end{align*}
\begin{figure}[t]
\centering
\includegraphics{figures/taylorPolyApprox.pdf}
\caption{Schematic of the approximations around the interstitial point. Red: left polynomial approximation $\P{f}(x,\V{\alpha})$; dotted red: extrapolation of $\P{f}(x,\V{\alpha})$ to the RHS; blue: right polynomial approximation, $\P{g}(x,\V{\beta})$; dotted blue: extrapolation of $\P{g}(x,\V{\beta})$ to the LHS; $\epsilon_i$ is the vertical distance between the extrapolated value and the observation. The approximation is constrained with the conditions: $\P{f}(0,\V{\alpha}) = \P{g}(0,\V{\beta})$ and $\P{f}'(0,\V{\alpha}) = \P{g}'(0,\V{\beta})$.}
\label{fig:taylorPolyApprox}
\end{figure}
\subsection{Combined Error}
The basic concept, which can be seen in Figure~\ref{fig:taylorPolyApprox}, is as follows: the left polynomial $\P{f}\left(x,\V{\alpha}\right)$, which approximates over the values $\V{x}_L$, is extended to the right and evaluated at the points $\V{x}_R$. Analogously, the right polynomial $\P{g}\left(x,\V{\beta}\right)$ is evaluated at the points $\V{x}_L$. If there is no $C^n$ discontinuity in the system, the polynomials $\P{f}$ and $\P{g}$ must be equal and consequently the extrapolated values won't differ significantly from the approximated values.
\subsubsection{Analytical Combined Error}
The extrapolation error in a continuous case, i.e. between the two polynomial models, can be computed with the following $2$-norm,
\begin{equation}
\epsilon_x =
\int_{x_{min}}^{x_{max}}
\left\{ \P{f}(x,\V{\alpha}) - \P{g}(x,\V{\beta})
\right\}^2 \, \md x.
\end{equation}
Given, the constraints which ensure that $\alpha_i = \beta_i \, i \in [0,\ldots,n-1]$, we obtain,
\begin{equation}
\epsilon_x =
\int_{x_{min}}^{x_{max}}
\left\{ (\alpha_{n} - \beta_{n} ) \, x^{n}
\right\}^2 \, \md x.
\end{equation}
Expanding and performing the integral yields,
\begin{equation}
\epsilon_x = (\alpha_{n} - \beta_{n})^2 \, \left\{
\frac{x_{max}^{2n + 1} - x_{min}^{2n + 1}}
{2n + 1} \right\}
\end{equation}
Given fixed values for $x_{min}$ and $x_{max}$ across a single computation implies that the factor,
\begin{equation}
k =
\frac{x_{max}^{2n + 1} - x_{min}^{2n + 1}}
{2n + 1}
\end{equation}
is a constant. Consequently, the extrapolation error is directly proportional to the square of the difference in the Taylor coefficients,
\begin{equation}
\epsilon_x \propto \, \left(\alpha_n - \beta_n\right)^2 \propto \, \left\{\Delta t_{\P{fg}}^{\left(n\right)}\right\}^2.
\end{equation}
\subsubsection{Numerical Combined Error}
In the discrete case, one can write the errors of $\P{f}(x,\V{\alpha})$ and $\P{g}(x,\V{\beta})$ as
\begin{equation}
\V{e}_\P{f} = \V{y}-\P{f}(\V{x},\V{\alpha}) \hspace{2ex} \text{and} \hspace{2ex}
\V{e}_\P{g} = \V{y}-\P{g}(\V{x},\V{\beta})
\end{equation}
respectively. Consequently, one could define an error function as
\begin{align}
&E_{\P{f}\P{g}} = \|\V{e}_\P{f}-\V{e}_\P{g}\|_2^2 =
\|(a_{n} - b_{n} ) \, \V{z}\|_2^2 =
(a_{n} - b_{n} )^2 \VT{z} \V{z}^{n} =
(a_{n} - b_{n} )^2\sum {x_i^{n}}
\end{align}
where $\V{z} \defas \V{x} .\, \hat{} \,\, n$. From these calculations it is clear that in the discrete case the error is also directly proportional to the square of the difference in the Taylor coefficients and that $ E_{\P{f}\P{g}} \propto \epsilon_x$. This proves that the numerical computation is consistent with the analytical continuous error.
\subsection{Extrapolation Error}
One could also define a different kind of error, based just on the extrapolative properties of the polynomials. Namely, using the notation from the beginning of Section~\ref{CoupledPoly}, one defines
\begin{equation*}
\V{r}_{e\P{f}} = \V{y}_L-\P{g}(\V{x}_L,\V{\beta}) = \V{y}_L-\M{V}_L\V{\beta}
\hspace{2ex} \text{and} \hspace{2ex}
\V{r}_{e\P{g}} = \V{y}_R-\P{f}(\V{x}_R,\V{\alpha}) = \V{y}_R-\M{V}_R\V{\alpha}
\end{equation*}
and then calculates the error as
\begin{align*}
&E_{e} = \VT{r}_{e\P{f}}\V{r}_{e\P{f}} + \VT{r}_{e\P{g}}\V{r}_{e\P{g}}\\
=&\left(\V{y}_L-\M{V}_L\V{\beta}\right)^\mathrm{T} \left(\V{y}_L-\M{V}_L\V{\beta}\right) + \left(\V{y}_R-\M{V}_R\V{\alpha}\right)^\mathrm{T} \left(\V{y}_R-\M{V}_R\V{\alpha}\right)\\
=&\VT{y}\V{y} -2\VT{\beta}\MT{V}_L \V{y}_L + \VT{\beta}\MT{V}_L\M{V}_L\V{\beta} - 2\VT{\alpha}\MT{V}_R\V{y}_R + \VT{\alpha}\MT{V}_R\M{V}_R\V{\alpha}.
\end{align*}
In the example in section~\ref{sec.examples}, it will be seen that there is no significant numerical difference between these two errors.
\section{Numerical Testing}
\label{sec.examples}
The numerical testing is performed with: synthetic data from a piecewise polynomial, where the locations of the $C^n$ discontinuities are known; and with real sensor data emanating from the monitoring of heavy machinery.
\subsection{Synthetic Data}
In the literature on splines, functions of the type $y\left(x\right) = e^{-x^2}$ are commonly used. However, this function is analytic and $C^{\infty}$ continuous; consequently it was not considered a suitable function for testing. In Figure~\ref{PieceWisePoly} a piecewise polynomial with a similar shape is shown; however, this curve has $C^2$ discontinuities at known locations.
\begin{figure}[h]
\centering
\includegraphics{figures/PieceWisePolyCn.pdf}
\caption{A piecewise polynomial of degree $d=2$, created from the
knots sequence $\V{x}_k = [0, 0.3, 0.7, 1]$ with the corresponding
values $\V{y}_k = [0, 0.3, 0.7, 1]$. The end points are clamped with
$y'(x)_{0,1} = 0$. Gaussian noise is added with $\sigma = 0.05$. Top:
the circles mark the known points of $C^2$ discontinuity; the blue and
red lines indicate the detected discontinuities; additionally the data
has been approximated by the b-spline (red) using the detected
discontinuities as knots. Bottom: shows $\Delta t^{(n)}_{\P{fg}} = t^{(n)}_\P{f} -
t^{(n)}_\P{g}$, together with the two identified peaks.}
\label{PieceWisePoly}
\end{figure}
The algorithm was applied to the synthetic data from the piecewise polynomial, with added noise with $\sigma = 0.05$ and the results for a single case can be seen in Figure~\ref{PieceWisePoly}. Additionally, a Monte Carlo simulation with $m=10000$ iterations was performed and the results of the algorithm were compared to the true locations of the two known knots. The mean errors in the location of the knots are: $\mu_1 = (5.59 \pm
2.05) \times 10^{-4}$ with $95 \%$ confidence, and $\mu_2 = (-4.62 \pm 1.94)\times 10^{-4}$. Errors in the scale of $10^{-4}$, in a support with a range $[0,\,1]$, and $5 \%$ noise amplitude in the curve can be considered a highly satisfactory result.
\subsection{Sensor Data}
The algorithm was also applied to a set of real-world sensor data\footnote{For confidentiality reasons the data has been anonymized.} emanating from the monitoring of heavy machinery. The original data set can be seen in Figure~\ref{fig:taylorCnDisKeller} (top). It has many local peaks and periods of little or no change, so the algorithm was used to detect discontinuities in the first derivative, in order to determine the peaks and phases.
\begin{figure}[h]
\centering
\includegraphics{figures/taylorCnDisKeller.pdf}
\caption{The top-most graph shows a function $y(x)$, together with the detected $C^1$ discontinuity points. The middle graph shows the difference in the Taylor polynomials $\Delta t_{\P{fg}}^{\left(n\right)}$ calculated at every interstitial point. The red and blue circles mark the relevant local maxima and minima of the difference respectively. According to this, the red and blue lines are drawn in the top-most graph. The bottom graph shows the approximation error evaluated at every interstitial point.}
\label{fig:taylorCnDisKeller}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics{figures/taylorCnDisKellerErrors.pdf}
\caption{The two error functions, $E_e$ and $E_{\P{f}\P{g}}$ as defined in Section \ref{Residuals}, for the example from Fig.~\ref{fig:taylorCnDisKeller}. One can see that the location of the peaks doesn't change, and the two errors don't differ significantly.}
\label{fig:taylorCnDisKellerErrors}
\end{figure}
The peaks in the Taylor differences were used in combination with the peaks of the extrapolation error to determine the points of discontinuity. A peak in the Taylor differences means that the Taylor coefficients are significantly different at that interstitial point, compared to other interstitial points in the neighbourhood. However, if there is no peak in the extrapolation errors at the same location, then the peak found by the Taylor differences is deemed insignificant, since one polynomial could model both the left and right values and as such the peak isn't a discontinuity.
Additionally, it can be seen in Figure~\ref{fig:taylorCnDisKellerErrors} that both the extrapolation error and the combined error, as defined in Section~\ref{Residuals}, have peaks at the same locations, and as such the results they provide do not differ significantly.
\section{Conclusion and Future Work}
It may be concluded, from the results achieved, that the coupled constrained polynomial approximation yield a good method for the detection of $C^n$ discontinuities in discrete observational data of continuous systems.
Local peaks in the square of the difference of the Taylor polynomials provide a relative measure as a means of determining the locations of discontinuities.
Current investigations indicate that the method can be implemented directly as a convolutional operator, which will yield a computationally efficient solution. The use of discrete orthogonal polynomials~\cite{Persson2003Smoothing,oleary2010Discrete} is being tested as a means of improving the sensitivity of the results to numerical perturbations.
\section*{Acknowledgements}
This work was partially funded by:
\begin{enumerate}
\item The COMET program within the K2 Center “Integrated Computational Material, Process and Product Engineering (IC-MPPE)” (Project No 859480). This program is supported by the Austrian Federal Ministries for Transport, Innovation and Technology (BMVIT) and for Digital and Economic Affairs (BMDW), represented by the Austrian research funding association (FFG), and the federal states of Styria, Upper Austria and Tyrol.
\item The European Institute of Innovation and Technology (EIT), a body of the European Union which receives support from the European Union's Horizon 2020 research and innovation programme. This was carried out under Framework Partnership Agreement No. 17031 (MaMMa - Maintained Mine \& Machine).
\end{enumerate}
The authors gratefully acknowledge this financial support.
\bibliographystyle{splncs04}
\bibliography{bibliography}
\end{document} | 215,763 |
TITLE: How do you prove that $\int_0^\frac{\pi}{2}\frac{\sin^mx}{\sin^mx+\cos^mx}\mathrm dx=\frac{\pi}{4}$
QUESTION [4 upvotes]: How do you prove that for all real $m$, $$\int_0^\frac{\pi}{2}\frac{\sin^mx}{\sin^mx+\cos^mx}\mathrm dx=\frac{\pi}{4}$$
I've let $y=\frac{\pi}{2}-x$ and got
$$\int_0^\frac{\pi}{2}\frac{\sin^mx}{\sin^mx+\cos^mx}\mathrm dx= \int_\frac{\pi}{2}^0\frac{\cos^my}{\cos^my+\sin^my}\mathrm dy = \int_\frac{\pi}{2}^0\frac{\cos^mx}{\cos^mx+\sin^mx}\mathrm dx $$
This is as far as I got. I know you're supposed to add the 2 results together, but the limits of integrations aren't the same. Adding both together I get,
$$\int_0^\frac{\pi}{2}\frac{\sin^mx-\cos^mx}{\sin^mx+\cos^mx}\mathrm dx$$ which doesn't simplify at all.
Thanks for all the help in advance.
REPLY [5 votes]: Your substiution is pretty close to the one needed. Consider $\frac{\pi}{2}-x=y\rightarrow dx=-dy\,$. The new bounds are $y_1=\frac{\pi}{2}-0=\frac{\pi}{2} \,$and $y_2=\frac{\pi}{2}-\frac{\pi}{2}=0$ $$I=\int^0_\frac{\pi}{2}\frac{\cos^my}{\cos^my+\sin^my}\mathrm (-dy)=\int_0^\frac{\pi}{2}\frac{\cos^my}{\cos^my+\sin^my}\mathrm dy$$
Add this with the initial integral in order to get: $$2I=\int_0^\frac{\pi}{2}\frac{\sin^mx+\cos^{m}x}{\sin^mx+\cos^mx}\mathrm dx\rightarrow I=\frac12\int_0^\frac{\pi}{2}dx=\frac{\pi}{4}$$ | 84,233 |
Stop Norton Antivirus Blocking Sites – Norton Antivirus shields your framework from the personality or information burglary while utilizing the web. You can secure you by hindering some noxious sites while utilizing the web. Likewise, you can include a few special cases in the Smart Firewall and Norton family and give them access to these sites. Here in this article, we will talk about the strategy to Stop Norton Antivirus Blocking Sites.
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TITLE: Probabilities of Survival
QUESTION [0 upvotes]: So I have these estimation of survival probabilities for men:
The probability that a man lives at least $70$ years is $80\%$.
The probability that a man lives at least $80$ years is $60\%$.
The probability that a $80$ years old man lives at least $90$ years is $40\%$.
The first question was what is the probability that a man lives at least $80$ if he has just celebrated his $70^{th}$ birthday?
What I did : I let $A$ be a man who reached $70$, and let $B$ be a man who reached $80$
$$P(A)=0.80$$
$$P(B)=0.60$$
$$P(B|A)=\frac{P(A \cap B)}{P(A)}=\frac{P(B)}{P(A)}=\frac{0.60}{0.80}=0.75$$
My question is what is the probability that this man lives at least
$90$ years?
REPLY [0 votes]: Following your notation, let $C$ be the unconditional probability of living to $90$ years. Then you are given $$\Pr[C \mid B] = 0.4,$$ that is to say, given that a man has lived to $80$ years, the probability of living to $90$ is $0.4$. Then what you want to find is $$\Pr[C \mid A] = \frac{\Pr[C \cap A]}{\Pr[A]} = \frac{\Pr[C]}{\Pr[A]}.$$ But $$\Pr[C \mid B] = \frac{\Pr[C \cap B]}{\Pr[B]} = \frac{\Pr[C]}{\Pr[B]},$$ so $$\Pr[C] = \Pr[C \mid B]\Pr[B]$$ and $$\Pr[C \mid A] = \frac{\Pr[C]}{\Pr[A]} = \frac{\Pr[C \mid B]\Pr[B]}{\Pr[A]}.$$ Now you can substitute all the values you have to get the answer. | 90,674 |
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Former Southampton defender Francis Benali has told talkSPORT Virgil van Dijk is good enough to play for any club in the world, and that he’d be a huge loss for the Saints if he leaves this summer.
Centre-back Van Dijk is one of the most in-demand players in the Premier League ahead of the summer transfer window.
The likes of Liverpool, Manchester United, Manchester City, Chelsea and Arsenal have all been linked with the 25-year-old Dutchman as they all look to strengthen in defence.
Saints boss Claude Puel has recently hinted that Van Dijk could end up staying at St Mary’s next season, with club bosses keen to keep him and under no pressure to sell.
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And Benali says Southampton would be left with a gaping hole in their squad if the ‘flawless’ defender becomes the latest Saints star to move to Anfield.
Speaking on the Alan Brazil Sports Breakfast, the former full-back said: “It’s been a familiar picture at Southampton for a few seasons now, and Virgil is the obvious player who is being talked about now.
“He is absolutely class. He’s a player I could see slotting into any team in the world, with the performances I’ve seen. There’s nothing he can’t do.
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As may become clear in the next post, (I was employed as a journalist once upon a time, so I know the merit of a snappy intro or even in this case A Snappy Nappy intro.) I have been pondering upon those occasions where through contact with an artefact, a battle field visit, or a work of art one brushes against the past in an almost tangible form.
I'm sorry, I'll start again.
Most of my wargaming life has been taken up with the Napoleonic Wars through reading, painting and playing games. As a result I have a strong though possibly slanted perception of that period, but I have been trying to lay that aside and consider in what ways that history impinged on me as a child living in a small Welsh market town. In effect, pre wargaming.
It is constructed entirely from carved mutton bones presumably retrieved from the soup or stew. It is a working model and as soon as I grasped its purpose I was fascinated by it. What the unknown prisoner was thinking when he spent so many hours carving it is not recorded. Possibly he was dreaming of the day when the real thing would be set up on the square outside the Royal Oak and bring Equality, Liberty and Fraternity to Mid Wales.
Much more civilised is the next, a story told to me by my Grandmother. How the French Officers were allowed to wonder about the town having given their parole, the only stipulation was that they had to swear not to pass beyond the parish boundary marker. They resolved this by walking freely where they wished but they were always preceeded by a hired local carrying the uprooted boundary maker in a wheel barrow. I suspect this story is so good that it is told about other parts of the UK Nevertheless it was told me by Gran who was not a lady widely read in Napoleonic memoirs.
Then there was the Sergeants Row, these canal side cottages were still standing in my infancy. They were made available by public subscription as housing for NCO's who were present at the Battle of Waterloo. This was spoken matter of factly as if the Sergeants had just slipped out for a pint and would soon be back home.
So without banging on about Methodism or free trade laws or God forbid, demographics, it is surprising that these little contacts still existed 150 years later. Which will bring me on to what I am really thinking about currently, wargaming the Great War.
2 comments:
I remember borrowing 'A Near Run Thing' from the school library ,it was probably the first grown up history book I read !
I can still remember the smell of that book from when I used to borrow it from my little local library as a youngster... at the time Napoleonic (and WWII to be fair) were my staple diet.... happy days.... | 321,256 |
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Diana Scharf-Hunt once said, “goals are dreams with deadlines.” Well, we know your goal is to learn SwiftUI because that’s why you’re here, so it’s time to add a deadline too: your mission today is to complete three animation challenges that will really put your skills – and your creativity! – to the test.
As you’ll see, one of the challenges today specifically asks you to be creative and try something yourself. You have all the tools now to create a huge variety of animations and transitions, so all that’s holding you back is the opportunity to practice.
Well, this is it – good luck!
Today you should work through the wrap up chapter for project 6, complete its review, then work through all three of its challenges.
I look forward to seeing what you create with the third challenge – make a video and share it on Twitter, making sure to add @twostraws somewhere in there so I see it!
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Deschamps gets lucky as France beat Australia but piles on the pressure
MOSCOW -- France did win vs. Australia, but Didier Deschamps lost.
Les Bleus got the three points they wanted with a 2-1 victory Saturday despite their manager getting pretty much everything wrong. Deschamps' ability to guide his very talented squad all the way through this World Cup had been questioned a lot before the start of the competition, and this is without a doubt the worst start he could have hoped for on a personal level.
Before facing the Socceroos, the pressure was on Deschamps almost more than on his players, which doesn't happen to him too often. Deschamps felt it as well. Deschamps led his team as it entered Kazan Arena, as he always does, but you could see the tension on his face as he arrived, almost as if he knew already then that the plan he had put in place for this match was not the right one.
The pressure was on and it got the best of Deschamps. Instead of pushing away doubts over his management, the French coach put more scrutiny on it.
A first game at a World Cup is never easy. It's the biggest stage of all and vital to start on the right foot. However, when you make it even more complicated by changing your tactics a couple of weeks before and then actually play in a different way on the day of the game, and when you decide to give a World Cup debut to seven of your 11 starters, you're not helping yourself, either.
In short, Deschamps' risky choices backfired on him. He tried to justify it after the game regarding his decision to start a front three of Antoine Griezmann, Ousmane Dembele and Kylian Mbappe by saying that they did well together against Italy two weeks ago. Yes, they did: in a friendly, at home in Lyon, at 9 p.m., against a team with a new manager and still traumatised by its World Cup playoff debacle.
Playing Australia in Kazan, under the midday sun and on a dry pitch against a determined, well-organised and solid team is not the same at all. Even Griezmann admitted after the victory that the lack of game time between the three was an issue. Yet Deschamps thought it was the right call to make.
For three weeks, the players worked on a new 4-3-1-2 formation, replacing the 4-4-2 they used in qualifying. They played their three friendly games like that, with Nabil Fekir and then Griezmann in the No. 10 role. On Saturday, Griezmann started wide in a 4-3-3. There was no No. 10 anymore; instead, it was a bit of a mess up front as Mbappe, Griezmann and Dembele ended up making the same runs, stepping on each other's toes and being totally transparent.
Hugo Lloris hinted after the game that the game plan was not very clear, either. He wasn't sure whether the instructions were to press or to drop deep. As a result, France did neither in the first half. It got a bit better after the break, but overall, it was simply not good enough at this level.
Deschamps had a shocker regarding his team selection. Dembele was certainly not ready to play in a game of this calibre. Corentin Tolisso, preferred to Blaise Matuidi, had a stinker in midfield. Benjamin Pavard, who played at right-back, struggled. The lack of experience of the team -- the average age was 24 years and 6 months, with only three players 26 or older -- was telling, but again, Deschamps went for it. Even his substitutions were disappointing. Fekir and Matuidi didn't bring anything to the game, unlike Olivier Giroud, who set up Paul Pogba's winner.
Lloris, whose analysis of his team's performance was spot-on after the game on Saturday, also asked each player to raise their game now that the World Cup is underway. It could not be truer for Deschamps as well. He has to do better.
Everything is not his fault, of course. The way the French players underestimated Australia is not down to him. He warned his men that the Aussies would be well organised and aggressive, which they were. Samuel Umtiti's stupid handball to gift Australia a way back in this game is not Deschamps' fault, either.
Ultimately, though, the France head coach had a lucky escape and can't afford to make more mistakes like these. Entering Thursday's game against Peru, the pressure on him is greater than. | 257,331 |
Was that dream mostly positive or negative?
Did your positive dream help you in any way to do what you practically could, in order to make that dream come true?
Did your negative dream at the very least warn you of what lay ahead, so that you could better prepare for it and minimize its impact?
Sweet dreams,
Brian
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\begin{document}
\hbox{}
\begin{center}
{\large\bf Inequivalent Vacuum States in Algebraic Quantum Theory}
\bigskip
\bigskip
G. Sardanashvily
\medskip
Department of Theoretical Physics, Moscow State University,
Moscow, Russia
\end{center}
\bigskip
\bigskip
\begin{abstract} The Gelfand--Naimark--Sigal representation
construction is considered in a general case of topological
involutive algebras of quantum systems, including quantum fields,
and inequivalent state spaces of these systems are characterized.
We aim to show that, from the physical viewpoint, they can be
treated as classical fields by analogy with a Higgs vacuum field.
\end{abstract}
\bigskip
\tableofcontents
\bigskip
\section{Introduction}
No long ago, one thought of vacuum in quantum field theory (QFT)
as possessing no physical characteristics, and thus being
invariant under any symmetry transformation. It is exemplified by
a particleless Fock vacuum (Section 8). Contemporary gauge models
of fundamental interactions however have arrived at a concept of
the Higgs vacuum (HV). By contrast to the Fock one, HV is equipped
with nonzero characteristics, and consequently is non-invariant
under transformations.
For instance, HV in Standard Model of particle physics is
represented by a constant background classical Higgs field, in
fact, inserted by hand into a field Lagrangian, whereas its true
physical nature still remains unclear. In particular, somebody
treats it as a {\it sui generis} condensate by analogy with the
Cooper one, and its appearance is regarded as a phase transition
characterized by some symmetry breakdown \cite{banks,bend,kawati}.
Thus, we come to a concept of different inequivalent and, in
particular, non-invariant vacua \cite{sard08a}. Here, we consider
some models of these vacua in the framework of algebraic quantum
theory (AQT). We aim to show that, from the physical viewpoint,
their characteristics are classical just as we observe in a case
of the above-mentioned Higgs vacuum.
In AQT, a quantum system is characterized by a topological
involutive algebra $A$ and a family of continuous positive forms
on $A$. Elements of $A$ are treated as quantum objects, and we
call $A$ the quantum algebra. In this framework, values of
positive forms on $A$ are regarded as numerical averages of
elements of $A$. In the spirit of Copenhagen interpretation of quantum theory, one
can think of positive forms on $A$ as being classical objects.
A corner stone of AQT is the following Gelfand--Naimark--Segal (GNS)
representation theorem \cite{book05,hor,schmu}.
\begin{theorem} \label{gns} \mar{gns}
Let $A$ be a unital topological involutive algebra and $f$ a
positive continuous form on $A$ such that $f(\bb)=1$ (i.e., $f$ is
a state). There exists a strongly cyclic Hermitian representation
$(\pi_f,\thh_f)$ of $A$ in a Hilbert space $E_f$ with a cyclic
vector $\thh_f$ such that
\mar{gns2}\beq
f(a)=\lng \pi(a)\thh_\f|\thh_\f\rng, \qquad a\in A. \label{gns2}
\eeq
\end{theorem}
It should be emphasized that a Hilbert space $E_f$ in Theorem
\ref{gns} is a completion of the quotient of an algebra $A$ with
respect to an ideal generated by elements $a\in A$ such that
$f(aa^*)=0$, and the cyclic vector $\thh_f$ is the image of the
identity $\bb\in A$ in this quotient. Thus, a carrier space of a
representation of $A$ and its cyclic vector in Theorem \ref{gns}
comes from a quantum algebra $A$, and they also can be treated as
quantum objects.
Since $\thh_f$ (\ref{gns2}) is a cyclic vector, we can think of it
as being a vacuum vector and, accordingly, a state $f$ as being a
vacuum of an algebra $A$. Let us note that a vacuum vector
$\thh_f$ is a quantum object, whereas a vacuum $f$ of $A$ is the
classical one. In particular, a quantum algebra $A$ acts on
quantum vectors, but not on its vacua (states).
Since vacua are classical objects, they are parameterized by
classical characteristics. A problem is that different vacua of
$A$ define inequivalent cyclic representations of a quantum
algebra $A$ in general. In this case, they are called
inequivalent.
We say that a quantum algebra $A$ performs a transition between
its vacua $f$ and $f'$ if there exist elements $b,b'\in A$ such
that $f'(a)=f(b^+ab)$ and $f(a)=f'(b'^+ab')$ for all $a\in A$. In
this case cyclic representations $\pi_f$ and $\pi_{f'}$ and,
accordingly, vacua $f$ and $f'$ are equivalent (Theorem
\ref{spr474'}). A problem thus is to characterize inequivalent
vacua of a quantum system.
One can say something in the following three variants.
(i) If a quantum algebra $A$ is a unital $C^*$-algebra, Theorem
\ref{gns} comes to well-known GNS (Theorem \ref{spr471}), and we
have a cyclic representation of $A$ by bounded operators in a
Hilbert space (Section 2). This is a case of quantum mechanics.
(ii) A quantum algebra $A$ is a nuclear involutive algebra
(Theorem \ref{gns36}). In particular, this is just the case of
quantum field theory (Sections ?9 and 10).
(iii) Given a group $G$ of automorphisms of a quantum algebra $A$,
its vacuum is $f$ invariant only under a proper subgroup of $G$.
This is the case of spontaneous symmetry breaking in a quantum
system (Sections 12 and 13).
If a quantum algebra $A$ is a unital $C^*$-algebra, one can show
that a set $F(A)$ of states of $A$ is a weakly$^*$-closed convex
hull of a set $P(A)$ of pure states of $A$, and it is weakly*
compact (Theorem \ref{gns17}). A set $P(A)$ of pure states of $A$,
in turn, is a topological bundle over the spectrum $\wh A$ of $A$
whose fibres are projective Hilbert space. The spectrum $\wh A$ of
$A$ is a set of its nonequivalent irreducible representations
provided with the inverse image of the Jacobson topology. It is
quasi-compact.
In accordance with Theorem \ref{spr474} a unital $C^*$-algebra
$A$ of a quantum system performs invertible transitions between
different vacua iff they are equivalent. At the same time, one can
enlarge an algebra $A$ to some algebra $B(E_F)$ so that all states
of $A$ become equivalent states of $B(E_F)$ (Theorem \ref{gns23}).
Moreover, this algebra contains the superselection operator $T$
(\ref{gns24}) which belongs to the commutant of $A$ and whose
distinct eigenvalues characterize different vacua of $A$.
In Section 4, an infinite qubit system modelled on an arbitrary
set $S$ is studied. Its quantum $C^*$-algebra $A_S$ possesses pure
states whose set is a set of maps $\si$ (\ref{gns91}) of a set $S$
to the unit sphere in $\mathbb C^2$. They are equivalent iff the
relation (\ref{gns67}) is satisfied and, in particular, if maps
$\si$ and $\si'$ differ from each other on a finite subset of $S$.
By analogy with a Higgs vacuum, one can treat the maps $\si$
(\ref{gns91}) as classical vacuum fields.
In Section 5, we consider an example of a locally compact group
$G$ and its group algebra $L^1_{\mathbb C}(G)$ of equivalence
classes of complex integrable functions on $G$. This is a Banach
involutive algebra with an approximate identity. There is
one-to-one correspondence between the representations of this
algebra and the strongly continuous unitary representations of a
group $G$ (Theorem \ref{gns61}). Continuous positive forms on
$L^1_{\mathbb C}(G)$ and, accordingly, its cyclic representations
are parameterized by continuous positive-definite functions $\psi$
on $G$ as classical vacuum fields (Theorem \ref{qm539}). If $\psi$
is square-integrable, the corresponding cyclic representation of
$L^1_{\mathbb C}(G)$ is contained in the regular representation
(\ref{qm344}). In this case, distinct square integrable continuous
positive-definite functions $\psi$ and $\psi'$ on $G$ define
inequivalent irreducible representations if they obey the
relations (\ref{qm345}).
However, this is not the case of unnormed topological
$*$-algebras. In order to say something, we restrict our
consideration to nuclear algebras (Section 6 and 7).
This technique is applied to the analysis of inequivalent
representations of infinite canonical commutative relations
(Section 8) and, in particular, free quantum fields, whose states
characterized by different masses are inequivalent.
Section 10 addresses the true functional integral formulation of
Euclidean quantum theory (Section 10). These integrals fail to be
translationally invariant that enables one to model a Higgs vacuum
a translationall inequivalent state (Section 11).
Sections 12 and 13 are devoted to the phenomenon of spontaneous
symmetry breaking when a state of a quantum algebra $A$ fails to
be stationary only with respect to some some proper subgroup $H$
of a group $G$ of automorphisms of $A$. Then a set of inequivalent
states of these algebra generated by these automorphisms is a
subset of the quotient $G/H$.
In particular, just this fact motivates us to describe classical
Higgs fields as sections of a fibre bundle with a typical fibre
$G/H$ \cite{higgs,sard08a,tmp}.
\section{GNS construction. Bounded operators}
We start with a GNS representation of a topological involutive
algebra $A$ by bounded operators in a Hilbert space. This is the
case of Banach involutive algebras with an approximate identity
(Theorem \ref{spr471}). Without a loss of generality, we however
restrict our consideration to GNS representations of
$C^*$-algebras because any involutive Banach algebra $A$ with an
approximate identity defines the enveloping $C^*$-algebra
$A^\dagger$ such that there is one-to-one correspondence between
the representations of $A$ and those of $A^\dagger$ (Remark
\ref{env}).
Let us recall the standard terminology \cite{dixm,book05}. A
complex associative algebra $A$ is called involutive (a
$^*$-algebra) if it is provided with an involution $*$ such that
\be
(a^*)^*=a, \quad (a +\la b)^*=a^* + \ol\la b^*, \quad
(ab)^*=b^*a^*, \quad a,b\in A, \quad \la\in \mathbb C.
\ee
An element $a\in A$ is normal if $aa^*=a^*a$, and it is Hermitian
or self-adjoint if $a^*=a$. If $A$ is a unital algebra, a normal
element such that $aa^*=a^*a=\bb$ is called the unitary one.
A $^*$-algebra $A$ is called the normed algebra (resp. the Banach
algebra) if it is a normed (resp. complete normed) vector space
whose norm $\|.\|$ obeys the multiplicative conditions
\be
\| ab\| \leq\| a\|\| b\|, \qquad \| a^*\|=\| a\|, \qquad a,b\in A.
\ee
A Banach $^*$-algebra $A$ is said to be a $C^*$-algebra if
$\|a\|^2= \| a^* a\|$ for all $a\in A$. If $A$ is a unital
$C^*$-algebra, then $\|\bb\|=1$. A $C^*$-algebra is provided with
a normed topology, i.e., it is a topological $^*$-algebra.
\begin{remark}
It should be emphasized that by a morphism of normed algebras is
meant a morphism of the underlying $^*$-algebras, without any
condition on the norms and continuity. At the same time, an
isomorphism of normed algebras means always an isometric morphism.
Any morphism $\f$ of $C^*$-algebras is automatically continuous
due to the property
\mar{spr410}\beq
\|\f(a)\|\leq \| a\|, \qquad a\in A. \label{spr410}
\eeq
\end{remark}
Any $^*$-algebra $A$ can be extended to a unital algebra $\wt
A=\mathbb C\oplus A$ by the adjunction of the identity $\bb$ to
$A$. The unital extension of $A$ also is a $^*$-algebra with
respect to the operation
\be
(\la\bb+a)^*=(\ol\la\bb+a^*), \qquad \la\in\mathbb C, \quad a\in
A.
\ee
If $A$ is a $C^*$-algebra, a norm on $A$ is uniquely prolonged to
the norm
\be
\|\la\bb +a\|= \op\sup_{\|a'\|\leq 1}\|\la a'+aa'\|
\ee
on $\wt A$ which makes $\wt A$ a $C^*$-algebra.
One says that a Banach algebra $A$ admits an approximate identity
if there is a family $\{u_\iota\}_{\iota\in I}$ of elements of
$A$, indexed by a directed set $I$, which possesses the following
properties:
$\bullet$ $\| u_\iota\|< 1$ for all $\iota\in I$,
$\bullet$ $\| u_\iota a-a\|\to 0$ and $\| au_\iota-a\|\to 0$ for
every $a\in A$.
\noindent It should be noted that the existence of an approximate
identity is an essential condition for many results (see, e.g.,
Theorems \ref{gns5} and \ref{spr471}).
For instance, a $C^*$-algebra has an approximate identity.
Conversely, any Banach $^*$-algebra $A$ with an approximate
identity admits the enveloping $C^*$-algebra $A^\dagger$ (Remark
\ref{env}) \cite{dixm,book05}.
An important example of $C^*$-algebras is an algebra $B(E)$ of
bounded (and, equivalently, continuous) operators in a Hilbert
space $E$ (Section 14.2). Every closed $^*$-subalgebra of $B(E)$
is a $C^*$-algebra and, conversely, every $C^*$-algebra is
isomorphic to a $C^*$-algebra of this type (Theorem \ref{spr424}).
An algebra $B(E)$ is endowed with the operator norm
\mar{spr427}\beq
\|a\|= \op\sup_{\|e\|_E=1} \|ae\|_E, \qquad a\in B(E).
\label{spr427}
\eeq
This norm brings the $^*$-algebra $B(E)$ of bounded operators in a
Hilbert space $E$ into a $C^*$-algebra. The corresponding topology
on $B(E)$ is called the normed operator topology.
One also provides $B(E)$ with the strong and weak operator
topologies, defined by the families of seminorms
\be
&& \{p_e(a)=\|ae\|, \quad e\in E\}, \\
&& \{p_{e,e'}(a)=|\lng ae|e'\rng|, \quad e,e'\in E\},
\ee
respectively. The normed operator topology is finer than the
strong one which, in turn, is finer than the weak operator
topology. The strong and weak operator topologies on a subgroup
$U(E)\subset B(E)$ of unitary operators coincide with each other.
It should be emphasized that $B(E)$ fails to be a topological
algebra with respect to strong and weak operator topologies.
Nevertheless, the involution in $B(E)$ also is continuous with
respect to the weak operator topology, while the operations
\be
B(E)\ni a\to aa'\in B(E),\qquad B(E)\ni a\to a'a\in B(E),
\ee
where $a'$ is a fixed element of $B(E)$, are continuous with
respect to all the above mentioned operator topologies.
\begin{remark} \label{w410} \mar{w410}
Let $N$ be a subset of $B(E)$. The commutant $N'$ of $N$ is a set
of elements of $B(E)$ which commute with all elements of $N$. It
is a subalgebra of $B(E)$. Let $N''=(N')'$ denote the bicommutant.
Clearly, $N\subset N''$. A $^*$-subalgebra $B$ of $B(E)$ is called
the von Neumann algebra if $B=B''$. This property holds iff $B$ is
strongly (or, equivalently, weakly) closed in $B(E)$ \cite{dixm}.
For instance, $B(E)$ is a von Neumann algebra. Since a strongly
(weakly) closed subalgebra of $B(E)$ also is closed with respect
to the normed operator topology on $B(E)$, any von Neumann algebra
is a $C^*$-algebra.
\end{remark}
\begin{remark} \label{w90} \mar{w90}
A bounded operator in a Hilbert space $E$ is called completely
continuous if it is compact, i.e., it sends any bounded set into
a set whose closure is compact. An operator $a\in B(E)$ is
completely continuous iff it can be represented by the series
\mar{spr462}\beq
a(e)=\op\sum_{k=1}^\infty\la_k\lng e|e_k\rng e_k, \label{spr462}
\eeq
where $e_k$ are elements of a basis for $E$ and $\la_k$ are
positive numbers which tend to zero as $k\to\infty$. For instance,
every degenerate operator (i.e., an operator of finite rank which
sends $E$ onto its finite-dimensional subspace) is completely
continuous. A completely continuous operator $a$ is called the
Hilbert--Schmidt operator if the series
\be
\|a\|^2_{\mathrm{HS}}= \op\sum_k \la^2_k
\ee
converges. Hilbert--Schmidt operators make up an involutive Banach
algebra with respect to this norm, and it is a two-sided ideal of
an algebra $B(E)$. A completely continuous operator $a$ in a
Hilbert space $E$ is called a nuclear operator if the series
\be
||a||_{\mathrm{Tr}}=\op\sum_k \la_k
\ee
converges. Nuclear operators make up an involutive Banach algebra
with respect to this norm, and it is a two-sided ideal of an
algebra $B(E)$. Any nuclear operator is the Hilbert--Schmidt one.
Moreover, the product of arbitrary two Hilbert--Schmidt operators
is a nuclear operator, and every nuclear operator is of this type.
\end{remark}
Let us consider representations of $^*$-algebras by bounded
operators in Hilbert spaces \cite{dixm,ped}. It is a morphism
$\pi$ of a $^*$-algebra $A$ to an algebra $B(E)$ of bounded
operators in a Hilbert space $E$, called the carrier space of
$\pi$. Representations throughout are assumed to be
non-degenerate, i.e., there is no element $e\neq 0$ of $E$ such
that $Ae=0$ or, equivalently, $AE$ is dense in $E$.
\begin{theorem} \label{spr424} \mar{spr424}
If $A$ is a $C^*$-algebra, there exists its exact (isomorphic)
representation.
\end{theorem}
\begin{theorem} \label{gns4} \mar{gns4}
A representation $\pi$ of a $^*$-algebra $A$ is uniquely prolonged
to a representation $\wt \pi$ of the unital extension $\wt A$ of
$A$.
\end{theorem}
Let $\{\pi^\iota\}$, $\iota\in I$, be a family of representations
of a $^*$-algebra $A$ in Hilbert spaces $E^\iota$. If the set of
numbers $\|\pi^\iota(a)\|$ is bounded for each $a\in A$, one can
construct a bounded operator $\pi(a)$ in a Hilbert sum $\oplus
E^\iota$ whose restriction to each $E^\iota$ is $\pi^\iota(a)$.
\begin{theorem} \label{sum} \mar{sum}
This is the case of a $C^*$-algebra $A$ due to the property
(\ref{spr410}). Then $\pi$ is a representation of $A$ in $\oplus
E^\iota$, called the Hilbert sum
\mar{0315}\beq
\pi=\op\oplus_I \pi^\iota \label{0315}
\eeq
of representations $\pi^\iota$.
\end{theorem}
Given a representation $\pi$ of a $^*$-algebra $A$ in a Hilbert
space $E$, an element $\thh\in E$ is said to be a cyclic vector
for $\pi$ if the closure of $\pi(A)\thh$ is equal to $E$.
Accordingly, $\pi$ (or a more strictly a pair $(\pi,\thh)$) is
called the cyclic representation.
\begin{theorem} \label{spr411} \mar{spr411}
Every representation of a $^*$-algebra $A$ is a Hilbert sum of
cyclic representations.
\end{theorem}
\begin{remark} \label{gns29} \mar{gns29}
It should be emphasized, that given a cyclic representation
$(\pi,\thh)$ of a $^*$-algebra $A$ in a Hilbert space $E$, a
different element $\thh'$ of $E$ is a cyclic for $\pi$ iff there
exist some elements $b, b'\in A$ such that $\thh'=\pi(b)\thh$ and
$\thh=\pi(b')\thh'$.
\end{remark}
Let $A$ be a $^*$-algebra, $\pi$ its representation in a Hilbert
space $E$, and $\thh$ an element of $E$. Then a map
\mar{spr470}\beq
\om_\thh: a\to \lng \pi(a)\thh|\thh\rng \label{spr470}
\eeq
is a positive form on $A$. It is called the vector form defined by
$\pi$ and $\thh$.
Therefore, let us consider positive forms on a $^*$-algebra $A$.
Given a positive form $f$, a Hermitian form
\mar{spr472}\beq
\lng a|b\rng = f(b^*a), \qquad a,b\in A, \label{spr472}
\eeq
makes $A$ a pre-Hilbert space. If $A$ is a normed $^*$-algebra,
continuous positive forms on $A$ are provided with a norm
\mar{0215}\beq
\|f\|=\op\sup_{\|a\|=1}|f(a)|, \qquad a\in A. \label{0215}
\eeq
\begin{theorem} \label{gns6} \mar{gns6} Let $A$ be a unital Banach
$^*$-algebra such that $\|\bb\|=1$. Then any positive form on $A$
is continuous.
\end{theorem}
In particular, positive forms on a $C^*$-algebra always are
continuous. Conversely, a continuous form $f$ on an unital
$C^*$-algebra is positive iff $f(\bb)=\|f\|$. It follows from this
equality that positive forms on a unital $C^*$-algebra $A$ obey a
relation
\mar{gns12}\beq
\|f_1+f_2\|=\|f_1\|+ \|f_2\|. \label{gns12}
\eeq
Let us note that a continuous positive form on a topological
$^*$-algebra $A$ admits different prolongations onto the unital
extension $\wt A$ of $A$. Such a prolongation is unique in the
following case \cite{dixm}.
\begin{theorem} \label{gns5} \mar{gns5}
Let $f$ be a positive form on a Banach $^*$-algebra $A$ with an
approximate identity. It is extended to a unique positive form
$\wt f$ on the unital extension $\wt A$ of $A$ such that $\wt
f(\bb)=\|f\|$.
\end{theorem}
A key point is that any positive form on a $C^*$-algebra equals a
vector form defined by some cyclic representation of $A$ in
accordance with the following GNS representation construction
\cite{dixm,book05}.
\begin{theorem} \label{spr471} \mar{spr471}
Let $f$ be a positive form on a Banach $^*$-algebra $A$ with an
approximate identity and $\wt f$ its continuous positive
prolongation onto the unital extension $\wt A$ (Theorems
\ref{gns6} and \ref{gns5}). Let $N_f$ be a left ideal of $\wt A$
consisting of those elements $a\in A$ such that $\wt f(a^*a)=0$.
The quotient $\wt A/N_f$ is a Hausdorff pre-Hilbert space with
respect to the Hermitian form obtained from $\wt f(b^*a)$
(\ref{spr472}) by passage to the quotient. We abbreviate with
$E_f$ the completion of $\wt A/N_f$ and with $\thh_f$ the
canonical image of $\bb\in \wt A$ in $\wt A/N_f\subset E_f$. For
each $a\in \wt A$, let $\tau(a)$ be an operator in $\wt A/N_f$
obtained from the left multiplication by $a$ in $\wt A$ by passage
to the quotient. Then the following holds.
(i) Each $\tau(a)$ has a unique extension to a bounded operator
$\pi_f(a)$ in a Hilbert space $E_f$.
(ii) A map $a\to \pi_f(a)$ is a representation of $A$ in $E_f$.
(iii) A representation $\pi_f$ admits a cyclic vector $\thh_f$.
(iv) $f(a)=\lng \pi(a)\thh_f|\thh_f\rng$ for each $a\in A$.
\end{theorem}
The representation $\pi_f$ and the cyclic vector $\thh_f$ in
Theorem \ref{spr471} are said to be defined by a form $f$, and
a form $f$ equals the vector form defined by $\pi_f$ and
$\thh_f$.
As was mentioned above, we further restrict our consideration of
the GNS construction in Theorem \ref{spr471} to unital
$C^*$-algebras in view of the following \cite{dixm,book05}.
\begin{remark} \label{env} \mar{env}
Let $A$ be an involutive Banach algebra $A$ with an approximate
identity, and let $P(A)$ be the set of pure states of $A$ (Remark
\ref{gns60}). For each $a\in A$, we put
\mar{spr542}\beq
\|a\|'=\op\sup_{f\in P(A)} f(aa^*)^{1/2}, \qquad a\in A.
\label{spr542}
\eeq
It is a seminorm on $A$ such that $\|a\|'\leq \|a\|$. If $A$ is a
$C^*$-algebra, $\|a\|'= \|a\|$ due to the relation (\ref{spr410})
and the existence of an isomorphic representation of $A$. Let
$\cI$ denote the kernel of $\|.\|'$. It consists of $a\in A$ such
that $\|a\|'=0$. Then the completion $A^\dagger$ of the factor
algebra $A/\cI$ with respect to the quotient of the seminorm
(\ref{spr542}), is a $C^*$-algebra, called the enveloping
$C^*$-algebra of $A$. There is the canonical morphism $\tau:A\to
A^\dagger$. Clearly, $A=A^\dagger$ if $A$ is a $C^*$-algebra. The
enveloping $C^*$-algebra $A^\dagger$ possesses the following
important properties.
$\bullet$ If $\pi$ is a representation of $A$, there is exactly
one representation $\pi^\dagger$ of $A^\dagger$ such that
$\pi=\pi^\dagger\circ\tau$. Moreover, the map $\pi\to\pi^\dagger$
is a bijection of a set of representations of $A$ onto a set of
representations of $A^\dagger$.
$\bullet$ If $f$ is a continuous positive form on $A$, there
exists exactly one positive form $f^\dagger$ on $A^\dagger$ such
that $f=f^\dagger\circ\tau$. Moreover, $\|f^\dagger\|=\|f\|$. The
map $f\to f^\dagger$ is a bijection of a set of continuous
positive forms on $A$ onto a set of positive forms on $A^\dagger$.
\end{remark}
Moreover, the cyclic vector $\thh_f$ in Theorem \ref{spr471}
defined by a positive form $f$ is the image of the identity under
the quotient $\wt A\to\wt A/N_f$, and thus the GNS construction
necessarily is concerned with unital algebras. In view of Theorems
\ref{gns4} and \ref{gns5}, we therefore can restrict our
consideration to unital $C^*$-algebras.
\section{Inequivalent vacua}
Let $A$ be a unital $C^*$-algebra of a quantum systems. As was
mentioned above, positive forms on a $C^*$ algebra are said to be
equivalent if they define its equivalent cyclic representations.
\begin{remark} \label{gns122} \mar{gns122}
Let us recall that two representations $\pi_1$ and $\pi_2$ of a
$^*$-algebra $A$ in Hilbert spaces $E_1$ and $E_2$ are equivalent
if there is an isomorphism $\g:E_1\to E_2$ such that
\mar{gns10}\beq
\pi_2(a)=\g\circ\pi_1(a)\circ \g^{-1}, \qquad a\in A.
\label{gns10}
\eeq
In particular, if representations are equivalent, their kernels
coincide with each other.
\end{remark}
Given two positive forms $f_1$ and $f_2$ on a unital $C^*$-algebra
$A$, we meet the following three variants.
(i) If $f_1=f_2$, there is an isomorphism $\g$ of the
corresponding Hilbert spaces $\g:E_1\to E_2$ such that the
relation (\ref{gns10}) holds, and moreover
\mar{gns11}\beq
\thh_2=\g(\thh_1). \label{gns11}
\eeq
(ii) Let positive forms $f_1$ and $f_2$ be equivalent, but
different. Then their equivalence morphism $\g$ fails to obey the
relation (\ref{gns11}).
(iii) Positive forms $f_1$ and $f_2$ on $A$ are inequivalent.
In particular, let $\pi$ be a representation of $A$ in a Hilbert
space $E$, and let $\thh$ be an element of $E$ which defines the
vector form $\om_\thh$ (\ref{spr470}) on $A$. Then a
representation $\pi$ contains a summand which is equivalent to the
cyclic representation $\pi_{\om_\thh}$ of $A$ defined by a vector
form $\om_\thh$.
There are the following criteria of equivalence of positive forms.
\begin{theorem} \label{gns123} \mar{gns123}
Positive forms on a unital $C^*$-algebra are equivalent only if
their kernels contain a common largest closed two-sided ideal.
\end{theorem}
\begin{proof}
The result follows from the fact that the kernel of a cyclic
representation defined by a positive form on a unital
$C^*$-algebra is a largest closed two-sided ideal of the kernel of
this form \cite{dixm}
\end{proof}
\begin{theorem} \label{spr474} \mar{spr474}
Positive forms $f$ and $f'$ on a unital $C^*$-algebra $A$ are
equivalent iff there exist elements $b,b'\in A$ such that
\be
f'(a)=f(b^+ab), \qquad f(a)=f'(b'^+ab'), \qquad a\in A.
\ee
\end{theorem}
\begin{proof} Let a positive form $f$ define a cyclic representation $(\pi_f,\thh_f)$ of $A$
in $E_f$. Let us consider an element $\pi_f(b)\thh_f\in E_f$. In
accordance with Remark \ref{gns29}, this element is a cyclic
element for a representation $\pi_f$. It provides a positive form
$\om_{\pi_f(b)\thh_f}$ on $A$ such that $\om_{\pi_f(b)\thh_f}=f'$.
Then a positive form $f'$ defines a cyclic representation
$(\pi_{f'},\thh_{f'})$ of $A$ in a Hilbert space $E_{f'}$ which is
isomorphic as $\g:E_{f'}\to E_f$ to a cyclic representation
$(\pi_f,\pi_f(b)\thh_f)$ in $E_f$ such that the relation
(\ref{gns11}) holds. Conversely, let positive forms $f$ and $f'$
be equivalent. Then a positive form $f'$ defines an isomorphic
cyclic representation $(\pi_f, \thh')$ in $E_f$, but with a
different cyclic vector $\thh'$. Then the result follows from
Remark \ref{gns29}.
\end{proof}
In particular, it follows from Theorem \ref{spr474} that given a
positive form $f$ on $A$, the state $f(\bb)^{-1}f$ of $A$ is
equivalent to $f$. Speaking on equivalent positive forms on $A$,
we therefore can restrict our consideration to states.
For instance, any cyclic representation of a $C^*$-algebra $A$ is
a summand of the Hilbert sum (\ref{0315}):
\mar{gns13}\beq
\pi_F=\op\oplus_{F(A)} \pi_f, \label{gns13}
\eeq
of cyclic representations of $A$ where $f$ runs through a set
$F(A)$ of all states of $A$. Since for any element $a\in A$ there
exists a state $f$ such that $f(a)\neq 0$, the representation
$\pi_F$ is injective and, consequently, isometric and isomorphic.
A space of continuous forms on a $C^*$-algebra $A$ is the
(topological) dual $A'$ of a Banach space $A$. It can be provided
both with a normed topology defined by the norm (\ref{0215}) and a
weak$^*$ topology (Section 14.1). It follows from the relation
(\ref{gns12}), that a subset $F(A)\subset A'$ of states is convex
and its extreme points are pure states.
\begin{remark} \mar{gns60} \label{gns60}
Let us recall that a positive form $f'$ on a $^*$-algebra $A$ is
said to be dominated by a positive form $f$ if $f-f'$ is a
positive form \cite{book05,dixm}. A non-zero positive form $f$ on
a $^*$-algebra $A$ is called pure if every positive form $f'$ on
$A$ which is dominated by $f$ reads $\la f$, $0\leq \la\leq 1$.
\end{remark}
A key point is the following \cite{dixm}
\begin{theorem} \label{spr480} \mar{spr480}
The cyclic representation of $\pi_f$ of a $C^*$-algebra $A$
defined by a positive form $f$ on $A$ is irreducible iff $f$ is a
pure form \cite{dixm}
\end{theorem}
In particular, any vector form defined by a vector of a carrier
Hilbert space of an irreducible representation is a pure form.
\begin{remark}
Let us note that a representation $\pi$ of a $^*$-algebra $A$ in a
Hilbert space $E$ is called topologically irreducible if the
following equivalent conditions hold:
$\bullet$ the only closed subspaces of $E$ invariant under
$\pi(A)$ are 0 and $E$;
$\bullet$ the commutant of $\pi(A)$ in $B(E)$ is a set of scalar
operators;
$\bullet$ every non-zero element of $E$ is a cyclic vector for
$\pi$.
\noindent At the same time, irreducibility of $\pi$ in the
algebraic sense means that the only subspaces of $E$ invariant
under $\pi(A)$ are 0 and $E$. If $A$ is a $C^*$-algebra, the
notions of topologically and algebraically irreducible
representations are equivalent. It should be emphasized that a
representation of a $C^*$-algebra need not be a Hilbert sum of the
irreducible ones.
\end{remark}
An algebraically irreducible representation $\pi$ of a
$^*$-algebra $A$ is characterized by its kernel $\Ker\pi\subset
A$. This is a two-sided ideal, called primitive. Certainly,
algebraically irreducible representations with different kernels
are inequivalent, whereas equivalent irreducible representations
possesses the same kernel. Thus, we have a surjection
\mar{spr736}\beq
\wh A\ni \pi\to \Ker\pi \in \mathrm{Prim}(A) \label{spr736}
\eeq
of a set $\wh A$ of equivalence classes of algebraically
irreducible representations of a $^*$-algebra $A$ onto a set
Prim$(A)$ of primitive ideals of $A$.
A set Prim$(A)$ is equipped with the so called Jacobson topology
\cite{dixm}. This topology is not Hausdorff, but obeys the
Fr\'echet axiom, i.e., for any two distinct points of Prim$(A)$,
there is a neighborhood of one of them which does not contain the
other. Then a set $\wh A$ is endowed with the coarsest topology
such that the surjection (\ref{spr736}) is continuous. Provided
with this topology, $\wh A$ is called the spectrum of a
$^*$-algebra $A$. In particular, one can show the following.
\begin{theorem} \label{spr413} \mar{spr413}
If a $^*$-algebra $A$ is unital, its spectrum $\wh A$ is
quasi-compact, i.e., it satisfies the Borel--Lebesgue axiom, but
need not be Hausdorff.
\end{theorem}
\begin{theorem} \label{spr414} \mar{spr414}
The spectrum $\wh A$ of a $C^*$-algebra $A$ is a locally
quasi-compact space.
\end{theorem}
It follows from Theorems \ref{spr413} and \ref{spr414} that the
spectrum of a unital $C^*$-algebra is quasi-compact.
\begin{example} \label{spr770} \mar{spr770} A $C^*$-algebra is said to be
elementary if it is isomorphic to an algebra $T(E)\subset B(E)$ of
compact operators in some Hilbert space $E$ (Example \ref{w90}).
Every non-trivial irreducible representation of an elementary
$C^*$ algebra $A\cong T(E)$ is equivalent to its isomorphic
representation by compact operators in $E$ \cite{dixm}. Hence, the
spectrum of an elementary algebra is a singleton set.
\end{example}
By analogy with Theorem \ref{spr474}, one can state the following
relations between equivalent pure states of a $C^*$-algebra.
\begin{theorem} \label{spr481} \mar{spr481}
Pure states $f$ and $f'$ of a unital $C^*$-algebra $A$ are
equivalent iff there exists a unitary element $U\in A$ such that
the relation
\mar{gns51'}\beq
f'(a)=f(U^*aU), \qquad a\in A. \label{gns51'}
\eeq
holds.
\end{theorem}
\begin{proof}
A key point is that, if $f$ is a pure state of a unital
$C^*$-algebra, a pseudo-Hilbert space $A/N_f$ in Theorem
\ref{spr471} is complete, i.e., $E_f=A/N_f$.
\end{proof}
\begin{corollary} \label{gns33} \mar{gns33}
Let $\pi$ be an irreducible representation of a unital
$C^*$-algebra $A$ in a Hilbert space $E$. Given two distinct
elements $\thh_1$ and $\thh_2$ of $E$ (they are cyclic for $\pi$),
the vector forms on $A$ defined by $(\pi,\thh_1)$ and $(\pi,
\thh_2)$ are equal iff there exists $\la\in\mathbb C$, $|\la|=1$,
such that $\thh_1=\la\thh_2$.
\end{corollary}
\begin{corollary} \label{gns34} \mar{gns34}
There is one-to-one correspondence between the pure states of a
unital $C^*$-algebra $A$ associated to the same irreducible
representation $\pi$ of $A$ in a Hilbert space $E$ and the
one-dimensional complex subspaces of $E$, i.e, these pure states
constitute a projective Hilbert space $PE$.
\end{corollary}
There is an additional important criterion of equivalence of pure
states of a unital $C^*$-algebra \cite{glimm}.
\begin{theorem} \label{gns18} \mar{gns18}
Pure states $f$ and $f'$ of a unital $C^*$-algebra are equivalent
if $\|f-f'\|<2$.
\end{theorem}
Let $P(A)$ denote a set of pure states of a unital $C^*$-algebra
$A$. Theorem \ref{spr481} implies a surjection $P(A)\to \wh A$.
One can show that, if $P(A)\subset A'$ is provided with a relative
weak$^*$ topology, this surjection is continuous and open, i.e.,
it is a topological fibre
bundle whose fibres are projective
Hilbert spaces \cite{dixm}.
Turning to a set $F(A)$ of states of a unital algebra
$C^*$-algebra $A$, we have the following.
\begin{theorem} \label{gns17} \mar{gns17}
A set $F(A)$ is a weakly$^*$-closed convex hull of a set $P(A)$
of pure states of $A$. It is weakly* compact \cite{dixm}.
\end{theorem}
Herewith, by virtue of Theorem \ref{gns17}, any set of mutually
inequivalent pure states of a unital $C^*$-algebra is totally
disconnected in a normed topology, i.e., its connected components
are points only.
By virtue of Theorem \ref{spr474}, elements of a quantum algebra
$A$ can not perform invertible transitions between its
inequivalent states. At the same time, one can show the following.
\begin{theorem} \label{gns23} \mar{gns23}
There exists a wider unital $C^*$-algebra such that inequivalent
states of $A$ become its equivalent ones.
\end{theorem}
\begin{proof}
Let us consider the Hilbert sum $\pi_F$ (\ref{gns13}) of cyclic
representations of $A$ whose carrier space is a Hilbert sum
\mar{gns21}\beq
E_F=\op\oplus_{F(A)}E_f. \label{gns21}
\eeq
Let $B(E_F)$ be a unital $C^*$ algebra of bounded operators in
$E_F$ (\ref{gns21}). Since the representation $\pi_F$ of $A$ is
exact, an algebra $A$ is isomorphic to a subalgebra of $B(E_F)$.
Any state $f$ of $A$ is equivalent to a vector state
$\om_{\thh_f}$ of $\pi_f(A)$ which also is that of $B(E_F)$. Since
all vector states of $B(E_F)$ are equivalent, all states of $A$
are equivalent as those of $B(E_F)$.
\end{proof}
An algebra $B(E_F)$ contains the projectors $P_f$ onto summands
$E_f$ of $E_F$ (\ref{gns21}). Let $r(f)$ be some real function on
$F(A)$. Then there exists a bounded operator in $E_F$
(\ref{gns21}), which we denote
\mar{gns24}\beq
T=\op\sum_{F(A)}r(f)P_f, \label{gns24}
\eeq
such that its restriction to each summand $E_f$ of $E_F$ is
$r(f)P_f$. Certainly, this operator belongs to the commutant
$\pi_F(A)'$ of $\pi_F(A)$ in $B(E_F)$.
One can think of $T$ (\ref{gns24}) as being a superselection
operator of a quantum system which distinguish its states
\cite{hor}.
\section{Example. Infinite qubit systems}
Let $Q$ be a two-dimensional complex space $\mathbb C^2$ equipped
with the standard positive non-degenerate Hermitian form
$\lng.|.\rng_2$. Let $M_2$ be an algebra of complex $2\times
2$-matrices seen as a $C^*$-algebra. A system of $m$ qubits
is usually described by a Hilbert space
$E_m=\op\ot^m Q$ and a $C^*$-algebra $A_m=\op\ot^m M_2$, which
coincides with the algebra $B(E_m)$ of bounded operators in $E_m$
\cite{keyl}. One can straightforwardly generalize this description
to an infinite set $S$ of qubits by analogy with a spin lattice
\cite{emch,book05,qubit}. Its algebra $A_S$ admits inequivalent
irreducible representations.
We follow the construction of infinite tensor products of Hilbert
spaces and $C^*$-algebras in \cite{emch}. Let $\{Q_s,s\in S\}$ be
a set of two-dimensional Hilbert spaces $Q_s=\mathbb C^2$. Let
$\op\times_SQ_s$ be a complex vector space whose elements are
finite linear combinations of elements $\{q_s\}$ of the Cartesian
product $\op\prod_S Q_s$ of the sets $Q_s$. A tensor product
$\op\ot_S Q_s$ of complex vector spaces $Q_s$ is the quotient of
$\op\times_S Q_s$ with respect to a vector subspace generated by
elements of the form:
$\bullet$ $\{q_s\} + \{q'_s\} -\{q''_s\}$, where $q_r + q'_r
=q''_r$ for some element $r\in S$ and $q_s = q'_s =q''_s$ for all
the others,
$\bullet$ $\{q_s\} - \la\{q'_s\}$, $\la\in\mathbb C$, where $q_r=
\la q'_r$ for some element $r\in S$ and $q_s = q'_s$ for all the
others.
Given a map
\mar{gns91}\beq
\si:S\to Q, \qquad , \lng\si(s)|\si(s)\rng_2=1, \label{gns91}
\eeq
let us consider an element
\mar{gns92}\beq
\thh_\si=\{\thh_s=\si(s)\}\in \op\prod_S Q_s. \label{gns92}
\eeq
Let us denote $\ot^\si Q_s$ the subspace of $\op\ot_S Q_s$ spanned
by vectors $\ot q_s$ where $q_s\neq \thh_s$ only for a finite
number of elements $s\in S$. It is called the $\thh_\si$-tensor
product of vector spaces $Q_s$, $s\in S$. Then $\ot^\si Q_s$ is a
pre-Hilbert space with respect to a positive non-degenerate
Hermitian form
\be
\lng \ot^\si q_s|\ot^\si q'_s\rng=\op\prod_{s\in S} \lng
q_s|q'_s\rng_2.
\ee
Its completion $Q_S^\si$ is a Hilbert space whose orthonormal
basis consists of the elements $e_{ir}=\ot q_s$, $r\in S$,
$i=1,2$, such that $q_{s\neq r}=\thh_s$ and $q_r=e_i$, where
$\{e_i\}$ is an orthonormal basis for $Q$.
Let now $\{A_s,s\in S\}$ be a set of unital $C^*$-algebras
$A_s=M_2$. These algebras are provided with the operator norm
\be
\|a\|=(\la_0\ol\la_0 +\la_1\ol\la_1 +\la_2\ol\la_2
+\la_3\ol\la_3)^{1/2}, \qquad a=i\la_0\bb
+\op\sum_{i=1,2,3}\la_i\tau^i,
\ee
where $\tau^i$ are the Pauli matrices. Given the family
$\{\bb_s\}$, let us construct the $\{\bb_s\}$-tensor product $\ot
A_s$ of vector spaces $A_s$. One can regard its elements as tensor
products of elements of $a_s\in A_s$, $s\in K$, for finite subsets
$K$ of $S$ and of the identities $\bb_s$, $s\in S\setminus K$. It
is easily justified that $\ot A_s$ is a normed $^*$-algebra with
respect to the operations
\be
(\ot a_s)(\ot a'_s)=\ot (a_sa'_s), \qquad (\ot a_s)^*=\ot a^*_s
\ee
and a norm
\be
\|\ot a_s\|=\op\prod_s\|a_s\|.
\ee
Its completion $A_S$ is a $C^*$-algebra treated as a quantum
algebra of a qubit system modelled over a set $S$. Then the
following holds \cite{emch}.
\begin{theorem} \label{gns63} \mar{gns63}
Given the element $\thh_\si=\{\thh_s\}$ (\ref{gns92}), the natural
representation $\pi^\si$ of a $^*$-algebra $\ot A_s$ in the
pre-Hilbert space $\ot^\si Q_s$ is extended to an irreducible
representation of a $C^*$-algebra $A_S$ in the Hilbert space
$Q_S^\si$ such that $\pi^\si(A_S)=B(Q_S^\si)$ is an algebra of all
bounded operators in $Q_S^\si$. Conversely, all irreducible
representations of $A_S$ are of this type.
\end{theorem}
An element $\thh_\si\in Q_S^\si$ in Theorem \ref{gns63} defines a
pure state $f_\si$ of an algebra $A_S$. Consequently, a set of
pure states of this algebra is a set of maps $\si$ (\ref{gns91}).
\begin{theorem}
Pure states $f_\si$ and $f_{\si'}$ of an algebra $A_S$ are
equivalent iff
\mar{gns67}\beq
\op\sum_{s\in S} ||\lng\si(s)|\si'(s)\rng_2|-1|<\infty.
\label{gns67}
\eeq
\end{theorem}
In particular, the relation (\ref{gns67}) holds if maps $\si$ and
$\si'$ differ from each other on a finite subset of $S$.
By analogy with a Higgs vacuum, one can treat the maps $\si$
(\ref{gns91}) as classical vacuum fields.
\section{Example. Locally compact groups}
Let $G$ be a locally compact group provided with a Haar measure
(Section 14.4). A space $L^1_\mathbb C(G)$ of equivalence classes
of complex integrable functions (or, simply, complex integrable
functions) on $G$ is an involutive Banach algebra (Section 14.3)
with an approximate identity. As was mentioned above, there is
one-to-one correspondence between the representations of this
algebra and the strongly continuous unitary representations of a
group $G$ (Theorem \ref{gns61}). Thus, one can employ the GNS
construction in order to describe these representations of $G$
\cite{dixm,book05}.
Let a left Haar measure $dg$ on $G$ hold fixed, and by an
integrability condition throughout is meant the
$dg$-integrability.
A uniformly (resp. {strongly) continuous unitary representation}
of a locally compact group $G$ in a Hilbert space $E$ is a
continuous homeomorphism $\pi$ of $G$ to a subgroup $U(E)\subset
B(E)$ of unitary operators in $E$ provided with the normed (resp.
strong) operator topology. A uniformly continuous representation
is strongly continuous. However, the uniform continuity of a
representation is rather rigorous condition. For instance, a
uniformly continuous irreducible unitary representation of a
connected locally compact real Lie group is necessarily
finite-dimensional. Therefore, one usually studies strongly
continuous representations of locally compact groups.
In this case, any element $\xi$ of a carrier Hilbert space $E$
yields the continuous map $G\ni g\to \pi(g)\xi\in E$. Since strong
and weak operator topologies on a unitary group $U(E)$ coincide,
we have a bounded continuous complex function
\mar{spr530}\beq
\vf_{\xi,\eta}(g)=\lng \pi(g)\xi|\eta\rng \label{spr530}
\eeq
on $G$ for any fixed elements $\xi,\eta\in E$. It is called the
coefficient of a representation $\pi$. There is an obvious
equality
\be
\vf_{\xi,\eta}(g)=\ol{\vf_{\eta,\xi}(g^{-1})}.
\ee
The Banach space $L^1_\mathbb C(G)$ of integrable complex
functions on $G$ is provided with the structure of an involutive
Banach algebra with respect to the contraction $f_1*f_2$
(\ref{spr531}) and the involution
\be
f(g)\to f^*(g)=\Delta(g^{-1})\ol{f(g^{-1})},
\ee
where $\Delta$ is the modular function of $G$. It is called the
group algebra of $G$. A map $f\to f(g)dg$ defines an isometric
monomorphism of $L^1_\mathbb C(G)$ to a Banach algebra $M^1(G,
\mathbb C)$ of bounded complex measures on $G$ provided with the
involution $\m^*=\ol{\m^{-1}}$. Unless otherwise stated,
$L^1_\mathbb C(G)$ will be identified with its image in
$M^1(G,\mathbb C)$. In particular, a group algebra $L^1_\mathbb
C(G)$ admits an approximate identity which converges to the Dirac
measure $\ve_\bb\in M^1(G,\mathbb C)$.
\begin{remark} \label{spr541} \mar{spr541}
The group algebra $L^1_\mathbb C(G)$ is not a $C^*$-algebra. Its
enveloping $C^*$-algebra $C^*(G)$ is called the $C^*$-algebra of a
locally compact group $G$.
\end{remark}
Unitary representations of a locally compact group $G$ and
representations of a group algebra $L^1_\mathbb C(G)$ are related
as follows \cite{dixm}.
\begin{theorem} \label{gns61} \mar{gns61}
There is one-to-one correspondence between the (strongly
continuous) unitary representations $\pi$ of a locally compact
group $G$ and the representations $\pi^L$ (\ref{spr567}) of its
group algebra $L^1_\mathbb C(G)$.
\end{theorem}
\begin{proof}
Let $\pi$ be a (strongly continuous) unitary representation of $G$
in a Hilbert space $E$. Given a bounded positive measure $\m$ on
$G$, let us consider the integrals
\be
\vf_{\xi,\eta}(\m)=\int \lng \pi(g)\xi|\eta\rng \m
\ee
of the coefficient functions $\vf_{\xi,\eta}(g)$ (\ref{spr530})
for all $\xi,\eta \in E$. There exists a bounded operator
$\pi(\m)\in B(E)$ in $E$ such that
\be
\lng \pi(\m)\xi|\eta\rng=\vf_{\xi,\eta}(\m), \qquad \xi,\eta \in
E.
\ee
It is called the operator-valued integral of $\pi(g)$ with respect
to the measure $\m$, and is denoted by
\mar{qm340}\beq
\pi(\m)=\int \pi(g)\m(g). \label{qm340}
\eeq
The assignment $\m\to\pi(\m)$ provides a representation of a
Banach $^*$-algebra $M^1(G,\mathbb C)$ in $E$. Its restriction
\mar{spr567}\beq
\pi^L(f)=\int \pi(g)f(g)dg\in B(E) \label{spr567}
\eeq
to $L^1_\mathbb C(G)$ is non-degenerate. One says that the
representations (\ref{qm340}) of $M^1(G,\mathbb C)$ and
(\ref{spr567}) of $L^1_\mathbb C(G)$ are determined by a unitary
representation $\pi$ of $G$. Conversely, let $\pi^L$ be a
representation of a Banach $^*$-algebra $L^1_\mathbb C(G)$ in a
Hilbert space $E$. There is a monomorphism $g\to\ve_g$ of a group
$G$ onto a subgroup of Dirac measures $\ve_g$, $g\in G$, of an
algebra $M^1(G,\mathbb C)$. Let $\{u_\iota(q)\}_{\iota\in I}$ be
an approximate identity in $L^1_\mathbb C(G)$. Then
$\{\pi^L(u_\iota)\}$ converges to an element of $B(E)$ which can
be seen as a representation $\pi^L(\ve_\bb)$ of the unit element
$\ve_\bb$ of $M^1(G,\mathbb C)$. Accordingly,
$\{\pi^L(\g(g)u_\iota)\}$ converges to $\pi^L(\ve_g)$. Thereby, we
obtain the (strongly continuous) unitary representation
$\pi(g)=\pi^L(\ve_g)$ of a group $G$ in a Hilbert space $E$.
Moreover, the representation (\ref{spr567}) of $L^1_\mathbb C(G)$
determined by this representation $\pi$ of $G$ coincides with the
original representation $\pi^L$ of $L^1_\mathbb C(G)$.
\end{proof}
Moreover, $\pi$ and $\pi^L$ have the same cyclic vectors and
closed invariant subspaces. In particular, a representation
$\pi^L$ of $L^1_\mathbb C(G)$ is topologically irreducible iff the
associated representation $\pi$ of $G$ is so. It should be
emphasized that, since $L^1_\mathbb C(G)$ is not a $C^*$-algebra,
its topologically irreducible representations need not be
algebraically irreducible. By irreducible representations of a
group $G$, we will mean only its topologically irreducible
representations.
Theorem \ref{gns61} enables us to apply the GNS construction
(Theorem \ref{spr471}) in order to characterize unitary
representations of $G$ by means of positive continuous forms on
$L^1_\mathbb C(G)$.
In accordance with Remark \ref{spr520}, a continuous form on a
group algebra $L_\mathbb C^1(G)$ is defined as
\mar{qm341}\beq
\f(f)=\int \psi(g)f(g)dg \label{qm341}
\eeq
by an element $\psi$ of a space $L^\infty_\mathbb C(G)$ of
infinite integrable complex functions on $G$ (Remark
\ref{spr520}). However, a function $\psi$ should satisfy the
following additional condition in order that the form
(\ref{qm341}) to be positive.
A continuous complex function $\psi$ on $G$ is called
positive-definite if
\be
\op\sum_{i,j}\psi(g_j^{-1}g_i)\ol\la_i\la_j\geq 0
\ee
for any finite set $g_1,\ldots,g_m$ of elements of $G$ and any
complex numbers $\la_1,\ldots,\la_m$. In particular, if $m=2$ and
$g_1=\bb$, we obtain
\be
\psi(g^{-1})=\ol{\psi(g)}, \qquad \nm\psi(g)\leq\psi(\bb), \qquad
g\in G,
\ee
i.e., $\psi(\bb)$ is bounded.
\begin{lemma} \label{1080} \mar{1080}
The continuous form (\ref{qm341}) on $L^1_\mathbb C(G)$ is
positive iff $\psi\in L^\infty_\mathbb C(G)$ locally almost
everywhere equals a continuous positive-definite function.
\end{lemma}
Then cyclic representations of a group algebra $L^1_\mathbb C(G)$
and the unitary cyclic representations of a locally compact group
$G$ are defined by continuous positive-definite functions on $G$
in accordance with the following theorem.
\begin{theorem} \label{qm539} \mar{qm539}
Let $\pi_\psi$ be a representation of $L^1_\mathbb C(G)$ in a
Hilbert space $E_\psi$ and $\thh_\psi$ a cyclic vector for
$\pi_\psi$ which are determined by the form (\ref{qm341}). Then
the associated unitary representation $\pi_\psi$ of $G$ in
$E_\psi$ is characterized by a relation
\mar{qm342}\beq
\psi(g)=\lng \pi(g)\thh_\psi|\thh_\psi\rng. \label{qm342}
\eeq
Conversely, a complex function $\psi$ on $G$ is continuous
positive-definite iff there exists a unitary representation
$\pi_\psi$ of $G$ and a cyclic vector $\thh_\psi$ for $\pi_\psi$
such that the equality (\ref{qm342}) holds.
\end{theorem}
By analogy with a Higgs vacuum, one can think of functions $\psi$
in Theorem \ref{qm539} as being classical vacuum fields.
\begin{example} \label{x10} \mar{x10}
Let a group $G$ acts on a Hausdorff topological space $Z$ on the
left. Let $\m$ be a quasi-invariant measure on $Z$ under a
transformation group $G$, i.e., $\g(g)\m=h_g\m$ where $h_g$ is the
Radon--Nikodym derivative in Theorem \ref{spr510}. Then there is a
representation
\mar{x11}\beq
G\ni g: f\to \Pi(g)f, \qquad (\Pi(g)f)(z)=h_g^{1/2}(z)f(gz)
\label{x11}
\eeq
of $G$ in a Hilbert space $L_\mathbb C^2(Z,\m)$ of square
$\m$-integrable complex functions on $Z$ \cite{dao}. It is a
unitary representation due to the equality
\be
\|f\|_\m=\int |f(z)|\m(z)=\int |f(g(z)|^2 \m(g(z))= \int
h_g(z)|f(g(z)|^2 \m(z)=\|\Pi(g)f\|^2_\m.
\ee
A group $G$ can be equipped with the coarsest topology such that
the representation (\ref{x11}) is strongly continuous. For
instance, let $Z=G$ be a locally compact group, and let $\m=dg$ be
a left Haar measure. Then the representation (\ref{x11}) comes to
the left-regular representation
\mar{spr570}\beq
(\Pi(g)f)(q)=f(g^{-1}q), \qquad f\in L^2_\mathbb C(G), \qquad q\in
G, \label{spr570}
\eeq
of $G$ in a Hilbert space $L^2_\mathbb C(G)$ of square integrable
complex functions on $G$. Note that the above mentioned coarsest
topology on $G$ is coarser then the original one, i.e., the
representation (\ref{spr570}) is strongly continuous.
\end{example}
Let us consider unitary representations of a locally compact group
$G$ which are contained in its left-regular representation
(\ref{spr570}). In accordance with the expression (\ref{qm340}),
the corresponding representation $\Pi(h)$ of a group algebra
$L^1_\mathbb C(G)$ in $L^2_\mathbb C(G)$ reads
\mar{qm344}\beq
(\Pi(h)f)(q)=\int h(g)(\Pi(g)f)(q)dg=\int
h(g)f(g^{-1}q)dg=(h*f)(q). \label{qm344}
\eeq
Let $G$ be a unimodular group. There is the following criterion
that its unitary representation is contained in the left-regular
one.
\begin{theorem} \label{qm360} \mar{qm360}
If a continuous positive-definite function $\psi$ on a unimodular
locally compact group $G$ is square integrable, then the
representation $\pi_\psi$ of $G$ determined by $\psi$ is contained
in the left-regular representation $\Pi$ (\ref{spr570}) of $G$.
Conversely, let $\pi$ be a cyclic unitary representation of $G$
which is contained in $\Pi$, and let $\thh$ be a cyclic vector for
$\pi$. Then a continuous positive-definite function
$\lng\pi(g)\thh|\thh\rng$ on $G$ is square integrable.
\end{theorem}
The representation $\pi_\psi$ in Theorem \ref{qm360} is
constructed as follows. Given a square integrable continuous
positive-definite function $\psi$ on $G$, there exists a
positive-definite function $\thh\in L^2_\mathbb C(G)$ such that
\be
\psi=\thh*\thh=\thh*\thh^*=\thh^**\ol\thh^*.
\ee
This is a cyclic vector for $\pi_\psi$. The coefficients
(\ref{spr530}) of a representation $\pi_\psi$ read
\be
\vf_{\xi,\eta}(g)=\ol{(\eta*\xi^*)(g)}.
\ee
In particular, if representations $\pi_\psi$ and $\pi_{\psi'}$,
determined by square integrable continuous positive-definite
functions $\psi$ and $\psi'$ on $G$, are irreducible and
inequivalent, then the corresponding cyclic vectors $\thh_\psi$
and $\thh_{\psi'}$ are orthogonal in $L^2_\mathbb C(G)$, while the
functions $\psi$ and $\psi'$ fulfil the relations
\mar{qm345}\beq
\int\psi'(g)\ol\psi(g)dg=0, \qquad \psi*\psi'=0.\label{qm345}
\eeq
Now, let $G$ be a connected locally compact (i.e.,
finite-dimensional) real Lie group. Any unitary representation of
$G$ yields a representation of its right Lie algebra $\cG$ as
follows. In particular, a finite-dimensional unitary
representation of $G$ in a Hilbert space $E$ is analytic, and a
Lie algebra $\cG$ is represented by bounded operators in $E$.
If $\pi$ is an infinite-dimensional (strongly continuous) unitary
representation of $G$ in a Hilbert space $E$, a representation of
a Lie algebra $\cG$ fails to be defined everywhere on $E$ in
general. To construct a carrier space of $\cG$, let us consider a
space $\cK^\infty(G,\mathbb C)\subset L^1_\mathbb C(G)$ of smooth
complex functions on $G$ of compact support and the vectors
\mar{spr586}\beq
e_f=\pi^L(f)e=\int \pi(g)f(g)edg, \qquad e\in E, \qquad f\in
\cK^\infty(G,\mathbb C), \label{spr586}
\eeq
where $\pi^L$ is the representation (\ref{spr567}) of a group
algebra $L^1_\mathbb C(G)$ \cite{hurt}. The vectors $e_f$
(\ref{spr586}) exemplify smooth vectors of the representation
$\pi$ because, for any $\eta\in E$, the coefficients
$\varphi_{e_f,\eta}(g)$ of $\pi$ are smooth functions on $G$. The
vectors $e_f$ (\ref{spr586}) for all $e\in E$ and $f\in
\cK^\infty(G,\mathbb C)$ constitute a dense vector subspace
$E_\infty$ of $E$. Let $u_a$ be a right-invariant vector field on
$G$ corresponding to an element $a\in \cG$. Then the assignment
\be
\pi_\infty(a): e_f\to \pi^L(u_a\rfloor df)e
\ee
provides a representation of a Lie algebra $\cG$ in $E_\infty$.
\section{GNS construction. Unbounded operators}
There are quantum algebras (e.g., of quantum fields) whose
representations in Hilbert spaces need not be normed. Therefore,
generalizations of the conventional GNS representation of
$C^*$-algebras (Theorem \ref{spr471}) to some classes of unnormed
topological $^*$-algebras has been studied
\cite{book05,hor,schmu}.
In a general setting, by an operator in a Hilbert (or Banach)
space $E$ is meant a linear morphism $a$ of a dense subspace
$D(a)$ of $E$ to $E$. The $D(a)$ is called the domain of an
operator $a$. One says that an operator $b$ on $D(b)$ is an
extension of an operator $a$ on $D(a)$ if $D(a)\subset D(b)$ and
$b|_{D(a)}=a$. For the sake of brevity, we will write $a\subset
b$. An operator $a$ is said to be
bounded on $D(a)$ if there exists a real number $r$ such that
\be
\|ae\|\leq r\|e\|, \qquad e\in D(a).
\ee
If otherwise, it is called unbounded. Any bounded operator on a
domain $D(a)$ is uniquely extended to a bounded operator
everywhere on $E$.
An operator $a$ on a domain $D(a)$ is called closed if the
condition that a sequence $\{e_i\}\subset D(a)$ converges to $e\in
E$ and that the sequence $\{ae_i\}$ does to $e'\in E$ implies that
$e\in D(a)$ and $e'=ae$. Of course, any operator defined
everywhere on $E$ is closed. An operator $a$ on a domain $D(a)$ is
called closable if it can be extended to a closed operator. The
closure of a closable operator $a$ is defined as the minimal
closed extension of $a$.
Operators $a$ and $b$ in $E$ are called adjoint if
\be
\lng ae|e'\rng=\lng e|be'\rng, \quad e\in D(a), \quad e'\in D(b).
\ee
Any operator $a$ has a maximal adjoint operator $a^*$, which is
closed. Of course, $a\subset a^{**}$ and $b^*\subset a^*$ if
$a\subset b$. An operator $a$ is called symmetric if it is
adjoint to itself, i.e., $a\subset a^*$. Hence, a symmetric
operator is closable. One can obtain the following chain of
extensions of a symmetric operator:
\be
a\subset\ol a\subset a^{**}\subset a^*=\ol a^*=a^{***}.
\ee
In particular, if $a$ is a symmetric operator, so are $\ol a$ and
$a^{**}$. At the same time, the maximal adjoint operator $a^*$ of
a symmetric operator $a$ need not be symmetric. A symmetric
operator $a$ is called self-adjoint if $a=a^*$, and it is called
essentially self-adjoint if $\ol a=a^*=\ol a^*$. It should be
emphasized that a symmetric operator $a$ is sometimes called
essentially self-adjoint if $a^{**}=a^*$. We here follow the
terminology of \cite{pow1}. If $a$ is a closed operator, the both
notions coincide. For bounded operators, the notions of symmetric,
self-adjoint and essentially self-adjoint operators coincide.
Let $E$ be a Hilbert space. A pair $(B,D)$ of a dense subspace $D$
of $E$ and a unital algebra $B$ of (unbounded) operators in $E$ is
called the $Op^*$-algebra ($O^*$-algebra in the terminology of
\cite{schmu}) on a domain $D$ if, whenever $b\in B$, we have
\cite{hor,pow1}: (i) $D(b)=D$ and $bD\subset D$, (ii) $D\subset
D(b^*)$, (iii) $b^*|_D\subset B$. An algebra $B$ is provided with
the involution $b\to b^+=b^*|_D$, and its elements are closable.
A representation $\pi(A)$ of a $^*$-algebra $A$ in a Hilbert space
$E$ is defined as a homomorphism of $A$ to an $Op^*$-algebra
$(B,D(\pi))$ of (unbounded) operators in $E$ such that
$D(\pi)=D(\pi(a))$ for all $a\in A$ and this representation is
Hermitian, i.e., $\pi(a^*)\subset \pi(a)^*$ for all $a\in A$. In
this case, one also considers the representations
\be
&& \ol\pi: a \to \ol\pi(a)=\ol{\pi(a)}|_{D(\ol\pi)},
\qquad D(\ol\pi)=\op\bigcap_{a\in A} D(\ol{\pi(a)}),\\
&& \pi^*: a \to \pi^*(a)=\pi(a^*)^*|_{D(\pi^*)},
\qquad D(\pi^*)=\op\bigcap_{a\in A} D(\pi(a)^*),
\ee
called the closure of a representation $\pi$, and the adjoint
representation, respectively. There are the representation
extensions $\pi\subset\ol\pi\subset\pi^*$, where $\pi_1\subset
\pi_2$ means $D(\pi_1)\subset D(\pi_2)$. The representations
$\ol\pi$ and $\pi^{**}$ are Hermitian, while $\pi^*=\ol\pi^*$. A
Hermitian representation $\pi(A)$ is said to be closed if
$\pi=\ol\pi$, and it is self-adjoint if $\pi=\pi^*$. Herewith, a
representation $\pi(A)$ is closed (resp. self-adjoint) if one of
operators $\pi(A)$ is closed (resp. self-adjoint).
A representation domain $D(\pi)$ is endowed with the
graph-topology. It is generated by neighborhoods of the origin
\be
U(M,\ve)=\{x\in D(\pi)\,:\,\op\sum_{a\in M} \|\pi(a)x\|<\ve\},
\ee
where $M$ is a finite subset of elements of $A$. All operators of
$\pi(A)$ are continuous with respect to this topology. Let us note
that the graph-topology is finer than the relative topology on
$D(\pi)\subset E$, unless all operators $\pi(a)$, $a\in A$, are
bounded \cite{schmu}.
Let $\ol N^g$ denote the closure of a subset $N\subset D(\pi)$
with respect to the graph-topology. An element $\thh\in D(\pi)$ is
called strongly cyclic (cyclic in the terminology of
\cite{schmu}) if
\be
D(\pi)\subset \ol{(\pi(A)\thh)}^g.
\ee
Then the GNS representation Theorem \ref{spr471} can be
generalized to Theorem \ref{gns} \cite{hor,schmu}.
Similarly to Remark \ref{gns123}, we say that representations
$\pi_1$ and $\pi_2$ of a $^*$-algebra $A$ are equivalent if there
exists an isomorphism $\g$ of their carrier spaces such that
\be
D(\pi_1)=\g(D(\pi_2)), \qquad \pi_1(a)=\g\circ\pi_2(a)\circ
\g^{-1}, \qquad a\in A.
\ee
In particular, if representations are equivalent, their kernels
coincide with each other.
Accordingly, states $f$ and $f'$ of a unital topological
$^*$-algebra $A$ in Theorem \ref{gns} are called equivalent if
they define equivalent representations $\pi_f$ and$\pi_{f'}$.
By analogy with Theorem \ref{spr474}, one can show the following.
\begin{theorem} \mar{spr474'} \label{spr474'}
Positive continuous forms $f$ and $f'$ on a unital topological
$^*$-algebra $A$ are equivalent if there exist elements $b,b'\in
A$ such that
\be
f'(a)=f(b^+ab), \qquad f(a)=f'(b'^+ab'), \qquad a\in A.
\ee
\end{theorem}
We point out the particular class of nuclear barreled
$^*$-algebras. Let $A$ be a locally convex topological
$^*$-algebra whose topology is defined by a set of multiplicative
seminorms $p_\iota$ which satisfy the condition
\be
p_\iota(a^*a)=p_\iota(a)^2, \qquad a\in A.
\ee
It is called a $b^*$-algebra. A unital $b^*$-algebra as like as a
$C^*$-algebra is regular and symmetric, i.e., any element $(\bb
+a^*a)$, $a\in A$, is invertible and, moreover, $(\bb +a^*a)^{-1}$
is bounded \cite{all,igu}.
The
$b^*$-algebras are related to $C^*$-algebras as follows.
\begin{theorem}
Any $b^*$-algebra is the Hausdorff projective limit of a family of
$C^*$-algebras, and {\it vice versa} \cite{igu}.
\end{theorem}
In particular, every $C^*$-algebra $A$ is a barreled
$b^*$-algebra, i.e., every absorbing balanced closed subset is a
neighborhood of the origin of $A$.
Let us additionally assume that $A$ is a nuclear algebra, i.e., a
nuclear space (Section 14.2). Then we have the following variant
of the GNS representation Theorem \ref{gns} \cite{igu}
\begin{theorem} \mar{gns36} \label{gns36}
Let $A$ be a unital nuclear barreled $b^*$-algebra and $f$ a
positive continuous form on $A$. There exists a unique (up to
unitary equivalence) cyclic representation $\pi_f$ of $A$ in a
Hilbert space $E_f$ by operators on a common invariant domain
$D\subset E_f$. This domain can be topologized to conform a rigged
Hilbert space such that all the operators representing $A$ are
continuous on $D$.
\end{theorem}
\begin{example} \mar{gns90} \label{gns90}
The following is an example of nuclear barreled $b^*$-algebras
which is very familiar from quantum field theory
\cite{sard91,ccr,axiom}. Let $Q$ be a nuclear space (Section
14.2). Let us consider a direct limit
\mar{x2}\beq
\wh\ot Q =\mathbb C \oplus Q \oplus Q\wh\ot Q\oplus \cdots\oplus
Q^{\wh\ot n}\oplus\cdots \label{x2}
\eeq
of vector spaces
\mar{x2a}\beq
\wh\ot^{\leq n} Q =\mathbb C \oplus
Q \oplus Q\wh\ot Q\oplus\cdots \oplus Q^{\wh\ot n}, \label{x2a}
\eeq
where $\wh\ot$ is the topological tensor product with respect to
Grothendieck's topology (which coincides with the $\ve$-topology
on the tensor product of nuclear spaces \cite{piet}). The space
(\ref{x2}) is provided with the inductive limit topology, the
finest topology such that the morphisms $\wh\ot^{\leq n} Q\to
\wh\ot Q$ are continuous and, moreover, are imbeddings
\cite{trev}. A convex subset $V$ of $\wh\ot Q$ is a neighborhood
of the origin in this topology iff $V\cap \wh\ot^{\leq n} Q$ is so
in $\wh\ot^{\leq n} Q$. Furthermore, one can show that $\wh\ot Q$
is a unital nuclear barreled LF-algebra with respect to a tensor
product \cite{belang}. The LF-property implies that a linear form
$f$ on $\wh\ot Q$ is continuous iff the restriction of $f$ to each
$\wh\ot^{\leq n} Q$ is so \cite{trev}. If a continuous conjugation
$*$ is defined on $Q$, the algebra $\wh\ot Q$ is involutive with
respect to the operation
\mar{x3}\beq
*(q_1\ot\cdots \ot q_n)=q_n^*\ot\cdots \ot q_1^* \label{x3}
\eeq
on $Q^{\ot n}$ extended by continuity and linearity to $Q^{\wh\ot
n}$. Moreover, $\wh\ot Q$ is a $b^*$-algebra as follows. Since $Q$
is a nuclear space, there is a family $\|.\|_k$, $k\in\mathbb
N_+$, of continuous norms on $Q$. Let $Q_k$ denote the completion
of $Q$ with respect to the norm $\|.\|_k$. Then one can show that
the tensor algebra $\ot Q_k$ is a $C^*$-algebra and that $\wh\ot
Q$ (\ref{x2}) is a projective limit of these $C^*$-algebras with
respect to morphisms $\ot Q_{k+1}\to \ot Q_k$ \cite{igu}. Since
$\wh\ot Q$ (\ref{x2}) is a nuclear barreled $b^*$-algebra, it
obeys GNS representation Theorem \ref{gns36}. Herewith, let us
note that, due to the LF-property, a positive continuous form $f$
on $\wh\ot Q$ is defined by a family of its restrictions $f_n$ to
tensor products $\wh\ot^{\leq n} Q$. One also can restrict a form
$f$ and the corresponding representation $\pi_f$ to a tensor
algebra
\mar{gns100}\beq
A_Q=\ot Q\subset \wh\ot Q \label{gns100}
\eeq
of $Q$. However, a representation $\pi_f(A_Q)$ of $A_Q$ in a
Hilbert space $E_f$ need not be cyclic.
\end{example}
In quantum field theory, one usually choose $Q$ the Schwartz space
of functions of rapid decrease (Sections 9 and 10).
\section{Example. Commutative nuclear groups}
Following Example \ref{gns90} in a case of a real nuclear space
$Q$, let us consider a commutative tensor algebra
\mar{gns101}\beq
B_Q=\mathbb R \oplus Q\oplus Q\vee Q\oplus\cdots\oplus \op\vee^n
Q\oplus \cdots. \label{gns101}
\eeq
Provided with the direct sum topology, $B_Q$ becomes a unital
topological $^*$-algebra. It coincides with the universal
enveloping algebra of the Lie algebra $T_Q$ of an additive Lie
group $T(Q)$ of translations in $Q$. Therefore, one can obtain the
states of an algebra $B_Q$ by constructing cyclic strongly
continuous unitary representations of a nuclear Abelian group
$T(Q)$.
\begin{remark} Let us note that, in contrast to that studied in
Section 5, a nuclear group $T(Q)$ is not locally compact, unless
$Q$ is finite-dimensional.
\end{remark}
A cyclic strongly continuous unitary representation $\pi$ of
$T(Q)$ in a Hilbert space $(E,\lng.|.\rng_E)$ with a normalized
cyclic vector $\theta\in E$ yields a complex function
\be
Z(q)=\lng \pi(T(q))\theta|\theta\rng_E
\ee
on $Q$. This function is proved to be continuous and
positive-definite, i.e., $Z(0)=1$ and
\be
\op\sum_{i,j} Z(q_i-q_j)\ol\la_i\la_j\geq 0
\ee
for any finite set $q_1,\ldots,q_m$ of elements of $Q$ and
arbitrary complex numbers $\la_1,\ldots,\la_m$.
In accordance with the well-known Bochner theorem for nuclear
spaces (Theorem \ref{spr525}), any continuous positive-definite
function $Z(q)$ on a nuclear space $Q$ is the Fourier transform
\mar{qm545}\beq
Z(q)=\int\exp[i\lng q,u\rng]\m(u) \label{qm545}
\eeq
of a positive measure $\m$ of total mass 1 on the dual $Q'$ of
$Q$. Then the above mentioned representation $\pi$ of $T(Q)$ can
be given by the operators
\mar{q2}\beq
T_Z(q)\rho(u)=\exp[i\langle q,u\rangle]\rho(u) \label{q2}
\eeq
in a Hilbert space $L_\mathbb C^2(Q',\m)$ of equivalence classes
of square $\m$-integrable complex functions $\rho(u)$ on $Q'$. A
cyclic vector $\thh$ of this representation is the
$\m$-equivalence class $\thh\ap_\m 1$ of a constant function
$\rho(u)=1$. Then we have
\mar{ccr1}\beq
Z(q)=\lng T_Z(q)\thh|\thh\rng_\m=\int \exp[i\langle q,u\rangle]\m.
\label{ccr1}
\eeq
Conversely, every positive measure $\m$ of total mass 1 on the
dual $Q'$ of $Q$ defines the cyclic strongly continuous unitary
representation (\ref{q2}) of a group $T(Q)$. By virtue of the
above mentioned Bochner theorem, it follows that every continuous
positive-definite function $Z(q)$ on $Q$ characterizes a cyclic
strongly continuous unitary representation (\ref{q2}) of a nuclear
Abelian group $T(Q)$. We agree to call $Z(q)$ the generating
function of this representation.
\begin{remark}
The representation (\ref{q2}) need not be topologically
irreducible. For instance, let $\rho(u)$ be a function on $Q'$
such that a set where it vanishes is not a $\m$-null subset of
$Q'$. Then the closure of a set $T_Z(Q)\rho$ is a $T(Q)$-invariant
closed subspace of $L_\mathbb C^2(Q',\m)$.
\end{remark}
Different generating functions $Z(q)$ determine inequivalent
representations $T_Z$ (\ref{q2}) of $T(Q)$ in general. One can
show the following \cite{gelf64}.
\begin{theorem} \label{gns77} \mar{gns77}
Distinct generating functions $Z(q)$ and $Z'(q)$ determine
equivalent representations $T_Z$ and $T_{Z'}$ (\ref{q2}) of $T(Q)$
in Hilbert spaces $L^2_\mathbb C(Q',\m)$ and $L^2_\mathbb
C(Q',\m')$ iff they are the Fourier transform of equivalent
measures on $Q'$.
\end{theorem}
Indeed, let
\mar{qm581}\beq
\m'= s^2\m, \label{qm581}
\eeq
where a function $s(u)$ is strictly positive almost everywhere on
$Q'$, and $\m(s^2)=1$. Then the map
\mar{qm580}\beq
L^2_\mathbb C(Q',\m')\ni\rho(u)\to s(u)\rho(u)\in L^2_\mathbb C
(Q',\m) \label{qm580}
\eeq
provides an isomorphism between the representations $T_{Z'}$ and
$T_Z$.
Similarly to the case of finite-dimensional Lie groups (Section
4), any strongly continuous unitary representation (\ref{q2}) of a
nuclear group $T(Q)$ implies a representation of its Lie algebra
by operators
\mar{gns102}\beq
\f(q)\rho(u)=\lng q,u\rng\rho(u) \label{gns102}
\eeq
in the same Hilbert space $L^2_\mathbb C(Q',\m)$. Their mean
values read
\mar{gns103}\beq
\lng\f(q)\rng=\om_\thh(\f(q))=\lng\f(q)\rng=\int \lng q,u\rng
\mu(u). \label{gns103}
\eeq
The representation (\ref{gns102}) is extended to that of the
universal enveloping algebra $B_Q$ (\ref{gns101}).
\begin{remark} \label{gns109} \mar{gns109}
Let us consider representations of $T(Q)$ with generating
functions $Z(q)$ such that $\mathbb R\ni t\to Z(tq)$ is an
analytic function on $\mathbb R$ at $t=0$ for all $q\in Q$. Then
one can show that a function $\lng q|u\rng$ on $Q'$ is square
$\m$-integrable for all $q\in Q$ and that, consequently, the
operators $\f(q)$ (\ref{gns102}) are bounded everywhere in a
Hilbert space $L^2_\mathbb C(Q',\m)$. Moreover, the corresponding
mean values of elements of $B_Q$ can be computed by the formula
\mar{ccr8}\beq
\lng \f(q_1)\cdots\f(q_n)\rng=i^{-n} \frac{\dr}{\dr\al^1}
\cdots\frac{\dr}{\dr\al^n}Z(\al^iq_i)|_{\al^i=0}= \int\lng
q_1,u\rng\cdots\lng q_n, u \rng \mu(u). \label{ccr8}
\eeq
\end{remark}
\section{Infinite canonical commutation relations}
The canonical commutation relations (CCR) are of central
importance in AQT as a method of canonical quantization. A
remarkable result about CCR for finite degrees of freedom is the
Stone--von Neumann uniqueness theorem which states that all
irreducible representations of these CCR are unitarily equivalent
\cite{petz}. On the contrary, CCR of infinite degrees of freedom
admit infinitely many inequivalent irreducible representations
\cite{flor}.
One can provide the comprehensive description of representations
of CCR modelled over an infinite-dimensional nuclear space $Q$
\cite{gelf64,book05,ccr}.
Let $Q$ be a real nuclear space endowed with a non-degenerate
separately continuous Hermitian form $\lng.|.\rng$. This Hermitian
form brings $Q$ into a Hausdorff pre-Hilbert space. A nuclear
space $Q$, the completion $\wt Q$ of a pre-Hilbert space $Q$, and
the dual $Q'$ of $Q$ make up the rigged Hilbert space $Q\subset
\wt Q\subset Q'$ (\ref{spr450}).
Let us consider a group $G(Q)$ of triples $g=(q_1,q_2,\la)$ of
elements $q_1$, $q_2$ of $Q$ and complex numbers $\la$, $|\la|=1$,
which are subject to multiplications
\mar{qm541}\beq
(q_1,q_2,\la)(q'_1,q'_2,\la')=(q_1+q'_1,q_2+q'_2, \exp[i\lng
q_2,q'_1\rng] \la\la'). \label{qm541}
\eeq
It is a Lie group whose group space is a nuclear manifold modelled
over a vector space $W=Q\oplus Q\oplus \mathbb R$. Let us denote
$T(q)=(q,0,0)$, $P(q)=(0,q,0)$. Then the multiplication law
(\ref{qm541}) takes a form
\mar{qm543}\beq
T(q)T(q')=T(q+q'), \quad P(q)P(q')=P(q+q'), \quad
P(q)T(q')=\exp[i\lng q|q'\rng]T(q')P(q). \label{qm543}
\eeq
Written in this form, $G(Q)$ is called the nuclear Weyl CCR group.
The complexified Lie algebra of a nuclear Lie group $G(Q)$ is the
unital Heisenberg CCR algebra $\ccG(Q)$. It is generated by
elements $I$, $\f(q)$, $\pi(q)$, $q\in Q$, which obey the
Heisenberg CCR
\mar{qm540}\beq
[\f(q),I]=\pi(q),I]=[\f(q),\f(q')]=[\pi(q),\pi(q')]=0, \qquad
[\pi(q),\f(q')]=-i\lng q|q'\rng I. \label{qm540}
\eeq
There is the exponential map
\be
T(q)=\exp[i\f(q)], \qquad P(q)=\exp[i\pi(q)].
\ee
Due to the relation (\ref{1085}), the normed topology on a
pre-Hilbert space $Q$ defined by a Hermitian form $\lng.|.\rng$ is
coarser than the nuclear space topology. The latter is metric,
separable and, consequently, second-countable. Hence, a
pre-Hilbert space $Q$ also is second-countable and, therefore,
admits a countable orthonormal basis. Given such a basis $\{q_i\}$
for $Q$, the Heisenberg CCR (\ref{qm540}) take a form
\be
[\f(q_j),I]=\pi(q_j),I]=[\f(q_j),\f(q_k)]=[\pi(q_k),\pi(q_j)]=0,
\qquad [\pi(q_j),\f(q_k)]=-i\dl_{jk}I.
\ee
A Weyl CCR group $G(Q)$ contains two nuclear Abelian subgroups
$T(Q)$ and $P(Q)$. Following the representation algorithm in
\cite{gelf64}, we first construct representations of a nuclear
Abelian group $T(Q)$. Then these representations under certain
conditions can be extended to representations of a Weyl CCR group
$G(Q)$.
Following Section 7, we treat a nuclear Abelian group $T(Q)$ as
being a group of translations in a nuclear space $Q$. Let us
consider its cyclic strongly continuous unitary representation
$T_Z$ (\ref{q2}) in a Hilbert space $L^2_\mathbb C(Q',\m)$ of
equivalence classes of square $\m$-integrable complex functions
$\rho(u)$ on the dual $Q'$ of $Q$ which is defined by the
generating function $Z$ (\ref{qm545}). This representation can be
extended to a Weyl CCR group $G(Q)$ if a measure $\m$ possesses
the following property.
Let $u_q$, $q\in Q$, be an element of $Q'$ given by the condition
$\lng q',u_q\rng=\lng q'|q\rng$, $q'\in Q$. These elements form
the range of a monomorphism $Q\to Q'$ determined by a Hermitian
form $\lng.|.\rng$ on $Q$. Let a measure $\m$ in the expression
(\ref{qm545}) remain equivalent under translations
\be
Q'\ni u\to u+u_q \in Q', \qquad u_q\in Q\subset Q',
\ee
in $Q'$, i.e.,
\mar{qm547}\beq
\m(u+u_q)=a^2(q,u)\m(u), \qquad u_q\in Q\subset Q', \label{qm547}
\eeq
where a function $a(q,u)$ is square $\m$-integrable and strictly
positive almost everywhere on $Q'$. This function fulfils the
relations
\mar{qm555}\beq
a(0,u)=1, \qquad a(q+q',u)=a(q,u)a(q',u+u_q). \label{qm555}
\eeq
A measure on $Q'$ obeying the condition (\ref{qm547}) is called
translationally quasi-invariant, though it does not remain
equivalent under an arbitrary translation in $Q'$, unless $Q$ is
finite-dimensional.
Let the generating function $Z$ (\ref{qm545}) of a cyclic strongly
continuous unitary representation of a nuclear group $T(Q)$ be the
Fourier transform of a translationally quasi-invariant measure
$\m$ on $Q'$. Then one can extend the representation (\ref{q2})
of this group to a unitary strongly continuous representation of a
Weyl CCR group $G(Q)$ in a Hilbert space $L^2_\mathbb C(Q',\m)$ by
the operators (\ref{x11}) in Example \ref{x10}. These operators
read
\mar{qm548}\beq
P_Z(q)\rho(u)=a(q,u)\rho(u+u_q). \label{qm548}
\eeq
Herewith, the following is true.
\begin{theorem} \label{gns78} \mar{gns78}
Equivalent representations of a group $T(Q)$ are extended to
equivalent representations of a Weyl CCR group $G(Q)$.
\end{theorem}
\begin{proof}
Let $\m'$ (\ref{qm581}) be a $\m$-equivalent positive measure of
total mass 1 on $Q'$. The equality
\be
\m'(u+u_q)=s^{-2}(u)a^2(q,u)s^2(u+u_q)\m'(u)
\ee
shows that it also is translationally quasi-invariant. Then the
isomorphism (\ref{qm580}) between representations $T_Z$ and
$T_{Z'}$ of a nuclear Abelian group $T(Q)$ is extended to the
isomorphism
\be
P_{Z'}(q)= s^{-1}P_Z(q)s: \rho(u)\to
s^{-1}(u)a(q,u)s(u+u_q)\rho(u+u_q)
\ee
of the corresponding representations of a Weyl CCR group $G(Q)$.
\end{proof}
Similarly to the case of finite-dimensional Lie groups (Section
4), any strongly continuous unitary representation $T_Z$
(\ref{q2}), $P_Z$ (\ref{qm548}) of a nuclear Weyl CCR group $G(Q)$
implies a representation of its Lie algebra $\ccG(Q)$ by
(unbounded) operators in the same Hilbert space $L^2_\mathbb
C(Q',\m)$ \cite{book05,ccr}. This representation reads
\mar{qm549,62}\ben
&& \f(q)\rho(u)=\lng q,u\rng\rho(u), \qquad
\pi(q)\rho(u)=-i(\dl_q+\eta(q,u))\rho(u), \label{qm549}\\
&& \dl_q\rho(u)=\op\lim_{\al\to 0}\al^{-1}[\rho(u+\al u_q)-\rho(u)],
\qquad \al\in\mathbb R,\nonumber\\
&& \eta(q,u)=\op\lim_{\al\to 0}\al^{-1}[a(\al q,u)-1].\label{qm562}
\een
It follows at once from the relations (\ref{qm555}) that
\be
&& \dl_q\dl_{q'}=\dl_{q'}\dl_q, \qquad
\dl_q(\eta(q',u))=\dl_{q'}(\eta(q,u)),\\
&& \dl_q=-\dl_{-q}, \qquad \dl_q(\lng q',u\rng)=\lng q'|q\rng,\\
&& \eta(0,u)=0, \quad u\in Q',\qquad
\dl_q\thh=0, \quad q\in Q.
\ee
Then it is easily justified that the operators (\ref{qm549})
fulfil the Heisenberg CCR (\ref{qm540}). The unitarity condition
implies the conjugation rule
\be
\lng q,u\rng^*=\lng q,u\rng, \qquad \dl_q^*=-\dl_q -2\eta(q,u).
\ee
Hence, the operators (\ref{qm549}) are Hermitian.
The operators $\pi(q)$ (\ref{qm549}), unlike $\f(q)$, act in a
subspace $E_\infty$ of all smooth complex functions in
$L^2_\mathbb C (Q',\m)$ whose derivatives of any order also
belongs to $L^2_\mathbb C(Q',\m)$. However, $E_\infty$ need not be
dense in a Hilbert space $L^2_\mathbb C(Q',\m)$, unless $Q$ is
finite-dimensional.
A space $E_\infty$ also is a carrier space of a representation of
the universal enveloping algebra $\ol\ccG(Q)$ of a Heisenberg CCR
algebra $\ccG(Q)$. The representations of $\ccG(Q)$ and
$\ol\ccG(Q)$ in $E_\infty$ need not be irreducible. Therefore, let
us consider a subspace $E_\thh=\ol\ccG(Q)\thh$ of $E_\infty$,
where $\thh$ is a cyclic vector for a representation of a Weyl CCR
group in $L^2_\mathbb C(Q',\m)$. Obviously, a representation of a
Heisenberg CCR algebra $\ccG(Q)$ in $E_\thh$ is algebraically
irreducible.
One also introduces creation and annihilation operators
\mar{qm552}\beq
a^\pm(q)=\frac{1}{\sqrt 2}[\f(q)\mp i\pi(q)]=\frac{1}{\sqrt
2}[\mp\dl_q \mp\eta(q,u) + \lng q,u\rng]. \label{qm552}
\eeq
They obey the conjugation rule $(a^\pm(q))^*=a^\mp(q)$ and the
commutation relations
\be
[a^-(q), a^+(q')]=\lng q|q'\rng\bb, \qquad
[a^+(q),a^+(q')]=[a^-(q),a^-(q')]=0.
\ee
The particle number operator $N$ in a carrier space $E_\thh$ is
defined by conditions
\be
[N,a^\pm(q)]=\pm a^\pm(q)
\ee
up to a summand $\la\bb$. With respect to a countable orthonormal
basis $\{q_k\}$, this operator $N$ is given by a sum
\mar{qm591}\beq
N=\op\sum_k a^+(q_k)a^-(q_k), \label{qm591}
\eeq
but need not be defined everywhere in $E_\thh$, unless $Q$ is
finite-dimensional.
Gaussian measures given by the Fourier transform (\ref{spr523})
exemplify a physically relevant class of translationally
quasi-invariant measures on the dual $Q'$ of a nuclear space $Q$.
Their Fourier transforms obey the analiticity condition in Remark
\ref{gns109}.
Let $\m_K$ denote a Gaussian measure on $Q'$ whose Fourier
transform is a generating function
\mar{qm563}\beq
Z_K=\exp[-\frac12 M_K(q)] \label{qm563}
\eeq
with the covariance form
\mar{qm560}\beq
M_K(q)=\lng K^{-1}q|K^{-1}q\rng, \label{qm560}
\eeq
where $K$ is a bounded invertible operator in the Hilbert
completion $\wt Q$ of $Q$ with respect to a Hermitian form
$\lng.|.\rng$. The Gaussian measure $\m_K$ is translationally
quasi-invariant, i.e.,
\be
\m_K(u+u_q)=a_K^2(q,u)\m_K(u).
\ee
Using the formula (\ref{ccr8}), one can show that
\mar{qm561}\beq
a_K(q,u)= \exp[-\frac14 M_K(Sq)- \frac12\lng Sq,u\rng],
\label{qm561}
\eeq
where $S=KK^*$ is a bounded Hermitian operator in $\wt Q$.
Let us construct a representation of a CCR algebra $\ccG(Q)$
determined by the generating function $Z_K$ (\ref{qm563}).
Substituting the function (\ref{qm561}) into the formula
(\ref{qm562}), we obtain
\be
\eta(q,u)= -\frac12\lng Sq,u\rng.
\ee
Hence, the operators $\f(q)$ and $\pi(q)$ (\ref{qm549}) take a
form
\mar{qm565}\beq
\f(q)=\lng q,u\rng, \qquad \pi(q)=-i(\dl_q-\frac12\lng Sq,u\rng).
\label{qm565}
\eeq
Accordingly, the creation and annihilation operators (\ref{qm552})
read
\mar{qm566}\beq
a^\pm(q)=\frac{1}{\sqrt 2}[\mp\dl_q \pm \frac12\lng Sq,u\rng +
\lng q,u\rng]. \label{qm566}
\eeq
They act on the subspace $E_\thh$, $\thh\ap_{\m_K}1$, of a Hilbert
space $L^2_\mathbb C(Q',\m_K)$, and they are Hermitian with
respect to a Hermitian form $\lng.|.\rng_{\m_K}$ on $L^2_\mathbb
C(Q',\m_K)$.
\begin{remark} \label{ccr10} \mar{ccr10}
If a representation of CCR is characterized by the Gaussian
generating function (\ref{qm563}), it is convenient for a
computation to express all operators into the operators $\dl_q$
and $\f(q)$, which obey commutation relations
\be
[\dl_q,\f(q')]=\lng q'|q\rng.
\ee
For instance, we have
\be
\pi(q)=-i\dl_q -\frac{i}{2}\f(Sq).
\ee
The mean values $\lng\f(q_1)\cdots\f(q_n)\dl_q\rng$ vanish, while
the meanvalues $\lng\f(q_1)\cdots\f(q_n)\rng$, defined by the
formula (\ref{ccr8}), obey the Wick theorem relations
\mar{ccr20}\beq
\lng\f(q_1)\cdots\f(q_n)\rng =\sum \lng\f(q_{i_1})\f(q_{i_2})\rng
\cdots \lng\f(q_{i_{n-1}})\f(q_{i_n})\rng, \label{ccr20}
\eeq
where the sum runs through all partitions of a set $1,\ldots,n$ in
ordered pairs $(i_1<i_2),\ldots(i_{n-1}<i_n)$, and where
\be
\lng\f(q)\f(q')\rng=\lng K^{-1}q|K^{-1}q'\rng.
\ee
\end{remark}
In particular, let us put $K=\sqrt2\cdot\bb$. Then the generating
function (\ref{qm563}) takes a form
\mar{qm567}\beq
Z_{\mathrm F}(q)=\exp[-\frac14\lng q|q\rng], \label{qm567}
\eeq
and defines the Fock representation of a Heisenberg CCR algebra
$\ccG(Q)$:
\mar{gns81}\ben
&& \f(q)=\lng q,u\rng, \qquad \pi(q)=-i(\dl_q-\lng q,u\rng), \label{gns81}\\
&& a^+(q)=\frac{1}{\sqrt 2}[-\dl_q + 2\lng q,u\rng], \qquad
a^-(q)=\frac{1}{\sqrt 2}\dl_q. \nonumber
\een
Its carrier space is the subspace $E_\thh$, $\thh\ap_{\m_{\mathrm
F}}1$
of the Hilbert space
$L^2_\mathbb C(Q',\m_{\mathrm F})$, where $\m_{\mathrm F}$ denotes
a Gaussian measure whose Fourier transform is (\ref{qm567}). We
agree to call it the Fock measure.
The Fock representation (\ref{gns81}) up to an equivalence is
characterized by the existence of a cyclic vector $\thh$ such that
\mar{qm570}\beq
a^-(q)\thh=0, \qquad q\in Q. \label{qm570}
\eeq
For the representation in question, this is $\thh\ap_{\m_{\mathrm
F}}1$. An equivalent condition is that the particle number
operator $N$ (\ref{qm591}) exists and its spectrum is lower
bounded. The corresponding eigenvector of $N$ in $E_\thh$ is
$\thh$ itself so that $N\thh=0$. Therefore, it is treated a
particleless vacuum.
A glance at the expression (\ref{qm566}) shows that the condition
(\ref{qm570}) does not hold, unless $Z_K$ is $Z_{\mathrm F}$
(\ref{qm567}). For instance, the particle number operator in the
representation (\ref{qm566}) reads
\be
&& N=\op\sum_j a^+(q_j)a^-(q_j)= \op\sum_j[-\dl_{q_j}\dl_{q_j}
+S^j_k\lng q_k,u\rng\dr_{q_j} + \\
&& \qquad
(\dl_{km}-\frac14 S^j_kS^j_m)\lng q_k,u\rng\lng q_m,u\rng -
(\dl_{jj}-\frac12 S^j_j)],
\ee
where $\{q_k\}$ is an orthonormal basis for a pre-Hilbert space
$Q$. One can show that this operator is defined everywhere on
$E_\thh$ and is lower bounded only if the operator $S$ is a sum
of the scalar operator $2\cdot\bb$ and a nuclear operator in $\wt
Q$, in particular, if
\be
\mathrm{Tr}(\bb -\frac12 S)<\infty.
\ee
This condition also is sufficient for measures $\m_K$ and $\m_F$
(and, consequently, the corresponding representations) to be
equivalent \cite{gelf64}. For instance, a generating function
\be
Z_c(q)=\exp[-\frac{c^2}{2}\lng q|q\rng], \qquad c^2\neq \frac12,
\ee
defines a non-Fock representation of nuclear CCR.
\begin{remark} \label{gns79} \mar{gns79}
Since the Fock measure $\m_{\mathrm F}$ on $Q'$ remains equivalent
only under translations by vectors $u_q\in Q\subset Q'$, the
measure
\be
\m_\si(u) =\m_{\mathrm F}(u+\si), \qquad \si\in Q'\setminus Q,
\ee
on $Q'$ determines a non-Fock representation of nuclear CCR.
Indeed, this measure is translationally quasi-invariant:
\be
\m_\si(u+u_q)=a^2_\si(q,u)\m_\si(u),\qquad a_\si(q,u)=a_{\mathrm
F}(q,u-\si),
\ee
and its Fourier transform
\be
Z_\si(q)=\exp[i\lng q,\si\rng]Z_{\mathrm F}(q)
\ee
is a positive-definite continuous function on $Q$. Then the
corresponding representation of a CCR algebra is given by
operators
\mar{qm597}\beq
a^+(q)=\frac{1}{\sqrt 2}(-\dl_q + 2\lng q,u\rng -\lng
q,\si\rng),\qquad a^-(q)=\frac{1}{\sqrt 2}(\dl_q + \lng
q,\si\rng). \label{qm597}
\eeq
In comparison with the all above mentioned representations, these
operators possess non-vanishing vacuum mean values
\be
\lng a^\pm(q)\thh|\thh\rng_{\m_{\mathrm F}}=\lng q,\si\rng.
\ee
If $\si\in Q\subset Q'$, the representation (\ref{qm597}) becomes
equivalent to the Fock representation (\ref{gns81}) due to a
morphism
\be
\rho(u)\to \exp[-\lng \si,u\rng]\rho(u+\si).
\ee
\end{remark}
\begin{remark} \label{gns80} \mar{gns80}
Let us note that the non-Fock representation (\ref{qm565}) of the
CCR algebra (\ref{qm540}) in a Hilbert space $L^2_\mathbb
C(Q',\m_K)$ is the Fock representation
\be
\f_K(q)=\f(q)=\lng q,u\rng,\qquad \pi_K(q)=\pi(S^{-1}q) =
-i(\dl^K_q -\frac12\lng q,u\rng), \qquad \dl^K_q =\dl_{S^{-1}q},
\ee
of a CCR algebra $\{\f_K(q),\pi_K(q),I\}$, where
\be
[\f_K(q),\pi_K(q)]=i\lng K^{-1}q|K^{-1}q'\rng I.
\ee
\end{remark}
\section{Free quantum fields}
There are two main algebraic formulation of QFT. In the framework
of the first one, called local QFT, one associates to a certain
class of subsets of a Minkowski space a net of von Neumann, $C^*$-
or $Op^*$-algebras which obey certain axioms
\cite{araki,buch,haag,halv,hor}. Its inductive limit is called
either a global algebra (in the case of von Neumann algebras) or a
quasilocal algebra (for a net of $C^*$-algebras). This
construction is extended to non-Minkowski spaces, e.g., globally
hyperbolic spacetimes \cite{brun,brun2,ruzz}.
In a different formulation of algebraic QFT with reference to the
field-particle dualism, realistic quantum field models are
described by tensor algebras, as a rule.
Let $Q$ be a nuclear space. Let us consider the direct limit
$\wh\otimes Q$ (\ref{x2}) of the vector spaces $\wh\ot^{\leq n} Q$
(\ref{x2a}) where $\wh\ot$ is the topological tensor product with
respect to Grothendieck's topology. As was mentioned above,
provided with the inductive limit topology, the tensor algebra
$\wh\otimes Q$ (\ref{x2}) is a unital nuclear $b^*$-algebra
(Example \ref{gns90}). Therefore, one can apply GNS representation
Theorem \ref{gns36} to it. A state $f$ of this algebra is given by
a tuple $\{f_n\}$ of continuous forms on the tensor algebra $A_Q$
(\ref{gns100}). Its value $f(q^1\cdots q^n)$ are interpreted as
the vacuum expectation of a system of fields $q^1,\ldots,q^n$.
In algebraic QFT, one usually choose by $Q$ the Schwartz space of
functions of rapid decrease.
\begin{remark} \label{spr451} \mar{spr451}
By functions of rapid decrease on an Euclidean space $\mathbb R^n$
are called complex smooth functions $\psi(x)$ such that the
quantities
\mar{spr453}\beq
\|\psi\|_{k,m}=\op\max_{|\al|\leq k} \op\sup_x(1+x^2)^m|D^\al
\psi(x)| \label{spr453}
\eeq
are finite for all $k,m\in \mathbb N$. Here, we follow the
standard notation
\be
D^\al=\frac{\dr^{|\al|}}{\dr^{\al_1} x^1\cdots\dr^{\al_n}x^n},
\qquad |\al|=\al_1+\cdots +\al_n,
\ee
for an $n$-tuple of natural numbers $\al=(\al_1,\ldots,\al_n)$.
The functions of rapid decrease constitute a nuclear space
$S(\mathbb R^n)$ with respect to the topology determined by the
seminorms (\ref{spr453}). Its dual is a space $S'(\mathbb R^n)$ of
tempered distributions \cite{bog,gelf64,piet}. The corresponding
contraction form is written as
\be
\lng \psi,h\rng=\op\int \psi(x) h(x) d^nx, \qquad \psi\in
S(\mathbb R^n), \qquad h\in S'(\mathbb R^n).
\ee
A space $S(\mathbb R^n)$ is provided with a non-degenerate
separately continuous Hermitian form
\be
\lng \psi|\psi'\rng=\int \psi(x)\ol{\psi'(x)}d^nx.
\ee
The completion of $S(\mathbb R^n)$ with respect to this form is a
space $L^2_C(\mathbb R^n)$ of square integrable complex functions
on $\mathbb R^n$. We have a rigged Hilbert space
\be
S(\mathbb R^n)\subset L^2_C(\mathbb R^n) \subset S'(\mathbb R^n).
\ee
Let $\mathbb R_n$ denote the dual of $\mathbb R^n$ coordinated by
$(p_\la)$. The Fourier transform
\mar{spr460,1}\ben
&& \psi^F(p)=\int \psi(x)e^{ipx}d^nx, \qquad px=p_\la x^\la,
\label{spr460}\\
&& \psi(x)=\int \psi^F(p)e^{-ipx}d_np, \qquad d_np=(2\pi)^{-n}d^np,
\label{spr461}
\een
defines an isomorphism between the spaces $S(\mathbb R^n)$ and
$S(\mathbb R_n)$. The Fourier transform of tempered distributions
is given by the condition
\be
\int h(x)\psi(x)d^nx=\int h^F(p)\psi^F(-p)d_np,
\ee
and it is written in the form (\ref{spr460}) -- (\ref{spr461}). It
provides an isomorphism between the spaces of tempered
distributions $S'(\mathbb R^n)$ and $S'(\mathbb R_n)$.
\end{remark}
For the sake of simplicity, we here restrict our consideration to
real scalar fields and choose by $Q$ the real subspace $RS^4$ of
the Schwartz space $S(\mathbb R^4)$ of smooth complex functions of
rapid decrease on $\mathbb R^4$ \cite{ccr}. Since a subset
$\op\ot^nS(\mathbb R^k)$ is dense in $S(\mathbb R^{kn})$, we
henceforth identify the tensor algebra $A_{RS^4}$ (\ref{gns100})
of a nuclear space $RS^4$ with the algebra
\mar{qm801}\beq
A=\mathbb R\oplus RS^4\oplus RS^8\oplus\cdots, \label{qm801}
\eeq
called the Borchers algebra \cite{borch,hor,ccr}. Any state $f$ of
this algebra is represented by a collection of tempered
distributions $\{W_k\in S'(\mathbb R^{4k})\}$ by the formula
\be
f(\psi_k)= \int
W_k(x_1,\ldots,x_k)\psi_k(x_1,\ldots,x_k)d^4x_1\cdots d^4x_k,
\qquad \psi_k\in RS^{4k}.
\ee
For instance, the states of scalar quantum fields in a Minkowski
space $\mathbb R^4$ are described by the Wightman functions
$W_n\subset S'(\mathbb R^{4k})$ in the Minkowski space which obey
the Garding--Wightman axioms of axiomatic QFT \cite{bog,wigh,zin}.
Let us mention the Poincar\'e covariance axiom, the condition of
the existence and uniqueness of a vacuum $\thh_0$, and the
spectrum condition. They imply that: (i) a carrier Hilbert space
$E_W$ of Wightman quantum fields admits a unitary representation
of a Poinar\'e group, (ii) a space $E_W$ contains a unique (up to
scalar multiplications) vector $\psi_0$, called the vacuum vector,
invariant under Poincar\'e transformations, (iii) the spectrum of
an energy-momentum operator lies in the closed positive light
cone. In particular, the Poincar\'e covariance condition implies
the translation invariance and the Lorentz covariance of Wightman
functions. Due to the translation invariance of Wightman functions
$W_k$, there exist tempered distributions $w_k\in S'(\mathbb
R^{4k-4})$, also called Wightman functions, such that
\mar{1278}\beq
W_k(x_1,\ldots,x_k)= w_k(x_1-x_2,\ldots,x_{k-1}-x_k). \label{1278}
\eeq
Note that Lorentz covariant tempered distributions for one
argument only are well described \cite{bog,zin07}. In order to
modify Wightman's theory, one studies different classes of
distributions which Wightman functions belong to
\cite{solov,thom}.
Let us here focus on states of the Borchers algebra $A$
(\ref{qm801}) which describe free quantum scalar fields of mass
$m$ \cite{ccr,axiom}.
Let us provide a nuclear space $RS^4$ with a positive complex
bilinear form
\mar{qm802,x5}\ben
&& (\psi|\psi')_m=\frac{2}{i}\int \psi(x)D^-_m(x-y)\psi'(y)d^4xd^4y=\int
\psi^F(-\om,-\op p^\to)\psi'^F(\om,\op p^\to)\frac{d_3p}{\om},
\label{qm802}\\
&& D^-_m(x)=i(2\pi)^{-3}\int \exp[-ipx]\thh(p_0)\dl(p^2-m^2)d^4p,
\label{x5}\\
&& \om=({\op p^\to}^2 +m^2)^{1/2}, \nonumber
\een
where $p^2$ is the Minkowski square, $\thh(p_0)$ is the Heaviside
function, and $D^-_m(x)$ is the negative frequency part of the
Pauli--Jordan function
\mar{gns120}\beq
D_m(x)=i(2\pi)^{-3}\int
\exp[-ipx](\thh(p_0)-\thh(-p_0))\dl(p^2-m^2)d^4p. \label{gns120}
\eeq
Since a function $\psi(x)$ is real, its Fourier transform
(\ref{spr460}) satisfies an equality $\psi^F(p)=\ol\psi^F(-p)$.
The bilinear form (\ref{qm802}) is degenerate because the
Pauli--Jordan function $D^-_m(x)$ obeys a mass shell equation
\be
(\Box +m^2)D^-_m(x)=0.
\ee
It takes non-zero values only at elements $\psi^F\in RS_4$ which
are not zero on a mass shell $p^2=m^2$. Therefore, let us consider
the quotient space
\mar{gns121}\beq
\g_m:RS^4\to RS^4/J, \label{gns121}
\eeq
where
\be
J=\{\psi\in RS^4\, :\, (\psi|\psi)_m=0\}
\ee
is the kernel of the square form (\ref{qm802}). The map $\g_m$
(\ref{gns121}) assigns the couple of functions $(\psi^F(\om,\op
p^\to),\psi^F(-\om,\op p^\to))$ to each element $\psi\in RS^4$
with a Fourier transform $\psi^F(p_0,\op p^\to)\in RS_4$. Let us
equip the factor space $RS^4/J$ with a real bilinear form
\mar{qm803}\ben
&& (\g\psi|\g\psi')_L=\mathrm{Re}(\psi|\psi')= \label{qm803}\\
&& \qquad \frac12\int [\psi^F(-\om,-\op p^\to)\psi'^F(\om,\op p^\to)
+\psi^F(\om,-\op p^\to) \psi'^F(-\om,\op p^\to)]\frac{d_3\op
p^\to}{\om}. \nonumber
\een
Then it is decomposed into a direct sum $RS^4/J=L^+\oplus L^-$ of
subspaces
\be
L^\pm=\{\psi^F_\pm(\om,\op p^\to)=\frac12(\psi^F(\om,\op p^\to)
\pm\psi^F(-\om,\op p^\to))\},
\ee
which are mutually orthogonal with respect to the bilinear form
(\ref{qm803}).
There exist continuous isometric morphisms
\be
\g_+:\psi^F_+(\om,\op p^\to) \to q^F(\op p^\to)=\om^{-1/2}\psi^F_+
(\om,\op p^\to),\qquad \g_-:\psi^F_-(\om,\op p^\to) \to q^F(\op
p^\to)=-i\om^{-1/2}\psi^F_- (\om,\op p^\to)
\ee
of spaces $L^+$ and $L^-$ to a nuclear space $RS^3$ endowed with a
non-degenerate separately continuous Hermitian form
\mar{qm807}\beq
\lng q|q'\rng=\int q^F(-\op p^\to)q'^F(\op p^\to)d_3p.
\label{qm807}
\eeq
It should be emphasized that the images $\g_+(L^+)$ and
$\g_-(L^-)$ in $RS^3$ are not orthogonal with respect to the
scalar form (\ref{qm807}). Combining $\g_m$ (\ref{gns121}) and
$\g_\pm$, we obtain continuous morphisms $\tau_\pm: RS^4\to RS^3$
given by the expressions
\be
&& \tau_+(\psi)=\g_+(\g_m\psi)_+=\frac{1}{2\om^{1/2}}\int[\psi^F(\om,\op
p^\to) + \psi^F(-\om,\op p^\to)]\exp[-i\op p^\to\op x^\to]d_3p,\\
&& \tau_-(\psi)=\g_-(\g_m\psi)_-=\frac{1}{2i\om^{1/2}}\int[\psi^F(\om,\op
p^\to) - \psi^F(-\om,\op p^\to)]\exp[-i\op p^\to\op x^\to]d_3p.
\ee
Now let us consider a Heisenberg CCR algebra
\mar{1110}\beq
\ccG(RS^3)=\{(\f(q),\pi(q)), I,\,q\in RS^3\} \label{1110}
\eeq
modelled over a nuclear space $RS^3$, which is equipped with the
Hermitian form (\ref{qm807}) (Section 8). Using the morphisms
$\tau_\pm$, let us define a map
\mar{1111}\beq
\G_m: RS^4\ni \psi \to \f(\tau_+(\psi)) -\pi(\tau_-(\psi))\in
\ccG(RS^3). \label{1111}
\eeq
With this map, one can think of (\ref{1110}) as being the algebra
of instantaneous CCR of scalar fields on a Minkowski space
$\mathbb R^4$. Owing to the map (\ref{1111}), any representation
of the Heisenberg CCR algebra $\ccG(RS^3)$ (\ref{1110}) defined
by a translationally quasi-invariant measure $\m$ on $S'(\mathbb
R^3)$ induces a state
\mar{qm805}\beq
f_m(\psi^1\cdots\psi^n)=\lng\f(\tau_+(\psi^1))
+\pi(\tau_-(\psi^1))] \cdots [\f(\tau_+(\psi^n))
+\pi(\tau_-(\psi^n))]\rng \label{qm805}
\eeq
of the Borchers algebra $A$ (\ref{qm801}). Furthermore, one can
justify that the corresponding distributions $W_n$ fulfil the mass
shell equation and that the following commutation relation holds:
\be
W_2(x,y) -W_2(y,x)=-iD_m(x-y),
\ee
where $D_m(x-y)$ is the Pauli--Jordan function (\ref{gns120}).
Thus, the state (\ref{qm805}) of the Borchers algebra $A$
(\ref{qm801}) describes quantum scalar fields of mass $m$.
For instance, let us consider the Fock representation $Z_{\mathrm
F}(q)$ (\ref{qm567}) of the Heisenberg CCR algebra $\ccG(RS^3)$
(\ref{1110}). Using the formulae in Remark \ref{ccr10} where a
form $\lng q|q'\rng$ is given by the expression (\ref{qm807}), one
observes that the state $f_m$ (\ref{qm805}) satisfies the Wick
theorem relations
\mar{1263}\beq
f_m(\psi^1\cdots\psi^n)=\op\sum_{(i_1\ldots i_n)}
f_2(\psi^{i_1}\psi^{i_2})\cdots f_2(\psi^{i_{n-1}}\psi^{i_n}),
\label{1263}
\eeq
where a state $f_2$ is given by the Wightman function
\mar{ccr21}\beq
W_2(x,y)=\frac{1}{i}D^-_m(x-y). \label{ccr21}
\eeq
Thus, the state $f_m$ (\ref{1263}) describes free quantum scalar
fields of mass $m$.
Similarly, one can obtain states of the Borchers algebra $A$
(\ref{qm801}) generated by non-Fock representations (\ref{qm563})
of the instantaneous CCR algebra $\ccG(RS^3)$, e.g., if
$K^{-1}=c\bb\neq 2^{-1/2}\bb$. These states fail to be defined by
Wightman functions.
It should be emphasized that, given a different mass $m'$, we have
a different map $\G_{m'}$ (\ref{1111}) of the Borchers algebra $A$
(\ref{qm801}) to the Heisenberg CCR algebra $\ccG(RS^3)$
(\ref{1110}). Accordingly, the Fock representation $Z_{\mathrm
F}(q)$ (\ref{qm567}) of the Heisenberg CCR algebra $\ccG(RS^3)$
(\ref{1110}) yields the state $f_{m'}$ (\ref{1263}) where a state
$f_2$ is given by the Wightman function
\mar{ccr21'}\beq
W_2(x,y)=\frac{1}{i}D^-_{m'}(x-y). \label{ccr21'}
\eeq
If $m\neq m'$, the states $f_m$ and $f_{m'}$ (\ref{1263}) the
Borchers algebra $A$ (\ref{qm801}) are inequivalent because its
representations $\G_m$ and $\G_{m'}$ (\ref{1111}) possess
different kernels.
\section{Euclidean QFT}
In QFT, interacting quantum fields created at some instant and
annihilated at another one are described by complete Green
functions. They are given by the chronological functionals
\mar{1260,030}\ben
&&f^c(\psi_k)= \int
W_k^c(x_1,\ldots,x_k)\psi_k(x_1,\ldots,x_k)d^4x_1\cdots d^4x_k,
\qquad \psi_k\in RS^{4k}, \label{1260}\\
&& W^c_k(x_1,\ldots,x_k)= \op\sum_{(i_1\ldots
i_k)}\thh(x^0_{i_1}-x^0_{i_2})
\cdots\thh(x^0_{i_{k-1}}-x^0_{i_n})W_k(x_1,\ldots,x_k),
\label{030}
\een
where $W_k\in S'(\mathbb R^{4k})$ are tempered distributions, and
the sum runs through all permutations $(i_1\ldots i_k)$ of the
tuple of numbers $1,\ldots,k$ \cite{bog2}.
A problem is that the functionals $W^c_k$ (\ref{030}) need not be
tempered distributions. For instance, $W^c_1\in S'(\mathbb R)$ iff
$W_1\in S'(\mathbb R_\infty)$, where $\mathbb R_\infty$ is the
compactification of $\mathbb R$ by means of a point
$\{+\infty\}=\{-\infty\}$ \cite{bog}. Moreover, chronological
forms are not positive. Therefore, they do not provide states of
the Borchers algebra $A$ (\ref{qm801}) in general.
At the same time, the chronological forms (\ref{030}) come from
the Wick rotation of Euclidean states of the Borchers algebra
\cite{sard91,ccr,axiom}. As is well known, the Wick rotation
enables one to compute the Feynman diagrams of perturbed QFT by
means of Euclidean propagators. Let us suppose that it is not a
technical trick, but quantum fields in an interaction zone are
really Euclidean. It should be emphasized that the above mentioned
Euclidean states differ from the well-known Schwinger functions in
the Osterwalder--Shraded Euclidean QFT
\cite{bog,ost,ccr,axiom,schlin,zin}. The Schwinger functions are
the Laplace transform of Wightman functions, but not chronological
forms.
Since the chronological forms (\ref{030}) are symmetric, the
Euclidean states of a Borchers algebra $A$ can be obtained as
states of the corresponding commutative tensor algebra $B_{RS^4}$
(\ref{gns101}) \cite{sard91,ccr,axiom}. Provided with the direct
sum topology, $B_{RS^4}$ becomes a topological involutive algebra.
It coincides with the enveloping algebra of the Lie algebra of an
additive Lie group $T(RS^4)$ of translations in $RS^4$. Therefore,
one can obtain states of an algebra $B_{RS^4}$ by constructing
cyclic strongly continuous unitary representations of a nuclear
Abelian group $T(RS^4)$ (Section 7). Such a representation is
characterized by a continuous positive-definite generating
function $Z$ on $SR^4$. By virtue of Bochner Theorem \ref{spr525},
this function is the Fourier transform
\mar{031}\beq
Z(\phi)=\int \exp[i \langle\phi,y\rangle]d\mu(y) \label{031}
\eeq
of a positive measure $\mu$ of total mass 1 on the dual $(RS^4)'$
of $RS^4$. Then the above mentioned representation of $T(RS^4)$
can be given by operators
\mar{qq2}\beq
\wh\phi \rho(y)=\exp[i\langle \f,y\rangle]\rho(y) \label{qq2}
\eeq
in a Hilbert space $L_{\mathbb C}^2((RS^4)',\m)$ of the
equivalence classes of square $\m$-integrable complex functions
$\rho(y)$ on $(RS^4)'$. A cyclic vector $\thh$ of this
representation is the $\m$-equivalence class $\thh\ap_\m 1$ of the
constant function $\rho(y)=1$.
Conversely, every positive measure $\m$ of total mass 1 on the
dual $(RS^4)'$ of $RS^4$ defines the cyclic strongly continuous
unitary representation (\ref{qq2}) of a group $T(RS^4)$. Herewith,
distinct generating functions $Z$ and $Z'$ characterize equivalent
representations $T_Z$ and $T_{Z'}$ (\ref{qq2}) of $T(RS^4)$ in the
Hilbert spaces $L^2_{\mathbb C}((RS^4)',\m)$ and $L^2_{\mathbb
C}((RS^4)',\m')$ iff they are the Fourier transform of equivalent
measures on $(RS^4)'$ (Theorem \ref{gns77}).
If a generating function $Z$ obeys the analiticity condition in
Remark \ref{gns109}, a state $f$ of $B_{RS^4}$ is given by the
expression
\mar{w0}\beq
f_k(\phi_1\cdots\phi_k)=i^{-k}\frac{\dr}{\dr \al^1}
\cdots\frac{\dr}{\dr\alpha^k}Z(\alpha^i\phi_i)|_{\alpha^i=0}=\int\langle
\phi_1,y\rangle\cdots\langle \phi_k,y \rangle d\mu(y). \label{w0}
\eeq
Then one can think of $Z$ (\ref{031}) as being a generating
functional of complete Euclidean Green functions $f_k$ (\ref{w0}).
For instance, free Euclidean fields are described by Gaussian
states. Their generating functions are of the form
\mar{a10}\beq
Z(\phi)=\exp(-\frac12M(\f,\f)), \label{a10}
\eeq
where $M(\f,\f)$ is a positive-definite Hermitian bilinear form on
$RS^4$ continuous in each variable. They are the Fourier transform
of some Gaussian measure on $(RS^4)'$. In this case, the forms
$f_k$ (\ref{w0}) obey the Wick relations (\ref{ccr20}) where
\be
f_1=0, \qquad f_2(\f,\f')=M(\f,\f').
\ee
Furthermore, a covariance form $M$ on $RS^4$ is uniquely
determined as
\mar{1270}\beq
M(\f_1,\f_2)=\int W_2(x_1,x_2)\f_1(x_1)\f_2(x_2) \label{1270}
d^nx_1d^nx_2.
\eeq
by a tempered distribution $W_2\in S'(\mathbb R^8)$.
In particular, let a tempered distribution $M(\f,\f')$ in the
expression (\ref{1270}) be Green's function of some positive
elliptic differential operator $\cE$, i.e.,
\be
\cE_{x_1}W_2(x_1,x_2)=\dl(x_1-x_2),
\ee
where $\dl$ is Dirac's $\dl$-function. Then the distribution $W_2$
reads
\mar{1272}\beq
W_2(x_1,x_2)=w(x_1-x_2), \label{1272}
\eeq
and we obtain a form
\be
&& f_2(\f_1\f_2)=M(\f_1,\f_2)=\int w(x_1-x_2)\f_1(x_1)\f_2(x_2)
d^4x_1 d^4x_2=\\
&& \qquad \int w(x)\f_1(x_1)\f_2(x_1-x)d^4x d^4x_1=\int w(x)\vf(x)d^4x=
\int w^F(p)\vf^F(-p) d_4p, \\
&& x=x_1-x_2, \qquad \vf(x)=\int \f_1(x_1)\f_2(x_1-x)d^4x_1.
\ee
For instance, if
\be
\cE_{x_1} =-\Delta_{x_1}+m^2,
\ee
where $\Delta$ is the Laplacian, then
\mar{1271}\beq
w(x_1-x_2)=\int\frac{\exp(-iq(x_1-x_2))}{p^2+m^2}d_4p,
\label{1271}
\eeq
where $p^2$ is the Euclidean square, is the propagator of a
massive Euclidean scalar field. Note that, restricted to the
domain $(x^0_1-x^0_2)<0$, it coincides with the Schwinger function
$s_2(x_1-x_2)$.
Let $w^F$ be the Fourier transform of the distribution $w$
(\ref{1272}). Then its Wick rotation is the functional
\be
\wt w(x)=\thh(x)\op\int_{\ol Q_+}w^F(p)\exp(-px)d_4p +
\thh(-x)\op\int_{\ol Q_-}w^F(p)\exp(-px)d_4p
\ee
on scalar fields on a Minkowski space \cite{ccr,axiom}. For
instance, let $w(x)$ be the Euclidean propagator (\ref{1271}) of a
massive scalar field. Then due to the analyticity of
\be
w^F(p)=(p^2+m^2)^{-1}
\ee
on the domain $\im p\cdot \re p>0$, one can show that $\wt
w(x)=-iD^c(x)$ where $D^c(x)$ is familiar causal Green's function.
A problem is that a measure $\m$ in the generating function $Z$
(\ref{031}) fails to be written in an explicit form.
At the same time, a measure $\mu$ on $(RS^4)'$ is uniquely defined
by a set of measures $\mu_N$ on the finite-dimensional spaces
$\mathbb R_N=(RS^4)'/E$ where $E\subset (RS^4)'$ denotes a
subspace of forms on $RS^4$ which are equal to zero at some
finite-dimensional subspace $\mathbb R^N\subset RS^4$. The
measures $\mu_N$ are images of $\mu$ under the canonical mapping
$(RS^4)'\to\mathbb R_N$. For instance, every vacuum expectation
$f_n(\f_1\cdots\f_n)$ (\ref{w0}) admits the representation by an
integral
\mar{S8}\beq
f_k(\f_1\cdots\f_k)=\int\langle w,\f_1\rangle\cdots\langle w,\f_k
\rangle d\mu_N(w) \label{S8}
\eeq
for any finite-dimensional subspace $\mathbb R^N$ which contains
$\f_1,\ldots,\f_k$. In particular, one can replace the generating
function (\ref{031}) by the generating function
\be
Z_N(\la_ie^i)=\int\exp(i\la_iw^i)\mu_N(w^i)
\ee
on $\mathbb R^N$ where $\{e^i\}$ is a basis for $\mathbb R^N$ and
$\{w^i\}$ are coordinates with respect to the dual basis for
$\mathbb R_N$. If $f$ is a Gaussian state, we have the familiar
expression (\ref{qm610}):
\mar{S9}\beq
d\mu_N=(2\pi\det[M^{ij}])^{-N/2}\exp[-\frac12(M^{-1})_{ij}w^iw^j]
d^Nw, \label{S9}
\eeq
where $M^{ij}=M(e^i,e^j)$ is a non-degenerate covariance matrix.
The representation (\ref{S8}) however is not unique, and the
measure $\mu_N$ depends on the specification of a
finite-dimensional subspace $\mathbb R^N$ of $RS^4$.
\begin{remark} \label{gns133} \mar{gns133}
Note that an expression
\mar{gns134}\beq
\exp(-\int L(\f)d^4x)\op\prod_x[d\f(x)] \label{gns134}
\eeq
conventionally used in perturbed QFT is a symbolic functional
integral, but not a true measure \cite{glimm,john,schmit}. In
particular, it is translationally invariant, i.e.,
\be
[d\f(x)]=[d(\f(x)+ \mathrm{const}.)],
\ee
whereas there is no (translationally invariant) Lebesgue measure
on infinite-dimensional vector space as a rule (see \cite{versh}
for an example of such a measure).
\end{remark}
\section{Higgs vacuum}
In contrast to the formal expression (\ref{gns134}) of perturbed
QFT, the true integral representation (\ref{031}) of generating
functionals enables us to handle non-Gaussian and inequivalent
Gaussian representations of the commutative tensor algebra $B_Q$
(\ref{gns101}) of Euclidean scalar fields. Here, we describe one
of such a representation as a model of a Higgs vacuum
\cite{NC91,sard91}.
In Standard Model of particle physics, Higgs vacuum is represented
as a constant background part $\si_0$ of a Higgs scalar field
$\si$ \cite{nov,SM}. In algebraic QFT, one can describe free Higgs
field similar to matter fields by a commutative tensor algebra
$B_\Si$ where $\Si$ is a real nuclear space.
Let $Z(\wh\si)$ be the generating function (\ref{a10}) of a
Gaussian state of $B_\Si$, and let $\m$ be the corresponding
Gaussian measure on the dual $\Si'$ of $\Si$. In contrast with a
finite-dimensional case, Gaussian measures on infinite-dimensional
spaces fail to be quasi-invariant under translations as a rule.
The introduction of a Higgs vacuum means a translation
\be
\g:\Si'\ni \si\to \si+\si_0\in\Si', \qquad \si_0\in\Si'
\ee
in a space $\Si'$ such that an original Gaussian measure $\m(\si)$
is replaced by a measure $\mu_{\si_0}(\si)=\m(\si+\si_0)$
possessing the Fourier transform
\be
Z_{\si_0}(\wh\si)=\exp(i\lng\wh\si,\si_0\rng Z(\wh\si)
\ee
The measures $\mu$ and $\mu_{\si_0}$ are equivalent iff a vector
$\si_0\in\Si'$ belongs to the canonical image of $\Si$ in $\Si'$
with respect to the scalar form $\langle|\rangle=M(,)$ (Remark
\ref{gns79}). Then the measures $\mu$ and $\mu_{\si_0}$ define the
equivalent states (\ref{w0}) of an algebra $B_\Si$. This
equivalence is performed by the unitary operator
\be
\rho(\si)\to \exp(-\langle \si|\si_0\rangle)\rho(\si+\si_0),
\qquad \rho(\si)\in L^2(\Si',\mu).
\ee
This operator fails to be constructed if $\si_0\in
\Si'\setminus\Si$, and the measures $\m$ and $\m_{\si_0}$ are
inequivalent.
Following the terminology of Standard Model, let us call $\si_0\in
\Si'\setminus\Si$ the Higgs vacuum field and $\si\in\Si\subset
\Si'$ the Higgs boson fields. Then we can say the following.
(i) A Higgs vacuum field $\si_0$ and Higgs boson fields $\si$
belong to different classes of functions. For instance, one
usually chooses a constant Higgs vacuum field $\si_0$ in QFT. If
$\Si=RS^4$, a constant function is an element of $(SR^4)'\setminus
SR^4$. At the same time, since $\Si$ is dense in $\Si'$, the
elements $\si_0$ and $\si$ can be arbitrarily closed to each other
with respect to a topology in $\Si'$. However, a covariance form
$M$ and some other functions being well defined at points $\si$
become singular at points $\si_0$.
(ii) One can think of a Higgs vacuum field $\si_0$ as being the
classical one in the sense that $\si_0\in \Si'\setminus\Si$,
whereas Higgs boson fields $\si\in \Si\subset \Si'$ are quantized
fields because the possess quantum partners $\wh\si\in\Si\subset
B_\Si$.
(iii) States of Higgs boson fields in the presence of in
equivalent Higgs vacua are inequivalent.
(iv) Let the generating function $Z$ and $Z_{\si_0}$ be restricted
to some finite-dimensional subspace $\mathbb R^N\subset\Si$. Then
there exists an element $\si_{0N}\in\mathbb R_N$ such that
$\lng\wh\si,\si_0\rng=\lng\wh\si|\si_{0N}\rng$ for any
$\wh\si\in\mathbb R^N$. As a consequence, a generating function
$Z_{\si_0}$ takes a form
\be
Z_{\si_0N}(\la_i\wh\si^i)=
(2\pi\det[M^{ij}])^{-N/2}\int\exp(\la_i\si^i)
\exp[-\frac12(M^{-1})_{ij}(\si-\si_{0N})^i(\si -\si_{0N})^j]
d^N\si,
\ee
where $M^{ij}$ is the covariance matrix of $Z_N$, $\si^i$ denote
coordinates in $\mathbb R_N$, and $\si^i_{0N}$ are coordinates of
a vector $\si_{0N}$ in $\mathbb R_N$. It follows that, if a number
of quantum Higgs boson fields $\\wh\si$ is finite, their
interaction with a classical Higgs vacuum field $\si_0$ reduced to
an interaction with some quantum fields $\wh\si_{0N}$ by
perturbation theory.
In Standard Model, a Higgs vacuum is responsible for spontaneous
symmetry breaking. Let us study this phenomenon (Sections 12 and
13).
\section{Automorphisms of quantum systems}
In order to say something, we mainly restrict our analysis to
automorphisms of $C^*$ algebras.
We consider uniformly and strongly continuous one-parameter groups
of automorphisms of $C^*$-algebras. Let us note that any weakly
continuous one-parameter group of endomorphism of a $C^*$-algebra
also is also strongly continuous and their weak and strong
generators coincide with each other \cite{brat,book05}.
\begin{remark} \label{w416} \mar{w416}
There is the following relation between morphisms of a
$C^*$-algebra $A$ and a set $F(A)$ of its states which is a convex
subset of the dual $A'$ of $A$ (Theorem \ref{gns17}). A linear
morphism $\g$ of a $C^*$-algebra $A$ as a vector space is called
the Jordan morphism if relations
\be
\g(ab+ba)=\g(a)\g(b)+\g(b)\g(a), \qquad \f(a^*)=\g(a)^*, \qquad
a,b\in A.
\ee
hold. One can show the following \cite{emch}. Let $\g$ be a Jordan
automorphism of a unital $C^*$-algebra $A$. It yields the dual
weakly$^*$ continuous affine bijection $\g'$ of $F(A)$ onto
itself, i.e.,
\be
\g'(\la f+(1-\la)f')=\la\g'(f) +(1-\la)\g'(f'), \qquad f,f',\in
F(A), \qquad \la\in [0,1].
\ee
Conversely, any such a map of $F(A)$ is the dual to some Jordan
automorphism of $A$. However, if $G$ is a connected group of
weakly continuous Jordan automorphisms of a unital $C^*$-algebra
$A$ is a weakly (and, consequently, strongly) continuous group of
automorphisms of $A$.
\end{remark}
A topological group $G$ is called the strongly (resp. uniformly)
continuous group of automorphisms of a $C^*$-algebra $A$ if there
is its continuous monomorphism to the group $\Is(A)$ of
automorphisms of $A$ provided with the strong (resp. normed)
operator topology, and if its action on $A$ is separately
continuous.
One usually deals with strongly continuous groups of automorphisms
because of the following reason. Let $G(\mathbb R)$ be
one-parameter group of automorphisms of a $C^*$-algebra $A$. This
group is uniformly (resp. strongly) continuous if it is a range
of a continuous map of $\mathbb R$ to the group $\Is(A)$ of
automorphisms of $A$ which is provided with the normed (resp.
strong) operator topology and whose action on $A$ is separately
continuous. A problem is that, if a curve $G(\mathbb R)$ in
$\Is(A)$ is continuous with respect to the normed operator
topology, then a curve $G(\mathbb R)(a)$ for any $a\in A$ is
continuous in a $C^*$-algebra $A$, but the converse is not true.
At the same time, a curve $G(\mathbb R)$ is continuous in $\Is(A)$
with respect to the strong operator topology iff a curve
$G(\mathbb R)(a)$ for any $a\in A$ is continuous in $A$. By this
reason, strongly continuous one-parameter groups of automorphisms
of $C^*$-algebras are most interesting. However, the infinitesimal
generator of such a group fails to be bounded, unless this group
is uniformly continuous.
\begin{remark} \label{w431} \mar{w431}
If $G(\mathbb R)$ is a strongly continuous one-parameter group of
automorphisms of a $C^*$-algebra $A$, there are the following
continuous maps \cite{brat}:
$\bullet$ $\mathbb R\ni t\to \lng G_t(a), f\rng \in \mathbb C$ is
continuous for all $a\in A$ and $f\in A'$;
$\bullet$ $A\ni a\to G_t(a) \in A$ is continuous for all
$t\in\mathbb R$;
$\bullet$ $\mathbb R\ni t\to G_t(a)\in A$ is continuous for all
$a\in A$.
\end{remark}
Without a loss of generality, we further assume that $A$ is a
unital $C^*$-algebra. Infinitesimal generators of one-parameter
groups of automorphisms of $A$ are derivations of $A$.
By a derivation $\dl$ of $A$ throughout is meant an (unbounded)
symmetric derivation of $A$ (i.e., $\dl(a^*)=\dl(a)^*$, $a\in A$)
which is defined on a dense involutive subalgebra $D(\dl)$ of $A$.
If a derivation $\dl$ on $D(\dl)$ is bounded, it is extended to a
bounded derivation everywhere on $A$. Conversely, every derivation
defined everywhere on a $C^*$-algebra is bounded \cite{dixm}. For
instance, any inner derivation $\dl(a)=i[b,a]$, where $b$ is a
Hermitian element of $A$, is bounded. A space of derivations of
$A$ is provided with the involution $u\to u^*$ defined by the
equality
\mar{w160}\beq
\dl^*(a)= -\dl(a^*)^*, \qquad a\in\cA. \label{w160}
\eeq
There is the following relation between bounded derivations of a
$C^*$-algebra $A$ and uniformly continuous one-parameter groups of
automorphisms of $A$ \cite{brat}.
\begin{theorem} \label{spr591} \mar{spr591}
Let $\dl$ be a derivation of a $C^*$-algebra $A$. The following assertions
are equivalent:
$\bullet$ $\dl$ is defined everywhere and, consequently, is bounded;
$\bullet$ $\dl$ is the infinitesimal generator of a uniformly
continuous one-parameter group $G(\mathbb R)$ of automorphisms of
a $C^*$-algebra $A$.
\noindent Furthermore, for any representation $\pi$ of $A$ in a
Hilbert space $E$, there exists a bounded self-adjoint operator
$\cH\in \pi(A)''$ in $E$ and the unitary uniformly continuous
representation
\mar{spr579'}\beq
\pi(G_t)=\exp(-it\cH), \qquad t\in\mathbb R, \label{spr579'}
\eeq
of the group $G(\mathbb R)$ in $E$ such that
\mar{spr592,3}\ben
&& \pi(\dl(a))=-i[\cH,\pi(a)], \qquad a\in A, \label{spr592}\\
&& \pi(G_t(a))=e^{-it\cH}\pi(a)e^{it\cH}, \qquad t\in\mathbb
R.\label{spr593}
\een
\end{theorem}
A $C^*$-algebra need not admit non-zero bounded derivations. For
instance, no commutative $C^*$-algebra possesses bounded
derivations.
The
following is the relation between (unbounded) derivations of a
$C^*$-algebra $A$ and strongly continuous
one-parameter groups of automorphisms of $A$ \cite{brat75,pow}.
\begin{theorem} \label{spr595} \mar{spr595}
Let $\dl$ be a closable derivation of a $C^*$-algebra $A$. Its
closure $\ol\dl$ is an infinitesimal generator of a strongly
continuous one-parameter group of automorphisms of $A$ iff
(i) the set $(\bb +\la\dl)(D(\dl)$ for any $\la\in\mathbb
R\setminus\{0\}$ is dense in $A$,
(ii)
$\|(\bb +\la\dl)(a)\|\geq \|a\|$ for any $\la\in\mathbb R$ and any $a\in
A$.
\end{theorem}
It should be noted that, if
$A$ is a unital algebra
and $\dl$ is its closable derivation, then $\bb\in D(\dl)$.
Let us mention a more convenient sufficient condition of a
derivation of a $C^*$-algebra to be an infinitesimal generator of
a strongly continuous one-parameter group of its automorphisms. A
derivation $\dl$ of a $C^*$-algebra $A$ is called well-behaved if,
for each element $a\in D(\dl)$, there exists a state $f$ of $A$
such that $f(a)=\|a\|$ and $f(\dl(a))=0$. If $\dl$ is a
well-behaved derivation, it is closable \cite{kish}, and obeys the
condition (ii) in Theorem \ref{spr595} \cite{brat75,pow}. Then we
come to the following.
\begin{theorem} \label{spr596} \mar{spr596}
If $\dl$ is a well-behaved derivation of a $C^*$-algebra $A$ and
it obeys the condition (i) in Theorem \ref{spr595}, its closure
$\ol\dl$ is an infinitesimal generator of a strongly continuous
one-parameter group of automorphisms of $A$.
\end{theorem}
For instance, a derivation $\dl$ is well-behaved if it is
approximately inner, i.e., there exists a sequence of self-adjoint
elements $\{b_n\}$ in $A$ such that
\be
\dl(a)=\op\lim_n i[b_n,a], \qquad a\in A.
\ee
In contrast with a case of a uniformly continuous one-parameter
group of automorphisms of a $C^*$-algebra $A$, a representation of
$A$ does not imply necessarily a unitary representation
(\ref{spr579'}) of a strongly continuous one-parameter group of
automorphisms of $A$, unless the following.
\begin{theorem} \label{spr759} \mar{spr759}
Let $G(\mathbb R)$ be a strongly continuous one-parameter group of
automorphisms of a $C^*$-algebra $A$ and $\dl$ its infinitesimal
generator. Let $A$ admit a state $f$ such that
\mar{w420}\beq
|f(\dl(a))|\leq\la[f(a^*a) + f(aa^*)]^{1/2} \label{w420}
\eeq
for all $a\in A$ and a positive number $\la$, and let
$(\pi_f,\thh_f)$ be a cyclic representation of $A$ in a Hilbert
space $E_f$ defined by $f$. Then there exist a self-adjoint
operator $\cH$ on a domain $D(\cH)\subset A\thh_f$ in $E_f$ and
the unitary strongly continuous representation (\ref{spr579'}) of
$G(\mathbb R)$ in $E_f$ which fulfils the relations (\ref{spr592})
-- (\ref{spr593}) for $\pi=\pi_f$.
\end{theorem}
It should be emphasized that the condition (\ref{w420}) in Theorem
\ref{spr759} is sufficient in order that a derivation $\dl$ to be
closable \cite{kish}.
Note that there is a general problem of a unitary representation
of an automorphism group of a $C^*$-algebra $A$. Let $\pi$ be a
representation of $A$ in a Hilbert space $E$. Then an automorphism
$\rho$ of $A$ possesses a unitary representation in $E$ if there
exists a unitary operator $U_\rho$ in $E$ such that
\mar{081}\beq
\pi(\rho(a))=U_\rho\pi(a)U_\rho^{-1}, \qquad a\in A. \label{081}
\eeq
A key point is that such a representation is never unique. Namely,
let $U$ and $U'$ be arbitrary unitary elements of the commutant
$\pi(A)'$ of $\pi(A)$. Then $UU_\rho U'$ also provides a unitary
representation of $\rho$. For instance, one can always choose
phase multipliers $U=\exp(i\al)\bb\in U(1)$. A consequence of this
ambiguity is the following.
Let $G$ be a group of automorphisms of an algebra $A$ whose
elements $g\in G$ admit unitary representations $U_g$ (\ref{081}).
The set of operators $U_g$, $g\in G$, however need not be a group.
In general, we have
\be
U_g U_{g'}=U(g,g')U_{gg'}U'(g,g'), \qquad U(g,g'),U'(g,g')\in
\pi(A)'.
\ee
If all $U(g,g')$ are phase multipliers, one says that the unitary
operators $U_g$, $g\in G$, form a projective representation
$U(G)$:
\be
U_g U_{g'}=k(g,g')U_{gg'}, \qquad g,g'\in G,
\ee
of a group $G$ \cite{cass,varad}. In this case, a set $U(1)\times
U(G)$ becomes a group which is a central $U(1)$-extension
\mar{084}\beq
\bb\ar U(1)\ar U(1)\times U(G)\ar G\ar \bb \label{084}
\eeq
of a group $G$. Accordingly, the projective representation
$\pi(G)$ of $G$ is a splitting of the exact sequence (\ref{084}).
It is characterized by $U(1)$-multipliers $k(g,g')$ which form a
two-cocycle
\mar{085}\beq
k(\bb,g)=k(g,\bb)=\bb, \qquad
k(g_1,g_2g_3)k(g_2,g_3)=k(g_1,g_2)k(g_1g_2,g_3) \label{085}
\eeq
of the cochain complex of $G$ with coefficients in $U(1)$
\cite{book05,mcl}. A different splitting of the exact sequence
(\ref{084}) yields a different projective representation $U'(G)$
of $G$ whose multipliers $k'(g,g')$ form a cocycle equivalent to
the cocycle (\ref{085}). If this cocycle is a coboundary, there
exists a splitting of the extension (\ref{084}) which provides a
unitary representation of a group $G$ of automorphisms of an
algebra $A$ in $E$.
For instance, let $B(E)$ be a $C^*$-algebra of bounded operators
in a Hilbert space $E$. All its automorphisms are inner. Any
(unitary) automorphism $U$ of a Hilbert space $E$ yields an inner
automorphism
\mar{spr750}\beq
a\to UaU^{-1}, \qquad a\in B(E), \label{spr750}
\eeq
of $B(E)$. Herewith, the automorphism (\ref{spr750}) is the
identity iff $U=\la\bb$, $|\la|=1$, is a scalar operator in $E$.
It follows that the group of automorphisms of $B(E)$ is the
quotient $U(E)/U(1)$ of a unitary group $U(E)$ with respect to a
circle subgroup $U(1)$. Therefore, given a group $G$ of
automorphisms of the $C^*$-algebra $B(E)$, the representatives
$U_g$ in $U(E)$ of elements $g\in G$ constitute a group up to
phase multipliers, i.e.,
\be
U_gU_{g'}=\exp[i\al(g,g')]U_{gg'}, \qquad \al(g,g')\in\mathbb R.
\ee
Nevertheless, if $G$ is a one-parameter weakly$^*$ continuous
group of automorphisms of $B(E)$ whose infinitesimal generator is
a bounded derivation of $B(E)$, one can choose the multipliers
$\exp[i\al(g,g')]=1$.
In a general setting, let $G$ be a group and $\cA$ a commutative
algebra. An $\cA$-multiplier of $G$ is a map $\xi: G\times G\to
\cA$ such that
\be
\xi(\bb_G,g)=\xi(g,\bb_G)=\bb_\cA, \qquad
\xi(g_1,g_2g_3)\xi(g_2,g_3)=\xi(g_1,g_2)\xi(g_1g_2,g_3), \qquad
g,g_i\in G.
\ee
For instance, $\xi:G\times G\to\bb_\cA\in \cA$ is a multiplier.
Two $A$-multipliers $\xi$ and $\xi'$ are said to be equivalent if
there exists a map $f:G\to \cA$ such that
\be
\xi(g_1,g_2)=\frac{f(g_1g_2)}{f(g_1)f(g_2)}\xi'(g_1,g_2), \qquad g_i\in
G.
\ee
An $\cA$-multiplier is called exact if it is equivalent to the
multiplier $\xi=\bb_\cA$. A set of $\cA$-multipliers is an Abelian
group with respect to the pointwise multiplication, and the set of
exact multipliers is its subgroup. Let $HM(G,\cA)$ be the
corresponding factor group.
If $G$ is a locally compact topological group and $\cA$ a Hausdorff
topological algebra,
one additionally requires that multipliers $\xi$ and equivalence maps
$f$ are measurable maps. In this case, there is a natural topology on
$HM(G,\cA)$ which is locally quasi-compact, but need not be Hausdorff
\cite{moor}.
\begin{theorem} \label{w711} \mar{w711} \cite{cass}.
Let $G$ be a simply connected locally compact Lie group. Each
$U(1)$-multiplier $\xi$ of $G$ is brought into a form
$\xi=\exp{i\al}$, where $\al$ is an $\mathbb R$-multiplier.
Moreover, $\xi$ is exact iff $\al$ is well. Any $\mathbb
R$-multiplier of $G$ is equivalent to a smooth one.
\end{theorem}
Let $G$ be a locally compact group of strongly continuous
automorphisms of a $C^*$-algebra $A$. Let $M(A)$ denote a
multiplier algebra of $A$, i.e., the largest $C^*$-algebra
containing $A$ as an essential ideal, i.e., if $a\in M(A)$ and
$ab=0$ for all $b\in A$, then $a=0$ \cite{woron94}). For instance,
$M(A)=A$ if $A$ is a unital algebra. Let $\xi$ be a multiplier of
$G$ with values in the center of $M(A)$. A $G$-covariant
representation $\pi$ of $A$ \cite{dopl,naud} is a representation
$\pi$ of $A$ (and, consequently, $M(A)$) in a Hilbert space $E$
together with a projective representation of $G$ by unitary
operators $U(g)$, $g\in G$, in $E$ such that
\be \pi(g(a)) = U(g)\pi(a)U^*(g),
\qquad U(g)U(g')=\pi(\xi(g,g'))U(gg').
\ee
\section{Spontaneous symmetry breaking}
Given a topological $^*$-algebra $A$ and its state $f$, let $\rho$
be an automorphism of $A$. Then it defines a state
\mar{0810}\beq
f_\rho(a)=f(\rho(a)), \qquad a\in A, \label{0810}
\eeq
of $A$. A state $f$ is said to be stationary with respect to an
automorphism $\rho$ of $A$ if
\mar{082}\beq
f(\rho(a))=f(a), \qquad a\in A. \label{082}
\eeq
One speaks about spontaneous symmetry breaking if a state $f$ of a
quantum algebra $A$ fails to be stationary with respect to some
automorphisms of $A$.
We can say something if $A$ is a $C^*$-algebra and its GNS
representations are considered \cite{brat,dixm,book05}.
\begin{theorem} \mar{t1} \label{t1} Let $f$ be a state of a $C^*$-algebra $A$ and
$(\pi_f,\thh_f,E_f)$ the corresponding cyclic representation of
$A$. An automorphism $\rho$ of $A$ defines the state $f_\rho$
(\ref{0810}) of $A$ such that a carrier space $E_{\rho f}$ of the
corresponding cyclic representation $\pi_{\rho f}$ is isomorphic
to $E_f$.
\end{theorem}
It follows that the representations $\pi_{\rho f}$ can be given by
operators $\pi_{\rho f}(a)=\pi_f(\rho(a))$ in the carrier space
$E_f$ of the representation $\pi_f$, but these representations
fail to be equivalent, unless an automorphism $\rho$ possesses the
unitary representation (\ref{081}) in $E_f$. In this case, a state
$f$ is stationary relative to $\rho$. The converse also is true.
\begin{theorem} \mar{t2} \label{t2}
If a state $f$ of a $C^*$-algebra $A$ is the stationary state
(\ref{082}) with respect to an automorphism $\rho$ of $A$, there
exists a unique unitary representation $U_\rho$ (\ref{0810}) of
$\rho$ in $E_f$ such that
\mar{087}\beq
U_\rho\thh_f=\thh_f. \label{087}
\eeq
\end{theorem}
It follows from Theorem \ref{spr591} that, since any uniformly
continuous one-parameter group of automorphisms of a $C^*$-algebra
$A$ admits a unitary representation, each state $f$ of $A$ is
stationary for this group. However, this is not true for an
arbitrary uniformly continuous group $G$ of automorphisms of $A$.
For instance, let $B(E)$ be the $C^*$-algebra of all bounded
operators in a Hilbert space $E$. Any automorphisms of $B(E)$ is
inner and, consequently, possesses a unitary representation in
$E$. Since the commutant of $B(E)$ reduces to scalars, the group
of automorphisms of $B(E)$ admits a projective representation in
$E$, but it need not be unitary.
It follows from Theorem \ref{spr759}) that, if a state $f$ of a
$C^*$-algebra $A$ is stationary under a strongly continuous group
$G(\mathbb R)$ of automorphisms of $A$, i.e., $f(]dl(a))=0$, there
exists unitary representation of this group in $E_f$. However,
this condition is sufficient, but not necessary.
Moreover, one can show the following \cite{dixm}.
\begin{theorem} \mar{t4} \label{t4}
Let $G$ be a strongly continuous group of automorphisms of a
$C^*$-algebra $A$, and let a state $f$ of $A$ be stationary for
$G$. Then there exists a unique unitary representation of $G$ in
$E_f$ whose operators obey the equality (\ref{087}).
\end{theorem}
Let now $G$ be a group of strongly or uniformly continuous group
of automorphisms of a $C^*$-algebra $A$, and let $f$ be a state of
$A$. Let us consider a set of states $f_g$ (\ref{0810}), $g\in G$,
of $A$ defined by automorphisms $g\in G$. Let $f$ be stationary
with respect to a proper subgroup $H$ of $G$. Then a set of
equivalence classes of states $f_g$, $g\in G$, is a subset of the
factor space $G/H$, but need not coincide with $G/H$.
This is just the case of spontaneous symmetry breaking in Standard
Model where A Higgs vacuum is a stationary state with respect to
some proper subgroup of a symmetry group \cite{nov,SM}.
In axiomatic QFT, the spontaneous symmetry breaking phenomenon is
described by the Goldstone theorem \cite{bog}.
Let $G$ be a connected Lie group of internal symmetries
(automorphisms of the Borchers algebra $A$ over $\id \mathbb R^4$)
whose infinitesimal generators are given by conserved currents
$j^k_\m$. One can show the following \cite{bog}.
\begin{theorem} \mar{t5} \label{t5}
A group $G$ of internal symmetries possesses a unitary
representation in $E_W$ iff the Wightman functions are
$G$-invariant.
\end{theorem}
\begin{theorem} \mar{t6} \label{t6}
A group $G$ of internal symmetries admits a unitary representation
if a strong spectrum condition holds, i.e., there exists a mass
gap.
\end{theorem}
As a consequence, we come to the above mentioned Goldstone
theorem.
\begin{theorem} \mar{t7} \label{t7}
If there is a group $G$ of internal symmetries which are
spontaneously broken, there exist elements $\f\in E_W$ of zero
spin and mass such that $\lng \f| j^k_\m\psi_0\rng\neq 0$ for some
generators of $G$.
\end{theorem}
These elements of unit norm are called Goldstone states. It is
easily observed that, if a group $G$ of spontaneously broken
symmetries contains a subgroup of exact symmetries $H$, the
Goldstone states carrier out a homogeneous representation of $G$
isomorphic to the quotient $G/H$.
This fact attracted great attention to such kind representations
and motivated to describe classical Higgs fields as sections of a
fibre bundle with a typical fibre $G/H$ \cite{higgs,sard08a,tmp}.
\section{Appendixes}
This Section summarizes some relevant material on topological
vector spaces and measures on non-compact spaces.
\subsection{Topological vector spaces}
There are several standard topologies introduced on an
(infinite-dimensional) complex or real vector space and its dual
\cite{book05,rob}. Topological vector spaces throughout are
assumed to be locally convex. Unless otherwise stated, by the dual
$V'$ of a topological vector space $V$ is meant its topological
dual, i.e., the space of continuous linear maps of $V\to \mathbb
R$.
Let us note that a topology on a vector space $V$ often is
determined by a set of seminorms. A non-negative real function $p$
on $V$ is called the seminorm if it satisfies the conditions
\be
p(\la x)=|\la|p(x), \qquad p(x+y)\leq p(x) +p(y), \qquad x,y\in V,
\quad \la\in\mathbb R.
\ee
A seminorm $p$ for which $p(x)=0$ implies $x=0$ is called the
norm. Given any set $\{p_i\}_{i\in I}$ of seminorms on a vector
space $V$, there is the coarsest topology on $V$ compatible with
the algebraic structure such that all seminorms $p_i$ are
continuous. It is a locally convex topology whose base of closed
neighborhoods consists of sets
\be
\{x\, :\, \op\sup_{1\leq i\leq n} p_i(x)\leq \ve\}, \qquad \ve>0,
\qquad n\in\mathbb N_+.
\ee
Let $V$ and $W$ be two vector spaces whose Cartesian product
$V\times W$ is provided with a bilinear form $\lng v,w\rng$ which
obeys the following conditions:
$\bullet$ for any element $v\neq 0$ of $V$, there exists an
element $w\in W$ such that $\lng v, w\rng\neq 0$;
$\bullet$ for any element $w\neq 0$ of $W$, there exists an
element $v\in V$ such that $\lng v, w\rng\neq 0$.
\noindent Then one says that $(V,W)$ is a dual pair. If $(V,W)$ is
a dual pair, so is $(W,V)$. Clearly, $W$ is isomorphic to a vector
subbundle of the algebraic dual $V^*$ of $V$, and $V$ is a
subbundle of the algebraic dual of $W$.
Given a dual pair $(V,W)$, every vector $w\in W$ defines a
seminorm $p_w=|\lng v,w\rng|$ on $V$. The coarsest topology
$\si(V,W)$ on $V$ making all these seminorms continuous is called
the weak topology determined by $W$ on $V$. It also is the
coarsest topology on $V$ such that all linear forms in $W\subset
V^*$ are continuous. Moreover, $W$ coincides with the dual $V'$ of
$V$ provided with the weak topology $\si(V,W)$, and $\si(V,W)$ is
the coarsest topology on $V$ such that $V'=W$. Of course, the
weak topology is Hausdorff.
For instance, if $V$ is a Hausdorff topological vector space with
the dual $V'$, then $(V,V')$ is a dual pair. The weak topology
$\si(V,V')$ on $V$ is coarser than the original topology on $V$.
Since $(V',V)$ also is a dual pair, the dual $V'$ of $V$ can be
provided with the weak$^*$ topology topology $\si(V',V)$. Then $V$
is the dual of $V'$, equipped with the weak$^*$ topology.
The weak$^*$ topology is the coarsest case of a topology of
uniform convergence on $V'$. A subset $M$ of a vector space $V$ is
said to absorb a subset $N\subset V$ if there is a number $\e\geq
0$ such that $N\subset \la M$ for all $\la$ with $|\la|\geq \e$.
An absorbent set is one which absorbs all points. A subset $N$ of
a topological vector space $V$ is called bounded if it is absorbed
by any neighborhood of the origin of $V$. Let $(V,V')$ be a dual
pair and $\cN$ some family of weakly bounded subsets of $V$. Every
$N\subset\cN$ yields a seminorm
\be
p_N(v')=\op\sup_{v\in N} |\lng v, v'\rng|
\ee
on the dual $V'$ of $V$. A topology on $V'$ defined by a set of
seminorms $p_N$, $N\in\cN$, is called the topology of uniform
convergence on the sets of $\cN$. When $\cN$ is a set of all
finite subsets of $V$, we have the coarsest topology of uniform
convergence which is the above mentioned weak$^*$ topology
$\si(V',V)$. The finest topology of uniform convergence is
obtained by taking $\cN$ to be a set of all weakly bounded subsets
of $V$. It is called the strong topology. The dual $V''$ of $V'$,
provided with the strong topology, is called the bidual. One says
that $V$ is reflexive if $V=V''$.
Since $(V',V)$ is a dual pair, a vector space $V$ also can be
provided with the topology of uniform convergence on subsets of
$V'$, e.g., the weak$^*$ and strong topologies. Moreover, any
Hausdorff locally convex topology on $V$ is a topology of uniform
convergence. The coarsest and finest topologies of them are the
weak$^*$ and strong topologies, respectively. There is the
following chain
\be
\mathrm{weak}^* \,<\, \mathrm{weak}\,<\, \mathrm{original}\,<\,
\mathrm{strong}
\ee
of topologies on $V$, where $<$ means "to be finer".
For instance, let $V$ be a normed space. The dual $V'$ of $V$ also
is equipped with a norm
\mar{spr446}\beq
\|v'\|'=\op\sup_{\|v\|=1}|\lng v,v'\rng|, \qquad v\in V, \qquad
v'\in V'. \label{spr446}
\eeq
Let us consider a set of all balls $\{v\,:\, \|v\|\leq\e,
\,\e>0\}$ in $V$. The topology of uniform convergence on this set
coincides with strong and normed topologies on $V'$ because weakly
bounded subsets of $V$ also are bounded by a norm. Normed and
strong topologies on $V$ are equivalent. Let $\ol V$ denote the
completion of a normed space $V$. Then $V'$ is canonically
identified to $(\ol V)'$ as a normed space, though weak$^*$
topologies on $V'$ and $(\ol V)'$ are different. Let us note that
both $V'$ and $V''$ are Banach spaces. If $V$ is a Banach space,
it is closed in $V''$ with respect to the strong topology on $V''$
and dense in $V''$ equipped with the weak$^*$ topology. One
usually considers the weak$^*$, weak and normed (equivalently,
strong) topologies on a Banach space.
It should be noted that topology on a finite-dimensional vector
space is locally convex and Hausdorff iff it is determined by the
Euclidean norm.
Let us say a few words on morphisms of topological vector spaces.
A linear morphism between two topological vector spaces is called
the weakly continuous morphism if it is continuous with respect to
the weak topologies on these vector spaces. In particular, any
continuous morphism between topological vector spaces is weakly
continuous \cite{rob}.
A linear morphism between two topological vector spaces is called
bounded if the image of a bounded set is bounded. Any continuous
morphism is bounded. A topological vector space is called the
Mackey space if any bounded endomorphism of this space is
continuous (we follow the terminology of \cite{rob}). Metrizable
and, consequently, normed spaces are of this type.
Any linear morphism $\g: V\to W$ of topological vector spaces
yields the dual morphism $\g': W'\to V'$ of the their topological
duals such that
\be
\lng v,\g'(w)\rng= \lng \g(v),w\rng, \qquad v\in V, \qquad w\in W.
\ee
If $\g$ is weakly continuous, then $\g'$ is weakly$^*$ continuous.
If $V$ and $W$ are normed spaces, then any weakly continuous
morphism $\g: V\to W$ is continuous and strongly continuous. Given
normed topologies on $V'$ and $W'$, the dual morphism $\g':W'\to
V'$ is continuous iff $\g$ is continuous.
\subsection{Hilbert, countably Hilbert and nuclear spaces}
Let us recall the relevant basics on pre-Hilbert and Hilbert
spaces \cite{bourb,book05}.
A Hermitian form on a complex vector space $E$ is defined as a
sesquilinear form $\lng.|.\rng$ such that
\be
\lng e|e'\rng=\ol{\lng e'|e\rng}, \qquad \lng \la e|e'\rng=\lng
e|\ol\la e'\rng=\la \lng e|e'\rng, \quad e,e'\in E, \quad \la\in
\mathbb C.
\ee
\begin{remark}
There exists another convention where $\lng e|\la e'\rng=\la \lng
e|e'\rng$.
\end{remark}
A Hermitian form $\lng.|.\rng$ is said to be positive if $\lng
e|e\rng\geq 0$ for all $e\in E$. All Hermitian forms throughout
are assumed to be positive. A Hermitian form is called
non-degenerate if the equality $\lng e|e\rng= 0$ implies $e=0$. A
complex vector space endowed with a Hermitian form is called the
pre-Hilbert space. Morphisms of pre-Hilbert spaces, by definition,
are isometric.
A Hermitian form provides $E$ with the topology defined by a
seminorm $\| e\|=\lng e|e\rng^{1/2}$. Hence, a pre-Hilbert space
is Hausdorff iff a Hermitian form $\lng.|.\rng$ is non-degenerate,
i.e., a seminorm $\| e\|$ is a norm. In this case, it is called
the scalar product.
A complete Hausdorff pre-Hilbert space is called the Hilbert
space. Any Hausdorff pre-Hilbert space can be completed to a
Hilbert space. .
The following are the standard constructions of new Hilbert spaces
from the old ones.
$\bullet$ Let $(E^\iota, \lng.|.\rng_{E^\iota})$ be a set of
Hilbert spaces and $\sum E^\iota$ denote a direct sum of vector
spaces $E^\iota$. For any two elements $e=(e^\iota)$ and
$e'=(e'^\iota)$ of $\sum E^\iota$, a sum
\mar{spr403}\beq
\lng e|e'\rng_\oplus = \op\sum_\iota \lng
e^\iota|e'^\iota\rng_{E^\iota} \label{spr403}
\eeq
is finite, and defines a non-degenerate Hermitian form on $\sum
E^\iota$. The completion $\oplus E^\iota$ of $\sum E^\iota$ with
respect to this form is a Hilbert space, called the Hilbert sum of
$E^\iota$.
$\bullet$ Let $(E,\lng.|.\rng_E)$ and $(H,\lng.|.\rng_H)$ be
Hilbert spaces. Their tensor product $E\ot H$ is defined as the
completion of a tensor product of vector spaces $E$ and $H$ with
respect to the scalar product
\be
&& \lng w_1|w_2\rng_\ot=\op\sum_{\iota,\bt}\lng e_1^\iota | e_2^\bt\rng_E
\lng h_1^\iota | h_2^\bt\rng_H,\\
&& w_1=\op\sum_\iota e^\iota_1\ot h^\iota_1, \quad
w_2=\op\sum_\bt e^\bt_2\ot h^\bt_2, \quad e_1^\iota,e_2^\bt\in E,
\quad h_1^\iota,h_2^\bt\in H.
\ee
$\bullet$ Let $E'$ be the topological dual of a Hilbert space $E$.
Then the assignment
\mar{spr400}\beq
e\to \ol e(e')= \lng e'|e\rng, \qquad e,e'\in E, \label{spr400}
\eeq
defines an antilinear bijection of $E$ onto $E'$, i.e., $\ol{\la
e}=\ol\la\ol e$. The dual $E'$ of a Hilbert space is a Hilbert
space provided with the scalar product $\lng \ol e|\ol e'\rng'=
\lng e'|e\rng$ such that the morphism (\ref{spr400}) is isometric.
The $E'$ is called the dual Hilbert space, and is denoted by $\ol
E$.
Physical applications of Hilbert spaces are limited by the fact
that the dual of a Hilbert space $E$ is anti-isomorphic to $E$.
The construction of a rigged Hilbert space describes the dual
pairs $(E,E')$ where $E'$ is larger than $E$ \cite{gelf64}.
Let a complex vector space $E$ have a countable set of
non-degenerate Hermitian forms $\lng.|.\rng_k$, $k\in\mathbb N_+,$
such that
\be
\lng e|e\rng_1\leq \cdots\leq \lng e|e\rng_k\leq\cdots
\ee
for all $e\in E$. The family of norms
\mar{spr445}\beq
\|.\|_k=\lng.|.\rng^{1/2}_k, \qquad k\in\mathbb N_+,
\label{spr445}
\eeq
yields a Hausdorff topology on $E$. A space $E$ is called the
countably Hilbert space if it is complete with respect to this
topology \cite{gelf64}. For instance, every Hilbert space is a
countably Hilbert space where all Hermitian forms $\lng.|.\rng_k$
coincide.
Let $E_k$ denote the completion of $E$ with respect to the norm
$\|.\|_k$ (\ref{spr445}). There is the chain of injections
\mar{1086}\beq
E_1\supset E_2\supset \cdots E_k\supset \cdots \label{1086}
\eeq
together with a homeomorphism $E=\op\cap_k E_k$. The dual spaces
form the increasing chain
\mar{1087}\beq
E'_1\subset E'_2\subset \cdots \subset E'_k \subset \cdots,
\label{1087}
\eeq
and $E'=\op\cup_k E'_k$. The dual $E'$ of $E$ can be provided with
the weak$^*$ and strong topologies. One can show that a countably
Hilbert space is reflexive.
Given a countably Hilbert space $E$ and $m\leq n$, let $T^n_m$ be
a prolongation of the map
\be
E_n\supset E\ni e\to e\in E\subset E_m
\ee
to a continuous map of $E_n$ onto a dense subset of $E_m$. A
countably Hilbert space $E$ is called the nuclear space if, for
any $m$, there exists $n$ such that $T^m_n$ is a nuclear map,
i.e.,
\be
T^n_m(e)=\op\sum_i\la_i\lng e|e^i_n\rng_{E_n} e^i_m,
\ee
where: (i) $\{e^i_n\}$ and $\{e_m^i\}$ are bases for the Hilbert
spaces $E_n$ and $E_m$, respectively, (ii) $\la_i\geq 0$, (iii)
the series $\sum \la_i$ converges \cite{gelf64}.
An important property of nuclear spaces is that they are perfect,
i.e., every bounded closed set in a nuclear space is compact. It
follows immediately that a Banach (and Hilbert) space is not
nuclear, unless it is finite-dimensional. Since a nuclear space is
perfect, it is separable, and the weak$^*$ and strong topologies
(and, consequently, all topologies of uniform convergence) on a
nuclear space $E$ and its dual $E'$ coincide.
Let $E$ be a nuclear space, provided with still another
non-degenerate Hermitian form $\lng.|.\rng$ which is separately
continuous, i.e., continuous with respect to each argument. It
follows that there exist numbers $M$ and $m$ such that
\mar{1085}\beq
\lng e|e\rng\leq M\|e\|_m, \qquad e\in E. \label{1085}
\eeq
Let $\wt E$ denote the completion of $E$ with respect to this
form. There are the injections
\mar{spr450}\beq
E\subset \wt E\subset E', \label{spr450}
\eeq
where $E$ is a dense subset of $\wt E$ and $\wt E$ is a dense
subset of $E'$, equipped with the weak$^*$ topology. The triple
(\ref{spr450}) is called the rigged Hilbert space. Furthermore,
bearing in mind the chain of Hilbert spaces (\ref{1086}) and that
of their duals (\ref{1087}), one can convert the triple
(\ref{spr450}) into the chain of spaces
\mar{1088}\beq
E\subset\cdots\subset E_k\subset\cdots E_1\subset\wt E\subset
E'_1\subset \cdots\subset E'_k\subset\cdots\subset E'.
\label{1088}
\eeq
\begin{remark}
Real Hilbert, countably Hilbert, nuclear and rigged Hilbert spaces
are similarly described.
\end{remark}
\subsection{Measures on locally compact spaces}
Measures on a locally compact space $X$ are defined as continuous
forms on spaces of continuous real (or complex) functions of
compact support on $X$, and they are extended to a wider class of
functions on $X$ \cite{bourb6,book05}.
Let $\mathbb C^0(X)$ be a ring of continuous complex functions on
$X$. For each compact subset $K$ of $X$, we have a seminorm
\mar{spr490}\beq
p_K(f)=\op\sup_{x\in K} |f(x)| \label{spr490}
\eeq
on $C^0(X)$. These seminorms provide $\mathbb C^0(X)$ with the
Hausdorff topology of compact convergence. We abbreviate with
$\cK(X,\mathbb C)$ the dense subspace of $\mathbb C^0(X)$ which
consists of continuous complex functions of compact support on
$X$. It is a Banach space with respect to a norm
\mar{spr508}\beq
\|f\|=\op\sup_{x\in X} |f(x)|. \label{spr508}
\eeq
Its normed topology, called the topology of uniform convergence,
is finer than the topology of compact convergence, and these
topologies coincide if $X$ is a compact space.
A space $\cK(X,\mathbb C)$ also can be equipped with another
topology, which is especially relevant to integration theory. For
each compact subset $K\subset X$, let $\cK_K(X,\mathbb C)$ be a
vector subspace of $\cK(X,\mathbb C)$ consisting of functions of
support in $K$. Let $\cU$ be a set of all absolutely convex
absorbent subsets $U$ of $\cK(X,\mathbb C)$ such that, for every
compact $K$, a set $U\cap \cK_K(X,\mathbb C)$ is a neighborhood of
the origin in $\cK_K(X,\mathbb C)$ under the topology of uniform
convergence on $K$. Then $\cU$ is a base of neighborhoods for a
(locally convex) topology, called the inductive limit topology, on
$\cK(X,\mathbb C)$ \cite{rob}. This is the finest topology such
that an injection $\cK_K(X,\mathbb C)\to \cK(X,\mathbb C)$ is
continuous. The inductive limit topology is finer than the
topology of uniform convergence, and these topologies coincide if
$X$ is a compact space. Unless otherwise stated, referring to a
topology on $\cK(X,\mathbb C)$, we will mean the inductive limit
topology.
A complex measure on a locally compact space $X$ is defined as a
continuous form $\m$ on a space $\cK(X,\mathbb C)$ of continuous
complex functions of compact support on $X$. The value $\m(f)$,
$f\in \cK(X,\mathbb C)$, is called the integral $\int f\m$ of $f$
with respect to the measure $\m$. The space $M(X,\mathbb C)$ of
complex measures on $X$ is the dual of $\cK(X, \mathbb C)$. It is
provided with the weak$^*$ topology.
Given a complex measure $\m$, any continuous complex function
$h\in \mathbb C^0(X)$ on $X$ defines the continuous endomorphism
$f\to hf$ of the space $\cK(X,\mathbb C)$ and yields a new complex
measure $h\m(f)=\m(hf)$. Hence, the space $M(X,\mathbb C)$ of
complex measures on $X$ is a module over the ring $\mathbb
C^0(X)$.
Let $\cK(X)\subset \cK(X,\mathbb C)$ denote a vector space of
continuous real functions of compact support on $X$. The
restriction of a complex measure $\m$ on $X$ to $\cK(X)$ is a
continuous complex form on $\cK(X)$, equipped with the inductive
limit topology. Any complex measure $\m$ is uniquely determined as
\mar{spr512}\beq
\m(f)=\m(\re f) + i\m(\im f), \qquad f\in\cK(X,\mathbb C),
\label{spr512}
\eeq
by its restriction to $\cK(X)$. A complex measure $\m$ on $X$ is
called a real measure if its restriction to $\cK(X)$ is a real
form. A complex measure $\m$ is real iff $\m=\ol\m$, where $\ol\m$
is the conjugate measure given by the condition
$\ol\m(f)=\ol{\m(\ol f)}$, $f\in \cK(X,\mathbb C)$.
A measure on a locally compact space $X$ is defined as a
continuous real form on a space $\cK(X)$ of continuous real
functions of compact support on $X$ provided with the inductive
limit topology. Any real measure on $\cK(X,\mathbb C)$ restricted
to $\cK(X)$ is a measure. Conversely, each measure $\m$ on
$\cK(X)$ is extended to the real measure (\ref{spr512}) on
$\cK(X,\mathbb C)$. Thus, measures on $\cK(X)$ and real measures
on $\cK(X,\mathbb C)$ can be identified.
A measure $\m$ on a locally compact space $X$ is called positive
if $\m(f)\geq 0$ for all positive functions $f\in \cK(X)$. Any
measure $\m$ defines the positive measure $|\m|(f)=|\m(f)|$, and
can be represented by the combination
\be
\m=\frac12(|\m|+\m)-\frac12(|\m|-\m)
\ee
of two positive measures.
A complex measure $\m$ on a locally compact space $X$ is called
bounded if there is a positive number $\la$ such that $|\m(f)|\leq
\la\|f\|$ for all $f\in \cK(X, \mathbb C)$. A complex measure $\m$
is bounded iff it is continuous with respect to the topology of
uniform convergence on $\cK(X,\mathbb C)$. Hence, a space
$M^1(X,\mathbb C)\subset M(X,\mathbb C)$ of bounded complex
measures is the dual of $\cK(X,\mathbb C)$, provided with this
topology. It is a Banach space with respect to a norm
\mar{spr565}\beq
\|\m\|=\sup\{ |\m(f)|\,:\, \|f\|=1\},\, f\in \cK(X,\mathbb C)\}.
\label{spr565}
\eeq
Of course, any complex measure on a compact space is bounded. If
$\m$ is a bounded complex measure and $h$ is a bounded continuous
function on $X$, the complex measure $h\m$ is bounded.
Similarly, a Banach space $M^1(X)$ of bounded measures on $X$ is
defined.
\begin{example} \label{spr492} \mar{spr492}
Given a point $x\in X$, the assignment $\ve_x:f\to f(x)$, $f\in
\cK(X)$, defines the Dirac measure on $X$. Any finite linear
combination of Dirac measures is a measure, called a point
measure. The Dirac measure $\ve_x$ is bounded, and $\|\ve_x\|=1$.
\end{example}
Now we extend a class of integrable functions as follows. Let
$\mathbb R_{\pm\infty}$ denote the extended real line, obtained
from $\mathbb R$ by the adjunction of points $\{+\infty\}$ and
$\{-\infty\}$. It is a ring such that $0\cdot\infty=0$ and
$\infty-\infty=0$. Let $J_+$ be a space of positive lower
semicontinuous functions on $X$ which take their values in the
extended real line $\mathbb R_{\pm\infty}$. These functions
possess the following important properties:
$\bullet$ the upper bound of any set of elements of $J_+$ and the
lower bound of a finite set of elements of $J_+$ also are elements
of $J_+$;
$\bullet$ any function $f\in J_+$ is an upper bound of a family of
positive functions $h\in\cK(X)$ such that $h\leq f$.
The last fact enables one to define the upper integral of a
function $f\in J_+$ with respect to a positive measure $\m$ on $X$
as the element
\mar{spr493}\beq
\m^*(f)=\int^* f\m=\sup\{\m(h)\,:\, h\in\cK(X),\,0\leq h\leq f\}
\label{spr493}
\eeq
of $\mathbb R_{\pm\infty}$. Of course, $\m^*(f)=\m(f)$ if $f\in
\cK(X)$.
\begin{example}
Let $U$ be an open subset of $X$ and $\vf_U$ its characteristic
function. It is readily observed that $\vf_U\in J_+$. Given a
positive measure $\m$ on $X$, the upper integral
$\m(U)=\m^*(\vf_U)$ is called the outer measure of $U$. For
instance, the outer measure of a relatively compact set $U$ (i.e.,
$U$ is a subset of a compact set) is finite. The (finite or
infinite) number $\m(X)=\m^*(1)$ is called the total mass of a
measure $\m$. In particular, a measure on a locally compact space
is bounded iff it has a finite total mass.
\end{example}
Let $f$ be an arbitrary positive $\mathbb R_{\pm\infty}$-valued
function on a locally compact space $X$ (not necessarily lower
semicontinuous). There exist functions $g\in J_+$ such that $g\geq
f$ (e.g., $g=+\infty$). Then the upper integral of $f$ with
respect to a positive measure $\m$ on $X$ is defined as
\mar{spr495}\beq
\m^*(f)=\inf\{\m^*(h)\,:\, h\in J_+,\, h\geq f\}. \label{spr495}
\eeq
\begin{example}
The outer measure $\m^*(V)=\m^*(\vf_V)$ of an arbitrary subset $V$
of $X$ exemplifies the upper integral (\ref{spr495}). In
particular, one says that $V\subset X$ is a $\m$-null set if
$\m(V)=0$. Two $\mathbb R_{\pm\infty}$-valued functions $f$ and
$f'$ on a locally compact space are called $\m$-equivalent if they
differ from each other only on a $\m$-null set; then
$\m^*(f)=\m^*(f')$. Two positive measures $\m$ and $\m'$ are said
to be equivalent if any compact $\m$-null set also is a $\m'$-null
set, and {\it vice versa}. They coincide if $\m(K)=\m'(K)$ for any
compact set $K\subset X$.
\end{example}
\begin{example}
A real function $f$ on a subset $V\subset X$ is said to be defined
almost everywhere with respect to a positive measure $\m$ on $X$
if the complement $X\setminus V$ of $V$ is a $\m$-null set. For
instance, an $\mathbb R_{\pm\infty}$-valued function $f$ which is
finite almost everywhere on $X$ exemplifies a real function
defined almost everywhere on $X$. Conversely, one can think of a
positive function defined almost everywhere on $X$ as being
$\m$-equivalent to some positive $\mathbb R_{\pm\infty}$-valued
function on $X$.
\end{example}
The following classes of integrable functions (and maps) are
usually considered.
Let $f$ be a map of a locally compact space $X$ to a Banach space
$F$, provided with a norm $|.|$ (e.g., $F$ is $\mathbb R$ or
$\mathbb C$). Given a positive measure $\m$ on $X$, let us define
the positive (finite or infinite) number
\mar{spr496}\beq
N_p(f)=\left[\int^* |f|^p \m\right]^{1/p}, \qquad 1\leq p<\infty.
\label{spr496}
\eeq
Clearly, $N_p(f)=N_p(f')$ if $f$ and $f'$ are $\m$-equivalent maps
on $X$, i.e., if they differ on a $\m$-null subset of $X$. There
is the Minkowski inequality
\mar{spr497}\beq
N_p(f+f')\leq N_p(f) + N_p(f'). \label{spr497}
\eeq
$\bullet$ Let $R^p_F(X,\m)$ be a space of maps $X\to F$ such that
$N_p(f)<+\infty$. In accordance with the Minkowski inequality
(\ref{spr497}), it is a vector space and $N_p$ (\ref{spr496}) is a
seminorm on $R^p_F(X,\m)$. Provided with the corresponding
topology, $R^p_F(X,\m)$ is a complete space, but not necessarily
Hausdorff. A space $\cK(X,F)$ of continuous maps $X\to F$ of
compact support belongs to $R^p_F(X,\m)$.
$\bullet$ A space $\cL^p_F(X,\m)$ is defined as the closure of
$\cK(X,F)\subset R^p_F(X,\m)$. Elements of $\cL^p_F(X,\m)$ are
called integrable $F$-valued functions of degree $p$. In
particular, elements of $\cL^1_F(X,\m)$ are called integrable
$F$-valued functions, while those of $\cL^2_F(X,\m)$ are square
integrable $F$-valued functions. Any element of $R^p_F(X,\m)$
which is $\m$-equivalent to an element of $\cL^p_F(X,\m)$ belongs
to $\cL^p_F(X,\m)$. An $F$-valued map defined almost everywhere on
$X$ also is called integrable if it is $\m$-equivalent to an
element of $\cL^p_F(X,\m)$.
$\bullet$ A space $L^p_F(X,\m)$ consists of classes of
$\m$-equivalent integrable $F$-valued maps of degree $p$. One
usually treat elements of this space as $F$-valued functions
without fear of confusion, and call them integrable $F$-valued
functions of degree $p$, too. The $L^p_F(X,\m)$ is a Banach space
with respect to the norm (\ref{spr496}).
There are the following important relations between the spaces
$L^p_F(X,\m)$, $1\leq p<+\infty$.
If $f\in \cL^p_F(X,\m)$, then $|f|^{(p/q)-1}f$ belongs to
$\cL^q_F(X,\m)$ for any $1\leq q<+\infty$, and {\it vice versa}.
Moreover, $f\to |f|^{(1/q)-1}f$ provides a homeomorphism between
topological spaces $\cL^1_F(X,\m)$ and $\cL^q_F(X,\m)$.
Let the numbers $1< p <+\infty$ and $1< q <+\infty$ obey the
condition
\mar{spr503}\beq
p^{-1} + q^{-1} =1. \label{spr503}
\eeq
If $f\in L^p_\mathbb C(X,\m)$ is an integrable complex function on
$X$ of degree $p$ and $f'\in L^q_\mathbb C(X,\m)$ is that of
degree $q$, then $f\ol f'$ is integrable, i.e., belongs to
$L^1_\mathbb C(X,\m)$. In particular, a space $L^2_\mathbb
C(X,\m)$ of square integrable complex functions on a locally
compact space $X$ is a separable Hilbert space with respect to a
scalar product
\mar{spr500}\beq
\lng f|f'\rng= \int f\ol f'\m. \label{spr500}
\eeq
One can say something more in the case of real functions. Let
numbers $p$ and $q$ obey the condition (\ref{spr503}). Any
integrable real function $f\in \cL^q_\mathbb R(X,\m)$ on $X$ of
degree $q$ defines a continuous real form
\mar{spr505}\beq
\f_f:f'\to \int ff'\m \label{spr505}
\eeq
on a space $L^p_\mathbb R(X,\m)$ such that $N_q(f)=\|\f_f\|$.
Conversely, each continuous real form on $L^p_\mathbb R(X,\m)$ is
of type (\ref{spr505}) where $f$ is an element of $\cL^q_\mathbb
R(X,\m)$ whose equivalence class in $L^q_\mathbb R(X,\m)$ is
uniquely defined. As a consequence, there is an isomorphism
between Banach spaces $L^q_\mathbb R(X,\m)$ and $(L^p_\mathbb
R(X,\m))'$, and a Banach space $L^p_\mathbb R(X,\m)$ is reflexive.
\begin{remark} \label{spr520} \mar{spr520}
One can define a space $L^\infty_\mathbb C(X,\m)$ of complex
infinite integrable functions on $X$ as the dual of a Banach space
$L^1_\mathbb C(X,\m)$. In particular, any bounded continuous
function belongs to $L^\infty_\mathbb C(X,\m)$. Let us note that a
space $L^1_\mathbb C(X,\m)$ is not reflexive, i.e., the dual of
$L^\infty_\mathbb C(X,\m)$, provided with the strong topology,
does not coincide with $L^1_\mathbb C(X,\m)$.
\end{remark}
We now turn to the relation between equivalent measures. Let $f$
be a positive $\mathbb R_{\pm\infty}$-valued function on a locally
compact space $X$. Given a positive measure $\m$ on $X$, a
quantity
\mar{spr521}\beq
\wt\m(f)=\op\sup_{K\subset X} \m^*(\vf_Kf), \label{spr521}
\eeq
where $K$ runs through a set of all compact subsets of $X$, is
called the essential upper integral of $f$. Since $\vf_Kf\leq f$
for any compact subset $K$, the inequality $\wt\m(f)\leq \m^*(f)$
holds. In particular, if $V$ is a subset of $X$ and $\wt
\m(\vf_V)=0$, one says that $V$ is a locally $\m$-null set,
i.e., any point $x\in X$ has a neighborhood $U$ such that $U\cap
V$ is a $\m$-null set. Essential upper integrals coincide with the
upper ones if $X$ is a locally compact space countable at
infinity.
One says that a function on a subset $V$ of $X$ is defined locally
almost everywhere if the complement of $V$ is a locally null set.
A real function $f$ defined locally almost everywhere on $X$ is
called locally $\m$-integrable if any point $x\in X$ has a
neighborhood $U$ such that $\vf_Uf$ is a $\m$-integrable function
or, equivalently, if $hf$ is a $\m$-integrable function for any
positive function $h\in \cK(X)$ of compact support.
Let $h$ be a locally $\m$-integrable function which is defined and
non-negative almost everywhere on $X$. There is a positive measure
$h\m$ on $X$ which obeys the relation $h\m(f)=\m(hf)$ for any
$f\in \cK(X)$. One says that $h\m$ is a measure with the basis
$\m$ and the density $h$. For instance, if $h$ is a
$\m$-integrable function, then $h\m$ is a bounded measure on $X$.
This construction also is extended to complex functions and
complex measures, seen as compositions of the positive real ones.
\begin{theorem} \label{spr510} \mar{spr510}
Positive measures $\m$ and $\m'$ on a locally compact space $X$
are equivalent iff $\m'=f\m$, where $f$ is a locally
$\m$-integrable function such that $f>0$ locally almost everywhere
on $X$.
\end{theorem}
The function $f$ in Theorem \ref{spr510} is called the
Radon--Nikodym derivative. Of course, $\wt\m'(f')=\wt\m(ff')$ for
any positive integrable function $f'$ on $X$.
\subsection{Haar measures}
Let us point out the peculiarities of measures on locally compact
groups \cite{bourb6,book05}.
Let $G$ be a topological group acting continuously on a locally
compact space $X$ on the left, i.e., a map
\mar{spr455}\beq
\g(g): X\ni x\to gx\in X, \qquad g\in G, \label{spr455}
\eeq
is continuous for any $g\in G$, and so is a map $G\ni g\to gx\in
X$ for any $x\in X$. It should be emphasized that a map
\be
G\times X\ni (g,x)\to gx\in X
\ee
need not be continuous.
Let $f$ be a real function on $X$ and $\m$ a measure on $X$. A
group $G$ acts on $f$ and $\m$ by the laws
\be
(\g(g)f)(x)=f(g^{-1}x),\qquad (\g(g)\m)(f)=\m(\g(g^{-1})f).
\ee
A measure $\g(g)\m$ is the image of a measure $\m$ with respect to
the map (\ref{spr455}).
A measure $\m$ on $X$, subject to the action of a group $G$, is
said to be:
$\bullet$ invariant if $\g(g)\m=\m$ for all $g\in G$;
$\bullet$ relative invariant, if there is a strictly positive
number $\chi(g)$ such that $\g(g)\m=\chi(g)^{-1}\m$ for each $g\in
G$;
$\bullet$ quasi-invariant, if measures $\m$ and $\g(g)\m$ are
equivalent for all $g\in G$.
\noindent A strictly positive function $g\to \chi(g)$ yields a
representation of $G$ in $\mathbb R$. It is called the multiplier
of a measure $\m$.
Let a topological group $G$ act continuously on a locally compact
space $X$ on the right, i.e.,
\be
\tau(g): X\ni x\to xg^{-1}\in X.
\ee
The corresponding transformations of functions and measures on $X$
read
\be
(\tau(g)f)(x)=f(xg),\qquad (\tau(g)\m)(f)=\m(\tau(g^{-1})f).
\ee
Then invariant, relative invariant, and quasi-invariant measures
on $X$ are defined similarly to the case of $G$ acting on $X$ on
the left.
Now let $G$ be a locally compact group acting on itself by left
and right multiplications
\mar{B10}\ben
&& \g(g): q\to gq, \qquad \tau(g):q\to qg^{-1}, \qquad q\in
G,\nonumber\\
&& \g(g_1)\tau(g_2)=\tau(g_2)\g(g_1). \label{B10}
\een
Accordingly, left- and right-invariant measures, relative left-
and right-invariant measures, left- and right-quasi-invariant
measures on a group $G$ are defined. Each measure $\m$ on $G$ also
yields the inverse measure $\m^{-1}$ given by a relation
\be
\int f(g)\m^{-1}(g)=\int f(g^{-1})\m(g), \qquad
f\in \cK(G).
\ee
A positive (non-vanishing) left-invariant measure on a locally
compact group $G$ is called the left Haar measure (or, simply, the
{Haar measure). Similarly, the right Haar measure} is defined.
\begin{theorem}
A locally compact group $G$ admits a unique Haar measure with
accuracy to a number multiplier. The total mass $\m(G)$ of a Haar
measure $\m$ on $G$ is finite iff $G$ is a compact group.
\end{theorem}
Let us choose, once and for all, a left Haar measure $dg$ on a
locally compact group $G$. If $G$ is a compact group, $dg$ is
customarily the Haar measure of total mass 1.
\begin{example}
The Lebesgue measure $dx$ is a Haar measure on the additive group
$G=\mathbb R$. Its inverse is $-dx$.
\end{example}
The equality (\ref{B10}) shows that, if $dg$ is a Haar measure on
$G$, a measure $\tau(g')dg$ for any $g'\in G$ also is
left-invariant. Therefore, there exists a unique continuous
strictly positive function $\Delta(g')$ on $G$ such that
$\tau(g')dg=\Delta(g')dg$, $g'\in G$. It is called the modular
function of $G$. If $dg$ is a left Haar measure, its inverse
$(dg)^{-1}$ is a right Haar measure. There is a relation
$(dg)^{-1} =\Delta(g)^{-1}dg$. If $\Delta(g)=1$, a group $G$ is
called unimodular. Left and right Haar measures on a unimodular
group differ from each other in a number multiplier. For instance,
compact, commutative, and semisimple groups are unimodular. There
is the following criterion of a unimodular group. If the unit
element of a locally compact group has a compact neighborhood
invariant under inner automorphisms, this group is unimodular.
Measures $\m_1,\ldots,\m_n$ on a locally compact group $G$ are
called mutually contractible if there exists a measure
$\op*_i\m_i$ on $G$ given by a relation
\mar{spr540}\beq
\int f(g)\op*_i\m_i(g)=\int f(g_1\cdots g_n) \m_1(g_1)\cdots
\m_n(g_n), \qquad f\in \cK(G). \label{spr540}
\eeq
It is an image of the product measure $\m_1\cdots\m_n$ on
$\op\times^nG$ with respect to a map
\be
\op\times^nG\ni (g_1,\ldots, g_n) \to g_1\cdots g_n\in G.
\ee
Let $\ve_g$, $g\in G$, be the Dirac measure on $G$. The following
relations hold for all $x,y,z\in G$:
$\bullet$ $\ve_x*\ve_y=\ve_{xy}$;
$\bullet$ $\ve_x*\m=\g(x)\m$ and $\m*\ve_x=\tau(x^{-1})\m$;
$\bullet$ if measures $\la,\m,\nu$ are contractible, the pairs of
measures $\la$ and $\m$, $\m$ and $\nu$, $\la*\m$ and $\nu$, $\la$
and $\m*\nu$ also are contractible, and we have
\be
\la*\m*\nu=(\la*\m)*\nu=\la*(\m*\nu).
\ee
For instance, any two bounded measures on $G$ are contractible,
and a space $M^1(G)$ of these measures is a unital Banach algebra
with respect to the contraction $*$ (\ref{spr540}),
where the Dirac measure $\ve_\bb$ is the unit element.
One also defines:
$\bullet$ the contraction of a measure $\nu$ and a $dg$-integrable
function $f$ on $G$ as the density of the contraction $\nu*(fdg)$
with respect to a Haar measure $dg$ on $G$;
$\bullet$ the contraction of $dg$-integrable functions $f_1$ and
$f_2$ on $G$ as the density of the contraction $(f_1dg)*(f_2dg)$
with respect to a Haar measure $dg$ on $G$, i.e.,
\mar{spr531}\beq
(f_1*f_2)(g)=\int f_1(q)f_2(q^{-1}g)dq, \qquad q,g\in G.
\label{spr531}
\eeq
\subsection{Measures on infinite-dimensional vector spaces}
Throughout this Section, $E$ denotes a real Hausdorff topological
vector space. Infinite-dimensional topological vector spaces need
not be locally compact, and measures on them are defined as
follows \cite{bourb6,gelf64,book05}. All measures are assumed to
be positive.
Let $N(E)$ denote a set of closed vector subspaces of $E$ of
finite codimension, i.e., a vector subspace $V$ of $E$ belongs to
$N(E)$ iff there exists a finite set $y_1,\ldots,y_n$ of elements
of the dual $E'$ of $E$ such that $V$ consists of $x\in E$ which
obey the equalities $\lng x,y_i\rng=0$, $i=1,\ldots,n$.
A quasi-measure (or a cylinder set measure in the terminology of
\cite{gelf64})
on $E$ is defined as family
$\m=\{\m_V, V\in N(E)\}$ of bounded measures $\m_V$ on
finite-dimensional vector spaces $E/V$ such that if $W\subset V$,
the measure $\m_V$ is the image of a measure $\m_W$ with respect
to the canonical morphism $E/W\to E/V.$
For instance, each bounded measure on $E$ yields a quasi-measure
$\{\m_V, V\in N(E)\}$, where $\m_V$ is the image of a measure $\m$
with respect to the canonical morphism $r_V:E\to E/V$. There is
one-to-one correspondence between the bounded measures on $E$ and
the quasi-measures on $E$ which obey the following condition. For
any $\ve>0$, there exists a compact subset $K\subset E$ such that
\be
\m_V(E/V-r_V(K))\leq\ve, \qquad V\in N(E).
\ee
Clearly, any quasi-measure on a finite-dimensional vector space is
a measure.
Let $\g:E\to F$ be a continuous morphism of topological vector
spaces. For any $W\in N(F)$, a subspace $V=\g^{-1}(W)$ of $E$
belongs to $N(E)$, and $\g$ yields a morphism $\g^W: E/V\to F/W$.
Let $\m=\{\m_V,V\in N(E)\}$ be a quasi-measure on $E$. Then one
can assign the measure
\be
\nu_W=\g^W_*(\m_{\g^{-1}(W)})
\ee
to each $W\in N(F)$. It is readily observed that the family
$\nu=\{\nu_W, W\in N(F)\}$ is a quasi-measure on $F$. It is
called the image of a quasi-measure $\m$ with respect to a
continuous morphism $\g$.
In particular, let $F=\mathbb R$, and let $y\in E'$ be a
continuous form on $E$. The image $\m_y$ of a quasi-measure $\m$
on $E$ with respect to a form $y$ is a measure on $\mathbb R$. The
Fourier transform of a quasi-measure $\m$ on $E$ is defined as a
complex function
\mar{G0}\beq
Z(y)=\int_{\mathbb R} e^{it}\m_y(t) \label{G0}
\eeq
on the dual $E'$ of $E$. If $\m$ is a bounded measure on $E$, its
Fourier transform reads
\mar{G1}\beq
Z(y)=\op\int_E\exp[i\lng x,y\rng]\m(x). \label{G1}
\eeq
Let us point out the following variant of the well-known Bochner
theorem. A complex function $Z$ on a topological vector space $F$
is called positive-definite if
\be
\op\sum_{i,j} Z(y_i-y_j)\ol c_ic_j \geq 0
\ee
for any finite set $y_1,\ldots,y_m$ of elements of $F$ and any
complex numbers $c_1,\ldots,c_m$.
\begin{theorem} \label{box} \mar{box}
The Fourier transform (\ref{G1}) provides a bijection of the set
of quasi-measures on a Hausdorff topological vector space $E$ to
the set of positive-definite functions on the dual $E'$ of $E$
whose restriction to any finite-dimensional subspace of $E'$ is
continuous.
\end{theorem}
For instance, let $M(y)$ be a seminorm on $E'$. Then a function
\mar{spr523}\beq
Z(y)=\exp\left[-\frac12 M(y)\right] \label{spr523}
\eeq
on $E'$ is positive-definite. By virtue of Theorem \ref{box},
there is a unique quasi-measure $\m_M$ on $E$ whose Fourier
transform is $Z(y)$ (\ref{spr523}). It is called the Gaussian
quasi-measure with a covariance form $M$.
\begin{example}
Let $E=\mathbb R^n$ be a finite-dimensional vector space,
coordinated by $(x^i)$, and let $M$ be a norm on the dual of $E$.
A Gaussian measure on $E$ with a covariance form $B$ is equivalent
to the Lebesgue measure on $E$, and reads
\mar{qm610}\beq
\m_M= \frac{\mathrm{det}[M]^{1/2}}{(2\pi)^{n/2}}
\exp\left[-\frac12 (M^{-1})_{ij}x^ix^j\right]d^nx. \label{qm610}
\eeq
\end{example}
\begin{example} \label{gaus} \mar{gaus}
Let $E$ be a Banach space and $E'$ its dual, provided with the
norm $\|.\|'$ (\ref{spr446}). A Gaussian quasi-measure on $E$ with
the covariance form $\|.\|'$ is called canonical. One can show
that this quasi-measure fails to be a measure, unless $E$ is
finite-dimensional. Let $T$ be a continuous operator in $E'$. Then
\mar{spr524}\beq
y\to \|Ty\| \label{spr524}
\eeq
is a seminorm on $E'$. A Gaussian quasi-measure on $E$ with the
covariance form (\ref{spr524}) is proved to be a measure iff $T$
is a Hilbert--Schmidt operator.
\end{example}
Let $E$ be a real nuclear space and $E'$ its dual, equipped with
the topology of uniform convergence. Let us recall that all
topologies of uniform convergence (including weak$^*$ and strong
topologies) on $E'$ coincide, and $E$ is reflexive. A
quasi-measure on $E'$ is a measure iff its Fourier transform on
$E$ (which is the dual of $E'$) is continuous. A variant of the
Bochner theorem for nuclear spaces states the following
\cite{gelf64,book05}.
\begin{theorem} \mar{spr525} \label{spr525}
The Fourier transform
\be
Z(x)=\int\exp[i\lng x,y\rng]\m(y)
\ee
provides a bijection of the set of measures on the dual $E'$ of a
real nuclear space $E$ to the set of continuous positive-definite
functions on $E$
\end{theorem}
\begin{remark} \label{spr526} \mar{spr526}
Let $E\subset\wt E\subset E'$ be a real rigged Hilbert space,
defined by a norm $\|.\|$ on $E$. Let $T$ be a nuclear operator in
$\wt E$ and $\|.\|_T$ the restriction of the seminorm
$y\to\|Ty\|$, $y\in \wt E$, (\ref{spr524}) on $\wt E$ to $E$. Then
the Gaussian measures $\m$ and $\m_T$ on $E'$ with the covariance
forms $\|.\|$ and $\|.\|_T$ are not equivalent. The Gaussian
measures $\m$ and $\m_T$ are equivalent if $T$ is a sum of the
identity and a nuclear operator. In particular, all Gaussian
measures on a finite-dimensional vector space are equivalent.
\end{remark} | 191,292 |
I got back in the water in Feb after a two month unintentional break, and most of my workouts have been moderate to hard aerobic, 5 or 6 days a week. Today I thought I would try something different.
Goal swim, break 1:00 in the 100M. First attempt is a straight 100, second attempt is 2x50, third attempt is 4x25s. My prediction was that I would do it on the 25s.
Lifetime lap pool was full, so spent about 10 minutes in the kiddie pool playing around with streamlines and SDK. Then all 5 lanes cleared out of the lap pool!
Warm up
- 200 swim, 200 back, 200 kick ez
Goal Set
- 1x100 all out (1:03)
-- at this point I thought I had a shot at making the goal set at the 50s.
- 200 ez back + lots of wall rest
- 2x50 all out (:29, :29-)
-- 50 ez between plus tons of rest; ~2:30 between
-- DONE
Cool down
- maybe 400 ez mixed back/free
Finis Swimsense: I bought a swimsense, a watch thing that keeps stroke count, lap count, splits etc. This toy should be really helpful, but it isn't accurate enough to be useful to competitive pool swimmers in my opinion. Ex: Yesterday I did an 100 IM set and my back splits was something like a 33 while my breast splits were around 23... I am a much better at back than breast. Another problem I have with the watch is that the times only resolve to 1 second, aka you split is 23, not a 22.6. In theory, this is not a big deal, but the way it has been implemented is that the splits can sum to more than the interval. The error seems to be +/- 2s, which is a huge error for fast swims in a short course pool. What irritates me the most about the watch would annoy any swimmer during a short rest set, or any swimmer wanting a time off a fast effort swim, the watch has to be manually started and stopped to record the interval. When you are trying to remember to breath, it takes a while to remember to press the button, and on short rest sets, especially near the end, having time to press the button twice might not actually exist.
The watch is accurate for recording distance, so for fitness swimmers who are interested in overall distance swam and stroke counts, it is a decent device. It just seems a little expensive for something that is limited to stroke and lap counting.
A freakin' book on the Lezak strength training plan at the end.
Didn't want to get out of bed this morning, I was tired and had a slight headache. Showed up at the pool and the water was cloudy and at that point I had enough excuses to goof off for a while and cut the workout short.
Warm up
- 400 swim
- 50 kick no fins on back
- 300 pull no paddles
- 2x100 kick with fins on back
- 2x100 drill (right, left, catchup, fist)
- 50 kick no fins on back
Transition
- 10x25 Focus on streamline and break out
-- 6 powerful strokes off the wall, no breath, then ez
Main Set
- 5x50 on 1:00 (31,30,29,30,30)
-- these need to be 27s if I am going to break 1:50 (just an aside, not the goal of this particular set)
Cool down
- 200 ez
Patrick thinks he can (should?) break :50 and 1:50 in the 100 and 200 free while swimming a 23 mid in the 50. I think I need to be sub 23 to break :50 and 1:50. If we are both right, I wonder if that means I should pay more attention or less attention to how he trains...
Lezak Plan: I have never been a fan of the Lezak lifting plan, but as I have been thinking more and more about rest and how it can be used to enhance training, and I have come up with why the Lezak plan is generally a good lifting plan for swimmers. This time last year, I was following a fairly traditional strength building plan, where you increase your weight every workout until you fail, drop back a little and start progressing again. This time last year was I doing 5 sets of 5 reps of back squats between 250 and 300 lbs (250 being the drop back, and 300 being the max) 3 times a week. I lifted MWF and used to dread Friday because it didn't really matter where I was in the cycle for each exercise, there would be at least one exercise that would be near max combined with the exhaustion of the lifting and swimming from the week made me want to cry. Monday's were always the easiest lifting days because the weekend usually included a full day of no exercise or at least two full days off lifting.
This year, I really have not spent much time lifting seriously. I couldn't lift in Jan-Feb because I hurt my ribs, it didn't make much sense to start in Mar then leave for Asia in April, nor start after Asia with Nationals in May. Since then, lifting has been sporadic and more core work than strength building.
As a side effect, I have had much more energy in the pool. Because of this extra energy, I have been thinking more and more about rest: be it sleep, rest built into a set, workout frequency or workout intensity.
I came to the conclusion that the Lezak plan wasn't a great strength building plan a long time ago. Too many exercises, too many reps and no serious progression of weight. The limiting factor in the Lezak plan is endurance more than strength or glycogen instead of muscle break down.
So why do people keep using it, including Lezak, and why does it seem to be effective?
Swimmers are used to endurance being the limiting factor in training, and a serious strength training program would quickly lead to over training since complete days of rest are infrequent. If you tear down the muscle, but don't allow it to heal, and keep repeating the process over training happens quick.
The Lezak plan results in minimal muscle damage, because the quantity of exercises, reps and sets requires keeping weights low so the entire workout can be completed. After years of descend and 'hold pace' sets, the Lezak plan fits right in with the standard swimmer mentality for training.
The Lezak plan is taking the long road to strength, but by doing so, swim training is only marginally impacted. Slowly gaining strength and being able to train in the water at a higher intensity is going to be more beneficial to the average swimmer when compared to gaining strength fast but not being able to train in the water at a consistently high intensity.
That concludes my theory of the day.
Updated September 27th, 2010 at 01:49 PM by qbrain
Warm up
- 400 swim
- 200 kick
- 300 back
Transition
- 6x50 catchup/fist
Main Set
8x
- 25 ez on :30
- 50 build on :50
- 25 fast (2x :14, 6x :13)
- 150 back
Cool Down
- Chris Stevenson Special
Each cycle of the main set took a little less than 5:00, but I wasn't paying too much attention to the exact times.
I synced up with the other fast lanes when they started the main set, and this is when my time dropped from 14 to 13. Coincidence?
Turnover speed was a problem this morning. Since the battery died in my tempotrainer, I will need to buy a new one if I want to use that to help work on this problem.
If I can swim a 23 mid, I should be able to swim 12s from a push, right?
It didn't feel like much got accomplished this morning, 8x25s isn't much, but I do feel a little tight in the chest and arms, so it must have been a decent 200s of work.
This was way over on the sprint side of the spectrum from where I normally train. I hope it was a good use of time.
Warm up
- 6x50 drill swim*
- 4x75 kick with fins (did an extra 50)
- 4x150 ?? (don't remember what the end of warm up was)
Main Set
- 10x100 on 1:20 (first 1:07, rest 1:09/1:10, goal was sub 1:10, so I was a little slow)
- 4x50 kick with fins on :45
- 4x50 back on :50
- 200 reverse IM
- 2x150 on 2:00 (1:40, 1:40)
Cool down
- 200 back ez
* Swimming 50s during warm up is like pouring icewater down my pants. I don't do it because it is too cold. Swam non-stop drill swim.
Swam with Mark, Gerald and Anna on the main set today. Mark said he was just going to try to survive 1:20s and Anna made funny faces. All made it (Gerald did 75s).
SCY - Had the lane to myself until the last 400.
Warm up
- 400 swim
- 200 kick
Transition
- 4x50 odd back/free, even drill/swim
Main Set
- 4x150 on 1:50 (about 5s rest)
- 300 back
- 3x400 on 5:00* (4:48, 4:49, 4:42)
Cool down
- 200 ez
* Took a minute extra rest between #2 and #3 trying to break 4:40. Marshall dove into the lane as I was coming into the wall at the 200 and I got washed into the lane line. I should have swam faster.
This workout took me 44.5 minutes from the time I dove in until I started cool down. Dennis asked me if workout was short, since it only took him 48 minutes to get to the cool down and I pointed out that there is almost no rest in today's workout. We are also used to doing about 3100 LCM, so today's workout would have felt short even if there were more short distance sets adding to our rest.
Updated August 25th, 2010 at 12:39 PM by qbrain
LCM
Warm up
- 300 swim
- 200 choice (back)
- 300 pull (did back)
Transition
- 4x50 free/back
Main Set
- 300 free
- 4x150 free on 2:45
- 300 pull with buoy and paddles
- 2x150 free on 2:45 (2:00, 1:56)
-- more but had to get out
Cool down
- 300
The lane was a cluster, speed and endurance levels were all over the place, so I made the 300s start on the finish of the 3rd person in the lane and the 150s on 2:45. There was actually a slower lane ahead of us, but this should be the last LCM practice until next summer.
Since the lane was in disarray, my goal was to maintain my pace at an uncomfortable level the entire main set. Only on the last 2 150s did I get my time, and I don't know if they were faster or slower than the rest of the set.
Speed and Training
LindsayNB posted this on one of the swimming with math threads
It is an article that Budd Termin wrote about a training method he used on his swimmers at the University of Buffalo. The focus of the training method is training swimmers to swim at their optimal stroke rate while maintaining their optimal dps.
Swimming Speed Simplified: There are two ways to get faster. Increase the length of the stroke or increase the turn over.
Focusing on this, Termin worked on increasing dps early in the season and then worked on increasing turnover the rest of the season. This is very very simple, and required very little training time to accomplish. Training consisted mainly of lots of short distance just trying to swim at a target turnover while maintaining dps. There were two types of practices, high velocity and recovery, with high velocity being the practices working on increasing turnover lasting 60 minutes and recovery focusing on easy technique for 30 minutes, plus warm up and cool down time. There were 2 high velocity days followed by 1 recovery day, 6 days a week, one practice a day. A surprisingly light load.
The study lasted for 4 years, and the cumulative improvement was 10% in the 100 and 200 free, the two events he had the largest samples of swimmers during the study. All strokes and distances improved, but it looked like they improved about half as well.
How can I use this? I have no desire to just copy the training plan, but the idea of tracking dps and turnover seems like a good idea. Identifying and training for a specific turnover for my 200 sounds like a great idea. Lacking Termin's setup, use a poolmate watch?
I understand very little about effective training, so I do suffer from the 'Oh shiny!' effect when I read about a new training methodology.
The entire training program is laid out in detail starting on the 4th page of the pdf at the bottom left with the heading Training paradigms.
Warm up
- 400 free
- 300 kick with fins and board
Transition
- 4x50 drill swim
Main Set x4
- 4x100 descend on (2:00 back, 1:50 free)
- 2x50 descend on (1:00 back, :50 free)
Cool down
- 200 ez
The main set was supposed to be swam so that the first 50 was faster than half the last 100. I didn't go a very good job at this.
The first round of the main set we did back, but the guy behind me, a technically better backstroker*, misunderstood and thought we were doing 1x100 back. He tagged up about 15M from the finish, which threw both of us off because I didn't expect him to go that fast on the easy one, and he didn't think I would go that slow on the only back. That round was a cluster after that, and I don't think I got a single time that made sense. Pretty much a loss.
The last 3 rounds were free and originally I wanted to go sub 1:20 on all the 100s while still descending. It ended up 1:21 descended to 1:16 the first round, 1:24 descended to 1:15 the second and 1:24 descend to 1:13 the last round. The 50s were piss poor (39-40) on each round until the last when I went 37, 35.
Good: (1:16, 1:15, 1:13)
Crap: (39, 39, 40,40, 37, 35)
LCM is almost over, but I think I would like to be breaking 1:10 on a set like this followed by 33,32 on the 50s.
It took until the first fast 100 free, about halfway through practice for my free to feel comfortable and my back never felt good.
I have been tired lately and I am not sure why. Plenty of sleep (~8 hours/night consistent), less frequent workouts than last season, easier weight workouts, so I am at a loss.
* He lacks my endurance
We didn't have to put the long course lane ropes in! Of course, the pool is closed, so no one has been in the pool since our practice Friday
LCM
Warm up
- 400 swim
- 200 kick
- 100 back
- 100 waitforpeopletokindacatchup free
Transition
- 2x100 drill
Main Set
- 400 free (5:26)
- 2x100 kick with fins and board
- 2x200 free on 3:30 (2:33, 2:37)
- 4x100 pull with paddles and buoy on 1:40 (all sub 1:20, 30/31 SPL)
- 2x50 on 1:00 (supposed to be 8) (:37, :36)
Cool down
- 2x50 ez
Meh, long course today.
Warm up
- 200 swim
- 300 back
Transition
- 4x50 drill/swim
Main Set
-- 1:40 base
- 500 free
- 5x100 free (was supposed to be pull)
- 400 free
- 4x100 back kick with fins on 2:00
- 300 free
-- See a pattern? I had to get out because of time.
Cool down
- 100 back
My pace was between 1:19 and 1:22 on the hundreds and probably sub 1:25 during the entire (free) main set. Not very fast considering the amount of rest I had.
Did 33 pushups.
No compliments today, Anna made fun of me, but I forgot the details.
Rant
I was highly annoyed last night when I found out that we would be long course again so the triathletes could get some more long course before the tri sprint hosted at the pool next weekend. All summer there would be no ropes in when we got there and I would usually unroll 5 of the 7 ropes, so the thought of pulling out 16 ropes and putting in 7 ropes had me considering skipping. We have between 30 and 40 people show up for practice but about 6-10 people either unrolled ropes or swam the ropes down the pool all summer. Last night the coach asked that people show up a little early to help switch out the lane ropes, and I was surprised, people showed up and helped.
Still, this cuts into the time I can spend swimming, so I dislike it. It isn't a problem short course, since people swim their own lane ropes, and they are rarely removed in the first place.
While I am in the middle of a rant, why were people wearing huge drag suits in my lane today? There were 4 people in the lane today, the 2 with drag suits that were board short sized, and those two didn't make the intervals. What benefit did the drag suits provide? I don't actually care, and am not upset that they did, it just seems dumb.
Monday should be the last day of long course nonsense. | 397,468 |
TITLE: Inverse of $x - \tanh(x)$
QUESTION [0 upvotes]: I am trying to find out the inverse of function $f:\mathbb R\to\mathbb R, f(x) = x - \tanh(x),\forall\in\mathbb R.$
What I tried:
Since $f(x)$ is invertible, so using $f(f^{-1}(x)) = x,$ I get $x = f^{-1}(x) - \tanh(f^{-1}(x)).$
Expanding or opening up $\tanh(x)$ leads me nowhere. I also tried the property $f^{-1'}(x) = \frac{1}{f^{'}(f^{-1}(x))}.$ But I couldn't get the inverse equation.
REPLY [3 votes]: 1.)
The inverse of your function isn't an elementary function:
$$f(x)=x-\tanh(x)$$
$$x-\tanh(x)=y$$
For algebraic values $y=a$, we have
$$x-\tanh(x)=a.\tag{1}$$
Each elementary standard function (like $\tanh$) can be represented in an exp-ln form:
$$x-\frac{(e^x)^2-1}{(e^x)^2+1}=a.$$
$$x(e^x)^2-a(e^x)^2-(e^x)^2+x-a+1=0\tag{2}$$
This is an irreducible polynomial equation over $\overline{\mathbb{Q}}$ in dependence of $x$ and $e^x$. The main theorems in [Lin 1983] and [Chow 1999] state that those kinds of equations don't have solutions except $0$ in the Elementary numbers or Explicit elementary numbers respectively. Because equation (2) and (1) have the same set of solutions, the same conclusion holds for equation (1).
And it follows with the theorem in Proof Check: Non-existence of the inverse function in a given class of functions that your function $f$ cannot have an inverse that is an elementary function.
2.)
$$-\frac{(x-1-y)}{x+1-y}e^{2x}=1$$
We see, your function doesn't have an inverse in terms of Lambert W. But the equation can be solved in terms of Generalized Lambert W:
$$x=\frac{1}{2}W(^{2y+2}_{2y-2};1)=-\frac{1}{2}W(^{-2y+2}_{-2y-2};1)$$
$-$ see [Mező 2017], [Mező/Baricz 2017], [Castle 2018].
$\ $
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018 | 41,540 |
\begin{document}
\title[Brieskorn manifolds, generalised Sieradski groups and coverings]{Brieskorn manifolds, generalised Sieradski groups, and coverings of lens spaces}
\author{Tatyana Kozlovskaya}
\address{Magadan Institute of Economics, 685000, Magadan, Russia}
\email{konus$\[email protected]}
\author{Andrei Vesnin}
\address{Sobolev Institute of Mathematics, 630090, Novosibirsk, Russia}
\email{[email protected]}
\thanks{The research was done under partial financial support by RFBR (grant 15-01-07906).}
\subjclass[2000]{57M05; 20F05; 57M50}
\keywords{Three-dimensional manifold, Brieskorn manifold, cyclically presented group, Sieradski group, lens space, branched covering.}
\dedicatory{To Sergei Matveev on the occasion of his seventieth birthday}
\maketitle
\begin{abstract}
The Brieskorn manifolds $B(p,q,r)$ are the $r$-fold cyclic coverings of the 3-sphere $S^{3}$ branched over the torus knot $T(p,q)$. The generalised Sieradski groups $S(m,p,q)$ are groups with $m$-cyclic pre\-sen\-tation $G_{m}(w)$, where defining word $w$ has a special form, depending of $p$ and $q$. In particular, $S(m,3,2) = G_{m}(w)$ is the group with $m$ generators $x_{1}, \ldots, x_{m}$ and $m$ defining relations $w(x_{i}, x_{i+1}, x_{i+2})=1$, where $w(x_{i}, x_{i+1}, x_{i+2}) = x_{i} x_{i+2} x_{i+1}^{-1}$. Presentations of $S(2n,3,2)$ in a certain form $G_{n}(w)$ were investigated by Howie and Williams. They proved that the $n$-cyclic presentations are geometric, i.e., correspond to the spines of closed orientable 3-manifolds. We establish an analogous result for the groups $S(2n,5,2)$. It is shown that in both cases the manifolds are $n$-fold cyclic branched coverings of lens spaces. To classify some of constructed manifolds we used the Matveev's computer program ``Recognizer''.
\end{abstract}
\tableofcontents
\section{Three-manifolds with cyclic symmetry} \label{sec1}
\subsection{Examples of manifolds with cyclic symmetry}
In this paper we discuss a relation between 3-dimensional closed orientable Brieskorn manifolds and the generalised Sieradski groups. The Brieskorn manifolds $B(p,q,r)$, firstly introduced in~\cite{Brieskorn}, are the $r$-fold cyclic coverings of the 3-sphere $S^{3}$ branched over the torus knots $T(p,q)$. The generali\-sed Sieradski groups $S(m,p,q)$, introduces in \cite{CavicchioliHegenbarthKim} for positive integers $p$ and $q$, where $p = 1 + d q$ and $d \geq 1$, are groups with cyclic presentation $G_{m}(w)$, where defining word $w$ will be described in subsection~\ref{subsection2.3}. In particular, $S(m,3,2) = G_{m}(w)$ is the group with $m$ generators $x_{1}, \ldots, x_{m}$ and $m$ defining relations $w(x_{i}, x_{i+1}, x_{i+2})=1$, where $w(x_{i}, x_{i+1}, x_{i+2}) = x_{i} x_{i+2} x_{i+1}^{-1}$.
Cyclic presentations of the groups $S(2n,3,2)$ (the case $q=2$ and $d=1$) in the form $G_{n}(w)$ were investigated by Howie and Williams in~\cite{Howie}. Namely, they proved that $n$-cyclic presentations of $S(2n,3,2) = G_{n} (x_{i} x_{i+2} x_{i+1}^{-1})$ are \emph{geometric}, i.e., correspond to spines of closed orientable 3-manifolds. Here we establish the analogous result for the groups $S(2n,5,2)$ (the case $q=2$ and $d=2$). More exactly, theorem~\ref{teo4} states that $n$-cyclic presentations
$$
S(2n,5,2) = G_{n} ( x_{i} x_{i+1} x_{i+2}^{2} x_{i+3} x_{i+4} x_{i+3}^{-1} x_{i+2}^{-1} x_{i+1} x_{i+2} x_{i+3} x_{i+2}^{-1} x_{i+1}^{-1})
$$
are geometric. Propositions~\ref{ass1} and~\ref{ass2} state that in both cases manifolds are $n$-fold cyclic branched coverings of lens spaces.
For last decades topological and geometrical properties of 3-manifolds are intensively studying from various points of view. An approach to constructing and describing a manifold depends on a problem posted. The approach can be based on a triangulation of a manifold, its Heegaard splitting, surgery description, branched covering description, etc. Any app\-ro\-ach can be useful in a suitable context. For basic definitions and facts of 3-manifold theory we refer to Hempel's book~\cite{Hempel}. One more useful approach is based on presenting 3-manifolds by spines. The theory of spines of 3-manifolds was developed in papers by Matveev, his students and followers. Main results the theory of spines are presented in his book~\cite{MatveevBook}. Methods and results on tabulating of 3-manifolds are also presented in his survey~\cite{MatveevSurvey}. The presentation of 3-manifolds by their spines is realised in the program tools \emph{3-manifold Recognizer}~\cite{Recognizer}, which was created and developed by Matveev and his group. The tools contain the original software for distinguishing and investiga\-ting 3-manifolds as well as the huge database of 3-manifolds, presented by their spines. In particular, the developed software is useful for studying symmetry groups of 3-manifolds and for understanding covering properties of 3-manifolds in a lot of concrete cases.
In this paper we will consider connected closed orientable 3-manifolds with cyclic symmetry which acts with fixed points. Moreover, we will interested in cases when the symmetry corresponds to a presentation of the manifold as a cyclic branched covering of the 3-sphere $S^{3}$ or of the lens spaces. Examples of such manifolds are well known in the literature. Let us recall some of them.
\begin{itemize}
\item The spherical and hyperbolic dodecahedral spaces constructed by Weber and Seifert in 1931 in the paper~\cite{WeberSeifert}: the first is the $3$-fold cyclic covering of $S^{3}$ branched over the trefoil knot, and the second is a $5$-fold cyclic covering of $S^{3}$ branched over the 2-component Whitehead link.
\item The smallest volume closed orientable hyperbolic 3-manifold, disco\-ve\-r\-ed by Matveev and Fomenko~\cite{MatveevFomenko} and independently by Weeks~\cite{Weeks}, is the $3$-fold cyclic covering of $S^{3}$ branched over the 2-bridge knot~$7/3$~\cite{MednykhVesnin1}.
\item The Fibonacci manifolds, constructed by Helling, Kim and Mennicke~\cite{HellingKimMennicke}, are the $n$-fold cyclic coverings of $S^{3}$ branched over the figure-eight knot.
\item The Sieradski manifolds, defined by Sieradski in~\cite{S} and investigated by Cavicchioli, Hegenbarth and Kim in~\cite{CavicchioliHegenbarthKim}, are the $n$-fold cyclic coverings of $S^{3}$ branched over the trefoil knot.
\end{itemize}
As one knows, the trefoil knot belongs to the family of torus knots (it has the corresponding notation $T(3,2)$). Thus, the Sieradski manifolds from~\cite{CavicchioliHegenbarthKim, S} belong to the bigger class of manifolds, known as the Brieskorn manifolds.
\subsection{Brieskorn manifolds}
Recall that Brieskorn~\cite{Brieskorn} (see also Milnor's book~\cite{Milnor68}) initiated a study of the following objects. For positive integer $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n+1} \geq 2$ consider the polynomial
$$
f(z_{1}, z_{2}, \ldots, z_{n+1}) = (z_{1})^{\alpha_{1}} + (z_{2})^{\alpha_{2}} + \cdots + (z_{n+1})^{\alpha_{n+1}}.
$$
\emph{A Brieskorn manifold} $\mathcal B(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n+1})$ is defined as the intersection of the complex hyperplane $V = f^{-1} (0)$ with the $(2n+1)$--dimensional sphere of unit radius
$$
S^{2n+1} = \{ (z_{1}, z_{2}, \ldots, z_{n+1}) \in \mathbb C^{n+1} \, : \, |z_{1}|^{2} + |z_{2}|^{2} + \cdots + |z_{n+1}|^{2} = 1 \}.
$$
Brieskorn manifolds are smooth manifolds of dimension $2n-1$. The interest to these manifolds is motivated, in particular, by the following surprising fact, proved by Milnor~~\cite{Milnor68}: Brieskorn manifolds, corresponding to polynomials
$$
f(z_1, z_2, z_3, z_4, z_5) = z_1^{6k-1} + z_2^3 + z_3^2 + z_4^2 + z_5^2
$$
where $k=1, 2, \ldots, 28$, are $28$ pairwise non-diffeomorphic exotic spheres and each of them is homeomorphic to the usual 7-dimensional sphere.
We will interested in the case when Brieskorn manifolds are 3-dimensional. Similar to Milnor's notation from~\cite{Milnor75}, for positive integers $p,q,r \geq 2$ we denote
$$
\mathcal B(p,q,r) = \{ (z_{1}, z_{2}, z_{3}) \in \mathbb C^{3} \, : \, z_{1}^{p} + z_{2}^{q} + z_{3}^{r} = 0, \quad |z_{1}|^{2} + |z_{2}|^{2} + |z_{3}|^{2} = 1 \}.
$$
According to lemma~1.1 from~\cite{Milnor75}, the manifold $\mathcal B(p,q,r)$ is the $r$-fold cyclic covering of the sphere $S^{3}$, branched over the torus link $T(p,q)$. Since parameters $p$, $q$, $r$ can be permuted in the definition of $\mathcal B(p,q,r)$, this manifold is also the $q$-fold cyclic covering of $S^{3}$, branched over $T(r,p)$, as well as the $p$-fold cyclic covering of $S^{3}$, branched over $T(q,r)$. Recall that the torus link $T(p,q)$ can be defined as the set of points $(z_1, z_2)$ on the unit sphere which satisfy the equation $z_1^p + z_2^q = 0$. Such a link has $d$ components, where $d = \gcd (p,q)$. An $n$-th component, $1 \leq n \leq d$, can be parametrized as the following:
$$
z_1 = \exp(2\pi i t/ p), \qquad z_2 =\exp(2 \pi i (t+n+ 1/2) /q )
$$
for $0 \leq t \leq pq/d$.
Another definition of the torus links can be done in terms of braids. Let $B_{p}$ be the group of geometrical braids on $p$ strings with standard generators $\sigma_{1}, \sigma_{2}, \ldots, \sigma_{p-1}$. Then the closure of the braid $(\sigma_{1} \sigma_{2} \cdots \sigma_{p-1})^{q}$ is the torus link $T(p,q)$.
It was shown in~\cite{Milnor75} that the Brieskorn manifold $\mathcal B(p,q,r)$ is spherical if $1/p + 1/q + 1/r > 1$; nilpotent if $1/p + 1/q + 1/r =1$; and $\widetilde{\operatorname{SL}}(2, \mathbb R)$-manifold if $1/p + 1/q + 1/r < 1$.
\section{Cyclically presented groups} \label{sec2}
\subsection{Cyclic presentation and defining word}
Let $\mathbb F_{n}$ be the free group of rank $m \geq 1$ with generators $x_{1}, x_{2}, \ldots, x_{m}$ and let $w = w(x_{1}, x_{2}, \ldots , x_{m})$ be a cyclically reduced word in $\mathbb F_{m}$. Let $\eta : \mathbb F_{m} \to F_{m}$ be an automorphism given by $\eta (x_i) = x_{i+1}$, $i=1,\dots,m-1$, and $\eta (x_m) = x_1$. The presentation
$$
G_m (w) \, = \, \langle x_1, \dots , x_m \ | \ w=1, \, \eta (w)=1, \, \dots, \, \eta^{m-1} (w)=1 \rangle ,
$$
is called an $m$-\emph{cyclic presentation} with \emph{defining word} $w$. A group $G$ is said to be \emph{cyclically presented group} if $G$ is isomorphic $G_{m}(w)$ for some $m$ and $w$~\cite{Johnson}. Beside purely algebraic questions (finiteness, commensurability, arithmeticity, etc.), the following two questions about cyclically presented groups were posted and investigated by many authors (see \cite{CavicchioliHegenbarthKim, Howie} for example).
\begin{question}
What cyclically presented groups are isomorphic to the fundamental groups of closed connected orientable 3-manifolds?
\end{question}
A 2-dimensional subpolyhedron $P$ of a closed connected 3-manifold $M$ is called a \emph{spine} of $M$ if $M \setminus \operatorname{Int} B^{3}$ collapses to $P$, where $B^{3}$ is a 3-ball in $M$ \cite{MatveevBook}. We shall say that a finite presentation $\mathcal P$ corresponds to a spine of a 3-manifold if its presentation complex $K_{\mathcal P}$ is a spine of a 3-manifold.
\begin{question}
What cyclic presentations $G_{m} (w)$ of groups are geometric, i.e., correspond to the spines of closed connected orientable 3-manifolds?
\end{question}
Let us recall some examples of cyclically presented groups which are fundamental groups of 3-manifolds, as well as demonstrate that some families of cyclically presented groups can not be realised as groups of hyperbolic 3-orbifolds (in particular, hyperbolic 3-manifolds) of finite volume.
\subsection{Fibonacci groups}
We start with groups, corresponding to the defining word $w(x_{1}, x_{2}, x_{3}) = x_{1} x_{2} x_{3}^{-1}$. Cyclically presented groups
$$
F(2,m) \ = \ \langle x_1, \dots , x_m \ | \ x_i x_{i+1} \, = \, x_{i+2}, \quad i = 1, \dots , m \rangle,
$$
where subscripts are taken by $\operatorname{mod} m$, are known as \emph{the Fibonacci groups}. It was shown by Helling, Kim and Mennicke~\cite{HellingKimMennicke} that if number of generators is even, $m=2n$, $n\geq 4$, then the groups $F(2,2n)$ are the fundamental groups of hyperbolic 3-manifolds. These manifolds were called \emph{the Fibonacci manifolds}. Thus, for even $m \geq 8$ the group $F(2,m)$ is infinite and torsion-free. For odd $m$ situation is different. If $m$ is odd, then $F(2,m)$ has elements of finite order. More exactly, the product $v = x_{1} x_{2} \ldots, x_{m}$ has order two (see \cite{BardakovVesnin} for details). Therefore, for odd $m$ the group $F(2,m)$ can not be the fundamental group of a hyperbolic 3-manifold. Moreover, it was shown by Maclachlan~\cite{Maclachlan}, that $F(2,m)$ can not be the group of hyperbolic 3-orbifold of finite volume. Recently, Howie and Williams~\cite{Howie} proved that for odd $m \geq 3$ the group $F(2,m)$ is the fundamental group of a 3-manifold if and only if $m=3,5,7$. In all this cases the group is finite cyclic.
A straightforward generalisation of the groups $F(2,m)$, namely, the groups $F(r,m)$ for $r \ge 2$ and $m \ge 3$ were defined in~\cite{JWW} as follows:
$$
F(r,m) \, = \, \langle x_1, \dots , x_m \ | \ x_i x_{i+1} \, \dots x_{i+r-1}= \, x_{i+r}, \quad i = 1, \dots , m \rangle ,
$$
where subscripts are taken by $\operatorname{mod} m$. The groups $F(r,m)$ are usually called~\emph{Fi\-bo\-nac\-ci groups} too. A survey of results about finiteness of these groups is given in~\cite{Johnson}. Using the same ideas as in~\cite{Maclachlan}, Szczepanski~\cite{Szcz} proved the following result: if $r$ is even and $m \ge r$ is odd, then the Fibonacci group $F(r,m)$ can not be the fundamental group of hyperbolic 3-manifold of finite volume.
\subsection{Sieradski groups} \label{subsection2.3}
One more interesting class of cyclically presented groups corresponds to the defining word $w(x_{1}, x_{2}, x_{3}) = x_{1} x_{3} x_{2}^{-1}$. Cyclically presented groups
$$
S(m) = \langle x_1, x_2, \ldots , x_{m} \, \mid \, x_{i} x_{i+2} = x_{i + 1}, \quad i=1, \ldots m \rangle,
$$
where all subscripts are taken by $\operatorname{mod} m$, were introduced by Sieradski~\cite{S}. In~\cite{CavicchioliHegenbarthKim} these groups were called \emph{the Sieradski groups} as well as a more general class of groups was introduced. The groups
$$
\begin{gathered}
S(m,p,q) = \langle x_1, \ldots , x_m \, \mid \, \hfill \\
{x_i} \, x_{i + q} \, \cdots \, x_{i + (q - 1)d q - q} \, x_{i + (q - 1)d q} \, = \, x_{i + 1} \, x_{i+q+1} \, \cdots \, x_{i + (q - 1)d q - q + 1} , \\ \hfill
i=1, \ldots, m \rangle ,
\end{gathered}
$$
are called \emph{the generalised Sieradski groups}. As above, all subscripts are taken by $\operatorname{mod} m$. Parameters $p$ and $q$ are co-prime integers such that $p = 1 + d q$, $d \in \mathbb{Z}$. Cavicchioli, Hegenbarth and Kim proved the following result.
\begin{theorem} \cite{CavicchioliHegenbarthKim} \label{theorem1}
The cyclic presentation $S(m,p,q)$ corresponds to a spine of the $m$-fold cyclic covering of the 3-sphere $S^{3}$ branched over the torus knot $T(p,q)$, i.e., the Brieskorn manifold $\mathcal B(m,p,q)$.
\end{theorem}
In particular, the Sieradski groups $S(m) = S(m,3,2)$ correspond to spines of manifolds $\mathcal B(m,3,2)$ which are $m$-fold cyclic coverings of $S^{3}$ branched over the trefoil knot $T(3,2)$.
Below we will interested in generalised Sieradski groups with parameter $q=2$. In this case $p=1 + 2d$ and we get the following presentation:
$$
\begin{gathered}
S(m,2d+1,2) = \langle x_{1}, x_{2}, \ldots, x_{m} \ | \ x_{i} x_{i+2} \cdots x_{i+2d} \ = \ x_{i+1} x_{i+3} \cdots x_{i+2d-1}, \\ \hfill \quad i=1, \ldots, m \rangle.
\end{gathered}
$$
\subsection{Generalised Fibonacci groups}
Consider two families of groups which were introduced in~\cite{C-R} and called \emph{the generalised Fibonacci groups}. The first family consists of the groups $F(r,m,k)$, where $r \ge 2$, $m \ge 3$, $k \ge 1$, given by
$$
F(r,m,k) = \langle x_1, \dots , x_m \ | \ x_i x_{i+1} \cdots x_{i+r-1} = x_{i+r-1+k}, \quad i = 1, \dots , m \rangle ,
$$
where all subscripts are taken by $\operatorname{mod} m$. Obviously, $F(r,m,1)=F(r,m)$.
The following statement generalises preceding results on groups $F(2,m)$ and $F(r,m)$, obtained in~\cite{Maclachlan} and~\cite{Szcz}, respectively.
\begin{theorem} \cite{Sz-Ves3} \label{theorem2}
Assume that $r$ is even, $m$ is odd and $(m, r+2k-1)=1$. Then the generalised Fibonacci group $F(r,m,k)$ can not be realised as the group of hyperbolic 3-orbifold of finite volume.
\end{theorem}
Let us discuss some groups $F(r,m,k)$ which do not satisfy to conditions of theorem~\ref{theorem2}. Suppose
\begin{equation}
m = r + 2k - 1 . \label{eqn_cyclic4}
\end{equation}
For $k=1$ in (\ref{eqn_cyclic4}) we have $m=r+1$. Corresponding groups $F(m-1,m,1)$ are the Fibonacci groups $F(m-1,m)$ with the following cyclic presentation:
$$
F(m-1,m) = \langle x_{1}, x_{2}, \ldots, x_{m} \ | \ x_{i} x_{i+1} \cdots x_{i+m-2} = x_{i+m-1}, \quad i=1, \ldots, m \rangle.
$$
Recall the definition of \emph{the generalised Neuwirth groups} $\Gamma^{\ell}_m$ from~\cite{Sz-Ves3}:
$$
\Gamma^{\ell}_m = \langle x_1, \dots , x_m \ | \ x_i x_{i+1} \ldots x_{i+m-2} = x_{i+m-1}^{\ell}, \quad i=1, \ldots, m \rangle ,
$$
where $m \geq 3$ and $\ell \geq 1$. These groups generalise the groups introduced by Nuewirth in~\cite{N}. In particular, $\Gamma^1_m = F(m-1, m)$.
\begin{proposition} \cite{Sz-Ves3}
The groups $\Gamma^{\ell}_m$ are the fundamental groups of the fibered Seifert spaces $\Sigma_{m}^{\ell}$ with parameters
$$
\Sigma_m^{\ell} \ = \ ( 0 \ {\small o} \ 0 \ | \ -1 ; \ \underbrace{(\ell+1,1), \ (\ell+1,1) , \ \dots , (\ell+1,1)}_{m~{\rm times}} ) .
$$
\end{proposition}
For $r=2$ in (\ref{eqn_cyclic4}) we have
$$
F(2,2k+1,k) = \langle x_1, \dots , x_{2k+1} \ | \ x_i x_{i+1} = x_{i+1+k}, \quad i=1, \dots , 2k+1 \rangle .
$$
It is easy to see~\cite{Sz-Ves2} that the groups $F(2,2k+1,k)$ and the Sieradski groups
$$
S(2k+1) = \langle a_{1}, a_{2}, \ldots, a_{2k+1} \, | \, a_{i} a_{i+2} = a_{i+1}, \quad i = 1, \ldots, 2k+1 \rangle
$$
are isomorphic under the following correspondence of generators:
$$
\left(
\begin{array}{llllllllll}
x_1 & x_2 & x_3 & \dots & x_k & x_{k+1} & x_{k+2} & \dots & x_{2k} & x_{2k+1} \\
a_1 & a_3 & a_5 & \dots & a_{2k-1} & a_{2k+1} & a_2 & \dots &
a_{2k-2} & a_{2k}
\end{array}
\right) .
$$
Theorem~\ref{theorem1} implies that the groups $F(2,2k+1,k)$ are the fundamental groups of 3-manifolds which are $(k+1)$-fold cyclic coverings of $S^{3}$ branched over the trefoil knot.
The second family from~\cite{C-R} consists of the groups $H(r,m,k)$, where $r \ge 2$, $m \ge 3$, $k \ge 1$, given by
$$
H(r,m,k) = \langle x_1, \dots , x_m \ | \ x_i x_{i+1} \cdots x_{i+r-1} = x_{i+r} \cdots x_{i+r-1+k}, \quad i = 1, \dots , m \rangle ,
$$
where all subscripts are taken by $\operatorname{mod} m$. In particular, $H(r,m,1) = F(r,m)$. Let us compare the groups $H(r,m,k)$ with the generalised Sieradski groups $S(m,2k-1,2)$:
$$
\begin{gathered}
S(m,2k-1,2) = \langle a_1, \dots , a_m \ | \ a_i a_{i+2} \ \dots \ a_{i+2k-2} = a_{i+1} a_{i+3} \ \dots \ a_{i+2k-3}, \\ \hfill \quad i=1 \dots n \rangle .
\end{gathered}
$$
It is easy to see~\cite{Sz-Ves2} that $H(k,2k-1,k-1)$ and $S(2k-1,2k-1,2)$ are isomorphic under the following correspondence of generators:
$$
\left(
\begin{array}{lllllllll}
x_1 & x_2 & x_3 & \dots & x_k & x_{k+1} & x_{k+2} & \dots & x_{2k-1} \\
a_1 & a_3 & a_5 & \dots & a_{2k-1} & a_{2} & a_4 & \dots & a_{2k-2}
\end{array}
\right) .
$$
Theorem~\ref{theorem1} implies that for $k \ge 2$ the group $H(k,2k-1,k-1) \cong S(2k-1,2k-1,2)$ is the fundamental group of the closed 3-manifold which is the $(2k-1)$-fold cyclic covering of the 3-sphere branched over the torus knot $T(2k-1,2)$, i.e., the Brieskorn manifold $\mathcal B(2k-1, 2k-1, 2)$.
\subsection{Johnson~-- Mawdesley groups}
In~\cite{J-M} Johnson and Mawdesley defined the class of groups with the following cyclic presentations:
$$
G_{n}(m,k) = \langle x_{1}, x_{2}, \ldots, x_{n} \, | \, x_{i} x_{i+m} = x_{i+k}, \quad i = 1, \ldots, n \rangle.
$$
Obviously, this class contains the Fibonacci groups and the Sieradski groups, namely $G_{n}(1,2) = F(2,n)$ and $G_{n} (2,1) = S(n,3,2)$. Structure of the groups $G_{n}(m,k)$ was studied in \cite{BardakovVesnin} and properties of the groups $G_{n}(m,1)$ were considered in \cite{G-H}. In \cite{Howie} there was investigated the following question: when $G_{n}(m,k)$ is the fundamental group of a 3-manifold? The answer was done for all group with only two exceptional cases: for the groups $G_{9}(4,1)$ and $G_{9}(7,1)$ the question is still open.
\section{Geometric group presentations} \label{sec3}
\subsection{Cyclic presentations of $S(2n,3,2)$ with $n$ generators}
It was men\-tioned above that the Fibonacci group $F(2,2n)$, $n \geq 2$, is the fundamental group of a 3-manifold which is the $n$-fold cyclic cover of the 3-sphere $S^{3}$ branched over the figure-eight knot. This manifold is spherical if $n=2$, Euclidean if $n=3$, and hyperbolic if $n \geq 4$. The covering is related to symmetry of order $n$ corresponding to the permutation $x_{i} \to x_{i+2}$ on the generators $x_{1}, x_{2}, \ldots, x_{2n}$ of $F(2,2n)$, but not to the permutation $x_{i} \to x_{i+1}$, which corresponds to the cyclic presentation of the group. It was observed in~\cite{Kim-Vesnin} that the $n$-fold cyclic covering corresponds to the following $n$-cyclic presentation:
$$
\begin{gathered}
F(2,2n) = G_{n}(y_{1}^{-1} y_{2}^{2} y_{3}^{-1} y_{2}) = \langle y_{1}, y_{2}, \ldots y_{n} \ | \ y_{i}^{-1} y_{i+1}^{2} y_{i+2}^{-1} y_{i+1} =1,
\ i=1, \ldots, n \rangle,
\end{gathered}
$$
where $y_{i} = x_{2i}$, $i=1, \ldots, n$. The correspondence $\sum_{i} y_{i}^{k_{i}} \to \sum_{i} k_{i} t^{i}$ sent the defining word $y_{0}^{-1} y_{1}^{2} y_{2}^{-1} y_{1}$ to the polynomial $-(t^{2} - 3t + 1)$, which is equivalent to the Alexander polynomial $\Delta(t) = t^{2} - 3t + 1$ of the figure-eight knot.
Analogously, one can consider the generalised Sieradski group $S(2n,3,2)$ with even number of generators. This group admits a cyclic presentation with $n$ gene\-ra\-tors:
\begin{eqnarray*}
S(2n,3,2) & = & G_{2n} (x_{1} x_{3} x_{2}^{-1}) \\
& = & \langle x_{1}, x_{2}, \ldots, x_{2n} \ | \ x_{i} x_{i+2} = x_{i+1}, \quad i=1, \ldots, 2n \rangle \\
& = & \langle x_{1}, x_{2}, \ldots, x_{2n} \ | \ x_{2j} x_{2j+2} = x_{2j+1}, \quad x_{2j+1} x_{2j+3} = x_{2j+2}, \\
& & \qquad \qquad \qquad \qquad \qquad \qquad j=1, \ldots, n \rangle \\
& = & \langle x_{2}, x_{4}, \ldots, x_{2n} \ | \ (x_{2j} x_{2j+2}) (x_{2j+2} x_{2j+4}) = x_{2j+2}, \quad j=1, \ldots, n \rangle \\
& = & \langle y_{1}, y_{2}, \ldots, y_{n} \ | \ y_{j} y_{j+1}^{2} y_{j+2} = y_{j+1}, \quad j = 1, \ldots, n \rangle \\
& = & G_{n} (y_{1} y_{2}^{2} y_{3} y_{2}^{-1}).
\end{eqnarray*}
Theorem~\ref{theorem1} implies that $S(2n,3,2)$ is the fundamental group of the 3-manifold $\mathcal B(2n,3,2)$, which is the $2n$-fold cyclic covering of $S^{3}$ branched over the trefoil knot. Moreover, the cyclic presentation $S(2n,3,2)$ is geometric.
It is natural to ask: if the cyclic presentation $G_{n} (x_{1} x_{2}^{2} x_{3} x_{2}^{-1})$ is geometric too?
Answering in affirmative on the question, Howie and Williams constructed a corresponding spine in~\cite{Howie}. Following notations from~\cite{Howie}, we denote the ge\-ne\-ra\-tors by $x_{0}, x_{1}, \ldots, x_{n-1}$. Consider the 2-complex $\mathcal P_{n}$, presented in figure~\ref{fig1}.
\begin{figure}[h]
\centering{
\includegraphics[totalheight=6.cm]{fig1.pdf}
\caption{The complex $\mathcal P_{n}$.} \label{fig1}
}
\end{figure}
It has $2n$ pentagonal 2-cells denoted by $F^{+}_{i}$, $F^{-}_{i}$, where $i=0, \ldots, n-1$. Let us orient edges and mark them by $x_{0}, x_{2}, \ldots, x_{n-1}$ as in the figure, and denote vertices by $N$, $S$, $u_{i}$, $v_{i}$, $w_{i}$ for $i=0, \ldots, n-1$.
Let us define pairing $F_{i}$ of the 2-cells of $\mathcal P_{n}$ in such a way that for each $i=0, \ldots, n-1$ the cells $F^{-}_{i}$ and $F^{+}_{i}$ are identified according to the following order of vertices:
$$
F_{i} : F_i^- = (N u_{i+1} v_{i+1} w_{i+1} u_{i+2}) \longrightarrow F_i^+ = (v_{i} w_{i} u_{i+1} v_{i+1} S).
$$
The pairing of 2-cells induces splitting of edges in classes of equivalent. In particular, for the complex, presented in figure~\ref{fig1}, the class of edges, marked by ${x_3}$, consists of the following edges, equivalent under pairings $F_{i}$ and their inverses:
$$
x_3 \, : \, [N, u_{4}] \stackrel{F_2}{\to} [v_{2}, S] \stackrel{F_1^{-1}}{\to} [w_{2}, u_{3}] \stackrel{F_2^{-1}}{\to} [u_{3}, v_{3}] \stackrel{F_{2}^{-1}}{\to} [v_{3}, w_{3}] \stackrel{F_{3}^{-1}}{\to} [N, u_{4}],
$$
where edges are presented by initial and terminal vertices and labels of arrows indicate gluings.
The following property holds.
\begin{theorem} \cite{Howie} \label{teo3}
Cyclic presentation $G_{n} (x_{0} x_{1}^{2} x_{2} x_{1}^{-1})$ is geometric, i.e., corres\-ponds to a spine of a closed 3-manifold.
\end{theorem}
Spine $\mathcal P_{n}$ has a cyclic symmetry of order $n$ which corresponds to the cyclic presenta\-tion $G_{n} (x_{0} x_{1}^{2} x_{2} x_{1}^{-1})$. This symmetry induces an order $n$ cyclic symmetry of the manifold. We will describe the corresponding quotient space.
\begin{proposition} \label{ass1}
For any $n$ the manifold from theorem~\ref{teo3} is an $n$-fold cyclic branched covering of the lens space~$L(3,1)$.
\end{proposition}
\begin{proof}
Let $\mathcal S(2n,3,2)$ be the closed 3-manifold, constructed from the spine $P_{n}$. To prove the statement we will use the standard way to move from a spine to a Heegaard diagram of the manifold. As the result, we get the Heegaard diagram, presented in figure~\ref{fig2}.
\begin{figure}[h]
\begin{center}
\includegraphics[totalheight=4.cm]{fig2.pdf}
\caption{Heegaard diagram of $\mathcal S(2n,3,2)$.}\label{fig2}
\end{center}
\end{figure}
This diagram admits a rotational symmetry of order $n$ which permutes discs cyclically: $F_{i}^{-} \to F_{i+1}^{-}$ and $F_{i}^{+} \to F_{i+1}^{+}$. Denote the symmetry by $\rho$ and its axe by $\ell$. The quotient space, corresponding to $\rho$, is a 3-orbifold and the image of $\ell$ is a singular set of it. By Singer's move of type IB \cite{Singer} we get the standard Heegaard diagram of the lens space $L(3,1)$, see figure~\ref{zin}. Thus, the underlying space of the 3-orbifold $\mathcal S(2n,3,2) / \rho$ is the lens space $L(3,1)$.
\begin{figure}[h]
\begin{center}
\includegraphics[totalheight=5.cm]{fig3.pdf}
\caption{Simplifying a Heegaard diagram of $\mathcal S(2n,3,2) / \rho$.} \label{zin}
\end{center}
\end{figure}
Therefore, the manifold $\mathcal S(2n,3,2)$ is an $n$-fold cyclic branched covering of the lens space $L(3,1)$.
\end{proof}
\subsection{Cyclic presentations of $S(2n,5,2)$ with $n$ generators}
Now we consider the generalised Sieradski groups $S(2n,5,2)$ with even number of generators. From cyclic presentation with $2n$ generators we move to cyclic presentation with $n$ generators:
\begin{eqnarray*}
S(2n,5,2) & = & G_{2n} (x_{1} x_{3} x_{5} x_{4}^{-1} x_{2}^{-1}) \\
& = & \langle x_{1}, x_{2}, \ldots, x_{2n} \, | \, x_{i+1} x_{i+3} x_{i+5} = x_{i+2} x_{i+4}, \quad i=1, \ldots, 2n \rangle \\
& = & \langle x_{1}, x_{2}, \ldots, x_{2n} \, | \, x_{j+1} x_{j+3} x_{j+5} = x_{j+2} x_{j+4}, \quad \\
& & \qquad \qquad \qquad \quad x_{j+2} x_{j+4} x_{j+6} = x_{j+3} x_{j+5}, \quad j=2,4 \ldots, 2n \rangle \\
& = & \langle x_{1}, x_{2}, \ldots, x_{2n} \, | \, x_{j+5} = (x_{j} x_{j+2} x_{j+4})^{-1} x_{j+2} x_{j+4}, \quad \\
& & \qquad \qquad \qquad \quad x_{j} x_{j+2} x_{j+4} = x_{j+1} x_{j+3}, \quad j=2,4 \ldots, 2n \rangle \\
& = & \langle x_{2}, x_{4}, \ldots, x_{2n} \, | \, x_{j} x_{j+2} x_{j+4} \left( x_{j+2}^{-1} x_{j}^{-1} x_{j-2} x_{j} x_{j+2} \right) \\
& & \qquad \qquad \qquad \cdot \left( x_{j}^{-1} x_{j-2}^{-1} x_{j-4} x_{j-2} x_{j} \right) = 1, \quad j=2, 4 \ldots, 2n \rangle \\
& = & \langle y_{1}, y_{2}, \ldots, y_{n} \, | \, y_{i} y_{i+1} y_{i+2} y_{i+1}^{-1} y_{i}^{-1} y_{i-1} y_{i} y_{i+1} y_{i}^{-1} y_{i-1}^{-1} y_{i-2} y_{i-1} y_{i} =1, \\
& & \qquad \qquad i = 1, \ldots, n \rangle \\
& = & G_{n} (y_{3} y_{4} y_{5} y_{4}^{-1} y_{3}^{-1} y_{2} y_{3} y_{4} y_{3}^{-1} y_{2}^{-1} y_{1} y_{2} y_{3} ) \\
& = & G_{n} (y_{1} y_{2} y_{3} y_{3} y_{4} y_{5} y_{4}^{-1} y_{3}^{-1} y_{2} y_{3} y_{4} y_{3}^{-1} y_{2}^{-1} ).
\end{eqnarray*}
We will show that obtained cyclic presentation is geometric. To formulate the subsequence statement we will renumerate generators similar to theorem~\ref{teo3}, namely, we put $x_i = y_{i+1}$, where $i = 0,1, \ldots, n-1$.
\begin{theorem} \label{teo4}
The cyclic presentation $G_{n} ( x_{0} x_{1} x_{2} x_{2} x_{3} x_{4} x_{3}^{-1} x_{2}^{-1} x_{1} x_{2} x_{3} x_{2}^{-1} x_{1}^{-1})$ is geometric, i.e., it corresponds to a spine of a closed 3-manifold.
\end{theorem}
\begin{proof} To prove the statement we will construct a $2$-complex $Q_{n}$ with $2n$ 2-cells and oriented edges marked by $x_{i}^{\pm1}$. Each 2-cell will be an 13-gon with the property that reading marks along its boundary gives some defining relation of the group. Moreover, for each defining relation there are exactly two 2-cell and their orientations are opposite.
Since the defining relations are quite long, we will demonstrate the constructi\-on for a small example. Let us consider the 2-complex $Q_{4}$, presented in figure~\ref{new}. $Q_{4}$ has eight 2-cells. Looking at the figure we keep in mind that edges on the left side should be identified in pairs with edges on the right side. And, also, that all vertical lines going upstairs meet in one point, as well as all vertical lines going downstairs meet in one point. The complex $Q_{4}$ corresponds to the cyclic presentation for $n=4$. Indeed, in this case we have four generators, say $x_{0}$, $x_{1}$, $x_{2}$ and $x_{3}$, and four defining relations, which we can write in the following form:
$$
\begin{gathered}
x_{0} x_{1} x_{2} x_{2} x_{3} x_{0} x_{3}^{-1} x_{2}^{-1} x_{1} x_{2} x_{3} x_{2}^{-1} x_{1}^{-1} = 1, \cr
x_{1} x_{2} x_{3} x_{3} x_{0} x_{1} x_{0}^{-1} x_{3}^{-1} x_{2} x_{3} x_{0} x_{3}^{-1} x_{2}^{-1} = 1, \cr
x_{2} x_{3} x_{0} x_{0} x_{1} x_{2} x_{1}^{-1} x_{0}^{-1} x_{3} x_{0} x_{1} x_{0}^{-1} x_{3}^{-1} = 1, \cr
x_{3} x_{0} x_{1} x_{1} x_{2} x_{3} x_{2}^{-1} x_{1}^{-1} x_{0} x_{1} x_{2} x_{1}^{-1} x_{0}^{-1} = 1. \cr
\end{gathered}
$$
It is easy to check that all four words in these relations appear in the process of reading marks along boundaries of the 2-cell in figure~\ref{new}. Indeed, we will get the first word by reading along the boundary of the 2-cell $F_{1}$ in the counterclockwise direction, as well as along the boundary of the 2-cell $\overline{F}_{1}$ in the clockwise direction. Analogously, other words correspond to pairs $F_{2}$ and $\overline{F}_{2}$, $F_{3}$ and $\overline{F}_{3}$, $F_{4}$ and $\overline{F}_{4}$, respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[totalheight=6.0cm]{fig4.pdf}
\caption{The complex $\mathcal Q_{4}$.} \label{new}
\end{center}
\end{figure}
Let us identify pairs of the 2-cells of $Q_{4}$ with equal words by following marks on their boundaries. These pairings induce splittings of 1-cells and 0-cells in classes of equivalent. The resulting quotient space is a closed 3-dimensional pseudo-manifold. It is ealy to check directly that its Euler characteristic vanishes. Therefore, complex $Q_{4}$ is a spine of a closed orientable 3-manifold~\cite{Seifert}. The described construction and above arguments work in the same way for arbitrary $n$.
\end{proof}
Let us denote by $\mathcal S(2n,5,2)$ the closed orientable 3-manifolds, corresponding to the spine $Q_{n}$. For small $n$ manifolds $\mathcal S(2n,5,2)$ can be distinguished by using \emph{3-manifold recognizer}~\cite{Recognizer}. For example, the computations for the case $n=4$ give us that $\mathcal S(8,5,2)$ is the Seifert manifold $(S^2,(4,1),(5,2),(5,2),(1,-1) )$.
Analogously to proposition~\ref{ass1}, the following result holds.
\begin{proposition} \label{ass2}
For each $n$ the manifold from theorem~\ref{teo4} is an $n$-fold cyclic covering of the lens space $L(5,1)$.
\end{proposition}
\begin{proof}
Because of the cyclic symmetry, it is enough to consider an example with small parameters. Consider the Heegaard diagram of the manifold $\mathcal S(8,5,2)$ which is obtained from the complex $Q_{4}$, see figure~\ref{fig5}. By the construction, the diagram has a rotational symmetry $\rho$ of order $4$.
\begin{figure}[h]
\begin{center}
\includegraphics[totalheight=4.cm]{fig5.pdf}
\caption{Heegaard diagram for the case $n=4$.} \label{fig5}
\end{center}
\end{figure}
Analogously to the proof of proposition~\ref{ass1}, it is easy to see that the quotient space $\mathcal S(8,5,2) / \rho$ is the 3-orbifold whose underlying space is the lens space $L(5,1)$.
\end{proof} | 20,424 |
Rationale
The Management of Chelsea House is committed to ensuring that the health and safety of all children, staff and visitors in the Centre is protected.
Objectives
- To ensure the 1998 Education (Early Childhood Centres) Regulations and all other relevant Acts are adhered to.
- To provide a safe and healthy working environment for all employees.
- To ensure equipment and materials are being used correctly.
- To ensure policies, procedures and systems, which maintain a healthy environment, are adhered to.
Procedures
Centre Manager and Staff will ensure that safe practices carried out to maintain the safety and good health of all who participate in the Centres day to day operation. The Centre will provide a copy of the Health and Safety Policy for visitors to read and acknowledge by signing.
Injury Prevention Strategies
- The doors at the front entrance, laundry, kitchen, maintenance storage and resource rooms will be kept closed at all times.
- All chemicals and cleaning products will be stored in marked containers and kept out of the reach of children at all times.
- No glass containers will be used by the children.
- Children will remain seated while eating and be supervised by a staff member.
- Children may not climb gates, sit on tables or jump off furniture.
- Educators will be conscious of potential hazards eg. water spills, toys on floors etc.
- Where practical educators will wear gloves when dealing with children’s body fluids.
- Hot drinks are to be consumed in staff only areas.
Accidents
- All qualified educators will hold current First Aid Certificates.
- All accidents will be recorded in the Accident Book in the child’s classroom.
- First Aid will be administered immediately. If necessary an Ambulance will be called.
- Where necessary the parents will be contacted and the child taken to a doctor.
- A staff member will stay with the child until the parent (or designated Person) arrives.
Staff Injury / Accident
- Administer first Aid immediately.
- If necessary call an ambulance and contact next of kin.
- Ensure staff member is able to get home safely
- Complete occupational Safety and Health form.
Illness and infectious disease procedures
It is the policy of Chelsea House that Management and all persons will endeavour to prevent the spread of infectious disease in accordance with the Early Childhood regulations (1998) and Ministry of Health information.
- Children and adults who are sick will not be at the Centre.
- Conditions including vomiting, diarrhoea, high temperatures, conjunctivitis, impetigo and other contagious conditions require exclusion from the Centre. (Reg 28).
Administering Medicines procedure ( refer to policy)
- Staff at Chelsea House will take all reasonable steps to ensure that correct medicine and medication is given to children.
- On arrival at the Centre, all medication must be handed to a staff member. The amount, date, time to be given and type of medicine must be written into the medication book.
- Only medication authorised by parents will be given to children. The medication should have a chemist label attached to the bottle with the child’s name, date, and doctor’s name, amount of medicine to be given and the length of time the medicine is to be given. No medicine will be given if it has been prescribed to another member of the family or is outdated.
- Permanent staff only, are to administer medicines to children and must sign medicine book when medicine is administered. Another staff member will witness and sign.
First Aid Cabinet
It will be the responsibility of the staff to inform the Centre Manager of any requirements for the First Aid Kits. It is then the responsibility of the Centre manager to ensure the kits are refurbished on a regular basis.
- The First Aid Kits will be kept in the bathroom, next to the changing area.
Sun Safety Procedure (refer to policy)
- Staff will follow the recommendations of the Cancer Society to protect children from harmful UV rays of the sun.
- Parents are to provide a suitable sunhat for their child/ren.
- The centre will provide sunscreen
- Staff will apply sunscreen prior to the children going outside and then at hourly – intervals.
- Permission must be given by parents prior to sunscreen being applied. Parents can provide their child’s own cream/lotion.
- All children will be required to wear a hat when they are outside.
- A cover over the sandpit will be maintained to ensure there is adequate shade.
- Staff will act as role models by wearing hats.
Supervision Plan
- Management will ensure adequate staff are present at all times to meet or better our licensing requirements, and to adequately supervise all children in attendance.
- Staff will be rostered to ensure the main playrooms, outdoor playground and toileting areas have staff rostered to adequately supervise children in these areas as required to meet the needs of the children and promote their wellbeing, health and safety.
- The Educators, adults visiting or working in this service, are well supervised and visible in the activities they perform with children. This includes having open door policies for private spaces where intimate care taking of children is required.
- Breaks for educators will be scheduled to enable proper supervision of children.
- Educators involved in supervision must not leave their children unsupervised under any circumstances unless relieved by another educator.
Hand washing Procedure
- Instruction will be given to children regarding washing of hands. Liquid soap and paper towels are available for children and staff.
- Children will wash their hands after toileting, before eating and after handling pets.
- Staff must wash their hands after any interaction with children that involves contact with body fluids. Staff will wash their hands before they handle food or drink, and after bathroom duties.
- All hand washing will take place in the bathroom area.
Cleaning Procedure
- A cleaner is employed 3 hours per day and the tasks are clearly set out.
- All cleaning fluids will be kept out of the reach of children at all times.
- Staff will keep the centre as clean as is practical during the day, with particular attention to the bathroom area. (Refer to the centre cleaning and maintenance check).
- Gloves are to be worn when cleaning toilets.
Animal Care Procedure
- Staff will ensure that when animals are kept at the centre, they will have a healthy living environment, which will include weekends and holidays.
- The cages will be cleaned regularly.
- Animal food will be stored away from other food and be clearly marked.
- Children must wash their hands after handling animals.
- All sick animals will be kept away from children and adequate care and Veterinary attention given to them.
Emergency Procedures
- Fire and Earthquake drills will be held at least once per month.
- The Manager or senior staff in the case of an earthquake drill will record the date and time.
- The Manager will be the Chief Fire Warden.
- The procedures will be on the wall and will be followed by all staff.
Maintenance Procedure
- The Centre Manager and Senior Educators will ensure all equipment is maintained in a safe condition.
- Assigned educators will complete a safety check in the outdoors every morning and complete a written record in the maintenance book or communication book, of any repairs to be carried out. (Refer to Osh checklist).
- It will be the role of the Centre Manager to ensure that the repairs are carried out.
Occupational Safety and Health
- The Licensee will ensure that the centre complies with the requirements of the Occupational Safety and Health Act. A daily check is carried out to ensure all areas are safe. (Refer Osh checklist).
- Staff training on occupational safety and health will be provided.
- Staff will identify hazards in the Centre.
- Strategies will be in place to:
Eliminate the hazard wherever possible.
Minimise the risks of the hazard.
Isolate the hazard to ensure no harm comes to a person as a Result of the hazard.
Review November 2013 Next Review: November 2014 | 117,437 |
Latest insights from our experts
The Government consults on repealing section 21 and abolishing Assured Shorthold Tenancies (AST)
Earlier this year the Government said that it intended to consult on its proposal to end a landlord’s ability to terminate an Assured Shorthold Tenancy (AST) in England by serving a notice under section 21 of the Housing Act 1988. This section allowed a landlord to end an AST on two months’ notice without giving a reason and without the tenant being at fault.
The Government published its consultation on 21 July 2019 seeking views on its proposals to end the Assured Shorthold Tenancy regime altogether.
The Government is encouraging all private landlords, which will include those in the parks sector, to take part in the consultation which is open until 12 October 2019.
Further details about the proposals and the consultation can be found here.
If you have any queries about the proposals. | 31,023 |
Weakness to Fire
Weakness to Fire is an effect that will render enemies weak to elemental damage depending on the strength of the potion. Ingredients that have the Weakness to Fire effect include:
What Links HereIngredients | Bleeding Crown | Frost Salts | Ice Wraith Teeth | Moon Sugar | Alchemy Ingredients Effects | Juniper Berries | Powdered Mammoth Tusk | | 5,489 |
TITLE: How to prove the relationship between Stern's diatomic series and Lucas sequence $U_n(x,1)$ over the field GF(2)?
QUESTION [7 upvotes]: I found the bit count of Lucas sequence $U_n(x,1)$ over the field GF(2) is Stern's diatomic series, I want to know the reason?
https://oeis.org/A002487 : Stern's diatomic series
https://oeis.org/A168081 : Lucas sequence $U_n(x,1)$ over the field GF(2)
REPLY [7 votes]: I always work mod 2. Let $f(n)$ be the bit count (number of nonzero
coefficients) of $U_n(x,1)$. $U_n(x,1)$ satisfies
$$ U_{2n}(x,1) = xU_n(x,1)^2 $$
$$ U_{2n+1}(x,1) = (U_n(x,1)+U_{n+1}(x,1))^2. $$
From the first equation $f(2n)=f(n)$, and only odd powers of $x$
can have nonzero coefficients. From the second equation, only even
powers of $U_{2n+1}$ can have nonzero coefficients. Hence no
cancellation occurs when we add $U_n(x,1)$ and $U_{n+1}(x,1)$, so
$f(2n+1)=f(n)+f(2n+1)$. Thus $f(n)$ satisfies the same recurrence and
same initial conditions $f(0)=0$ and $f(1)=1$ as Stern's diatomic sequence.
Addendum. The two formulas above can be proved by induction on
$n$. Namely, writing just $U_n$ for $U_n(x,1)$,
$$ U_{2n}=xU_{2n-1}+U_{2n-2} = x(U_{n-1}+U_n)^2+xU_{n-1}^2
=xU_n^2 $$
and
$$ U_{2n+1} = xU_{2n}+U_{2n-1} = x^2U_n^2+(U_{n-1}+U_n)^2
=U_n^2+(xU_n+U_{n-1})^2=U_n^2+U_{n+1}^2. $$ | 74,426 |
Home > About Us > Faculty & Staff Directory > Hen Energy Technology Innovation Policy project,. Additionally, he has worked with private and public organizations, including the InterAmerican Development Bank, the International Finance Corporation, the State of Sao Paulo, the U.S. Departments of Energy and Interior, the National Research Council, the Intercontinental Energy Corporation, General Electric, the Massachusetts Department of Conservation and Recreation, and the U.S. EPA. His recent research interests focus on energy and transportation, China's energy policy, and public infrastructure projects in developing countries. Mr. Lee is the author of recent papers on both the U.S. and China, the economic viability of electric vehicles, as well as case studies on tariffs to promote solar energy, Iceland's green energy agenda, Liberia's electricity sector and health reform in Lesotho.
Henry Le. | 193,558 |
Liv Tyler In New York City
September 2nd, 2008: The paps caught up with Liv Tyler in New York City Tuesday afternoon. Her response to the attention reminded me of the very different atmosphere of New York, as opposed to Los Angeles. "Okay guys, this is really embarrassing," she pleaded before the paps quit shooting. | 85,366 |
Why Set & Save?
Just click "Add to Set & Save" when placing items in your cart. softgels daily with food or as directed by a healthcare professional.
Store in a cool, dry place. | 235,470 |
PANDAProject reference: 616419
Funded under :
Phylogenetic ANalysis of Diversification Across the tree of life
Project details
Total cost:EUR 1 857 856
EU contribution:EUR 1 857 856
Coordinated in:France
Topic(s):ERC-CG-2013-LS8 - ERC Consolidator Grant - Evolutionary, Population and Environmental Biology
Call for proposal:ERC-2013-CoGSee other projects for this call
Funding scheme:ERC-CG - ERC Consolidator Grants
Objective
"Explaining the tremendous disparity in the number of species among different taxonomic groups and geographic regions is one of the greatest challenges in biodiversity research. This project aims to significantly advance our fundamental understanding of the ecological and evolutionary processes that explain how species richness is distributed on earth and in the tree of life. The first part of the project consists in significantly improving phylogenetic approaches to diversification. This implies developing stochastic lineage-based birth-death models that account for species interactions, stochastic lineage-based birth-death models integrating phylogenetic and fossil information, and stochastic individual-based birth-death-speciation models. The second part of the project consists in implementing these models, as well as other recently developed models, in efficient and user-friendly software packages. Finally, the third part of the project consists in applying these phylogenetic tools to large datasets spanning the tree of life in order to understand how and why speciation and extinction rates vary over evolutionary time, geographical space, and species groups. We will use our new phylogenetic comparative methods to uncover diversity-through-time curves for a wide variety of taxonomic groups, evaluate the relative role of the biotic and the abiotic environment in driving diversification, shed new light on the latitudinal diversity gradient, and consider microbial diversity with a macroevolutionary perspective. The proposed research will provide the scientific community with novel, user-friendly modeling tools for understanding biodiversity, and will yield unprecedented insights into the dynamics and determinants of biodiversity." | 92,567 |
Thank you for your interest in supporting PRAY LOVE HEAL. This non-profit Christian ministry is funded exclusively by the generous gifts of supporters like you.
When you give to PRAY LOVE HEAL you are:
- strengthening a global community of praying people
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You can support PRAY LOVE HEAL with one-time gifts of any amount or easily set up automated monthly donations, which are especially helpful.
* PRAY LOVE HEAL is a missions project of The Brushfires Foundation, a 501(c)3 non-profit ministry. All donations are tax-deductible as allowed by law.
I WANT TO BLESS PEOPLE IN NEED
Your gifts contribute to this mission in the following ways:
- A $25 gift allows us to distribute incoming prayer requests to our prayer teams for a day
- A $50 gift pays for a month of email and hosting costs
- A $100 gift covers half of our monthly design costs
- A $750 gift pays our prayer coordinator for two weeks
- A $2,000 gift supports an entire month of ministry (This includes funding the prayer coordinator’s stipend for that month, all graphic design costs, email and hosting services, and administrative costs.)
Prayer Leaders and Site Sponsors
We have a special program for individuals and organizations that want to become praying leaders. Joining at this level opens up a library of resources to guide your own local or church prayer outreach. This is also a greater way to strengthen our outreach to the broken. | 228,813 |
TITLE: Differentiability of $|x|^p$?
QUESTION [0 upvotes]: Let $p > 0$, and let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined piecewise by $f(x)= |x|^p$ if $x \in \mathbb{Q}$ and $f(x)=0$ if $x \in \mathbb{R} \setminus \mathbb{Q}$. For what values of $p$ is $f$ differentiable?
Do I have to show the left/right hand limits exist and agree using the definition of absolute value (i.e. $|x|=x$ if $x \geq 0$ and $|x|=-x$ if $x < 0$)? But how do I proceed? By fixing $p$ and then solving for it somehow?
Also, since both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense, we can pick a sequence $(x_n) $of rationals and $(y_n)$ of irrationals that both converge to 0. Then, by the sequential definition of continuity, we must have $(f(x_n)) \rightarrow 0$ and $(f(y_n)) \rightarrow 0$ (since before being differentiable, a function has to be continuous).
That means that $f$ will be differentiable only for $p=0$?
Thanks!
REPLY [2 votes]: As you have observed the function is not continous at $x$ if $x \neq 0$ so it is not differentiable either. For differentiability at $x=0$ consider $\lim_{x \to 0} \frac {|x|^{p}} x$. If $p>1$ this limit is $0$ because $|\frac {|x|^{p}} x|=|x|^{p-1} \to 0$. For $p<1$ it is clear that $|\frac {|x|^{p}} x|=|x|^{p-1} \to \infty$. For $p=1$ consider right hand and left hand limits to see that the function is not differentiable | 182,903 |
CASES; For Elderly, A Little Fall, A Big Worry
By ANNE MARIE VALINOTI, M.D
Published: July 29, 2008
''She's 88 and tripped and fell yesterday,'' the emergency room doctor told me when I answered the page. ''She was on the floor about 15 hours when her family found her.''
''What did she break?'' I asked.
''Nothing,'' he said, as I recall. ''But she's too weak to stand, and we can't send her home since she lives alone.''
The ''little old lady found on the floor'' is a staple of hospital admissions, and Martha, a patient in our group practice, was hardly the first I had cared for. The causes of falls in the elderly are myriad, as are the consequences. But frankly, they're at the bottom of my list of fascinating new cases.
Often, my role as an internist consists mainly of arranging for physical therapy in the hospital and social work intervention if needed to ensure a safe discharge plan. It is much less exciting than, say, managing bacterial sepsis or a pulmonary embolism. Resigned, I drove over to the hospital to meet Martha.
In the emergency room, Martha's nurse handed me the chart. ''Her family is here,'' he told me. ''And I just got her lab results -- her CPK is 32,000.''
This was interesting. Creatine phosphokinase, or CPK, is an enzyme found in muscle cells in the body. If a muscle is injured or inflamed, the level will rise. Small elevations are not usually worrisome, but large ones signify extensive muscle injury.
That injury, called rhabdomyolysis, was first described in survivors of the London blitz in World War II who, though rescued after a limb had been crushed or pinned under fallen masonry, later died of kidney failure. A British physician, Eric Bywaters, determined that the crushed muscle cells leaked the protein myoglobin into the blood, where it was then able to poison the kidneys. He published his findings in 1941.
The level of CPK in Martha's blood was through the roof. Though she had not been pulled from rubble, by quietly lying on her bedroom floor for so many hours, her leg pinned beneath her, she risked complications as severe as those of any victim of a building collapse.
I went to examine Martha and found her surrounded by her concerned family. She was an alert, elderly woman who, according to her daughter, lived independently without difficulty. Her family lived nearby, saw her frequently and called daily. In fact, they had called her twice while she lay on the floor, unable to get up. Twice, Martha had answered the phone and said nothing about her predicament. ''Typical,'' her daughter told me. ''She didn't want to worry us.''
There was plenty to worry about now. Large intravenous infusions of sodium bicarbonate will sometimes prevent kidney damage in rhabdomyolysis, but this is easier said than done in an 88-year-old. The treatment could overwhelm her heart, resulting in a buildup of fluid in the lungs, which itself can be fatal. Hospital-acquired infections, delirium and blood clots were other lurking dangers. The potential for complications was daunting, as it so often is with elderly patients.
This was not going to be easy to communicate to her family. Martha looked well -- tired after her ordeal but otherwise none the worse for wear. The tea-colored urine in the catheter bag was the only ominous sign of the assault on her kidneys. Her son and daughter had been relieved that there were no fractures. Before the lab tests came back, this was going to be a simple hospital admission for physical therapy.
The family listened as I explained the diagnosis, the treatment and the risks involved. Martha would be admitted to an intensive care unit where she could be monitored, especially for signs of heart failure.
Fortunately, Martha's renal function remained stable after her hospital admission. This was in stark contrast to my last patient with rhabdomyolysis, who spent weeks in the hospital enduring uncomfortable dialysis until his failing kidneys recovered. Last year a patient in our practice died from complications of rhabdomyolysis.
But Martha was able to leave the hospital a week later, having avoided the complications of her injury and its treatment. Although the care she received was critical to her good outcome, I can't help but think that pure luck was at play, too.
According to the Centers for Disease Control and Prevention, falls are the leading cause of death from injury among people over 65. An older person may seem independent, but friends and relatives should routinely survey homes and apartments for slippery surfaces, clutter or other fall-inducing conditions. Something as seemingly innocuous as a loose rug can lead to a dire medical outcome.
PHOTO (PHOTO ILLUSTRATION BY VIVIENNE FLESHER)(pg. F7) | 220,207 |
\begin{document}
\title{Renormalization of singular elliptic stochastic PDEs using flow equation}
\author{Pawe{\l} Duch
\\
Faculty of Mathematics and Computer Science \\
Adam Mickiewicz University in Pozna\'n\\
ul. Uniwersytetu Pozna\'nskiego 4, 61-614 Pozna\'n, Poland\\
[email protected]
}
\date{\today}
\maketitle
\begin{abstract}
We develop a solution theory for singular elliptic stochastic PDEs with fractional Laplacian, additive white noise and cubic non-linearity. The method covers the whole sub-critical regime. It is based on the Wilsonian renormalization group theory and the Polchinski flow equation.
\end{abstract}
\tableofcontents
\section{Introduction}
A general technique that allows to renormalize and prove universality of parabolic singular SPDEs with fractional Laplacian, additive noise and polynomial non-linearity was developed in~\cite{duch}. The goal of this paper is to give a different application of this technique. We present a self-contained construction of solutions of the following non-local singular elliptic SPDEs
\begin{equation}
\big(1+(-\Delta)^{\sigma/2}\big)\varPhi(x) = \xi(x)+\lambda\varPhi(x)^3 - \infty\varPhi(x),
\quad
x\in\bR^\rdim,
\end{equation}
where $\rdim\in\{1,\ldots,6\}$, $\sigma\in(\rdim/3,\rdim/2]$, $(-\Delta)^{\sigma/2}$ is the fractional Laplacian, $\lambda\in\bR$ is sufficiently small and $\xi$ is the periodization of the white noise on $\bR^\rdim$ with period $2\pi$. Recall that the regularity of the noise $\xi$ is slightly worse than $-\rdim/2$ and the expected regularity of the solution is slightly worse than $\sigma-\rdim/2$. For $\sigma>\rdim/2$ the above equation is not singular and can be solved using classical PDE theory. For $\sigma\leq\rdim/2$ the solution is not a function but only a distribution. As a result, the cubic term is ill-defined and has to be renormalized by subtracting appropriate mass counterterm (for $\rdim>6$ other counterterms are needed even if one takes into account all the symmetries of the equation). The renormalization problem is tractable only if the equation is sub-critical (super-renormalizable). This is the case if the expected regularity of the renormalized non-linearity is better than the regularity of $\xi$, i.e. $3(\sigma-\rdim/2)>-\rdim/2$. Let us remark that for $\rdim=5$ and $\sigma=2$ the above equation is sometimes called the elliptic quantization equation of the $\varPhi^4_3$ model (provided $\xi$ is replaced by the white noise on $\bR^\rdim$).
Let $G\in L^1(\bR^\rdim)$ be the fundamental solution for the pseudo-differential operator $\fQ:=1+(-\Delta)^{\sigma/2}$ and let $G_\uv$ be the smooth approximation of $G$ with a spatial UV cutoff of order $[\uv]:=\uv^{1/\sigma}$ introduced in Def.~\ref{dfn:kernel_G}. We rewrite the above singular SPDE in the following regularized mild form
\begin{equation}\label{eq:intro_mild}
\varPhi = G_\uv \ast F_\uv[\varPhi],\qquad \uv\in(0,1/2].
\end{equation}
The functional $F_\uv[\varphi]$, called the force, is defined by
\begin{equation}\label{eq:force}
F_\uv[\varphi](x):= \xi(x)
+\lambda\, \varphi^3(x)
+\sum_{i=1}^{i_\sharp}\lambda^i\, c_{\uv}^{[i]} \varphi(x),
\end{equation}
where $i_\sharp=\floor{\sigma/(3\sigma-\rdim)}$ and the parameters $c_{\uv}^{[i]}\in\bR$ depending on the UV cutoff $\uv$ are called the counterterms. Let us state our main result.
\begin{thm}
There exists a choice of counterterms and a random variable $\lambda_0$ such that $\bE(\lambda_0^{-n})<\infty$ for every $n\in\bN_+$ and for every random variable \mbox{$\lambda\in[-\lambda_0,\lambda_0]$} and $\uv\in(0,1/2]$ Eq.~\eqref{eq:intro_mild} has a periodic solution \mbox{$\varPhi_\uv\in C^\infty(\bR^\rdim)$} and for any $\beta<\sigma-\rdim/2$ the limit $\lim_{\uv\searrow0}\varPhi_\uv$ exists almost surely in the Besov space $\sC^\beta(\bR^\rdim)$.
\end{thm}
\begin{proof}
We first establish bounds for cumulants of the enhanced noise introduced in Def.~\ref{dfn:modified_coefficients}. The bounds are stated in Theorem~\ref{thm:cumulants} and hold true for an appropriate choice of the counterterms. Using these bounds and a Kolmogorov type argument we deduce bounds for the enhanced noise stated in Theorem~\ref{thm:probabilistic_bounds}. This together with the deterministic result of Theorem~\ref{thm:solution} implies the statement.
\end{proof}
\begin{rem}
There exists a general technique developed in~\cite{hairer2014structures,chandra2016bphz,bruned2019algebraic,bruned2021renormalising} based on the theory of regularity structures that allows to systematically renormalize virtually all sub-critical singular SPDEs with local differential operators. However, due to the slow decay of the kernel $G$ at infinity, this technique does not cover Eq.~\eqref{eq:intro_mild}.
\end{rem}
\section{Effective force}\label{sec:intro_flow}
The basic object of the flow equation approach is the effective force functional
\begin{equation}
\sS(\bM)\ni\varphi\mapsto F_{\uv,\sIR}[\varphi]\in \sS'(\bM)
\end{equation}
depending on the UV cutoff $\uv\in(0,1/2]$ and the flow parameter $\sIR\in[0,1]$, where $\sS(\bM)$ and $\sS'(\bM)$ denote the space of Schwartz functions and Schwartz distributions on $\bM=\bR^\rdim$, respectively.
\begin{dfn}\label{dfn:kernel_G}
Fix $\chi\in C^\infty(\bR)$ such that $\chi(r)=0$ for $|r|\leq1$ and $\chi(r)=1$ for $|r|>2$ and let $\chi_\sIR(r) := \chi(r(1-\sIR)/\sIR)$ for $\sIR\in(0,1]$. Let $G\in L^1(\bM)$ be the fundamental solution for the pseudo-differential operator $\fQ:=1+(-\Delta)^{\sigma/2}$. For $\uv\in(0,1/2]$ and $\sIR\in(0,1]$ the smooth kernels $G_\uv,G_{\uv,\sIR}\in L^1(\bM)$ are defined by
\begin{equation}
G_{\uv}(x) :=
\chi_\uv(|x|^\sigma)\,
G(x),
\qquad G_{\uv,\sIR}(x) :=
\chi_\uv(|x|^\sigma)\,
\chi_\sIR(|x|^\sigma)\,
G(x).
\end{equation}
\end{dfn}
\begin{rem}
Since $\chi_\uv(r)\chi_\sIR(r)=\chi_\uv(r)$ for $\sIR\leq\uv/2$ it holds $G_{\uv,\sIR}=G_\uv$ for $\sIR\leq\uv/2$ and $G_{\uv,\sIR}=G_\sIR$ for $\sIR\geq 2\uv$. Moreover, $\lim_{\uv\searrow0}G_\uv=G$ in $L^1(\bM)$ and $G_{\uv,1}=0$ for all $\uv\in(0,1/2]$.
\end{rem}
By definition, the effective force satisfies the following flow equation
\begin{equation}\label{eq:intro_flow_eq}
\langle\partial_\sIR F_{\uv,\sIR}[\varphi],\psi\rangle
=
-\langle \rD F_{\uv,\sIR}[\varphi,\partial_\sIR G_{\uv,\sIR}\ast F_{\uv,\sIR}[\varphi]],\psi\rangle
\end{equation}
with the boundary condition $F_{\uv,0}[\varphi]=F_\uv[\varphi]$, where $\varphi,\psi\in \sS(\bM)$ and the force $F_\uv[\varphi]$ is defined by Eq.~\eqref{eq:force}. The paring between a distribution $V$ and a test function $\psi$ is denoted by $\langle V,\psi\rangle$ and $\langle\rD V[\varphi,\zeta],\psi\rangle$ is the derivative of the functional $\sS(\bM)\ni\varphi\mapsto \langle V[\varphi],\psi\rangle\in\bR$ in the direction $\zeta\in \sS(\bM)$. In contrast to the force $F_\uv[\varphi]$ the effective force $F_{\uv,\sIR}[\varphi]$ is generically a non-local functional of $\varphi$.
We claim that $\varPhi_\uv := G_\uv\ast F_{\uv,1}[0]$ is a solution of Eq.~\eqref{eq:intro_stationary_relation}. The above statement is a consequence of the equalities $G_{\uv,\sIR}=G_\uv$ and $F_{\uv,\sIR}[\varphi]=F_\uv[\varphi]$ that hold for all $\sIR\in[0,\uv/2]$ and the identity
\begin{equation}\label{eq:intro_stationary_relation}
F_{\uv,1}[0] = F_{\uv,\sIR}[G_{\uv,\sIR}\ast F_{\uv,1}[0]]
\end{equation}
that holds for all $\sIR\in[0,1]$. In order to prove the last identity we use the flow equation to show that the difference between the LHS and RHS of Eq.~\eqref{eq:intro_stationary_relation}, denoted by $g_{\uv,\sIR}$, satisfies the following linear ODE
\begin{equation}
\partial_\sIR g_{\uv,\sIR}
=
-\rD F_{\uv,\sIR}[G_{\uv,\sIR}\ast F_{\uv,1}[0],\partial_\sIR G_{\uv,\sIR}\ast g_{\uv,\sIR}]
\end{equation}
with the boundary condition $g_{\uv,1}=0$. This implies that $g_{\uv,\sIR}=0$ for all $\sIR\in[0,1]$.
\begin{rem}
The effective force plays also a central role in the approach to singular SPDEs proposed earlier by Kupiainen~\cite{kupiainen2016rg,kupiainen2017kpz} and applied to the dynamical $\varPhi^4_3$ model and the KPZ equation. The method developed by Kupiainen is based on the Wilsonian discrete renormalization group theory~\cite{wilson1971}. In this approach one uses the fact that for $\sIR\geq\uIR$ the effective force satisfies the following equation
\begin{equation}
F_{\uv,\sIR}[\varphi] = F_{\uv,\uIR}[(G_{\uv,\uIR}-G_{\uv,\sIR})\ast F_{\uv,\sIR}[\varphi]+\varphi].
\end{equation}
Given $F_{\uv,\uIR}[\varphi]$ the above equation is viewed as an equation for $F_{\uv,\sIR}[\varphi]$ that can be solved using the Banach fixed point theorem. One defines recursively the effective force $F_{\uv,\sIR}[\varphi]$ for $\uv=2L^{-N}$ and $\sIR\in\{L^{-N},\ldots,1\}$, where $L>1$, starting with $F_{2L^{-N}}[\varphi]=F_{2L^{-N},L^{-N}}[\varphi]$ and finishing with $F_{2L^{-N},1}[\varphi]$. In order to prove uniform bounds for $F_{2L^{-N},1}[\varphi]$ one has to appropriately adjust the counterterms which are the coefficients of the force $F_{2L^{-N}}[\varphi]$. The fine-tuning problem becomes exceedingly difficult for equations close to criticality. The flow equation provides a different, more efficient and simpler method of constructing the effective force that allows to treat equations arbitrarily close to criticality.
\end{rem}
\section{Construction of effective force coefficients}
The starting point of the construction of the effective force is the formal ansatz
\begin{equation}\label{eq:intro_ansatz}
\langle F_{\uv,\sIR}[\varphi],\psi\rangle
:=\sum_{i=0}^\infty \sum_{m=0}^\infty \lambda^i\,\langle F^{i,m}_{\uv,\sIR},\psi\otimes\varphi^{\otimes m}\rangle,
\end{equation}
where $\lambda$ is the prefactor of the cubic non-linearity in the original equation. The distributions $F^{i,m}_{\uv,\sIR}\in\sS'(\bM^{1+m})$, $i,m\in\bN_0$, are called the effective force coefficients. By definition the expression $\langle F^{i,m}_{\uv,\sIR},\psi\otimes\varphi_1\otimes\ldots\otimes\varphi_m\rangle$ is invariant under permutations of the test functions $\varphi_1,\ldots,\varphi_m\in\ \sS(\bM)$. The coefficients $F^{i,m}_\uv$ of the force $F_\uv[\varphi]$ are defined by an equality analogous to Eq.~\eqref{eq:intro_ansatz}.
\begin{rem}
Let us list the non-vanishing force coefficients $F^{i,m}_{\uv}$:
\begin{gather}
F^{0,0}_\uv(x)=\xi(x),\qquad
F^{1,3}_\uv(x;x_1,x_2,x_3)=\delta_\bM(x-x_1)\delta_\bM(x-x_2)\delta_\bM(x-x_3),
\\
F^{i,1}_\uv(x;x_1)=c_{\uv}^{[i]} \delta_\bM(x-x_1),
\quad i\in\{0,\ldots,i_\sharp\}.
\end{gather}
\end{rem}
The flow equation~\eqref{eq:intro_flow_eq} for the effective force $F_{\uv,\sIR}[\varphi]$ formally implies that the effective force coefficients $F^{i,m}_{\uv,\sIR}$ satisfy the following flow equation
\begin{multline}\label{eq:intro_flow_eq_i_m}
\langle \partial_\sIR^{\phantom{i}} F^{i,m}_{\uv,\sIR}\,,
\,\psi\otimes\varphi^{\otimes m}\rangle
\\
=
-\sum_{j=0}^i\sum_{k=0}^m
\,(1+k)
\,\big\langle F^{j,1+k}_{\uv,\sIR}\otimes F^{i-j,m-k}_{\uv,\sIR},
\psi\otimes \varphi^{\otimes k}
\otimes \fV\partial_\sIR G_{\uv,\sIR}
\otimes \varphi^{\otimes(m-k)}
\big\rangle
\end{multline}
with the boundary condition $F^{i,m}_{\uv,0}=F^{i,m}_{\uv}$, where $\fV\partial_\sIR G_{\uv,\sIR}\in C^\infty(\bM\times\bM)$ is defined by $\fV\partial_\sIR G_{\uv,\sIR}(x,y):=\partial_\sIR G_{\uv,\sIR}(x-y)$.
The basic idea behind the flow equation approach is a recursive construction of the effective force coefficients $F^{i,m}_{\uv,\sIR}$:
\begin{enumerate}
\item[(0)] We set $F^{0,0}_{\uv,\sIR}=\xi$ and $F^{i,m}_{\uv,\sIR}= 0$ if $m>3i$.
\item[(I)] Assuming that all $F^{i,m}_{\uv,\sIR}$ with $i<i_\circ$, or $i=i_\circ$ and $m>m_\circ$ were constructed we define $\partial_\sIR F^{i,m}_{\uv,\sIR}$ with $i=i_\circ$ and $m=m_\circ$ with the use of Eq.~\eqref{eq:intro_flow_eq_i_m}.
\item[(II)] Subsequently, $F^{i,m}_{\uv,\sIR}$ is defined by $F^{i,m}_{\uv,\sIR} = F^{i,m}_{\uv} + \int_0^\sIR \partial_\uIR F^{i,m}_{\uv,\uIR}\,\rd\uIR$.
\end{enumerate}
Using this procedure we construct all the effective force coefficients $F^{i,m}_{\uv,\sIR}$ for arbitrary $\uv\in(0,1/2]$, $\sIR\in[0,1]$.
\begin{rem}\label{rem:explicit_coefficients}
One easily shows that the effective force coefficients $F^{i,m}_{\uv,\sIR}$ actually vanish if $i=0$ and $m>0$, or $i>0$ and $m>2(i-1)+3$. The only non-zero coefficients $F^{i,m}_{\uv,\sIR}$ which are independent of the value of the flow parameter $\sIR$ are $F^{0,0}_{\uv,\sIR}(x)=\xi(x)$ and $F^{1,3}_{\uv,\sIR}(x;x_1,x_2,x_3)=F^{1,3}_{\uv}(x;x_1,x_2,x_3)$. These coefficients happen to be independent of the UV cutoff $\uv$. Let us give some further examples:
\begin{equation}
\begin{gathered}
F^{1,2}_{\uv,\sIR}(x;x_1,x_2)=3\varPsi_{\uv,\sIR}(x)\, \delta_\bM(x-x_1)\delta_\bM(x-x_2),
\\
F^{1,1}_{\uv,\sIR}(x;x_1)=(3\varPsi_{\uv,\sIR}^2(x)+c_{\uv}^{[1]})\, \delta_\bM(x-x_1),
\qquad
F^{1,0}_{\uv,\sIR}(x)=\varPsi_{\uv,\sIR}^3(x)+c_{\uv}^{[1]}\,\varPsi_{\uv,\sIR}(x),
\end{gathered}
\end{equation}
where $\varPsi_{\uv,\sIR}:=G_{\uv\shortparallel\sIR}\ast \xi$ and $G_{\uv\shortparallel\sIR}:=G_\uv-G_{\uv,\sIR}$ is the so-called fluctuation propagator. The coefficient $F^{2,5}_{\uv,\sIR}(x;x_1,\ldots,x_5)$ is obtained from the distribution
$\delta_\bM(x-x_1)\delta_\bM(x-x_2)\,G_{\uv\shortparallel\sIR}(x-x_3)\, \delta_\bM(x_3-x_4)\delta_\bM(x_3-x_5)$ by symmetrization in variables $x_1,\ldots,x_5$. We also have
\begin{multline}
F^{2,1}_{\uv,\sIR}(x;x_1) =
(3\varPsi_{\uv,\sIR}^2(x)+c_{\uv}^{[1]})\,
G_{\uv\shortparallel\sIR}(x-x_1)\,
(3\varPsi_{\uv,\sIR}^2(x_1)+c_{\uv}^{[1]})
\\
+\left(\int_{\bM}6\varPsi_{\uv,\sIR}(x)\,G_{\uv\shortparallel\sIR}(x-x_2)\,
(\varPsi_{\uv,\sIR}^3(x_2)+c_{\uv}^{[1]}\varPsi_{\uv,\sIR}(x_2))\,\rd x_2
+ c_{\uv}^{[2]}\right)\delta_\bM(x-x_1),
\end{multline}
\begin{equation}
F^{2,0}_{\uv,\sIR}(x) =
(3\varPsi_{\uv,\sIR}^2(x)+c_{\uv}^{[1]})\,
G_{\uv\shortparallel\sIR}(x-x_1)\,
(\varPsi_{\uv,\sIR}^3(x_1)+c_{\uv}^{[1]}\varPsi_{\uv,\sIR}(x_1))
+ c_{\uv}^{[2]}\, \varPsi_{\uv,\sIR}(x).
\end{equation}
For the sake of brevity, we did not give expressions for the coefficients $F^{2,4}_{\uv,\sIR}$, $F^{2,3}_{\uv,\sIR}$, $F^{2,2}_{\uv,\sIR}$, which should be constructed after $F^{2,5}_{\uv,\sIR}$ and before $F^{2,1}_{\uv,\sIR}$.
\end{rem}
\begin{rem}
The effective force is an analog of the effective potential in QFT. A recursive construction of the effective potential coefficients based on the flow equation is the backbone of a very simple proof of perturbative renormalizability of QFT models proposed by Polchinski~\cite{polchinski1984} (see~\cite{muller2003} for a review).
\end{rem}
The solution of Eq.~\eqref{eq:intro_mild} is formally given by the following sum
\begin{equation}\label{eq:solution_series}
\varPhi_\uv = G_\uv\ast F_{\uv,1}[0] := G_\uv\ast \sum_{i=0}^\infty \lambda^i F^{i,0}_{\uv,1}.
\end{equation}
Assuming that $\lambda$ is sufficiently small in Sec.~\ref{sec:solution} we prove that the above series converges absolutely and $\varPhi_\uv$, as defined above, solves Eq.~\eqref{eq:intro_mild} and converges almost surely as $\uv\searrow0$ in the Besov space $\mathscr{C}^\beta(\bM)$ for every $\beta<\sigma-\rdim/2$. To this end, we will establish certain bounds for $\partial_\uv^r\partial_\sIR^s F_{\uv,\sIR}^{i,m}$ which are stated in Sec.~\ref{sec:deterministic}. The bounds involve a regularizing kernel $K_\sIR$, which is introduced in Sec.~\ref{sec:kernels}, and a norm $\|\Cdot\|_{\sV^m}$, which is introduced in Sec.~\ref{sec:topology}.
\section{Regularizing kernels}\label{sec:kernels}
\begin{dfn}
For $n\in\bN_+$ let $\cK^n\subset\sS'(\bM^n)$ be the space of signed measures $K$ on $\bM^n$ with finite total variation $|K|$. We set $\|K\|_{\cK^n} = \int_{\bM^n} |K(\rd x_1\ldots\rd x_n)|$. If $n=1$, then we write $\cK^1=\cK\subset\sS'(\bM)$.
\end{dfn}
\begin{rem} It holds $\|K\|_{\cK^n}=\|K\|_{L^1(\bM^n)}$ for any $K\in L^1(\bM^n)\subset\cK^n$. Note also that $\delta_{\bM^n}\in\cK^n$, where $\delta_{\bM^n}$ is the Dirac delta in $\bM^n$.
\end{rem}
\begin{dfn}
For $\sIR\geq 0$ the kernel $K_\sIR\in\cK$ is the unique solution of the equation $\fP_\sIR K_\sIR=\delta_\bM$, where $\fP_\sIR:=1-[\sIR]^2\Delta_x$ and $[\sIR]:=\sIR^{1/\sigma}$. We set $K_\sIR^{\ast0}:=\delta_\bM$, $\fP^0_\sIR:=1$ and $K^{\ast(\oo+1)}_\sIR:=K_\sIR\ast K^{\ast\oo}_\sIR$, $\fP^{\oo+1}_\sIR:=\fP_\sIR\, \fP^{\oo}_\sIR$ for $\oo\in\bN_0$. We omit the index $\sIR$ if $\sIR=1$.
\end{dfn}
\begin{rem}
It holds $K_0=\delta_{\bM}$ and $K_\sIR\in L^1(\bM)$ for $\sIR>0$. For $\oo\in\bN_0$ and $\sIR\geq 0$ the kernel $K_\sIR^{\ast\oo}$ is a positive measure with total mass $\|K_\sIR^{\ast\oo}\|_\cK=1$. We have $\fP^\oo_\sIR K^{\ast\oo}_\sIR=\delta_\bM$. The fact that the regularizing kernel $K_\sIR^{\ast\oo}$ is an inverse of a differential operator simplifies the analysis in Sec.~\ref{sec:taylor}.
\end{rem}
\begin{lem}\label{lem:kernel_u_v}
For any $\sIR>0$ and $\uIR\geq0$ there exists $K_{\sIR,\uIR}\in\cK$ such that
\begin{equation}\label{eq:kernel_u_v}
\begin{gathered}
K_\sIR = K_{\sIR,\uIR} \ast K_\uIR,
\qquad
\|K_{\sIR,\uIR}\|_\cK= 1\vee(2[\uIR/\sIR]^2-1).
\end{gathered}
\end{equation}
In particular, if $\sIR\geq\uIR$, then $\|K_{\sIR,\uIR}\|_\cK=1$.
\end{lem}
\begin{proof}
Note that $K_{\sIR,\uIR} = [\uIR/\sIR]^2\delta_\bM + (1-[\uIR/\sIR]^2) K_\sIR$ and $\|\delta_\bM\|_\cK=\|K_\sIR\|_{\cK}=1$.
\end{proof}
\begin{dfn}\label{dfn:periodization}
Let $\bT=\bM/(2\pi\bZ)^\rdim$. For $V\in L^1(\bM)$ we define $\fT V\in L^1(\bT)$ by
\begin{equation}
\fT V(x):= \sum_{y\in(2\pi\bZ)^\rdim} V(x+y).
\end{equation}
\end{dfn}
\begin{rem}\label{rem:periodic}
For $K\in L^1(\bM)$ and periodic $f\in C(\bT)$ it holds $K\ast f = \fT K \star f$, where $\ast$ and $\star$ are the convolutions in $\bM$ and $\bT$, respectively.
\end{rem}
\begin{lem}\label{lem:kernel_simple_fact}
Let $\oo\in\bN_0$, $a\in\bN_0^{\rdim}$ and $n\in[1,\infty]$. The following is true:
\begin{enumerate}
\item[(A)]
If $|a|\leq\oo$, then $\|\partial^a K^{\ast\oo}_\sIR\|_\cK
\lesssim [\sIR]^{-|a|}$ uniformly in $\sIR>0$.
\item[(B)]
$\|\fT K^{\ast\rdim}_\sIR\|_{L^n(\bT)}\lesssim [\sIR]^{-\rdim (n-1)/n}$ uniformly in $\sIR\in(0,1]$.
\item[(C)]
It holds $\|\partial_\sIR K_\sIR\|_{\cK} \lesssim [\sIR]^{-\sigma}$ uniformly in $\sIR>0$.
\end{enumerate}
\end{lem}
\begin{rem}
Let $f\in C(\bM)$, $\oo\in\bN_0$ and $a\in\bN_0^\rdim$ be such that $|a|\leq\oo$. Then $K^{\ast\oo}_{\sIR}\ast f\in C^\oo(\bM)$ and $\|\partial^a K^{\ast\oo}_{\sIR}\ast f\|_{L^\infty(\bM)}\lesssim [\sIR]^{-|a|}\|f\|_{L^\infty(\bM)}$ uniformly over $\sIR>0$ and $f\in C(\bM)$.
\end{rem}
\section{Function spaces for coefficients}\label{sec:topology}
\begin{dfn}
For $\alpha<0$ and $\phi\in C^\infty_{\rb}(\bM)$ we define
\begin{equation}
\|\phi\|_{\sC^\alpha(\bM)}:= \sup_{\sIR\in(0,1]} [\sIR]^{-\alpha}\,
\|K_\sIR^{\ast\oo}\ast \phi\|_{L^\infty(\bM)},
\qquad
\oo=\ceil{-\alpha}\in\bN_+.
\end{equation}
The space $\sC^\alpha(\bM)$ is the completion of $C^\infty_\rb(\bM)$ equipped with the norm $\|\Cdot\|_{\sC^\alpha(\bM)}$.
\end{dfn}
\begin{dfn}\label{dfn:sV}
Let $m\in\bN_0$. The vector space $\sV^m$ consists of $V\in C(\bM^{1+m})$ such that
\begin{equation}
\|V\|_{\sV^m}
:=
\\
\sup_{x\in\bM} \int_{\bM^m}
|V(x;y_1,\ldots,y_m)|\,\rd y_1\ldots\rd y_m
\end{equation}
is finite and the function $x\mapsto V(x;y_1+x,\ldots,y_n+x)$ is $2\pi$ periodic for every $y_1,\ldots,y_m\in\bM$. For $\oo\in\bN_0$ the space $\cD^{m}_\oo$ consists of $V\in \sS'(\bM^{1+m})$ such that it holds $K^{\ast\oo,\otimes (1+m)}\ast V\in\sV^{m}$. The space $\cD^{m}$ is the union of the spaces $\cD^{m}_\oo$, $\oo\in\bN_0$. We also set $\sV^0=\sV$, $\cD^0_\oo=\cD_\oo$ and $\cD^0=\cD$.
\end{dfn}
\begin{rem}
Observe that $\sV=\cD_0=C(\bT)$. Moreover, we have $\|v\|_{\sV}=\|v\|_{L^\infty(\bM)}$ for any $v\in\sV$. Since $K$ is the inverse of a differential operator $\cD=\sS'(\bT)$.
\end{rem}
\begin{dfn}\label{dfn:map_Y}
The permutation group of the set $\{1,\ldots,n\}$ is denoted by $\cP_n$. For $m\in\bN_0$ and $V\in\cD^m$ and $\pi\in\cP_m$ we define $\fY_\pi V\in\cD^m$ by
\begin{equation}
\big\langle \fY_\pi V,\psi \otimes
\botimes_{q=1}^m \varphi_q\big\rangle
:=
\big\langle V,\psi\otimes
\botimes_{q=1}^{m} \varphi_{\pi(q)}\rangle,
\end{equation}
where $\psi,\varphi_1,\ldots,\varphi_m\in\ \sS(\bM)$.
\end{dfn}
\begin{rem}\label{rem:fY_pi}
The map $\fY_\pi:\sV^{m}\to\sV^{m}$ is well defined and has norm one.
\end{rem}
\begin{dfn}\label{dfn:map_B}
Let $m\in\bN_0$, $k\in\{0,\ldots,m\}$. We define the trilinear map $\fB\,:\,\sS(\bM)\times \sV^{1+k}\times\sV^{m-k}\to\sV^m$ by
\begin{multline}\label{eq:fB1_dfn}
\fB(G,W,U)(x;y_1,\ldots,y_m)
\\
:=
\int_{\bM^2} W(x;y_0,\ldots,y_k)
\,G(y_0-z)\,
U(z;y_{1+k},\ldots,y_m)\,\rd y_0\rd z.
\end{multline}
\end{dfn}
\begin{rem}
The RHS of Eq.~\eqref{eq:intro_flow_eq_i_m} can be written compactly using the map $\fB$.
\end{rem}
\begin{lem}\label{lem:fB1_bound}
It holds
\begin{equation}\label{eq:fB1_bound}
\|\fB(G,W,U)\|_{\sV^{m}}
\leq
\|G\|_\cK\,\|W\|_{\sV^{1+k}} \|U\|_{\sV^{m-k}}.
\end{equation}
\end{lem}
\begin{proof}
We have
\begin{multline}
\|\fB(G,W,U)\|_{\sV^{m}}=
\sup_{x\in\bM} \int_{\bM^{m}} |\fB(G,W,U)(x;y_1,\ldots,y_m)|\,\rd y_1\ldots\rd y_m
\\
\leq
\sup_{x\in\bM}\int_{\bM^{2+m}} |W(x;y_0,\ldots,y_k)|\,
|G(y_0-z)|\, |U(z;y_{1+k},\ldots,y_{m})|\, \rd z\rd y_0\ldots\rd y_m.
\end{multline}
It is easy to see that the last line is bounded by the RHS of~\eqref{eq:fB1_bound}.
\end{proof}
\begin{rem}\label{rem:fB_Ks}
The fact that $P_\sIR^\oo K_\sIR^{\ast\oo}=\delta_\bM$ implies that for any $\sIR>0$ it holds
\begin{equation}
\begin{gathered}
K_\sIR^{\ast\oo,\otimes(1+m)}\ast\fB(G,W,U)=
\fB\big(\fP^{2\oo}_\sIR G,
K_\sIR^{\ast\oo,\otimes(2+k)}\ast W,
K_\sIR^{\ast\oo,\otimes(1+m-k)}\ast U\big).
\end{gathered}
\end{equation}
This allows to define $\fB(G,W,U)\in\cD^m$ for all $W\in\cD^{1+k}$, $U\in\cD^{m-k}$.
\end{rem}
\section{Deterministic bounds for irrelevant coefficients}\label{sec:deterministic}
\begin{dfn}
Recall that $\sigma\in(\rdim/3,\rdim/2]$, $[\sIR]=\sIR^{1/\sigma}$ and set
\begin{equation}
\dim(\xi):=\rdim/2>0,\qquad
\dim(\varPhi):=\rdim/2-\sigma \geq 0,\qquad
\dim(\lambda):=3\sigma-\rdim>0.
\end{equation}
\end{dfn}
\begin{dfn}\label{dfn:relevant_irrelevant}
For $\varepsilon\geq0$ and $i,m\in\bN_0$ we define
\begin{equation}
\varrho_\varepsilon(i,m) :=
-\dim(\Xi)-\varepsilon
+ m\, (\dim(\Phi)+2\varepsilon)
+ i\, (\dim(\lambda)-6\varepsilon)\in \bR.
\end{equation}
We omit $\varepsilon$ if $\varepsilon=0$. The effective force coefficients $F^{i,m}_{\uv,\sIR}$ such that $\varrho(i,m)\leq 0$ are called relevant. The remaining coefficients are called irrelevant.
\end{dfn}
\begin{rem}
The number of relevant coefficients $F^{i,m}_{\uv,\sIR}$ such that $m\leq3i$ is always finite (recall that $F^{i,m}_{\uv,\sIR}=0$ if $m>3i$). For example, in the case $\rdim=5$ and $\sigma=2$ the relevant coefficients are $F^{0,0}_{\uv,\sIR}$, $F^{1,0}_{\uv,\sIR}$, $F^{1,1}_{\uv,\sIR}$, $F^{1,2}_{\uv,\sIR}$, $F^{1,3}_{\uv,\sIR}$, $F^{2,0}_{\uv,\sIR}$, $F^{2,1}_{\uv,\sIR}$. For explicit expressions for these coefficients see Remark~\ref{rem:explicit_coefficients}.
\end{rem}
\begin{rem}\label{rem:rho}
For arbitrary $\varepsilon>0$ and $i,m\in\bN_0$ such that $m\leq 3i$ it holds $\varrho_\varepsilon(i,m)<\varrho(i,m)$. Let $i_\diamond$ be the smallest integer such that $\varrho(i_\diamond+1,0)>0$. It holds $i_\diamond\in\bN_+$. Moreover, let $\varrho_\diamond>0$ be the minimum of $\varrho(i,m)+\ro$ for $i\in\{0,\ldots,i_\diamond\}$, $m\in\{0,\ldots,3i\}$, $\ro\in\bN_+$ such that $\varrho(i,m)+\ro>0$. Define $\varepsilon_\diamond:=\dim(\xi)/3\wedge\dim(\lambda)/9\wedge\varrho_\diamond/(7+6i_\diamond)\wedge\sigma$. We claim that for all $\varepsilon\in(0,\varepsilon_\diamond)$ and all $i,m,\ro\in\bN_0$ it holds \mbox{$\varrho_\varepsilon(i,m)+\ro>0$} if \mbox{$\varrho(i,m)+\ro>0$}. In what follows, we fix some $\varepsilon\in(0,\varepsilon_\diamond/3)$.
\end{rem}
\begin{lem}\label{lem:kernel_G}
For all $\oo\in\bN_0$, $r\in\bN_0$ it holds $\partial_\uv^r \partial_\sIR G_{\uv,\sIR}\in C^\infty_\rc(\bM)$ and
\begin{equation}\label{eq:bound_G}
\|\fP^{\oo}_\sIR \partial_\uv^r \partial_\sIR G_{\uv,\sIR}\|_\cK \lesssim [\uv]^{(\varepsilon-\sigma)r}\,[\sIR]^{-\varepsilon r}
\end{equation}
uniformly in $\uv\in(0,1/2]$, $\sIR\in(0,1]$, where $G_{\uv,\sIR}$ was introduced in Def.~\ref{dfn:kernel_G}.
\end{lem}
\begin{proof}
Let $a\in\bN_0^\rdim$. It holds $|\partial^a G(x)|\lesssim |x|^{\sigma-\rdim-|a|}$ uniformly for $|x|\leq 2$. If $\sigma\in2\bN_+$, then $\partial^a G$ decays fast at infinity. In general, $|\partial^a G(x)|\lesssim |x|^{-\sigma-\rdim-|a|}$ uniformly for $|x|> 1$. Moreover, we have
\begin{equation}
|\partial^a \partial_\uv^r \chi_\uv(|x|^\sigma)|\lesssim [\uv]^{(\varepsilon-\sigma)r} |x|^{(\sigma-\varepsilon)r-|a|},
\qquad
|\partial^a \partial_\sIR\chi_\sIR(|x|^\sigma)|\lesssim |x|^{\sigma-|a|}/\sIR^2
\end{equation}
uniformly in $\uv\in(0,1/2]$, $\sIR\in(0,1]$ and $x\in\bM$. Furthermore, $|\partial^a \partial_\sIR\chi_\sIR(|x|^\sigma)|$ vanishes unless $\sIR<(1-\sIR)|x|^\sigma\leq 2\sIR$. Using the above properties and considering separately $\sIR\in(0,1/2]$ and $\sIR\in(1/2,1]$ we obtain $\|\partial^a \partial_\uv^r \partial_\sIR G_{\uv,\sIR}\|_\cK \lesssim [\uv]^{(\varepsilon-\sigma)r}[\sIR]^{-\varepsilon r-|a|}$. This implies the lemma since $\fP_\sIR^\oo = (1-[\sIR]^2\Delta_x)^\oo$.
\end{proof}
\begin{rem}\label{rem:tilde_R}
Given $\oo\in\bN_0$ there exists $\tilde R>0$ such that
\begin{equation}
\frac{(1+m)\,\sigma}{\varrho_\varepsilon(i,m)}\leq \tilde R^{1/2},
\qquad
\|\fP^{2\oo}_\sIR \partial_\uv^r \partial_\sIR G_{\uv,\sIR}\|_\cK \leq 1/6\,\tilde R^{1/2}\, [\uv]^{(\varepsilon-\sigma)r}\,[\sIR]^{-\varepsilon}
\end{equation}
for all $i,m\in\bN_0$ such that $\varrho(i,m)>0$ and all $r\in\{0,1\}$, $\uv\in(0,1/2]$, $\sIR\in(0,1]$. The first of the above bounds implies, in particular, that $1/(\dim(\varPhi)+2\varepsilon)\leq\tilde R^{1/2}$.
\end{rem}
\begin{thm}\label{thm:deterministic_convergence}
Fix $\oo\in\bN_0$. Let $\tilde R>1$ as in Remark~\ref{rem:tilde_R}. Assume that for $r\in\{0,1\}$, $s=0$, all $i,m\in\bN_0$ such that $\varrho(i,m)\leq0$ and all $\uv\in(0,1/2]$, $\sIR\in(0,1]$ the following bound holds
\begin{equation}\label{eq:bound_deterministic_main}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\sIR^s F_{\uv,\sIR}^{i,m}\|_{\sV^m}\leq \frac{\tilde R^{1-s/2+2(3i-m)}}{4(1+i)^2\,4(1+m)^{2-s}} [\uv]^{(\varepsilon-\sigma)r}\, [\sIR]^{\varrho_\varepsilon(i,m)-\sigma s}.
\end{equation}
Then the above bound holds for all $r,s\in\{0,1\}$, $i,m\in\bN_0$, $\uv\in(0,1/2]$, $\sIR\in(0,1]$.
\end{thm}
\begin{rem}
By Theorems~\ref{thm:cumulants},~\ref{thm:probabilistic_bounds}~\ref{thm:deterministic_taylor}~\ref{thm:deterministic_long_range} (applied in this order), there exists a choice of counterterms such that the assumption of the above theorem is satisfied for some random $\tilde R>0$ such that $\bE \tilde R^n <\infty$ for any $n\in\bN_+$.
\end{rem}
\begin{rem}
If the bound~\eqref{eq:bound_deterministic_main} holds for $\oo=\ooo$, then it holds for all $\oo>\ooo$.
\end{rem}
\begin{rem}
Let us comment on the assumption in two simple cases. For $i=0$, $m=0$ the bound~\eqref{eq:bound_deterministic_main} says that \mbox{$\|K_\sIR^{\ast\oo}\ast\xi\|_{\sV}\leq \tilde R/4^2\,[\sIR]^{-\dim(\xi)-\varepsilon}$}, which is known to be true for $\oo=\rdim$ and some random $R>0$ such that $\bE \tilde R^n<\infty$ for all $n\in\bN_+$. For $i=1$, $m=3$ the bound~\eqref{eq:bound_deterministic_main} says that $\|K_\sIR^{\ast\oo,\otimes4}\ast F^{1,3}_{\uv,\sIR}\|_{\sV^3}\leq \tilde R/4^5$. It is easy to see that the above bound is satisfied for $\tilde R=4^5$ using the fact that $F^{1,3}_{\uv,\sIR}(x;x_1,x_2,x_3)=\delta_\bM(x-x_1)\delta_\bM(x-x_2)\delta_\bM(x-x_3)$.
\end{rem}
\begin{rem}
For $\lambda\leq \tilde R^{-6}$ the bounds~\eqref{eq:bound_deterministic_main} imply convergence of the series~\eqref{eq:solution_series} defining $\varPhi_\uv$, which is our candidate for the solution of Eq.~\eqref{eq:intro_mild}. In fact, one easily proves that $\|K^{\ast\oo}\ast\partial_\uv^r \varPhi_\uv\|_{\sV}\leq R\,[\uv]^{(\varepsilon-\sigma)r}$ for $r\in\{0,1\}$ and all $\uv\in(0,1]$. This implies the existence of the limit $\lim_{\uv\searrow0}\varPhi_\uv$ in $\sS'(\bM)$. In Sec.~\ref{sec:solution} we will prove that $\varPhi_\uv$ solves Eq.~\eqref{eq:solution_series} and $\lim_{\uv\searrow0}\varPhi_\uv$ exists in $\sC^{-\dim(\varPhi)-2\varepsilon}(\bM)$.
\end{rem}
\begin{proof}
Fix some $i_\circ\in\bN_+$, $m_\circ\in\bN_0$ and assume that the statement with $s=0$ is true for all $i,m\in\bN_+$ such that either $i<i_\circ$ or $i=i_\circ$ and $m>m_\circ$. We shall prove the statement for $i=i_\circ$, $m=m_\circ$. We first consider the case $s=1$. Using the flow equation~\eqref{eq:intro_flow_eq_i_m} and the notation introduced in Sec.~\ref{sec:topology} we obtain
\begin{multline}\label{eq:flow_deterministic_i_m}
\partial_\sIR \partial_\uv^r F^{i,m}_{\uv,\sIR}
=
-\frac{1}{m!}
\sum_{\pi\in\cP_m}\sum_{j=0}^i\sum_{k=0}^m\sum_{u+v+w=r}
(1+k)\,
\\
\times \fY_\pi\fB\big(\partial_\uv^u\partial_\sIR G_{\uv,\sIR},\partial_\uv^v F^{j,1+k}_{\uv,\sIR},\partial_\uv^wF^{i-j,m-k}_{\uv,\sIR}\big),
\end{multline}
where $r,u,v,w\in\{0,1\}$. By Remark~\ref{rem:fY_pi}, Lemma~\ref{lem:fB1_bound} and Remark~\ref{rem:fB_Ks} we get
\begin{multline}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast\partial_\sIR \partial_\uv^r F^{i,m}_{\uv,\sIR}\|_{\sV^m}
\leq
(1+m)~\sum_{u+v+w=r} \|\fP_\sIR^\oo\partial_\uv^u\partial_\sIR G_{\uv,\sIR}\|_\cK
\\
\times \sum_{j=0}^i\sum_{k=0}^m\|K_{\sIR}^{\ast\oo,\otimes(2+k)}\ast \partial_\uv^v F^{j,1+k}_{\uv,\sIR}\|_{\sV^{1+k}}~
\|K_{\sIR}^{\ast\oo,\otimes(1+m-k)}\partial_\uv^wF^{i-j,m-k}_{\uv,\sIR}\|_{\sV^{m-k}}.
\end{multline}
The statement of the theorem with $s=1$ follows now from the induction hypothesis, Remark~\ref{rem:tilde_R},
the equality
\begin{equation}
\varrho_\varepsilon(i,m)-\sigma = \varrho_\varepsilon(j,1+k) + \varrho_\varepsilon(i-j,m-k)-\varepsilon
\end{equation}
and the bound
\begin{equation}
\sum_{j=0}^i\sum_{k=0}^m \frac{\tilde R^{1+2(3j-k-1)}}{4(1+j)^2\,4(2+k)^2}
\frac{\tilde R^{1+2(3i-3j-m+k)}}{4(1+i-j)^2\,4(1+m-k)^2}
\leq \frac{\tilde R^{2(3i-m)}}{4(1+i)^2\,4(1+m)^2},
\end{equation}
which is a consequence of the following inequality
\begin{equation}
\sum_{j=0}^i \frac{1}{4(1+j)^2\,4(1+i-j)^2} \leq \frac{1}{4(1+i)^2}.
\end{equation}
In order to prove the statement of the theorem for $s=0$ and $i,m\in\bN_0$ such that $\varrho(i,m)>0$ we use the identity
\begin{equation}\label{eq:determinisitc_intagrate}
F^{i,m}_{\uv,\sIR} = F^{i,m}_{\uv} + \int_0^\sIR \partial_\uIR F^{i,m}_{\uv,\uIR}\,\rd\uIR.
\end{equation}
We first observe that $F^{i,m}_{\uv,0}=F^{i,m}_\uv=0$ if $\varrho(i,m)>0$. Next, we note that
\begin{equation}\label{eq:determinisitc_bound_u_v}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast
\partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m}\|_{\sV^m}\leq\|K_{\uIR}^{\ast\oo,\otimes(1+m)}\ast
\partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m}\|_{\sV^m}
\end{equation}
for $\uIR\leq\sIR$ by Lemma~\ref{lem:kernel_u_v}. The statement of the theorem with $s=0$ follows now from the statement with $s=1$ and the bounds
\begin{equation}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r F_{\uv,\sIR}^{i,m}\|_{\sV^m}
\\
\leq
\int_{0}^{\sIR}
\|K_{\uIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m}\|_{\sV^m}\,\rd\uIR
\end{equation}
and
\begin{equation}
\int_{0}^\sIR [\uIR]^{\varrho_\varepsilon(i,m)-\sigma}\,\rd\uIR \leq \sigma/\varrho_\varepsilon(i,m)~ [\sIR]^{\varrho_\varepsilon(i,m)}
\leq \tilde R^{1/2}/(1+m)\,[\sIR]^{\varrho_\varepsilon(i,m)}
\end{equation}
We stress that the first inequality in the last bound is valid only if $\varrho_\varepsilon(i,m)> 0$, which holds provided $\varrho(i,m)>0$ by Remark~\ref{rem:rho}.
\end{proof}
\section{Generalized effective force coefficients}\label{sec:modified_coefficients}
In order to prove the bound~\eqref{eq:bound_deterministic_main} for all relevant effective force coefficients we will use generalized coefficients introduced in this section.
\begin{dfn}\label{dfn:modified_coefficients}
We denote the set of multi-indices by $\frM=\bN_0^\rdim$. For for $m\in\bN_+$ and $a=(a_1,\ldots,a_m)\in\frM^m$ we define $|a|:=|a_1|+\ldots+|a_m|$. We also set $\frM^0\equiv\{0\}$. For $a\in\frM$ we define $\cX^a\in C^\infty(\bM)$ by $\cX^a(x):= \frac{1}{a!}x^a$ and for $a\in\frM^m$ we define $\cX^{m,a}\in C^\infty(\bM^{1+m})$ by
\begin{equation}
\cX^{m,a}(x;y_1,\ldots,y_m)
:=
\frac{1}{a!}\,(x-y_1)^{a_1} \ldots (x-y_m)^{a_m}.
\end{equation}
For $i,m\in\bN_0$, $a\in\frM^m$ we define $F^{i,m,a}_{\uv,\sIR}\in\sS'(\bM^{1+m})$ and $f^{i,m,a}_{\uv,\sIR}\in\sS'(\bM)$ by
\begin{equation}
F^{i,m,a}_{\uv,\sIR}:= \cX^{m,a} F^{i,m}_{\uv,\sIR},
\qquad
\langle f^{i,m,a}_{\uv,\sIR},\psi\rangle:=\langle F^{i,m,a}_{\uv,\sIR},\psi\otimes 1_\bM^{\otimes m}\rangle,
\end{equation}
where $\psi\in\sS(\bM)$ and $1_\bM(x)=1$ for all $x\in\bM$. The force coefficients $F^{i,m,a}_\uv$ and $f^{i,m,a}_\uv$ are defined analogously. The effective force coefficients $F^{i,m,a}_{\uv,\sIR}$ or $f^{i,m,a}_{\uv,\sIR}$ such that $\varrho(i,m)+|a|\leq 0$ are called relevant. The remaining coefficients are called irrelevant. The finite collection of all relevant coefficients $f^{i,m,a}_{\uv,\sIR}$ such that $m\leq 3i$ is called the enhanced noise (recall that $f^{i,m,a}_{\uv,\sIR}=0$ if $m>3i$).
\end{dfn}
\begin{rem}
Let us list the non-zero force coefficients $f^{i,m,a}_{\uv\phantom{0}}=f^{i,m,a}_{\uv,0}\in\sS'(\bM)$:
\begin{equation}
f^{0,0,0}_\uv = \xi(x),\quad
f^{1,3,0}_\uv = 1,\quad
f^{i,1,0}_\uv = c^{[i]}_\uv,
\quad i\in\{1,\ldots,i_\sharp\}.
\end{equation}
\end{rem}
\begin{rem}
For example, in the case $\rdim=5$ and $\sigma=2$ the enhanced noise consists of $f^{0,0,0}_{\uv,\sIR}=\xi$, $f^{1,0,0}_{\uv,\sIR}$, $f^{1,1,0}_{\uv,\sIR}$, $f^{1,2,0}_{\uv,\sIR}$, $f^{1,3,0}_{\uv,\sIR}=1$, $f^{2,0,0}_{\uv,\sIR}$, $f^{2,1,0}_{\uv,\sIR}$. Explicit expressions for all these coefficients can be easily obtained using Remark~\ref{rem:explicit_coefficients}.
\end{rem}
\begin{rem}\label{rem:taylor}
Given $i,m\in\bN_0$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\in(-\ro,1-\ro]$ for some $\ro\in\bN_+$ the relevant coefficient $F^{i,m,a}_{\uv,\sIR}$ can be expressed in terms of irrelevant coefficients $F^{i,m,b}_{\uv,\sIR}$, $|b|=\ro$, and relevant coefficients $f^{i,m,b}_{\uv,\sIR}$, $|b|<\ro$. The above fact plays a crucial role in the proof of bounds for relevant coefficients given in Sec.~\ref{sec:deterministic_relavant}. In Sec.~\ref{sec:taylor} we show that it is a consequence of the Taylor theorem.
\end{rem}
\begin{rem}\label{rem:polynomials}
For a list of multi-indices $a=(a_1,\ldots,a_m)\in\frM^m$ and a permutation $\pi\in\cP_n$ we set $\pi(a):=(a_{\pi(1)},\ldots,a_{\pi(m)})$. Note that it holds
\begin{equation}\label{eq:poly_perm}
\cX^{m,a}(x;y_1,\ldots,y_m)
\\
=
\cX^{m,\pi(a)}(x;y_{\pi(1)},\ldots,y_{\pi(m)})
\end{equation}
We claim that
\begin{multline}\label{eq:poly_binom}
\cX^{m,a}(x;y_1,\ldots,y_m)
=
\sum_{b,c,d}
\cX^{1+k,(b_{1+k}+\ldots+b_m,b_1,\ldots,b_k)}(x;y_0,y_1,\ldots,y_k)
\\
\cX^{1,c_{1+k}+\ldots+c_m}(y_0;z)
\,\cX^{m-k,(d_{1+k},\ldots,d_m)}(z;y_{1+k},\ldots,y_m),
\end{multline}
where the sum is over all $b_p,c_p,d_p$ with $p\in\{1,\ldots,m\}$ such that $b_p=a_p$, $c_p=0$, $d_p=0$ for $p\in\{1,\ldots,k\}$ and $b_p+c_p+d_p=a_p$ for $p\in\{1+k,\ldots,m\}$. Throughout the paper we use the following schematic notation for the sums of the above type
\begin{multline}\label{eq:poly_binom_notation}
\cX^{m,a}(x;y_1,\ldots,y_m)
\\
=\sum_{b+c+d=a} \cX^{1+k,b}(x;y_0,y_1,\ldots,y_k)
\cX^{c}(y_0-z) \cX^{m-k,d}(z;y_{1+k},\ldots,y_m).
\end{multline}
Let us note that the formula~\eqref{eq:poly_binom} above was obtained by writing $(y_p-x)^{a_p}$, $p\in\{1+k,\ldots,m\}$, as
\begin{equation}
(y_p-x)^{a_p}
=\sum_{\substack{b_p,c_p,d_p\\b_p+c_p+d_p=a_p}}\frac{a_p!}{b_p!c_p!d_p!}~(x-y_0)^{b_p}(y_0-z)^{c_p}(z-y_p)^{d_p}.
\end{equation}
\end{rem}
Using the flow equation~\eqref{eq:intro_flow_eq_i_m} and the identity~\eqref{eq:poly_binom} we show that the effective force coefficients $F^{i,m,a}_{\uv,\sIR}$ satisfy the following flow equation
\begin{multline}\label{eq:flow_deterministic_i_m_a}
\partial_\sIR^{\phantom{i}}\partial_\uv^r F^{i,m,a}_{\uv,\sIR}
=
-\frac{1}{m!}
\sum_{\pi\in\cP_m}\sum_{j=0}^i\sum_{k=0}^m\sum_{u+v+w=r}
\sum_{b+c+d=\pi(a)}(1+k)\,
\\
\times\fY_\pi\fB\big(\cX^c\partial_\uv^u\partial_\sIR G_{\uv,\sIR},\partial_\uv^v F^{j,1+k,b}_{\uv,\sIR},\partial_\uv^w F^{i-j,m-k,d}_{\uv,\sIR}\big),
\end{multline}
where $r\in\bN_0$, $u,v,w\in\{0,\ldots,r\}$ and the sum over multi-indices $b,c,d$ is restricted to the set specified below Eq.~\eqref{eq:poly_binom}.
\section{Taylor polynomial and remainder}\label{sec:taylor}
\begin{dfn}
For $a\in\frM^m$, $V\in C^\infty(\bM^{1+m})$ we define $\partial^a V\in C^\infty(\bM^{1+m})$ by $\partial^a V(x;x_1,\ldots,x_m):=
\partial^{a_1}_{x_1} \ldots\partial^{a_m}_{x_m}
V(x;x_1,\ldots,x_m)$.
The above map extends in an obvious way to $\sS'(\bM^{1+m})\supset\cD^m$.
\end{dfn}
\begin{dfn}\label{dfn:map_L_m}
For $v\in\cD$ we define $\fL^m v\in\cD^m$ by the equality
\begin{equation}
\langle\,\fL^m v\,,\,\psi\otimes\varphi_1\otimes\ldots\otimes\varphi_m\,\rangle= \langle v\,,\,\psi\varphi_1\ldots\varphi_m\rangle,
\end{equation}
where $\psi,\varphi_1,\ldots,\varphi_m\in \sS(\bM)$ are arbitrary.
\end{dfn}
\begin{dfn}\label{dfn:map_Z}
For $\tau>0$ and $V\in C(\bM^{1+m})$ we define $\fZ_\tau V\in C(\bM^{1+m})$ by the equality
\begin{equation}
\fZ_\tau V(x;y_1,\ldots,y_m)
:=\tau^{-\rdim m}\,V(x;x+(y_1-x)/\tau,\ldots,x+(y_m-x)/\tau).
\end{equation}
The above map extends in an obvious way to $\sS'(\bM^{1+m})\supset\cD^m$.
\end{dfn}
\begin{dfn}\label{dfn:map_I}
The linear map $\fI\,:\,\sV^m\to\sV$ is defined by
\begin{equation}
\fI V(x):= \int_{\bM^m} V(x;y_1,\ldots,y_m)\,\rd y_1\ldots\rd y_m.
\end{equation}
The map $\fI$ is extended to $V\in\cD^m_\oo$ by the formula
\begin{equation}
\langle \fI V,\psi\rangle := \langle K^{\ast\oo,\otimes(1+m)}\ast V,\fP^\oo \psi\otimes 1^{\otimes m}_\bM\rangle,
\end{equation}
where $\psi\in \sS(\bM)$.
\end{dfn}
\begin{lem}\label{lem:map_I}
The map $\fI$ is well defined and has the following properties.
\begin{enumerate}
\item[(A)] $\fI\big( K_\sIR^{\ast\oo,\otimes(1+m)} \ast V\big)=K^{\ast\oo}_\sIR\ast\fI V$
for any $V\in\sV^m$, $\oo\in\bN_0$ and $\sIR>0$.
\item[(B)] $\|\fI V\|_\sV \leq \|V\|_{\sV^m}$ for $V\in\sV^m$.
\end{enumerate}
\end{lem}
\begin{proof}
The well definedness and Part (A) follow from $\int_\bM K_\sIR(x)\,\rd x=1$. Part~(B) is a consequence of Def.~\ref{dfn:sV} of the norm~$\sV^m$.
\end{proof}
\begin{dfn}\label{dfn:map_X}
For $\ro\in\bN_+$, $a\in\frM^m$ such that $|a|<\ro$ and two collections of distributions: $v^b\in\cD$, $b\in\frM^m$, $|b|<\ro$, and $V^b\in\cD^m$, $b\in\frM^m$, $|b|=\ro$, the distribution $\fX_\ro^a(v^b,V^b)\in \cD^m$ is defined by the equality
\begin{multline}
\fX_\ro^a(v^b,V^b)
:=
\sum_{|a+b|<\ro} \,(-1)^{|b|}\,\tbinom{a+b}{a}
\partial^b \fL^m(v^{a+b})
\\
+
\sum_{|a+b|=\ro} |b| \,(-1)^{|b|}\,\tbinom{a+b}{a}
\int_0^1 (1-\tau)^{|b|-1}\,
\partial^b \fZ_\tau(V^{a+b})\,\rd\tau,
\end{multline}
where the sums above are over $b\in\frM^m$.
\end{dfn}
\begin{thm}\label{thm:taylor_bounds}
Let $\ro\in\bN_+$ and $\oo\in\bN_0$. There exists a constant $c>0$ such that the following statement is true. Let $V^b\in\sV^m$, $v^b\in\sV$ be as in Def.~\ref{dfn:map_X} and $\sIR\in(0,1]$. Assume that there exists a constant $C>0$ such that for $b\in\frM$
\begin{equation}\label{eq:taylor_ass_V}
\|K_\sIR^{\ast\oo,\otimes(1+m)}\ast V^b\|_{\sV^m} \leq C\, [\sIR]^{[b]},\quad |b|=\ro,
\quad\quad
\|K_\sIR^{\ast\oo}\ast v^b\|_\sV \leq C\, [\sIR]^{[b]}.
\quad |b|<\ro,
\end{equation}
Then for $a\in\frM$ such that $|a|<\ro$ it holds
\begin{equation}\label{eq:taylor_thm_bound}
\| K_\sIR^{\ast(6\oo+\ro),\otimes(1+m)} \ast \fX_\ro^a(v^b,V^b)\|_{\sV^m} \leq c\, C\, [\sIR]^{|a|}.
\end{equation}
\end{thm}
\begin{proof}
The theorem follows from Def.~\ref{dfn:map_X} of the map $\fX^a_\ro$, Lemma~\ref{lem:taylor_bounds_aux} and the bound $\|\partial^c K^{\ast\ro}_\sIR\|_\cK\lesssim |\sIR|^{-|c|}$, $c\in\frM$, $|c|\leq\ro$, proved in Lemma~\ref{lem:kernel_simple_fact}~(A).
\end{proof}
\begin{thm}\label{thm:taylor}
Let $\ro\in\bN_+$ and $V\in\cD^m$ such that $\cX^{m,b} V \in \cD^m$ for any $b\in\frM^m$, $|b|\leq\ro$. Then
$
\cX^{m,a} V = \fX_\ro^a(\fI(\cX^{m,b} V),\cX^{m,b} V)
$
for all $a\in\frM^m$, $|a|<\ro$.
\end{thm}
\begin{proof}
First recall that for any $\varphi\in C^\infty(\bR^N)$, $N\in\bN_+$, it holds
\begin{multline}
\varphi(y) = \sum_{|b|<\ro} \frac{1}{b!}(y-x)^b\,(\partial^b \varphi)(x)
\\
+ \sum_{|b|=\ro} \frac{1}{b!}\,(y-x)^b \int_0^1 (1-\tau)^{\ro-1}\,(\partial^b \varphi)(x+\tau(y-x))\,\rd\tau
\end{multline}
by the Taylor theorem, where $b\in\bN_0^N$. Consequently, for any $\varphi\in \sS(\bM^{1+m})$ we have
\begin{multline}
(\cX^{m,a}\varphi)(x;y_1,\ldots,y_m) = \sum_{|a+b|<\ro} \tbinom{a+b}{b}\,\cX^{a+b}(x;y_1,\ldots,y_m)\, (\partial^b \varphi)(x;x,\ldots,x)
\\
+ \sum_{|a+b|=\ro} \tbinom{a+b}{b}\,\cX^{a+b}(x;y_1,\ldots,y_m) \int_0^1 (1-\tau)^{\ro-1}\,(\fZ_{\tau}^\dagger\partial^b\varphi)(x;y_1,\ldots,y_m)\,\rd\tau,
\end{multline}
where the map $\fZ_{\tau}^\dagger:=\tau^{\rdim m}\fZ_{1/\tau}$ is a formal dual to the map $\fZ_\tau$ and the sums are over $b\in\frM$. To complete the proof we test $\cX^{m,a} V$ with $\varphi\in \sS(\bM^{1+m})$, apply the above formula and use the definitions of the maps $\fX_\ro^a$ and $\fI$.
\end{proof}
\begin{lem}\label{lem:taylor_bounds_aux}
Let $\oooo\in\bN_0$. The following bounds:
\begin{enumerate}
\item[(A)] $\|K^{\ast\oooo,\otimes(1+m)}_\sIR\ast \fL^m v\|_{\sV^m} \lesssim \|K^{\ast\oooo}_\sIR\ast v\|_\sV$,
\item[(B)]
$\|K^{\ast2\oooo,\otimes(1+m)}_\sIR\ast\fZ_\tau V\|_{\sV^m}
\lesssim
\|K^{\ast\oooo,\otimes(1+m)}_\sIR\ast V\|_{\sV^m}$
\end{enumerate}
hold uniformly in $\tau,\sIR\in(0,1]$ and $v\in\sV$, $V\in\sV^m$.
\end{lem}
\begin{proof}
By the exact scaling of the norms and kernels it is enough to prove the lemma for $\sIR=1$. Using $\fP^{\oooo} K^{\ast\oooo}=\delta_\bM$ we obtain
\begin{multline}
(K^{\ast\oooo,\otimes(1+m)}\ast\fL^m v)(x;y_1,\ldots,y_m)
\\
=\int_\bM
(K^{\ast\oooo}\ast v)(z)~
\fP^{\oooo}(\partial_z) K^{\ast\oooo}(x-z)K^{\ast\oooo}(y_1-z)\ldots K^{\ast\oooo}(y_m-z)
\,\rd z
\\
=\int_\bM
(K^{\ast\oooo}\ast v)(z)~
H^{\ast\oooo}_0(x-z,y_1-z,\ldots,y_m-z)
\,\rd z,
\end{multline}
where $H_0\in\cK^{1+m}$ is defined in the lemma below and $\fP^{\oooo}(r)=(1-r^2)^\oooo$. This proves the bound (A) since
$
K^{\ast\oooo,\otimes(1+m)}\ast \fL^m v = H_0^{\ast\oooo}\ast \fL^m(K^{\ast\oooo}\ast v).
$
The bound~(B) follows from the identities
\begin{equation}\label{eq:lem_taylor_useful_identities}
\begin{gathered}
K^{\otimes(1+m)}
=(K\otimes\check K^{\otimes m}_\tau) \ast \fZ_\tau(\delta_\bM\otimes K^{\otimes m}),
\\
K^{\otimes (1+m)}
= H_\tau\ast\fZ_\tau(K\otimes\delta_\bM^{\otimes m}),
\end{gathered}
\end{equation}
where the kernels $\check K_\tau$ and $\check H_\tau$ are defined in the lemma below. Indeed, applying the above equalities recursively and using $\fZ_\tau H\ast \fZ_\tau V=\fZ_\tau(H\ast V)$ we arrive at
\begin{equation}\label{eq:taylor_S_hat_property}
K^{\ast2\oooo,\otimes(1+m)}\ast\fZ_\tau V
=
\check H^{\ast\oooo}_\tau\ast \fZ_\tau(K^{\ast\oooo,\otimes(1+m)}\ast V),
\quad
\check H_\tau:=(K\otimes \check K^{\otimes m}_\tau) \ast H_\tau,
\end{equation}
which implies Part (B) since $\|\check H_\tau\|_{\cK^{1+m}}\lesssim 1$ uniformly in $\tau\in(0,1]$ by the lemma below. It remains to establish the identities~\eqref{eq:lem_taylor_useful_identities}. The first one follows from the fact that $\fZ_\tau(\delta_\bM\otimes K^{\otimes m}) = \delta_\bM\otimes\breve K_\tau^{\otimes m}$, where $\breve K_\tau(x):=\tau^{-1-\rdim}K(x/\tau)$. To show the second one we observe that by the definition of $\check H_\tau$ given in the lemma below
\begin{equation}
\fZ_{1/\tau}(H_\tau)(x;y_1,\ldots,y_m)
=
\tau^{\rdim m}\, \fP(\partial_x)\, K(x)
K(x+\tau(y_1-x)) \ldots K(x+\tau(y_1-x)).
\end{equation}
After convolving both sides with the kernel $K\otimes\delta_\bM^{\otimes m}$ we obtain
\begin{equation}
\fZ_{1/\tau}(H_\tau)\ast\big(K\otimes\delta_\bM^{\otimes m}\big)
=
\fZ_{1/\tau}(K^{\otimes (1+m)})
\end{equation}
The second of the identities \eqref{eq:lem_taylor_useful_identities} follows after applying the map $\fZ_\tau$ to both sides of the above equality.
\end{proof}
\begin{lem}\label{lem:taylor_aux}
The distributions $\check K_\tau\in\sS'(\bM)$ and $H_\tau\in\sS'(\bM^{1+m})$ defined by
\begin{equation}
\begin{gathered}
\check K_\tau(x):= \fP(\tau \partial_x) K(x) = \tau^2 \delta_\bM + (1-\tau^2) K,
\\
H_\tau(x;y_1,\ldots,y_m)
:=
\fP(\partial_x+(1-\tau)\partial_y) \, K(x) K(y_1) \ldots K(y_m),
\end{gathered}
\end{equation}
where $\partial_y:=\partial_{y_1}+\ldots+\partial_{y_m}$ and $\fP(r)=1-r^2$, are polynomials in $\tau>0$ of degree~$2$ whose coefficients belong to $\cK$ and $\cK^{1+m}$, respectively.
\end{lem}
\section{Deterministic bounds for relevant coefficients}\label{sec:deterministic_relavant}
\begin{lem}
For every $\oo\in\bN_0$, $a\in\frM$ and $r\in\bN_0$ the following bound
\begin{equation}
\|\fP^{\oo}_\sIR \cX^a\partial_\uv^r \partial_\sIR G_{\uv,\sIR}\|_\cK \lesssim [\uv]^{(\varepsilon-\sigma)r} [\sIR]^{|a|-\varepsilon r}
\end{equation}
holds uniformly in $\uv,\sIR\in(0,1/2]$.
\end{lem}
\begin{rem}
For $\sigma\notin2\bN_+$ the bound stated in the lemma does not hold uniformly for all $\sIR\in(0,1]$ because of the slow decay of the kernel $G(x)$ at infinity. The above lemma can be easily proved using the method of the proof of Lemma~\ref{lem:kernel_G}.
\end{rem}
\begin{thm}\label{thm:deterministic_taylor}
Fix $R>1$. Assume that for all $i,m\in\bN_0$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\leq0$ there exists $\oo\in\bN_0$ such that
\begin{equation}
\|K^{\ast\oo}_\sIR\ast \partial_\uv^r f^{i,m,a}_{\uv,\sIR}\|_\sV \leq R\,[\uv]^{(\varepsilon-\sigma) r}[\sIR]^{\varrho_\varepsilon(i,m)+|a|}
\end{equation}
for all \mbox{$r\in\{0,1\}$} and $\uv,\sIR\in(0,1/2]$. Then for all $i,m\in\bN_0$, $a\in\frM^m$ there exists $\oo\in\bN_0$ such that
\begin{equation}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\sIR^s F_{\uv,\sIR}^{i,m,a}\|_{\sV^m}\lesssim R^{1+3i -m}\,[\uv]^{(\varepsilon-\sigma)r}\, [\sIR]^{\varrho_\varepsilon(i,m)+|a|-\sigma s}
\end{equation}
for all $r,s\in\{0,1\}$ uniformly in $\uv,\sIR\in(0,1/2]$. The constants of proportionality in the above bounds depend only on $i,m\in\bN_+$, $a\in\frM^m$ and are otherwise universal.
\end{thm}
\begin{rem}
Let us remind the reader that the coefficients $F^{i,m,a}_{\uv,\sIR}$ and $f^{i,m,a}_{\uv,\sIR}$ such that $\varrho(i,m)+|a|\leq0$ are called relevant and $F^{i,m,a}_{\uv,\sIR}$ and $f^{i,m,a}_{\uv,\sIR}$ vanish unless $m\leq 3i$. There only finitely many non-zero relevant coefficients $F^{i,m,a}_{\uv,\sIR}$ or $f^{i,m,a}_{\uv,\sIR}$. Recall that the collection of the relevant coefficients $f^{i,m,a}_{\uv,\sIR}$ such that $m\leq 3i$ is called the enhanced noise.
\end{rem}
\begin{rem}
The above theorem together with Theorem~\ref{thm:deterministic_long_range} imply that under the assumption of the above theorem there exists $\oo\in\bN_0$ and a universal constant $c>1$ such that the assumption of Theorem~\ref{thm:deterministic_convergence} is satisfied with $\tilde R=cR$.
\end{rem}
\begin{proof}
{\it The base case:}
We observe that $F^{0,0,0}_{\uv,\sIR}=f^{0,0,0}_{\uv,\sIR}=\xi$. Hence, for $i=0$, $m=0$ the statement follows from the assumption.
{\it The inductive step:} Fix some $i_\circ\in\bN_+$, $m_\circ\in\bN_0$ and assume that the statement with $s=0$ is true for all $i,m\in\bN_+$ such that either $i<i_\circ$ or $i=i_\circ$ and $m>m_\circ$. We shall prove the statement for $i=i_\circ$, $m=m_\circ$. As in the proof of Theorem~\ref{thm:deterministic_convergence} the induction hypothesis, the flow equation~\eqref{eq:flow_deterministic_i_m_a}, and the identities
\begin{equation}
\varrho_\varepsilon(i,m)+|a|-\sigma = \varrho_\varepsilon(j,1+k)+|b| + |c| + \varrho_\varepsilon(i-j,m-k)+|d| -\varepsilon,
\end{equation}
\begin{equation}
R^{1+3j-k-1} R^{1+3(i-j)-m+k} = R^{1+3i-m}
\end{equation}
imply the statement for $s=1$.
In order to prove the statement for $s=0$ let us first assume that $a\in\frM^m$ is such that \mbox{$\varrho(i,m)+|a|>0$}. Then $F^{i,m,a}_{\uv,0}=F^{i,m,a}_\uv=0$. Consequently, the analogs of Eq.~\eqref{eq:determinisitc_intagrate} and the bound~\eqref{eq:determinisitc_bound_u_v} imply that
\begin{equation}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast\partial_\uv^r F_{\uv,\sIR}^{i,m,a}\|_{\sV^m}
\\
\leq
\int_{0}^{\sIR}
\|K_{\uIR}^{\ast\oo,\otimes(1+m)}\ast
\partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m,a}\|_{\sV^m}\,\rd\uIR.
\end{equation}
The statement follows now from the bound for $\partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m,a}$ and the fact that \mbox{$\varrho_\varepsilon(i,m)+|a|-\varepsilon>0$} if \mbox{$\varrho(i,m)+|a|>0$} by Remark~\ref{rem:rho}.
Now let us suppose that $a\in\frM^m$ is such that \mbox{$\varrho(i,m)+|a|\leq0$}. Then our bound for $\|K_{\uIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m,a}\|_{\sV^m}$ is not integrable at $\uIR=0$. Consequently, the strategy used in the previous paragraph does not work. If $m=0$, then $a=0$ and $\partial_\uv^r F^{i,0,0}_{\uv,\sIR}=\partial_\uv^r f^{i,0,0}_{\uv,\sIR}$. As a result, the statement follows from the assumption. If $m\in\bN_+$, then we use the identity
\begin{equation}\label{eq:thm_deterministic_relevant}
\partial_\uv^r F^{i,m,a}_{\uv,\sIR}
=
\fX^a_\ro(\partial_\uv^r f^{i,m,b}_{\uv,\sIR},
\partial_\uv^r F^{i,m,b}_{\uv,\sIR}),
\end{equation}
which follows from Theorem~\ref{thm:taylor}. We choose $\ro\in\bN_+$ to be the smallest positive integer such that $\varrho(i,m)+\ro>0$. With this choice of $\ro$ the RHS of the above equality involves only the coefficients $f^{i,m,b}_{\uv,\sIR}$ such that $\varrho(i,m,b)\leq0$, which satisfy the assumed bound, and the coefficients $F^{i,m,b}_{\uv,\sIR}$ such that $\varrho(i,m,b)>0$, for which the statement of the theorem has already been established. To finish the proof of the inductive step we apply Theorem~\ref{thm:taylor_bounds} with $C\lesssim [\uv]^{(\varepsilon-\sigma)r}[\sIR]^{\varrho_\varepsilon(i,m)-\varepsilon r}$.
\end{proof}
\section{Deterministic bounds for long-range part of coefficients}
\begin{thm}\label{thm:deterministic_long_range}
Fix $R>1$. Assume that for all $i,m\in\bN_0$ there exists $\oo\in\bN_0$ such that
\begin{equation}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\sIR^s F_{\uv,\sIR}^{i,m}\|_{\sV^m}\lesssim R^{1+3i-m}\,[\uv]^{(\varepsilon-\sigma)r}\, [\sIR]^{\varrho_\varepsilon(i,m)-\sigma s}
\end{equation}
for $r,s\in\{0,1\}$ uniformly in $\uv,\sIR\in(0,1/2]$. Then for all $i,m\in\bN_0$ there exists $\oo\in\bN_0$ such that the above bound is true for $r,s\in\{0,1\}$ uniformly in $\uv\in(0,1/2]$, $\sIR\in(0,1]$. The constants of proportionality in the above bounds depend only on $i,m\in\bN_+$ and are otherwise universal.
\end{thm}
\begin{proof}
{\it The base case:}
We observe that $F^{0,0}_{\uv,\sIR}=f^{0,0,0}_{\uv,\sIR}=\xi$ is independent of $\uv,\sIR$. The assumed bound and Lemma~\ref{lem:kernel_u_v} imply that
\begin{equation}
\|K_\sIR^{\ast\oo}\ast\xi\|_\sV\leq \|K_{\sIR/2}^{\ast\oo}\ast\xi\|_\sV \lesssim [\sIR]^{\varrho_\varepsilon(0,0)}
\end{equation}
uniformly in $\sIR\in(0,1]$. This proves the statement for $i=0$ and $m=0$.
{\it The induction step:} Fix some $i_\circ\in\bN_+$, $m_\circ\in\bN_0$ and assume that the statement with $s=0$ is true for all $i,m\in\bN_+$ such that either $i<i_\circ$ or $i=i_\circ$ and $m>m_\circ$. We shall prove the statement for $i=i_\circ$, $m=m_\circ$. The induction hypothesis and the flow equation imply the statement for $s=1$. In order to prove the statement for $s=0$ we use the identity
\begin{equation}
F^{i,m}_{\uv,\sIR} = F^{i,m}_{\uv,1/2} + \int_{1/2}^\sIR \partial_\uIR F^{i,m}_{\uv,\uIR}\,\rd\uIR,
\end{equation}
which by Lemma~\ref{lem:kernel_u_v} implies that
\begin{multline}
\|K_{\sIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r F_{\uv,\sIR}^{i,m}\|_{\sV^m}
\\
\leq
\|K_{1/2}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r F_{\uv,1/2}^{i,m}\|_{\sV^m}
+
\int_{1/2}^{\sIR}
\|K_{\uIR}^{\ast\oo,\otimes(1+m)}\ast \partial_\uv^r\partial_\uIR F_{\uv,\uIR}^{i,m}\|_{\sV^m}\rd\uIR.
\end{multline}
This completes the proof of inductive step.
\end{proof}
\section{Deterministic construction of solution}\label{sec:solution}
Assuming the bound~\eqref{eq:bound_deterministic_main} for all $r,s\in\{0,1\}$, $i,m\in\bN_0$, $\uv\in(0,1/2]$, $\sIR\in(0,1]$ with fixed $\oo\in\bN_0$, $\tilde R>1$ we show how to construct a solution of Eq.~\eqref{eq:intro_mild}.
\begin{dfn}
The families of sets $\sV_{\sIR}$, $\sB_{\sIR}$, $\sIR\in(0,1]$, are defined as follows
\begin{equation}
\begin{aligned}
\sV_{\sIR}&:=\{K^{\ast\oo}_\sIR\ast\varphi\in\sV\,|\,\varphi\in\sV\},
\\
\sB_{\sIR}&:=\{K^{\ast\oo}_\sIR\ast\varphi\in\sV\,|\,\varphi\in\sV,\,\|\varphi\|_\sV< \tilde R^2\,[\sIR]^{-\dim(\varPhi)-2\varepsilon}\}.
\end{aligned}
\end{equation}
By Lemma~\ref{lem:kernel_u_v}, if $\varphi\in\sB_{\sIR}$, then $\varphi\in\sB_{\uIR}$ for any $\uIR$ in a sufficiently small neighbourhood of $\sIR$. For $\uv\in(0,1]$, $\sIR\in[0,1]$ and $\lambda\leq \tilde R^{-6}$ we define the functionals
\begin{equation}
F_{\uv,\sIR}\,:\, \sB_{\sIR}\to \cD,
\qquad
\rD F_{\uv,\sIR}\,:\, \sB_{\sIR}\times \sV_{\sIR} \to \cD
\end{equation}
by the following formulas
\begin{equation}\label{eq:deterministic_eff_force_series}
\langle F_{\uv,\sIR}[\varphi],\psi\rangle
:=\sum_{i=0}^\infty \sum_{m=0}^\infty \lambda^i\,\langle F^{i,m}_{\uv,\sIR},\psi\otimes\varphi^{\otimes m}\rangle,
\end{equation}
\begin{equation}\label{eq:deterministic_D_eff_force_series}
\langle \rD F_{\uv,\sIR}[\varphi,\zeta],\psi\rangle
:=\sum_{i=0}^\infty \sum_{m=0}^\infty \lambda^i\,(1+m)\,
\langle F^{i,1+m}_{\uv,\sIR},\psi\otimes\zeta\otimes\varphi^{\otimes m}\rangle,
\end{equation}
where $\psi\in \sS(\bM)$, $\varphi\in\sB_{\sIR}$ and $\zeta\in\sV_{\sIR}$.
\end{dfn}
\begin{lem}\label{lem:functionals}
For $\lambda\leq \tilde R^{-6}$ the functionals $F_{\uv,\sIR}$, $\rD F_{\uv,\sIR}$ are well defined and satisfy the following bounds
\begin{equation}
\begin{gathered}
\|K_\sIR^{\ast\oo}\ast \partial_\uv^r F_{\uv,\sIR}[\varphi]\|_\sV \leq \tilde R\, [\uv]^{(\varepsilon-\sigma)r}\,[\sIR]^{-\dim(\xi)-\varepsilon}
\\
\|K_\sIR^{\ast\oo}\ast \partial_\uv^r \rD F_{\uv,\sIR}[\varphi,\zeta]\|_\sV \leq \tilde R\, [\uv]^{(\varepsilon-\sigma)r}\,[\sIR]^{-\dim(\xi)-\varepsilon}\,\|\fP^\oo_\sIR\zeta\|_\sV
\end{gathered}
\end{equation}
for $r\in\{0,1\}$ and $\uv\in(0,1/2]$, $\sIR\in(0,1]$ and $\varphi\in\sB_\sIR$, $\zeta\in\sV_\sIR$. The functional $\rD F_{\uv,\sIR}$ is the directional derivative of $F_{\uv,\sIR}$. Moreover, given $\uIR\in(0,1]$ and $\varphi\in\sB_\uIR$ the flow equation~\eqref{eq:intro_flow_eq} holds for all $\sIR$ in a sufficiently small neighbourhood of $\uIR$.
\end{lem}
\begin{proof}
The proof is straightforward. For example, using Def.~\ref{dfn:sV} of the space $\sV_m$ and the bounds for the effective force coefficients we obtain
\begin{multline}
\|K_\sIR^{\ast\oo}\ast F_{\uv,\sIR}[K_\sIR^{\ast\oo}\ast\varphi]\|
\\
\leq \sum_{i=0}^\infty\sum_{m=0}^\infty
\frac{\lambda^i \tilde R^{1+2(3i-m)}}{4(1+i)^2\,4(1+m)^{2}}\, [\uv]^{(\varepsilon-\sigma)r}\, [\sIR]^{\varrho_\varepsilon(i,m)} ~\tilde R^{2m}\, [\sIR]^{-m(\dim(\varPhi)+2\varepsilon)}
\end{multline}
for $\varphi\in\sV$ such that $\|\varphi\|_\sV\leq \tilde R^2\,[\sIR]^{-\dim(\varPhi)-2\varepsilon}$, which implies the desired bound. The flow equation~\eqref{eq:intro_flow_eq} follows from the flow equation~\eqref{eq:intro_flow_eq_i_m}.
\end{proof}
\begin{lem}\label{lem:solution}
For $\lambda\leq \tilde R^{-6}$ and $\uv\in(0,1/2]$, $\sIR\in(0,1]$ it holds
\begin{itemize}
\item[(A)] $\|K_\sIR^{\ast\oo}\ast \partial_\uv^r F_{\uv,1}[0]\|_{\sV} \leq \tilde R\,[\uv]^{r(\varepsilon-\sigma)}[\sIR]^{-\dim(\xi)-\varepsilon}$,
\item[(B)] $\|\fP_\sIR^\oo G_{\uv,\sIR}\ast\partial_\uv^r F_{\uv,1}[0]\|_{\sV} \leq 1/3\,\tilde R^2\,[\uv]^{r(\varepsilon-\sigma)}[\sIR]^{-\dim(\varPhi)-2\varepsilon}$,
\item[(C)] $F_{\uv,1}[0] = F_{\uv,\sIR}[G_{\uv,\sIR}\ast F_{\uv,1}[0]]$,
\item[(D)] $\partial_\uv F_{\uv,1}[0]
=
(\partial_\uv F_{\uv,\sIR})[G_{\uv,\sIR}\ast F_{\uv,1}[0]]
+ \rD F_{\uv,\sIR}[G_{\uv,\sIR}\ast F_{\uv,1}[0],\partial_\uv (G_{\uv,\sIR}\ast F_{\uv,1}[0])]$.
\end{itemize}
\end{lem}
\begin{proof}
By Lemma~\ref{lem:functionals} and the equality $G_{\uv,1}=0$ the statement holds true for $\sIR=1$. Chose $\nu\in(0,1]$ and assume that the statement holds true for all $\sIR\in[\nu,1]$. Then by Lemma~\ref{lem:kernel_u_v} we have
\begin{equation}
\|K_{\uIR}^{\ast\oo}\ast \partial_\uv^r F_{\uv,1}[0]\|_{\sV}
\leq
2\, \|K_{\sIR}^{\ast\oo}\ast \partial_\uv^r F_{\uv,1}[0]\|_{\sV}
\leq 2\tilde R\,[\uv]^{r(\varepsilon-\sigma)}[\sIR]^{-\dim(\xi)-\varepsilon}
\end{equation}
for all $\uIR\in[\tau\sIR,1]$, $\sIR\in[\nu,1]$ for some $\tau\in(0,1)$ depending only on $\tilde R$. As a result,
\begin{equation}
\|K_{\sIR}^{\ast\oo}\ast \partial_\uv^r F_{\uv,1}[0]\|_{\sV} \leq 2\tilde R\,[\uv]^{r(\varepsilon-\sigma)}[\sIR]^{-\dim(\xi)-\varepsilon}
\end{equation}
for all $\sIR\in[\tau\nu,1]$. This together with Remark~\ref{rem:tilde_R} and the estimate
\begin{equation}
\|\fP_\sIR^\oo G_{\uv,\sIR}\ast\partial_\uv^r F_{\uv,1}[0]\|_{\sV} \leq
\int_\sIR^1 \|\fP_\uIR^{2\oo}\partial_\uIR G_{\uv,\uIR}\|_\cK\,
\|K_{\uIR}^{\ast\oo}\ast \partial_\uv^r F_{\uv,1}[0]\|_{\sV}
\end{equation}
imply Part (B) for all $\sIR\in[\tau\nu,1]$ and shows that $G_{\uv,\sIR}\ast F_{\uv,1}[0]\in\sB_\sIR$. Using Lemma~\ref{lem:functionals} and the reasoning presented in Sec.~\ref{sec:intro_flow} below Eq.~\eqref{eq:intro_stationary_relation} we prove that Parts (C), (D) hold for all $\sIR\in[\tau\uv,1]$. Part (A) for all $\sIR\in[\tau\uv,1]$ follows now from Lemma~\ref{lem:functionals}. To complete the proof of the theorem it is enough to apply the above reasoning recursively with $\nu=\tau^n$, $n\in\bN_0$.
\end{proof}
\begin{thm}\label{thm:solution}
Fix $R>1$. Assume that for all $i,m\in\bN_+$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\leq0$ there exists $\oo\in\bN_0$ such that
\begin{equation}
\|K^{\ast\oo}_\sIR\ast \partial_\uv^r f^{i,m,a}_{\uv,\sIR}\|_\sV \leq R\,[\uv]^{(\varepsilon-\sigma) r}[\sIR]^{\varrho_\varepsilon(i,m)+|a|}
\end{equation}
for all \mbox{$r\in\{0,1\}$} and $\uv\in(0,1/2]$, $\sIR\in(0,1]$. There exists $\oo\in\bN_0$ and a universal constant $c>1$ such that for $\tilde R=c\,R$, $\lambda<\tilde R^{-6}$ and $\uv\in(0,1/2]$ the function $\varPhi_\uv:=G_\uv\ast F_{\uv,\sIR}[0]$ is well defined, solves Eq.~\eqref{eq:intro_mild} and satisfies the bound
\begin{equation}
\|K_\sIR^{\ast\oo}\ast \partial_\uv^r \varPhi_\uv\|_{\sV} \leq \tilde R^2\, [\uv]^{(\varepsilon-\sigma) r}[\sIR]^{-\dim(\varPhi)-2\varepsilon}
\end{equation}
for all $\uv,\sIR\in(0,1]$.
\end{thm}
\begin{proof}
By Theorems~\ref{thm:deterministic_convergence},~\ref{thm:deterministic_taylor} and~\ref{thm:deterministic_long_range} the assumption implies that there exists $\oo\in\bN_0$ and a universal constant $c>1$ such that the bound~\eqref{eq:bound_deterministic_main} holds true for all $r,s\in\{0,1\}$, $i,m\in\bN_0$ and $\uv\in(0,1/2]$, $\sIR\in(0,1]$ with $\tilde R=c\,R$. The function $\varPhi_\uv$ is well-defined by Lemma~\ref{lem:functionals}. To conclude, we observe that
\begin{equation}
\|K_\sIR^{\ast\oo}\ast \partial_\uv^r \varPhi_\uv\|_{\sV}
\leq
\sum_{u+v=r} \int_0^1 \|\fP^\oo_\uIR \partial_\uv^u \partial_\uIR G_{\uv,\uIR}\|_\cK\,
\|K_\uIR^{\ast\oo}\ast K_\sIR^{\ast\oo}\ast \partial_\uv^v F_{\uv,1}[0]\|_{\sV}\,\rd \uIR,
\end{equation}
use Remark~\ref{rem:tilde_R} and apply Lemma~\ref{lem:solution} with $\sIR$ replaced by $\uIR\vee\sIR$.
\end{proof}
\section{Cumulants of effective force coefficients}\label{sec:cumulants_estimates}
In the remaining part of the paper we show that the assumption of Theorem~\ref{thm:solution} is satisfied for some random $R$ such that $\bE R^{n}<\infty$ for all $n\in\bN_+$. This will follow from the bounds for the joint cumulants of the effective force coefficients $F^{i,m,a}_{\uv,\sIR}$ proved in Sec.~\ref{sec:cumulants_uniform_bounds} and the probabilistic estimates proved in Sec.~\ref{sec:probabilistic}. The bounds for the joint cumulants of $F^{i,m,a}_{\uv,\sIR}$ involve the norm $\|\Cdot\|_{\sV^{\mathsf{m}}}$ introduced in Sec.~\ref{sec:topology_cumulants}. We prove the above-mentioned bounds using a certain flow equation given in Sec.~\ref{sec:flow_equation_cumulants}. We consider joint cumulants of $F^{i,m,a}_{\uv,\sIR}$ instead of $f^{i,m,a}_{\uv,\sIR}$ because there is no flow equation for the joint cumulants of $f^{i,m,a}_{\uv,\sIR}$.
\begin{dfn}
Let $p\in\bN_+$, $I=\{1,\ldots,p\}$ and $\zeta_q$, $q\in I$, be random variables. The joint cumulant of $(\zeta_q)_{q\in I}$ is defined by the formula
\begin{equation}
\llangle \zeta_1,\ldots,\zeta_p\rrangle
\equiv
\llangle(\zeta_q)_{q\in I}\rrangle
=
(-\ri)^p \partial_{t_1}\ldots\partial_{t_p}\log\bE \exp(\ri t_1 \zeta_1+\ldots+\ri t_p \zeta_p) \big|_{t_1=\ldots=t_p=0}.
\end{equation}
In particular, $\llangle \zeta\rrangle\equiv\bE\zeta$, $\llangle \zeta_1,\zeta_2\rrangle=\llangle \zeta_1\zeta_2\rrangle-\llangle \zeta_1\rrangle\llangle \zeta_2\rrangle$. The joint cumulant $\llangle(\zeta_q)_{q\in I}\rrangle$ does not depend on the order of elements in the list $(\zeta_q)_{q\in I}$.
\end{dfn}
\begin{lem}\label{lem:cumulants}
Let $p\in\bN_+$, $I=\{1,\ldots,p\}$ and $\zeta_1,\ldots,\zeta_p,\Phi,\Psi$ be random variables. It holds
\begin{equation}\label{eq:expectation_cumulants}
\llangle\zeta_1\ldots\zeta_p\rrangle
=\sum_{r=1}^p
\sum_{\substack{I_1,\ldots,I_r\subset I,\\I_1\cup\ldots\cup I_r=I\\I_1,\ldots,I_r\neq \emptyset}}
\llangle(\zeta_q)_{q\in I_1}\rrangle
\ldots
\llangle(\zeta_q)_{q\in I_r}\rrangle,
\end{equation}
\begin{equation}\label{eq:cumulants_product}
\llangle(\zeta_q)_{q\in I},\Phi\Psi\rrangle
=
\llangle(\zeta_q)_{q\in I},\Phi,\Psi\rrangle
+
\sum_{\substack{I_1,I_2\subset I\\I_1\cup I_2= I}}
\llangle(\zeta_q)_{q\in I_1},\Phi\rrangle
~
\llangle(\zeta_q)_{q\in I_2},\Psi\rrangle.
\end{equation}
\end{lem}
\begin{rem}
For the proof of the above lemma see Proposition~3.2.1 in~\cite{peccati2011wiener}.
\end{rem}
\begin{dfn}\label{dfn:notation_cumulants_distributions}
Given $n\in\bN_+$, $I=\{1,\ldots,n\}$, $m_1,\ldots,m_n\in\bN_0$ and random distributions \mbox{$\zeta_q\in\sS'(\bM^{1+m}_q)$}, $q\in I$, we define the deterministic distribution
$\llangle (\zeta_q)_{q\in I}\rrangle
\equiv
\llangle \zeta_1,\ldots,\zeta_n\rrangle
\in
\sS'(\bM^n\times\bM^{m_1+\ldots+m_n})$ by the equality
\begin{equation}
\langle\llangle \zeta_1,\ldots,\zeta_n\rrangle,\psi_1\otimes\ldots\otimes\psi_n\otimes\varphi_1\otimes\ldots\otimes\varphi_n\rangle
:=
\llangle
\langle\zeta_1,\psi_1\otimes\varphi_1\rangle,\ldots,\langle\zeta_n,\psi_n\otimes\varphi_n\rangle\rrangle,
\end{equation}
where $\psi_q\in \sS(\bM)$, $\varphi_q\in \sS(\bM^{m_q})$, $q\in I$, are arbitrary.
\end{dfn}
\begin{dfn}\label{dfn:cumulants_eff_force}
A list $(i,m,a,s,r)$, where $i,m\in\bN_0$, $a\in\frM^m$ and $s\in\{0,1\}$, $r\in\{0,1,2\}$ is called an index. Let $n\in\bN_+$ and
\begin{equation}\label{eq:list_indices}
\vI\equiv ((i_1,m_1,a_1,s_1,r_1),\ldots,(i_n,m_n,a_n,s_n,r_n))
\end{equation}
be a list of indices. We set $\rn(\vI):=n$, $\rri(\vI):=i_1+\ldots+i_n$, $\mathsf{m}(\vI):=(m_1,\ldots,m_n)$, $\rrm(\vI):=m_1+\ldots+m_n$, $\ra(\vI):=|a_1|+\ldots+|a_n|$, $\rs(\vI):=s_1+\ldots+s_n$ and $\rr(\vI):=r_1+\ldots+r_n$. We use the following notation for the joint cumulants of the effective force coefficients
\begin{equation}
E^\vI_{\uv,\sIR}:=
\llangle \partial_\sIR^{s_1}\partial_\uv^{r_1} F^{i_1,m_1,a_1}_{\uv,\sIR} ,\ldots,
\partial_\sIR^{s_n}\partial_\uv^{r_n} F^{i_n,m_n,a_n}_{\uv,\sIR}\rrangle\in \sS'(\bM^{\rn(\vI)}\times\bM^{\rrm(\vI)}).
\end{equation}
\end{dfn}
\begin{dfn}\label{dfn:varrho_I}
For $\varepsilon\geq0$ and a list of indices $\vI$ of the form~\eqref{eq:list_indices} we define
\begin{equation}
\varrho_\varepsilon(\vI) := \varrho_\varepsilon(i_1,m_1)+|a_1| +\ldots +\varrho_\varepsilon(i_n,m_n)+|a_n|\in\bR.
\end{equation}
We also set $\varrho(\vI):=\varrho_0(\vI)$. The cumulant $E^\vI_{\uv,\sIR}$ such that $\varrho(\vI)+(n-1)\rdim\leq 0$ are called relevant. The remaining cumulants are called irrelevant.
\end{dfn}
\begin{rem}
If $\ri(\vI)=0$, then either $E^\vI_{\uv,\sIR}=0$ or $\rn(\vI)=2$, $\rrm(\vI)=0$, $\ra(\vI)=0$, $\rs(\vI)=0$, $\rr(\vI)=0$. In the later case $E^\vI_{\uv,\sIR}$ is relevant and coincides with the covariance of the white noise. For $\ri(\vI)>0$ the only relevant cumulants are the expectations of the relevant force coefficients.
\end{rem}
\begin{rem}\label{rem:rho2}
For $\varepsilon>0$ and any list of indices $\vI$ such that $\rrm(\vI)\leq 3\rri(\vI)$ it holds $\varrho_\varepsilon(\vI)<\varrho(\vI)$. Moreover, \mbox{$\varrho_\varepsilon(\vI)+(\rn(\vI)-1)\rdim>0$} for $\varepsilon\in(0,\varepsilon_\diamond)$ and lists of indices $\vI$ such that $\varrho(\vI)+(\rn(\vI)-1)\rdim>0$, where $\varepsilon_\diamond$ was introduced in Remark~\ref{rem:rho}. Recall that $\varepsilon\in(0,\varepsilon_\diamond/3)$ is fixed (see Remark~\ref{rem:rho}).
\end{rem}
\section{Probabilistic analysis}\label{sec:probabilistic}
\begin{thm}\label{thm:probabilistic_bounds}
Fix $n\in2\bN_+$ such that $\rdim/n<\varepsilon$ and $i,m\in\bN_0$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\leq0$. For $s\in\{0,1\}$, $r\in\{0,1,2\}$ we define the list of indices $\vI=\vI(s,r)=((i,m,a,s,r),\ldots,(i,m,a,s,r))$, $\rn(\vI)=n$. Assume that there exists $\oo\in\bN_0$ such that for $s\in\{0,1\}$, $r\in\{0,1,2\}$ the following bound
\begin{multline}\label{eq:probabilistic_thm_assumption}
\int_{\bT^{n-1}}[K^{\ast\oo,\otimes(n+nm)}_\sIR\ast E^\vI_{\uv,\sIR}](x_1,\ldots,x_n)\,\rd x_2\ldots\rd x_n
\\
\lesssim [\uv]^{(\varepsilon-\sigma)\rr(\vI)}
[\sIR]^{\varrho_{3\varepsilon}(\vI)-\sigma\rs(\vI)+(n-1)\rdim}
\end{multline}
holds uniformly in $x_1\in\bM$, $\uv,\sIR\in(0,1/2]$, where
\begin{multline}
[K^{\ast\oo,\otimes(n+nm)}_\sIR\ast E^\vI_{\uv,\sIR}](x_1,\ldots,x_n)\\
:=\int_{\bM^{nm}} |(K^{\ast\oo,\otimes(n+nm)}_\sIR\ast E^\vI_{\uv,\sIR})(x_1,\ldots,x_n;\ry_1,\ldots,\ry_n)|\, \rd \ry_1\ldots \rd \ry_n.
\end{multline}
Then there exists $\oo\in\bN_0$ and a random variable $R>0$ such that $\bE R^n<\infty$ and for $r\in\{0,1\}$ the following bound
\begin{equation}
\|
K_\sIR^{\ast\oo}
\ast \partial_\uv^r f^{i,m,a}_{\uv,\sIR}\|_{\sV} \leq R\,[\uv]^{(\varepsilon-\sigma)r}\,[\sIR]^{\varrho_\varepsilon(i,m)+|a|}.
\end{equation}
holds uniformly in $\uv,\sIR\in(0,1/2]$.
\end{thm}
\begin{rem}
The function $[K^{\ast\oo,\otimes(n+nm)}_\sIR\ast E^\vI_{\uv,\sIR}]$ is $2\pi$ periodic in all variables and the integral on the LHS of the bound~\eqref{eq:probabilistic_thm_assumption} is an integral over one period.
\end{rem}
\begin{rem}
The assumption of the above theorem is verified in Theorem~\ref{thm:cumulants}.
\end{rem}
\begin{proof}
By Remark~\ref{rem:periodic} and Lemma~\ref{lem:kernel_simple_fact} (B), which says that \mbox{$\|\fT K^{\ast\rdim}_\sIR\|_{\sV}\lesssim [\sIR]^{-\rdim}$}, the assumption of the theorem implies that for $\ooo=\oo+\rdim$ the following bound
\begin{equation}\label{eq:probabilistic_thm_bound}
[K^{\ast\ooo,\otimes(n+nm)}_\sIR\ast E^\vI_{\uv,\sIR}](x_1,\ldots,x_n)
\lesssim [\uv]^{(\varepsilon-\sigma)\rr(\vI)}
[\sIR]^{\varrho_{3\varepsilon}(\vI)-\sigma\rs(\vI)}
\end{equation}
holds uniformly in $x_1,\ldots,x_n\in\bM$, $\uv,\sIR\in(0,1/2]$. Using the above bound with $x_1=\ldots=x_n=x$, the relation between the expectation of a product of random variables and their joint cumulants given by Eq.~\eqref{eq:expectation_cumulants}, the equalities $\varrho_{3\varepsilon}(\vI)=n(\varrho_{3\varepsilon}(i,m)+|a|)$, $\rs(\vI)=n s$, $\rr(\vI)=nr$ and
\begin{equation}
(K^{\ast\ooo}_\sIR \ast \partial_\sIR^s\partial_\uv^r f^{i,m,a}_{\uv,\sIR})(x)=\int_{\bM^m}\!(K^{\ast\ooo,\otimes(1+m)}_\sIR \ast \partial_\sIR^s\partial_\uv^r F^{i,m,a}_{\uv,\sIR})(x;y_1,\ldots,y_m)\,\rd y_1\ldots\rd y_m
\end{equation}
we obtain for $r\in\{0,1,2\}$, $s\in\{0,1\}$ the following bound
\begin{equation}
\sup_{x\in\bM} \bE (K^{\ast\ooo}_\sIR \ast \partial_\sIR^s\partial_\uv^r f^{i,m,a}_{\uv,\sIR}(x))^n
\lesssim [\uv]^{n(\varepsilon-\sigma)r}\,[\sIR]^{n(\varrho_{3\varepsilon}(i,m)+|a|-\sigma s)}
\end{equation}
uniform in $x\in\bM$, $\uv,\sIR\in(0,1/2]$. Lemma~\ref{lem:expectation_sup} and Lemma~\ref{lem:kernel_simple_fact} (C) imply
\begin{equation}
\bE \|\partial_\sIR^s\partial_\uv^r (K^{\ast\oooo}_\sIR \ast f^{i,m,a}_{\uv,\sIR})\|^n_{\sV}
\lesssim [\uv]^{n(\varepsilon-\sigma)r}\,[\sIR]^{n(\varrho_{3\varepsilon}(i,m)+|a|-\sigma s-\varepsilon)}
\end{equation}
with $\oooo=\ooo+\rdim$. The theorem follows now from Lemma~\ref{lem:probabilistic_estimate} applied with
\begin{equation}
\zeta_{2\uv,2\sIR}= K^{\ast\oooo}_\sIR \ast \partial_\uv^r f^{i,m,a}_{\uv,\sIR},\qquad r\in\{0,1\},
\end{equation}
and $-\omega=(\varepsilon-\sigma)r$ and $-\rho=\varrho_{3\varepsilon}(i,m)+|a|-2\varepsilon \geq \varrho_{\varepsilon}(i,m)+|a|$, where the last inequality holds for $m\leq3i$.
\end{proof}
\begin{rem}
The advantage of the bound of the form~\eqref{eq:probabilistic_thm_assumption} over~\eqref{eq:probabilistic_thm_bound} is that the former bound contains the extra factor $[\sIR]^{(n-1)\rdim}$ on the RHS, and consequently can be more easily proved by induction using the flow equation for cumulants stated in Sec.~\ref{sec:flow_equation_cumulants} and the equality
$E^\vI_{\uv,\sIR}=E^\vI_{\uv,0} + \int_0^\sIR \partial_\uIR E^{\vI}_{\uv,\uIR}\,\rd\uIR$.
\end{rem}
\begin{lem}\label{lem:expectation_sup}
Let $n\in2\bN_+$. There exists a constant $C>0$ such that for all random fields $\zeta\in L^n(\bT)$ and $\sIR\in(0,1/2]$ it holds
\begin{equation}
\bE
\|K_\sIR^{\ast\rdim} \ast \zeta\|^n_{L^\infty(\bT)}
\leq C\, [\sIR]^{-\rdim}\,
\bE \|\zeta\|^n_{L^n(\bT)}.
\end{equation}
\end{lem}
\begin{proof}
Note that $K_\sIR^{\ast\rdim} \ast \zeta = \fT K_\sIR^{\ast\rdim} \star \zeta$, where $\star$ is the convolution in $\bT$ and $\fT K_\sIR^{\ast\rdim}$ is the periodization of $K_\sIR^{\ast\rdim}$ (see Def.~\ref{dfn:periodization}). Using the Young inequality for convolutions we obtain
\begin{equation}
\bE\|K_\sIR^{\ast\rdim}\ast \zeta\|^n_{L^\infty(\bT)}
\leq
\|\fT K_\sIR^{\ast\rdim}\|_{L^{n/(n-1)}(\bT)}^n\,
\bE\|\zeta\|^n_{L^n(\bT)}.
\end{equation}
The lemma follows now from Lemma~\ref{lem:kernel_simple_fact}~(B).
\end{proof}
\begin{lem}\label{lem:probabilistic_estimate}
Fix $n\in\bN_+$. There exists a universal constant $c>0$ such that if
\begin{equation}
\bE \|\partial_\sIR^s\partial_\uv^{u} \zeta_{\uv,\sIR}^{\phantom{l}}\|^n_{L^\infty(\bT)}
\\
\leq C\,[\uv]^{-n(\omega + \sigma u-\varepsilon u)} [\sIR]^{-n(\rho+\sigma s-\varepsilon s)}
\end{equation}
for some differentiable random function $\zeta:(0,1]^2\to L^\infty(\bT)$, some $C>0$, $\omega,\rho\leq0$ and all $u,s\in\{0,1\}$ and $\uv,\sIR\in(0,1]$, then
\begin{equation}
\bE\Big( \sup_{\uv,\sIR\in(0,1]}
[\uv]^{n\omega}[\sIR]^{n\rho}\,\|\zeta_{\uv,\sIR}\|^n_{L^\infty(\bT)}\Big) \leq c\,C.
\end{equation}
\end{lem}
\begin{proof}
For all $\uv,\sIR\in(0,1]$ we have
\begin{multline}
[\uv]^{\omega} [\sIR]^{\rho}\,\|\zeta_{\uv,\sIR}\|_{L^\infty(\bT)}
\leq
\|\zeta_{1,1}\|_{L^\infty(\bT)}
\\
+
\int_\sIR^1 [\uIR]^{\rho}\,\|\partial_\uIR\zeta_{1,\uIR}\|_{L^\infty(\bT)}\,\rd\uIR
+
\int_\sIR^1 \int_\uv^1 [\nu]^{\omega}[\uIR]^{\rho}\,\|\partial_\uIR\partial_\nu \zeta_{\nu,\uIR}\|_{L^\infty(\bT)} \,\rd\nu\rd\uIR
\end{multline}
By the Minkowski inequality we get the bound
\begin{multline}\label{eq:probabilistic_proof_ieq}
\bE\Big( \sup_{\uv,\sIR\in(0,1]}
[\uv]^{n\omega}[\sIR]^{n\rho}\,\|\zeta_{\uv,\sIR}\|^n_{L^\infty(\bT)}\Big)^{\!\frac{1}{n}}
\leq
\bE\Big(\|\zeta_{1,1}\|^n_{L^\infty(\bT)}\Big)^{\!\frac{1}{n}}
\\
+
\int_0^1 [\sIR]^\rho \,\bE\Big( \|\partial_\sIR \zeta_{1,\sIR}\|^n_{L^\infty(\bT)}\Big)^{\!\frac{1}{n}}\,\rd\sIR
+
\int_0^1\int_0^1 [\uv]^\omega[\sIR]^\rho\, \bE\Big(\|\partial_\sIR\partial_\uv \zeta_{\uv,\sIR}\|^n_{L^\infty(\bT)}\Big)^{\!\frac{1}{n}} \,\rd\uv\rd\sIR,
\end{multline}
which implies the statement.
\end{proof}
\section{Function spaces for cumulants}\label{sec:topology_cumulants}
\begin{dfn}\label{dfn:sVm}
Let $n\in\bN_+$, \mbox{$\mathsf{m}=(m_1,\ldots,m_n)\in\bN_0^n$} and $m=m_1+\ldots+m_n$. The vector space $\sV^{\mathsf{m}}$ consists of functions $V\in C(\bM^n\times\bM^m)$ such that
\begin{equation}
\|V\|_{\sV^{\mathsf{m}}}
:=
\\
\sup_{x_1\in\bM} \int_{\bT^{n-1}}
[V](x_1,\ldots,x_n)\,\rd x_2\ldots\rd x_n,
\end{equation}
where
\begin{equation}
[V](x_1,\ldots,x_n):=\int |V(x_1,\ldots,x_n;y_1,\ldots,y_m)|\,\rd y_1\ldots\rd y_m,
\end{equation}
is finite and the function $\fU^{\mathsf{m}} V\in C(\bM^n\times\bM^m)$ defined by
\begin{equation}
(\fU^{\mathsf{m}} V)(x_1,\ldots,x_n;y_1,\ldots,y_m):=V(x_1,\ldots,x_n;\mathrm y_1+x_1,\ldots,\mathrm y_n+x_n)
\end{equation}
is $2\pi$ periodic in variables $x_1,\ldots,x_n$ for every
\begin{equation}
(y_1,\ldots,y_m)=(\mathrm y_1,\ldots,\mathrm y_n)\in\bM^{m_1}\times\ldots\times\bM^{m_n}=\bM^m,
\end{equation}
where for arbitrary $x\in\bM$ and $\ry=(y_1,\ldots,y_m)\in\bM^m$, $m\in\bN_0$, we use the following notation $\ry+x:=(y_1+x,\ldots,y_n+x)\in\bM^m$. For $\oo\in\bN_0$ the space $\cD^{\mathsf{m}}_\oo$ consists of distributions $V\in \sS'(\bM^n\times\bM^m)$ such that $K^{\ast\oo,\otimes (n+m)}_\sIR\ast V\in\sV^{\mathsf{m}}$. The space $\cD^{\mathsf{m}}$ is the union of the spaces $\cD^{\mathsf{m}}_\oo$, $\oo\in\bN_0$.
\end{dfn}
\begin{rem}
For $V\in\sV^{\mathsf{m}}$, the function $[V]$ is $2\pi$ periodic. Note also that for $n=1$ and $\mathsf{m}=m$ we have $\sV^{\mathsf{m}}=\sV^m$, $\cD^{\mathsf{m}}=\cD^m$, $\cD^{\mathsf{m}}_\oo=\cD^m_\oo$, where the spaces $\sV^m$, $\cD^m$, $\cD^m_\oo$ were introduced in Def.~\ref{dfn:sV}.
\end{rem}
\begin{dfn}\label{dfn:map_Ym}
For $V\in\cD^{\mathsf{m}}$ and $\pi\in\cP_{m_1}$ we define $\fY_\pi V\in\cD^{\mathsf{m}}$ by
\begin{equation}
\big\langle \fY_\pi V,\botimes_{q=1}^n \psi_q\otimes
\botimes_{q=1}^m \varphi_q\big\rangle
:=
\big\langle V,\botimes_{q=1}^n \psi_q\otimes
\botimes_{q=1}^{m_1} \varphi_{\pi(q)}\otimes
\botimes_{q=m_1+1}^{m} \varphi_q \big\rangle,
\end{equation}
where $\psi_1,\ldots,\psi_n,\varphi_1,\ldots,\varphi_m\in\ \sS(\bM)$. For $V\in\cD^{\mathsf{m}}$ and $\omega\in\cP_{n}$ we define $\fY^\omega V\in\cD^{\omega(\mathsf{m})}$, where $\omega(\mathsf{m}):=(m_{\omega(1)},\ldots,m_{\omega(n)})$, by
\begin{equation}
\big\langle \fY^\omega V,\botimes_{q=1}^n \psi_q\otimes
\botimes_{q=1}^n \varphi_q\big\rangle
:=
\big\langle V,\botimes_{q=1}^n \psi_{\omega(q)}\otimes
\botimes_{q=1}^{n} \varphi_{\omega(q)}\big\rangle,
\end{equation}
where and $\psi_q\in\ \sS(\bM)$, $\varphi_q\in \sS(\bM^{m_q})$ for $q\in\{1,\ldots,n\}$.
\end{dfn}
\begin{rem}
Recall that $\cP_n$ is the permutation group. The map $\fY_\pi:\sV^{\mathsf{m}}\to\sV^{\mathsf{m}}$ is well defined and bounded with norm one. The map $\fY^\omega:\sV^{\mathsf{m}}\to\sV^{\mathsf{m}}$ is also well-defined but in general not bounded because of the distinguished role of the variable $x_1$ in Def.~\ref{dfn:sVm} of the norm $\|\Cdot\|_{\sV^{\mathsf{m}}}$.
\end{rem}
\begin{dfn}
Let $n\in\bN_+$ and $V\in\sS'(\bM^n)$. We say that $V$ is translationally invariant iff
$
\langle V,\psi_1\otimes\ldots\otimes\psi_n\rangle
=
\langle V,\psi_1(\Cdot+x)\otimes\ldots\otimes\psi_n(\Cdot+x)\rangle
$
for any $x\in\bM$ and $\psi_1,\ldots,\psi_n\in \sS(\bM)$.
\end{dfn}
\begin{rem}\label{rem:translational_permutation}
$\|\fY^\omega V\|_{\sV^{\mathsf{m}}}\leq \|V\|_{\sV^{\mathsf{m}}}$ for translationally-invariant $V\in\sV^{\mathsf{m}}$ since
\begin{equation}
\|V\|_{\sV^{\mathsf{m}}}
=
(2\pi)^{-\rdim} \int_{\bT^n}
[V](x_1,\ldots,x_n)\,\rd x_1\ldots\rd x_n.
\end{equation}
\end{rem}
\begin{dfn}\label{dfn:maps_A_B}
Fix $n\in\bN_0$, $\hat n\in\{1,\ldots,n\}$, $m_1,\ldots,m_{n+1}\in\bN_0$ and
\begin{equation}
\begin{gathered}
\mathsf{m}=(m_1+m_{n+1},m_2,\ldots,m_n)\in\bN_0^n,
\qquad
\tilde{\mathsf{m}}=(m_1+1,m_2,\ldots,m_{n+1})\in\bN_0^{n+1},
\\
\hat{\mathsf{m}}=(m_1+1,m_2,\ldots,m_{\hat n})\in\bN_0^{\hat n},
\qquad
\check{\mathsf{m}}=(m_{\hat n+1},\ldots,m_{n+1})\in\bN_0^{n-\hat n+1}.
\end{gathered}
\end{equation}
The bilinear map $\fA\,:\,\sS(\bM) \times\sV^{\tilde{\mathsf{m}}}\to\sV^{\mathsf{m}}$ is defined by
\begin{multline}\label{eq:fA_dfn}
\fA(G,V)(x_1,\ldots,x_n;\ry_1,\ry_{n+1},\ry_2,\ldots,\ry_n)
\\
:=
\int_{\bM^2} V(x_1,\ldots,x_{n+1}; y,\ry_1\ldots\ry_{n+1})\,G(y-x_{n+1})\,\rd y\rd x_{n+1}.
\end{multline}
The trilinear map $\fB\,:\,\sS(\bM)\times \sV^{\hat{\mathsf{m}}}\times\sV^{\check{\mathsf{m}}}\to\sV^{\mathsf{m}}$ is defined by
\begin{multline}\label{eq:fB_dfn}
\fB(G,W,U)(x_1,\ldots,x_{n};\ry_1,\ry_{n+1},\ry_2,\ldots,\ry_{n})
:=
\int_{\bM^2} W(x_1,\ldots,x_{\hat n};y,\ry_1,\ldots,\ry_{\hat n})
\\
\times\,G(y-x_{n+1})\,
U(x_{\hat n+1},\ldots,x_{n+1};\ry_{\hat n+1},\ldots,\ry_{n+1})\,\rd y\rd x_{n+1}.
\end{multline}
In the above equations $\ry_j\in\bM^{m_j}$, $j\in\{1,\ldots,n+1\}$.
\end{dfn}
\begin{rem}
The map $\fB$ is a generalization of the map $\fB$ introduced in Def.~\ref{dfn:map_B}. The maps $\fA$ and $\fB$ appear on the RHS of the flow equation for the cumulants of the effective force coefficients introduced in Sec.~\ref{sec:flow_equation_cumulants}.
\end{rem}
\begin{lem}\label{lem:fA_fB_bounds}
If $V$, $W$ and $U$ are translationally invariant, then $\fA(G,V)$ and $\fB(G,W,U)$ are also translationally invariant and it holds
\begin{equation}\label{eq:fA_ieq}
\|\fA(G,V)\|_{\sV^{\mathsf{m}}}
\leq
\|\fT G\|_{\sV}\,
\|V\|_{\sV^{\tilde{\mathsf{m}}}},
\end{equation}
\begin{equation}\label{eq:fB_ieq}
\|\fB(G,W,U)\|_{\sV^{\mathsf{m}}}
\leq
\|G\|_\cK\,
\|W\|_{\sV^{\check{\mathsf{m}}}}\,
\|U\|_{\sV^{\hat{\mathsf{m}}}},
\end{equation}
where $\fT G$ is the periodization of $G$ introduced in Def.~\ref{dfn:periodization}.
\end{lem}
\begin{proof}
The translational invariance follows easily from the definition of the maps $\fA$ and $\fB$. Using Def.~\ref{dfn:sVm}, Def.~\ref{dfn:periodization} and Remark~\ref{rem:periodic} we obtain:
\begin{equation}\label{eq:proof_fA_fB}
\begin{gathered}
[\fA(G,V)](x_1,\ldots,x_n)
\leq
\|\fT G\|_{L^\infty(\bM)}\,
\,\int_\bT [V](x_1,\ldots,x_n,x)\,\rd x,
\\
[\fB(G,W,U)](x_1,\ldots,x_n)
\leq
\|G\|_{L^1(\bM)}
\,[W](x_1,\ldots,x_{\hat n})\,
\sup_{x\in\bM}\,[U](x_{\hat n+1},\ldots,x_n,x).
\end{gathered}
\end{equation}
The statement follows now from Remark~\ref{rem:translational_permutation}.
\end{proof}
\begin{rem}\label{lem:fA_fB_Ks}
The fact that $P_\sIR^\oo K_\sIR^{\ast\oo}=\delta_\bM$ implies that for any $\sIR>0$ it holds
\begin{equation}
\begin{gathered}
K_\sIR^{\ast\oo,\otimes(n+m)}\ast\fA(G,V)=
\fA\big(\fP^{2\oo}_\sIR G,
K_\sIR^{\ast\oo,\otimes (n+m+2)}\ast V\big),
\\
K_\sIR^{\ast\oo,\otimes(n+m)}\ast\fB(G,W,U)=
\fB\big(\fP^{2\oo}_\sIR G,
K_\sIR^{\ast\oo,\otimes(\hat n+\hat m+1)}\ast W,
K_\sIR^{\ast\oo,\otimes(n-\hat n+\check m+1)}\ast U\big),
\end{gathered}
\end{equation}
where \mbox{$m=m_1+\ldots+m_{n+1}$}, \mbox{$\hat m=m_1+\ldots+m_{\hat n}$} and \mbox{$\check m=m_{\hat n+1}+\ldots+m_{n+1}$}. Note also that by Lemma~\ref{lem:kernel_simple_fact} (B)
\begin{equation}
\|\fT \fP^{2\oo}_\sIR G\|_\sV\leq \|\fT K_\sIR^{\ast\rdim}\|_\sV \|\fP^{2\oo+\rdim}_\sIR G\|_\cK\lesssim [\sIR]^{-\rdim}\,\|\fP^{2\oo+\rdim}_\sIR G\|_\cK
\end{equation}
uniformly in $\sIR\in(0,1]$.
\end{rem}
\section{Flow equation for cumulants}\label{sec:flow_equation_cumulants}
\begin{lem}\label{lem:flow_E_general}
Let $n\in\bN_+$, $i_1\in\bN_0$, $m_1,\ldots,m_n\in\bN_0$, $a_1\in\frM^{m_1}$, $r_1\in\{0,1,2\}$ and $I\equiv \{2,\ldots,n\}$. For any random distributions $\zeta_q\in\cD^{m_q}$, $q\in I$, the following flow equation holds
\begin{multline}\label{eq:flow_E_general}
\llangle
\partial_\sIR \partial_\uv^{r_1} F^{i_1,m_1,a_1}_{\uv,\sIR},
(\zeta_q)_{q\in I}\rrangle
=
-\frac{1}{m_1!}
\sum_{\pi\in\cP_{m_1}}\sum_{j=0}^{i_1}\sum_{k=0}^{m_1}\sum_{u+v+w=r_1}
\sum_{b+c+d=\pi(a_1)}(1+k)\,
\\\times\Bigg(
\sum_{\substack{I_1,I_2\subset I\\I_1\cup I_2= I}}
\fY_\pi\fB\big(\cX^c\partial_\uv^u\partial_\sIR G_{\uv,\sIR},
\llangle
\partial_\uv^v F^{j,1+k,b}_{\uv,\sIR},(\zeta_q)_{q\in I_1}\rrangle
,\llangle
(\zeta_q)_{q\in I_2},\partial_\uv^{w} F^{i_1-j,m_1-k,d}_{\uv,\sIR}
\rrangle\big)
\\+
\fY_\pi\fA\big(\cX^c\partial_\uv^u\partial_\sIR G_{\uv,\sIR},
\llangle
\partial_\uv^v F^{j,1+k,b}_{\uv,\sIR},
(\zeta_q)_{q\in I},
\partial_\uv^w F^{i_1-j,m_1-k,d}_{\uv,\sIR}
\rrangle\big)
\Bigg),
\end{multline}
where we used the notation introduced in Def.~\ref{dfn:notation_cumulants_distributions}.
\end{lem}
\begin{proof}
The statement follows immediately from Eqs.~\eqref{eq:flow_deterministic_i_m_a}, \eqref{eq:cumulants_product}.
\end{proof}
\begin{lem}\label{lem:flow_E_form_bound}
Let $\vJ\equiv (\vJ_1,\ldots,\vJ_n)=((i_1,m_1,a_1,s_1,r_1),\ldots,(i_n,m_n,a_n,s_n,r_n))$ be a list of indices such that $s_l=1$ for some $l\in\{1,\ldots,n\}$.
\begin{enumerate}
\item[(A)]
The distribution $E^\vJ_{\uv,\sIR}$ can be expressed as a linear combination of distributions of the form
\begin{equation}
\fY^\omega\fY_\pi\fA\big(\cX^c\partial_\uv^u\partial_\sIR G_{\uv,\sIR},E^{\vK}_{\uv,\sIR}\big)
\qquad
\textrm{or}
\qquad
\fY^\omega\fY_\pi\fB\big(\cX^c\partial_\uv^u\partial_\sIR G_{\uv,\sIR},
E^{\vL}_{\uv,\sIR},
E^{\vM}_{\uv,\sIR}
\big),
\end{equation}
where $u\in\{0,\ldots,r_l\}$, $c\in\frM$ is some multi-index, $\vK$, $\vL$, $\vM$ are some lists of indices and $\omega\in\cP_n$, $\pi\in\cP_{m_l}$ are some permutation.
\item[(B)] The lists of indices $\vK$, $\vL$, $\vM$ satisfy the following conditions
\begin{equation}
\begin{split}
&\rn(\vK)=\rn(\vJ)+1,\\
&\rri(\vK)=\rri(\vJ),\\
&\rrm(\vK)=\rrm(\vJ)+1,\\
&\ra(\vK)+|c|=\ra(\vJ),\\
&\rs(\vK)=\rs(\vJ)-1,\\
&\rr(\vK)=\rr(\vJ)-r,\\
&\varrho_{3\varepsilon}(\vJ) -\sigma = \varrho_{3\varepsilon}(\vK)-\rdim -3\varepsilon,
\end{split}
~\textrm{or}\qquad
\begin{split}
&\rn(\vL)+\rn(\vM)=\rn(\vJ)+1,\\
&\rri(\vL)+\rri(\vM)=\rri(\vJ),\\
&\rrm(\vL)+\rrm(\vM)=\rrm(\vJ)+1,\\
& \ra(\vL)+\ra(\vM)+|c|=\ra(\vJ),\\
&\rs(\vL)+\rs(\vM)=\rs(\vJ)-1,\\
&\rr(\vL)+\rr(\vM)=\rr(\vJ)-r,\\
&\varrho_{3\varepsilon}(\vJ) -\sigma = \varrho_{3\varepsilon}(\vL)+\varrho_{3\varepsilon}(\vM)-3\varepsilon.
\end{split}
\end{equation}
\item[(C)] Fix some $\oo\in\bN_0$. Suppose that the bound
\begin{equation}
\big\|K^{\ast\oo,\otimes(\rn(\vI)+\rrm(\vI))}_\sIR\ast E^\vI_{\uv,\sIR}\big\|_{\sV^{\mathsf{m}(\vI)}}
\\
\lesssim [\uv]^{(\varepsilon-\sigma)\rr(\vI)}
[\sIR]^{\varrho_{3\varepsilon}(\vI)-\sigma\rs(\vI)+(\rn(\vI)-1)\rdim}
\end{equation}
holds uniformly in $\uv,\sIR\in(0,1/2]$ for all lists of indices \mbox{$\vI\in\{\vK,\vL,\vM\}$}, where $\vK,\vL,\vM$ are arbitrary lists of indices satisfying the conditions specified in Part~(B) given the list of indices $\vJ$. Then the above bound holds uniformly in $\uv,\sIR\in(0,1/2]$ for $\vI=\vJ$.
\end{enumerate}
\end{lem}
\begin{proof}
Without loss of generality we can assume that $l=1$. Part (A) of the theorem follows immediately from Lemma~\ref{lem:flow_E_general} applied with
\begin{equation}
\zeta_q\equiv \partial_\sIR^{s_q}\partial_\uv^{r_q} F^{i_q,m_q,a_q}_{\uv,\sIR},
\qquad q\in\{2,\ldots,n\}.
\end{equation}
The multi-index $c\in\frM$, $u\in\{0,\ldots,r_1\}$ and the permutation $\pi\in\cP_{m_1}$ in the statement coincide with the respective objects in Eq.~\eqref{eq:flow_E_general}. The permutation $\omega\in\cP_n$ is trivial because of the assumption $l=1$. Moreover, it holds
\begin{equation}
\begin{gathered}
\vK=((j,1+k,b,0,v),\,\vJ_2,\ldots,\vJ_n,\,(i_1-j,m_1-k,d,0,w))
\\
\vL\sqcup\vM = ((j,1+k,b,0,v),\,\vJ_2,\ldots \vJ_n,\,(i_1-j,m_1-k,d,0,w)),
\end{gathered}
\end{equation}
where $\vL\sqcup\vM$ denotes the concatenation of the lists $\vL$, $\vM$ and $j\in\{1,\ldots,i_1\}$, $k\in\{0,\ldots,m_1\}$, $b\in\frM^{1+k}$, $d\in\frM^{m-k}$, $v,w\in\{0,\ldots,r_1\}$ coincide with the respective objects in Eq.~\eqref{eq:flow_E_general}. This implies that the lists $\vK$, $\vL$, $\vM$ satisfy the conditions given in Part (B). The last condition follows from Def.~\eqref{dfn:varrho_I} and $\dim(\Xi)+\dim(\Phi)=\rdim-\sigma$. To prove Part (C) we use Parts (A), (B), Lemma~\ref{lem:fA_fB_bounds}, Remark~\ref{lem:fA_fB_Ks} and Lemma~\ref{lem:kernel_G} applied with $r\in\{0,1,2\}$.
\end{proof}
\section{Uniform bounds for cumulants}\label{sec:cumulants_uniform_bounds}
\begin{thm}\label{thm:cumulants}
There exists a unique choice of the counterterms $c^{[i]}_\uv$ in Eq.~\eqref{eq:force} such that:
$\bE f^{1,3,0}_{\uv,1/2}=1$, $\bE f^{i,1,0}_{\uv,1/2}=\mathfrak{f}^{[i]}$, $i\in\{1,\ldots,i_\sharp\}$, and $\bE f^{i,m,a}_{\uv,1/2} = 0$ for all other $i\in\bN_+$, $m\in\bN_0$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\leq 0$. Fix arbitrary \mbox{$\mathfrak{f}^{[i]}\in\bR$}, $i\in\{1,\ldots,i_\sharp\}$. For all list of indices
$\vI$ there exists \mbox{$\oo\in\bN_0$} such that the bound
\begin{equation}\label{eq:thm_cumulants}
\|K^{\ast\oo,\otimes(n+m)}_{\sIR}\ast E^\vI_{\uv,\sIR}\|_{\sV^{\mathsf{m}}}
\lesssim
[\uv]^{(\varepsilon-\sigma)\rr(\vI)}
[\sIR]^{\varrho_{3\varepsilon}(\vI)-\sigma\rs(\vI)+(n-1)\rdim}
\end{equation}
holds uniformly in $\uv,\sIR\in(0,1/2]$, where $n=\rn(\vI)$, $\mathsf{m}=\mathsf{m}(\vI)$, \mbox{$m=\rrm(\vI)$}. Moreover, the following condition is satisfied
\begin{equation}
E^\vI_{\uv,\sIR}(x_1,\ldots,x_n;\ry_1,\ldots,\ry_m) = (-1)^{\ra(\vI)}
E^\vI_{\uv,\sIR}(-x_1,\ldots,-x_n;-\ry_1,\ldots,-\ry_m),
\end{equation}
where $x_j\in\bM$, $\ry_j\in\bM^{m_j}$ for $j\in\{1,\ldots,n\}$ and $E^\vI_{\uv,\sIR}=0$ unless $m+n\in2\bN_0$.
\end{thm}
\begin{rem}
By stationarity $\bE f^{i,m,a}_{\uv,\sIR}(x)$ is a constant. Since $\partial_\sIR F^{1,3}_{\uv,\sIR}=0$ it holds $\bE f^{1,3,0}_{\uv,1/2}=f^{1,3,0}_{\uv,0}=1$. There is no distinguished value of $\bE f^{i,1,0}_{\uv,1/2}=\mathfrak{f}^{[i]}$, $i\in\{1,\ldots,i_\sharp\}$, ultimately, because there is no distinguished function $\chi$ in Def.~\ref{dfn:kernel_G} of the cutoff propagator $G_\uv$. The vanishing of $\bE f^{i,m,a}_{\uv,1/2}$ for all other $i\in\bN_+$, $m\in\bN_0$, $a\in\frM^m$ such that $\varrho(i,m)+|a|\leq 0$ is enforced by the properties of the cumulants given in the last sentence of the theorem (here we use the assumption $\rdim\in\{1,\ldots,6\}$). Observe that these properties are consequences of the following symmetries of Eq.~\eqref{eq:intro_mild}: $\varPhi(x)\mapsto-\varPhi(x)$, $\xi(x)\mapsto-\xi(x)$ and $\varPhi(x)\mapsto\varPhi(-x)$, $\xi(x)\mapsto\xi(-x)$, which in particular preserve the law of $\xi$. The counterterms $c^{[i]}_\uv$ are related to the renormalization parameters $\mathfrak{f}^{[i]}$ by the following formula
\begin{equation}
c^{[i]}_\uv:=f^{i,1,0}_{\uv}=f^{i,1,0}_{\uv,0} = \mathfrak{f}^{[i]} -\int_0^{1/2} \bE\,\partial_\sIR f^{i,1,0}_{\uv,\sIR}\,\rd\sIR,\qquad i\in\{1,\ldots,i_\sharp\}.
\end{equation}
The constants $\mathfrak{f}^{[i]}\in\bR$, $i\in\{1,\ldots,i_\sharp\}$ parameterize the class of solutions of Eq.~\eqref{eq:intro_mild} constructed in the paper (generically this is an over-parametrization).
\end{rem}
\begin{proof}
\textit{The base case:}
We first prove the theorem for list of indices $\vI$ such that \mbox{$\rri(\vI)=0$}. In this case the cumulants $E^\vI_{\uv,\sIR}$ coincide with the cumulants of the white noise $\xi$. The only non-vanishing cumulant is the covariance which corresponds to $\rn(\vI)=2$, $\mathsf{m}(\vI)=(0,0)$, $\rrm(\vI)=0$, $\ra(\vI)=0$, $\rr(\vI)=0$ and $\rs(\vI)=0$. It holds
\begin{equation}
\|\llangle K_\sIR^{\ast\oo}\ast\xi,K_\sIR^{\ast\oo}\ast\xi\rrangle\|_{\sV^{\mathsf{m}}} \leq \sup_{x_2\in\bT}\int_{\bT}|\bE(\xi(\rd x_1)\xi(x_2))| = 1.
\end{equation}
This finishes the proof of the base case.
\textit{Induction step:}
Fix $i\in\bN_+$ and $m\in\bN_0$. Assume that the theorem is true for all lists of indices $\vI$ such that either $\rri(\vI)<i$, or $\rri(\vI)=i$ and $\rrm(\vI)>m$. We shall prove the theorem for all $\vI$ such that $\rri(\vI)=i$ and $\rrm(\vI)=m$.
Consider the case $\rs(\vI)> 0$. In this case by Lemma~\ref{lem:flow_E_form_bound}~(A), (B) the cumulants $E^\vI_{\uv,\sIR}$ can be expressed in terms of the cumulants for which the statement of the theorem has already been established. As a result, the bound~\eqref{eq:thm_cumulants} with $\rs(\vI)> 0$ follows from the inductive assumption and Lemma~\ref{lem:flow_E_form_bound}~(C).
Now consider $\vI = ((i_1,m_1,a_1,0,r_1),\ldots,(i_n,m_n,a_n,0,r_n))$, $\rs(\vI)=0$. It follows from Def.~\ref{dfn:cumulants_eff_force} of the cumulants $E^\vI_{\uv,\sIR}$ that
\begin{equation}\label{eq:thm_cumulants_ind_step}
E^\vI_{\uv,\sIR}
=
E^\vI_{\uv,0} + \sum_{q=1}^n \int_0^\sIR E^{\vI_q}_{\uv,\uIR}\,\rd\uIR,
\quad
E^\vI_{\uv,\sIR}
=
E^\vI_{\uv,1/2} - \sum_{q=1}^n \int_\sIR^{1/2} E^{\vI_q}_{\uv,\uIR}\,\rd\uIR,
\end{equation}
where $\vI_q = ((i_1,m_1,a_1,0,r_1),\ldots, (i_q,m_q,a_q,1,r_q),\ldots,(i_n,m_n,a_n,0,r_n))$. Note that $\rs(\vI_q)=1$, hence the bound~\eqref{eq:thm_cumulants} has already been established for $E^{\vI_q}_{\uv,\uIR}$. We will use the first of Eqs.~\eqref{eq:thm_cumulants_ind_step} to bound the irrelevant cumulants $E^\vI_{\uv,\sIR}$, i.e. those with $\vI$ such that $\varrho(\vI)+(\rn(\vI)-1)\rdim>0$. The second equality will be used to bound certain contributions to the relevant cumulants $E^\vI_{\uv,\sIR}$, i.e. those with $\vI$ such that $\varrho(\vI)+(\rn(\vI)-1)\rdim\leq0$.
First, let us analyse the irrelevant contributions. If $\rn(\vI)>1$, then $E^\vI_{\uv,0}$ is a joint cumulant of a list of at least two random distributions. Since $\rri(\vI)=i>0$ one of the elements of this list is a deterministic distribution of the form $\partial^r_\uv F^{i,m,a}_{\uv,0}$. Hence, the cumulant vanishes. If $\rn(\vI)=1$ and $\varrho(\vI)>0$, then $E^\vI_{\uv,0}$ coincides with $\partial^r_\uv F^{i,m,a}_{\uv,0}=\partial^r_\uv F^{i,m,a}_{\uv\phantom{0}}$ for some $i,m,a$ such that $\varrho(i,m)+|a|>0$ and consequently vanishes. To prove the bound for $E^\vI_{\uv,0}$ we use the first of Eqs.~\eqref{eq:thm_cumulants_ind_step}. As we argued above, the first term on the RHS of this equation vanishes. The claim of the theorem is a consequence of the bound
\begin{equation}\label{eq:K_bound_s_u}
\|K_\sIR^{\ast\oo,\otimes(n+m)}\ast E^\vI_{\uv,\sIR}\|_{\sV^{\mathsf{m}}}\leq
\sum_{q=1}^n\int_0^\sIR \|K_\uIR^{\ast\oo,\otimes(n+m)}\ast E^{\vI_q}_{\uv,\uIR}\|_{\sV^{\mathsf{m}}}\,\rd\uIR.
\end{equation}
We used the fact $\|K_\sIR^{\ast\oo,\otimes(n+m)}\ast E^{\vI_q}_{\uv,\uIR}\|_{\sV^{\mathsf{m}}}\leq \|K_\uIR^{\ast\oo,\otimes(n+m)}\ast E^{\vI_q}_{\uv,\uIR}\|_{\sV^{\mathsf{m}}}$ for $\uIR\leq\sIR$ which follows from Lemma~\ref{lem:kernel_u_v}.
Next, let us analyse the relevant contributions. We note that the inequality $\varrho(\vI)+(\rn(\vI)-1)\rdim\leq0$ implies \mbox{$\rn(\vI)=1$}. Consequently, $\vI=(i,m,a,0,r)$ for some $r\in\{0,1,2\}$ and $a\in\frM$ such that $\varrho(i,m)+|a|\leq0$. Hence, $E^\vI_{\uv,\sIR}=\bE\,\partial^r_\uv F^{i,m,a}_{\uv,\sIR}$. We first study
\begin{equation}
\bE\, \partial_\uv^r f^{i,m,a}_{\uv,\sIR}
=\fI( \bE\,\partial_\uv^r F^{i,m,a}_{\uv,\sIR})=
\fI(K_\sIR^{\ast\oo,\otimes(n+m)}\ast E^\vI_{\uv,\sIR})\in\bR,
\end{equation}
where the map $\fI$ was introduced in Def.~\ref{dfn:map_I}. Note that by the translational invariance $\bE\, f^{i,m,a}_{\uv,\sIR}$ is a constant function. The application of the map $\fI$ to both sides of the second of Eqs.~\eqref{eq:thm_cumulants_ind_step} yields
\begin{equation}\label{eq:thm_cumulants_f_ren}
\bE\, \partial_\uv^r f^{i,m,a}_{\uv,\sIR}
=
\bE\, \partial_\uv^r f^{i,m,a}_{\uv,1/2}
-
\sum_{q=1}^n \int_\sIR^{1/2} \fI(K_\sIR^{\ast\oo,\otimes(n+m)}\ast E^{\vI_1}_{\uv,\sIR} )\,\rd\uIR,
\end{equation}
where $E^{\vI_1}_{\uv,\sIR} = \bE\, \partial_\uIR\partial_\uv^r F^{i,m,a}_{\uv,\uIR}$. Recalling that $\sV^m\equiv\sV^{\mathsf{m}}$ for $n=1$ and using Lemma~\ref{lem:map_I} we arrive at
\begin{equation}\label{eq:thm_cumulants_f_ren_bound}
|\bE\, \partial_\uv^r f^{i,m,a}_{\uv,\sIR}|
\leq
|\bE\, \partial_\uv^r f^{i,m,a}_{\uv,1/2}|
+
\sum_{q=1}^n \int_\sIR^{1/2} \|K^{\ast\oo,\otimes(n+m)}_{\uIR}\ast E^{\vI_1}_{\uv,\uIR}\|_{\sV^m}\,\rd\uIR.
\end{equation}
By assumption, $\bE\, f^{i,m,a}_{\uv,1/2}$ is independent of $\uv$ and $\bE\, \partial_\uv f^{i,m,a}_{\uv,1/2}=0$. Hence, using the bound~\eqref{eq:thm_cumulants_f_ren_bound} and the bound~\eqref{eq:thm_cumulants} applied to $E^{\vI_1}_{\uv,\sIR}$ we obtain
\begin{multline}\label{eq:thm_cumulants_relevant}
|\bE\, \partial_\uv^r f^{i,m,a}_{\uv,\sIR}|=|\fI(\bE\,\partial_\uv^r F^{i,m,a}_{\uv,\sIR})|
\lesssim 1+\int_\sIR^{1/2} [\uv]^{(\varepsilon-\sigma)r}
[\uIR]^{\varrho_{3\varepsilon}(i,m)+|a|-\sigma-\varepsilon r}\,\rd\uIR
\\
\lesssim
[\uv]^{(\varepsilon-\sigma)r}
[\sIR]^{\varrho_{3\varepsilon}(i,m)+|a|-\varepsilon r}.
\end{multline}
If $m=0$, then $a=0$ and $E^\vI_{\uv,\sIR} = \bE\, \partial^r_\uv F^{i,0,0}_{\uv,\sIR}=\bE\,\partial_\uv^r f^{i,0,0}_{\uv,\sIR}$. Hence, in this case the statement of the theorem follows from the bound~\eqref{eq:thm_cumulants_relevant}. To prove the case $m>0$ we first recall that we have already proved the following bounds
\begin{equation}\label{eq:cumulants_proof_rel1}
\|K_\sIR^{\ast\ooo,\otimes(1+m)}\ast \bE\, \partial_\uv^r F^{i,m,b}_{\uv,\sIR}\|_{\sV^m} \lesssim [\uv]^{(\varepsilon-\sigma)r}
[\sIR]^{\varrho_{3\varepsilon}(i,m)+|b|-\varepsilon r}
\end{equation}
for all irrelevant coefficients $F^{i,m,b}_{\uv,\sIR}$ and the following bounds
\begin{equation}\label{eq:cumulants_proof_rel2}
|\fI(\bE\, \partial_\uv^r F^{i,m,b}_{\uv,\sIR})|\lesssim [\uv]^{(\varepsilon-\sigma)r}
[\sIR]^{\varrho_{3\varepsilon}(i,m)+|b|-\varepsilon r}
\end{equation}
for all relevant coefficients $F^{i,m,b}_{\uv,\sIR}$. Let $\ro\in\bN_+$ be the smallest positive integer such that $\varrho(i,m)+\ro>0$ and note that by Theorem~\ref{thm:taylor} it holds
\begin{equation}
\bE\, \partial_\uv^r F^{i,m,a}_{\uv,\sIR}
=
\fX^a_\ro(\fI(\bE\, \partial_\uv^r F^{i,m,b}_{\uv,\sIR}),
\bE\, \partial_\uv^r F^{i,m,b}_{\uv,\sIR}).
\end{equation}
The arguments of the map $\fX^a_\ro$ above satisfy the bounds~\eqref{eq:cumulants_proof_rel2} and~\eqref{eq:cumulants_proof_rel1}. This together with Theorem~\ref{thm:taylor_bounds} applied with $C\lesssim [\uv]^{(\varepsilon-\sigma)r}[\sIR]^{\varrho_{3\varepsilon}(i,m)-\varepsilon r}$ implies the statement of the theorem.
\end{proof}
\section*{Acknowledgments}
I wish to thank Markus Tempelmayr and Pavlos Tsatsoulis for discussions. The main part of this work was carried out when the author was affiliated at the Max-Planck Institute for Mathematics in the Sciences in Leipzig. The financial support of the Max-Planck Society, grant Proj. Bez. M.FE.A.MATN0003 and partial support from
the National Science Centre, Poland, grant `Sonata Bis' 2019/34/E/ST1/00053 are gratefully acknowledged. | 210,073 |
TITLE: Continuous $f:[x_1,x_2]\to\mathbb{R}$ is not "1-1" if there are local maxima at interval ends
QUESTION [1 upvotes]: Let $f:[x_1,x_2]\to\mathbb{R}$ be a continuous function such that $f$ has a local maximum at $x_1$ and at $x_2$. Show that $f$ is not "1-1".
I have thought of using EVT, but I seem unable to prove that $f$ cannot be "1-1". It is sure, however, that reductio ad absurdum has to be used.
REPLY [0 votes]: Since $f$ has a local maxima at both $x_1,x_2$ so $\exists \delta_1,\delta_2>0$ such that $f(x_1)>f(x)\forall x\in [x_1,x_1+\delta) $ and $f(x_2)>f(x)\forall x\in (x_2-\delta ,x_2] $.
So $f(x)$ first decreases and then increases.
Graphically: The graph of $f $ first decreases attains a minimum value and then again increases to attain its local maximum value.(Note that it attains the minimum value because of its continuity).
So there must exist two points $x_1<a<b<x_2$ such that $f(a)=f(b). $ | 89,776 |
TITLE: Characteristic Exponent of Levy Process
QUESTION [1 upvotes]: I'm reading a chapter of a book on Levy processes and it states:
"Any Levy process $X$ enjoys the following property: For all $t\geq0$
$$\mathbb{E}[e^{iu X_t}]=e^{t\phi(u)},$$
where $\phi$ is the characterstic function of $X$ of the process $X$".
I read somewhere else that this is a consequence of stationary increment property of Levy processes. Can somoneone provide me with more insight on this? I want to understand why this holds
REPLY [0 votes]: If you denote the left side by $f(t)$ then $f(t+s)=f(t)f(s)$. This is an easy consequence of the fact that the process has stationary independent increments. Since $f$ is a continuous function and $f(0)=1$ we can show that $f$ must of the type specified. | 140,069 |
Excerpt:
For our birthday party, we decided to keep it pretty low key! We invited Sean Sullivan of the Maine Brewers’ Guild on to chat Maine beer with us. We have learned a lot in a year and can’t wait to see what the next year brings for us and for the Maine beer scene.
Guide:
Sean Sullivan; Maine Brewers’ Guild
Website:
Twitter:
Instagram:
Subject of Study:
The Maine Brewers’ Guild & Being in the Podcasting space for a year.
Entry:
Further Field Notes:
We let go of our usual format and just drink beers. We chat with Sean about why no one knows EXACTLY how many breweries there are in Maine right now, as well as, how much things have changed in the last 3 years.
Additional Notes:
The Maine Brewers’ Guild puts on a fabulous beer festival: Summer Session. It takes place on July 23 and is going to be awesome. Get your tickets here:
Add Comment | 161,646 |
\begin{document}
\title[Identities for a parametric Weyl algebra]{Identities for a parametric Weyl algebra over a ring}
\thanks{The first author was supported by CNPq 313358/2017-6, FAEPEX 2054/19 and FAEPEX 2655/19}
\author{Artem Lopatin}
\address{Artem Lopatin\\
State University of Campinas, 651 Sergio Buarque de Holanda, 13083-859 Campinas, SP, Brazil}
\email{[email protected] (Artem Lopatin)}
\author{Carlos Arturo Rodriguez Palma}
\address{Carlos Arturo Rodriguez Palma\\
State University of Campinas, 651 Sergio Buarque de Holanda, 13083-859 Campinas, SP, Brazil; Industrial University of Santander, Cl.~9 \#Cra 27, Ciudad Universit\'aria,
Bucaramanga, Santander, Colombia}
\email{[email protected] (Carlos Arturo Rodriguez Palma)}
\begin{abstract}
In 2013 Benkart, Lopes and Ondrus introduced and studied in a series of papers the infinite-dimensional unital associative algebra $\A_h$ generated by elements $x,y,$ which satisfy the relation $yx-xy=h$ for some $0\neq h\in \FF[x]$. We generalize this construction to $\A_h(\B)$ by working over the fixed $\FF$-algebra $\B$ instead of $\FF$. We describe the polynomial identities for $\A_h(\B)$ over the infinite field $\FF$ in case $h\in\B[x]$ satisfies certain restrictions.
\noindent{\bf Keywords: } polynomial identities, matrix identities, Weyl algebra, Ore extensions, positive characteristic.
\noindent{\bf 2020 MSC: } 16R10; 16S32.
\end{abstract}
\maketitle
\section{Introduction}
Assume that $\FF$ is a field of arbitrary characteristic $p=\Char\FF\geq0$. All vector spaces and algebras are over $\FF$ and all algebras are associative, unless other\-wise is stated. For the fixed $\FF$-algebra $\B$ with unity we write $\B\LA x_1,\ldots,x_m\RA$ for the $\FF$-algebra of non-commutative $\B$-polynomials in variables $x_1,\ldots,x_m$, i.e., $\B\LA x_1,\ldots,x_m\RA$ is a free left (and a free right) $\B$-module with the basis given by the set of all non-commutative monomials in $x_1,\ldots,x_m$, where we assume that $\be x_i = x_i \be$ for all $\be\in \B$ and $1\leq i\leq m$. The unity $1$ of $\B\LA x_1,\ldots,x_m\RA$ corresponds to the empty monomial. In case the variables are $x_1,x_2,\dotsc$ the algebra of non-commutative $\B$-polynomials is denoted by $\B\LA X\RA$. Similarly, we define the algebra of commutative $\B$-polynomials $\B[x_1,\ldots,x_m]$ as a free left (and a free right) $\B$-module with the basis given by the set of all monomials in $x_1^{i_1}\cdots x_m^{i_m}$ with $i_1,\ldots,i_m\geq0$, where we assume that $\be x_i = x_i \be$ and $x_i x_j=x_j x_i$ for all $\be\in \B$ and $1\leq i,j\leq m$. Note that $\B\LA x\RA = \B[x]$.
\subsection{Parametric Weyl algebra $\A_h(\B)$ as the Ore extension}
We study the polynomial identities for the following family of infinite-dimensional unital algebras $\A_h(\B)$, which are parametrized by a polynomial $h$ from the center of $\B[x]$:
\begin{defin}\label{Def Ah}
For $h\in Z(\B)[x]$, the {\it parametric Weyl algebra} $\A_h(\B)$ {\it over the ring} $\B$ is the unital associative algebra over $\FF$ generated by $\B$ and letters $x$, $y$ commuting with $\B$ subject to the defining relation $yx=xy+h$ (equivalently, $[y,x]=h$, where $[y,x]=yx-xy$), i.e., $$\A_h(\B)=\Bxy/\id{yx-xy-h}.$$
\end{defin}
\medskip
For short, we denote $\A_h=\A_h(\FF)$. The partial cases of the given construction are the Weyl algebra $\A_1$, the polynomial algebra $\A_0=\FF[x,y]$, and the universal enveloping algebra $\A_x$ of the two-dimensional nonabelian Lie algebra. For $h\in \FF[x]$, the following isomorphism of $\FF$-algebras holds:
\begin{eq}\label{eq_tensor}
\A_h(\B) \simeq \B \otimes_{\FF} \A_h.
\end{eq}
Note that in general the isomorphism~\Ref{eq_tensor} does not hold because $A_h$ is not well-defined in case $h\not\in\FF[x]$. Given a polynomial $f= \eta_d x^d + \eta_{d-1} x^{d-1} + \cdots + \eta_0$ of $\B[x]$ with $d\geq0$, we say that $\eta_d$ is the {\it leading coefficient} of $f$ and the product $\eta_d x^d$ is the {\it leading term} of $f$.
The algebra $\A_h$ was introduced and studied by Benkart, Lopes, Ondrus~\cite{Benkart_Lopes_Ondrus_I, Benkart_Lopes_Ondrus_II, Benkart_Lopes_Ondrus_III} as a natural object in the theory of Ore extensions. In particular, they determined automorphisms of $\A_h$ over an arbitrary field $\FF$ and the invariants of $\A_h$ under the automorphisms, completely described the simple modules and derivations of $\A_h$ over any field. Then Lopes and Solotar~\cite{Lopes_Solotar_2019} described the Hochschild cohomology ${\rm HH}^{\bullet}(\A_h)$ over a field of arbitrary characteristic. Over an algebraically closed field of zero characteristic simple $\A_h$-modules were independently classified by Bavula~\cite{Bavula_2020}. In recent preprints~\cite{Bavula_2021_zero},~\cite{Bavula_2021_prime} Bavula continued the study of the automorphism group of $\A_h$.
Let us recall that an Ore extension of $R$ (or, equivalently, a skew polynomial ring over $R$) $A=\R[y,\sigma,\delta]$ is given by a unital associative (not necessarily commutative) algebra $\R$ over a field $\FF$, an $\FF$-algebra endomorphism $\sigma:\R\rightarrow \R$, and a $\sigma$-derivation $\delta:\R\rightarrow \R$, i.e., $\delta$ is $\FF$-linear map and $\delta(ab)=\delta(a)b+\sigma(a)\delta(b)$ for all $a,b\in \R$. Then $A=\R[y,\sigma,\delta]$ is the unital algebra generated by $y$ over $\R$ subject to the relation $$ya=\sigma(a)y+\delta(a) \quad \text{ for all } a\in \R.$$
Assume that $\R=\B[x]$, $\sigma={\rm id}_{\R}$ is the identity automorphism on $\R$, and $\delta:\R\rightarrow \R$ is given by $\delta(a)=a'h$ for all $a\in \R$, where $a'$ stands for the usual derivative of a $\B$-polynomial $a$ with respect to the variable $x$. Since $h\in Z(\R)$, $\de$ is a derivation of $\R$. Using the linearity of derivative and induction on the degree of $a\in\B[x]$ it is easy to see that
\begin{eq}
[y,a]=a'h \text{ holds in }\A_h(\B) \text{ for all }a\in \B[x].
\end{eq}
\noindent{}Hence $\A_h(\B)=\R[y,\sigma,\delta]$ is an Ore extension. The following lemma is a corollary of Observation 2.1 from~\cite{AVV} proven by Awami, Van den Bergh and Van Oystaeyen (see also Proposition 3.2 of \cite{AD2} and Lemma 2.2 of \cite{Benkart_Lopes_Ondrus_I}).
\begin{lemma} Assume that $A=\R[y,\si,\de]$ is an Ore extension of $\R=\FF[x]$, where $\si$ is an automorphism of $\R$. Then $A$ is isomorphic to one of the following algebras:
\begin{enumerate}
\item[$\bullet$] a quantum plane, i.e., $A\simeq\Fxy/\id{yx-q xy}$ for some $q\in\FF^{*}=\FF\backslash \{0\}$;
\item[$\bullet$] a quantized Weyl algebra, i.e., $A\simeq\Fxy/\id{yx-q xy-1}$ for some $q\in\FF^{*}$;
\item[$\bullet$] an algebra $\A_h$ for some $h\in\FF[x]$.
\end{enumerate}
\end{lemma}
\medskip
\noindent{}Note that by Theorem 9.3 of~\cite{Benkart_Lopes_Ondrus_I} the algebra $\A_h$ is not a generalized Weyl algebra over $\FF[x]$ in the sense of Bavula~\cite{Bavula_1992} in case $h\not\in\FF$.
Since the algebra of $\B$-polynomials $\B[x,y]$ is well studied, in what follows we assume that $h$ is non-zero. Moreover, we assume that the following restriction holds:
\begin{conv}\label{conv1} The leading coefficient of $h\in Z(\B)[x]$ is not a zero divisor.
\end{conv}
\medskip
\subsection{Polynomial identities}
A polynomial identity for a unital $\FF$-algebra $\algA$ is an element $f(x_1,\ldots,x_m)$ of $\FF\LA X\RA$ such that $f(a_1,\ldots,a_m)=0$ in $\algA$ for all $a_1,\ldots,a_m\in \algA$. The set $\Id{\algA}$ of all polynomial identities for $\algA$ is a T-ideal, i.e., $\Id{\algA}$ is an ideal of $\FX$ such that $\phi(\Id{\algA})\subset \Id{\algA}$ for every endomorphism $\phi$ of $\FX$. An algebra that satisfies a nontrivial polynomial identity is called a PI-algebra. A T-ideal $I$ of $\FF\LA X\RA$ generated by $f_1,\ldots,f_k\in \FF\LA X\RA$ is the minimal T-ideal of $\FF\LA X\RA$ that contains $f_1,\ldots,f_k$. We say that $f\in \FF\LA X\RA$ follows from $f_1,\ldots,f_k$ if $f\in I$. Given a monomial $w\in \FF\LA x_1,\ldots,x_m\RA$, we write $\deg_{x_i}(w)$ for the number of letters $x_i$ in $w$ and $\mdeg(w)$ for the multidegree $(\deg_{x_1}(w),\ldots,\deg_{x_m}(w))$ of $w$. An element $f\in\FX$ is called multihomogeneous if it is a linear combination of monomials of the same multidegree. We say that algebras $\algA$, $\algB$ are called PI-equivalent and write $\algA \eqPI \algB$ if $\Id{\algA} =\Id{\algB}$.
Denote the $n^{\rm th}$ Weyl algebra by
$$\AW_n=\FF\LA x_1,\ldots,x_n,y_1,\ldots,y_n\RA/I,$$
where the ideal $I$ is generated by $[y_i,x_j]=\de_{ij}$, $[x_i,x_j]=0$, $[y_i,y_j]=0$ for all $1\leq i,j\leq n$. Note that $\A_1=\AW_1$.
Assume that $p=0$. It is well-known that the algebra $\AW_n$ does not have nontrivial polynomial identities. Nevertheless, some subspaces of $\AW_n$ satisfy certain polynomial identities. Namely, denote by $\AW_n^{(1,1)}$ the $\FF$-span of $x_i y_j$ in $\AW_n$ for all $1\leq i,j \leq n$ and by $\AW_n^{(-,r)}$ the $\FF$-span of $a y_{j_1}\cdots y_{j_r}$ in $\AW_n$ for all $1\leq j_1,\ldots, j_r\leq n$ and $a\in \FF[x_1,\ldots,x_n]$. Dzhumadil'daev~\cite{Askar_2004, Askar_2014} studied the standard polynomial
$${\rm St}_N(t_{1},\ldots,t_{N})=\sum_{\sigma\in S_{N}}(-1)^{\sigma}t_{\sigma(1)}\cdots t_{\sigma(N)}$$
over some subspaces of $\AW_n$. Namely, he showed that
\begin{enumerate}
\item[$\bullet$] ${\rm St}_N$ is a polynomial identity for $\AW_n^{(-,1)}$ in case $N\geq n^2 + 2n$;
\item[$\bullet$] ${\rm St}_N$ is not a polynomial identity for $\AW_n^{(-,1)}$ in case $N< n^2 + 2n - 1$;
\item[$\bullet$] ${\rm St}_N$ is a polynomial identity for $\AW_1^{(-,r)}$ if and only if $N>2r$;
\item[$\bullet$] the minimal degree of nontrivial identity in $\AW_1^{(-,r)}$ is $2r+1$.
\end{enumerate}
Using graph--theoretic combinatorial approach Dzhumadil'daev and Yeliussizov~\cite{Askar_Yeliussizov_2015} established that
\begin{enumerate}
\item[$\bullet$] ${\rm St}_{2n}$ is a polynomial identity for $\AW_n^{(1,1)}$ if and only if $n=1,2,3$.
\end{enumerate}
Note that the space $\AW_n^{(-,1)}$ together the multiplication given by the Lie bracket is the $n^{\rm th}$ Witt algebra $W_n$, which is a simple infinitely dimensional Lie algebra. The polynomial identities for the Lie algebra $W_n$ were studied by Mishchenko~\cite{Mishchenko_1989}, Razmyslov~\cite{Razmyslov_book} and others. The well-known open conjecture claims that all polynomial identities for $W_1$ follow from the standard Lie identity
$$\sum_{\sigma\in S_{4}}(-1)^{\sigma}[[[[t_0,t_{\si(1)}]t_{\si(2)}]t_{\si(3)}]t_{\si(4)}].$$
\noindent{}$\ZZ$-graded identities for $W_1$ were described by Freitas, Koshlukov and Krasilnikov~\cite{W1_2015}.
\subsection{Results}
In Theorem~\ref{Teorema Principal} we prove that over an infinite field $\FF$ of positive characteristic $p$ the algebra $\A_h(\B)$ is PI-equivalent to the algebra of $p\times p$ matrices over $\B$ in case $h(\al)$ is not a zero divisor for some $\al\in Z(\B)$. On the other hand, over a finite field the similar result does not hold in case $\B=\FF$ (see Theorem~\ref{theo_finite}). As about the case of zero characteristic, in Theorem~\ref{theo0} we prove that similarly to $\A_1$, the algebra $\A_h(\B)$ does not have nontrivial polynomial identities.
In Section~\ref{section_example} we consider the algebra $\A_{h}(\B)=\A_{\zeta}(\FF^2)$ such that $h=\zeta$ does not satisfy Convention~\ref{conv1} and the statements of Theorems \ref{theo0} and ~\ref{Teorema Principal} do not hold for $\A_{\zeta}(\FF^2)$. We describe polynomial identities for $\A_{\zeta}(\FF^2)$ and compare them with the polynomial identities for the Grassmann unital algebra of finite rank.
\section{Properties of $\A_h(\B)$}
Many properties of an Ore extension $A=\R[y,\sigma,\delta]$ are inherited from an underlying algebra $\R$. Namely, it is well-known that when $\sigma$ is an automorphism, then:
\begin{itemize}
\item $A$ is a free right and a free left $\R$-module with the basis $\{y^i \tq i\geq 0\}$ (see Proposition 2.3 of~\cite{K. R. Goodearl});
\item in case $R$ is left (right, respectively) noetherian we have that $A$ is left (right, respectively) noetherian (see Theorem 2.6 of~\cite{K. R. Goodearl});
\item in case $\R$ is a domain we have that $A$ is a domain (see Exercise 2O~of \cite{K. R. Goodearl}).
\end{itemize}
\noindent{}In case $\B=\FF$ the algebra $\A_h=\A_h(\B)$ is a noetherian domain, but in general case $\A_h(\B)$ lacks these properties, since $\B\subset \A_h(\B)$ (see also Example~\ref{ex2} below).
In order to distinguish the generators for the algebras $\A_h(\B)$ and $\A_1(\B)$, we will use the following
\begin{conv}\label{conv2} The generators of $\A_h(\B)$ are denoted by $x,\y, 1$ and the generators of $\A_1(\B)$ are denoted by $x,y,1$.
\end{conv}
\medskip
\begin{lemma}\label{lemma_basis}
The sets $\{x^{i}\y^{j}\tq i,j\geq 0\}$ and $\{\y^{j}x^{i}\tq i,j\geq 0\}$ are $\B$-bases for $\A_h(B)$.
\end{lemma}
\begin{proof}
Obviously, $\A_h(\B)$ is the $\B$-span of each of the sets from the lemma. On the other hand, $\B$-linear independence of these sets follows from the fact that $\A_h(\B)$ is a free right and a free left $\B[x]$-module with the basis $\{\y^i \tq i\geq 0\}$.
\end{proof}
Introduce the following lexicographical order on $\ZZ^2$:
$(i,j)<(r,s)$ in case $j<s$ or $j=s,\;i<r$. Denote the multidegree of a monomial $w=x^i\y^j$ of $\A_h(\B)$ by $\mdeg(w)=(i,j)$. Given an arbitrary element
$a=\sum_{i,j\geq0}\be_{ij} x^i \y^j$ of $\A_h(\B)$, where only finitely many $\be_{ij}\in\B$ are non-zero, define its {\it multidegree} $\mdeg(a)=(d_x,d_y)$ as the maximal multidegree of its momomials, i.e. as the maximal element of the set $\{(i,j) \tq \be_{ij}\neq0\}$. By Lemma~\ref{lemma_basis} the multidegree is well-defined. As above, the coefficient $\be_{d_x,d_y}$ is called the {\it leading coefficient} of $a$ and the product $\be_{d_x,d_y}\,x^{d_x} \y^{d_y}$ is called the {\it leading term} of $a$. In case $a\in\B$ we set $\mdeg(a)=(0,0)$ and the leading coefficient as well as the leading term of $a$ is $a$.
\begin{lemma}\label{lemma_mult}
Assume $i,j,r,s\geq 0$. Then
\begin{enumerate}
\item[(a)] the leading term of $x^i \y^j\cdot x^r \y^s$ is $x^{i+r}\, \y^{j+s}$;
\item[(b)] in case $h\in\B$ we have
$$x^i \y^j\cdot x^r \y^s = \sum_{k=0}^{\min\{j,r\}} k! \binom{j}{k} \binom{r}{k} x^{i+r-k} h^k\, \y^{j+s-k}.$$
\end{enumerate}
\end{lemma}
\begin{proof}
Recall that $\de(a)=a'h$ for each $a\in\B[x]$. Since $[\y,a]=\de(a)$, the induction on $j$ implies that
\begin{eq}\label{eq1_lemma_mult}
\y^j x^r = \sum_{k=0}^{j} \binom{j}{k} \de^k(x^r) \y^{j-k}
\end{eq}
(cf.~Lemma 5.2 of~\cite{Benkart_Lopes_Ondrus_I}). Taking $k=0$ in equality~\Ref{eq1_lemma_mult}, we obtain that the leading term of $\y^j x^r$ is $x^r\y^j$. Similarly we conclude the proof of part (a). Part (b) follows from~\Ref{eq1_lemma_mult} and
$$\de^k(x^r) =
\left\{
\begin{array}{cl}
\frac{r!}{(r-k)!} x^{r-k} h^k, & \text{if } k\leq r \\
0, & \text{if } k> r \\
\end{array}
\right..
$$
\end{proof}
\begin{lemma}\label{lemma_deg}
If the leading coefficient of one of non-zero elements $a,b\in \A_h(\B)$ is not a zero divisor, then $\mdeg(ab) = \mdeg(a) + \mdeg(b)$. In particular, $ab$ is not zero.
\end{lemma}
\begin{proof} Consider $a=\sum_{i=1}^m \be_{i} x^{r_i} \y^{s_i}$ and $b=\sum_{j=1}^n \ga_{j} x^{k_j} \y^{l_j}$ for $m,n\geq1$ and non-zero $\be_i,\ga_j\in\B$, where we assume that elements of each of the sets $\{(r_i,s_i)\tq 1\leq i\leq m\}$ and $\{(k_j,l_j)\tq 1\leq j\leq n\}$ are pairwise different. Assume that $\mdeg(a)=(r_1,s_1)$ and $\mdeg(b)=(k_1,l_1)$. Part (a) of Lemma~\ref{lemma_mult} implies that
$$\mdeg(x^{r_1} \y^{s_1} x^{k_1} \y^{l_1})=(r_1\!+\!k_1, s_1\!+\!l_1) \;\text{ and }\; \mdeg(x^{r_i} \y^{s_i} x^{k_j} \y^{l_j})<(r_1\!+\!k_1, s_1\!+\!l_1)$$ if $(i,j)\neq (1,1)$. Since $\be_1 \ga_1\neq 0$, we obtain $\mdeg(ab) = (r_1+k_1, s_1 + l_1)$ and the proof is concluded.
\end{proof}
\begin{lemma}\label{lemma_embedding}
\begin{enumerate}
\item[(a)] The $\B$-linear homomorphism of $\FF$-algebras $\psi: \A_h(\B)\to \A_1(\B)$, defined by
$$1\to 1,\quad x\to x,\quad \y\to y h,$$
is an embedding $\A_h(\B)\subset \A_1(\B)$.
\item[(b)] $\{x^{i}h^{j}y^{j}\tq i,j\geq 0\}$ and $\{y^{j}h^{j}x^{i}\tq i,j\geq 0\}$ are $\B$-bases for $\A_h(\B)\subset \A_1(\B)$.
\end{enumerate}
\end{lemma}
\begin{proof}
\noindent{\bf (a)} Since $\psi([\y,x]-h)=([y,x]-1)h=0$ in $\A_1(\B)$, the homomorphism $\psi$ is well-defined. Assume that $\psi$ is not an embedding, i.e., there exists non-zero finite sum $a=\sum_{i,j\geq0}\be_{ij} x^i \y^j$ with $\be_{ij}\in \B$ such that
$$
\psi(a)=\sum_{i,j\geq0}\be_{ij} x^i (y h)^j =0 \quad\text{ in }\quad\A_1(\B).
$$
Denote $\mdeg(a)=(r,s)$. If $(r,s)=(0,0)$, then $a\in \B$ and $\psi(a)=a$ is not zero; a contradiction. Therefore, $(r,s)\neq (0,0)$.
Since $\mdeg(x^i (y h)^j)= (i + j \deg(h), j)$ by Lemma~\ref{lemma_deg} and Convention~\ref{conv1}, we obtain that $\mdeg(\psi(a))= (r + s \deg(h), s)$ is not zero; a contradiction.
\medskip
\noindent{\bf (b)} Since $h$ lies in the center of $\B[x]$, repeating the proof of Lemma 3.4 from~\cite{Benkart_Lopes_Ondrus_I} for $\A_h(\B)$ we can see that
$$\A_h(\B)=\bigoplus_{j\geq 0}\B[x]h^{j}y^{j}=\bigoplus_{j\geq 0}y^{j}h^{j}\B[x].$$
Similarly to part (a), we conclude the proof by the reasoning with multidegree.
\end{proof}
\begin{example}\label{ex2} Assume $\B$ is the $\FF$-algebra of double numbers, i.e., $\B$ has an $\FF$-basis $\{1,\zeta\}$ with $\zeta^2=0$. Then the ideal $I=\FF\text{-span}\{\zeta x^i \y^j \tq i,j\geq 0\}$ is a proper nilpotent ideal of $\A_h(\B)$. In particular, the algebra $\A_h(\B)$ is not semi-prime.
\end{example}
\medskip
\section{$\A_h(\B)$ as the algebra of differential operators}\label{section_diff}
Denote by ${\rm Map}(\B[z])$ the algebra of all $\FF$-linear maps over $\B[z]$ with respect to composition. Assume that $\Diff_h(\B[z])$ is the subalgebra of ${\rm Map}(\B[z])$ generated by the following maps: the multiplication $\be\,{\rm Id}$ by an element $\be$ of $\B$, i.e., $(\be\,{\rm Id})(f)=\be f$, the multiplication $\chi$ by $z$, i.e., $\chi(f)=zf$, and the derivation $\de$ given by $\de(f)=f'h(z)$ for all $f\in\B[z]$. Note that $\de= h(\chi)\, \pa$, where $\pa$ stands for the operator of the usual derivative. Obviously, maps $\chi$, $h(\chi)$ and $\pa$ are $\B$-linear. For short, we write $\chi^0$ for ${\rm Id}$.
\begin{prop}\label{prop_diff}
\begin{enumerate}
\item[(a)] $\{\chi^{i} h(\chi)^j \pa^{j}\tq i,j\geq 0\}$ is an $\B$-basis for $\Diff_h(\B[z])$ in case $p=0$.
\item[(b)] $\{\chi^{i} h(\chi)^j \pa^{j}\tq 0\leq i,\; 0\leq j<p \}$ is an $\B$-basis for $\Diff_h(\B[z])$ in case $p>0$.
\item[(c)]$\A_h(\B) /\id{h^p y^p} \simeq \Diff_h(\B[z])$ for each $p\geq 0$.
\end{enumerate}
\end{prop}
\begin{proof}
Consider the $\B$-linear homomorphism of $\FF$-algebras $\Phi: \A_1(\B) \to {\rm Map}(\B[z])$ given by
$1\to {\rm Id}$, $x\to \chi$, $y\to \pa$. Since $\Phi([y,x]-1)(f)=(\pa \chi - \chi \pa - {\rm Id})(f) = f + z f' - z f' - f = 0$ for all $f\in\B[z]$, the map $\Phi$ is a well-defined. Applying $\Phi$ to parts (a) and (b) of Lemma~\ref{lemma_embedding} we obtain that $\Diff_h(\B[z]) = \Phi(\A_h(\B))$ is an $\B$-span of $\{\chi^{i} h(\chi)^j \pa^{j}\tq i,j\geq 0\}$.
Let $p=0$. Assume that some non-zero finite sum $\pi=\sum_{i,j\geq 0} \be_{ij} \chi^{i} h(\chi)^j \pa^{j}$ with $\be_{ij}\in\B$ belongs to the kernel of $\Phi$. Denote by $j_0$ the minimal $j\geq0$ with $\be_{ij}\neq 0$ for some $i$ and denote by $i_0$ the maximal $i$ with $\be_{ij_0}\neq 0$. Then $\pi(z^{j_0}) = j_0!\, h(z)^{j_0} \sum_{0\leq i\leq i_0} \be_{i,j_0}\, z^i =0$ in $\B[z]$. Thus Convention~\ref{conv1} together with $j_0!\,\be_{i_0,j_0} \neq 0$ implies a contradiction. Part~(a) is proven.
Assume that $p>0$ and some non-trivial finite sum
$$\pi=\sum\limits_{0\leq i,\; 0\leq j<p } \be_{ij} \chi^{i} h(\chi)^j \pa^{j}$$
with $\be_{ij}\in\B$ belongs to the kernel of $\Phi$. As above, we obtain a contradiction. Since $\pa^p=0$, we conclude the proof of part (b).
Parts (a) and (b) together with part (b) of Lemma~\ref{lemma_embedding} conclude the proof of part (c).
\end{proof}
\begin{theo}\label{theo0}
In case $p=0$ the algebra $\A_h(\B)$ does not have nontrivial polynomial identities.
\end{theo}
\begin{proof}
Assume that $\FF$-algebra $\A_h(\B)$ has a nontrivial polynomial identity. Since $p=0$, there exists $N>0$ such that $\A_h(\B)$ satisfies a nontrivial multilinear identity $f(x_1,\ldots,x_N)=\sum_{\si\in S_N}\al_{\si} x_{\si(1)}\cdots x_{\si(N)}$ with $\al_{\si}\in \FF$. Moreover, we can assume that $\al_{\rm Id}\neq0$ for the identity permutation ${\rm Id}$. Given $j>0$, we write $F_j$ for a $\B$-linear map $\chi^{2j} h(\chi)^j \pa^{j}$ from $\Diff_h(\B[z])$. Denote by $d\geq0$ the degree of $h$ and we write $\eta$ for the leading coefficient of $h$. Recall that $\eta$ is not a zero divisor by Convention~\ref{conv1}. Note that
\begin{eq}\label{eq_theo0}
F_{j}(z^m)=\left\{
\begin{array}{cc}
0, & \text{ in case }m<j\\
\frac{m!}{(m-j)!} \, z^{m+j} h(z)^j, & \text{ in case }m\geq j.\\
\end{array}
\right.
\end{eq}
In particular, $\deg(F_{j}(z^j))=j(d+2)$ and the leading coefficient of $F_{j}(z^j)$ is $j!\,\eta^j$, which is not a zero divisor.
By parts (a) and (c) of Proposition~\ref{prop_diff}, the equality $f(F_{j_{1}},\ldots,F_{j_N})=0$ holds in $\Diff_h(\B[z])$ for all $j_1,\ldots,j_{N}>0$. Consider $j_k=(d+2)^{N-k}$ for all $0\leq k\leq N$.
Note that $1=j_N<j_{N-1}<\cdots < j_0$. We claim that for any $\si\in S_N$ we have
\begin{eq}\label{claim_theo0}
F_{j_{\si(1)}}\circ \cdots \circ F_{j_{\si(N)}}(z) \neq 0 \text{ if and only if } \si= {\rm Id}.
\end{eq}
\begin{eq}\label{claim2_theo0}
\text{The leading term of } F_{j_1}\circ \cdots \circ F_{j_N}(z) \text{ is } j_{1}!\cdots j_N!\,\eta^{j_{1}+\cdots +j_N} z^{j_{0}}.
\end{eq}
Let us prove these claims. Assume that $F_{j_{\si(1)}}\circ \cdots \circ F_{j_{\si(N)}}(z) \neq 0$ for some $\si\in S_N$.
Since $F_{j_{\si(N)}}(z)\neq 0$, then equality~\Ref{eq_theo0} implies that $\si(N)=N$, $j_{\si(N)}=1$, $\deg(F_{j_{\si(N)}}(z))=d+2=j_{N-1}$ and the leading coefficient of $F_{j_{\si(N)}}(z)$ is $\eta$, which is not a zero divisor.
Similarly, assume that for $1\leq l< N$ with $\si(l)\leq l$ the inequality $F_{j_{\si(l)}}(g)\neq 0$ holds for some $g\in \B[z]$ with the leading term $j_{l+1}!\cdots j_N!\,\eta^{j_{l+1}+\cdots + j_N} z^{j_l}$. Then equality~\Ref{eq_theo0} implies that $\si(l)=l$ and $\deg(F_{j_{\si(l)}}(g))=j_{l-1}$. Moreover, the leading term of $F_{j_{\si(l)}}(g)$ is $j_{l}!\cdots j_N!\,\eta^{j_{l}+\cdots +j_N} z^{j_{l-1}}$. Consequently applying this reasoning to $l=N-1,\, l=N-2,\ldots, l=1$, we conclude the proof of claims~\Ref{claim_theo0} and~\Ref{claim2_theo0}.
Claims~\Ref{claim_theo0} and~\Ref{claim2_theo0} imply that $0=f(F_1,\ldots,F_N)(z)=\al_{\rm Id}\, F_{j_1}\circ \cdots \circ F_{j_N}(z)\neq 0$ by Convention~\ref{conv1}; a contradiction.
\end{proof}
\section{Polynomial Identities for $\A_h(\B)$ in positive characteristic}\label{section_positive}
We write $M_n=M_n(\FF)$ for the algebra of all $n\times n$ matrices over $\FF$ and denote by $\M_n$ the algebra of all $n\times n$ matrices over $\B[x,y]$. Denote by $I_n$ the identity $n\times n$ matrix and by $E_{ij}\in M_n$ the matrix such that the $(i,j)^{\rm th}$ entry is equal to one and the rest of entries are zeros. Consider the properties of the next two matrices of $M_p$:
$$A_{0}=\sum_{i=1}^{p-1}E_{i+1,i}\;\;\text{ and }\;\;B_{0}=\sum_{i=1}^{p-1}i\cdot E_{i,i+1}.$$
\begin{lemma}\label{lemma_A0_B0}
\begin{enumerate}
\item[(a)] For all $0\leq k<p$ we have that
$$A_{0}^{k}=\sum_{i=1}^{p-k}E_{k+i,i}\;\;\text{ and }\;\; B_{0}^{k}=\sum_{i=1}^{p-k}\frac{(k+i-1)!}{(i-1)!}E_{i,k+i},$$
\noindent{}where $A_0^0$ and $B_0^0$ stand for $I_p$.
\item[(b)] $B_{0}A_{0}-A_{0}B_{0}=I_{p}$.
\end{enumerate}
\end{lemma}
\begin{proof} The formula for $A_{0}^{k}$ is trivial. We prove the formula for $B_{0}^{k}$ by induction on $k$. For $k=1$ the claim holds. Assume that the claim is valid for some $k<p-1$. Then
\begin{align*}
B_{0}^{k+1}&=\left(\sum_{i=1}^{p-k}\frac{(k+i-1)!}{(i-1)!}E_{i,k+i}\right)\left(\sum_{r=1}^{p-1}r\cdot E_{r,r+1} \right)=\sum_{i=1}^{p-(k+1)}\frac{(i+k)!}{(i-1)!}E_{i,k+1+i}
\end{align*}
and the required is proven. Part (b) is straightforward.
\end{proof}
Define the $\B$-linear homomorphism $\varphi:\B\LA x,y \RA \rightarrow \M_p$ of algebras by
$$x\mapsto A,\;\; y\mapsto B,\;\; 1\mapsto I_p,$$
where $A=xI_{p}+A_{0}$ and $B=yI_{p}+B_{0}$. Since $\be A=A \be$ and $\be B = B\be$ for each $\be\in B$, the homomorphism $\varphi$ is well-defined.
\begin{lemma}\label{A1 subset Map}
The homomorphism $\varphi$ induces the injective $\B$-linear homomorphism $\ov{\varphi}:\A_1(\B)\rightarrow \M_p$. In particular, the restriction of $\ov{\varphi}$ to $\A_h(\B)\subset \A_1(\B)$ is the injective $\B$-linear homomorphism $\A_h(\B)\rightarrow \M_p$.
\end{lemma}
\begin{proof}
By part (b) of Lemma \ref{lemma_A0_B0} we have that $\varphi(yx-xy-1)=BA-AB-I_{p}=0$. Therefore, $\varphi$ induces a $\B$-linear homomorphism $\ov{\varphi}:\A_1(\B)\rightarrow \M_p$ of algebras.
Assume that there exists a nonzero $a\in \A_1(\B)$ such that $\ov{\varphi}(a)=0$. Since $\{x^{i}y^{j}\tq i,j\geq 0\}$ is an $\B$-basis for $\A_1(\B)$ by Lemma~\ref{lemma_basis}, we have $a=\sum_{i,j\geq 0}\be_{ij}x^{i}y^{j}$ for a finite sum with $\be_{ij}\in \B$. Thus
$0= \ov{\varphi}(a) = \sum_{i,j\geq 0} \be_{ij} A^i B^j$. The equalities
$$A^{i}=\begin{pmatrix}
x^{i} & 0 &\cdots & 0 \\
* & * &\cdots & * \\
\vdots &\vdots&&\vdots\\
* & * &\cdots & * \\
\end{pmatrix}
\;\; \text{and} \;\;
B^{j}=\begin{pmatrix}
y^{j} & * &\cdots & * \\
0 & * &\cdots & * \\
\vdots &\vdots&&\vdots\\
0 & * &\cdots & * \\
\end{pmatrix}$$
imply that the $(1,1)^{\rm th}$ entry of $A^{i}B^{j}$ is $(A^{i}B^{j})_{1,1}=x^{i}y^{j}$. Hence
$0 = (\varphi(a))_{1,1} = \sum_{i,j\geq 0} \be_{ij} x^{i}y^{j}$ in $\B[x,y]$.
Hence $\be_{ij}=0$ for all $i,j\geq 0$ and $a=0$; a contradiction. Therefore $\ov{\varphi}$ is injective.
\end{proof}
Given $1\leq i,j\leq p$ and $k\geq 1$, we write $z_{ij}(k)$ for $x_{i+p(j-1)+p^2(k-1)}\in \FF\LA X\RA$. The {\it generic $p\times p$ matrix $X_k$ with non-commutative elements} is the matrix $X_k=(z_{ij}(k))_{1\leq i,j\leq p}$.
\begin{cor}\label{Id(Mp) subset Id(Ah)}
$\Id{M_p(\B)}\subset \Id{\A_h(\B)}$, if the field $\FF$ is infinite.
\end{cor}
\begin{proof}
Lemma~\ref{A1 subset Map} implies that $\Id{\M_p}\subset \Id{\A_h(\B)}$.
Since $B\subset B[x,y]$, we have $\Id{\M_p}\subset \Id{M_p(\B)}$. On the other hand, assume that $f=f(x_1,\ldots,x_n)$ is a polynomial identity for $M_p(\B)$. Then $f(X_1,\ldots,X_n)= (f_{ij})_{1\leq i,j\leq n}$ for some $f_{ij}\in \FF\LA X\RA$ with $f_{ij}\in \Id{\B}$. It is well-known that for an infinite field $\FF$ and a commutative unital $\FF$-algebra $\algC$ the polynomial identities for a unital $\FF$-algebra $\algB$ and $\algC\otimes_{\FF} \algB$ are the same (for example, see Lemma 1.4.2 of~\cite{A.Gia}). Since $\B[x,y]=\FF[x,y]\otimes_{\FF} \B$, we obtain $f_{ij}\in \Id{\B[x,y]}$ and $f\in \Id{\M_p}$. The required is proven.
\end{proof}
For each $\al\in Z(\B)$ consider the evaluation $\B$-linear homomorphism $\epsilon_{\al}:\B[x,y]\rightarrow \B$ of unital $\FF$-algebras defined by
$$x\mapsto \al, \;\; \ y\mapsto 0$$
and extend it to the evaluation homomorphism $\varepsilon_{\al}:\M_p\rightarrow M_p(\B)$. Since $\A_h(\B)$ is a subalgebra of $\M_p$ by means of embedding $\ov{\varphi}$ (see Lemma \ref{A1 subset Map}), we can consider the images of $x,\y\in \A_h(\B)$ in $M_p(\B)$, which we denote by $C_{\al}$ and $D_{\al}$, respectively:
$$\begin{array}{cll}
C_{\al}=\varepsilon_{\al}(\ov{\varphi}(x))& =\varepsilon_{\al}(A) & = \al I_p + A_0,\\
D_{\al}=\varepsilon_{\al}(\ov{\varphi}(\y))& =\varepsilon_{\al}(Bh(A)) & = B_{0}\varepsilon_{\al}(h(A)).
\end{array}
$$
\noindent{}Obviously, $\be C_{\al} = C_{\al} \be$ and $\be D_{\al} = D_{\al} \be$ for each $\be\in B$. To obtain the explicit description of the matrix $D_{\al}$ we calculate $h(A)$. For $r\geq1$ denote the $r^{\rm th}$ derivative of $h\in\B[x]$ by $h^{(r)}=\frac{d^{r}h}{dx^{r}}$ and write $h^{(0)}$ for $h$. Note that $h^{(r)}(\al)$ lies in the center of $\B$.
\begin{lemma}\label{lemma_hA}
$$h(A) =\sum_{i=1}^{p}\sum_{j=1}^{i} \frac{1}{(i-j)!} h^{(i-j)} E_{ij}.$$
\end{lemma}
\begin{proof}
We start with the case of $h=x^{k}\in\B[x]$ for some $k\geq0$. Obviously, the claim of the lemma holds for $h=1$. Therefore, we assume that $k\geq1$. Since $A_{0}^{r}=0$ for all $r\geq p$, we have
$$h(A) = A^{k} = (xI_{p}+A_{0})^{k} = \sum_{r=0}^{\min\{k,p-1\}}\binom{k}{r}x^{k-r}A_{0}^{r}.$$
\noindent{}Part (a) of Lemma \ref{lemma_A0_B0} implies
$$h(A) = \sum_{r=0}^{\min\{k,p-1\}}\binom{k}{r}x^{k-r} \left(\sum_{i=1}^{p-r}E_{r+i,i}\right).$$
\noindent{}Regrouping the terms we obtain
\begin{eq}\label{eq_lemma_hA}
h(A)=\sum_{i=1}^{p}\; \sum_{j=\max\{1,i-k\}}^{i} \binom{k}{i-j}x^{k-(i-j)}E_{ij}.
\end{eq}
\noindent{}Note that for $0\leq r< p$ we can rewrite
$$\frac{1}{r!} h^{(r)} = \left\{
\begin{array}{rl}
\binom{k}{r}x^{k-r}, & \text{ if } r\leq k \\
0, & \text{ otherwise } \\
\end{array}
\right.,
$$
where $r!$ is not zero in $\FF$. Hence equality~\Ref{eq_lemma_hA} implies that the claim holds for $h=x^{k}$.
The general case follows from the proven partial case and the $\B$-linearity of derivatives.
\end{proof}
Lemma~\ref{lemma_hA} together with the definition of $D_{\al}$ immediately implies the next corollary.
\begin{cor}\label{cor_D0}
$$D_{\al} = \sum_{i=1}^{p-1} \sum_{j=1}^{i+1} \frac{i}{(i-j+1)!} h^{(i-j+1)}(\al) E_{ij}.$$
\end{cor}
\bigskip
For short, denote the $(i,j)^{\rm th}$ entry of $D_{\al}$ by $\D{i}{j}\in Z(\B)$ and for all $1\leq k < p$ define
\begin{eq}\label{eq_Dak}
D_{\al,k} = D_{\al} - \sum_{r=0}^{k-1}\D{k}{k-r}A_0^{r} = D_{\al} - \sum_{r=0}^{k-1} \sum_{i=1}^{p-r} \D{k}{k-r}E_{r+i,i}.
\end{eq}
\noindent{}We apply the following technical lemma in the proof of key Proposition~\ref{Id(Ah) subset Id(Mp)} (see below).
\begin{lemma}\label{Epk.Dk}
For all $1\leq r\leq p$ and $1\leq k < p$ we have
$$E_{rk}\, D_{\al,k}=k\, h(\al) E_{r,k+1}.$$
\end{lemma}
\begin{proof}
We have
\begin{align*}
E_{rk}D_{\al,k}&=\sum_{i=1}^{p-1} \sum_{j=1}^{i+1} \D{i}{j}E_{rk}E_{ij} - \sum_{j=0}^{k-1}\sum_{i=1}^{p-j} \D{k}{k-j} E_{rk} E_{i+j,i}\\
&= \sum_{j=1}^{k+1} \D{k}{j}E_{rj} - \sum_{j=1}^{k} \D{k}{j} E_{rj}\\
&= \D{k}{k+1}E_{r,k+1}.\\
\end{align*}
Equality $\D{k}{k+1} = k\, h(\al)$ concludes the proof.
\end{proof}
\begin{prop}\label{Id(Ah) subset Id(Mp)}
\begin{enumerate}
\item[(a)] Assume $\al\in Z(\B)$. Then $\varepsilon_{\al}(\A_h(\B))$ contains $h(\al)^{2(p-1)} M_p(\B)$.
\item[(b)] Assume $h(\al)$ is invertible in $\B$ for some $\al\in Z(\B)$. Then $\Id{\A_h(\B)}\subset \Id{M_p(\B)}$.
\item[(c)] Assume $h(\al)$ is not a zero divisor for some $\al\in Z(\B)$ and $\FF$ is infinite. Then $\Id{\A_h(\B)}\subset \Id{M_p(\B)}$.
\end{enumerate}
\end{prop}
\begin{proof} For short, we write $\be$ for $h(\al)\in Z(\B)$.
\medskip
\noindent{\bf (a)}
Denote by $\La = \varepsilon_{\al}(\A_h(\B)) = \alg_{\B}\{I_p,C_{\al}, D_{\al}\}$ the $\FF$-algebra generated by $\B I_p$, $C_{\al}$, $D_{\al}$. Since $A_0=C_{\al} - \al I_p$, we obtain that
$$A_0^{k}=\sum_{i=1}^{p-k}E_{k+i,i}\in \La$$
for all $0\leq k<p$. In particular, $E_{p1}=A_0^{p-1}\in\La$. Equality~\Ref{eq_Dak} implies that $D_{\al,k}\in\La$ for all $1\leq k < p$.
The statement of part (a) follows from the following claim:
\begin{eq}\label{claim1}
\{ \be^{p+k-r-1} E_{rk} \tq 1\leq r,k \leq p\}\subset \La.
\end{eq}
\noindent{}To prove the claim we use descending induction on $r$.
Assume $r=p$. We have $E_{p1}\in\La$. Lemma~\ref{Epk.Dk} implies that $E_{p1} D_{\al,1} = \be E_{p2}$. Since $E_{p1}$, $D_{\al,1}$ belong to $\La$, we can see that $\be E_{p2}\in\La$. Similarly, the equality $\be E_{p2}D_{\al,2} = 2 \be^2 E_{p3}$ implies $\be^2 E_{p3}\in\La$. Repeating this reasoning we obtain that $\be^{k-1} E_{pk}\in\La$ for all $1\leq k\leq p$.
Assume that for some $1\leq r< p$ claim~\Ref{claim1} holds for all $r'>r$, i.e., for every $1\leq k\leq p$ we have $\be^{p+k-r'-1} E_{r'k}\in\La$. Since
$$\be^{p-r}\left(A_0^{r-1} - \sum_{k=2}^{p-r+1} E_{(r-1)+k,k}\right) = \be^{p-r} E_{r1},$$
we obtain $\be^{p-r} E_{r1}\in\La$. Lemma~\ref{Epk.Dk} implies that $\be^{p-r} E_{r1} D_{\al,1} = \be^{p+1-r} E_{r2}$. Hence $\be^{p+1-r} E_{r2}\in\La$. Repeating this reasoning we obtain that $\be^{p+k-r-1} E_{rk} \in\La$ for all $1< k\leq p$, since $\be^{p+k-r-2} E_{r,k-1} D_{\al,k-1} = (k-1)\, \be^{p+k-r-1} E_{r,k}$. Claim~\Ref{claim1} is proven.
\medskip
\noindent{\bf (b)} Since $h(\al)$ is invertible in $\B$, part (a) implies that $\varepsilon_{\al}(\A_h(\B))=M_p(\B)$. Since $\varepsilon_{\al}$ is a homomorphism of $\FF$-algebras, the required is proven.
\medskip
\noindent{\bf (c)} Consider a polynomial identity $f\in \FF\LA x_1,\ldots,x_m\RA$ for $\A_h(\B)$. Since $\FF$ is infinite, without loss of generality we can assume that $f$ is homogeneous with respect to the natural grading of $\FF\LA x_1,\ldots,x_m\RA$ by degrees, i.e., each monomial of $f$ has one and the same degree $t>0$.
Part (a) implies that for every $A_1,\ldots,A_m$ from $M_p(\B)$ there exist $a_1,\ldots,a_m$ from $A_h(\B)$ such that
$$\be^{2(p-1)t} f(A_1,\ldots,A_m) = f(\be^{2(p-1)} A_1,\ldots,\be^{2(p-1)} A_m) = f(\varepsilon_{\al}(a_1),\ldots,\varepsilon_{\al}(a_m)).$$
Since $\varepsilon_{\al}$ is a homomorphism of $\FF$-algebras, we have $f(\varepsilon_{\al}(a_1),\ldots,\varepsilon_{\al}(a_m))=0$. Therefore $f$ is a polynomial identity for $\A_h(\B)$ because $\be$ is not a zero divisor.
\end{proof}
To illustrate the proof of part (a) of Proposition~\ref{Id(Ah) subset Id(Mp)}, we repeat it in the partial case of $p=3$ in the following example.
\begin{example}\label{ex1}
Assume $p = 3$ and $h(\al)\neq0$ for some $\al\in Z(\B)$. For short, denote $\be=h(\al)$, $\be'=h'(\al)$ and $\be''=h''(\al)$. Then $A_0=E_{21} + E_{32}$,
$$ C_{\al}=\begin{pmatrix}
\al& 0 & 0\\
1 & \al & 0\\
0 & 1 & \al \\
\end{pmatrix},\quad\text{ and }\quad
D_{\al}=\begin{pmatrix}
\be'&\be&0\\
\be''& 2\be' & 2\be\\
0&0&0
\end{pmatrix}.
$$
To show that $\La = \alg_{\B}\{I_p,C_{\al}, D_{\al}\}$ contains $\be^4 M_3(\B)$, we consider the following elements of $\La$:
$$ D_{\al,1}=D_{\al} - \be' I_3=
\begin{pmatrix}
0 & \be & 0\\
\be'' & \be' & 2 \be\\
0 & 0 & -\be' \\
\end{pmatrix},
$$
$$ D_{\al,2}=D_{\al} - 2 \be' I_3 - \be'' A_0 =
\begin{pmatrix}
-\be' & \be & 0\\
0 & 0 & 2 \be\\
0 & -\be'' & -2 \be' \\
\end{pmatrix}.
$$
\noindent{}Note that $A_0=C_{\al} - \al I_3$ and $A_0^2=E_{31}$ belong to $\La$. Since
$$E_{31} D_{\al,1} = \be E_{32}\;\;\text{ and }\;\; \be E_{32} D_{\al,2} = 2 \be^2 E_{33},$$
we obtain that $ \be E_{32},\, \be^2E_{33}\in\La$. Thus $ \be(A_0-E_{32})= \be E_{21}$ lies in $\La$. Since
$$\be E_{21} D_{\al,1} = \be^2 E_{22}\;\;\text{ and }\;\; \be^2 E_{22} D_{\al,2} = 2 \be^3 E_{23},$$
we obtain
that $ \be^2 E_{22},\, \be^3 E_{23}\in\La$. Hence $ \be^2(I_3 - E_{22} - E_{33}) = \be^2 E_{11}$ lies in $\La$. Since
$$\be^2 E_{11} D_{\al,1} = \be^3 E_{12}\;\;\text{ and }\;\; \be^3 E_{12} D_{\al,2} = 2 \be^4 E_{13},$$ we obtain
that $\be^3 E_{12},\, \be^4 E_{13}\in\La$. Therefore, $\La$ contains $\be^4 M_3(\B)$.
\end{example}
\medskip
\begin{theo}\label{Teorema Principal}
Assume that $\FF$ is an infinite field of characteristic $p>0$.
\begin{enumerate}
\item[(a)] If $h(\al)$ is not a zero divisor for some $\al\in Z(\B)$, then $\A_h(\B)\eqPI M_p(\B)$.
\item[(b)] If $\B=\FF$, then $\A_h\eqPI M_p$.
\end{enumerate}
\end{theo}
\begin{proof} Part (a) follows from Corollary \ref{Id(Mp) subset Id(Ah)} and part (c) of Proposition \ref{Id(Ah) subset Id(Mp)}.
Assume that $\B=\FF$. Then there exists $\al\in\FF$ with $h(\al)\neq0$, because $h\in\FF[x]$ is not zero and $\FF$ is infinite. Part (a) concludes the proof of part (b).
\end{proof}
\medskip
\begin{cor}\label{cor1}
Assume that $\FF$ is an infinite field of characteristic $p>0$ and for $h= \eta_d x^d + \eta_{d-1} x^{d-1} + \cdots + \eta_0$ from $Z(\B)[x]$ we have that $\eta_d$ and $\eta_0$ are not zero divisors. Then $\A_h(\B)\eqPI M_p(\B)$.
\end{cor}
\bigskip
\section{$\A_h$ over finite fields}\label{section_finite}
In this section we assume that $\B=\FF$ is the field of finite of order $q=p^k$ and $\FF\subset \KK$ for an infinite field $\KK$. Since $h\in\FF[x]$, Convention~\ref{conv1} is equivalent to the inequality $h\neq0$. As above, we write $M_p$ for $M_p(\FF)$ and $\A_h$ for $\A_h(\FF)$. Given a $\KK$-algebra $\algA$, we write $\IdK{\algA}$ for the ideal of $\KK\LA X\RA$ of polynomial identities for $\algA$ over $\KK$. In this section we proof the next result.
\begin{theo}\label{theo_finite}
\begin{enumerate}
\item[(a)] $\IdK{M_p(\KK)} \bigcap \FF\LA X\RA \subset \Id{\A_h}$.
\item[(b)] $\Id{\A_h} \subset \Id{M_p}$, if $h(\al)\neq0$ for some $\al\in\FF$.
\item[(c)] $\A_h \not\eqPI M_p$.
\end{enumerate}
\end{theo}
\begin{proof}
Since $\A_h\subset \A_h(\KK)= \A_h \otimes_{\FF}\KK$ as $\FF$-algebras, we can see that
$$\Id{\A_h \otimes_{\FF}\KK}\; = \; \IdK{\A_h(\KK)} \cap \FF\LA X\RA \; \subset \; \Id{\A_h}.$$
Part (b) of Theorem~\ref{Teorema Principal} concludes the proof of part (a). Part (b) follows from part (b) of Proposition \ref{Id(Ah) subset Id(Mp)}.
Consider $F_{p,q}(x,y)=G_{p,q}(x)\, R_{p,q}(x,y)\, (y^{q}-y)$ of $\FF\LA x,y\RA$, where
\begin{align*}
G_{p,q}(x)&=(x^{q^{2}}-x)(x^{q^{3}}-x)\cdots(x^{q^{p}}-x),\\
R_{p,q}(x,y)&= \left(1-(y\,({\rm ad}\, x)^{p-1})^{q-1}\right)
\left(1-(y\,({\rm ad}\, x)^{p-2})^{q-1}\right) \cdots
\left(1-(y\,{\rm ad}\, x)^{q-1}\right)
\end{align*}
for $y\,{\rm ad}\,x=[y,x]$.
Genov~\cite{G. Genov} proved that $F_{p,q}(x,y)$ is a polynomial identity for $M_{p}$.
Since $x\,{\rm ad}\,x = [x,x] = 0$, for $x\in\A_h$ we have $R_{p,q}(x,x)=1$ and $$F_{p,q}(x,x)=(x^{q}-x)(x^{q^{2}}-x)(x^{q^{3}}-x)\cdots(x^{q^{p}}-x).$$
By part (b) of Lemma~\ref{lemma_embedding} elements $x,x^2,x^3,\ldots $ are linearly independent in $\A_h$. Therefore, $F_{p,q}(x,x)\neq0$ in $\A_h$; part (c) is proven.
\end{proof}
\begin{conj}\label{conj}
$\Id{M_p(\KK)} \bigcap \FF\LA X\RA = \Id{\A_h}$.
\end{conj}
\medskip
\section{Counterexample}\label{section_example}
In this section we consider a counterexample to show that without Convention~\ref{conv1} Theorems~\ref{theo0} and~\ref{Teorema Principal} do not hold. Namely, we consider the commutative algebra $\B\simeq\FF^2$ of double numbers from Example~\ref{ex2}, i.e., $\B$ has an $\FF$-basis $\{1,\zeta\}$ with $\zeta^2=0$, and set $h=\zeta$. Note that Convention~\ref{conv1} does not hold for $h$. Then the statements of Theorems~\ref{theo0} and \ref{Teorema Principal} are not valid for $\A_{\zeta}(\FF^2)=\A_h(\B)$ (see Proposition~\ref{prop_ex} below).
\begin{remark}\label{remark_ex}
If Convention~\ref{conv1} does not hold for $h$, then Lemmas~\ref{lemma_basis} and \ref{lemma_mult} are still valid for $\A_h(\B)$.
\end{remark}
\bigskip
Part (b) of Lemma~\ref{lemma_mult} together with Remark~\ref{remark_ex} implies that for all $i,j,r,s\geq 0$ we have
\begin{eq}\label{eq1_ex}
x^i \y^j\cdot x^r \y^s = x^{i+r}\, \y^{s+j} + \zeta\, jr \, x^{i+r-1}\, \y^{s+j-1},
\end{eq}
where we use conventions that $x^{-1}=0$ and $y^{-1}=0$. Then
\begin{eq}\label{eq2_ex}
[x^i \y^j, x^r \y^s] = \zeta\, (jr-is) \, x^{i+r-1}\, \y^{s+j-1} \;\text{ in }\; \A_{\zeta}(\FF^2).
\end{eq}
The unital finite dimensional Grassmann algebra $\G_k$ of rank $k$ has an $\FF$-basis
$$\{1,e_{i_1}\cdots e_{i_m} \tq 1\leq i_1<\cdots < i_m\leq k\}$$
and satisfies the defining relations $e_i^2=0$ and $e_i e_j = - e_j e_i$ for all $1\leq i,j\leq k$. The polynomial identities for $\G_k$ were described by Di Vincenzo~\cite{DiVincenzo_1991} for $p=0$ and by Giambruno, Koshlukov~\cite{GiambrunoKoshlukov_2001} for any infinite field.
\begin{prop}\label{prop_ex} Assume that $\FF$ is an infinite field.
\begin{enumerate}
\item[(a)] The T-ideal of identities $\Id{\A_{\zeta}(\FF^2)}$ is generated by $$ f_1=[[x_1,x_2],x_3],\quad f_2=[x_1,x_2]\, [x_3,x_4].$$
\item[(b)] $\A_{\zeta}(\FF^2)\not\eqPI M_t(\C)$ for every $t\geq 2$ and every $\FF$-algebra $\C$ with unity.
\item[(c)] $\A_{\zeta}(\FF^2)\eqPI \G_k$ if and only if $k\in\{2,3\}$.
\end{enumerate}
\end{prop}
\begin{proof}
\noindent{\bf (a)} By $\FF$-linearity formula~\Ref{eq2_ex} implies that $[a,b]$ belongs to $\zeta\, \A_{\zeta}(\FF^2)$ for all $a,b\in \A_{\zeta}(\FF^2)$. Then $f_1,f_2\in \FF\LA X\RA$ are nontrivial polynomial identities for $\A_{\zeta}(\FF^2)$, since $\zeta^2=0$. Note that
$$f_3 = [x_1,x_2]\,x_3\, [x_4,x_5] = [[x_1,x_2],x_3] [x_4,x_5] + x_3 [x_1,x_2] [x_4,x_5]\in \Id{\A_{\zeta}(\FF^2)}$$
follows from $f_1,f_2$. Denote by $I$ the T-ideal generated by $f_1,f_2$.
Assume that $f=\sum_k \al_k w_k$ is a nontrivial identity for $\A_{\zeta}(\FF^2)$, where $\al_k\in\FF$ and $w_k\in\FF\LA x_1,\ldots,x_m\RA$ is a monomial. Since $\FF$ is infinite, we can assume that $f$ is multihomogeneous. i.e., there exists $\un{d}\in\NN^m$ with $\mdeg(w_k)=\un{d}$ for each $k$. We apply equalities
$$u x_j x_i v = u x_i x_j v - u [x_i,x_j] v,$$
$$u [x_i,x_j] v = [x_i,x_j] u v - [[x_i,x_j], u] v,$$
where monomials $u,v$ can be empty and $i<j$, to monomials $\{w_k\}$ and then repeat this procedure. Since $f_1,f_3\in \Id{\A_{\zeta}(\FF^2)}$, we finally obtain that there exist $g\in I$, $\al_0, \al_{ij}\in\FF$ such that
$$f = g + \al_0 x_1^{d_1}\cdots x_m^{d_m} + \sum_{1\leq i<j\leq m} \al_{ij} [x_i,x_j] x_1^{d_1}\cdots x_i^{d_i-1} \cdots x_j^{d_j-1} \cdots x_m^{d_m} \;\text{ in }\;\FF\LA X\RA.$$
Since $f(1,\ldots,1) = g(1,\ldots,1) + \al_0$, we obtain that $\al_0=0$. Consider $i<j$ with $d_i,d_j\geq1$. Making substitutions $x_i\to x$, $x_j\to \y$, $x_l\to 1$ for each $l$ different from $i$ and $j$, we can see that $0=0 + \al_{ij} [x,\y] x^{d_i-1}\y^{d_j-1}$ in $\A_{\zeta}(\FF^2)$. Thus $-\al_{ij} \zeta x^{d_i-1}\y^{d_j-1}=0$ in $\A_{\zeta}(\FF^2)$. Lemma~\ref{lemma_basis} together with Remark~\ref{remark_ex} implies that $\al_{ij}=0$. Therefore, $f=g$ lies in $I$.
\medskip
\noindent{\bf (b)} Since $\FF\subset \C$, every polynomial identity for $M_t(\C)$ lies in $\Id{M_t(\FF)}$. By Amitsur--Levitzki Theorem~\cite{Amitsur_Levitzki} the minimal degree of a polynomial identity for $M_t(\FF)$ is $2t$. In particular, $f_1$ is not an identity for $M_t(\C)$.
\medskip
\noindent{\bf (c)} Since $\G_k$ is commutative in case $p=2$ or $k=1$, we can assume that $p\neq 2$ and $k\geq2$. Note that
\begin{eq}\label{eq3_ex}
f_2(e_1,e_2,e_3,e_4)=4e_1 e_2 e_3 e_4 \neq 0 \;\text{ in }\; \G_k \;\text{ for }\; k\geq4.
\end{eq}
Thus we can assume that $k\in\{2,3\}$. The T-ideal $\Id{\G_k}$ is generated by
\begin{enumerate}
\item[$\bullet$] $f_1,f_2$ in case $p=0$ or $p=k=3$.
\item[$\bullet$] $f_1,{\rm St}_4$ in case $p>k$, where $k\in\{2,3\}$.
\end{enumerate}
Since
$${\rm St}_4(x_1,x_2,x_3,x_4) = [x_1,x_2]\circ [x_3,x_4] - [x_1,x_3]\circ [x_2,x_4] + [x_1,x_4] \circ [x_2,x_3],$$
where $u\circ v$ stands for $uv+vu$, part (a) implies that ${\rm St}_4$ lies in $\Id{\A_{\zeta}(\FF^2)}$. On the other hand, we can see that $f_2$ is a polynomial identity for $\G_k$ when $k\in\{2,3\}$. Part (c) is proven.
\end{proof} | 177,122 |
In this week’s fashion roundup, clearly everyone got over the cold and decided to throw those freakum dresses back on. Much of that has to do with the last week being Grammy weekend and this whole week leading up to the SuperBowl. Whatever, let’s just be happy there’s not a single coat to be seen in this gallery. Don’t get too cocky about it, though. Who knows when this polar vortex will be back to troll us.
VH1 Fashions Of The Week: The Coats Are Off
View Photo Gallery | 226,442 |
According insecticidal treated nets (LLINs)..” | 318,585 |
TITLE: How can I find such a $z$? (Injectivity of $T(z) = \lambda z + \mu\bar{z}$)
QUESTION [2 upvotes]: We consider $T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\lambda z+\mu\overline{z}$, where $\lambda ,\mu\in\mathbb{C}$. I want to prove that if $T$ is injective, then $\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}$.
Clearly, if $\mu =0$, then $\lambda\neq 0$ because $T$ is injective.
Now, if $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ and $\mu\neq 0$, then I wanted to find a $z\in\mathbb{C}\setminus\{0\}$ such that $z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}$, because we would have that:
$T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0$
('Cause this would prove that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ implies $T$ is not injective.)
So, the question is, can we always find such a $z\in\mathbb{C}\setminus\{0\}$?
REPLY [1 votes]: Here's an alternative proof.
Suppose that $T$ is injective. Note that $\mu \neq \bar{\lambda}$, otherwise $T(z) = \lambda z + \overline{\lambda z}$ in which case $T(z) = 0$ for any $z \in \mathbb{C}$ such that $\Re(\lambda z) = 0$. Now note that $T(\bar{\lambda}) = |\lambda|^2 + \mu\lambda$ and $T(\mu) = \lambda\mu + |\mu|^2$. Therefore $T(\bar{\lambda}) - T(\mu) = |\lambda|^2 - |\mu|^2$; as $\mu \neq \bar{\lambda}$ and $T$ is injective, $|\lambda|^2 \neq |\mu|^2$. | 199,223 |
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\begin{document}
\begin{abstract}
We construct two models for the classifying space for the family of infinite cyclic subgroups of the fundamental group of the Klein bottle.
These examples do not fit in general constructions previously done, for example, for hyperbolic groups.
\end{abstract}
\maketitle
\section{Introduction}
Let $G$ be a discrete group. A \emph{family}, $\f$, of subgroups of $G$ is a nonempty set of subgroups of $G$
which is closed under conjugation and taking subgroups.
A model, $E_{\f}(G)$, for the \textit{classifying space of the fami\-ly $\f$} is
a $G$-CW-complex $X$, such that all of its isotropy groups belong to $\f$ and if
$Y$ is a $G$-CW-complex with isotropy groups belonging to $\f$, there is precisely one
$G$-map $Y \rightarrow X$ up to $G$-homotopy. Classifying spaces for families appear frequently in mathematics, notably, in various Assembly Isomorphism conjectures
such as the Baum--Connes and the Farrell--Jones Conjectures, see \cite{luck}.
A model for $E_{\vcy } (\z\times \z)$, the classifying space for the family of virtually
cyclic subgroups of $\z \times \z$, may be constructed as countable join $*_{i \in \z}\r_i$
with each $\r_i \simeq \r$ see \cite{JL}, with a suitable action of $\z\times\z$.
This space is built by using the fact that each nontrivial virtually cyclic subgroup $H$
of $\z \times \z$ is normal in $\z\times \z$. We cannot build a model for $E_{\vcy}(\zz) $
in the same way because this is not the case in $\zz$. In this note, we present two models
for $E_{\vcy}(\zz )$ (which are $\zz$-homotopy equivalent). Our principal results are Theorems \ref{thm-join} and \ref{second-model}, and Proposition \ref{prop-homology}.
The constructions follow the
work of W. L\"uck and M. Weiermann in \cite{luck-weiermann}, and of D. Farley
in \cite{farley}. The latter uses the fact that $\zz$ is a $CAT(0)$ group as it acts
by isometries on the plane, and the former follows a general construction.
We thank the referee for valuable suggestions.
\section{Classifying Spaces for Families}
Let $G$ be a discrete group. A \emph{family} $\f$ of subgroups of $G$ is a nonempty set of subgroups of $G$
which is closed under conjugation and taking subgroups. Some examples are:
$ \{1\}$, the family consisting of the trivial subgroup in $G$,
$\fin$, the family of finite subgroups of $G$,
$\vcy$ the family of virtually cyclic subgroups of $G$ and
$\all$, the family of all subgroups of $G$.
Let $H$ be a subgroup of $G$, and $\f$ be a family of subgroups of $G$, then $\f$ defines a family of $H$ as follows
\begin{equation*}
\f (H)=\{K\subseteq H| K\in \f \}. \label{fsubg}
\end{equation*}
\begin{defi}Let $\f$ be a family of subgroups of $G$. A model for the
\textit{classifying space $E_{\f}(G)$ of the family $\f$} is a $G$-CW-complex $X$,
such that all of whose isotropy groups belong to $\f$. If $Y$ is a $G$-CW-complex
with isotropy groups belonging to $\f$, there is precisely one map $G$-map $Y \rightarrow X$ up
to $G$-homotopy. We denote by $B_{\f}(G)$ be the quotient of $X$ by the action of $G$.
\end{defi}
In other words, $X$ is a terminal object in the category of $G$-CW complexes with isotropy groups belonging to $\f$.
In particular, two models for $E_{\f}(G)$ are $G$-homotopy equivalent, and then we denote $X$
by $E_{\f}(G)$.
\begin{rem}\label{incfam}
Given two families $\f_1\subseteq \f_2$ of subgroups of $G$, since $E_{\f_2} (G)$ is a terminal object in the category of $G$-CW complexes with isotropy groups belonging to $\f_2$, there exists precisely one $G$-map up to $G$-homotopy
$$E_{\f_1}(G) \rightarrow E_{\f_2}(G).$$
\end{rem}
\begin{defi}
Let $G$ be a group, $H\subseteq G$ and $X$ a $G$-set. The \emph{fixed point set} $X^H$ is defined as
\begin{align}
X^H =\{x\in X \mid \forall h\in H, \; hx =x \; \}.
\end{align}
\end{defi}
\begin{thm}\cite[Thm. 1.9]{luck} \label{luck-caract}
A $G$-CW-complex $X$ is a model for $E_{\f}(G)$ if and only if the $H$-fixed point set $X^H$ is contractible for $H \in \f$ and is empty for $H \not\in \f$.
\end{thm}
A model for $E_{\mathcal{ALL}}(G)$ is $G/G$, a model for $E_{\{1\}}(G)$ is the same as a model for $EG$, the total space of the universal $G$-principal bundle $EG \rightarrow BG$, \cite{milnor}.
We write $\efin$ for $E_{\mathcal{FIN}(G)}$, this is known as the \textit{universal} $G$-CW-\textit{complex for proper $G$-actions}, and we write $\evyc$ for $E_{\mathcal{VCY}}(G)$.
\subsection{Constructing models from models for smaller families} \label{section-build}
Let $\f$ and $\g$ be two families of a group $G$, with $\f \subseteq \g$, such that we
know a model for $E_{\f}(G)$. In \cite{luck-weiermann} W. L\"uck and M. Weiermann build a model
$E_{\g}(G)$ from $E_{\f}(G)$, and in \cite{farley} Farley builds another model for groups
acting on CAT(0)-spaces. In this Section we present these models.
Let $\f=\fin$ and $\g=\vcy$. Following \cite{luck-weiermann}, define an equivalence relation $\sim$ on $\fo:= \vcy - \fin$ as
\begin{equation} \label{rel-eq}
V\sim W \:\:\Longleftrightarrow |V \cap W |= \infty ,
\end{equation}
for $V$ and $W$ in $\fo$, where
$|\star|$ denotes the cardinality of the set $\star$.
Let $[\fo]$ denote the set of equivalence classes under the above relation
and let $[H] \in [\fo ]$ be the equivalence class of $H \in \fo$.
For $H\in \f$
\begin{align}\label{nh}
\nh :=&\{ g\in G \mid [g^{-1} Hg]=[ H] \} \nonumber
\\ =&\{ g\in G \mid |g^{-1} H g \cap H|= \infty \}.
\end{align}
This is the isotropy group of $[H]$ under the $G$-action on $[\fo]$ induced by conjugation.
Note that $\nh$ is the commensurator of $H$ in $G$.
Define a family of subgroups $\gh$ of $\nh$ by
\begin{align} \label{gh}
\gh:=\{K \subseteq \nh \mid K\in \fo, |K\cap H|= \infty \} \cup \{ \fin \cap \nh \}.
\end{align}
The method to build a model of $E_{\g}(G)$ from one of $E_{\f}(G)$ is with the following theorem.
\begin{thm}\cite[Thm. 2.3]{luck-weiermann}\label{thm-luck-weierm}
Let $\f \subseteq \g$ and $\sim$ as above. Let $I$ be a complete system of representatives $[H]$ of the $G$-orbits in $[\g-\f]$ under the $G$-action coming from conjugation. Choose arbitrary $N_G[H]$-CW-models for $E_{\f\cap N_G[H]}(N_G[H])$, $E_{\g[H]}(N_G[H])$ and an arbitrary $G$-CW-model for $E_\f(G)$. Define $X$ a $G$-CW-complex by the cellular $G$-pushout
\begin{align*}
\xymatrix{
\coprod_{[H]\in I} G \times_{N_G[H]} E_{\f\cap N_G[H]}(N_G[H])
\ar[d]^{\coprod_{[H]\in I} id_G \times_{N_G[H]}f_{[H]}}
\ar[r]^(0.75){i}
&E_{\f}(G)
\ar[d]
\\
\coprod_{[H]\in I} G \times_{N_G[H]} E_{\g[H]}(N_G[H])
\ar[r]
& X
}
\end{align*}
such that $f_{[H]}$ is a cellular $N_G[H]$-map for every $[H]\in I$ and $i$ is an inclusion of $G$-CW-complexes, or such that every map $f_{[H]}$ is an inclusion of $N_G[H]$-CW-complexes for every $[H]\in I$ and $i$ is a cellular $G$-map.
Then $X$ is a model for $E_{\g}(G).$
\end{thm}
The maps in Theorem \ref{thm-luck-weierm} are given by the universal property of classifying spaces for families and inclusions of families of subgroups (see Remark (\ref{incfam})). \\
The following is Definition 2.2, in \cite{farley}.
\begin{defi}\label{def-i-g-g}
Let $\f \subseteq \g $ be families of subgroups of a group $G$. We say that a $G$-CW complex $X$ is an \emph{$I_{\g -\f} G$-complex} if
\begin{enumerate}
\item[(i)] whenever $H \in \g-\f $, $X^H$ is contractible;
\item[(ii)] whenever $H\not\in \g $, $X^H=\emptyset $.
\end{enumerate}
\end{defi}
Observe that if all isotropy groups are in $\g$ then (ii) holds. Since the trivial subgroup
is not in $\g- \f$, $X$ is not necessarily contractible.
\begin{thm} \cite[Prop. 2.4]{farley} \label{join-g-f}
If $G$ is a group and $\f \subseteq \g$ are families of subgroups of $G$, then the join
$$(E_{\f})*(I_{\g -\f}G)$$
is a model for the classifying space $E_{\g}G$.
\end{thm}
\section{First model for $\underline{\underline{E}}({\z\rtimes \z})$} \label{section-modelcat}
Let $\mathcal{K}:=(\zz )\backslash \r^2$ be the klein bottle. Following \cite{farley} we construct a model for $\underline{\underline{E}}({\z\rtimes \z})$ using the fact that $\zz $ acts by isometries on $\r^2$. This action is given by deck transformations of the universal covering $p\colon \r^2 \to \mathcal{K}$ of the Klein bottle as $\zz\cong \pi_1(\mathcal{K})$.
\subsection{Virtually cyclic subgroups of $\zz$}
Let $\zz$ be the \emph{Klein bottle group} with multiplication
\begin{align*}
(n_1, m_1)(n_2, m_2)=(n_1+(-1)^{m_1}n_2, m_1+m_2),
\end{align*}
inverse element
$$(n,m)^{-1}=((-1)^{1-m}n,-m)$$
and the neutral element is $(0,0)$.
\begin{rem} \label{remconj}
For $(t_1,t_2), (n,m) \in \zz$ we have that
\begin{align}
(t_1,t_2)(n,m)(t_1,t_2)^{-1} &=(t_1,t_2)(n,m)((-1)^{1-t_2} t_1, -t_2) \nonumber \\
&= (t_1,t_2) (n+(-1)^m (-1)^{1-t_2} t_1, \; m-t_2 ) \nonumber \\
&= (t_1 + (-1)^{t_2} [n + (-1)^{(m+1-t_2)} t_1] , \; m ) \nonumber \\
&=( (-1)^{t_2}n + t_1 + (-1)^{m+1} t_1, \; m) \label{conj}
\end{align}
\end{rem}
In $\zz$ the families $\fin$ and $\vcy$ are
\begin{align*}
\fin &=\{1\}\\
\vcy &=\{C\subseteq \zz \mid \, C\;\text{is infinite cyclic}\}\cup\{1 \}.
\end{align*}
\begin{rem}\label{rem-clasif-subg}
We classify all infinite cyclic subgroups of the family $\vcy$ in $\zz$. Let $n,m\in \z$, $k \in \mathbb{N}$. Observe that
\begin{align*}
&(n,m)^k= ([1+(-1)^m +(-1)^{2m}+ \cdots (-1)^{(k-1)m }] n, \: km) ;\\
&(n,m)^{-k}=([(-1)^{1-m} + (-1)^{1-2m}+ \cdots + (-1)^{1-km} ] n, \; -km).
\end{align*}
Therefore infinite cyclic subgroups in $\vcy$ are of the following form
\begin{align}
& \langle(n,2m')\rangle =\{(kn,2km') \mid k\in \z\}, \;\; \text{where} \;\; (n,2m')\neq (0,0) ; \label{power-e} \\
& \langle(n,2m'+1)\rangle = \{ (n,2m'+1)^k \mid k\in \z \}, \;\;\; \text{where} \label{power-o-e} \\
&\;\;\;\;\; (n,2m'+1)^k =\left \{ \begin{array}{cc} (0,k(2m'+1)), & \;\; \text{if $k$ is even }
\\ (n,k(2m'+1)) & \;\; \text{if $k$ is odd}.
\end{array} \right. \nonumber
\end{align}
\end{rem}
\subsection{A model for $\underline{\underline{E}}({\z\rtimes \z})$}
The group $\zz$ acts on $\r^2$ by deck transformations of the universal covering
$p\colon \r^2 \to \mathcal{K}$ of the Klein bottle. Explicitly, the action is as follows: let $(n,m)\in \zz$ and $(t,r)\in \r^2$, then
\begin{align}\label{action-r2}
(n,m)(t,r)=(n+(-1)^m t, m+r).
\end{align}
\begin{rem} \label{r2-free}
A model for $E(\zz)$ is $\r^2$, because $\r^2$ is contractible and the action (\ref{action-r2}) is free and properly discontinuous.
\end{rem}
Let $\l(a,b)$ denote the geodesic line in $\r^2$ determined by $a,b \in \r$ as
\begin{align}\l(a,b):=\{(x, ax +b) \mid x \in \r \},
\end{align}
and let
\begin{align}\l(\infty, b):=\{ (b, y) \mid y \in \r \}, \end{align}
denote the geodesic line parallel to $y$-axis, determined by $b$. Let $\L$ be the \emph{ space of lines} in $\r^2$:
\begin{align}
\L:=\{\l(a,b) \mid a \in \r\cup\{\infty \}, \;\; b \in \r \}.
\end{align}
The space of lines is a metric space with the following distance
\begin{align*}
d(\l_1,\l_2)= \left \{
\begin{array}{cc}
\frac{k}{1+k} & \text{if $\l_1$ and $\l_2$ bound a flat strip of width }k; \\
1 & \text{if $\l_1$ and $\l_2$ are not parallel.}
\end{array}
\right.
\end{align*}
Then we have that $\L= \coprod_{a\in \r \cup \{\infty \}} \r_a $, where
$\r_a:=\{\l(a,b)\mid b\in \r\},$ and the metric in each connected component $\r_a$ is given by $d$.
Since the action of $\zz$ in $\r^2$ sends geodesic lines to geodesic lines, it induces an action of $\zz$ on $\L$. For $(n,m)\in \zz$ and $\l(a,b)\in \L$, this action is as follows,
\begin{align}
&(n,m)\l(a,b)=\l((-1)^m a, b+m-(-1)^m an ), \; \text{ if } a \in \r ; \label{action-lines} \\
&(n,m)\l(\infty,b)= \l(\infty, n+ (-1)^m b ). \label{action-lines2}
\end{align}
\begin{defi}
Let $(n,m)\neq (0,0)$ in $\zz$. We say that a line $\l \subset \r^2$ is an \emph{axis} for
$(n,m)$ if
$(n,m)\l =\l $ and $(n,m) $ acts by translation on $\l $.
\emph{The axis space}, for elements of $\zz$ in $\r^2$, is defined as follows
\begin{align*}
\a& :=\{\l \in \L \mid \l \;\: \text{is an axis for some }\; (n,m)\in \zz-(0,0) \} .
\end{align*}
\end{defi}
By (\ref{action-lines2}) all the lines $\l(\infty,b)$ are axes. And by (\ref{action-lines}) we have
\begin{align}
(n,m) \l(a,b) =\l(a,b) \; \;\;& \text{iff } \;\; \l((-1)^m a, b+m-(-1)^m an )= \l(a,b) ;
\nonumber \\
& \text{iff } \; \; m \; \text{even and } \; a=\tfrac{m}{n} . \label{arterisco}
\end{align}
Observe that if $(n,m)$ fixes $\l(a,b)$, then it acts on $\l(a,b)$ by translation.
Therefore $\l(a,b) \in \a $ if only if $a \in \q\cup \infty $, and then
\begin{align}
\a &= \{ \l(a,b) \in \L \mid \; a\in \q \cup \infty , \; b \in \r \} \nonumber \\
&= \coprod_{a\in \q \cup \{\infty \}} \r_a. \label{axis-coprod}
\end{align}
\begin{prop}\label{axis}
The axis space $\a$ of $\r^2$ is an $I_{\vcy-\{1\}}(\zz)$-complex.
\end{prop}
The proof of Proposition \ref{axis} is given in the next Subsection.
\begin{thm}\label{thm-join}
A model for the classifying space for the family of virtually cyclic subgroups of $\zz$ is the join
\begin{align} \label{e-vcy-klein}
\underline{\underline{E}}(\zz) = \r^2 * \coprod_{a\in \q \cup \{\infty \}} \r.
\end{align}
Therefore, the quotient by the action is
\begin{align}\label{b-vcy-klein}
\underline{\underline{B}}(\zz) = \mathcal{K} * \coprod_{a \in \q_{\geq 0} \cup \{\infty\} }\s^1.
\end{align}
\end{thm}
\begin{proof}
By Proposition \ref{axis}, the axis space of $\r^2$, is an $I_{\vcy-\{1\}}(\zz)$-complex.
Thus, by Theorem \ref{join-g-f} and because
$\r^2$ is a model for $E(\zz)$, we have (\ref{e-vcy-klein}). On the other hand,
(\ref{b-vcy-klein}) easily follows by looking the action of $\zz$ on $\mathcal{K}$ and $\a$,
since the action of $\zz$ on $\a$ is by translation.
\end{proof}
\subsection{Proof of Proposition \ref{axis}}
In this Section we will prove that the axis space $\a$ is an
$I_{\vcy-\{1\}}(\zz)$-complex, which follow from Lemmas \ref{isotropy} and \ref{fixed}, below.
We denote by $\Iso(a,b)$ the isotropy subgroup of $\l(a,b)\in \a$, where $a\in \q \cup \{\infty\}$, and $b\in \r$, that is,
\begin{align*}
\Iso(a,b):=\{ (n,m)\in \zz \mid (n,m)\l(a,b)=\l(a,b)\}.
\end{align*}
\begin{lem} \label{isotropy}
All isotropy subgroups of the action of $\zz$ in the axis space $\a$ are in the set $\vcy-\{1\}$.
\end{lem}
\begin{proof}
We compute the isotropy subgroups of the action of $\zz$ on $\a$. See Remark \ref{rem-clasif-subg} about the family $\vcy$ of $\zz$.
\begin{itemize}
\item[(i)] If $a \in \q- \{0\}$ and $b \in \r$, by (\ref{arterisco}), we have
\begin{align*}
(n,m) \in \Iso(\l(a,b)) & \; \text{iff} \; m \;\text{is even and } \; a= \tfrac{m}{n}.
\end{align*}
Suppose $a=\tfrac{a_1}{a_2}$ with $gcd(a_1,a_2)=1$, then:
\begin{enumerate}
\item[1.] If $a_1$ is even then
$$ \Iso(a,b)=\langle (a_2,a_1) \rangle \in \vcy-\{1\}.$$
\item[2.] If $a_1$ is odd then
$$\Iso(a,b)=\langle (2a_2,2a_1) \rangle \in \vcy-\{1\}.$$
\end{enumerate}
\item[(ii)] Now if $a=0$, by (\ref{action-lines}) we have
$$ \Iso(0,b)=\langle (1,0) \rangle \in \vcy-\{1\}.$$
\item[(iii)] Lastly, if $a=\infty$, by (\ref{action-lines2}) we have
\begin{align*}
(n,m)\in \zz \in \Iso(\infty,b) \; & \text{ iff }
n=(1-(-1)^m) b.
\end{align*}
If $m$ is even, then $n=0$, and if $m $ is odd, $n=2b $, but $b \in \r$ and $n \in \z$, then we conclude that
\begin{enumerate}
\item[1)] if $2b \in \z$ then by (\ref{power-o-e})
\begin{align*}
\Iso(\infty, b)&=\langle (0,2) \rangle \cup \{(2b,2m'+1) \mid m' \in \z\} \\
&=\langle (2b,1) \rangle \in \vcy-\{1\};
\end{align*}
\item[2)] if $b\in \r$ and $2b \not\in \z$ then by (\ref{power-e})
\begin{align*}
\Iso(\infty,b) = \langle (0,2) \rangle \in \vcy-\{1\}.
\end{align*}
\end{enumerate}
\end{itemize}
\end{proof}
\begin{lem}\label{fixed}
If $H\in \vcy - \{1\}$, the fixed point set $\a^H$ is contractible.
\end{lem}
\begin{proof}
Let $H=\langle (n,m)\rangle \in \vcy-\{1\}$, a infinite cyclic subgroup of $\zz$ (see (\ref{power-e}) and (\ref{power-o-e})), we will describe
$$\a^H= \{\l \in \a \mid H \l =\l \},$$
along the following lines.
\begin{itemize}
\item[(i)] Suppose $m$ is even, and $n \neq 0$, then by (\ref{power-e}), $$H=\{(kn,km) \mid k \in \z \}.$$
Observe by (\ref{action-lines2}) that
$$(kn,km)\l(\infty,b)=\l(\infty, kn+b).$$ Since $n \neq 0$, then $\l(\infty, b) \not \in \a^H$ for every $b \in \r$. \\
Now if $a \neq \infty$ and $b\in \r$, by (\ref{arterisco}), we have
\begin{align*}
\l(a,b) \in \a^H \;\; & \text{iff } a=\frac{km}{kn}=\frac{m}{n}.
\end{align*}
Therefore
$$ \a ^H = \r_{\frac{m}{n}},$$
which is contractible.
\item[(ii)] Let $m\neq 0$ even and $K=\langle (0,m)\rangle =\{(0,km ) \mid k\in \z \}$. By (\ref{action-lines})
$$(0,km )\l(a,b)= \l(a, b + 2km ),$$
we have that $\l(a,b) \not \in \a^K$ whenever $a \in \q$. \\
On the other side, by (\ref{action-lines2}),
$$(0,km)\l(\infty,b)= \l(\infty, b), \;\:\forall k \in \z,\text{ and }\forall b \in \r .$$
Therefore,
$$\a^K= \r_{\infty},$$
which is contractible.
\item[(iii)] If $m$ is odd and $R=\langle (n,m)\rangle $, recall from (\ref{power-o-e}) that
$$(n,m)^k= \left \{ \begin{array}{cc} (0,km) & \;\;\;\text{if $k $ is even} \\ (n,km)& \;\;\; \text{if $k$ is odd} . \end{array} \right.$$
Let $a \in \q $ and $b \in \r$. Since $m$ is odd, by (\ref{arterisco}), no $(n,m) \in R$ fixes $\l(a,b)$.
\\
Now, if $a=\infty$ and $b \in \r$, by (\ref{action-lines2}) we have
\begin{align*}
\l(\infty,b) \in \a^R
& \;\;\text{iff} \;\; b=\tfrac{n}{2}.
\end{align*}
Therefore
$$ \a^R =\{\l(\infty,\tfrac{n}{2})\},$$
the space with a single point in $\r_{\infty}$.
\end{itemize}
\end{proof}
\section{Building a second model for $\underline{\underline{E}}(\zz)$} \label{chapter-build}
In this Section we build a model for $\underline{\underline{E}}(\zz)$ following \cite[Sec. 2.1]{luck-weiermann}. We recall
that subgroups in $\vcy$ are of the form (\ref{power-e}) or (\ref{power-o-e}), and $\fin=\{1\}$.
In $\fo=\vcy - \{1\} $, we have the following equivalences:
(i) Define $H:=\langle(1,0)\rangle$. Let $n\in \z- \{0\}$. By (\ref{power-e}), $$\langle(n,0)\rangle \subseteq \langle(1,0)\rangle ,$$
hence by (\ref{rel-eq})
\begin{align*}
\langle(n,0)\rangle \sim \langle(1,0)\rangle.
\end{align*}
In fact these subgroups of $\fo$ are the unique subgroups that intersect $H$ in an infinite set. Denote this class by
\begin{align} \label{h}
[H]=[ \langle(1,0)\rangle]=[\langle(n,0)\rangle],
\;\;\;\: \forall n \in \z- \{0\}.
\end{align}
(ii) Fix $n,m \in \z- \{0\}$ and
\begin{align} \label{r}
R= \langle(n,2m)\rangle.
\end{align}
Then there is a maximal subgroup in $\fo$ which contains $R$.\\
Let $s:=gcd(m,n)$, we define a subgroup $R'$ as follows,
\begin{align*}
R'=\langle (\frac{n}{s}, \frac{2m}{s})\rangle.
\end{align*}
By (\ref{power-e}), it is easy to see that $R'$ is the maximal subgroup in $\fo$ which contains $R$, then by (\ref{rel-eq}),
\begin{align*}
R \sim \; R'.
\end{align*}
Observe by (\ref{power-e}) that the only subgroups of $\fo$ related to $R'$ are the subgroups of $\fo$ which are contained in $R'$.
Denote by $[R]$ this class in $[\fo].$
(iii) We have the following inclusions by (\ref{power-e}) and (\ref{power-o-e})
\begin{align*}
\langle (0,2k) \rangle& \subseteq \langle ( 0,2)\rangle, \;\;\;\; \text{for every }\; k \in \z- \{0\};
\\
\langle ( 0,2(2s+1))\rangle & \subseteq \langle (r,2s+1 )\rangle, \;\;\;\; \text{for every } \;r ,s\in \z ;\text{ and}
\\
\langle ( t,2u+1)\rangle &\subseteq \langle (t,1 )\rangle \;\;\;\; \text{for every }\; t,u \in \z.
\end{align*}
By the previous inclusions and (\ref{rel-eq}) we have the following relations
\begin{align*}
& \langle(0,2(2m+1))\rangle \sim \langle(0,2)\rangle \sim \langle(r,1)\rangle \text{, and }\\
&\langle(n,2m+1)\rangle \sim \langle(n,1)\rangle \sim \langle(0,2)\rangle \sim \langle(r,1)\rangle \sim \langle(r,2s+1)\rangle
\end{align*}
for every $n,m,r,s \in \z$.
Denote this class by
\begin{align}\label{k}
[K]=[\langle (n,2m+1)\rangle].
\end{align}\\
We have in $[\fo]$ the classes $[H]$, $[K]$ and infinitely many countable classes of type $[R]$, as many as maximal subgroups in $\fo$ of the form $\langle (n,2m)\rangle$ there are, with $n,m\neq 0$ relatively prime.
Also $\zz$ acts on $[\fo]$ by conjugation. The classes $[H]$ and $[K]$ are fixed by conjugation and the classes of type $[R]$ are permuted.
\subsection{Explicit models } \label{models}
We describe models for $E(\zz)$, $E_{\g[\star]}(N_{\zz} [\star])$ and $E(N_{\zz}[\star])$, with $N_G[\star] $, $\mathcal{G}[\star]$ defined in (\ref{nh}),(\ref{gh}) and $[\star] \in [\fo]$.
Let $g=(t_1,t_2)\in \zz$.
\begin{enumerate}
\item[1.]\label{freeaction-r2}
A model for $E(\zz)$ is ${\r}^2$.
\item[2.] A model for $E(\z \times \z)$ is $\r^2$, since $\z \times \z $ acts freely on $\r^2$ by translation.
\item[3.]
(a) Let $[H] \in [\fo] $, where $H$ is as in (\ref{h}), by (\ref{conj}) note: if $$(t_1,t_2)(1,0)(t_1,t_2)^{-1}=((-1)^{t_2} , 0 ),$$
then $g^{-1}Hg=H$, and therefore
\begin{align*}
N_{\zz}[H] &= \{ g\in \zz \mid |g^{-1} H g \cap H|= \infty \} \nonumber \\
&\cong \zz
\end{align*}
So by (\ref{gh}):
$
\gh = \vcy (H). $
\\
(b) We claim that a model for $E_{\vcy(H)}(\zz)$ is $\r$. Define the action of $\zz$ on $\r$ as follows
$$(t_1,t_2) x = t_2 +x, \;\;\;\; (t_1,t_2)\in \zz, \;\; x\in \r. $$
Observe that any point $x\in \r$ is fixed by $(t_1,t_2)$ if only if $t_2=0$. If $S$ is a subgroup not in $\vcy(H)$, then the $S$-fixed point set $R^S $ is the empty set. \\
Let $S\in \vcy(H)$, then $S=\langle (n,0)\rangle$ for some $n \in \z -\{0\}$. Then the $S$-fixed point set $R^S = \r$ is contractible. Therefore, by Proposition \ref{luck-caract},
we have the claim.
\item[4.]
(a) Let $R=\langle(n,2m) \rangle$, with $n,m $ fixed in $\z- \{0\}$, and suppose $R$ is maximal in $\fo$.
By (\ref{conj}) we have that
\begin{align*}
(t_1,t_2)(n,2m)(t_1,t_2)^{-1}= ((-1)^{t_2}n, 2m ).
\end{align*}
Then we have two possibilities,
\begin{align*}
\text{ i) } &gRg^{-1}=R \; \;\text{if only if } \: t_2 \text{ is even }, \\
\text{ ii) } &gRg^{-1}=\langle (-n,2m) \rangle \;\; \text{if only if } \: t_2 \; \text{is odd},
\\
&\text{therefore}\;\;\; gRg^{-1}\cap R=\{1\}.
\end{align*}
We conclude by (\ref{nh}) that
\begin{align*}
N_{\zz}[R]=\{(t_1,2t_2) \mid t_1,t_2 \in \z\} = \z \times \z,
\end{align*}
and so by (\ref{gh}) and (\ref{power-e}):
\begin{align*}
\mathcal{G}[R]&= \{ \langle(ln,2lm) \rangle \mid \; l \in \z \} \nonumber \\
&= \vcy(R) \label{gr}
\end{align*}
(b) A model for $E_{\vcy(R)}(\z\times \z)$ is $\r$ by Proposition \ref{luck-caract}. Observe that the normalizer $N_{\zz}(R)$
of $R$ is equal to $N_{\zz}[R]$, therefore
$R$ is a normal subgroup of $N_{\zz}[R]=\z\times \z$ and we have the following exact sequence
\begin{align*}
0\rightarrow R \xrightarrow{i} \z \times \z \xrightarrow{\phi} \z \rightarrow 0
\end{align*}
where $i$ is the inclusion and $\phi$ is the projection onto the quotient $(\z \times \z) / R \cong \z .$
Let $(t_1,t_2) \in \z \times \z$ and $x \in \r$, define the action of $\z \times \z$ in $\r$ as follows:
\begin{align*}
(t_1,t_2) \cdot x= \phi(t_1,t_2)+ x.
\end{align*}
So, $ \phi(t_1,t_2) +x =x \;\;\;\text{if only if} \;\;\; (t_1,t_2)\in \ker \phi=R,$
therefore the isotropy groups are in $\vcy(R)$.
Furthermore, $\r^{R}=\r$ is contractible, and if $S\not\in \vcy(R)$ then $\r^S=\emptyset $.
\item[5.] (a) Let $[K]$ be as in (\ref{k}) and $(t_1,t_2)\in \zz$. By equation (\ref{conj}), it follows that
$$(t_1,t_2)(0,2)(t_1,t_2)^{-1}=(0,2). $$
Since $[\langle(0,2)\rangle]=[K]$, we conclude by (\ref{nh}) that
\begin{equation*} \label{nk}
N_{\zz}[K]=\zz,
\end{equation*}
so by (\ref{nh}), (\ref{power-e}) and (\ref{power-o-e}), note:
\begin{align*}
\mathcal{G}[K]&= \{D \subseteq \zz \mid D\in \fo, \: |D\cap K|=\infty \} \cup \{1\} \nonumber \\
&= \{ \langle (n,2m+1)\rangle \mid n,m \in \z\} \cup\{1\}. \label{gk}
\end{align*}
\item[(b)] A model for $E_{\g[K]}\zz$ is as follows: \\
By (\ref{power-o-e}), note that $ \langle (n,2m+1)\rangle \subseteq \langle (n,1)\rangle$ for all $n,m \in \z$.
Let $K_n=\langle (n,1)\rangle $ and
consider a point $k_n$ for each $K_n$. Observe for $g=(t_1,t_2)\in \zz$,
$gK_ng^{-1}=K_{m}$ that $m=(-1)^{t_2}n+2t_1$ by Remark \ref{remconj}. Then $\zz$ acts by permuting the subgroups
$K_n$, $n\in \z$.
\\
We define an action of $\zz$ on $X:=\{k_n \mid n\in \z\}$ as follows
$$g \cdot k_n =k_{m} \, \text{ iff } \, gK_ng^{-1}=K_{m} .$$
\\
By the above observation and (\ref{power-o-e}), note:
$$g.k_n=k_n \text{ iff } g\in N_{\zz}(K_n)=K_n.$$
Therefore the $\zz$-set $X$ is a model for $I_{\g[K]-\{1\}}$.
Since a model for $E(\zz)$ is $\r^2$, by Theorem \ref{join-g-f} we conclude that a model for
$E_{\g[K]}(\zz)$ is the join $X * \r^2$.
\end{enumerate}
\subsection{A second model for $\underline{\underline{E}}(\zz)$ }
We are now ready to apply Theorem \ref{thm-luck-weierm} and obtain a model for $\underline{\underline{E}}(\zz)$ by the following $(\zz)$-pushout.
Let $G=\zz$ and $A=\z\times \z$, then
\begin{align}
\xymatrix{
G \times_{G} EG \;
\coprod \; G\times_{G} EG \;
\coprod_{l\in I}\; G\times_{A} EA
\ar[r]^(0.7){i}
\ar[d]|{id \times_{G} p
\coprod id \times_{G} g 677
\coprod_{k \in I} id \times_{A} f_l }
& \r^2
\ar[d]
\\
G \times_G E_{\vcy(H)}G \;
\coprod \; G\times_G E_{\g[K]}G \;
\coprod_{l\in I} \; G\times_A E_{\vcy(R)}A
\ar[r]
& \underline{\underline{E}}(\zz),
}
\end{align}
where $I$ is a complete system of representatives $[R]$ of the $G$-orbits under conjugation over the classes of subgroups of type $[R]=[ \langle (n,2m)\rangle ]$, $(n,m) \in \zz- \{(0,0)\}.$
Applying models given in Section \ref{models} and the fact
$G \times_G Y = Y$,
we have
\begin{align}\label{pushout}
\xymatrix{
\r^2 \;
\coprod \; \r^2 \;
\coprod_{l\in I}\; G\times_{A} \r^2
\ar[d]|{ p
\coprod g
\coprod_{k \in I} id \times_{A} f_l }
\ar[r]^(0.7){i}
& \r^2
\ar[d]
\\
\r \;
\coprod \; (\{k_n\}_{n\in \z} \ast \r^2 )\;
\coprod_{l \in I} \; G\times_A \r
\ar[r]
& \underline{\underline{E}}(\zz)
}
\end{align}
The maps are given by the universal property of classifying spaces, applied with inclusions of families of subgroups, these are as follows:
\begin{enumerate}
\item[1.] $p\colon \r^2 \to \r$ is the projection on the $y$-axis, $(t,s)\mapsto s $. Since $G$ acts on the $y$-axis of $\r^2$ by translation and the action of $G$ on $\r$ is also by translation, then $p$ is a $G$-map, which is cellular. By Remark \ref{incfam} $p$ is unique up to $G$-homotopy.
\item[2.] The map $g \colon \r^2 \to \{k_n\}_{n\in \z} \ast \r^2$ is the inclusion, because the $G$-action is the same on $\r^2$, we have that $g$ is a $G$-map.
\item[3.] Let $l \in I$, $f_l \colon \r^2 \to \r $ are the quotient map of $\r^2$ on the line through $(n,2m)$ and the origin. It follows the map is $G$-equivariant.
\item[4.] The map $i$ is the identity on the first two $G$-spaces corresponding to the disjoint union, and it is the natural $G$-map on the third $G$-space.
\end{enumerate}
\begin{thm}\label{second-model}
Let $G=\zz$, $A=\z\times \z$ and $I$ be a complete system of representatives $[R]$ of the $G$-orbits under conjugation over the classes of subgroups of type $[R]=[ \langle (n,2m)\rangle ]$, $(n,m) \in \zz- \{(0,0)\}.$ From the $G$-pushout (\ref{pushout}) we have
\begin{align*}
\underline{\underline{E}} (\zz)= \frac{\r \coprod \{k_n\}_{n\in \z}\ast \r^2 \coprod_{l\in I} G\times_A \r^2 }{\forall x\in \r^2: p(x)\sim g(x)\sim [1_G,f_l(x)]},
\end{align*}
where the maps $p$, $g$ and $f_l$ are as before.
\end{thm}
\section{Homology}
In this Section we compute the homology groups of the model given in Theorem \ref{thm-join},
$\underline{\underline{B}} (\zz)=(\coprod_{J} \s^1 )* \mathcal{K}$, where $J$ is an
countably infinite set. To simplify notation, denote $X= \coprod_{J}\s^1$.
Since the Klein bottle is path-connected, then the join $X*\mathcal{K}$ is simply connected, \cite[Sec. 7.2]{homotopical}. Therefore $H_0(X*\mathcal{K})= \z $ and $ H_1(X*\mathcal{K})=0.$
From the well known short exact sequence of the join, see \cite[Ch. 8]{munkres}, we have: let $n>0$,
\begin{align*}
\xymatrix{0 \ar[r] &\widetilde{H}_{n+1} (X * \mathcal{K} ) \ar[r]& \widetilde{H}_n(X\times \mathcal{K})
\ar[r]^(0.4){\pi}& \widetilde{H}_n (X) \oplus \widetilde{H}_n (\mathcal{K}) \ar[r]& 0
}
\end{align*}
where the homomorphism $\pi $ is given by $a \mapsto (\pi_X (a), -\pi_\mathcal{K}(a))$. (Here $\pi_X$
and $\pi_\mathcal{K}$ are the homomorphisms induced in homology by the projections of $X\times \mathcal{K} $
on $X$ and $\mathcal{K}$ respectively.)
Then we have the following exact sequence:
\begin{align}\label{homology}
0 \rightarrow \widetilde{H}_{n+1} (X * \mathcal{K}) \rightarrow \bigoplus_J \widetilde{H}_n(\s^1\times \mathcal{K})
\rightarrow \left(\bigoplus_J \widetilde{H}_n (\s^1)\right) \oplus \widetilde{H}_n (\mathcal{K})
\rightarrow 0.
\end{align}
Thus, by the CW structure of $\mathcal{K} $ and $\s^1$, we use the K\"unneth theorem to obtain the homology groups of the product $\s^1 \times \mathcal{K}$:
\begin{align*}
H_i(\s^1 \times \mathcal{K})=\left \{ \begin{array}{cc} \z, \;\;\;& i=0;
\\ \z \oplus \z\oplus \z_2,\;\; &i=1;
\\ \z\oplus \z_2, \; & i=2;
\\ 0, & i>2.
\end{array}\right.
\end{align*}
Then for $n=1$:
\begin{align*}
0 \rightarrow \widetilde{H}_{n+1} (X * \mathcal{K} ) \xrightarrow{} \bigoplus_J (\z \oplus \z\oplus \z_2 )
\xrightarrow{\pi} (\bigoplus_J \z ) \oplus \z\oplus \z_2
\rightarrow 0,
\end{align*}
and we conclude that
\begin{align*}
H_2(X * \mathcal{K})= \bigoplus_{J'} (\z\oplus \z_2),
\end{align*}
where $J'$ is a countably infinite set. If $n=2$ in (\ref{homology}), then
$$H_3(X*\mathcal{K})= H_2(X\times \mathcal{K})=\bigoplus_J(\z\oplus \z_2).$$ And at last, for $i>3$,
$H_i(X*\mathcal{K})=0$.
\begin{prop}\label{prop-homology}
Homology groups of $\underline{\underline{B}} (\zz)=\big(\coprod_{J} \s^1 \big)* \mathcal{K},$ where $J$ is an
countably infinite set (Thm. \ref{thm-join}), are the following:
\begin{align*}
H_i\big(\big(\coprod_{J} \s^1 \big)\ast \mathcal{K}\big)=\left \{ \begin{array}{ccc} \z , \;\;\;& i=0;\\
0, \;\;\;& i=1; \\
\bigoplus_{J'} (\z\oplus \z_2), \;\;\;& i=2;
\\
\bigoplus_{J} (\z\oplus \z_2), \;\;\;& i=3;
\\ 0, & i>3,
\end{array}\right.
\end{align*}
where $J'$ is a countable infinite set.
\end{prop} | 45,661 |
Review: Shin Megami Tensei Devil Summoner Soul Hackers 3DS
Soul Hackers delivers a first-person, dungeon-crawling RPG experience set in a future where technology and otherworldly forces meet in a fusion of cyberpunk futurism and gothic horror.
Pros:
- Graphics are nice from a distance and some of the maps are really good with some amazing artwork gone into them along with the backdrops.
- FMV cutscenes bring the retro feel with an almost VHS quality to them.
- First person dungeon crawler RPG where you summon demons.
- Top screen is the game screen with the bottom screen used for menu management and showing off the map in the dungeons.
- Turn based combat that is menu driven.
- You can analyze enemies in battle which helps identify them along with any weaknesses.
- The game has a real horror meets the 90s set in the future vibe about it all. By that I mean it has retro gameplay at its core but is set in a very futuristic town with ancient indian influenced horror thrown in like images of coyotes giving heeds of warning and sending messages of doom.
- Random encounters through out.
- Menus are detailed and within them when dungeon crawling, Allows you to do things like set up party formations, assign secial moves or movesets and do some basic item management.
- Controls are really simple with full d pad support, Using the bumpers will do a side step in game and the menu is brought up via the Y button, Attacking/confirming and basically everything else is done with the B button.
- Can state save at any point and triggering it will take you back to the main screen. You cannot use another save or start a new game with state save, Instead you are only allowed to continue your state save.
- Earn xp, items and money from enemies.
- When you level up you get given a set number of upgrade points for each character and you can oost their base stats.
- You can unlock the ability to use play coins to buy better demons and is an extremely handy option for new players or people finding the game a bit hard.
- 3D is implemented but is not compulsory to enjoy the game.
- Use in game money to maintain and buy new demons for/in your party.
- Multiple save slots, 3 to be precise.
- You can talk to enemies and can even have the chance to recruit them to your party.
- You can in game, Hack the computer as it were to make the game easier, harder.
Cons:
- Slow start as it sets the story and introduces the characters.
- No proper tutorial to explain everything you can do.
- The game itself is a slow paced affair with a lot of text screen breaks in between fights.
- It is a port and for that it does show in close ups on the graphics, The FMV while cool dooes show its age.
- Game has a lot of difficulty spikes throughout.
- Loads of menus to navigate and needless menu changing/hopping.
- Huge gaps between actual proper save points.
In summary, The game is very much aimed at the hardcore crowd. New players will be able to get into it but you really need to attack this game with a huge 90s vibe going through your head. Back in the day this game was way ahead of its time in many ways and by this time it is a bit dated in places, clunky menus and slow paced combat, But the story is a well written one and has many twists and turns. RPG fans of yester year will relish the chance to replay one of the classics on a modern console but also love the familiarity of the old way of doing RPG games. It took me a fair few hours but I got there and I am enjoying it so stick to it and think about the 90s. Not too much though as I cannot be held responsible for any depression or the inability to find cans of Tab Clear or jolly ranchers. | 199,966 |
\begin{document}
\begin{center}
{{\large\bf The invariance principle and the large deviation for the biased random walk on $\mathbb{Z}^d$}\footnote{The project is supported partially by CNNSF (No.~11671216).}}
\end{center}
\vskip 2mm
\begin{center}
Yuelin Liu$^a$, Vladas Sidoravicius$^b$, Longmin Wang$^a$, Kainan Xiang$^a$
\vskip 1mm
\footnotesize{$^a$School of Mathematical Sciences, LPMC, Nankai University}\\
\footnotesize{Tianjin City, 300071, P. R. China}\\
\footnotesize{$^b$Courant Institute of Mathematical Sciences, New York, NY 10012, USA}\\
\footnotesize{\& NYU-ECNU Institute of Mathematical Sciences at NYU Shanghai}\\
\footnotesize{Emails: [email protected]} (Liu)\\
\footnotesize{[email protected]} (Sidoravicius)\\
\footnotesize{[email protected]} (Wang)\\
\footnotesize{[email protected]} (Xiang)
\end{center}
\begin{abstract}
In this paper, we establish the invariance principle and the large deviation for the biased random walk $\RW_\lambda$ with $\lambda\in [0,1)$ on $\mathbb{Z}^d, d\geq 1.$\\
\noindent{\bf AMS 2010 subject classifications}. 60J10, 60F05, 60F10.\\
\noindent{\bf Key words and phrases.} Biased random walk, invariance principle, large deviation principle.
\end{abstract}
\section{Introduction}
\setcounter{equation}{0}
\noindent The biased random walk $\RW_\lambda$ with parameter $\lambda\in [0,\infty)$ was introduced to design a Monte-Carlo algorithm for the self-avoiding walk by Berretti and Sokal \cite{BS1985}. The idea was refined and developed in \cite{LS1988,SJ1989, RD1994}. Lyons, and Lyons, Pemantle and Peres produced a sequence of remarkable papers on $\RW_\lambda$ (\cite{LR1990, LR1992, LR1995, LPP1996a, LPP1996b}).
$\RW_\lambda$ was studied in a number of works later (see for example \cite{BHOZ2013, AE2014, BFS2014, HS2015, BF2014} and the references therein). \cite{SSSWX2018} gave a spectral radius and several additional properties for biased random walk on infinite graphs. In \cite{SSSWX2018b}, it was shown that there is a phase transition in relation of the tree number in the uniform spanning forest on Euclidean lattice equipped with a network corresponding to biased random walk. Bowditch \cite{BA2018} proved a quenched invariance principle for $\{\vert X_n\vert\}_{n=0}^{\infty}$ when $\{X_n\}_{n=0}^{\infty}$ is a $\RW_\lambda$ on supercritical Galton-Watson tree, and showed that the corresponding scaling limit is a one dimensional Brownian motion.
Our paper aims to study the invariance principle (IP) and the large deviation principle (LDP) for $\RW_\lambda $ with $\lambda\in [0,1)$ on $d$-dimensional integer lattice $\mathbb{Z}^d\ (d\geq 1).$ Main results of this paper are Theorems \ref{thm2.1} and \ref{thm3.1}.
We define $\RW_\lambda\ (\lambda\geq 0)$ on $\mathbb{Z}^d$ as follows: Let $\mathbf{0}=(0,\cdots,0)\in \mathbb{Z}^d$ and
$$\vert x\vert=\sum\limits_{i=1}^d\vert x_i\vert,\ x=(x_1,\cdots,x_d)\in\mathbb{Z}^d$$
which is the graph distance between $x$ and $\mathbf{0}.$ Write $\mathbb{N}$ for the set of natural numbers, and let $\mathbb{Z}_{+}=\mathbb{N}\cup\{0\}.$ For any $n\in\mathbb{Z}_{+},$ define
$$B(n)=\left\{x\in \mathbb{Z}^d:\ \vert x\vert\leq n\right\},\ \partial B(n)=\left\{x\in \mathbb{Z}^d:\ \vert x\vert=n\right\}.$$
If the edge $e=\{x,y\}$ is at graph distance $n$ from $\mathbf{0},$ namely $\vert x\vert\wedge\vert y\vert=n,$ then let its conductance to be $\lambda^{-n}.$ Denote by
$\RW_\lambda\ (X_n)_{n=0}^\infty$ the random walk associated to the above conductances and call it the biased random walk with parameter $\lambda.$ $\RW_{\lambda}$ $(X_n)_{n=0}^{\infty}$ has the following transition probability:
\begin{eqnarray}\label{(1.1)}
p(v,u):=p_{\lambda}(v,u)=\left\{\begin{array}{cl}
1/d_v &{\rm if}\ v=\mathbf{0},\\
\ \\
\frac{\lambda}{d_v+\left(\lambda-1\right)d_v^-} &{\rm if}\ u\in \partial B(|v|-1)\ \mbox{and}\ v\neq \mathbf{0},\\
\ \\
\frac{1}{d_v+\left(\lambda-1\right)d_v^-} &{\rm otherwise}.
\end{array}
\right.
\end{eqnarray}
Here $d_v=2d$ is the degree of the vertex $v;$ and $d_v^-$ (resp. $d_v^+$) is the number of edges connecting $v$ to $\partial B(|v|-1)$ (resp. $\partial B(|v|+1)$). Note that
$\RW_1$ $(X_n)_{n=0}^{\infty}$ is the simple random walk (SRW) on $\mathbb{Z}^d;$ and
\begin{eqnarray*}
d_v^+=d+\kappa (v),\ d_v^-=d-\kappa (v),\ v=(v_1,\cdots,v_d)\in\mathbb{Z}^d,
\end{eqnarray*}
where $\kappa(v)=\#\{i:\ v_i=0\}$ (with $\# A$ being the cardinality of a set $A$). For any $n\in\mathbb{Z}_+,$ write $X_n=\left(X_n^1,\cdots,X_n^d\right).$
It is known that on $\Z^d\ (d\geq1)$, $\RW_{\lambda}$ $(X_n)_{n=0}^{\infty}$ is transient for $\lambda<1$ and positive recurrent for $\lambda>1$ (R. Lyons \cite{LR1995}, R. Lyons and Y. Peres \cite[Theorem~3.10]{LP2017}).
Let
\[\sX = \left\{ x=(x_1,\cdots,x_d) \in \Z^d:\ x_i = 0 \text{ for some } 1 \leq i \leq d \right\}. \]
Then from \cite[Lemma 2.3 and Theorem 2.4]{SSSWX2018b}, the next two results were stated for $\lambda\in (0,1)$. In fact, for any $0\leq \lambda<1,$ with probability $1$,
\begin{eqnarray}\label{facts}
\left\{\begin{array}{ll}
\RW_{\lambda}\ (X_n)_{n=0}^{\infty}\ \mbox{visits}\ \sX \ \mbox{only finitely many times;}\\
\ \ \ \\
\frac{1}{n} \left(\left|X_n^1\right|, \cdots, \left|X_n^d\right|\right) \rightarrow \frac{1- \lambda}{1+ \lambda} \left(\frac{1}{d},\cdots,\frac{1}{d}\right)\ \mbox{as}\ n\rightarrow\infty.
\end{array}
\right.
\end{eqnarray}
And when $\lambda>1,$ $(X_n)_{n=0}^{\infty}$ is positive recurrent and ergodic with ${\bf 0}$ speed:
\[\lim\limits_{n\rightarrow\infty}\frac{\left(\left|X_n^1\right|, \cdots, \left|X_n^d\right|\right)}{n}=\mathbf{0}\] almost surely. The related
central limit theorem (CLT) and IP can be derived from \cite{JG2004} and \cite{MPU2006} straightforwardly.
Thus a natural question is to study CLT, IP with $\lambda\in [0,1)$ and LDP for $\RW_\lambda$ on $\mathbb{Z}^d.$ The LDP for $\RW_\lambda$ $(X_n)_{n=0}^{\infty}$ means LDP for the scaled reflected biased random walk $\left\{\frac{1}{n}\left(\left\vert X_n^1\right\vert,\cdots,\left\vert X_n^d\right\vert\right)\right\}_{n=1}^{\infty}$. And CLT, IP and LDP for $\RW_0$ are interesting
only for $d\geq 2.$ In this paper, Theorem \ref{thm2.1} proves CLT and IP for $\RW_\lambda, \lambda\in[0,1)$. However the proof of the Theorem \ref{thm2.1} shows that the scaling limit in this case is not a $d$-dimensional Brownian motion. For Theorem \ref{thm3.1}, we derive LDP for scaled reflected $\RW_\lambda, \lambda\in[0,1)$. The rate function of scaled reflected $RW_{\lambda}$ differs from that of drifted random walk, though there is a strong connection between them. Here, we only can calculate the rate function $\Lambda^\ast$ out when $d\in\{1,2\}$ and $\lambda\in (0,1)$, $d\geq 2$ and $\lambda=0$. To obtain an explicit rate function when $d\geq 3$ and $\lambda\in (0,1)$ still remains as an open problem. And in the recurrent case when $\lambda>1$, we hope to establish a LDP with proper normalization in the future work.
\section{CLT and Invariance Principle for $\RW_\lambda$ with $\lambda\in [0,1)$}\label{sec2}
\setcounter{equation}{0}
\noindent In this section, we fix $\lambda\in [0,1),$ then use the matingale's CLT for $\mathbb{R}^d$-valued martingales and the martingale characterization of Markov chains
to prove the IP for reflected $\RW_\lambda$
(Theorem \ref{thm2.1}). And CLT for reflected $\RW_\lambda$ is a consequence of corresponding IP.
To describe our main result, we need to introduce some notations. For any nonnegative definite symmetric $d\times d$ matrix $A$, let $\mathcal{N}(\mathbf{0},A)$ be the normal distribution with mean $\mathbf{0}$ and covariance matrix $A.$ Define the following positive definite symmetric $d\times d$ matrix $\Sigma=(\Sigma_{ij})_{1\leq i,j\leq d}:$
$$\Sigma_{ii}=\frac{1}{d}-\frac{(1-\lambda)^2}{d^2(1+\lambda)^2},\ \Sigma_{ij}=-\frac{(1-\lambda)^2}{d^2(1+\lambda)^2},\ 1\leq i\not=j\leq d.$$
For a random sequence $\{Y_n\}_{n=1}^{\infty}\subset\mathbb{R}^d,$ $Y_n\rightarrow \mathcal{N}(\mathbf{0},A)$ means that as $n\rightarrow\infty,$ $Y_n$ converges in distribution to the normal distribution $\mathcal{N}(\mathbf{0},A)$ in Skorohod space $D\left([0,\infty),\mathbb{R}^d\right)$.
For any $a\in\mathbb{R},$ let $\lfloor a\rfloor$ be the integer part of $a.$ Put
$$v=\left(\frac{1-\lambda}{d(1+\lambda)},\cdots,\frac{1-\lambda}{d(1+\lambda)}\right)\in\mathbb{R}^d.$$
\vskip 3mm
\begin{thm}[IP and CLT]\label{thm2.1}
Let $0 \leq \lambda < 1$ and $(X_m)_{m=0}^{\infty}$ be $\RW_{\lambda}$ on $\mathbb{Z}^d$ starting at $x$, $x\in \mathbb{Z}^d$.
Then, on $D\left([0,\infty),\mathbb{R}^d\right),$
\[\left(\frac{\left(\left|X_{\lfloor nt\rfloor}^1\right|,\cdots, \left|X_{\lfloor nt\rfloor}^d\right|\right)-nvt}{\sqrt{n}}\right)_{t\geq 0}\
\]
converges in distribution to $\frac{1}{\sqrt{d}}\left[I- \frac{1-\rho_\lambda}{d}E\right]W_t$ as $n\rightarrow\infty$, where $(W_t)_{t\geq 0}$ is the $d$-dimensional Brownian motion starting at $\mathbf{0}$, $I$ is identity matrix and $E$ denotes the $d\times d$ matrix whose entries are all equal to $1$ and $\rho_\lambda = 2 \sqrt{\lambda}/(1+\lambda)$ is the spectral radius of $X_n$ (see \cite[Theorem~1.1]{SSSWX2018b}).
In particular, we have
\begin{eqnarray*}
\frac{(|X_n^1|,\cdots,|X_n^d|)-nv}{\sqrt{n}}\rightarrow \mathcal{N}(\mathbf{0},\Sigma)\
\end{eqnarray*}
in distribution as $n \to \infty$.
\end{thm}
\vskip 2mm
\begin{remark} Note $\Sigma=0$ when $d=1$ and $\lambda=0.$ Hence for $d=1$ and $\lambda=0,$ both
$$\frac{|X_n|-nv}{\sqrt{n}}=\frac{\vert X_0\vert+n-nv}{\sqrt{n}}\rightarrow \mathcal{N}(\mathbf{0},\Sigma)$$
and $\left(\frac{\left|X_{\lfloor nt\rfloor}\right|-nvt}{\sqrt{n}}\right)_{t\geq 0}$ converging in distribution to $Y=(Y_t)_{t\geq 0}$
are not interesting.
From Theorem \ref{thm2.1}, the following holds: For $\RW_{\lambda}$ $(X_m)_{m=0}^{\infty}$ on $\mathbb{Z}^d$ starting at any fixed vertex with $\lambda<1,$
on $D\left([0,\infty),\mathbb{R}\right),$ as $n\rightarrow\infty,$ $\left(\frac{\left|X_{\lfloor nt\rfloor}\right|-n\frac{1-\lambda}{1+\lambda}t}{\sqrt{n}}\right)_{t\geq 0}$
converges in distribution to $\left(\rho_{\lambda} B_t\right)_{t\geq 0}$ and $(B_t)_{t\geq 0}$ is the $1$-dimensional Brownian motion starting at $0.$
\end{remark}
\vskip 2mm
\noindent{\bf Proof of Theorem \ref{thm2.1}.}
We only prove IP for $\left(\frac{\left(\left|X_{\lfloor nt\rfloor}^1\right|,\cdots, \left|X_{\lfloor nt\rfloor}^d\right|\right)-nvt}{\sqrt{n}}\right)_{t\geq 0}.$
Let
$$\mathcal{F}_k=\sigma(X_0,X_1,\cdots,X_k),\ k\in\mathbb{Z}_{+}.$$
Give any $1\leq i\leq d,$ and define $f_{i}: ~\mathbb{Z}^d\rightarrow \mathbb{R}$ as follows: for any $x=(x_1,\cdots,x_d)\in\mathbb{Z}^d,$
\begin{eqnarray}\label{generator}
f_{i}(x) = \left\{\begin{array}{cl}
\frac{2}{d+\kappa_x+\lambda \left(d-\kappa_x\right)} &{\rm if} \ x_i= 0 ,\\
\ \\
\frac{1- \lambda}{d+\kappa_x+\lambda \left(d-\kappa_x\right)} &{\rm if} \ x_i\neq 0 ,\\
\end{array}
\right.
\end{eqnarray}
then for any $k\in\mathbb{N},$
$$f_i\left(X_{k-1}^i\right)=\mathbb{E}\left(\left.\left|X_{k}^i\right|-\left|X_{k-1}^i\right|\ \right| \mathcal{F}_{k-1}\right).$$
By martingale characterization theorem of Markov chain $\{|X_k^i|\}_k$, \[\left\{\left|X_k^i\right|-\left|X_{k-1}^i\right|-f_i(X_{k-1})\right\}_{k\geq 1}\] is an $\mathcal{F}_k$-adapted martingale-difference sequence, and so is
$$\left\{\xi_{n, k}^i:= \frac{1}{\sqrt{n}}\left(\left|X_k^i\right|-\left|X_{k-1}^i\right|-f_i(X_{k-1})\right)\right\}_{k= 1}\ \mbox{for any}\ n\in\mathbb{N}.$$
For each random sequence $\left\{\xi_{n,k}:=\left(\xi_{n,k}^1,\cdots,\xi_{n,k}^d\right)\right\}_{k\geq 1},$ define $M^n=(M^n_t)_{t\geq 0}$ and $A_n=(A_n(t))_{t\geq 0}$ as follows
\begin{eqnarray*}
M_t^n=\left(M_t^{n,1},\cdots,M_t^{n,d}\right)=\sum\limits_{k= 1}^{\lfloor nt\rfloor} \xi_{n,k}, \ A_n(t)=\left(A_n^{i,j}(t)\right)_{1\leq i,j\leq d}=\sum\limits_{k= 1}^{\lfloor nt\rfloor}\left(\xi_{n,k}^i\xi_{n,k}^j\right)_{1\leq i,j\leq d},\ t\geq 0.
\end{eqnarray*}
Then each $\left(M_t^{n,i}\right)_{t\geq 0}$ is an $\mathcal{F}_{\lfloor nt\rfloor}$-martingale with $M^{n,i}_0=0,$
and further each $M^n$ is a $\mathbb{R}^d$-valued $\mathcal{F}_{\lfloor nt\rfloor}$-martingale with $M_0=\mathbf{0}.$
Note that for any $t>0,$
$$\left\vert M_t^n-M^n_{t-}\right\vert\leq \left\vert \xi_{n,\lfloor nt\rfloor}\right\vert =\sum\limits_{i=1}^d\left\vert\xi_{n,\lfloor nt\rfloor}^i\right\vert\leq \frac{2d}{\sqrt{n}},$$
and for any $T\in (0,\infty),$
\begin{eqnarray}\label{bound}
\lim\limits_{n\rightarrow\infty}\mathbb{E}\left[\sup_{t\leq T}\left|M^n_t-M^n_{t-}\right|\right]\leq\lim\limits_{n \rightarrow \infty} \frac{2d}{\sqrt n} = 0.\end{eqnarray}
Fix any $1\leq i,j\leq d$ and $0\leq s<t<\infty.$ Then by the martingale property,
\begin{eqnarray*}
\mathbb{E}\left[\left. M_s^{n,j}\left(M_t^{n,i}-M_s^{n,i}\right)\right\vert \mathcal{F}_{\lfloor ns\rfloor}\right]=M_s^{n,j}
\mathbb{E}\left[\left. M_t^{n,i}-M_s^{n,i}\right\vert \mathcal{F}_{\lfloor ns\rfloor}\right]=0,
\end{eqnarray*}
and similarly $\mathbb{E}\left[\left. M_s^{n,i}\left(M_t^{n,j}-M_s^{n,j}\right)\right\vert \mathcal{F}_{\lfloor ns\rfloor}\right]=0.$ Additionally
\begin{eqnarray*}
&&\mathbb{E}\left[\left.\left(M_t^{n,i}-M_s^{n,i}\right)\left(M_t^{n,j}-M_s^{n,j}\right)\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]\\
&&=\mathbb{E}\left[\left.\sum\limits_{k=\lfloor ns\rfloor+1}^{\lfloor nt\rfloor}\xi_{n,k}^i\xi_{n,k}^j\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]+
\mathbb{E}\left[\left.\sum\limits_{\lfloor ns\rfloor+1\leq r<k\leq \lfloor nt\rfloor}\xi_{n,k}^i\xi_{n,r}^j\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]\\
&&\ \ \ \ \ + \mathbb{E}\left[\left.\sum\limits_{\lfloor ns\rfloor+1\leq r<k\leq \lfloor nt\rfloor}\xi_{n,k}^j\xi_{n,r}^i\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]\\
&&=\mathbb{E}\left[\left.\sum\limits_{k=\lfloor ns\rfloor+1}^{\lfloor nt\rfloor}\xi_{n,k}^i\xi_{n,k}^j\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]+
\sum\limits_{\lfloor ns\rfloor+1\leq r<k\leq \lfloor nt\rfloor}\mathbb{E}\left[\left.\mathbb{E}\left[\left.\xi_{n,k}^i\xi_{n,r}^j\right\vert\mathcal{F}_r\right]
\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]\\
&&\ \ \ \ \ + \sum\limits_{\lfloor ns\rfloor+1\leq r<k\leq \lfloor nt\rfloor}\mathbb{E}\left[\left.\mathbb{E}\left[\left.\xi_{n,k}^j\xi_{n,r}^i\right\vert\mathcal{F}_r\right]
\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]\\
&&=\mathbb{E}\left[\left.\sum\limits_{k=\lfloor ns\rfloor+1}^{\lfloor nt\rfloor}\xi_{n,k}^i\xi_{n,k}^j\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right]=
\mathbb{E}\left[\left. A_n^{i,j}(t)-A_n^{i,j}(s)\right\vert\mathcal{F}_{\lfloor ns\rfloor}\right].
\end{eqnarray*}
Since
\begin{eqnarray*}
&&M_t^{n,i}M_t^{n,j}-A_n^{i,j}(t)\\
&&=M_s^{n,i}M_s^{n,j}-A_n^{i,j}(s)+M_s^{n,i}\left(M_t^{n,j}-M_s^{n,j}\right)+M_s^{n,j}\left(M_t^{n,i}-M_s^{n,i}\right)\\
&&\ \ \ +\left(M_t^{n,i}-M_s^{n,i}\right)\left(M_t^{n,j}-M_s^{n,j}\right)-\left(A_n^{i,j}(t)-A_n^{i,j}(s)\right),
\end{eqnarray*}
we see that
\begin{eqnarray}\label{martingale}
\mathbb{E}\left[\left. M_t^{n,i}M_t^{n,j}-A_n^{i,j}(t)\right\vert \mathcal{F}_{\lfloor ns\rfloor}\right]=M_s^{n,i}M_s^{n,j}-A_n^{i,j}(s),
\end{eqnarray}
which implies that $A_n^{i,j}= \left[M_n^i, M_n^j\right]$ with $[X, Y]$ being the cross-variation process of $X$ and $Y$.
For any $1\leq i\neq j\leq d,$ by (\ref{facts}) and (\ref{generator}), almost surely,
\begin{eqnarray*}
f_i\left(X_{k-1}\right)=f_j\left(X_{k-1}\right)=\frac{1-\lambda}{d(1+\lambda)}\ \mbox{for large enough}\ k,
\end{eqnarray*}
and as $n\rightarrow\infty,$
\begin{eqnarray*}
&&\frac{1}{n} \sum\limits_{k=1}^{\lfloor nt \rfloor}\left(\left|X_{k}^j\right|-\left|X_{k-1}^j\right|\right)=\frac{1}{n}\left(\left|X_{\lfloor nt \rfloor}^j\right|-\left|X_{0}^j\right|\right)\rightarrow
\frac{(1-\lambda)t}{d(1+\lambda)}\\
\end{eqnarray*}
which also holds with $j$ replaced by $i$. Thus
\[
\begin{split}
&\sum_{k=1}^{\lfloor nt \rfloor} \xi_{n,k}^{i}\xi_{n, k}^{j}= \sum_{k=1}^{\lfloor nt \rfloor} \frac{(|X_{k}^i|-|X_{k-1}^i|-f_{i}(X_{k-1}))}{\sqrt{n}} \frac{(|X_{k}^j|-|X_{k-1}^j|-f_{j}(X_{k-1}))}{\sqrt{n}} \\
&\ \ \ \ =\frac{1}{n} \sum\limits_{k=1}^{\lfloor nt \rfloor} \left\{\left(|X_{k}^i|-|X_{k-1}^i|\right)(|X_{k}^j|-|X_{k-1}^j|)-
(|X_{k}^i|-|X_{k-1}^i|) f_j(X_{k-1})\right.\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left. (|X_{k}^j|-|X_{k-1}^j|) f_i(X_{k-1})+ f_i(X_{k-1}) f_j(X_{k-1})\right\}\\
&\ \ \ \ =\frac{1}{n} \sum\limits_{k=1}^{\lfloor nt \rfloor} \left\{-
(|X_{k}^i|-|X_{k-1}^i|) f_j(X_{k-1}) -(|X_{k}^j|-|X_{k-1}^j|) f_i(X_{k-1})+ f_i(X_{k-1}) f_j(X_{k-1})\right\}\\
&\ \ \ \ \rightarrow -\frac{(1-\lambda)^2}{d^2 (1+\lambda)^2}t,
\end{split}
\]
where we use the fact that $(|X_{k}^i|-|X_{k-1}^i|)(|X_{k}^j|-|X_{k-1}^j|)=0,\ k\geq 1.$
On the other hand, for any fixed $1\leq i\leq d,$ construct the following martingale-difference sequence $\left(\zeta_{n, k}^i\right)_{k\geq 1}:$
$$\zeta_{n, k}^i=\left(\sqrt{n}\xi_{n, k}^i\right)^2- \mathbb{E}\left[\left.\left(\sqrt{n} \xi_{n, k}^i\right)^2\right| \mathcal{F}_{k-1}\right],\ k\in\mathbb{N}.$$
Then for any $1\leq k<\ell<\infty,$ $\mathbb{E}\left[\zeta_{n,k}^i\right]=\mathbb{E}\left[\zeta_{n,\ell}^i\right]=0,$ and
$$\mathbb{E}\left[\zeta_{n, k}^i\zeta_{n, \ell}^i\right]=\mathbb{E}\left[\zeta_{n,k}^i\mathbb{E}\left[\left.\zeta_{n, \ell}^i\right\vert \mathcal{F}_{\ell-1}\right]\right]=0,$$
which implies $\left(\zeta_{n, k}^i\right)_k$ is a sequence of uncorrelated random variables. Notice that for any $k\in\mathbb{N},$
$$\left\vert \zeta_{n,k}^i\right\vert\leq 2\ \mbox{and hence}\ {\rm Var}\left(\zeta_{n,k}^i\right)\leq 4.$$
By the strong law of large numbers for uncorrelated random variables (\cite[Theorem~13.1]{LP2017}), we have that almost surely, as $n\rightarrow\infty,$
\begin{eqnarray}\label{lln}
\frac{1}{n}\sum\limits_{k=1}^{\lfloor nt\rfloor}\zeta^i_{n,k}=\sum\limits_{k=1}^{\lfloor nt\rfloor}\left\{
\left(\xi_{n, k}^i\right)^2- \mathbb{E}\left[\left.\left(\xi_{n,k}^i\right)^2\right| \mathcal{F}_{k-1}\right]\right\}\rightarrow 0.
\end{eqnarray}
Due to (\ref{facts}) and (\ref{generator}),
almost surely, as $n\rightarrow\infty,$
\begin{eqnarray*}
&&\sum\limits_{k=1}^{\lfloor nt \rfloor}\mathbb{E}\left[\left.\left(\xi_{n, k}^i\right)^2\right| \mathcal{F}_{k-1}\right]= \sum\limits_{k=1}^{\lfloor nt \rfloor}
\mathbb{E}\left[\left.\left(\frac{\left|X_{k}^i\right|-\left|X_{k-1}^i\right|- f_{i}(X_{k-1})}{\sqrt{n}}\right)^2\right| \mathcal{F}_{k-1}\right] \\
&&=\frac{1}{n}\sum\limits_{k=1}^{\lfloor nt \rfloor}\left\{\mathbb{E}\left[\left.\left(\left|X_{k}^i\right|-\left|X_{k-1}^i\right|\right)^2\right| \mathcal{F}_{k-1}\right]-
2\mathbb{E}\left[\left.\left(\left|X_{k}^i\right|-\left|X_{k-1}^i\right|\right)f_i(X_{k-1}^i)\right| \mathcal{F}_{k-1}\right]\right.\\
&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.+ \mathbb{E}\left[\left.\left(f_i\left(X_{k-1}^i\right)\right)^2\right| \mathcal{F}_{k-1}\right]\right\}\\
&&=\frac{1}{n} \sum\limits_{k=1}^{\lfloor nt \rfloor}\left\{\mathbb{E}\left[\left.\left(\left|X_{k}^i\right|-\left|X_{k-1}^i\right|\right)^2\right| \mathcal{F}_{k-1}\right]-\left(f_i\left(X_{k-1}^i\right)\right)^2\right\}\\
&&\rightarrow t \frac{1}{d }-t \frac{(1-\lambda)^2}{d^2 (1+\lambda)^2}.
\end{eqnarray*}
Together with (\ref{lln}), we have
\begin{eqnarray}\label{crossvariation}
\lim\limits_{n\rightarrow \infty}A_n(t)=t\Sigma.
\end{eqnarray}
Therefore (\ref{bound}), (\ref{martingale}) and (\ref{crossvariation}) implies that (\cite[Theorem~1.4]{EK2005}) on $D\left([0,\infty),\mathbb{R}^d\right),$ as $n\rightarrow\infty,$
\[\left(\frac{\left(\left|X_{\lfloor nt\rfloor}^1\right|,\cdots, \left|X_{\lfloor nt\rfloor}^d\right|\right)-nvt}{\sqrt{n}}\right)_{t\geq 0}\]
converges in distribution to a $\mathbb{R}^d$-valued process $Y :Y_t=\frac{1}{\sqrt{d}}\left[I- \frac{1-\rho_\lambda}{d}E\right]W_t$ with independent Gaussian increments such that $Y_0=\mathbf{0}$ and $Y_{t+s}-Y_s$ has the law $\mathcal{N}(\mathbf{0},t\Sigma)$
for any $0\leq s, t<\infty.$ \qed
\section{LDP for scaled reflected ${\rm RW}_\lambda$ with $\lambda\in [0,1)$}\label{sec3}
\setcounter{equation}{0}
\noindent In this section, we fix $\lambda\in [0,1)$ when $d\geq 2$ and $\lambda\in (0,1)$ when $d=1,$
then prove that the sequence $\left\{\frac{1}{n}\left(\left|X_n^1\right|, \cdots ,\left|X_n^d\right|\right)\right\}_{n=1}^{\infty}$
satisfy the LDP with a good rate function as $n\rightarrow\infty$ (Theorem \ref{thm3.1}).
Write $\mathbb{R}_+=[0,\infty).$ Let
$$s_0:=s_0(\lambda)=\frac{1}{2}\ln\lambda,\ \rho_\lambda=\frac{2\sqrt{\lambda}}{1+\lambda}.$$
Here $\rho_\lambda$ is the spectral radius of $\RW_\lambda$ as in Theorem \ref{thm2.1} and by \cite[Theorem~2.1]{SSSWX2018b}, for any $\lambda\in [0,1)$,
\begin{eqnarray*}
p_{\lambda}^{(2n)} (\mathbf{0},\mathbf{0})=\mathbb{P}\left[X_{2n}=\mathbf{0}\vert X_0=\mathbf{0}\right]\asymp \rho_\lambda^{2n} \frac{1}{n^{3d/2}}.
\end{eqnarray*}
Here for two nonnegative sequences $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty},$ $a_n\asymp b_n$ means that there are two positive constants $c_1$ and $c_2$ such that
$c_1b_n\leq a_n\leq c_2b_n$ for large enough $n.$ For any $s=(s_1, \cdots, s_d)\in\mathbb{R}^d,$ let
\begin{eqnarray}\label{psi}
\psi(s)=N(s)\frac{\rho _\lambda}{d}+\frac{1}{d(1+ \lambda)}\sum\limits_{i=1}^d\left\{\lambda e^{-s_i}+ e^{s_i}\right\}I_{\{s_i\geq s_0\}},\ \mbox{where}\ N(s)=\sum\limits_{i=1}^dI_{\{s_i<s_0\}}.
\end{eqnarray}
\begin{thm}[LDP]\label{thm3.1} Fix $\lambda\in [0,1)$ when $d\geq 2,$ and $\lambda\in (0,1)$ when $d=1.$ Assume $\RW_\lambda$ $\{X_n\}_{n=1}^{\infty}$ starts at any fixed point in $\mathbb{Z}^d.$
{\bf (i)} For any $d\in\mathbb{N},$
$\left\{\frac{1}{n}\left(\left|X_n^1\right|, \cdots ,\left|X_n^d\right|\right)\right\}_{n=1}^{\infty}$ satisfies the LDP with the following good rate function:
\begin{eqnarray*}
\Lambda^{\ast}(x)= \sup_{s\in \mathbb{R}^d}\left\{(s,x)-\ln\psi (s)\right\},\ x\in\mathbb{R}^d.
\end{eqnarray*}
In addition, when $\lambda\in (0,1),$
$$\mathcal{D}_{\Lambda^\ast}:=\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)<\infty\right\}=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d_{+}:\ 0\leq \sum\limits_{i=1}^dx_i\leq 1\right\},$$
$$\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)=0\right\}=\left\{\left(\frac{1-\lambda}{d(1+\lambda)},\cdots,\frac{1-\lambda}{d(1+\lambda)}\right)\right\};$$
and when $d\geq 2$ and $\lambda=0,$
$$\mathcal{D}_{\Lambda^\ast}(0):=\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)<\infty\right\}=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d_{+}:\ \sum\limits_{i=1}^dx_i=1\right\},$$
$$\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)=0\right\}=\left\{\left(\frac{1}{d},\cdots,\frac{1}{d}\right)\right\}.$$
{\bf (ii)} In particular, when $d=1, \lambda \in (0, 1),$
\begin{eqnarray*}
\Lambda^{\ast}(x)=\left\{\begin{array}{cl}
\frac{x}{2}\ln\lambda - \ln \rho_\lambda+ (1+x)\ln \sqrt{1+ x}+ (1-x)\ln\sqrt{1-x}, & x\in [0,1],\\
\ \\
+\infty, & otherwise;
\end{array}
\right.
\end{eqnarray*}
and when $d= 2, \lambda \in (0, 1),$
\begin{eqnarray*}
\Lambda^{\ast}(x)=\left\{\begin{array}{cl}
\frac{1}{2}(x_1+x_2)\ln\lambda-\ln\rho_\lambda +\overline{\Lambda^*}(x), & (x_1, x_2)\in\mathcal{D}_{\Lambda^*},\\
\ \\
+\infty, & otherwise,
\end{array}
\right.
\end{eqnarray*}
where for any $x=(x_1,x_2)\in\mathbb{R}^2_+$ with $x_1+x_2\leq 1,$
\begin{eqnarray*}
&&\overline{\Lambda^*}(x)=x_1\ln \left[\frac{2x_1+\sqrt{x_1^2-x_2^2+1}}{\sqrt{(x_1^2-x_2^2)^2+1-2(x_1^2+x_2^2)}}\right]+x_2\ln \left[\frac{2x_2+\sqrt{x_2^2-x_1^2+1}}{\sqrt{(x_1^2-x_2^2)^2+1-2(x_1^2+x_2^2)}}\right]\\
\ \\
&&\ \ \ \ \ \ \ \ \ \ \ - \ln \left[\frac{\sqrt{x_1^2-x_2^2+1}+\sqrt{x_2^2-x_1^2+1}}{\sqrt{(x_1^2-x_2^2)^2+1-2(x_1^2+x_2^2)}}\right]+\ln 2.
\end{eqnarray*}
In addition, when $d\geq 2, \lambda=0,$
\begin{eqnarray*}
&&\Lambda^\ast (x)=\left\{\begin{array}{cl}
\ln d+\sum\limits_{i=1}^dx_i\ln x_i,\ &x\in\mathcal{D}_{\Lambda^\ast}(0),\\
\ \\
+\infty, \ & othwise.
\end{array}
\right.
\end{eqnarray*}
\end{thm}
\vskip 2mm
\begin{remark}
{\bf (i)} When $d\geq 3$ and $\lambda\in (0,1),$ we can not calculate $\Lambda^\ast$ out to get an explicit expression.
But recall that the rate function of the Cram\'{e}r theorem for SRW is
$$\overline{\Lambda^*}(x)=\sup\limits_{y\in\mathbb{R}^d}\ln\left\{\frac{e^{\sum\limits_{i=1}^dy_ix_i}}{\frac{1}{2d}\sum\limits_{i=1}^d(e^{-y_i}+e^{y_i})}\right\},\ x\in\mathbb{R}^d.$$
By (\ref{rate1}), for any $x=(x_1,\cdots,x_d)\in\mathbb{R}_+^d,$
$$\Lambda^*(x)=\frac{1}{2}\sum\limits_{i=1}^dx_i\ln\lambda-\ln\rho_\lambda+\overline{\Lambda^*}(x).$$
To obtain an explicit rate function when $d\geq 3$ and $\lambda\in (0,1)$, it remains as an open problem.
{\bf (ii)} Assume $\lambda\in [0,1)$ if $d\geq 2$ and $\lambda\in (0,1)$ if $d=1.$ The following sample path large deviation (Mogulskii type theorem) holds for the reflected $\RW_\lambda$ starting at $\mathbf{0}.$ For any $0\leq t\leq 1,$ let
\begin{eqnarray*}
&&Z_n(t)=\frac{1}{n}\left(\left\vert X_{\lfloor nt\rfloor}^1\right\vert,\cdots,\left\vert X_{\lfloor nt\rfloor}^d\right\vert\right),\\
&&\widetilde{Z}_n(t)=Z_n(t)+\left(t-\frac{\lfloor nt\rfloor}{n}\right)\left(\left(\left\vert X_{\lfloor nt\rfloor+1}^1\right\vert,\cdots,\left\vert X_{\lfloor nt\rfloor+1}^d\right\vert\right)-\left(\left\vert X_{\lfloor nt\rfloor}^1\right\vert,\cdots,\left\vert X_{\lfloor nt\rfloor}^d\right\vert\right)\right).
\end{eqnarray*}
Write $L_{\infty}([0,1])$ for the $\mathbb{R}^d$-valued $L_{\infty}$ space on interval $[0,1],$ and $\mu_n$ (resp. $\widetilde{\mu}_n$) the law of $Z_n(\cdot)$ (resp. $\widetilde{Z}_n(\cdot)$) in $L_{\infty}([0,1]).$ From \cite[Lemma~5.1.4]{DZ1998}, $\mu_n$ and $\tilde{\mu}_n$ are exponentially equivalent.
Denote by $\mathcal{AC}^{'}$ the space of non-negative absolutely continuous $\mathbb{R}^d$-valued functions $\phi$ on $[0,1]$ such that
$$||\phi||\leq 1,\ \phi(0)=\mathbf{0}$$
where $\Vert\cdot\Vert$ is the supremum norm.
Let
$C_{\mathbf{0}}([0,1])=\left\{\left. f:\ [0,1]\rightarrow\mathbb{R}^d\right\vert f\ \mbox{is continuous},\ f(0)=\mathbf{0}\right\},$
and
$$\mathcal{K}=\left\{f\in C_{\mathbf{0}}([0,1])\left\vert \sup\limits_{0\leq s<t\leq 1}\frac{\left\vert f(t)-f(s)\right\vert}{t-s}\leq 1,\
\sup\limits_{0\leq t\leq 1}\vert f(t)\vert\leq 1\right.\right\}.$$
By the Arzel\`{a}-Ascoli theorem, $\mathcal{K}$ is compact in $\left(C_{\mathbf{0}}([0,1]),\Vert\cdot\Vert\right)$. Note that each $\widetilde{\mu}_n$ concentrates on $\mathcal{K}$ which implies exponential tightness in $C_0[0, 1]$ equipped with supremum topology. Given that the finite dimensional LDPs for $\mu_n\circ p_j^{-1}$ in $(\R^d)^{|j|}$ follows from Theorem \ref{thm3.1}, $\mu_n\circ p_j^{-1}$ and $\tilde{\mu}_n\circ p_j^{-1}$ are exponential equivalent, $\tilde{\mu}_n$ satisfy the LDP in $L_{\infty}[0,1]$ with pointwise convergence topology by Dawson-G\"{a}rtner theorem (\cite{DZ1998}). Together with exponential tightness in $C_0[0, 1]$, we can lift LDP of $\tilde{\mu}_n$ to $L_{\infty}[0,1]$ with supremum topology with the good rate function \[
I(\phi)=\left\{\begin{array}{cl}
\int_0^1\Lambda^*\left(\dot{\phi}(t)\right)\ {\rm d}t &{\rm if}\ \phi\in\mathcal{AC}^{'},\\
\ \\
\infty & {\rm otherwise}.
\end{array}
\right.
\]
The same holds for $\mu_n$ due to exponential equivalence. The proof of the above sample path LDP is similar to that of \cite[Theorem~5.1.2]{DZ1998}.
\end{remark}
\vskip 2mm
For any $n\in\mathbb{N},$ $s=(s_1,\cdots,s_d)\in \mathbb{R}^d$ and $x\in\mathbb{Z}^d,$ let
\[
\Lambda_n (s,x)= \ln\mathbb{E}_x\left[\exp\left\{\left(n s, \left(\frac{|X^1_n|}{n},\cdots,\frac{\vert X^d_n\vert}{n}\right)\right)\right\}\right]
=\ln\mathbb{E}_x\left[\exp\left\{\sum\limits_{i=1}^d s_i|X_n^i|\right\} \right].
\]
To prove Theorem \ref{thm3.1}, we firstly prove the following lemmas.
\begin{lem}
\label{L:Lambda0}
For each $s \in \R^d$ and $x \in \Z^d$, there exists a constant $C > 0$ such
that
\begin{equation}
\label{e:Lambda0}
\Lambda_{n-|x|}(s,\, x) - C \leq \Lambda_n(s,\, \zero) \leq \Lambda_{n+|x|}(s,\, x) + C, \quad n > |x|.
\end{equation}
\end{lem}
\begin{proof}
By the Markov property, we have
\[ \E_\zero \left[ e^{\sum_{i=1}^d s_i |X_n^i|} \right]
\geq
\E_0 \left[ I_{\{X_{|x|}=x\}} e^{\sum_{i=1}^d s_i |X_n^i|} \right]
\geq
\P_{\zero} \left( X_{|x|} = x \right) \E_x \left[ e^{\sum_{i=1}^d s_i
|X_{n-|x|}^i|} \right],
\]
and the first inequality in \eqref{e:Lambda0} follows. The second inequality
is proved similarly.
\end{proof}
Let $e_1$, $\ldots$, $e_d$ be the standard unit vectors in $\mathbb{Z}^d$. Suppose
$\left\{Z_n=\left(Z_n^1,\,\cdots,\,Z_n^d\right)\right\}_{n=0}^{\infty}$ is a
drifted random walk on $\mathbb{Z}^d$ such that for any $1\leq i\leq d$ and
$n\in\mathbb{Z}_+$,
\begin{equation}
\label{e:Zn}
\mathbb{P}\left (Z_{n+1}=Z_n+e_i \,\big|\, Z_0,\,\cdots,\, Z_n\right)=\frac{1}{d(1+ \lambda)},\quad
\mathbb{P}\left(Z_{n+1}=Z_n-e_i \,\big|\, Z_0,\,\cdots,\,Z_n\right)=\frac{\lambda}{d(1+ \lambda)}.
\end{equation}
\begin{lem}
\label{L:hkupper}
We have that for $k \in \Z_+^d$,
\begin{equation*}
\label{e:hkupper}
\P_\zero \left( X_n = k \right) \leq \P \left( Z_n = k \,\big|\, Z_0 = \zero \right), \quad n \geq 0.
\end{equation*}
\end{lem}
\begin{proof}
For $x$, $k \in \Z^d$ and $n \in \N$, let $\Gamma_n(x,\, k)$ be the set of all
nearest-neighbor paths in $\Z^d$ from $x$ to $k$ with length $n$. For a path $\gamma =
\gamma_0\gamma_1\cdots\gamma_n \in \Gamma_n(x,\, k)$, let
\[ p(\gamma) = p(x,\, \gamma_1) p(\gamma_1,\, \gamma_2) \cdots
p(\gamma_{n-1},\, k). \]
Consider the first $n$ steps of $\RW_{\lambda}$ along the path $\gamma \in
\Gamma_n(0,\, k)$. Each
time the transition probability for the walk is either $\frac{1}{d + m +
(d-m)\lambda}$ (with $0 \leq m \leq d$) or $\frac{\lambda}{d+m +
(d-m)\lambda}$ (with $0 \leq m \leq d-1$). The total number of probability
terms of the forms $\frac{1}{d + m +
(d-m)\lambda}$ (resp. $\frac{\lambda}{d + m +
(d-m)\lambda}$) is exactly $\frac{n + |k|}{2}$ (resp. $\frac{n - |k|}{2}$).
Note that $d+m + (d-m)\lambda \geq d(1+\lambda)$. Therefore, we have for
$\gamma \in \Gamma_n(\zero,\, k)$,
\[ p(\gamma) \leq \left( \frac{1}{d(1+\lambda)} \right)^{\frac{n+|k|}{2}}
\left( \frac{\lambda}{d(1+\lambda)} \right)^{\frac{n-|k|}{2}}. \]
As a consequence,
\[ \P_{\zero}(X_n = k) = \sum_{\gamma \in \Gamma_n(\zero,\, k)} p(\gamma) \leq
\sum_{\gamma \in \Gamma_n(\zero,\, k)} \left( \frac{1}{d(1+\lambda)} \right)^{\frac{n+|k|}{2}}
\left( \frac{\lambda}{d(1+\lambda)} \right)^{\frac{n-|k|}{2}} = \P \left(
Z_n = k \, \big|\, Z_0 = \zero \right). \]
\end{proof}
\begin{lem}
\label{L:hklower}
For every $k \in \Z_+^d$ and $z \in \Z_+^d \setminus \mathcal{X}$,
we have that
\begin{equation*}
\label{e:hklower}
\P_z \left( X_n = k \right) \geq n^{-d} \P \left( Z_n = k \,\big|\, Z_0 = z \right).
\end{equation*}
\end{lem}
\begin{proof}
Recall that $\mathcal{X}$ is the boundary of $\Z_+^d$. Define
\[ \sigma = \inf \left\{ n:\ X_n \in \mathcal{X} \right\}, \quad \tau = \inf
\left\{ n:\ Z_n \in \mathcal{X} \right\}. \]
Starting at $z \in \Z_+^d \setminus \mathcal{X}$, the process $\left( X_n
\right)_{0 \leq n \leq \sigma}$ has the same distribution as the drifted
random walk $(Z_n)_{0 \leq n \leq \tau}$. Then we have for $k \in \Z_+^d$,
\begin{equation}
\label{e:XnZn}
\P_z \left( X_n = k \right) \geq \P \left( Z_n = k,\, n \leq \tau
\,\big|\, Z_0 = z \right).
\end{equation}
For $\alpha$, $\beta \in \Z$ and $n \in \N$, denote by $P_{n,\,
\beta}(\alpha)$ the number of
paths $\gamma = \gamma_0 \gamma_1 \cdots \gamma_n$ in $\Z$ with $\gamma_0 = \alpha$
and $\gamma_n = \beta$, and by $Q_{n,\, \beta}(\alpha)$ the number of those
paths with
additional property that $\gamma_i \geq \alpha \wedge \beta$ for $0 \leq i
\leq n$. By \cite[Theorem~4.3.2 and Lemma~4.3.3]{DR2010}, we have that
\begin{equation}
\label{e:PnQn}
Q_{n,\, \beta}(\alpha) \geq \frac{|\alpha-\beta| \vee 1}{n} P_{n,\, \beta}(\alpha).
\end{equation}
For $k = (k_1,\, \cdots,\, k_d) \in \Z_+^d$ and $z = (z_1,\, \cdots,\, z_d)
\in \Z_+^d \setminus \sX$, write $a = \frac{n + |k| - |z|}{2}$ and $b =
\frac{n - |k| + |z|}{2}$. By \eqref{e:PnQn}, we obtain that
\begin{align*}
& \P \left( Z_n = k,\, n \leq \tau \vert Z_0= z\right) \\
\geq &
\sum_{m_1+\cdots+m_d = n} {n \choose m_1,\, \cdots,\, m_d} d^{-n} \left( \frac{\lambda}{1+\lambda} \right)^b \left( \frac{1}{1+\lambda} \right)^a \prod_{m_j > 0} Q_{m_j,\, k_j}(z_j) \\
\geq &
\sum_{m_1+\cdots +m_d = n} {n \choose m_1,\, \cdots,\, m_d} d^{-n} \left( \frac{\lambda}{1+\lambda} \right)^b \left( \frac{1}{1+\lambda} \right)^a \prod_{m_j > 0} \frac{|k_j-z_j|\vee 1}{m_j} P_{m_j,\, k_j}(z_j) \\
\geq &
n^{-d} \sum_{m_1+\cdots+m_d=n} {n \choose m_1,\, \cdots,\, m_d} d^{-n} \left( \frac{\lambda}{1+\lambda} \right)^b \left( \frac{1}{1+\lambda} \right)^a \prod_{m_j > 0} P_{m_j,\, k_j}(z_j) \\
= &
n^{-d} \P \left( Z_n = k \,\big|\, Z_0 = z \right),
\end{align*}
which together with \eqref{e:XnZn} proves this lemma.
\end{proof}
Recall that $s_0 = \frac{1}{2} \ln \lambda$.
For fixed $s \in \R^d$, let $I_1 = \left\{ 1 \leq i \leq d:\ s_i < s_0 \right\}$
and $I_2 = \left\{ 1,\, \cdots,\, d \right\} \setminus I_1$. Define $s^{I_1} =
\left( s_i \right)_{i \in I_1}$ and $s^{I_2}$, $Z_n^{I_1}$, $Z_n^{I_2}$ are
defined similarly.
\begin{lem}
\label{L:orthant}
Let $y_0 = \zero^{I_1}$ if $n$ is even and $y_0 = (1,\, 0,\, \cdots,\, 0) \in \R^{N(s)}$
otherwise.
For any $z \in \Z^d$ and $s \in
\R^d$,
we have that
\begin{equation*}
\label{e:orthant}
\E \left[ e^{\sum_{i=1}^d s_i Z_n^i} I_{\{Z_n^i \geq z_i\}} \,\big|\, Z_0 = z \right]
\geq
2^{-|I_2|} \left( e^{s_1} \wedge 1 \right) \E \left[ e^{\sum_{i\in I_2} s_i Z_n^i} I_{\{Z_n^{I_1} = z^{I_1} + y_0\}} \,\big|\, Z_0 = z \right].
\end{equation*}
\end{lem}
\begin{proof}
Without loss of generality, we assume that $z = \zero$.
As in the proof of Lemma~\ref{L:hklower}, for $\alpha$, $\beta \in \Z$, we
denote by $P_{n,\,\beta}(\alpha)$ the number of nearest-neighbor paths in $\Z$
from $\alpha$ to $\beta$. Then we have, for $k \in \Z^d$,
\begin{align}\label{orthant1}
\nonumber& \P \left( Z_n = k \,\big|\, Z_0 = \zero \right) \\
=& \sum_m {n \choose m_1,\, \cdots,\, m_d} d^{-n} \left( \frac{\lambda}{1+\lambda}
\right)^{\sum\limits_{i=1}^d \frac{m_i-k_i}{2}} \left( \frac{1}{1+\lambda}
\right)^{\sum\limits_{i=1}^d \frac{m_i + k_i}{2}} \prod_{m_j>0} P_{m_j,\, k_j}(0), \\
=\nonumber& \lambda^{\frac{n - \sum_{i=1}^d k_i}{2}} \sum_m {n \choose m_1,\, \cdots,\, m_d} \left(\frac{1}{d(1+\lambda)}\right)^n \prod_{m_j>0} P_{m_j,\, k_j}(0),
\end{align}
where the sum is over all tuples $m = (m_1,\, \cdots,\, m_d) \in \Z^d$ such that
$|m| = n$ and $m_j \geq |k_j|$ for $1 \leq j \leq d$.
Note that
\[
\E \left[ e^{\sum_{i=1}^d s_i Z_n^i} I_{\{Z_n \in \Z_+^d\}} \,\big|\, Z_0 =
\zero \right]
\geq
\left( e^{s_1} \wedge 1 \right) \E \left[ e^{\sum_{i \in I_2} s_i Z_n^i} I_{\{
Z_n^{I_1} = y_0,\, Z_n^{I_2} \in \Z_+^{I_2} \}} \,\big|\, Z_0 = \zero \right]. \]
For $\epsilon = \left( \epsilon_i \right)_{i \in I_2} \in \left\{ -1,\, 1
\right\}^{I_2}$, let
\[ \sO_{\epsilon} = \left\{ y = \left( y_i \right)_{i \in I_2} \in \Z^{I_2} :\
\epsilon_i y_i \geq 0,\ i \in I_2 \right\}. \]
Note that for every $k \in \Z^d$, $P_{m_j,\, k_j}(0) = P_{m_j,\,
\epsilon_j k_j}(0)$. Therefore, by (\ref{orthant1})
\[ \P \left( Z_n = k \,\big|\, Z_0 = \zero \right)
=
\lambda^{\sum_{i=1}^d \frac{1}{2} \left( \epsilon_i - 1 \right) k_i}
\P \left( Z_n = \left( \epsilon_1 k_1,\, \cdots,\, \epsilon_d k_d \right)
\,\big|\, Z_0 = 0 \right).
\]
Applying the fact that $e^{s_i} \lambda^{-1/2} \geq 1$ for $i \in I_2$, we
obtain that for every $\epsilon \in \left\{ -1,\, 1 \right\}^{I_2}$,
\begin{align*}
& \E \left[ e^{\sum_{i \in I_2} s_i Z_n^i} I_{\{
Z_n^{I_1} = y_0,\, Z_n^{I_2} \in \Z_+^{I_2} \}} \,\big|\, Z_0 = \zero \right] \\
\geq&
\sum_{k \in \Z_+^{I_2}}
\left( e^{s_i} \lambda^{-1/2} \right)^{\sum_{i \in I_2} \left( 1 - \epsilon_i \right) k_i}e^{\sum_{i\in I_2} s_i \epsilon_i k_i} \P \left( Z_n^{I_1} = y_0,\, Z_n^{I_2} = \left( \epsilon_i k_i\right)_{i\in I_2} \,\big|\, Z_0 = \zero \right) \\
\geq &
\E \left[ e^{\sum_{i\in I_2} s_i Z_n^i} I_{\{ Z_n^{I_1} = y_0, Z_n^{I_2} \in \sO_{\epsilon}\}} \,\big|\, Z_0 = \zero \right].
\end{align*}
By taking sum over all $\epsilon \in \{-1,\, 1\}^{I_2}$, we complete the proof of
this lemma.
\end{proof}
Recall the definition of $\psi$ from \eqref{psi}.
\begin{lem}
\label{L:Lambdau}
For every $s \in \R^d$ and $x \in \Z^d$ we have that
\[ \limsup_{n \to \infty} \frac{1}{n} \Lambda_n(s,\, x) \leq \ln \psi(s). \]
\end{lem}
\begin{proof}
By Lemma~\ref{L:Lambda0}, we only need to consider the case $x = \zero$.
Recall that $s_0:= \frac{1}{2} \ln \lambda$.
For $s \in \R^d$, let $\tilde{s} := \left( s_1\vee s_0,\, \cdots, \, s_d
\vee s_0 \right)$. Since $\Lambda_n(s,\, 0)$ is increasing in each
coordinate $s_i$, we have that $\Lambda_n(s,\, 0) \leq \Lambda_n(\tilde{s},\,
0)$. Let $(Z_n)$ be the drifted random walk defined by \eqref{e:Zn}. Then, by
Lemma~\ref{L:hkupper},
\begin{align*}
& \E_{\zero} \left[ e^{\sum_{i=1}^d \tilde{s}_i |X_n^i|} \right] \leq 2^d \sum_{k \in \Z_+^d} e^{\sum_{i=1}^d \tilde{s}_i k_i}
\P \left( Z_n = k \,\big|\, Z_0 = \zero \right) \leq 2^d \E \left[
e^{\sum_{i=1}^d \tilde{s}_i Z_n^i} \,\big|\, Z_0 = \zero \right] \\
=& 2^d
\left( \sum_{i=1}^d \frac{\lambda e^{-\tilde{s}_i} +
e^{\tilde{s}_i}}{d(1+\lambda)} \right)^n = 2^d \left( \psi(s) \right)^n.
\end{align*}
The lemma follows immediately.
\end{proof}
\begin{lem}
\label{L:Lambdal}
For every $s \in \R^d$ and $x \in \Z^d$ we have that
\[ \liminf_{n \to \infty} \frac{1}{n} \Lambda_n(s,\, x) \geq \ln \psi(s). \]
\end{lem}
\begin{proof}
It suffices to prove only for $n$ even.
We use the same notations as in Lemma~\ref{L:orthant}. Fix $x \in \Z_+^d
\setminus \sX$. By Lemma~\ref{L:hklower} and~\ref{L:orthant}, we have
\begin{align*}
\E_x \left[ e^{\sum_{i=1}^d s_i |X_{2n}^i|} \right]
\geq&
n^{-d} \sum_{k \in \Z_+^d} e^{\sum_{i=1}^d s_i k_i} \P \left( Z_{2n} = k \,\big|\, Z_0 = x \right) \\
=& n^{-d} \E \left[ e^{\sum_{i=1}^d s_i Z_{2n}^i} I_{\{Z_{2n} \in \Z_+^d\}} \,\big|\, Z_0 =x \right] \\
\geq &
c n^{-d} \E \left[ e^{\sum_{i \in I_2} s_i Z_{2n}^i} I_{\{Z_{2n}^{I_1} = x^{I_1} \}} \,\big|\, Z_0 = x \right]
\end{align*}
for some constant $c > 0$. Let $\Gamma_1(2n)$ be the
number of nearest-neighbor paths of length $2n$ in $\Z^{I_1}$ from $\zero^{I_1}$ to
$\zero^{I_1}$. Similarly, for $k \in \Z^{I_2}$, let $\Gamma_2(2n,\, k)$ be
the number of paths of
length $2n$ in $\Z^{I_2}$ from $\zero^{I_2}$ to $k$. Then we have for $k \in
\Z^{I_2}$,
\[
\P \left( Z_{2n}^{I_1} = x^{I_1}, \, Z_{2n}^{I_2} = x^{I_2} + k \,\big|\,
Z_0 = x \right)
=
\sum_{m=0}^{n} {2n \choose 2m} \frac{\lambda^{n - \sum_{i \in I_2}
k_i}}{\left[ d \left( 1+\lambda \right) \right]^{2n}} \Gamma_1(2m)
\Gamma_2(2n-2m,\, k).
\]
Recall that $N(s) = |I_1|$.
Let $(W_n)$ be the drifted random walk in $\Z^{I_2}$ starting at $\zero$, that
is, the transition
probability is given by \eqref{e:Zn} with $d$ replaced by $d - N(s)$. Since
$\Gamma_1(2m) \geq c m^{-N(s)/2} \left( 2 N(s) \right)^{2m}$, we obtain that
\begin{align*}
& \P \left( Z_{2n}^{I_1} = x^{I_1}, \, Z_{2n}^{I_2} = x^{I_2} + k \,\big|\,
Z_0 = x \right) \\
\geq&
c n^{-N(s)/2} d^{-2n} \sum_{m=0}^{n} {2n \choose 2m} \left( N(s) \rho_{\lambda} \right)^{2m}
\left( d - N(s) \right)^{2n-2m} \P \left( W_{2n-2m} = k \right).
\end{align*}
Therefore,
\begin{align*}
& \E_x \left[ e^{\sum_{i=1}^d s_i |X_{2n}^i|} \right] \\
\geq&
n^{-3d/2} d^{-2n}\sum_{m=0}^n \sum_{k \in \Z^{I_2}} e^{\sum_{i \in I_2} s_i (k_i+ x_i)} {2n \choose 2m} \left(
N(s) \rho_{\lambda} \right)^{2m} \left( d-N(s) \right)^{2n-2m} \P \left(
W_{2n-2m} = k \right) \\
= & c n^{-3d/2} d^{-2n}\sum_{m=0}^n {2n \choose 2m} \left(
N(s) \rho_{\lambda} \right)^{2m} \left( d-N(s) \right)^{2n-2m} \E \left[ e^{\sum_{i \in I_2} s_i W_{2n-2m}^i} \right] \\
= & c n^{-3d/2} d^{-2n} \sum_{m=0}^n {2n \choose 2m} \left(
N(s) \rho_{\lambda} \right)^{2m} \left( d-N(s) \right)^{2n-2m} \left( \frac{\sum_{i \in I_2} \left( \lambda e^{-s_i} + e^{s_i} \right)}{(d-N(s)) (1+\lambda)} \right)^{2n-2m} \\
\geq & \frac{c}{3} n^{-3d/2} \left( \psi(s) \right)^{2n}.
\end{align*}
The last inequality holds since for any positive real number $a, b,$
\[\frac{\sum\limits_{m=0}^{n} \binom{2n}{2m} a^{2n-2m} b^{2m}}{\left(a+ b\right)^{2n}}\rightarrow \frac{1}{2}, \ \ n\rightarrow \infty.
\]
The proof is finished.
\end{proof}
Combining Lemma~\ref{L:Lambdau} and~\ref{L:Lambdal}, we get the following
result.
\begin{cor}
\label{C:limit}
For every $s \in \R^d$ and $x \in \Z^d$, we have
\begin{equation*}
\label{e:limit}
\lim_{n \to \infty} \frac{1}{n} \Lambda_n(s,\, x) = \ln \psi(s).
\end{equation*}
\end{cor}
Define $\Lambda(s) := \ln \psi(s)$ and let $\Lambda^{\ast}(x) = \sup\limits_{s \in \R^d}
\left\{ \left( s,\, x \right) - \Lambda(s) \right\}$ be Fenchel-Legendre
transform of $\Lambda$.
\begin{lem}\label{property0-1} Give any $d\geq 1$ and $\lambda\in (0,1).$ The effective domain of $\Lambda^{\ast}(\cdot)$ is
$$\mathcal{D}_{\Lambda^{\ast}}(\lambda):=\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)<\infty\right\}=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d:\ \sum\limits_{i=1}^dx_i\leq 1,\ 0\leq x_i\leq 1,\ 1\leq i\leq d\right\}.$$
Furthermore, $\Lambda^{\ast}(\cdot)$ is strictly convex in
$x=(x_1,\cdots,x_d)\in \mathcal{D}_{\Lambda^{\ast}}(\lambda)$ with
$\sum\limits_{i=1}^dx_i<1,$ and $\Lambda^{\ast}(x) = 0$ if and only if $x =
\left(\frac{1-\lambda}{d(1+\lambda)},\cdots,\frac{1-\lambda}{d(1+\lambda)}\right)$.
\end{lem}
\pf
Note that $s_0=\frac{1}{2}\ln\lambda.$
Firstly, let $$\mathcal{D}=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d: 0\leq x_i\leq \sum\limits_{j=1}^{d}x_j\leq 1,\ 1\leq i\leq d\right\}.$$
Then for any $x\in \mathcal{D}$, we have
\begin{eqnarray}\label{rate1}
\nonumber \Lambda^*(x)&=&\sup\limits_{s_i\geq s_0,1\leq i\leq d}\ln\left\{\frac{1}{2}(1+\lambda)\frac{e^{\sum\limits_{i=1}^ds_ix_i}}{\frac{1}{2d}\sum\limits_{i=1}^d(\lambda e^{-s_i}+e^{s_i})}\right\}\\
\nonumber&=&\sup\limits_{y_i\geq 0,1\leq i\leq d}\ln\left\{\frac{1}{2}(1+\lambda)\lambda^{-\frac{1}{2}+\frac{1}{2}\sum\limits_{i=1}^dx_i}\frac{e^{\sum\limits_{i=1}^dy_ix_i}}{\frac{1}{2d}\sum\limits_{i=1}^d(e^{-y_i}+e^{y_i})}\right\}\\
&=&\frac{1}{2}\sum\limits_{i=1}^d x_i \ln (\lambda)- \ln(\rho_{\lambda})+\sup\limits_{y_i\geq 0,1\leq i\leq d}\ln\left\{\frac{e^{\sum\limits_{i=1}^dy_ix_i}}{\frac{1}{2d}\sum\limits_{i=1}^d(e^{-y_i}+e^{y_i})}\right\}< \infty.
\end{eqnarray}
In fact, for any $s=(s_1,\cdots,s_d)\in\mathbb{R}^d,$ $\sum\limits_{i=1}^ds_ix_i$ is increasing in each $s_i$ and
$\psi(s)=\psi\left(\tilde{s}\right)$, where $\tilde{s}$ is defined in Lemma \ref{L:Lambdau}. And for $x \in \mathcal{D}^c,$ it's easy to verify that $\Lambda^*(x)= \infty.$
By \cite[Lemma~2.3.9]{DZ1998}, $\Lambda^*$ is a good convex rate function. It's obvious that the Hessian matrix of $\Lambda(s)$ is positive-definite which implies strict concavity of $s\cdot x- \Lambda(s)$, thus the local maximum of $s\cdot x- \Lambda(s)$ exists uniquely and is attained at a finite solution $s=s(x)$, i.e.
\begin{eqnarray*}\label{solution}
\Lambda^{\ast}(x)=s(x)\cdot x- \Lambda(s(x)).
\end{eqnarray*}
Then according to implicit function theorem, $\Lambda^{\ast}(\cdot)$ in $\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}_+^d:\ \sum\limits_{i=1}^{d} x_i< 1\right\}$ is strictly convex.
Finally, due to the strict convexity of $\Lambda^{*}(\cdot)$ and (\ref{rate1}), $
\left(\frac{1-\lambda}{d(1+\lambda)},\cdots,\frac{1-\lambda}{d(1+\lambda)}\right)$ is the unique solution of $\Lambda^{*}(x)= 0$.
\qed
\vskip 3mm
Assume $d=1, 2 \mbox{ and } \lambda\in (0,1)$, we can get explicit formula of $\Lambda^{*}$ by calculating rate function $\overline{\Lambda^{*}}$ of SRW which is omitted here. For $\lambda=0$, we have the following explicit expression.\\
\begin{lem}\label{property0}
For any $d\geq 2$ and $\lambda=0,$
\begin{eqnarray*}
&&\mathcal{D}_{\Lambda^{\ast}}(0):=\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)<\infty\right\}=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d_+:\ \sum\limits_{i=1}^dx_i=1\right\},\\
&&\Lambda^\ast (x)=\left\{\begin{array}{cl}
\ln d+\sum\limits_{i=1}^dx_i\ln x_i,\ &x\in\mathcal{D}_{\Lambda^\ast}(0),\\
\ \\
+\infty, \ & otherwise,
\end{array}
\right.\\
&&\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)=0\right\}=\left\{\left(\frac{1}{d},\cdots,\frac{1}{d}\right)\right\}.
\end{eqnarray*}
\end{lem}
\pf Clearly
$$\mathcal{D}_{\Lambda^{\ast}}(0)\subseteq \left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d_+:\ \sum\limits_{i=1}^dx_i=1\right\}.$$
Assume firstly $x_i> 0, 1\leq i\leq d.$ Let $y_i=e^{s_i},\ 1\leq i\leq d,$ then by the Jensen inequality,
\begin{eqnarray*}
&&s\cdot x-\ln\left(\frac{1}{d}\sum\limits_{i=1}^de^{s_i}\right)=\sum\limits_{i=1}^dx_i\ln\frac{y_i}{x_i}-\ln\left(\sum\limits_{i=1}^dy_i\right)+ \sum\limits_{i=1}^dx_i\ln x_i+\ln d \\
&&\ \ \ \ \leq \ln\left(\sum\limits_{i=1}^d x_i\frac{y_i}{x_i}\right)-\ln\left(\sum\limits_{i=1}^dy_i\right)+\sum\limits_{i=1}^dx_i\ln x_i+\ln d\\
&&\ \ \ \ =\sum\limits_{i=1}^dx_i\ln x_i+\ln d,
\end{eqnarray*}
and the inequality in the second line becomes equality only if each $s_i=\ln x_i.$
If there exists some $i$ such that $x_i=0,$ we still have that
\begin{eqnarray*}
s\cdot x-\ln\left(\frac{1}{d}\sum\limits_{i=1}^de^{s_i}\right)\leq \sum\limits_{i=1}^dx_i\ln x_i+\ln d,
\end{eqnarray*}
then $\sup\limits_{s\in\mathbb{R}^d}\left\{s\cdot x-\ln\left(\frac{1}{d}\sum\limits_{i=1}^de^{s_i}\right)\right\}=\sum\limits_{i=1}^dx_i\ln x_i+\ln d$ by lower semi-continuity of $\Lambda^*(\cdot).$
Hence
\begin{eqnarray*}
&&\mathcal{D}_{\Lambda^{\ast}}(0)=\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}^d_+:\ \sum\limits_{i=1}^dx_i=1\right\},\\
&&\left\{x\in\mathbb{R}^d:\ \Lambda^\ast(x)=0\right\}=\left\{\left(\frac{1}{d},\cdots,\frac{1}{d}\right)\right\}.
\end{eqnarray*}
\qed
\vskip 3mm
Denote by $\mathbb{F}$ the set of exposed points of $\Lambda^{\ast}(\cdot)$ whose exposing hyperplane belong to $\mathcal{D}_{\Lambda}^{o}$.
Here $y\in\mathbb{R}^d$ is an exposed point of $\Lambda^\ast$ if for some $s\in\mathbb{R}^d,$
$$(y,s)-\Lambda^\ast(y)>(x,s)-\Lambda^\ast(x),\ \forall x\in\mathbb{R}^d\setminus\{y\};$$
and we call the above $s$ an exposing hyperplane.
\begin{lem}\label{infimumattained}
For any open set $G$ of $\mathbb{R}^d,$
\[\inf_{x\in G\cap{\mathbb{F}}} \Lambda^{\ast}(x)=\inf_{x\in G} \Lambda^{\ast}(x)\].
\end{lem}
\pf Assume $\lambda\in (0,1)$ and $d\geq 1.$
By Duality lemma \cite[Lemma~4.5.8]{DZ1998}, we have that
$$\left\{x=(x_1,\cdots,x_d)\in\mathbb{R}_+^d:\ \sum\limits_{i=1}^d x_i< 1\right\}\subseteq \mathbb{F}.$$
By strict convexity of $\Lambda^*(\cdot)$ and (\ref{rate1}), for any $x=(x_1,\cdots,x_d)\in\mathbb{R}_+^d$ with $\sum\limits_{i=1}^dx_i=1,$
$$\Lambda^*(x)\geq\limsup\limits_{n\rightarrow\infty}\Lambda^*\left[x-\frac{1}{n}\left(x-\left(\frac{1-\lambda}{d(1+\lambda)},\cdots, \frac{1-\lambda}{d(1+\lambda)}\right)\right)\right]$$
which leads to the conclusion.
\vskip 2mm
Assume $\lambda=0$ and $d\geq 2.$ We have that
$$\mathbb{F}= \left\{x=(x_1,\cdots,x_d)\in\mathbb{R}_+^d:\ \sum\limits_{i=1}^d x_i = 1,\mbox{and}\ x_i >0, 1\leq i\leq d\right\}.$$
Then by Lemma \ref{property0}, for any open set $G,$
\[\inf_{x\in G\cap \mathbb{F}} \Lambda^{\ast}(x)=\inf_{x\in G} \Lambda^{\ast}(x).\]
\qed
\vskip 3mm
\noindent{\bf Proof of Theorem \ref{thm3.1}.} Let $\mu_n$ be the law of $\left(\frac{\left\vert X_n^1\right\vert}{n},\cdots,\frac{\left\vert X_n^d\right\vert}{n}\right)$ for any $n\in\mathbb{N}.$ From
Corollary~\ref{C:limit}, Lemma~\ref{property0-1} and \ref{property0}, the logarithmic moment generating function exists with $\Lambda(s)= \ln \psi(s)$ and
\[\mathcal{D}_\Lambda=\left\{s\in\mathbb{R}^d:\ \Lambda(s)<\infty\right\}=\mathbb{R}^d=\mathcal{D}_\Lambda^o.\]
By $\bf{(a)}$ and $\bf{(b)}$ of the Gartner-Ellis theorem and Lemma \ref{infimumattained}, we have that
for any closed set $F\subseteq\mathbb{R}^d,$
\[\limsup\limits_{n \rightarrow \infty} \frac{1}{n} \ln\mu_n(F)\leq-\inf\limits_{x\in F}\Lambda^{\ast}(x);\]
and for any open set $G$ of $\mathbb{R}^d,$
\[\liminf\limits_{n \rightarrow \infty} \frac{1}{n} \ln \mu_n(G)\geq-\inf\limits_{x\in G} \Lambda^{\ast}(x).\]
The proof is done.
\qed
\bigskip
\noindent{\bf Acknowledgements.} Y. Liu, L. Wang and K. Xiang thank NYU Shanghai - ECNU Mathematical Institute for hospitality and financial support. V. Sidoravicius thanks Chern Institute of Mathematics for
hospitality and financial support. | 111,446 |
Hopes are rising that despite some devastating property loss, Tasmania's bushfires will prove to be fatality free.
Five days after a bushfire tore through communities east of Hobart, police acting commissioner Scott Tilyard said police had not received any official missing persons reports.
"At this point, more than 2,200 individuals have been identified as safe and well," Mr Tilyard said on Wednesday.
"My advice is there are no reports of missing persons in circumstances that cause us to have grave fears for their safety at this time."
Emergency services screened 731 properties in and around the worst-hit town of Dunalley, including 126 damaged or destroyed by fire, without finding any human remains.
Searches are continuing with the aid of a contingent of Victoria Police sworn in as special constables, and the preliminary screening is expected to be completed later on Wednesday.
As losses are counted, the damage bill is expected to be sharply higher than an initial $42 million total, according to the Insurance Council of Australia.
Many property owners have been unable to return to homes in the fire-affected region, and police said a major priority would be to expedite entry to intact residences. There have been no reports of looting, police confirmed.
Animal welfare advocates warned that a crisis was growing in fire-affected regions, with landowners prohibited access to check on animals in their care.
“It is imperative that the animal feed so generously donated by the public gets to its intended recipients immediately” said Chris Simcox, spokesman for Against Animal Cruelty Tasmania.
"Authorities are currently preventing people from providing proper care to their animals, and this is not good enough,” Mr Simcox said.
As the state government began to plan recovery, concerns grew for local tourism operators on the Tasman Peninsula dependent on the current high season.
“Many operators who survived the Victorian Black Saturday bushfires subsequently went to the wall because people simply assumed the entire region had been burnt and stayed away," said Greens senator Peter Whish-Wilson.
"The Tasman Peninsula and east coast are an integral part of our tourism economy, with international drawcards including Freycinet National Park and Port Arthur," Senator Whish-Wilson said. "We need to get the message out that these areas are open for business."
The story Police optimistic over Tasmanian fires first appeared on The Sydney Morning Herald. | 311,896 |
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\begin{document}
\title[Arithmetic statistics and Diophantine Stability]{Arithmetic statistics and Diophantine Stability for Elliptic Curves}
\author[A.~Ray]{Anwesh Ray}
\address[Ray]{Department of Mathematics\\
University of British Columbia\\
Vancouver BC, Canada V6T 1Z2}
\email{[email protected]}
\subjclass[2010]{11G05, 11R23}
\keywords{Arithmetic statistics, Iwasawa theory, Selmer groups, elliptic curves.}
\begin{abstract}
We study the growth and stability of the Mordell-Weil group and Tate-Shafarevich group of an elliptic curve defined over the rationals, in various cyclic Galois extensions of prime power order. Mazur and Rubin introduced the notion of diophantine stability for the Mordell-Weil group an elliptic curve $E_{/\Q}$ at a given prime $p$. Inspired by their definition of stability for the Mordell-Weil group, we introduce an analogous notion of stability for the Tate-Shafarevich group, called $\Sha$-stability. Using methods in arithmetic statistics and Iwasawa theory, we study the diophantine stability of elliptic curves on average. First, we prove results for a fixed elliptic curve $E$ and varying prime $p$. It is shown that any non-CM elliptic curve of rank 0 defined over the rationals is diophantine stable and $\Sha$-stable at $100\%$ of primes $p$. Next, we show that standard conjectures on rank distribution give lower bounds for the proportion of rational elliptic curves $E$ that are diophantine stable at a fixed prime $p\geq 11$. Related questions are studied for rank jumps and growth of ranks Tate-Shafarevich groups on average in prime power cyclic extensions.
\end{abstract}
\maketitle
\section{Introduction}
\par In \cite{mazurrubin}, B.~Mazur and K.~Rubin introduced the notion of \emph{diophantine stability} for an irreducible variety $V$ defined over a number field $K$. Let $p$ be a prime number. Given a number field extension $L/K$, $V$ is said to be diophantine stable in $L$ if $V(L)=V(K)$. The variety $V$ is said to be diophantine stable at the prime $p$ if there are abundantly many cyclic $p$-extensions of $p$-power order in which $V$ is diophantine stable. More precisely, $V$ is diophantine stable at $p$ if for any choice of $n\in \Z_{\geq 1}$ and finite set of primes $\Sigma$ of $K$, $V$ is diophantine stable in infinitely many $\Z/p^n\Z$-extensions $L/K$ in which the primes of $\Sigma$ split. Mazur and Rubin showed that if $A$ is a simple abelian variety defined over $K$ such that all $\bar{K}$-endomorphisms of $A$ are defined over $K$, then, $A$ is diophantine stable at a postive density set of primes of $K$. A similar result is proved for any curve of genus $\geq 1$ defined over a number field.
\par We study diophantine stability for non-CM elliptic curves $E$ defined over the rationals with good ordinary reduction at $p$. Given a pair $(E,p)$, we ask if $E$ is diophantine stable at $p$. From a statistical point of view, we are interested in the following questions.
\begin{question}
\begin{enumerate}
\item Given an elliptic curve $E_{/\Q}$, what can be said about proportion of primes $p$ at which $E$ is diophantine stable at $p$?
\item Given a prime $p$, what can be said about proportion of elliptic curves $E_{/\Q}$ ordered by height that are diophantine stable at $p$?
\end{enumerate}
\end{question}
Inspired by the notion of diophantine stability (for the Mordell-Weil group), we introduce an analogous notion for the $p$-primary part of the Tate-Shafarevich group, see Definition \ref{shastab}. Given an elliptic curve $E_{/\Q}$, we shall assume throughout this paper that $\Sha(E/\Q)$ is finite. Let $p$ be a prime number and $L/\Q$ a cyclic $p$-extension. We say that $E$ is $\Sha$-stable in $L/\Q$ if $\#\Sha(E/L)[p^\infty]=\#\Sha(E/\Q)[p^\infty]$. Further, $E$ is said to be $\Sha$-stable at $p$ if for any $(n, \Sigma)$, there are infinitely many $\Z/p^n\Z$-extensions in which the primes of $\Sigma$ split and in which $E$ is $\Sha$-stable.
\begin{lthm}[Theorem \ref{thm section 4 main}]\label{thmA}
Let $E_{/\Q}$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)=0$. Then, $E$ is diophantine stable and $\Sha$-stable at $100\%$ of the primes $p$.
\end{lthm}
Given an elliptic curve $E$ satisfying the above conditions, it is shown in earlier work (see \cite[Theorem A]{BKR}) that for $100\%$ of the primes $p$, $E(L)=E(\Q)$ for infinitely many $\Z/p\Z$-extensions $L/\Q$. The above result is much stronger, in fact for any pair $(n, \Sigma)$ the method gives an explicit recipe to construct extensions for which $E(L)=E(\Q)$, see Remark \ref{remark}. One is able to prove a refinement of Theorem \ref{thmA} when $E(\Q)_{\op{tors}}\neq 0$.
\begin{lthm}[Theorem \ref{4.15}]Let $E_{/\Q}$ be an elliptic curve without complex multiplication such that $E(\Q)_{\op{tors}}\neq 0$ and $\op{rank} E(\Q)=0$. Then $E$ is diophantine stable \emph{and} $\Sha$-stable at all but a finite set of primes $p$ at which it has good ordinary reduction.
\end{lthm}
Furthermore, the finite set of primes outside which the results apply is made explicit. For instance, the elliptic curve $X_0(14)$ is diophantine and $\Sha$-stable at all primes $p\geq 11$ at it has good ordinary reduction.
\par Next, we study the growth of ranks and Tate-Shafarevich groups in cyclic $p$-extensions. Independent results are proved for elliptic curves of rank 0 and positive rank.
\begin{lthm}[Theorem \ref{5.1 main theorem}]
Let $E$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)=0$. Then, for $100\%$ of the primes $p$, there are abundantly many $p$-cyclic extensions in which the $p$-primary Selmer group grows. More precisely, for any $n\in \Z_{\geq 1}$ and a finite set of primes $\Sigma$, there are infinitely many $\Z/p^n\Z$-extensions $L/\Q$ such that
\begin{enumerate}
\item all primes of $\Sigma$ split in $L$,
\item $\op{Sel}_{p^\infty}(E/\Q)=0$ and $\op{Sel}_{p^\infty}(E/\Q)\neq 0$.
\end{enumerate}
\end{lthm}
Given an elliptic curve $E_{/\Q}$ without complex multiplication, let $\mathcal{P}_E$ be the set of primes satisfying the conditions of Definition \ref{def 5.1}. These are precisely the set of primes $p$ at which $E$ has good ordinary reduction for which the $\mu$-invariant vanishes and the $\lambda$-invariant is equal to $\op{rank}E(\Q)$. It is expected that $\mathcal{P}_E$ consists of $100\%$ of the primes, according to \cite[Conjecture 1.1]{kunduray1}.
\begin{lthm}[Theorem \ref{5.1 theorem pos rank}]
Let $E$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)>0$ and let $p\in \mathcal{P}_E$. Let $L$ be \textit{any} cyclic $p$-extension. Then, at least one of the following assertions hold:
\begin{enumerate}
\item $\op{rank}E(L)\geq [L:\Q]\op{rank} E(\Q)$,
\item the $p$-adic regulator over $\Q$ is a unit in $\Z_p$, and over $L$, it is not a unit in $\Z_p$,
\item
$\Sha(E/\Q)[p^\infty]=0$ and $\Sha(E/L)[p^\infty]\neq 0$.
\end{enumerate}
\end{lthm}
Next, we prove stability results for a fixed prime $p\geq 11$ and varying elliptic curve $E_{/\Q}$ of Mordell-Weil rank 0. We assume a version of the rank distribution conjecture. Delaunay's heuristics \cite{delaunay} for the Tate-Shafarevich group give a precise estimate for the proportion of elliptic curves $E_{/\Q}$ such that $E$ has rank 0 and $p|\#\Sha(E/\Q)$.
\begin{lthm}[Theorem \ref{last thm}]
Let $p\geq 11$ be a prime and assume that Conjectures \ref{RDC} and \ref{Del} are satisfied. Let $\mathscr{E}_p$ be the set of isomorphism classes of elliptic curves $E_{/\Q}$ satisfying the following conditions:
\begin{enumerate}
\item $\op{rank} E(\Q)=0$,
\item $E[p]$ is irreducible as a Galois module,
\item $E$ has good reduction at $2,3$,
\item $E$ has good ordinary reduction at $p$,
\item $E$ is diophantine stable and $\Sha$-stable at $p$.
\end{enumerate}
Then, for the effective constant $c_1>0$ (see Lemma \ref{lemma 6.5}) we have that the lower density of $\mathscr{E}_p$ (in the set of all elliptic curves $E_{/\Q}$) satisfies the lower bound
\[\geq \frac{1}{6}-\frac{1}{p}-\frac{1}{2}\left(1-\prod_i \left(1-\frac{1}{p^{2i-1}}\right)\right)-(\zeta(p)-1)-\zeta(10) c_1\frac{\log p\cdot (\log \log p)^2}{\sqrt{p}}.\]
\end{lthm}
\par In particular, the lower density approaches $\frac{1}{6}$ as $p\rightarrow \infty$.
\par We apply results in Iwasawa theory to prove results on the rank jumps and growth of Tate-Shafarevich groups in cyclic $p$-extensions. The initial conception of this idea can be traced back to the work of T.~Dokchitser, see \cite{Dok}, in which the rank jump of elliptic curves in cubic extensions is studied. This is done by applying a formula of Y.~Hachimori and K.~Matsuno (see \cite{hachimorimatsuno}) for the growth of the Iwasawa $\lambda$-invariant in cyclic $p$-extensions.
\par We extend this approach to show that there is a close relationship between behaviour of Iwasawa invariants and diophantine stability. In previous work \cite{kunduray1}, results are proved for Iwasawa invariants on average. These methods are significantly extended to prove the results in this paper.
\subsection*{Acknowledgements} I would like to thank Debanjana Kundu, Ravi Ramakrishna and Tom Weston for helpful discussions.
\section{Preliminaries}
\label{section: preliminaries}
\subsection{}
\par Let $E_{/\Q}$ be an elliptic curve of conductor $N$ and $p$ an odd prime. We introduce main object of study in Iwasawa study, namely the $p$-primary Selmer group over the cyclotomic $\Z_p$-extension. For each integer, $n\in \Z_{\geq 1}$, denote by $E[n]$ the subgroup of $n$-torsion points of $E(\bar{\Q})$. Set $E[p^\infty]$ to denote the $p$-primary torsion subgroup of $E(\bar{\Q})$ given by the union
\[E[p^\infty]:=\bigcup_{k\geq 1} E[p^k].\]
Let $S$ be a set of primes containing those dividing $Np$, and $\Q_S\subset \bar{\Q}$ the maximal extension of $\Q$ in which all primes $\ell\notin S$ are unramified. Note that $E[p^\infty]$ admits an action of $\op{Gal}(\Q_S/\Q)$. Given a number field extension $F$ of $\Q$ contained in $\Q_S$, we set \[H^1(\Q_S/F, E[p^\infty]):=H^1(\op{Gal}(\Q_S/F), E[p^\infty]).\]
The $p$-primary Selmer group for $E$ over $F$ consists of global cohomology classes subject to local conditions, defined as follows
\[\op{Sel}_{p^\infty}(E/F):=\op{ker}\left(H^1(\Q_S/F, E[p^\infty])\longrightarrow \bigoplus_{v} H^1(F_v, E)[p^\infty]\right),\]where in the above sum, $v$ ranges over all finite primes of $F$.
\par For $n\geq 0$, let $\Q_n$ be the subfield of $\Q(\mu_{p^{n+1}})$ degree $p^n$ over $\Q$. Let $F_n$ denote the composite $F_n:=F\cdot \Q_n$, and note that $F_n$ is contained in $F_{n+1}$. Let $F_{\infty}$ be the union \[F_{\infty}:=\bigcup_{n\geq 0} F_n\]and set $\Gamma_F:=\op{Gal}(F_{\infty}/F)$. Note that there are isomorphisms of topological groups \[\op{Gal}(F_{\infty}/F)\xrightarrow{\sim} \varprojlim_n\op{Gal}(F_{n}/F)\xrightarrow{\sim} \Z_p.\] The extension $F_{\infty}$ is the cyclotomic $\Z_p$-extension of $F$ and $F_n$ is its \textit{$n$-th layer}. Choose a topological generator $\gamma_F\in \Gamma_F$ and fix an isomorphism $\Z_p\xrightarrow{\sim} \Gamma_F$ sending $a$ to $\gamma_F^a$. The Iwasawa algebra $\Lambda(\Gamma_F)$ is defined as the following inverse limit
\[\Lambda(\Gamma_F):=\varprojlim_n \Z_p[\op{Gal}(F_n/F)].\] Fix an isomorphism of $\Lambda(\Gamma_F)$ with the ring of formal power series $\Z_p\llbracket x\rrbracket$, by identifying $\gamma_F-1$ with $x$.
\par Assume that $F$ is a number field extension of $\Q$ such that $F\cap \Q_\infty=\Q$. Then, there is a natural isomorphism $\Gamma_\Q\xrightarrow{\sim} \Gamma_F$ and thus we may identify $\Lambda(\Gamma_F)$ with $\Lambda:=\Lambda(\Gamma_\Q)$. Consider the direct limit taken with respect to restriction maps
\[\op{Sel}_{p^\infty} (E/F_\infty):=\varinjlim_n \op{Sel}_{p^\infty}(E/F_n).\]
The Pontryagin dual $\op{Sel}_{p^\infty} (E/F_\infty)^{\vee}$ is a finitely generated $\Lambda$-module and is the main object of interest in the Iwasawa theory of elliptic curves. When $E$ has good ordinary reduction at $p$, it is a deep result of Kato \cite{katozeta} that the Pontryagin dual \[\op{Sel}_{p^\infty} (E/F_\infty)^{\vee}:=\op{Hom}_{\op{cnts}}\left(\op{Sel}_{p^\infty} (E/F_\infty), \Q_p/\Z_p\right)\] is a torsion $\Lambda$-module.
\subsection{}
\par Let $M$ be a cofinitely generated cotorsion $\Z_p\llbracket x\rrbracket$-module, i.e., the Pontryagin-dual $M^{\vee}:=\op{Hom}(M, \Q_p/\Z_p)$ is a finitely generated and torsion $\Z_p\llbracket x\rrbracket$-module. Recall that a polynomial $f(x)\in \Zx$ is said to be \textit{distinguished} if it is a monic polynomial whose non-leading coefficients are all divisible by $p$. Note that all height $1$ prime ideals of $\Z_p\llbracket x\rrbracket$ are principal ideals $(a)$, where $a=p$ or $a=f(x)$, where $f(x)$ is an irreducible distinguished polynomial.
According to the structure theorem for $\Zx$-modules (see \cite[Theorem 13.12]{washington1997}), $M^{\vee}$ is pseudo-isomorphic to a finite direct sum of cyclic $\Zx$-modules, i.e., there is a map
\[
M^{\vee}\longrightarrow \left(\bigoplus_{i=1}^s \Zx/(p^{\mu_i})\right)\oplus \left(\bigoplus_{j=1}^t \Zx/(f_j(T)) \right)
\]
with finite kernel and cokernel.
Here, $\mu_i>0$ and $f_j(T)$ is a distinguished polynomial.
The characteristic ideal of $M^\vee$ is (up to a unit) generated by
\[
f_{M}^{(p)}(T) = f_{M}(T) := p^{\sum_{i} \mu_i} \prod_j f_j(T).
\]
The $\mu$-invariant of $M$ is defined as the power of $p$ in $f_{M}(T)$.
More precisely,
\[
\mu_p(M):=\begin{cases}
\sum_{i=1}^s \mu_i & \textrm{ if } s>0\\
0 & \textrm{ if } s=0.
\end{cases}
\]
The $\lambda$-invariant of $M$ is the degree of the characteristic element, i.e.,
\[
\lambda_p(M) :=\begin{cases}
\sum_{i=1}^s \deg f_i & \textrm{ if } s>0\\
0 & \textrm{ if } s=0.
\end{cases}
\]
\subsection{} In the rest of this section, assume that $E$ is an elliptic curve defined over $\Q$ with good ordinary reduction at an odd prime $p$. Let $L/\Q$ be a cyclic extension with $\op{Gal}(L/\Q)\simeq \Z/p^n\Z$ for some integer $n\in \Z_{\geq 1}$. We shall refer to such an extension of $\Q$ as a $p$-cyclic extension of $\Q$. Kato showed that the Selmer group $\op{Sel}_{p^\infty}(E/\Q_\infty)$ is a cofinitely generated and cotorsion $\Lambda$-module. Let $\mu_p(E/\Q)$ and $\lambda_p(E/\Q)$ denote the Iwasawa $\mu$ and $\lambda$-invariants of $\op{Sel}_{p^\infty}(E/\Q_\infty)$ respectively. Likewise, let $\mu_p(E/L)$ and $\lambda_p(E/L)$ be the $\mu$ and $\lambda$-invariants of $\op{Sel}_{p^\infty}(E/L_\infty)$ respectively. The following conjecture of R.Greenberg is wide open.
\begin{Conjecture}\label{mu=0 conjecture} \cite[Conjecture 1.11]{greenbergITEC}
Let $E_{/\Q}$ be an elliptic curve and $p$ a prime at which $E$ has good ordinary reduction. Assume that $E[p]$ is irreducible as a module over $\op{Gal}(\bar{\Q}/\Q)$, then, $\mu_p(E/\Q)=0$.
\end{Conjecture}The hypothesis that $E[p]$ is irreducible is indeed necessary, since examples for which $\mu_p(E/\Q)>0$ were constructed by Mazur in \cite{mazurrationalpoints}, when $E[p]$ is reducible as a Galois module. In \cite{raymuinvariant}, it is shown that the Galois modules for which $E[p]$ is reducible may be classified into two types, namely \textit{aligned} and \textit{skew}. In the case when $E[p]$ is aligned, it is shown that $\mu_p(E/\Q)>0$ and in the case when $E[p]$ is skew, examples indicate that $\mu_p(E/\Q)=0$.
\par From a statistical point of view, one is interested in how often the Iwasawa $\mu$-invariant vanishes. The following implication of Greenberg's conjecture follows from well known results.
\begin{Proposition}
Assume Greenberg's conjecture for all pairs $(E,p)$, where $E$ is an elliptic curve defined over $\Q$ and $p$ is a prime at which $E$ has good ordinary reduction. Then, the following assertions hold.
\begin{enumerate}
\item If $E_{/\Q}$ is a semistable elliptic curve without complex multiplication, then, $\mu_p(E/\Q)=0$ for all primes $p\geq 13$ at which $E$ has good ordinary reduction.
\item For a fixed elliptic curve $E_{/\Q}$ without complex-multiplication, for all but finitely many primes $p$ at which $E$ has good ordinary reduction, $\mu_p(E/\Q)=0$.
\item For $100\%$ of elliptic curves $E_{/\Q}$, the $\mu$-invariant $\mu_p(E/\Q)=0$ for all primes $p$ at which $E$ has good ordinary reduction.
\end{enumerate}
\end{Proposition}
\begin{proof}
\par Let $E_{/\Q}$ be a semistable elliptic curves without complex mutliplication. Mazur showed that $E[p]$ is irreducible as a Galois module for all primes $p\geq 13$.
\par Let $E$ be a fixed elliptic curve without complex-multiplication. It follows from Serre's renowned open image theorem \cite{serreopenimage} that the residual representation \[\bar{\rho}_{E,p}:\op{Gal}(\bar{\Q}/\Q)\rightarrow \op{GL}_2(\F_p)\] for the action of the absolute Galois group on $E[p]$ is surjective for all but finitely many primes. As a consequence, the Galois representation on $E[p]$ is irreducible for all but finitely many primes, and hence, Greenberg's conjecture imples that $\mu_p(E/\Q)=0$ for all but finitely many primes $p$.
\par It is shown by W.Duke \cite{duke} that for $100\%$ of elliptic curves defined over $\Q$, the residual representation on $E[p]$ is irreducible for all but finitely many primes. Hence, Greenberg's conjecture implies that $\mu_p(E/\Q)=0$ for all but finitely many primes $p$ at which $E$ has good ordinary reduction.
\end{proof}
According to the next result of Hachimori-Matsuno \cite{hachimorimatsuno}, the $\mu=0$ property is stable in $p$-cyclic extensions. Furthermore, the $\lambda$-invariants $\lambda_p(E/\Q)$ and $\lambda_p(E/L)$ are related by a formula which generalizes the formula of Kida in \cite{kida}. Let $E_{/\Q}$ be an elliptic curve with good ordinary reduction at $p$ and $L/\Q$ a $p$-cyclic extension. In what follows $\tilde{w}$ will be a prime of $L_\infty$ and $\tilde{\ell}$ the prime below it in $\Q_\infty$. Set $e(\tilde{w}/\tilde{\ell})$ to denote the ramification index of $\tilde{w}$ over $\tilde{\ell}$. Let $P_1$ (resp. $P_2$) be the set of primes $\tilde{w}$ at which $E$ has split multiplicative reduction (resp. $E(L_{\infty, \tilde{w}})$ has a point of order $p$).
\begin{Th}[Y.Hachimori-K.Matsuno]\label{kida thm}
Let $E_{/\Q}$ be an elliptic curve and $p$ an odd prime at which $E$ has good ordinary reduction. Let $L/\Q$ be a $p$-cyclic extension and that $L\cap \Q_\infty= \Q$. Assume $\mu_p(E/\Q)=0$, then the Selmer group $\op{Sel}_{p^\infty}(E/L_\infty)$ is cotorsion as a $\Lambda$-module and $\mu_p(E/L)=0$. Furthermore, the $\lambda$-invariants are related according to the following formula
\begin{equation}\label{kidaformula}\lambda_p(E/L)=[L:\Q]\lambda_p(E/\Q)+\sum_{\tilde{w}\in P_1} \left(e(\tilde{w}/\tilde{\ell})-1\right)+2\sum_{\tilde{w}\in P_2} \left(e(\tilde{w}/\tilde{\ell})-1\right).\end{equation}
\end{Th}
\begin{proof}
The above result is \cite[Theorem 3.1]{hachimorimatsuno}.
\end{proof}
\section{The truncated Euler characteristic}
\par In this section, we recall the notion of the truncated Euler characteristic and its relationship with Iwasawa invariants. The reader may consult \cite{csslinks, zerbes, ray1, raymulti} for a more detailed introduction to the topic. Let $E_{/\Q}$ be an elliptic curve and $p$ an odd prime at which $E$ has good ordinary reduction. Let $L/\Q$ be a $p$-cyclic extension and assume that $L\cap \Q_\infty=\Q$. Note that there is a canonical isomorphism $\Gamma_\Q\simeq \Gamma_L$, with respect to which $\Lambda=\Lambda(\Gamma_\Q)$ is identified with $\Lambda(\Gamma_F)$. According to a result of K.Kato and K.Rubin, the Selmer group $\op{Sel}_{p^\infty}(E/L_\infty)$ is cotorsion as a $\Lambda$-module. In this section, let $\Gamma:=\Gamma_\Q$ and identify $\Gamma$ with $\Gamma_L$. Let $F$ be either $\Q$ or $L$, and consider the natural map
\[\Phi: \op{Sel}_{p^\infty} (E/F_\infty)^\Gamma\rightarrow \op{Sel}_{p^\infty} (E/F_\infty)_{\Gamma}\]
sending an element $x\in \op{Sel}_{p^\infty} (E/F_\infty)^\Gamma$ to its residue class in $\Phi(x)=\bar{x}\in \op{Sel}_{p^\infty} (E/F_\infty)_{\Gamma}$. If $\op{rank} E(F)\leq 1$, the kernel and cokernel of $\Phi$ are both finite. Moreover, when $\op{rank} E(\Q)>1$, there is an explicit criterion for the kernel and cokernel of $\Phi$ to be finite, see \cite[Lemma 3.3]{raymulti}.
\begin{Definition}
The truncated Euler characteristic $\chi_t(\Gamma_F, E[p^\infty])$ is defined as the following quotient
\[\chi_t(\Gamma_F, E[p^\infty]):=\frac{\# \op{ker}\Phi}{\#\op{cok} \Phi},\] provided the kernel and cokernel of $\Phi$ are both finite.
\end{Definition}
Let $f(T)$ denote the characteristic element of $\op{Sel}_{p^\infty}(E/F_\infty)$. Let $r\in \Z_{\geq 0}$ denote the order of vanishing of $f(T)$ at $T=0$, and write $f(T)=T^r\cdot g(T)$. Note that $g(0)\neq 0$. Let $\op{Reg}_p(E/F)$ denote the $p$-adic regulator of $E$ over $F$, defined as the determinant of the $p$-adic height pairing, see \cite{schneiderhp1}. Note that the $p$-adic regulator is conjectured to be non-zero, see \cite{Schneider85}. Let $\mathcal{R}_p(E/F)$ denote the normalized regulator, defined as $\mathcal{R}_p(E/F):=p^{-\op{rank}E(F)}\op{Reg}_p(E/F)$. Given $a, b\in \Q_p$, we write $a\sim b$ if $a=ub$ for a $p$-adic unit $u\in \Z_p^{\times}$.
\par The following result gives the formula for the truncated Euler characteristic of the $p$-primary Selmer group when it is defined.
In the CM case, this was proven by B. Perrin-Riou (see \cite{PR82}) and in the general case by P. Schneider (see \cite{Schneider85}).
\begin{Th}\label{th ECF}
Let $E$ be an elliptic curve over a number field $F$ and $p$ an odd prime. Assume that the following conditions are satisfied:
\begin{enumerate}
\item\label{th ECF c1} $E$ has good ordinary reduction at all primes $v|p$,
\item\label{th ECF c2} the $p$-primary part of the Tate-Shafarevich group $\Sha(E/F)[p^\infty]$ is finite,
\item\label{th ECF c3} the $p$-adic regulator $\op{Reg}_p(E/F)$ is non-zero,
\item\label{th ECF c4} $E(F)[p]=0$.
\end{enumerate}
Then, the order of vanishing of $f(T)$ at $T=0$ is equal to $\op{rank} E(F)$ and
\[g(0)\sim \mathcal{R}_p(E/F) \times \# \Sha(E/F)[p^\infty] \times \prod_{v\nmid p} c_{v}^{(p)}(E/F) \times \left(\# \widetilde{E}(\kappa_v)[p^\infty]\right)^2,\]where we recall that $\mathcal{R}_p(E/F):=\frac{\op{Reg}_p(E/F)}{p^{\op{rank}E(F)}}$.
\end{Th}
\begin{Remark}
When the rank of $E(F)$ is 0, the term $\mathcal{R}_p(E/F)=1$ by definition, hence there are only three contributing terms to the above formula in this case. Note that if $E$ is an elliptic curve over $\Q$ such that $E(\Q)[p]=0$, then $E(L)[p]=0$ for any $p$-cyclic extension $L/\Q$. Thus the condition \eqref{th ECF c4} is satisfied for $E_{/L}$.
\end{Remark}
\begin{Proposition}
Let $E_{/F}$ be such that the conditions of Theorem \ref{th ECF} are satisfied and furthermore, the kernel and cokernel of the map $\Phi$ are finite. Then, we have that
\[\chi_t(\Gamma_F, E[p^\infty])\sim g(0).\]
\end{Proposition}
The above result motivates the following definition.
\begin{Definition}
Let $E_{/F}$ be an elliptic curve satisfying the conditions of Theorem \ref{th ECF}. Suppose that $\op{rank} E(F)>1$ and that the truncated Euler characteristic is not defined. Then, we simply set $\chi_t(\Gamma_F, E[p^\infty])$ to denote $|g(0)|_p^{-1}$.
\end{Definition}
Thus, from here on in, the truncated Euler characteristic is always defined and equals $|g(0)|_p^{-1}$. Putting it all together, we have the following result.
\begin{Corollary}\label{cor ECF}
Let $E$ be an elliptic curve over a number field $F$ and $p$ an odd prime. Assume that the conditions of Theorem \ref{th ECF} and satisfied. Then, the truncated Euler characteristic is given (up to a $p$-adic unit) by
\[\chi_t(\Gamma_F, E[p^\infty])\sim \mathcal{R}_p(E/F) \times \# \Sha(E/F)[p^\infty] \times \prod_{v\nmid p} c_{v}^{(p)}(E/F) \times \left(\# \widetilde{E}(\kappa_v)[p^\infty]\right)^2.\]
\end{Corollary}
The truncated Euler characteristic $\chi_t(\Gamma_F, E[p^\infty])$ is of the form $p^n$, where $n\in \Z_{\geq 0}$. Thus, either $\chi_t(\Gamma_F, E[p^\infty])=1$ or divisible by $p$. When $\chi_t(\Gamma_F, E[p^\infty])=1$, the Iwasawa invariants $\mu_p(E/F)$ and $\lambda_p(E/F)$ are fully understood. The following result gives a relationship between the Iwasawa invariants and the (truncated) Euler characteristic.
\begin{Lemma}\label{lemma ECF mulambda}
Let $E_{/F}$ be an elliptic curve satisfying the conditions of Theorem \ref{th ECF}. Then, the following are equivalent
\begin{enumerate}
\item $\mu_p(E/F)=0$ and $\lambda_p(E/F)=\op{rank} E(F)$,
\item $\chi_t(\Gamma_F, E[p^\infty])=1$.
\end{enumerate}
\end{Lemma}
\begin{proof}
The result follows from \cite[Lemma 3.4]{kunduray1}.
\end{proof}
\section{Stability for a fixed elliptic curve of rank 0 and varying prime $p$}\label{s4}
\par In this section, we fix an odd prime number $p$ and an elliptic curve $E_{/\Q}$ with good ordinary reduction at $p$. Recall that a cyclic $p$-extension $L/\Q$ is a Galois extension for which $\op{Gal}(L/\Q)\simeq \Z/p^n\Z$ for some positive integer $n$. We shall also refer to such an extension as a $\Z/p^n\Z$-extension. Let $\op{Cyc}_p$ denote the set of all cyclic $p$-extensions $L/\Q$ and for $n\in \Z_{\geq 1}$ denote by $\op{Cyc}_p^n\subset \op{Cyc}_p$ the subset of cyclic $p$-extensions $L/\Q$ such that $\op{Gal}(L/\Q)\simeq \Z/p^n\Z$. Given a finite set of prime numbers $\Sigma$, let $\op{Cyc}_p^{\Sigma, n}$ consist of $L\in \op{Cyc}_p^{n}$ such that all primes $\ell \in \Sigma$ split in $L$.
\begin{Definition}\label{def diophantine stab}
Let $V$ be a variety defined over $\Q$. For a number field extension $L/\Q$, $V$ is said to be \textit{diophantine stable} w.r.t. $L$ if $V(L)=V(\Q)$. Let $p$ be a prime number, $V$ is said to be diophantine stable at $p$ if for every positive integer $n$ and every finite set of primes $\Sigma$, there are infinitely many cyclic $p$-extensions $L\in \op{Cyc}_p^{\Sigma, n}$ such that $V$ is diophantine stable w.r.t. $L$.
\end{Definition}
We note that the additional requirement that any finite set of primes $\Sigma$ should be split is quite a strong one. One is naturally interested in the following question in arithmetic statistics.
\begin{question}
Given a variety $V_{/\Q}$, what is the proportion of primes $p$ such that it can be shown that $V$ is diophantine stable at $p$?
\end{question}
For a simple abelian variety $A_{/\Q}$ whose $\bar{\Q}$-endomorphisms are all defined over $\Q$, then Mazur and Rubin show that there is a set of primes $p$ of positive density at which $A$ is diophantine stable at $p$. It is shown in Theorem \ref{thm section 4 main} that if $E_{/\Q}$ is a non-CM elliptic curve with Mordell-Weil rank 0, then $E$ is diophantine stable at $p$ for consists of $100\%$ of the primes $p$. This particular result should be contrasted to an earlier result \cite[Theorem A]{BKR}, in which it is shown that for $100\%$ of the primes $p$, there are infinitely many $\Z/p\Z$-extensions $L/\Q$ in for which $E(\Q)=E(L)$. However, the result proved in this section is much stronger since it gives more control on the splitting behaviour of primes in $p$-cyclic $\Z/p^n\Z$-extensions. According to Remark \ref{remark after thm}, the set of all extensions $L\in \op{Cyc}_p^{\Sigma, n}$ w.r.t. which $E$ is diophantine stable has a an explicit description, which makes them very abundant. Furthermore, Theorem \ref{thm section 4 main} asserts that the stability of the $p$-primary part of the Tate-Shafarevich can also be controlled for $100\%$ of the primes $p$. Inspired by the definition of Mazur and Rubin for diophantine stability of the Mordell Weil group, we introduce the notion of $\Sha$-stability, for the $p$-primary part of the Tate-Shafarevich group.
\begin{Definition}\label{shastab}
Let $E_{/\Q}$ be an elliptic curve for which $\Sha(E/\Q)$ is finite. Let $p$ be a prime number. We say that $E$ is $\Sha$-stable at $p$ if for every $n\in \Z_{\geq 1}$ and finite set of primes $\Sigma$, there are infinitely many $L\in \op{Cyc}_p^{n,\Sigma}$ such that \begin{equation}\label{sha stable}\#\Sha(E/L)[p^\infty]=\#\Sha(E/\Q)[p^\infty].\end{equation}
\end{Definition}
\par Note that since $\Sha(E/\Q)$ is assumed to be finite, it follows that $\Sha(E/\Q)[p^\infty]=0$ for all but finitely many primes $p$. Thus for all but finitely many primes $p$, \eqref{sha stable} translates to the condition $\Sha(E/L)[p^\infty]=0$.
\begin{Definition}\label{def PE}For an elliptic curve $E_{/\Q}$ for which $\op{rank} E(\Q)=0$, let $\mathcal{P}_E$ consist of the primes $p$ such that
\begin{enumerate}
\item $p$ is odd,
\item $E[p]$ is irreducible as a $\op{Gal}(\bar{\Q}/\Q)$-module,
\item $E$ has good ordinary reduction at $p$,
\item $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$.
\end{enumerate}
\end{Definition}
The following result will be used at various points in in this paper.
\begin{Lemma}\label{NSW lemma}
Let $G$ and $M$ be finite abelian groups of $p$-power order such that $G$ acts on $M$. Suppose that $M^G=0$, then $M=0$.
\end{Lemma}
\begin{proof}
The result follows from \cite[Proposition 1.6.12]{NSW}.
\end{proof}
\begin{Remark}\label{remark}
For $p\in \mathcal{P}_E$, we have that $E(\Q)[p]=0$ since $E[p]$ is irreducible as a Galois module. Since $\Q_\infty$ is a pro-$p$ extension of $\Q$, it follows from Lemma \ref{NSW lemma} that $E(\Q_\infty)[p^\infty]=0$. Therefore, $\op{Sel}_{p^\infty}(E/\Q)$ injects into $\op{Sel}_{p^\infty}(E/\Q_\infty)$. On the other hand, $\op{Sel}_{p^\infty}(E/\Q)$ fits into a short exact sequence
\[0\rightarrow E(\Q)\otimes \Q_p/\Z_p\rightarrow \op{Sel}_{p^\infty}(E/\Q)\rightarrow \Sha(E/\Q)[p^\infty]\rightarrow 0.\] Hence, for $p\in \mathcal{P}_E$, we have that $\Sha(E/\Q)[p^\infty]=0$.
\end{Remark}
For $x>0$, let $\mathcal{P}_E^c(x)$ consist of all primes $p\notin \mathcal{P}_E$ such that $E$ has good ordinary reduction at $p$ and $p\leq x$.
\begin{Th}\label{100 percent}
The set $\mathcal{P}_E$ consists of $100\%$ of the primes $p$. Furthermore, we have the following growth estimate for the complement of $\mathcal{P}_E$
\[\# \mathcal{P}_E^c(x)\ll \frac{x (\log \log x)^2}{(\log x)^2}.\]
\end{Th}
\begin{proof}
The result essentially follows is due to R.Greenberg \cite[Theorem 5.1]{greenbergITEC} and the asymptotic estimate follows from the main result of V.K.Murty \cite{murty}. We provide details here for completeness. It follows from Serre's open image theorem that there $E[p]$ is irreducible as a Galois module for all but finitely many primes. Serre showed that $E$ has good ordinary reduction for $100\%$ of primes $p$, see \cite{serrechebotarev}. It follows from \cite[Corollary 3.6]{kunduray1} and the Euler characteristic formula that the following conditions are equivalent for a prime $p$ at which $E$ has good ordinary reduction and $E[p]$ is irreducible as a Galois module
\begin{enumerate}
\item $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$,
\item $\mu_p(E/\Q)=0$ and $\lambda_p(E/\Q)=0$,
\item $\chi_t(\Gamma_\Q, E[p^\infty])=1$,
\item $p\nmid \#\Sha(E/\Q)$, $p\nmid c_\ell(E)$ for all primes $\ell\neq p$ and $p\nmid \#\widetilde{E}(\F_p)$.
\end{enumerate}
Clearly, $p\nmid \#\Sha(E/\Q)$ and $p\nmid c_\ell(E)$ for all but finitely many primes $p$. It follows from \cite{murty} that the proportion of primes $p$ such that $p\mid \#\widetilde{E}(\F_p)$ is $\ll \frac{x(\log \log x)^2}{(\log x)^2}$. The result follows from this.
\end{proof}
The main result of this section is the following.
\begin{Th}\label{thm section 4 main}
Let $E_{/\Q}$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)=0$. Then $E$ is diophantine stable \emph{and} $\Sha$-stable at every prime $p\geq 11$ such that $p\in \mathcal{P}_E$. As a result, $E$ is diophantine stable and $\Sha$-stable at $100\%$ of the primes $p$.\end{Th}
\begin{Remark}\label{remark after thm}
In fact the method produces many extensions $L/\Q$ with respect to which $E$ is both diophantine and $\Sha$-stable. Let $p\geq 11$ be a prime in $\mathcal{P}_E$ and $n\in \Z_{\geq 1}$. Then, there is a positive density set of primes $\mathfrak{S}_n$ such that given a finite set of primes $\Sigma$, and any set of primes $\ell_1, \dots, \ell_{\#\Sigma+1}\in \mathfrak{S}_n$, there is a $\Z/p^n\Z$-extension $L/\Q$ such that all of the following conditions hold:
\begin{enumerate}
\item the primes that ramify in $L$ are a subset of $\{\ell_1, \dots, \ell_{\#\Sigma+1}\}$,
\item all primes of $\Sigma$ split in $L$,
\item $E(L)=E(\Q)$,
\item $\Sha(E/L)[p^\infty]=0$.
\end{enumerate}
Furthermore, according to Lemma \ref{s4 positive density}, the density of $\mathfrak{S}_n$ is equal to $\frac{p^2-p-1}{p^{n-1} (p+1)(p-1)^2}$.
\end{Remark}
We now make preparations for the proof of Theorem \ref{thm section 4 main}. Given an elliptic curve with good reduction at a prime $\ell$, let $\widetilde{E}$ denote the reduced curve at $\ell$, and $\widetilde{E}(\F_\ell)$ the group of $\F_\ell$-points of $E$.
\begin{Definition}
Let $E_{/\Q}$ be an elliptic curve and $p$ an odd prime. Let $Q_1=Q_1(E,p)$ be the set of primes $\ell\neq p$ at which $E$ has bad reduction and $Q_2=Q_2(E,p)$ the set of primes $\ell\neq p$ at which $E$ has good reduction and $p$ divides $\#\widetilde{E}(\F_\ell)$.
\end{Definition}
Note that the set of primes $Q_1$ is finite, since the set of primes at which $E$ has bad reduction is finite. Given a cyclic $p$-extension $L/\Q$, let $\Sigma_L$ be the set of primes $\ell\neq p$ that ramify in $L$.
\begin{Lemma}\label{stability lemma}
Let $E_{/\Q}$ be an elliptic curve with $\op{rank} E(\Q)=0$ and $p$ an odd prime such that
\begin{enumerate}
\item $E$ has good ordinary reduction at $p$,
\item $E[p]$ is irreducible as a Galois module,
\item $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$.
\end{enumerate}
Let $L/\Q$ be a cyclic $p$-extension such that
\begin{enumerate}
\item $L\cap \Q_\infty=\Q$,
\item $\Sigma_L$ and $Q_1\cup Q_2$ are disjoint.
\end{enumerate} Then, the following assertions hold
\begin{enumerate}
\item\label{stability: 1} $\op{rank} E(L)=0$,
\item\label{stability: 2} $\Sha(E/L)[p^\infty]=\Sha(E/\Q)[p^\infty]=0$.
\item\label{stability: 3} If $p\geq 11$, then $E$ is diophantine-stable w.r.t. $L$, i.e., $E(L)=E(\Q)$.
\end{enumerate}
\end{Lemma}
\begin{proof}
First, we prove \eqref{stability: 1}. Let $\ell\in \Sigma_L$, and $\tilde{w}$ a prime of $L_\infty$ above $\ell$. Since $\ell\notin Q_1$, the elliptic curve $E$ has good reduction at $\ell$. Therefore, $\tilde{w}$ is not contained in $P_1$. Since $\ell\notin Q_2$, it follows that $\widetilde{E}(\F_\ell)[p]=0$. Let $k_{\tilde{w}}$ denote the residue field at $\tilde{w}$, since $k_w/\F_\ell$ is pro-$p$ it follows that $\widetilde{E}(k_{\tilde{w}})[p]=0$. Note that the reduction map
\[E(L_{\infty,\tilde{w}})\rightarrow \widetilde{E}(k_{\tilde{w}})\] is surjective (since $E$ has good reduction at $\ell$) and its kernel is pro-$\ell$. It follows that the map on $p$-torsion
\[E(L_{\infty,\tilde{w}})[p]\rightarrow \widetilde{E}(k_{\tilde{w}})[p]\]is an isomorphism. Therefore, $E(L_{\infty,\tilde{w}})$ has no $p$-torsion point and we have shown that $\tilde{w}\notin P_1\cup P_2$. Since the Selmer group $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$, we have that
\[\mu_p(E/\Q)=0\text{ and }\lambda_p(E/\Q)=0.\] Therefore, it follows from \eqref{kidaformula} that $\lambda_p(E/L)=[L:\Q]\lambda_p(E/\Q)=0$. On the other hand, according to \cite[Theorem 1.9]{greenbergITEC}
\[\op{rank} E(L)\leq \lambda_p(E/L),\] and we deduce that $\op{rank} E(L)=0$.
\par Next, we prove \eqref{stability: 2}. Since the Selmer group is $0$, we have that $\mu_p(E/\Q)=0$. It follows from Theorem \ref{kida thm} that $\mu_p(E/L)=0$ as well. On the other hand, it has been shown in the previous paragraph that $\lambda_p(E/L)=0$. Note that since $\op{rank} E(L)=0$, we have that $\mathcal{R}_p(E/L)=1$, and thus the hypotheses of Lemma \ref{lemma ECF mulambda} are satisfied. Therefore, it follows from Lemma \ref{lemma ECF mulambda} that $\chi_t(\Gamma_L, E[p^\infty])=1$. Since $E(\Q)[p]=0$ and $L/\Q$ is a $p$-cyclic extension, it follows from Lemma \ref{NSW lemma} that $E(F)[p]=0$. Therefore, according to the Euler characteristic formula
\[\chi_t(\Gamma_L, E[p^\infty])\sim \# \Sha(E/L)[p^\infty] \times \prod_{v\nmid p} c_{v}^{(p)}(E/L) \times \prod_{v|p}\left(\# \widetilde{E}(\kappa_v)[p^\infty]\right)^2.\]
Since the Euler characteristic is $1$, it follows that each of the terms in the above formula are equal to $1$. In particular, we have shown that $\Sha(E/L)[p^\infty]=0$.
\par It has been shown that $\op{rank} E(L)=0$. Therefore, in order to prove \eqref{stability: 3}, it suffices to observe that $E(L)_{\op{tors}}=E(\Q)_{\op{tors}}$. This assertion follows from \cite[Theorem 7.2]{GJN}.
\end{proof}
\begin{Lemma}\label{s4 splitting}
Let $\Sigma$ be a finite set of rational primes and $n\in \Z_{\geq 1}$. Set $t:=\# \Sigma$. Let $\ell_1, \dots, \ell_{t+1}$ be primes such that $\ell_i\notin \Sigma$ and $\ell_i\equiv 1\mod{p^n}$ for all $i$ and $m:=\prod_{i=1}^{t+1} \ell_i$. There is a $\Z/p^n\Z$-extension $L/\Q$ contained in $\Q(\mu_m)$ in which all primes $q\in \Sigma$ split.
\end{Lemma}
\begin{proof}
Note that there is a Galois extension $E/\Q$ contained in $\Q(\mu_m)$ such that
\[\op{Gal}(E/\Q)\simeq \left(\Z/p^n \Z \right)^{t+1}.\] Consider the subgroup $H$ of $\op{Gal}(E/\Q)$ generated by the Frobenius elements at the $t$ primes in $\Sigma$. We show that there is a quotient of $\op{Gal}(E/\Q)/H$ isomorphic to $\Z/p^n\Z$. From this, it shall follow that there is a $\Z/p^n\Z$-extension $L/\Q$ contained in $E$ in which all primes $q\in \Sigma$ split.
\par Set $G:=\op{Gal}(E/\Q)\simeq \left(\Z/p^n \Z \right)^{t+1}$ and for $k>0$, let $G_k$ be the subgroup generated by $p^k$-th multiples of the generators of $G$. In other words, $G_k$ corresponds to the subgroup $\left(p^k\Z/p^n \Z \right)^{t+1}$ of $\left(\Z/p^n \Z \right)^{t+1}$. Set $H_k:=G_k\cap H$, and note that $H_{n-1}$ is a $\Z/p\Z$-vector space of dimension $\leq t$. On the other hand, $G_{n-1}$ is a $\Z/p\Z$-vector space of dimension $t+1$. Let $N$ be a subspace of $G_{n-1}$ of dimension $t$ containing $H_{n-1}$, and let $\tilde{N}$ be the subgroup of $G$ consisting of all elements $g\in G$ such that $p^{n-1} g$ is contained in $N$. It is easy to see that $\tilde{N}$ contains $H$ and the quotient $G/\tilde{N}$ is isomorphic to $\Z/p^n\Z$.
\end{proof}
\begin{Definition}\label{definition sn} Let $E_{/\Q}$ be an elliptic curve of conductor $N$ and $p$ a prime. For $n\in \Z_{\geq 1}$, and $x>0$, let $\mathfrak{S}_n$ be the set of primes $\ell\nmid N$ such that
\begin{enumerate}
\item $\ell\equiv 1\mod{p^n}$,
\item $p\nmid \# \widetilde{E}(\F_\ell)$.
\end{enumerate}
For $x>0$, let $\mathfrak{S}_n(x)$ be the set of primes $\ell\in \mathfrak{S}_n$ such that $\ell\leq x$.
\end{Definition}
\begin{Lemma}\label{s4 positive density}
Let $E_{/\Q}$ be an elliptic curve and $p$ a prime such that $p\geq 5$. Assume that the residual representation $\bar{\rho}_{E,p}:\op{Gal}(\bar{\Q}/\Q)\rightarrow \op{GL}_2(\F_p)$ on $E[p]$ is surjective. For $n\in \Z_{\geq 1}$, let $\mathfrak{S}_n$ be as above.
Then, the set of primes $\mathfrak{S}_n$ has positive density, and furthermore
\[\lim_{x\rightarrow \infty} \frac{\#\mathfrak{S}_n(x)}{\pi(x)}=\frac{p^2-p-1}{p^{n-1} (p+1)(p-1)^2}.\]
\end{Lemma}
\begin{proof}
Given a prime $\ell\nmid Np$, set $a_\ell:=\ell+1-\#\widetilde{E}(\F_\ell)$. Let $\Q(E[p])$ be the extension of $\Q$ which is fixed by the kernel of $\bar{\rho}_{E,p}$. The residual representation gives a natural isomorphism
\[\op{Gal}(\Q(E[p])/\Q)\xrightarrow{\sim} \op{GL}_2(\F_p).\] Note that the determinant of $\bar{\rho}=\bar{\rho}_{E,p}$ is the mod-$p$ cyclotomic character, hence, \begin{itemize}
\item $\Q(E[p])$ contains $\Q(\mu_p)$,
\item $\bar{\rho}$ induces an isomorphism
\[\op{Gal}(\Q(E[p])/\Q(\mu_p))\xrightarrow{\sim} \op{SL}_2(\F_p).\]
\end{itemize}The prime $\ell$ is unramified in $\Q(E[p])$ and let $\sigma_\ell$ is the Frobenius in $\op{Gal}(\Q(E[p])/\Q)$. The trace and determinant of $\bar{\rho}(\sigma_\ell)$ are as follows,
\[\op{trace}\left(\bar{\rho}(\sigma_\ell)\right)=a_\ell\text{ and }\op{det}\left(\bar{\rho}(\sigma_\ell)\right)=\ell.\]
Let $\mathcal{S}'$ be the subset of $\op{Gal}(\Q(E[p])/\Q)$ consisting of $\sigma$ such that
\[\op{trace}\left(\bar{\rho}(\sigma)\right)\neq 2\text{ and }\op{det}\left(\bar{\rho}(\sigma)\right)=1.\]
Note that $\sigma_\ell\in \mathcal{S}'$ if and only if $\ell\equiv 1\mod{p}$ and $p\nmid \#\widetilde{E}(\F_\ell)$. On the other hand, C.Jordan showed that $\op{PSL}_2(\F_p)$ is simple, see \cite{jordan}. As a result, $\op{SL}_2(\F_p)$ does not contain any normal subgroup $N$ such that $p$ divides $[\op{SL}_2(\F_p):N]$. Hence the intersection $\Q(E[p])\cap \Q(\mu_{p^n})$ is equal to $\Q(\mu_p)$. Note that for $\sigma\in \mathcal{S}'$, the determinant of $\bar{\rho}(\sigma)$ is equal to $1$, and as a result, $\mathcal{S}'$ is a subset of $\op{Gal}(\Q(E[p])/\Q(\mu_p))$. Let $\Q(E[p], \mu_{p^n})$ be the composite $\Q(E[p])\cdot \Q(\mu_{p^n})$ and note that
\[\op{Gal}(\Q(E[p], \mu_{p^n})/\Q(\mu_p))\simeq \op{Gal}\left(\Q(E[p])/\Q(\mu_p)\right)\times \op{Gal}\left(\Q(\mu_{p^n})/\Q(\mu_p)\right).\]
Let $\mathcal{S}''$ be the subset of $\op{Gal}(\Q(E[p], \mu_{p^n})/\Q(\mu_p))$ such that the projection to $\op{Gal}\left(\Q(E[p])/\Q(\mu_p)\right)$ lies in $\mathcal{S}'$ and the projection to $\op{Gal}\left(\Q(\mu_{p^n})/\Q(\mu_p)\right)$ is trivial. Note that $\sigma_\ell\in \mathcal{S}''$ if and only if
\[\ell\equiv 1\mod{p^n}\text{ and } p\nmid \#\widetilde{E}(\F_\ell).\]
Note that $\#\mathcal{S}''=\mathcal{S}'$. Thus according to the Chebotarev density theorem, we have that
\[\lim_{x\rightarrow \infty} \frac{\#\mathfrak{S}_n(x)}{\pi(x)}=\frac{\#\mathcal{S}'}{p^{n-1}\left(\#\op{GL}_2(\F_p)\right)}>0.\]
In order to compute $\#\mathcal{S}'$, we need to compute the cardinality of the set of all matrices
\[\mtx{a}{b}{c}{d}\in \op{SL}_2(\F_p)\] with trace $2$ and determinant $1$. An elementary matrix calculation shows that \[\#\mathcal{S}'=\# \op{SL}_2(\F_p)-p^2=(p^2-1)p-p^2=p^3-p^2-p.\]
Putting it all together we have
\[\lim_{x\rightarrow \infty} \frac{\#\mathfrak{S}_n(x)}{\pi(x)}=\frac{p^2-p-1}{p^{n-1} (p^2-1)(p-1)}.\]
\end{proof}
\begin{proof}[proof of Theorem \ref{thm section 4 main}]
\par We begin by stating what needs to be proved. We are given a prime $p\geq 11$ contained in $\mathcal{P}_E$. Recall that according to Definition \ref{def PE}, this means that the following conditions hold:
\begin{enumerate}
\item $E[p]$ is irreducible as a $\op{Gal}(\bar{\Q}/\Q)$-module,
\item $E$ has good ordinary reduction at $p$,
\item $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$.
\end{enumerate}
Given $n\geq 1$ and a finite set of primes $\Sigma$, we show that there are infinitely many $\Z/p^n\Z$-extensions $L/\Q$ such that \begin{enumerate}
\item all primes in $\Sigma$ split in $L$,
\item $E(L)=E(\Q)$,
\item $\Sha(E/L)[p^\infty]=\Sha(E/\Q)[p^\infty]=0$.
\end{enumerate}
Let $t:=\# \Sigma$ and $\mathfrak{S}_n$ be as in Definition \ref{definition sn}. By Lemma \ref{s4 positive density}, the set $\mathfrak{S}_n$ has positive density. All that is required to prove the result is that $\mathfrak{S}_n$ is infinite. Let $\ell_1, \dots, \ell_{t+1}$ be a set of primes in $\mathfrak{S}_n$, and set $m:=\prod_{i=1}^{t+1} \ell_i$. According to Lemma \ref{s4 splitting}, there is a $\Z/p^n\Z$ extension $L/\Q$ such that
\begin{enumerate}
\item all the primes in $\Sigma$ split completely,
\item $L/\Q$ is ramified outside $\{\ell_i\mid i=1,\dots, t+1\}$.
\end{enumerate}Given a set of primes $\Omega=\{\ell_1, \dots, \ell_{t+1}\}\subset \mathfrak{S}_n$, let $L_{\Omega}$ be a choice of such a $\Z/p^n\Z$-extension satisfying the above properties. Clearly, there are infinitely many choices of mutually disjoint sets $\Omega$, and hence infinitely fields $L_{\Omega}$. The set of primes that ramify in $L_{\Omega}$ is contained in $\Omega$ which are primes of good reduction of $E$. Said differently, all the primes of $Q_1$ are unramified in $L_{\Omega}$. Furthermore, for each prime $\ell\in \mathfrak{S}_n$, we have arranged that $p\nmid \#\widetilde{E}(\F_\ell)$. Hence, each prime of $Q_2$ is also unramified in $L_{\Omega}$. Hence, it follows from Lemma \ref{stability lemma} that for $L=L_{\Omega}$, both of the following conditions are satisfied
\begin{enumerate}
\item $E(L)=E(\Q)$,
\item $\Sha(E/L)[p^\infty]=\Sha(E/\Q)[p^\infty]=0$.
\end{enumerate}
This completes the proof.
\end{proof}
We conclude this section with a refinement of Theorem \ref{thm section 4 main} for elliptic curves $E_{/\Q}$ for which $E(\Q)_{\op{tors}}\neq 0$.
\begin{Lemma}\label{boring lemma}
Let $E_{/\Q}$ be an elliptic curve without CM and assume that
\begin{enumerate}
\item $E(\Q)_{\op{tors}}\neq 0$,
\item $\Sha(E/\Q)$ is finite.
\end{enumerate} Then $\mathcal{P}_E$ consists of all primes $p$ at which $E$ has good ordinary reduction \emph{except} a finite set of primes $\mathcal{P}_E'$. The set $\mathcal{P}_E'$ consists of a subset of the primes $p$ for which the following conditions are satisfied
\begin{enumerate}
\item\label{c1} $p\leq 5$ or $p\mid N$,
\item $E[p]$ is reducible as a Galois module,
\item\label{c3} $p\mid \#\Sha(E/\Q)$.
\item\label{c4} $\# E(\Q)_{\op{tors}}$ is of $p$-power order.
\end{enumerate}
\end{Lemma}
\begin{proof}
We refer to the discussion in the proof of Theorem \ref{100 percent}. A prime $p$ at which $E$ has good ordinary reduction is contained in $\mathcal{P}_E$ if and only if $p\nmid \# \Sha(E/\Q)$, $p\nmid c_\ell(E)$ for all primes $\ell\neq p$ and $p\nmid \#\widetilde{E}(\F_p)$. Suppose that $p\notin \mathcal{P}_E$ does not satisfy \eqref{c1}-\eqref{c3}, then, $p$ divides $\#\widetilde{E}(\F_p)$. Since $p\geq 7$, it follows from the Hasse bound that $\#\widetilde{E}(\F_p)=p$. Suppose that there is a prime $\ell\neq p$ that divides $\#E(\Q)_{\op{tors}}$. Denote by $\chi_\ell$ the mod-$\ell$ cyclotomic character.
Since $E(\Q)[\ell]\neq 0$, the mod-$\ell$ representation is of the form \[\bar{\rho}_{E,\ell}=\mtx{1}{\ast}{0}{\chi_{\ell}}.\]
Let $\sigma_p$ be the Frobenius at $p$.
We find that
\[
1+p-\#\widetilde{E}(\F_p)\equiv \op{trace}\left(\bar{\rho}_{E,\ell}(\sigma_p)\right)\equiv 1+\chi_\ell(p)=1+p\pmod{\ell}.
\]
Therefore, $\ell$ divides $\# \widetilde{E}(\F_p)$, a contradiction. Therefore, \eqref{c4} is satisfied, hence the result.
\end{proof}
The following Theorem follows immediately from Theorem \ref{thm section 4 main} and Lemma \ref{boring lemma}.
\begin{Th}\label{4.15}
Let $E_{/\Q}$ be an elliptic curve without complex multiplication such that $E(\Q)_{\op{tors}}\neq 0$ and $\op{rank} E(\Q)=0$. Then $E$ is diophantine stable \emph{and} $\Sha$-stable at all but a finite set of primes $p$ at which it has good ordinary reduction.
\end{Th}
\par \emph{Example:} Let us illustrate the above result through an explicit example. Consider the elliptic curve $E$ with Cremona label \href{https://www.lmfdb.org/EllipticCurve/Q/14a1/}{14a1}, this is the modular curve $X_0(14)$. Let us note down a few properties of $E$,
\begin{enumerate}
\item $E$ is a non-CM elliptic curve with rank 0.
\item At all primes $p\geq 5$ the Galois representation on $E[p]$ is surjective (hence irreducible).
\item The Tate Shafarevich group $\Sha(E/\Q)=0$.
\item The group $E(\Q)=\Z/6\Z$, hence, is not of prime power order.
\end{enumerate}
It follows from Lemma \ref{boring lemma} and Corollary \ref{4.15} that $E$ is diophantine stable at all primes $p\geq 11$ at which $E$ has good ordinary reduction.
\section{Rank jumps and Growth of the Tate-Shafarevich group on average}\label{s5}
\par In the previous section, it was shown that a rational elliptic curve of rank zero without complex multiplication is diophantine stable at $100\%$ of the primes. Furthermore, a similar result was proven for stability of the $p$-primary part of the Tate-Shafarevich group as $p$ varies. In this section, we study the average growth of ranks and Tate-Shafarevich groups in cyclic $p$-extensions. As in the previous section, results are proved for a fixed elliptic curve $E_{/\Q}$ and varying prime $p$.
\par First, we extend Definition \ref{def PE} to elliptic curves of arbitrary rank. Throughout, it is assumed that the $\Sha(E/\Q)$ is finite and that the $p$-adic regulator $\mathcal{R}_p(E/\Q)$ is non-zero.
\begin{Definition}\label{def 5.1}
For an elliptic curve $E_{/\Q}$, let $\mathcal{P}_E$ consist of the primes $p$ such that
\begin{enumerate}
\item $p$ is odd,
\item $E[p]$ is irreducible as a $\op{Gal}(\bar{\Q}/\Q)$-module,
\item $E$ has good ordinary reduction at $p$,
\item the equivalent conditions of Lemma \ref{lemma ECF mulambda} are satisfied.
\end{enumerate}
\end{Definition}
Note that when $\op{rank} E(\Q)=0$ and $E(\Q)[p]=0$, the Selmer group $\op{Sel}_{p^\infty}(E/\Q_\infty)$ does not contain any proper finite index submodules, see \cite[Proposition 4.14]{greenbergITEC}. If the $\mu$ and $\lambda$-invariants vanish, then, $\op{Sel}_{p^\infty}(E/\Q_\infty)$ is finite, and hence $0$. Consequently, the above definition coincides with Definition \ref{def PE} when $\op{rank} E(\Q)=0$. It is expected that even in the positive rank setting, the set $\mathcal{P}_E$ consists of $100\%$ of primes, see \cite[Conjecture 1.1]{kunduray1}. Indeed, this is the case if the normalized $p$-adic regulator $\mathcal{R}_p(E/\Q)$ is a $p$-adic unit for $100\%$ of primes $p$, and there is computational evidence for this (see Theorem 3.13 and the subsequent discussion in \emph{loc. cit.}).
\par The main result in this section for elliptic curves of rank $0$ states that for $100\%$ of the primes $p$, the $p$-primary Selmer group exhibits growth for base-change by a large number of $p$-cyclic extensions.
\begin{Th}\label{5.1 main theorem}
Let $E$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)=0$ and let $p\in \mathcal{P}_E$. Assume furthermore that the residual representation
\[\bar{\rho}:\op{Gal}(\bar{\Q}/\Q)\rightarrow\op{GL}_2(\F_p)\] is surjective. Then, for any $n\in \Z_{\geq 1}$ and a finite set of primes $\Sigma$, there are infinitely many $\Z/p^n\Z$-extensions $L/\Q$ such that
\begin{enumerate}
\item all primes of $\Sigma$ split in $L$,
\item $\op{Sel}_{p^\infty}(E/\Q)=0$ and $\op{Sel}_{p^\infty}(E/\Q)\neq 0$.
\end{enumerate}
\end{Th}
\par We obtain an independent result for elliptic curves of positive rank as well. Note that this is the first time in this paper that elliptic curves of positive rank are studied.
\begin{Th}\label{5.1 theorem pos rank}
Let $E$ be an elliptic curve without complex multiplication such that $\op{rank} E(\Q)>0$ and let $p\in \mathcal{P}_E$. Let $L$ be \textit{any} cyclic $p$-extension. Then, at least one of the following assertions hold:
\begin{enumerate}
\item\label{5.3 c1} $\op{rank}E(L)\geq [L:\Q]\op{rank} E(\Q)$,
\item\label{5.3 c2} $\mathcal{R}_p(E/\Q)\in\Z_p^\times$ and $\mathcal{R}_p(E/L)\notin \Z_p^\times$,
\item\label{5.3 c3}
$\Sha(E/\Q)[p^\infty]=0$ and $\Sha(E/L)[p^\infty]\neq 0$.
\end{enumerate}
\end{Th}
Thus for any elliptic curve $E_{/\Q}$ without complex multiplicative any triple $(p, n, \Sigma)$, with $p\in \mathcal{P}_E$, there are infinitely many cyclic extensions such that the Selmer group grows. This implies that either the rank increases or the Tate-Shafarevich group grows. The following special case of the conjecture of David-Fearnley-Kisilevsky (see \cite[Conjecture 1.2]{DFK}) predicts that given an elliptic curve over $\Q$, the rank rarely jumps in $\Z/p\Z$-extensions with $p\geq 7$.
\begin{Conjecture}[David-Fearnley-Kisilevsky]\label{DFK conjecture}
Let $p\geq 7$ be a prime and $E$ an elliptic curve over $\Q$. Then, there are only finitely many $
\Z/p\Z$-extensions $L/\Q$ such that $\op{rank} E(L)>\op{rank} E(\Q)$.
\end{Conjecture}
The following is an immediate Corollary to Theorem \ref{5.1 main theorem}. The result shows that if the Conjecture of David-Fearney-Kisilevski is true, then indeed there are many $\Z/p\Z$-extensions in which the size of the Tate-Shafarevich group increases. In fact, infinitely many extensions may be constructed that in which a given finite set of primes split.
\begin{Corollary}
Let $E$ be an elliptic curve without complex multiplication and let $p\in \mathcal{P}_E$. Assume that $p\geq 7$ and that Conjecture \ref{DFK conjecture} is satisfied for $E$ at $p$. Suppose $\op{rank} E(\Q)=0$, then, for any finite set of primes $\Sigma$, there are infinitely many $\Z/p\Z$-extensions $L/\Q$ in which all primes of $\Sigma$ split and
\[\Sha(E/L)[p^\infty]\neq 0\text{ and }\Sha(E/\Q)[p^\infty]=0.\]
Suppose that $\op{rank} E(\Q)>0$,
then, for any finite set of primes $\Sigma$, there are infinitely many $\Z/p\Z$-extensions $L/\Q$ in which all primes of $\Sigma$ split and at least one of the the following conditions is satisfied:
\begin{enumerate}
\item $\mathcal{R}_p(E/\Q)\in\Z_p^\times$ and $\mathcal{R}_p(E/L)\in p\Z_p$,
\item
$\Sha(E/\Q)[p^\infty]=0$ and $\Sha(E/L)[p^\infty]\neq 0$.
\end{enumerate}
\end{Corollary}
\begin{proof}
\par First, consider the case when $\op{rank}E(\Q)=0$. According to Theorem \ref{5.1 main theorem}, there are an infinitude of $\Z/p\Z$-extensions $L/\Q$ in which the primes of $\Sigma$ split and such that $\op{Sel}_{p^\infty}(E/L)\neq 0$ and $\op{Sel}_{p^\infty}(E/\Q)=0$. If Conjecture \ref{DFK conjecture} is true, then for all but finitely many of the extensions $L/\Q$, the rank of $E(L)$ is zero. Suppose $L/\Q$ is a $\Z/p\Z$-extension such that $\op{rank} E(L)=0$. Note that since $p\in \mathcal{P}_E$, it is assumed that $E[p]$ is irreducible as a Galois module. As a consequence, we have that $E(\Q)[p]=0$. Since $L/\Q$ is a $p$-extension, it follows that $E(L)[p]=0$ as well, see Lemma \ref{NSW lemma}. As a result, the $p$-primary Mordell-Weil group $E(L)\otimes \Q_p/\Z_p=0$, and hence, the Selmer group $\op{Sel}_{p^\infty}(E/L)$ coincides with $\Sha(E/L)[p^\infty]$. Since $\op{Sel}_{p^\infty}(E/L)\neq 0$, the result follows.
\par When $\op{rank} E(\Q)>0$, the result is a direct consequence of Theorem \ref{5.1 theorem pos rank}. \end{proof}
\begin{Lemma}\label{boringlemma}
Let $p\geq 5$ be a prime and $L/\Q$ a $p$-cyclic extension, and $\ell\neq p$ be a prime. Make the following assumptions:
\begin{enumerate}
\item $c_\ell^{(p)}(E/\Q)=1$,
\item if $\ell$ is ramified in $L$, then $E$ has good reduction at $\ell$.
\end{enumerate}Then, we have that $\prod_{v\mid\ell} c_v^{(p)}(E/L)=1$, where $v$ ranges over all primes of $L$ above $\ell$.
\end{Lemma}
\begin{proof}
Since $\ell\neq p$, and $L/\Q$ is a pro-$p$, it follows that $\ell$ is tamely ramified in $L$, and the results for base-change of Tamagawa numbers in \cite[Table 1, pp. 556-557]{Kida} apply. We consider two cases.
\begin{enumerate}
\item First consider the case where $\ell$ is unramified in $L$. Fix a prime $v\mid \ell$. Since $p\geq 5$, the Tamagawa number $c_v^{(p)}(E/L)\neq 1$ if and only if the Kodaira type of $E_{/L_v}$ is $\op{I}_n$ for an integer $n\in \Z_{\geq 1}$ that is divisible by $p$ (see \cite[p. 448]{Sil09}). However, according to \cite[Table 1, pp. 556-557]{Kida}, the only way this is possible is if if the Kodaira type of $E_{/\Q_\ell}$ is $\op{I}_n$ to begin with. However, this is not the case, since $c_\ell^{(p)}(E/\Q)= 1$.
\item Next, consider the case when $\ell$ ramifies in $L$. In this case, it is assumed that $E$ has good reduction at $\ell$, and the result is clear.
\end{enumerate}
\end{proof}
\begin{Proposition}\label{boring prop}
Let $E_{/\Q}$ be an elliptic curve and $p$ an odd prime at which $E$ has good ordinary reduction. Assume that $p\in \mathcal{P}_E$ and that $\op{rank} E(\Q)=0$.
Let $L/\Q$ be a $\Z/p^n\Z$-extension of $\Q$ for which the following conditions are satisfied:
\begin{enumerate}
\item\label{boring prop c1} $E$ has good reduction at all primes $\ell\neq p$ which ramify in $L$,
\item\label{boring prop c2} there exists a prime $\ell\neq p$ which ramifies in $L$, we have that $p|\#\widetilde{E}(\F_\ell)$.
\end{enumerate}
Then at least one of the following holds
\begin{enumerate}
\item $\op{rank} E(L)>0$,
\item $\Sha(E/L)[p^\infty]\neq 0$ and $\Sha(E/\Q)[p^\infty]=0$.
\end{enumerate}
\end{Proposition}
\begin{proof}
Recall that since $\op{rank} E(\Q)=0$, the (normalized) regulator $\mathcal{R}_p(E/\Q)=1$. By definition, since $p\in \mathcal{P}_E$, the following equivalent conditions are satisfied:
\begin{enumerate}
\item $\mu_p(E/\Q)=0$ and $\lambda_p(E/\Q)=0$,
\item\label{29 aug c2} $\chi(\Gamma_\Q, E[p^\infty])=1$,
\item\label{29 aug c3} $\Sha(E/\Q)[p^\infty]=0$, $p\nmid c_\ell(E/\Q)=1$ for all $\ell\neq p$ and $p\nmid \#\widetilde{E}(\F_p)$.
\end{enumerate}The equivalence of \eqref{29 aug c2} and \eqref{29 aug c3} follows from the Euler characteristic formula. Assume by way of contradiction that both the conditions \eqref{boring prop c1} and \eqref{boring prop c2} of the proposition are not satisfied. In other words, assume that $\op{rank} E(L)=0$ and $\Sha(E/L)[p^\infty]=0$. Note that since $E[p]$ is irreducible as a Galois module, it follows that $E(\Q)[p]=0$. Since $L/\Q$ is a $p$-cyclic extension, it follows from Lemma \ref{NSW lemma} that $E(L)[p]=0$. Thus, we have that
\[\chi_t(\Gamma_L, E[p^\infty])\sim \# \Sha(E/L)[p^\infty] \times \prod_{v\nmid p} c_{v}^{(p)}(E/L) \times \prod_{v|p}\left(\# \widetilde{E}(\kappa_v)[p^\infty]\right)^2.\]
We show that $\chi_t(\Gamma_L, E[p^\infty])$ by showing that the contributing factors are all $1$. It follows from Lemma \ref{boringlemma} that the Tamagawa product $\prod_{v\nmid p} c_v^{(p)}(E/L)=1$. Finally, for each prime $v|p$, $[k_v:\F_p]$ is either $1$ or a power of $p$. Hence, according to
\[\#\widetilde{E}(\F_p)[p]=0\Rightarrow \#\widetilde{E}(k_v)[p]=0.\] Putting everything together, we conclude that $\chi_t(\Gamma_L, E[p^\infty])=1$. Hence, according to Lemma \ref{lemma ECF mulambda}, this implies that $\mu_p(E/L)=0$ and $\lambda_p(E/L)=0$. On the other hand, since there is a prime $\ell\neq p$ which is ramified in $L$ such that $p|\#\widetilde{E}(\F_\ell)$, we have that $\ell\in Q_2$. Therefore, it follows from Kida's formula that $\lambda_p(E/L)>0$, a contradiction.
\end{proof}
\begin{proof}[proof of Theorem \ref{5.1 main theorem}]
Note that since $p\in \mathcal{P}_E$, the Selmer group $\op{Sel}_{p^\infty}(E/\Q_\infty)=0$. As argued before, the irreducibility of $E[p]$ implies that the Selmer group $\op{Sel}_{p^\infty}(E/\Q)$ injects into $\op{Sel}_{p^\infty}(E/\Q_\infty)$, hence is $0$ as well. Let $\mathfrak{T}_n$ be the set of primes $\ell\nmid Np$ such that
\begin{enumerate}
\item $\ell\equiv 1\mod{p^n}$,
\item $p\mid \# \widetilde{E}(\F_\ell)$.
\end{enumerate}
The same arguments as in the proof of Lemma \ref{s4 positive density} show that $\mathfrak{T}_n$ has positive density, hence, is infinite. Let $t:=\#\Sigma$, and let $\Omega$ be a set of $t+1$ primes in $\mathfrak{T}_n$. There is a $\Z/p^n\Z$-extension $L_\Omega$ with ramification contained in $\Omega$ and in which $\Sigma$ is split. The extension $L_\Omega$ is ramified at at least one prime $\ell\in \Omega$. Since $p$ divides $\#\widetilde{E}(\F_\ell)$, the result follows from Proposition \ref{boring prop}.
\end{proof}
\begin{proof}[proof of Theorem \ref{5.1 theorem pos rank}]
\par The proof of this result is similar to that of Proposition \ref{boring prop}. Assume that the conditions \eqref{5.3 c2} and \eqref{5.3 c3} are not satisfied. Then, we show that \eqref{5.3 c1} is satisfied, i.e., $\op{rank}E(L)\geq p\op{rank} E(\Q)$. Recall that since $p\in \mathcal{P}_E$, $\chi_t(\Gamma_\Q, E[p^\infty])=1$. Since $E[p]$ is irreducible as a Galois module, the Euler characteristic formula implies that $\mathcal{R}_p(E/\Q)\in \Z_p^\times$. Since \eqref{5.3 c2} is not satisfied, $\mathcal{R}_p(E/L)\in \Z_p^\times$. Moreover $\Sha(E/\Q)[p^\infty]=0$, and since \eqref{5.3 c3} is not satisfied, it follows that $\Sha(E/L)[p^\infty]=0$. By the same reasoning as in the proof of Proposition \ref{boring prop}, we have that $\chi_t(\Gamma_L, E[p^\infty])=1$. Therefore, by Lemma \ref{lemma ECF mulambda}, $\lambda_p(E/L)=\op{rank} E(L)$. On the other hand, since $p\in \mathcal{P}_E$, $\lambda_p(E/\Q)=\op{rank} E(\Q)$. Therefore, according to \eqref{kidaformula},
\[\op{rank} E(L)\geq [L:\Q] \op{rank} E(\Q),\] hence, condition \eqref{5.3 c1} is satisfied.
\end{proof}
\section{Stability for Elliptic curves on average at a fixed prime $p$}
\par Recall that in sections \ref{s4} and \ref{s5}, results were proved for a fixed elliptic curve and varying prime $p$. In this section, we study the dual questions.
\begin{question}\label{question 6.1}
Let $p$ be a prime. What can one say about the proportion of rational elliptic curves that are diophantine stable at $p$?
\end{question}
Our results on diophantine stability only apply to $p\geq 11$, hence, we make this assumption throughout. Recall that any elliptic curve $E_{/\Q}$ admits a unique Weierstrass equation
\begin{equation}\label{weier}
E:Y^2 = X^3 + aX + b
\end{equation}
where $a, b$ are integers and $\gcd(a^3 , b^2)$ is not divisible by any twelfth power.
Since $p\geq 5$, such an equation is minimal.
Recall that the \emph{height of} $E$ satisfying the minimal equation $\eqref{weier}$ is given by $H(E) := \op{max}\left(|a|^3, b^2\right)$.
Let $\mathscr{E}$ be the set of isomorphism classes of elliptic curves defined over $\Q$, and for any subset $\mathcal{S}\subset \mathscr{E}$, let $\mathcal{S}(x)$ consist of all $E\in \mathcal{S}$ such that $H(E)<x$.
\par In order to provide some answers to the above question, we assume some well known conjectures.
\begin{Conjecture}[Rank distribution conjecture]
For elliptic curves ordered by height, the proportion of elliptic curves with rank $0$ is $50\%$, the proportion with rank $1$ is $50\%$.
\end{Conjecture}
Thus, the proportion of elliptic curves $E_{/\Q}$ with $\op{rank} E(\Q)\geq 2$ is expected to be $0\%$. Denote by $\mathscr{E}$ the set of all isomorphism classes of rational elliptic curves. Let $\mathscr{E}'$ be the set of isomorphism classes of rational elliptic curves of rank 0 with good reduction at the primes $2$ and $3$. At any prime $\ell$, the proportion of elliptic curves with good reduction is $(1-\ell^{-1})$, see the proof of \cite[Proposition 4.2 (1)]{stats2}. It is reasonable to make the following conjecture.
\begin{Conjecture}[RDC+]\label{RDC}
The lower density of $\mathscr{E}'$ satisfies
\[\liminf_{x\rightarrow \infty} \frac{\# \mathscr{E}'(x)}{\# \mathscr{E}(x)}= \frac{1}{2}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=\frac{1}{6}.\]
\end{Conjecture}
Along with the above rank distribution conjecture, one also makes the following assumption due to Delaunay, see \cite{delaunay}. \begin{Conjecture}[Del-p]\label{Del}
The proportion of all elliptic curves such that $E$ has rank 0 for which $p\mid \# \Sha(E/\Q)$ is given by
\[\frac{1}{2}\left(1-\prod_{i\geq 1} \left(1-\frac{1}{p^{2i-1}}\right)\right).\]
\end{Conjecture}
Delaunay predicts that the proportion of rank 0 elliptic curves with $p$ dividing the order of the Tate-Shafarevich group is given by \[1-\prod_{i\geq 1} \left(1-\frac{1}{p^{2i-1}}\right)= \frac{1}{p}+\frac{1}{p^3}-\frac{1}{p^4}+\dots.\] The term $1/2$ is expected due to rank distribution.
\par For $(a,b)\in \F_p\times \F_p$ with $\Delta:=4a^3+27b^2$ non-zero, associate the elliptic curve $E_{a,b}$ defined by the Weierstrass equation
\[
E_{a,b}:Y^2=X^3+aX+b.
\]
Let $\mathfrak{T}_p$ be the set of pairs $(a,b)\in \F_p\times \F_p$ such that $\#E_{a,b}(\F_p)\in \{p,p+1\}$. For $t\in \{0,1\}$, the number of isomorphism classes of elliptic curves over $\F_p$ with $\#E(\F_p)=p+1-t$ is given by $H(t^2-4p)$, where, $H(\cdot)$ denotes the Kronecker class number (see \cite[p. 184]{Schoof87}).
\begin{Lemma}\label{lemma 6.5}
Assume that $p\geq 5$ is a prime. There is an explicit positive constant $c_1$ such that
\[
\# \mathfrak{T}_p \leq c_1 p^{\frac{3}{2}} \log p \left( \log \log p\right)^2.
\]
\end{Lemma}
\begin{proof}
It is well known that $E_{a,b}$ is isomorphic to $E_{a',b'}$ over $\F_p$ if and only if
\[a'=c^4 a\text{ and }b'=c^6 b\] for some element $c\in \F_p^\times$. Thus, the number of curves $E_{a',b'}$ that are isomorphic to $E_{a,b}$ is at most $\frac{p-1}{2}$, hence, the first inequality. The inequality follows from \cite[Proposition 1.9]{Lenstra_annals}.
\end{proof}
\par The following result provides a conditional answer to Question \ref{question 6.1}.
\begin{Th}\label{last thm}
Let $p\geq 11$ be a prime and assume that Conjectures (RDC+) and (Del-p) are satisfied. Let $\mathscr{E}_p$ be the set of isomorphism classes of elliptic curves $E_{/\Q}$ satisfying the following conditions:
\begin{enumerate}
\item\label{6.5 c1} $\op{rank} E(\Q)=0$,
\item $E[p]$ is irreducible as a Galois module,
\item $E$ has good reduction at $2,3$,
\item\label{6.5 c4} $E$ has good ordinary reduction at $p$,
\item\label{6.5 c5} $E$ is diophantine stable and $\Sha$-stable at $p$.
\end{enumerate}
Then, for the effective constant $c_1>0$ of Lemma \ref{lemma 6.5}, we have that
\[\begin{split}\liminf_{x\rightarrow \infty} \frac{\# \mathscr{E}_p(x)}{\#\mathscr{E}(x)} \geq &\frac{1}{6}-\frac{1}{p}-\frac{1}{2}\left(1-\prod_i \left(1-\frac{1}{p^{2i-1}}\right)\right) \\
& -(\zeta(p)-1)-\zeta(10) c_1\frac{\log p\cdot (\log \log p)^2}{\sqrt{p}}.\\
\end{split}\]
In particular, we find that
\[\liminf_{p\rightarrow \infty}\left(\liminf_{x\rightarrow \infty} \frac{\# \mathscr{E}_p(x)}{\#\mathscr{E}(x)}\right)=\frac{1}{6}.\]
\end{Th}
\begin{proof}
Suppose that the conditions \eqref{6.5 c1}-\eqref{6.5 c4} are satisfied for $(E,p)$. Then, if $p\in \mathcal{P}_E$ it follows from Theorem \ref{thm section 4 main} that $E$ is diophantine stable and $\Sha$-stable at $p$.
For $i=1,\dots, 5$ let $\mathscr{E}_p^{(i)}$ be the subset of $\mathscr{E}$ defined as follows:
\begin{enumerate}
\item $\mathscr{E}_p^{(1)}$ is the subset of all elliptic curves of rank 0 such that $p|\#\Sha(E/\Q)$. According to (Del-p),
\[\limsup_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(1)}(x)}{\#\mathscr{E}(x)}=\frac{1}{2}\left(1-\prod_i \left(1-\frac{1}{p^{2i-1}}\right)\right).\]
\item Let $\mathscr{E}_p^{(2)}$ be the subset of all elliptic curves such that $p\mid \prod_{\ell\neq p} c_\ell(E/\Q)$. It follows from \cite[Corollary 8.8]{HKR} that
\[\limsup_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(2)}(x)}{\#\mathscr{E}(x)}\leq (\zeta(p)-1) \]
\item Let $\mathscr{E}_p^{(3)}$ be the subset of all elliptic curves such that at least of the following conditions hold:
\begin{enumerate}
\item $p\mid \#\widetilde{E}(\F_p)$,
\item $p$ has good supersingular reduction at $p$.
\end{enumerate} Note that since $p\geq 11$, it follows from the Hasse bound that this is the case precisely when $\#\widetilde{E}(\F_p)=p+1-t$, where $t\in \{0,1\}$. We have that
\[\limsup_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(3)}(x)}{\#\mathscr{E}(x)}\leq \zeta(10)\frac{\#\mathfrak{T}_p}{p^2}, \]see the proof of \cite[Theorem 4.14]{kunduray1}. It follows from Lemma \ref{lemma 6.5} that
\[\#\mathfrak{T}_p\leq c_1 p^{\frac{3}{2}} \log p \cdot (\log \log p)^2.\]
Therefore, we have that
\[\limsup_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(3)}(x)}{\#\mathscr{E}(x)}\leq \frac{\zeta(10) c_1\log p\cdot (\log \log p)^2}{\sqrt{p}}. \]
\item Let $\mathscr{E}_p^{(4)}$ be the subset of all elliptic curves such that $E[p]$ is reducible as a Galois module. It follows from the main result of \cite{duke} that \[\limsup_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(4)}(x)}{\#\mathscr{E}(x)}=0. \]
\item Let $\mathscr{E}_p^{(5)}$ be the subset of all elliptic curves such $E$ has bad reduction at $p$, it follows from \cite[Proposition 4.2 (1)]{stats2} that \[\lim_{x\rightarrow \infty} \frac{\#\mathscr{E}_p^{(5)}(x)}{\#\mathscr{E}(x)}= \frac{1}{p}. \]
\end{enumerate}
Note that $\mathscr{E}_p$ is contained in $\mathscr{E}'\backslash \left(\bigcup_{i=1}^5 \mathscr{E}_p^{(i)}\right)$. Therefore,
\[\liminf_{x\rightarrow \infty} \frac{\# \mathscr{E}_p(x)}{\#\mathscr{E}(x)}\geq \liminf_{x\rightarrow \infty} \frac{\# \mathscr{E}'(x)}{\#\mathscr{E}(x)}-\sum_{i=1}^5 \limsup_{x\rightarrow \infty} \frac{\# \mathscr{E}_p^{(i)}(x)}{\#\mathscr{E}(x)}.\]
The result follows.
\end{proof} | 196,957 |
As part of our general membership meeting we will be hosting a Professional Development Workshop titled “Business of Politics: Understanding Your World of Politics” to give our members and young professionals in South Florida the opportunity to learn and discuss different avenues of politics. Additionally, with the upcoming elections we feel that it is imperative to inform the public about the electoral process, voting rights etc. Our distinguished panelists will discuss pertinent topics on the intersection of politics and business in the 21st century. | 196,729 |
TITLE: Linear map help needed, explain the theorem
QUESTION [0 upvotes]: Suppose that (v1,…,vn) is a basis of V and (w1,…,wm) is a basis of W. Then M is an invertible linear map between L(V,W) and Mat(m,n,F).
But all invertible matrices are square matrices, therefore, it means that m=n in this case, V and W have same dimensions?
If so, does it mean that V=W?
REPLY [2 votes]: No. You misunderstood what $M$ is.
$M$ is a linear transformation, which map linear transformation to its matrix according to the basis. $M$ is NOT a linear transformation from $V$ to $W$. $M$ is in fact a linear transformation from a space of dimension $mn$ to another space of dimension $mn$. Hence if you represent $M$ as a matrix, you will end up with a matrix $(mn)\times(mn)$. So all you can conclude from this is that $mn=mn$.
So what this means is that $L(V,W)$ is isomorphic to $M(m,n,F)$. | 114,288 |
\begin{document}
\title{\bf Shifted Matroid Optimization}
\author{
Asaf Levin
\thanks{\small Technion - Israel Institute of Technology, Haifa, Israel.
Email: [email protected]}
\and
Shmuel Onn
\thanks{\small Technion - Israel Institute of Technology, Haifa, Israel.
Email: [email protected]}
}
\date{}
\maketitle
\begin{abstract}
We show that finding lexicographically minimal $n$ bases in a matroid can
be done in polynomial time in the oracle model. This follows from a more general result
that the shifted problem over a matroid can be solved in polynomial time as well.
\vskip.2cm
\noindent {\bf Keywords:} integer programming, combinatorial optimization,
unimodular, matroid, spanning tree, matching
\end{abstract}
\section{Introduction}
\label{introduction}
Let $G$ be a connected graph and let $n$ be a positive integer. Given $n$ spanning
trees in $G$, an edge is {\em vulnerable} if it is used by all trees.
We wish to find $n$ spanning trees with minimum number of vulnerable edges.
One motivation for this problem is as follows. We need to make a sensitive
broadcast over $G$. In the planning stage, $n$ trees are chosen and prepared.
Then, just prior to the actual broadcast, one of these trees is randomly chosen and used.
An adversary, trying to harm the broadcast and aware of the prepared trees
but not of the tree finally chosen, will try to harm a vulnerable edge,
used by all trees. So we protect each vulnerable edge with high cost,
and our goal is to choose $n$ spanning trees with minimum number of vulnerable edges.
Here we consider the following harder problem. For $k=1,\dots,n$,
call an edge {\em $k$-vulnerable} if it is used by at least $k$ of the $n$ trees.
We want to find $n$ {\em lexicographically minimal} trees,
that is, which first of all minimize the number of $n$-vulnerable edges,
then of $(n-1)$-vulnerable edges, and so on. More precisely, given $n$ trees,
define their {\em vulnerability vector} to be $f=(f_1,\dots,f_n)$ with $f_k$ the
number of $k$-vulnerable edges. Then $n$ trees with vulnerability vector $f$ are better
than $n$ trees with vulnerability vector $g$ if the last nonzero entry of $g-f$ is positive.
(We remark that this order is often used in the symbolic computation literature, where it is called
{\em reverse lexicographic}, but for brevity we will simply call it here {\em lexicographic}.)
As a byproduct of our results we show how to find in polynomial time $n$ lexicographically
minimal spanning trees, which in particular minimize the number of vulnerable edges.
This problem can be defined for any combinatorial optimization set as follows.
For matrix $x\in\R^{d\times n}$ let $x^j$ be its $j$-th column.
Define the {\em $n$-product of} a set $S\subseteq\R^d$ by
$$S^n\ :=\ \times_n S\ :=\ \{x\in\R^{d\times n}\ :\ x^j\in S\,,\ j=1,\dots,n \}\ .$$
Call two matrices $x,y\in\R^{d\times n}$ {\em equivalent} and write $x\sim y$ if
each row of $x$ is a permutation of the corresponding row of $y$.
The {\em shift} of matrix $x\in\R^{d\times n}$ is the unique matrix ${\la x}\in\R^{d\times n}$
which satisfies ${\la x}\sim x$ and $x^1\geq \cdots\geq x^n$, that is, the unique matrix
equivalent to $x$ with each row nonincreasing. Let $|x^j|:=\sum_{i=1}^d |x_{i,j}|$
and $|x|:=\sum_{j=1}^n|x^j|$ be the sums of absolute values of the components of $x^j$ and $x$,
respectively. The {\em vulnerability vector} of $x\in\{0,1\}^{d\times n}$ is
$$(|{\la x}^1|,\dots,|{\la x}^n|)\ .$$
We then have the following nonlinear combinatorial optimization problem.
\vskip.2cm\noindent{\bf Lexicographic Combinatorial Optimization.}
Given $S\subseteq\{0,1\}^d$ and $n$, solve
\begin{equation}\label{lexicographic_problem}
\lexmin\{(|{\la x}^1|,\dots,|{\la x}^n|)\ :\ x\in S^n\}\ .
\end{equation}
\vskip.2cm\noindent
The complexity of this problem depends on the presentation of the set $S$.
In \cite{KOS} it was shown that it is polynomial time solvable
for $S=\{z\in\{0,1\}^d\,:\,Az=b\}$ for any totally unimodular $A$ and any integer $b$.
Here we solve the problem for matroids.
\bt{matroid_theorem}
The lexicographic combinatorial optimization problem \eqref{lexicographic_problem}\,over the
bases of any matroid given by an independence oracle and any $n$ is polynomial time solvable.
\et
Our spanning tree problem is the special case with $S$ the set of indicators of spanning
trees in a given connected graph with $d$ edges, and hence is polynomial time solvable.
We note that Theorem \ref{matroid_theorem} provides a solution of a nonlinear optimization
problem over matroids, adding to available solutions of other nonlinear optimization
problems over matroids and independence systems in the literature,
see e.g. \cite{BLMORWW,BLOW,LOW,Onn} and the references therein.
We proceed as follows. In Section 2 we discuss the
{\em shifted combinatorial optimization problem} and its relation to the lexicographic
combinatorial optimization problem. In section 3 we solve the shifted problem
over matroids in Theorem \ref{matroid_theorem_2} and conclude Theorem \ref{matroid_theorem}.
In Section 4 we discuss matroid intersections, partially solve the shifted problem over
the intersection of two strongly base orderable matroids (which include gammoids)
in Theorem \ref{matroid_intersection_theorem}, and leave open the complexity
of the problem for the intersection of two arbitrary matroids.
We conclude in Section 5 with some final remarks about the polynomial time
solvability of the shifted and lexicographic problems over totally unimodular systems
from \cite{KOS}, and show that these problems are NP-hard over matchings already for $n=2$
and cubic graphs.
\section{Shifted Combinatorial Optimization}
\label{shifted}
Lexicographic combinatorial optimization can be reduced to the following problem.
\vskip.2cm\noindent{\bf Shifted Combinatorial Optimization.}
Given $S\subseteq\{0,1\}^d$ and $c\in\Z^{d\times n}$, solve
\begin{equation}\label{shifted_problem}
\max\{{\ra c}\,{\la x}\ :\ x\in S^n\}\ .
\end{equation}
The following lemma was shown in \cite{KOS}.
We include the proof for completeness.
\bl{reduction_1}\cite{KOS}
The Lexicographic Combinatorial Optimization problem \eqref{lexicographic_problem} can be reduced
in polynomial time to the Shifted Combinatorial Optimization problem \eqref{shifted_problem}.
\el
\bpr
Define the following $c\in\Z^{d\times n}$, and note that it satisfies ${\ra c}=c$,
$$c_{i,j}\ :=\ -(d+1)^{j-1}\,,\quad i=1,\dots,d\,,\quad j=1,\dots,n\ .$$
Consider any two vectors $x,y\in S^{n}$, and suppose that the vulnerability vector
$(|{\la x}^1|,\dots,|{\la x}^n|)$ of $x$ is lexicographically smaller than the
vulnerability vector $(|{\la y}^1|,\dots,|{\la y}^n|)$ of $y$. Let $r$ be the largest
index such that $|{\la x}^r|\neq |{\la y}^r|$. Then $|{\la y}^r|\geq |{\la x}^r|+1$.
We then have
\begin{eqnarray*}
{\ra c}\,{\la x}-{\ra c}\,{\la y}\ =\ c\,{\la x}-c\,{\la y}
& = & \sum_{j=1}^n(d+1)^{j-1}\left(|{\la y}^j|-|{\la x}^j|\right) \\
& \geq & \sum_{j<r}(d+1)^{j-1}\left(|{\la y}^j|-|{\la x}^j|\right)+(d+1)^{r-1} \\
& \geq & (d+1)^{r-1} -\sum_{j<r}d(d+1)^{j-1}\ >\ 0\ .
\end{eqnarray*}
Thus, an optimal solution $x$ for problem \eqref{shifted_problem}
is also optimal for problem \eqref{lexicographic_problem}.
\epr
We proceed to reduce the Shifted Combinatorial Optimization problem \eqref{shifted_problem}
in turn to two yet simpler auxiliary problems.
For a set of matrices $U\subseteq\{0,1\}^{d\times n}$ let $[U]$ be the set of
matrices which are equivalent to some matrix in $U$,
$$[U]\ :=\ \{x\in\{0,1\}^{d\times n}\ :\ \exists\ y\in U\,,\ x\sim y \}\ .$$
Consider the following two further algorithmic problems over a given $S\subseteq\{0,1\}^d$:
\begin{equation}\label{auxiliary_problem_1}
\mbox{{\bf Shuffling.} Given $c\in\Z^{d\times n}$, solve $\max\{cx\,:\,x\in[S^n]\}$.}
\end{equation}
\begin{equation}\label{auxiliary_problem_2}
\mbox{{\bf Fiber.} Given $x\in[S^n]$, find $y\in S^n$ such that $x\sim y$.}
\end{equation}
\bl{reduction_2}
The Shifted Combinatorial Optimization problem \eqref{shifted_problem} can be reduced in polynomial
time to the Shuffling and Fiber problems \eqref{auxiliary_problem_1} and \eqref{auxiliary_problem_2}.
\el
\bpr
First solve the Shuffling problem \eqref{auxiliary_problem_1} with profit matrix $\ra c$ and
let $x\in[S^n]$ be an optimal solution. Next solve the Fiber problem \eqref{auxiliary_problem_2}
for $x$ and find $y\in S^n$ such that $x\sim y$.
We claim that $y$ is optimal for the Shifted Combinatorial Optimization problem
\eqref{shifted_problem}. To prove this, we consider any $z$ which is feasible in
\eqref{shifted_problem}, and prove that the following inequality holds,
$${\ra c}\,{\la y}\ =\ {\ra c}\,{\la x}\ \geq\ {\ra c}\,x\ \geq\ {\ra c}\,{\la z}\ .$$
Indeed, the first equality follows since $x\sim y$ and therefore we have ${\la y}={\la x}$.
The middle inequality follows since ${\ra c}$ is nonincreasing. The last inequality follows since
$z\in S^n$ implies that $\la z\in[S^n]$ and hence $\la z$ is feasible in \eqref{auxiliary_problem_1}.
So $y$ is indeed an optimal solution for problem \eqref{shifted_problem}.
\epr
\section{Matroids}
\label{matroids}
Define the {\em $n$-union} of a set $S\subseteq\{0,1\}^E$ where $E$ is any finite set to be the set
$$\vee_n S\ :=\ \{x\in\{0,1\}^E\ :\ \exists x_1,\dots,x_n\in S\,,\ x=\sum_{k=1}^n x_k\}\ .$$
Call $S\subseteq\{0,1\}^E$ a matroid if it is the set of indicators
of independent sets of a matroid over $E$. We will have matroids over
$E=[d]:=\{1,\dots,d\}$ and $E=[d]\times[n]$.
The following facts about $n$-unions of matroids are well known, see e.g. \cite{Sch}.
\bp{partition_matroid}
For any matroid $S$ and any $n$ we have that $\vee_n S$ is also a matroid.
Given an independence oracle for $S$, it is possible in polynomial time to realize an
independence oracle for $\vee_n S$, and if $x\in\vee_n S$,
to find $x_1,\dots,x_n\in S$ with $x=\sum_{k=1}^nx_k$.
\ep
Define the {\em $n$-lift} of a set of vectors $S\subseteq\{0,1\}^d$
to be the following set of matrices,
$$\uparrow_n\!S\ :=\ \{x\in\{0,1\}^{d\times n}\ :\ \sum_{j=1}^n x^j\in S\}\ .$$
We need two lemmas.
\bl{partition_1}
For any set $S\subseteq\{0,1\}^d$ and any $n$ we have that
$[S^n]=\vee_n \uparrow_n\!S$ in $\{0,1\}^{d\times n}$.
\el
\bpr
Consider $x\in [S^n]$. Then $x\sim y$ for some $y\in S^n$,
and thus $y^j\in S$ for $j=1,\dots,n$. Since $x\sim y$,
each row of $x$ is a permutation of the corresponding row of $y$.
Assume that the $i$-th row of $x$ is given by the permutation $\pi_i$
of the corresponding row of $y$. That is, $x_{i,j}=y_{i,\pi_i(j)}$.
For $k=1,\dots,n$, we define a matrix $z_k\in \{0,1\}^{d\times n}$
whose column sum satisfies $\sum_{j=1}^n z_k^j=y^k$, by $(z_k)_{i,j}:=x_{i,j}$ if $\pi_i(j)=k$,
and otherwise $(z_k)_{i,j}:=0$. Since $y^k\in S$, we conclude that $z_k\in \uparrow_n\!S$.
Since the supports of the $z_k$ are pairwise disjoint, we have
that $\sum_{k=1}^n z_k \in \vee_n \uparrow_n\!S$.
However, $\sum_{k=1}^n z_k=x$ by definition, and thus $x\in \vee_n \uparrow_n\!S$.
In the other direction, assume that $x\in \vee_n \uparrow_n\!S$. Then there are
$x_1,\dots,x_n\in \uparrow_n\!S$ such that $x=\sum_{k=1}^n x_k$.
That is, there are $x_1,\dots,x_n \in \{0,1\}^{d\times n}$ such that for each
$k=1,\dots,n$, we have $\sum_{j=1}^n x_k^j \in S$, and $x=\sum_{k=1}^n x_k$.
Let $y$ be a matrix whose $k$-th column is $y^k=\sum_{j=1}^n x_k^j$.
Then $y^k\in S$, and therefore $y\in S^n$. The matrices $x$ and $y$ are
$0-1$ matrices and $\sum_{j=1}^n x^j=\sum_{j=1}^n y^j$, and therefore $x\sim y$.
Thus, $x\in [S^n]$, as required.
\epr
\bl{uparrow_matroid}
Let $S\subseteq\{0,1\}^d$ be a matroid given by an independence oracle. Then we have:
\begin{enumerate}
\item
$\uparrow_n\!S\subseteq\{0,1\}^{d\times n}$ is also a matroid,
for which an independence oracle can be realized.
\item The rank of the matroid $\uparrow_n\!S$ equals to the rank of the matroid $S$.
\item The rank of the matroid $[S^n]=\vee_n \uparrow_n S$ equals $n$
times the rank of the matroid $S$.
\end{enumerate}
\el
\bpr
We begin with the first claim. Let $z$ be the $0$ matrix (where $z_{i,j}=0$ for all $i,j$).
Then $\sum_{j=1}^n z^j$ is the zero vector that belongs to $S$ since $S$ is a matroid, and
therefore $z\in\uparrow_n\!S$. Let $x\geq y$ be a pair of $0-1$ matrices such that
$x\in \uparrow_n\!S$. Let $\hat{x}=\sum_{j=1}^n x^j$, and $\hat{y}=\sum_{j=1}^n y^j$.
Then $\hat{x}\geq \hat{y}$. Since $x\in \uparrow_n\!S$, $\hat{x}\in S$, and because $S$ is
a matroid $\hat{y}\in S$, and thus $y\in \uparrow_n\!S$. Last, assume that $x,y\in\uparrow_n\!S$
and $|x|>|y|$. Let $\hat{x}=\sum_{j=1}^n x^j$ and $\hat{y}=\sum_{j=1}^n y^j$. Since $|x|>|y|$
we conclude that $|\hat{x}|>|\hat{y}|$. Since $x,y\in\uparrow_n\!S$, $\hat{x},\hat{y}\in S$
and since $S$ is a matroid there is $i$ such that $\hat{x}_i=1$, $\hat{y}_i=0$ and changing
$\hat{y}_i$ to $1$ results in an indicating vector of an independent set of $S$.
Consider this value of $i$, and let $j$ be such that $x_{i,j}=1$. Then, by changing $y_{i,j}$
to $1$, we get a larger matrix whose column sum is the vector resulted from $\hat{y}$ by
changing its $i$-th component to $1$, and thus the new matrix is in $\uparrow_n\!S$.
Therefore, $\uparrow_n\!S$ is a matroid. To present an independence oracle of $\uparrow_n\!S$,
assume that we would like to test the independence of a $0-1$ matrix $x$, then we compute
$\sum_{j=1}^n x^j$ and test if the resulting vector is independent in $S$.
Next we show the second claim. Let $x$ be an indicating matrix of an independent set
of the matroid $\uparrow_n\!S$, then its column sum is in $S$ and its rank in $S$ equals
to the rank of $x$ in $\uparrow_n\!S$. In the other direction, let $\hat{x}\in S$.
Define a matrix $x$ whose first column is $\hat{x}$ and its all other columns are zero columns,
then $\sum_{j=1}^n x^j=\hat{x}\in S$, and therefore $x\in \uparrow_n\!S$. The rank of $x$
in $\uparrow_n\!S$ equals the rank of $\hat{x}$ in $S$, and thus the claim follows.
Last, consider the remaining claim. Given a matroid $T$ it is always the case that the rank
of the ground set according to $\vee_n T$ is at most $n$ times the rank of the ground set
according to $T$. Thus, to show the claim using the second claim, it suffices to show that given
a base $z$ of $S$, there is an independent set of $\vee_n \uparrow_n S$ whose rank
(in $\vee_n \uparrow_n S$) is $n$ times larger than the rank of $z$ (in $S$).
With a slight abuse of notation, assume that $z$ is the indicator of the base $z$.
Let $x$ be a matrix such that for $j=1,\dots,n$, $x^j=z$. Then $x\in S^n$, and therefore it is
an indicator of an independent set of $\vee_n \uparrow_n S$ whose rank is $|x|=n|z|$ that
equals $n$ times the rank of the base $z$ (according to the rank function of the matroid $S$).
\epr
\bt{matroid_theorem_2}
The shifted combinatorial optimization problem \eqref{shifted_problem} over any matroid given
by an independence oracle or over its set of bases and any $n$ is polynomial time solvable.
\et
\bpr
We begin with the problem over (the independent sets of) a matroid $S$.
Consider problem \eqref{auxiliary_problem_1} over $S$. Since $[S^n]=\vee_n \uparrow_n\!S$
by Lemma \ref{partition_1}, it follows from Proposition \ref{partition_matroid} and
Lemma \ref{uparrow_matroid} that $[S^n]$ is a matroid for which we can realize an independence
oracle. So the greedy algorithm over $[S^n]$ solves problem \eqref{auxiliary_problem_1} over $S$.
Next consider problem \eqref{auxiliary_problem_2} over $S$. Let $x\in[S^n]$ be given.
By Proposition \ref{partition_matroid} and Lemma \ref{partition_1} and Lemma \ref{uparrow_matroid}
again, we can find $x_1,\dots,x_n\in\uparrow_n\!S$ with $x=\sum_{k=1}^nx_k$.
Let $y$ be the matrix with $y^k=\sum_{j=1}^nx_k^j$ for $k=1,\dots,n$. Then
$$\sum_{j=1}^nx^j\ =\ \sum_{j=1}^n\sum_{k=1}^nx_k^j\ =\
\sum_{k=1}^n\sum_{j=1}^nx_k^j\ =\ \sum_{k=1}^ny^k\ .$$
This implies that $x\sim y$. Now, since $x_k\in\uparrow_n\!S$ and
$y^k=\sum_{j=1}^nx_k^j$, we have that $y^k\in S$ for all $k$.
So we can find $y\in S^n$ with $x\sim y$ and therefore solve problem \eqref{auxiliary_problem_2}
over $S$ as well. It now follows from Lemma \ref{reduction_2} that we can
indeed solve problem \eqref{shifted_problem} over $S$.
To solve problem \eqref{shifted_problem} over the set of bases of $S$ proceed as follows.
Define a new profit matrix $w$ by $w_{i,j}:=c_{i,j}+2|c|+1$ for all $i,j$.
Solve problem \eqref{auxiliary_problem_1} over (the independent sets of) $S$ with
profit $w$ by the greedy algorithm over $[S^n]$. Let $r$ be the rank of $S$
so that by Lemma \ref{uparrow_matroid} the rank of $[S^n]$ is $nr$.
Consider any $x,y,z\in[S^n]$ with $x,y$ bases and $z$ not a basis. Then
$$wx\ =\ cx+(2|c|+1)|x|\ =\ cx+(2|c|+1)nr\ \geq\ -|c|+(2|c|+1)nr\ ,$$
$$wz\ =\ cz+(2|c|+1)|z|\ \leq\ |c|+(2|c|+1)(nr-1)\ .$$
So $wx-wz>0$ and hence the $w$-profit of any basis $x$ is larger than that
of any independent set $z$ which is not a basis, so the algorithm will output a basis.
Also, $wx-wy=cx-cy$ and hence the $w$-profit of basis $x$ is larger
than that of basis $y$ if and only if the $c$-profit of $x$ is larger than that
of $y$, so the algorithm will output a basis $x$ of $[S^n]$ with largest $c$-profit.
Now solve problem \eqref{auxiliary_problem_2} and find $y\in S^n$ with $x\sim y$ as before.
Since $S$ has rank $r$, each column $y^k\in S$ satisfies $|y^k|\leq r$.
Since $nr=|x|=|y|=\sum_{k=1}^n|y^k|$ we must have in fact $|y^k|=r$ so that $y^k$ is a basis
of $S$ for all $k$. Therefore $y$ is an optimal solution for problem
\eqref{shifted_problem} over the bases.
\epr
\vskip.2cm\noindent{\em Proof of Theorem \ref{matroid_theorem}.}
This follows at once from Lemma \ref{reduction_1} and Theorem \ref{matroid_theorem_2}.
\epr
\section{Matroid Intersections}
\label{matroid intersections}
We next consider the Shifted Combinatorial Optimization problem over matroid intersections.
We can only provide a partial solution for the class of {\em strongly base orderable} matroids,
introduced in \cite{Bru}, which strictly includes the class of {\em gammoids},
see \cite[Section 42.6c]{Sch}.
Let $S\subseteq\{0,1\}^d$ be a matroid and let ${\cal B}$ be the set of subsets of $[d]$
which are supports of bases of $S$. Then $S$ is {\em strongly base orderable} if for every pair
$B_1,B_2\in{\cal B}$ there is a bijection $\pi:B_1\rightarrow B_2$ such that
for all $I\subseteq B_1$ we have $\pi(I)\cup(B_1\setminus I)\in {\cal B}$
(where $\pi(I)=\{ \pi(i): i\in I\}$).
We need the following elegant result of \cite{DM}, see also \cite[Theorem 42.13]{Sch}.
\bp{intersection}
For any two strongly base orderable matroids $S_1$ and $S_2$ we have
$$(\vee_n S_1)\cap(\vee_n S_2)\ =\ \vee_n(S_1\cap S_2)\ .$$
\ep
We also need the following lemma.
\bl{strongly_based_lemma}
If $S$ is a strongly base orderable matroid, then $\uparrow_n\!S$ is also a strongly
base orderable matroid. Further, for any $S_1,S_2\subseteq\{0,1\}^d$ we have that
$\uparrow_n\!S_1\cap \uparrow_n\!S_2=\uparrow_n\!(S_1\cap S_2)$.
\el
\bpr
We begin with the first part. Let $S$ be a strongly base orderable matroid.
Then by Lemma \ref{uparrow_matroid}, $\uparrow_n\!S$ is also a matroid.
Let $x,y\in\{ 0,1\}^{d\times n}$ whose supports $X,Y$ are two bases of $\uparrow_n\!S$.
By Lemma \ref{uparrow_matroid}, the supports $\hat{X}$ and $\hat{Y}$ of $\hat{x}=\sum_{j=1}^n x^j$
and $\hat{y}=\sum_{j=1}^n y^j$ respectively are two bases of $S$. Since $S$ is a
strongly base orderable matroid there is a bijection $\pi:\hat{X}\rightarrow \hat{Y}$
such that $\pi(\hat{I})\cup (\hat{X}\setminus \hat{I})$ is a base of $S$ for every
$\hat{I} \subseteq \hat{X}$. We define a bijection $\pi':X\rightarrow Y$ as follows.
For every $(i,j)\in X$, we let $\pi'(i,j)=(\pi(i),j')$ where $j'$ is the unique column
containing $1$ in the $\pi(i)$ row of $y$ (there is such a column as $\pi(i)\in \hat{Y}$).
A support of a $0-1$ matrix is a base in $\uparrow_n\!S$ if and only if its column sum is a
base in $S$. Thus, the required properties of $\pi'$ follow from these properties of $\pi$.
Next we prove the second claim. Let $x\in \uparrow_n\!S_1\cap \uparrow_n\!S_2$ and denote
its column sum by $\hat{x}=\sum_{j=1}^n x^j$. Since $x\in \uparrow_n\!S_1$, we have
$\hat{x}\in S_1$, and similarly since $x\in \uparrow_n\!S_2$, we have $\hat{x}\in S_2$.
Thus, $\hat{x}\in S_1\cap S_2$, and thus $x \in \uparrow_n\!(S_1\cap S_2)$.
In the other direction, let $x \in \uparrow_n\!(S_1\cap S_2)$ and denote
$\hat{x}=\sum_{j=1}^n x^j$. Then, $\hat{x} \in S_1\cap S_2$. Thus, $\hat{x} \in S_1$,
and therefore $x\in \uparrow_n\!S_1$, and similarly $\hat{x} \in S_2$, and therefore
$x\in\uparrow_n\!S_2$. Therefore, $x\in \uparrow_n\!S_1\cap \uparrow_n\!S_2$.
\epr
\bt{matroid_intersection_theorem}
The optimal objective function value of the shifted combinatorial optimization problem
\eqref{shifted_problem} over the intersection of any two strongly base orderable
matroids given by independence oracles and any $n$ can be computed in polynomial time.
\et
\bpr
Let $S_1,S_2\subseteq\{0,1\}^d$ be strongly base orderable matroids and let $S:=S_1\cap S_2$.
By Lemma \ref{strongly_based_lemma}, $\uparrow_n S_1$ and $\uparrow_n S_2$ are strongly base
orderable matroids. Applying Proposition \ref{intersection} to $\uparrow_n\!S_1$,
$\uparrow_n\!S_2$, and using Lemmas \ref{partition_1} and \ref{strongly_based_lemma},
we get $$[S_1^n]\cap[S_2^n]\ =\ (\vee_n \uparrow_n\!S_1)\cap(\vee_n \uparrow_n\!S_2)
\ =\ \vee_n(\uparrow_n\!S_1\cap \uparrow_n\!S_2)\ =\ \vee_n\uparrow_n\!(S_1\cap S_2)=[S^n]\ .$$
By Proposition \ref{partition_matroid} applied to the matroids $\uparrow_n\!S_1$ and
$\uparrow_n\!S_2$, and using Lemma \ref{partition_1}, we can realize in polynomial time independence
oracles for $[S_1^n]$ and $[S_2^n]$. So we can maximize a given profit matrix $c$ over $[S^n]$
by maximizing $c$ over the intersection of the matroids $[S_1^n]$ and $[S_2^n]$.
This solves problem \eqref{auxiliary_problem_1} over $S$ and gives the optimal value
of the shifted optimization problem \eqref{shifted_problem} over $S$ via
the equality $\max\{{\ra c}\,{\la x}\,:\,x\in S^n\}=\max\{cx\,:\,x\in [S^n]\}$.
\epr
We raise the following natural questions. Is it possible to actually find an optimal solution
(and not only the optimal value) for the shifted problem over the intersection of
strongly base orderable matroids in polynomial time? What is the complexity of
shifted combinatorial optimization over the intersection of two arbitrary matroids?
The answer to the first question is yes for transversal matroids since then the Fiber
problem \eqref{auxiliary_problem_2} over the intersection can also be solved.
This implies in particular that the shifted problem
over matchings in bipartite graphs is polynomial time solvable.
However, if the Fiber problem cannot be solved efficiently, that is given $x\in [S^n]$,
it is hard to find $y\in S^n$ such that $x\sim y$, then there are matrices $c$ for which
it is hard to solve the Shifted Combinatorial Optimization problem as well.
To see this, let $c$ be such that for all $i,j$ we have $c_{i,j}=1$ if $x_{i,j}=1$
and $c_{i,j}=-1$ if $x_{i,j}=0$. Then, the optimal profit will be $|x|$, and every optimal
solution $y$ to the Shifted Combinatorial Optimization problem must satisfy $y\in S^n$
and $y\sim x$, and thus solves the Fiber problem as well. Thus, the Fiber problem is
not harder than the Shifted Combinatorial Optimization problem.
\section{Remarks}
\label{remarks}
It was shown in \cite{KOS} that problem \eqref{shifted_problem} and hence
also problem \eqref{lexicographic_problem} are polynomial time solvable
for $S=\{z\in\{0,1\}^d\,:\,Az=b\}$ with $A$ totally unimodular and $b$ integer.
We briefly give a new interpretation of this by showing how the new auxiliary
problems \eqref{auxiliary_problem_1} and \eqref{auxiliary_problem_2} over such $S$
can also be efficiently solved, so Lemma \ref{reduction_2} can be applied.
\bp{unimodular}
For $S=\{z\in\{0,1\}^d:Az=b\}$ with $A$ totally unimodular and $b$ integer,
\begin{eqnarray}\label{unimodular_equation}
[S^n]\ =\ \{x\in\{0,1\}^{d\times n}\ :\ A\sum_{j=1}^nx^j=nb\}\ .
\end{eqnarray}
Moreover, problems \eqref{auxiliary_problem_1} and \eqref{auxiliary_problem_2}
hence \eqref{shifted_problem} and \eqref{lexicographic_problem}
over $S$ are polynomial time solvable.
\ep
\bpr
If $x$ is in the left hand side of \eqref{unimodular_equation} then there is a
$y\in S^n$ with $x\sim y$. Then $A\sum_{j=1}^nx^j=A\sum_{j=1}^ny^j=nb$ so $x$
is also in the right hand side. If $x$ is in the right hand side, then a decomposition
theorem of \cite{BT} implies that there is a $y\in S^n$ with $\sum_{j=1}^ny^j=\sum_{j=1}^nx^j$.
Then $x\sim y$ so $x$ is also in the left hand side of \eqref{unimodular_equation}.
Moreover, by an algorithmic version of \cite{KOS}
of this decomposition theorem, such a $y$ can be found in polynomial time,
solving problem \eqref{auxiliary_problem_2} over $S$. By the equality in
\eqref{unimodular_equation}, to solve problem \eqref{auxiliary_problem_1} over $S$ with a given
profit $c$, we can maximize $c$ over the right hand side of \eqref{unimodular_equation},
and this can be done in polynomial time by linear programming since the defining matrix
of this set is $[A,...,A]$, which is totally unimodular since $A$ is.
Lemmas \ref{reduction_1} and \ref{reduction_2} now imply that problems \eqref{shifted_problem}
and \eqref{lexicographic_problem} over $S$ can also be solved in polynomial time.
\epr
This in particular implies that lexicographic combinatorial optimization over $s-t$ paths
in digraphs or perfect matchings in bipartite graphs is polynomial time doable.
\vskip.2cm
The shifted combinatorial optimization problem over $S\subseteq\{0,1\}^d$ in the special
case $n=1$ is just the standard combinatorial optimization problem $\max\{cx:x\in S\}$
over $S$. So a solution of the former implies a solution of the latter. Is the converse true?
The above results show that it is true for matroids, certain matroid intersections, and unimodular
systems including $s-t$ paths in digraphs and perfect matchings in bipartite graphs.
Unfortunately, as we next show, the answer in general is negative.
\bp{matching}
Let $S\subseteq\{0,1\}^d$ be the set of perfect matchings in a cubic graph.
Then the lexicographic problem \eqref{lexicographic_problem} and shifted
problem \eqref{shifted_problem} over $S$ are NP-hard already for fixed $n=2$.
\ep
\bpr
By Lemma \ref{reduction_1} it is enough to show that solving
$\lexmin\{(|{\la x}^1|,|{\la x}^2|)\ :\ x\in S^2\}$ is NP-hard.
We claim that deciding if there is an $x\in S^2$ with $|{\la x}^2|=0$ is NP-complete.
Indeed, there is such an $x$ if and only if there are two edge disjoint perfect matchings
in the given graph. Now this holds if and only if there are three
pairwise edge disjoint perfect matchings in the given graph, since it is cubic.
This is equivalent to the existence of a $3$ edge coloring, which is NP-complete
to decide on cubic graphs \cite{Hol}.
\epr
We point out that the set of perfect matchings in a graph can be written in the
form $S=\{z\in\{0,1\}^d\,:\,Az=b\}$ with $A$ the vertex-edge incidence matrix of
the graph and $b$ the all $1$ vector in the vertex space. Thus, Proposition \ref{matching}
shows, in contrast with Proposition \ref{unimodular}, that if $A$ is not totally unimodular,
then the shifted problem is NP hard over such sets $S$, even if we can efficiently do
linear optimization over $S$, as is the case for perfect matchings.
\vskip.2cm
The shifted optimization problem $\max\{{\sh c}\,{\sh x}:x\in S^n\}$
can be defined for any set $S\subset\Z^d$, and its complexity
will depend on the structure and presentation of $S$.
Consider sets of bounded cardinality. Note that even with $|S|=2$,
the number of feasible solutions is $|S^n|=2^n$, so exhaustive search
is not polynomial. However, we do have the following simple statement.
\bp{fixed} For any fixed positive integer $s$, the shifted optimization problem
over any set $S\subset\Z^d$ satisfying $|S|\leq s$, any $n$,
and any $c\in Z^{d\times n}$, can be solved in polynomial time.
\ep
\bpr
Let $S=\{z^1,\dots,z^m\}$ be the list of elements of $S$, with $m\leq s$ and $z^i\in\Z^d$ for all
$i$. Given $x\in S^n$, let $n_i:=|\{k:x^k=z^i\}|$ for $i=1,\dots,m$. Let $y\in S^n$ be the matrix
with first $n_1$ columns equal to $z^1$, next $n_2$ columns equal to $z^2$, and so on, with last
$n_m$ columns equal to $z^m$. Then $y\sim x$ (in fact, $y$ is obtained from $x$ by applying
the same permutation to each row), and therefore ${\sh c}\,{\sh x}={\sh c}\,{\sh y}$.
So the objective value of $x$ can be computed from $n_1,\dots,n_m$. Now, any integers
$0\leq n_i\leq n$ with $\sum_{i=1}^m n_i=n$ give a feasible $y\in S^n$ with these counts.
So, it is enough to go over all such $n_1,\dots,n_m$, for each compute the corresponding $y$
and the value ${\sh c}\,{\sh y}$, and pick the best. The number of such tuples is at most
$(n+1)^{m-1}=O(n^{s-1})$, since $n_m=n-\sum_{i=1}^{m-1} n_i$ is determined by the others,
which is polynomial in $n$ for fixed $s$.
\epr | 151,968 |
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TITLE: Computing the exponential of a $2 \times 2$ matrix using trace $0$ matrices
QUESTION [3 upvotes]: It is an easily proved fact that for a $2\times 2$ traceless matrix $A$,
$$ e^A = \cos\left(\sqrt{\det(A)}\right)I + \frac{\sin\left(\sqrt{\det(A)}\right)}{\sqrt{\det(A)}}A$$
Problem 2.7 of Lie Groups, Lie Algebras, and Representations by Bryan Hall asks to use this fact to compute $\exp(X)$, where
$$ X = \begin{pmatrix}
4 & 3\\
-1 & 2
\end{pmatrix}$$
In other words, I have to write $X$ in terms of traceless matrices, and employ the above fact. My question is: is there a systematic way to do this?
My idea to solve this problem is to write $X = X_1 + X_2$, where $X_1$ is traceless, $X_2$ is diagonal or nilpotent, and $[X_1, X_2] = 0$, and compute the exponent using $e^{X_1 + X_2} = e^{X_1}e^{X_2}$. For example, I tried the most obvious thing:
$$X = \begin{pmatrix} -2 & 3\\-1 & 2\end{pmatrix} + \begin{pmatrix} 6 & 0\\0 & 0\end{pmatrix},$$
but the two matrices above do not commute.
REPLY [2 votes]: You were close.
Try $X=X_1+X_2$, where $X_1=X-3I$ and $X_2=3I$. | 136,296 |
TITLE: If functions agree at all but finitely many points then the integrals are the same
QUESTION [2 upvotes]: Exercise 3-2 from Calculus on Manifolds by Spivak:
Let $A\subset R^n,\ f:A\rightarrow R$ an integrable (in the sense of
Darboux) function. Let $g=f$ except at finitely many points. Prove
that $g$ is also integrable and $\int_Af=\int_A g$.
Note that there is a similar question show that $g$ is integrable, with $f=g$ except in finite set and $f$ integrable but the answers assume the knowledge of measure theory whereas Spivak doesn't.
I guess I need to use the criterion saying that $f$ is integrable iff there is a partition $P$ of $A$ such that $U(f,P)-L(f,P)< \epsilon$ for any $\epsilon < 0$. But I don't know how to apply it to both functions. I thought about considering $f-g$ (which should be integrable except finitely many points), but Spivak doesn't even state that the sum of two integrable functions is integrable, so perhaps I'm not supposed to use this. (Even if I consider $f-g$, I don't know how to proceed).
REPLY [2 votes]: Your idea of considering the difference is a good one, since your problem is then seen to be equivalent to proving that a function which is $0$ except at finitely many points is integrable and must have integral equal to $0$ (try to understand why if this isn't clear)
In order to prove this statement, taking partitions with small diameter will show that you can let your upper and lower sums be as near zero as you want (they will be bounded by $\pm N \cdot\mathrm{diam}(P)^n\cdot\max |f|$, where $N$ is the number of points where the function is not zero).
Now, to see why $g$ is integrable, notice that $g=(g-f)+f$.
Therefore, $\int g=\int(g-f)+\int f=\int f$.
You mention in the question that the book does not say that sum of integrable functions is integrable before this point. If you are uncomfortable with using this fact, try to adapt the proof above to avoid using it. The core idea is in the second paragraph of this answer. | 36,353 |
\begin{document}
\title{Power Allocation in Multi-user Cellular Networks With Deep Q Learning Approach}
\author{Fan~Meng, Peng~Chen and Lenan~Wu
\thanks{Fan~Meng, and Lenan~Wu are with the School of Information Science and
Engineering, Southeast University, Nanjing 210096, China (e-mail: [email protected], [email protected]).}
\thanks{
Peng Chen is with the State Key Laboratory of Millimeter Waves, Southeast University, Nanjing 210096, China (e-mail: [email protected]).}
}
\maketitle
\begin{abstract}
The model-driven power allocation (PA) algorithms in the wireless cellular networks with interfering multiple-access channel (IMAC) have been investigated for decades. Nowadays, the data-driven model-free machine learning-based approaches are rapidly developed in this field, and among them the deep reinforcement learning (DRL) is proved to be of great promising potential. Different from supervised learning, the DRL takes advantages of exploration and exploitation to maximize the objective function under certain constraints. In our paper, we propose a two-step training framework. First, with the off-line learning in simulated environment, a deep Q network (DQN) is trained with deep Q learning (DQL) algorithm, which is well-designed to be in consistent with this PA issue. Second, the DQN will be further fine-tuned with real data in on-line training procedure. The simulation results show that the proposed DQN achieves the highest averaged sum-rate, comparing to the ones with present DQL training. With different user densities, our DQN outperforms benchmark algorithms and thus a good generalization ability is verified.
\end{abstract}
\begin{IEEEkeywords}
Deep reinforcement learning, deep Q learning, interfering multiple-access channel, power allocation.
\end{IEEEkeywords}
\section{Introduction}\label{sec:intro}
Data transmitting in wireless communication networks has experienced explosively growth in recent decades and will keep rising in the future. The user density is greatly increasing, resulting in critical demand for more capacity and spectral efficiency. Therefore, both intra-cell and inter-cell interference managements are significant to improve the overall capacity of a cellular network system. The problem of maximizing a generic sum-rate is studied in this paper, and it is non-convex, NP-hard and cannot be solved efficiently.
Various model-driven algorithms have been proposed in the present papers for PA problems, such as fractional programming (FP)~\cite{Shen2018Fractional}, weighted MMSE (WMMSE)~\cite{Shi2011An} and some others~\cite{Chiang2008Power, 5770666}. Excellent performance can be observed through theoretical analysis and numerical simulations, but serious obstacles are faced in practical deployments~\cite{Liye}. First, these techniques highly rely on tractable mathematical models, which are imperfect in real communication scenarios with the specific user distribution, geographical environment, etc. Second, the computational complexities of these algorithms are high.
In recent years, the machine learning (ML)-based approaches have been rapidly developed in wireless communications~\cite{8054694}. These algorithms are usually model-free, and are compliant with optimizations in practical communication scenarios. Additionally, with developments of graphic processing unit (GPU) or specialized chips, the executions can be both fast and energy-efficient, which brings in solid foundations for massive applications.
Two main branches of ML, supervised learning and reinforcement learning (RL)~\cite{Lecun2015Deep}, are briefly introduced here. With supervised learning, a deep neural network (DNN) is trained to approximate some given optimal (or suboptimal) objective algorithms, and it has been realized in some applications~\cite{8444648, 8454504, 8052521}. However, the target algorithm is usually unavailable and the performance of DNN is bounded by the supervisor. Therefore, the RL has received widespread attention, due to its nature of interacting with an unknown environment by exploration and exploitation. The Q learning method is the most well-studied RL algorithm, and it is exploited to cope with power allocation (PA) in~\cite{DBLP:journals/corr/abs-1803-06760, 7997440, 8466370}, and some others~\cite{8412128}. The DNN trained with Q learning is called deep Q network (DQN), and it is proposed to address the distributed downlink single-user PA problem~\cite{DBLP:DRL}.
In our paper, we extend the work in~\cite{DBLP:DRL}, and the PA problem in cellular cells with multiple users is investigated. The design of the DQN model is discussed and introduced. Simulation results show that our DQN outperforms the present DQNs and the benchmark algorithms. The contributions of this work are summarized as follows:
\begin{itemize}
\item A model-free two-step training framework is proposed. The DQN is first off-line trained with DRL algorithm in simulated scenarios. Second, the learned DQN can be further dynamically optimized in real communication scenarios, with the aid of transfer learning.
\item The PA problem using deep Q learning (DQL) is discussed, then a DQN enabled approach is proposed to be trained with current sum-rate as reward function, including no future reward. The input features are well-designed to help the DQN get closer to the optimal solution.
\item After centralized training, the proposed DQN is tested by distributed execution. The averaged rate-sum of DQN outperforms the model-driven algorithms, and also shows good generalization ability in a series of benchmark simulation tests.
\end{itemize}
The remainder of this paper is organized as follows. Section~\ref{sec:system} outlines the PA problem in the wireless cellular network with IMAC. In Section~\ref{sec:dqn} our proposed DQN is introduced in detail. Then, this DQN is tested in distinct scenarios, along with benchmark algorithms, and the simulation results are analyzed in Section~\ref{sec:sim}. Conclusions and discussion are given in Section~\ref{sec:con}.
\section{System Model}\label{sec:system}
The problem of PA in the cellular network with interfering multiple-access channel (IMAC) is considered. In a system with $ N $ cells, at the center of each cell a base stations (BS) simultaneously serves $ K $ users with sharing frequency bands. A simple network example is shown in Fig.~\ref{fig:cellular}. At time slot $ t $, the independent channel coefficient between the $ n $-th BS and the user $ k $ in cell $ j $ is denoted by $ g^t_{n,j,k} $, and can be expressed as
\begin{equation}\label{equ:g}
g_{n,j,k} = |h^t_{n,j,k}|^2\beta_{n,j,k},
\end{equation}
where $ h^t_{n,j,k} $ is the small scale complex flat fading element, and $ \beta_{n,j,k} $ is the large scale fading component taking account of both the geometric attenuation and the shadow fading. Therefore, the signal to interference plus noise ratio (SINR) of this link can be described by
\begin{equation}\label{equ:sinr}
\textup{sinr}^t_{n,k} = \frac{g^t_{n,n,k} p^t_{n,k}}{\sum_{k' \neq k} g^t_{n,n,k} p^t_{n,k'} + \sum_{n' \in D_n} g^t_{n',n,k} \sum_{j} p^t_{n',j} + \sigma^2},
\end{equation}
where $ D_n $ is the set of interference cells around the $ n $-th cell, $ p $ is the emitting power of BS, and $ \sigma^2 $ denotes the additional noise power. With normalized bandwidth, the downlink rate of this link is given as
\begin{equation}\label{equ:C}
C^t_{n,k} = \log_2\left(1 + \textup{sinr}^t_{n,k}\right),
\end{equation}
The optimization target is to maximize this generic sum-rate objective function under maximum power constraint, and it is formulated as
\begin{equation}\label{equ:tar}
\begin{split}
&\max_{\textbf{p}^t}\quad \sum_n\sum_k C^t_{n,k}\\
&\textup{s.t.}\quad 0 \leq p^t_{n,k} \leq P_{\textup{max}}, \;\forall n,k,
\end{split}
\end{equation}
where $ \textbf{p}^t = \{p^t_{n,k}|\forall n,k\} $, and $ P_{\textup{max}} $ denotes the maximum emitting power. We also define sum-rate $ C^t = \sum_n\sum_k C^t_{n,k} $, $ \textbf{C}^t = \{C^t_{n,k}|\forall n,k\} $, and channel state information (CSI) $ \textbf{g}^t = \{g^t_{n,j,k}|\forall n,j,k\} $. This problem is non-convex and NP-hard, so we propose a data-driven learning algorithm based on the DQN model in the following section.
\begin{figure}
\centering
\includegraphics[width=3.6in]{cellular.pdf}
\caption{An illustrative example of a multi-user cellular network with $ 9 $ cells. In each cell, a BS serves $ 2 $ users simultaneously.}
\label{fig:cellular}
\end{figure}
\section{Deep Q Network}\label{sec:dqn}
\subsection{Background}
Q learning is one of the most popular RL algorithms aiming to deal with the Markov decision process (MDP) problems~\cite{Mnih2015Human}. At time instant $ t $, by observing the state $ s^t \in S $, the agent takes action $ a^t \in A $ and interacts with the environment, and then get the reward $ r^t $ and the next state $ s^{t+1} $ is obtained. The notations $ A $ and $ S $ are the action set and the state set, respectively. Since $ S $ can be continuous, the DQN is proposed to combine Q learning with a flexible DNN to settle infinite state space. The cumulative discounted reward function is given as
\begin{equation}\label{equ:Rt}
R^t = \sum_{\tau = 0}^{\infty} \gamma^{\tau} r^{t+\tau+1},
\end{equation}
where $ \gamma \in [0, 1) $ is a discount factor that trades off the importance of immediate and future rewards, and $ r $ denotes the reward. Under a certain policy $ \pi $, the Q function of the agent with an action $ a $ in state $ s $ is given as
\begin{equation}\label{equ:value}
Q_{\pi}(s,a;\boldsymbol{\theta}) = \mathbb{E}_{\pi}\left[R^t | s^t = s, a^t = a\right],
\end{equation}
where $ \boldsymbol{\theta} $ denotes the DQN parameters, and $ \mathbb{E}\left[\cdot\right] $ is the expectation operator. Q learning concerns with how agents ought to interact with an unknown environment so as to maximize the Q function. The maximization of~\eqref{equ:value} is equivalent to the Bellman optimality equation~\cite{Introduction}, and it is describe as
\begin{equation}\label{equ:yt}
y^t = r^{t} + \gamma \max_{a'} Q(s^{t+1},a';\boldsymbol{\theta}^t),
\end{equation}
where $ y^t $ is the optimal Q value. The DQN is trained to approximate the Q function, and the standard Q learning update of the parameters $ \boldsymbol{\theta} $ is described as
\begin{equation}\label{equ:update}
\boldsymbol{\theta}^{t+1} = \boldsymbol{\theta}^{t} + \eta \left(y^t - Q(s^t,a^t;\boldsymbol{\theta}^t)\right)\nabla Q(s^t,a^t;\boldsymbol{\theta}^t),
\end{equation}
where $ \eta $ is the learning rate. This update resembles stochastic gradient descent, gradually updating the current value $ Q(s^{t},a^{t};\boldsymbol{\theta}^t) $ towards the target $ y^t $. The experience data of the agent is loaded as $ \left(s^t, a^t, r^t, s^{t+1}\right) $. The DQN is trained with recorded batch data randomly sampled from the experience replay memory, which is a first-in first-out queue.
\subsection{Discussion on DRL}\label{sec:dis}
In many applications such as playing video games~\cite{Mnih2015Human}, where current strategy has long-term impact on cumulative reward, the DQN achieve remarkable results and beat humans. However, the discount factor is suggested to be zero in this PA problem. The DQL aims to maximize the Q function. Let $ \gamma = 0 $, from~\eqref{equ:value} we have
\begin{equation}\label{equ:q1}
\max Q = \max_{a \in A} \mathbb{E}_{\pi}\left[r^t | s^t = s, a^t = a\right].
\end{equation}
For a PA problem, clearly that $ s = \textbf{g}^t $, $ a = \textbf{p}^t $. Then we let $ r^t = C^t $ and get that
\begin{equation}\label{equ:q2}
\max Q = \max_{\textbf{0} \preceq \textbf{p}^t \preceq \textbf{p}_{\max}} \mathbb{E}_{\pi}\left[C^t | \textbf{g}^t, \textbf{p}^t\right].
\end{equation}
In the execution period the policy is deterministic, and thus~\eqref{equ:q2} can be written as
\begin{equation}\label{equ:q3}
\max Q = \max_{\textbf{0} \preceq \textbf{p}^t \preceq \textbf{p}_{\max}} C^t \left(\textbf{g}^t, \textbf{p}^t\right),
\end{equation}
which is a equivalent form of~\eqref{equ:tar}. In this inference process we assume that $ \gamma = 0 $ and $ r^t = C^t $, indicating that the optimal solution to~\eqref{equ:tar} is identical to that of~\eqref{equ:value}, under these two conditions.
As shown in Fig.~\ref{fig:opt}, it is well-known that the optimal solution $ \textbf{p}^{t*} $ of~\eqref{equ:tar} is only determined by current CSI $ \textbf{g}^t $, and the sum-rate $ \textbf{C}^t $ is calculated with $ (\textbf{g}^t, \textbf{p}^t) $. Theoretically the optimal power $ \textbf{p}^{t*} $ can be obtained using a DQN with input being just $ \textbf{g}^t $. In fact, the performance of this designed DQN is poor, since it is non-convex and the optimal point is hard to find. Therefore, we propose to utilize two more auxiliary features: $ \textbf{C}^{t-1} $ and $ \textbf{p}^{t-1} $. Since that the channel can be modeled as a first-order Markov process, the solution of last time period can help the DQN get closer to the optimum, and~\eqref{equ:q3} can be rewritten as
\begin{equation}\label{equ:q4}
\max Q = \max_{\textbf{0} \preceq \textbf{p}^t \preceq \textbf{p}_{\max}} C^t\left(\textbf{g}^t, \textbf{p}^t, \textbf{C}^{t-1}, \textbf{p}^{t-1}\right).
\end{equation}
\begin{figure}
\centering
\includegraphics[width=3.6in]{opt.pdf}
\caption{The solution of DQN is determined by CSI $ \textbf{g}^t $, along with downlink rate $ \textbf{C}^{t-1} $ and transmitting power $ \textbf{p}^{t-1} $.}
\label{fig:opt}
\end{figure}
Once $ \gamma = 0 $ and $ r^t = C^t $,~\eqref{equ:yt} is simplified to be $ y^t = C^t $, and the replay memory is also reduced to be $ \left(s^t, a^t, r^t\right) $. The DQN works as an estimator to predict the current sum-rate of corresponding power levels with a certain CSI. These discussions provide good guidance for the following DQN design.
\subsection{DQN Design in Cellular Network}
In our proposed model-free two-step training framework, the DQN is first off-line pre-trained with DRL algorithm in simulated wireless communication system. This procedure is to reduce the on-line training stress, due to the large data requirement of data-driven algorithm by nature. Second, with the aid of transfer learning, the learned DQN can be further dynamically fine-tuned in real scenarios. Since the practical wireless communication system is dynamic and influenced by unknown issues, the data-driven algorithm is believed to be a promising technique. We just discuss the two-step framework here, and the first training step is mainly focused in the following manuscript.
In a certain cellular network, each BS-user link is regarded as an agent and thus a multi-agent system is studied. However, multi-agent training is difficult since it needs much more learning data, training time and DNN parameters. Therefore, centralized training is considered, and only one agent is trained by using all agents' experience replay memory. Then, this agent's learned policy is shared in the distributed execution period. For our designed DQN, components of the replay memory are introduced as follows.
\subsubsection{State}
The state design for a certain agent $ (n,k) $ is important, since the full environment information is redundant and irrelevant elements must be removed. The agent is assumed to have corresponding perfect instant CSI information in~\eqref{equ:sinr}, and we define logarithmic normalized interferer set $ \boldsymbol{\Gamma}_{n,k}^t $ as
\begin{equation}\label{equ:gamma_g}
\boldsymbol{\Gamma}_{n,k}^t = \left\{ \underbrace{{1,\cdots,1}}_{K-1}, \left\{\log_2\left(1 + \frac{g^t_{n',j,k}}{g^t_{n,k,k}} \right)\bigg| n'\in D_n, \forall j \right\}\right\}.
\end{equation}
The channel amplitude of interferers are normalized by that of the needed link, and the logarithmic representation is preferred since the amplitudes of channel often vary by orders of magnitude. The cardinality of $ \boldsymbol{\Gamma}_{n,k}^t $ is $ (|D_n|+1)K-1 $. To further decrease the input dimension and reduce the computational complexities, the elements in $ \boldsymbol{\Gamma}_{n,k}^t $ are sorted in decrease turn and only the first $ C $ elements remain. As we discussed in~\ref{sec:dis}, these remained components' and this link's corresponding downlink rate $ \textbf{C}_{n,k}^{t-1} $ and transmitting power $ \textbf{p}_{n,k}^{t-1} $ at last time slot, are the additional two parts of the input to our DQN. Therefore, the state is composed of three features: $ s^t_{n,k} = \{ \boldsymbol{\Gamma}_{n,k}^t, \textbf{C}_{n,k}^{t-1}, \textbf{p}_{n,k}^{t-1} \} $. The cardinality of state, i.e., the input dimension for DQN is $ |S| = 3C+2 $.
\subsubsection{Action}
In~\eqref{equ:tar} the downlink power is a continuous variable, and is only constrained by maximum power constraint. Since the action space of DQN must be finite, the possible emitting power is quantized in $ |A| $ levels. The allowed power set is given as
\begin{equation}\label{equ:p_set}
A = \left \{0, P_{\textup{min}}, P_{\textup{min}}\left(\frac{P_{\textup{max}}}{P_{\textup{min}}}\right)^{\frac{1}{|A|-2}}, \cdots, P_{\textup{max}} \right\},
\end{equation}
where $ P_{\textup{min}} $ is the non-zero minimum emitting power.
\subsubsection{Reward}
In some manuscripts the reward function is elaborately designed to improve the agent's transmitting rate and also mitigate the interference influence. However, most of these reward functions are suboptimal approaches to the target function of~\eqref{equ:tar}. In our paper, the $ C^t $ is directly used as the reward function, and it is shared by all agents. In the training simulations with small or medium scale cellular network, this simple method proves to be feasible.
\section{Simulation Results}\label{sec:sim}
\subsection{Simulation Configuration}
A cellular network with $ N = 25 $ cells is simulated. At center of each cell, a BS is deployed to synchronously serve $ K = 4 $ users which are located uniformly and randomly within the cell range $ r \in [R_{\min}, R_{\min}] $, where $ R_{\min} = 0.01 $ km and $ R_{\min} = 1 $ km are the inner space and half cell-to-cell distance, respectively. The small-scale fading is simulated to be Rayleigh distributed, and the Jakes model is adopted with Doppler frequency $ f_d = 10 $ Hz and time period $ T = 20 $ ms. According to the LTE standard, the large-scale fading is modeled as $ \beta = 120.9 + 37.6 \log_{10}(d) + 10 \log_{10}(z) $ dB, where $ z $ is a log-normal random variable with standard deviation being $ 8 $ dB, and $ d $ is the transmitter-to-receiver distance (km). The AWGN power $ \sigma^2 $ is $ -114 $ dBm, and the emitting power constraints $ P_{\textup{min}} $ and $ P_{\textup{max}} $ are $ 5 $ and $ 38 $ dBm, respectively.
A four-layer feed-forward neural network (FNN) is chosen as DQN, and the neuron numbers of two hidden layers are $ 128 $ and $ 64 $, respectively. The activation function of output layer is linear, and the ReLU is adopted in the hidden layers. The cardinality of adjacent cells is $ |D_n| = 18, \forall n $, the first $ C = 16 $ interferers remain and power level number $ |A| = 10 $. Therefore, the input and output dimensions are $ 50 $ and $ 10 $, respectively.
In the off-line training period, the DQN is first randomly initialized and then trained epoch by epoch. In the first $ 100 $ episodes, the agents only take actions stochastically, then they follow by adaptive $ \epsilon $-greedy learning strategy~\cite{Introduction} to step in the following exploring period. In each episode, the large-scale fading is invariant, and thus the number of training episode must be large enough to overcome the generalization problem. There are $ 50 $ time slots per episode, and the DQN is trained with $ 256 $ random samples in the experience replay memory every $ 10 $ time slots. The Adam algorithm~\cite{Kingma2014Adam} is adopted as the optimizer in our paper, and the learning rate $ \eta $ exponentially decays from $ 10^{-3} $ to $ 10^{-4} $. All training hyper-parameters are listed in Tab.\ref{table:hyperparameter} for better illustration. In the following simulations, these default hyper-parameters will be clarified once changed.
The FP algorithm, WMMSE algorithm, maximum PA and random PA schemes are treated as benchmarks to evaluate our proposed DQN-based algorithm. The perfect CSI of current moment is assumed to be known for all schemes. The simulation code will be available after formal publication.
\begin{table}
\caption{Hyper-parameters setup of DQN training}
\centering
\begin{tabular}{c|c||c|c}
\hline
\textbf{Parameter} & \textbf{Value} & \textbf{Parameter} & \textbf{Value}\\
\hline
Number of $ T $ per episode & $ 50 $ & Initial $ \eta $ & $ 10^{-3} $\\
Observe episode number & $ 100 $ & Final $ \eta $ & $ 10^{-4} $\\
Explore episode number & $ 9900 $ & Initial $ \epsilon $ & $ 0.2 $\\
Train interval & 10 & Final $ \epsilon $ & $ 10^{-4} $\\
Memory size & $ 50000 $ & Batch size & $ 256 $\\
\hline
\end{tabular}
\label{table:hyperparameter}
\end{table}
\subsection{Discount Factor}\label{gamma}
In this subsection, the performance of different discount factor $ \gamma $ is studied. We set $ \gamma \in \{0.0,0.1,0.3,0.7,0.9\} $, and the average rate $ \bar{C} $ over the training period is shown in Fig.~\ref{fig:gamma_train}. At the same time slot, obviously the values of $ \bar{C} $ with higher $ \gamma \in \{0.7,0.9\} $ are lower than the rest with lower $ \gamma $ values. The trained DQNs are then tested in three cellular networks with different cell numbers. As shown in Fig.~\ref{fig:gamma_test} shows that DQN with $ \gamma = 0.0 $ achieves the highest $ \bar{C} $ score, while the lowest value is obtained by the one with highest $ \gamma $ value. The simulation result shows that the non-zero $ \gamma $ has a negative influence on the performance of DQN, which is consistent with the analysis in \ref{sec:dis}. Therefore, a zero or low discount factor value is recommended.
\begin{figure}
\centering
\includegraphics[width=3.6in]{gamma_train.pdf}
\caption{With different $ \gamma $ values, the recorded average rate during training period (Curves smoothed by averaged window).}
\label{fig:gamma_train}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=3.6in]{gamma_test.pdf}
\caption{The average rate $ \bar{C} $ versus cellular network scalability for trained DQNs with different $ \gamma $ values.}
\label{fig:gamma_test}
\end{figure}
\subsection{Algorithm Comparison}\label{vs}
The DQN trained with zero $ \gamma $ is used, and the four benchmark algorithms stated before are tested as comparisons. In real cellular network, the user density is changing over time, and the DQN must have good generalization ability against this issue. The user number per cell $ K $ is assumed to be in set $ \{1,2,4,6\} $. The averaged simulation results are obtained after $ 500 $ repeats. As shown in Fig.~\ref{fig:rate_user}, the DQN achieves the highest $ \bar{C} $ in all testing scenarios. Although it is trained with $ K=4 $, the DQN still outperforms the other algorithms in the other cases. We also note that the gap between random/maximum PA schemes and the rest optimization algorithms is increased when $ K $ becomes larger. This can be mainly attributed that the intra-cell interference gets stronger with increased user density, which indicating that the optimization of PA is more significant in the cellular networks with denser users.
We also give an example result of one testing episode here ($ K = 4 $). In comparison with the averaged sum-rate values in Fig.~\ref{fig:rate_user}, in Fig.~\ref{fig:rate_time} the performance of three PA algorithms (DQN, FP, WMMSE) is not stable, especially depending on the specific large-scale fading effects. Additionally, in some episodes the DQN can not be better than the other algorithms over the time (not shown in this paper), which means that there is still potential to improve the DQN performance.
In terms of computation complexity, the time cost of DQN is in linear relationship with layer numbers, with the utilization of GPU. Meanwhile, both FP and WMMSE are iterative algorithms, and thus the time cost is not constant, depending on the stopping criterion condition, initialization and CSI.
\begin{figure}
\centering
\includegraphics[width=3.6in]{rate_user.pdf}
\caption{The average rate $ \bar{C} $ versus user number per cell. Five power allocation schemes are tested.}
\label{fig:rate_user}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=3.6in]{rate_time.pdf}
\caption{Comparisons of all five power allocation schemes over $ 1000 $ time slots (Curves smoothed by averaged window).}
\label{fig:rate_time}
\end{figure}
\section{Conclusions}\label{sec:con}
The PA problem in the cellular network with IMAC has been investigated, and the data-driven model-free DQL has been applied to solve this issue. To be in consistent with the PA optimization target, the current sum-rate is used as reward function, including no future reward. This designed DQL algorithm is proposed, and the DQN simply works as an estimator to predict the current sum-rate under all power levels with a certain CSI. Simulation results show that the DQN trained with zero $ \gamma $ achieves the highest average sum-rate. Then in a series of different scenarios, the proposed DQN outperforms the benchmark algorithms, indicating that the designed DQN has good generalization abilities. In our two-step training framework, we have realized the off-line centralized learning with simulated communication networks, and the learned DQN is tested by distributed executions. In our future work, the on-line learning will be further studied to accommodate the real scenarios with specific user distributions and geographical environments.
\section{Acknowledgments}
This work was supported in part by the National Natural Science Foundation of China (Grant No. 61801112, 61471117, 61601281), the Natural Science Foundation of Jiangsu Province (Grant No. BK20180357), the Open Program of State Key Laboratory of Millimeter Waves (Southeast University, Grant No. Z201804).
\input{revised.bbl}
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\begin{document}
\title[Bellman approach to the one-sided bumping]{Bellman approach to the one-sided bumping for weighted estimates of Calder\'on--Zygmund operators}
\author{Fedor Nazarov}
\address{Department of Mathematics, University of Wisconsin-Madison and Kent Sate University}
\email{[email protected]}
\thanks{Work of F.~Nazarov is supported by the NSF grant}
\author{Alexander Reznikov}
\address{Department of Mathematics, Michigan State University, East
Lansing, MI 48824, USA}
\email{[email protected]}
\author{Alexander Volberg}
\thanks{Work of A.~Volberg is supported by the NSF under the grant DMS-0758552.
}
\address{Department of Mathematics, Michigan State University, East
Lansing, MI 48824, USA}
\email{[email protected]}
\urladdr{http://sashavolberg.wordpress.com}
\makeatletter
\@namedef{subjclassname@2010}{
\textup{2010} Mathematics Subject Classification}
\makeatother
\subjclass[2010]{42B20, 42B35, 47A30}
\keywords{\cz operators, dyadic shift, Orlicz spaces,
weighted estimates}
\begin{abstract}
We give again a proof of weighted estimate of any \cz operator. This is under a universal sharp sufficient condition that is weaker than the so-called bump condition. Bump conjecture was recently solved independently and simultaneously by A. Lerner and Nazarov--Reznikov--Treil--Volberg. The latter paper uses the Bellman approach. Immediately a very natural and seemingly simple question arises how to to strengthen the bump conjecture result by weakening its assumptions in a natuarl symmetric way. This is what we are dealing with here. However we meet an unexpected and, in our opinion, deep obstacle, that allows us to make only partial result.
Our proof consists of two main parts: reduction to a simple model operator, construction of Bellman function for estimating this simple operator. The newer feature is that the domain of definition of our Bellman function is infinitely dimensional.
\end{abstract}
\date{}
\maketitle
\section{Introduction}
In this paper we consider a question about the sufficiency of certain ``bump'' conditions for the boundedness of all Calder\'on-Zygmund operators. Precisely, we consider two functions $u, v$, positive almost everywhere, and ask a question:
\begin{equation}
\label{mainquestion}
\begin{aligned}
&\text{ When there exists a constant}\; C,\;\text{ such that for every function}
\\
&\qquad\qquad\qquad f\in C_{0}^{\infty} \quad \|T(fu)\|_{L^2(v)}\leqslant C \|f\|_{L^2(u)}?
\end{aligned}
\end{equation}
Of course the constant $C$ is assumed to be independent of $f$.
The famous ``joint $A_2$'' condition, necessary but not sufficient, was introduced by D. Sarason. It looked like this:
$$
\mbox{There exists a constant $C$, such that for any interval $I$ the following holds:}
$$
$$
\frac{1}{|I|}\ili_I u\,dx\; \;\frac{1}{|I|}\ili_I v\,dx \leqslant C\,.
$$
We would like to rewrite this condition in the following way:
\begin{equation}
\label{a2}
\|u\|_{L^1(I, \frac{dx}{|I|})} \cdot \|v\|_{L^1(I, \frac{dx}{|I|})} \leqslant C\,.
\end{equation}
It is well known that this condition is not sufficient for the boundedness of $T$ for interesting $T$ (like the Hilbert transform, or a dyadic shift). So, we want to consider a bigger left-hand side, to make the condition stronger. Thus, instead of the $L^1$-norm, we would like to put something bigger. This brings us to the notion of Orlicz norms.
\subsection{Orlicz norms}
Consider a function $\Phi$ that is increasing, and convex. Then define
$$
\|u\|_{L^\Phi_I} = \inf\{\la \colon \frac{1}{|I|}\ili_I \Phi\left(\frac{f(t)}{\la}\right) dt \leqslant 1\}.
$$
Notice that $\Phi(t)=t$ gives the normalized $L^1$ norm, and $\Phi(t)=t^p$ gives the normalized $L^p$ norm.
\subsection{History of the question}
\label{history}
An interesting ``bump" conjecture was open for quite a while, and it was recently
solved independently and at the same time by two quite distinct (but having some
fundamental similarity) methods . One solution, due to Andrei Lerner, uses {\it local sharp maximal function} approach, see \cite{Le}. Another solution, due to
Nazarov--Reznikov--Treil--Volberg \cite{NRTV} used a Bellman function technique,
but with a new twist, the Bellman function of \cite{NRTV} depends on infinitely many variables. This bump conjecture is the statement that replaces
the left hand side of \eqref{a2} by its ``bumped-up" version, conjecturing that this version is now sufficient for the boundedness of all \cz operators. Thus, it introduces the bumped $A_2$ condition:
\begin{equation}
\label{ba2}
\|u\|_{L^\Phi_I} \cdot \|v\|_{L^\Phi_I} \leqslant C\,.
\end{equation}
Of course $\Phi(t)=t$ is just the same as \eqref{a2}, so we need some condition (preferably sharp) on $\Phi$
to ensure the boundedness of interesting (actually of all) \cz operators.
This condition (and this is known to be sharp) was invented by Carlos P\'erez and David Cruz-Uribe, \cite{CUP99}, \cite{CUPpisa}, and it is
\begin{equation}
\label{intPhi}
\Phi \;\text{is convex increasing function such that}\;\int_1^\infty \frac{dt}{\Phi(t)} <\infty\,.
\end{equation}
The bump conjecture itself (see \cite{CUP99}, \cite{CUPpisa}, \cite{CUP00b}, \cite{CUMaP07}, \cite{CUMaP12}, \cite{CUMaPbook}) reads now: given that two weights $u, v$ satisfy \eqref{ba2} and $\Phi$ in \eqref{ba2} satisfies \eqref{intPhi}, prove that
any \cz operator is bounded from $L^2(u)$ into $L^(v)$ in the sense \eqref{mainquestion} stated above.
This has been proved, as we already mentioned in \cite{Le} and in \cite{NRTV}.
However, a very natural conjecture is that \eqref{ba2} can be weakening even more.
Namely, we want to bump only one weight at a time. We get the following quite natural one-sided bump assumption:
\begin{equation}
\label{1sba2}
\begin{aligned}
&\mbox{There exists a constant $C$, such that for any interval $I$ the following holds:}
\\
&\|u\|_{L^\Phi_I} \cdot \|v\|_{L^1(I, \frac{dx}{|I|})} \leqslant C, \\
&\mbox{and}\\
&\|u\|_{L^1(I, \frac{dx}{|I|})} \cdot \|v\|_{L^\Phi_I} \leqslant C.
\end{aligned}
\end{equation}
And now one-sided bump conjecture is the following statement:
suppose \eqref{1sba2} holds for all intervals (cubes), and suppose $\Phi$
satisfies integrability condition \eqref{intPhi}, then any \cz operator is bounded from $L^2(u)$ into $L^2(v)$ in the sense \eqref{mainquestion} stated above.
The attempt to prove this has been done in \cite{CURV}. But we could do this only for {\it some} $\Phi$.
The present article is another attempt. It is sort of different in technique, it uses a Bellman function method unlike \cite{CURV} that used a stopping time argument.
The present approach is slightly more propitious, because Bellman technique is ``reversible".
We actually manage to prove by Bellman technique only slightly more general result than in \cite{CURV}.
However, this ``reversibility" feature probably indicates that the one sided bump conjecture in full generality (for all $\Phi$ subject to \eqref{intPhi}) might be actually wrong.
Recently M. Lacey \cite{L} using a parallel corona argument generalized the results of this paper to the case $p\not=2$ and to a more general bump condition.
\section{A construction from ~\cite{NRTV}}
\label{co}
To formulate the main result we use a certain language.
For that we need the following construction. Define a function $\Psi$ in the following parametric way:
$$
\begin{cases}
s=\frac{1}{\Phi(t)\Phi'(t)} \\
\Psi(s):=\Phi'(t).
\end{cases}
$$
Of course, we define $\Psi$ in this way near $s=0$.
We give the following definition.
\begin{defin}\label{regbumps}
A function $\Phi$ is called \textit{regular bump}, if for any function $u$ there holds
$$
\|u\|_{L^\Phi_I}\geqslant C\int N_I(t) \Psi(N_I(t))dt.
$$
\end{defin}
\begin{zamech}[\cite{NRTV}]
An example of regular bump is the following: $\Phi(t)=t\rho(t)$, and
$$
t\frac{\rho'(t)}{\rho(t)}\log\rho(t) \to 0, \; \mbox{as} \; t\to \infty.
$$
\end{zamech}
The important result is the following.
\begin{lemma}
The function $s\mapsto\Psi(s)$ is decreasing; the function $s\mapsto s\Psi(s)$ is increasing; the function $\frac{1}{s\Psi(s)}$ is integrable near $0$.
Moreover, the following inequality is true with a uniform constant $C$ (which may depend only on $\Phi$):
$$
C\|u\|_{L^\Phi_I}\geqslant \int N_I(t) \Psi(N_I(t))dt,
$$
where
$$
N_I(t)=\frac{1}{|I|} |\{x\in I\colon u(x)\geqslant t\}|.
$$
Further, for ``regular'' functions $\Phi$ we have that
$$
\|u\|_{L^\Phi_I}\sim \int N_I(t) \Psi(N_I(t))dt.
$$
\end{lemma}
\section{The main results. Boundedness and weak boundedness.}
\label{main}
Given a function $\Phi$, satisfying \eqref{intPhi}, build the corresponding function $\Psi$ as in Section \ref{co}. We prove the following theorems. Regularity conditions are not very important, but the last condition in the statement of the theorem is actually an important restriction. This is the restriction one would wish to get rid of. Or to prove that it is actually needed. Lately we believe that one cannot get rid of it.
We give a non-standard definition.
\begin{defin}\label{weaklyconc}
A function $f$ is ``weakly concave'' on its domain, if for any numbers $x_1, \ldots, x_n$ and $\la_1, \ldots, \la_n$, such that $0\leqslant \la_j\leqslant 1$,
and $\sli \la_j = 1$, the following inequality holds:
$$
f(\sli \la_j x_j)\geqslant C \sli \la_j f(x_j),
$$
where the constant $C$ does not depend on $n$.
\end{defin}
\begin{theorem}
\label{propep}
Suppose there exists a function $\Phi_0$ with corresponding $\Psi_0$, such that:
\begin{itemize}
\item $\Phi_0$ satisfies \eqref{intPhi};
\item $\Phi$ and $\Phi_0$ are regular bumps;
\item There is a function $\ep$, such that $\Psi_0(s)\leqslant C \Psi(s) \ep(\Psi(s))$;
\item The function $t\mapsto t\ep(t)$ is weakly concave, in the sense of the Definition \ref{weaklyconc};
\item The function $t\mapsto t\ep(t)$ is strictly increasing near $\infty$;
\item The function $t\mapsto t\ep(t)$ is concave near $\infty$;
\item The function $t\mapsto \frac{\ep(t)}{t}$ is integrable at $\infty$.
\end{itemize}
Suppose that there exists a constant $C$, such that a one-sided bump condition \eqref{1sba2} holds.
Then any \cz operator is bounded from $L^2(u)$ into $L^{2}(v)$ in the sense of \eqref{mainquestion}.
\end{theorem}
\begin{theorem}
Suppose the function $\Phi$ satisfies all conditions from the theorem above. Suppose that there exists a constant $C$, such that
$$
\|u\|_{L^1(I, \frac{dx}{|I|})} \cdot \|v\|_{L^\Phi_I} \leqslant C.
$$
Then any Calderon-Zygmund operator is weakly bounded from $L^2(u)$ into $L^{2, \infty}(v)$, i.e. there exists a constant $C$, such that for any function $f\in C_0^\infty$ there holds
\begin{equation}\label{weakbdd}
\|T(fu)\|_{L^{2, \infty}(v)} \leqslant C \|f\|_{L^2(u)}.
\end{equation}
\end{theorem}
\section{Examples of $\Phi$ satisfying the restrictions of the main results: the cases from \cite{CURV}}
\label{examples}
The biggest difference of the above results with those of \cite{CURV} is that here we gave the integral condition on the corresponding bump function $\Phi$.
To compare with \cite {CURV} we notice that in \cite{CURV} theorems above were proved in two cases:
\begin{enumerate}
\item $\Phi(t)=t\log^{1+\sigma}(t)$;
\item $\Phi(t)=t\log(t)\log\log^{1+\sigma}(t)$, for sufficiently big $\sigma$.
\end{enumerate}
We show that these results are covered by our theorems.
First, suppose $\Phi(t)=t\log^{1+\sigma}(t)$. Then $\Psi(s)\asymp\log^{1+\sigma}(\frac1s)$. We put $\Phi_0(s)=t\log^{1+\frac\sigma 2}(t)$, and then $\ep(t)=t^{-\frac{\sigma}{2(1+\sigma)}}$.
Then, clearly, all properties of $\ep$ from our theorem are satisfied.
\bigskip
Next, suppose $\Phi(t)=t\log(t)\log\log^{1+\sigma}(t)$. Then $\Psi(s)\asymp\log(\frac1s)\log\log^{1+\sigma}(\frac1s)$. We put $\Phi_0(t)=t\log(t)\log\log^{1+\delta\sigma}(t)$, $\delta<1$ which gives $\ep(t)=\log^{-(1-\delta)\sigma}(t)$.
Then, the integral $\int^\infty \frac{\ep(t)}{t}\,dt$ converges if $\sigma>1$, and we choose $\delta$ to be very small.
\bigskip
Moreover, examining the proof of Theorem 5.1 from \cite{CURV}, we get the result from our paper but with a condition
$$
\mbox{The function $t\mapsto \frac{\sqrt{\ep(t)}}{t}$ is integrable at $\infty$.}
$$
We notice that for regular functions we have $\ep(t)\to 0$ when $t\to \infty$, and so $\ep(t)<\sqrt{\ep(t)}$. Thus, our results work for more function $\ep$ and, thus, bumps $\Phi$.
\section{Preliminary results}
\label{prelim}
In this section we state two helpful results. They are important building blocks in our proof. The first result is due to A. Lerner. It treats the so called {\it Banach function spaces}, see the definition in \cite{Le}.
\begin{theorem}[Theorem 1.1 from \cite{Ler}]\label{Ler}
Suppose $X$ is a {\it Banach function space} over $\R$ equipped with Lebesgue measure. Then, for any appropriate $f$,
$$
\|Tf\|_X \leqslant C(T, n) \sup_{\mathcal{D}, \{a_I\}}\|T_{\mathcal{D}, \{a_I\}}|f|\|_X,
$$
where the supremum is taken over all dyadic lattices $\mathcal{D}$ and all sequences $\{a_I\}$, such that $\sli_{I\subset J, I\in \mathcal{D}}a_I|I|\leqslant 2|J|$ for any $J\in \mathcal{D}$. Here
$$
T_{\mathcal{D}, \{a_I\}}f = \sli_I a_I \av{f}{I}\chi_I.
$$
\end{theorem}
Here
$$
\av{f}{I} := \frac{1}{|I|} \ili_I f dx\,.
$$
The next result is the famous testing conditions type theorem. We state it in the way we will use it. First let us introduce the small notation
$$
u(J) := \int_J u\, dx\,.
$$
Notice that
$$
u(J)= \|1_J\|_{L^2(u)}^2\,.
$$
\begin{theorem}\label{Laceyandco}
Suppose there exists a constant $C$, such that for any dyadic interval $J\in \mathcal{D}$ there holds
\begin{equation}
\label{test}
\begin{aligned}
&\|\chi_J T_{\mathcal{D}, \{a_I\}}(u\chi_J)\|_{L^2(v)}^2 \leqslant C\, u(J),
\\
&\|\chi_J T_{\mathcal{D}, \{a_I\}}(v\chi_J)\|_{L^2(u)}^2 \leqslant C\, v(J)\,.
\end{aligned}
\end{equation}
Then the operator $T_{\mathcal{D}, \{a_I\}}$ is bounded in the sense of \eqref{mainquestion}.
Moreover, if the weights $u$ and $v$ satisfy the joint $A_2$ condition, meaning that for any interval $J$ there holds $\av{u}{J}\av{v}{J}\leqslant C$, and there holds
$$
\|\chi_J T_{\mathcal{D}, \{a_I\}}(v\chi_J)\|_{L^2(u)}^2 \leqslant C\, v(J)\,,
$$
then the operator $T_{\mathcal{D}, \{a_I\}}$ is weakly bounded in the sense of \eqref{weakbdd}.
\end{theorem}
The first part can be found in \cite{NTV-mrl}. The last statement follows from the Theorem 4.3 of \cite{HLMORSUT} and Corollary 3.2 of \cite{HyLa}. It also needs the known fact that the maximal function is weakly bounded if weights satisfy the joint $A_2$ condition.
\subsection{Self improvements of Orlicz norms.}
In this section we prove a technical result, which has the following ``hand-waving'' explanation: suppose we take a function $\Phi$ and a smaller function $\Phi_0$. We explain how small can be the quotient $\frac{\|u\|_{L^{\Phi_0}_I}}{\|u\|_{L^{\Phi}_I}}$ in terms of smallness of $\frac{\Phi_0}{\Phi}$. In what follows we consider only ``regular bumps'' functions in the sense of the Definition \ref{regbumps}.
Suppose we have two functions $\Phi$ and $\Phi_0$, and we have built functions $\Psi$ and $\Psi_0$. We suppose that
$$
\Psi_0(s)\leqslant C \Psi(s) \ep(\Psi(s)).
$$
The following theorem holds.
\begin{theorem}
\label{selfimp}
Let $I$ be an arbitrary interval (cube). If a function $t\mapsto t\ep(t)$ is weakly concave, then
$$
\|u\|_{L_I^{\Phi_0}}\leqslant C \|u\|_{L_I^{\Phi}} \;\ep\Big(\frac{\|u\|_{L_I^\Phi}}{\av{u}{I}}\Big)\,.
$$
\end{theorem}
To do that we need the following easy lemma:
\begin{lemma}
For weakly concave functions the Jensen inequality holds with a constant:
$$
\ili f(g(t))d\mu(t) \leqslant C f(\ili g(t)d\mu(t)).
$$
\end{lemma}
\begin{proof}
This is true since if $g$ is a step function, then this is just a definition. Then we pass to the limit.
Here we essentially used that we can take a convex combination of $n$ points, and the constant in the definition above does not depend on $n$.
\end{proof}
\begin{proof}[Proof of the Theorem]
In the proof we omit the index $I$.
Since for regular bumps we know that
$$
\|u\|_{L^\Phi} \sim \int \Psi(N(t))N(t)dt,
$$
we simply need to prove that
$$
\int \Psi_0(N(t))N(t)dt \leqslant C \int \Psi(N(t))N(t)dt\; \ep\Big(\frac{\int \Psi(N(t))N(t)dt}{\int N(t)dt}\Big)
$$
Our first step is the obvious estimate of the left-hand side:
$$
\int \Psi_0(N(t))N(t)dt\leqslant C \int \Psi(N(t))\ep(\Psi(N(t))N(t)dt\,.
$$
Denote $a(t)=t\ep(t)$. Then we need to prove that
$$
\int a(\Psi(N(t)) N(t)dt \leqslant C \int N(t)dt \; a\Big(\frac{\int \Psi(N(t))N(t)dt}{\int N(t)dt}\Big)\,.
$$
We denote
$$
d\mu = \frac{N(t)}{\int N(t)dt}dt,
$$
it is a probability measure. Moreover, by assumption, $t\mapsto a(t)$ is concave. Therefore, by Jensen's inequality (from the Lemma),
$$
\int a(f(t))d\mu(t) \leqslant C a\Big(\int f(t)d\mu(t)\Big)\,.
$$
Take $f(t)=\Psi(N(t))$, and the result follows.
\end{proof}
\subsection{Examples}
\subsubsection{$\log$-bumps}
First, if $\Phi(t)=t\log^{1+\sigma}(t)$, then $\Psi(s)=\log^{1+\sigma}(1/s)$, and
$$
\frac{\Psi_0(s)}{\Psi(s)}=\log^{-\frac{\sigma}{2}}(1/s) = \Psi^{-\frac{\sigma}{2(1+\sigma)}}.
$$
Thus, $\ep(t)=t^{-\frac{\sigma}{2(1+\sigma)}}$, and everything is fine.
\bigskip
\subsubsection{$\log\log$-bumps}\label{exloglog}
Next example is with double logs. In fact, when
$$
\Psi(s)=log(1/s)(\log\log(1/s))^{1+\sigma}, \;\;\;\;
\Psi_0(s)=log(1/s)(\log\log(1/s))^{1+\sigma/2}
$$
then
$$
\frac{\Psi_0(s)}{\Psi(s)}=\log\log^{-\sigma/2}(1/s) \sim (\log(\Psi(s)))^{-\sigma/2}.
$$
Thus, $\ep(t)=(\log t)^{-\frac{\sigma}2}$. Everything would be also fine, except for one little thing: the function $t\mapsto t\ep(t)$ is concave on infinity, but not near $1$. However, $t\mapsto t\ep(t)$ is weakly concave on $[2, \infty)$, and this is enough for our goals as without loss of generality, $\Psi(s)\geqslant 2$.
So let us prove that $a(t)= t\ep(t)$ is weakly concave on $[2, \infty)$.
Let $\varkappa:= \frac{\sigma}2$. The function $a$ has a local minimum at $e^{\varkappa}$ and its concavity changes at $e^{\varkappa+1}$. We now take $x_j$, $\la_j$ and $x=\sli \la_j x_j$. We first notice that if $x>e^{\varkappa+1}$, the we are done, because then $(x,\sli \la_j a(x_j))$ lies under the graph of $a$.
If $2\le x<e^{\varkappa+1}$, then $a(x)>\min_{[2, e^{\varkappa+1}]}a = c(\varkappa)$. Moreover, if $\ell$ is a line tangent to graph of $a$, starting at $(2, a(2))$, and $\ell$ ``kisses'' the graph at a point $(r, a(r))$, then $\sli \la_j a(x_j)\leqslant a(r)=c_1(\varkappa)$. This follows from the picture: a convex combination of $a(x_j)$ can not be higher than this line.
Therefore,
$$
a(\sli \la_j x_j)\geqslant c(\varkappa)\geqslant C c_1(\varkappa) \geqslant \sli \la_j a(x_j)\,.
$$
This finishes our proof.
\subsection{Proof of the main result: notation and the first reduction.}
We fix a dyadic grid $\mathcal{D}$. Theorems \ref{Ler} and \ref{Laceyandco} show that to prove our main results it is enough to show that the following implication holds:
$$
\mbox{if for all}\, \,\,I \, \; \|u\|_{L^\Phi_I} \cdot \|v\|_{L^1(I, \frac{dx}{|I|})} \leqslant B_{u,v} \; \mbox{then} \; \|\chi_J T_{\mathcal{D}, \{a_I\}}(u\chi_J)\|_{L^2(v)}^2\leqslant C\, u(J) ,
$$
where $C$ does not depend neither on the grid, nor on the sequense $\{a_I\}$. It can, of course, depend on $B_{u,v}$. This will prove the weak bound $T: L^2(v)\rightarrow L^{2,\infty}(u)$.
For simplicity, we denote $T_a = T_{\mathcal{D}, \{a_I\}}$. It is an easy calculation that, under the joint $A_2$ condition (which is definitely satisfied under the bump condition), it is enough to get an estimate of the following form:
\begin{equation}
\label{glav}
\frac{1}{|J|}\sli_{J\subset I} a_I \cdot \av{u}{I} \cdot \frac{1}{|I|}\sli_{K\subset I}a_K \av{u}{K}\av{v}{K}|K| \cdot |I|\leqslant C\, u(J)\,.
\end{equation}
\begin{zamech}
By the rescaling argument it is clear that we can assume $B_{u,v}$ as small as we need (where ``smallness'', of course, depends only on the function $\Phi$). We need this remark, since all behaviors of our function $\ep$ are studied near $0$.
\end{zamech}
\begin{zamech}
Everything is reduced to \eqref{glav}. We concentrate on proving \eqref{glav}. Clearly, by scale invariance, it looks very tempting to make \eqref{glav} a Bellman function statement. This will be exactly our plan from now on.
\end{zamech}
\section{Bellman proof of \eqref{glav}: introducing the ``main inequality''}
\label{mainineq}
We start this Section with the following notation. We fix two weights $u$ and $v$, and a Carleson sequense $\{a_I\}$. We denote
$$
u_I = \av{u}{I}, \; \; v_I=\av{v}{I}; \; \; \; N_I(t)=\frac1{|I|} |\{x\colon u(x)\geqslant t\}|;
$$
$$
A_I = \frac{1}{|I|}\sli_{J\subset I}a_J |J|;
$$
$$
L_I = \frac{1}{|I|}\sli_{J\subset I}a_J \av{u}{J}\av{v}{J} |J|.
$$
We proceed with two theorems that prove our main result. Everywhere in the future we use that $\av{u}{I}\av{v}{I}=u_I v_I \leqslant \delta < 1$ for any $I$. We can do it due to simple rescaling.
\begin{theorem}
\label{mainineq}
Suppose that
$$
\frac{\Psi_0}{\Psi}\leqslant \ep (\Psi),
$$
where $\ep$ satisfies properties of Theorem \ref{main}, from which the main one is
\begin{equation}
\label{intep}
\int^\infty \frac{\ep(t)}{t}\, dt<\infty\,.
\end{equation}
Let $\delta$ be small enough, and
$$
\Om_1 = \{(N, A) \colon 0\leqslant N \leqslant 1; \; 0\leqslant A \leqslant 1\}
$$
and for some constant $P$
$$
\Om_2 = \{(u,v,L,A)\colon 0\leqslant A \leqslant 1; \; u,v,L \geqslant 0; \; uv\leqslant \delta; \; L\leqslant P\cdot \sqrt{uv}\}.
$$
Suppose we have found a function $B_1$, defined on $\Om_1$, and a function $B_2$, defined on $\Om_2$, such that:
\begin{align}
& 0\leqslant B_1\leqslant N;\\
& (B_1)'_A \geqslant 10\frac{N}{\Psi_0(N)};\\
& -d^2 \,B_1 \geqslant 0;\\
& 0\leqslant B_2\leqslant u;\\
& (B_2)'_A \geqslant 0\\
&(B_2)'_A\geqslant c\cdot u\cdot L, \; \mbox{when} \; P\cdot \sqrt{uv}\ge L\geqslant \frac{uv}{\ep(\frac{1}{uv})}; \\
&uv(B_2)'_L \geqslant -\delta_1 uL, \; \mbox{for sufficiently small $\delta_1$ in the whole of $\Omega_2$}; \\
& -d^2\,B_2 \geqslant 0.
\end{align}
Then for the function of an interval $\mathcal{B}(I):= B_2(u_I, v_I, L_I, A_I)+\ili_0^\infty B_1(N_I(t), A_I)dt$ the following holds:
\begin{align}
& 0\leqslant \mathcal{B}(I) \leqslant 2u_I\\
&\mathcal{B}(I)-\frac{\mathcal{B}(I_+)+\mathcal{B}(I_-)}{2} \geqslant C\, a_I\cdot u_I\cdot L_I. \label{diff}
\end{align}
\end{theorem}
Next, we state
\begin{theorem}
\label{thenglav}
If such two functions $B_1$ and $B_2$ exist, then \eqref{glav} holds, namely
$$
\frac{1}{|I|}\sli_{J\subset I} a_I \cdot \av{u}{J} \cdot \frac{1}{|J|}\sli_{K\subset J}a_K \av{u}{K}\av{v}{K}|K| \cdot |I|\leqslant R^2 \ili_I u.
$$
\end{theorem}
\begin{proof}[Proof of the Theorem \ref{thenglav}]
This is a standard Green's formula applied to function $\mathcal{B}(I)$ on the tree of dyadic intervals. Let us explain the details.
Since the function $\mathcal{B}$ is non-negative, we have that
$$
2|I| u_I \geqslant |I|\mathcal{B}(I) \geqslant |I|\mathcal{B}(I) - \sli_{k=1}^{2^n} |I_{n,k}| \mathcal{B}(I_{n,k}).
$$
Here $n$ is fixed, and $I_{n,k}$ are $n-th$ generation descendants of $I$. Clearly, all $|I_{n,k}|$ are equal to $2^{-n}$.
Let us denote $\Delta(J) = |J|\mathcal{B}(J) - |J_+|\mathcal{B}(J_+) - |J_-|\mathcal{B}(J_-)$, where $J_\pm$ are children of $J$. By the property \eqref{diff} we know that $\Delta(J)\geqslant C|J|\, a_J u_J L_J$.
By the telescopic cancellation, we get that
$$
|I|\mathcal{B}(I) - \sli_{k=1}^{2^n} |I_{n,k}| \mathcal{B}(I_{n,k}) =
\sli_{m=0}^{n-1} \sli_{k=1}^{2^m}\Delta(I_{m,k}).
$$
Combining our estimates, we get
$$
2|I| u_I \geqslant C \sli_{m=0}^{n-1} \sli_{k=1}^{2^m} |I_{m,k}| a_{I_{m,k}}u_{I_{m,k}}L_{I_{m,k}} = C \sli_{J\subset I, |J|\geqslant 2^{-n}|I|}|J|a_J u_J L_J.
$$
This is true for every $n$, with the constant $C$ independent of $n$. Thus,
$$
u_I\geqslant C \frac{1}{|I|}\sli_{J\subset I} a_J u_J L_J |J|.
$$
The result follows from the definition of $L_J$.
\end{proof}
In the future we use the following variant of Sylvester criterion of positivity of matrix.
\begin{lemma}
\label{Sy}
Let $M=(m_{ij})_{i, j=1}^3$ be a $3\times 3$ real symmetric matrix such that $m_{11}<0$, $m_{11} m_{22}-m_{12}m_{21} >0$, and $\det \,M=0$. Then $M$ is nonpositive definite.
\end{lemma}
\begin{proof}
Let $E$ be a matrix with all entries being $0$ except for $e_{33}=1$. Consider $t>0$ and $A:=A(t):=M+tE$. It is easy to see that $a_{11}<0$, $a_{11} a_{22}-a_{12}a_{21} >0$, and $\det\, A=t\cdot (m_{11} m_{22}-m_{12}m_{21} )>0$ when $t>0$. By Sylvester criterion, matrices $A(t)$, $t>0$, are all negatively definite. Therefore, tending $t$ to $0+$, we obtain, that $M$ is nonpositive definite.
\end{proof}
We need the following lemma, which is in spirit of \cite{VaVo}.
\begin{lemma}
Let $L_I$ be given by
$$
L_I = \frac{1}{|I|}\sli_{J\subset I}a_J \av{u}{J}\av{v}{J} |J|.
$$
Let $A_I$ given by
$A_I = \frac{1}{|I|}\sli_{J\subset I}a_J |J|$. Suppose that it is bounded by $1$ for any dyadic $I$ (Carleson condition).
If for any dyadic interval $I$ we have that $\av{u}{I} \av{v}{I} \leqslant 1$, then it holds that for any dyadic interval $I$ we have $L_I \leqslant P \sqrt{\av{u}{I} \av{v}{I}}$.
\end{lemma}
\begin{proof}
It is true since the function $T(u, v, A)=100\sqrt{uv}-\frac{uv}{A+1}$ is concave enough in the domain $G:=\{0\leqslant A\leqslant 1, \; uv < 1, \; u,v\geqslant 0\}$. One can adapt the proof from \cite{VaVo}.
First, we need to check that the function $T(x,y,A)$ is concave in $G$. Clearly, $T''_{A,A} < 0$. Next,
\begin{equation}
\label{detT}
\det \begin{pmatrix} T''_{A,A} & T''_{A,v} \\ T''_{A, v} & T''_{v,v}
\end{pmatrix} = \frac{x}{y(A+1)^4} \cdot (50(A+1)\sqrt{xy}-xy) > 0\,.
\end{equation}
This expression is non-negative, because $A+1\geqslant 1$, and $\sqrt{uv}\leqslant 1$.
Finally,
$$
\det \begin{pmatrix} T''_{A,A} & T''_{A,v} & T''_{A,u} \\ T''_{A, v} & T''_{v,v} & T''_{v,u} \\ T''_{A,u} & T''_{v,u} & T''_{u,u}
\end{pmatrix} = 0.
$$
Therefore, by Lemma \ref{Sy} we conclude that $T(u,v, A)$ is a concave function.
Next,
$$
T'_A = \frac{uv}{(A+1)^2} \geqslant \frac{1}{4} uv.
$$
Thus, if we fix three points $(u, v, A)$, $(u_\pm, v_\pm, A_\pm)$, such that $u=\frac{u_+ + u_-}{2}$, $v=\frac{v_+ + v_-}{2}$, and $A=\frac{A_+ + A_-}{2} + a$, we get by the Taylor formula:
$$
T(u,v,A) - \frac{T(u_+, v_+, A_+)+T(u_-, v_-, A_-)}{2} \geqslant a T'_A(u,v,A) \geqslant C\, a\cdot uv.
$$
This requires the explanation. The Taylor formula we used has a remainder with the second derivative at the intermediate point $P_\pm$ on segments $S_+ :=[(u,v, \frac{A_-+A_+}{2}), (u,v, A_+)],$ $S_-:= [(u,v, \frac{A_-+A_+}{2}), (u,v, A_-)]$. One of this segments definitely lies inside domain $G$, where $T$ is concave, and this remainder will have the right sign. However the second segment can easily stick out of domain $G$, because $G$ itself is not convex. But notice that if, for example, $S_+$ is not inside $G$, still $(x,y, B)\in S_+$ implies that one of the coordinates, say $x$, must be smaller than $u$. Then $y$ can be bigger than $v$, but not much. In fact,
$$
v_+- v= v-v_-\,\Rightarrow \, v_+ \le 2v-v_-\le 2v\,.
$$
Therefore, $y\le v_+\le 2v$. Then we have that $xy\le 2uv \le 2$. Let us consider $\widetilde{G}:=\{(x,y, A): 0\le A\le 1,\; x, y\ge 0,\; 0\le xy \le 2\}$. Now come back to the proof that $T$ is concave in $G$. In \eqref{detT} we used that if $(x, y, A)\in G$, then $xy\le 1$ and the corresponding determinant is non-negative. But the same non-negativity in \eqref{detT} holds under slightly relaxed assumption $(x,y, A)\in \widetilde{G}$.
We notice that our $u_I=\av{u}{I}$, $v_I=\av{v}{I}$, and $A_I= \frac{1}{|I|}\sli_{J\subset I}a_I |I|$ have the dynamics above. The rest of the proof reads exactly as the proof of the Theorem \ref{thenglav}.
\end{proof}
\begin{proof}[Proof of the Theorem \ref{mainineq}]
We start with the following corollary from the Taylor expansion.
Suppose we have three tuples $(N, A)$, $(N_\pm, A_\pm)$, such that:
$$
N = \frac{N_+ + N_-}{2}; \; \; \; A = \frac{A_+ + A_-}{2} + m.
$$
Moreover, suppose there are $(u, v, L)$, $(u_\pm, v_\pm, L_\pm)$, such that
$$
u=\frac{u_+ + u_-}{2}; \; \; \; v=\frac{v_+ + v_-}{2};\;\;\; L=\frac{L_+ + L_-}{2} + m\cdot uv.
$$
Then, since $d^2B_1 \leqslant 0$, we write
$$
B_1(N_+, A_+) \leqslant B_1(N, A)+(B_1)'_N (N, A)(N_+-N) + (B_1)'_A(N, A)(A_+-A).
$$
Thus,
$$
B_1(N,A) - \frac{B_1(N_+, A_+)+B_1(N_-, A_-)}{2}\geqslant (B_1)'_A(N,A) \cdot (A-\frac{A_+ + A_-}{2}) = m\cdot (B_1)'_A(N,A)\geqslant m \frac{N}{\Psi_0(N)}.
$$
Similarly,
$$
B_2(u,v,L,A) - \frac{B_2(u_+, v_+, L_+, A_+)+B_2(u_-, v_-, L_-, A_-)}{2}\geqslant m\cdot ((B_2)'_A(u,v, L, A) + uv(B_2)'_L)
$$
First, suppose that $L_I\leqslant \frac{u_I v_I}{\ep(\frac{1}{u_I v_I})}$.
Then, using $m=a_I$ we get
\begin{multline}
\mathcal{B}(I)-\frac{\mathcal{B}(I_+)+\mathcal{B}(I_-)}{2} \geqslant \\ \geqslant \ili \left(B_1(N_I(t), A_I) - \frac{B_1(N_{I_+}(t), A_{I_+})+B_1(N_{I_-}(t), A_{I_-})}{2}\right)dt + \\+
\left( B_2(u_I,v_I,L_I,A_I) - \frac{B_2(u_{I_+}, v_{I_+}, L_{I_+}, A_{I_+})+B_2(u_{I_-}, v_{I_-}, L_{I_-}, A_{I_-})}{2}\right)
\\ \geqslant
a_I\left((B_2)'_A(u_I,v_I,L_I,A_I) + u_I v_I (B_2)'_L(u_I,v_I,L_I,A_I)\right)+a_I \left( \ili (B_1)'_A (N_I(t), A_I) dt \right) \geqslant \\
a_I \left(\ili \frac{N_I(t)}{\Psi_0(N_I(t))} dt - \delta_1 u_I L_I\right).
\end{multline}
The last inequality is true, since $(B_2)'_A \geqslant 0$ and $uv(B_2)'_L\geqslant -\delta_1 uL$ on the domain of $B_2$.
We use H\"older's inequality (and that $\int N_I(t)dt = u_I$) to get:
\begin{multline}
\ili \frac{N_I(t)}{\Psi_0(N_I(t))} dt\geqslant \frac{u_I^2}{\ili N_I(t)\Psi_0(N_I(t))dt} \geqslant C \frac{u_I^2}{\ili N_I(t)\Psi(N_I(t))dt\;\ep\!\left(\frac{\ili N_I(t)\Psi(N_I(t))dt}{u_I}\right)}.
\end{multline}
Last inequality is Theorem \ref{selfimp}.
Therefore, we get that
\begin{multline}
\ili \frac{N_I(t)}{\Psi_0(N_I(t))} dt \geqslant
u_I \cdot \frac{u_I}{\|u\|_{L^\Phi_I}}\cdot \frac{1}{\ep\left(\frac{\|u\|_{L^\Phi_I}}{u_I}\right)}= u_I \frac{u_Iv_I}{\|u\|_{L^\Phi_I}v_I}\cdot \frac{1}{\ep\left(\frac{\|u\|_{L^\Phi_I}v_I}{u_Iv_I}\right)}.
\end{multline}
We are going to use the one-sided bump condition $\|u\|_{L^\Phi_I}v_I\leqslant B_{u,v}\leqslant 1$. Thus,
$$
u_Iv_I \leqslant \frac{u_I v_I}{\|u\|_{L^\Phi_I}v_I}.
$$
Since the function $x\mapsto \frac{x}{\ep(\frac1x)}$ is increasing near $0$ (on $[0, c_\ep]$) and bounded from below between $c_\ep$ and $1$, we get
$$
\frac{u_Iv_I}{v_I\,\|u\|_{L^\Phi_I}}\cdot \frac{1}{\ep\left(\frac{v_I\,\|u\|_{L^\Phi_I}}{u_Iv_I}\right)}\geqslant C\cdot u_I v_I \frac{1}{\ep(\frac{1}{u_Iv_I})}.
$$
Therefore,
$$
\ili \frac{N_I(t)}{\Psi_0(N_I(t))} dt \geqslant C u_I \frac{u_I v_I}{\ep(\frac{1}{u_I v_I})}\geqslant C u_I L_I.
$$
The last inequality follows from our assumption that $L_I\leqslant \frac{u_I v_I}{\ep(\frac{1}{u_I v_I})}$.
Putting everything together, we get
$$
\mathcal{B}(I)-\frac{\mathcal{B}(I_+)+\mathcal{B}(I_-)}{2} \geqslant a_I u_I L_I (C-\delta_1)\geqslant C_1 \cdot a_I u_I L_I.
$$
\bigskip
We proceed to the case $L_I\geqslant \frac{u_I v_I}{\ep(\frac{1}{u_I v_I})}$. Then we write
$$
\mathcal{B}(I)-\frac{\mathcal{B}(I_+)+\mathcal{B}(I_-)}{2} \geqslant B_2(u_I, v_I, L_I, A_I) - \frac{B_2(u_{I_+}, v_{I_+}, L_{I_+}, A_{I_+})+B_2(u_{I_-}, v_{I_-}, L_{I_-}, A_{I_-})}{2}.
$$
This is obviously true, since $(B_1)'_A\geqslant 0$ everywhere and $B_1$ is a concave function. Next, we use
\begin{multline}
B_2(u_I, v_I, L_I, A_I) - \frac{B_2(u_{I_+}, v_{I_+}, L_{I_+}, A_{I_+})+B_2(u_{I_-}, v_{I_-}, L_{I_-}, A_{I_-})}{2}\geqslant \\
a_I\left((B_2)'_A + uv (B_2)'_L\right)\geqslant ca_I\cdot u_I L_I,
\end{multline}
by the property of $B_2$. Therefore, we are done.
\end{proof}
\section{Fourth step: building the function $B_2$}
\label{B2}
In order to finish the proof, we need to build functions $B_1$ and $B_2$. In this section we will present the function $B_2$. Denote
$$
\vf(x)=\frac{x}{\ep(\frac1x)}.
$$
This function is increasing (by regularity assumptions on $\ep$ in Theorem \ref{main}), therefore, there exists $\vf^{-1}$. We introduce
$$
B_2(u,v,L,A)= Cu - \frac{L^2}{v} \ili_{\frac{A+1}{L}}^{\infty} \vf^{-1}\Big(\frac{1}{x}\Big)\,dx.
$$
Let us explain why the integral is convergent. In fact, using change of variables, we get
$$
\ili_1^{\infty}\vf^{-1}\Big(\frac{1}{x}\Big)\,dx = \ili_0^{\vf^{-1}(1)} \frac{\ep(\frac 1t) - t\frac{d}{dt}(\ep(\frac 1t))}{t}\,dt,
$$
which converges at $0$ by assumption \eqref{intep}.
Therefore, since $L_I\leqslant C \sqrt{u_I v_I}$, we get
$$
0\leqslant B(u_I, v_I, L_I, A_I) \leqslant Cu_I.
$$
\bigskip
Next,
\begin{multline}
(B_2)'_A + uv(B_2)'_L = \frac{L}{v}\vf^{-1}\Big(\frac L{A+1}\Big) - u(A+1)\vf^{-1}\Big(\frac L{A+1}\Big) - 2uL\ili_{\frac{A+1}{L}}^{\infty}\vf^{-1}\Big(\frac 1x\Big)\,dx =
\\
uL \cdot\left(\frac{1}{uv}\vf^{-1}\Big(\frac L{A+1}\Big) - \frac{A+1}{L}\vf^{-1}\Big(\frac L{A+1}\Big) - 2\ili_{\frac{A+1}{L}}^{\infty}\vf^{-1}\Big(\frac 1x\Big)\,dx \right)
\end{multline}
We use that $L\geqslant \frac{uv}{\ep\big(\frac{1}{uv}\big)}=\vf(uv)$. Then $\vf^{-1}(L)\geqslant uv$, and, since $A+1\sim 1$, we get
$$
\frac{1}{uv}\vf^{-1}\Big(\frac L{A+1}\Big) \geqslant C_1.
$$
Moreover, since $uv\leqslant \delta$ is a small number, we get that $L$ is small enough for the integral $\ili_{\frac{A+1}{L}}^{\infty}\vf^{-1}(\frac 1x)dx$ to be less than a small number $c_2$. Finally, let us compare $\frac{A+1}{L}\vf^{-1}(\frac L{A+1})$ with a small number $c_3$. Since $L$ is small, we can write
$$
\ep\Big(\frac 1{c_3 L}\Big) \leqslant c_3.
$$
We do it, since $c_3$ is fixed from the beginning (say, $c_3=\frac{1}{10}$). Thus,
$$
L\leqslant \vf(c_3L).
$$
This implies
$$
\vf^{-1}(L)\leqslant c_3 L,
$$
thus
$$
\frac1L \vf^{-1}(L)\leqslant c_3.
$$
Since $A+1\sim 1$, we get the desired.
Therefore, if $L\geqslant \frac{uv}{\ep(\frac{1}{uv})}=\vf(uv)$ then $(B_2)'_A + uv(B_2)'_L \geqslant cuL$.
Moreover, in the whole domain of $B_2$ we get, since $(B_2)'_A \geqslant 0$,
$$
(B_2)'_A + uv(B_2)'_L \geqslant uv (B_2)'_L \geqslant -(c_2+c_3)uL
$$
with small $c_2+c_3$. This is a penultimate inequality in the statement of Theorem \ref{mainineq}.
Now we shall prove the concavity of $B_2$. For this it is enough to prove the concavity of the function of three variables: $B(v, L, A):= B_2(u,v, L, A) -Cu$. Clearly, $(B)''_{vv} < 0$, which is obvious.
Also, it is a calculation that
$$
\det \begin{pmatrix}
&(B)''_{vv} &&(B)''_{vA} &&&(B)''_{vL}\\
&(B)''_{vA} &&(B)''_{AA} &&&(B)''_{AL}\\
&(B)''_{vL} &&(B)''_{AL} &&&(B)''_{LL}
\end{pmatrix}
=0.
$$
Thus, we need to consider the matrix
$$
\begin{pmatrix}
&(B)''_{vv} &&(B)''_{vA} \\
&(B)''_{vA} &&(B)''_{AA}
\end{pmatrix}
$$
and to prove that its determinant is positive.
We denote $f(t)=\vf^{-1}(t)$, to simplify the next formula.
The calculation shows that the determinant above is equal to
$$
g\Big(\frac{L}{A+1}\Big):=-f\Big(\frac{L}{A+1}\Big)^2 + 2\Big(\frac{L}{A+1}\Big)^2\cdot f'\Big(\frac{L}{A+1}\Big) \ili_{\frac{A+1}{L}}^{\infty} f\Big(\frac1x\Big)\,dx.
$$
We need to prove that $g$ is positive near $0$. First, $g(0)=0$. Next,
\begin{multline}
g'(s) = -2 f(s)f'(s) + 4s f'(s) \ili_{\frac1s}^{\infty} f\Big(\frac1x\Big)\,dx + 2s^2 f''(s)\ili_{\frac1s}^{\infty} f\Big(\frac1x\Big)\,dx + 2 f'(s) f(s) =
\\
4s f'(s) \ili_{\frac1s}^{\infty} f\Big(\frac1x\Big)\,dx + 2s^2 f''(s)\ili_{\frac1s}^{\infty} f\Big(\frac1x\Big)\,dx.
\end{multline}
We notice that $f'$ is positive, since $\vf^{-1}$ is increasing near $0$. Moreover, by the fact that $\vf$ is strictly monotonous, and by concavity of $t\ep(t)$ (see Theorem \ref{propep}), we get that $\vf$ is {\it strictly} convex, hence $\vf^{-1}$ is strictly convex near $0$ as well. That is, $f''$ is also positive.
Therefore, $g'(s)>0$, and so $g(s)> g(0)=0$. The application of Lemma \ref{Sy} finishes the proof of concavity of $B$ (and therefore of the concavity of $B_2$). We are done.
\begin{zamech}
We can always think that the bump constant $B_{u,v}\leqslant C_\ep$, where $C_\ep$ is such that $L_I\leqslant c_\ep$. Then we can use the monotonicity and concavity of the function $\vf$ near $0$.
\end{zamech}
\section{Fifth step: building the function $B_1$}
\label{B1}
We present the function from \cite{NRTV}.
$$
B_1(N, A) = CN - N\ili_{0}^{\frac NA}\frac{ds}{s\Psi_0(s)}
$$
\section{Why function $\ep$ was so needed?}
\label{whyB2}
Integrability condition \eqref{intep} on function $\ep$ was used in constructing $B_2$ in a very essential way. A natural question arises, why not to get rid of $\ep$?
Suppose we can build function $B$ in the domain $\widehat\Omega$ such that
$$
\widehat\Om = \{(u,v,L,A)\colon 0\leqslant A \leqslant 1; \; u,v,L \geqslant 0; \; uv\leqslant \delta; \; L\leqslant P\cdot \sqrt{uv}\}.
$$
\begin{equation}
\label{hypoB}
\begin{aligned}
& 0\leqslant B\leqslant u;\\
& (B)'_A \geqslant 0\\
&(B)'_A\geqslant c\cdot u\cdot L, \; \mbox{when} \; P \sqrt{uv}\ge L\geqslant uv; \\
&uv(B)'_L \geqslant -\delta_1 uL, \; \mbox{for sufficiently small $\delta_1$ in the whole of $\widehat\Omega$}; \\
& -d^2\,B \geqslant 0.
\end{aligned}
\end{equation}
Looking at the proof of Theorem \ref{mainineq} we immediately see that this $B$ can replace our $B_2$ in this proof and, thus, give us \eqref{glav}
{\it without} any extra conditions on $\Phi$ or corresponding $\Psi$ apart a necessary condition of integrability: $\int^\infty\frac{dt}{\Phi(t)}<\infty$.
Hence, the existence of such a function would prove (as we have explained before) the one-sided bump conjecture in full generality.
\medskip
\begin{zamech}
However, we are inconclusive whether $B$ as in \eqref{hypoB} exists. We ``almost" prove below that it does not exist.
\end{zamech}
\medskip
Put
$$
\Om := \{(u,v, L, A)\colon 0\le A\le 1;\; u, v, L\ge 0;\; uv \le 1; P\sqrt{uv} \ge L\ge uv\}\,,
$$
and
$$
\,\Om_0: = \{(u,v, A)\colon 0\le A\le 1;\; u, v \ge 0;\; uv \le 1\}\,.
$$
\begin{theorem}
\label{shep}
It is impossible to find a smooth function $B$, defined on $\Om$, such that:
\begin{align}
& 0\leqslant B \leqslant u;\\
&B'_A \geqslant c\cdot u\cdot L, \; \mbox{when} \; P\cdot \sqrt{uv}\ge L\geqslant uv;
\\
& -d^2\,B \geqslant 0;
\end{align}
which satisfies one extra condition:
\begin{equation}
\label{concB0}
B_0(u,v, A) := \sup_{L\colon uv \le L\le P\sqrt{uv}} B(u,v, L, A) \;\;\;\text{is concave in}\;\;\;\Om_0 \,.
\end{equation}
\end{theorem}
The proof will consist of two parts. First we show that if function $B$ in Theorem \ref{shep} exists, then a certain other function must exist.
Only then we come to a contradiction with the existence of this new function built in the lemma that now follows.
\begin{lemma}
\label{B0}
Given a smooth function $B$ from Theorem \ref{shep} one can build a function $B_0$ such that it is defined in $\Om_0$
and such that
\begin{align}
& 0\leqslant B_0 \leqslant u;\\
&(B_0)'_A \geqslant c\cdot u^2\cdot v ;
\\
& -d^2\,B_0 \geqslant 0.
\end{align}
\end{lemma}
\begin{proof}
Given $B$ in $\Omega$ consider a new function defined on $\Omega_0$:
$$
B_0(u, v, A) :=\max_{L\colon\;(u,v, A, L)\in \Omega}B(u,v, L, A) = \max_{L\colon \; uv \le L\le P\,\sqrt{uv}}B(u,v, L, A) \,.
$$
Call the point, where the maximum is attained $L(u,v, A)$. Fix $A, u$. Let set $I_{A, u}$ be the set of $v\in [0, 1/u]$
such that this maximum is attained strictly inside: $L(u, v, A)\in (uv, P\sqrt{uv})$.
Then for $v\in I_{A,u}$ we have
$ \frac{\pd B}{\pd L} (u, v, L(u,v, A), A) =0$. Consequently
\begin{equation}
\label{rav}
\frac{\pd B_0}{\pd A} = \frac{\pd B}{\pd A} (u, v, L(u,v, A), A) + \frac{\pd B}{\pd L}(u, v, L(u,v, A), A)\cdot \frac{\pd L}{\pd A}(u,v, A) \,.
\end{equation}
Using the middle property of $B$ from Theorem \ref{shep} we get
\begin{equation}
\label{oce}
\frac{\pd B_0}{\pd A} \ge c \cdot u\cdot uv = c\cdot u^2v
\end{equation}
to be satisfied on the closure of $I_{A,u}$ for all $A, u$. If this closure is not the whole $[0,1/u]$,
then its complement contains a sub-interval of $[0,1/u]$. Call this sub-interval $S$. On $S$ maximum in the definition of
$B_0$ is attained either for $L(u,v, A)= uv$ or $L(u,v, A)=P\,\sqrt{uv}$. It is easy to see that we can
fix one of this choices (may be by making $S$ smaller).
So suppose maximum is attained for $v\in S_{A,u}$ for $L(u,v, A)= uv$. Let us start to vary $A,u$ a bit.
We will see that the set of $A,u$, where the closure of $I_{A,u}$ is not the whole $[0,1/u]$ is an open set $G$.
Foliating $\{(u, v, \,A)\colon (A,u)\in G, v\in S_{A,u}\}$ by surfaces $uv=const$ we see that $\frac{\pd L(u,v, A)}{\pd A} =0$ on $G$.
Then again $\frac{\pd B}{\pd L}(u, v, L(u,v, A), A)\cdot \frac{\pd L}{\pd A}(u,v, A) =0$, and \eqref{oce} is thus valid because of equality \eqref{rav}.
We are left to prove the concavity of $B_0$: $-d^2 B_0 \ge 0$. But this concavity in guaranteed by our extra requirement \eqref{concB0}.
\end{proof}
\begin{zamech}
Function $B_0$ is a supremum of concave functions, and as such is not {\it automatically} concave. Only the infimum of concave functions is automatically concave. Therefore our extra requirement \eqref{concB0} seems very ad hoc and too strong. In fact, it is ``almost" no requirement at all. We cannot get rid of the word ``almost" though. However, let us explain that there is a very generic common situation when the supremum of the family of concave functions is indeed concave.
It is really easy to see that when one has a concave function of several variables given in the {\it convex} domain, and one forms the supremum of it over varying one of its variables with all other variables fixed, then one gets a concave function again.
We are doing ``almost" that in our construction of $B_0$ from $B$. The difference is that $\Omega$ is not convex anymore! And this is the only obstacle for deleteing the word ``almost" above. If not for that small obstacle the next lemma would prove that the set of functions having properties \eqref{hypoB} is empty.
\end{zamech}
Now we prove
\begin{lemma}
\label{B0nonexi}
Function $B_0$ such that $-d^2\,B_0 \geqslant 0$, $(B_0)'_A \geqslant c\cdot u^2\cdot v, c >0$, $ 0\leqslant B_0 \leqslant u$ in $\Omega_0=\{(u,v, A)\colon 0\le A\le 1;\; u, v \ge 0;\; uv \le 1\}$ does not exist.
\end{lemma}
\begin{proof}
Suppose it does exist. Take any sequence $\{\al_I\}, 0\le \al_I\le 1,$ enumerated by
dyadic lattice of $I_0:=[0,1]$ such that for any $J\in \cD$
\begin{equation}
\label{carse}
A_J:=\frac1{|J|}\sum_{I\in\cD\colon I\subset J} \al_I |I| \le 1\,.
\end{equation}
(Carleson property.)
Take any two functions $u, v$ on $[0,1]$ such that
\begin{equation}
\label{avav}
\av{u}{I}\cdot \av{v}{I} \le 1,\;\forall I\in\cD\,.
\end{equation}
Consider
$$
\cB(I):= B_0 (\av{u}{I}, \av{v}{I}, A_I)\,.
$$
It is then easy to see using the properties of $B_)$ and Taylor's formula that
\begin{equation}
\label{ind10}
|I| \cB(I) - (|I_+| \cB(I_+) + |I_-| \cB(I_-) )\ge c\cdot\al_I |I|\cdot \av{u}{I}^2\cdot \av{v}{I} \,.
\end{equation}
Summing this up, we get
\begin{equation}
\label{ind11}
c\sum_{I\in \cD, \, I\subset I_0} \av{u}{I}^2\cdot \av{v}{I}\al_I |I| \le\cB(I_0) \le \int_0^1 u \,dx\,.
\end{equation}
\medskip
Now we construct functions $u,v$ and Carleson sequence $\al_I$ such that they satisfy the properties just mentioned, but such that
\eqref{ind11} fails. To do that we choose $u\ge 0$ on $I_0$ whose specifications will be made later. Let
$\cG_n$ be the family of maxima dyadic intervals inside $I_0$ such that
$$
\av{u}{I}\ge 3^n\,.
$$
Here are two facts, firstly:
\begin{equation}
\label{sv}
\forall I\in \cG_n\;\; \av{u}{I} \le 2\cdot 3^n\,,
\end{equation}
and secondly, if for any $J\in \cG_{n-1}$ we denote $\cG_n(J)$ those $I\in \cG_n$ that lie inside $J$, then
\begin{equation}
\label{sn}
\frac{|J\setminus\cup_{I\in \cG_n(J)} I |} {|J|} \ge \frac13\,.
\end{equation}
Inequality \eqref{sv} is obvious by maximality of intervals. Inequality \eqref{sn} the followslike that:
$$
\frac{\cup_{I\in \cG_n(J)}| I | \cdot 3^n }{|J|} \le \frac{\sum_{I\in \cG_n(J)}\int_I u\,dx }{|J|} \le \frac{\int_J u\,dx}{|J|} =\av{u}{J} \le 2\cdot 3^{n-1}\,.
$$
Therefore,
$$
\frac{\cup_{I\in \cG_n(J)}| I | }{|J|}\le \frac23\,,
$$
and \eqref{sn} is proved.
Before choosing $u$ we will now choose $v$. We build $v$ ``from bottom to top".
Choose a large $n$, and on each $I\in \cG_n$ choose $v$ to be the same constant $3^{-n}$. At this moment \eqref{avav} is of course satisfied.
Now we consider $J\in \cG_{n-1}$, and we want to keep \eqref{avav} for this $J$.
If we would keep $v$ to be $3^{-n}$ on all $I\in\cG_n(J)$ and we put $v=3^{-n}$ on $J\setminus\cup_{I\in \cG_n(J)} I $,
then notice that $\av{u}{J}$ drops $3$ times with respect to $\av{u}{I}$, but $\av{v}{J}$ does not drop
with respect to $\av{v}{I}$, we would see that the product drops $3$ times. However, we want it not to drop at all. So we keep
$v$ to be $3^{-n}$ on all $I\in\cG_n(J)$ and we put $v= 9\cdot 3^{-n}$ on $J\setminus\cup_{I\in \cG_n(J)} I $.
This portion is at least $\frac13 |J|$. Therefore, even without extra help from $v$ on all $I\in\cG_n(J)$ we would have at this moment $\av{u}{J}\av{v}{J}\ge 1$.
We want exactly $1$. So we choose $c\in (1,9)$ such that if $v$ on all $I\in\cG_n(J)$ is $3^{-n}$ and on $J\setminus\cup_{I\in \cG_n(J)} I $ it is $c\cdot 3^{-n}$, then
\begin{equation}
\label{avJ}
\av{u}{J}\av{v}{J} =1\,.
\end{equation}
On the top of that we have an absolute estimate for all $L\in\cD$ such that $L\subset J$ and $L$ not a subset of $\cup_{I\in \cG_n(J)} I $. In fact,
for such $L$ we have $\av{u}{L} \le 3^n, \av{v}{L} \le 9\cdot 3^{-n}$, hence
\begin{equation}
\label{avL}
\av{u}{L}\av{v}{L} \le 9\,.
\end{equation}
Now we already built $v$ on every $J\in \cG_{n-1}$. The passage from $\cG_{n-1}$ to $\cG_n$ repeats those steps. Finally we will be finishing with $v$ such that
\eqref{avJ} holds for all $J\in \bigcup_{k=1}^n\cG_k$, and \eqref{avL} holds for all $L\in\cD$ such that $L$ is not equal to or being inside of any of the intervals $I\in \cG_n$.
Making $n\to\infty$ we have $v$ such that
\eqref{avJ} holds for all $J\in \bigcup_{k=1}^\infty\cG_k$, and \eqref{avL} holds for all $L\in\cD$.
Now we notice two things: we can multiply $v$ by $1/9$ to have $1$ in the right hand side of \eqref{avL}, and we can put
$$
\al_I := \begin{cases} 1/3\,,\;\; I \in \bigcup_{k=1}^\infty\cG_k
\\
0\,,\;\;\text{otherwise}\,.
\end{cases}
$$
Then by \eqref{sn} it is a Carleson sequence in the sense of \eqref{carse}. Now let us disprove \eqref{ind11} by the choice of $u$. The left hand side of \eqref{ind11} now can be written (by looking at \eqref{avJ}) as
$$
\sum_{I \in \bigcup_{k=1}^\infty\cG_k}\av{u}{I}^2\av{v}{I} \al_I |I| = \frac1{27} \sum_{k=1}^\infty\sum_{I \in \cG_k}\av{u}{I} |I|\,.
$$
Let for those $I$ consider $E_I:= I\setminus \bigcup_{\ell\in \cG_{k+1}}\ell$. Let $M^d$ means dyadic maximal operator. Then on those $I$ we have $M^d u\le 3^{k+1}$, $\av{u}{I}\ge 3^k$, and $E_I$ tile $I_0$.
Hence
$$
\sum_{k=1}^\infty\sum_{I \in \cG_k}\av{u}{I} |I|\ge \sum_{k=1}^\infty\sum_{I \in \cG_k}\av{u}{I} |E_I|\ge \frac13 \int_{I_0} (M^d u)(x)\, dx\,.
$$
Here is the choice of $u$ that finally violates \eqref{ind11}: $u$ must be such that $\int_0^2 M^d u\, dx =\infty, \int_0^1 u\, dx <\infty$. Lemma is proved.
\end{proof}
Theorem \ref{shep} is proved.
\section{Appendix: translating from \cite{CURV} to our language}
\label{appendix}
For simplicity we focus only on the case $p=2$. In \cite{CURV} the authors considered a bump funtcion $A(t)$ and studied the norm $\|u^{\frac12}\|_A$. In our language, $\Phi(t)=A(t^{\frac12})$. Then we notice that
$$
\|u^{\frac12}\|_A = \|u\|_{\Phi}^\frac12.
$$
This follows from the definition of the Orlitz norm.
Suppose now that we chose a function $\Phi_0$ with $\Psi_0(s)\leqslant \Psi(s)\ep(\Psi(s))$. We have peoved that it implies the following improvement of Orlitz norm:
$$
\|u\|_{\Phi_0}\leqslant C \|u\|_{\Phi} \cdot \ep\bigg(\frac{\|u\|_{\Phi}}{\ave{u}}\bigg).
$$
Translating it to the language of \cite{CURV}, we get (after taking square roots of both sides)
$$
\|u^\frac12\|_{A_0} \leqslant C \|u^{\frac12}\|_A \cdot \left(\ep\bigg(\frac{\|u^{\frac12}\|_{A}^2}{\|u^{\frac12}\|_{L^2}^2}\bigg)\right)^{\frac12}.
$$
Thus, the Orlisz norms for $A$ and $A_0$ improve in the following way:
$$
\|u^\frac12\|_{A_0} \leqslant C \|u^{\frac12}\|_A \cdot \ep_{A_0, A}\bigg(\frac{\|u^{\frac12}\|_{A}}{\|u^{\frac12}\|_{L^2}}\bigg),
$$
where
$$
\ep_{A_0, A}(t) = \sqrt{\ep(t^2)}.
$$
Thus, our integral condition on $\ep$ gives the following condition on $\ep_{A_0, A}$:
$$
\ili^{\infty} \frac{\ep_{A_0, A}(\sqrt{t})^2}{t}dt\leqslant \infty.
$$
We denote $y=\sqrt{t}$, thus $dy/y = dt/t$. And we get
$$
\ili^{\infty} \frac{\ep_{A_0, A}(y)^2}{y}dt <\infty.
$$
However, the proof of the results from \cite{CURV} gives a condition
$$
\ili^{\infty} \frac{\ep_{A_0, A}(y)}{y}dt <\infty.
$$
We notice that $\ep_{A_0, A}(y)$ is small at infinity, and thus $\ep_{A_0, A}(y)^2 < \ep_{A_0, A}$. Therefore, our result gives the result of \cite{CURV} and improves it. | 129,981 |
TITLE: Help with Proof for Connect Open Space implying Finitely path connectedness
QUESTION [0 upvotes]: I would appreciate help with my (sketch of a) proof for the following problem:
Question:
Let $E\subset \mathbb{R^n}$ be an open connected open set. Show that for any $x,y\in E$ there is a finite sequence
$$x_0=x,x_1,...,x_n=y\in E$$
such that the line segment from $x_i$ to $x_{i+1}$ is contained in E.
What I did:
It's quite rough yet, but what I did was saying that if one of these segments is no fully in E, then there would be a point in E arbitrarily close to something not in E, which would mean that E is not open, reaching thus a contradiction.
It doesn't seem quite right though.
REPLY [1 votes]: Choose some $x_0\in E$ and let $F$ be the set of $y\in E$ that can be reached from $x_0$ on a path $P_y$ consisting of finitely many line segments that lie in $E.$
(i). For any $y\in F,$ there is an open ball $B(y,r)\subset E$ with radius $r>0,$ and for any $z\in B(y,r),$ the union of $P_y$ with the line-segment from $y$ to $z$ is a suitable $P_z.$ So $B(y,r),$ which is open in the space $E,$ is a subset of $F$.
Therefore $F$ is open in the space $E$.
(ii).For any $y$ in the closure of $F$ in the space $E,$ we have $y\in E,$ so consider the same $B(y,r)$ as in (i). $ B(y,r)$ is a nbhd of $y$ in the space $E,$ and $y\in Cl_E(F),$ so there must exist $w\in F\cap B(y,r).$ Now the union of $P_w$ with the segment from $w$ to $y$ lies in in $E,$ so $y\in F$.
Therefore $F$ is closed in the space $E.$
(iii). By (i) and (ii), $F$ is a non-empty open-and-closed subset of the connected space $E,$ so $F=E.$ Now for any $x,y\in E,$ consider the union of the path $P_x$ from $x$ to $x_0$ with the path $P_y$ from $x_0$ to $y.$ | 197,801 |
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