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Hours of Infinity: Recording the Imperfect Eternal
Hours of Infinity: Recording the Imperfect Eternal is a book I wrote that was published by The Kelsey Museum of Archaeology in 2012. The book's forward was written by sound theorist and critic Marc Weidenbaum, with an introduction by Egyptologist T.G. Wilfong. The book is held in the collections of the ZKM Library in Karlsruhe and the Herskovits Library of African Studies at Northwestern University in Chicago.
The description of the book from the publisher's website:
."
• Hours of Infinity entry at Goodreads | 96,595 |
\begin{document}
\maketitle
\begin{center}
$^{1}$Department of Mathematics and Statistics, Texas Tech University\\
Box 41042, Lubbock, TX 79409-1042, U.S.A.\\
Email address: \texttt{[email protected]}
\medskip
$^{2}$Mathematics Department, Tulane University\\
6823 St. Charles Ave, New Orleans, LA 70118, U.S.A.\\
Email address: \texttt{[email protected]}
\end{center}
\begin{abstract}
We study the long-time behavior of spatially periodic solutions of the Navier-Stokes equations in the three-dimensional space.
The body force is assumed to possess an asymptotic expansion or, resp., finite asymptotic approximation, in Sobolev-Gevrey spaces, as time tends to infinity, in terms of polynomial and decaying exponential functions of time.
We establish an asymptotic expansion, or resp., finite asymptotic approximation, of the same type for the Leray-Hopf weak solutions.
This extends previous results that were obtained in the case of potential forces, to the non-potential force case, where the body force may have different levels of regularity and asymptotic approximation.
This expansion or approximation, in fact, reveals precisely how the structure of the force influences the asymptotic behavior of the solutions.
\end{abstract}
\section{Introduction}\label{intro}
We study the Navier-Stokes equations (NSE) for a viscous, incompressible fluid in the three-dimensional space, $\R^3$.
Let $\vecx\in \R^3$ and $t\in\R$ denote the space and time variables, respectively.
Let the (kinematic) viscosity be denoted by $\nu>0$, the velocity vector field by $\vecu(\vecx,t)\in\R^3$, the pressure by $p(\vecx,t)\in\R$, and the body force by $\mathbf f(\vecx,t)\in\R^3$. The NSE which describe the fluid's dynamics are given by
\begin{align}\label{nse}
\begin{split}
&\bds \frac{\partial \vecu}{\partial t}\eds + (\vecu\cdot\nabla)\vecu -\nu\Delta \vecu = -\nabla p+\mathbf f \quad\text{on }\R^3\times(0,\infty),\\
&\textrm{div } \vecu = 0 \quad\text{on }\R^3\times(0,\infty).
\end{split}
\end{align}
The initial condition is
\beq\label{ini}
\vecu(\vecx,0) = \vecu^0(\vecx),
\eeq
where $\vecu^0(\vecx)$ is a given divergence-free vector field.
In this paper, we focus on solutions $\vecu(\vecx, t)$ and $p(\vecx, t)$ which are $L$-periodic for some $L>0$.
Here, a function $g(\vecx)$ is $L$-periodic if
\beqs
g(\vecx+L\vece_j)=g(\vecx)\quad \textrm{for all}\quad \vecx\in \R^3,\ j=1,2,3,\eeqs
where $\{\vece_1,\vece_2,\vece_3\}$ is the standard basis of $\R^3$.
By the remarkable Galilean transformation, we can assume also that $\vecu(\vecx,t)$, for all $t\ge 0$, has zero average over the domain $\Omega=(-L/2, L/2)^3$.
A function $g(\vecx)$ is said to have zero average over $\Omega$ if
\beq \label{Zacond}
\int_\Omega g(\vecx)d\vecx=0.
\eeq
By rescaling the spatial and time variables, we assume throughout, without loss of generality, that $L=2\pi$ and $\nu =1$.
\begin{notation} We will use the following standard notation.
\begin{enumerate}[label={\rm (\alph*)}]
\item In studying dynamical systems in infinite dimensional spaces, we denote, regarding \eqref{nse} and \eqref{ini}, $u(t)=\vecu(\cdot,t)$, $f(t)=\mathbf f(\cdot,t)$, and $u^0=\vecu^0(\cdot)$.
\item For non-negative functions $h(t)$ and $g(t)$, we write
\beqs
h(t)=\mathcal{O}(g(t))\ \text{as $t\to\infty$}\quad \text{if there exist $T,C>0$ such that }\ h(t)\leq Cg(t),\ \forall t>T.
\eeqs
\item The Sobolev spaces on $\Omega$ are denoted by $H^m(\Omega)$ for $m=0,1,2,\ldots$, each consists of functions on $\Omega$ with distributional derivatives up to order $m$ belonging to $L^2(\Omega)$.
\end{enumerate}
\end{notation}
The type of asymptotic expansion that we study here is defined, in a general setting, as follows.
\begin{definition}\label{expanddef}
Let $X$ be a real vector space.
{\rm (a)} An $X$-valued polynomial is a function $t\in \R\mapsto \sum_{n=1}^d a_n t^n$,
for some $d\ge 0$, and $a_n$'s belonging to $X$.
{\rm (b)} When $(X,\|\cdot\|)$ is a normed space, a function $g(t)$ from $(0,\infty)$ to $X$ is said to have the asymptotic expansion
\beq \label{gexpand}
g(t) \sim \sum_{n=1}^\infty g_n(t)e^{-nt} \text{ in } X,
\eeq
where $g_n(t)$'s are $X$-valued polynomials, if for all $N\geq1$, there exists $\varep_N>0$ such that
\beq \label{grem}
\Big\|g(t)- \sum_{n=1}^N g_n(t)e^{-nt}\Big\|=\mathcal O(e^{-(N+\varep_N)t})\ \text{as }t\to\infty.
\eeq
\end{definition}
This article aims at studying the asymptotic behavior of the solution $\vecu(\vecx,t)$ as $t\to\infty$ for a certain class of forces $\mathbf f(\vecx,t)$.
The case when $\mathbf f$ is a potential force, i.e., $\mathbf f(\vecx,t)=-\nabla \phi(\vecx,t)$, for some scalar function $\phi$, has been well-studied. For instance, it is well-known that any Leray-Hopf weak solution becomes regular eventually and decays in the $H^1(\Omega)$-norm exponentially.
For more precise asymptotic behavior, Dyer and Edmunds \cite{DE1968} proved that a non-trivial, regular solution is also bounded below by an exponential function.
Foias and Saut \cite{FS84a} then proved that in bounded or periodic domains, a non-trivial, regular solution decays exponentially at an exact rate which is an eigenvalue of the Stokes operator.
Furthermore, they established in \cite{FS87} that for such a solution $u(t)$, the following asymptotic expansion holds, in the sense of Definition \ref{expanddef}, in Sobolev spaces $H^m(\Omega)^3$, for all $m\ge 0$:
\beq \label{expand}
u(t) \sim \sum_{n=1}^\infty q_n(t)e^{-nt},
\eeq
where $q_n(t)$'s are unique polynomials in $t$
with trigonometric polynomial values.
Recently, in \cite{HM1}, it was shown by the authors that the expansion in fact holds in Gevrey spaces.
More precisely, for any $\sigma>0$ and $m\in\N$, there exists an $\varep_N>0$, for each integer $N>0$, such that
\beq\label{vHM}
\Big\|e^{\sigma (-\Delta)^{1/2}} \Big(u(t)- \sum_{n=1}^N q_n(t)e^{-nt}\Big)\Big\|_{H^m(\Omega)^3}=\mathcal O\big(e^{-(N+\varepsilon_N)t}\big)\textrm{ as }t\to\infty.
\eeq
Note that the (Gevrey) norm in estimate \eqref{vHM} is much stronger than the standard Sobolev norm in $H^m(\Omega)$.
More importantly, the simplified approach in \cite{HM1} allows the proof to be applied to wider classes of equations; that approach will be adopted in this paper.
Regarding the case of potential forces, the interested reader is referred to \cite{FS83,FS84a,FS84b,FS87,FS91} for deeper studies on the asymptotic expansion, its associated normalization map, and invariant nonlinear manifolds; for the associated (Poincar\'e-Dulac) normal form, see \cite{FHOZ1,FHOZ2,FHS1}; for its applications to statistical solutions of the NSE, decaying turbulence, and analysis of helicity, see \cite{FHN1,FHN2}; for a result in the whole space $\mathbb R^3$, see \cite{KukaDecay2011}.
\medskip
The main goal of this paper is to establish \eqref{expand} when $\mathbf f$ is \textit{not} a potential function.
To understand the result without calling for technical details, we state it here as a `meta theorem'; the rigorous version will be provided by Theorem \ref{mainthm}, whose proof will then be presented in section \ref{pfsec}.
\begin{theorem}[Meta Theorem]\label{meta}
Assume that the body force has an asymptotic expansion
\beq\label{fexpand}
f(t)\sim \sum_{n=1}^\infty f_n(t)e^{-nt},
\eeq
in some appropriate functional space.
Then any Leray-Hopf weak solution $u(t)$ of \eqref{nse} and \eqref{ini} admits an asymptotic expansion of the form \eqref{expand} in the same space.
\end{theorem}
On the other hand, when $f(t)$ satisfies a finitary version of \eqref{fexpand}, then we obtain a corresponding finite asymptotic approximation result in Theorem \ref{finitetheo}.
More specifically, if the right-hand side of \eqref{fexpand} is a finite sum, then we show that the corresponding solution, $u(t)$, admits a finite sum approximation of the same type analogous to \eqref{expand}.
Theorem \ref{finitetheo} is also suitable for the case when the force is not smooth, but belongs only to a pre-specified Sobolev class.
Theorem \ref{meta} develops one of the possible avenues of extending the Foias-Saut theory \cite{FS87} for the NSE to the non-potential force case. The finite sum version, Theorem \ref{finitetheo}, which is a new feature only for the non-potential force case, allows asymptotic analysis of the solution even when the force has restricted regularity and only limited information on the force's asymptotic behavior is known. Our analysis, moreover, explicitly indicates how each term in the expansion of the force integrates into the expansion of the solution.
We emphasize that the assumptions that we do impose on the force are very natural. This point is not so obvious because if one directly adopts the argument of Foias-Saut in \cite{FS87}, one would then be led to impose conditions on \textit{time-derivatives of all orders} on the force. We, nevertheless, are able to avoid these conditions and ultimately establish the claimed expansion by applying the refined approach, in the case of periodic domains, initiated in \cite{HM1}.
We also remark that we obtain a welcome technical improvement over \cite{HM1} in obtaining the expansion in Sobolev spaces, when Gevrey regularity is not available; it requires a more elaborate bootstrapping process which is carried out in Part II of the proof of Proposition \ref{theo23}.
The results obtained in this paper are the first steps of the larger program of understanding the relation between the asymptotic expansion and the external body force. On the other hand, in spite of assuming the force to have regular modes of decay, further understanding of the solution's resulting expansion and its consequences will shed insights into the nonlinear structure of the NSE, and decaying turbulence theory.
It is worth mentioning at this point that the global existence of regular solutions and the uniqueness of the global weak solutions of the 3D NSE remain outstanding open problems.
However, our result details the asymptotic behavior of any Leray-Hopf weak solution, regardless of the resolution to either of these issues.
Moreover, to the best of the authors' knowledge, there have been no numerical studies made to determine the polynomials $q_n(t)$ in \eqref{expand} as of yet. By extending the expansion to accommodate non-potential forces, our result should facilitate the formulation and testing of possible numerical algorithms for their computation. Indeed, one can attempt to compute the expansion of \emph{explicit} solutions, particularly, those for which the nonlinear term in NSE does not vanish. These solutions are quite easy to generate when certain, specific forces are used at the onset, but harder to find
when the projected force in the functional equation \eqref{fctnse} is \emph{given and fixed}, albeit zero, as in the case of potential forces (see \eqref{potential}).
For the structure of this paper, Section \ref{bkgmain} lays the necessary background for formulating the result.
The main theorems are Theorems \ref{mainthm} and \ref{finitetheo}, while Corollary \ref{Vcor} emphasizes the scenario of finite-dimensional polynomial coefficients in \eqref{fexpand}, and the construction of the corresponding polynomials appearing \eqref{expand} as solutions to finite-dimensional, linear ordinary differential equations.
Section \ref{Gdecay} contains explicit estimates that ultimately furnish ``eventual" exponential decay for the weak solutions in Sobolev and Gevrey spaces, see Propositions \ref{theo22} and \ref{theo23}. They are used crucially in section \ref{pfsec}, which is devoted to the proofs of our main results.
\section{Background and main results}\label{bkgmain}
The space $L^2(\Omega)^3$ of square (Lebesgue) integrable vector fields on $\Omega$ is a Hilbert space with the standard inner product $\inprod{\cdot,\cdot}$ and norm $|\cdot
|$ defined by
$$
\inprod{u,v}=\int_\Omega \vecu(\vecx)\cdot \vecv(\vecx) d\vecx
\quad\text{and}\quad |u|
=\inprod{u,u}^{1/2}\quad\text{for } u=\vecu(\cdot),\ v=\vecv(\cdot).$$
We note that $|\cdot|$ is also used to denote the absolute value, modulus, and, more generally, the Euclidean norm in $\R^n$ and $\C^n$, for $n\in\N$. Nonetheless, its meaning will be made clear by the context.
Let $\mathcal{V}$ be the set of all $L$-periodic trigonometric polynomial vector fields which are divergence-free and have zero average over $\Omega$.
Define
$$H, \text{ resp. } V\ =\text{ closure of }\mathcal{V} \text{ in }
L^2(\Omega)^3, \text{ resp. } H^1(\Omega)^3.$$
We use the following embeddings and identification
$$V\subset H=H'\subset V',$$
where each space is dense in the next one, and the embeddings are compact.
Let $\mathcal{P}$ denote the orthogonal (Leray) projection in $L^2(\Omega)^3$ onto $H$. Explicitly,
\begin{align*}
\mathcal{P}\Big(\sum_{\veck\ne \mathbf 0}
\widehat \vecu(\veck)e^{i\veck\cdot \vecx}\Big) = \sum_{\veck\ne \mathbf 0} \Big\{\widehat\vecu(\veck)-\Big(\widehat\vecu(\veck)\cdot \frac{\veck}{|\veck|}\Big)\frac{\veck}{|\veck|}\Big\}e^{i\veck\cdot \vecx} .
\end{align*}
We define the Stokes operator $A:V\to V'$ by
\beqs
\inprod{A\vecu,\vecv}_{V',V}=
\doubleinprod{\vecu,\vecv}
\eqdef \sum_{i=1}^3 \inprod{ \frac{\partial \vecu}{\partial x_i} , \frac{\partial \vecv}{\partial x_i} },\quad \text{for all } \vecu,\vecv\in V.
\eeqs
As an unbounded operator on $H$, the operator $A$ has the domain $\mD(A)=V\cap H^2(\Omega)^3$, and
\beqs A\vecu = - \mathcal{P}\Delta \vecu=-\Delta \vecu\in H, \quad \textrm{for all}\quad \vecu\in\mD(A).
\eeqs
The last identity is due to the periodic boundary conditions.
It is known that the spectrum of the Stokes operator $A$ is $\sigma(A)=\{\lambda_j:j\in \N\}$,
where $\lambda_j$ is strictly increasing in $j$, and is an eigenvalue of $A$.
In fact, for each $j\in\N$, $\lambda_j=|\veck|^2$ for some $\veck\in\Z^3\setminus \{\mathbf 0\}$. Note that since $\sigma(A)\subset \N$ and $\lambda_1=1$, the additive semigroup generated by $\sigma(A)$ is equal to $\N$.
\medskip
For $\alpha,\sigma \in \R$ and $\vecu=\sum_{\veck\ne \mathbf 0}
\widehat \vecu(\veck)e^{i\veck\cdot \vecx}$, define
$$A^\alpha \vecu=\sum_{\veck\ne \mathbf 0} |\veck|^{2\alpha} \widehat \vecu(\veck)e^{i\veck\cdot
\vecx},$$
$$A^\alpha e^{\sigma A^{1/2}} \vecu=\sum_{\veck\ne \mathbf 0} |\veck|^{2\alpha}e^{\sigma
|\veck|} \widehat \vecu(\veck)e^{i\veck\cdot
\vecx}.$$
We then define the Gevrey spaces by
\beqs
G_{\alpha,\sigma}=\mD(A^\alpha e^{\sigma A^{1/2}} )\eqdef \{ \vecu\in H: |\vecu|_{\alpha,\sigma}\eqdef |A^\alpha
e^{\sigma A^{1/2}}\vecu|<\infty\},
\eeqs
and the domain of the fractional operator $A^\al$ by
\beqs
\mD(A^\alpha)=G_{\alpha,0}=\{ \vecu\in H: |A^\alpha \vecu|=|\vecu|_{\alpha,0}<\infty\}.
\eeqs
Thanks to the zero-average condition \eqref{Zacond}, the norm $|A^{m/2}\vecu|$ is equivalent to $\|\vecu\|_{H^m(\Omega)^3}$ on the space $\mathcal D(A^{m/2})$ for $m=0,1,2,\ldots$
Note that $\mD(A^0)=H$, $\mD(A^{1/2})=V$, and $\|\vecu\|\eqdef |\nabla \vecu|$ is equal to $|A^{1/2}\vecu|$ for $\vecu\in V$.
Also, the spaces $G_{\alpha,\sigma}$ are decreasing in $\alpha$ and $\sigma$.
Denote for $\sigma\in\R$ the space
\beqs
E^{\infty,\sigma}=\bigcap_{\alpha\ge 0} G_{\alpha,\sigma}=\bigcap_{m\in\N } G_{m,\sigma}.
\eeqs
We will say that an asymptotic expansion \eqref{gexpand} holds in $E^{\infty,\sigma}$ if it holds in $G_{\alpha,\sigma}$ for all $\alpha\ge 0$.
Let us also denote by $\mathcal{P}^{\alpha,\sigma}$ the space of $G_{\alpha,\sigma}$-valued polynomials in case $\alpha\in \R$, and
the space of $E^{\infty,\sigma}$-valued polynomials in case $\alpha=\infty$.
\medskip
We define the bilinear mapping $B:V\times V\to V'$, which is associated with
the nonlinear term in the NSE, by
\beqs
\inprod{B(\vecu,\vecv),\vecw}_{V',V}=b(\vecu,\vecv,\vecw)\eqdef \int_\Omega ((\vecu\cdot \nabla) \vecv)\cdot \vecw\, d\vecx, \quad \textrm{for all}\quad \vecu,\vecv,\vecw\in V.
\eeqs
In particular,
\beqs
B(\vecu,\vecv)=\mathcal{P}((\vecu\cdot \nabla) \vecv), \quad \textrm{for all}\quad \vecu,\vecv\in\mD(A).
\eeqs
More precisely, for $\vecu=\sum_{\veck\ne 0}
\widehat \vecu(\veck)e^{i\veck\cdot \vecx}$ and $\vecv=\sum_{\veck\ne 0}
\widehat \vecv(\veck)e^{i\veck\cdot \vecx}$,
\beqs
B(\vecu,\vecv)
= \sum_{\veck\ne \mathbf 0} \Big\{\widehat\vecb(\veck)-\Big(\widehat\vecb(\veck)\cdot \frac{\veck}{|\veck|}\Big)\frac{\veck}{|\veck|} \Big\}e^{i\veck\cdot
\vecx},
\quad\text{where }\widehat\vecb(\veck)=\sum_{\vecm+\vecl=\veck}
i (\widehat\vecu(\vecm)\cdot \vecl)\widehat\vecv(\vecl).
\eeqs
It is clear that
\beq\label{BVV}
B(\mathcal V,\mathcal V)\subset \mathcal V.
\eeq
By applying the Leray projection $\mathcal{P}$ to \eqref{nse} and \eqref{ini},
we rewrite the initial value problem for NSE in the functional form as
\beq\label{fctnse}
\frac{du(t)}{dt} + Au(t) +B(u(t),u(t))=\mathcal P f(t) \quad \text{ in } V' \text{ on } (0,\infty),
\eeq
with the initial data
\beq\label{uzero}
u(0)=u^0\in H.
\eeq
(See e.g. \cite{LadyFlowbook69,CFbook,TemamAMSbook} for more details.)
Because of the projection $\mathcal P$ on the right-hand side of \eqref{fctnse}, it is convenient to assume, without loss of generality that the force belongs to $H$. Then, we have
$$\mathcal P f(t)=f(t)\text{ in \eqref{fctnse}}.$$
In the case $\mathbf f(\vecx,t)$ is a potential force, then, by the Helmholtz-Leray decomposition,
\beq \label{potential}
\mathcal P f(t)\equiv 0 \text{ in the functional equation \eqref{fctnse}}.
\eeq
In dealing with weak solutions of \eqref{fctnse}, we follow the presentation in \cite{FMRTbook} and use the results there.
\begin{definition}\label{lhdef}
Let $f\in L^2_{\rm loc}([0,\infty),H)$.
A \emph{Leray-Hopf weak solution} $u(t)$ of \eqref{fctnse} is a mapping from $[0,\infty)$ to $H$ such that
\beq\label{lh:wksol}
u\in C([0,\infty),H_{\rm w})\cap L^2_{\rm loc}([0,\infty),V),\quad u'\in L^{4/3}_{\rm loc}([0,\infty),V'),
\eeq
and satisfies
\beq\label{varform}
\ddt \inprod{u(t),v}+\doubleinprod{u(t),v}+b(u(t),u(t),v)=\inprod{f(t),v}
\eeq
in the distribution sense in $(0,\infty)$, for all $v\in V$, and the energy inequality
\beq\label{Lenergy}
\frac12|u(t)|^2+\int_{t_0}^t \|u(\tau)\|^2d\tau\le \frac12|u(t_0)|^2+\int_{t_0}^t \langle f(\tau),u(\tau)\rangle d\tau
\eeq
holds for $t_0=0$ and almost all $t_0\in(0,\infty)$, and all $t\ge t_0$.
We will say that a Leray-Hopf weak solution $u(t)$ is \emph{regular} if $u\in C([0,\infty),V)$.
\end{definition}
Above, $H_{\rm w}$ is the topological vector space $H$ with the weak topology.
This definition of the Leray-Hopf weak solutions with the choice of the energy inequality \eqref{Lenergy} is, in fact, equivalent to the weak solutions used in \cite[Chapter II, section 7]{FMRTbook}, see e.g. Remark 1(e) of \cite{FRT2010} for the explanations.
\begin{assumption}
It is assumed throughout the paper that the function $f(t)$ belongs to $L^\infty_{\rm loc}([0,\infty),H)$.
\end{assumption}
This assumption serves to guarantee the existence of the Leray-Hopf weak solutions for any $u^0\in H$, see e.g. \cite{FMRTbook}. We note that, later in the paper, we will further impose that the force, $f(t)$, decays in time, for instance, in Sobolev norms.
Regarding the problem of finding asymptotic expansions for solutions of the NSE, it is natural, at this stage, to study the class of functions $f(t)$ which have similar asymptotic behavior as $u(t)$ in \eqref{expand}.
We specify the condition on $f(t)$ more precisely in the next theorem, which is our first main result.
\begin{theorem}[Asymptotic expansion] \label{mainthm}
Assume that there exist a number $\sigma_0\geq0$ and polynomials $f_n\in\mathcal{P}^{\infty,\sigma_0}$, for all $n\ge 1$, such that $f(t)$ has the asymptotic expansion
\beq\label{forcexpand}
f(t)\sim \sum_{n=1}^\infty f_n(t)e^{-nt}\quad \text{in }E^{\infty,\sigma_0} .
\eeq
Let $u(t)$ be a Leray-Hopf weak solution of \eqref{fctnse} and \eqref{uzero}.
Then there exist polynomials $q_n\in\mathcal{P}^{\infty,\sigma_0}$, for all $n\ge 1$, such that $u(t)$ has the asymptotic expansion
\beq\label{uexpand}
u(t)\sim \sum_{n=1}^\infty q_n(t) e^{-nt}\quad \text{in }E^{\infty,\sigma_0} .
\eeq
Moreover, the mappings
\beq\label{uF}
u_n(t)\eqdef q_n(t) e^{-nt} \quad\text{and}\quad F_n(t)\eqdef f_n(t)e^{-nt},
\eeq
satisfy the following ordinary differential equations in the space $E^{\infty,\sigma_0}$
\beq\label{unODE}
\ddt u_n(t) + Au_n(t) +\sum_{\stackrel{k,m\ge 1}{k+m=n}}B(u_k(t),u_m(t))= F_n(t),\quad t\in\R,
\eeq
for all $n\ge 1$.
\end{theorem}
Regarding equation \eqref{unODE}, when $n=1$, the sum on its left-hand side is empty, hence the equation reads as
\beq\label{u1ODE}
\ddt u_1(t) + Au_1(t) = F_1(t).
\eeq
\begin{remark}\label{betterem}
Observe that since the expansion \eqref{forcexpand} is an infinite sum, it immediately implies the following remainder estimate:
\begin{align*}
\Big|f(t)-\sum_{n=1}^N f_n(t)e^{-nt}\Big|_{\alpha,\sigma_0}
&\le \Big|f_{N+1}(t)e^{-(N+1)t}\Big|_{\alpha,\sigma_0}+\Big|f(t)-\sum_{n=1}^{N+1} f_n(t)e^{-nt}\Big|_{\alpha,\sigma_0}\\
&=\mathcal O(e^{-(N+\varep)t})+\mathcal O(e^{-(N+1+\delta_{N+1},\alpha)t}),
\end{align*}
which holds for all $N\geq1$, $\alpha\ge 0$, $\varep\in(0,1)$ and some $\delta_{N+1,\alpha}\in(0,1)$. Therefore, we have for all $N\geq1$, $\alpha\ge0$ that
\beq\label{fep}
\Big|f(t)-\sum_{n=1}^N f_n(t)e^{-nt}\Big|_{\alpha,\sigma_0}
=\mathcal O(e^{-(N+\varep)t}) \quad\text{as }t\to\infty,\quad \forall\varep\in(0,1).
\eeq
Similarly, the expansion \eqref{uexpand} implies for any $N\ge 1$ and $\alpha\ge0$ that
\beq\label{uqa}
\Big |u(t)-\sum_{n=1}^N q_n(t) e^{-nt}\Big |_{\alpha,\sigma_0} =\mathcal O(e^{-(N+\varep)t} )\quad\text{as } t\to\infty,\quad \forall\varep\in(0,1).
\eeq
(In fact, the same argument applies to the general expansion \eqref{gexpand} so that $\varep_N$ in \eqref{grem} can be taken arbitrarily in $(0,1)$.)
\end{remark}
\begin{remark}\label{mainrmk} The following additional remarks on Theorem \ref{mainthm} are in order.
\begin{enumerate}[label={\rm (\alph*)}]
\item We do not require the time derivative $d^m f/dt^m$, \emph{for all} $m\in\N$, to have any kind of expansion.
Indeed, this rather stringent requirement would have been imposed if one adapts Foias-Saut's original proof.
Our relaxation of this condition is owed to the higher regularity of the solutions for large time, in the particular case of periodic domains, either in Sobolev or Gevrey spaces.
In the case of Sobolev spaces ($\sigma_0=0$), it requires a bootstrapping scheme to gradually increase the regularity to any needed level.
In the case of the Gevrey spaces ($\sigma_0>0$), the effect is immediate, hence the proof is shorter.
(For the related Gevrey norm techniques, see \cite{FT-Gevrey, HM1, OliverTiti2000} and references therein.)
\item The equations \eqref{unODE} determine the polynomials $q_n(t)$'s and indicate the interactions on all scales between the body force and the nonlinear terms in NSE.
Even though the solution $u(t)$ decays to zero, such interactions are complicated.
\item Condition \eqref{forcexpand} is easily satisfied for any finite sum
\beqs
f(t)=\sum_{n=1}^N f_n(t)e^{-nt},
\eeqs
for some fixed $N\ge 1$, and polynomials $f_n(t)$'s belonging to $\mathcal V$. Even in this case, the result in Theorem \ref{mainthm} is new and the expansion \eqref{uexpand} for $u(t)$ can still be an infinite sum.
\item If the expansion \eqref{forcexpand} for $f(t)$ holds in $G_{0,\sigma_1}$ for some $\sigma_1>0$, then it holds in $E^{\infty,\sigma_0}$ for any $\sigma_0\in (0,\sigma_1)$, and hence, by Theorem \ref{mainthm}, the solution $u(t)$ admits the expansion \eqref{uexpand} in $E^{\infty,\sigma_0}$ for any $\sigma_0\in (0,\sigma_1)$.
\item The equations \eqref{unODE} in fact are {\it linear} systems of ordinary differential equations (ODEs) in infinite-dimensional spaces. They form an integrable system in the sense that it can be recursively solved by the variation of constants formula. Moreover, each solution $u_n(t)$ of the form \eqref{uF} is uniquely determined provided that $R_n u_n(0)=\xi_n\in R_nH$ is given.
\end{enumerate}
\end{remark}
In the following simple scenario, the ODEs \eqref{unODE} are finite-dimensional systems, which make them more accessible for deeper study, and, in particular, for numerical computations.
\begin{corollary}\label{Vcor}
If all $f_n(t)$'s in Theorem \ref{theo22} are $\mathcal V$-valued polynomials,
then so are the polynomials $q_n(t)$'s in the expansion \eqref{uexpand}, and consequently, the equations \eqref{unODE} are systems of linear ODEs in finite-dimensional spaces.
\end{corollary}
Next is the paper's second main result which deals with the case when $f(t)$ does not possess an asymptotic expansion \eqref{forcexpand}, but rather a \emph{finite asymptotic approximation}. Then it is proved that the weak solution also admits a finite asymptotic approximation of the same type.
\begin{theorem}[Finite asymptotic approximation]\label{finitetheo}
Suppose there exist an integer $N_*\ge1$, real numbers $\sigma_0\ge 0$, $\mu_*\ge \alpha_*\ge N_*/2$, and, for any $1\le n\le N_*$, numbers $\delta_n\in(0,1)$ and polynomials $f_n\in\mathcal{P}^{\mu_n,\sigma_0}$, such that
\beq \label{ffinite}
\Big|f(t)-\sum_{n=1}^N f_n(t)e^{-nt}\Big|_{\alpha_N,\sigma_0}=\mathcal O(e^{-(N+\delta_N)t}) \quad\text{as }t\to\infty,
\eeq
for $1\le N\le N_*$, where
\beqs
\mu_n=\mu_*-(n-1)/2,\quad \alpha_n=\alpha_*-(n-1)/2.
\eeqs
Let $u(t)$ be a Leray-Hopf weak solution of \eqref{fctnse} and \eqref{uzero}.
\begin{enumerate}[label={\rm{(\roman*)} }]
\item Then there exist polynomials $q_n\in\mathcal{P}^{\mu_n+1,\sigma_0}$, for $1\le n\le N_*$,
such that one has for $1\le N\le N_*$ that
\beq \label{ufinite}
\Big |u(t)-\sum_{n=1}^N q_n(t)e^{-nt}\Big|_{\alpha_N,\sigma_0}=\mathcal O(e^{-(N+\varep)t})\quad\text{as }t\to\infty,\quad\forall\varep\in(0,\delta_N^*),
\eeq
where
$ \delta_N^*=\min\{\delta_1,\delta_2,\ldots,\delta_N\}$.
Moreover, the ODEs \eqref{unODE} hold in the corresponding space $G_{\mu_n,\sigma_0}$ for $1\le n\le N_*$.
\item In particular, if all $f_n(t)$'s belong to $\mathcal V$, resp., $E^{\infty,\sigma_0}$, then so do all $q_n(t)$'s, and the ODEs \eqref{unODE} hold in $\mathcal V$, resp., $E^{\infty,\sigma_0}$.
\end{enumerate}
\end{theorem}
Regarding the finite approximation \eqref{ffinite}, in addition to its sum having only finitely many terms ($N\le N_*$), the force $f(t)$ now has much less regularity compared to the expansion \eqref{forcexpand}. This, therefore, determines the regularity of the solution $u(t)$ and its asymptotic approximation \eqref{ufinite}. Such dependence was worked out in detail in the above theorem.
It is worth noticing that Theorem \ref{finitetheo} is stronger than Theorem \ref{mainthm}. Nevertheless, the proof of Theorem \ref{finitetheo}, in the current presentation, is adapted from that of Theorem \ref{mainthm}.
\section{Exponential decay in Gevrey and Sobolev spaces}\label{Gdecay}
This section prepares for the proofs in section \ref{pfsec}. Particularly, we will derive the exponential decay for weak solutions in both Gevrey and Sobolev spaces.
First, we have a few basic inequalities concerning the Sobolev and Gevrey norms.
It is elementary to see that
\beqs
\max_{x\ge 0} (x^{2\alpha}e^{-\sigma x})=\Big(\frac{2\alpha}{e\sigma }\Big)^{2\alpha} \text{ for any } \sigma,\alpha>0.
\eeqs
Applying this inequality, one easily obtains
\beq\label{als}
|A^\alpha u|=|(A^\alpha e^{-\sigma A^{1/2}})e^{\sigma A^{1/2}}u| \le \Big(\frac{2\alpha}{e\sigma}\Big)^{2\alpha} |e^{\sigma A^{1/2}}u|\quad \forall \alpha,\sigma>0.
\eeq
We recall a key estimate for the Gevrey norm of the bilinear form (cf. Lemma 2.1 in \cite{HM1}), which is based mainly on the work of Foias and Temam \cite{FT-Gevrey}.
\begin{lemma}
\label{nonLem}
Let $\sigma\ge 0$ and $\alpha\ge 1/2$. There exists an absolute constant $K>1$, independent of $\alpha,\sigma$, such that
\beq\label{AalphaB}
|B(u,v)|_{\alpha ,\sigma }\le K^\alpha |u|_{\alpha +1/2,\sigma } \, |v|_{\alpha +1/2,\sigma}\quad \forall u,v\in G_{\alpha+1/2,\sigma}.
\eeq
\end{lemma}
Although inequality \eqref{AalphaB} is not sharp, it is very convenient for our calculations below and will be sufficient for our purposes.
As a consequence of Lemma \ref{nonLem}, we have
\beq\label{BGG}
B(G_{\alpha+1/2,\sigma},G_{\alpha+1/2,\sigma})\subset G_{\alpha,\sigma}\quad\text{for } \alpha\ge 1/2,\ \sigma\ge 0,
\eeq
\beq\label{BEE}
B(E^{\infty,\sigma},E^{\infty,\sigma})\subset E^{\infty,\sigma}\quad\text{for }\sigma\ge 0.
\eeq
Next, we prove a small data result, which establishes the global existence of the solution in Gevrey spaces and its exponential decay as time goes to infinity.
\begin{proposition} \label{theo22}
Let $\delta\in (0,1), \lambda\in(1-\delta,1]$ and $\sigma\ge 0, \alpha\ge1/2$.
Define the positive numbers $C_0=C_0(\alpha,\delta)$ and $C_1=C_1(\alpha,\delta,\lambda)$ by
\beqs
\begin{cases}
C_0=\frac\delta{6K^\alpha},\
C_1=\frac2{\sqrt3}\,\sqrt{\delta(\lambda-1+\delta)}\, C_0&\text{if }\sigma>0,\\
C_0=\frac\delta{4K^\alpha},\
C_1=\sqrt{2}\,\sqrt{\delta(\lambda-1+\delta)}\, C_0&\text{if }\sigma=0.
\end{cases}
\eeqs
Suppose
\beq\label{usmall}
|A^\alpha u^0|\le C_0,
\eeq
and
\beq\label{fta}
|f(t)|_{\alpha-1/2,\sigma}\le C_1e^{-\lambda t},\quad\forall t\ge0.
\eeq
Then there exists a unique solution $u(t)$ of \eqref{fctnse} and \eqref{uzero} that satisfies
$$u\in C([0,\infty),\mathcal D(A^\alpha))$$ and
\beq\label{uest}
|u(t)|_{\alpha,\sigma}\le \sqrt{2}C_0e^{-(1-\delta)t},\quad \forall t\ge t_*,
\eeq
where $t_*=6\sigma/\delta$.
Moreover, one has for all $t\ge t_*$ that
\beq\label{intAa}
\int_t^{t+1} |u(\tau)|_{\alpha+1/2,\sigma}^2d\tau
\le \frac{3C_0^2}{2(1-\delta)}e^{-2(1-\delta)t}.
\eeq
\end{proposition}
\begin{proof}
While the estimates below are formal, they can be justified by performing them at the level of the Galerkin approximation and then passing to the limit. The estimates will hold for the unique, regular solution $u(t)$.
\textbf{Part I: case $\sigma>0$.} Let $\varphi(t)$ be a function in $C^\infty(\R)$ such that
$$\varphi((-\infty,0])=\{0\},\quad
\varphi([0,t_*])=[0,\sigma],\quad
\varphi([t_*,\infty))=\{\sigma\},$$
and
$$0< \varphi'(t)< 2\sigma/t_*=\delta/3\quad \text{for all }t\in(0,t_*).$$
From equation \eqref{fctnse}, we have
\beq\label{daeu}
\ddt (A^{\alpha}e^{\varphi(t) A^{1/2}}u(t)) =A^{\alpha}e^{\varphi(t) A^{1/2}}(-Au-B(u,u)+f)
+ \varphi'(t)A^{1/2}A^{\alpha}e^{\varphi(t) A^{1/2}} u.
\eeq
Taking inner product of the equation \eqref{daeu} with $A^{\alpha}e^{\varphi(t) A^{1/2}}u(t)$ gives
\begin{align*}
&\frac12\ddt |u|_{\alpha,\varphi(t)}^2 + |A^{1/2}u|_{\alpha,\varphi(t)}^2
=\varphi'(t)\langle A^{2\alpha+1/2}e^{2\varphi(t) A^{1/2}}u,u\rangle\\
&\quad -\langle A^{\alpha}e^{\varphi(t) A^{1/2}}B(u,u),A^{\alpha}e^{\varphi(t) A^{1/2}}u\rangle+ \langle A^{\alpha-1/2}e^{\varphi(t) A^{1/2}}f,A^{\alpha+1/2}e^{\varphi(t) A^{1/2}}u\rangle.
\end{align*}
Applying the Cauchy-Schwarz inequality, then Lemma \ref{nonLem} to the second term on the right-hand side, we obtain
\begin{align*}
&\frac12\ddt |u|_{\alpha,\varphi(t)}^2 + |A^{1/2}u|_{\alpha,\varphi(t)}^2\\
& \le \varphi'(t) |u|_{\alpha+1/2,\varphi(t)}^2+K^\alpha |A^{1/2}u|_{\alpha,\varphi(t)}^2 |u|_{\alpha,\varphi(t)}
+ |f(t)|_{\alpha-1/2,\varphi(t)}|u|_{\alpha+1/2,\varphi(t)}, \\
&\le \frac\delta3 |u|_{\alpha+1/2,\varphi(t)}^2+K^\alpha |A^{1/2}u|_{\alpha,\varphi(t)}^2 |u|_{\alpha,\varphi(t)}
+ \frac3{4\delta}|f(t)|_{\alpha-1/2,\varphi(t)}^2 +\frac\delta3|u|_{\alpha+1/2,\varphi(t)}^2.
\end{align*}
This implies
\beq\label{s1}
\frac12\ddt |u|_{\alpha,\varphi(t)}^2 + \Big(1-\frac{2\delta}3 -K^\alpha |u|_{\alpha,\varphi(t)}\Big)|A^{1/2}u|_{\alpha,\varphi(t)}^2 \le \frac3{4\delta}|f(t)|_{\alpha-1/2,\sigma}^2.
\eeq
Let $T\in(0,\infty)$. Note that $|u(0)|_{\alpha,\varphi(0)}=|A^\alpha u^0|<2C_0$. Assume that
\beq\label{uT}
|u(t)|_{\alpha,\varphi(t)}\le 2C_0,\quad \forall t\in[0,T).
\eeq
Then for $t\in (0,T)$, we have from \eqref{s1} and \eqref{fta} that
\beq\label{s2}
\ddt |u|_{\alpha,\varphi(t)}^2 + 2(1-\delta)|A^{1/2}u|_{\alpha,\varphi(t)}^2 \le \frac3{2\delta}|f(t)|_{\alpha-1/2,\sigma}^2\leq \frac{3C_1^2}{2\delta} e^{-2\lambda t}.
\eeq
Applying Gronwall's inequality in \eqref{s2} yields for all $t\in(0,T)$ that
\begin{align*}
|u(t)|_{\alpha,\varphi(t)}^2
&\le e^{-2(1-\delta)t}|u^0|_{\alpha,0}^2+\frac{3C_1^2}{2\delta}e^{-2(1-\delta)t}\int_0^t e^{2(1-\delta)\tau} \cdot e^{-2\lambda\tau} d\tau\\
&\le e^{-2(1-\delta)t}|u^0|_{\alpha,0}^2+\frac{3C_1^2}{4\delta (\lambda-1+\delta)} e^{-2(1-\delta)t}\\
&= \Big(|u^0|_{\alpha,0}^2 + C_0^2\Big)e^{-2(1-\delta)t}.
\end{align*}
Combining this with condition \eqref{usmall} for the initial data, we obtain
\beqs
|u(t)|_{\alpha,\varphi(t)}^2
\le 2C_0^2e^{-2(1-\delta)t} ,
\eeqs
which gives
\beq\label{s4}
|u(t)|_{\alpha,\varphi(t)}
\le \sqrt{2} C_0 e^{-(1-\delta)t}, \quad \forall t\in(0,T).
\eeq
In particular, letting $t\to T^-$ in \eqref{s4} yields
\beqs
\lim_{t\to T^-}|u(t)|_{\alpha,\varphi(t)}
\le \sqrt{2}C_0<2C_0.
\eeqs
By the standard contradiction argument, we have that the inequality in \eqref{uT} holds for all $t>0$, and consequently, so does \eqref{s4}. Since $\varphi(t)=\sigma$ for $t\ge t_*$, the desired estimate \eqref{uest} follows \eqref{s4}.
For $t\ge t_*$, integrating \eqref{s2} from $t$ to $t+1$ gives
\begin{align*}
& 2(1-\delta)\int_t^{t+1} |A^{1/2}u(\tau)|_{\alpha,\sigma}^2d\tau
\le |u(t)|_{\alpha,\sigma}^2+\frac{3C_1^2}{2\delta}\int_t^{t+1}e^{-2\lambda \tau}d\tau\\
&\le 2C_0^2 e^{-2(1-\delta)t}+\frac{3C_1^2}{4\delta\lambda}e^{-2\lambda t}
\le C_0^2 e^{-2(1-\delta)t}\Big(2+\frac{\lambda-1+\delta}{\lambda}\Big)\\
&\le 3C_0^2 e^{-2(1-\delta)t}.
\end{align*}
Then estimate \eqref{intAa} follows.
\textbf{Part II: case $\sigma=0$.} The proof is similar to Part I without using the function $\varphi(t)$.
Here, we perform necessary calculations.
First, using Sobolev norms, we have
\beq\label{dtAa}
\frac12\ddt |A^\alpha u|^2+\Big(1-\frac\delta2-K^\alpha|A^\alpha u|\Big)|A^{\alpha+1/2}u|^2\le \frac1{2\delta}|A^{\alpha-1/2}f|^2.
\eeq
As long as $|A^\alpha u(t)|\le 2C_0=\delta/(2K^\alpha)$ in $[0,T)$ for some $T\in(0,\infty]$, we have
for $t\in(0,T)$ that
\begin{align*}
|A^\alpha u(t)|^2
&\le |A^\alpha u^0|^2 e^{-2(1-\delta)t} + \frac{C_1^2 e^{-2(1-\delta)t}}{\delta}
\int_0^t e^{-2(\lambda-\delta+1)\tau}d\tau\\
&\le \Big(C_0^2 +\frac{C_1^2}{2\delta(\lambda-\delta+1)}\Big) e^{-2(1-\delta)t}= 2C_0^2 e^{-2(1-\delta)t}.
\end{align*}
This implies $T=\infty$ and then also proves \eqref{uest}.
Now, using \eqref{uest} for the second term on the left-hand side of \eqref{dtAa}, and then integrating in time gives
\begin{align*}
2(1-\delta)\int_t^{t+1} |A^{\alpha+1/2}u|^2d\tau
&\le |A^\alpha u(t)|^2+ \frac1\delta\int_t^{t+1}|A^{\alpha-1/2}f|^2d\tau\\
&\le 2C_0^2 e^{-2(1-\delta)t} + \frac{C_1^2 e^{-2\lambda t}}{2\delta\lambda}
\le 3C_0^2 e^{-2(1-\delta)t}.
\end{align*}
This implies \eqref{intAa}, and the proof is complete.
\end{proof}
\begin{remark}
We point out that Proposition \ref{theo22} is essentially an application of Leray-Hopf energy inequality and the well-known fact (cf. \cite{DoeringTiti}) that having control on the growth of the rate of global energy dissipation, $\epsilon:=\sup_{t\geq0}\epsilon(t)$, where $\epsilon(t):=\nu\lVert{\nabla u(t)}\rVert_{L^2}^2$, can sustain exponential decay in the spectrum of the corresponding solution. The need for proving Proposition \ref{theo22} comes from the fact that we require detailed information of the decay rates of the solution and the particular effect on it from the body force, which is not immediately available from \cite{DoeringTiti}, where the case of having a non-potential force is not treated.
\end{remark}
Considering decaying forces, we assume at the moment up to Proposition \ref{theo23} that there are numbers $M_*,\kappa_0>0$ such that
\beq\label{fkappa}
|f(t)|\le M_* e^{-(1+\kappa_0)t/2},\quad \forall t\ge 0.
\eeq
We recall estimate (A.39) of \cite[Chap. II]{FMRTbook} for Leray-Hopf weak solutions (under the Basic Assumption,)
\beqs
|u(t)|^2\le e^{-t}|u_0|^2 + e^{-t}\int_0^t e^\tau |f(\tau)|^2d\tau,\quad \forall t>0.
\eeqs
It then follows from \eqref{fkappa} that
\beq\label{uenerM}
|u(t)|^2\le e^{-t}(|u_0|^2 + M_*^2/\kappa_0),\quad \forall t>0.
\eeq
By applying the Cauchy-Schwarz, Cauchy, and Poincar\'e inequalities to the last term on the right-hand side of \eqref{Lenergy}, upon simplifying we obtain
\beqs
|u(t)|^2+\int_{t_0}^t \|u(\tau)\|^2d\tau\le |u(t_0)|^2+\int_{t_0}^t |f(\tau)|^2\ d\tau,
\eeqs
for $t_0=0$ and almost all $t_0\in(0,\infty)$, and all $t\ge t_0$.
Using \eqref{uenerM} for $|u(t_0)|^2$, and, again, \eqref{fkappa} yields
\beq\label{uVest}
\int_{t_0}^{t_0+1}\|u(\tau)\|^2d\tau
\le e^{-t_0}\Big(|u_0|^2 + \frac{M_*^2}{\kappa_0}\Big) + \frac1{1+\kappa_0}M_*^2 e^{-(1+\kappa_0)t_0}
\le e^{-t_0}\Big(|u_0|^2 + \frac{2 M_*^2}{\kappa_0}\Big).
\eeq
For any $t\ge 0$, let $\{t_n\}_{n=1}^\infty$ be a sequence in $(0,\infty)$ converging to $t$ such that \eqref{uVest} holds for $t_0=t_n$. Then letting $n\to\infty$ gives
\beq\label{tt1}
\int_t^{t+1}\|u(\tau)\|^2d\tau
\le e^{-t}\Big(|u_0|^2 + \frac{2 M_*^2}{\kappa_0}\Big).
\eeq
\begin{proposition}\label{theo23}
Assume \eqref{fkappa} and, additionally, that there are $\sigma\ge 0$, $\alpha\ge 1/2$ and $\lambda_0\in(0,1)$ such that
\beq\label{falphaonly}
|f(t)|_{\alpha,\sigma}=\mathcal O(e^{-\lambda_0t})\quad\text{as }t\to\infty.
\eeq
Let $u(t)$ be a Leray-Hopf weak solution of \eqref{fctnse}.
Then for any $\delta\in (1-\lambda_0,1)$, there exists $T_*>0$
such that $u(t)$ is a regular solution of \eqref{fctnse} on $[T_*,\infty)$, and one has for all $t\ge 0$ that
\beq\label{preus0}
|u(T_*+t)|_{\alpha+1/2,\sigma} \le K^{-\alpha-1/2}e^{-(1-\delta)t},
\eeq
\beq\label{preBlt}
|B(u(T_*+t),u(T_*+t))|_{\alpha,\sigma}\le K^{-\alpha-1}e^{-2(1-\delta)t},
\eeq
where $K$ is the constant in Lemma \ref{nonLem}.
\end{proposition}
\begin{proof}
First, we note that \eqref{preBlt} is a direct consequence of \eqref{preus0}. Indeed, applying Lemma \ref{nonLem} with the use of \eqref{preus0}, we have for $t\ge 0$,
\beqs
|B(u(T_*+t),u(T_*+t))|_{\alpha,\sigma}\le K^\alpha |u(T_*+t))|_{\alpha+1/2,\sigma}^2\le K^\alpha \Big(\frac{e^{-(1-\delta)t}}{K^{\alpha+1/2}}\Big)^2,
\eeqs
which yields \eqref{preBlt}.
We focus on proving \eqref{preus0} now.
Define
\beqs
\lambda=\frac{1-\delta+\lambda_0}{2}\in(1-\delta,\lambda_0).
\eeqs
We consider each case $\sigma>0$ and $\sigma=0$ separately.
\textbf{(i) Case $\sigma>0$.}
\textit{Step 1.}
By \eqref{tt1} and \eqref{falphaonly}, there exists $t_0>0$ such that
\beqs
|A^{1/2}u(t_0)|<C_0(1/2,\delta),
\eeqs
\beqs
|f(t_0+t)|_{0,\sigma}\le C_1(1/2,\delta,\lambda)e^{-\lambda t},\quad \forall t\ge 0.
\eeqs
Applying Proposition \ref{theo22} to $u(t_0+\cdot)$, $f(t_0+\cdot)$, $\alpha=1/2$ results in
\beqs
|u(t_0+t)|_{1/2,\sigma}\le \sqrt 2 C_0(1/2,\delta) e^{-(1-\delta)t}\le K^{-1/2}e^{-(1-\delta)t},\quad \forall t\ge t_*=6\sigma/\delta.
\eeqs
Then by \eqref{als}, we have for all $t\ge t_*$ that
\begin{align}
|A^{\alpha+1/2} u(t_0+t)|
&\le \Big(\frac{2\alpha+1}{e\sigma}\Big)^{2\alpha+1} |e^{\sigma A^{1/2}} u(t_0+t)|
\le \Big(\frac{2\alpha+1}{e\sigma}\Big)^{2\alpha+1} |u(t_0+t)|_{1/2,\sigma} \notag \\
&\le \Big(\frac{2\alpha+1}{e\sigma}\Big)^{2\alpha+1} K^{-1/2} e^{-(1-\delta)t}.\label{Aaut0}
\end{align}
\textit{Step 2.}
From \eqref{Aaut0} and \eqref{falphaonly} we deduce that there is a sufficiently large $T>t_0+t_*$ so that
\beqs
|A^{\alpha+1/2} u(T)|\le C_0(\alpha+1/2,\delta),
\eeqs
\beqs
|f(T+t)|_{\alpha,\sigma}\le C_1(\alpha+1/2,\delta,\lambda)e^{-\lambda t} \quad \forall t\ge 0.
\eeqs
Applying Proposition \ref{theo22} again to $u(T+\cdot)$, $\alpha:=\alpha+1/2$, we obtain that there is $T_*>T+t_*$ such that
\beqs
|u(T_*+t)|_{\alpha+1/2,\sigma}\le \sqrt 2 C_0(\alpha+1/2,\delta)e^{-(1-\delta)t}\le \frac1{K^{\alpha+1/2}}e^{-(1-\delta)t}\quad \forall t\ge 0.
\eeqs
This yields \eqref{preus0} and completes the proof of Case $(i)$.
\textbf{(ii) Case $\sigma=0$.}
We will apply Proposition \ref{theo22} recursively to gain the exponential decay for $u(t)$ in higher Sobolev norms.
For $j\in \N$, suppose
\beq\label{intAj}
\lim_{t\to\infty}\int_t^{t+1}|A^{j/2}u(\tau)|^2d\tau=0,
\eeq
and
\beq\label{fj}
|A^{(j-1)/2}f(t)|=\mathcal O(e^{-\lambda_0 t})\quad\text{as }t\to\infty.
\eeq
Then there is $T>0$ so that
\beqs
|A^{j/2} u(T)|\le C_0(j/2,\delta),
\eeqs
\beqs
|A^{j/2-1/2}f(T+t)|\le C_1(j/2,\delta,\lambda)e^{-\lambda t} \quad \forall t\ge 0.
\eeqs
Applying Proposition \ref{theo22} to $u(T+\cdot)$, $\alpha:=j/2$, $\sigma:=0$, we obtain
\beqs
|A^{j/2}u(T+t)|\le \sqrt 2 C_0(j/2,\delta)e^{-(1-\delta)t}\le \frac1{K^{j/2}}e^{-(1-\delta)t}\quad \forall t\geq0,
\eeqs
and
\beq\label{intAj1}
\int_t^{t+1}|A^{(j+1)/2}u(\tau)|^2d\tau=\mathcal O(e^{-2(1-\delta)t})\text{ as }t\to\infty.
\eeq
Note, by \eqref{tt1}, that \eqref{intAj} holds true for $j=1$.
Let $m\in\N\cup\{0\}$ be given such that
\beq \label{mal}
\alpha \le m/2<\alpha+1/2.
\eeq
Since $\alpha\ge1/2$, condition \eqref{mal} gives $m\ge1$. Also, observe that \eqref{mal} implies $(m-1)/2<\alpha$. Hence, by \eqref{falphaonly}, condition \eqref{fj} is satisfied for $j=1,2,\ldots,m$.
Now we repeat the argument from \eqref{intAj} to \eqref{intAj1} for $j=1,2,\ldots,m$.
Particularly, when $j=m$ we obtain from \eqref{intAj1} that
\beqs
\int_t^{t+1}|A^{(m+1)/2}u(\tau)|^2d\tau=\mathcal O(e^{-2(1-\delta)t}) \text{ as }t\to\infty.
\eeqs
Since $\alpha\le m/2$, this yields
\beq\label{intAm}
\int_t^{t+1}|A^{\alpha+1/2}u(\tau)|^2d\tau=\mathcal O(e^{-2(1-\delta)t}) \text{ as }t\to\infty.
\eeq
Using \eqref{intAm} in place of \eqref{Aaut0}, we can proceed as in Step 2 of part (i) and obtain \eqref{preus0}. The proof is complete.
\end{proof}
\section{Proofs of main results}\label{pfsec}
We will use the following elementary identities: for $\beta>0$, integer $d\ge 0$, and any $t\in \R$,
\beq\label{id1}
\int_{-\infty}^t \tau^d e^{\beta \tau}\ d\tau=\frac{e^{\beta t}}{\beta}\sum_{n=0}^{d}\frac{(-1)^{d-n} d!}{n!\beta^{d-n}}t^n,
\eeq
\beq\label{id2}
\int_t^\infty \tau^d e^{-\beta \tau}\ d\tau=\frac{e^{-\beta t}}{\beta}\sum_{n=0}^{d}\frac{d!}{n!\beta^{d-n}}t^n.
\eeq
The next lemma is a building block of the construction of the polynomials $q_n(t)$'s. It summarizes and reformulates the facts
used in \cite{FS87} and \cite[Lemma 3.2]{FS91}, see also \cite{HM1}.
\begin{definition}
Let $X$ be a Banach space with its dual $X'$. Let $u(t)$ and $g(t)$ be functions in $L^1_{\rm loc}([0,\infty),X)$.
We say $g(t)$ is the $X$-valued distribution derivative of $u(t)$, and denote $g=u'$, if
\beq\label{disprimet}
\ddt \inprod{u(t),v}=\inprod{g(t),v}\text{ in the distribution sense on $(0,\infty)$}, \forall v\in X',
\eeq
where $\inprod{\cdotp,\cdotp}$ in \eqref{disprimet} denotes the usual duality pairing between an element of $X$ and $X'$.
\end{definition}
\begin{lemma}\label{polylem}
Let $(X,\|\cdot\|)$ be a Banach space. Suppose $y(t)$ is a function in $C([0,\infty),X)$ that solves the following ODE
\beqs
y'(t)+ \beta y(t) =p(t)+g(t)
\eeqs
in the $X$-valued distribution sense on $(0,\infty)$.
Here, $\beta\in \R$ is a fixed constant, $p(t)$ is an $X$-valued polynomial in $t$, and $g\in L^1_{\rm loc}([0,\infty),X)$ satisfies
\beqs
\|g(t)\|\le Me^{-\delta t} \quad \forall t\ge 0, \quad \text{for some } M,\delta>0.
\eeqs
Define $q(t)$ for $t\in \R$ by
\beq\label{qdef}
q(t)=
\begin{cases}
e^{-\beta t}\int_{-\infty}^t e^{\beta\tau }p(\tau) d\tau&\text{if }\beta >0,\\
y(0) +\int_0^\infty g(\tau)d\tau + \int_0^t p(\tau)d\tau &\text{if }\beta =0,\\
-e^{-\beta t}\int_t^\infty e^{\beta\tau }p(\tau) d\tau&\text{if }\beta <0.
\end{cases}
\eeq
Then $q(t)$ is an $X$-valued polynomial of degree at most $\deg(p)+1$ that satisfies
\beq\label{pode2}
q'(t)+\beta q(t) = p(t),\quad t\in \R,
\eeq
and the following estimates hold:
\begin{enumerate}[label={\rm (\roman*)}]
\item
If $\beta>0$ then
\beq\label{g1b1}
\|y(t)-q(t)\|\le \Big(\|y(0)-q(0)\| + \frac{M}{|\beta-\delta|}\Big)e^{-\min\{\delta,\beta\} t}, \quad t\ge 0,\text{ for } \beta\ne \delta,
\eeq
and
\beq\label{g1b1equal}
\|y(t)-q(t)\|\le (\|y(0)-q(0)\| + M t )e^{-\delta t}, \quad t\ge 0,\text{ for } \beta=\delta.
\eeq
\item If $\beta=0$ then
\beq\label{g1b2}
\|y(t)-q(t)\|\le \frac{M}{\delta}e^{-\delta t},\quad t\ge 0.
\eeq
\item If $\beta<0$ and
\beq\label{yexpdec}
\lim_{t\to\infty} (e^{\beta t}y(t))=0,
\eeq
then
\beq\label{g1b3}
\|y(t)-q(t)\|\le \frac{M}{|\beta|+\delta}e^{-\delta t}, \quad t\ge 0.
\eeq
\end{enumerate}
\end{lemma}
\begin{proof}
The fact that $q(t)$ is a polynomial in $t$ follows the identities \eqref{id1} and\eqref{id2}.
The equation \eqref{pode2} obviously results from the definition \eqref{qdef} of $q(t)$. It remains to prove estimates \eqref{g1b1}, \eqref{g1b2} and \eqref{g1b3}.
Let $z(t)=y(t)-q(t)$, then
\beqs
z'(t)+\beta z(t)=g(t)\text{ in the $X$-valued distribution sense on } (0,\infty).
\eeqs
Multiplying this equation by $e^{\beta t}$ yields
\beq\label{eaz}
(e^{\beta t} z(t))'=e^{\beta t} g(t) \text{ in the $X$-valued distribution sense on } (0,\infty).
\eeq
For $t_0\ge 0$, it follows \eqref{eaz} and \cite[Ch. III, Lemma 1.1]{TemamAMSbook}
that
\beq\label{prevoc}
e^{\beta t} z(t)=\xi+\int_{t_0}^t e^{\beta \tau}g(\tau)d\tau,
\eeq
for some $\xi\in X$ and almost all $t\in(t_0,\infty)$.
Since $e^{\beta t} z(t)$ is continuous on $[0,\infty)$ and $e^{\beta t} g(t)\in L^1_{\rm loc}([0,\infty))$, we have $\xi=e^{\beta t_0} z(t_0)$ and equation \eqref{prevoc} holds for all $t\ge t_0$. Hence, we obtain the standard variation of constant formula
\beq\label{voc}
z(t)=e^{-\beta(t-t_0)} z(t_0)+e^{-\beta t} \int_{t_0}^t e^{\beta\tau}g(\tau)d\tau \quad \forall t\ge t_0.
\eeq
(i) Case $\beta>0$.
Setting $t_0=0$ in \eqref{voc}, we estimate
\begin{align*}
\|z(t)\|
&\le e^{-\beta t}\|z(0)\|+ e^{-\beta t} \int_0^t e^{\beta \tau}\|g(\tau)\| d\tau
\le e^{-\beta t}\|z(0)\| + e^{-\beta t} \int_0^t e^{\beta\tau} Me^{-\delta\tau} d\tau.
\end{align*}
Since the last term is $M(e^{-\delta t}-e^{-\beta t})/(\beta-\delta)$ if $\beta\ne \delta$, and is $M t e^{-\delta t}$ if $\beta=\delta$, we easily obtain \eqref{g1b1} and \eqref{g1b1equal}.
(ii) Case $\beta=0$. Note from \eqref{qdef} that $z(0)=y(0)-q(0)=-\int_0^\infty g(\tau)d\tau$.
Letting $t_0=0$ in \eqref{prevoc} gives
\beqs
z(t)=z(0) +\int_0^t g(\tau)d\tau = - \int_t^\infty g(\tau)d\tau.
\eeqs
Hence
\beqs
\|z(t)\|\le \int_t^\infty \|g(\tau)\|d\tau \le \int_t^\infty Me^{-\delta \tau}d\tau=\frac{M}\delta e^{-\delta t},
\eeqs
which proves \eqref{g1b2}.
(iii) Case $\beta<0$. By \eqref{yexpdec} and the fact $q(t)$ is a polynomial, we have $e^{\beta t}z(t)\to 0$ as $t\to\infty$.
Then letting $t\to\infty$ in \eqref{prevoc} and setting $t_0=t$ yield
\beqs
z(t)=-e^{-\beta t} \int_t^\infty e^{\beta\tau} g(\tau) d\tau.
\eeqs
It follows that
\beqs
\|z(t)\|\le e^{-\beta t} \int_t^\infty M e^{(\beta-\delta)\tau} d\tau = e^{-\beta t} \frac{Me^{(\beta-\delta)t}}{-\beta+\delta}
=\frac{M}{|\beta|+\delta}e^{-\delta t},
\eeqs
which proves \eqref{g1b3}.
\end{proof}
The remainder of this paper is focused on the proofs of main results, and will use the following notation.
\begin{notation} If $n\in\sigma (A)$, we define $R_n$ to be the orthogonal projection in $H$ on the eigenspace of $A$ corresponding to $n$. In case $n\notin \sigma (A)$, set $R_n=0$.
For $n\in \N$, define $P_n=R_1+R_2+ \cdots +R_n$. Note that each vector space $P_nH$ is finite dimensional.
\end{notation}
\subsection{Proof of Theorem \ref{mainthm}}
We start by obtaining some additional properties for the force $f(t)$ and solution $u(t)$ which we will make use of later.
By the expansion \eqref{forcexpand} of $f(t)$ in $E^{\infty,\sigma_0}$, for each $N\in\N$ and $\alpha\geq0$, there exists a number $\delta_{N,\alpha}\in(0,1)$ such that
\beq\label{Fcond}
\Big|f(t)-\sum_{n=1}^N f_n(t)e^{-nt}\Big|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+\delta_{N,\alpha})t})\quad \text {as }t\to\infty.
\eeq
Observe that we have the following immediate consequences:
\begin{enumerate}[label={\rm (\alph*)}]
\item The relation \eqref{Fcond} implies for each $\alpha\ge 0$ that $f(t)$ belongs to $G_{\alpha,\sigma_0}$ for $t$ large.
\item Note that when $N=1$, the function $f(t)$ itself satisfies
\beqs
|f(t)- f_1(t) e^{-t}|_{\alpha,\sigma_0}=\mathcal O(e^{-(1+\delta_{1,\alpha}) t} ).
\eeqs
Since $f_1(t)$ is a polynomial, it follows that
\beq\label{fsure}
|f(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-\lambda t}),\quad \forall \lambda\in (0,1),\quad\forall \alpha\ge 0.
\eeq
Consequently, for any $\varep>0$, $\alpha\ge 0$, and $\lambda\in(0,1)$, applying \eqref{fsure} with $(\lambda+1)/2$ replacing $\lambda$, it follows that there is $T>0$ such that
\beq\label{fdecay}
|f(T+t)|_{\alpha,\sigma_0}\le \varep e^{-\lambda t}\quad \forall t\ge 0.
\eeq
\item Combining \eqref{fsure} for $\alpha=0$, with the Basic Assumption, we assume, without loss of generality, for each $\lambda\in (0,1)$ that
\beq\label{fh}
|f(t)|\le M_\lambda e^{-\lambda t},\quad \forall t\ge 0, \text{ for some }M_\lambda>0.
\eeq
\end{enumerate}
For the solution $u(t)$, we summarize the key estimates in section \ref{Gdecay} into the following.
\textbf{Claim.} For any $\alpha\geq0$ and $\delta\in (0,1)$, there exists a positive number $T_*>0$
such that $u(t)$ is a regular solution on $[T_*,\infty)$, and one has for $t\ge 0$ that
\beq\label{us0}
|u(T_*+t)|_{\alpha+1/2,\sigma_0} \le e^{-(1-\delta) t},
\eeq
\beq\label{Blt}
|B(u(T_*+t),u(T_*+t))|_{\alpha,\sigma_0}\le e^{-2(1-\delta) t}.
\eeq
\textit{Proof of {Claim}.}
We apply Proposition \ref{theo23}. By \eqref{fdecay}, we have that \eqref{falphaonly} holds for $\sigma=\sigma_0$, any $\alpha\ge 1/2$ and any $\lambda_0\in(0,1)$.
Also, \eqref{fkappa} is satisfied because of \eqref{fh}.
Therefore \eqref{preus0} and \eqref{preBlt} hold for $\sigma=\sigma_0$, any $\alpha\ge 1/2$ and any $\delta\in(0,1)$. These directly yield \eqref{us0} and \eqref{Blt}.
\bigskip
Returning to the main proof, it suffices to prove that there exist polynomials $q_n$'s for all $n\ge 1$ such that for each $N\ge 1$ the following properties ($\mathcal H1$), ($\mathcal H2$), and ($\mathcal H3$) hold true:
\begin{enumerate}
\item[($\mathcal H1$)] $q_N\in \mathcal P^{\infty,\sigma_0}$.
\item[($\mathcal H2$)] For $\alpha\ge 1/2$,
\beq\label{remdelta}
\Big |u(t)-\sum_{n=1}^N q_n(t) e^{-nt}\Big |_{\alpha,\sigma_0} =\mathcal O(e^{-(N+\varep)t} )\quad\text{as } t\to\infty,\quad\forall\varep\in(0,\delta_{N,\alpha}^*),
\eeq
where the numbers $\delta_{n,\alpha}^*$'s, for $\alpha\ge 1/2$, are defined recursively by
\beqs
\delta_{n,\alpha}^*=\begin{cases}
\delta_{1,\alpha},&\text{for }n=1,\\
\min\{\delta_{n,\alpha},\delta^*_{n-1,\alpha+1/2}\},&\text{for }n\ge 2.
\end{cases}
\eeqs
\item[($\mathcal H3$)] The ODE \eqref{unODE} holds in $E^{\infty,\sigma_0}$ for $n=N$.
\end{enumerate}
\bigskip
We prove these statements by constructing the polynomials $q_N(t)$'s recursively.
\medskip
\noindent {\bf Base case: $N=1$.}
Let $k\geq1$. By taking $v\in R_kH$ in the the weak formulation \eqref{varform}, we have
\beq\label{rku}
\ddt R_ku + k R_k u = R_k (f(t) - B(u(t),u(t)))
\eeq
in the $R_kH$-valued distribution sense on $(0,\infty)$.
(Since $R_kH$ is finite dimensional, ``$R_kH$-valued distribution sense," is simply the same as ``distribution sense''.)
Let $w_0(t)=e^t u(t)$ and $w_{0,k}(t)=R_k w_0(t)$. By virtue of the $H_{\rm w}$-continuity of $u(t)$ (see \eqref{lh:wksol}), we have $w_{0,k}\in C([0,\infty),R_kH)$. It follows from \eqref{rku} that
\beq\label{wj}
\ddt w_{0,k} + (k-1) w_{0,k} = R_k f_1 + R_kH_0(t),
\eeq
where
\beq\label{H0def}
H_0(t)=e^t (f(t)-F_1(t) - B(u(t),u(t))),
\eeq
and $F_1$ is defined in \eqref{uF}. Note that $R_k f_1(t)$ is an $R_kH$-valued polynomial in $t$.
Let $\alpha\ge 1/2$ be fixed.
Using \eqref{Fcond} for $N=1$ and applying \eqref{Blt} with $\delta=(1-\delta_{1,\alpha})/4$, there are $T_0>0$ and $D_0\ge 1$ such that for $t\ge 0$,
\beq\label{ch1}
e^t |f(T_0+t)-F_1(T_0+t)|_{\alpha,\sigma_0}\le D_0e^{-\delta_{1,\alpha} t},
\eeq
\beq\label{ch2}
e^t |B(u(T_0+t),u(T_0+t))|_{\alpha,\sigma_0}\le e^{(2\delta-1)t}=e^{-(1+\delta_{1,\alpha})t/2}\le e^{-\delta_{1,\alpha} t}.
\eeq
Then, by setting $D_1=D_0+1$, we have
\beq\label{ch3}
|H_0(T_0+t)|_{\alpha,\sigma_0} \le D_1e^{-\delta_{1,\alpha} t},\quad\forall t\ge0.
\eeq
We will now identify the components of the desired polynomial, $q_1(t)$, belonging to each eigenspace $R_kH$.
\medskip
\noindent\textit{\underline{Case: $k=1$}.}
Applying Lemma \ref{polylem}(ii) to equation \eqref{wj} with $X=R_1H$, $\|\cdot\|=|\cdot|_{\alpha,\sigma_0}$, $\beta=0$,
$$y(t)=w_{0,1}(T_0+t),\quad p(t)=R_1f_1(T_0+t), \quad g(t)=R_1H_0(T_0+t),$$
we infer that there is an $R_1H$-valued polynomial $q_{1,1}(t)$ such that for any $t\ge 0$
\beqs
|w_{0,1}(T_0+t)-q_{1,1}(t)|_{\alpha,\sigma_0} \le \frac{D_1}{\delta_{1,\alpha}}e^{-\delta_{1,\alpha} t},
\eeqs
thus,
\beq\label{q11}
|R_1 w_0(t)-q_{1,1}(t-T_0)|_{\alpha,\sigma_0} \le \frac{D_1e^{\delta_{1,\alpha} T_0}}{\delta_{1,\alpha}} e^{-\delta_{1,\alpha} t},\quad\forall t\geq T_0.
\eeq
In fact,
\beq\label{q11def}
q_{1,1}(t)= \xi_1+\int_0^t R_1f_1(\tau+T_0)d\tau\quad \text{for some } \xi_1\in R_1H.
\eeq
\medskip
\noindent\textit{\underline{Case: $k\geq2$}.} We apply Lemma \ref{polylem}(i) to equation \eqref{wj} with
$$y(t)=w_{0,k}(T_0+t),\quad p(t)=R_kf_1(T_0+t),\quad g(t)=R_kH_0(T_0+t),$$
where $\beta=k-1>\delta_{1,\alpha}$, and the norm $\|\cdot\|$ being $|\cdot|_{\alpha,\sigma_0}$ on the space $X=R_kH$.
In particular, there is an $R_kH$-valued polynomial, $q_{1,k}(t)$ such that for any $t\ge 0$
\beqs
|w_{0,k}(T_0+t)- q_{1,k}(t)|_{\alpha,\sigma_0} \le e^{-\delta_{1,\alpha} t}\Big(|w_{0,k}(T_0)|_{\alpha,\sigma_0} +|q_{1,k}(0)|_{\alpha,\sigma_0} + \frac {D_1}{k-1-\delta_{1,\alpha}}\Big),
\eeqs
which implies for all $t\ge T_0$ that
\beq\label{q1j}
|w_{0,k}(t)- q_{1,k}(t-T_0)|_{\alpha,\sigma_0} \le e^{-\delta_{1,\alpha} (t-T_0)}\Big(|w_k(T_0)|_{\alpha,\sigma_0} +|q_{1,k}(0)|_{\alpha,\sigma_0} + \frac {D_1}{k-1-\delta_{1,\alpha}}\Big).
\eeq
In fact, for $k\ge 2$
\beq\label{q1k}
q_{1,k}(t)=-e^{(1-k)t}\int_{-\infty}^t e^{(k-1)\tau}R_kf_1(T_0+\tau)d\tau.
\eeq
\noindent{\bf Polynomial $q_1(t)$.}
Define
\beq\label{q1def}
q_1(t)=\sum_{k=1}^\infty q_{1,k}(t-T_0),\quad t\in\R.
\eeq
We next prove that $q_1\in\mathcal{P}^{\infty,\sigma_0}$. Write
\beqs
f_1(t+T_0)=\sum_{d=0}^m a_d t^d,\quad \text{for some }a_d\in E^{\infty,\sigma_0}.
\eeqs
Clearly, by \eqref{q11def}, $R_1q_1(t+T_0)=q_{1,1}(t)$ is a $\mathcal V$-valued polynomial, and hence,
\beq\label{R1q1poly}
\text{the mapping }t\mapsto R_1q_1(t+T_0) \text{ belongs to } \mathcal{P}^{\infty,\sigma_0}.
\eeq
We consider the remaining part $(I-R_1)q_1(t+T_0)$.
Using the integral formula \eqref{id1},
\begin{align*}
(I-R_1)q_1(t+T_0)
&=\sum_{k=2}^\infty q_{1,k}(t)
=\sum_{k=2}^\infty -e^{(1-k)t} \int_{-\infty}^t e^{(k-1)\tau}\Big(\sum_{d=0}^m R_k a_d \tau^d\Big) d\tau\\
&=\sum_{k=2}^\infty \frac{1}{k-1}\sum_{d=0}^m \sum_{n=0}^{d}\frac{(-1)^{d-n} d!}{n!(k-1)^{d-n}}t^n R_k a_d\\
&=\sum_{k=2}^\infty \frac{1}{k-1}\sum_{n=0}^d\Big(\sum_{d=n}^{m}\frac{(-1)^{d-n} d!}{n!(k-1)^{d-n}}R_k a_d\Big)t^n.
\end{align*}
Thus,
\beq\label{IRq1}
(I-R_1)q_1(t+T_0)=\sum_{n=0}^d b_nt^n,
\eeq
where
the coefficient $b_n$, for $0\le n\le d$, is
\begin{align*}
b_n=\sum_{k=2}^\infty \frac{1}{k-1}\Big(\sum_{d=n}^{m}\frac{(-1)^{d-n} d!}{n!(k-1)^{d-n}}R_k a_d\Big).
\end{align*}
For any $\mu\ge 0$, we have
\begin{align*}
|b_n|_{\mu+1,\sigma_0}^2=|Ab_n|_{\mu,\sigma_0}^2
&=\sum_{k=2}^\infty \Big|\frac{1}{k-1}\sum_{d=n}^m \frac{(-1)^{d-n} d!}{n!(k-1)^{d-n}} k \cdot R_k a_d\Big |_{\mu,\sigma_0}^2\\
&\le \sum_{k=2}^\infty \frac{k^2}{(k-1)^2} \Big(\sum_{d=n}^m \frac{ m!}{n!} |R_k a_d|_{\mu,\sigma_0}\Big)^2\\
&\le 4\sum_{k=2}^\infty \frac{ (m!)^2 (m-n+1)^2 }{(n!)^2} |R_ka_d|_{\mu,\sigma_0}^2.
\end{align*}
(Above, we simply used $k/(k-1)\le 2$ for the last inequality.)
Thus,
\beq\label{Abn}
|b_n|_{\mu+1,\sigma_0}^2
\le \frac{ 4(m!)^2 (m-n+1)^2 }{(n!)^2} |(I-R_1)a_d|_{\mu,\sigma_0}^2<\infty.
\eeq
Therefore, $b_n\in E^{\infty,\sigma_0}$ for all $0\le n\le d$,
and, by \eqref{IRq1}, the mapping $t\mapsto (I-R_1)q_1(t+T_0)$ belongs to $\mathcal{P}^{\infty,\sigma_0}$.
This, together with \eqref{R1q1poly}, implies that $t\mapsto q_1(t+T_0)$ belongs to $\mathcal{P}^{\infty,\sigma_0}$ and, ultimately, that $t\mapsto q_1(t)$ belongs to $\mathcal{P}^{\infty,\sigma_0}$, as desired.
\noindent{\bf Remainder estimate.} We estimate $|u(t)-q_1(t)e^{-t}|_{\alpha,\sigma_0}$ now.
Firstly, inequality \eqref{q11} yields
\beq\label{R1wq1}
|R_1(w_0(t)-q_1(t))|_{\alpha,\sigma_0}=\mathcal O( e^{-\delta_{1,\alpha} t}).
\eeq
Secondly, we have from \eqref{q1j} that
\begin{align*}
&\sum_{k= 2}^\infty |R_k(w_0(t)-q_1(t))|_{\alpha,\sigma_0}^2 \\
&\le 3 e^{2\delta_{1,\alpha} T_0} e^{-2\delta_{1,\alpha} t}\sum_{k=2}^\infty\Big (|w_{0,k}(T_0)|_{\alpha,\sigma_0}^2 +|R_kq_1(T_0)|_{\alpha,\sigma_0}^2+\frac {D_1^2}{(k-1-\delta_{1,\alpha})^2}\Big)\\
&\le D_2^2 e^{-2\delta_{1,\alpha} t}
\end{align*}
for all $t\ge T_0$, where
\beqs
D_2^2 =3 e^{2\delta_{1,\alpha} T_0}\Big\{|(I-R_1)w_0(T_0)|_{\alpha,\sigma_0}^2 + |(I-R_1)q_1(T_0)|_{\alpha,\sigma_0}^2+D_1^2\sum_{k=2}^\infty \frac {1 }{(k-1-\delta_{1,\alpha})^2})\Big\}<\infty.
\eeqs
This implies
\beq\label{wq1}
|(I-R_1)(w_0(t)-q_1(t))|_{\alpha,\sigma_0}\le D_2 e^{-\delta_{1,\alpha} t},\quad \forall t\ge T_0.
\eeq
Combining \eqref{R1wq1} with \eqref{wq1} gives
\beqs
|w_0(t)-q_1 (t)|_{\alpha,\sigma_0}= \mathcal O (e^{-\delta_{1,\alpha} t}),
\eeqs
and consequently,
\beq\label{rem1}
|u(t)-q_1 (t) e^{-t}|_{\alpha,\sigma_0}= \mathcal O (e^{-(1+\delta_{1,\alpha})t}).
\eeq
Thanks to \eqref{rem1}, the polynomial $q_1(t)$ is independent of $\alpha$. Hence the same $q_1$ satisfies \eqref{rem1} for all $\alpha\ge 1/2$, which proves ($\mathcal H2$) for $N=1$.
This proves \eqref{remdelta} for $N=1$.
\noindent{\bf Establishing the ODE \eqref{u1ODE}.}
By \eqref{pode2} in Lemma \ref{polylem}, the polynomial $q_1(t)$ satisfies
\beq\label{pode3}
\ddt R_k q_1(T_0+t)+(k-1)R_kq_1(T_0+t)=R_kf_1(T_0+t),\quad \forall k\ge 1,\quad \forall t\in\R.
\eeq
For each $\mu\ge 0$, we have $Aq_1(T_0+t)$ and $f_1(T_0+t)$ belong to $G_{\mu,\sigma_0}$. Hence,
we can sum over $k$ in \eqref{pode3} and obtain
\beqs
\ddt q_1(t)+(A-1)q_1(t)=f_1(t) \quad\text{ in } G_{\mu,\sigma_0},\quad \forall t\in\R,
\eeqs
which implies that the differential equation \eqref{u1ODE} holds in $E^{\infty,\sigma_0}$.
Therefore, $q_1$ satisfies ($\mathcal H1$), ($\mathcal H2$), and ($\mathcal H3$) for $N=1$.
\medskip
\noindent{\bf Recursive step.} Let $N\ge1$. Suppose that there already exist $q_1,q_2,\ldots,q_N\in \mathcal P^{\infty,\sigma_0}$ that satisfy ($\mathcal H2$),
and the ODE \eqref{unODE} holds in $E^{\infty,\sigma_0}$ for each $n=1,2,\ldots,N$.
We will construct a polynomial $q_{N+1}(t)$ that satisfies ($\mathcal H1$), ($\mathcal H2$), and ($\mathcal H3$) with $N+1$ replacing $N$.
Let $\alpha\ge 1/2$ be given and $\varep_*$ be arbitrary in $(0,\delta_{N+1,\alpha}^*)$.
Define
$$\bar u_N=\sum_{n=1}^N u_n\quad\text{and}\quad v_N=u-\bar u_N.$$
Assumption ($\mathcal H2$) particularly yields
\beq\label{vNrate}
| v_N(t)|_{\alpha+1/2,\sigma_0}=\mathcal O(e^{-(N+\varep)t}),\quad \forall \varep\in(0,\delta_{N,\alpha+1/2}^*).
\eeq
Subtracting \eqref{unODE} for $n=1,2,\ldots,N$ from \eqref{fctnse}, we have
\beq\label{uminusuN}
\frac d{dt}v_N+Av_N +B(u,u)- \sum_{m+j\le N} B(u_m,u_j)=f-\sum_{n=1}^N F_n,
\eeq
where the functions $F_n$'s are defined in \eqref{uF}.
We reformulate equation \eqref{uminusuN} as
\beq\label{vNeq}
\frac d{dt}v_N+Av_N +\sum_{m+j=N+1} B(u_m,u_j)=F_{N+1}+h_N,
\eeq
where
\begin{align*}
h_N&=-B(u,u)+\sum_{\stackrel{1\le m,j\le N}{m+j\le N+1}} B(u_m,u_j)+f-\sum_{n=1}^{N+1} F_n\\
&=-\Big\{B(u,u)-B(\bar u_N,\bar u_N)\Big\}- \Big\{B(\bar u_N,\bar u_N)-\sum_{\stackrel{1\le m,j\le N}{m+j\le N+1}} B(u_m,u_j)\Big\}+\Big\{f-\sum_{n=1}^{N+1} F_n\Big\}.
\end{align*}
With this way of grouping, we rewrite $h_N$ as
\beq
\label{hdef}
h_N=-B(v_N,u)-B(\bar u_N,v_N)-\sum_{\stackrel{1\le m,j\le N}{m+j\ge N+2}} B(u_m,u_j)+\tilde F_{N+1},
\eeq
where
\beqs
\tilde F_{N+1}(t)=f(t)-\sum_{n=1}^{N+1} F_n(t).
\eeqs
Note in case $N=1$ that neither of the following terms
$$ \sum_{m+j\le N} B(u_m,u_j)\text{ in \eqref{uminusuN}
nor }\sum_{\stackrel{1\le m,j\le N}{m+j\ge N+2}} B(u_m,u_j)\text{ in \eqref{hdef} will appear.}$$
\noindent{\bf Estimate of $h_N(t)$.}
By \eqref{Fcond} and Remark \ref{betterem}, we have
\beq \label{Ftil}
|\tilde F_{N+1}(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+1+\delta_{N+1,\alpha})t})
=\mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
It is also obvious that
\beq \label{Bmj}
\sum_{\stackrel{1\le m,j\le N}{m+j\ge N+2}} |B(u_m,u_j)|_{\alpha,\sigma_0}
= \sum_{\stackrel{1\le m,j\le N}{m+j\ge N+2}} e^{-(m+j)t}|B(q_m,q_j)|_{\alpha,\sigma_0}
=\mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
Take $\varep\in(\varep_*,\delta_{N+1,\alpha}^*)\subset (0,\delta_{N,\alpha+1/2}^*)$ in \eqref{vNrate}, and set $\delta=\varep-\varep_*\in(0,1)$.
Then we have from the definition of $u_n(t)$ and \eqref{us0} that
\beq \label{ubu}
|\bar u_N(t)|_{\alpha+1/2,\sigma_0},|u(t)|_{\alpha+1/2,\sigma_0}=\mathcal O(e^{-(1-\delta)t}).
\eeq
By Lemma \ref{nonLem} and estimates \eqref{vNrate}, \eqref{ubu}, it follows that
\beq \label{BvNu}
|B(v_N,u)|_{\alpha,\sigma_0},|B(\bar u_N,v_N)|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+\varep+1-\delta)t})=\mathcal O(e^{-(N+1+\varep_*)t}).\eeq
Therefore, by \eqref{hdef}, \eqref{Ftil}, \eqref{Bmj} and \eqref{BvNu},
\beq\label{hNo}
|h_N(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
{\flushleft{\bf Construction of $q_{N+1}(t)$.}}
Using the weak formulation of \eqref{vNeq}, which is similar to \eqref{varform}, and then taking the test function, $v$, to be in $R_kH$, we obtain
\beq\label{vNeq1}
\frac d{dt}R_kv_N+kR_k v_N +\sum_{m+j=N+1} R_k B(u_m,u_j)=R_k F_{N+1}+ R_k h_N \text{ on } (0,\infty).
\eeq
Let $w_{N}(t)=e^{(N+1)t} v_N(t)$ and $w_{N,k}=R_k w_N(t)$.
Using \eqref{vNeq1}, we write the ODE for $w_{N,k}$ as
\beq\label{wNeq}
\frac d{dt}w_{N,k}+(k-(N+1))w_{N,k}=\Big(R_k{f}_{N+1}-\sum_{m+j=N+1} R_k B(q_m,q_j)\Big)+ H_{N,k},
\eeq
with
$H_{N,k}(t)= e^{(N+1)t}R_k h_N(t)$.
Note from \eqref{hNo} that
\beqs
|H_{N,k}(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-\varep_* t}).
\eeqs
Then there exist $T_N>0$ and $D_3>0$ such that
\beq\label{HNkD3}
|H_{N,k}(T_N+t)|_{\alpha,\sigma_0}\le D_3 e^{-\varep_* t},\quad \forall t\geq0.
\eeq
By the first property in \eqref{lh:wksol}, each $w_{N,k}(t)$ is continuous from $[0,\infty)$ to $R_kH$.
We apply Lemma \ref{polylem} to equation \eqref{wNeq} on $(T_N,\infty)$ with space $X=R_kH$, norm $\|\cdot\|=|\cdot|_{\alpha,\sigma_0}$,
solution $y(t)=w_{N,k}(T_N+t)$, constant $\beta=k-(N+1)$, polynomial
\begin{align*}
p(t)&=R_k{f}_{N+1}(T_N+t)-\sum_{m+j=N+1} R_k B(q_m(T_N+t),q_j(T_N+t)),
\end{align*}
and function $g(t)=H_{N,k}(T_N+t)$ which satisfies the estimate \eqref{HNkD3}.
\medskip
\noindent\textit{\underline{Case $k=N+1$}.}
We have $\beta=0$, then Lemma \ref{polylem}(ii) implies that there is a polynomial $q_{N+1,N+1}(t)$ valued in $R_{N+1}H$ such that
\beqs
|w_{N,N+1}(T_N+t)- q_{N+1,N+1}(t)|_{\alpha,\sigma_0}= \mathcal O (e^{-\varep_* t}).
\eeqs
Thus,
\beq \label{form0}
|R_{N+1}w_{N}(t)- q_{N+1,N+1}(t-T_N)|_{\alpha,\sigma_0}= \mathcal O (e^{-\varep_* t}).
\eeq
\medskip
\noindent\textit{\underline{Case $k\le N$}.} Note that $\beta<0$ and by \eqref{vNrate} we have
$$\lim_{t\to\infty}( e^{\beta t}y(t) ) = \lim_{t\to\infty}e^{\beta (t-T_N)} w_{N,k}(t)= e^{-\beta T_N}\lim_{t\to\infty}e^{kt}R_kv_N(t)=0.$$
Then, by applying Lemma \ref{polylem}(iii), there is a polynomial $q_{N+1,k}(t)$ valued in $R_kH$ such that
\beqs
|w_{N,k}(T_N+t)- q_{N+1,k}(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-\varep_* t}),
\eeqs
hence,
\beq\label{form1}
|R_k w_{N}(t)- q_{N+1,k}(t-T_N)|_{\alpha,\sigma_0}=\mathcal O(e^{-\varep_* t}).
\eeq
\medskip
\noindent\textit{\underline{Case $k\ge N+2$}.}
Similarly, $\beta=k-N-1>\varep_*$ and, by using Lemma \ref{polylem}(i), there is a polynomial $q_{N+1,k}(t)$ valued in $R_kH$ such that for $t\ge 0$,
\beqs
|w_{N,k}(T_N+t)
- q_{N+1,k}(t)|_{\alpha,\sigma_0}
\le \Big(|R_k v_N(T_N)|_{\alpha,\sigma_0} + |q_{N+1,k}(0)|_{\alpha,\sigma_0} + \frac{D_3}{k-(N+1)-\varep_*}\Big) e^{-\varep_* t}.
\eeqs
Thus
\begin{multline}\label{form2}
|R_kw_{N}(t)
- q_{N+1,k}(t-T_N)|_{\alpha,\sigma_0} \\
\le e^{\varep_* T_N} \Big(|R_k v_N(T_N)|_{\alpha,\sigma_0} + |q_{N+1,k}(0)|_{\alpha,\sigma_0} + \frac{D_3}{k-(N+1)-\varep_*}\Big) e^{-\varep_* t},
\quad\forall t\ge T_N.
\end{multline}
\medskip
We define $$q_{N+1}(t)=\sum_{k=1}^\infty q_{N+1,k}(t-T_N), \quad t\in \R.$$
It follows from \eqref{forcexpand} that $f_{N+1}\in \mathcal{P}^{\infty,\sigma_0}$, and from the recursive assumptions that
$q_m, q_j\in \mathcal{P}^{\infty,\sigma_0}$ for $1\le m,j\le N$; then by \eqref{BEE}, we have $f_{N+1}-\sum_{m+j=N+1} B(q_m,q_j)\in \mathcal{P}^{\infty,\sigma_0}$. Following the same proof that shows $q_1\in\mathcal{P}^{\infty,\sigma_0}$, we argue similarly to show that $q_{N+1}\in\mathcal{P}^{\infty,\sigma_0}$.
\medskip
\noindent{\bf Estimate of $v_{N+1}(t)$.}
From \eqref{form0} and \eqref{form1}, we immediately have
\beq\label{PNw}
|P_{N+1}(w_N(t)-q_{N+1}(t))|_{\alpha,\sigma_0} = \mathcal O(e^{-\varep_* t}).
\eeq
Squaring \eqref{form2} and summing over $k\ge N+2$, we obtain for $t\ge T_N$ that
\begin{align*}
&|(I-P_{N+1})(w_N(t)-q_{N+1}(t))|_{\alpha,\sigma_0}^2 = \sum_{k=N+2}^\infty |R_k (w_N(t)-q_{N+1}(t))|_{\alpha,\sigma_0}^2 \\
&\le 3 e^{2\varep_* T_N} \Big(\sum_{k=N+2}^\infty|R_k v_N(T_N)|_{\alpha,\sigma_0}^2 + \sum_{k=N+2}^\infty|R_kq_{N+1}(T_N)|_{\alpha,\sigma_0}^2 \\
&\quad + \sum_{k=N+2}^\infty\frac{D_3^2}{(k-(N+1)-\varep_*)^2}\Big)e^{-2\varep_* t}.
\end{align*}
Since the last three sums are convergent, we deduce
\beq\label{QNw}
|(I-P_{N+1})(w_N(t)-q_{N+1}(t))|_{\alpha,\sigma_0} = \mathcal O(e^{-\varep_* t}).
\eeq
From \eqref{PNw} and \eqref{QNw}, we have
\beq\label{wNqN1}
|w_N(t)-q_{N+1}(t)|_{\alpha,\sigma_0}= \mathcal O(e^{-\varep_*t}).
\eeq
Therefore,
\beq \label{vN1ep}
|v_{N+1}(t)|_{\alpha,\sigma_0}
=|v_N(t)-e^{-(N+1)t} q_{N+1}(t)|_{\alpha,\sigma_0}= \mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
Thanks to \eqref{wNqN1}, the polynomial $q_{N+1}(t)$ is independent of $\alpha$ and $\varep_*$. Therefore, \eqref{vN1ep} holds for any $\alpha\ge1/2$ and $\varep_*\in(0,\delta_{N+1}^*)$, which proves ($\mathcal H2$) with $N+1$ replacing $N$.
\medskip
\noindent{\bf Establishing the ODE \eqref{unODE} for $n=N+1$.}
By our construction of the polynomials $q_{N+1,k}(t)$ above, and
by \eqref{pode2} in Lemma \ref{polylem}, the polynomial $q_{N+1}(t)$ satisfies
\begin{align*}
&\ddt R_k q_{N+1}(T_N+t)+(k-(N+1))R_kq_{N+1}(T_N+t) \\
&\quad =R_kf_{N+1}(T_N+t)- \sum_{m+j=N+1} R_k B(q_m(T_N+t),q_j(T_N+t))
\quad \forall k\ge 1,\quad \forall t\in\R.
\end{align*}
This yields, for each $k\ge 1$,
\beq\label{Rkueq}
\ddt R_k u_{N+1}(t)+AR_ku_{N+1}(t) + \sum_{m+j=N+1} R_k B(u_m(t),u_j(t)) =R_kF_{N+1}(t),\quad \forall t\in\R.
\eeq
For any $\mu\ge 0$, since $Au_{N+1}(t)$, $\sum_{m+j=N+1} B(u_m(t),u_j(t))$, $F_{N+1}(t)$ each is a $G_{\mu,\sigma_0}$-valued polynomial, then summing up equation \eqref{Rkueq} in $k$ gives
\beqs
\ddt u_{N+1}+Au_{N+1}+ \sum_{m+j=N+1} B(u_m,u_j)=F_{N+1} \quad \text{in } G_{\mu,\sigma_0}.
\eeqs
Thus, the ODE \eqref{unODE} holds in $E^{\infty,\sigma_0}$ for $n=N+1$.
We have established the existence of the desired polynomial, $q_{N+1}(t)$, which completes the recursive step, and hence, the proof of Theorem \ref{mainthm}.
\qed
\begin{remark}
By using the extra information \eqref{fep} in Remark \ref{betterem}, we can prove directly the remainder estimate \eqref{remdelta} for any $\varep\in(0,1)$, which, in fact, is expected by \eqref{uqa}.
Nonetheless, the above proof with specific $\varep\in(0,\delta^*_{N,\alpha})$ is more flexible and will be easily adapted in section \ref{sec43} below to serve the proof of Theorem \ref{finitetheo}.
\end{remark}
\subsection{Proof of Corollary \ref{Vcor}}\label{sec42}
We follow the proof of Theorem \ref{mainthm}. Since $f_1\in \mathcal V$, there is $N_1\ge 1$ such that $f_1\in P_{N_1}H$. As a consequence, we see from \eqref{q1k} that $q_{1,k}=0$ for $k>N_1$.
Hence, $q_1(t)=\sum_{k=1}^{N_1} q_{1,k}(t-T_0)$ is a polynomial in $P_{N_1}H$.
For the recursive step, the functions $f_{N+1}$, $q_m$ and $q_j$ ($1\le m,j\le N$), in this case, are $\mathcal{V}$-valued polynomials.
Hence, by \eqref{BVV}, so is $f_{N+1}-\sum_{m+j=N+1} B(q_m,q_j)$. It follows that there are at most finitely many $k$ such that $q_{N+1,k}$ is nonzero. Since each $q_{N+1,k}$ is an $\mathcal{V}$-valued polynomial, clearly $q_{N+1}$, as a finite sum of those, is a $\mathcal{V}$-valued polynomial.\qed
\subsection{Proof of Theorem \ref{finitetheo}}\label{sec43}
We follow the proof of Theorem \ref{mainthm} closely and make necessary modifications.
We prove part (i), while the proof of part (ii) is entirely similar to that in Theorem \ref{mainthm} and thus omitted.
The same notation $u_n(t)$, $F_n(t)$, $v_n(t)$, $\bar u_n(t)$ as in Theorem \ref{mainthm} is used here.
By \eqref{ffinite} with $N=1$,
\beq\label{ff1}
e^t |f(t)-F_1(t)|_{\alpha_*,\sigma_0}=\mathcal O(e^{-\delta_{1,\alpha_*} t}).
\eeq
Also,
\beq\label{ffirst}
|f(t)|_{\alpha_*,\sigma_0}\le |f_1(t)|_{\alpha_*,\sigma_0}e^{-t}+|f(t)-f_1(t)e^{-t}|_{\alpha_*,\sigma_0}=\mathcal O(e^{-\lambda t}),\quad \forall \lambda\in(0,1).
\eeq
Using \eqref{ffirst} and by applying Proposition \ref{theo23}, we have
\beq \label{newua}
|u(t)|_{\alpha_*+1/2,\sigma_0}= \mathcal O(e^{-(1-\delta) t}),\quad \forall \delta\in(0,1),
\eeq
\beq\label{newBu}
|B(u(t),u(t))|_{\alpha_*,\sigma_0}=\mathcal O(e^{-2(1-\delta) t}),\quad \forall \delta\in(0,1).
\eeq
\medskip
\noindent{\bf Base case: $N=1$.}
Let $\alpha=\alpha_*$ and $\mu=\mu_*$.
In estimating $H_0(t)$ defined by \eqref{H0def}, the estimate \eqref{ch1}, resp., \eqref{ch2}, comes from \eqref{ff1}, resp. \eqref{newBu}.
Hence we obtain the bound \eqref{ch3} for $|H_0(T_0+t)|_{\alpha,\sigma_0}$.
Then the existence and definition of $q_1(t)$ are the same as in \eqref{q11def}, \eqref{q1k} and \eqref{q1def}.
Since $f_1\in \mathcal P^{\mu,\sigma_0}$, then the same proof gives
$q_1\in\mathcal P^{\mu+1,\sigma_0}$, see \eqref{R1q1poly}, \eqref{IRq1} and \eqref{Abn}.
The remainder estimate \eqref{rem1} still holds true, which, for the current value $\alpha=\alpha_*$, proves \eqref{ufinite} for $N=1$.
Also, the ODE \eqref{u1ODE} holds in $G_{\mu,\sigma_0}$ (for the current $\mu=\mu_*$).
If $N_*=1$, then the proof is finished here. We consider $N_*\ge 2$ now.
\medskip
\noindent{\bf Recursive step.} Let $1\le N\le N_*-1$.
Assume there already exist $q_n\in \mathcal P^{\mu_n,\sigma_0}$ for $1\le n\le N$, such that
\beq\label{farate0}
| v_N(t)|_{\alpha_N,\sigma_0}=\mathcal O(e^{-(N+\varep)t}),\quad\forall \varep\in(0,\delta_N^*),
\eeq
and \eqref{unODE} holds in $ G_{\mu_N,\sigma_0}$ for $n=1,2,3,\ldots,N$.
Let
\beq\label{almu}
\alpha=\alpha_{N+1}=\alpha_N-1/2\ge 1/2, \quad \mu=\mu_{N+1}=\mu_N-1/2\ge \alpha\ge 1/2.
\eeq
Note for $n=1,2,\ldots,N$ that $\mu_n\ge \alpha_n\ge 1/2$ and both $\mu_n$, $\alpha_n$ are decreasing,
hence,
\beq \label{uqn}
u_n(t),q_n(t)\in G_{\mu_n,\sigma_0} \subset G_{\mu_N,\sigma_0}=G_{\mu+1/2,\sigma_0} \subset G_{\alpha_N,\sigma_0}=G_{\alpha+1/2,\sigma_0},\quad\forall t\in\R.\eeq
Rewrite \eqref{farate0} as
\beq\label{favNrate}
| v_N(t)|_{\alpha+1/2,\sigma_0}=\mathcal O(e^{-(N+\varep)t}),\quad \forall \varep\in(0,\delta_N^*).
\eeq
We now construct a polynomial $q_{N+1}\in\mathcal P^{\mu+1,\sigma_0}$ such that
\eqref{ufinite} holds true with $N+1$ replacing $N$, and the ODE \eqref{unODE}, with $n=N+1$, holds in $G_{\mu_{N+1},\sigma_0}=G_{\mu,\sigma_0}$.
We proceed with the construction of $q_{N+1}(t)$ as in the proof of Theorem \ref{mainthm} using the specific values of $\alpha$ and $\mu$ in \eqref{almu}.
Note that equation \eqref{vNeq} for $v_N$ still holds in the weak sense as in Definition \ref{lhdef}.
\medskip
$\bullet$ We check the estimate \eqref{hNo} for the function $h_N(t)$ defined by \eqref{hdef}.
Let $\varep_*\in(0,\delta^*_{N+1})$.
By \eqref{ffinite} with $N+1$ replacing $N$, we have
\beq \label{faFtil}
|\tilde F_{N+1}(t)|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+1+\delta_{N+1})t})
=\mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
Thanks to \eqref{uqn} and Lemma \ref{nonLem}, estimate \eqref{Bmj} stays the same as
\beq \label{faBmj}
\sum_{\stackrel{1\le m,j\le N}{m+j\ge N+2}} |B(u_m,u_j)|_{\alpha,\sigma_0}
=\mathcal O(e^{-(N+1+\varep_*)t}).
\eeq
Again, take $\varep\in(\varep_*,\delta^*_{N+1})\subset (0,\delta^*_N)$ in \eqref{favNrate} and $\delta=\varep-\varep_*\in(0,1)$.
Then we have
\beq \label{faubu}
|\bar u_N(t)|_{\alpha+1/2,\sigma_0}=|\bar u_N(t)|_{\alpha_N,\sigma_0}=\mathcal O(e^{-(1-\delta)t}),
\eeq
and by \eqref{newua}
\beqs
|u(t)|_{\alpha+1/2,\sigma_0}\le |u(t)|_{\alpha_*,\sigma_0}=\mathcal O(e^{-(1-\delta)t}).
\eeqs
By Lemma \ref{nonLem} and estimates \eqref{favNrate}, \eqref{faubu}, it follows that
$$|B(v_N,u)|_{\alpha,\sigma_0},|B(\bar u_N,v_N)|_{\alpha,\sigma_0}=\mathcal O(e^{-(N+\varep+1-\delta)t})=\mathcal O(e^{-(N+1+\varep_*)t}).$$
Therefore, by \eqref{faFtil}, \eqref{faBmj} and \eqref{faubu}, we again obtain \eqref{hNo}.
\medskip
$\bullet$ The same construction of $q_{N+1}(t)$ now goes through. Indeed, since $f_{N+1}\in \mathcal P^{\mu,\sigma_0}$, by \eqref{uqn} and the fact that $q_m, q_j\in \mathcal P^{\mu+1/2,\sigma_0}$ for $1\le m,j\le N$, it then follows from \eqref{BGG} that
$$f_{N+1}-\sum_{m+j=N+1} B(q_m,q_j)\in \mathcal P^{\mu,\sigma_0}.$$
The same proof as for the case of $q_1$ then yields that $q_{N+1}\in\mathcal P^{\mu+1,\sigma_0}$.
\medskip
$\bullet$
For the estimate of $v_{N+1}(t)$, same arguments yield
\beqs
|v_{N+1}(t)|_{\alpha,\sigma_0}
=|v_N(t)-e^{-(N+1)t} q_{N+1}(t)|_{\alpha,\sigma_0}= \mathcal O(e^{-(N+1+\varep_*)t}).
\eeqs
Since this holds for any $\varep_*\in(0,\delta_{N+1}^*)$, the remainder estimate \eqref{ufinite} holds true with $N+1$ replacing $N$.
\medskip
$\bullet$ As for the ODE \eqref{unODE} with $n=N+1$, the proof is unchanged from that of Theorem \ref{mainthm}, noting that the ODE now holds in the corresponding space $G_{\mu,\sigma_0}$.
\medskip
We have proved that the polynomial $q_{N+1}$ has the desired properties. This completes the recursive step, and hence, the proof of Theorem \ref{finitetheo}.\qed
\section*{}
\noindent\textbf{\large Acknowledgements.}
The authors would like to thank Peter Constantin for prompting the question of extending the Foias-Saut theory to the case of non-potential forces. The authors are also grateful to Ciprian Foias for his encouragement towards this work and helpful discussions,
as well as to Animikh Biswas for stimulating discussions. L.H. gratefully acknowledges the support for his research by the NSF grant DMS-1412796
\bibliography{paperbaseall}{}
\bibliographystyle{abbrv}
\end{document} | 159,638 |
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Apple Faces Lawsuit for Text Disruption Caused by their iMessage Service
Apple’s iMessage is a free messaging service. This service was announced in 2011, and it along with other free messaging apps have been blamed for the steady decline in the usage of SMS service. There was though a major flaw in iMessage that has got Apple another visit to the courthouse.
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- Sandy | 404,269 |
TITLE: Looking for a function $f$ that is $n$-differentiable, but $f^{(n)}$ is not continuous
QUESTION [1 upvotes]: I am looking for a real valued function of real variable that is $n$-differentiable, but whose $n$th derivative is not continuous.
This is my function:
$f_n(x) = x^{n+1} \cdot \sin{\frac{1}{x}}$, if $x \neq 0$ and 0, if $x=0$. $n\in \{0,1,2,\ldots\}$ For example, if $n=0$, $f_0$ it's continuous and non diferentiable in $0$.
REPLY [5 votes]: Let $n$ be a positive integer and let
$$f(x) = x^{2n} \cdot \sin\left(\frac{1}{x}\right)$$
$$f(0) = 0$$
Then mathematical induction can be used to prove: (a) The $n$th derivative of $f(x)$ exists for each value of $x$. (b) The $n$th derivative $f(x)$ is not continuous at $x = 0$.
Let $n$ be a positive integer and let
$$g(x) = x^{2n+1} \cdot \sin\left(\frac{1}{x}\right)$$
$$g(0) = 0$$
Then mathematical induction can be used to prove: (a) The $n$th derivative of $g(x)$ is continuous at each value of $x$. (b) The $(n+1)$st derivative of $g(x)$ does not exist at $x = 0$.
More generally, let $a$, $b$ be positive real numbers, let $n$ be a positive integer, and define $h(x)$ by:
$$h(x) = x^a \cdot \sin\left(\frac{1}{x^b}\right)$$
$$h(0) = 0$$
The $n$th derivative of $h(x)$ exists for all values of $x$ if and only if $a > n + (n-1)b$.
The $n$th derivative of $h(x)$ is bounded on every bounded interval if and only if $a \geq n + nb$.
The $n$th derivative of $h(x)$ is continuous at each point if and only if $a > n + nb$.
To prove these statements, you can use mathematical induction to prove that the $n$th derivative of $h(x)$ has the form
$$P_{n}(x) \cdot \cos\left(\frac{1}{x^b}\right) + Q_{n}(x) \cdot \sin\left(\frac{1}{x^b}\right),$$
where $P_n$ and $Q_n$ are polynomials such that at least one of them has a lowest degree term that is a NONZERO multiple of $x^{a-n-nb}$ and neither has a lower degree term. [The added emphasis on nonzero is because this becomes vital for the "only if" halves of the statements above.] To prove the "if" halves, you may want to incorporate into the induction statement the fact that the $n$th derivative at $x = 0$ is zero. | 39,811 |
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- From I 20 take Mountain Creek Parkway north. The boat ramp is on your left.
- One-lane ramp with parking for 29 vehicles
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Beautiful river-front lot located in upscale Black Water Cove. Close to Conway and Highway 22 for access to the beach. Shaftsbury Glen golf course is a short walk. Build your dream home on the river with the builder of your choice. Riverside park with pavilion and community boat dock. This lot can be purchased with Lot 12, MLS #1823255, next to it making a total of .59 acres with 145' of riverfront. | 159,291 |
\section{The {\sc Maple} Code for computing Symmetries of Binary Forms}
The \Maple\ code consists of two main programs --- {\tt symm} and {\tt matrices} ---
and two auxiliary functions --- {\tt simple} and {\tt l\_f}.
The programs are available st the web site at {\tt http://www.math.umn.edu/~berchenk}.
The program {\tt symm} is the main function.
The input consists of a complex-valued polynomial $ f(p)$ considered as the
projective version of homogeneous binary form $Q(x,y)$, and
the degree $n=deg(Q)$. The program
computes the invariants $J$ and $K$ in reduced form, determines the dimension of the symmetry
group, and, in the case of a finite symmetry group, applies the
\Maple\ command {\tt solve} to solve the two polynomial symmetry equations \eq{FG} to find
explicit form of symmetries. The
output of {\tt symm} consists of the projective index of the form and the explicit formulae for its
discrete projective symmetries. The program also notifies the user if the
symmetry group is not discrete, or is in the maximal discrete symmetry class.
The program works well when applied to very simple forms, but experienced
difficulties simplifying complicated rational algebraic formulae into the basic linear
fractional form.
\mapskip
{\tt > with(linalg):}
{\tt > symm:=proc(form,n)
global tr,error;
local Q,Qp,Qpp,Qppp,Qpppp,H,T,V,U,J,K,j,k, Eq1,Eq2,i,eqtr,
ans;
\lone{ tr:='tr':
Q:=form(p);
Qp:=diff(Q,p);
Qpp:=diff(Qp,p);
Qppp:=diff(Qpp,p);
Qpppp:=diff(Qppp,p);
H:=n*(n-1)*(Q*Qpp-(n-1)/n*Qp\^{}2);
if H=0 then }
\ltwo { ans:=`Hessian is zero: two-dimensional symmetry group`}
\lone { else }
\ltwo
{
T:=-n\^{}2*(n-1)*(Q\^{}2*Qppp-3*(n-2)/n*Q*Qp*Qpp
+2*(n-1)*(n-2)/n\^{}2*Qp\^{}3);
V:=Q\^{}3*Qpppp-4*(n-3)/n*Q\^{}2*Qp*Qppp+6*(n-2)*(n-3)/n\^{}2
*Q*Qp\^{}2*Qpp-3*(n-1)*(n-2)*(n-3)/n\^{}3*Qp\^{}4;
U:=n\^{}3*(n-1)*V-3*(n-2)/(n-1)*H\^{}2;
J:=simple(T\^{}2/H\^{}3); K:=simple(U/H\^{}2);
j:=subs(p=P,J);k:=subs(p=P,K);
Eq1:=simplify(numer(J)*denom(j)-numer(j)*denom(J));
Eq2:=simplify(numer(K)*denom(k)-numer(k)*denom(K));
if Eq1=0 then}
\lthree { ans:=`Form has a one-dimensional symmetry group`;}
\ltwo{ else }
\lthree{ if Eq2=0 then }
\lfour{ print (` Form has the maximal possible discrete
symmetry group`);
eqtr:= [solve(Eq1,P)];
tr:=map(radsimp,map(allvalues,eqtr));}
\lthree{ else}
\lfour{ eqtr:=[solve(\{Eq1,Eq2\},P)];
tr:= [];
for i from 1 to nops(eqtr) do}
\lfive { tr:=map(radsimp,[op(tr),allvalues(rhs(eqtr[i][1]))]);}
\lfour { od}
\lthree { fi;
print(`The number of elements in the symmetry group`
=nops(tr));
ans:=map(l\_{}f,tr);
if error=1 then }
\lfour{ print(`ERROR: Some of the transformations are not
linear-fractional`) }
\lthree{ else }
\lfour{ if error=2 then }
\lfive { print(`WARNING: Some of the transformations are not
written in the form polynomial over polynomial`)}
\lfour{ fi;}
\lthree{ fi;}
\ltwo{ fi;}
\lone{ fi;
ans }
end:
}
\mapskip
\noindent
The program {\tt matrices} determines the matrix
symmetry corresponding to a given (list of) projective symmetries. As discussed in the text, this only
requires determining an overall scalar multiple, which can be found by substituting the projective
symmetry into the form. The output consists of each projective symmetry, the scalar factor $\mu $,
and the resulting matrix symmetry.
\mapskip
{\tt
> matrices:=proc(form,n,L::list)
local Q,ks,ksi,i,Sf,M;
\lone {ksi:='ksi';
for i from 1 to nops(L) do }
\ltwo {Sf:=simplify(denom(L[i])\^{}n*form(L[i]));
ks:=quo(Sf,form(p),p);
ksi:=simplify(ks\^{}(1/n),radical,symbolic);
M[i]:=matrix(2,2,[coeff(numer(L[i]),p)/ksi,
coeff(numer(L[i]),p,0)/ksi,coeff(denom(L[i]),p)/ksi,
coeff(denom(L[i]),p,0)/ksi]);
print(L[i], mu=ksi, map(simplify,M[i]))}
\lone { od;}
end:
}
\mapskip
\noindent
The auxiliary function {\tt simple} helps to simplify rational
expressions by manipulating the numerator and denominator separately.
The simplified rational expression is returned.
\mapskip
{\tt
> simple:=proc(x)
local nu,de,num,den;
\lone{ nu:=numer(x);
de:=denom(x);
num:=(simplify((nu,radical,symbolic)));
den:=(simplify((de,radical,symbolic)));
simplify(num/den);}
end:
}
\mapskip
\noindent
The auxiliary function {\tt l\_f} uses polynomial division to
reduce rational expressions to linear fractional form (when possible).
\mapskip
{ \tt
> l\_{}f:=proc(x)
local A,B,C,S,de,nu,r,R;
global error;error:='error';
\lone { nu:=numer(x);
de:=denom(x);
if type(nu,polynom(anything,p))
and type(de,polynom(anything,p)) then}
\ltwo{ if degree(nu,p)+1=degree(de,p) then}
\lthree{ A:=quo(de,nu,p,'B');
S:=1/A; R:=0}
\ltwo { else }
\lthree { A:=quo(nu,de,p,'B');
if B=0 then }
\lfour { S:=A; R:=0; }
\lthree{ else }
\lfour { C:=quo(de,B,p,'r'); R:=simple(r);
S:=simplify(A+1/C) }
\lthree{ fi;}
\ltwo{ fi;
if R=0 then }
\lthree { collect(S,p)}
\ltwo { else }
\lthree { error:=1; x }
\ltwo { fi;}
\lone { else }
\ltwo { error:=2; x }
\lone {fi;}
end:}
\section{ Cubic Forms.}
We now present the results of applying the function {\tt symm} and {\tt matrices} to cubic forms.
We begin with simple cases, ending with a cubic whose formulae required extensive manipulation.
\medskip
\noindent\iz{2}. {Cubics with one triple root:}
\smallskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^3;}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{3}
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{symm(f,3);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{Hessian\ is\ zero:\ two-dimensional\ symmetry\ group}
\]
\end{maplelatex}
\end{maplegroup}
\smallskip
\noindent\iz{2}. {Cubics with one double root and one single root:}
\smallskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p;}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{symm(f,3);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{Form\ has\ a\ one-dimensional\ symmetry\ group}
\]
\end{maplelatex}
\end{maplegroup}
\smallskip
\noindent\iz{3}. {Cubics with three simple roots:}
\smallskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^3+1;}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{3} + 1
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{S:=symm(f,3);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{\ Form\ has\ the\ maximal\ possible\ discrete\ symmetry\
group}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{The\ number\ of\ elements\ in\ the\ symmetry\ group}=6
\]
\end{maplelatex}
\begin{maplelatex}
\[
S := \left[ \; p, \,{\displaystyle \frac {1}{p}} , \,
{\displaystyle \frac { - {\displaystyle \frac {1}{2}} +
{\displaystyle \frac {1}{2}} \,I\,\sqrt{3}}{p}} , \,
{\displaystyle \frac { - {\displaystyle \frac {1}{2}} -
{\displaystyle \frac {1}{2}} \,I\,\sqrt{3}}{p}} , \,( -
{\displaystyle \frac {1}{2}} + {\displaystyle \frac {1}{2}} \,I
\,\sqrt{3})\,p, \,( - {\displaystyle \frac {1}{2}} -
{\displaystyle \frac {1}{2}} \,I\,\sqrt{3})\,p \; \right]
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{matrices(f,3,[S[2],S[4]]);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
{\displaystyle \frac {1}{p}} , \quad \mu =1, \quad \left[
{\begin{array}{rr}
0 & 1 \\
1 & 0
\end{array}}
\right]
\]
\end{maplelatex}
\begin{maplelatex}
\[
{\displaystyle \frac { - {\displaystyle \frac {1}{2}} -
{\displaystyle \frac {1}{2}} \,I\,\sqrt{3}}{p}} , \quad \mu =2, \quad
\left[
{\begin{array}{rc}
0 & - {\displaystyle \frac {1}{2}} - {\displaystyle \frac {1}{2
}} \,I\,\sqrt{3} \\ [2ex]
1 & 0
\end{array}}
\right]
\]
\end{maplelatex}
\end{maplegroup}
\smallskip
\noindent\iz{4}. A more complicated cubic example.
\smallskip
All cubics with a discrete symmetry group are complex
equivalent to $x^3+y^3$ and have projective index $6$.
However, when we apply the same code to a cubic not in canonical form. The initial \Maple\ result
is not in the correct linear fractional form. We must simplify the rational algebraic expressions ``by
hand'' to put them in the form of a projective linear fractional transformation.
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^3+p+1;}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{3} + p + 1
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{S:=symm(f,3);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{\ Form\ has\ the\ maximal\ possible\ discrete\ symmetry\
group}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{The\ number\ of\ elements\ in\ the\ symmetry\ group}=6
\]
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{\mathit{WARNING:\ Some\ of\ the\ transformations\ are\
not\ written\ \ \ \ in\ the\ form\ polynomial\ \backslash }} \\
& & \mathit{over\ polynomial}\mbox{\hspace{280pt}}
\end{eqnarray*}
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{S := [p, \,{\displaystyle \frac {( - 9 + I\,\sqrt{31})\,
p + 2}{9 + I\,\sqrt{31} + 6\,p}} , \, - {\displaystyle \frac {(9
+ I\,\sqrt{31})\,p - 2}{9 - I\,\sqrt{31} + 6\,p}} ,
{\displaystyle \frac {1}{18}} ((54\,p^{4} + 9\,2^{(2/3)}\,3^{(1/3
)}\,\mathrm{\%1}^{(1/3)}\,p^{3}} \\
& & \mbox{} + 324\,p^{3} + 3\,2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}
^{(2/3)}\,p^{2} + 450\,p^{2} - 108\,p + 9\,2^{(1/3)}\,3^{(2/3)}\,
\mathrm{\%1}^{(2/3)}\,p \\
& & \mbox{} + 9\,2^{(2/3)}\,3^{(1/3)}\,\mathrm{\%1}^{(1/3)}\,p
- 2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)} + 6 + 9\,2^{(2/3)}
\,3^{(1/3)}\,\mathrm{\%1}^{(1/3)})3^{(2/3)} \\
& & 2^{(1/3)}) \left/ {\vrule height0.56em width0em depth0.56em}
\right. \! \! (\mathrm{\%1}^{(1/3)}\,(27\,p^{3} - 9\,p^{2} - 1)
), - {\displaystyle \frac {1}{36}} ((54\,I\,\sqrt{3}\,p^{4} + 54
\,p^{4} + 324\,p^{3} \\
& & \mbox{} + 324\,I\,\sqrt{3}\,p^{3} - 18\,2^{(2/3)}\,3^{(1/3)}
\,\mathrm{\%1}^{(1/3)}\,p^{3} + 450\,p^{2} + 450\,I\,\sqrt{3}\,p
^{2}
\end{eqnarray*}
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
& & \mbox{} - 9\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}\,
p^{2} + 3\,2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)}\,p^{2} - 27
\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}\,p \\
& & \mbox{} - 108\,p - 18\,2^{(2/3)}\,3^{(1/3)}\,\mathrm{\%1}^{(
1/3)}\,p + 9\,2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)}\,p - 108
\,I\,\sqrt{3}\,p \\
& & \mbox{} + 3\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}
+ 6\,I\,\sqrt{3} - 18\,2^{(2/3)}\,3^{(1/3)}\,\mathrm{\%1}^{(1/3)
} - 2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)} + 6 \\
& & )3^{(2/3)}\,2^{(1/3)}) \left/ {\vrule
height0.56em width0em depth0.56em} \right. \! \! (\mathrm{\%1}^{
(1/3)}\,(27\,p^{3} - 9\,p^{2} - 1)), {\displaystyle \frac {1}{36}
} ((54\,I\,\sqrt{3}\,p^{4} - 54\,p^{4} - 324\,p^{3} \\
& & \mbox{} + 324\,I\,\sqrt{3}\,p^{3} + 18\,2^{(2/3)}\,3^{(1/3)}
\,\mathrm{\%1}^{(1/3)}\,p^{3} - 450\,p^{2} + 450\,I\,\sqrt{3}\,p
^{2} \\
& & \mbox{} - 9\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}\,
p^{2} - 3\,2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)}\,p^{2} - 27
\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}\,p \\
& & \mbox{} + 108\,p + 18\,2^{(2/3)}\,3^{(1/3)}\,\mathrm{\%1}^{(
1/3)}\,p - 9\,2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)}\,p - 108
\,I\,\sqrt{3}\,p \\
& & \mbox{} + 3\,I\,3^{(1/6)}\,2^{(1/3)}\,\mathrm{\%1}^{(2/3)}
+ 6\,I\,\sqrt{3} + 18\,2^{(2/3)}\,3^{(1/3)}\,\mathrm{\%1}^{(1/3)
} + 2^{(1/3)}\,3^{(2/3)}\,\mathrm{\%1}^{(2/3)} - 6 \\
& & )3^{(2/3)}\,2^{(1/3)}) \left/ {\vrule
height0.56em width0em depth0.56em} \right. \! \! (\mathrm{\%1}^{
(1/3)}\,(27\,p^{3} - 9\,p^{2} - 1))] \\
& & \mathrm{\%1} := 9 + 18\,p - 81\,p^{2} + 261\,p^{3} + 27\,
\sqrt{31}\,\sqrt{3}\,p^{3} - 9\,\sqrt{31}\,\sqrt{3}\,p^{2} -
\sqrt{31}\,\sqrt{3}
\end{eqnarray*}
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
The first three components of {\tt S} are in the proper linear fractional form. The problem with the
other expressions is that \Maple\ does not automatically factor polynomials under a radical.
One approach to simplification is to first do the required factorization:
\end{maplegroup}
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{
n1:=factor(9+18*p-81*p^2+261*p^3+27*sqrt(31)*sqrt(3)*p^3\break
-9*sqrt(31)*sqrt(3)*p^2-sqrt(31)*sqrt(3));
}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{n1} := - {\displaystyle \frac {1}{24}} \,(29 + 3\,\sqrt{
31}\,\sqrt{3})\,( - 6\,p - 9 + \sqrt{31}\,\sqrt{3})^{3}
\]
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
Substituting {\tt n1} into the fourth rational algebraic expression in {\tt S} above,
we can now force \Maple\ to take the cube root and obtain the actual linear fractional
formula for this symmetry:
\end{maplegroup}
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{
simp1:=radsimp(1/18*((54*p^4+9*2^(2/3)*3^(1/3)*(n1)^(1/3)*p^3+324*p^3
\break
+3*2^(1/3)*3^(2/3)*(n1)^(2/3)*p^2+450*p^2+9*2^(1/3)*3^(2/3)*(n1)^(2/3)\break
*p+9*2^(2/3)*3^(1/3)*(n1)^(1/3)*p-108*p-2^(1/3)*3^(2/3)*(n1)^(2/3)+6
\break
+9*2^(2/3)*3^(1/3)*(n1)^(1/3))*3^(2/3)*2^(1/3))/((n1)^(1/3)\break
*(27*p^3-9*p^2-1)));
}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{simp2:=l_f(simp1);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}
{simp3:=collect(expand(numer(simp2))/expand(denom(simp2)),p);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{\mathit{simp3} := (( - 226\,2^{(1/3)} - 20\,\mathrm{\%2}
- 20\,2^{(2/3)}\,\mathrm{\%1}^{(2/3)} - 22\,\mathrm{\%3} - 208\,
\mathrm{\%1}^{(1/3)})\,p - 8\,\mathrm{\%3}} \\
& & \mbox{} - 58\,2^{(2/3)}\,\mathrm{\%1}^{(2/3)} - 566\,
\mathrm{\%1}^{(1/3)} - 58\,\mathrm{\%2} - 116\,2^{(1/3)}) \left/
{\vrule height0.56em width0em depth0.56em} \right. \! \! ( \\
& & ( - 6\,2^{(2/3)}\,\mathrm{\%1}^{(2/3)} - 174\,\mathrm{\%1}^{
(1/3)} - 1686\,2^{(1/3)} - 174\,\mathrm{\%3} - 6\,\mathrm{\%2})\,
p \\
& & \mbox{} + 20\,2^{(2/3)}\,\mathrm{\%1}^{(2/3)} + 208\,
\mathrm{\%1}^{(1/3)} + 226\,2^{(1/3)} + 22\,\mathrm{\%3} + 20\,
\mathrm{\%2}) \\
& & \mathrm{\%1} := 29 + 3\,\sqrt{31}\,\sqrt{3} \\
& & \mathrm{\%2} := \mathrm{\%1}^{(1/3)}\,\sqrt{31}\,\sqrt{3} \\
& & \mathrm{\%3} := 2^{(1/3)}\,\sqrt{31}\,\sqrt{3}
\mbox{\hspace{303pt}}
\end{eqnarray*}
\end{maplelatex}
\end{maplegroup}
\mapskip
The linear fractional formulae for the other symmetries are derived in a similar fashion.
\section { The Octahedral Symmetry Group.}
As we remarked in the text, the sextic polynomial $Q(p) = p^5 + p$ has an octahedral symmetry group.
Here we illustrate how the symmetries are computed using our \Maple\ program.
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^5+p;
}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{5} + p
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{symm(f,6);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{\ Form\ has\ the\ maximal\ possible\ discrete\ symmetry\
group}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{The\ number\ of\ elements\ in\ the\ symmetry\ group}=24
\]
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{\mathit{WARNING:\ Some\ of\ the\ transformations\ are\
not\ written\ in\ the\ form
}} \\
& & \mathit{polynomial\ over\ polynomial}\mbox{\hspace{246pt}}
\end{eqnarray*}
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{\left [\ {\displaystyle \frac {1}{p}} , \,p, \, - p, \, -
{\displaystyle \frac {1}{p}} , \,{\displaystyle \frac {I}{p}} ,
\, - {\displaystyle \frac {I}{p}} , \,I\,p, \, - I\,p, \,
{\displaystyle \frac { - 2\,p^{3} + 2\,I\,p + \mathrm{\%3}}{p^{4}
+ 1}} , \,{\displaystyle \frac { - 2\,p^{3} + 2\,I\,p - \mathrm{
\%3}}{p^{4} + 1}} ,\right . } \\
& & {\displaystyle \frac { - 2\,p^{3} - 2\,I\,p + \mathrm{\%4}}{
p^{4} + 1}} , \,{\displaystyle \frac { - 2\,p^{3} - 2\,I\,p -
\mathrm{\%4}}{p^{4} + 1}} , \,{\displaystyle \frac {2\,p^{3} + 2
\,I\,p + \mathrm{\%4}}{p^{4} + 1}} , \,{\displaystyle \frac {2\,p
^{3} + 2\,I\,p - \mathrm{\%4}}{p^{4} + 1}} , \\
& & {\displaystyle \frac {2\,p^{3} - 2\,I\,p + \mathrm{\%3}}{p^{
4} + 1}} , \,{\displaystyle \frac {2\,p^{3} - 2\,I\,p - \mathrm{
\%3}}{p^{4} + 1}} , \,{\displaystyle \frac { - 2\,p + 2\,I\,p^{3}
+ \mathrm{\%1}}{p^{4} + 1}} , \,{\displaystyle \frac { - 2\,p +
2\,I\,p^{3} - \mathrm{\%1}}{p^{4} + 1}} , \\
& & {\displaystyle \frac { - 2\,p - 2\,I\,p^{3} + \mathrm{\%2}}{
p^{4} + 1}} , \,{\displaystyle \frac { - 2\,p - 2\,I\,p^{3} -
\mathrm{\%2}}{p^{4} + 1}} , \,{\displaystyle \frac {2\,p + 2\,I\,
p^{3} + \mathrm{\%2}}{p^{4} + 1}} , \,{\displaystyle \frac {2\,p
+ 2\,I\,p^{3} - \mathrm{\%2}}{p^{4} + 1}} , \\
& & \left .{\displaystyle \frac {2\,p - 2\,I\,p^{3} + \mathrm{\%1}}{p^{
4} + 1}} , \,{\displaystyle \frac {2\,p - 2\,I\,p^{3} - \mathrm{
\%1}}{p^{4} + 1}} \ \right ] \\
& & \mathrm{\%1} := \sqrt{ - 4\,p^{6} + 4\,p^{2} + I\,p^{8} - 6
\,I\,p^{4} + I} \\
& & \mathrm{\%2} := \sqrt{ - 4\,p^{6} + 4\,p^{2} - I\,p^{8} + 6
\,I\,p^{4} - I} \\
& & \mathrm{\%3} := \sqrt{4\,p^{6} - 4\,p^{2} + I\,p^{8} - 6\,I
\,p^{4} + I} \\
& & \mathrm{\%4} := \sqrt{4\,p^{6} - 4\,p^{2} - I\,p^{8} + 6\,I
\,p^{4} - I}
\end{eqnarray*}
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
Again, \Maple\ has failed to simplify the expressions $\%1,\%2,\%3,\%4$, and we need to make it
take the square root. In the case of symmetries numbers $9,11,13,15,17,19,21,23$ this is done as
follows. The others are handled in a similar fashion, and, for brevity, we omit the formulae here.
\end{maplegroup}
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{
for j in [9,11,13,15,17,19,21,23] do \break
sq:=sqrt(factor(op(op(numer(tr[j]))[3])[1],I),symbolic):\break
s[j]:=l_f((op(numer(tr[j]))[1]+op(numer(tr[j]))[2]+sq)/denom(tr[j]));\break
print(s.j=s[j]);\break
od: }{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{s9}= - {\displaystyle \frac {(\sqrt{2} + I\,\sqrt{2})\,p
- 2}{ - \sqrt{2} + I\,\sqrt{2} - 2\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s11}= - {\displaystyle \frac {( - \sqrt{2} + I\,\sqrt{2})
\,p + 2}{I\,\sqrt{2} + \sqrt{2} + 2\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s13}={\displaystyle \frac {( - \sqrt{2} + I\,\sqrt{2})\,p
- 2}{I\,\sqrt{2} + \sqrt{2} - 2\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s15}={\displaystyle \frac {(\sqrt{2} + I\,\sqrt{2})\,p +
2}{ - \sqrt{2} + I\,\sqrt{2} + 2\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s17}= - {\displaystyle \frac {( - \sqrt{2} + I\,\sqrt{2})
\,p - 2}{ - \sqrt{2} + I\,\sqrt{2} - 2\,I\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s19}= - {\displaystyle \frac {I\,((\sqrt{2} + I\,\sqrt{2}
)\,p + 2)}{ - \sqrt{2} + I\,\sqrt{2} + 2\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s21}={\displaystyle \frac {(\sqrt{2} + I\,\sqrt{2})\,p -
2}{\sqrt{2} + I\,\sqrt{2} + 2\,I\,p}}
\]
\end{maplelatex}
\begin{maplelatex}
\[
\mathit{s23}={\displaystyle \frac {( - \sqrt{2} + I\,\sqrt{2})\,p
+ 2}{ - \sqrt{2} + I\,\sqrt{2} + 2\,I\,p}}
\]
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
As we remarked in the text, the octahedral symmetry group has two generators.
The matrix form of these generators is computed as follows:
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{matrices(f,6,[tr[7],s[9]]);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
I\,p, \qquad \mu =(-1)^{(1/12)}, \qquad \left[
{\begin{array}{cc}
(-1)^{(5/12)} & 0 \\
0 & - (-1)^{(11/12)}
\end{array}}
\right]
\]
\end{maplelatex}
\begin{maplelatex}
\[
- {\displaystyle \frac {(\sqrt{2} + I\,\sqrt{2})\,p - 2}{ -
\sqrt{2} + I\,\sqrt{2} - 2\,p}} , \qquad \mu =2\,\sqrt{2}, \qquad \left[
{\begin{array}{cc}
- {\displaystyle \frac {1}{2}} - {\displaystyle \frac {1}{2}}
\,I & {\displaystyle \frac {1}{2}} \,\sqrt{2} \\ [2ex]
- {\displaystyle \frac {1}{2}} \,\sqrt{2} & - {\displaystyle
\frac {1}{2}} + {\displaystyle \frac {1}{2}} \,I
\end{array}}
\right]
\]
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
We end with two further examples. We already know that the following octavic polynomial also has an
octahedral symmetry group. In this case, {\tt symm} produces the projective symmetries directly:
\end{maplegroup}
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^8+14*p^4+1;
}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{8} + 14\,p^{4} + 1
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{S:=symm(f,8);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{The\ number\ of\ elements\ in\ the\ symmetry\ group}=24
\]
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{S := [ - {\displaystyle \frac {1}{p}} , \, -
{\displaystyle \frac {p - 1}{p + 1}} , \, - {\displaystyle
\frac {p + 1}{p - 1}} , \,p, \, - p, \,{\displaystyle \frac {p +
1}{p - 1}} , \,{\displaystyle \frac {p - 1}{p + 1}} , \,
{\displaystyle \frac {1}{p}} , \,{\displaystyle \frac {I\,(p - 1)
}{p + 1}} , \, - {\displaystyle \frac {I\,(p - 1)}{p + 1}} , \,
{\displaystyle \frac {I\,(p + 1)}{p - 1}} , } \\
& & - {\displaystyle \frac {I\,(p + 1)}{p - 1}} , \,
{\displaystyle \frac {I}{p}} , \, - {\displaystyle \frac {I}{p}}
, \,I\,p, \, - I\,p, \,{\displaystyle \frac { - 1 + I\,p}{ - I +
p}} , \, - {\displaystyle \frac {1 + I\,p}{I + p}} , \,
{\displaystyle \frac {1 + I\,p}{I + p}} , \, - {\displaystyle
\frac { - 1 + I\,p}{ - I + p}} , \,{\displaystyle \frac { - 1 + I
\,p}{1 + I\,p}} , \\
& & {\displaystyle \frac {1 + I\,p}{ - 1 + I\,p}} , \, -
{\displaystyle \frac {1 + I\,p}{ - 1 + I\,p}} , \, -
{\displaystyle \frac { - 1 + I\,p}{1 + I\,p}} ]
\end{eqnarray*}
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{matrices(f,8,[S[11],S[15]]);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
{\displaystyle \frac {I\,(p + 1)}{p - 1}} , \qquad \mu =\sqrt{2}, \qquad
\left[
{\begin{array}{cc}
{\displaystyle \frac {1}{2}} \,I\,\sqrt{2} & {\displaystyle
\frac {1}{2}} \,I\,\sqrt{2} \\ [2ex]
{\displaystyle \frac {1}{2}} \,\sqrt{2} & - {\displaystyle
\frac {1}{2}} \,\sqrt{2}
\end{array}}
\right]
\]
\end{maplelatex}
\begin{maplelatex}
\[
I\,p, \qquad \mu =1, \qquad \left[
{\begin{array}{cr}
I & 0 \\
0 & 1
\end{array}}
\right]
\]
\end{maplelatex}
\end{maplegroup}
\mapskip
\begin{maplegroup}
Finally, for illustrative purposes, we present a higher order example given by a binary form of degree
$12$.
\end{maplegroup}
\mapskip
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{f:=p->p^12-33*p^8-33*p^4+1;}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
f := p\rightarrow p^{12} - 33\,p^{8} - 33\,p^{4} + 1
\]
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{S:=symm(f,12);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
\mathit{The\ number\ of\ elements\ in\ the\ symmetry\ group}=24
\]
\end{maplelatex}
\begin{maplelatex}
\begin{eqnarray*}
\lefteqn{S := [p, \, - p, \, - {\displaystyle \frac {1}{p}} , \,
{\displaystyle \frac {1}{p}} , \, - {\displaystyle \frac {p - 1}{
p + 1}} , \, - {\displaystyle \frac {p + 1}{p - 1}} , \,
{\displaystyle \frac {p + 1}{p - 1}} , \,{\displaystyle \frac {p
- 1}{p + 1}} , \,{\displaystyle \frac {I}{p}} , \, -
{\displaystyle \frac {I}{p}} , \,I\,p, \, - I\,p, \,
{\displaystyle \frac { - 1 + I\,p}{ - I + p}} , \, -
{\displaystyle \frac {1 + I\,p}{I + p}} , } \\
& & {\displaystyle \frac {1 + I\,p}{I + p}} , \, -
{\displaystyle \frac { - 1 + I\,p}{ - I + p}} , \,{\displaystyle
\frac {I\,(p - 1)}{p + 1}} , \, - {\displaystyle \frac {I\,(p - 1
)}{p + 1}} , \,{\displaystyle \frac {I\,(p + 1)}{p - 1}} , \, -
{\displaystyle \frac {I\,(p + 1)}{p - 1}} , \,{\displaystyle
\frac { - 1 + I\,p}{1 + I\,p}} , \,{\displaystyle \frac {1 + I\,p
}{ - 1 + I\,p}} , \\
& & - {\displaystyle \frac {1 + I\,p}{ - 1 + I\,p}} , \, -
{\displaystyle \frac { - 1 + I\,p}{1 + I\,p}} ]
\end{eqnarray*}
\end{maplelatex}
\end{maplegroup}
\begin{maplegroup}
\begin{mapleinput}
\mapleinline{active}{1d}{matrices(f,12,[S[11],S[19]]);}{
}
\end{mapleinput}
\mapleresult
\begin{maplelatex}
\[
I\,p, \qquad \mu =1, \qquad \left[
{\begin{array}{cr}
I & 0 \\
0 & 1
\end{array}}
\right]
\]
\end{maplelatex}
\begin{maplelatex}
\[
{\displaystyle \frac {I\,(p + 1)}{p - 1}} , \qquad \mu =(-1)^{(1/12)}
\sqrt{2}, \qquad \left[
{\begin{array}{cc}
{\displaystyle \frac {1}{2}} \,(-1)^{(5/12)}\,\sqrt{2} &
{\displaystyle \frac {1}{2}} \,(-1)^{(5/12)}\,\sqrt{2} \\ [2ex]
- {\displaystyle \frac {1}{2}} \,(-1)^{(11/12)}\,\sqrt{
2} & {\displaystyle \frac {1}{2}} \,(-1)^{(11/12)}\,
\sqrt{2}
\end{array}}
\right]
\]
\end{maplelatex}
\end{maplegroup} | 143,208 |
\begin{document}
\title[Complete bounded $\lambda$-hypersurfaces]
{Complete bounded $\lambda$-hypersurfaces in the weighted volume-preserving mean curvature flow}
\author[Yecheng Zhu]{Yecheng Zhu}
\address{
1. Department of Mathematics, University of Science and Technology of China,
230026, Hefei, Anhui Province, People's Republic of China \\
2. Department of Applied Mathematics, Anhui University of Technology,
243002, Maanshan, Anhui Province, People's Republic of China }
\email{
[email protected] }
\thanks{This work was supported by the National Natural Science Foundation of China (No. 11271343). }
\author{Yi Fang}
\address{Department of Applied Mathematics, Anhui University of Technology,
243002, Maanshan, Anhui Province, People's Republic of China}
\email{[email protected]}
\author{Qing Chen}
\address{Department of Mathematics, University of Science and Technology of China,
230026, Hefei, Anhui Province, People's Republic of China}
\email{[email protected]}
\subjclass{Primary 53C42; Secondary 53C44}
\keywords{Volume comparison theorem, Topology, Second fundamental form, $\infty$ - Bakry - Emery Ricci tensor, Mean curvature flow}
\date{September 24, 2016}
\begin{abstract}
In this paper, we study the complete bounded $\lambda$-hypersurfaces in weighted volume-preserving mean curvature flow. Firstly, we investigate the volume comparison theorem of complete bounded $\lambda$-hypersurfaces with $|A|\leq\alpha$ and get some applications of the volume comparison theorem. Secondly, we consider the relation among $\lambda$, extrinsic radius $k$, intrinsic diameter $d$, and dimension $n$ of the complete $\lambda$-hypersurface, and we obtain some estimates for the intrinsic diameter and the extrinsic radius. At last, we get some topological properties of the bounded $\lambda$-hypersurface with some natural and general restrictions.
\end{abstract}
\maketitle
\section{Introduction}
A hypersurface $X : M^n \rightarrow R^{n+1}$ is said to be a self-shrinker in $R^{n+1}$ if it satisfies the following equation (see [10]) for the mean curvature and the normal
\begin{equation}
H-\frac{<X, N>}{2}=0.
\end{equation}
Self-shrinkers play an important role in the study of the mean curvature flow. Not only they correspond to the self-shrinking solutions to mean
curvature flow, but also they describe all possible blow ups at a given singularity points of the mean curvature flow. The simplest self-shrinkers are $R^n$, the sphere of radius $\sqrt{2n}$ and more generally cylindrical products $S^k \times R^{n-k}$ (where $S^k$ has radius $\sqrt{2k}$). All of these examples are mean
convex. Without the assumption on mean convexity, there are expected to be many more examples of self-shrinkers in $R^3$. In particular, Angenent constructed a self-shrinking torus (¡°shrinking donut¡±) of revolution in [2], and there is numerical evidence for a number of other examples (see [3], [9], [24]). We refer the readers to [10, 11, 15, 16] and references therein for more information on self-shrinkers and singularities of mean curvature flow.
As generalizations of self-shrinkers, $\lambda$-hypersurfaces were first introduced by Cheng and Wei in [8], where they
proved that $\lambda$-hypersurfaces are critical points of the weighted area functional for the weighted
volume-preserving variations. Furthermore, they classified the complete $\lambda$-hypersurfaces with polynomial volume growth and studied F-stability of $\lambda$-hypersurfaces, which are generalizations of the results due to Huisken [15] and Colding-Minicozzi [10]. Guang proved some gap theorems and Bernstein type theorems for complete $\lambda$-hypersurfaces with polynomial volume growth in terms of the norm of the second fundamental form in [14].
More results on $\lambda$-hypersurfaces can be found in [6, 19, 25, etc.].
We follow the notations of [14, 19] and call a hypersurface $X: M^n \rightarrow R^{n+1}$ a $\lambda$-hypersurface if
it satisfies the curvature condition
\begin{equation}
H-\frac{<N,X>}{2}=\lambda,
\end{equation}
where $\lambda$ is a constant, $N$ is the unit normal vector of $X$ and $H$ is the mean curvature of $M$. One can prove that $\lambda$-hypersurface is a hypersurface with constant mean curvature $\lambda$ in $R^{n+1}$ with respect to the metric $g_{ij} = e^{-\frac{|X|^2}{4}}\delta_{ij}$.
In this paper, we study the volume comparison theorem and topology of complete bounded $\lambda$-hypersurfaces in the weighted volume-preserving mean curvature flow. The organization of this article is as follows:
In section 2, we recall some backgrounds and derive some formulas for $\lambda$-hypersurfaces.
In section 3, we investigate the volume comparison theorem of complete bounded $\lambda$-hypersurfaces with $|A|\leq\alpha$.
In section 4, we give some applications of the volume comparison theorem of $\lambda$-hypersurfaces.
In section 5, we study the relation among $\lambda$, the radius $k$ and the dimension $n$, for the complete $\lambda$-hypersurfaces with controlled intrinsic
volume growth contained in the Euclidean closed ball $\overline{\mathcal{B}}^{n+1}_k(0)$.
In section 6, we generalize the well-known Myers' theorem on a complete and connected $\lambda$-hypersurface with $Ric_f \geq \frac{1-\lambda^2-3\alpha^2}{2} >0$.
In section 7, we obtain some properties on the topology at the infinity of a bounded $\lambda$-hypersurface with $Ric_f\geq 0$.
In section 8, we get some natural and general restrictions that force the $\lambda$-hypersurface to be compact.
\section{ Preliminaries}
Throughout this paper, the Einstein convention of summing over repeated indices from
1 to $n$ will be adopted.
Let $X:M^n \rightarrow R^{n+1}$ be an $n$-dimensional complete hypersurface of
Euclidean space $R^{n+1}$. We choose a local orthonormal frame field $\{e_\mathcal{\epsilon}\}^{n+1}_{\mathcal{\epsilon}=1}$ in $R^{n+1}$
with dual coframe field $\{\omega_\mathcal{\epsilon}\}^{n+1}_{\mathcal{\epsilon}=1}$ , such that, restricted to $M^n$, $e_1, \cdots , e_n$ are tangent
to $M^n$, and $e_{n+1}$ is the unit normal vector $N$. The coefficients of the second fundamental form $A$ are defined to be
\begin{equation}
h_{ij} = <\nabla_{e_i}e_j, N> .
\end{equation}
In particular, we have
\begin{equation}
\nabla_{e_i}N = -h_{ij}e_j.
\end{equation}
Since $<\nabla_NN, N>=0$, then the mean curvature
\begin{equation}
H = <\nabla_{e_i}N, e_i>=-h_{ii}.
\end{equation}
The Riemann curvature tensor and the Ricci tensor are
given by Gauss equation
\begin{equation}
R_{ijkl} = h_{ik}h_{jl} - h_{il}h_{jk},
\end{equation}
\begin{equation}
R_{ij} = -Hh_{ij} - h_{il}h_{lj}.
\end{equation}
Let $f = \frac{|X|^2}{4}$,
and denote by $d vol_f$ the corresponding weighted volume measure of $M$,
\begin{equation}
dvol_f = e^{-f} dvol.
\end{equation}
Thus, $M = (M, dvol_f)$ is a smooth metric measure space. There is a natural drifted Laplacian on $(M, dvol_f)$ defined by
\begin{equation}
\triangle_f = e^f div(e^{-f}\nabla) =\triangle -<\nabla ,\nabla f>.
\end{equation}
The $\infty$ - Bakry - Emery Ricci tensor $Ric_f$ of $(M, dvol_f)$ is defined by
\begin{equation}
Ric_f = Ric + Hess(f).
\end{equation}
Next we look at the $\infty$ - Bakry - Emery Ricci tensor $Ric_f$ of $\lambda$-hypersurface. For simplicity, we choose a frame such that $\nabla^T_{e_i}e_j=0$, then
\begin{eqnarray}
& &(Ric_f)_{ij}=R_{ij}+\nabla_{e_i}\nabla_{e_j} f\nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =R_{ij}+ \frac{1}{2}(<\nabla_{e_i}X^T,e_j>+ <X^T,\nabla_{e_i}e_j>) \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =R_{ij}+ \frac{1}{2}(<\nabla_{e_i}X,e_j>- <\nabla_{e_i}(<X,N>N),e_j> \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ + <X,\nabla_{e_i}e_j>-<X,N><N,\nabla_{e_i}e_j> )\nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =R_{ij}+ \frac{1}{2}(\delta_{ij}-<X,N><\nabla_{e_i}N,e_j> \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ +<X,<\nabla_{e_i}e_j,N>N>-<X,N><\nabla_{e_i}e_j,N>) \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ =R_{ij}+ \frac{1}{2}\delta_{ij}-\frac{1}{2}<X,N><\nabla_{e_i}N,e_j> \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ =R_{ij}+ \frac{1}{2}\delta_{ij}+\frac{1}{2}<X,N>h_{ij} \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ =-Hh_{ij} - h_{il}h_{lj}+\frac{1}{2}\delta_{ij}+(H-\lambda)h_{ij} \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\delta_{ij}-\lambda h_{ij}- h_{il}h_{lj} \nonumber\\
& &\ \ \ \ \ \ \ \ \ \ \ \geq\frac{1}{2}\delta_{ij}-\frac{1}{2}(\lambda^2+h^2_{ij})- h_{il}h_{lj}. \nonumber\\
\end{eqnarray}
Hence, we get the following lower bound for the $\infty$ - Bakry - Emery Ricci tensor $Ric_f$ of $\lambda$-hypersurface,
\begin{equation}
Ric_f \geq \frac{1-\lambda^2}{2}-\frac{3}{2}|A|^2.
\end{equation}
\section{ Volume comparison theorem of $\lambda$-hypersurfaces}
The classical volume comparison theorem shows that the volume of any ball is
bounded above by the volume of the corresponding ball in the model space, validating the intuitive picture: the bigger the curvature, the smaller the volume. Moreover, this is much less intuitive, if the volume of a big ball has a lower bound, then all the smaller balls also have lower bounds. It enjoys many geometric and topological applications.
In this section, we will investigate the volume comparison theorem of the complete bounded $\lambda$-hypersurface $M^n$ with $|A|\leq\alpha$, that is, the $\infty$ - Bakry - Emery Ricci tensor
\begin{equation}
Ric_f \geq \frac{1-\lambda^2-3\alpha^2}{2},
\end{equation}
where $\alpha$ is an arbitrary nonnegative constant.
Firstly, we fix a point $p\in M^n$, and let $r(x) = d(p, x)$
be the intrinsic distance from $p$ to $x$. This defines a Lipschitz function on the $\lambda$-hypersurface, which is
smooth except the cut locus of $p$. In geodesic polar coordinates, the volume element
$d vol = dr \wedge A_{\vartheta}(r) d\vartheta$, where $d\vartheta$ is the volume form of the standard $S^{n-1}$.
Let $B(p,R)$ be the geodesic ball of $M^n$ with radius $R$ centered at $p$, the
volume of $B(p,R)$ is defined by
\begin{equation}
Vol(B(p,R))=\int_0^R dr\int_{S^{n-1}(p,r)}A_{\vartheta}(r) d\vartheta.
\end{equation}
where $S^{n-1}(p,r)=\{x\in M| d(p,x)=r\}$.
Let $H(r)$ denote the mean curvature of the geodesic sphere at $p$ with outer normal vector $N$, then we have
\begin{equation}
\triangle r=H(r)=\frac{\partial}{\partial r}log A_\vartheta(r).
\end{equation}
Let $\omega_\alpha(t)$ be the solution to
\begin{equation}
\omega_\alpha^{''} + \frac{1-\lambda^2-3\alpha^2}{2(n-1)}\omega_\alpha = 0
\end{equation}
such that $\omega_\alpha(0) = 0$ and $\omega_\alpha^{'}(0) = 1$, i.e. $\omega_\alpha$ are the coefficients of the Jacobi
fields of the simply connected model space $M^n_{\alpha,\lambda}$ with constant curvature $\frac{1-\lambda^2-3\alpha^2}{2(n-1)}$, and
\begin{equation}
\omega_\alpha(t) =
\left\{
\begin{array}{ll}
\frac{\sqrt{2(n-1)}}{\sqrt{1-\lambda^2-3\alpha^2}}sin(\sqrt{\frac{1-\lambda^2-3\alpha^2}{2(n-1)}} \ t),\ \lambda^2+3\alpha^2<1 \\
t, \ \lambda^2+3\alpha^2=1 \\
\frac{\sqrt{2(n-1)}}{\sqrt{|1-\lambda^2-3\alpha^2|}}sinh(\sqrt{\frac{|1-\lambda^2-3\alpha^2|}{2(n-1)}}\ t), \ \lambda^2+3\alpha^2>1\\
\end{array}
\right.
\end{equation}
Let
$d vol_\alpha = dr \wedge A_{\vartheta_\alpha}(r) d\vartheta_\alpha$ be the volume element of model space $M^n_\alpha$, and denote by $H_{\alpha}$ the mean curvature of the geodesic sphere, then we have
\begin{equation}
H_\alpha(r)=\frac{A^{'}_{\vartheta_\alpha}(r)}{A_{\vartheta_\alpha}(r)}=(n-1)\frac{\omega_\alpha^{'}(r)}{\omega_\alpha(r) }.
\end{equation}
For real numbers $\alpha, \lambda, k$ and $n$, let
\begin{equation}
V_{\alpha,\lambda,k}(r)= vol(S^{n-1}(1))\int_0^r (A_{\vartheta_\alpha}(t))^{(1+\frac{k^2}{2(n-1)})}dt.
\end{equation}
Then we have the following volume comparison theorem for complete bounded $\lambda$-hypersurfaces.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq\alpha$, where $\alpha$ is an arbitrary nonnegative constant, and $\overline{\mathcal{B}}^{n+1}_k(0)$ denotes the Euclidean closed ball with center $0$ and radius $k$. Then for any point $p\in M^n$, and $ 0 < R_1\leq R_2$, we have
\begin{equation}
\frac{Vol(B(p,R_2))}{Vol(B(p,R_1))}\leq e^{\frac{k^2}{4}}\frac{V_{\alpha,\lambda,k}(R_2)}{V_{\alpha,\lambda,k}(R_1)},
\end{equation}
where we assume $R_2 \leq \frac{\sqrt{2(n-1)}\pi}{4\sqrt{1-\lambda^2-3\alpha^2}}$, if $\lambda^2+3\alpha^2<1$. \\
\end{thm}
\begin{proof}
(of Theorem 3.1)
Fix $p \in M$ as a base point, and let $\gamma : [0, r] \rightarrow M$ be a minimizing unit speed
geodesic from $p$. Let $\{E_i(t)\}_{i=1}^{n-1}$ be parallel orthonormal vector fields along
$\gamma(t) $ which are orthogonal to $\dot{\gamma}$. Constructing vector fields $\{X_i(t)= \frac{\omega_\alpha(t)}{|\omega_\alpha(r)|} E_i(t)\}_{i=1}^{n-1}$
along $\gamma $, then by the second variation formula, we have
\begin{eqnarray}
& &\triangle r\leq \int^r_0 \sum_{i=1}^{n-1}(|\nabla_{\dot{\gamma}}X_i|^2-<X_i,R_{X_i,\dot{\gamma}}\dot{\gamma}>)dt \nonumber \\
& &\ \ \ \ \ =\frac{1}{\omega_\alpha^2(r)}\int^r_0 ((n-1)(\omega^{'}_\alpha(t))^2-\omega_\alpha^2(t) Ric(\dot{\gamma},\dot{\gamma}))dt. \nonumber \\
\end{eqnarray}
By (2.8), (2.10) and the assumption $|A|\leq\alpha$, we have
\begin{eqnarray}
& &\triangle r\leq \frac{1}{\omega_\alpha^2(r)}\int^r_0 ((n-1)(\omega^{'}_\alpha(t))^2-\omega_\alpha^2(t)Ric_f(\dot{\gamma},\dot{\gamma}) )dt \nonumber \\
& &\ \ \ \ \ \ \ +\frac{1}{\omega_\alpha^2(r)}\int^r_0\omega_\alpha^2(t)(Hess(f)(\dot{\gamma},\dot{\gamma}))dt \nonumber \\
& &\ \ \ \ \ \leq\frac{1}{\omega_\alpha^2(r)}\int^r_0 ((n-1)((\omega^{'}_\alpha(t))^2-\frac{1-\lambda^2-3\alpha^2}{2(n-1)}\omega_\alpha^2(t)))dt \nonumber \\
& &\ \ \ \ \ \ \ +\frac{1}{\omega_\alpha^2(r)}\int^r_0(\omega_\alpha^2(t) \frac{d^2}{d t^2}(f\circ \gamma))dt. \nonumber \\
\end{eqnarray}
On the other hand, by (3.5) and (3.6), we can get
\begin{equation}
(\omega_\alpha^2(t)H_{\alpha}(t))^{'} = (n-1)((\omega^{'}_\alpha(t))^2-\frac{1-\lambda^2-3\alpha^2}{2(n-1)}\omega_\alpha^2(t)).
\end{equation}
Thus, the inequality (3.10) becomes
\begin{eqnarray}
& &\triangle r\leq \frac{1}{\omega_\alpha^2(r)}\int^r_0 (\omega_\alpha^2(t)H_{\alpha}(t))^{'}dt +\frac{1}{\omega_\alpha^2(r)}\int^r_0 (\omega_\alpha^2(t)\frac{d^2}{d t^2}(f\circ \gamma))dt \nonumber \\
& &\ \ \ \ \ =H_{\alpha}(r)+ \frac{1}{\omega_\alpha^2(r)}\int^r_0 (\omega_\alpha^2(t)\frac{d^2}{d t^2}(f\circ \gamma))dt, \nonumber \\
\end{eqnarray}
where we have used $\omega_\alpha(0) = 0$. By integration by parts on the last term, the expression (3.12) can be written as
\begin{eqnarray}
& &\triangle r\leq H_{\alpha}(r)- \frac{1}{\omega_\alpha^2(r)}\int^r_0 (\omega_\alpha^2(t))^{'}\frac{d}{d t}(f\circ \gamma) dt+ \frac{d}{d t}(f\circ \gamma)|_{t=r} \nonumber \\
& &\ \ \ \ \ =H_{\alpha}(r)+ \frac{1}{\omega_\alpha^2(r)}(\int^r_0 (\omega_\alpha^2(t))^{''}(f\circ \gamma) dt- (\omega_\alpha^2(r))^{'}(f\circ \gamma)(r))\nonumber \\
& &\ \ \ \ \ \ \ \ + \frac{d}{d t}(f\circ \gamma)|_{t=r}. \nonumber \\
\end{eqnarray}
Since the $\lambda$-hypersurface is contained in the ball $\mathcal{B}^{n+1}_k(0)$, then $f = \frac{|X|^2}{4} \leq \frac{k^2}{4}$. On the other hand, we assume $r\in(0, \frac{\sqrt{2(n-1)}\pi}{4\sqrt{1-\lambda^2-3\alpha^2}}]$, if $\alpha<\sqrt{\frac{1-\lambda^2}{3}}$. This implies that $(\omega_\alpha^2)^{'}\geq 0$ and $(\omega_\alpha^2)^{''}\geq 0$, for all $\alpha\geq 0$. Then
\begin{eqnarray}
& &\triangle r\leq H_{\alpha}(r)+ \frac{1}{\omega_\alpha^2(r)}(\frac{k^2}{4}\int^r_0 (\omega_\alpha^2(t))^{''}dt- (\omega_\alpha^2(r))^{'}(f\circ \gamma)(r))\nonumber \\
& &\ \ \ \ \ \ \ \ + \frac{d}{d t}(f\circ \gamma)|_{t=r} \nonumber \\
& &\ \ \ \ \ = H_{\alpha}(r)+ \frac{(\omega_\alpha^2(r))^{'}}{\omega_\alpha^2(r)}(\frac{k^2}{4} - (f\circ \gamma)(r))+ \frac{d}{d t}(f\circ \gamma)|_{t=r} \nonumber \\
& &\ \ \ \ \ \leq H_{\alpha}(r)+ \frac{k^2}{4}\frac{(\omega_\alpha^2(r))^{'}}{\omega_\alpha^2(r)}+\frac{d}{d t}(f\circ \gamma)|_{t=r} \nonumber \\
& &\ \ \ \ \ = H_{\alpha}(r)+ \frac{k^2}{2(n-1)}H_{\alpha}(r)+\frac{d}{d t}(f\circ \gamma)|_{t=r} \nonumber \\
& &\ \ \ \ \ = (1+ \frac{k^2}{2(n-1)})H_{\alpha}(r)+\frac{d}{d t}(f\circ \gamma)|_{t=r}, \nonumber \\
\end{eqnarray}
here in the fifth line we have used (3.6). By (3.3) and (3.6), the expression (3.14) can be written as
\begin{equation}
\frac{\partial}{\partial r}log A_\vartheta(r) \leq (1+ \frac{k^2}{2(n-1)})\frac{d}{d r}log A_{\vartheta_\alpha}(r)+\frac{d}{d t}(f\circ \gamma)|_{t=r}.
\end{equation}
Integrating it from $s_1$ to $s_2$, together with $f = \frac{|X|^2}{4} \leq \frac{k^2}{4}$, we have
\begin{equation}
log \frac{A_\vartheta(s_2)}{A_\vartheta(s_1)} \leq (1+ \frac{k^2}{2(n-1)})log \frac{A_{\vartheta_\alpha}(s_2)}{A_{\vartheta_\alpha}(s_1)}+\frac{k^2}{4}.
\end{equation}
Then
\begin{equation}
\frac{A_\vartheta(s_2)}{A_\vartheta(s_1)} \leq e^{\frac{k^2}{4}}(\frac{A_{\vartheta_\alpha}(s_2)}{A_{\vartheta_\alpha}(s_1)})^{(1+ \frac{k^2}{2(n-1)})},
\end{equation}
that is,
\begin{equation}
A_\vartheta(s_2) \cdot (A_{\vartheta_\alpha}(s_1))^{(1+ \frac{k^2}{2(n-1)})} \leq e^{\frac{k^2}{4}} \cdot A_\vartheta(s_1) \cdot (A_{\vartheta_\alpha}(s_2))^{(1+ \frac{k^2}{2(n-1)})}.
\end{equation}
Integrating it from $0$ to $R_1$ with respect to $s_1$, and from $0$ to $R_2$ with respect to $s_2$, we have
\begin{eqnarray}
& &\ \ \ \ \int_0^{R_2}A_\vartheta(s_2)d s_2 \cdot \int_0^{R_1}(A_{\vartheta_\alpha}(s_1))^{(1+ \frac{k^2}{2(n-1)})}d s_1 \nonumber \\
& &\leq e^{\frac{k^2}{4}} \cdot \int_0^{R_1}A_\vartheta(s_1)d s_1 \cdot \int_0^{R_2}(A_{\vartheta_\alpha}(s_2))^{(1+ \frac{k^2}{2(n-1)})}d s_2 . \nonumber \\
\end{eqnarray}
Integration along the sphere direction gives
\begin{equation}
Vol(B(p,R_2))\cdot V_{\alpha,\lambda,k}(R_1)\leq e^{\frac{k^2}{4}}\cdot Vol(B(p,R_1)) \cdot V_{\alpha,\lambda,k}(R_2).
\end{equation}
Then the result follows.
\end{proof}
Note that the $\lambda$-hypersurface is a self-shrinker of the mean curvature flow when $\lambda= 0$, then we have the following corollary.
\begin{cor}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete self-shrinker with $|A|\leq\alpha$. Then for any $p\in M^n$, $0 < R_1\leq R_2$, we have
\begin{equation}
\frac{Vol(B(p,R_2))}{Vol(B(p,R_1))}\leq e^{\frac{k^2}{4}}\frac{V_{\alpha,0,k}(R_2)}{V_{\alpha,0,k}(R_1)},
\end{equation}
where $R_2 \leq \frac{\sqrt{2(n-1)}\pi}{4\sqrt{1-\lambda^2-3\alpha^2}}$ if $\lambda^2+3\alpha^2<1$.
\end{cor}
Since $(A_{\vartheta_\alpha}(t))^{(1+\frac{k^2}{2(n-1)})}=(\omega_\alpha(t))^{(n-1)(1+\frac{k^2}{2(n-1)})}=(\omega_\alpha(t))^{(n+\frac{k^2}{2}-1)}$,
then $dr \wedge A_{\vartheta_\alpha}(r)^{(1+\frac{k^2}{2(n-1)})} d\vartheta_\alpha$ can be considered as the volume element of simply connected model space $M^{n+\frac{k^2}{2}}_{\alpha,\lambda}$ of dimension $n+\frac{k^2}{2}$ with constant curvature $\frac{1-\lambda^2-3\alpha^2}{2(n-1)}$. Now by $(3.18)$, we can obtain the following volume comparison for balls.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq\alpha$. Then for any $p\in M^n$, $ 0 < R_1\leq R_2$, we have
\begin{equation}
\frac{Vol(B(p,R_2))}{Vol(B(p,R_1))}\leq e^{\frac{k^2}{4}}\frac{Vol_{\alpha,\lambda}^{n+\frac{k^2}{2}}(R_2)}{Vol_{\alpha,\lambda}^{n+\frac{k^2}{2}}(R_1)},
\end{equation}
where $Vol_{\alpha,\lambda}^{n+\frac{k^2}{2}}(r)$ is the volume of the ball with radius $r$ in model space $M^{n+\frac{k^2}{2}}_{\alpha,\lambda}$, and $R_2 \leq \frac{\sqrt{2(n-1)}\pi}{4\sqrt{1-\lambda^2-3\alpha^2}}$ if $\lambda^2+3\alpha^2<1$.
\end{thm}
Actually, it is not easy to figure out the relevant conclusions by Theorem 3.1 and Theorem 3.3. In particular, for the complete $\lambda$-hypersurfaces with $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$(i.e. $Ric_f \geq 0$ and $|\lambda|\leq 1$ ), we obtain the following interesting result.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq\sqrt{\frac{1-\lambda^2}{3}}$. Then for any $p\in M^n$, $ 0 < R_1\leq R_2$, we have
\begin{equation}
\frac{Vol(B(p,R_2))}{Vol(B(p,R_1))}\leq e^{\frac{3k^2}{4}}\frac{V(R_2)}{V(R_1)},
\end{equation}
where $V(r)$ is the volume of the ball with radius $r$ in Euclidean space $R^{n}$. Moreover, we can get
\begin{equation}
Vol(B(p,R_2))\leq e^{\frac{3k^2}{4}} V(R_2).
\end{equation}
\end{thm}
\begin{proof}
(of Theorem 3.4) Since $|A| \leq\sqrt{\frac{1-\lambda^2}{3}}$ implies $Ric_f \geq 0$, then the expression (3.13) can be written as
\begin{eqnarray}
& &\triangle r\leq \frac{n-1}{r}+\frac{1}{r^2}(\int^r_0 (t^2)^{''}(f\circ \gamma)dt-((t^2)^{'} f\circ \gamma)|_{t=r})+ \frac{d}{d t}(f\circ \gamma)dt|_{t=r} \nonumber \\
& &\ \ \ \ \ =\frac{n-1}{r}+\frac{2}{r^2}\int^r_0(f\circ \gamma)dt-\frac{2}{r} f\circ \gamma(r)+ \frac{d}{d t}(f\circ \gamma)dt|_{t=r}. \nonumber \\
\end{eqnarray}
That is,
\begin{equation}
\frac{\partial}{\partial r}log A_\vartheta(r) \leq \frac{d}{d r}log r^{n-1}+\frac{2}{r^2}\int^r_0(f\circ \gamma)dt-\frac{2}{r} f\circ \gamma(r)+ \frac{d}{d t}(f\circ \gamma))dt|_{t=r}.
\end{equation}
Integrating from $s_1$ to $s_2$, together with $f = \frac{|X|^2}{4} \leq \frac{k^2}{4}$, we have
\begin{eqnarray}
& &log \frac{A_\vartheta(s_2)}{A_\vartheta(s_1)} \leq log \frac{s_2^{n-1}}{s_1^{n-1}}+\int^{s_2}_{s_1}(\frac{2}{r^2}\int^r_0(f\circ \gamma)dt)dr-\int^{s_2}_{s_1}(\frac{2}{r} f\circ \gamma(r))dr \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +f\circ \gamma|_{s_1}^{s_2} \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ = log \frac{s_2^{n-1}}{s_1^{n-1}}-\frac{2}{r}\int^r_0(f\circ \gamma)dt|_{s_1}^{s_2}+f\circ \gamma|_{s_1}^{s_2} \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \leq log \frac{s_2^{n-1}}{s_1^{n-1}}+\frac{k^2}{2}+\frac{k^2}{4} \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ = log \frac{s_2^{n-1}}{s_1^{n-1}}+\frac{3k^2}{4}. \nonumber \\
\end{eqnarray}
Now the result is obvious.
\end{proof}
\section{Some applications of volume comparison theorem}
The classical volume comparison theorem is a powerful tool in studying the manifolds with lower Ricci
curvature bound(See [36]). In this section, we will give some applications of
the volume comparison of $\lambda$-hypersurfaces with lower $\infty$ - Bakry - Emery Ricci
curvature bound.
Firstly, we obtain the lower bound and upper bound on
volume growth for $\lambda$-hypersurface with $Ric_f \geq 0$ (i.e. $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$).
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$. Then, for any $p \in M$,
\begin{equation}
C_1 R^{n} \geq Vol(B(p,R))\geq C_2 R.
\end{equation}
\end{thm}
\begin{proof}
(of Theorem 4.1) Since $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$, for any $p \in M$,
by Theorem 3.4, we have
\begin{equation}
Vol(B(p,R))\leq C_1 R^{n}. \\
\end{equation}
On the other hand, for $T_1\leq T_2, R_1\leq R_2, T_1\leq R_1, T_2\leq R_2,$ and $q \in M$, we let $A_q(R_1,R_2)$ be the set of $x \in M$ such that $R_1\leq r(x) \leq R_2$ and $V(T_1,T_2)= vol(S^{n-1}(1))\int_{T_1}^{T_2} t^{n-1}dt$, where $r(x)=d(q,x)$. Then
by (3.27), we can get
\begin{equation}
\frac{Vol(A_q(R_1,R_2))}{Vol(A_q(T_1,T_2))}\leq e^{\frac{3k^2}{4}}\frac{V(R_1,R_2)}{V(T_1,T_2)}.
\end{equation}
Let $\Gamma$ be a geodesic based at $p$ in $M$, by $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$ and the annulus relative volume comparison
(4.3) to annuli centered at $\Gamma(t)$, we have
\begin{eqnarray}
& &\frac{Vol(A_{\Gamma(t)}(t-1,t+1))}{Vol(B(\Gamma(t),t-1))}\leq e^{\frac{3k^2}{4}}\frac{(t+1)^{n}-(t-1)^{n}}{(t-1)^{n}} \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\tilde{c}(n,k)}{t}. \nonumber \\
\end{eqnarray}
By $B(\Gamma(0),1)\subset A_{\Gamma(t)}(t-1,t+1)$, we get
\begin{eqnarray}
& &Vol(B(\Gamma(t),t-1))\geq \frac{t}{\tilde{c}(n,k)}Vol(A_{\Gamma(t)}(t-1,t+1)) \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \geq \frac{t}{\tilde{c}(n,k)}Vol(B(p,1)) \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tilde{\tilde{c}} t. \nonumber \\
\end{eqnarray}
Therefore,
\begin{equation}
Vol(B(p,R))\geq Vol(B(\Gamma(\frac{R}{2}),\frac{R}{2}-1))\geq C_2 R.
\end{equation}
Combining (4.2) and (4.6), the result follows easily.
\end{proof}
In [21], Milnor observed that polynomial volume growth on the universal cover of a manifold restricts the structure of its fundamental group. Thus Theorem 3.4 also implies the following extension of Milnor's Theorem.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$. Then any finite generated subgroup of the fundamental group
of $M$ has polynomial
growth of order at most $n$.
\end{thm}
Note that the relation between the fundamental group and the first Betti number given by the Hurewicz Theorem
[34], Ricci curvature can also give control on the first Betti number. By the same assertions as in M. Gromov [13], we can prove the following theorem by Theorem 3.1.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq \alpha$, $diam_M\leq d$. Then $b_1(M) \leq c(n,k,\alpha,d^2)$.
\end{thm}
Theorem 3.1 also implies the following extensions
theorems of Anderson [1]. We will leave it for readers.
\begin{thm}
For the class of $n$-dimensional $\lambda$-hypersurfaces $M$ with $|A|\leq \alpha, vol\geq v>0, diam \leq d,$ which contained in the compact ball $\overline{\mathcal{B}}^{n+1}_k(0) \subset R^{n+1}$, there are only finitely many isomorphism classes of the fundamental group of $M$.
\end{thm}
The volume comparison has many other geometric applications, such as, in the Gromov-Hausdorff convergence theory, in the rigidity and pinching theory. We will leave these statements to the interested readers.
\section{Estimate of the exterior radius}
The $n$-dimensional sphere $S^{n}(k)$ with radius $k=\sqrt{\lambda^2+2n}+\lambda$ is a compact $\lambda$-hypersurface and contained in the compact closed ball $\overline{\mathcal{B}}^{n+1}_k(0)\subset R^{n+1}$.
Our first remark is that if a complete $\lambda$-hypersurface with controlled intrinsic
volume growth is contained in some Euclidean closed ball $\overline{\mathcal{B}}^{n+1}_k(0)$ with center $0$ and radius $k$, then there is an obvious
relation among $\lambda$, the radius $k$ and the dimension $n$. To prove this, we need the following elementary lemma.
\begin{lem}
(see[27])Let $(M, dvol_f)$ be a geodesically complete weighted manifold satisfying the volume growth condition
\begin{equation}
\frac{R}{log(Vol_f B(p,R))}
\notin L^1(+\infty),
\end{equation}
Then the weak maximum principle at infinity for the $f$-Laplacian holds on $M$.
\end{lem}
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface whose intrinsic volume growth satisfies
\begin{equation}
\frac{R}{log(Vol B(p,R))}
\notin L^1(+\infty),
\end{equation}
where $B(p,R)$ is the geodesic ball of $\lambda$-hypersurface $M^n$ with radius $R$ centered at $p$, and $\mathcal{B}^{n+1}_k(0)$ denotes the Euclidean ball with center $0$ and radius $k$. Then
\begin{equation}
k\geq \sqrt{\lambda^2+2n}-|\lambda|.
\end{equation}
\end{thm}
\begin{proof}
(of Theorem 5.2)
Since $\triangle X = - H N$ on the hypersurface and $H-\frac{<N,X>}{2}=\lambda$ on the $\lambda$-hypersurface, then
\begin{eqnarray}
& &\triangle |X|^2=2|\nabla X|^2+2<\triangle X, X> \nonumber \\
& &\ \ \ \ \ \ \ \ \ =2|\nabla X|^2-2H<N, X> \nonumber \\
& &\ \ \ \ \ \ \ \ \ =2|\nabla X|^2-2(\frac{<N,X>}{2}+\lambda)<N, X> \nonumber \\
& &\ \ \ \ \ \ \ \ \ =2n-2\lambda<N, X>- <N,X>^{2}. \nonumber \\
\end{eqnarray}
Also note that $\nabla|X|^2 = 2 X^T$, where $X^T$ is the tangential projection of $X$, we can get
\begin{eqnarray}
& &\triangle_f |X|^2=\triangle |X|^2- <\nabla |X|^2,\nabla f> \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =\triangle |X|^2- <\nabla |X|^2,\nabla \frac{|X|^2}{4}> \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =\triangle |X|^2- \frac{1}{4}<2 X^T,2 X^T> \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =2n-2\lambda<N, X>- <N,X>^{2}- |X^T|^2 \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ =2n-2\lambda<N, X>- |X|^2 \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \geq 2n-2|\lambda||X|- |X|^2. \nonumber \\
\end{eqnarray}
On the other hand, since $c^{-1} d vol_f \leq d vol \leq c d vol_f$
for a large enough constant $c > 1$, then
\begin{equation}
R \rightarrow \frac{R}{log(Vol_f B(p,R))}
\notin L^1(+\infty),
\end{equation}
which implies that on the $\lambda$-hypersurface the weak maximum
principle holds at infinity for the drifted Laplacian $\triangle_f$ (Lemma 5.1). Therefore
\begin{equation}
0 \geq 2n- 2|\lambda|sup_M|X|- sup_M|X|^2\geq 2n- 2|\lambda|k- k^2,
\end{equation}
and the claimed lower estimate on $k$ follows. This completes the proof of Theorem 5.2.
\end{proof}
By(4.1), we can specialize Theorem 5.2 to the following
\begin{cor}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$, then
\begin{equation}
k\geq \sqrt{\lambda^2+2n}-|\lambda|.
\end{equation}
\end{cor}
\section{Estimate of the intrinsic diameter}
The purpose of this section is to generalize the well-known Myers' theorem [23]
on a complete and connected $\lambda$-hypersurface with $Ric_f \geq \frac{1-\lambda^2-3\alpha^2}{2} >0$ (i.e. $|A|\leq\alpha<\sqrt{\frac{1-\lambda^2}{3}}$). In particular, we obtain the following
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with $|A|\leq \alpha < \sqrt{\frac{1-\lambda^2}{3}}$. Then $M$ is compact and the intrinsic diameter satisfies
\begin{equation}
diam(M) \leq \sqrt{\frac{2(n-1)(n+k^2-1)}{1-\lambda^2-3\alpha^2}}\pi.
\end{equation}
\end{thm}
\begin{proof}
(of Theorem 6.1) The proof goes by contradiction. If $\gamma: [0,l] \rightarrow M$ is a unit speed geodesic of length $l >\sqrt{\frac{2(n-1)(n+k^2-1)}{1-\lambda^2-3\alpha^2}}\pi$.
Let $\{E_i(t)\}_{i=1}^{n-1}$ be parallel orthonormal vector fields along
$\gamma $ which are orthogonal to $\dot{\gamma}$. Using vector fields $\{X_i(t)= \sin(\frac{\pi}{l}t) E_i(t)\}_{i=1}^{n-1}$
along $\gamma $, then we get the index form
\begin{equation}
I(X_i(t),X_i(t))=\int^l_0 (|\nabla_{\dot{\gamma}}X_i|^2-<X_i,R_{X_i,\dot{\gamma}}\dot{\gamma}>)dt.
\end{equation}
Then by $|A|\leq \alpha < \sqrt{\frac{1-\lambda^2}{3}}$, $f = \frac{|X|^2}{4} \leq \frac{k^2}{4}$, $l >\sqrt{\frac{2(n-1)(n+k^2-1)}{1-\lambda^2-3\alpha^2}}\pi$, (2.8) and (2.10), we have
\begin{eqnarray}
& & \ \ \ \ \sum_{i=1}^{n-1}I(X_i(t),X_i(t)) \nonumber \\
& &=\int^l_0 \sum_{i=1}^{n-1}(|\nabla_{\dot{\gamma}}X_i|^2-<X_i,R_{X_i,\dot{\gamma}}\dot{\gamma}>)dt \nonumber \\
& &=\int^l_0((n-1)(\frac{\pi}{l}cos(\frac{\pi}{l}t))^2- (\sin(\frac{\pi}{l}t))^2Ric(\dot{\gamma},\dot{\gamma}))dt\nonumber \\
& & \leq\int^l_0((n-1)(\frac{\pi}{l}cos(\frac{\pi}{l}t))^2- (\sin(\frac{\pi}{l}t))^2(\frac{1-\lambda^2-3\alpha^2}{2(n-1)}))dt\nonumber \\
& & \ \ \ \ +\int^l_0(\sin(\frac{\pi}{l}t))^2(Hess(f)(\dot{\gamma},\dot{\gamma}))dt \nonumber \\
& &=\frac{(n-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)}+\int^l_0(\sin(\frac{\pi}{l}t))^2\frac{\partial^2}{\partial t^2}(f\circ\gamma) dt\nonumber \\
& &=\frac{(n-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)}-\frac{\pi}{l}
\int^l_0\sin(\frac{2\pi}{l}t)\frac{\partial}{\partial t}(f\circ\gamma) dt\nonumber \\
& &=\frac{(n-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)}+ \frac{2\pi^2}{l^2}
\int^l_0\cos(\frac{2\pi}{l}t)(f\circ\gamma) dt\nonumber \\
& &\leq \frac{(n-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)}+ \frac{2\pi^2}{l^2}
\int^l_0\ |(f\circ\gamma)| dt\nonumber \\
& &\leq \frac{(n-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)}+ \frac{\pi^2k^2}{2l}
\nonumber \\
& &=\frac{(n+k^2-1)\pi^2}{2l}-\frac{(1-\lambda^2-3\alpha^2)l}{4(n-1)} \nonumber \\
& &< 0.\nonumber \\
\end{eqnarray}
This implies $I(X_i(t),X_i(t))< 0$, for some $1\leq i\leq n-1$. Namely, the index form is not positive semi-definite. It is a contradiction. So we finish the proof.
\end{proof}
\section{Topology at infinity of $\lambda$-hypersurfaces }
In this section, by the following Cheeger-Gromoll-Lichnerowicz splitting theorem, we obtain a bit of information on the topology at infinity of a bounded $\lambda$-hypersurfaces with $Ric_f\geq 0$.
\begin{lem}
(see[17])Let $(M^n, dvol_f)$ be a geodesically complete weighted manifold with $Ric_f\geq 0$ for some
bounded function $f$ and $M^n$ contains a line, then $M^n = N^{n-1} \times R$ and $f$ is constant along the line.
\end{lem}
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete non-compact $\lambda$-hypersurface with $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$. Then $M$ does not contain a line. In particular, $M$ is connected at infinity, i.e., $M$ has only one end.
\end{thm}
\begin{proof}
(of Theorem 7.2) The proof goes by contradiction. By $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$, we have $Ric_f\geq 0$. If $M$ contains a line, by Lemma 7.1, $M$ can be split isometrically as the Riemannian product $(N^{n-1} \times R, g_N + dt\otimes dt)$, and $f$ is constant along the line. Hence,
\begin{equation}
Hess(f)(\frac{\partial}{\partial t}, \frac{\partial}{\partial t}) = 0.
\end{equation}
On the other hand, $H-\frac{<N,X>}{2}=\lambda$ implies that
\begin{equation}
H_i=-\frac{1}{2}\sum_jh_{ij}<X,e_j>,
\end{equation}
and
\begin{equation}
H_{ik}=-\frac{1}{2}(\sum_j(h_{ijk}<X,e_j>+h_{ik}+\sum_jh_{ij}h_{kj}(2H-2\lambda)).
\end{equation}
Therefore,
\begin{eqnarray}
& &\triangle h_{ij}=-|A|^2h_{ij}-H h_{ik}h_{kj}- H_{ij} \nonumber \\
& &\ \ \ \ \ \ \ =-|A|^2h_{ij}-H h_{ik}h_{kj}+\frac{1}{2}(\sum_k(h_{ikj}<X,e_k>+h_{ij} \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ +\sum_k h_{ik}h_{jk}(2H-2\lambda)) \nonumber \\
& &\ \ \ \ \ \ \ =-|A|^2h_{ij}+\sum_kh_{ikj}<\frac{X}{2},e_k>+\frac{1}{2}h_{ij}-\lambda\sum_k h_{ik}h_{jk}. \nonumber \\
\end{eqnarray}
Then it follows that
\begin{equation}
\triangle_f h_{ij}=(\frac{1}{2}-|A|^2)h_{ij}-\lambda\sum_k h_{ik}h_{jk}.
\end{equation}
By $\frac{1}{2}\triangle_f\eta^2=|\nabla \eta|^2+\eta\triangle_f\eta$, we have
\begin{equation}
\frac{1}{2}\triangle_f |A|^2=\sum_{ijk}h_{ijk}^2+(\frac{1}{2}-|A|^2)|A|^2-\lambda\sum_{ijk} h_{ij}h_{jk}h_{ki}.
\end{equation}
Then, by $|A|\leq \sqrt{\frac{1-\lambda^2}{3}}$, we have
\begin{eqnarray}
& &\frac{1}{2}\triangle_f |A|^2\geq\sum_{ijk}h_{ijk}^2+(\frac{1}{2}-|A|^2)|A|^2-|\lambda||A|^3 \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ =\sum_{ijk}h_{ijk}^2+(\frac{1}{2}-|\lambda||A|-|A|^2)|A|^2 \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \geq\sum_{ijk}h_{ijk}^2+(\frac{1-\lambda^2}{2}-\frac{3}{2}|A|^2)|A|^2 \nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ \ \geq0. \nonumber \\
\end{eqnarray}
Hence, the strong maximum principle for the drifted Laplacian yields that either $|A| < \sqrt{\frac{1-\lambda^2}{3}} $ or $|A| \equiv \sqrt{\frac{1-\lambda^2}{3}}$.\\
Case 1: If $|A| < \sqrt{\frac{1-\lambda^2}{3}},$ then $|A|=0$ and
\begin{equation}
Hess(f)(\frac{\partial}{\partial t}, \frac{\partial}{\partial t})=Ric_f(\frac{\partial}{\partial t}, \frac{\partial}{\partial t})\geq \frac{(1-\lambda^2-3|A|^2)}{2} >0 ,
\end{equation}
which contradicts (7.1).\\
Case 2: If $|A| \equiv \sqrt{\frac{1-\lambda^2}{3}},$ then $h_{ijk}=0.$ By the classical Lawson's classification theorem, $M$ is a cylindrical
product $S^k\times R^{n-k}$. Since the $\lambda$-hypersurface is bounded,
we conclude that $M=S^n$, which contradicts the assumption that $M$ is not compact.
Then it completes the proof.
\end{proof}
\section{Compactness of $\lambda$-hypersurfaces}
In this section, we will follow the notations and conclusions of [4]. Then the technique of the Feller property combining with the stochastic completeness, will enable us to get natural and general restrictions that force the $\lambda$-hypersurface to be compact.
\begin{thm}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete non-compact $\lambda$-hypersurface. Then we have
\begin{equation}
\lim_{r\rightarrow \infty}\sup_{M\backslash B(r)}|A|\geq \sqrt{\frac{1-\lambda^2}{3}},
\end{equation}
where $B(r)$ is the geodesic ball of $\lambda$-hypersurface $M^n$ with radius $r$.
\end{thm}
\begin{proof}
(of Theorem 8.1) By contradiction. Suppose that $\displaystyle\lim_{r\rightarrow \infty}\sup_{M\backslash B(r)}|A|< \sqrt{\frac{1-\lambda^2}{3}}$, we have $|A| \in L^\infty(M)$. Since $|\nabla f|=\frac{1}{2}|X^T|\leq\frac{1}{2}|X|<\frac{1}{2}k$,
then by (2.10), Theorem 7 and Theorem 8 in $[4]$, $M$ is stochastically complete and Feller with respect to $\triangle_f$.
By the Simons type equation (7.7), we have, for some $\alpha$,
\begin{equation}
\frac{1}{2}\triangle_f |A|^2\geq(\frac{1}{2}-|\lambda|\alpha-\alpha^2)|A|^2,
\end{equation}
outside a smooth domain $\Omega\subset\subset M$. Then Theorem 2 in [4] gives
\begin{equation}
|A|(x) \rightarrow 0, \mbox{\ as\ } x \rightarrow \infty.
\end{equation}
By (8.3) and $Hess(f)_{ij}=\frac{1}{2}(\delta_{ij}+<X,N>h_{ij})$, we have
\begin{equation}
\frac{d^2}{dt^2}(f\circ \gamma)(t)=Hess(f)(\dot{\gamma}, \dot{\gamma})\geq \frac{1}{4},
\end{equation}
for some ray $\gamma: [0, +\infty) \rightarrow M$, and $t >>1$.
It follows by integration that $|X|^2\rightarrow +\infty$ along $\gamma$, therefore, $M$ is
unbounded. This is a contradiction. So we finish the proof.
\end{proof}
By Theorem 8.1, the following corollary is obvious.
\begin{cor}
Let $X : M \rightarrow \overline{\mathcal{B}}^{n+1}_{k} (0) \subset R^{n+1}$ be an $n$-dimensional complete $\lambda$-hypersurface with
\begin{equation}
\lim_{r\rightarrow \infty}\sup_{M\backslash B(r)}|A|< \sqrt{\frac{1-\lambda^2}{3}},
\end{equation}
where $B(r)$ is the geodesic ball of $\lambda$-hypersurface $M^n$ with radius $r$. Then $M$ is compact.
\end{cor} | 73,679 |
\section{Motivic homological algebra}
All schemes are assumed to be noetherian
and of finite Krull dimension.
We will say `$S$-scheme' for `separated scheme of finite type over $S$'.
If $\A$ is an abelian category, we let $\Comp(\A)$, $\Htp(\A)$ and
$\Der(\A)$ be respectively the category of unbounded cochain
complexes of $\A$, the same category modulo cochain homotopy equivalence,
and the unbounded derived category of $\A$.
\subsection{$\AA^1$-invariant cohomology}
\begin{paragr}\label{axiomesVV}
We suppose given a scheme $S$.
We consider a full subcategory $\V$ of the category $\sm/S$
of smooth $S$-schemes satisfying the following
properties\footnote{In practice, the category $\V$
will be $\sm/S$ itself or the full subcategory of
smooth affine $S$-schemes.}.
\begin{itemize}
\item[(a)] $\AA^n_S$ belongs to $\V$ for $n\geq 0$.
\item[(b)] If $X'\To X$ is an \'etale morphism
and if $X$ is in $\V$, then there exists a Zariski
covering $Y\To X'$ with $Y$ in $\V$.
\item[(c)] For any pullback square of $S$-schemes
$$\xymatrix{
X'\ar[r]\ar[d]_{u'}&X\ar[d]^u\\
Y'\ar[r]&Y}$$
in which $u$ is smooth, if $X$,$Y$ and $Y'$ are in $\V$,
so is $X'$.
\item[(d)] If $X$ and $Y$ are in $\V$, then their disjoint
union $X\amalg Y$ is in $\V$.
\item[(e)] For any smooth $S$-scheme $X$, there exists
a Nisnevich covering $Y\To X$ of $X$
with $Y$ in $\V$.
\end{itemize}
We recall that a Nisnevich covering is a surjective and completely
decomposed \'etale morphism. This defines the
Nisnevich topology on $\V$; see e.g. \cite{nis,KS,TT,MV}.
The last property (e) ensures that
the category of sheaves on $\V$ is equivalent to the category
of sheaves on the category of smooth $S$-schemes as far as we consider
sheaves for the Nisnevich topology (or any stronger one).
\end{paragr}
\begin{paragr}\label{defidsisqu}
A \emph{distinguished square} is a pullback square of schemes
\begin{equation}\label{distsquare}\begin{split}\xymatrix{
W\ar[r]^i\ar[d]_g&V\ar[d]^f\\
U\ar[r]_j&X}
\end{split}\end{equation}
where $j$ is an open immersion and $f$ is an \'etale morphism
such that the induced map from $f^{-1}((X-U)_{\mathit{red}})$ to
$(X-U)_{\mathit{red}}$ is an isomorphism. For such a distinguished
square, the map $(j,f) : U\amalg V\To X$ is a Nisnevich
covering. A very useful property of the Nisnevich topology is that
any Nisnevich covering can be refined by a covering coming from
a distinguished square (as far as we work with noetherian schemes).
This leads to the following characterization of the Nisnevich sheaves.\\
\indent A presheaf $F$ on $\V$ is a sheaf for the Nisnevich topology
if and only if for any distinguished square of shape \eqref{distsquare},
we obtain a pullback square
\begin{equation}\label{distsquare2}\begin{split}\xymatrix{
F(X)\ar[r]^{f^*}\ar[d]_{j^*}&F(V)\ar[d]^{i^*}\\
F(U)\ar[r]_{g^*}&F(W)}
\end{split}\end{equation}
This implies that Nisnevich sheaves are stable by filtering colimits
in the category of presheaves on $\V$. In other words,
if $I$ is a small filtering category,
and if $F$ is a functor from $I$ to the category
of presheaves on $\V$ such that $F_{i}$ is a Nisnevich sheaf
for all $i\in I$, then the presheaf $\limind_{i\in I}F_{i}$
is a Nisnevich sheaf.
\end{paragr}
\begin{paragr}\label{nissheavespremodcat}
We fix a commutative ring $R$.
Let $\mathit{Sh}(\V,R)$ be the category of Nisnevich sheaves
of $R$-modules on $\V$. For a presheaf (of $R$-modules) $F$,
we denote by $F_\nis$ the Nisnevich sheaf associated to $F$.
We can form its derived category
$$\Der(\V,R)=\Der(\mathit{Sh}(\V,R)) \ .$$
More precisely, the category $\Der(\V,R)$ is obtained
as the localization of the category $\Comp(\V,R)$ of
(unbounded) complexes of the Grothendieck abelian category
$\mathit{Sh}(\V,R)$ by the class of
quasi-isomorphisms. As we have an equivalence of categories
$$\mathit{Sh}(\V,R)\simeq\mathit{Sh}(\sm/S,R) \ , $$
we also have a canonical equivalence of categories
\begin{equation}\label{Nispremodcat1}
\Der(\V,R)\simeq\Der(\sm/S,R) \ .
\end{equation}
We have a canonical functor
\begin{equation}
R:\V\To\mathit{Sh}(\V,R)\quad , \qquad X\longmapsto R(X)
\end{equation}\label{Nispremodcat2}
where $R(X)$ denotes the Nisnevich sheaf associated to the
presheaf
$$Y\longmapsto\text{free $R$-module generated by $\Hom^{}_\V(Y,X)$.}$$
Note that according to \ref{defidsisqu},
for any $X$ in $\V$, and any small filtering system $(F_i)_{i \in I}$
of $\mathit{Sh}(\V,R)$, the canonical map
\begin{equation}\label{filtglobsection}
\limind_{i\in I}\Hom_{\mathit{Sh}(\V,R)}(R(X),F_{i})\To
\Hom_{\mathit{Sh}(\V,R)}(R(X),\limind_{i\in I}F_{i})
\end{equation}
is an isomorphism (we can even take
$X$ to be any smooth $S$-scheme according to the previous equivalence).
For a complex $K$ of presheaves of $R$-modules on $\V$, we have
a canonical isomorphism
\begin{equation}\label{Nispremodcat3}
\hypercoh^n_\nis(X,K_\nis)=\Hom^{}_{\Der(\V,R)}(R(X),K_\nis[n])
\end{equation}
where $X$ is an object of $\V$, $n$ is an integer, and
$\hypercoh^n_\nis(X,K_\nis)$ is the
Nisnevich hypercohomology with coefficients in $K$.
\end{paragr}
\begin{paragr}\label{predefcmfnis}
For a sheaf of $R$-modules $F$, and an integer $n$,
we denote by $D^nF$ the complex concentrated in degrees $n$
and $n+1$ whose only non trivial differential is the identity of $F$.
We write $S^{n}F$ for the sheaf $F$ seen as a complex concentrated
in degree $n$. We have a canonical inclusion of $S^{n+1}F$ in $D^nF$.
We say that a morphism of complexes of sheaves of $R$-modules
is a \emph{$\V$-cofibration} if it is contained in the smallest
class of maps stable by pushout, transfinite composition and retract
that contains the maps of the form $S^{n+1}R(X)\To D^nR(X)$
for any integer $n$ and any $X$ in $\V$. For example, for any $X$ in $\V$,
the map $0\To R(X)$ is a $\V$-cofibration (where $R(X)$ is seen
as a complex concentrated in degree $0$). A complex of presheaves $K$ is
\emph{$\V$-cofibrant} if $0\To K$ is a $\V$-cofibration.
A complex of presheaves of $R$-modules $K$ on $\V$
is \emph{$\V_\nis$-local} if for any $X$ in $\V$, the canonical map
$$H^n(K(X))\To \hypercoh^n_\nis(X,K_\nis)$$
is an isomorphism of $R$-modules.
A morphism $p:K\To L$ of complexes of presheaves of $R$-modules on $\V$
is $\V$-surjective if for any $X$ in $\V$, the map
$K(X)\To L(X)$ is surjective.
\end{paragr}
\begin{prop}\label{existmodcatlocalnis}
The category of complexes of Nisnevich sheaves of $R$-modules on $\V$
is a proper Quillen closed model category structure whose weak equivalences are the
quasi-isomorphisms, whose cofibrations are the
$\V$-cofibrations and whose fibrations are the $\V$-surjective
morphisms with $\V_{\nis}$-local kernel.
In particular, for any $X$ in $\V$, $R(X)$ is $\V$-cofibrant.
\end{prop}
\begin{proof}
If $\X$ is a simplicial object of $\V$, we denote by
$R(\X)$ the associated complex. If $\X$ is a Nisnevich hypercovering
of an object $X$ of $\V$, we have a canonical morphism
from $R(\X)$ to $R(X)$ and we define
$$\widetilde{R}(\X)={\mathrm{Cone}}\big(R(\X)\To R(X)\big) \ . $$
Let $\G$ be the collection of the $R(X)$'s for $X$ in $\V$,
and $\H$ the class of the $\widetilde{R}(\X)$'s
for all the Nisnevich hypercoverings of any object $X$ of $\V$.
Then $(\G,\H)$ is a descent structure
on $\mathit{Sh}(\V,R)$ as defined in \cite[Definition 1.4]{HCD},
so that we can apply \cite[Theorem 1.7 and Corollary 4.9]{HCD}.
\end{proof}
\begin{paragr}
The model structure above will be called the \emph{$\V$-local
model structure}.
\end{paragr}
\begin{cor}\label{flasqueresolution}
For any complex of Nisnevich sheaves of $R$-modules $K$,
there exists a quasi-isomorphism $K\To L$
where $L$ is $\V_\nis$-local.
\end{cor}
\begin{proof}
We just have to choose a factorization of
$K\To 0$ into a quasi-isomorphism $K\To L$
followed by a fibration $L\To 0$ for the above
model structure.
\end{proof}
\begin{prop}\label{exactdistsquare}
For any distinguished square
$$\xymatrix{
W\ar[r]^i\ar[d]_g&V\ar[d]^f\\
U\ar[r]_j&X}$$
the induced commutative square of sheaves of $R$-modules
$$\xymatrix{
R(W)\ar^{i*}[r]\ar_{g_*}[d]&R(V)\ar^{f_*}[d]\\
R(U)\ar_{j_*}[r]&R(X)}$$
is exact; this means that it is cartesian and cocartesian,
or equivalently that it gives rise to a short exact
sequence in the category of sheaves of $R$-modules
$$0\To R(W)
\xrightarrow{g_*-i_*} R(U)\oplus R(V)
\xrightarrow{(j_*,f_*)} R(X)\To 0 \ .$$
\end{prop}
\begin{proof}
The characterization of Nisnevich sheaves given in \ref{defidsisqu}
implies that the sequence
$$0\To R(W)\To R(U)\oplus R(V)\To R(X)\To 0$$
is right exact. So the result comes from the injectivity of the
map from $R(W)$ to $R(V)$ induced by $i$.
\end{proof}
\begin{paragr}
Let $K$ be a complex of presheaves of $R$-modules on $\V$.
A \emph{closed pair} will be a couple $(X,Z)$ such that
$X$ is a scheme in $\V$, $Z \subset X$ is a closed subset
and $X-Z$ belongs to $\V$. Let $j$ be the immersion
of $X-Z$ in $X$. We put
$$
K_Z(X)=\mathrm{Cone}\big(K(X) \xrightarrow{j^*} K(X-Z)\big)[-1].
$$
A morphism of closed pairs
$f:(Y,T) \rightarrow (X,Z)$ is a morphism of schemes
$f:Y \rightarrow X$ such that $f^{-1}(Z) \subset T$.
The morphism of closed pairs $f$ will be called \emph{excisive}
when the induced square
$$\xymatrix{
Y-T\ar[r]^/4pt/i\ar[d]_g&Y\ar[d]^f\\
X-Z\ar[r]_/4pt/j&X}$$
is distinguished.
The complex $K_Z(X)$ is obviously functorial with respect
to morphisms of closed pairs. We will say that $K$
has the \emph{excision property on $\V$}
if for any excisive morphism $f:(Y,T) \rightarrow (X,Z)$
the map $K_T(Y) \rightarrow K_Z(X)$ is a quasi-isomorphism.
We will say that $K$ has the \emph{Brown-Gersten property on $\V$
with respect to the Nisnevich topology}, or the \emph{B.-G.-property} for short,
if for any distinguished square
$$\xymatrix{
W\ar[r]^i\ar[d]_g&V\ar[d]^f\\
U\ar[r]_j&X}$$
in $\V$, the square
$$\xymatrix{
K(X)\ar[r]^{f^*}\ar[d]_{j^*}&K(V)\ar[d]^{i^*}\\
K(U)\ar[r]_{g^*}&K(W)}$$
is a homotopy pullback (or equivalently a
homotopy pushout) in the category of complexes of $R$-modules.
The latter condition means that the commutative square
of complexes of $R$-modules
obtained from the distinguished square above by applying $K$
leads canonically to a long exact sequence ``\`a la Mayer-Vietoris''
$$H^n(K(X))
\xrightarrow{j^*+f^*} H^n(K(U))\oplus H^n(K(V))
\xrightarrow{g^*-i^*} H^n(K(W)) \To H^{n+1}(K(X))$$
The complexes satisfying the B.-G.-property are in fact
the fibrant objects of the model structure of Proposition \ref{existmodcatlocalnis}.
This is shown by the following result which is essentially due to
Morel and Voevodsky.
\end{paragr}
\begin{prop}\label{bgtilde}
Let $K$ be a complex of presheaves of $R$-modules on $\V$.
Then the following conditions are equivalent.
\begin{itemize}
\item[(i)] The complex $K$ has the B.-G.-property.
\item[(i$^{\prime}$)] The complex $K$ has the excision property.
\item[(ii)] For any $X$ in $\V$, the canonical map
$$H^n(K(X))\To \hypercoh^n_\nis(X,K_\nis)$$
is an isomorphism of $R$-modules (\emph{i.e.} $K$ is $\V_\nis$-local).
\end{itemize}
\end{prop}
\begin{proof}
The equivalence of (i) and (i$^{\prime}$) follows from the definition
of a homotopy pullback.
As any short exact sequence of sheaves of $R$-modules
defines canonically a distinguished triangle in $\Der(\V,R)$,
the fact that (ii) implies (i) follows easily from
proposition \ref{exactdistsquare}.
To prove that (i) implies (ii), we need a little more
machinery. First, we can choose a monomorphism of complexes $K\To L$
which induces a quasi-isomorphism between $K_\nis$ and $L_\nis$,
and such that $L$ is $\V_\nis$-local. For this, we first choose
a quasi-isomorphism $K_\nis\To M$ where $M$ is $\V_\nis$-local
(which is possible by Corollary \ref{flasqueresolution}).
We have a natural embbeding of $K$ into
the mapping cone of its identity $\mathrm{Cone}(1_K)$.
But $\mathrm{Cone}(1_K)$ is obviously $\V_\nis$-local
as it is already acyclic as a complex of presheaves.
This implies that the direct sum $L=\mathrm{Cone}(1_K)\oplus M$
is also $\V_\nis$-local. Moreover, as
$K$ and $L$ both satisfy the B.-G.-property,
one can check easily that the quotient presheaf $L/K$ also
has the B.-G.-property. Hence it is sufficient to prove that
$H^n(L(X)/K(X))=0$ for any $X$ in $\V$ and any integer $n$.\\
\indent Let us fix an object $X$ of $\V$ and an integer $n$.
One has to consider for any $q \geq 0$ the presheaf $T_q$ on the small
Nisnevich site $X_\nis$ of $X$ defined by
$$T_q(Y)=H^{n-q}(L(Y)/K(Y)) \ . $$
These are B.-G.-functors as defined by Morel and
Voevodsky \cite[proof of Prop. 1.16, page 101]{MV},
and for any integer $q\geq 0$, the Nisnevich sheaf associated to $T_q$
is trivial (this is because $K_\nis\To L$ is a quasi-isomorphism
of complexes of Nisnevich sheaves by construction of $L$).
This implies by virtue of \cite[Lemma 1.17, page 101]{MV}
that $T_q=0$ for any $q\geq 0$. In particular,
we have $H^n(L(X)/K(X))=0$. Therefore $L/K$ is an acyclic
complex of presheaves over $\V$, and $K\To L$ is a quasi-isomorphism
of complexes of presheaves. This proves that $K$ is $\V_\nis$-local
if and only if $L$ is, hence that $K$ is $\V_\nis$-local.
\end{proof}
\begin{cor}\label{filtcolimNiscoh}
Let $I$ be a small filtering category, and
$K$ a functor from $I$ to the category of complexes of
Nisnevich sheaves of $R$-modules. Then for any smooth
$S$-scheme $X$, the canonical maps
$$\limind_{i\in I}\hypercoh^n_\nis(X,K_{i})\To
\hypercoh^n_\nis(X,\limind_{i\in I}K)$$
are isomorphisms for all $n$.
\end{cor}
\begin{proof}
We can suppose that $K_{i}$ is $\V_{\nis}$-local for all $i\in I$
(we can take a termwise fibrant replacement of $K$ with respect to the
model structrure of Proposition \ref{existmodcatlocalnis}).
It then follows from Proposition \ref{bgtilde}
that $\limind_{i\in I}K_{i}$ is still $\V_{\nis}$-local:
it follows from the fact that the filtering colimits are exact
that the presheaves with the B.-G.-property are
stable by filtering colimits. The map
$$\limind_{i\in I}H^n(K_{i}(X))\To
H^n(\limind_{i\in I}K_{i}(X))$$
is obviously an isomorphism for any $X$ in $\V$.
As we are free to take $\V=\sm/S$, this proves the assertion.
\end{proof}
\begin{paragr}\label{defNiscompact}
Remember that if $\T$ is a triangulated category with
small sums, an object $X$ of $\T$ is \emph{compact}
if for any small family $(K_{\lambda})_{\lambda\in \Lambda}$
of objects of $\T$, the canonical map
$$\bigoplus_{\lambda\in\Lambda}\Hom_{\T}\big(X, K_{\lambda}\big)
\To\Hom_{\T}\big(X,\bigoplus_{\lambda\in\Lambda} K_{\lambda}\big)$$
is bijective (as this map is always injective, this is
equivalent to say it is surjective). One denotes by $\T_{c}$
the full subcategory of $\T$ that consists of compact objects.
It is easy to see that $\T_{c}$ is a thick subcategory of $\T$
(which means that $\T_{c}$ is a triangulated subcategory of $\T$ stable by direct factors).
\end{paragr}
\begin{cor}\label{filtcolimNiscoh2}
For any smooth $S$-scheme $X$, $R(X)$
is a compact object of the derived category
of Nisnevich sheaves of $R$-modules.
\end{cor}
\begin{proof}
As any direct sum is a filtering colimit of
finite direct sums, this follows from \eqref{Nispremodcat3}
and Corollary \ref{filtcolimNiscoh}.
\end{proof}
\begin{paragr}\label{defA1equiv}
Let $\Der$ be a triangulated category.
Remember that a \emph{localizing subcategory} of $\Der$
is a full subcategory $\TDer$ of $\Der$ with the following properties.
\begin{itemize}
\item[(i)] $A$ is in $\TDer$ if and only if $A[1]$ is in $\TDer$.
\item[(ii)] For any distinguished triangle
$$A'\To A\To A''\To A'[1] \ , $$
if $A'$ and $A''$ are in $\TDer$, then $A$ is in $\TDer$.
\item[(iii)] For any (small) family $(A_i)_{i\in I}$
of objects of $\TDer$, $\bigoplus_{i\in I}A_i$ is in $\TDer$.
\end{itemize}
If $\T$ is a class of objects of $\Der$, the \emph{localizing
subcategory of $\Der$ generated by $\T$} is the smallest
localizing subcategory of $\Der$ that contains $\T$ (\emph{i.e.}
the intersection of all the localizing subcategories of $\Der$
that contain $\T$).\\
\indent Let $\T$ be the class of complexes of shape
$$\dots\To 0\To R(X\times_S\AA^1_S)\To R(X)\To 0\To\dots$$
with $X$ in $\V$ (the non trivial differential is induced
by the canonical projection). Denote by $\TDer(\V,\AA^1,R)$ the localizing
subcategory of $\Der(\V,R)$ generated by $\T$. We define the
triangulated category \smash{$\Der^{\mathit{eff}}_{\AA^1}(\V,R)$}
as the Verdier quotient of $\Der(\V,R)$ by $\TDer(\V,\AA^1,R)$.
$$\Der^{\mathit{eff}}_{\AA^1}(\V,R)=\Der(\V,R)/\TDer(\V,\AA^1,R)$$
We know that $\Der(\V,R)\simeq\Der(\sm/S,R)$,
and an easy Mayer-Vietoris argument for the Zariski topology
shows that the essential image of $\TDer(\V,\AA^1,R)$ in $\Der(\sm/S,R)$
is precisely $\TDer(\sm/S,\AA^1,R)$. Hence we get a canonical equivalence
of categories
\begin{equation}\label{eqdmdmV}
\Der^{\mathit{eff}}_{\AA^1}(\V,R)\simeq\Der^{\mathit{eff}}_{\AA^1}(\sm/S,R) \ .
\end{equation}
We simply put:
$$\DMte(S,R)=\Der^{\mathit{eff}}_{\AA^1}(\sm/S,R) \ .$$
According to F. Morel insights, the category $\DMte(S,R)$
is called the \emph{triangulated category of effective real
motives}\footnote{This terminology comes from the fact
$\DMte(S,R)$ give quadratic informations on $S$, which implies
it is bigger than Voevodsky's triangulated category of
mixed motives; see \cite{pizero,ratmot}.
The word `real' is meant here as opposed to `complex'.}
(with coefficients in $R$). In the sequel of this paper, we will
consider the equivalence \eqref{eqdmdmV} as an
equality\footnote{The role of the category $\V$ is only to define
model category structures on the category of complexes of
$\mathit{Sh}(\V,R)\simeq\mathit{Sh}(\sm/S,R)$ which
depend only on the local behaviour of
the schemes in $\V$ (e.g. the smooth
affine shemes over $S$).}.
We thus have a canonical localization functor
\begin{equation}\label{deflocdmequat}
\gamma : \Der(\V,R)\To\DMte(S,R) \ .
\end{equation}
We will say that a morphism of complexes of $\mathit{Sh}(\V,R)$
is an \emph{$\AA^1$-equivalence} if its image
in \smash{$\DMte(S,R)$}
is an isomorphism.\\
\indent A complex of presheaves of $R$-modules $K$ over $\V$
is \emph{$\AA^1$-homotopy invariant} if for any $X$ in $\V$,
the projection of $X\times_S\AA^1_S$ on $X$ induces a quasi-isomorphism
$$K(X)\To K(X\times_S\AA^1_S) \ .$$
\end{paragr}
\begin{prop}\label{A1cmfeff}
The category of complexes of $\mathit{Sh}(\V,R)$ is endowed with
a proper Quillen model category structure whose weak equivalences are the $\AA^1$-equivalences,
whose cofibrations are the $\V$-cofibrations, and whose fibrations
are the $\V$-surjective morphisms with $\AA^1$-homotopy
invariant and $\V_{\nis}$-local kernel.
In particular, the fibrant objects of this model structure are exactly
the $\AA^1$-homotopy invariant and $\V_\nis$-local complexes.
The corresponding homotopy category is the triangulated
category of effective real motives
\smash{$\DMte(S,R)$}.
\end{prop}
\begin{proof}
This is a direct application of \cite[Proposition 3.5 and Corollary 4.10]{HCD}.
\end{proof}
\begin{paragr}
Say that a complex $K$ of presheaves of $R$-modules
on $\V$ is \emph{$\AA^1$-local}
if for any $X$ in $\V$, the projection of $X\times_S\AA^1_S$
on $X$ induces isomorphisms in Nisnevich hypercohomology
$$\hypercoh^n_\nis(X,K_\nis)\simeq
\hypercoh^n_\nis(X\times_S\AA^1_S,K_\nis) \ . $$
It is easy to see that a $\V_\nis$-local complex
is $\AA^1$-local if and only if it is $\AA^1$-homotopy
invariant. In general, a complex of sheaves $K$
is $\AA^1$-local if and only if, for any quasi-isomorphism
$K\To L$, if $L$ is $\V_\nis$-local, then $L$ is $\AA^1$-homotopy
invariant. We deduce from this the following result.
\end{paragr}
\begin{cor}
The localization functor $\Der(\V,R)\To\DMte(S,R)$
has a right adjoint that is fully faithful and whose essential
image consists of the $\AA^1$-local complexes. In other words,
\smash{$\DMte(S,R)$} is canonically equivalent
to the full subcategory of $\AA^1$-local complexes in $\Der(\V,R)$.
\end{cor}
\begin{proof}
For any complex of sheaves $K$, one can produce functorially
a map $K\To L^{}_{\AA^1}K$ which is an $\AA^1$-equivalence
with $L^{}_{\AA^1}K$ a $\V_\nis$-local and $\AA^1$-homotopy
invariant complex (just consider a functorial fibrant resolution
of the model category of \ref{A1cmfeff}). Then the functor
$L^{}_{\AA^1}$ takes $\AA^1$-equivalences to quasi-isomorphisms
of complexes of (pre)sheaves, and induces a functor
$$L^{}_{\AA^1} : \DMte(S,R)\To\Der(\V,R)$$
which is the expected right adjoint of the localization
functor.
\end{proof}
\begin{ex}\label{exampleR}
The constant sheaf $R$ is $\AA^1$-local (if it is considered as a complex
concentrated in degree $0$).
\end{ex}
\begin{paragr}\label{changeringcoeff}
Let $R'$ be another commutative ring, and $R\To R'$
a morphism of rings. The functor $K\longmapsto K\otimes^{}_{R}R'$
is a symmetric monoidal left Quillen functor from $\Comp(\V,R)$ to $\Comp(\V,R')$
for the model structures of Proposition \ref{A1cmfeff}. Hence it has a total
left derived functor
$$\DMte(S,R)\To\DMte(S,R')\qquad ,
\quad K\longmapsto K\otimes^\derL_{R}R'$$
whose right adjoint is the obvious forgetful functor.
I.e. for a complex of sheaves of $R$-modules $K$
and a complex of sheaves of $R'$-modules $L$, we have a canonical
isomorphism
$$\Hom_{\DMte(S,R)}(K,L)\simeq
\Hom_{\DMte(S,R')}(K\otimes^\derL_{R}R',L) \ .$$
\end{paragr}
\begin{ex}\label{exampleGm}
Let $\GG_{m}=\AA^1_{S}-\{0\}$ be the multiplicative group.
It can be considered as a presheaf of groups on $\sm/S$,
and one can check that it is a Nisnevich sheaf. Moreover,
for any smooth $S$-scheme $X$, one has
\begin{equation}\label{computeGmcoh}
H^{i}_{\nis}(X,\GG_{m})=\begin{cases}
\bundle O^*(X)&\text{if $i=0$,}\\
\pic(X)&\text{if $i=1$,}\\
0&\text{otherwise.}
\end{cases}
\end{equation}
As $S$ is assumed to be regular,
$\GG_{m}$ is $\AA^1$-local as a complex concentrated in degree $0$.
We deduce that we have the formula
\begin{equation}\label{GminDmeffk}
H^{i}_{\nis}(X,\GG_{m})=\Hom_{\DMte(k,\ZZ)}(\ZZ(X),\GG_{m}[i]) \ .
\end{equation}
In particular, it follows from \ref{changeringcoeff} that for any smooth
$S$-scheme $X$, one has a canonical morphism of abelian groups
\begin{equation}\label{PicDmeffkR}
\pic(X)\To
\Hom_{\DMte(S,R)}(R(X),\GG_{m}\otimes^\derL_{\ZZ}R[1]) \ .
\end{equation}
If moreover $R$ is flat over $\ZZ$, we get the formula
\begin{equation}\label{PicDmeffkRflat}
\pic(X)\otimes^{}_{\ZZ}R\simeq
\Hom_{\DMte(S,R)}(R(X),\GG_{m}\otimes^\derL_{\ZZ}R[1]) \ .
\end{equation}
\end{ex}
\begin{prop}\label{A1filtcolimNiscoh}
Let $I$ be a small filtering category, and
$K$ a functor from $I$ to the category of complexes of
Nisnevich sheaves of $R$-modules. Then for any smooth
$S$-scheme $X$, the canonical map
$$\limind_{i\in I}\Hom_{\DMte(S,R)}(R(X),K_{i})\To
\Hom_{\DMte(k,R)}(R(X),\limind_{i\in I}K)$$
is an isomorphism.
\end{prop}
\begin{proof}
This follows from Corollary \ref{filtcolimNiscoh}
once we see that $\AA^1$-homotopy invariant
complexes are stable by filtering colimits.
\end{proof}
\begin{cor}\label{repcompactdmtildeeff}
For any smooth $S$-scheme $X$, $R(X)$
is a compact object of $\DMte(S,R)$.
\end{cor}
\subsection{Derived tensor product and derived $\Hom$}
\begin{paragr}
We consider a full subcategory $\V$ of $\sm/S$ as in \ref{axiomesVV}
and a commutative ring $R$.
The category of sheaves of $R$-modules on $\V$ has a tensor product $\otimes^{}_R$
defined in the usual way: if $F$ and $G$ are two sheaves of $R$-modules,
$F\otimes^{}_R G$ is the Nisnevich sheaf associated to the presheaf
$$X\longmapsto F(X)\otimes^{}_R G(X) \ . $$
the unit of this tensor product is $R=R(S)$.
This makes the category $\mathit{Sh}(\V,R)$ a closed symmetric
monoidal category. For two objects $X$ and $Y$ of $\V$, we have a canonical
isomorphism
$$R(X\times_S Y)\simeq R(X)\otimes^{}_R R(Y) \ . $$
Finally, an important property of this tensor product is that
for any $X$ in $\V$, the sheaf $R(X)$ is flat, by which we mean
that the functor
$$F\longmapsto R(X)\otimes^{}_R F$$
is exact. This implies that the family of the sheaves
$R(X)$ for $X$ in $\V$ is \emph{flat} in the sense of \cite[2.1]{HCD}.
Hence we can apply
Corollary 2.6 of \emph{loc. cit.} to get that
the $\V_\nis$-local model structure of \ref{existmodcatlocalnis}
is compatible with the tensor product
in a very (rather technical but also)
gentle way: define the tensor product of two complexes
of sheaves of $R$-modules $K$ and $L$ on $\V$ by the formula
$$(K\otimes^{}_R L)^n=\bigoplus_{p+q=n}K^p\otimes^{}_R L^q$$
with differential $d(x\otimes y)=dx\otimes y+(-1)^{\mathrm{deg}(x)} x\otimes dy$.
This defines a structure of symmetric monoidal category
on $\Comp(\mathit{Sh}(\V,R))$
(the unit is just $R$ seen as complex
concentrated in degree $0$, and the symmetry rule is given by
the usual formula
$x\otimes y\longmapsto (-1)^{\mathrm{deg}(x)\mathrm{deg}(y)}y\otimes x$).
A consequence of \emph{loc. cit.} Corollary 2.6 is that the functor
$(K,L)\longmapsto K\otimes^{}_R L$
is a left Quillen bifunctor, which implies in particular that it has
a well behaved total left derived functor
$$\Der(\V,R)\times\Der(\V,R)\To\Der(\V,R)\quad ,
\qquad (K,L)\longmapsto K\otimes^\derL_R L \ .$$
Moreover, for a given complex $L$, the functor
$$\Der(\V,R)\To\Der(\V,R)\quad ,
\qquad K\longmapsto K\otimes^\derL_R L$$
is the total left derived functor of the functor
$K\longmapsto K\otimes^{}_R L$ (see Remark 2.9 of \emph{loc. cit.}).
This means that for
any $\V$-cofibrant complex $K$ (that is, a complex such that
the map $0\To K$ is a $\V$-cofibration), the canonical map
$$K\otimes^\derL_R L\To K\otimes^{}_R L$$
is an isomorphism in $\Der(\V,R)$ for any complex $L$. In particular,
if $F$ is a direct factor of some $R(X)$ (with $X$ in $\V$),
then for any complex of sheaves $L$, the map
$$F\otimes^\derL_R L\To F\otimes^{}_R L$$
is an isomorphism in $\Der(\V,R)$.
This derived tensor product makes $\Der(\V,R)$ a
closed symmetric monoidal triangulated category.
This means that for two objects $L$ and $M$ of
$\Der(\V,R)$, there is an object $\derR\sHom(L,M)$ of $\Der(\V,R)$
that is defined by the universal property
$$\forall K\in\Der(\V,R)\ , \quad
\Hom_{\Der(\V,R)}(K\otimes^\derL_{R}L,M)\simeq
\Hom_{\Der(\V,R)}(K,\derR\sHom(L,M)).$$
The functor $\derR\sHom$ can also be characterized as the
total right derived functor of the internal Hom of the category
of complexes of Nisnevich sheaves of $R$-moddules on $\V$.
If $L$ is $\V$-cofibrant and if $M$ if $\V_{\nis}$-local, then
$\derR\sHom(L,M)$ can be represented by the complex
of sheaves
$$X\longmapsto\mathrm{Tot}\big [\Hom_{\mathit{Sh}(\V,R)}(R(X)\otimes^{}_{R}L,M)\big ].$$
The derived tensor product on $\Der(\V,R)$
induces a derived tensor product on $\DMte(S,R)$
as follows.
\end{paragr}
\begin{prop}\label{A1derivedtensor}
The tensor product of complexes $\otimes^{}_R$ has a total
left derived functor
$$\DMte(S,R)\times\DMte(S,R)
\To\DMte(S,R)\quad ,
\qquad (K,L)\longmapsto K\otimes^\derL_R L $$
that makes $\DMte(S,R)$ a closed symmetric
tensor triangulated category.
Moreover, the localization functor
$\Der(\V,R)\To\DMte(S,R)$
is a triangulated symmetric monoidal functor.
\end{prop}
\begin{proof}
This follows easily from \cite[Corollary 3.14]{HCD} applied to the classes $\G$
and $\H$ defined in the proof of \ref{existmodcatlocalnis}
and to the class $\T$ of complexes of shape
$$\cdots\To 0\To R(X\times_S\AA^1_S)\To R(X)\To 0\To\cdots$$
with $X$ in $\V$.
\end{proof}
\begin{paragr}\label{A1compactRHomeffprep}
It follows from Proposition \ref{A1derivedtensor} that
the category $\DMte(S,R)$ has an internal Hom that
we still denote by $\derR\sHom$. Hence for three objects $K$, $L$ and $M$
in $\DMte(S,R)$, we have a canonical isomorphism
\begin{equation}\label{A1RHomeff}
\Hom_{\DMte(S,R)}(K\otimes^\derL_{R}L,M)\simeq
\Hom_{\DMte(S,R)}(K,\derR\sHom(L,M)).
\end{equation}
If $L$ is $\V$-cofibrant and $M$ is $\V_{\nis}$-local and $\AA^1$-local,
then
\begin{equation}\label{A1RHomeff2}
\derR\sHom(L,M)=\mathrm{Tot}\big [ \Hom_{\mathit{Sh}(\V,R)}(R(-)\otimes^{}_{R}L,M)\big ].
\end{equation}
\end{paragr}
\begin{prop}\label{A1compactRHomeff}
If $L$ is a compact object of $\DMte(S,R)$, then
for any small family $(K_{\lambda})_{\lambda\in \Lambda}$
of objects of $\DMte(S,R)$, the canonical map
$$\bigoplus_{\lambda\in\Lambda}\derR\sHom\big(L,K_{\lambda}\big)\To
\derR\sHom\big(L,\bigoplus_{\lambda\in\Lambda} K_{\lambda}\big)$$
is an isomorphism in $\DMte(S,R)$.
\end{prop}
\begin{proof}
Once the family $(K_{\lambda})_{\lambda\in \Lambda}$ is fixed, this map defines
a morphism of triangulated functors from the triangulated category of compact objects of
$\DMte(S,R)$ to $\DMte(S,R)$.
Therefore, it is sufficient to check this property when $L=R(Y)$
with $Y$ in $\V$. This is equivalent to say that for any $X$ in $\V$,
the map
$$\Hom
\big(R(X),\bigoplus_{\lambda\in\Lambda}\derR\sHom\big(R(Y),K_{\lambda}\big)\big)\To
\Hom
\big(R(X),\derR\sHom\big(R(Y),\bigoplus_{\lambda\in\Lambda} K_{\lambda}\big)\big)$$
is bijective.
As $R(X)$ is compact (\ref{repcompactdmtildeeff}), we have
$$\Hom
\big(R(X),\bigoplus_{\lambda\in\Lambda}\derR\sHom\big(R(Y),K_{\lambda}\big)\big)
\simeq\bigoplus_{\lambda\in\Lambda}
\Hom
\big(R(X),\derR\sHom\big(R(Y),K_{\lambda}\big)\big),$$
and as $R(X\times_{S}Y)\simeq R(X)\otimes^\derL_{R}R(Y)$ is compact as well,
we have
$$\begin{aligned}
\Hom
\big(R(X),\derR\sHom\big(R(Y),\bigoplus_{\lambda\in\Lambda} K_{\lambda}\big)\big)&\simeq
\Hom
\big(R(X)\otimes^\derL_{R}R(Y),\bigoplus_{\lambda\in\Lambda} K_{\lambda}\big)\\
&\simeq\bigoplus_{\lambda\in\Lambda}
\Hom\big(R(X)\otimes^\derL_{R}R(Y),K_{\lambda}\big)\\
&\simeq\bigoplus_{\lambda\in\Lambda}
\Hom\big(R(X),\derR\sHom(R(Y),K_{\lambda})\big).
\end{aligned}$$
This implies our claim immediately.
\end{proof}
\begin{paragr}\label{quillenconstantsheaf}
Let $R\Mod$ be the category of $R$-modules.
If $M$ is an $R$-module, we still denote by $M$
the constant Nisnevich sheaf of $R$-modules on $\V$
generated by $M$.
This defines a symmetric monoidal functor from
the category of (unbounded) complexes of $R$-modules $\Comp(R)$
to the category $\Comp(\V,R)$
\begin{equation}\label{quillenconstantsheaf1}
\Comp(R)\To\Comp(\V,R)\quad , \qquad
M\longmapsto M.
\end{equation}
This functor is a left adjoint to the global sections functor
\begin{equation}\label{quillenconstantsheaf2}
\Gamma : \Comp(\V,R)\To\Comp(R)\quad , \qquad
M\longmapsto \Gamma(M)=\Gamma(S , M).
\end{equation}
The category $\Comp(R)$ is a Quillen model category
with the quasi-isomorphims as weak equivalences and the
degreewise surjective maps as fibrations (see \emph{e.g.} \cite[Theorem 2.3.11]{Hov}).
We call this model structure the \emph{projective model structure}.
This implies that the constant sheaf functor \eqref{quillenconstantsheaf1}
is a left Quillen functor for the model structures of
Propositions \ref{existmodcatlocalnis} and \ref{A1cmfeff} on
$\Comp(\V,R)$. Therefore, the global sections functor \eqref{quillenconstantsheaf2}
is a right Quillen functor and has total right derived functor
\begin{equation}\label{quillenconstantsheaf3}
\derR\Gamma : \DMte(S,R)\To\Der(R)
\end{equation}
where $\Der(R)$ denotes the derived category of $R$.
For two objects $M$ and $N$ of $\DMte(S,R)$,
we define
\begin{equation}\label{quillenconstantsheaf4}
\derR\Hom(M,N)=\derR\Gamma\big(\derR\sHom(M,N)\big).
\end{equation}
We invite the reader to check that $\derR\Hom$ is the derived Hom
of $\DMte(S,R)$. In particular, for any integer $n$, we have
a canonical isomorphism
\begin{equation}\label{quillenconstantsheaf5}
H^n\big(\derR\Hom(M,N)\big)\simeq\Hom_{\DMte(S,R)}(M,N[n]).
\end{equation}
\end{paragr}
\subsection{Tate object and purity}
\begin{paragr}
Let $\GG_m=\AA^1_S-\{0\}$ be the multiplicative group scheme over $S$.
The unit of $\GG_m$ defines a morphism $R=R(S)\To R(\GG_m)$,
and we define the \emph{Tate object} $R(1)$ as the cokernel
$$R(1)=\mathrm{coker}\big(R\To R(\GG_m)\big)[-1]$$
(this definition makes sense in the category of complexes of
$\mathit{Sh}(\V,R)$ as well as in $\Der(\V,R)$
or in $\DMte(S,R)$ as we take the cokernel
of a split monomorphism). By definition, $R(1)[1]$
is a direct factor of $R(\GG_m)$, so that $R(1)$
is $\V$-cofibrant. Hence for any integer $n\geq 0$,
$R(n)=R(1)^{\otimes n}$ is also $\V$-cofibrant.
For a complex $K$ of $\mathit{Sh}(\V,R)$,
we define $K(n)=K\otimes^{}_R R(n)$.
As $R(n)$ is $\V$-cofibrant, the map
$$K\otimes^\derL_RR(n)\To K\otimes^{}_RR(n)=K(n)$$
is an isomorphism in $\DMte(k,R)$. Another
description of $R(1)$ is the following.
\end{paragr}
\begin{prop}\label{motivicsphere}
The inclusion of $\GG_{m}$ in $\AA^{1}$ induces
a canonical split distinguished triangle in $\DMte(S,R)$
$$R(\GG_{m})\To R(\AA^1_{S})\overset{0}{\To} R(1)[2]\To R(\GG_{m})[1]$$
that gives the canonical decomposition $R(\GG_{m})=R\oplus R(1)[1]$.
\end{prop}
\begin{proof}
This follows formally from the definition
of $R(1)$ and from the fact that $R(\AA^1)=R$ in $\DMte(S,R)$.
\end{proof}
\begin{paragr}\label{link_homotopy_cat}
Let $\op\Delta\mathit{Sh(\sm/S)}$ be the category of
Nisnevich sheaves of simplicial sets on $Sm/S$.
Morel and Voevodsky defined in \cite{MV} the $\AA^1$-homotopy
theory in $\op\Delta\mathit{Sh(\sm/S)}$. In particular, we have a notion
of $\AA^1$-weak equivalences of simplicial sheaves
that defines a proper model category structure (with the
monomorphisms as cofibrations).
Furthermore, we have a canonical functor
$$\op\Delta\mathit{Sh}(\sm/S)\To\Comp(\sm/S,R)\quad ,
\qquad \X\longmapsto R(\X)$$
which has the following properties; see e.g. \cite{A11,A12}.
\begin{itemize}
\item[(1)] The functor $R$ above preserves colimits.
\item[(2)] The functor $R$ preserves monomorphisms.
\item[(3)] The functor $R$ sends $\AA^1$-weak equivalences
to $\AA^1$-equivalences.
\end{itemize}
We deduce from these properties that the functor $R$
sends homotopy pushout squares of $\op\Delta\mathit{Sh(\sm/S)}$
to homotopy pushout squares of $\Comp(\sm/S,R)$ and induces
a functor
$$R:\mathscr H(S)\To\DMte(S,R)$$
where $\mathscr H(S)$ denotes the localization of $\op\Delta\mathit{Sh(\sm/S)}$
by the $\AA^1$-weak equivalences.
This implies that all the results of \cite{MV} that are formulated
in terms of $\AA^1$-weak equivalences (or isomorphisms
in $\H(S)$) and in terms of homotopy pushout have their counterpart
in $\DMte(S,R)$. We give below the results
we will need that come from this principle.
\end{paragr}
\begin{paragr}\label{defthomspaces}
Let $X$ be a smooth $S$-scheme
and $\bundle V$ a vector bundle over $X$.
Consider the open immersion $j:\bundle V^\times \rightarrow \bundle V$
of the complement of the zero section of $\bundle V/X$.
We define the \emph{Thom space of $\bundle V$} as the quotient
\begin{equation}\label{defthomspacescoker}
R(\thom\bundle V)
=\mathrm{coker}\big(R(\bundle V^\times)
\xrightarrow{j_*} R(\bundle V)\big) \ .
\end{equation}
We thus have a short exact sequence of sheaves of $R$-modules
\begin{equation}\label{defthomspacesshort}
0\To R(\bundle V^\times)\To R(\bundle V)\To R(\thom\bundle V)\To 0 \ .
\end{equation}
\end{paragr}
\begin{prop}\label{trivialthom99}
Let $\bundle O^n$ be the trivial vector bundle of dimension $n$
on a smooth $S$-scheme $X$. Then we have a canonical
isomorphism in $\DMte(S,R)$:
$$R(\thom \bundle O^n)\simeq R(X)(n)[2n] \ .$$
\end{prop}
\begin{proof}
This follows from Proposition \ref{motivicsphere}
and from the second statement of
\cite[Proposition 2.17, page 112]{MV}.
\end{proof}
For a given vector bundle $\bundle V$, over a $S$-scheme $X$,
we will denote by $\PP(\bundle V)\To X$ the corresponding projective
bundle.
\begin{prop}\label{projthom}
Let $\bundle V$ be a vector bundle on a smooth $S$-scheme $X$.
Then we have a canonical distinguished triangle in $\DMte(S,R)$
$$R(\PP(\bundle V))\To R(\PP(\bundle V\oplus \bundle O))\To R(\thom\bundle V)
\To R(\PP(\bundle V))[1] \ .$$
\end{prop}
\begin{proof}
This follows from Proposition \ref{motivicsphere}
and from the third statement of
\cite[Proposition 2.17, page 112]{MV}.
\end{proof}
\begin{cor}\label{trivialprojthom}
We have a canonical distinguished triangle in $\DMte(S,R)$
$$R(\PP^n_{S})\To R(\PP^{n+1}_{S})\To R(n+1)[2n+2]\To R(\PP^n_{S})[1] \ .$$
Moreover, this triangle splits canonically for $n=0$ and gives the decomposition
$$R(\PP^1_{S})=R\oplus R(1)[2] \ .$$
\end{cor}
\begin{proof}
This is a direct consequence of Propositions \ref{trivialthom99}
and \ref{projthom}. The splitting of the case $n=0$ comes obviously from
the canonical map from $\PP^1_{S}$ to $S$.
\end{proof}
\begin{paragr}\label{defProjinfty}
The inclusions $\PP^n_{S}\subset\PP^{n+1}_{S}$ allow us to define the
Nisnevich sheaf of sets
\begin{equation}\label{eqdefprojinfty}
\PP^\infty_{S}=\textstyle\limind_{\, n\geq 0}\PP^n_{S}
\end{equation}
We get a Nisnevich sheaf of $R$-modules
\begin{equation}\label{eqdefRprojinfty}
R(\PP^\infty_{S})=\textstyle\limind_{\, n\geq 0}R(\PP^n_{S})\ .
\end{equation}
For a complex $K$ of sheaves of $R$-modules, we define
the hypercohomology of $\PP^\infty_{S}$ with coefficients in $K$
to be
\begin{equation}\label{defhypercohprojinfty}
\hypercoh^i_\nis(\PP^\infty_{S},K)=\Hom_{\Der(\sm/S,R)}(R(\PP^\infty_{S}),K[i]) \ .
\end{equation}
\end{paragr}
\begin{prop}\label{MilnorProj}
There is a short exact sequence
$$\textstyle 0\To\limproj^1_{\, n\geq 0}\hypercoh^{i-1}_\nis(\PP^n_{S},K)
\To\hypercoh^i_\nis(\PP^\infty_{S},K)\To
\limproj_{\, n\geq 0}\hypercoh^i_\nis(\PP^n_{S},K)\To 0 \ .$$
\end{prop}
\begin{proof}
As the filtering colimits are exact in $\mathit{Sh}(\sm/S,R)$
we have an isomorphism
$\smash{\hocolim R(\PP^n_{S})\simeq R(\PP^\infty_{S})}$
in $\Der(\sm/S,R)$. This result is thus a direct application
of the Milnor short exact sequence applied to this homotopy
colimit (see e.g. \cite[Proposition 7.3.2]{Hov}).
\end{proof}
\begin{prop}[Purity Theorem]\label{purity}
Let $i:Z\To X$ a closed immersion of smooth $S$-schemes,
and $U=X-i(Z)$. Denote by $\bundle N_{X,Z}$ the normal vector bundle of $i$.
Then there is a canonical distinguished triangle in $\DMte(S,R)$
$$\xymatrix{R(U)\To R(X)\To R(\thom\bundle N_{X,Z})\To R(U)[1] \ .}$$
\end{prop}
\begin{proof}
This follows from \cite[Theorem 2.23, page 115]{MV}.
\end{proof}
\begin{cor}\label{anmoinszerosphere}
There is a canonical decomposition $R(\AA^n_{S}-\{0\})=R\oplus R(n)[2n-1]$
in $\DMte(S,R)$.
\end{cor}
\begin{proof}
The Purity Theorem and Proposition \ref{trivialthom99}
give a distinguished triangle
$$\xymatrix{R(\AA^n_{S}-\{0\})\To R(\AA^n_{S})\To R(n)[2n]\To R(\AA^n_{S}-\{0\})[1] \ .}$$
But this triangle is isomorphic to the distinguished triangle
$$\xymatrix{R(\AA^n_{S}-\{0\})\To R\To Q[1]\To R(\AA^n_{S}-\{0\})[1]}$$
where $Q$ is the kernel of the obvious
map $R(\AA^n_{S}-\{0\})\To R$, which shows that these triangle
split.
\end{proof}
\subsection{Tate spectra}
\begin{paragr}
We want the derived tensor product by $R(1)$
to be an equivalence of categories. As this is not the case
in $\DMte(S,R)$, we will modify
the category $\DMte(S,R)$ and construct
the triangulated category of real motives
${\DMt}(S,R)$ in which this will occur by definition.
For this purpose, we will define the model category of
symmetric Tate spectra. We will give only
the minimal definitions we will need to
work with. We invite the interested reader
to have look at \cite[Section 6]{HCD}
for a more complete account. The main
properties of ${\DMt}(S,R)$ are listed in \ref{proprietesDMt}.
We consider given a category of smooth $S$-schemes $\V$
as in \ref{axiomesVV}.
\end{paragr}
\begin{paragr}
A \emph{symmetric Tate spectrum} (in $\mathit{Sh}(\V,R)$)
is a collection
$E=(E_{n},\s_{n})_{n\geq 0}$, where for each integer $n\geq 0$,
$E_{n}$ is a complex of Nisnevich sheaves on $\V$
endowed with an action of the symmetric group $\mathfrak S_{n}$,
and $\s_n : R(1)\otimes^{}_{R}E_{n}\To E_{n+1}$ is a morphism
of complexes, such that the induced maps obtained by composition
$$R(1)^{\otimes m}\otimes^{}_{R} E_{n}\To R(1)^{\otimes m-1}\otimes^{}_{R} E_{n+1}\To\cdots\To
R(1)\otimes^{}_{R} E_{m+n-1}\To E_{m+n}$$
are $\mathfrak S_{m}\times\mathfrak S_{n}$-equivariant.
We have to define the actions to be precise: $\mathfrak S_{m}$
acts on $R(m)=R(1)^{\otimes m}$ by permutation,
and the action on $E_{m+n}$ is induced by the diagonal
inclusion $\mathfrak S_{m}\times\mathfrak S_{n}\subset \mathfrak S_{m+n}$.
A morphism of symmetric Tate spectra
$u:(E_{n},\s_{n})\To(F_{n},\tau_{n})$ is a collection of $\mathfrak S_{n}$-equivariant maps
$u_{n}:E_{n}\To F_{n}$ such that the squares
$$\xymatrix{
R(1)\otimes^{}_{R} E_{n}\ar[r]^{\s_{n}}\ar[d]_{R(1)\otimes u_{n}}&E_{n+1}\ar[d]^{u_{n+1}}\\
R(1)\otimes^{}_{R} F_{n}\ar[r]_{\tau_{n}}&F_{n+1}
}$$
commute. We denote by $\Spt(\V,R)$ the category of
symmetric Tate spectra.
If $A$ is a complex of sheaves of $R$-modules
on $\V$, we define its \emph{infinite suspension} $\Sigma^\infty(A)$
as the symmetric Tate spectrum that consists of the collection
$(A(n),1_{A(n+1)})_{n\geq 0}$ where $\mathfrak S_{n}$ acts on
$A(n)=R(1)^{\otimes n}\otimes^{}_{R}A$
by permutation on $R(n)=R(1)^{\otimes n}$. This defines
the infinite suspension functor
\begin{equation}\label{definftysusptate}
\Sigma^\infty : \Comp(\V,R)\To\Spt(\V,R)
\end{equation}
This functor has a right adjoint
\begin{equation}\label{defomegainftytate}
\Omega^\infty : \Spt(\V,R)\To\Comp(\V,R)
\end{equation}
defined by $\Omega^\infty(E_{n},\s_{n})_{n\geq 0}=E_{0}$.
According to \cite[6.14 and 6.20]{HCD}
we can define a ($R$-linear) tensor product of symmetric spectra
$E\otimes^{}_{R}F$ satisfying the following properties (and these
properties determine this tensor product up to a canonical isomorphism).
\begin{itemize}
\item[(1)] This tensor product makes the category of symmetric
Tate spectra a closed symmetric monoidal category
with $\Sigma^\infty(R(S))$ as unit.
\item[(2)] The infinite suspension functor \eqref{definftysusptate}
is a symmetric monoidal functor.
\end{itemize}
Say that a map of symmetric Tate spectra $u:(E_{n},\s_{n})\To(F_{n},\tau_{n})$
is a \emph{quasi-isomorphism} if the map $u_{n}:E_{n}\To F_{n}$
is a quasi-isomorphism of complexes of Nisnevich sheaves of $R$-modules
for any $n\geq 0$. We define the \emph{Tate derived category
of $\mathit{Sh}(\V,R)$} as the localization of $\Spt(\V,R)$ by the class
of quasi-isomorphisms. We will write $\Der_{\tate}(\V,R)$ for this
``derived category''. One can check that $\Der_{\tate}(\V,R)$
is a triangulated category (according to \cite[Remark 6.19]{HCD},
this is the homotopy category of a stable model category)
and that the functor induced by $\Sigma^\infty$
is a triangulated functor (because this is a left Quillen
functor between stable model categories).
A symmetric Tate spectrum $E=(E_{n},\s_{n})_{n\geq 0}$ is
a \emph{weak $\Omega^\infty$-spectrum} if for any integer $n\geq 0$,
the map $\s_{n}$ induces an isomorphism
$E_{n}\simeq\derR\sHom(R(1),E_{n+1})$ in $\DMte(S,R)$.
A symmetric Tate spectrum $E=(E_{n},\s_{n})_{n\geq 0}$ is
a \emph{$\Omega^\infty$-spectrum} if it is a weak $\Omega^\infty$-spectrum
and if, for any integer $n\geq 0$, the complex $E_{n}$ is $\V_{\nis}$-local
and $\AA^1$-homotopy invariant.
A morphism of symmetric Tate spectra $u:A\To B$ is a
\emph{stable $\AA^1$-equivalence} if for any weak
$\Omega^\infty$-spectrum $E$, the map
$$u^* : \Hom_{\Der_{\tate}(\V,R)}(B,E)\To\Hom_{\Der_{\tate}(\V,R)}(A,E)$$
is an isomorphism of $R$-modules.
A morphism of Tate spectra is a \emph{stable $\AA^1$-fibration} if it is termwise $\V_{\nis}$-surjective and if its kernel is a
$\Omega^\infty$-spectrum.
A morphism of Tate spectra is a \emph{stable $\V$-cofibration} if it has the left lifting property with respect to the stable $\AA^1$-fibrations
which are also stable $\AA^1$-equivalences. A symmetric Tate spectrum $E$ is \emph{stably $\V$-cofibrant} if the map $0\To E$ is a stable
$\V$-cofibration.
\end{paragr}
\begin{prop}\label{cmfA1stable}
The category of symmetric Tate spectra is a stable proper symmetric
monoidal model category with the stable $\AA^1$-equivalences
as the weak equivalences, the stable $\AA^1$-fibrations as fibrations
and the stable $\V$-cofibrations as cofibrations. The
infinite suspension functor is a symmetric left Quillen functor
that sends the $\AA^1$-equivalences to the stable
$\AA^1$-equivalences. Moreover, the tensor product by
any stably $\V$-cofibrant symmetric Tate spectrum
preserves the stable $\AA^1$-equivalences.
\end{prop}
\begin{proof}
The first assertion is an application of \cite[Proposition 6.15]{HCD}.
The fact that the functor $\Sigma^\infty$ preserves weak
equivalences comes from \emph{loc. cit.}, Proposition 6.18.
The last assertion follows from \emph{loc. cit.}, Proposition 6.35.
\end{proof}
\begin{paragr}\label{proprietesDMt}
The proposition above means the following.
Define the \emph{triangulated category of real
mixed motives} ${\DMt}(S,R)$ as the localization of
the category $\Spt(\V,R)$ by the class of stable
$\AA^1$-equivalences. Then $\DMt(S,R)$ is a triangulated category
with infinite direct sums and products. To be more precise,
any short exact sequence in $\Spt(\V,R)$
gives rise canonically to an exact triangle in $\DMt(S,R)$,
and any distinguished triangle is isomorphic to
an exact triangle that comes from a short exact sequence.
Furthermore, this triangulated category does not depend on the
category $\V$: the category $\V$ is only a technical tool
to define a model category structure that is well
behaved with the tensor product and Nisnevich descent in $\V$.
The infinite suspension functor sends $\AA^1$-equivalences to
stable $\AA^1$-equivalences and thus induces a functor
\begin{equation}\label{DMtsigmainfty}
\Sigma^\infty : \DMte(S,R)\To\DMt(S,R) \ .
\end{equation}
The right adjoint of the infinite suspension functor has a total right
derived functor
\begin{equation}\label{DMtomegainfty}
\derR\Omega^\infty : \DMt(S,R)\To\DMte(S,R) \ .
\end{equation}
For a (weak) $\Omega^\infty$-spectrum $E$, one has
\begin{equation}\label{DMtomegaspectra}
\derR\Omega^\infty(E)=E_{0} \ .
\end{equation}
The tensor product on $\Spt(\V,R)$ has a total left derived functor
\begin{equation}\label{DMttensor}
\DMt(S,R)\times\DMt(S,R)\To\DMt(S,R)
\quad , \qquad (E,F)\longmapsto E\otimes^\derL_{R}F \ .
\end{equation}
If $E$ is stably $\V$-cofibrant, then the canonical map
$E\otimes^\derL_{R}F\To E\otimes^{}_{R}F$ is an isomorphism in $\DMt(S,R)$.
Moreover, the functor \eqref{DMtsigmainfty} is symmetric monoidal.
In particular, for two complexes of Nisnevich sheaves of $R$-modules $A$ and $B$
we have a canonical isomorphism
\begin{equation}\label{DMttensor2}
\Sigma^\infty(A\otimes^\derL_{R}B)\simeq
\Sigma^\infty(A)\otimes^\derL_{R}\Sigma^\infty(B)\ .
\end{equation}
The category $\DMt(S,R)$ has also an internal Hom that we denote by
$\derR\sHom(E,F)$.
We will write $\tR=\Sigma^\infty(R(S))$,
and for a smooth $S$-scheme $X$, we define
$\tR(X)$ to be $\Sigma^\infty(R(X))$.
We also define $\tR(n)=\Sigma^\infty(R(n))$
for $n\geq 0$. Note that $\tR$ is the unit of the (derived) tensor product.
We define the symmetric Tate spectrum $\tR(-1)$ by
the formula $\tR(-1)_{n}=R(n+1)$ with the action of $\mathfrak S_{n}$
defined as the action by permutations on the first $n$ factors of
$R(n+1)=R(1)^{\otimes n}\otimes^{}_{R} R(1)$. The maps $\tR(-1)_{n}\otimes R(1)\To \tR(-1)_{n+1}$
are just the identities. One can check that $\tR(-1)$ is $\V$-cofibrant.
\end{paragr}
\begin{prop}\label{TateinverseDMt}
The object $\tR(1)$ is invertible in $\DMt(S,R)$ and
we have an isomorphism $\tR(-1)\simeq\tR(1)^{-1}$.
In other words, there are isomorphisms
$$\tR(1)\otimes^{\derL}_{R}\tR(-1)\simeq\tR\quad
\text{and}\quad\tR(-1)\otimes^\derL_{R}\tR(1)\simeq\tR \ . $$
\end{prop}
\begin{proof}
This follows from \cite[Proposition 6.24]{HCD}.
\end{proof}
\begin{paragr}
For an integer $n\geq 0$, we define $\tR(-n)=\tR(-1)^{\otimes n}$.
For an integer $n$, and a symmetric Tate spectrum $E$, we define
$$E(n)=E\otimes\tR(n) \ .$$
As $\tR(n)$ is $\V$-cofibrant, the canonical maps
$E\otimes^{\derL}_{R}\tR(n)\To E\otimes^{}_{R}\tR(n)=E(n)$
are isomorphisms in $\DMt(S,R)$.
We will say that a symmetric Tate spectrum
$E=(E_{n},\s_{n})_{n\geq 0}$ is a \emph{weak
$\Omega^\infty$-spectrum} if for any integer $n\geq 0$,
the map $\s_{n}$ induces by adjunction an isomorphism
$$E_{n}\simeq\derR\sHom(R(1),E_{n+1})$$
in $\DMte(S,R)$.
\end{paragr}
\begin{prop}\label{suspDMt}
Let $E$ be a weak $\Omega^\infty$-spectrum.
Then for any integer $n\geq 0$ and any complex of Nisnevich
sheaves of $R$-modules $A$, there is a canonical isomorphism of
$R$-modules
$$\Hom_{\DMt(S,R)}(\Sigma^\infty(A),E(n))\simeq
\Hom_{\DMte(S,R)}(A,E_{n}) \ .$$
In particular, for any smooth $S$-scheme $X$, one has isomorphisms
$$\hypercoh^{i}_{\nis}(X,E_{n})\simeq
\Hom_{\DMt(S,R)}(\tR(X),E(n)[i]) \ .$$
\end{prop}
\begin{proof}
This is an application of \cite[Proposition 6.28]{HCD}.
\end{proof}
\begin{cor}
A morphism of weak $\Omega^\infty$-spectra $E\To F$
is a stable $\AA^1$-equivalence if and only if
the map $E_{n}\To F_{n}$ is a $\AA^1$-equivalence
for all $n\geq 0$.
\end{cor}
\begin{prop}\label{compactdmtilde}
For any smooth $S$-scheme $X$ and any integer $n$,
$\tR(X)(n)$ is a compact object of $\DMt(S,R)$.
\end{prop}
\begin{proof}
Let $(E_{\lambda})_{\lambda\in\Lambda}$ be a small family of
Tate spectra. We want to show that the map
$$\bigoplus_{\lambda\in\Lambda}\Hom_{\DMt(S,R)}\big(\tR(X)(n),E_{\lambda}\big)
\To\Hom_{\DMt(S,R)}\big(\tR(X)(n),\bigoplus_{\lambda\in\Lambda}E_{\lambda}\big)$$
is bijective. Replacing the spectra $E_{\lambda}$ by the spectra $E_{\lambda}(-n)$,
we can suppose that $n=0$. Furthermore, we can assume that the spectra
$E_{\lambda}$ are weak $\Omega^\infty$-spectra.
As $R(1)$ is a compact object of $\DMte(S,R)$
(this is by definition a direct factor of the object $R(\GG_{m})$
which is compact by \ref{repcompactdmtildeeff}),
it follows from Proposition \ref{A1compactRHomeff}
that $\bigoplus_{\lambda\in\Lambda}E_{\lambda}$
is a weak $\Omega^\infty$-spectrum as well. Therefore, by Proposition \ref{suspDMt},
we have
$$\begin{aligned}
\bigoplus_{\lambda\in\Lambda}\Hom_{\DMt(S,R)}\big(\tR(X),E_{\lambda}\big)&\simeq
\bigoplus_{\lambda\in\Lambda}\Hom_{\DMte(S,R)}\big(R(X),E_{0,\lambda}\big)\\
&\simeq
\Hom_{\DMte(S,R)}\big(R(X),\bigoplus_{\lambda\in\Lambda}E_{0,\lambda}\big)\\
&\simeq
\Hom_{\DMte
(S,R)}\big(R(X),\big(\bigoplus_{\lambda\in\Lambda}E_{\lambda}\big)_{0}\big)\\
&\simeq
\Hom_{\DMt(S,R)}\big(\tR(X),\bigoplus_{\lambda\in\Lambda}E_{\lambda}\big).
\end{aligned}$$
This proves the result.
\end{proof}
\begin{paragr}\label{constantsheafnoneff}
The functor
\begin{equation}\label{constantsheafnoneff1}
\Comp(R)\To\Spt(\V,R)\quad , \qquad
M\longmapsto \Sigma^\infty(M).
\end{equation}
is a left Quillen functor from the projective model structure on $\Comp(R)$
(see \ref{quillenconstantsheaf}) to the model structure of Proposition \ref{cmfA1stable}.
The right adjoint of \eqref{constantsheafnoneff1}
\begin{equation}\label{constantsheafnoneff2}
\Spt(\V,R)\To\Comp(R)\quad , \qquad
M\longmapsto \Gamma(\Omega^\infty(M))
\end{equation}
is a right Quillen functor. The corresponding total right derived functor
is canonically isomorphic to the composed
functor $\derR\Gamma\circ\derR\Omega^\infty$.
For two objects $E$ and $F$ of $\DMt(S,R)$, we define
\begin{equation}\label{constantsheafnoneff3}
\derR\Hom(E,F)=\derR\Gamma\Big(\derR\Omega^\infty\big(\derR\sHom(E,F)\big)\Big).
\end{equation}
For any integer $n$, we have
a canonical isomorphism
\begin{equation}\label{qconstantsheafnoneff4}
H^n\big(\derR\Hom(E,F)\big)\simeq\Hom_{\DMt(S,R)}(E,F[n]).
\end{equation}
\end{paragr}
\subsection{Ring spectra}
\begin{paragr}\label{defringspectrum}
A \emph{ring spectrum} is a monoid object in the category
of symmetric Tate spectra. A ring spectrum is
\emph{commutative} if it is commutative as a monoid
object of $\Spt(\V,R)$.
Given a ring spectrum $\ste$,
one can form the category of left $\ste$-modules.
These are the symmetric Tate spectra $M$ endowed with
a left action of $\ste$
$$\ste\otimes^{}_{R} M\To M$$
satisfying the usual associativity and unit properties.
We denote by $\Spt(\V,\ste)$
the category of left $\ste$-modules. There is a \emph{base change functor}
\begin{equation}\label{basechangestetate}
\Spt(\V,R)\To\Spt(\V,\ste)\quad , \qquad F\longmapsto \ste\otimes^{}_{R}F
\end{equation}
which is a left adjoint of the forgetful functor
\begin{equation}\label{forgetstetate}
\Spt(\V,\ste)\To\Spt(\V,R)\quad , \qquad M\longmapsto M \ .
\end{equation}
If $\ste$ is commutative, the category $\Spt(\V,\ste)$
is canonically endowed with a closed symmetric
monoidal category structure such that the functor \eqref{basechangestetate}
is a symmetric monoidal functor. We denote by $\otimes^{}_{\ste}$
the corresponding tensor product. The unit of this monoidal structure
is $\ste$ seen as an $\ste$-module.
A morphism of $\ste$-modules is
a \emph{stable $\AA^1$-equivalence}
(resp. a \emph{stable $\AA^1$-fibration})
if it is so as a morphism of symmetric Tate spectra.
A morphism of $\ste$-modules is a stable $\V$-cofibration
if it is has the left lifting property with respect to the
stable $\AA^1$-fibrations which are also
stable $\AA^1$-equivalences.
\end{paragr}
\begin{prop}\label{cmftatestemod}
For a given ring spectrum $\ste$, the category
$\Spt(\V,\ste)$ is endowed with a stable proper
model category structure with the
stable $\AA^1$-equivalences as weak equivalences,
the stable $\AA^1$-fibrations as fibrations, and the
stable $\V$-co\-fibra\-tions as cofibrations.
The base change functor \eqref{basechangestetate}
is a left Quillen functor. Moreover, if $\ste$ is commutative,
then this model structure is symmetric monoidal.
\end{prop}
\begin{proof}
See \cite[Corollary 6.39]{HCD}.
\end{proof}
\begin{paragr}\label{defringspectrum2}
Let $\ste$ be a commutative ring spectrum.
We define $\DMt(S,\ste)$ to be the localization
of the category $\Spt(\V,\ste)$ by the class of
stable $\AA^1$-equivalences. It follows from the
proposition above that this category is canonically endowed with
a triangulated category structure. The base change functor
has a total left derived functor
\begin{equation}\label{derbasechangestetate}
\DMt(S,R)\To\DMt(S,\ste)\quad , \qquad F\longmapsto \ste\odmt_{R}F
\end{equation}
which is a left adjoint of the forgetful functor
\begin{equation}\label{derforgettate}
\DMt(S,\ste)\To\DMt(S,R)\quad , \qquad M\longmapsto M \ .
\end{equation}
The forgetful functor \eqref{derforgettate} is conservative
(which means that an object of $\DMt(S,\ste)$
is null if and only if it is null in $\DMt(S,R)$.
There is a derived tensor product
\begin{equation}\label{dertensorstemod}
\DMt(S,\ste)\times\DMt(S,\ste)\To\DMt(S,\ste)
\quad , \qquad (M,N)\longmapsto M\odmt_{\ste}N
\end{equation}
that turns $\DMt(S,\ste)$ into a symmetric monoidal
triangulated category (by applying \cite[Theorem 4.3.2]{Hov}
to the model structure of Proposition \ref{cmftatestemod}).
The derived base change
functor \eqref{derbasechangestetate} is of course
a symmetric monoidal functor. The category
$\DMt(S,\ste)$ also has an internal Hom that we denote by
$\derR\sHom_{\ste}(M,N)$. We thus have the formula
\begin{equation}\label{internalhomste}
\Hom_{\DMt(S,\ste)}(L\otimes^\derL_{\ste}M,N)\simeq
\Hom_{\DMt(S,\ste)}(L,\derR\sHom_{\ste}(M,N)).
\end{equation}
It follows from \ref{constantsheafnoneff} that the functor
\begin{equation}\label{constantsheafste1}
\Comp(R)\To\Spt(\V,\ste)\quad , \qquad
M\longmapsto\ste\otimes^{}_{R}\Sigma^\infty(M).
\end{equation}
is a left Quillen functor. The right derived functor of
its right adjoint is the composition of the forgetful functor \eqref{derforgettate}
with the functor $\derR\Gamma\circ\derR\Omega^\infty$.
For two objects $M$ and $N$ of $\DMt(S,\ste)$, we define
\begin{equation}\label{constantsheafste2}
\derR\Hom_{\ste}(M,N)=
\derR\Gamma\Big(\derR\Omega^\infty\big(\derR\sHom_{\ste}(M,N)\big)\Big).
\end{equation}
For any integer $n$, we have
a canonical isomorphism
\begin{equation}\label{constantsheafste3}
H^n\big(\derR\Hom_{\ste}(M,N)\big)\simeq\Hom_{\DMt(S,\ste)}(M,N[n]).
\end{equation}
For a smooth $S$-scheme $X$, we define
the free $\ste$-module generated by $X$ as
$$\ste(X)=\ste\otimes^{}_{R}\tR(X)=\ste\otimes^{}_{R}\Sigma^\infty(R(X)) \ .$$
As $\tR(X)$ is stably $\V$-cofibrant, the canonical map
$$\ste\otimes^{\derL}_{R}\tR(X)\To\ste\otimes^{}_{R}\tR(X)=\ste(X)$$
is an isomorphism in $\DMt(S,R)$ (hence in $\DMt(S,\ste)$
as well). This implies that for any $\ste$-module $M$,
we have canonical isomorphisms
\begin{equation}\label{calculcohstemod}
\Hom_{\DMte(S,R)}(R(X),\derR\Omega^\infty(M))\simeq
\Hom_{\DMt(S,\ste)}(\ste(X),M) \ .
\end{equation}
Note that as the forgetful functor \eqref{derforgettate} preserves direct sums,
Proposition \ref{compactdmtilde} implies that $\ste(X)(n)$
is a compact object of $\DMt(S,\ste)$ for all smooth $S$-scheme $X$
and integer $n$.
\end{paragr} | 206,514 |
On Thu, Jun 9, 2011 at 02:17, Jonathan Schleifer <js-pkgsrc%webkeks.org@localhost> wrote: > asterisk18 has more than 2 security problems for several months now. For all > of them, patches were released, yet none was incorporated into NetBSD. In the > meantime, the number of security problems increased, yet no update to pkgsrc > at all. Currently, it has 5 security issues - patches exist for all of them. > > So, I suggest to either maintain it again or remove the package, as a package >. --Benny. | 278,931 |
TITLE: Application of Martingale representation theorem to exponential martingale
QUESTION [4 upvotes]: I am trying to solve the following problem from Oksendal textbook.
Let $M_t$ be a martingale defined in the following way
$$M_t:=\mathbb{E}\left[e^{\sigma W_T}|\mathcal{F}_t\right]$$ By the
martingale representation theorem, we have
$$M_t=\mathbb{E}\left[M_0\right] + \int_0^t g\left(s,\omega\right)dW_s$$ for $g\left(s,\omega\right)$ such that the Ito integral be defined... Find said function !
I can derive (but maybe it is wrong) the following intermediate results $$M_t:=\mathbb{E}\left[e^{\sigma W_T}|\mathcal{F}_t\right]=e^{\frac{\sigma^2}{2}\left(T-t\right)+\sigma W_t}=\int_{0}^{t}e^{\frac{\sigma^2}{2}\left(T-s\right)+\sigma W_s}\sigma dW_s$$ $$\mathbb{E}\left[M_0\right]=e^{\frac{\sigma^2}{2}T} $$
This leads to $$\int _0^tg\left(s,\omega\right)dW_s=\int_0^te^{\frac{\sigma^2}{2}\left(T-s\right)+\sigma W_s}\sigma dW_s-e^{\frac{\sigma^2}{2}T}$$ then I am trapped since I do not see how to bring back the right most term under the ito integral.
Would appreciate any hints, thanks !
Following Saz advice
Taking $t\leftarrow T$, we have $$\int_0^T g\left(s,\omega\right)dW_s=e^{\sigma W_T}-e^{\frac{\sigma^2T}{2}}=e^{\frac{\sigma^2T}{2}}\left(e^{\sigma W_T-\frac{\sigma^2T}{2}}-1\right)$$ the parenthesis content is of the form $\left(X_T-X_0\right)$ in the equality $X_T = X_0 + \int_0^TdX_s$. Also $X_T$ is clearly an exponential martingale so that $dX_s=\sigma e^{-\frac{\sigma^2s}{2}+\sigma W_s}dW_s$. This leads to
$$\int_0^T g\left(s,\omega\right)dW_s=e^{\frac{\sigma^2T}{2}}\int_0^T \sigma e^{-\frac{\sigma^2s}{2}+\sigma W_s}dW_s \Rightarrow g\left(s,\omega\right)\equiv \sigma e^{\frac{\sigma^2}{2}\left(T-s\right)+\sigma W_s}$$
REPLY [2 votes]: Hints:
Show (or recall) that it suffices to find a function $g$ such that $$e^{\sigma W_T} = M_T = \mathbb{E}(M_0) + \int_0^T g(s) \, dW_s. \tag{1}$$
Applying Itô's formula show that $$e^{\sigma W_T-T \sigma^2/2} -1 = \sigma \int_0^T e^{\sigma W_s-s \sigma^2/2} \, dW_s.$$ Use this to find a representation of the form $(1)$.
Remark: Suppose that $X$ is an $\mathcal{F}_T$-measurable random variable which is of the form $X = g(W_T)$ for some nice function $g$. In order to find the representation $(1)$, the standard approach is to find a deterministic function $f$ such that $f(t) g(W_t)$ is a martingale. Usually, Itô's formula is quite useful to determine $f$. | 24,152 |
\begin{document}
\newcommand{\HDS}{\vrule width0pt height2.3ex depth1.05ex\displaystyle}
\def\f#1#2{{{\HDS #1}\over{\HDS #2}}}
\def\lpravilo#1{ \makebox[-.5em][r]{\mbox{\it #1}} {\mbox{\hspace{0.5em}}}}
\def\str{\rightarrow}
\newcommand{\qed}{\hfill $\Box$}
\def\k#1#2{\stackrel{\raisebox{-2pt}{\mbox{\tiny $#1$}}}{k}^
{\raisebox{-8pt}{\scriptsize #2}}}
\def\G{\mbox{\it Gen}}
\def\R{\mbox{\it Rel}}
\def\mj{\mbox{$\boldmath 1$}}
\def\cl{\mbox{\rm Cl}}
\title{\bf Generality of Proofs \\ and its Brauerian Representation}
\author{{\sc Kosta Do\v sen} and {\sc Zoran Petri\' c}
\\[.05cm]
\\Mathematical Institute, SANU \\
Knez Mihailova 35, p.f. 367 \\
11001 Belgrade, Yugoslavia \\
email: \{kosta, zpetric\}@mi.sanu.ac.yu}
\date{}
\maketitle
\begin{abstract}
\noindent The generality of a derivation is an equivalence relation on
the set of occurrences of variables in its premises and conclusion such
that two occurrences of the same variable are in this relation if and
only if they must remain occurrences of the same variable in every
generalization of the derivation. The variables in question are
propositional or of another type. A generalization of the derivation
consists in diversifying variables without changing the rules of
inference.
This paper examines in the setting
of categorial proof theory the conjecture that two derivations
with the same premises and conclusions stand for the
same proof if and only if they have the same generality. For that purpose
generality is defined within a category whose arrows are
equivalence relations on finite
ordinals, where composition is rather complicated.
Several examples are given of deductive systems of derivations covering
fragments of logic, with the associated map into the category of
equivalence relations of generality.
This category is isomorphically represented in the category whose arrows are
binary
relations between finite ordinals, where composition is the usual
simple composition of relations.
This representation is related to a classical representation result of
Richard Brauer.
\vspace{0.3cm}
\noindent{\it Mathematics Subject Classification} ({\it 2000}): 03F07,
03G30, 18A15, 16G99 \\[.2cm]
{\it Keywords}: identity criteria for proofs, generality of proof,
categories of proofs, Brauer algebras, representation
\end{abstract}
\section{Introduction}
Up to now, two conjectures
were made concerning identity criteria for proofs.
The first, and dominating conjecture, which we call the {\it Normalization
Conjecture}, was made by Prawitz in \cite{P71}. There Prawitz raised the
question of identity criteria for proof as one of the central questions
of general proof theory, and conjectured that two derivations in
natural deduction should stand for the same proof iff they both reduce to
the same normal form. (For details and further references concerning
the Normalization Conjecture see \cite{D02}.) Via the Curry-Howard
correspondence, the Normalization Conjecture says that identity of
proofs is caught by equality between typed lambda terms, and this shows
that for fragments of intuitionistic logic the equivalence relation
on derivations envisaged by Prawitz is a well-grounded mathematical
notion. A further corroboration for this conjecture is brought by
the categorial equivalence between typed lambda calculuses and
cartesian closed categories, which was demonstrated by Lambek
(see \cite{L74} and \cite{LS86}, Part I). Cartesian closed categories
provide an alternative description of proofs in intuitionistic
conjunctive-implicative logic, and the equivalence relation between
proofs given by identity of arrows in cartesian closed categories is in
accordance with the Normalization Conjecture.
The second conjecture concerning identity criteria for proofs, which we call
the {\it Generality Conjecture}, was suggested by Lambek in
\cite{L68} (pp. 287, 316) and \cite{L69} (p. 89). According to
this conjecture, two derivations with the
same premises and conclusions should stand for the same proof iff for
every generalization of one of them, which consists in diversifying
the variables without changing the rules of inference, there is a
generalization of the other derivation so that in the two generalizations
we have the same premises and conclusions. For example, let
$\k{\wedge}{1}_{p,p}:p\wedge p\str p$ and
$\k{\wedge}{2}_{p,p}:p\wedge p\str p$
be obtained respectively by the first rule and second rule of conjunction
elimination, let $\langle \k{\wedge}{1}_{p,p},
\k{\wedge}{2}_{p,p}\rangle:p\wedge p\str
p\wedge p$ be obtained from $\k{\wedge}{1}_{p,p}$ and
$\k{\wedge}{2}_{p,p}$ by
conjunction introduction, and let $1_{p\wedge p}:p\wedge p\str p\wedge p$
be the identity derivation from $p\wedge p$ to $p\wedge p$. For every
generalization of $\langle \k{\wedge}{1}_{p,p},
\k{\wedge}{2}_{p,p}\rangle$---and there
is only one sort of generalization here, namely,
$\langle\k{\wedge}{1}_{q,r},
\k{\wedge}{2}_{q,r}\rangle:q\wedge r\str q\wedge r$---we have
a generalization of $1_{p\wedge p}$, namely,
$1_{q\wedge r}:q\wedge r\str q\wedge r$.
So, by the Generality Conjecture,
$\langle \k{\wedge}{1}_{p,p},\k{\wedge}{2}_{p,p}\rangle$ and
$1_{p\wedge p}$ should stand
for the same proof, which is in accordance with the Normalization Conjecture
in its beta-eta version. On the other hand, $\k{\wedge}{1}_{p,p}$ can be
generalized to $\k{\wedge}{1}_{q,r}:q\wedge r\str q$,
and there is no generalization
of $\k{\wedge}{2}_{p,p}$ of the type $q\wedge r\str q$. So, by the Generality
Conjecture, $\k{\wedge}{1}_{p,p}$ and $\k{\wedge}{2}_{p,p}$
would not stand for the same proof;
this is again in accordance with the Normalization Conjecture.
(We will see in Section 3 that when we go beyond the conjunctive fragment
of logic, the two conjectures need not accord with each other any more.)
The notion of generality of derivation, which is implicit in the Generality
Conjecture, will be made precise in the context of category theory. A proof
will be an arrow in a category, and a derivation will be represented by an
arrow term. The source of the arrow is the premise and the
target the conclusion of the proof, and analogously for arrow terms and
derivations. The generality of an arrow term will be an equivalence relation
on the set of occurrences of variables in the source and target of this
arrow term. If two occurrences in this set are equivalent, then they
must be occurrences of the same variable (but not vice versa). The
Generality Conjecture will say that two arrow terms with the same
source and target stand for the same arrow iff they have the same
generality.
Note that Lambek's formal notion of generality
from \cite{L68} and \cite{L69}, with which he became dissatisfied in
\cite{L72} (p. 65), is not the notion we will define
in this paper. We take from Lambek just the inspiration of his
intuitive remarks about generality of proofs.
In the next section, we will introduce a category called \G, in
which we will define formally the generality of derivations.
We may conceive this category as obtained by abstraction from the category
whose arrows are proofs and whose objects are formulae by replacing formulae
with sequences of
occurrences of variables. From arrows that stood for proofs we keep
only the equivalence relation holding between occurrences of
variables that
must in every generalization remain occurrences of the same variable.
Since sequences of occurrences of variables are finite we represent
them by finite ordinals. The variables that motivate the introduction
of \G\ are of an arbitrary type: they may be propositional
variables, or individual variables, or of any other type of
variable in a language from which we take our formulae. We
may also deal simultaneously with several types of variable.
In Section 3, we will give several examples of deductive systems,
covering fragments of logic, and of generality obtained
via maps into \G\ defined for these deductive systems, which induces identity
of proofs according to the Generality Conjecture.
The definition of composition of arrows in \G\ is rather involved, and we
will not attempt to give a direct formal proof of the associativity of this
composition. An indirect proof of this associativity will be provided
by an isomorphic representation of \G\ in the category \R\ whose arrows are
binary
relations between finite ordinals, where composition is just
composition of relations. We devote Sections 4-6 to this matter. In Section
4 we first deal with some very general questions about representing
equivalence relations, while in Section 5 we give our representation
of \G\ in \R, and prove the isomorphism of this representation.
In Section 6 we show that our representation of \G\ in \R\ is
related to a classical result of representation theory
due to Richard Brauer.
The representation of \G\ in \R\ does not serve only to prove the
associativity of the composition of \G, but it may be of an independent
interest. Although in passing from \G\ to \R\ we have a functor
involving an exponential function, \R\ may be helpful in
computations, since composing binary relations is like multiplying
matrices. The representation of \G\ in \R\ sheds also some light on
Brauer's representation result, which it generalizes to a certain
extent.
In a companion to this paper \cite{DP03} we will consider
a category more general than \G\ obtained by replacing equivalence relations
by preordering, i.e. reflexive and transitive, relations.
This category will also be isomorphically represented in \R,
in a manner that generalizes further Brauer's representation.
\section{The category \G}
The objects of the category \G\ are finite ordinals, and its arrows
$R:n\str m$ are equivalence relations $R\subseteq (n+m)^2$.
As usual, we write sometimes $xRy$ for $(x,y) \in R$.
If $X\subseteq \omega$, then $X^{+k}=_{\mbox{\scriptsize\it def}}
\{n+k\mid n\in X\}$. For $n,m\in \omega$,
the sum $n+m$ is by definition $n\cup m^{+n}$. For an arbitrary
relation $R\subseteq (n\cup m^{+n})^2$ and $k\in\omega$, let $R^{+k}
\subseteq (n^{+k}\cup m^{+n+k})^2$ be the relation defined by
\[x R^{+k} y \quad {\mbox{\rm iff}}\quad (x-k) R (y-k).\]
Note that if $R$ is an equivalence relation in \G, then for $k\neq 0$ the
equivalence relation $R^{+k}$ is not an arrow of \G.
The identity arrow $\mj_n$ in \G\ is the equivalence relation
$\mj_n\subseteq (n+n)^2$ defined by
\[(x,y) \in \mj_n \quad {\mbox{\rm iff}}
\quad x=y\;({\makebox{\rm mod }}n).\]
We define composition of arrows in \G\ in the following manner.
Let $\cl(R)$ be the transitive closure of the binary relation $R$.
Suppose
we have the equivalence relations $R_1:n\str m$ and $R_2:m\str k$; that is,
$R_1\subseteq(n\cup m^{+n})^2$ and $R_2\subseteq(m\cup k^{+m})^2$.
Then
$R_1\cup R_2^{+n}\subseteq (n\cup m^{+n}\cup k^{+m+n})^2$,
and it is easy to check that
$\cl(R_1\cup R_2^{+n})$ is an equivalence relation.
For $x\in n\cup k^{+n}$ and $m\in\omega$ let
\[x'=\left\{\begin{array}{lll}
x & {\mbox{\rm if}} & x\in n
\\
x+m & {\mbox{\rm if}} & x\in k^{+n}.
\end{array}
\right.
\]
Then for an arbitrary relation $R\subseteq (n\cup k^{+m+n})^2$ let
$R^\dagger\subseteq(n\cup k^{+n})^2$ be defined by
\[ x R^\dagger y \quad {\mbox{\rm iff}} \quad x' R y'.\]
It is easy to see that if $R$ is an equivalence relation, then
so is $R^\dagger$.
Then we define the composition $R_2\ast R_1:n\str k$ of the arrows
$R_1$ and $R_2$ above by
\[
R_2\ast R_1 =_{\mbox{\scriptsize\it def}} ((\cl(R_1\cup R_2^{+n})\cap
(n\cup k^{+m+n})^2)^\dagger.
\]
It can be checked that $R_2\ast R_1$ is indeed an equivalence relation on
$n\cup k^{+n}$. It is more involved to check formally that $\ast$ is
associative, though this is rather clear if we represent this
operation geometrically (as this is done in \cite{EK66}, and also
in categories of tangles; see \cite{K95}, Chapter 12, and references therein).
For example, for $R_1\subseteq (3\cup 9^{+3})^2=12^2$ that
corresponds to the partition
$\{\{0,3\}$, $\{4,5\}$, $\{1,6\}$, $\{7,8\}$, $\{2,9\}$, $\{10,11\}\}$, and
$R_2\subseteq (9\cup 1^{+9})^2=10^2$ that corresponds to the partition
$\{\{0,1\}$, $\{2,9\}$, $\{3,4\}$, $\{5,6\}$, $\{7,8\}\}$, the composition
$R_2\ast R_1\subseteq (3\cup 1^{+3})^2=4^2$ that corresponds
to the partition $\{\{0,3\}$, $\{1,2\}\}$ is obtained from the following
drawing
\begin{center}
\begin{picture}(160,120)
\put(3,23){\line(1,1){34}}
\put(0,63){\line(0,1){34}}
\put(57,63){\line(-1,1){34}}
\put(117,63){\line(-2,1){74}}
\put(0,20){\circle*{2}}
\put(0,60){\circle*{2}}
\put(20,60){\circle*{2}}
\put(40,60){\circle*{2}}
\put(60,60){\circle*{2}}
\put(80,60){\circle*{2}}
\put(100,60){\circle*{2}}
\put(120,60){\circle*{2}}
\put(140,60){\circle*{2}}
\put(160,60){\circle*{2}}
\put(0,100){\circle*{2}}
\put(20,100){\circle*{2}}
\put(40,100){\circle*{2}}
\put(10,57){\oval(20,20)[b]}
\put(30,63){\oval(20,20)[t]}
\put(70,57){\oval(20,20)[b]}
\put(110,57){\oval(20,20)[b]}
\put(90,63){\oval(20,20)[t]}
\put(150,57){\oval(20,20)[b]}
\put(150,63){\oval(20,20)[t]}
\put(0,17){\makebox(0,0)[t]{\scriptsize$0$}}
\put(-5,60){\makebox(0,0)[r]{\scriptsize$0$}}
\put(15,60){\makebox(0,0)[r]{\scriptsize$1$}}
\put(35,60){\makebox(0,0)[r]{\scriptsize$2$}}
\put(55,60){\makebox(0,0)[r]{\scriptsize$3$}}
\put(75,60){\makebox(0,0)[r]{\scriptsize$4$}}
\put(95,60){\makebox(0,0)[r]{\scriptsize$5$}}
\put(115,60){\makebox(0,0)[r]{\scriptsize$6$}}
\put(135,60){\makebox(0,0)[r]{\scriptsize$7$}}
\put(155,60){\makebox(0,0)[r]{\scriptsize$8$}}
\put(0,103){\makebox(0,0)[b]{\scriptsize$0$}}
\put(20,103){\makebox(0,0)[b]{\scriptsize$1$}}
\put(40,103){\makebox(0,0)[b]{\scriptsize$2$}}
\end{picture}
\end{center}
\noindent In this example all members of partitions have two elements, but
this is by no means necessary.
The complexity of the definition of $\ast$ in \G\ motivates the introduction
of an isomorphic representation of \G, considered in Section 5, in
which composition will be defined in an elementary way. The proof that
we have
there an isomorphic representation will provide an indirect proof
that $\ast$ in \G\ is associative, and that \G\ is indeed a category.
\section{Generality of derivations in fragments of logic}
The language $\cal L$ of conjunctive logic is built from a
nonempty set of propositional variables $\cal P$ with the binary connective
$\wedge$ and the propositional constant, i.e. nullary connective, $\top$
(the exact cardinality of $\cal P$ is not important here). We use the
schematic letters $A,B,C,\ldots$ for formulae of $\cal L$.
We have the following axiomatic derivations for every $A$ and $B$ in
$\cal L$:
\[
\begin{array}{l}
1_A:A\str A,
\\
\k{\wedge}{1}_{A,B}:A\wedge B\str A,
\\
\k{\wedge}{2}_{A,B}:A\wedge B\str B,
\\
\k{\wedge}{}_A:A\str\top,
\end{array}
\]
and the following inference rules for generating derivations:
\[
\f{f:A\str B \quad g:B\str C}{g\circ f:A\str C}
\]
\[
\f{f:C\str A \quad g:C\str B}{\langle f,g\rangle:C\str A\wedge B}
\]
This defines the deductive system $\cal D$ of conjunctive logic (both
intuitionistic and classical). In this system $\top$ is included as an
``empty conjunction''. For $f:A\str B$ and $g:C\str D$ let $f\wedge
g:A\wedge C\str B\wedge D$ abbreviate $\langle
f\circ\k{\wedge}{1}_{A,C}, g\circ\k{\wedge}{2}_{A,C}\rangle$.
We define now a function $G$ from $\cal L$ to the objects of \G\ by
taking that $G(A)$ is the number of occurrences of propositional
variables in $A$. Next we define inductively a function, also denoted
by $G$, from the derivations of $\cal D$ to the arrows of \G:
\[
\begin{array}{ll}
G(1_A)=\mj_{G(A)}, &
\\[.1cm]
(x,y)\in G(\k{\wedge}{1}_{A,B}) & {\mbox{\rm iff}} \quad
x=y\; ({\mbox{\rm mod }}G(A\wedge B)),
\\[.1cm]
(x,y)\in G(\k{\wedge}{2}_{A,B}) & {\mbox{\rm iff}} \quad
(x,y < G(A) \;{\mbox {\it and }} x=y)
\;{\mbox {\it or }}
(x,y \geq G(A) \;{\mbox {\it and }}
\\ & \hspace{5cm} x=y\; ({\mbox{\rm mod }} G(B))),
\\[.1cm]
(x,y)\in G(\k{\wedge}{}_{A}) & {\mbox{\rm iff}} \quad x=y,
\\[.2cm]
{\makebox[1em][l]{$G(g\circ f)=G(g)\ast G(f);$}} &
\end{array}
\]
for $x\in G(C)\cup G(B)^{+G(A)+G(C)}$ let
\[
x''=\left\{\begin{array}{ll}
x & {\mbox{\rm if }} x\in G(C)
\\
x-G(A) & {\mbox{\rm if }} x\in G(B)^{+G(A)+G(C)},
\end{array}
\right.
\]
and let $(x,y)\in G(g)^{\dagger\dagger}$ iff $(x'',y'')\in G(g)$; then
\[
\makebox[30em][l]{$G(\langle f,g\rangle)=\cl (G(f)\cup G(g)^
{\dagger\dagger}).$}
\]
Then, in accordance with the Generality Conjecture, we stipulate that
two derivations $f,g:A\str B$ of $\cal D$ are equivalent iff
$G(f)=G(g)$. A proof of $\cal D$ is then the equivalence class of a
derivation with respect to this equivalence relation.
It is then clear that proofs make the arrows of a category $\cal C$, with
the obvious sources and targets,
and that $G$
gives rise to a faithful functor from $\cal C$ to \G. The category
$\cal C$ happens to be the free cartesian category generated by the set of
propositional variables $\cal P$ as the generating set of objects (this
set may be conceived as a discrete category).
This fact about $\cal C$ follows from the coherence result for cartesian
categories treated in \cite{DP01} and \cite{P02}.
(Cartesian categories are
categories with all finite products, including the empty product, i.e.
terminal object. The category $\cal C$ can be equationally presented;
see \cite{LS86}, Chapter I.3, or \cite{DP01}.)
The language of disjunctive logic is dual to the language we had above:
instead of $\wedge$ and $\top$
we have $\vee$ and $\bot$ in $\cal L$, and instead
of $\k{\wedge}{\it i}$,
$\k{\wedge}{}$ and $\langle\; ,\; \rangle$ we have
in the corresponding deductive system $\cal D$
\[
\begin{array}{l}
\k{\vee}{1}_{A,B}:A\str A\vee B,
\\
\k{\vee}{2}_{A,B}:B\str A\vee B,
\\
\k{\vee}{}_A:\bot\str A,
\end{array}
\]
\[
\f{f:A\str C \quad g:B\str C}{[ f,g]:A\vee B\str C}
\]
Dually to what we had above, we obtain that equivalence of derivations
induced by generality makes of $\cal D$ the free category with all
finite coproducts generated by $\cal P$.
Let us now assume we have in $\cal D$ both $\wedge$ and $\vee$, but
without $\top$ and $\bot$, and let us introduce the category of proofs
$\cal C$ induced by generality as we did above. The category $\cal C$
will, however, not be the free category with nonempty finite products and
coproducts generated by $\cal P$, in spite of the coherence result of
\cite{DP02}. The problem is that in $\cal C$ we don't have the equation
\[\k{\wedge}{1}_{A,B}\circ(1_A\wedge[1_B,1_B])=\;\k{\wedge}{1}_{A,B\vee B}.\]
The graphs with respect to which coherence is proved in \cite{DP02}
don't correspond to the equivalence relations of \G. The graphs of
\cite{DP01}, with respect to which coherence can be proved for
cartesian categories are also not the equivalence relations of \G, but
in this case one easily passes to \G\ from the category whose arrows
are these graphs.
The coherence result of \cite{KML71}
for symmetric monoidal closed categories without
the unit object I can be used to show that generality in
the appropriate deductive system,
which corresponds to the
implication-tensor fragment of intuitionistic linear logic,
gives rise to the free symmetric
monoidal closed category without I. The graphs
of \cite{KML71} correspond to the equivalence relations of \G.
If we add intuitionistic implication to conjunctive or
conjunctive-disjunctive
logic, the equivalence of derivations induced by generality does not
give rise to free cartesian closed or free bicartesian
closed categories (for counterexamples see \cite{DP02}, Section 1, and
\cite{S75}).
Up to now, we considered generality of derivations in fragments of
logic with respect to propositional variables. In the next example this
generality will be considered with respect to individual variables.
Let now $\cal L$ be the language generated from a nonempty set of
individual variables $\{x,y,z,\ldots\}$, whose exact
cardinality is again not important, with the binary relational symbol
$=$, the binary connective $\wedge$ and the propositional constant $\top$.
As axioms and inference rules for our deductive system $\cal D$ we now
have whatever we had for conjunctive logic in the preceding section
plus all axioms of the form
\vspace{.3cm}
\[
\begin{array}{l}
r_x:\top\str x=x,
\\
s_{x,y}:x=y\str y=x,
\\
t_{x,y,z}:x=y\wedge y=z\str x=z.
\end{array}
\]
\vspace{.2cm}
The function $G$ from $\cal L$ to the objects of \G\ is now defined by
taking that $G(A)$ is the number of occurrences of individual variables
in $A$. We define inductively the function $G$ from the derivations of
$\cal D$ to the arrows of \G\ as we did in the preceding section for
conjunctive logic with the following additional clauses, where $X^+$ is
the reflexive and symmetric closure of $X$,
\[
\begin{array}{l}
G(r_x)=\{(0,1)\}^+,
\\
G(s_{x,y})=\{(0,3),(1,2)\}^+,
\\
G(t_{x,y,z})=\{(0,4),(1,2),(3,5)\}^+.
\end{array}
\]
Graphically, these clauses correspond to
\begin{center}
\begin{picture}(180,80)
\put(60,20){\line(1,2){20}}
\put(80,20){\line(-1,2){20}}
\put(140,20){\line(-1,2){20}}
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\put(10,20){\oval(20,20)[t]}
\put(150,60){\oval(20,20)[b]}
\put(10,17){\makebox(0,0)[t]{$x=x$}}
\put(70,17){\makebox(0,0)[t]{$y=x$}}
\put(150,17){\makebox(0,0)[t]{$x=z$}}
\put(70,65){\makebox(0,0)[b]{$x=y$}}
\put(130,65){\makebox(0,0)[b]{$x=y$}}
\put(170,65){\makebox(0,0)[b]{$y=z$}}
\put(10,65){\makebox(0,0)[b]{$\top$}}
\put(150,65){\makebox(0,0)[b]{$\wedge$}}
\put(-5,40){\makebox(0,0)[r]{$r_x$}}
\put(55,40){\makebox(0,0)[r]{$s_{x,y}$}}
\put(125,40){\makebox(0,0)[r]{$t_{x,y,z}$}}
\end{picture}
\end{center}
Then in the category $\cal C$ of proofs defined via generality we find,
for example, the following equations:
\[
\begin{array}{l}
t_{x,y,y}\circ(1_{x=y}\wedge r_y)=\; \k{\wedge}{1}_{x=y,\top},
\\
t_{x,x,y}\circ(r_x\wedge 1_{x=y})=\; \k{\wedge}{2}_{\top,x=y},
\\
s_{y,x}\circ s_{x,y}=1_{x=y},
\\
s_{x,x}\circ r_x=r_x.
\end{array}
\]
\vspace{.2cm}
\section{Representing equivalence relations by sets of functions}
In this section we will consider some general matters
concerning the representation of arbitrary equivalence relations.
This will serve for demonstrating in the next section the isomorphism of
our representation of \G\ in the category whose arrows are
binary relations between finite ordinals.
Let $X$ be an arbitrary set, and let $R\subseteq X^2$.
Let $p$ be a set such that for $p_0 \neq p_1$ we have
$p_0,p_1\in p$. Let $S\subseteq p^2$, and for $i,j\in\{0,1\}$
let $p_i S p_j$ iff $i\leq j$. Consider the following set of functions:
\[
{\cal F}^S(R)=_{\mbox{\scriptsize\it def}} \{f:X\str p\mid
(\forall x,y\in X)(xRy\Rightarrow (f(x),f(y))\in S)\},
\]
and for $x,y\in X$ let $f_x:X\str p$ be defined as follows:
\[
f_x(y)=\left\{\begin{array}{ll}
p_1 & {\mbox{\rm if $xRy$}}
\\
p_0 & {\mbox{\rm if not $xRy$.}}
\end{array}
\right.
\]
The function $f_x$ is the characteristic function of the $R$-cone
over $x$. We can then prove the following proposition.
\vspace{.3cm}
\noindent {\sc Proposition} 1.\quad {\it The relation $R$ is transitive
iff $(\forall x\in X)f_x\in{\cal F}^S(R)$.}
\vspace{.2cm}
\noindent {\it Proof.}\quad $(\Rightarrow)$ Suppose $yRz$.
If $f_x(y)=p_0$, then $(f_x(y),f_x(z))\in S$. If $f_x(y)=p_1$, then
$f_x(z)=p_1$ by the transitivity of $R$.
\vspace{.1cm}
$(\Leftarrow)$ If $yRz\Rightarrow (f_x(y),f_x(z))\in S$, then
$yRz\Rightarrow (f_x(y)=p_0\;{\mbox{\it or}}\; f_x(z)=p_1)$, which
means $yRz\Rightarrow(xRy\Rightarrow xRz)$.
\qed
\vspace{.2cm}
With the definition
\[
{\cal F}^=(R)=_{\mbox{\scriptsize\it def}}\{f:X\str p\mid
(\forall x,y\in X)(xRy\Rightarrow f(x)=f(y))\}
\]
we can prove the following proposition.
\vspace{.3cm}
\noindent {\sc Proposition} 2.\quad {\it If $S$ is reflexive and
antisymmetric, and $R$ is symmetric, then ${\cal F}^=(R)={\cal F}^S(R)$.}
\vspace{.2cm}
\noindent {\it Proof.}\quad If $f\in {\cal F}^=(R)$,
then $f\in {\cal F}^S(R)$ by the
reflexivity of $S$; and if $f\in {\cal F}^S(R)$, then $f\in {\cal F}^=(R)$
by the symmetry of $R$ and the antisymmetry of $S$.
\qed
\vspace{.2cm}
Suppose now the relation $S$ is reflexive and antisymmetric.
We can then prove the following proposition.
\vspace{.3cm}
\noindent {\sc Proposition} 3.\quad {\it The relation $R$ is
an equivalence relation iff $(\forall x,y\in X)(xRy$ $\Leftrightarrow
(\forall f\in {\cal F}^=(R))f(x)=f(y))$.}
\vspace{.2cm}
\noindent {\it Proof.}\quad Note first that in
the equivalence on the right-hand
side the direction $xRy\Rightarrow (\forall f\in{\cal F}^=(R)) f(x)=f(y)$
is satisfied by definition.
\vspace{.1cm}
$(\Rightarrow)$
If $(\forall f\in {\cal F}^=(R))f(x)=f(y)$, then
$f_x(x)=f_x(y)$ by Propositions 1 and 2, and since
$f_x(x) = p_1$ by the reflexivity of $R$, we have $xRy$.
\vspace{.1cm}
$(\Leftarrow)$ From
\[(\ast)\quad (\forall x,y\in X)((\forall f\in {\cal F}^=(R)) f(x)=f(y)
\Rightarrow xRy)
\]
we obtain that $R$ is reflexive by taking that $x=y$.
For transitivity, suppose $xRy$ and $yRz$. Then for every
$f \in {\cal F}^=(R)$
we have $f(x)=f(y)=f(z)$, and hence $xRz$ by $(\ast)$.
For symmetry, suppose $xRy$. Then for every $f \in {\cal F}^=(R)$
we have $f(y)=f(x)$, and hence $yRx$ by $(\ast)$.
\qed
\vspace{.3cm}
\noindent {\sc Corollary.} \quad {\it If $R_1,R_2\subseteq X^2$ are
equivalence relations, then $R_1= R_2$ iff
${\cal F}^=(R_1)={\cal F}^= (R_2)$.}
\vspace{.2cm}
{\it Proof.}\quad We have, of course, that $R_1= R_2$ implies
${\cal F}^=(R_1)={\cal F}^= (R_2)$. The converse follows
from Proposition 3. \qed
\vspace{.2cm}
\section{Representing \G\ in the category \R}
Let \R\ be the category whose arrows are binary relations between finite
ordinals. Let $I_n \subseteq n\times n$ be the identity relation on $n$;
the composition $R_2\circ R_1\subseteq n\times k$
of $R_1\subseteq n\times m$ and $R_2\subseteq m\times k$ is
$\{(x,y)\mid (\exists z \in m)(xR_1z \;{\mbox {\it and}}\; zR_2y)\}$.
Then for $p\in \omega$ such that $p\geq 2$ we define a functor $F_p$
from \G\ to \R\ in the following manner. On objects $F_p$ is defined by
$F_p(n)=p^n$. Every element of the ordinal $p^n$ is identified by a
function $f:n\str p$ in the following way. Every $f:n\str p$
corresponds to a sequence of members of $p$ of length $n$. These
sequences can be ordered lexicographically, and $(p^n,\leq)$ is
isomorphic to the set of these sequences ordered lexicographically.
For $f_1: n\str p$ and $f_2:m\str p$, let $[f_1,f_2]:n\cup m^{+n}\str p$
be defined by
\[
[f_1,f_2](x)=\left\{\begin{array}{ll}
f_1(x) & {\mbox{\rm if }}x \in n
\\
f_2(x-n) & {\mbox{\rm if }}x \in m^{+n}.
\end{array}
\right.
\]
For $R:n\str m$ an arrow of \G, and for $f_1:n\str p$ and $f_2:m\str
p$, we define $F_p(R)$ by
\[(f_1,f_2)\in F_p(R)\quad{\mbox{\rm iff}}\quad [f_1,f_2]\in{\cal F}^=(R),
\]
where ${\cal F}^=(R)$ is the set of functions defined as in the preceding
section. Here $X$ is $n+m$, while for
the ordinal $p\geq 2$ we have that $p_0$ is $0$, $p_1$ is $1$ and
$S$ is $\leq$. It
remains to check that $F_p$ so defined is indeed a functor.
\vspace{.3cm}
\noindent {\sc Proposition} 4.\quad {\it $F_p$ is a functor.}
\vspace{.2cm}
\noindent {\it Proof.}\quad We show first that $F_p(\mj_n)=I_{p^n}$
by remarking that
\[
[f_1,f_2]\in {\cal F}^=(\mj_n)\quad{\mbox{\rm iff}}\quad f_1=f_2.
\]
Next we have to show that $F_p(R_2\ast R_1)=F_p(R_2)\circ F_p(R_1)$ for
$R_1:n\str m$ and $R_2: m\str k$. This amounts to showing that for
$f_1: n\str p$ and $f_2:k\str p$
\[
(\ast\ast)\quad (\forall x,y\in n\cup k^{+n})(x(R_2\ast R_1)y\Rightarrow
[f_1,f_2](x)=[f_1,f_2](y))
\]
is equivalent to the assertion that there exists an $f_3:m\str p$ such
that the following two statements are satisfied:
\[\begin{array}{l}
(\ast 1)\quad (\forall x,y\in n\cup m^{+n})(x R_1 y\Rightarrow
[f_1,f_3](x)=[f_1,f_3](y)),
\\[.1cm]
(\ast 2)\quad (\forall x,y\in m\cup k^{+m})(x R_2 y\Rightarrow
[f_3,f_2](x)=[f_3,f_2](y)).\end{array}
\]
First we show that $(\ast\ast)$ implies that for some $f_3:m\str p$ we have
$(\ast 1)$ and $(\ast 2)$. Before defining $f_3$ we check the
following facts that follow from $(\ast\ast)$:
\vspace{.2cm}
{\footnotesize
\[
\begin{array}{ll}
{\mbox {\rm (I)}} &
(\forall x,y\in n)((x,z),(y,z)\in\cl(R_1\cup R_2^{+n})\Rightarrow
f_1(x)=f_1(y)),
\\[.1cm]
{\mbox {\rm (II)}} &
(\forall x,y\in k^{+m+n})((x,z),(y,z)\in\cl(R_1\cup R_2^{+n})\Rightarrow
f_2(x-m-n)=f_2(y-m-n)),
\\[.1cm]
{\mbox {\rm (III)}} &
(\forall x\in n)(\forall y\in k^{+m+n})((x,z),(y,z)\in\cl(R_1\cup
R_2^{+n}) \Rightarrow f_1(x)=f_2(y-m-n)).
\end{array}
\]}
Then for $z\in m$ we define $f_3(z)\in p$ as follows. If for some
$x\in n\cup k^{+m+n}$ we have $(x,z)\in\cl(R_1\cup R_2^{+n})$, then
\[
f_3(z)=\left\{
\begin{array}{ll}
f_1(x) & {\mbox{\rm if }} x\in n
\\
f_2(x-m-n) & {\mbox{\rm if }} x\in k^{+m+n}.
\end{array}
\right.
\]
This definition is correct according to (I)-(III). Otherwise, if
there is no such $x$, we put $f_3(z)=0$.
Let us now demonstrate $(\ast 1)$. Suppose $xR_1 y$. If $x,y\in n$,
then $f_1(x)=f_1(y)$ by $(\ast\ast)$. If $x,y\in m^{+n}$, then
$f_3(x)=f_3(y)$ by the definition of $f_3$. If $x\in n$ and $y\in
m^{+n}$, then $f_1(x)=f_3(y)$ by the definition of $f_3$. In
every case we obtain $[f_1,f_3](x)=[f_1,f_3](y)$. We demonstrate
analogously $(\ast 2)$.
It remains to show that from the assumption that for some $f_3:m\str p$
we have $(\ast 1)$ and $(\ast 2)$ we can infer $(\ast\ast)$. Suppose
$x(R_2\ast R_1)y$ and $x,y\in n$. Then there is a sequence
$x_1,\ldots, x_{2l}$, $l\geq 1$, such that $x=x_1$, $y=x_{2l}$, for
every $i$ such that $1<i<2l$ we have $x_i\in m^{+n}$,
and
\[
x_1 R_1 x_2, \; x_2R_2^{+n} x_3, \; x_3 R_1 x_4,\ldots,
\; x_{2l-1} R_1 x_{2l}.
\]
Then by
applying $(\ast 1)$ and $(\ast 2)$ we obtain $f_1(x)=f_1(y)$,
and hence $[f_1,f_2](x)=[f_1,f_2](y)$. We proceed analogously when
$x,y\in k^{+n}$, or when $x\in n$ and $y\in k^{+n}$.
\qed
\vspace{.2cm}
Next we show that $F_p$ is faithful. Since $F_p$ is one-one on objects,
this amounts to showing that it is one-one on arrows.
\vspace{.3cm}
\noindent {\sc Proposition} 5.\quad {\it $F_p$ is faithful.}
\vspace{.2cm}
\noindent {\it Proof.}\quad Suppose $F_p(R_1)=F_p(R_2)$.
This means that for
every $f_1:n\str p$ and every $f_2: m\str p$ we have $[f_1,f_2]\in
{\cal F}^=(R_1)$ iff $[f_1,f_2]\in {\cal F}^=(R_2)$. But every function
$f:n+m\str p$ is of the form $[f_1,f_2]$ for some $f_1:n\str p$ and
some $f_2:m\str p$. Hence ${\cal F}^=(R_1)={\cal F}^=(R_2)$, and $R_1=R_2$
by the Corollary of the preceding section.
\qed
\vspace{.2cm}
So we have an isomorphic representation of \G\ in \R.
\section{Connection with Brauer algebras}
The representation of the category \G\ in the category \R,
which we have presented in the preceding section, is closely connected
to Brauer's representation of Brauer algebras, which is the orthogonal
group case of \cite{B37} (Section 5; see also \cite{W88}, Section 2,
\cite{J94}, Section 3, and \cite{DKP02}).
An $(n,n)$-{\it diagram} is an arrow $R:n\str n$ of \G\ such that
every member of the partition that corresponds to $R$ is a two-element
set. When defining the composition $R_2\ast R_1:n\str k$ of $R_1:n\str
m$ and $R_2:m\str k$ we didn't take into account the ``circles'', or
``closed loops'', in $\cl(R_1\cup R_2^{+n})$, namely, those members $X$
of the partition corresponding to $\cl(R_1\cup R_2^{+n})$ such that
$X\cap n=\emptyset$ and $X\cap k^{+m+n}=\emptyset$.
(In our example in Section 2 we have a circle involving 7 and 8
in the drawing.) Let $l(R_1,R_2)$ be
the number of those circles.
For $n\in \omega$ and $c$ a complex number,
the {\it Brauer algebra} $B(n,c)$ is the
algebra with basis the set of all $(n,n)$-diagrams, and multiplication
between two $(n,n)$-diagrams $R_1$ and $R_2$ defined as
$c^{l(R_1,R_2)}(R_2\ast R_1)$ (cf. \cite{J94}, Section 2).
Let $V$ be a vector space of dimension $p$ with basis $w_0,\ldots, w_{p-1}$.
For the $(n,n)$-diagram $R$ define $\beta(R)\in {\mbox{\rm
End}}(\otimes^n V)$ by the matrix (with respect to the basis
$\{w_{a_0}\otimes \ldots \otimes w_{a_{n-1}} \mid a_i\in\{0,\ldots,
p-1\}\}$ of $\otimes^n V$)
\[
\beta(R)_{a_0\ldots a_{n-1},a_n\ldots a_{2n-1}}
=_{\mbox{\scriptsize\it def}}
\prod_{\{i,j\}\in R}\delta(a_i,a_j)
\]
where $\delta$ is the Kronecker $\delta$ (cf. \cite{J94}, Definition
3.1). The sequences $a_0\ldots a_{n-1}$ and $a_n\ldots a_{2n-1}$
correspond to functions from $n$ to $p$. This defines a homomorphism of
the Brauer algebra $B(n,p)$ onto a subalgebra of ${\mbox{\rm
End}}(\otimes^n V)$ (see \cite{J94}, Lemma 3.2), which is Brauer's
representation of Brauer algebras in \cite{B37} (Section 5).
Every $n\times m$-matrix $M$ whose entries are only 0 and 1 may be
identified with a binary relation ${\cal R}_M\subseteq n\times m$ such that
$M(i,j)=1$ iff $(i,j)\in {\cal R}_M$. The matrix $M$ is the characteristic
function of ${\cal R}_M$. The $p^n\times p^n$-matrix $\beta(R)$ is a 0-1
matrix such that ${\cal R}_{\beta(R)}$ is precisely $F_p(R)$. So at the root of
Brauer's representation we find a particular case of our
representation of \G\ in \R.
One difference between our representation and Brauer's is that we deal
with arbitrary equivalence relations, whereas Brauer's equivalence
relations are more special---they correspond to partitions whose every
member is a two-element set. Another difference is that we have
composition of relations where Brauer has multiplication of matrices.
One passes from multiplication of matrices to our composition of
relations by disregarding ``circles'' and by taking that $1+1=1$.
\vspace{.5cm}
\noindent {\footnotesize {\it Acknowledgement.} The writing of this paper
was financed by the Ministry of Science, Technology and Development of
Serbia through grant 1630 (Representation of proofs with applications,
classification of structures and infinite combinatorics).}
\newpage | 186,935 |
TITLE: Convert $(1+i) ^ {1+i}$ to polar form
QUESTION [0 upvotes]: Can someone please help me understand the exponent/logarithm relationships to get through this problem?
Thank you.
REPLY [0 votes]: We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $\log z=\log |z|+it$.
We need to write $1+i$ in polar form: $1+i=\sqrt2\,e^{i\pi/4}$.
By definition, \begin{align}(1+i)^{1+i}&=\exp((1+i)\log(1+i))=\exp((1+i)(\log\sqrt2+i\tfrac\pi4)\\ \ \\
&=\exp(\tfrac12\,(1+i)(\log2+i\tfrac\pi2)=\exp(\tfrac12\,(-\tfrac\pi2+\log 2 + i(\tfrac\pi2+\log 2))\\ \ \\
&\exp(-\tfrac\pi4+\tfrac12\log 2 + i(\tfrac\pi4+\tfrac12\log 2))\\ \ \\
&=e^{-\pi/4}\sqrt2\,\exp\left[i\left(\tfrac\pi4+\log \sqrt2\right)\right].
\end{align}
REPLY [0 votes]: Hints to explore
Can you write $1+i$ in polar form $re^{i\theta}$ with $r$ being a positive real?
Can you write $1+i$ in the form $e^{s+i \theta}$ with $s=\log(r)$ being a real number?
Then consider $\left(e^{s+i \theta}\right)^{1+i} = e^{(s+i\theta)(1+i)} = e^{(s-\theta)+i(s+ \theta)}$
Then remember that $e^{x+iy} = e^x \cos(y) + i e^x\sin(y)$
That will get you one value. But complex powers are not so conveniently defined and $re^{i\theta} = re^{i(\theta+2n\pi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer | 195,052 |
Versuri Nothing Left To Mutilate Lyrics
Album: Cannibal Corpse - The Wretched Spawn
As I walk behind her, her scent trails to me
What is it that draws me neared,ils my plight
Searching for the one, 21, she will run from the gleam of my blade, she will pay
Nothing left to mutilate
Nothing left to mutilate
[Solo Jack Owen] | 41,626 |
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ah it ain't too bad, well i gonna go now, gotta get up in like 5 hours, wanna try to sleep a little.
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Not bad man, just chilling out. Hows yourself?
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woah, woah... you just woke up?!?! What time is it, where you are? My day was pretty much uneventful, just laid in bed browsing FJ and what not.
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Your getting up at 3 in the morning!!! Damn son!
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I know the feeling man, I bored 24/7!
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haha i was gonna post a picture like that, but didn't know where to find that 'feel' pic.
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It could be better.... could be better. There's nothing to do in my city and I'm broke and the money i do have needs to go towards college tuition. How's Florida???
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Where abouts do you live in Florida?
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Sweet deal, i use to live in Valrico in the Hillsborough County, just outside of Tampa.
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oh ya, crazy! its cool I can still find my old house on google earth and I remembered it was actually in bloomingdale, which just south of Valrico.
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Hmm well i'm in British Columbia on the Pacific coast, so its not that cold, like 25-32 F, sometimes colder. | 85,409 |
Episode #11
Paradise Kiss and Princess Jellyfish
It's coming to the end of the anime season, and I've already finished Recovery of an MMO Junkie and Juni Taisen, with more coming in the following weeks. I always like the one week between seasons. Gives me time to procrastinate on watching everything else.
This episode, we're talking about Ai Yazawa's Paradise Kiss published by Vertical Comics and Akiko Higashimura's Princess Jellyfish published by Kodansha.
That song you heard at the top of the episode was "Koko Dake no Hanashi"by Chatmonchy from the Princess Jellyfish anime. Also included in this episode are "♥Lonely in Gorgeous♥" by Tommy february6 from the Paradise Kiss anime and "Kimi no Kirei ni Kizuiteokure" by Sambomaster, the ending from the Princess Jellyfish:
1:46 - We talk about Paradise Kiss!
19:56 - We talk about Princess Jellyfish!
Next time, we'll be talking about Kitchen Princess and Delicious in Dungeon!
Published on December 20 2017 | 59,853 |
Zeppelin Model
$139 Final Sale
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In 1863, a ride in a hot-air balloon over the lakes and forests of Minnesota introduced a German military officer to the wonders of lighter-than-air flight. Inspired by that thrilling experience, Count Ferdinand Von Zeppelin became a pioneer in the field – his name synonymous with the steerable rigid airships he invented. Crafted from aluminum and mounted on a museum stand, our remarkably detailed model salutes those leviathans of the air, and the age of innovation that spawned them.
Show Details +
Dimensions
- 28"W x 8"D x 12"H
- Weight: 6.6 lbs.
Views
Zeppelin Model
$139 Final Sale
$125 Member | 38,860 |
Maize Farming Business Plan Get N Plan For Corn After Cover Crop Maize Farming Business Plan In South Africa
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How To Set Up the onX Hunt App for Deer Hunting
The onX Hunt App is the Best Deer Hunting App. Follow these simple steps to set up the Hunt App for Deer Hunting. Not yet a member?
The onX Hunt App is the Best Deer Hunting App. Follow these simple steps to set up the Hunt App for Deer Hunting. Not yet a member?
Step 1. Log in to access all available tools and layers.
If you’re new to onX start a free 7-day trial and create your account. If you already have an onX Membership, log in here.
TRY THE ONX HUNT APP TODAY
No credit card required.
Step 2. Choose your state and download Offline Maps.
onX Hunt is available for all 50 U.S. states. Choose your state and find the area in which you plan to hunt deer.
Save your Offline Map so you won’t need cell service to use every feature in the onX Hunt App. Offline maps can be saved at three different resolutions and widths, at five miles wide, 10 miles wide or 150 miles wide. Five-mile wide maps have the best resolution when zoomed in. Offline maps even work while you’re in airplane mode, so you can save your battery life when in the field.
When you want to use your saved Offline Map for deer hunting, click on Offline Maps at the bottom of the Hunt App, followed by Go Offline. All your saved maps will be visible at the same time, so you don’t need to do anything when you move from one saved map to another. Do you hunt in more than one state? It’s easy to switch states within the onX Hunt App.
Step 3. Turn on Layers.
To find all available layers for your state, click Map Layers at the bottom of the Hunt App. Toggle between My Layers and Layer Library to find all available Layers. Premium and Elite onX Members will have access to all available layers.
To create the best deer hunting app experience, we recommend turning on all these layers as a baseline:
- Private Lands Layer
- GMUs / Hunting Units Layer
- Trails Layer, especially if you’re hiking into your deer hunting area
- QDMA CWD Layer. If you’re concerned about the potential of CWD in your area, use this layer to identify counties where chronic wasting disease (CWD) is present.
The more layers you have turned on, the slower your device may run. You can turn layers on and off as you need. Move any layers you don’t use regularly into the Layer Library.
You should find all the information you’ll need for deer hunting since the onX Hunt App features 400,000+ miles of trails, 121 million private properties, 985 million acres of public land, 421 map overlays and up-to-date landowner names and property boundaries.
Step 4. Know the Map Tools in the onX Hunt App.
The onX Hunt App is one of the best tools to have with you when deer hunting because of these available features, all found under Map Tools at the bottom of the Hunt App:
- Waypoints. onX members can also save Colored Waypoints and Photo Waypoints, which are great for marking specific tree stands, trail cameras or water sources.
- Line Distance. Know the distance between two points on your map and save it. It’s perfect for measuring shooting lanes and distances between your deer stand and a food plot.
- Area Shape Tool. Use this tool to calculate the acreage of your deer hunting land or the size of a food plot you might plant.
Make a mistake marking a waypoint or drawing a line? It’s easy to edit your Markups.
Step 5. Before you go on your deer hunt.
After you’ve set up the onX Hunt App for deer hunting and you’re ready to head out for your hunt, there are few more important tools and features you can use.
- Check the weather and wind direction. The Hunt App provides live weather reports, weather forecasts, wind direction, sunrise / sunset times and barometric pressure all curated in one place.
- Share your waypoints. Whether you’re hunting with friends or want your family to know where you’ll be, waypoint sharing keeps you connected.
- Double-check your access. In addition to the public and private land boundaries in the Hunt App, users can add Walk-In Layers to find public access to private land. Learn more.
Set Up onX Hunt for Deer Hunting Today
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Identification
- leaves glabrous, lanceolate, with strong rank odor - lateral veins sunken above creating a pleated appearance - tubular cream-green flowers - shiny purple-black elliptic berries
Description: Tree, reproductive at 4-10 m tall, reaching 22 cm dbh, leaves to 5 x 13 cm, alternate and solitary, membranous, lanceolate, lateral veins sunken above creating a bullate appearance, strong rank odor even in undamaged leaves, tubular cream-green flowers 15-18 mm, on short axillary and terminal inflorescences, fruit 6-10 mm, a shiny purple-black elliptic berry.
Similar species
Local distribution: Pacific slope from 1000 to 1400m.
Species range: Belize to Brazil
Habitat: Roadside, forest edge, rarely in forest light gaps.
Herbivores
Phenology: Flower: mar-may. Fruit: apr-jun.
Pollinators: Moths
Seed dispersers: Birds
Comments: Notable for its extremely pungent rank odor. Leaves usually bent upward on the margins forming a V in cross-section.
Credits: Images and text copyright 2002-2006 by William A. Haber, | 396,992 |
I came across a slideshow by Copper Mobile Inc. that I thought provided some interesting statistics, as well as some good suggestions for what you might want to consider and prepare before you begin developing your enterprise mobile application.
- Let”s start with some interesting numbers. Technology Analysts at Gartner say, “by 2015, there will be more than 6.7 billion smartphones globally”. TEKsystems talked to more than 1500 IT leaders this past March about what technology trend will impact their organization most. Prioritizing their answers yielded these results. 1. Mobile 2. Cloud Computing 3. Consumerization of IT 4. Social Media.
- Morgan Stanley in their 2012 research report, stated that more users will likely connect to Internet through mobile devices than desktop PCs in 5 years. In my blog last Thursday, I shared that I hold the same opinion.
- An enterprise application lets you and your personnel, particularly those working remotely and in the field, interact with customers directly. Access your business”s database remotely, and manage your company processes, projects, and operational plans remotely from any mobile device. Why is this so important to the enterprise? Of the worlds 4 billion mobile phones in use, 1.08 billion are smartphones. As I”ve said before and also sited in the Morgan Stanley report, “In 2014, mobile Internet usage will overtake desktop usage” By then, mobile Internet should also have taken over desktop Internet usage.
- The next question to ask, is what platforms should I build an enterprise mobile app for? Employees are getting their smartphones & tablets to offices and organizations too are in favor of BYOD (Bring your own device culture). When there”s a single application and multiple devices, it is important to have cross-platform support. An Enterprise Mobile App Development Platform like Snappii is a great solution, letting you easily use the visual What You See Is What You Get (WYSIWYG) Drag and Drop Editor to make mobile business apps for iOS, Android and HTML5. Best of all, you don”t have to learn Objective-C or Java code. The Snappii platform will properly write the code as you build your app.
- You will want to understand that a mobile application is different from the mobile version of your website. There needs to be differences in:the user interface, the screen size the app is designed to appear on, navigation specifically designed for a mobile device, mobile device specific clickable areas, content, be it video, audio, or text designed to show well on mobile devices and tablets, and the list can continue.
- To ensure your app will work properly for your staff and customers, you”ll want to ensure you”ll have an opportunity to Test your App, Traditionally, App Testing can be time-consuming, expensive, and potentially require even more effort. Testing is important to ensure any potential bug threats are eliminated, there are no high fragmentation in the mobile OS that would increase the chances for errors. Device specific testing is also important, at least for native mobile business apps.
- This is another area where the Snappii Enterprise Mobile App Platform offers you tremendous value. With the Snappii Live Build Preview App, you are able to build your app, then test it on any mobile device or tablet you”d like. If you find areas you”d like to change or correct, you can easily make those changes in the Snappii Platform Editor and retest the app before uploading it to either the Apple or Google stores.
- While 60 to 80 percent of the application costs charged by nearly all other developers goes to maintenance and adding features after the initial release, this is another important feature of the Snappii Platform. Whether an OS (Operating System) changes its requirements or you would like to make more or new changes to your enterprise mobile app, you can ensure the Snappii Platform is kept up to date on any OS changes and you can easily make changes to your App and easily update all of your mobile business apps and test them as part of the services you receive as a Snappii customer.
- Finally, it”s also important to have a deep understanding of the enterprise app trends within your business marketplace. With a rich library of pre-built business specific templates, you will find a template that will help you ensure your enterprise mobile application offers the features needed to ensure your app meets every industry standard and, in many cases, exceeds those standards.
- Go to today and start building the enterprise mobile app that will grow your business and your customer loyalty in 2013. | 190,264 |
TITLE: The cyclic nature of group U(n)
QUESTION [0 upvotes]: For what possible values of n is U(n) cyclic?Is there any general result regarding this?If yes please could anyone state it and possibly give me a hint about its proof.
REPLY [2 votes]: The group is cyclic if and only if $n=1,2,4,p^k$ or $2p^k$ where $p$ is an odd prime.
Gauss knew this. It seems to me the result is fairly nontrivial, requiring the proof that there is a primitive element mod $n$ in those, and only those, cases. If memory serves this problem was at least partially considered in his Disquisitiones Arithmeticae of $1801$. | 33,455 |
TITLE: Proximal Operator / Mapping of $ g \left( x \right) = {\left\| x \right\|}_{1} $ ($ {L}_{1} $ Norm) in the Complex Domain
QUESTION [2 upvotes]: The $ \operatorname{Prox}_{g \left( \cdot \right)} $ operator is given by:
$$ \operatorname{Prox}_{\lambda g \left( \cdot \right)} \left( x \right) = \arg \min_{u} \left\{ g \left( u \right) + \frac{1}{2 \lambda} {\left\| u - x \right\|}^{2} \right\} $$
For $ g \left( x \right) = {\left\| x \right\|}_{1} $ in the $ {\mathbb{R}}^{n} $ the is given by:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} = \begin{cases}
{x}_{i} - \lambda & \text{ if } {x}_{i} \geq \lambda \\
0 & \text{ if } \left| {x}_{i} \right| \leq \lambda \\
{x}_{i} + \lambda & \text{ if } {x}_{i} \leq -\lambda \\
\end{cases} $$
My question is, what would be the result if the domain is $ {\mathbb{C}}^{n} $?
Namely:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} \left( x \right) = \arg \min_{u} \left\{ {\left\| x \right\|}_{1} + \frac{1}{2 \lambda} {\left\| u - x \right\|}^{2} \right\}, \; u ,x \in {\mathbb{C}}^{n} $$
Thank You.
REPLY [2 votes]: Let $\phi(u,x) = \|u\|_1 + {1 \over 2\lambda} \|x-u\|_2^2$. With a slight abuse
of notation, note that
$\phi(u,x) = \sum_k \phi(u_k,x_k)$, so we may as well assume that $u,x \in \mathbb{C}$.
If $x=0$, we see that $\operatorname{prox}_{\lambda {|\cdot|}} (x) = 0$, so assume $x \neq 0$.
Note that if $|\theta|=1$, then $\phi(u,x) = \phi(\theta u, \theta x)$ and
also $\phi(u,x) \ge \phi(\operatorname{re} u, \operatorname{re} x)$.
In particular,
$\phi(u,x) = \phi({\bar{x} \over |x|} u, |x|) \ge \phi(\operatorname{re}({\bar{x} \over |x|} u), |x|)$, from which we see that (i) we need only
optimise over real $u$ and (ii)
$\operatorname{prox}_{\lambda {|\cdot|}} (x) = {|x| \over \bar{x}} \operatorname{prox}_{\lambda {|\cdot|}} (|x|)$, where the latter is
computed over a real domain (we have
$\operatorname{prox}_{\lambda {|\cdot|}} (|x|) = \max(0,|x|-\lambda)$).
Notes:
(i) ${|x| \over \bar{x}} = { x \over |x| }$.
(ii) The real (one or higher dimension) problem
$\min_u \|u\|_1 + {1 \over 2 \lambda } \|x-u\|_2^2$ has a nice solution
via a minor extension of the von Neumann mimimax theorem.
We have $\|u\|_1 =\max_{\|h\|_\infty \le 1} \langle h, u \rangle$ and
the problem can be written as
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2$. Applying the aforementioned theorem (with a minor extension to
deal with the non compact domain for $u$) we have
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|^2 = \max_{\|h\|_\infty \le 1} \min_u \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2$ and solving the inner quadratic problem
gives
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2 = \max_{\|h\|_\infty \le 1} \langle h, x \rangle - {\lambda \over 2} \|h\|_2^2$ (with $\lambda h - (x-u) = 0$).
This is separable, and reduces to solving
$\max_{|h_k| \le 1} h_k x_k - {\lambda \over 2} h_k^2$, which
has solution $h_k = \operatorname{sgn} x_k \min(1,{|x_k| \over \lambda})$, and
substituting into $u_k= x_k-\lambda h_k$
gives the solution $u_k=\operatorname{sgn} x_k \max(0,|x_k|-\lambda)$ | 107,005 |
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ISLAMABAD - General (r) Pervez Musharraf treason trial episode took a new, yet anticipated, turn on Tuesday with a top defence official jumping into the fray in a bid to dissociate the military from the entire saga.“It is extremely surprising why speculations are on about the military’s involvement in this case,” said Defence Secretary Lt-Gen (r) Asif Yasin Malik in a brief and candid interaction with media people Tuesday. The secretary was leaving for the Ministry of Defence (MoD), Rawalpindi, after attending a scheduled session of National Assembly Standing Committee on Defence at the Parliament House when several journalists raised a set of queries, all concerning the former dictator.The defence secretary spoke about the treason trial case against Pervez Musharraf in the Special Court to which a medical report on Musharraf’s treatment at Armed Forces Institute of Cardiology (AFIC), Rawalpindi, was submitted. The four-page report, prepared by AFIC Commandant Maj-Gen Syed Muhammad Imran Majeed, recommended coronary angiography of Pervez Musharraf to “optimise the management and to ascertain the possibility of further interventions like coronary artery bypass surgery.”In his Tuesday’s interaction with journalists, Defence Secretary Asif Yasin Malik attempted to mince words when asked about Musharraf’s possible fate. Passing the buck on the judiciary, he said, “The matter is pending with the court. It’s for the court to decide the case, which is the rightful and authorised platform to do so.”The secretary, however, added, “I wonder why speculative reports keep running in the media about this issue. I come across so many news items that suggest that the military is behind him (Musharraf). Pakistan Army has nothing to do with this whole episode. As I said the matter is with the court, let the court decide on it.”Malik further said, “When courts are civilians, a trial is conducted by civilians and there are civilians everywhere (dealing with the matter), there’s absolutely no room and possibility for the military’s role whatsoever.”Seen as the military establishment’s perspective on the Musharraf issue, the secretary’s viewpoint came against the backdrop of intense purported discussions within the top military ranks on devising the army’s response. Lately, the military top bosses were discussing whether to issue a formal statement on the treason trial episode or informally brief journalists on the issue.Pressure was mounting on the military to speak on Musharraf’s trial after the ex-army chief had claimed in televised interviews that he had full backing of Pakistan Army, which, he said, was deeply concerned over his trial. The same issue was discussed in the corps commanders’ conference last week.The defence secretary also dispelled the impression that Musharraf had gone into hiding in the AFIC. “Everybody knows AFIC has the best heart physicians. It is one of the best hospitals for cardiac treatment. He is being treated there and doctors are taking good care of him,” he remarked.Meanwhile, a meeting of NA Standing Committee on Defence was held under the chairmanship of Sheikh Rohail Asghar to discuss Cantonment Laws (Amendments) Bill 2013 and Surveying and Mapping Ordinance 2013. Senior officials at MoD, Military Lands & Cantonments (ML&Cs) and Survey of Pakistan appeared before the panel to give related briefings. The Supreme Court is hearing a case regarding delay in local government elections in the cantonment boards and its proceedings were also held Tuesday.Published in The Nation newspaper on 08-Jan-2014
| 71,721 |
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Responsibilities:
Under the direct supervision of the Chief of the Japan Information Center
Monitors and analyzes newspapers, radio, TV, and other public and private media in 10 Midwest states.
Research on media relations in 10 states, including building a network of contacts for the JIC.
Ensure implementation of the JIC’s media strategy as it is related to media outreach, and promotion of priority issues and events, with a development focus.
Conduct a survey of media relations issues in 10 Midwest states.
General office responsibilities including Japan-related inquiries and correspondence.
Qualifications:
US citizen or permanent residency holder ( working permit holder)
Majoring in media relations, public relations, journalism, communications or related area will be highly desirable.
Excellent communications skills.
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Languages:
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TITLE: Every $W^{1,p}$ has a representative in ACL
QUESTION [1 upvotes]: Let $\Omega:=(0,1)^n$ and define $ACL_i(\Omega)$ as the set of all Borel functions $u:\Omega\to\mathbb{R}$ such that
$$ t\mapsto u(x_1,\dots,x_{i-1},t,x_{i+1},\dots,x_n) $$
is $AC$ for a.e. $(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)$. Let $ACL(\Omega):=\cap_i ACL_i(\Omega)$ and denote by $ACL^p(\Omega)$ the set of functions $u\in ACL(\Omega)$ having all partial derivatives in $L^p(\Omega)$.
It seems to be well-known that any element of $W^{1,p}$ has a representative in $ACL^p(\Omega)$, but all proofs I have found actually show that, for any $i$, one can find a representative in $ACL_i(\Omega)$.
How can one deduce that there exists a representative in $ACL(\Omega)$? Notice that, given $u\in W^{1,p}(\Omega)$, it is not (a priori) sufficient to subsequently modify $u$ on the negligible sets $\Sigma_1,\dots,\Sigma_n$ provided by the above weaker assertion.
REPLY [2 votes]: This is addressed in section 4.9.2 of the Evans and Gariepy's book on measure theory. The key is to show that the representative in $ACL_i$ is the precise representative of the function (defined in the book), so it does not depend on $i$. | 105,963 |
TITLE: Evaluate the double integral with sin/cos
QUESTION [1 upvotes]: $$\iint_\limits{D}\sin(x)\cos(y) dA$$ where $A$ is bounded by $y = 1 + x, y = 1 - x, y = 0$.
So this is obviously a triangle region, lets split it into 2 region such that $D_1 + D_2 = D$.
Lets calculate: $\iint_\limits{D_1}\sin(x)\cos(y) dA$ first.
If I fix $y$ from say $0 \to 1$, then I get $x_{\text{left}} = -1$ and $x_{\text{right}} = 0$.
So I should get
$$\iint_\limits{D_1}\sin(x)\cos(y) dA = \int_{0}^{1} \int_{-1}^{0} \sin(x)\cos(y) dx dy = \sin(1) \cdot (\cos(1) -1)$$
But this isnt right. How do I make sure I am getting the right limits?
REPLY [1 votes]: You let $y$ go from $0$ to $1$. When $y$ is at some value, say $y=1/2$, then the range for $x$ is not $-1$ to $0$. The left bound is the line $y=1+x$ and when $y=1/2,$ the left limit for $x$ is $-1/2$. And it's different for every value of $y$. So the $x$ limits can't be $-1$ to $0$. They have to depend on $y$.
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\begin{document}
\maketitle
\vspace{-1.1cm}
\begin{abstract}
For any cosmological constant $\Lambda=-3/\ell^2<0$ and any $\alpha<9/4$, we find a Kerr-AdS spacetime $(\MM,g_{\mathrm{KAdS}})$, in which the Klein-Gordon equation $\Box_{g_{\mathrm{KAdS}}}\psi+\alpha/\ell^2\psi=0$ has an exponentially growing mode solution satisfying a Dirichlet boundary condition at infinity. The spacetime violates the Hawking-Reall bound $r_+^2>|a|\ell$.
We obtain an analogous result for Neumann boundary conditions if $5/4<\alpha<9/4$.
Moreover, in the Dirichlet case, one can prove that, for any Kerr-AdS spacetime violating the Hawking-Reall bound, there exists an open family of masses $\alpha$ such that the corresponding Klein-Gordon equation permits exponentially growing mode solutions.
Our result adopts methods of Shlapentokh-Rothman developed in \citep{ShlapentokhGrowing} and provides the first rigorous construction of a superradiant instability for negative cosmological constant.
\end{abstract}
\tableofcontents
\section{Introduction}
\label{sec:IntroductionI}
\subsection{The Klein-Gordon equation in asymptotically anti-de\,Sitter spacetimes}
The Einstein vacuum equations
\begin{align}
\label{eqn:Einstein}
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=0
\end{align}
with cosmological constant $\Lambda$ can be understood as a system of second-order partial differential equations for the metric tensor $g$ of a four-dimensional spacetime $(\MM,g)$. Solutions with negative cosmological constant have drawn considerable attention in recent years, mainly due to the conjectured instability of these spacetimes. For more details, see \citep{AndersonUniqueness}, \citep{DafermosHolzegelInstability}, \citep{Bizon}, \citep{DiasGravitational}, \citep{DiasHorowitzMarolfSantos}, \citep{HolzegelLukSmuleviciWarnick} and references therein.
\begin{wrapfigure}{r}{.25\textwidth}
\hspace{0.2cm}
\def\svgwidth{80pt}
\input{KerrAdS.pdf_tex}
\caption*{\footnotesize Penrose diagram of the exterior of the Kerr-AdS spacetime \label{figure:Penrose}}
\end{wrapfigure}
In appropriate coordinates, (\ref{eqn:Einstein}) forms a system of non-linear wave equations. A first step in understanding the global dynamics of solutions to (\ref{eqn:Einstein}) -- and thus eventually answering the question of stability -- is the study of linear wave equations on a fixed background. For $\Lambda<0$, efforts have focused on understanding the dynamics of the Klein-Gordon equation
\begin{align}
\label{eqn:KG}
\Box_g\psi+\frac{\alpha}{\ell^2}\psi=0
\end{align}
for an asymptotically AdS metric $g$ with cosmological constant $\Lambda=-3/\ell^2$ and a mass term $\alpha$ satisfying the Breitenlohner-Freedman bound $\alpha<9/4$ \citep{BreitenlohnerFreedman}, which is required for well-posedness of the equation -- see \citep{WarnickMassive}, \citep{HolzegelWell} and \citep{Vasy}. The conformally coupled case $\alpha=2$ encompasses scalar-type metric perturbations around an exact AdS spacetime \citep{IshibashiWald3}.
For $g$ being the metric of an exact AdS spacetime, the massive wave equation (\ref{eqn:KG}) allows for time-periodic solutions due to the timelike nature of null and spacelike infinity $\mathcal I$; in particular, general solutions to (\ref{eqn:KG}), while remaining bounded, do not decay.
The behaviour of solutions to (\ref{eqn:KG}) on black-hole spacetimes is very different. Given a Kerr-AdS spacetime with parameters $\ell$, $M$ and $a$ satisfying $|a|<\ell$, define the Hawking-Reall Killing vector field
\begin{align*}
K:=T+\frac{a\Xi}{r_+^2+a^2}\Phi,
\end{align*}
where, using Boyer-Lindquist coordinates, $T=\del_t$ and $\Phi=\del_{\tilde\phi}$; see Section~\ref{subsec:KerrAdS} for definitions of $\Xi$ and $r_+$. The vector field $K$ is the (up to normalisation) unique Killing vector field that is null on the horizon $\HH$ and non-spacelike in a neigbourhood of $\HH$. It is globally timelike in the black hole exterior if the Hawking-Reall bound $r_+^2>|a|\ell$ is satisfied. If the bound is violated, $K$ becomes non-timelike far away from the horizon.
\begin{wrapfigure}{r}{.4\textwidth}
\hspace{0.2cm}
\def\svgwidth{250pt}
\input{Admissible_parameter_space.pdf_tex}
\caption*{\footnotesize For $r_+<\ell$, the two shaded regions represent the set of admissible parameters for $|a|/\ell$ and $r_+/\ell$. Within the plain gray area (bottom right), the Hawking-Reall bound is satisfied, whereas it is violated in the striped (intermediate) domain.\label{figure:parameters}}
\end{wrapfigure}
In \citep{HawkingReall}, Hawking and Reall use the existence of a globally causal $K$ for $r_+^2>|a|\ell$ to argue towards the stability of these spacetimes.
Indeed, uniform boundedness of solutions to (\ref{eqn:KG}) in the full regime $\alpha<9/4$ was proved for $r_+^2>|a|\ell$ in \citep{HolzegelMassive} and \citep{HolzegelWarnickBoundedness}. Moreover, in \citep{HolzegelSmuleviciDecay}, it was shown that solutions with the fastest radial decay (Dirichlet conditions at infinity) in fact decay logarithmically in time\footnote{\,\,\,Slightly stronger restrictions on $\alpha$ and the spacetime parameters were imposed in \citep{HolzegelSmuleviciDecay} for technical reasons, but the result is believed to hold in full generality by virtue of \citep{HolzegelWarnickBoundedness}.} and \citep{HolzegelSmuleviciQuasi} proves that this logarithmic bound is sharp.
For spacetimes violating the Hawking-Reall bound, the global behaviour of solutions to (\ref{eqn:KG}) has not been investigated rigorously, but it was argued in the physics literature -- see \citep{CardosoDias}, \citep{CardosoDiasLemosYoshida}, \citep{CardosoDiasYoshida} and \citep{DiasGravitational} -- that at least for small black holes, i.\,e. for $|a|\ll\ell$ and $|a|\ll r_+$, instability of solutions to (\ref{eqn:KG}) is to be expected if $r_+^2<|a|\ell$. As, in this regime, there is no Killing vector field which is globally timelike in the black hole exterior, this parallels the situation of asymptotically flat Kerr spacetimes, where superradiance
is present. For the present discussion, we will understand superradiance loosely as energy extraction from a rotating black hole. We will make this more precise in Lemma~\ref{lemma:negative_energy}.
\subsection{Unstable modes and superradiance in spacetimes with $\Lambda=0$}
The study of energy extraction from black holes in asymptotically flat spacetimes has a long history in the physics literature and two different, but related mechanisms have been proposed.
On the one hand, Press and Teukolsky \citep{PressTeukolsky} suggested that the leakage of energy through the horizon of a rotating black hole could be used to create a black hole bomb by placing a mirror around it. Superradiance would increase the radiation pressure on the mirror over time until it finally breaks, setting free all the energy at once.\footnote{\,\,\,In the asymptotically AdS case, infinity could serve as such a mirror due to the timelike character of spacelike and null infinity.} On the other hand, it was argued that energy could be extracted by the aid of massive waves acting as a natural mirror. This goes back to Zel'dovich \citep{Zeldovich} and was explored further by Starobinsky in \citep{Starobinskii}. Numerous heuristic and numerical studies on the superradiant behaviour of solutions to the Klein-Gordon equation followed, e.\,g. \citep{DamourDeruelleRuffini}, \citep{Zouros}, \citep{Detweiler}, \citep{Dolan2} and \citep{Dolan}. These studies found exponentially growing solutions to the massive wave equation on Kerr spacetimes.
Remarkably, this instability is not present at the level of the massless wave equation
\begin{align*}
\Box_g\psi=0,
\end{align*}
see \citep{DafermosRodnianskiShlapentokh}, where boundedness and decay for such solutions is proved in the full subextremal range $|a|<M$. Even though energy can potentially leak out of the black hole, superradiance can be overcome here as the superradiant frequencies in Fourier spaces are not trapped. In particular, in the context of scattering \citep{DafermosRodnianskiShlapentokhScattering}, a quantitative bound on the maximal superradiant amplification was shown.
In accordance with the above heuristic of massive waves acting as a natural mirror for a black hole bomb, this situation changes dramatically for the Klein-Gordon equation
\begin{align}
\label{eqn:KG_Yakov}
\Box_g\psi-\mu^2\psi=0
\end{align}
with scalar mass $\mu>0$.
A first rigorous construction of exponentially growing finite-energy solutions
in Kerr spacetimes was given by Shlapentokh-Rothman \citep{ShlapentokhGrowing}. The constructed solutions were modes. Mode solutions are solutions of the form
\begin{align}
\label{eqn:Mode}
\psi(t,r,\theta,\tilde\phi)=\e^{-\im\omega t}\e^{\im m\tilde\phi}S_{ml}(\cos\theta)R(r)
\end{align}
in Boyer-Lindquist coordinates $(t,r,\theta,\tilde\phi)$ for $\omega\in\CC$, $m\in\ZZ$ and $l\in\ZZ_{\geq|m|}$, where the smooth functions $S_{ml}$ and $R$ satisfy ordinary differential equations arising from the separability property of the wave equation in Boyer-Lindquist coordinates \citep{Carter_Separability}. We call a mode unstable if it is exponentially growing in time, i.\,e. if $\Im\omega>0$. Shlapentokh-Rothman showed that, for any given Kerr spacetime with $0<|a|<M$,
there is an open family of masses $\mu$ producing unstable modes with finite energy. The construction starts from proving existence of real modes and hence produces in particular periodic solutions. We will adopt this strategy.
\subsection{Unstable modes and superradiance in Kerr-AdS spacetimes}
Let us return to the Kerr-AdS case and
connect the existence of unstable modes to superradiance. Recall that the energy-momentum tensor for the Klein-Gordon equation (\ref{eqn:KG}) is given by
\begin{align*}
\mathbb{T}_{\mu\nu}:=\Re\left(\nabla_{\mu}\psi\overline{\nabla_{\nu}\psi}\right)-\frac{1}{2}g_{\mu\nu}\left(|\nabla\psi|^2-\frac{\alpha}{\ell^2}|\psi|^2\right)
\end{align*}
and that, for each vector field $X$, we obtain a current
\begin{align*}
J_{\mu}^X:=\mathbb{T}_{\mu\nu}X^{\nu}.
\end{align*}
While in Kerr spacetimes, $T=\del_t$ (see Section~\ref{subsec:KerrAdS}) is the (up to normalisation) unique timelike Killing field at infinity, the family of vector fields $T+\lambda\Phi$ with $\Phi=\del_{\tilde{\phi}}$ is timelike near infinity in Kerr-AdS spacetimes if and only if
\begin{align}
\label{eqn:Lambda_range}
-\ell^{-2}\left(\ell+a\right)<\lambda<\ell^{-2}\left(\ell-a\right).
\end{align}
Hence, in this range of values for $\lambda$, the conserved current $J_{\mu}^{T+\lambda\Phi}$ encapsulates the energy density of the scalar field measured by different (rotating) observers at infinity. The vector field $T+\lambda\Phi$ becomes spacelike or null at the horizon.
Recall that the Hawking-Reall vector field $K$ is tangent to the null generators of the horizon $\HH$. Therefore the energy density radiated through the horizon is measured by
\begin{align*}
J_{\mu}^{T+\lambda\Phi}K^{\mu}\Big\lvert_{\mathcal H}&=\Re\left(\left(T\psi+\lambda\Phi\psi\right)\overline{K\psi}\right)\Big\lvert_{\mathcal H}\\
&=\Re\left(\left(T\psi+\lambda\Phi\psi\right)\overline{\left(T\psi+\frac{a\Xi}{r_+^2+a^2}\Phi\psi\right)}\right)\bigg\lvert_{\mathcal H}
\end{align*}
since $g(T+\lambda\Phi,K)=0$ on the horizon.
For mode solutions (\ref{eqn:Mode}), this yields
\begin{align}
\label{eqn:ModeEnergy}
J_{\mu}^{T+\lambda\Phi}K^{\mu}\Big\lvert_{\HH}&=\left(|\omega|^2-\Re(\omega)\frac{ma\Xi}{r_+^2+a^2}+m\lambda\left(\frac{ma\Xi}{r_+^2+a^2}-\Re(\omega)\right)\right)|\psi|^2\bigg\lvert_{\HH}.
\end{align}
A non-trivial mode solution radiates energy away from the horizon if and only if the expression (\ref{eqn:ModeEnergy}) is negative
for all $\lambda$ in the range (\ref{eqn:Lambda_range}). The thusly characterised frequencies $\omega$ form the superradiant regime.
\begin{lemma}
\label{lemma:negative_energy}
Let $r_+^2<|a|\ell$. Let $\psi$ be a mode solution with $\omega(\epsilon)=\omega_R(\epsilon)+\im\epsilon$ for sufficiently small $\epsilon>0$, $\omega_R(\epsilon)\in\RR$ and $\omega_R(0)=ma\Xi/(r_+^2+a^2)$. If
\begin{align}
\label{eqn:superradiance}
\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)<0,
\end{align}
then $J_{\mu}^{T+\lambda\Phi}K^{\mu}\Big\lvert_{\HH}<0$ for sufficiently small $\epsilon>0$ and $\lambda$ in (\ref{eqn:Lambda_range}).
\end{lemma}
\begin{proof}
Since $J_{\alpha}^{T+\lambda\Phi}K^{\alpha}=0$ at the horizon for $\epsilon=0$, it suffices to differentiate (\ref{eqn:ModeEnergy}) with respect to $\epsilon$ and evaluate at $\epsilon=0$. We see that the derivative is negative if and only if
\begin{align*}
\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)-m\lambda\frac{\del\omega_R}{\del\epsilon}(0)<0.
\end{align*}
This, however, can be easily checked to hold using (\ref{eqn:Lambda_range}) and $r_+^2<|a|\ell$.
\end{proof}
\begin{rk}
If $r_+^2>|a|\ell$, then $K$ induces an energy density at infinity and $J_{\mu}^KK^{\mu}\geq 0$, in accordance with the intuition of not being in the superradiant regime if the Hawking-Reall bound is satisfied.
\end{rk}
We will show that our constructed growing mode solutions -- as the modes of \citep{ShlapentokhGrowing} -- satisfy the assumptions of Lemma~\ref{lemma:negative_energy}. This corroborates our interpretation that the unstable modes are a linear manifestation of the superradiant properties of Kerr-AdS spacetimes.
\subsection{The Kerr-AdS family}
\label{subsec:KerrAdS}
Before stating our results, we introduce the Kerr-AdS family of spacetimes. For a more exhaustive presentation, we refer the reader to \citep{HolzegelSmuleviciDecay}. Kerr-AdS spacetimes depend on three parameters $(\ell,M,a)$, where $\ell$ is related to the cosmological constant $\Lambda$ via $\Lambda=-3/\ell^2$. The parameter $M>0$ represents the mass of the black hole and $a$, the angular momentum per unit mass, is assumed to satisfy $|a|<\ell$. This condition guarantees for the metric to be regular. Let
\begin{align*}
\Delta_-(r):=(r^2+a^2)\left(1+\frac{r^2}{\ell^2}\right)-2Mr.
\end{align*}
The polynomial $\Delta_-$ has two real roots, denoted by $r_-< r_+$. We can write
\begin{align}
\label{eqn:Delta-}
\Delta_-(r)=\ell^{-2}(r-r_+)(r^3+r^2r_++r(r_+^2+a^2+\ell^2)-a^2\ell^2r_+^{-1}),
\end{align}
whence
\begin{align*}
\del_r\Delta_-(r_+)=\frac{1}{\ell^2}(3r_+^3+r_+a^2+r_+\ell^2-a^2\ell^2r_+^{-1}).
\end{align*}
This expression imposes some restrictions on the range of $|a|$ in terms of $r_+$ as shown in the following
\begin{lemma}
\label{lemma:r+}
If $r_+<\ell$,
\begin{align}
\label{eqn:rlessr}
a^2<r_+^2\frac{3\frac{r_+^2}{\ell^2}+1}{1-\frac{r_+^2}{\ell^2}}.
\end{align}
If $r_+\geq \ell$, $|a|$ can take any value in $[0,\ell)$.
\end{lemma}
\begin{proof}
These statements follow from $\del_r\Delta_-(r_+)>0$, which is a necessary condition for $r_-<r_+$. Note also that $r_+\geq \ell$ implies $r_+^2\geq|a|\ell$.
\end{proof}
Therefore, under the restriction of Lemma~\ref{lemma:r+}, there is a bijection between Kerr-AdS spacetimes with parameters $(\ell,M,a)$ and spacetimes with parameters $(\ell,r_+,a)$.
Henceforth we will use the shorthand notations $\MM_{\mathrm{KAdS}}(\ell,M,a)$ and $\MM_{\mathrm{KAdS}}(\ell,r_+,a)$ to denote Kerr-AdS spacetimes with parameters $(\ell,M,a)$ and $(\ell,r_+,a)$ respectively. The restriction of Lemma~\ref{lemma:r+} can be seen in the above figure.
Given $(\ell,M,a)$, a chart covering all of the domain of outer communication is given by Boyer-Lindquist coordinates $(t,r,\theta,\tilde{\phi})\in\RR\times(r_+,\infty)\times S^2$. The metric in these coordinates is
\begin{align*}
g_{\mathrm{AdS}}&=-\frac{\Delta_--\Delta_{\theta}a^2\sin^2\theta}{\Sigma}\,\dd t^2-2\frac{\Delta_{\theta}(r^2+a^2)-\Delta_-}{\Xi\Sigma}a\sin^2\theta\,\dd t\,\dd\tilde{\phi}+\frac{\Sigma}{\Delta_-}\,\dd r^2\\
&~~~~~~~~~~~~+\frac{\Sigma}{\Delta_{\theta}}\,\dd\theta^2+\frac{\Delta_{\theta}(r^2+a^2)^2-\Delta_-a^2\sin^2\theta}{\Xi^2\Sigma}\sin^2\theta\,\dd\tilde{\phi}^2,
\end{align*}
where
\begin{align*}
\Sigma=r^2+a^2\cos^2\theta,~~~~~~\Delta_{\theta}=1-\frac{a^2}{\ell^2}\cos^2\theta,~~~~~~\Xi=1-\frac{a^2}{\ell^2}.
\end{align*}
Since Boyer-Lindquist coordinates break down at $r=r_+$, we introduce Kerr-AdS-star coordinates $(\sta t,r,\theta,\phi)$. These are related to Boyer-Lindquist coordinates by
\begin{align*}
\sta t:=t+A(r)~~~\mathrm{and}~~~\phi:=\tilde{\phi}+B(r),
\end{align*}
where
\begin{align*}
\frac{\dd A}{\dd r}=\frac{2Mr}{\Delta_-(1+r^2/\ell^2)}\mathrm{~~~and~~~}\frac{\dd B}{\dd r}=\frac{a\Xi}{\Delta_-}.
\end{align*}
In these coordinates, the metric extends smoothly through $r=r_+$. One sees that the boundary $r=r_+$ of the Boyer-Lindquist patch is null and we shall call it the event horizon $\HH$.
Finally, we introduce the tortoise coordinate $\sta r$ which is related to $r$ by
\begin{align*}
\frac{\dd \sta r}{\dd r}=\frac{r^2+a^2}{\Delta_-(r)}
\end{align*}
with $\sta r(+\infty)=\pi/2$. We will denote the derivative with respect to $\sta r$ by $'$.
\subsection{Statement of the results}
The analysis in this paper yields two types of instability results:
\begin{itemize}
\item[A.] Given a cosmological constant $\Lambda$ and a mass $\alpha$, there is a Kerr-AdS spacetime for this $\Lambda$ in which (\ref{eqn:KG}) has a growing solution.
\item[B.] Given a Kerr-AdS spacetime violating the Hawking-Reall bound, there is a range for the scalar mass such that, in this spacetime, (\ref{eqn:KG}) has a growing solution.
\end{itemize}
To make this more precise, recall that mode solutions are Fourier modes that take the form
\begin{align*}
\psi(t,r,\theta,\tilde\phi)=\e^{-\im\omega t}\e^{\im m\tilde\phi}S_{ml}(\cos\theta)R(r)
\end{align*}
in Boyer-Lindquist coordinates $(t,r,\theta,\tilde\phi)$ for $\omega\in\CC$, $m\in\ZZ$ and $l\in\ZZ_{\geq|m|}$. Define $u(r):=(r^2+a^2)^{1/2}R(r)$. Use $\Ss_{\mathrm{mode}}(\alpha,\omega,m,l)$ to denote the set of all mode solutions with parameters $\omega,m,l$ to the Klein-Gordon equation with scalar mass $\alpha$. Set
\begin{align*}
\kappa^2:=9/4-\alpha.
\end{align*}
We require that all mode solutions are smooth. For the $S_{ml}$ this is ensured automatically by the definition -- see Section \ref{subsec:Spheroidal}. Hence we only need to impose a regularity condition on the function $u$, given parameters $\ell$, $r_+$, $a$, $m$ and $\omega$.
\begin{defn}[Horizon regularity condition]
\label{defn:HRC}
A smooth function $f:\,(r_+,\infty)\rightarrow\CC$ satisfies the horizon regularity condition if
$f(r)=(r-r_+)^{\xi}\rho$ for a smooth function $\rho$ as well as a constant
\begin{align}
\label{eqn:defn_xi}
\xi:=\im\frac{\Xi am-(r_+^2+a^2)\omega}{\del_r\Delta_-(r_+)}.
\end{align}
\end{defn}
Henceforth we will only call a mode $\psi$ a mode solution to (\ref{eqn:KG}) if its radial part $R$ (and hence $u$) satisfies the horizon regularity condition.
At infinity, we will study two different boundary conditions for $u$.
\begin{defn}[Dirichlet boundary condition]
\label{defn:BdyConds}
Given a mass $\alpha<9/4$ (i.\,e. a $\kappa>0$), a smooth function $f:\,(r_+,\infty)\rightarrow\CC$ satisfies the Dirichlet boundary condition if
\begin{align*}
r^{1/2-\kappa}f\rightarrow 0
\end{align*}
as $r\rightarrow\infty$.
We say that a $\psi\in\Ss_{\mathrm{mode}}(\alpha,\omega,m,l)$ satisfies Dirichlet boundary condition if its radial part $u$ satisfies the Dirichlet boundary condition.
\end{defn}
Mode solutions satisfying these boundary conditions are analogous to the modes considered in \citep{ShlapentokhGrowing}.
We are able to show the following result.
\begin{thm}
\label{thm:oldD}
Given a cosmological constant $\Lambda=-3/\ell^2$, a black hole radius $0<r_+<\ell$ and a scalar mass parameter $\alpha_0\in(-\infty,9/4)$, there are a spacetime parameter $a$ satisfying the regularity condition $|a|<\ell$, mode parameters $m$ and $l$ and a $\delta>0$ such that there are a smooth curve
\begin{align*}
(-\delta,\delta)\rightarrow\RR^2,~\epsilon\mapsto(\alpha(\epsilon),\omega_R(\epsilon))
\end{align*}
with
\begin{align}
\label{eqn:HRfrequency}
\alpha(0)=\alpha_0~~~\mathrm{and}~~~\omega_R(0)=\frac{\Xi am}{r_+^2+a^2}
\end{align}
and corresponding
mode solutions in $\Ss_{\mathrm{mode}}(\alpha(\epsilon),\omega_R(\epsilon)+\im\epsilon,m,l)$ satisfying the horizon regularity condition and Dirichlet boundary conditions.
For all $\epsilon\in(0,\delta)$, these modes satisfy
\begin{align*}
\frac{\dd\alpha}{\dd\epsilon}(0)>0 ~~~\mathrm{and}~~~\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)<0.
\end{align*}
\end{thm}
\begin{rk}
The $u$ in the theorem has finite energy and hence the spacetime parameters of the theorem must violate the Hawking-Reall bound as explained in the previous sections; this is explained further in Lemma~\ref{lemma:HR_positivity} and Remark~\ref{rk:HR_positivity}. By Lemma~\ref{lemma:r+}, we know that the $a$ can be located anywhere in the range
\begin{align*}
\frac{r_+^2}{\ell}<|a|<r_+\sqrt{\frac{3\frac{r_+^2}{\ell^2}+1}{1-\frac{r_+^2}{\ell^2}}}.
\end{align*}
We remark that our result does not restrict to small $|a|$. In fact, we can enforce $|a|$ to be as close to $\ell$ as we wish by choosing $r_+/\ell<1$ large.
\end{rk}
Lemma~\ref{lemma:negative_energy} implies that the constructed modes are superradiant and indicates that the instability is driven by energy leaking through the horizon.
Our next theorem builds on the first, but allows for the construction of an unstable superradiant mode with Dirichlet boundary conditions for each given $\alpha<9/4$.
\begin{thm}
\label{thm:new}
Let $\ell>0$ and $\alpha<9/4$. Then there is an $\MM_{\mathrm{KAdS}}(\ell,r_+,a)$ and a super\-radiant $\psi\in\Ss_{\mathrm{mode}}(\alpha,\omega_R+\im\epsilon,m,l)$ for an $\omega_R\in\RR$ and $\epsilon>0$ satisfying Dirichlet boundary conditions.
\end{thm}
The methods used in our proof also show the following statement:
\begin{cor}
\label{cor:new}
Let $\ell>0$, $\alpha<9/4$ and $0<r_+<\ell$. Then there is an $\epsilon>0$ such that for all $|a|\in(r_+^2/\ell,r_+^2/\ell+\epsilon)$, the Klein-Gordon equation with mass $\alpha$ has an exponentially growing mode solution in $\MM_{\mathrm{KAdS}}(\ell,r_+,a)$.
\end{cor}
\begin{rk}
These results also apply to the massless wave equation, which is an important difference to the asymptotically flat case.
\end{rk}
Furthermore, although this will not be pursued explicitly in this paper, one can also show the analogue of Shlapentokh-Rothman's result in our setting by only adapting the proof slightly.
\begin{thm}
\label{thm:analogue_Yakov}
Given a Kerr-AdS spacetime $\MM_{\mathrm{KAdS}}(\ell,r_+,a)$ satisfying, $|a|<\ell$, $r_+>0$ and $r_+^2<|a|\ell$ (and the restrictions of Lemma~\ref{lemma:r+}), there are mode parameters $m$ and $l$ as well as a $\delta>0$ such that, for each $\epsilon\in(-\delta,\delta)$, there is an open family of masses $\alpha(\epsilon)$ and a mode solution in $\Ss_{\mathrm{mode}}(\alpha(\epsilon),\omega_R(\epsilon)+\im\epsilon,m,l)$ satisfying Dirichlet boundary conditions with $\omega_R(0)$ as in (\ref{eqn:HRfrequency}).
\end{thm}
\begin{rk}
\begin{compactenum}
\item Conversely, in the asymptotically flat Kerr case of \citep{ShlapentokhGrowing}, it is also possible to prove an analogue of Theorem~\ref{thm:new} instead of only the analogue of Theorem~\ref{thm:oldD}, using our strategy explained in the next section.
\item To contrast our case to the asymptotically flat setting, we add three observations. First, in \citep{ShlapentokhGrowing}, the curve $\epsilon\mapsto(\mu(\epsilon),\omega_R(\epsilon)+\im\epsilon)$ must satisfy $\mu(0)^2>\omega_R(0)^2$. There is no equivalent condition for Kerr-AdS spacetimes as the instability is not driven by the interplay of frequency and mass, but by the violation of the Hawking-Reall bound. Second, in both cases, $\del\omega_R/\del\epsilon<0$ for small $\epsilon$, so $\omega_R(0)$ can be seen as the upper bound of the superradiant regime. Third, the result in Kerr holds for all $m\neq 0$, $l\geq |m|$. In contrast, our result is a statement about large $m=l$.
\end{compactenum}
\end{rk}
It is known -- see \citep{HolzegelWarnickBoundedness} and references therein -- that, for $0<\kappa<1$, i.\,e. $5/4<\alpha<9/4$, we also have well-posedness for different boundary conditions at infinity. This underlies the following
\begin{defn}[Neumann boundary condition]
\label{defn:BdyCondsNeumann}
Given a mass $5/4<\alpha<9/4$ (i.\,e. $0<\kappa<1$), a smooth function $f:\,(r_+,\infty)\rightarrow\CC$ satisfies the Neumann boundary condition if
\begin{align*}
r^{1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f\right)\rightarrow 0
\end{align*} as $r\rightarrow\infty$.
\end{defn}
Using the techniques of twisted derivatives, introduced in \citep{WarnickMassive}, we can prove versions of Theorems~\ref{thm:oldD} and \ref{thm:new} for Neumann boundary conditions.
\begin{thm}
\label{thm:oldN}
Given a cosmological constant $\Lambda=-3/\ell^2$, a black hole radius $0<r_+<\ell$ and a scalar mass parameter $\alpha_0\in(5/4,9/4)$, there are a spacetime parameter $a$ satisfying the regularity condition $|a|<\ell$, mode parameters $m$ and $l$ and a $\delta>0$ such that there is a smooth curve
\begin{align*}
(-\delta,\delta)\rightarrow\RR^2,~\epsilon\mapsto(\alpha(\epsilon),\omega_R(\epsilon))
\end{align*}
with (\ref{eqn:HRfrequency}). Moreover, there are corresponding mode solutions in $\Ss_{\mathrm{mode}}(\alpha(\epsilon),\omega_R(\epsilon)+\im\epsilon,m,l)$ satisfying Neumann boundary conditions. If $\epsilon\in(0,\delta)$, then the modes satisfy
\begin{align*}
\frac{\dd\alpha}{\dd\epsilon}(0)>0 ~~~\mathrm{and}~~~\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)<0.
\end{align*}
\end{thm}
\begin{thm}
\label{thm:newN}
Let $\ell>0$ and $5/4<\alpha<9/4$. Then there is an $\MM_{\mathrm{KAdS}}(\ell,r_+,a)$ and a super\-radiant $\psi\in\Ss_{\mathrm{mode}}(\alpha,\omega_R+\im\epsilon,m,l)$ for an $\omega_R\in\RR$ and $\epsilon>0$ satisfying Neumann boundary conditions.
\end{thm}
Let us conclude this section with a general remark on boundedness.
From \citep{HolzegelWarnickBoundedness}, we know that solutions to the Klein-Gordon equation with Dirichlet boundary conditions remain bounded for all $r_+^2>|a|\ell$. A similar statement holds for Neumann boundary conditions under more restrictive assumptions on the parameters. For $r_+^2=|a|\ell$, one can easily repeat the proof of the second theorem of \citep{HolzegelSmuleviciDecay} to see that there are no periodic solutions. One can potentially also extend the decay result of \citep{HolzegelSmuleviciDecay} to $r_+^2=|a|\ell$. Our results do not rule out boundedness in the entire parameter range in which $r_+^2<|a|\ell$ since we did not show that for \emph{any} given Kerr-AdS spacetime and \emph{any} $\alpha$, there are unstable mode solutions; they do, however, impose restrictions on the ranges of spacetime parameters and masses $\alpha$ in which boundedness could potentially hold. It is believed that, using more refined spectral estimates, our results can be shown to hold in the full regime $r_+^2<|a|\ell$, but we will not pursue this further.
\subsection{Outline of the proof}
The difficulty lies in the construction of the radial part $u$, for which we use the strategy of \citep{ShlapentokhGrowing}, which, as our present work shows, can be applied to more general settings than Kerr spacetimes. The technique contains two main steps.
\begin{itemize}
\item[I.] Construct $u$ corresponding to a real frequency $\omega_0\in\RR$.
\item[II.] Obtain a mode solution corresponding to a complex $\omega$ with $\Im\omega>0$ by varying spacetime and mode parameters.
\end{itemize}
We note that both steps are completely independent of each other, in particular step II does not rely on the method by which the periodic mode solution was constructed, but only requires existence of such a mode.
Let us first only deal with Dirichlet boundary conditions. To complete step I, $u$ needs to satisfy the radial ODE
\begin{align}
\label{eqn:angularODE_intro}
u''-(V-\omega_0^2)u=0
\end{align}
for the given boundary condition -- see Section~\ref{subsec:Spheroidal}. Lemma~\ref{thm:RestrReal} then already restricts $\omega_0$ to $\omega_+:=ma\Xi/(r_+^2+a^2)$. It is important to note that the boundary value problem does not admit nontrivial solutions in general.
\begin{lemma}
If $u$ satisfies the Dirichlet boundary condition for real $\omega_0$ and $V-\omega_0^2\geq 0$, then $u=0$.
\end{lemma}
\begin{proof}
Define $Q(r):=\Re(u'\overline u)$, note that $Q(r_+)=Q(\infty)=0$ and integrate $\dd Q/\dd r$.
\end{proof}
Hence, in a first step in Section~\ref{subsec:Potential}, we will find spacetime and mode parameters such that $V-\omega_0^2<0$ on some subinterval of $(r_+,\infty)$ for given $\ell$ and $\alpha_0$ by a careful analysis of the shape of the potential $V$ in Lemma~\ref{lemma:Vneg}. This requires proving an asymptotic estimate for the eigenvalues of the modified oblate spheroidal harmonics (Lemma~\ref{lemma:GroundState}). The spacetime parameters will necessarily violate the Hawking-Reall bound.
The radial ODE is the Euler-Lagrange equation of the functional
\begin{align}
\label{eqn:functional_intro}
\LL_a(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+(V-\omega^2)\frac{r^2+a^2}{\Delta_-}|f|^2\right)\,\dd r.
\end{align}
The functional is not bounded below, so we need to impose a norm constraint, which we choose to be $\norm{f/r}_{L^2(r_+,\infty)}=1$. Then Lemma~\ref{lemma:VariationInequalityD} gives a coercivity-type estimate. To carry out the direct method of the calculus of variations, we use the weighted Sobolev spaces that arise naturally from the functional -- see Section~\ref{sec:Real}. This setting of the minimisation problem then guarantees that the minimiser satisfies the correct boundary conditions. We remark that we will directly work with the functional (\ref{eqn:functional_intro}) instead of regularising first at the horizon and then taking the limit, as in \citep{ShlapentokhGrowing}.
Then, in Lemma~\ref{lemma:ELD}, we obtain an ODE
\begin{align*}
u''-(V-\omega_0^2)u+\nu_a\frac{u}{r^2}=0
\end{align*}
with a Lagrange multiplier $\nu_a\leq 0$ that depends continuously on the spacetime parameter $a$. By varying $a$, we find an $\hat a$ such that $\nu_{\hat a}=0$ (Proposition~\ref{propn:a_hat}) and hence a solution to the radial ODE.
To carry out step II, we need the asymptotic analysis of (\ref{eqn:angularODE_intro}) that is worked out in Section~\ref{subsec:LocalAna}. There are two branches that asymptote $r^{-1/2+\kappa}$ and $r^{-1/2-\kappa}$, respectively, at infinity. Let $h_1$ denote the branch with slow decay and $h_2$ the one with fast decay. Then
\begin{align*}
u(r,\alpha,\omega)=A(\alpha,\omega)h_1(r,\alpha,\omega)+B(\alpha,\omega)h_2(r,\alpha,\omega).
\end{align*} For the parameters from step I, $A(\alpha_0,\omega_0)=0$. By varying $\omega$ and $\alpha$ simultaneously in Section~\ref{subsec:PerturbingD}, the implicit function theorem yields a curve
\begin{align*}
\epsilon\mapsto(\omega_R(\epsilon)+\im\epsilon,\alpha(\epsilon))
\end{align*}
with $\omega_R(0)=\omega_0$ and $\alpha(0)=\alpha_0$ such that
\begin{align*}
A(\alpha(\epsilon),\omega(\epsilon))=0.
\end{align*}
along the curve. As $\Im\omega(\epsilon)>0$ for $\epsilon>0$, these modes grow exponentially whilst satisfying Dirichlet boundary conditions.
In Section~\ref{subsec:Crossing}, we show that
\begin{align}
\label{epsilon_dependence_outline}
\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)<0~~~\mathrm{and}~~~
\frac{\del\alpha}{\del\epsilon}(0)>0,
\end{align}
which proves Theorem~\ref{thm:oldD}. A careful analysis of the domain of the implicit function theorem in Section~\ref{subsec:Continuity} yields Theorem~\ref{thm:new}. Here, the analysis heavily exploits several continuity properties in the parameters. A difficulty is caused by $\hat a$ being defined as the infimum of an open set.
For Corollary~\ref{cor:new}, one observes that, by Lemma~\ref{lemma:Vneg}, once the Hawking-Reall bound is violated, one can always make the potential $V$ negative on some interval by choosing $|m|$ sufficiently large. This yields periodic modes for very small violation of the Hawking-Reall bound and hence growing modes by repeating the above argument.
The situation is more complicated if $u$ satisfies the Neumann boundary condition. Since, in this case, $u\sim r^{-1/2+\kappa}$ as $r\rightarrow\infty$, $\LL_a$ is not well-defined and hence cannot be used to produce periodic modes.
To carry out the construction of step I, we use twisted derivatives as introduced in \citep{WarnickMassive} and used extensively in \citep{HolzegelWarnickBoundedness}. To find the minimiser via the variational argument, we also need to modify our function spaces and use twisted weighted Sobolev spaces. All details are given in Section~\ref{sec:RealDN}.
The main technical problems, however, arise in the second part of the argument.
The underlying reason is that the proofs for step II rely severely on establishing monotonicity properties for the functional when varying $\alpha$. Since the twisting necessarily depends on $\alpha$, proving monotonicity in $\alpha$ is more involved and indeed the monotonicity properties shown in the Neumann case are weaker; nevertheless, the ideas introduced in Section~\ref{subsec:PerturbingN} are sufficiently robust not only to construct the growing modes, but also to be applicable to showing (\ref{epsilon_dependence_outline}) and to transition from Theorem~\ref{thm:oldN} to Theorem~\ref{thm:newN}. It is also in the Neumann case, where the independence of steps I and II -- alluded to above -- is exploited.
\section{Preliminaries}
\label{sec:Prelim}
\subsection{The modified oblate spheroidal harmonics}
\label{subsec:Spheroidal}
Following \citep{HolzegelSmuleviciDecay}, we define the $L^2(\sin\theta\,\dd\theta\,\dd\tilde{\phi})$-self adjoint operator $P$ acting on $H^1(S^2)$-complex valued functions as
\begin{align*}
-P(\omega,\ell,a)f&=\frac{1}{\sin\theta}\del_{\theta}(\Delta_{\theta}\sin\theta\del_{\theta}f)+\frac{\Xi^2}{\Delta_{\theta}}\frac{1}{\sin^2\theta}\del^2_{\tilde{\phi}}f\\&~~~~~~~~+\Xi\frac{a^2\omega^2}{\Delta_{\theta}}\cos^2\theta f-2\im a\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta\del_{\tilde{\phi}}f.
\end{align*}
We also define
\begin{align*}
P_{\alpha}(\omega,\ell,a,\alpha):=\begin{cases}
P(\omega,\ell,a)+\frac{\alpha}{\ell^2}a^2\sin^2\theta & \mathrm{if~}\alpha>0\\
P(\omega,\ell,a)-\frac{\alpha}{\ell^2}a^2\cos^2\theta & \mathrm{if~}\alpha\leq 0.
\end{cases}
\end{align*}
For equivalent definitions in Kerr spacetime see \citep{DafermosRodnianskiSmalla} and also \citep{FinsterSchmid} for a more detailed discussion.
From elliptic theory \citep[cf.][]{HolzegelSmuleviciDecay}, we can make the following definitions: $P(\omega,\ell,a)$ has eigenvalues $\tilde{\lambda}_{ml}(\omega,\ell,a)$ with eigenfunctions $\e^{\im m\tilde{\phi}}\tilde S_{ml}(\omega,\ell,a,\cos\theta)$; $P_{\alpha}(\omega,\ell,a,\alpha)$ has eigenvalues ${\lambda}(\omega,\ell,a,\alpha)$ with eigenfunctions $\e^{\im m\tilde{\phi}} S_{ml}(\omega,\ell,a,\alpha,\cos\theta)$. The eigenfunctions form an orthonormal basis of $L^2(\sin\theta\,\dd\theta\,\dd\tilde{\phi})$. Below we will suppress $(\omega,\ell,a,\alpha)$ in the notation.
If $\alpha\leq 0$, $S_{ml}$ satisfies the angular ODE
\begin{align}
\begin{aligned}
\label{eqn:AngularODE1}
&\frac{1}{\sin\theta}\del_{\theta}\left(\Delta_{\theta}\sin\theta\del_{\theta}S_{ml}(\cos\theta)\right)-\bigg(\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}-\frac{\Xi}{\Delta_{\theta}}a^2\omega^2\cos^2\theta\\&~~~~~~~~~-2ma\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta-\frac{\alpha}{\ell^2}a^2\cos^2\theta\bigg)S_{ml}(\cos\theta)+\lambda_{ml} S_{ml}(\cos\theta)=0
\end{aligned}
\end{align}
for $\lambda_{m\ell}(\omega,\alpha,a)\in\CC$. If $\alpha>0$, the angular ODE takes the form
\begin{align}
\begin{aligned}
\label{eqn:AngularODE2}
&\frac{1}{\sin\theta}\del_{\theta}\left(\Delta_{\theta}\sin\theta\del_{\theta}S_{ml}(\cos\theta)\right)-\bigg(\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}-\frac{\Xi}{\Delta_{\theta}}a^2\omega^2\cos^2\theta\\&~~~~~~~~~-2ma\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta+\frac{\alpha}{\ell^2}a^2\sin^2\theta\bigg)S_{ml}(\cos\theta)+\lambda_{ml} S_{ml}(\cos\theta)=0.
\end{aligned}
\end{align}
Using these modified oblate spheroidal harmonics, one obtains that, for fixed $m$ and $l$, $u:=\sqrt{r^2+a^2}R$ satisfies the radial ODE
\begin{align}
\label{eqn:radial_ODE_prelim}
u''(r)+(\omega^2-V(r))u(r)=0,
\end{align}
with
\begin{align*}
V(r)&=V_{+}(r)+V_{0}(r)+V_{\alpha}(r)\\
V_{+}(r)&=-\Delta_-^2\frac{3r^2}{(r^2+a^2)^4}+\Delta_-\frac{5\frac{r^4}{\ell^2}+3r^2\left(1+\frac{a^2}{\ell^2}\right)-4Mr+a^2}{(r^2+a^2)^3}\\
V_{0}(r)&=\frac{\Delta_-(\lambda_{ml}+\omega^2a^2)-\Xi^2a^2m^2-2m\omega a\Xi(\Delta_- -(r^2+a^2))}{(r^2+a^2)^2}\\
V_{\alpha}(r)&=-\frac{\alpha}{\ell^2}\frac{\Delta_-}{(r^2+a^2)^2}(r^2+\Theta(\alpha)a^2).
\end{align*}
Here $\Theta(x)=1$ if $x>0$ and zero otherwise.
We will use the shorthand $\tilde V:=V-\omega^2$.
Recall that $'$ denotes an $r^{\ast}$-derivative.
To indicate the dependence upon $a$, we will often write $V_a$ and $\tilde V_a$ for $V$ and $\tilde V$ respectively.
\subsection{Local analysis of the radial ODE}
\label{subsec:LocalAna}
To see which boundary conditions are appropriate for $u$, we perform a local analysis of the radial ODE near the horizon $r=r_+$ and at infinity, using the following theorem about regular singularities, which we cite from \citep{Teschl}, but it can also be found in \citep{ShlapentokhGrowing} or \citep{Olver}.
\begin{thm}
\label{thm:RegularSing}
Consider the complex ODE
\begin{align}
\label{eqn:CplxODE}
\frac{\dd^2H}{\dd z^2}+f(z,\nu)\frac{\dd H}{\dd z}+g(z,\nu)H=0.
\end{align}
Suppose $f$ and $g$ are meromorphic and have poles of order (at most) one and two, respectively, at $z_0\in\CC$. Let $f_0(\nu)$ and $g_0(\nu)$ be the coefficients of pole of order one and two, respectively, in the Laurent expansions.
Let $s_1(\nu)$ and $s_2(\nu)$ be the two solutions of the indicial equation
\begin{align*}
s(s-1)+f_0(\nu)s+g_0(\nu)=0
\end{align*}
with $\Re(s_1)\leq\Re(s_2)$.
If $s_2(\nu)-s_1(\nu)\notin\NN_0$, a fundamental system of solutions is given by
\begin{align*}
h_j(z,\nu)=(z-z_0)^{s_j(\nu)}\rho_j(z,\nu),
\end{align*}
where the functions $\rho_j$ are holomorphic and satisfy $\rho_j(z_0,\nu)=1$.
If $s_2(\nu)-s_1(\nu)=m\in\NN_0$, a fundamental system is given by
\begin{align*}
h_1&=(z-z_0)^{s_1}\rho_1+c\log(z)h_2\\
h_2&=(z-z_0)^{s_2}.
\end{align*}
The constant $c$ may be zero unless $m=0$.
In both cases, the radius of convergence of the power series of $\rho_j$ is at least equal to the minimum of the radii of convergence of the Laurent series of $f$ and $g$.
\end{thm}
\subsubsection{The horizon}
Adopting the notation of the previous section, and, after expressing the radial ODE (\ref{eqn:radial_ODE_prelim}) with $r$-derivatives, we have
\begin{align*}
f=\frac{\del_r\Delta_-}{\Delta_-}-\frac{2r}{r^2+a^2},~~~~~~~~~~g=\frac{(r^2+a^2)}{\Delta_-^2}(\omega^2-V).
\end{align*}
Thus we obtain
\begin{align*}
f_0&=\lim_{r\rightarrow r_+}(r-r_+)f=1\\
g_0&=\lim_{r\rightarrow r_+}(r-r_+)^2\frac{(r^2+a^2)^2}{\Delta_-^2}(\omega^2-V)=\lim_{r\rightarrow r_+}\frac{(r-r_+)^2}{\Delta_-^2}\left(\omega (r^2+a^2)-\Xi a m\right)^2
=-\xi^2
\end{align*}
with
\begin{align*}
\xi:=\im\frac{\Xi a m-\omega(r_+^2+a^2)}{\del_r\Delta_-(r_+)}
\end{align*}
as $\del_r\Delta_-(r_+)>0$. Thus, the indicial equation is solved by $s=\pm\xi$.
Therefore if $\xi\neq 0$, a local basis of solutions $u$ (or $R$) is given by
\begin{align*}
\{(\cdot-r_+)^{\xi}\phi_1,(\cdot-r_+)^{-\xi}\phi_2\}
\end{align*}
for holomorphic functions $\phi_i$ satisfying $\phi_i(r_+)=1$. For $\xi=0$, a local basis is given by
\begin{align*}
\{\phi_1,\phi_1\left(1+c\log(\cdot-r_+)\right)\}
\end{align*}
for $\phi_1(r_+)=1$ and some constant $c$.
\begin{lemma}
\label{lemma:SmoothHorizon}
If $u$ extends smoothly to the horizon, then there is a smooth function $\rho:\,[r_+,\infty)\rightarrow \CC$ such that
\begin{align*}
u=(\cdot-r_+)^\xi\rho.
\end{align*}
\end{lemma}
\begin{proof}
Boyer-Lindquist coordinates break down at the horizon, so we need to change to Kerr-star coordinates. Then the solution $\psi$ takes the form
\begin{align*}
\psi(\sta t,r,\sta\phi,\theta)=\e^{-\im\omega(t-A(r))}\e^{\im m(\sta\phi-B(r))}S_{ml}(a\omega,\cos\theta)\frac{u(r)}{(r^2+a^2)^{1/2}},
\end{align*}
where
\begin{align*}
\frac{\dd A}{\dd r}=\frac{(r^2+a^2)(1+r^2/\ell^2)-\Delta_-}{\Delta_-(1+r^2/\ell^2)},~~~~~~\frac{\dd B}{\dd r}=\frac{a(1-a^2/\ell^2)}{\Delta_-}.
\end{align*}
Hence $R$ extends smoothly to the horizon if there is a smooth function $f$ such that
\begin{align*}
u(r)=\e^{-\im(\omega A(r)-m B(r))}f(r).
\end{align*}
Therefore the claim reduces to showing that
\begin{align*}
\rho(r):=(r-r_+)^{-\xi}\e^{-\im(\omega A(r)-m B(r))}
\end{align*}
is smooth. Since
\begin{align*}
\frac{\dd}{\dd r}\left(-\im(\omega A(r)-m B(r))\right)=\frac{\xi}{r-r_+}+\OO(1),
\end{align*}
we have
\begin{align*}
\rho(r)=\e^{-\xi\log(r-r_+)}\e^{\xi\log(r-r_+)+\OO(r-r_+)},
\end{align*}
which proves the claim.
\end{proof}
\begin{cor}
\label{cor:horizon}
Assume $u$ satisfies the horizon regularity condition. Then a local basis of solutions to the ODE at the horizon is given by
\begin{align*}
(\cdot-r_+)^{\xi}\rho
\end{align*}
for a holomorphic function $\rho$ defined around $r=r_+$.
\end{cor}
This asymptotic analysis at the horizon motivates the horizon regularity condition of Definition~\ref{defn:HRC}.
\subsubsection{Infinity}
\label{subsec:local_infinity}
The radial ODE has a regular singularity at $r=\infty$. To analyse it using the Theorem~\ref{thm:RegularSing}, we rewrite equation (\ref{eqn:CplxODE}) by introducing $x:=1/z$. This yields
\begin{align*}
\frac{\dd^2 H}{\dd x^2}+\left(\frac{2}{x}-\frac{f}{x^2}\right)\frac{\dd H}{\dd x}+\frac{g}{x^4}H=0.
\end{align*}
For the radial ODE, we have $x=1/r$. We obtain
\begin{align*}
f(x=0)=0,~~~~\lim_{x\rightarrow 0}\frac{f}{x}=2,~~~~
g(x=0)=0,~~~~\lim_{x\rightarrow 0}\frac{g}{x}=0,~~~~\lim_{x\rightarrow 0}\frac{g}{x^2}=\alpha-2.
\end{align*}
The indicial equation becomes
\begin{align*}
s^2-s+\alpha=0,
\end{align*}
which is solved by $s_{\pm}=\frac{1}{2}\pm\sqrt{\frac{9}{4}-\alpha}$. Set
\begin{align*}
\mathcal E:=\left\{\frac{9-k^2}{4}\,:\,k\in\NN\right\}.
\end{align*}
Then, for $\alpha\notin\mathcal E$, a local basis of solutions near infinity is given by
\begin{align*}
\{r^{-1/2+\sqrt{9/4-\alpha}}\rho_1(r),r^{-1/2-\sqrt{9/4-\alpha}}\rho_2(r)\}
\end{align*}
with functions $\rho_1,\rho_2$, smooth at $\infty$ and satisfying $\rho_1(\infty)=\rho_2(\infty)=1$. For $\alpha\in\mathcal E$, a local basis is given by
\begin{align*}
\left\{C_3r^{-1/2-\kappa}\log\frac{1}{r}+r^{-1/2+\kappa}\rho_2,r^{-1/2-\sqrt{9/4-\alpha}}\rho_2(r)\right\}.
\end{align*}
If $u$ extends smoothly to $r=r_+$ and we specify a boundary value $u(r_+)$, then the arguments of Section~\ref{subsec:uniqueness_continuity} show that $C_3$ has to be zero.
\begin{lemma}
Let $u$ satisfy (\ref{eqn:radial_ODE_prelim}) on $(r_+,\infty)$ and extend smoothly to $r=r_+$, then, for large $r$, $u$ is a linear combination of
\begin{align*}
h_1(r,\alpha,\omega,a)&=r^{-1/2+\kappa}\rho_1(r,\alpha,\omega,a)\\
h_2(r,\alpha,\omega,a)&=r^{-1/2-\kappa}\rho_2(r,\alpha,\omega,a)
\end{align*}
for functions $\rho_1$ and $\rho_2$ holomorphic at $r=\infty$ and satisfying $\rho_1(\infty)=\rho_2(\infty)=1$.
\end{lemma}
\begin{cor}
If $u$ satisfies the horizon regularity condition and the Neumann boundary condition at infinity, then, for $5/4<\alpha<9/4$,
\begin{align*}
u=C_1h_1
\end{align*}
for a constant $C_1\in\CC$.
If $u$ satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity, then, for all $\alpha<9/4$,
\begin{align*}
u(r)=C_2h_2
\end{align*}
for a constant $C_2\in\CC$.
\end{cor}
{
\begin{rk}
The asymptotics near infinity do not change if we add $\nu(r^2+a^2)/(r^2\Delta_-)$ to $g$ as in Section~\ref{sec:Real}.
\end{rk}
\subsubsection{Uniqueness of solutions and dependence on parameters}
\label{subsec:uniqueness_continuity}
As one would expect, specifying one of the boundary conditions at infinity and choosing a value of $u$ at $r=r_+$ determines the solution to the radial ODE uniquely, which is being made more precise in the following standard lemma.
\begin{lemma}
\label{lemma:uniqueness_ODE}
Let $C_0\in\CC$. Then there is a unique classical solution to (\ref{eqn:radial_ODE_prelim}) on $(r_+,\infty)$ satisfying $u(r_+)=C_0$ and extending smoothly to $r=r_+$.
\end{lemma}
The continuous dependence of the solution $u$ on parameters is also well-known:
\begin{lemma}
\label{lemma:continuous_parameters}
Let $u_0$ be a unique solution to (\ref{eqn:radial_ODE_prelim}) for a certain set of parameters $(\alpha_0,\omega_0,a_0)$ with fixed $u(r_+)$ satisfying either the Dirichlet or Neumann boundary condition. Let there be a neighbourhood of these parameters such that for all $(\alpha,\omega,a)$ in said neighbourhood, there is a unique solution $u_{\alpha,\omega,a}$ with the same boundary conditions. Fix an $\hat r\in(r_+,\infty)$. Then
\begin{align*}
(\alpha,\omega,a)\mapsto u_{\alpha,\omega,a}(\hat r)
\end{align*}
is smooth.
\end{lemma}
Let $u$ be a solution to (\ref{eqn:radial_ODE_prelim}) that extends smoothly to the horizon. Fixing $u(r_+)$, we can uniquely define
reflection and transmission coefficients $A(\alpha,\omega,a)$ and $B(\alpha,\omega,a)$ via
\begin{align}
\label{eqn:reflection_transmission}
u(r,\alpha,\omega,a)= A(\alpha,\omega,a) h_1(r,\alpha,\omega,a)+B(\alpha,\omega,a) h_2(r,\alpha,\omega,a)
\end{align}
for large $r$. Here $h_1$ and $h_2$ are the local basis near infinity from Section~\ref{subsec:local_infinity}.
Let $W$ denote the Wronskian. {Then
\begin{align*}
A=\frac{W(u,h_2)}{W(h_1,h_2)}
\end{align*}
and similarly for $B$.
\begin{lemma}
\label{lemma:continuous_AB}
$A$ and $B$ are smooth in $\alpha$, $\omega$ and $a$.
\end{lemma}
\begin{proof}
Note that $A$ and $B$ are independent of $r$ and apply Lemma~\ref{lemma:continuous_parameters}.
\end{proof}
\subsection{Detailed analysis of the potential}
\label{subsec:Potential}
From the analysis in \citep{HolzegelSmuleviciDecay} we know that the angular ODE has countably many simple eigenvalues $\lambda_{ml}$, labelled by $l=|m|,|m|+1,\ldots$ for any given $m\in\ZZ$, and corresponding real-valued eigenfunction $S_{ml}$. For later use, we need a bound from below which can be found in \citep{HolzegelSmuleviciDecay}, where it is proved under the assumption of the Hawking-Reall bound. We give the slight extension to our regime.
\begin{lemma}
Let $\omega\in\RR$. For $|a|<\ell$, the eigenvalues satisfy
\begin{align}
\begin{aligned}
\label{eqn:bound_below}
\lambda_{ml}+a^2\omega^2&\geq\Xi^2|m|(|m|+1)\\
\lambda_{ml}+a^2\omega^2&\geq\Xi^2|m|(|m|+1)+a^2\omega_+^2-C_{\ell,a}|m||\omega-\omega_+|,
\end{aligned}
\end{align}
where $C_{\ell,a}>0$ depends on $\ell$ and $a$ only and
\begin{align*}
\omega_+(\ell,r_+,a,m):=\frac{ma\Xi}{r_+^2+a^2}
\end{align*}
\end{lemma}
\begin{proof}
We focus on the second inequality since the first one can be obtained similarly.
Let
\begin{align*}
\tilde P f:=-\frac{1}{\sin\theta}\del_{\theta}\left(\del_{\theta}\Delta_{\theta}\sin\theta\del_{\theta}f\right)+\Xi^2\frac{m^2}{\sin^2\theta}f.
\end{align*}
Then
\begin{align*}
\lambda_{ml}S_{ml}+a^2\omega^2S_{ml}&\geq\tilde PS_{ml}-\Xi^2\frac{m^2}{\sin^2\theta}S_{ml}+\Xi^2\frac{m^2}{\sin^2\theta}\frac{1}{\Delta_{\theta}}S_{ml}-\Xi\frac{a^2\omega^2}{\Delta_{\theta}}\cos^2\theta S_{ml}\\
&~~~~~~~~-2ma\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta S_{ml}+a^2\omega_+^2S_{ml}+a^2(\omega^2-\omega_+^2)S_{ml}\\
&=: \tilde PS_{ml}+P_cS_{ml}+a^2\omega_+^2S_{ml},
\end{align*}
where we have already used that the mass term is always nonnegative. We want to show that $P_c\geq 0$. We have the decomposition
\begin{align*}
P_c=P_c^++a^2(\omega^2-\omega_+^2)\left(1-\frac{\Xi}{\Delta_{\theta}}\cos^2\theta\right)-2ma\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}(\omega-\omega_+)\cos^2\theta,
\end{align*}
where $P_c^+$ is the $\omega_+$-part (i.\,e. the part for $\omega=\omega_+$) with
\begin{align*}
P_c^+&=\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}\left(\frac{a^2}{\ell^2}\cos^2\theta+\frac{a^4}{(r_+^2+a^2)^2}\sin^4\theta-2\frac{a^2}{r_+^2+a^2}\frac{a^2}{\ell^2}\sin^2\theta\cos^2\theta\right).
\end{align*}
Interpreting the bracket as a function in $\theta$, we see that it has critical points only at $\theta=0,\pi/2,\pi$. Hence $P_c^+\geq 0$.
Therefore, we know
\begin{align*}
P_c&\geq a^2(\omega^2-\omega_+^2)\left(1-\frac{\Xi}{\Delta_{\theta}}\cos^2\theta\right)-2ma\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}(\omega-\omega_+)\cos^2\theta\\
&\geq \frac{a^2}{\Delta_{\theta}} 2\omega_+(\omega-\omega_+)\sin^2\theta-2ma\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}(\omega-\omega_+)\cos^2\theta\\
&\geq -\left(2\frac{a^2}{\Delta_{\theta}}|\omega_+|\sin^2\theta+2|m||a|\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta\right)|\omega-\omega_+|
\end{align*}
To obtain the estimate, one only needs to integrate by parts on the sphere. The $\tilde P$ term yields
\begin{align*}
\int_{0}^{\pi}\tilde P S_{ml}\cdot\overline{S}_{ml}&=\int_{0}^{\pi}\left(\Delta_{\theta}|\del_{\theta}S_{ml}|^2+\Xi^2\frac{m^2}{\sin\theta}|S_{ml}|^2\right)\,\dd\theta\\
&\geq\Xi^2\int_{0}^{\pi}\left(|\del_{\theta}S_{ml}|^2+\frac{m^2}{\sin^2\theta}|S_{ml}|^2\right)\sin\theta\,\dd\theta\\
&\geq\Xi^2|m|(|m|+1),
\end{align*}
where we compared with spherical harmonics via the min-max principle.
\end{proof}
We will also need an asymptotic upper bound on the ground state eigenvalue $\lambda_{mm}$.
By the min-max principle, we know that
\begin{align*}
\lambda_{mm}&=\min_{u\in U,\norm{u}=1}\int_0^{\pi}\Bigg(\left[\Delta_{\theta}\left\lvert\frac{\dd u}{\dd\theta}\right\lvert^2+\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}|u|^2\right]-\frac{\Xi}{\Delta_{\theta}}a^2\omega^2\cos^2\theta|u|^2\\&~~~~~~~~~~-2ma\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}\cos^2\theta|u|^2-\frac{\alpha}{\ell^2}a^2\cos^2\theta|u|^2\Bigg)\sin\theta\,\dd\theta\\
&\leq \min_{u\in U,\,\norm{u}=1}\int_0^{\pi}\left(\Delta_{\theta}\left\lvert\frac{\dd u}{\dd \theta}\right\lvert^2+\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}|u|^2\right)\sin\theta\,\dd\theta+\left(|\alpha|+2|ma|\cdot|\omega-\omega_+|\right)\frac{a^2}{\ell^2}
\end{align*}
for $U=\{(\sin\theta)^{|m|}\rho(\theta)~:~\rho~\mathrm{analytic}\}$, which is the subspace of $L^2$ which contains all $S_{ml}$ -- see Appendix~\ref{sec:AngularODE}.
\begin{lemma}
Let $n\in\NN$. Then
\begin{align*}
\int_0^{\pi}\sin^n\theta\,\dd \theta=\sqrt{\pi}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\mathrm{~~~and~~~}
\int_0^{\pi}\sin^n\theta\cos^2\theta\,\dd\theta=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+4}{2}\right)},
\end{align*}
in particular
\begin{align*}
\int_0^{\pi}\sin^n\theta\,\dd \theta\sim\sqrt{2\pi}n^{-1/2}~~~\mathrm{and}~~~
\int_0^{\pi}\sin^n\theta\cos^2\theta\,\dd\theta\sim\sqrt{2\pi}n^{-3/2}
\end{align*}
as $n\rightarrow\infty$.
\end{lemma}
\begin{proof}
The expressions follow by induction and the Stirling formula.
\end{proof}
Define
\begin{align*}
u_m:=\left(\pi^{-1/2}\frac{\Gamma\left({(|m|+2)}/{2}\right)}{\Gamma\left({(|m|+3)}/{2}\right)}\right)^{1/2}\sin^{|m|}\theta.
\end{align*}
Then $u_m\in U$ and $\norm{u_m}_{L^2((0,\pi);\sin\theta\,\dd\theta)}$=1. As $1-\cos^2\theta\leq\Delta_{\theta}\leq 1$, we hence know that
\begin{align*}
\lambda_{mm}\leq\int_0^{\pi}\left(\left\lvert\frac{\dd u_m}{\dd \theta}\right\lvert^2+\Xi^2\frac{m^2}{\sin^4\theta}|u_m|^2\right)\sin\theta\,\dd\theta+\left(|\alpha|+2|ma|\cdot|\omega-\omega_+|\right)\frac{a^2}{\ell^2}.
\end{align*}
\begin{lemma}
\label{lemma:GroundState}
\begin{align*}
\lim_{m\rightarrow\infty}\frac{\lambda_{mm}}{m^2}=\Xi^2
\end{align*}
\end{lemma}
\begin{proof}
By equation (\ref{eqn:bound_below}),
we already have $\lim_{m\rightarrow\infty}\lambda_{mm}/m^2\geq \Xi^2$. To prove the result, we compute
\begin{align*}
\frac{1}{m^2}\int_0^{\pi}\left(\left\lvert\frac{\dd u_m}{\dd \theta}\right\lvert^2+\Xi^2\frac{m^2}{\sin^4\theta}|u_m|^2\right)\sin\theta\,\dd\theta&=\pi^{-1/2}\frac{\Gamma\left(\frac{|m|+2}{2}\right)}{\Gamma\left(\frac{|m|+3}{2}\right)}\times\\
&~~~~~\times \int_0^{\pi}\left(\cos^2\theta\sin^2\theta+\Xi^2\right)\sin^{2|m|-3}\,\dd\theta\\
&\sim m^{1/2}m^{-3/2}+\Xi^2m^{1/2}m^{-1/2}
\end{align*}
by the previous lemma. Hence $\lim_{m\rightarrow\infty}\lambda_{mm}/m^2\leq\Xi^2$.
\end{proof}
\begin{lemma}
\label{lemma:Vneg}
Let $N,L>0$. Then, given $\ell>0$ and $\alpha<9/4$. Moreover assume the spacetime parameters $r_+$ and $a$ satisfy
\begin{align}
\label{eqn:aRange}
\frac{r_+^4-a^2\ell^2}{(r_+^2+a^2)^2}<-N.
\end{align}
Then
there is an $m_0>0$ such that for all mode parameters $|m|\geq m_0$ and $l=m$, we have
\begin{align}
\label{eqn:V_above}
V-\omega^2\leq -N\frac{\Delta_-m^2\Xi^2}{(r^2+a^2)^2}
\end{align}
on an interval $(R_1,R_2)$ of length $L$ at $\omega=\omega_+(\ell,r_+,a,m)$.
\end{lemma}
\begin{proof}
Let us first rewrite the potential:
\begin{align}
\label{eqn:potential_rewritten}
V-\omega^2&=V_++V_{\alpha}-\omega^2\notag\\\nonumber
&~~~~~~~~~+\frac{\Delta_-}{(r^2+a^2)^2}\left(\Xi^2m^2-2\frac{a^2\Xi^2m^2}{r_+^2+a^2}+\frac{m^2a^4\Xi^2}{(r_+^2+a^2)^2}-\frac{m^2a^2\Xi^2}{\Delta_-}\frac{(r^2-r_+^2)^2}{(r_+^2+a^2)^2}\right)\\\notag
&~~~~~~~~~+\frac{\Delta_-}{(r^2+a^2)^2}(\lambda-\Xi^2m^2)\\\notag
&=V_++V_{\alpha}+\frac{\Delta_-}{(r^2+a^2)^2}(\lambda-\Xi^2m^2)+\frac{\Delta_-m^2\Xi^2}{(r^2+a^2)^2(r_+^2+a^2)^2}(r_+^4-a^2\ell^2)\\\notag
&~~~~~~~~~+\frac{\Delta_-m^2\Xi^2}{(r^2+a^2)^2(r_+^2+a^2)^2}\frac{a^2}{\Delta_-}(r-r_+)[r(2r_+^2+a^2+\ell^2)-a^2\ell^2r_+^{-1}+r_+^3]\\\notag
&=\frac{\Delta_-m^2\Xi^2}{(r^2+a^2)^2}\times\\\notag
&~~~~~~~\Big[\frac{2r^2}{\Xi^2m^2\ell^2}+\frac{\Delta_-}{(r^2+a^2)^2}\frac{a^2}{m^2\Xi^2}+\frac{r^2-a^2}{(r^2+a^2)^2}\frac{2Mr}{m^2\Xi^2}-\frac{\alpha}{\ell^2}\frac{1}{m^2\Xi^2}\left(r^2+\Theta(\alpha)a^2\right)\\
&~~~~~~+\left(\frac{\lambda}{\Xi^2m^2}-1\right)+\boxed{\frac{r_+^4-a^2\ell^2}{(r_+^2+a^2)^2}}\\\notag
&~~~~~~+\frac{a^2(r-r_+)}{\Delta_-}[r(2r_+^2+a^2+\ell^2)-a^2\ell^2r_+^{-1}+r_+^3]\Big]
\end{align}
Note that
\begin{align*}
[r(2r_+^2+a^2+\ell^2)-a^2\ell^2r_+^{-1}+r_+^3]|_{r=r_+}=\Delta'_-(r_+)>0.
\end{align*}
Moreover $\lambda-\Xi^2m^2>0$ by (\ref{eqn:bound_below}). Therefore, to obtain negativity, we will violate the Hawking-Reall bound in the boxed term. First we can choose $|m|$ large such that $\lambda/\Xi^2m^2-1$ is sufficiently small by Lemma~\ref{lemma:GroundState}. Since the term in the last line is decaying, we can find an $R_1$ such that the last term is bounded on $[R_1,R_1+L]$. By making $|m|$ possibly larger, the terms of the first line are also bounded on the interval.
Let
\begin{align*}
\frac{r_+^4-a^2\ell^2}{(r_+^2+a^2)^2}\leq-N-\epsilon_m
\end{align*}
Now we choose $m$ sufficiently large such that
\begin{align*}
\frac{\lambda_{mm}}{\Xi^2m^2}-1&<\frac{\epsilon_m}{2}.
\end{align*}
There is an $R_1$ such that
\begin{align*}
\frac{a^2(r-r_+)}{\Delta_-}[r(2r_+^2+a^2+\ell^2)-a^2\ell^2r_+^{-1}+r_+^3]<\frac{\epsilon_m}{2}
\end{align*}
for all $r\geq R_1$. Set $R_2:=R_1+L$ and choose $m$ such that
\begin{align*}
\frac{1}{m^2}\left(\frac{2r^2}{\Xi^2\ell^2}+\frac{\Delta_-}{(r^2+a^2)^2}\frac{a^2}{\Xi^2}+\frac{r^2-a^2}{(r^2+a^2)^2}\frac{2Mr}{\Xi^2}-\frac{\alpha}{\ell^2}\frac{r^2}{\Xi^2}\right)<\frac{\epsilon_m}{2}
\end{align*}
on $[R_1,R_2]$.
Putting everything together, the lemma follows.
\end{proof}
\begin{rk}
\label{rk:HR_bound}
The same proof yields the analogous negativity results for $V-\omega^2+F$, where $F$ is any continuous function on $(r_+,\infty)$ that is independent of $m$. This will be used in Section~\ref{sec:RealDN}.
\end{rk}
Define the functional
\begin{align*}
\LL_{\alpha,r_+,a}(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+\frac{(V-\omega^2)(r^2+a^2)}{\Delta_-}|f|^2\right)\,\dd r
\end{align*}
on $C_0^{\infty}(r_+,\infty)$. We often suppress some of the indices and write $\LL_a$ and $V_a$ in view of Section~\ref{sec:Real}.
\begin{lemma}
\label{lemma:FunctionalNegative}
Choose $(r_+,a,\ell)$ and $(m,l)$ as in Lemma~\ref{lemma:Vneg}. Then there is a function $f\in C_0^{\infty}(r_+,\infty)$ such that
\begin{align*}
\LL_{a}(f)<0.
\end{align*}
\end{lemma}
\begin{proof}
We have the following estimate for the functional if $f$ is supported in $(R_1,R_2)$:
\begin{align*}
\LL_{a}(f)\leq\int_{R_2}^{R_1}\left(\frac{\Delta_-(R_2)}{r_+^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2-N\frac{m^2\Xi^2}{R_2^2+a^2}|f|^2\right)\,\dd r
\end{align*}
Choose an $f$ such that $f$ is 1 on $[R_1+L/4,R_2-L/4]$ and 0 outside of $(R_1,R_2)$. Furthermore we require that
\begin{align*}
\left\lvert\frac{\dd f}{\dd r}\right\lvert\leq 2\frac{4}{L}.
\end{align*}
Hence
\begin{align*}
\LL_{a}(f)\leq\frac{64\Delta_-(R_2)}{(r_+^2+a^2)L^2}-N\frac{m^2\Xi^2L}{2(R_2^2+a^2)}.
\end{align*}
If necessary, we can increase $m$ further to make the expression negative.
\end{proof}
\begin{rk}
\label{rk:Choice}
Fix $\ell>0$ and $0<r_+<\ell$. Choose $\alpha_0<9/4$ and $a=a_0$ such that (\ref{eqn:aRange}) holds. Then there is a non-empty open interval $I\subseteq(-\infty,9/4)$ with $\alpha_0\in I$, a non-empty open Interval $I'$ around $a_0$ and an $m_0$ such that Lemma~\ref{lemma:FunctionalNegative} holds on $[R_1,R_2]$ for all $\alpha\in I$, $a\in I'$ and $|m|\geq m_0$.
\end{rk}
We want to conclude this section by showing that $\LL_a$ is always non-negative if the Hawking-Reall bound is satisfied. We borrow the following Hardy inequality from \citep{HolzegelSmuleviciDecay}.
\begin{lemma}
\label{lemma:Hardy}
For any $r_{\mathrm{cut}}\geq r_+$, we have for a smooth function $f$ with $fr^{1/2}=o(1)$ at infinity that
\begin{align*}
\frac{1}{4\ell^2}\int_{r_{\mathrm{cut}}}^{\infty}|f|^2\,\dd r\leq \int_{r_{\mathrm{cut}}}^{\infty}\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\,\dd r.
\end{align*}
\end{lemma}
\begin{proof}
We include a proof for the sake of completeness. Integrating by parts and applying the Cauchy-Schwarz inequality yields
\begin{align*}
\int_{r_{\mathrm{cut}}}^{\infty}\frac{\dd}{\dd r}(r-r_{\mathrm{cut}})|f|^2\,\dd r\leq 4\int_{r_{\mathrm{cut}}}^{\infty}(r-r_{\mathrm{cut}})^2\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\,\dd r.
\end{align*}
The lemma follows by estimating $r_{\mathrm{cut}}\geq r_+$.
\end{proof}
Thus we can prove the
\begin{lemma}
\label{lemma:HR_positivity}
Let $r_+^2\geq |a|\ell$. Then there is an $m_0$ such that, for all $|m|\geq m_0$ and $f\in C_0^{\infty}(r_+,\infty)$,
\begin{align*}
\LL_a(f)\geq 0.
\end{align*}
\end{lemma}
\begin{proof}
Let us first assume $r_+^2>|a|\ell$. Then, noting that
\begin{align*}
V_+=\frac{2\Delta_-}{(r^2+a^2)^2}\frac{r^2}{\ell^2}+\frac{\Delta_-}{(r^2+a^2)^4}\left(a^4\Delta_-+(r^2-a^2)2Mr\right),
\end{align*}
on sees from (\ref{eqn:potential_rewritten}) that
\begin{align*}
\tilde V_a>\frac{2-\alpha}{\ell^2}\frac{\Delta_-}{(r^2+a^2)^2}r^2>-\frac{1}{4\ell^2}\frac{\Delta_-}{(r^2+a^2)^2}r^2.
\end{align*}
Using Lemma~\ref{lemma:Hardy}, we conclude $\LL_a(f)>0$. By continuity, we obtain $\LL_a(f)\geq 0$ for $r_+^2\geq |a|\ell^2$.
\end{proof}
An analogue of Lemma~\ref{lemma:FunctionalNegative} can be proved for the twisted functional used in Section~\ref{sec:RealDN}. For $0<\kappa<1$, define
\begin{align*}
\tL_a(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f\right)\right\lvert^2+\tilde V_a^{h}\frac{r^2+a^2}{\Delta_-}|f|^2\right)\,\dd r
\end{align*}
with $\tilde V_a^{h}$ as in Section~\ref{sec:RealDN}.
\begin{lemma}
\label{lemma:FunctionalNegativeTwisted}
Choose $(r_+,a,\ell)$ and $(m,l)$ as in Lemma~\ref{lemma:Vneg}. Then there is a function $f\in C_0^{\infty}(r_+,\infty)$ such that
\begin{align*}
\tL_{a}(f)<0.
\end{align*}
\end{lemma}
\begin{proof}
For $f\in C_0^{\infty}(r_+,\infty)$, $\LL_a(f)=\tL_a(f)$ by choice of $\Vmod_a$.
\end{proof}
\subsection{Periodic mode solutions}
\label{sec:RestrReal}
\begin{lemma}
\label{thm:RestrReal}
Suppose we have a $\psi\in\Ss_{\mathrm{mod}}(\alpha,\omega,m,l)$ such that $\omega\in\RR$. Then the following statements are true:
\begin{compactenum}
\item[(i)] We have $m a\Xi-(r_+^2+a^2)\omega=0$, i.\,e. that $\omega=\omega_+(\ell,r_+,a,m)$.
\item[(ii)] We have $am\neq 0$.
\end{compactenum}
\end{lemma}
\begin{proof}
We wish to show $(i)$. First let us only deal with the Dirichlet branch.
Then $u$ is decaying at infinity. Define the microlocal energy current
\begin{align*}
Q_T:=\Im\left(u'\overline{u}\right).
\end{align*}
We have $Q_T(\infty)=0$. Moreover
\begin{align*}
\frac{\dd Q_T}{\dd r}&=\frac{\dd\sta r}{\dd r}\Im(u''\overline{u}+|u'|^2)=0
\end{align*}
by the radial ODE.
By Lemma~\ref{lemma:SmoothHorizon}, we obtain
\begin{align*}
u'&=\frac{\dd r}{\dd\sta r}\left(\frac{\xi}{r-r_+}u(r)+(r-r_+)^{\xi}\rho'(r)\right)=\frac{\Delta_-}{r^2+a^2}\left(\frac{\xi}{r-r_+}u(r)+(r-r_+)^{\xi}\rho'(r)\right)
\end{align*}
and so
\begin{align*}
u'(r_+)=\im\frac{\Xi am-(r_+^2+a^2)\omega}{r_+^2+a^2}u(r_+).
\end{align*}
We conclude
\begin{align}
\label{eqn:microlocal_horizon}
0=Q_T(r_+)=(r_+^2+a^2)\Im\left(\frac{\dd u}{\dd\sta r}(r_+)\overline{u(r_+)}\right)=(am\Xi-(r_+^2+a^2)\omega)|u(r_+)|^2.
\end{align}
If $u(r_+)=0$, then $u$ vanishes identically by Lemma~\ref{lemma:uniqueness_ODE}. Hence we conclude that
\begin{align*}
m a\Xi-(r_+^2+a^2)\omega=0.
\end{align*}
For the Neumann branch of the solution we observe that
\begin{align*}
Q_T=\Im\left(r^{-\frac{1}{2}+\kappa}\frac{\dd}{\dd\sta r}\left(r^{\frac{1}{2}-\kappa}u\right)\overline u\right).
\end{align*}
From the boundary condition, we immediately get $Q_T(\infty)=0$ as well and the rest follows as above.
Part (ii) follows immediately from $(r_+^2+a^2)\omega=\Xi am$.
\end{proof}
\section{Growing mode solutions satisfying Dirichlet boundary conditions}
\label{sec:Growing_Dirichlet}
\subsection{Existence of real mode solutions}
\label{sec:Real}
We now fix $\ell>0$, $\alpha<9/4$ and $0<r_+<\ell$.
Recall the variational functional
\begin{align*}
\LL_a(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+\tilde V_a\frac{r^2+a^2}{\Delta_-}|f|^2\right)\,\dd r.
\end{align*}
for $\omega=\omega_+=am\Xi/(r_+^2+a^2)$.
Define
\begin{align*}
\Aa:=\{a>0\,:\,\exists f\in C_0^{\infty}:\,\LL_a(f)<0\}.
\end{align*}
By Lemma~\ref{lemma:FunctionalNegative}, there is an $m_0$ such that $\Aa$ is non-empty for all $|m|\geq m_0$ and $l=m$. Fix $m$ and $l$ henceforth.
If the bound $r_+^2\geq |a|\ell$ is satisfied, then $\LL_a(f)\geq 0$ for all compactly supported $f$ by Lemma~\ref{lemma:HR_positivity}. Hence $\Aa$ is bounded below by a strictly positive infimum. Moreover $\Aa$ is open as $a\mapsto \LL_{a}(f)$ is continuous for any fixed $f$.
\begin{rk}
We restrict ourselves to $a>0$, but we could have defined the set $\Aa$ to also include negative values of $a$.
\end{rk}
Our aim is to show that $\LL_a$ has a minimiser for $a\in \Aa$. We will apply the natural steps of the direct method of the calculus of variations. First, we will specify an appropriate function space, then we will show that the functional obeys a coercivity condition and that the functional is weakly lower semicontinuous. The existence of a minimiser follows by an application of compactness results.
For $U\subseteq(r_+,\infty)$ define the weighted norm
\begin{align*}
\norm{f}^2_{\uL^2(U)}=\int_U\frac{1}{r^2}|f|^2\,\dd r
\end{align*}
and the space
\begin{align*}
\uL^2(U):=\{f\mathrm{~measurable~}:~\norm{f}_{\uL^2(U)}<\infty\}.
\end{align*}
This is clearly a Hilbert space with the natural inner product $(\cdot,\cdot)_{\uL^2(U)}$.
For $U\subseteq(r_+,\infty)$, we define the weighted Sobolev space $\uH^1$ via the norm
\begin{align*}
\norm{f}_{\uH^1(U)}^2:=\int_U\left(|f|^2+r(r-r_+)\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\right)\,\dd r
\end{align*}
Note that for $U\subseteq(r_+,\infty)$ compact, the $\uH^1$ norm is equivalent to the standard Sobolev norm. As usual, let $\uH^1_0(U)$ be the completion of $C_0^{\infty}(U)$ under $\norm{\cdot}_{\uH^1(U)}$.
\begin{lemma}
\label{lemma:trace}
Let $u\in\uH_0^1(r_+,\infty)$. Then is $u$ is also in $C(r_++1,\infty)$ (after possibly changing it on a set of measure zero) and
\begin{align*}
\lim_{r\rightarrow\infty } u(r)=\lim_{r\rightarrow\infty }r^{1/2-\kappa} u(r)=0
\end{align*}
for all $\kappa>0$.
\end{lemma}
\begin{proof}
Establishing the embedding $\uH_0^1(r_++1,\infty)\subseteq C(r_++1,\infty)$ is standard. Now take a sequence $(u_m)$ in $C_0^{\infty}$ such that $u_m\rightarrow u$ in $\uH^1_0$ and pointwise almost everywhere. Choose an $R$ such that $(u_m)$ converges pointwise there. For any $\beta<1/2$, we have
\begin{align*}
\left\lvert\lim_{r\rightarrow\infty}r^{\beta} u(r)\right\lvert&\leq R^{\beta}\left\lvert u-u_m\right\lvert(R)+\int_R^{\infty}\left\lvert\del_r\left(r^{\beta}( u-u_m)\right)\right\lvert\,\dd r\\
&\leq R^{\beta}\left\lvert u-u_m\right\lvert(R)+C''\norm{ u-u_m}_{\uH^1}.
\end{align*}
Therefore, the claim follows.
\end{proof}
To establish a coercivity-type inequality, we use the Hardy inequality of Lemma~\ref{lemma:Hardy}:
\begin{lemma}
\label{lemma:VariationInequalityD}
Let $a\in \Aa$ be fixed. There exist constants $r_+<B_0<B_1<\infty$ and constants $C_0,C_1,C_2>0$, such that, for sufficiently large $m$, we have for all smooth functions $f$ with $fr^{1/2}=o(1)$ at infinity that
\begin{align*}
\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+C_01_{[B_0,B_1]^c}|f|^2\right)\,\dd r\leq C_1\int_{B_0}^{B_1}|f|^2\,\dd r+C_2\LL_{a}(f).
\end{align*}
Here we can choose $C_2=1$ if $\alpha<2$.
\end{lemma}
\begin{rk}
Note that the dependence of the expression on $\alpha$ is via $\tilde V$ in $\LL_a$. Recall from Section~\ref{subsec:Spheroidal} that
\begin{align*}
\tilde V=V-\omega^2=V_++V_0+V_{\alpha}-\omega^2.
\end{align*}
\end{rk}
\begin{proof}
First, we have to study the potential again:
\begin{align*}
(r_+^2+a^2)\frac{V_+}{\Delta_-}(r_+)&=\frac{3\frac{r_+^4}{\ell^2}+r_+^2\left(1+\frac{a^2}{\ell^2}\right)-a^2}{(r_+^2+a^2)^2}\\
(r_+^2+a^2)\frac{V_{\alpha}}{\Delta_-}(r_+)&=-\frac{\alpha}{\ell^2}\frac{1}{r_+^2+a^2}(r_+^2+\Theta(\alpha)a^2)\\
(r_+^2+a^2)\frac{V_0-\omega_+^2}{\Delta_-}(r_+)&=\frac{\lambda+\omega_+^2a^2-2m\omega a\Xi}{r_+^2+a^2}\\
&\geq \frac{\Xi^2m^2}{(r_+^2+a^2)^3}r_+^4
\end{align*}
Thus for sufficiently large $|m|$, the expression is greater than zero. Furthermore, note the asymptotics
\begin{align}
\label{eqn:Potential_asymptotics}
(r^2+a^2)\frac{\tilde V}{\Delta_-}\rightarrow\ell^{-2}(2-\alpha)
\end{align}
as $r\rightarrow\infty$.
We will deal with the cases $\alpha<2$ and $\alpha\geq 2$ separately. First, let $\alpha<2$. The function $\frac{r^2+a^2}{\Delta_-}\tilde V$ is only nonpositive on an interval $[R_1,R_2]$. Choose constants such that $r_+<B_1<R_1<R_2<B_2<\infty$. Set $C_0$ to be the minimum of $\frac{r^2+a^2}{\Delta_-}\tilde V$ on $(r_+,\infty)\backslash[B_1,B_2]$ and set $-C_1$ to be its minimum on $[B_1,B_2]$. This immediately yields the result.
Now let $\alpha \geq 2$. There exist $R_1,R_2$ such that $\frac{r^2+a^2}{\Delta_-}\tilde V$ is positive on $(r_+,R_1)$ and
\begin{align*}
\frac{r^2+a^2}{\Delta_-}\tilde V>-\frac{1}{4\ell^2}(1-\epsilon)
\end{align*}
on $(R_2,\infty)$ for an $\epsilon>0$ because of (\ref{eqn:Potential_asymptotics}). Hence
\begin{align*}
\int_{R_2}^{\infty}\frac{r^2+a^2}{\Delta_-}\tilde V|f|^2\,\dd r&>-\frac{1-\epsilon/2}{4\ell^2}\int_{R_2}^{\infty}|f|^2\,\dd r+\frac{\epsilon}{8\ell^2}\int_{R_2}^{\infty}|f|^2\,\dd r\\
&\geq -\left(1-\frac{\epsilon}{2}\right)\int_{R_2}^{\infty}\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\,\dd r+\frac{\epsilon}{8\ell^2}\int_{R_2}^{\infty}|f|^2\,\dd r
\end{align*}
by Lemma~\ref{lemma:Hardy}. Choose $B_1,B_2$ as before. Let $C$ be the minimum of $\frac{r^2+a^2}{\Delta_-}\tilde V$ on $(r_+,B_1)$. Let $\epsilon C_0/2$ be the minimum of $C$ and $\epsilon/(8\ell^2)$. Moreover, set $-\epsilon C_1/2$ to be the minimum of $\frac{r^2+a^2}{\Delta_-}\tilde V$ on $[B_1,B_2]$, we obtain
\begin{align*}
\int_{r_+}^{\infty}\left(\epsilon \frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+\epsilon C_01_{[B_0,B_1]^c}|f|^2\right)\,\dd r\leq \epsilon C_1\int_{B_0}^{B_1}|f|^2\,\dd r+\LL_{a}(f)
\end{align*}
and hence the inequality.
\end{proof}
\begin{lemma}
\label{lemma:SemictsD}
The functional $\LL_a$ is weakly lower semicontinuous in $\uH^1(r_+,\infty)$ when restricted to functions of untit $\uL^2$ norm.
\end{lemma}
\begin{proof}
As the functional is convex in the derivative, the statement is standard and a proof can be extracted from \citep[][\textsection 8]{Evans}. We note that the boundedness from below comes from the norm constraint. The $r$ weight deals with $(r_+,\infty)$ having non-finite measure.
\end{proof}
\begin{lemma}
\label{lemma:RegMinD}
Let $a\in\Aa$. Then there exists an $f_{a}\in \uH_{0}^1(r_+,\infty)$ with unit $\uL^2(r_+,\infty)$ norm such that $\LL_{a}$ achieves its infimum over
\begin{align*}
\{f\in \uH_0^1(r_+,\infty)\,:\,\norm{f}_{\uL^2}=1\}
\end{align*}
on $f_{a}$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma:VariationInequalityD},
\begin{align}
\label{eqn:boundLmuD}
\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2+C_01_{[B_0,B_1]^c}|f|^2\right)\,\dd r\leq C_1\int_{B_0}^{B_1}|f|^2\,\dd r+C_2\LL_{a}(f)
\end{align}
holds for all $f\in \uH_0^{1}$.
From this, it is evident that
\begin{align*}
\LL_{a}(f)>-\infty
\end{align*}
if $\norm{f}_{\uL^2}=1$,
whence
\begin{align*}
\nu_{a} &=\inf\{\LL_{a}(f)\,:\,f\in \uH_0^{1},\,\norm{f}_{\uL^2}=1\}>-\infty.
\end{align*}
We can choose a minimising sequence of functions of compact support by density. Thus let $\{f_{a,n}\}$ be a sequence of smooth functions, compactly supported in $(r_+,\infty)$ with $\norm{f_{a,n}}_{\uL^2}=1$, such that
\begin{align*}
\LL_{a}(f_{a,n})\rightarrow\nu_{a}.
\end{align*}
The bound (\ref{eqn:boundLmuD}) implies that $\norm{f_{a,n}}_{\uH^1}$ is uniformly bounded. Thus by the Banach-Alaoglu theorem, it has a weakly convergent subsequence in $\uH_0^1(r_+,\infty)$.
Recall a simple version of Rellich-Kondrachov: $H^1[a,b]$ embeds compactly into $L^2[a,b]$. Hence by the equivalence of norms, the subsequence has a strongly in $L^2$ convergent subsequence on compact subsets of $(r_+,\infty)$.
Relabelling, we have a sequence $\{f_{a,n}\}$ that converges to $f_{a}$ weakly in $\uH_0^1$ and strongly in $L^2$ on compact subsets of $(r_+,\infty)$. The space $\uH_0^1$ is a linear (hence convex) subspace of $\uH^1$ that is norm-closed. Every convex subset that is norm closed is weakly closed. Therefore, $f_a\in \uH_0^1$.
We claim that $\norm{f_{a}}_{\uL^2}=1$. We have
\begin{align*}
\left\lvert\norm{f_a}_{\uL^2}-1\right\lvert&\leq\left\lvert \norm{f_a}_{\uL^2(r_++1/N,N)}-\norm{f_{a,n}}_{\uL^2(r_++1/N,N)}\right\lvert\\
&~~~~~~~~~+\left\lvert\norm{f_a}_{\uL^2(r_++1/N,N)^c}-\norm{f_{a,n}}_{\uL^2(r_++1/N,N)^c}\right\lvert
\end{align*}
Due to the $L^2$ convergence on compact subsets, the claim follows if
\begin{align*}
\lim_{N\rightarrow\infty}\lim_{n\rightarrow\infty}\norm{f_n}_{\uL^2((r_+,\infty)\backslash[r_++1/N,N]}=0.
\end{align*}
Suppose not.
Then there is a $\rho$ such that, for any $N$, there are infinitely many of the $f_{a,n}$ such that
\begin{align*}
\norm{f_{a,n}}_{\uL^2((r_+,\infty)\backslash [r_++1/N,N])}\geq \rho>0,
\end{align*}
i.\,e. the norm must concentrate either near the horizon or near infinity.
Suppose first that
\begin{align*}
\norm{f_{a,n}}_{\uL^2(r_+,r_++\delta)}\geq \rho_1>0
\end{align*}
for infinitely many $f_{a,n}$ and any $\delta>0$.
By (\ref{eqn:boundLmuD}), we have for $r\in(r_+,r_++1)$:
\begin{align*}
|f_{a,n}(r)|&\leq \int_{r}^{r_++1}\left\lvert\frac{\dd f_{a,n}}{\dd r'}\right\lvert\,\dd r'+\int_{r_++1}^{\infty}\left\lvert\frac{\dd f_{a,n}}{\dd r'}\right\lvert\,\dd r'\\
&\leq \left(\int_{r}^{r_++1}\frac{1}{r'-r_+}\,\dd r'\right)^{1/2}\left(\int_{r}^{r_++1}(r'-r_+)\left\lvert\frac{\dd f_{a,n}}{\dd r'}\right\lvert^2\,\dd r'\right)^{1/2}\\
&~~~~~~~~+\left(\int_{r_++1}^{\infty}\frac{1}{(r')^2}\,\dd r'\right)^{1/2}\left(\int_{r_++1}^{\infty}r^2\left\lvert\frac{\dd f_{a,n}}{\dd r'}\right\lvert^2\,\dd r'\right)^{1/2}\\
&\leq C\left(1+\sqrt{\log\frac{1}{r-r_+}}\right)
\end{align*}
for a constant $C>0$. Since $r\mapsto \sqrt{|\log (r-r_+)|}$ is integrable on compact subsets of $[r_+,\infty)$, we obtain $\norm{f_{a,n}}_{\uL^2(r_+,r_++\delta)}\rightarrow 0$ as $\delta\rightarrow 0$, a contradiction.
Hence we only need to exclude the case that the norm is bounded away from zero for large $r$. Thus, suppose that
\begin{align*}
\norm{f_{a,n}}_{\uL^2(R_0,\infty)}\geq \rho_2>0
\end{align*}
for infinitely many $f_{a,n}$ and any $R_0>0$. However,
\begin{align*}
R_0\rho_2\leq\norm{f_{a,n}}_{L^2(R_0,\infty)}\leq C'
\end{align*}
for a constant $C'>0$ by (\ref{eqn:boundLmuD}) and any $R_0$, a contradiction. This shows that
\begin{align*}
\nu_a=\inf\{\LL_a(f)\,:\,f\in\uH_0^1,\,\norm{f}_{\uL^2}=1\}.
\end{align*}
By the infimum property, we have
\begin{align*}
\nu_{a}\leq\LL_{a}(f_{a}).
\end{align*}
By Lemma~\ref{lemma:SemictsD}, we get
\begin{align*}
\LL_{a}(f_{a})\leq\liminf_{n\rightarrow\infty}\LL_{a}(f_{a,n}).
\end{align*}
As
\begin{align*}
\LL_{a}(f_{a,n})\rightarrow\nu_{a},
\end{align*}
the latter equals $\nu_{a}$.
Thus the minimum is attained by $f_{a}$.
\end{proof}
We would like to derive the Euler-Lagrange equation corresponding to this minimiser.
\begin{lemma}
\label{lemma:ELD}
The minimiser $f_{a}$ satisfies
\begin{align}
\label{eqn:ELregD}
\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+a^2}\frac{\dd f_{a}}{\dd r}\frac{\dd\psi}{\dd r}+\tilde V_a\frac{r^2+a^2}{\Delta_-}f_{a}\psi\bigg)\,\dd r=-\nu_a\int_{r_+}^{\infty}\frac{f_a}{r^2}\psi\,\dd r
\end{align}
for all $\psi\in\uH_0^1(r_+,\infty)$.
\end{lemma}
\begin{proof}
The proof can be extracted from \citep{Evans}. The analogous proof for twisted derivatives is given for Lemma~\ref{lemma:ELDN}.
\end{proof}
\begin{propn}
\label{propn:a_hat}
There is an $\hat a$ and a corresponding non-zero function $f_{\hat a}\in C^{\infty}(r_+,\infty)$ such that
\begin{align*}
\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd f_{\hat a}}{\dd r}\right)-\tilde V_{\hat a}f_{\hat a}=0
\end{align*}
and $f_{\hat a}$ satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity.
\end{propn}
\begin{proof}
First we would like to show that $\nu_a$ is continuous in $a$. We will use the notation $\Delta_-^a$ to denote the $\Delta_-$ corresponding to $a$. Given $a_1$ and $a_2$, we have
\begin{align*}
\nu_{a_1}&=\LL_{a_1}(f_{a_1})\\
&=\int_{r_+}^{\infty}\left(\frac{\Delta_-^{a_2}}{r^2+a_2^2}\left\lvert\frac{\dd f_{a_1}}{\dd r}\right\lvert^2+\tilde V_{a_2}\frac{r^2+a_2^2}{\Delta_-^{a_2}}|f_{a_1}|^2\right)\,\dd r\\
&~~~~+\int_{r_+}^{\infty}\left[\left(\frac{\Delta_-^{a_1}}{r^2+a_1^2}-\frac{\Delta_-^{a_2}}{r^2+a_2^2}\right)\left\lvert\frac{\dd f_{a_1}}{\dd r}\right\lvert^2+\left(\tilde V_{a_1}\frac{r^2+a_1^2}{\Delta_-^{a_1}}-\tilde V_{a_2}\frac{r^2+a_2^2}{\Delta_-^{a_2}}\right)|f_{a_1}|^2\right]\,\dd r.
\end{align*}
Due to the continuity of $\LL_a(f)$ in $a$, the first line is greater or equal than $\nu_{a_2}$ if $a_1$ is sufficiently close to $a_2$. Since the coefficients in the second line are continuously differentiable in $a$, we can use the mean value theorem to obtain
\begin{align*}
\nu_{a_1}\geq \nu_{a_2}-C|a_1-a_2|\int_{r_+}^{\infty}\left((r-r_+)\left\lvert\frac{\dd f_{a_1}}{\dd r}\right\lvert^2+|f_{a_1}|^2\right)\,\dd r
\end{align*}
for some constant $C>0$. We obtain an analogous inequality reversing the r\^{o}les of $a_1$ and $a_2$. Using (\ref{eqn:boundLmuD}) and $\norm{f_{a}}_{\uL^2}=1$ yields
\begin{align*}
|\nu_{a_1}-\nu_{a_2}|&\leq C|a_1-a_2|\int_{r_+}^{\infty}\left((r-r_+)\left\lvert\frac{\dd f_{a_1}}{\dd r}\right\lvert^2+|f_{a_1}|^2\right)\,\dd r\leq C'|a_1-a_2|.
\end{align*}
Since $\Aa\neq \emptyset$, we set
\begin{align*}
\hat a:=\inf\Aa.
\end{align*}
As stated in the introduction to this section, $\Aa$ is open, so $\hat a\notin\Aa$. By continuity of $\nu_a$, this implies that $\nu_{\hat a}=0$.
Now choose a sequence $a_n\rightarrow \hat a$ and corresponding minimisers $f_{a_n}\in\uH^1_0$ satisfying $\norm{f_{a_n}}_{L^2}=1$. Then, as in the proof of Lemma~\ref{lemma:RegMinD}, by Lemma~\ref{lemma:VariationInequalityD}, $f_{a_n}$ is bounded in $\uH^1$ and there is a subsequence (also denoted $(a_n)$) such that $f_{a_n}\rightarrow f_{\hat a}$ weakly in $\uH^1$ and strongly in $L^2$ on compact subsets for a $f_{\hat a}\in\uH^1_0$. Again by Lemma~\ref{lemma:VariationInequalityD} and the strong $L^2$ convergence on compact subsets, we see that $f_{\hat a}$ is non-zero. Moreover, we have sufficient decay towards infinity by Lemma~\ref{lemma:trace}. Hence we get the desired asymptotics.
From the weak convergence of $(f_{a_n})$, Lemma~\ref{lemma:ELD} yields that $f_{\hat a}$ satisfies
\begin{align*}
\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd f_{\hat a}}{\dd r}\frac{\dd\psi}{\dd r}+\tilde V_{\hat a}\frac{r^2+\hat a^2}{\Delta_-}f_{\hat a}\psi\bigg)\,\dd r=0
\end{align*}
for all $\psi\in\uH_0^1(r_+,\infty)$. For ordinary differential equations, weak solutions are classical solutions -- see for example \citep[][Ch.~1]{TaoDispersive}; so
\begin{align*}
\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd f_{\hat a}}{\dd r}\right)-\tilde V_{\hat a}f_{\hat a}=0,
\end{align*}
from which we obtain that $f_{\hat a}\in C^{\infty}$.
It remains to check the boundary condition at the horizon. The lower semi-continuity of convex functionals with respect to weak convergence implies that
\begin{align*}
\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a_n^2} \left\lvert\frac{\dd f_{a_n}}{\dd r}\right\lvert^2+\tilde V_{a_n}\frac{r^2+a_n^2}{\Delta_-}|f_{a_n}|^2\right)\,\dd r\leq\nu_{a_n},
\end{align*}
whence
\begin{align*}
\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+\hat a^2} \left\lvert\frac{\dd f_{\hat a}}{\dd r}\right\lvert^2+\tilde V_{\hat a}\frac{r^2+\hat a^2}{\Delta_-}|f_{\hat a}|^2\right)\,\dd r\leq 0.
\end{align*}
Hence
\begin{align}
\label{eqn:Boundfa}
\int_{r_+}^{\infty}\frac{\Delta_-}{r^2+\hat a^2}\left\lvert\frac{\dd f_{\hat a}}{\dd r}\right\lvert^2\,\dd r<\infty.
\end{align}
Near $r_+$, the local theory (Theorem~\ref{thm:RegularSing}) implies that there exist constants $A$, $B$ and non-zero analytic functions $\phi_i$ such that
\begin{align*}
f_{\hat a}=A\phi_1+B(\log(r-r_+)\phi_2+\phi_3).
\end{align*}
If $B\neq 0$, then
\begin{align*}
\int_{r_+}^{\infty}\frac{\Delta_-}{r^2+\hat a^2}\left\lvert\frac{\dd f_{\hat a}}{\dd r}\right\lvert^2\,\dd r=\infty,
\end{align*}
whence $B=0$. Hence $f_{\hat a}$ satisfies the horizon regularity condition.
\end{proof}
\begin{rk}
\label{rk:HR_positivity}
From Lemma~\ref{lemma:HR_positivity}, we already know that $|\hat a|\geq r_+^2/\ell$. In \citep{HolzegelSmuleviciDecay}, it is shown directly that, if the Hawking-Reall bound is satisfied, there are no periodic solutions. One can easily see that the proof generalises to the case when the Hawking-Reall bound is saturated. Thus we even obtain $|\hat a|>r_+^2/\ell$.
\end{rk}
\begin{cor}
Assume our choice of parameters, $a=\hat a$ and $\omega=\omega_+$. Let $C_0\in\CC$. Then the radial ODE (\ref{eqn:radial_ODE_prelim}) has a unique solution satisfying $u(r_+)=C_0$ and the Dirichlet boundary condition at infinity.
\end{cor}
\subsection{Perturbing the Dirichlet modes into the complex plane}
\label{subsec:PerturbingD}
We have shown that, for given $\ell>0$ and $\alpha<9/4$, there exists a real mode solution in a Kerr-AdS spacetime with parameters $(\ell,r_+,\hat a)$ and $\omega=\omega_R(0):=\Xi \hat am/(r_+^2+\hat a^2)$. Henceforth, we shall denote the chosen $\hat a$ simply by $a$.
Now we wish to vary $\omega$ and $\alpha$, keeping all the other parameters constant.
Keeping $u(r_+,\omega,\alpha)$ fixed, satisfying $|u|(r_+,\omega,\alpha)=1$, the local theory yields a unique solution to the radial ODE of the form
\begin{align*}
u(r,\alpha,\omega)=A(\alpha,\omega)h_1(r,\alpha,\omega)+B(\alpha,\omega)h_2(r,\alpha,\omega)
\end{align*}
for large $r$, cf. Lemma~\ref{lemma:uniqueness_ODE} and (\ref{eqn:reflection_transmission}).
The functions $A$ and $B$ are smooth in $\omega$ and $\alpha$. Finding a mode solution is equivalent to finding a zero of $A$. We already have $A(\alpha(0),\omega_R(0))=0$. Write $A=A_R+\im A_I$.
Recall
\begin{align*}
Q_T(r)=\Im(u'\overline u)
\end{align*}
and that
\begin{align*}
\frac{\dd Q_T}{\dd r}(r)&=\frac{r^2+a^2}{\Delta_-}\Im\left(V-\omega^2\right)\\
Q_T(r_+)&=\Xi am-\omega_R(r_+^2+a^2),
\end{align*}
where we have used $|u(r_+)|=1$.
We have
\begin{align*}
Q_T(r)&=|A|^2\frac{\Delta_-}{r^2+a^2}\Im\left(\frac{\dd h_1}{\dd r}\overline{h_1}\right)+\frac{\Delta_-}{r^2+a^2}\Im\left(A\frac{\dd h_1}{\dd r}\overline{B h_2}\right)\\
&~~~~~~~~~+\frac{\Delta_-}{r^2+a^2}\Im\left(B\frac{\dd h_2}{\dd r}\overline{A h_1}\right)+|B|^2\frac{\Delta_-}{r^2+a^2}\Im\left(\frac{\dd h_2}{\dd r}\overline{h_2}\right)
\end{align*}
and hence
\begin{align*}
Q_T(\infty)&=\frac{1}{\ell^2}\left(-\frac{1}{2}+\sqrt{\frac{9}{4}-\alpha}\right)\Im(A\overline B)+\frac{1}{\ell^2}\left(-\frac{1}{2}-\sqrt{\frac{9}{4}-\alpha}\right)\Im(B\overline A)\\
&=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\Im(A\overline B).
\end{align*}
due to the asymptotics of the $h_i$. We obtain
\begin{align*}
\Xi am-(r_+^2+a^2)\omega_R+\int_{r_+}^{\infty}\frac{r^2+a^2}{\Delta_-}\Im\left(V-\omega^2\right)\,\dd r=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\Im(A\overline B).
\end{align*}
Now we differentiate at $\omega_R=\omega_R(0)$ and $\alpha=\alpha_0$ with respect to $\omega_R$ and $\alpha$:
\begin{align}
\label{eqn:det1}
-(r_+^2+a^2)&=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\Im\left(\frac{\del A}{\del\omega_R}\overline B\right)
=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\left(\frac{\del A_I}{\del\omega_R}B_R-\frac{\del A_R}{\del\omega_R}B_I\right)\\
\label{eqn:det2}
0&=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\Im\left(\frac{\del A}{\del\alpha}\overline B\right)
=\frac{2}{\ell^2}\left(\frac{9}{4}-\alpha\right)^{1/2}\left(\frac{\del A_I}{\del\alpha}B_R-\frac{\del A_R}{\del\alpha}B_I\right)
\end{align}
To extend the coefficient $A(\alpha,\omega_R)=0$ to complex $\omega$, we want to appeal to the implicit function theorem establishing
\begin{align*}
\det\begin{pmatrix}
\frac{\del A_R}{\del\omega_R} & \frac{\del A_R}{\del\alpha}\\
\frac{\del A_I}{\del\omega_R} & \frac{\del A_I}{\del\alpha}
\end{pmatrix}
\neq 0.
\end{align*}
From equations (\ref{eqn:det1}) and (\ref{eqn:det2}) we see that this holds if
\begin{align*}
\frac{\del A}{\del\alpha}(\alpha(0),\omega_R(0))\neq 0.
\end{align*}
This is true indeed:
\begin{lemma}
\label{lemma:del_A__del_alphe}
\begin{align*} \frac{\del A}{\del\alpha}(\alpha(0),\omega_R(0))\neq 0.\end{align*}
\end{lemma}
\begin{proof}
Suppose $\del A/\del\alpha=0$. Then we have
\begin{align*}
\frac{\del u}{\del\alpha}(r,\omega_R(0),\alpha(0))&=\frac{\del B}{\del\alpha}(\omega_R(0),\alpha(0))h_2(r,\omega_R(0),\alpha(0))\\
&~~~~~~~+B(\alpha(0),\omega_R(0))\frac{\del h_2}{\del\alpha}(r,\omega_R(0),\alpha(0)).
\end{align*}
Thus $\del u/\del \alpha$ is polynomially decreasing at infinity as $r^{-1/2-\sqrt{9/4-\alpha(0)}}$ and extends smoothly to $r=r_+$.
Defining the derivative $u_{\alpha}:=\del u/\del\alpha$, we get from the radial ODE
\begin{align*}
\frac{\Delta_-}{r^2+a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u_{\alpha}}{\dd r}\right)-\tilde Vu_{\alpha}=\left[\frac{\Delta_-}{(r^2+a^2)^2}\frac{\del\lambda}{\del\alpha}-\frac{1}{\ell^2}\frac{\Delta_-}{(r^2+a^2)^2}(r^2+\Theta(\alpha)a^2)\right]u.
\end{align*}
Multiplying by $\overline u$ and integrating by parts, we obtain at $\omega_R(0)$ and $\alpha(0)$
\begin{align}
\label{eqn:IBPLambda}
\int_{r_+}^{\infty}\frac{\Delta_-}{(r^2+a^2)^2}\left(\frac{\del\lambda}{\del\alpha}-\frac{1}{\ell^2}(r^2+\Theta(\alpha)a^2)\right)|u|^2\,\dd r=0.
\end{align}
Now the two cases $\alpha\leq 0$ and $0<\alpha<9/4$ have to be treated separately. If $\alpha\leq 0$, then Proposition~\ref{propn:LambdaDeriv} readily gives $\del\lambda/\del\alpha<0$, so that $u$ would vanish identically.
For $\alpha>0$, we need to use the formula for $\del\lambda/\del\alpha$
from Proposition~\ref{propn:LambdaDeriv}.
Together with (\ref{eqn:IBPLambda}), this yields
\begin{align*}
\int_{r_+}^{\infty}\frac{\Delta_-}{(r^2+a^2)^2}\int_0^{\pi}\frac{1}{\ell^2}\left(-r^2-a^2\cos^2\theta\right)|S|^2\sin\theta|u|^2\,\dd\theta\,\dd r=0,
\end{align*}
whence we get the same contradiction.
\end{proof}
\subsection{Behaviour for small $\epsilon>0$ for Dirichlet boundary conditions}
\label{subsec:Crossing}
From the analysis of the previous section, we have a family of mode solutions $u(r,\epsilon)$ to the radial ODE parameters $(\omega(\epsilon),m,l,\alpha(\epsilon))$, where
\begin{align*}
\omega(\epsilon)=\omega_R(\epsilon)+\im\epsilon.
\end{align*}
The mode $u$ satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity.
This proves the first part of Theorem~\ref{thm:oldD}. To prove the second part, we would like to study the behaviour of $\omega(\epsilon)$ and $\alpha(\epsilon)$ for small $\epsilon>0$.
To obtain the following statements, we potentially need to make $|m|$ even larger than in the previous sections.
\begin{propn}
\label{lemma:DelOmega}
If $|m|$ is sufficiently large, we have
\begin{align*}
\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)< 0.
\end{align*}
\end{propn}
\begin{proof}
Define
\begin{align*}
\tilde Q_T:=\Im\left(u'\overline{\omega u}\right).
\end{align*}
Let $\epsilon>0$. We have $\tilde Q_T(\infty)=0$. Moreover
\begin{align*}
\tilde Q_T(r_+)=\Im\left(\frac{\xi}{r_+^2+a^2}\overline{\omega}\right)|u|^2(r_+)=0
\end{align*}
since $\xi$ has a positive real part (see (\ref{eqn:defn_xi})), $u\sim(r-r_+)^{\xi}$ and hence $|u|(r_+)=0$. Furthermore, using the radial ODE, one computes
\begin{align*}
\frac{\dd\tilde Q_T}{\dd r}=-\epsilon\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd u}{\dd r}\right\lvert^2+\frac{r^2+a^2}{\Delta_-}\Im\left((V_{a}-\omega^2)\overline{\omega}\right)|u|^2.
\end{align*}
Hence
\begin{align}
\label{eqn:int_exact_equality}
\int_{r_+}^{\infty}\left(\epsilon\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd u}{\dd r}\right\lvert^2-\frac{r^2+a^2}{\Delta_-}\Im\left((V_{a}-\omega^2)\overline{\omega}\right)|u|^2\right)\,\dd r=0
\end{align}
with
\begin{align*}
-\Im((V_{a}-\omega^2)\overline{\omega})&=\frac{\epsilon}{(r^2+a^2)^2} \Big(V_+(r^2+a^2)^2+|\omega|^2(r^2+a^2)^2-\Xi^2a^2m^2\\&~~~~~~-\frac{\alpha}{\ell^2}\Delta_-(r^2+a^2\Theta(\alpha))\Big)-\frac{\Delta_-}{(r^2+a^2)^2}\Im((\lambda+a^2\omega^2)\overline\omega).
\end{align*}
From Proposition~\ref{propn:ImOmega}, we know that $-\Im(\lambda\overline\omega)>0$. Hence
\begin{align}
\begin{split}
\label{eqn:Im_V}
-\Im((V_{a}-\omega^2)\overline{\omega})&>\frac{\epsilon}{(r^2+a^2)^2} \Big(V_+(r^2+a^2)^2+|\omega|^2(r^2+a^2)^2-\Xi^2a^2m^2\\&~~~~~~~-\frac{\alpha}{\ell^2}\Delta_-(r^2+a^2\Theta(\alpha))\Big)-\frac{\Delta_-}{(r^2+a^2)^2}a^2\epsilon|\omega|^2.
\end{split}
\end{align}
We set
\begin{align*}
K(r):=|\omega|^2(r^2+a^2)^2-\Xi^2a^2m^2-\Delta_-a^2|\omega|^2.
\end{align*}
We have
\begin{align}
\begin{split}
\label{eqn:derivative_K}
\frac{\dd}{\dd r}K(r)
&=|\omega|^2\left(4\left(1-\frac{a^2}{\ell^2}\right)r^3+2a^2M+2a^2\left(1-\frac{a^2}{\ell^2}\right)r\right)>0.
\end{split}
\end{align}
As already used in Section~\ref{sec:Real}, there is an $R>r_+$ such that, for $r\geq R$,
\begin{align*}
V_++V_{\alpha}>-\frac{1}{4\ell^2}\frac{\Delta_-}{r^2+a^2}.
\end{align*}
By an application of Lemma~\ref{lemma:Hardy}, we conclude
\begin{align}
\begin{split}
\label{eqn:int_K_large_R}
&\int_{R}^{\infty}\left(\epsilon\frac{\Delta_-}{r^2+a^2}\left\lvert\frac{\dd u}{\dd r}\right\lvert^2-\frac{r^2+a^2}{\Delta_-}\Im\left((V_{a}-\omega^2)\overline{\omega}\right)|u|^2\right)\,\dd r\\&~~~~~~~~~~~~>\int_R^{\infty}\frac{\epsilon}{(r^2+a^2)^2}K(r)|u|^2\,\dd r.
\end{split}
\end{align}
For the sake of contradition, suppose $K(r_+)\geq 0$. Then, by (\ref{eqn:derivative_K}), $K>0$ on $(r_+,\infty)$, whence we obtain strict positivity for (\ref{eqn:int_K_large_R}).
As
\begin{align*}
|\omega(0)|^2=\frac{m^2a^2\Xi^2}{(r_+^2+a^2)^2}
\end{align*}
and as, for fixed $r_+$, $\hat a$ is bounded away from zero for all $m$,
\begin{align*}
|\omega(0)|^2\geq Cm^2.
\end{align*}
Since $\epsilon\mapsto\omega(\epsilon)$ is continuous, $|\omega|^2$ scales as $m^2$ for small $\epsilon$, so $\dd K/\dd r$ can be chosen as large as possible at $r=r_+$, in particular, it can be used to overcome the potentially non-positive derivative of the remaining terms of the right hand side of (\ref{eqn:Im_V}) on $(r_+,R)$. Then, (\ref{eqn:int_exact_equality}) implies $u=0$, a contradiction.
Hence $K(r_+)<0$ which is equivalent to
\begin{align*}
\omega_R(\epsilon)^2+\epsilon^2<\left(\frac{am}{r_+^2+a^2}\right)^2.
\end{align*}
This in turn is equivalent to the claim.
\end{proof}
In the following, we will fix an $|m|\geq m_0$ such that Proposition~\ref{lemma:DelOmega} holds.
\begin{rk}
\label{rk:large_m}
The choice of $m$ could have been made right at the beginning as the choice of $m_0$ in Lemma~\ref{lemma:Vneg} is independent of the largeness required for Proposition~\ref{lemma:DelOmega}.
\end{rk}
The next proposition shows that the mass $\alpha$ is at first increasing along the curve obtained by the implicit function theorem. The proof requires a technical lemma which is given at the end of this section.
\begin{propn}
\label{propn:alpha_Dirichlet}
Let $\alpha(0)<9/4$. Then
\begin{align*}
\frac{\del\alpha}{\del\epsilon}(0)>0.
\end{align*}
\end{propn}
\begin{proof}
Define
\begin{align*}
u_{\epsilon}=\frac{\del u}{\del\epsilon}.
\end{align*}
Then
\begin{align*}
\frac{\del}{\del r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\del u_{\epsilon}}{\del r}\right)-\frac{(V-\omega^2)(r^2+a^2)}{\Delta_-}u_{\epsilon}=\frac{\del}{\del\epsilon}\left(\frac{r^2+a^2}{\Delta_-}\tilde V\right)u.
\end{align*}
We would like to multiply this equation by $\overline u$ and then integrate by parts, but, at $r=r_+$, $u_{\epsilon}$ does not satisfy the boundary conditions of a mode solution. However, using $u\sim (r-r_+)^{-1/2-\kappa(\epsilon)}$ for all $\epsilon$ by Section~\ref{subsec:PerturbingD}, $u_{\epsilon}\sim \log r(r-r_+)^{-1/2-\kappa}$ and hence satisfies the Dirichlet boundary condition at infinity.
We know that
\begin{align*}
f(r,\epsilon):=\exp\left(-\im\frac{\Xi am-(r_+^2+a^2)\omega(\epsilon)}{\del_r\Delta_-}\log(r-r_+)\right)u(r,\epsilon)
\end{align*}
is smooth,
whence
\begin{align*}
\frac{\del f}{\del\epsilon}&=\im(r_+^2+a^2)\frac{\log(r-r_+)}{\del_r\Delta_-(r_+)}\left(\frac{\del\omega_R}{\del\epsilon}+\im\right)f(r,\epsilon)\\&~~~~~~~+\exp\left(-\im\frac{\Xi am-(r_+^2+a^2)\omega(\epsilon)}{\del_r\Delta_-(r_+)}\log(r-r_+)\right)u_{\epsilon}
\end{align*}
and
\begin{align*}
u_{\epsilon}(r,0)=\frac{r_+^2+a^2}{\del_r\Delta_-(r_+)}\left(1-\im\frac{\del\omega_R}{\del\epsilon}\right)\log(r-r_+)u+\frac{\del f}{\del\epsilon}(r,0).
\end{align*}
We have
\begin{align*}
\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u_{\epsilon}}{\dd r}\right)\overline u&=\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u_{\epsilon}}{\dd r}\overline u\right)-\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}u_{\epsilon}\frac{\dd \overline u}{\dd r}\right)\\&~~~~~~~~~~+u_{\epsilon}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd\overline u}{\dd r}\right)
\end{align*}
and
\begin{align*}
\frac{\dd u_{\epsilon}}{\dd r}\overline u-u_{\epsilon}\frac{\dd\overline u}{\dd r}&=\frac{r_+^2+a^2}{\del_r\Delta_-(r_+)}\left(1-\im\frac{\del\omega_R}{\del\epsilon}\right)\frac{1}{r-r_+}|u|^2\\&~~~~~~~~~~+\frac{r_+^2+a^2}{\del_r\Delta_-(r_+)}\left(1-\im\frac{\del\omega_R}{\del\epsilon}\right)\log(r-r_+)\Im\left(\frac{\dd u}{\dd r}\overline u\right).
\end{align*}
We conclude that
\begin{align*}
\frac{\Delta_-}{r^2+a^2}\log(r-r_+)\Im\left(\frac{\dd u}{\dd r}\overline u\right)
\end{align*}
is zero at $r=r_+$.
Thus evaluating the radial ODE at $\epsilon=0$, multiplying it by $\overline u$, taking real parts and integrating by parts yields
\begin{align}
\label{eqn:IntV}
-|u(r_+)|^2&=\int_{r_+}^{\infty}\frac{r^2+a^2}{\Delta_-}\Re\left(\frac{\del\tilde V}{\del\epsilon}\right)\Bigg\lvert_{\epsilon=0}|u|^2\,\dd r
\end{align}
For $\alpha\neq 0$, the derivative is given by
\begin{align}
\begin{split}
\label{eqn:real_part_V_epsilon}
\Re\left(\frac{\del\tilde V}{\del\epsilon}\right)\Bigg\lvert_{\epsilon=0} &=\frac{\Delta_-}{(r^2+a^2)^2}\left[\Re\left(\frac{\del\lambda}{\del\epsilon}\right)+2a^2\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)\right]\\
&~~~~-\frac{\Delta_-}{(r^2+a^2)^2}2ma\Xi\frac{\del\omega_R}{\del\epsilon}(0)-2\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)\frac{r^2-r_+^2}{r^2+a^2}\\
&~~~~-\frac{1}{\ell^2}\frac{\del\alpha}{\del\epsilon}\frac{\Delta_-}{(r^2+a^2)^2}(r^2+\Theta(\alpha)a^2)\\
&=\frac{\Delta_-}{(r^2+a^2)^2}\left[\Re\left(\frac{\del\lambda}{\del\epsilon}-2r_+^2\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)\right)-\frac{1}{\ell^2}\frac{\del\alpha}{\del\epsilon}(r^2+\Theta(\alpha)a^2)\right]\\
&~~~~~~~~~~~-2\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)\frac{r^2-r_+^2}{r^2+a^2}.
\end{split}
\end{align}
Noting that $\int_0^{\pi}|S|^2\sin\theta\,\dd\theta=1$ and using Lemma~\ref{propn:IfThen} to eliminate the dependence on $\lambda$ and then Proposition~\ref{lemma:DelOmega}, we conclude that $\del\alpha/\del\epsilon(0)$ needs to be positive to make the integrand of (\ref{eqn:IntV}) negative. The restriction to $\alpha\neq 0$ can by removed by continuity of the reflection and transmission coefficients $A$ and $B$.
\end{proof}
\begin{lemma}
\label{propn:IfThen}
At $\epsilon=0$, for $\alpha\leq 0$,
\begin{align*}
&\int_0^{\pi}\Bigg(2\left[\Xi a^2+(r_+^2+a^2)\frac{a^2}{\ell^2}\right]\frac{\cos^2\theta}{\Delta_{\theta}}\omega_R\frac{\del\omega_R}{\del\epsilon}+\frac{a^2}{\ell^2}\cos^2\theta\frac{\del\alpha}{\del\epsilon}+\Re\left(\frac{\del\lambda}{\del\epsilon}\right)\Bigg)|S|^2\sin\theta\,\dd\theta=0
\end{align*}
and, for $\alpha>0$,
\begin{align*}
&\int_0^{\pi}\Bigg(2\left[\Xi a^2+(r_+^2+a^2)\frac{a^2}{\ell^2}\right]\frac{\cos^2\theta}{\Delta_{\theta}}\omega_R\frac{\del\omega_R}{\del\epsilon}-\frac{a^2}{\ell^2}\sin^2\theta\frac{\del\alpha}{\del\epsilon}+\Re\left(\frac{\del\lambda}{\del\epsilon}\right)\Big)|S|^2\sin\theta\,\dd\theta=0.
\end{align*}
\end{lemma}
\begin{proof}
Let $\alpha\geq 0$. Set $S_{\epsilon}:=\del S/\del\epsilon$. Then, differentiating the angular ODE with respect to $\epsilon$,
evaluating at $\epsilon=0$, multiplying by $\overline S$, taking the real part and integrating by part yields the claimed identity.
An analogous computation yields the result for $\alpha>0$.
\end{proof}
\subsection{A continuity argument}
\label{subsec:Continuity}
We now deduce Theorem~\ref{thm:new} from Theorem~\ref{thm:oldD}. In this section, we fix $\ell>0$ and $\alpha_0<9/4$. In the previous sections, we have produced a curve $\epsilon\mapsto\alpha_0(\epsilon)$ of masses with $\del\alpha_0(0)/\del\epsilon>0$. This means that the constructed mode solutions will solve a radial ODE with a different scalar mass.
This section formalises the intuitive idea of ``following up" the curves $\epsilon\mapsto\alpha(\epsilon)$ starting at an $\alpha$ close to $\alpha_0$ until one ``hits" the desired mass, which is made possible by $\del\alpha(0)/\del\epsilon>0$. The proof consists simply in establishing necessary continuity and carefully choosing neighbourhoods. This can be divided into two independent steps.
\begin{enumerate}
\item We show that the function mapping $\alpha$ to the corresponding $\hat a$ is left-continuous.
\item We show that, for $\alpha$ and corresponding $\hat a$ sufficiently close to $\alpha_0$ and the corresponding $\hat a_0$, the implicit function theorem guarantees a curve, starting at $\alpha$ and the corresponding $\hat a$ and real frequency $\omega_+$, which exists ``long enough" to ``hit" $\alpha_0$.
\end{enumerate}
Note that, for any $f\in C_0^{\infty}$,$(\alpha,r_+,a)\mapsto \LL_{\alpha,r_+,a}(f)$
defines a continuous function. For a given $f\in C_0^{\infty}$, define the family of sets
\begin{align*}
\Aa_{\alpha,r_+}(f):=\{a>0\,:\,\LL_{\alpha,r_+,a}(f)<0\}
\end{align*}
and
\begin{align*}
\Aa_{\alpha,r_+}:=&\bigcup_{f\in C_0^{\infty}}\Aa_{\alpha,r_+}(f)\\
=&\{a>0\,:\,\exists f\in C_0^{\infty}:\,\LL_{\alpha,r_+,a}(f)<0\}.
\end{align*}
\begin{rk}
$\Aa_{\alpha,r_+}$ corresponds to the set $\Aa$ from Section~\ref{sec:Real}.
\end{rk}
Define the function
\begin{align*}
\Phi:\,(-\infty,9/4)\times(0,\infty)\rightarrow(0,\infty),~\Phi(\alpha,r_+):=\inf\Aa_{\alpha,r_+}
\end{align*}
if $\Aa_{\alpha,r_+}\neq\emptyset$.
\begin{lemma}
\label{lemma:well-defined}
Let $0<r_+<\ell$. Then there is an interval $I\subseteq(-\infty,9/4)$ with $\alpha_0\in I$ and an $m_0$ such that $\Phi(\cdot,r_+)$ is well-defined for all $\alpha\in I$ and $|m|\geq m_0$ .
\end{lemma}
\begin{proof}
The set $\Aa_{\alpha,r_+}$ non-empty, open and bounded away from zero for all $\alpha\in I\subseteq (-\infty,9/4)$ by Remark~\ref{rk:Choice}, whence $\Phi(\cdot,r_+)$ is well-defined.
\end{proof}
We shall fix $r_+$ now. Moreover, we shall fix an $m\geq m_0>0$.
\begin{lemma}
\label{lemma:non-increasing}
The function $\Phi(\cdot,r_+)$ is non-increasing in $\alpha\in I$.
\end{lemma}
\begin{proof}
Suppose $\Phi(\cdot,r_+)$ was not non-increasing. Then
\begin{align*}
\inf\Aa_{\alpha,r_+}<\inf\Aa_{\alpha',r_+}
\end{align*}
for some $\alpha<\alpha'$. Hence there is an $a>0$ with
\begin{align*}
\inf\Aa_{\alpha,r_+}<a<\inf\Aa_{\alpha',r_+}
\end{align*}
and an $f\in C_0^{\infty}$ such that
\begin{align*}
\LL_{\alpha,r_+,a}(f)<0\leq\LL_{\alpha',r_+,a}(f).
\end{align*}
This contradicts that $\LL_{\alpha,r_+,a}(f)>\LL_{\alpha',r_+,a}(f)$ for all $f$ and $a$ if $\alpha<\alpha'$.
\end{proof}
\begin{lemma}
\label{lemma:CL}
The function $\Phi(\cdot,r_+)$ is left-continuous at $\alpha_0$, i.\,e.
\begin{align*}
\lim_{\alpha\uparrow\alpha_0}\Phi(\alpha,r_+)=\Phi(\alpha_0,r_+).
\end{align*}
\end{lemma}
\begin{proof}
Suppose $\Phi(\cdot,r_+)$ was not left-continuous at $\alpha_0$. Then there is an $\epsilon>0$ such that, for all $\delta>0$, there is an $\alpha<\alpha_0$ with
\begin{align*}
\alpha_0-\alpha<\delta
\end{align*}
and
\begin{align*}
\Phi(\alpha,r_+)-\Phi(\alpha_0,r_+)\geq \epsilon.
\end{align*}
Then there is an $a$ between $\Phi(\alpha_0,r_+)$ and $\Phi(\alpha,r_+)$ such that there is an $f$ with $a\in\Aa_{\alpha_0,r_+}(f)$,
but, for each $\delta$, there is an $\alpha$ with $a\notin\Aa_{\alpha,r_+}(f)$.
Since $\LL_{\alpha_0,r_+,a}(f)<0$ and due to the continuity of $\LL_{\cdot}(f)$, there is a $\delta>0$ such that for all $\alpha_0-\alpha<\delta$, we have $\LL_{\alpha,r_+,a}(f)<0$, i.\,e. $a\in\Aa_{\alpha,r_+}(f)$, a contradiction.
\end{proof}
For $\alpha\in I$, we define
\begin{align}
\label{eqn:defn_omega_+}
\Omega_R(\alpha):=\frac{m\Phi(\alpha,r_+)\left(1-\frac{\Phi(\alpha,r_+)^2}{\ell^2}\right)}{r_+^2+\Phi(\alpha,r_+)^2}.
\end{align}
As shown, this is left-continuous at $\alpha_0$.
Now we turn to the second step. Recall that, for all $\alpha\in I$, there is a periodic Dirichlet mode with frequency $\omega=\Omega_R(\alpha)\in\RR$ in a Kerr-AdS spacetime with parameters $(\ell,r_+,\Phi(\alpha,r_+))$ by Proposition~\ref{propn:a_hat}. Using Section~\ref{subsec:PerturbingD}, we can find unstable Dirichlet mode solutions with frequency $\omega=\omega_R+\im\omega_I=\omega_R(\epsilon)+\im\epsilon$ (where $\omega_R(0)=\Omega_R(\alpha)$) to the Klein-Gordon equation with mass $\alpha(\epsilon)$ (where $\alpha(0)=\alpha_0$). As by Lemma~\ref{lemma:well-defined} the results of Section~\ref{subsec:Crossing} hold, we know that
\begin{align}
\label{eqn:sign_omega_alpha}
\frac{\del\alpha}{\del\epsilon}(0)>0,~~~~~~~~\frac{\del\omega_R}{\del\epsilon}(0)<0.
\end{align}
In this section, $B_{\rho}(x)$ will denote an open $\ell^{\infty}$ ball of radius $\rho$ centered around $x\in\RR^4$, i.\,e.
\begin{align*}
B_{\rho}(x):=\left\{y\in\RR^4\,:\,\max_{j=1,\ldots,4}|x_j-y_j|<\rho\right\}.
\end{align*}
We view the column vectors $(\alpha,\omega_R,\omega_I,a)^t$ as points in $\RR^4$.
Consider
\begin{align*}
D:=\det\begin{pmatrix}
\frac{\del A_R}{\del\omega_R} & \frac{\del A_R}{\del\alpha}\\
\frac{\del A_I}{\del\omega_R} & \frac{\del A_I}{\del\alpha}
\end{pmatrix}.
\end{align*}
It was shown in Section~\ref{subsec:PerturbingD} that $D(\alpha,\Omega_R(\alpha),0,\Phi(\alpha,r_+))\neq 0$ for all $\alpha\in I$.
From Lemma~\ref{lemma:continuous_AB}, we know that $A$ is smooth in $\alpha$, $\omega$ and $a$. Hence there is an $L>0$ such that $D\neq 0$ in $B_L(\alpha,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$ and such that, for all values $(\alpha,\Omega_R(\alpha),0,\Phi(\alpha,r_+))\in B_L(\alpha,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$, we have $\alpha\in I$.
Hence in this neighbourhood, the vector field
\begin{align*}
W:=-\begin{pmatrix}
\frac{\del A_R}{\del\alpha} & \frac{\del A_R}{\del\omega_R} & 0 & 0\\
\frac{\del A_I}{\del \alpha} & \frac{\del A_I}{\del\omega_R} & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{pmatrix}^{-1}
\begin{pmatrix}
\frac{\del A_R}{\del\omega_I}\\
\frac{\del A_I}{\del\omega_I}\\
1\\
0
\end{pmatrix}
\end{align*}
is well-defined. It is this vector field whose integral curves describe the solutions given by the implicit function theorem as applied in Section~\ref{subsec:PerturbingD}. In particular, solving the ODE
\begin{align*}
\frac{\dd}{\dd \epsilon}
(\alpha(\epsilon),\omega_R(\epsilon),\omega_I(\epsilon),a(\epsilon))^t
=W(\alpha(\epsilon),\omega_R(\epsilon),\omega_I(\epsilon),a(\epsilon))
\end{align*}
with initial conditions $(\alpha,\Omega_R(\alpha),0,\Phi(\alpha,r_+))$ ($\alpha\in I$) gives the previously introduced $\alpha(\epsilon)$ and $\omega_R(\epsilon)$.
Set $W:=(W^{\alpha},W^{\omega_R},W^{\omega_I},W^a)^t$. By (\ref{eqn:sign_omega_alpha}),
\begin{align*}
W^{\alpha}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))>0
\end{align*}
and
\begin{align*}
W^{\omega_R}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))<0.
\end{align*}
Let $\delta>0$. Again by smoothness of $A$, there are $\rho>0$ and $L'\leq L$ such that
\begin{align*}
\norm{W-W(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))}_{\infty}<\delta~~\mathrm{and}~~
W^{\alpha}\geq\rho,~W^{\omega_R}\leq -\rho
\end{align*}
in $B_{L'}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$.
We now study integral curves of $W$ in $B_{L'}(\alpha_0,\omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$. Let
\begin{align*}
\tau\mapsto\gamma(\tau,p)
\end{align*}
be the integral curve of $W$ with $\gamma(0,p)=p\in B_{L'}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$.
Define the map
\begin{align*}
&T:\,B_{2L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))\rightarrow\RR,\\
&T(p)=\inf\left\{\tau>0\,:\,\gamma(\tau,p)\in\overline{B_{2L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))}^c\right\}.
\end{align*}
The set $\left\{\tau>0\,:\,\gamma(\tau,p)\in\overline{B_{2L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))}^c\right\}$ is non-empty since
\begin{align}
\label{eqn:del_omega_I}
\frac{\dd}{\dd \tau}\omega_I(\tau)=1.
\end{align}
Therefore $T$ is well-defined.
\begin{lemma}
T is continuous.
\end{lemma}
\begin{proof} For this proof use the abbreviation $B:=B_{2L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$.
Let $p_0\in B$, $\tau_0:=T(p_0)>0$. Let $0<\epsilon<\tau_0$ such that $\gamma(\tau,p_0)\in \overline{B}^c$ for $\tau_0+\epsilon\leq \tau\leq \tau_0+2\epsilon$, which exists by (\ref{eqn:del_omega_I}). Define
\begin{align*}
d_1:=\min\{\mathrm{dist}(\gamma(\tau,p_0),\del B)\,:\,0\leq \tau\leq \tau_0-\epsilon\}
\end{align*}
We claim that $d_1>0$. Suppose not. Then there is a $\tau'\in(0,\tau_0-\epsilon]$ such that $\mathrm{dist}(\gamma(\tau',p_0),\del B)=0$. Since
\begin{align}
\label{eqn:Vectorfield}
W^{\alpha}\geq\rho,~~~W^{\omega_R}\leq-\rho,~~~W^{\omega_I}=1,~~~W^{a}=0,
\end{align}
whence $W$ is not parallel to any side of the boundary $\del B$ of the $\ell^{\infty}$ ball, this would imply that $\gamma(\tau,p_0)\in\overline B^c$ for a range of $\tau$'s in a small neighbourhood of $\tau'$. This, however, contradicts $\tau_0=T(p_0)$. Hence $d_1>0$.
Furthermore define
\begin{align*}
d_2:=\min\{\mathrm{dist}(\gamma(\tau,p_0),\del B)\,:\,\tau_0+\epsilon\leq \tau\leq \tau_0+2\epsilon\}.
\end{align*}
Using (\ref{eqn:Vectorfield}), we can see again that $d_2>0$. Set $d:=\min(d_1,d_2)$.
Set
\begin{align*}
G:=\{\gamma(\tau,p_0)\,:\,\tau\in[0,\tau_0+2\epsilon]\backslash[\tau_0-\epsilon,\tau_0+\epsilon]\}
\end{align*}
Since the solutions of linear ODEs depend continuously on the initial data, there is a $\delta>0$ such that, for all $p\in B_{\delta}(p_0)$, $\gamma(\tau,p)$ is in a $d/2$-neighbourhood of $G$ for all $\tau\in[0,\tau_0+2\epsilon]\backslash(\tau_0-\epsilon,\tau_0+\epsilon)$. Thus for all $p\in B_{\delta}(p_0)$, $T(p)>\tau_0-\epsilon$ and $T(p)<\tau_0+\epsilon$, i.\,e. $|T(p)-T(p_0)|<\epsilon$.
\end{proof}
Hence there exists a $T_0\geq0$ such that
\begin{align*}
T(p)\geq T_0
\end{align*}
for all $p\in \overline{B_{L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))}$. As continuous functions attain their minimum on compact sets, $T_0$ can be chosen to be positive. This shows that all integral curves of $W$ starting in $B_{L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$ exist for $0\leq \tau\leq T_0$ and remain in $B_{L'}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$.
We can prove the following
\begin{lemma} Given $(\alpha(0),\omega_R(0),\omega_I(0),a(0))^t\in B_{L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))$ with
\begin{align*}
\alpha_0-T_0\rho\leq\alpha(0)<\alpha_0,
\end{align*}
let
\begin{align*}
s\mapsto (\alpha(s),\omega_R(s),\omega_I(s),a(s))^t
\end{align*}
be the integral curve starting at $(\alpha(0),\omega_R(0),\omega_I(0),a(0))^t$. Then there is a $\tau\in (0,T_0]$ such that $\alpha(\tau)=\alpha_0$.
\end{lemma}
\begin{proof} The ODE yields
\begin{align*}
\alpha(T_0)&=\alpha(0)+\int_0^{T_0}W^{\alpha}(\alpha(s),\omega_R(s),\omega_I(s),a(s))\,\dd s\\
&\geq \alpha(0)+T_0\rho.
\end{align*}
If $\alpha_0-T_0\rho\leq\alpha(0)<\alpha_0$, then $\alpha(T_0)\geq\alpha_0$ and, by the intermediate value theorem, there is a $\tau\in (0,T_0]$ such that $\alpha(\tau)=\alpha_0$.
\end{proof}
The function $\Phi(\cdot,r_+)$ induces the curve
\begin{align*}
\Gamma:\,\alpha\mapsto(\alpha,\Omega_R(\alpha),0,\Phi(\alpha,r_+))^t
\end{align*}
for $\alpha\in I$; it is continuous on the left at $\alpha_0$. The result of the previous section says that along this curve, the implicit function theorem produces parameter curves that correspond to superradiant modes; these parameter curves are exactly the integral curves of $W$ starting on a point of $\Gamma$. Since $\Gamma$ is left-continuous,
\begin{align*}
\Gamma\cap B_{L'/3}(\alpha_0,\Omega_R(\alpha_0),0,\Phi(\alpha_0,r_+))\cap \{\alpha_0-T_0\rho\leq\alpha(0)<\alpha_0\}\neq\emptyset.
\end{align*}
This shows Theorem~\ref{thm:new}
\section{Growing mode solutions satisfying Neumann boundary conditions}
\label{sec:Neumann_construction}
\subsection{Existence of real mode solutions}
\label{sec:RealDN}
In this section, we will construct growing mode solutions satisfying Neumann boundary conditions. Every result has a counterpart in Section~\ref{sec:Growing_Dirichlet}. In the following, whenever proofs will be short in detail, the reader can extract those from Section~\ref{sec:Growing_Dirichlet}. The two novel techniques in this section are the use of twisted derivatives with appropriately modified Sobolev spaces and a new Hardy inequality (Lemma~\ref{lemma:Hardy_twisted}).
Fix $\ell>0$ and $r_+$. In this section, we look at the range $5/4<\alpha<9/4$, i.\,e. $0<\kappa<1$, for Neumann boundary conditions.
To treat the Neumann case variationally, we need to modify the functional, so it becomes finite for Neumann modes. We achieve this by conjugating the derivatives by a power of r; more precisely, we consider the twisted derivative $h\frac{\dd}{\dd r}\left(h^{-1}\cdot\right)$, where $h=r^{-1/2+\kappa}$. This ``kills off" the highest order term of the Neumann branch. Moreover, squaring the twisted derivative term does not introduce any ``mixed terms" in $f$ and its derivative; it only produces a zeroth order term that also makes the potential finite.
Thus introduce the twisted variational functional
\begin{align*}
\tL_a(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}f\right)\right\lvert^2+\Vmod_a\frac{r^2+a^2}{\Delta_-}|f|^2\right)\,\dd r,
\end{align*}
where $\Vmod_a$ as in Appendix~\ref{sec:twisted_derivative}, i.\,e.
\begin{align*}
\Vmod_a=\tilde V_a+\left(\frac{1}{2}-\kappa\right)\frac{\Delta_-}{r^2+a^2}r^{\frac{1}{2}-\kappa}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}r^{-\frac{3}{2}+\kappa}\right).
\end{align*}
By Lemma~\ref{lemma:twisting_positive_g}, $\Vmod_a=\OO(1)$ and $\Vmod_a$ is positive near infinity for sufficiently large $|m|$, which shall be assumed henceforth. Moreover, $\Vmod_a$ is chosen such that the twisted variational problem leads to the same Euler-Lagrange equation as the untwisted one.
For $U\subseteq(r_+,\infty)$, we define the twisted Sobolev norm
\begin{align*}
\norm{f}_{\uH_{\kappa}^1(U)}^2:=\int_U\left(\frac{1}{r^2}|f|^2+r(r-r_+)h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}f\right)\right\lvert^2\right)\,\dd r.
\end{align*}
Note that for $U\subseteq(r_+,\infty)$ compact, the $\uH_{\kappa}^1$ norm is equivalent to the standard Sobolev norm.
For $U=(r_+,\infty)$, let $\uH^1_{\kappa}(U)$ be the completion of functions of the form
\begin{align}
\label{eqn:basis_twisting}
f(r)=r^{-\frac{1}{2}+\kappa}g(r)
\end{align}
under $\norm{\cdot}_{\uH_{\kappa}^1(U)}$, where $(x\mapsto g(1/x))\in C_0^{\infty}[0,1/r_+)$. Henceforth, we will sometimes refer to such a function $g$ as being ``compactly supported around infinity".
\begin{lemma}
\label{lemma:SobolevNeumann}
Let $f\in\uH^1_{\kappa}(r_+,\infty)$, then $f$ is also in $C(r_++1,\infty)$ and $r^{1/2-\kappa}f(r)$ is bounded.
\end{lemma}
\begin{proof}
The existence of a continuous version follows from Sobolev embedding as in Lemma~\ref{lemma:trace}. Then, there exists a sequence $(f_n)\in C^{\infty}$ as in the definition such that $f_n\rightarrow f$ in $\uH^1_{\kappa}$. Let $\tilde R>R>r_+$ and let $f_n(R)$ converge to $f(R)$:
\begin{align*}
\left\lvert \tilde R^{1/2-\kappa}(f_n(\tilde R)-f(\tilde R))\right\lvert&\leq \left\lvert r^{1/2-\kappa}(f_n(R)-f(R))\right\lvert+\int_R^{\tilde R}\left\lvert\frac{\dd}{\dd r}\left(r^{1/2-\kappa}\left(f_n-f\right)\right)\right\lvert\,\dd r\\
&\leq \left\lvert r^{1/2-\kappa}(f_n(R)-f(R))\right\lvert\\&~~~~~+\left(\int_R^{\infty}r(r-r_+)r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(f_n-f\right)\right\lvert^2\,\dd r\right)^{1/2}\times\\&~~~~~~~~~~\times\left(\int_R^{\infty}\frac{1}{r(r-r_+)r^{-1+2\kappa}}\,\dd r\right)^{1/2}
\end{align*}
Hence $r^{1/2-\kappa}\left(f_n(r)-f(r)\right)$ converges uniformly for all $r\geq R$. Hence we even have convergence at $r=\infty$. Since $\lim_{r\rightarrow\infty}r^{1/2-\kappa}f_n(r)\neq \infty$ for all $n$, we obtain the result.
\end{proof}
As in Section~\ref{sec:Real}, choose mode parameters such that the conditions for Lemma~\ref{lemma:FunctionalNegativeTwisted} are satisfied.
Let
\begin{align}
\label{eqn:defn_A}
\Aa:=\{a>0\,:\,\exists\,(x\mapsto g(1/x))\in C_0^{\infty}[0,1/r_+):\,\tilde{\LL}_a(r^{-1/2+\kappa}g)<0\}.
\end{align}
Note that $\Aa$ is non-empty, open and bounded below.
\begin{lemma}
\label{lemma:Hardy_twisted}
For $r_{\mathrm{cut}}> r_++1$, $0<\kappa<1$ and a smooth function $f$ with $fr^{1/2-\kappa}=\OO(1)$ at infinity, we have that
\begin{align*}
\int_{r_{\mathrm{cut}}}^{\infty}\frac{|f|^2}{r^2}\,\dd r&=\frac{1}{2(1-\kappa)}r_{\mathrm{cut}}^{-1}\left(\frac{c}{2}\int_{r_{\mathrm{cut}}-1}^{r_{\mathrm{cut}}}|f|^2\,\dd r+\frac{1}{2c}\int_{r_{\mathrm{cut}}-1}^{r_{\mathrm{cut}}}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\,\dd r\right)\\&~~~~~+\frac{1}{(1-\kappa)^2}\int_{r_{\mathrm{cut}}}^{\infty}r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{1/2-\kappa}f\right)\right\lvert^2\,\dd r
\end{align*}
for any $c>0$ sufficiently large.
\end{lemma}
\begin{proof}
We compute:
\begin{align*}
\int_{r_{\mathrm{cut}}}^{\infty}\frac{|f|^2}{r^2}\,\dd r&=\frac{1}{2(1-\kappa)}r_{\mathrm{cut}}^{-1}|f|^2(r_{\mathrm{cut}})\\&~~~~~+\frac{1}{1-\kappa}\int_{r_{\mathrm{cut}}}^{\infty}r^{2\kappa-2}\Re\left(r^{1/2-\kappa}f\frac{\dd}{\dd r}\left( r^{1/2-\kappa}\overline f\right)\right)\,\dd r.
\end{align*}
Using the Cauchy-Schwarz inequality, one easily sees that
\begin{align*}
\int_{r_{\mathrm{cut}}}^{\infty}\frac{|f|^2}{r^2}\,\dd r&\leq \frac{1}{1-\kappa}r_{\mathrm{cut}}^{-1}|f|^2(r_{\mathrm{cut}})\\&~~~~~+\frac{1}{(1-\kappa)^2}\int_{r_{\mathrm{cut}}}^{\infty}r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{1/2-\kappa}f\right)\right\lvert^2\,\dd r.
\end{align*}
Let $\chi\geq 0$ be a smooth function of compact support with $\chi(r_{\mathrm{cut}})=1$ and $\chi=0$ for $r\leq r_{\mathrm{cut}-1}$. Then
\begin{align*}
|f|^2(r_{\mathrm{cut}})&\leq|f|^2(r_{\mathrm{cut}})\chi(r_{\mathrm{cut}})\\
&\leq \frac{c}{2}\int_{r_{\mathrm{cut}}-1}^{r_{\mathrm{cut}}}|f|^2\,\dd r+\frac{1}{2c}\int_{r_{\mathrm{cut}}-1}^{r_{\mathrm{cut}}}\left\lvert\frac{\dd f}{\dd r}\right\lvert^2\,\dd r
\end{align*}
for any $c>0$ sufficiently large.
\end{proof}
\begin{lemma}
\label{lemma:VariationInequalityND}
Let $a\in \Aa$ be fixed. There exist constants $r_+<B_0<B_1<\infty$ and $C_0,C_1>0$, such that, for large enough $m$, we have for all smooth functions $f$ for which the following integrals are defined that
\begin{align*}
\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}f\right)\right\lvert^2+C_01_{[B_0,B_1]^c}\frac{|f|^2}{r^2}\right)\,\dd r\leq C_1\int_{B_0}^{B_1}|f|^2\,\dd r+2\tL_{a}(f).
\end{align*}
\end{lemma}
\begin{proof}
The proof follows the strategy of Lemma~\ref{lemma:VariationInequalityD}. The analysis of the potential goes through as in Section~\ref{sec:Real} as the twisting part of $\Vmod_a$ does not depend on $m$ and has the right asymptotics. Thus we know that there is an $R_1$ such that
\begin{align*}
\frac{r^2+a^2}{\Delta_-}\Vmod_a>0
\end{align*}
on $(r_+,R_1)$. Moreover, there is an $R_2>R_1$ such that
\begin{align*}
\frac{r^2+a^2}{\Delta_-}\Vmod_a>-\frac{C}{2r^2}~~~\mathrm{and}~~~
\frac{C}{(1-\kappa)^2}<\frac{\Delta_-}{r^2+a^2}
\end{align*}
for $r\geq R_2$.
Hence
\begin{align*}
\int_{R_2}^{\infty}\frac{r^2+a^2}{\Delta_-}\Vmod_a|f|^2\,\dd r&\geq -\frac{1}{2}\int_{R_2-1}^{\infty}\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{1/2-\kappa}f\right)\right\lvert^2\,\dd r-C'\int_{R_2-1}^{R_2}|f|^2\,\dd r
\end{align*}
for some large constant $C'>0$ by the Hardy inequality of Lemma~\ref{lemma:Hardy_twisted}. Choosing $B_0$,$ B_1$, $C_0$ and $C_1$ appropriately (as in the proof of Lemma~\ref{lemma:VariationInequalityD}), we obtain the inequality.
\end{proof}
\begin{lemma}
\label{lemma:Semicts}
The functional $\tL_a$ is weakly lower semicontinuous in $\uH_{\kappa}^1(r_+,\infty)$.
\end{lemma}
\begin{proof}
See the comments to Lemma~\ref{lemma:SemictsD}.
\end{proof}
\begin{lemma}
\label{lemma:RegMinim}
Let $a\in\Aa$.
There exists an $f_{a}\in\uH_{\kappa}^1(r_+,\infty)$ with norm $\norm{f_{a}}_{\uL^2(r_+,\infty)}=1$ such that $\tL_a$ achieves its infimum in
\begin{align*}
\uH_{\kappa}^1(r_+,\infty)\cap\{\norm{f}_{\uL^2(r_+,\infty)}=1\}
\end{align*}
on $f_{a}$.
\end{lemma}
\begin{proof}
The proof is similar to the one in Section~\ref{sec:Real}. We obtain a minimising sequence $(f_{a,n})$ that converges weakly in $\uH_{\kappa}^1$ and strongly in $L^2$ on compact subsets of $(r_+,\infty)$. In analogy to the Dirichlet, the $f_{a,n}$ are can be taken from a dense subset and can be chosen to be of the form $f_{a,n}=r^{-1/2+\kappa}g_{a,n}$ for $g_{a,n}$ smooth and compactly supported around infinity.
We will show that the norm is conserved. Suppose not.
Then, for any $N$, there are infinitely many of the $f_{a,n}$ such that
\begin{align*}
\norm{f_{a,n}}_{\uL^2((r_+,\infty)\backslash [r_++1/N,N])}\geq \rho>0.
\end{align*}
Suppose
\begin{align*}
\norm{f_{a,n}}_{\uL^2(r_+,r_++\delta)}\geq \rho_1>0
\end{align*}
for infinitely many $f_{a,n}$ and any $\delta>0$.
Because of the $L^2$ convergence on compact subsets, there is an $R$ such that $f_{a,n}(R)\rightarrow f_a(R)$ as $n\rightarrow\infty$, in particular $f_{a,n}(R)$ is bounded for all $n$.
By Lemma~\ref{lemma:VariationInequalityND}, we have for $r\in(r_+,R)$:
\begin{align*}
|r^{1/2-\kappa}f_{a,n}(r)|&\leq \int_{r}^{R}\left\lvert\frac{\dd }{\dd r'}\left(r'^{1/2-\kappa}f_{a,n}\right)\right\lvert\,\dd r'+R^{1/2-\kappa}f_{a,n}(R)\\
&\leq \left(\int_{r}^{R}\frac{1}{r'-r_+}\,\dd r'\right)^{1/2}\times\\&~~~~~~~~~~\times\left(\int_{r}^{R}(r'-r_+)r'^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r'}\left(r'^{1/2-\kappa}f_{a,n}\right)\right\lvert^2\,\dd r'\right)^{1/2}\\&~~~~~~~~~~~~~~~+R^{1/2-\kappa}f_{a,n}(R)\\
&\leq C\left(1+\sqrt{\log\frac{R-r_+}{r-r_+}}\right)
\end{align*}
for a constant $C>0$. Since $r\mapsto \sqrt{|\log (r-r_+)|}$ is integrable on compact subsets of $[r_+,\infty)$, we obtain $\norm{f_{a,n}}_{\uL^2(r_+,r_++\delta)}\rightarrow 0$ as $\delta\rightarrow 0$, a contradiction.
Hence we only need to exclude the case that the norm is bounded away from zero for large $r$. Thus, suppose that
\begin{align*}
\norm{f_{a,n}}_{\uL^2(R_0,\infty)}\geq \rho_2>0
\end{align*}
for infinitely many $f_{a,n}$ and any $R_0>0$.
Since $f_{a,n}(r_+)=0$, we have
\begin{align*}
r^{\frac{1}{2}-\kappa}|f_{a,n}|(r)&\leq \int_{r_+}^{r}\left\lvert\frac{\dd}{\dd r}\left(r'^{\frac{1}{2}-\kappa}f_{a,n}\right)\right\lvert\,\dd r'\\&\leq \left(\int_{r_+}^{\infty}\frac{1}{r^{1+2\kappa}}\,\dd r\right)^{1/2}\left(\int_{r_+}^{\infty}r^{1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{a,n}\right)\right\lvert\,\dd r\right)^{1/2},
\end{align*}
which is uniformly bounded for all $n$. Hence
\begin{align*}
\int_{R_0}^{\infty}\frac{|f_{a,n}|^2}{r^2}\,\dd r\leq C'\int_{R_0}^{\infty}r^{-3+2\kappa}\,\dd r\rightarrow 0
\end{align*}
as $R_0\rightarrow\infty$, a contradiction.
As in the proof of Lemma~\ref{lemma:RegMinD}, we have
\begin{align*}
\nu_a\leq\LL_a(f_a)\leq\liminf_{n\rightarrow\infty}\LL_a(f_{a,n})=\nu_a
\end{align*}
and the rest follows.
\end{proof}
We would like to derive the Euler-Lagrange equation corresponding to this minimiser.
\begin{lemma}
\label{lemma:ELDN}
The minimiser $f_{a}$ satisfies
\begin{align}
\begin{aligned}
\label{eqn:ELregNeumann}
\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}f_{a}\right)\frac{\dd}{\dd r}\left(h^{-1}\psi\right)+\Vmod_a\frac{r^2+a^2}{\Delta_-}f_{a}\psi\bigg)\,\dd r=-\nu_a\int_{r_+}^{\infty}\frac{f_{a}}{r^2}\psi\,\dd r
\end{aligned}
\end{align}
for all $\psi\in\uH_{\kappa}^1(r_+,\infty)$.
\end{lemma}
The proof of Lemma~\ref{lemma:ELDN} can be found in Appendix~\ref{sec:Twisted_Euler_Lagrange}.
\begin{propn}
There is an $\hat a$ and a corresponding non-zero function $f_{\hat a}\in C^{\infty}(r_+,\infty)$ such that
\begin{align*}
\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+\hat a^2}\frac{\dd f_{\hat a}}{\dd r}\right)-\tilde V_{\hat a}f_{\hat a}=0
\end{align*}
and $f_a$ satisfies the horizon regularity condition and the Neumann boundary condition at infinity.
\end{propn}
\begin{proof}
As in Proposition~\ref{propn:a_hat}, we find an $f_{\hat a}\in\uH_{\kappa}^1$ such that
\begin{align*}
\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+\hat a^2}r^{-1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{\hat a}\right)\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}\psi\right)+\Vmod_{\hat a}\frac{r^2+\hat a^2}{\Delta_-}f_{\hat a}\psi\bigg)\,\dd r=0.
\end{align*}
Choosing $\psi(r)=r^{-\frac{1}{2}+\kappa}g(r)$ with $g$ having compact support around infinity and integrating by parts, we obtain
\begin{align*}
\frac{\Delta_-}{r^2+\hat a^2}r^{-1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{\hat a}\right)g\rightarrow 0
\end{align*}
as $r=\infty$ for all $g$ as in (\ref{eqn:basis_twisting}). This yields the asymptotics.
Moreover, as in the proof of Proposition~\ref{propn:a_hat}, we retrieve the ODE.
The boundary condition at the horizon follows analogously to Section~\ref{sec:Real}.
\end{proof}
\subsection{Perturbing the Neumann modes into the complex plane}
\label{subsec:PerturbingN}
In Section~\ref{sec:RealDN}, we constructed real mode solutions for $5/4<\alpha<9/4$ satisfying Neumann conditions.
For the growing radial parts, we proceed as in Section~\ref{sec:Real} with the difference that here, finding a mode solution is equivalent to finding a zero of $B$. The present case is considerably more difficult than the Dirichlet case. A first manifest difference is the asymmetry in the definitions of Dirichlet and Neumann boundary conditions since a Dirichlet mode has more decay than required by Definition~\ref{defn:BdyConds}. This means that if a function satisfies the Dirichlet boundary condition for a mass $\alpha_1$, it also does so for every $\alpha_2$ sufficiently close to $\alpha_1$. As Definition~\ref{defn:BdyCondsNeumann} is tighter, this is not true in the Neumann case. Another difficulty stems from twisting as the dependence of the equations on $\alpha$ becomes more complicated.
We have already chosen $B(\alpha(0),\omega_R(0))=0$. Recall from Section~\ref{sec:RestrReal} that
\begin{align*}
Q_T=\Im\left(r^{-\frac{1}{2}+\kappa}\frac{\dd}{\dd\sta r}\left(r^{\frac{1}{2}-\kappa}u\right)\overline u\right),~~~~~ Q_T(r_+)=\Xi am-\omega_R(r_+^2+a^2).
\end{align*}
Hence, analogously to Section~\ref{subsec:PerturbingD}, the problem reduces to showing that
\begin{align*}
\frac{\del B}{\del\alpha}(\alpha(0),\omega_R(0))\neq 0.
\end{align*}
Again, for the sake of contradiction, suppose that this is not the case.
Then, near infinity, we have
\begin{align*}
u_{\alpha}(r,\alpha(0),\omega_R(0))&=\frac{\del A}{\del\alpha}(\alpha(0),\omega_R(0)) h_1(r,\alpha(0),\omega_R(0))\\
&~~~~~~~+A(\alpha(0),\omega_R(0))\frac{\del h_1}{\del\alpha}(r,\alpha(0),\omega_R(0)).
\end{align*}
By the horizon regularity condition, $u\sim(r-r_+)^{\xi}$ near the horizon, $u_{\alpha}$ is smooth at $r=r_+$. However, $u_{\alpha}$ does not satisfy the Neumann condition at infinity as the second term behaves as $r^{-1/2+\kappa}u\log r$.
Let $f:\,(r_+,\infty)\rightarrow\CC$ be $C^1$ and piecewise $C^2$. Then the function
\begin{align*}
v(r):=u_{\alpha}(r)-\frac{\del\kappa}{\del\alpha}f(r)u(r)=u_{\alpha}(r)+\frac{1}{2\kappa}f(r)u(r),
\end{align*}
does satisfy the Neumann boundary condition if, for large $r$, $f(r)=\log r+\OO(r^{-\gamma})$, where $\gamma>0$.
From the radial ODE, we obtain
\begin{align*}
&\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u_{\alpha}}{\dd r}\right)-\frac{r^2+a^2}{\Delta_-}\tilde V_au_{\alpha}=\frac{1}{r^2+a^2}\left[\frac{\del\lambda}{\del\alpha}-\frac{1}{\ell^2}(r^2+a^2)\right]u.
\end{align*}
Lemma~\ref{lemma:EL_twisted_g} yields a twisted version
\begin{align}
\begin{split}
\label{eqn:twisted_del_alpha_perturbation}
\frac{1}{h}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u_{\alpha}}{h}\right)\right)+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}u_{\alpha}=\frac{1}{r^2+a^2}\left[\frac{\del\lambda}{\del\alpha}-\frac{1}{\ell^2}(r^2+a^2)\right]u.
\end{split}
\end{align}
We will use the previous twisting, i.\,e. $h=r^{-1/2+\kappa}$.
For the second term of $v$, we compute:
\begin{align*}
\frac{1}{h}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(f\frac{u}{h}\right)\right) &=h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)(h^{-1}u)\\&~~~~~+2h^{-1}\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\frac{\dd}{\dd r}(h^{-1}u)\\&~~~~~+h^{-1}f\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}(h^{-1}u)\right)
\end{align*}
We add this to the equation to (\ref{eqn:twisted_del_alpha_perturbation}) multiplied by $2\kappa$. Then we multiply the resulting equation by $\overline u$ and integrate by parts, noting that $v$ satisfies the Neumann boundary condition. Hence we obtain
\begin{align}
\begin{split}
\label{eqn:Bzero_contradiction}
0&=\int_{r_+}^{\infty}\frac{2\kappa}{r^2+a^2}\left(\frac{\del\lambda}{\del\alpha}-\frac{a^2}{\ell^2}\right)|u|^2\,\dd r\\
&~~~~~-\int_{r_+}^{\infty}\left(\frac{2\kappa}{\ell^2}\frac{r^2}{r^2+a^2}|u|^2-\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)\left\lvert h^{-1}u\right\lvert^2\right)\,\dd r\\
&~~~~~-2\int_{r_+}^{\infty}f\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r.
\end{split}
\end{align}
Our aim is to show that the right hand side of (\ref{eqn:Bzero_contradiction}) is negative, which yields the desired contradiction.
From Section~\ref{subsec:PerturbingD}, we already know that
\begin{align*}
\int_{r_+}^{\infty}\frac{2\kappa}{r^2+a^2}\left(\frac{\del\lambda}{\del\alpha}-\frac{a^2}{\ell^2}\right)|u|^2\,\dd r=-\int_{r_+}^{\infty}\frac{2\kappa}{r^2+a^2}\int_0^{\pi}\frac{a^2}{\ell^2}\cos^2\theta|S|^2\sin\theta|u|^2\,\dd\theta\,\dd r
\end{align*}
has the right sign. We set
\begin{align}
\label{f_perturbation}
f(r):=\begin{cases}
\log r+\frac{1}{2\kappa}\frac{R^{2\kappa}}{r^{2\kappa}}, & r\geq R\\
\log R+\frac{1}{2\kappa}, & r<R
\end{cases}
\end{align}
for an $R>r_++1$ to be determined. Note that $f$ is continuously differentiable. For $r> R$,
\begin{align*}
\frac{\dd f}{\dd r}(r)=\frac{1}{r}\left(1-\frac{R^{2\kappa}}{r^{2\kappa}}\right)>0,
\end{align*}
whence $f$ is monotonic.
First, we choose $R$ sufficiently large such that $\tilde V^h_a>0$ for $r>R$ according to Lemma~\ref{lemma:twisting_positive_g}, whence
\begin{align*}
&\int_{r_+}^{\infty}f\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r\\=&\left(\log R+\frac{1}{2\kappa}\right)\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r\\
&~~~~~~~+\int_R^{\infty}\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\frac{R^{2\kappa}}{r^{2\kappa}}-1\right)\right)\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r,
\end{align*}
which is non-negative since the first integral with the constant coefficient is zero and the second integral is positive.
For $r>R$, one easily computes
\begin{align*}
\frac{2\kappa}{\ell^2}\frac{r^2}{r^2+a^2}-h^{-2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)
=-\frac{2\kappa a^2/\ell^2+(2\kappa-2)}{r^2}-2\frac{R^{2\kappa}}{r^{2\kappa}}\frac{1}{r^2}+\OO(r^{-3})
\end{align*}
Therefore, there is a $C_1>0$ such that
\begin{align*}
\int_R^{\infty}\left(\frac{2\kappa}{\ell^2}\frac{r^2}{r^2+a^2}|u|^2-h^{-2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)\right)|u|^2\,\dd r>-C_1\int_R^{\infty}\frac{|u|^2}{r^2}\,\dd r.
\end{align*}
We can prove the following Hardy inequality:
\begin{lemma}
Let $u$ satisfy the Neumann boundary condition at infinity and let $\beta>0$. Then
\begin{align*}
\int_R^{\infty}\frac{1}{r^{1+\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u\right\lvert^2\,\dd r&\leq \lim_{r\rightarrow\infty}\frac{2\beta}{R^{\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u(r)\right\lvert^2\\&~~~~~~+4\beta^2 \int_R^{\infty}r^{2-2\kappa-\beta}\left(1-\left(\frac{r}{R}\right)^{\beta}\right)^2h^2\left\lvert\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right\lvert^2\,\dd r.
\end{align*}
\end{lemma}
\begin{proof} We compute
\begin{align*}
\int_R^{\infty}\frac{1}{r^{1+\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u\right\lvert^2\,\dd r&=\int_R^{\infty}\del_r\left(-\frac{\beta}{r^{\beta}}+\frac{\beta}{R^{\beta}}\right)\left\lvert r^{\frac{1}{2}-\kappa}u\right\lvert^2\,\dd r\\
&\leq\frac{\beta}{R^{\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u\right\lvert^2(\infty)+\int_R^{\infty}\frac{1}{2}\frac{1}{r^{1+\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u\right\lvert^2\,\dd r\\&~~~~~~~~~~+2\beta^2 \int_R^{\infty}r^{1+\beta}\left(r^{-\beta}-R^{-\beta}\right)^2\left\lvert\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}u\right)\right\lvert^2\,\dd r,
\end{align*}
yielding the result.
\end{proof}
Since
\begin{align*}
C_1\lim_{r\rightarrow\infty}\frac{2\beta}{R^{\beta}}\left\lvert r^{\frac{1}{2}-\kappa}u(r)\right\lvert^2\rightarrow 0
\end{align*}
as $R\rightarrow\infty$,
by choosing $R$ possibly larger, we obtain
\begin{align*}
C_1\frac{2\beta}{R^{\beta}}\lim_{r\rightarrow\infty}\left\lvert r^{\frac{1}{2}-\kappa}u(r)\right\lvert^2<\int_{r_+}^{\infty}\frac{2\kappa}{r^2+a^2}\left(\frac{\del\lambda}{\del\alpha}-\frac{a^2}{\ell^2}\right)|u|^2\,\dd r.
\end{align*}
For convenience, set $C_2:=4\beta^2 C_1$. Since $\tilde V_a^h\sim r^{-2}$, we need to show that
\begin{align}
\begin{split}
\label{eqn:goal_perturbation}
C_2\int_R^{\infty}\frac{|u|^2}{r^2}\,\dd r<&\int_R^{\infty}\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\frac{R^{2\kappa}}{r^{2\kappa}}-1\right)\right)\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2\,\dd r.
\end{split}
\end{align}
We will deal with the two cases $0<\kappa\leq 1/2$ and $1/2<\kappa<1$ separately. Let us first consider $0<\kappa\leq 1/2$. We choose $\beta=2\kappa$. Note that in this case
\begin{align*}
\int_R^{\infty}\frac{|u|^2}{r^2}\,\dd r\leq \int_R^{\infty}\frac{1}{r^{1+2\kappa}}\left\lvert r^{1/2-\kappa}u\right\lvert^2\,\dd r.
\end{align*}
\begin{lemma}
Let $C>0$. There is an $R$ such that, for all $r>R$,
\begin{align*}
Cr^{2-4\kappa}\left(1-\left(\frac{r}{R}\right)^{2\kappa}\right)^2\leq r^2\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\left(\frac{R}{r}\right)^{2\kappa}-1\right)\right).
\end{align*}
\end{lemma}
\begin{proof}
It suffices to show that
\begin{align*}
Cr^{-4\kappa}\left(1-\left(\frac{r}{R}\right)^{2\kappa}\right)^2\leq \log\frac{r}{R}+\frac{1}{2\kappa}\left(\left(\frac{R}{r}\right)^{2\kappa}-1\right).
\end{align*}
As this holds at $r=R$, it suffices to show the statement for the derivatives. Substituting $x:=r^{2\kappa}$, we need to show
\begin{align*}
0\leq x^2-(R^{2\kappa}+4\kappa C R^{-2\kappa})x+4\kappa C=(x-R^{2\kappa})(x-4\kappa C R^{-2\kappa}).
\end{align*}
Therefore, the result holds if $R^{4\kappa}>4\kappa C$.
\end{proof}
This lemma immediately yields (\ref{eqn:goal_perturbation}) for $0<\kappa\leq 1/2$. Let us now turn to $1/2<\kappa<1$. Here we choose $\beta=2-2\kappa$.
\begin{lemma}
Let $C>0$ and $1/2<\kappa<1$. There is an $R$ such that, for all $r>R$,
\begin{align*}
C\left(1-\left(\frac{r}{R}\right)^{2-2\kappa}\right)^2\leq r^2\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\left(\frac{R}{r}\right)^{2\kappa}-1\right)\right).
\end{align*}
\end{lemma}
\begin{proof}
As equality holds for $r=R$, it suffices to consider the derivatives, i.\,e. we would like to establish
\begin{align*}
&2r\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\left(\frac{R}{r}\right)^{2\kappa}\right)\right)+r^2\left(\frac{1}{r}-\left(\frac{R}{r}\right)^{2\kappa}\frac{1}{r}\right)\\
&~~~~~~~~~~~~~~~~~~~-2C\left(\left(\frac{r}{R}\right)^{2-2\kappa}-1\right)(2-2\kappa)R^{-2+2\kappa}r^{1-2\kappa}\geq 0.
\end{align*}
This again holds for $r=R$, so, after dividing the inequality by $r$, it suffices to prove the corresponding inequality for the derivatives, i.\,e.
\begin{align*}
&\frac{1}{r}\left(2-2\left(1-\kappa\right)\left(\frac{R}{r}\right)^{2\kappa}\right)-4C\kappa(2-2\kappa)\frac{1}{R^2}\left(\frac{r}{R}\right)^{-1-2\kappa}\\&~~~~~~~~~~~+2C(2-2\kappa)(4\kappa-2)R^{-4-4\kappa}r^{1-4\kappa}\geq 0.
\end{align*}
The last term on the left hand side is always positive. Thus the left hand side is greater than
\begin{align*}
\frac{\kappa}{r}-4C\kappa(2-2\kappa)\frac{1}{R}\left(\frac{R}{r}\right)^{2\kappa}\geq \left(\frac{1}{2}-4C\frac{1}{R}\right)\frac{1}{r},
\end{align*}
which is positive for sufficienlty large $R$.
\end{proof}
Therefore, for both ranges of $\kappa$, the right hand side of (\ref{eqn:Bzero_contradiction}) is bounded below by
\begin{align*}
-\int_{r_+}^R\frac{1}{\ell^2}\frac{r^2}{r^2+a^2}|u|^2\,\dd r-2\int_R^{\infty}\left(\log\frac{r}{R}+\frac{1}{2\kappa}\left(\frac{R^{2\kappa}}{r^{2\kappa}}-1\right)\right)\tilde V_a^h\frac{r^2+a^2}{\Delta_-}|u|^2\,\dd r<0
\end{align*}
for non-trivial $u$,
a contradiction. Thus we have shown the following
\begin{lemma}
\begin{align*}
\frac{\del B}{\del\alpha}(\alpha(0),\omega_R(0))\neq 0.
\end{align*}
\end{lemma}
\subsection{Behaviour for small $\epsilon>0$ for Neumann boundary conditions}
\label{subsec:small_behaviour_Neumann}
The main new idea of this section can be found in the proof of Proposition~\ref{propn:alpha_decreasing_Neumann}, where the insights of Section~\ref{subsec:PerturbingN} are essential to overcome the difficulties outlined at the beginning of the previous section.
\begin{propn}
\label{propn:del_omega_Neumann}
For sufficiently large $|m|$,
\begin{align*}
\omega_R(0)\frac{\del\omega_R}{\del\epsilon}(0)< 0.
\end{align*}
\end{propn}
\begin{proof}
We define an appropriate modified microlocal energy current
\begin{align*}
\tilde Q_T:=\Im\left(r^{-\frac{1}{2}+\kappa}\left(r^{\frac{1}{2}-\kappa}\right)'\overline{\omega u}\right).
\end{align*}
Let $\epsilon>0$, then $\tilde Q_T(r_+)=\tilde Q_T(\infty)=0$. This yields
\begin{align}
\label{eqn:int_whole}
\int_{r_+}^{\infty}\left(\epsilon\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd }{\dd r}\left(h^{-1}u\right)\right\lvert^2-\frac{r^2+a^2}{\Delta_-}\Im(\Vmod_a\overline\omega)|u|^2\right)\,\dd r=0.
\end{align}
Similarly to Section~\ref{subsec:Crossing}, we obtain
\begin{align}
\begin{aligned}
\label{eqn:Im_ineq}
-\Im\left(\Vmod_a\overline\omega\right)&>\frac{\epsilon}{(r^2+a^2)^2}\left(K(r)+V_+(r^2+a^2)^2-\frac{\alpha}{\ell^2}\Delta_-(r^2+a^2)\right)\\&~~~~~~~~+\epsilon \left(\frac{1}{2}-\kappa\right)\frac{\Delta_-}{r^2+a^2}r^{1/2-\kappa}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}r^{-3/2+\kappa}\right)
\end{aligned}
\end{align}
with the additional term due to the twisting. Again
\begin{align*}
K(r)=|\omega|^2(r^2+a^2)^2-\Xi^2 a^2m^2-\Delta_-a^2|\omega|^2.
\end{align*}
Recall from Section~\ref{subsec:Crossing} that
\begin{align}
\label{eqn:K_growth}
\frac{\dd K}{\dd r}(r)=|\omega|^2\left(4\left(1-\frac{a^2}{\ell^2}\right)r^3+2a^2M+2a^2\left(1-\frac{a^2}{\ell^2}\right)\right)>0.
\end{align}
By Lemma~\ref{lemma:twisting_positive_g}, there is an $R>r_+$ such that
\begin{align*}
\frac{r^2+a^2}{\Delta_-}\left\lvert V_++V_{\alpha}+\left(\frac{1}{2}-\kappa\right)\frac{\Delta_-}{r^2+a^2}r^{1/2-\kappa}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}r^{-3/2+\kappa}\right)\right\lvert<\frac{C}{2r^2}
\end{align*}
for any $C>0$. Thus, by an application of Lemma~\ref{lemma:Hardy_twisted} as in the proof of Lemma~\ref{lemma:VariationInequalityND}, we have
\begin{align*}
\begin{split}
&\int_R^{\infty}\left(\epsilon\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd }{\dd r}\left(h^{-1}u\right)\right\lvert^2-\frac{r^2+a^2}{\Delta_-}\Im(\Vmod_a\overline\omega)|u|^2\right)\,\dd r\\&~~~~~~~~~~~~~~~~~~~~~~>\int_R^{\infty}\frac{\epsilon}{(r^2+a^2)^2}K(r)|u|^2-\int_{R-1}^R\epsilon C'|u|^2\,\dd r
\end{split}
\end{align*}
for sufficiently large $R$ and a large constant $C'>0$.
For the sake of contradiction, suppose that $K(r_+)\geq 0$. Then, of course,
\begin{align}
\begin{split}
\label{eqn:int_R}
&\int_R^{\infty}\left(\epsilon\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd }{\dd r}\left(h^{-1}u\right)\right\lvert^2-\frac{r^2+a^2}{\Delta_-}\Im(\Vmod_a\overline\omega)|u|^2\right)\,\dd r\\&~~~~~~~~>\int_R^{\infty}\frac{\epsilon}{(r^2+a^2)^2}K(r)|u|^2+\int_{R-1}^R\epsilon\left(\frac{K(r)}{(r^2+a^2)^2}-C'\right) |u|^2\,\dd r
\end{split}
\end{align}
By (\ref{eqn:K_growth}), this means that $K>0$ on $(r_+,\infty)$. Since $\epsilon\mapsto\omega(\epsilon)$ is continuous and
\begin{align*}
|\omega(0)|^2\geq Cm^2,
\end{align*}
$|\omega|^2$ scales as $m^2$, so $\dd K/\dd r$ can be chosen to be as large as possible by increasing $m^2$, in particular, it can be used to overcome the potentially negative derivative of the remaining terms of the right hand side of (\ref{eqn:Im_ineq}) on $(r_+,R)$ and in (\ref{eqn:int_R}) on $(R-1,R)$. Using (\ref{eqn:int_whole}) and (\ref{eqn:int_R}), we conclude $u=0$, a contradiction.
\end{proof}
From now on we fix $m$ -- see Remark~\ref{rk:large_m}.
\begin{propn}
\label{propn:alpha_decreasing_Neumann}
\begin{align*}
\frac{\del\alpha}{\del\epsilon}(0)>0.
\end{align*}
\end{propn}
\begin{proof}
The proof proceeds as in Section~\ref{subsec:Crossing}, adapting the idea used already in Section~\ref{subsec:PerturbingN}.
Set $u_{\epsilon}:=\del u/\del\epsilon$. For an $f$ as in (\ref{f_perturbation}) with an $R$ to be determined,
\begin{align*}
v(r):=u_{\epsilon}(r)-\frac{\del\kappa}{\del\epsilon}f(r)u(r)=u_{\epsilon}(r)+\frac{1}{2\kappa}\frac{\del\alpha}{\del\epsilon}f(r)u(r)
\end{align*}
satisfies the Neumann boundary condition at infinity. As $fu$ extends smoothly to the horizon, the behaviour of $v$ at $r=r_+$ is dominated by $u_{\epsilon}$.
Using the $h$ of Lemma~\ref{lemma:twisting_positive_g} yields the ODE
\begin{align*}
& h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd }{\dd r}\left(\frac{v}{h}\right)\right)-\frac{r^2+a^2}{\Delta_-}\tilde V^h_a v\\&~~~~~~~=\frac{r^2+a^2}{\Delta_-}\frac{\del\tilde V_a}{\del\epsilon}u+\frac{1}{2\kappa}\frac{\del\alpha}{\del\epsilon}h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)\left(h^{-1}u\right)\\
&~~~~~~~~~~~+2\frac{1}{2\kappa}\frac{\del\alpha}{\del\epsilon}h^{-1}\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\frac{\dd}{\dd r}\left(h^{-1}u\right)+\frac{1}{2\kappa}\frac{\del\alpha}{\del\epsilon}fh^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}u\right)\right).
\end{align*}
Observe that as in the proof of Proposition~\ref{propn:alpha_Dirichlet}
\begin{align*}
\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}u_{\epsilon}\right)\right)h^{-1}\overline u&=\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h\left[\frac{\dd}{\dd r}\left(h^{-1}u_{\epsilon}\right)\overline u-u_{\epsilon}\frac{\dd}{\dd r}\left(h^{-1}\overline u\right)\right]\right)\\
&~~~~~~~~+u_{\epsilon}h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}\overline u\right)\right).
\end{align*}
This yields
\begin{align}
\begin{split}
\label{eqn:integral_var_epsilon}
-2\kappa|u(r_+)|^2&=\int_{r_+}^{\infty}\left(\frac{r^2+a^2}{\Delta_-}2\kappa\Re\left(\frac{\del\tilde V}{\del\epsilon}\right)\Bigg\lvert_{\epsilon=0}+\frac{\del\alpha}{\del\epsilon}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)\left\lvert h^{-1}u\right\lvert^2\right)\,\dd r\\
&~~~~~-\frac{\del\alpha}{\del\epsilon}\int_{r_+}^{\infty}f\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r\\
&~~~~~-\frac{\del\alpha}{\del\epsilon}\int_{r_+}^{\infty}f\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2\,\dd r.
\end{split}
\end{align}
The expression for $\Re\left(\frac{\del\tilde V}{\del\epsilon}\right)\Big\lvert_{\epsilon=0}$ can be taken from (\ref{eqn:real_part_V_epsilon}). First one can eliminate the explicit $\lambda$ dependence via Lemma~\ref{propn:IfThen} and one obtains a lower bound on the right hand side using Proposition~\ref{propn:del_omega_Neumann}.
Then suppose for the sake of contradiction that $\del\alpha/\del\epsilon\leq 0$. It follows immediately from Section~\ref{subsec:PerturbingN} that the right hand side of (\ref{eqn:integral_var_epsilon}) is positive, a contradiction.
\end{proof}
\subsection{The continuity argument for Neumann boundary conditions}
\label{subsec:Continuity_Neumann}
To apply the continuity argument to the Neumann case, we need to take the two steps outlined in the introduction to Section~\ref{subsec:Continuity}. The second step merely relied on continuity properties of $A$ and the monotonicity properties of $\omega(\epsilon)$ and $\alpha(\epsilon)$ established in Sections~\ref{subsec:Crossing} and \ref{subsec:small_behaviour_Neumann}, respectively; in particular, it did not rely directly on properties of the functional. Hence this part of the argument can be carried out almost verbatim. Therefore, we only need to deal with the first step here.
We make the analogous definitions for $\Aa_{\alpha,r_+}$ and $\Phi$ as in Section~\ref{subsec:Continuity}. For $f=r^{-1/2+\kappa}g$, $x\mapsto g(1/x)\in C_0^{\infty}[0,1/r_+)$, we define
\begin{align*}
\Aa_{\alpha,r_+}(f):=\{a>0\,:\,\tL_{\alpha,r_+,a}(f)<0\}
\end{align*}
and
\begin{align*}
\Aa_{\alpha,r_+}:=&\bigcup_{g\in C_0^{\infty}[0,1/r_+)}\Aa_{\alpha,r_+}(r\mapsto r^{-1/2+\kappa}g(1/r)))\\
=&\{a>0\,:\,\exists (x\mapsto g(1/x))\in C_0^{\infty}[0,1/r_+):\,\tL_{\alpha,r_+,a}(r^{-1/2+\kappa}g)<0\}.
\end{align*}
Moreover, we define
\begin{align*}
\Phi:\,(5/4,9/4)\times(0,\infty)\rightarrow(0,\infty),~\Phi(\alpha,r_+):=\inf\Aa_{\alpha,r_+}
\end{align*}
if $\Aa_{\alpha,r_+}\neq\emptyset$.
Instead of showing monotonicity for $\Phi$, we will define a left-continuous function $\Psi$ that can play the r\^ole of $\Phi$ in the continuity argument. For each $\alpha\in (5/4,9/4)$, there will be a value $\Phi(\alpha,r_+)$ for $a$ such that there is a real mode solution satisfying the Neumann boundary condition for $\alpha$ and this $a$. The function $\Psi(\cdot,r_+)$ will essentially look like $\Phi(\cdot,r_+)$, but will be modified on potential jump points to achieve left-continuity. The arguments of Sections~\ref{subsec:PerturbingN} and~\ref{subsec:small_behaviour_Neumann} (which depend only on the existence of a Neumann mode solution) can be repeated for $\Psi(\alpha,r_+)$ instead of $\Phi(\alpha,r_+)$, thus we can substitue $\Phi$ by $\Psi$ in the remainder of the proof of Section~\ref{subsec:Continuity}.
\begin{lemma}
\label{lemma:Psi_for_Phi}
There is a left-continuous function $\Psi(\cdot,r_+)$ such that there is a real mode solution satisfying the Neumann boundary condition for each $5/4<\alpha<9/4$ and each $a=\Psi(\alpha,r_+)$.
\end{lemma}
To prove this lemma, we need a monotonicity result about the twisted functional. Note that for a fixed $g$ (where $(x\mapsto g(1/x))\in C_0^{\infty}[0,1/r_+))$, the function
\begin{align}
\label{eqn:continuity_functional_Neumann}
(\alpha,a)\mapsto\tL_{\alpha,r_+,a}(r^{-1/2+\kappa}g)
\end{align}
is continuous.
\begin{lemma}
\label{lemma:negative_derivative_twisted_functional}
Let $5/4<\kappa_0<9/4$. Fix all spacetime parameters. Let $u_0:=r^{-1/2+\kappa_0}g_0$ be a solution to the radial ODE at $\kappa_0$. Define $u(r,\kappa):=r^{-1/2+\kappa}g_0$. Then
\begin{align*}
\frac{\del}{\del\kappa}\tL_{\alpha,r_+,a}(u(r,\kappa))\bigg\lvert_{\kappa=\kappa_0}>0.
\end{align*}
\end{lemma}
\begin{proof}
We start from the identity
\begin{align*}
h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u=\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u}{\dd r}\right)-\tilde V_a\frac{r^2+a^2}{\Delta_-}u
\end{align*}
where we always take $h=r^{-1/2+\kappa}$. Set $u_{\kappa}:=\del u/\del\kappa$. Let $f$ be as in (\ref{f_perturbation}) with an $R>r_++1$ to be determined and set $v:=u_{\kappa}-f u$. Then we have
\begin{align*}
&\frac{\del}{\del\kappa}\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\\&~~~~~~~~=\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd u_{\kappa}}{\dd r}\right)-\tilde V_a\frac{r^2+a^2}{\Delta_-}u_\kappa-\frac{\del\tilde V_a}{\del\kappa}\frac{r^2+a^2}{\Delta_-}u
\\&~~~~~~~~=h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u_{\kappa}}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u_{\kappa}-\frac{\del\tilde V_a}{\del\kappa}\frac{r^2+a^2}{\Delta_-}u
\\&~~~~~~~~=h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{v}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}v
-\frac{\del\tilde V_a}{\del\kappa}\frac{r^2+a^2}{\Delta_-}u\\
&~~~~~~~~~~~+h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}fu\right)\right)-\tilde V^h_a\frac{r^2+a^2}{\Delta_-}fu.
\end{align*}
Multiplying by $\overline u$, integrating over $(r_+,\infty)$, integrating by parts as in Section~\ref{subsec:PerturbingN} and evaluating at $\kappa=\kappa_0$ yields:
\begin{align*}
&\int_{r_+}^{\infty}\frac{\del}{\del\kappa}\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\overline u\,\dd r\,\bigg\lvert_{\kappa=\kappa_0}\\
=&2\kappa\int_{r_+}^{\infty}\frac{1}{r^2+a^2}\left(\frac{\del\lambda}{\del\alpha}-\frac{a^2}{\ell^2}\right)|u|^2\,\dd r\,\bigg\lvert_{\kappa=\kappa_0}\\
&-\int_{r_+}^{\infty}\left(\frac{2\kappa}{\ell^2}\frac{r^2}{r^2+a^2}|u|^2-\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd f}{\dd r}\right)\left\lvert h^{-1}u\right\lvert^2\right)\,\dd r\,\bigg\lvert_{\kappa=\kappa_0}\\
&-2\int_{r_+}^{\infty}f\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\,\dd r\,\bigg\lvert_{\kappa=\kappa_0}.
\end{align*}
By repeating the proof of Section~\ref{subsec:PerturbingN}, one shows that the right hand side is negative. For the left hand side, we compute:
\begin{align*}
&\frac{\del}{\del\kappa}\left(\frac{\Delta_-}{r^2+a^2}h^2\left\lvert\frac{\dd}{\dd r}\left(h^{-1}u\right)\right\lvert^2+\tilde V^h_a\frac{r^2+a^2}{\Delta_-}|u|^2\right)\\
=&\frac{\del^2}{\del r\del\kappa}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}u\right)h^{-1}\overline u\right)\\
&~~~-\frac{\del}{\del\kappa}\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)\overline u-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\overline u\right)\\
=&-\frac{\del}{\del\kappa}\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\overline u\\
&~~~-\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\frac{\del\overline u}{\del\alpha}\\
&~~~+\frac{\del^2}{\del r\del\kappa}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}u\right)h^{-1}\overline u\right)
\end{align*}
Again we have
\begin{align*}
\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\,\bigg\lvert_{\kappa=\kappa_0}=0.
\end{align*}
Moreover,
\begin{align*}
\frac{\del}{\del\kappa}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}u\right)h^{-1}\overline u\right)\,\bigg\lvert_{\kappa=\kappa_0}\sim r^{1+2\kappa_0}\frac{\dd}{\dd r}\left(r^{1/2-\kappa_0}u_0\right)\log r.
\end{align*}
Therefore,
\begin{align*}
\frac{\del}{\del\kappa}\tL_{\alpha,r_+,a}(u(r,\kappa))\bigg\lvert_{\kappa=\kappa_0}=-\int_{r_+}^{\infty}\frac{\del}{\del\kappa}\left(h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(\frac{u}{h}\right)\right)-\tilde V_a^h\frac{r^2+a^2}{\Delta_-}u\right)\overline u\,\dd r,
\end{align*}
whence positivity.
\end{proof}
\begin{cor}
\label{cor:local_monotonicity_right}
For all $\alpha\in (5/4,9/4)$, there is a $\delta>0$ such that $\Phi(\cdot,r_+)$ is decreasing in $[\alpha,\alpha+\delta)$
\end{cor}
\begin{proof}
Lemma~\ref{lemma:negative_derivative_twisted_functional} shows that monotonicity is an open property.
\end{proof}
Now we can prove the main lemma.
\begin{proof}[Proof of Lemma~\ref{lemma:Psi_for_Phi}]
Let $\epsilon>0$. By Corollary~\ref{cor:local_monotonicity_right}, there is an indexing set $X$ and disjoint half-open intervals $I_{\beta}$, $\beta\in X$, containing their left endpoints, such that one has $\cup_{\beta\in X}I_{\beta}=[5/4+\epsilon,9/4)$ and $\Phi(\cdot,r_+)\big\lvert_{ I_\beta}$ is decreasing for all $\beta\in X$.
For $\alpha\in\cup_{\beta\in X}\mathring I_{\beta}$, set
\begin{align*}
\Psi(\alpha,r_+):=\Phi(\alpha,r_+).
\end{align*}
Let $\alpha_0\in\del I_{\beta_1}\cap I_{\beta_2}$. Choose a sequence $(\alpha_k)\subseteq I_{\beta_1}$ such that $\alpha_k\rightarrow\alpha_0$. As the sequence $(\Phi(\alpha_k,r_+))$ is monotonically decreasing and bounded below, it is convergent. We set $a_0:=\lim_k\Phi(\alpha_k,r_+)$. Let $f_{\alpha_k}$ be the unique solution to the radial ODE with parameters $\alpha_k$, $a_k=\Phi(\alpha_k,r_+)$ and $\omega_k=\Omega_R(\alpha_k)$ -- see definition (\ref{eqn:defn_omega_+}). Let $f_{\alpha_0}$ be the unique solution corresponding to the parameters $\alpha_0$ and $a_0$. Since all $f_{\alpha_k}$ satisfy the Neumann boundary condition, continuity of the reflection and transmission coefficients yields that $f_{\alpha_0}$ satisfies the Neumann boundary condition as well. We set
\begin{align*}
\Psi(\alpha_0,r_+):=a_0.
\end{align*}
As we can repeat this construction for all $\epsilon>0$ and all jump points $\alpha_0$, we obtain a function $\Psi(\cdot,r_+)$ defined in $(5/4,9/4)$, whose values correspond to parameters $a$ with periodic mode solutions.
Since we have left-continuity at the jump points by construction, it remains to show that $\Psi(\cdot,r_+)$ is left-continuous in $\alpha\in\cup_{\beta\in X}\mathring I_{\beta}$, which can be proved as Lemma~\ref{lemma:CL}: Suppose not.
Then there is an $\epsilon>0$ such that, for all $\delta>0$, there is an $\alpha'<\alpha$ with
\begin{align*}
\alpha-\alpha'<\delta
\end{align*}
and
\begin{align*}
\Psi(\alpha',r_+)-\Psi(\alpha,r_+)\geq \epsilon.
\end{align*}
Then there is an $a$ between $\Psi(\alpha,r_+)$ and $\Psi(\alpha',r_+)$ such that there is an $g$ with $a\in\Aa_{\alpha,r_+}(r^{-1/2+\kappa}g)$,
but, for each $\delta$, there is an $\alpha'$ with $a\notin\Aa_{\alpha',r_+}(r^{-1/2+\kappa}g)$.
Therefore, since $\tL_{\alpha,r_+,a}(r^{-1/2+\kappa}g)<0$ and due to the continuity (\ref{eqn:continuity_functional_Neumann}),
there is a $\delta>0$ such that for all $\alpha-\alpha'<\delta$, we have $\tL_{\alpha',r_+,a}(r^{-1/2+\kappa}g)<0$, i.\,e. $a\in\Aa_{\alpha',r_+}(r^{-1/2+\kappa}g)$, a contradiction.
\end{proof}
\section{Acknowledgements}
I would like to thank my supervisors Mihalis Dafermos and Gustav Holzegel for proposing this project, helpful discussions, guidance and advice on the exposition. I would also like to thank Yakov Shlapentokh-Rothman for useful conversations and advice. I gratefully acknowledge the financial support of EPSRC, the Cambridge Trust and the Studienstiftung des deutschen Volkes.
\begin{appendices}
\section{The angular ODE}
\label{sec:AngularODE}
Assume throughout the section that $m\neq0$. Recall equations (\ref{eqn:AngularODE1}) and (\ref{eqn:AngularODE2}).
We will only give details for $\alpha\leq 0$. The other case can be treated analogously.
Define $x:=\cos\theta$. Then the equation becomes
\begin{align*}
\frac{\dd}{\dd x}\left(\Delta_{\theta}(1-x^2)\frac{\dd S}{\dd x}\right)-&\left(\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{1-x^2}-\left(\frac{\Xi}{\Delta_{\theta}}a^2\omega^2-2ma\omega\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}-\frac{\alpha}{\ell^2}a^2\right)x^2\right)S+\lambda S=0.
\end{align*}
Set
\begin{align*}
K(x):=\frac{\dd}{\dd x}\left(\left(1-\frac{a^2}{\ell^2}x^2\right)\left(1-x^2\right)\right)=4\frac{a^2}{\ell^2}x^3-2x\left(1+\frac{a^2}{\ell^2}\right).
\end{align*}
Using the language of Theorem~\ref{thm:RegularSing}, we see that at $\pm 1$, we have
\begin{align*}
f_0=1,~~~~~~~~~~g_0=-m^2/4.
\end{align*}
Thus for $m\neq 0,1$, we have two zeros which do not differ by an integer. Then we know that solutions are linear combinations of $(x\mp 1)^{-|m|/2}$ and $(x\mp 1)^{|m|/2}$ near $\pm 1$.
\begin{propn}
Suppose that for some fixed $\omega_0,\alpha_0\in\RR$, we have an eigenvalue $\lambda_0$. Then, for $\kappa$ sufficiently close to $\kappa_0$, we can uniquely find a complex analytic function $\lambda(\omega,\alpha)$ of eigenvalues for the angular ODE with parameter $(\omega,\alpha)\in\CC\times\RR$ such that $\lambda_0=\lambda(\omega,\alpha)$.
\end{propn}
\begin{proof}
We can use the proof in \citep{ShlapentokhGrowing}.
If $S$ is an eigenfunction, we clearly must have
\begin{align*}
S\sim (1\mp x)^{|m|/2}
\end{align*}
as $x\rightarrow \pm 1$. For any $\omega,\alpha$ and $\lambda$, we can uniquely define a solution $S(\theta,\omega,\alpha,\lambda)$ by requiring that
\begin{align*}
S(x,\omega,\alpha,\lambda)(1+x)^{-|m|/2}
\end{align*}
is holomorphic at $x=-1$ and
\begin{align*}
\left(S(\cdot,\omega,\alpha,\lambda)(1+\cdot)^{-|m|/2}\right)(x=-1)=1.
\end{align*}
Then we have holomorphic functions $F(\omega,\alpha,\lambda)$ and $G(\omega,\alpha,\lambda)$ such that
\begin{align*}
S(x,\omega,\alpha,\lambda)\sim F(\omega,\alpha,\lambda)(1-x)^{-|m|/2}+G(\omega,\alpha,\lambda)(1-x)^{|m|/2}
\end{align*}
as $x\rightarrow 1$. Since $\lambda_0$ is an eigenvalue, we have $F(\omega_0,\alpha_0,\lambda_0)=0$. We want to appeal to the implicit function theorem and define our function $\lambda(\omega,\alpha)$ uniquely near $(\omega_0,\alpha_0)$. Suppose (for the sake of contradiction) that
\begin{align*}
\frac{\del F}{\del\lambda}(\omega_0,\alpha_0,\lambda_0)=0.
\end{align*}
Set $S_{\lambda}:=\del S/\del\lambda$. Since $\del F/\del\lambda=0$, $S_{\lambda}$ satisfies the boundary conditions of eigen\-functions. Moreover, we have
\begin{align*}
&\frac{\dd}{\dd x}\left(\Delta_{\theta}(1-x^2)\frac{\dd S_{\lambda}}{\dd x}\right)-\left(\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{1-x^2}-\left(\frac{\Xi}{\Delta_{\theta}}a^2\omega_0^2-2ma\omega_0\frac{\Xi}{\Delta_{\theta}}\frac{a^2}{\ell^2}-\frac{\alpha}{\ell^2}a^2\right)x^2\right)S_{\lambda}\\&~~~~~~~~~~~~~~~~~~~+\lambda_0 S=-S.
\end{align*}
Multiplying both sides by $\overline S$, integrating over $(0,\pi)$ with measure $\sin\theta\,\dd\theta$, integrating by parts and using that $\overline S$ satisfies the angular ODE implies
\begin{align*}
\int_0^{\pi}|S|^2\sin\theta\,\dd\theta=0,
\end{align*}
which is a contradiction. The proof for $\alpha<0$ proceeds similarly.
\end{proof}
\begin{propn}
\label{propn:ImOmega}
If $\omega_I>0$, then
\begin{align*}
-\Im(\lambda\overline{\omega})>0.
\end{align*}
\end{propn}
\begin{proof}
Let $\alpha\leq 0$. Multiplying the ODE by $\overline{\omega S}$, integrating by parts and taking imaginary parts gives
\begin{align*}
-\int_0^{\pi}\Im(\lambda\overline{\omega})\sin\theta\,\dd\theta
&=\int_0^{\pi}\omega_I\left(\Delta_{\theta}\left\lvert\frac{\dd S}{\dd\theta}\right\lvert^2+\left[\frac{\Xi^2}{\Delta_{\theta}}\frac{m^2}{\sin^2\theta}-\frac{\alpha}{\ell^2}a^2\cos^2\theta\right]|S|^2\right)\sin\theta\,\dd\theta\\
&~~~~~~~~~~+\int_0^{\pi}\frac{\Xi}{\Delta_{\theta}}\cos^2\theta\,\Im(a^2\omega^2\overline{\omega})|S|^2\sin\theta\,\dd\theta,
\end{align*}
which is positive for $\omega_I>0$. For $\alpha>0$, the proof proceeds almost verbatim.
\end{proof}
\begin{propn}
\label{propn:LambdaDeriv}
When $\omega$ is real, we have
\begin{align*}
\frac{\del\lambda}{\del\alpha}=-\frac{a^2}{\ell^2}\int_0^{\pi}\cos^2\theta|S|^2\sin\theta\,\dd\theta
\end{align*}
for $\alpha\leq 0$ and
\begin{align*}
\frac{\del\lambda}{\del\alpha}=\frac{a^2}{\ell^2}\int_0^{\pi}\sin^2\theta|S|^2\sin\theta\,\dd\theta
\end{align*}
for $\alpha>0$.
\end{propn}
\begin{proof}
Let $S_{\alpha}:=\del S/\del\alpha$. First, let $\alpha\leq 0$.
First one differentiates (\ref{eqn:AngularODE1}) with respect to $\alpha$, then multiplies by $\overline S$ and then integrates by part. Since $\omega\in\RR$, $\overline S$ satisfies the angular ODE, which yields the result. Similarly we obtain the result for $\alpha>0$.
\end{proof}
\section{Twisted derivatives and the modified potential}
\label{sec:twisted_derivative}
To deal with the slow decay or even growth of modes satisfying the Neumann boundary condition, we need to use renormalised derivatives
\begin{align*}
h\frac{\dd}{\dd r}\left(h^{-1}\cdot\right)
\end{align*}
with a sufficiently regular function $h$.
Defining the modified potential
\begin{align*}
\tilde V^h_a:=\tilde V-\frac{1}{h}\frac{\Delta_-}{r^2+a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd h}{\dd r}\right),
\end{align*}
we obtain a twisted expression for the radial ODE:
\begin{lemma}
\label{lemma:EL_twisted_g}
For all $f\in C^1$ that are piecewise $C^2$,
\begin{align*}
h^{-1}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}h^2\frac{\dd}{\dd r}\left(h^{-1}f\right)\right)-\tilde V^h_a\frac{r^2+a^2}{\Delta_-}f=\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd f}{\dd r}\right)-\tilde V_a\frac{r^2+a^2}{\Delta_-}f.
\end{align*}
\end{lemma}
By virtue of twisting, the modified potential can be chosen to be positive for large $r$:
\begin{lemma}
\label{lemma:twisting_positive_g}
Let $h:=r^{-1/2+\kappa}$.
If $|m|$ is sufficiently large, then there is an $R>r_+$ such that $\tilde V^h_a>0$ for $r>R$. The choice of $R$ is independent of $a$ and $\alpha$. Moreover $\tilde V_a^h=\OO(1)$ as $r\rightarrow\infty$.
\end{lemma}
\begin{proof}
We look at the asymptotic behaviour of the different parts of $\tilde V_a$:
\begin{align*}
V_0-\omega^2&\sim\frac{1}{\ell^2}\left(\lambda+a^2\omega^2-2ma\omega\Xi\right)>\frac{1}{\ell^2}m^2\Xi^2>0\\
V_+&=\frac{2\Delta_-}{(r^2+a^2)^2}\frac{r^2}{\ell^2}+\frac{\Delta_-}{(r^2+a^2)^4}\left(a^4\Delta_-+(r^2-a^2)2Mr\right)\\&\sim\frac{2\Delta_-}{(r^2+a^2)^2}\frac{r^2}{\ell^2}+\frac{a^4\Delta_-^2}{(r^2+a^2)^4}
\end{align*}
One easily computes that
\begin{align*}
v(r):=\frac{2-\alpha}{\ell^2}\frac{\Delta_-}{(r^2+a^2)^2}r^2-h^{-1}\frac{\Delta_-}{r^2+a^2}\frac{\dd}{\dd r}\left(\frac{\Delta_-}{r^2+a^2}\frac{\dd h}{\dd r}\right)=\OO(1),
\end{align*}
which yields the result.
\end{proof}
\section{The twisted Euler-Lagrange equation}
\label{sec:Twisted_Euler_Lagrange}
We give here the derivation of the weak twisted Euler-Lagrange equation.
\begin{proof}[Proof of Lemma~\ref{lemma:ELDN}]
The following proof can be extracted from \citep{Evans}. We give the extension to twisted derivatives here for the sake of completeness.
The minimiser $f_{a}$ is a minimiser of the functional
\begin{align*}
\tL_{a}(f):=\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\left\lvert\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f\right)\right\lvert^2+\Vmod_a\frac{r^2+a^2}{\Delta_-}|f|^2\right)\,\dd r
\end{align*}
under the constraint
\begin{align*}
\mathcal{J}(f)=0,
\end{align*}
where
\begin{align*}
\JJ(f)=\int_{r_+}^{\infty} G(r,f)\,\dd r,~~~~G(r,f)=\frac{1}{r^2}\left(|f|^2-r_+\right).
\end{align*}
Moreover, define $g(r,f):={2f}/{r^2}$.
Fix $\psi_1\in \uH_{\kappa}^{1}(r_+,\infty)$. We assume in a first step that $g(r,f_a)$ is not identically zero almost everywhere on $(r_+,\infty)$. Then we can find a $\psi_2\in \uH_{\kappa}^{1}(r_+,\infty)$ such that
\begin{align*}
\int_{r_+}^{\infty}g(r,f_{a})\psi_2(r)\,\dd r\neq 0.
\end{align*}
Define $j(\tau,\sigma):=\JJ(f_{a}+\tau\psi_1+\sigma\psi_2)$ for $\tau,\sigma\in\RR$. Clearly, $j(0,0)-0$. Since $\frac{\del g(r,f_a+\tau\psi_1+\sigma\psi_2)}{\del\tau}\psi_1$ and $\frac{\del g(r,f_a+\tau\psi_1+\sigma\psi_2)}{\del\tau}\psi_2$ are integrable on $(r_+,\infty)$, $j$ is in $C^1$. In particular, we have
\begin{align*}
\frac{\del j}{\del\sigma}(0,0)=\int_{r_+}^{\infty}g(r,f_{a})\psi_2(r)\,\dd r\neq 0.
\end{align*}
By the Implicit Function Theorem, there is a $\kappa:\,\RR\rightarrow\RR$ such that $\kappa(0)=0$ and
\begin{align*}
j(\tau,\kappa(\tau))=0.
\end{align*}
In other words, the function $f_{a}+\chi(\tau)$, where
\begin{align}
\label{eqn:implicit_curve}
\chi(\tau):=\tau\psi_1+\kappa(\tau)\psi_2,
\end{align}
satisfies the integral constraint. Thus, setting $i(\tau):=\tL_a(f_{a}+\chi(\tau))$,
we obtain $i'(0)=0$. Note here that $i$ is differentiable in $\tau$ since $f_{a}\in \uH_{\kappa}^1(r_+,\infty)$. We have
\begin{align*}
\frac{\dd i}{\dd\tau}\bigg\lvert_{\tau=0}&=2\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{a}\right)\left(\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}\psi_1\right)+\kappa'(0)\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}\psi_2\right)\right)\\&~~~~~~+\Vmod_a\frac{r^2+a^2}{\Delta_-}f_{a}(\psi_1+\kappa'(0)\psi_2)\bigg)\,\dd r.
\end{align*}
From (\ref{eqn:implicit_curve}), we deduce
\begin{align*}
\kappa'(0)=-\frac{\int_{r_+}^{\infty}g(r,f_{a})\psi_1\,\dd r}{\int_{r_+}^{\infty}g(r,f_{a})\psi_2\,\dd r}.
\end{align*}
Setting
\begin{align*}
\lambda:=2\frac{\int_{r_+}^{\infty}\left(\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{a}\right)\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}\psi_2\right)+\Vmod_a\frac{r^2+a^2}{\Delta_-}f_{a}\psi_2\right)\,\dd r}{\int_{r_+}^{\infty}g(r,f_{a})\psi_2\,\dd r}
\end{align*}
yields that
\begin{align*}
\begin{split}
\int_{r_+}^{\infty}\bigg(\frac{\Delta_-}{r^2+a^2}r^{-1+2\kappa}\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}f_{a}\right)\frac{\dd}{\dd r}\left(r^{\frac{1}{2}-\kappa}\psi_1\right)+\Vmod_a\frac{r^2+a^2}{\Delta_-}f_{a}\psi_1\bigg)\,\dd r=\lambda\int_{r_+}^{\infty}\frac{f_a}{r^2}\psi\,\dd r
\end{split}
\end{align*}
for all $\psi\in \uH_{\kappa}^1(r_+,\infty)$. We have $f_a\in \uH_{\kappa}^1(r_+,\infty)$, whence $\lambda=-\nu_a$.
It remains to deal with the case $g(r,f_a)=0$ a.\,e. This, however, would yield that $f=0$ in contradiction to the norm constraint.
\end{proof}
\end{appendices}
\bibliographystyle{alphadin}
\addcontentsline{toc}{section}{References}
{\footnotesize
\bibliography{literatureSuperradiance}}
\end{document} | 53,659 |
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\begin{document}
\title{On Solving the Cauchy Problem with Propagators}
\author{Henrik Stenlund\thanks{The author is grateful to Visilab Signal Technologies for supporting this work.}
\\Visilab Signal Technologies Oy, Finland}
\date{November 5, 2014}
\maketitle
\begin{abstract}
The abstract first order Cauchy problem is solved in terms of Taylor's series leading to a series of operators which is a propagator. It is found that higher order Cauchy problems can be solved in the same way. Since derivatives of order lower than the Cauchy problem are built-in to the problem, it suffices to solve the higher order derivatives in terms of the lower order ones, in a simple way. \footnote{Visilab Report \#2014-11}
\subsection{Keywords}
Cauchy problem, differential equations, initial value problems
\subsection{Mathematical Classification}
Mathematics Subject Classification 2010: 47D09, 34A12, 35E15, 35F10, 35F40, 35G10, 35G40
\end{abstract}
\subsection{}
Dedicated to my wife Anna-Maija.
\tableofcontents
\section{Introduction}
The first order Cauchy problem (CP) is the following differential equation, treated as an initial value problem. This is a general abstract case with no particular application in mind with ($t \in{R^+},x \in{R}$).
\begin{equation}
\frac{\partial{u(x,t)}}{\partial{t}}=L{u(x,t)} \label{eqn10}
\end{equation}
Here $L$ is a linear operator and independent of $t$ and $\frac{\partial}{\partial{t}}$. It is a function of $x$ and the differential operator $\frac{\partial}{\partial{x}}$ but can adopt many forms depending on the particular application.
\begin{equation}
L=L(x,\frac{\partial}{\partial{x}}) \label{eqn20}
\end{equation}
The initial value of the problem is
\begin{equation}
u(x,t_0)=u_0(x).
\end{equation}
The exponential function of $L$ is defined as a Taylor's series with $a \in {C}$, a parameter
\begin{equation}
e^{aL}=\sum^{\infty}_{n=0}\frac{a^n{L^n}}{n!} \label{eqn30}
\end{equation}
and $L$ commutes with it. The traditional way of solving the (CP) formally is to integrate equation (\ref{eqn10}), see e.g. \cite{Pechukas1966}, leading to
\begin{equation}
u(x,t)=e^{(t-t_o)L}{u_0}, \ \ \ t > t_0 \label{eqn50}
\end{equation}
It is not difficult to prove that this is a unique solution to the (CP). Many functions satisfy equation (\ref{eqn10}). For example, those do, formed from a solution $u$
\begin{equation}
\frac{\partial{u(x,t)}}{\partial{t}} \label{eqn60}
\end{equation}
and
\begin{equation}
h(L){u(x,t)} \label{eqn70}
\end{equation}
where $h()$ can be almost any function of $L$. Those functions do not, however, satisfy the initial value condition and therefore are not solutions to the (CP).
The (CP) can be treated in multiple dimensions and complex variables too. The (CP) is studied widely and the literature is extensive. The methodologies generally applied are various semigroup methods, regularization methods and abstract distribution methods. The Laplace transform can be applied successfully in many cases.
The motivation for this paper is to find a solution method for arbitrary order (CP) for solving physical problems. In the following chapter we present the first order (CP) solved with the new approach. Then we apply the method to the second order, to the third, fourth and arbitrary orders. Chapter 3 shows a simple example. We are working in an intuitive manner and are not stepping into rigorous formalism. We keep the presentation simple.
\section{Propagator Solutions}
\subsection{The First Order Cauchy Problem}
We assume well-posedness for the (CP) but do not intend to handle that issue since it has already been treated thoroughly, for example \cite{Pechukas1966}, \cite{Fattorini1983}, \cite{Melnikova2001} and \cite{Xiao1998} and references therein. We presume that the (CP) solution $u(x,t)$ can be expanded as a Taylor's power series, ($t,x \in{R}$) as
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\frac{(t-t_0)^{n}}{n!}[\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}]_{t=t_0}} \label{eqn1000}
\end{equation}
We could equally work in ($t,x \in{C}$) having the requirement of analyticity for the solution. At the end we could flatten our solution to ($t,x \in{R}$). For real variables the requirements are within the interval in question:
\itshape
- $u(x,t)$ is continuous as a function of $t, t > t_0$
- the series converges
- all the derivatives exist for any $n=1,2,3..$:
\begin{equation}
\left|\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}\right| < \infty \nonumber
\end{equation}
\normalfont
The initial value function is
\begin{equation}
u(x,t_0)=u_0(x).
\end{equation}
In order to solve the (CP) we can integrate equation (\ref{eqn10}) and place the expression (\ref{eqn1000}) to it, obtaining
\begin{equation}
u(x,t)=u_0+L\int^{t}_{t_0}{ds\cdot{u(x,s)}} \label{eqn1050}
\end{equation}
\begin{equation}
=u_0+L\sum^{\infty}_{n=0}{\frac{1}{n!}[\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}]_{t=t_0}}\int^{t}_{t_0}{ds(s-t_0)^{n}} \label{eqn1070}
\end{equation}
\begin{equation}
=u_0+L\sum^{\infty}_{n=0}{\frac{(t-t_0)^{n+1}}{(n+1)!}[\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}]_{t=t_0}} \label{eqn1090}
\end{equation}
Next we need to solve for all the derivatives of the function $u(x,t)$. In effect, we are able to do it by repeatedly differentiating equation (\ref{eqn10}) getting
\begin{equation}
\frac{\partial^n{u(x,t)}}{\partial{t^n}}=L^n{u(x,t)} \label{eqn1110}
\end{equation}
and thus we get after substitution
\begin{equation}
u(x,t)=u_0+\sum^{\infty}_{n=0}{\frac{L^{n+1}(t-t_0)^{n+1}}{(n+1)!}u_0}=e^{(t-t_0)L}u_0, \ \ \ t > t_0 \label{eqn1140}
\end{equation}
This is the traditional result. The operator $e^{(t-t_0)L}$ has become a propagator for this particular (CP) with $L$.
\subsection{The Second Order Cauchy Problem}
The second order (CP) is the following initial value problem
\begin{equation}
\frac{\partial^2{u(x,t)}}{\partial{t^2}}=M{u(x,t)} \label{eqn1210}
\end{equation}
with
\begin{equation}
u(x,t_0)=u_0(x) \ \ \ \label{eqn1214}
\end{equation}
and
\begin{equation}
[\frac{\partial{u(x,t)}}{\partial{t}}]_{t=t_0}=u_1(x) \ \ \ \label{eqn1216}
\end{equation}
being the initial values. $M$ is a linear operator and not a function of $t$ nor $\frac{\partial}{\partial{t}}$.
\begin{equation}
M=M(x,\frac{\partial}{\partial{x}}) \label{eqn1220}
\end{equation}
We attempt to generalize the procedure of the preceding section. We generate the derivatives in the same way as above, obtaining
\begin{equation}
\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}= \nonumber
\end{equation}
\begin{equation}
{M^{\frac{n}{2}}u(x,t), n \ even} \nonumber
\end{equation}
\begin{equation}
{M^{\frac{n-1}{2}}\frac{\partial{u(x,t)}}{\partial{t}}, n \ odd} \ \ \ \label{eqn1240}
\end{equation}
By applying the series in equation (\ref{eqn1000}) and integrating twice we get
\begin{equation}
u(x,t)=u_0+(t-t_0)u_1+M\sum^{\infty}_{n=0}{\frac{(t-t_0)^{n+2}}{(n+2)!}[\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}]_{t=t_0}} \label{eqn1250}
\end{equation}
By placing the derivatives and after changing the indices, we have
\newpage
\begin{equation}
u(x,t)=u_0+(t-t_0)u_1+\sum^{\infty}_{n=1}{{(\sqrt{M})}^{2n}\frac{(t-t_0)^{2n}}{(2n)!}u_0} \nonumber
\end{equation}
\begin{equation}
+\frac{1}{\sqrt{M}}\sum^{\infty}_{n=1}{{(\sqrt{M})}^{2n+1}\frac{(t-t_0)^{2n+1}}{(2n+1)!}u_1} \label{eqn1260}
\end{equation}
We condense the expressions to get
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\frac{{(\sqrt{M})}^{2n}(t-t_0)^{2n}}{(2n)!}u_0}+\frac{1}{\sqrt{M}}\sum^{\infty}_{n=0}{\frac{{(\sqrt{M})}^{2n+1}(t-t_0)^{2n+1}}{(2n+1)!}u_1}, \ \ \ t > t_0 \label{eqn1270}
\end{equation}
The series are identified as $sinh()$ and $cosh()$ and thus we have an optional representation for the result
\begin{equation}
u(x,t)=cosh[(t-t_0)\sqrt{M}]u_0+\frac{sinh[(t-t_0)\sqrt{M}]}{\sqrt{M}}u_1, \ \ \ t > t_0 \label{eqn1290}
\end{equation}
We have obtained something which is reminiscent of the integrated sine and cosine functions $S(x)$ and $C(x)$, \cite{Kisynski1972}, \cite{Arendt1989}, \cite{Zheng1996}.
\subsection{The Third Order Cauchy Problem}
Continuing the generalization process, the third order (CP) is the following initial value problem
\begin{equation}
\frac{\partial^3{u(x,t)}}{\partial{t^3}}=P{u(x,t)} , \label{eqn1410}
\end{equation}
\begin{equation}
u(x,t_0)=u_0
\end{equation}
\begin{equation}
[\frac{\partial{u(x,t)}}{\partial{t}}]_{t=t_0}=u_1
\end{equation}
\begin{equation}
[\frac{\partial^2{u(x,t)}}{\partial{t^2}}]_{t=t_0}=u_2
\end{equation}
as the initial values.
$P$ is a linear operator independent of $t$ and $\frac{\partial}{\partial{t}}$.
\begin{equation}
P=P(x,\frac{\partial}{\partial{x}}) \label{eqn1420}
\end{equation}
The derivatives are
\begin{equation}
\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}= \nonumber
\end{equation}
\begin{equation}
{P^{\frac{n}{3}}u(x,t), \ \ \ n=0,3,6,9...} \nonumber
\end{equation}
\begin{equation}
{P^{\frac{n-1}{3}}\frac{\partial{u(x,t)}}{\partial{t}}, \ \ \ n=1,4,7,10...} \nonumber
\end{equation}
\begin{equation}
{P^{\frac{n-2}{3}}\frac{\partial{^{2}u(x,t)}}{\partial{t^2}}, \ \ \ n=2,5,8,11...} \label{eqn1440}
\end{equation}
By applying the series in equation (\ref{eqn1000}) we get
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[3]{P})^{3n}}{(3n)!}}u_0 \nonumber
\end{equation}
\begin{equation}
+\frac{1}{P^{\frac{1}{3}}}\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[3]{P})^{3n+1}}{(3n+1)!}}u_1 \nonumber
\end{equation}
\begin{equation}
+\frac{1}{P^{\frac{2}{3}}}\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[3]{P})^{3n+2}}{(3n+2)!}}u_2, \ \ \ t > t_0 \label{eqn1460}
\end{equation}
\subsection{The Fourth Order Cauchy Problem}
Going further, the fourth order (CP) is the following initial value problem.
\begin{equation}
\frac{\partial^4{u(x,t)}}{\partial{t^4}}=K{u(x,t)}, \label{eqn1510}
\end{equation}
\begin{equation}
u(x,t_0)=u_0
\end{equation}
\begin{equation}
[\frac{\partial{u(x,t)}}{\partial{t}}]_{t=t_0}=u_1
\end{equation}
\begin{equation}
[\frac{\partial^2{u(x,t)}}{\partial{t^2}}]_{t=t_0}=u_2
\end{equation}
\begin{equation}
[\frac{\partial^3{u(x,t)}}{\partial{t^3}}]_{t=t_0}=u_3
\end{equation}
are the initial values. $K$ is a linear operator independent of $t$ and $\frac{\partial}{\partial{t}}$.
\begin{equation}
K=K(x,\frac{\partial}{\partial{x}}) \label{eqn1520}
\end{equation}
The derivatives are
\begin{equation}
\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}= \nonumber
\end{equation}
\begin{equation}
{K^{\frac{n}{4}}u(x,t), \ \ \ n=0,4,8,12...} \nonumber
\end{equation}
\begin{equation}
{K^{\frac{n-1}{4}}\frac{\partial{u(x,t)}}{\partial{t}}, \ \ \ n=1,5,9,13...} \nonumber
\end{equation}
\begin{equation}
{K^{\frac{n-2}{4}}\frac{\partial{^{2}u(x,t)}}{\partial{t^2}}, \ \ \ n=2,6,10,12...} \nonumber
\end{equation}
\begin{equation}
{K^{\frac{n-3}{4}}\frac{\partial{^{3}u(x,t)}}{\partial{t^3}}, \ \ \ n=3,7,11,15...} \label{eqn1540}
\end{equation}
By using the procedure shown we get
\newpage
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[4]{K})^{4n}}{(4n)!}}u_0 \nonumber
\end{equation}
\begin{equation}
+\frac{1}{K^{\frac{1}{4}}}\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[4]{K})^{4n+1}}{(4n+1)!}}u_1 \nonumber
\end{equation}
\begin{equation}
+\frac{1}{K^{\frac{2}{4}}}\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[4]{K})^{4n+2}}{(4n+2)!}}u_2 \nonumber
\end{equation}
\begin{equation}
+\frac{1}{K^{\frac{3}{4}}}\sum^{\infty}_{n=0}{\frac{((t-t_0)\sqrt[4]{K})^{4n+3}}{(4n+3)!}}u_3, \ \ \ t > t_0 \label{eqn1560}
\end{equation}
\subsection{The $N$th Order Cauchy Problem}
We are able to solve the abstract $N$th order (CP) below with the same assumptions as above.
\begin{equation}
\frac{\partial^N{u(x,t)}}{\partial{t^N}}=G{u(x,t)} \label{eqn2510}
\end{equation}
We mark the initial value functions as follows ($i=0,1,2...N-1$)
\begin{equation}
[\frac{\partial^i{u(x,t)}}{\partial{t^i}}]_{t=t_0}=u_i(x) \label{eqn2500}
\end{equation}
$G$ is a linear operator independent of $t$ and $\frac{\partial}{\partial{t}}$
\begin{equation}
G=G(x,\frac{\partial}{\partial{x}}) \label{eqn2520}
\end{equation}
The series is
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\frac{(t-t_0)^{n}}{n!}[\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}]_{t=t_0}} \label{eqn2525}
\end{equation}
Working as before we get the derivatives
\begin{equation}
\frac{\partial{^{n}u(x,t)}}{\partial{t^n}}= \nonumber
\end{equation}
\begin{equation}
{G^{\frac{n}{N}}u(x,t), \ \ \ n=0,N,2N,3N...} \nonumber
\end{equation}
\begin{equation}
{G^{\frac{n-1}{N}}\frac{\partial{u(x,t)}}{\partial{t}}, \ \ \ n=1,N+1,2N+1,3N+1...} \nonumber
\end{equation}
\begin{equation}
{G^{\frac{n-2}{N}}\frac{\partial{^{2}u(x,t)}}{\partial{t^2}}, \ \ \ n=2,N+2,2N+2,3N+2...} \nonumber
\end{equation}
...
\begin{equation}
{G^{\frac{n-N+1}{N}}\frac{\partial{^{N-1}u(x,t)}}{\partial{t^{N-1}}}, \ \ \ n=N-1,N+N-1,2N+N-1...} \label{eqn2540}
\end{equation}
The solution will be
\begin{equation}
u(x,t)=\sum^{\infty}_{n=0}{\sum^{N-1}_{j=0}{\frac{((t-t_0)G^{\frac{1}{N}})^{Nn+j}G^{\frac{-j}{N}}}{(Nn+j)!}}}u_j(x) \label{eqn2560}
\end{equation}
or
\begin{equation}
u(x,t)=\sum^{N-1}_{j=0}{(t-t_0)^j\sum^{\infty}_{n=0}{\frac{((t-t_0)^{N}G)^n}{(Nn+j)!}}}u_j(x), \ \ \ t > t_0 \label{eqn2580}
\end{equation}
Since the problem itself provides the derivatives of order ($0...N-1$) at $t_0$, it suffices to solve any higher order ($n>N-1$) derivatives in terms of lower order derivatives, equation (\ref{eqn2540}). The equation (\ref{eqn2580}) above shows that the fractional power operators disappear in the arbitrary order too and we get integer powers. The author believes this approach is new, working just as well for multidimensional problems.
\section{An Example}
\subsection{The Classic Wave Equation}
One can show that the following exponential propagator relation holds for $f(x)$ if it can be expanded as a Taylor's power series with $x,a \in {C}$.
\begin{equation}
e^{a\frac{\partial}{\partial{x}}}f(x)=f(x+a) \label{eqn2010}
\end{equation}
This propagator is a universal translation operator. As an application of equation (\ref{eqn1290}) we solve a simple second-order wave equation-like initial value problem, as in equation (\ref{eqn1210}) with
\begin{equation}
M=v^2{\frac{\partial^2}{\partial{x^2}}} \label{eqn2030}
\end{equation}
The initial values are
\begin{equation}
u(x,t_0)=f(x) \label{eqn2032}
\end{equation}
\begin{equation}
[\frac{\partial}{\partial{x}}u(x,t)]_{t=t_0}=g(x) \label{eqn2034}
\end{equation}
We use the formal solution and place the items and change the hyperbolic functions to exponential functions, getting
\begin{equation}
u(x,t)=\frac{1}{2}(e^{v(t-t_0){\frac{\partial}{\partial{x}}}}+e^{{-v(t-t_0){\frac{\partial}{\partial{x}}}}})f(x)+\frac{1}{2v\frac{\partial}{\partial{x}}}(e^{v(t-t_0){\frac{\partial}{\partial{x}}}}-e^{-v(t-t_0){\frac{\partial}{\partial{x}}}})g(x) \label{eqn2040}
\end{equation}
We can then apply equation (\ref{eqn2010}) and obtain
\begin{equation}
u(x,t)=\frac{1}{2}[f(x+v(t-t_0))+f(x-v(t-t_0))] \nonumber
\end{equation}
\begin{equation}
+\frac{1}{2v\frac{\partial}{\partial{x}}}[g(x+v(t-t_0))-g(x-v(t-t_0))] \label{eqn2060}
\end{equation}
The operator
\begin{equation}
\frac{1}{\frac{\partial}{\partial{x}}} \label{eqn2070}
\end{equation}
is an integration operator and the result will be
\begin{equation}
u(x,t)=\frac{1}{2}[f(x+v(t-t_0))+f(x-v(t-t_0))] \nonumber
\end{equation}
\begin{equation}
+\frac{1}{2v}\int_{x_0}^{x}{dx[g(x+v(t-t_0))-g(x-v(t-t_0))]}, \ \ \ t > t_0 \label{eqn2090}
\end{equation}
This is the d'Alembert's formula. It has two waves traveling to opposite directions. The interesting second part of it comes from the assumption of nonzero time derivative of the function.
\section{Discussion}
We have presented a new approach for solving the abstract first order Cauchy problem (CP) producing the familiar result (\ref{eqn1140}). It is based on expanding the solution as a Taylor's time power series. The series becomes an operator series being a propagator for time development turned to the initial value function.
The procedure extends itself naturally to higher orders, (\ref{eqn1290}), (\ref{eqn1460}) and (\ref{eqn1560}) and to arbitrary order (\ref{eqn2580}). The series become operator series and are propagators for the initial value functions. Identifying the propagator series as elementary functions is possible in some cases, becoming more difficult at higher orders. If the operator has any fractional derivatives, it may still have a formal solution according to this method. However, not all fractional (CP)'s have a solution, see \cite{Bazhlekova1998} for a treatment.
In an arbitrary order $N$ of (CP) the approach is based on solving the derivatives required in the Taylor's series in terms of lower order ($0...N-1$) derivatives in a simple cyclic pattern, equation (\ref{eqn2540}). The fractional operators in the derivatives disappear from the final expressions. The lower order derivatives are provided by the (CP) itself. The main result of this work is equation (\ref{eqn2580}). | 38,030 |
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In saying the rosary, it is hard to keep from distractions. What if you don't pay attention all the way through?
The rosary is mainly a mental prayer. While saying the Hail Mary's, like a chant or mantra in the background, we are supposed to be meditating and concentrating on the mysteries which are announced for each decade. As in all our prayers, we should struggle against deliberate distractions and try to focus our thoughts upon our communication with God. Involuntary distractions are often unavoidable, however, given the psychological make-up of the human mind. In the rosary, it would be impossible to concentrate on the words of each of the fifty Hail Mary's. We should not even try to do that. In our efforts to place ourselves and our thoughts in the presence of God when we pray, we should also depend on God's grace. God, after all, rewards effort in this area more than success. Read the Catechism of the Catholic Church, number 2729.
Reprinted from July 30, 1999
How can a Catholic say a novena without lapsing into superstition? Are there any guidelines for this? How can, we explain novenas to non-Catholics?
A novena is any series of prayers purposely offered in a series of nine - for nine consecutive weeks, etc. Novenas are always private devotions, that is, they are not part of the official liturgy of the Church, even though sometimes they are done publicly, led by a priest, etc. The number nine is always used because of the nine days that Mary and the apostles waited in prayer for the coming of the Holy Spirit (Acts 1:14; 2:1). Usually novenas are prayed for a particular intention, or in honor of the Person of Christ or in veneration of some saint. The best way to avoid superstition in their use is to use only novenas approved by Church authorities (i.e., a bishop) and always pray with total resignation to God's holy will, never claiming to be able to "manipulate God," which would be blasphemy. You can tell non-Catholics what is said here, but remember: "To those who believe, no explanation is necessary and to those who do not, no explanation is possible."
Reprinted from April 2, 1999
Why is receiving Holy Communion on First Fridays important?
In some private revelations to St. Margaret Mary Alacoque in the 17th century our Lord promised that those who would receive Holy Communion in honor of His Sacred Heart on nine consecutive First Fridays would not die without receiving all the help necessary for their eternal salvation, Devotion to the Sacred Heart of Jesus, the grand symbol of His infinite love, tenderness and mercy, is widely practiced in the Catholic Church. The Solemnity of the Sacred Heart in the United States is celebrated on the Friday after "Corpus Christi," the Solemnity of the Body and Blood of Christ (two weeks after Pentecost Sunday).
Reprinted from May 7, 1999
Why do Protestants add the words "for thine is the kingdom, etc." to the Lord's Prayer?
That doxology which is sometimes added to the Lord's Prayer is very ancient, but it is not part of that prayer itself, as contained in the Bible (Matthew 6:9-13; Luke 11:2-4). It was first found in the important Christian document called the "Didache," which dates probably from about the year A.D. 90. Protestantism was only in vented in the 16th century, so the doxology is actually a Catholic prayer. For more information about this matter consult the Catechism of the Catholic Church, number 2760.
Reprinted from May 14, 1999
Why do some prayers use such expressions as "thee" and "thou"?
In the English language the second person singular pronoun has become an archaic usage. The second person plural "you" is now used generally for both the singular and the plural. Prayers are usually the last group of writings and sayings to change when language changes. Also, sometimes the older expressions, consecrated by generations of use, lend a certain dignity and elegance to the expression, which it is desirable to maintain. This is also the case with some verbs and the "st" and "th" form, e.g. "givest," "sayeth," etc.
Reprinted from March 26, 1999 January 22, 1999
At a funeral I heard a priest say we should pray for the deceased for the remission of his temporal punishment. What is temporal punishment?
Temporal punishment is in contrast to eternal punishment. Even after the sins are forgiven, the damage which they do to the soul of the sinner, to the whole Church and other people remains. This must be repaired and restitution made. If this is not done in one's lifetime, it must be done in purgatory. The souls in purgatory can be helped by our prayers and their temporal punishment thereby remitted, mitigated and shortened. Read about this in the Catechism of the Catholic Church, numbers 1030-1032 and 1472-1479.
Reprinted from December 24, 1998
What is the temptation" that we, in the Lord's Prayer, ask God not to lead us into?
The English phrase is difficult to understand apart from the Greek original here. It is a request to God not to allow us, by His grace, to enter into sinful temptation and not to permit us to yield to temptation. It also is a wish that God not subject us to a certain kind of trial. This petition in the "Our Father" is treated in the Catechism of the Catholic Church, numbers 2846-2849. I suggest you study it there.
Reprinted from September 18, 1998 | 72,828 |
\begin{document}
\title{Symmetry breaking for toral actions in simple mechanical
systems}
\author{Petre Birtea, Mircea Puta, Tudor S. Ratiu, R\u azvan
Tudoran}
\date{November 3, 2003}
\maketitle
\begin{abstract}
For simple mechanical systems, bifurcating branches of relative
equilibria with trivial symmetry from a given set of relative
equilibria with toral symmetry are found. Lyapunov stability
conditions along these branches are given.
\end{abstract}
\tableofcontents
\section{Introduction}
This paper investigates the problem of symmetry breaking in the
context of simple mechanical systems with compact symmetry Lie
group $G$. Let $\mathbb{T}$ be a maximal torus of $G$ whose Lie
algebra is denoted by $\mathfrak{t}$. Denote by $Q $ the
configuration space of the mechanical system. Assume that every
infinitesimal generator defined by an element of
$\mathfrak{t}$ evaluated at a symmetric configuration $q_e \in Q$
whose symmetry subgroup $G_{q_e}$ lies in $\mathbb{T}$ is a relative
equilibrium. The goal of this paper is to give sufficient conditions
capable to insure the existence of points in this set from which
branches of relative equilibria with trivial symmetry will emerge.
Sufficient Lyapunov stability conditions along these branches
will be given if $G = \mathbb{T}$.
The strategy of the method can be roughly described
as follows. Denote by $\mathfrak{t} \cdot q_e$ the set of relative
equilibria described above. Take a regular element
$\mu\in \mathfrak{g}^{\ast }$ which happens to be the momentum
value of some relative equilibrium in $\mathfrak{t}\cdot q_{e}$.
Choose a one parameter perturbation $\beta(\tau, \mu) \in
\mathfrak{g}^\ast$ of $\mu$ that lies in
the set of regular points of $\mathfrak{g}^{\ast }$, for
small values of the parameter $\tau>0 $. Consider
the $G_{q_{e}}$-representation on the tangent space
$T_{q_{e}}Q$. Let $v_{q_e}$ be an element in the $\{e\}$-stratum
of the representation and also in the normal space to the tangent space at $q_{e}$ to the orbit
$G\cdot q_{e}$. Assume that its norm is small enough in order for
$v_{q_e}$ to lie in the open ball centered at the origin $0_{q_e}
\in T_{q_e}Q $ where the Riemannian exponential is a
diffeomorphism. The curve $\tau v_{q_e}$ projects by the
exponential map to a curve $q_e( \tau) $ in a neighborhood of
$q_e $ in $Q$ whose value at $\tau= 0$ is $q_e$. Note that the
isotropy subgroup at every point on this curve, except for $\tau=
0 $, is trivial. We shall search for relative equilibria in $TQ$
starting at points of
$\mathfrak{t}\cdot q_e $ such that their base curve in $Q $
equals $q_e(\tau)$ and their momentum values are $\beta(\tau,
\mu) $. To do this, we shall choose a curve $\xi(\tau,
v_{q_e}, \mu) \in \mathfrak{g} $ uniquely determined by $\beta(
\tau, \mu)$; as will be explained in the course of the
construction, $\xi(\tau, v_{q_e}, \mu)$ equals the value of
the inverse of the locked inertial tensor on $\beta(\tau, \mu)$
for $\tau \neq 0 $. If one can show that the limit of $\xi( \tau,
v_{q_e}, \mu) $ exists and belongs to $\mathfrak{t}$ for $\tau
\rightarrow 0$, then the infinitesimal generator of this value
evaluated at $q_e$ is automatically a relative equilibrium since
it belongs to $\mathfrak{t}\cdot q_e$. It will be also shown that
the infinitesimal generators of $\xi(\tau, v_{q_e}, \mu)$
evaluated at $q_e(\tau)$ are relative equilibria. This produces a
branch of relative equilibria starting at this specific point in
$\mathfrak{t}\cdot q_e$ which has trivial isotropy for $\tau>0 $
and which depends smoothly on the additional parameter $\mu \in
\mathfrak{g}^\ast$.
In this method, there are two key technical problems, namely, the
existence of the limit of $\xi( \tau,v_{q_e}, \mu)$ as $\tau
\rightarrow 0$ and the extension of the amended potential at
points with symmetry. The existence of the limit of
$\xi( \tau,v_{q_e}, \mu)$ as $\tau \rightarrow 0$ will be
shown using the Lyapunov-Schmidt procedure.
To extend the amended potential and its derivative at points with
symmetry, two auxiliary functions obtained by blow-up will
be introduced. The analysis breaks up in two problems on a space
orthogonal to the $G$-orbit.
The present paper can be regarded as a sequel to the
work of Hern\'andez and Marsden \cite{hm}. The main difference
is that one single hypothesis from
\cite{hm} has been retained, namely that all points of
$\mathfrak{t} \cdot q_e $ are relative equilibria. We have also
eliminated a strong nondegeneracy assumption in \cite{hm}. But
the general principles of the strategy of the proof having to do
with a regularization of the amended potential at points with
symmetry, where it is not a priori defined, remains the same. In
a future paper we shall further modify this method to deal with
bifurcating branches of relative equilibria that have a given
isotropy, different from the trivial one, along the branch.
The paper is organized as follows. In \S \ref{Lagrangian
mechanical systems} we quickly review the necessary material on
symmetric simple mechanical systems and introduce the notations
and conventions for the entire paper. Relative equilibria and
their characterizations for general symmetric mechanical systems
and for simple ones in terms of the augmented and amended
potentials are recalled in \S \ref{Relative equilibria}. Section
\S \ref{Some basic results from the theory of group actions}
gives a brief summary of facts from the theory of proper group
actions needed in this paper. After these short introductory
sections, \S \ref{Regularization of the amended potential
criterion} presents the main bifurcation result of the
paper. The existence of branches of relative
equilibria starting at certain
points in $\mathfrak{t}\cdot q_e$, depending on several
parameters and having trivial symmetry off
$\mathfrak{t}\cdot q_e$, is proved in Theorem \ref{principala},
the main result of this paper. In \S
\ref{stability section}, using a result of Patrick \cite{patrick
thesis}, Lyapunov stability conditions for these branches are
given if the symmetry group is a torus.
\bigskip
\section{Lagrangian mechanical systems}
\label{Lagrangian mechanical systems}
This section summarizes the key facts from the theory of
Lagrangian systems with symmetry and sets the notations and
conventions to be used throughout this paper. The references for
this section are \cite{f of m}, \cite{lm}, \cite{marsden 92}, \cite{ims}.
\subsection{Lagrangian mechanical systems with symmetry }
Let $Q$ be a smooth manifold, the configuration space of a
mechanical system. The {\bfi fiber derivative\/} or {\bfi
Legendre transform\/} $\mathbb{F}L:TQ\rightarrow T^{\ast }Q$ of
$L$ is a vector bundle map covering the identity defined by
\[
\langle \mathbb{F}L(v_{q}),w_{q}\rangle
= \left.\frac{d}{dt}\right |_{t=0}L(v_{q}+tw_{q})
\]
for any $v_q, w_q \in TQ$. The {\bfi energy\/} of $L $ is
defined by $E(v_{q})= \langle \mathbb{F}L(v_{q}),v_{q} \rangle
-L(v_{q})$, $v_{q}\in T_{q}Q$. The pull back by $\mathbb{F}L $ of
the canonical one-- and two--forms of $T ^\ast Q $ give the
{\bfi Lagrangian one\/} and {\bfi two-forms\/} $\Theta _{L}$ and
$\Omega_L$ on $TQ$ respectively, that have thus the expressions
\[
\langle\Theta _{L}(v_{q}),\delta v_{q}\rangle=\langle\mathbb{F}L(v_{q}),T_{v_{q}}\pi
_{Q}(\delta v_{q})\rangle,\quad v_{q}\in T_{q}Q,\quad \delta v_{q}\in T_{v_{q}}TQ,
\qquad \Omega_{L}=-\mathbf{d}\Theta _{L},
\]
where $\pi_Q : TQ \rightarrow Q$ is the tangent bundle projection.
The Lagrangian $L $ is called {\bfi regular\/} if $\mathbb{F}L $
is a local diffeomorphism, which is equivalent to $\Omega_L $
being a symplectic form on $TQ$. The Lagrangian $L $ is called
{\bfi hyperregular\/} if $\mathbb{F}L $ is a diffeomorphism and
hence a vector bundle isomorphism. The {\bfi Lagrangian vector
field\/} $X_E $ of $L$ is uniquely determined by the equality
\[
\Omega _{L}(v_{q})(X_{E}(v_{q}),w_{q})=\langle
\mathbf{d}E(v_{q}),w_{q}\rangle,
\quad \text{for} \quad v_{q}, \; w_{q}\in T_{q}Q.
\]
A {\bfi Lagrangian dynamical system\/}, or simply a {\bfi
Lagrangian system\/}, for $L$ is the dynamical system defined by
$X_{E}$, i.e., $\dot{v} =X_{E}(v)$. In standard coordinates $(q^i,
\dot{q}^i) $ the trajectories of $X_E $ are given by the second
order equations
\[
\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^{i}}-\frac{\partial L}{
\partial q^{i}}=0,
\]
which are the classical the Euler-Lagrange equations.
\medskip
Let $\Psi: G \times Q \rightarrow Q$ be a smooth left Lie group
action on $Q$ and let
$L:TQ\rightarrow \mathbb{R}$ be a Lagrangian that is invariant
under the lifted action of $G$ to $TQ$. Denote by $\mathfrak{g}$
the Lie algebra of $G$. From the definition of the fiber
derivative it immediately follows that $\mathbb{F}L $ is
equivariant relative to the lifted $G $--actions to $TQ$ and
$T^\ast Q$. The $G$-invariance of $L$ implies that
$X_{E}$ is $G$-equivariant, that is, $\Psi_g ^\ast X_E = X_E $
for any $g \in G $. The $G $--action on $TQ $ admits a momentum
map given by
\[
\langle\mathbf{J}_{L}\mathbf{(}v_{q}),\xi \rangle=\langle\mathbb{F}L(v_{q}),\xi
_{Q}(q)\rangle, \quad \text{for} \quad v_{q}\in T_{q}Q, \quad \xi
\in \mathfrak{g}.
\]
where $\xi _{Q}(q) : = d\exp (t\xi) \cdot q/dt|_{t=0}$ is the
{\bfi infinitesimal generator\/} of $\xi \in \mathfrak{g}$, where
$\mathfrak{g}$ denotes the Lie algebra of $G $. Recall that the
momentum map
$\mathbf{J}:T^{\ast}Q\rightarrow
\mathfrak{g}^{\ast }$ on $T^{\ast }Q$ is given by
\[
\langle\mathbf{J(}\alpha _{q}),\xi \rangle=\langle\alpha _{q},\xi _{Q}(q)\rangle,
\quad \text{for} \quad \alpha _{q}\in
T_{q}^{\ast }Q, \quad \xi \in \mathfrak{g}
\]
and hence $\mathbf{J}_{L}= \mathbf{J\circ }\mathbb{F}L$.
We shall denote by $g \cdot q: =\Psi(g,q)$ the action of the
element $g\in G $ on the point $q \in Q $. Similarly, the lifted
actions of $G $ on $TQ$ and $T^\ast Q $ are denoted by
\[
g \cdot v_q : = T_q \Psi_g(v_q) \quad \text{and} \quad
g \cdot \alpha_q : = T^\ast_{g\cdot q} \Psi_{g^{-1}}(\alpha_q)
\]
for $g\in G $, $v_q \in T_q Q $, and $\alpha_q \in T_q ^\ast Q $.
\subsection{Simple mechanical systems}
A {\bfi simple mechanical system} $(Q,\langle \!\langle \cdot ,\cdot \rangle \! \rangle
_{Q},V)$ consists of a Riemannian manifold $(Q,\langle \!\langle \cdot ,\cdot \rangle
\! \rangle _{Q})$ together with a potential function $V:Q\rightarrow \mathbb{R}$. These
elements define a Hamiltonian system on $(T^{\ast }Q,\omega )$ with Hamiltonian given
by $ H:T^{\ast }Q\rightarrow \mathbb{R}$, $H(\alpha _{q})= \frac{1}{2} \langle \!\langle
\alpha _{q},\alpha _{q}\rangle \! \rangle _{T^{\ast }Q}+V(q)$, where
$\alpha_q \in T_q ^\ast Q $ and $\langle \!\langle
\cdot ,\cdot \rangle
\! \rangle _{T^{\ast }Q}$ is the vector bundle metric on $T^{\ast
}Q$ induced by the Riemannian metric of $Q$. The Hamiltonian
vector field $X_H $ is uniquely given by the relation
$\mathbf{i} _{X_{H}}\omega =\mathbf{d}H$, where $\omega$ is the
canonical symplectic form on $ T^{\ast }Q$.
The dynamics of a simple mechanical system can also be described in terms of
Lagrangian mechanics, whose description takes place on $TQ$. The Lagrangian
for a simple mechanical system is given by $L:TQ\rightarrow \mathbb{R}$, $
L(v_{q})=\frac{1}{2}\langle\!\langle v_{q},v_{q}\rangle\!\rangle
_{Q}-V(q)$, where $v_q \in T_q Q $. The energy of $L $ is
$E(v_q) = \frac{1}{2}\langle\!\langle v_q, v_q \rangle\!\rangle +
V(q)$. Since the fiber derivative for a simple mechanical system
is given by
$\langle\mathbb{F}L(v_{q}),w_{q}\rangle=\langle\!\langle
v_{q},w_{q}\rangle\!\rangle _{Q}$, or in local coordinates
$\mathbb{F}L\left(\dot{q}^{i}\frac{
\partial }{\partial q^{i}}\right)=g_{ij}\dot{q}^{j}dq^{i}$, where
$g_{ij}$ is the local expression for the metric on $Q$, it follows
that $L $ is hyperregular.
The relationship between the Hamiltonian and the Lagrangian
dynamics is the following: the vector bundle
isomorphism $\mathbb{F}L$ bijectively maps the trajectories of
$X_E$ to the trajectories of $X_H$, $(\mathbb{F}L)^\ast X_H = X_E
$, and the base integral curves of $X_E $ and $X_H $ coincide.
\subsection{Simple mechanical systems with symmetry}
\label{Simple mechanical systems with symmetry}
Let $G$ act on the configuration manifold $Q$ of a simple
mechanical system $(Q,\langle\!\langle \cdot ,\cdot
\rangle\!\rangle _{Q},V)$ by isometries.
The {\bfi locked inertia tensor\/} $\mathbb{I}:Q\rightarrow
\mathcal{L(} \mathfrak{g},\mathfrak{g}^{\ast })$, where
$\mathcal{L(} \mathfrak{g},\mathfrak{g}^{\ast })$ denotes the
vector space of linear maps from $\mathfrak{g}$ to
$\mathfrak{g}^\ast$, is defined by
\[
\langle\mathbb{I(}q)\xi ,\eta \rangle=\langle\!\langle \xi
_{Q}(q),\eta _{Q}(q)\rangle\!\rangle _{Q}
\]
for any $q \in Q $ and any $\xi, \eta \in \mathfrak{g} $.
If the action is {\bfi locally free\/} at $q\in Q$, that is, the
isotropy subgroup $G_q $ is discrete, then
$\mathbb{I(}q)$ is an isomorphism and hence defines an inner
product on $\mathfrak{g}$. In general, the defining formula of
$\mathbb{I}(q)$ shows that $\ker \mathbb{I}(q) = \mathfrak{g}_q:
= \{\xi\in \mathfrak{g} \mid \xi_Q(q) = 0 \}$.
Suppose the action is locally free at every point $q \in Q $.
Then on can define the {\bfi mechanical connection\/}
$\mathcal{A} \in \Omega^1(Q; \mathfrak{g})$ by
\[
\mathcal{A}(q)(v_{q})=\mathbb{I(}q)^{-1}\mathbf{J}_{L}(v_{q}),
\quad v_{q}\in T_{q}Q.
\]
If the $G $--action is free and proper, so $Q \rightarrow Q/G $
is a $G $--principal bundle, then
$\mathcal{A}$ is a connection one--form on the principal bundle
$Q \rightarrow Q/G$, that is, it satisfies the following
properties:
\begin{itemize}
\item $\mathcal{A}(q) :T_q Q\rightarrow \mathfrak{g}$ is linear
and $G$-equivariant for every $q \in Q $, which means that
\[
\mathcal{A}(g \cdot q)(g\cdot v_q) =
\operatorname{Ad}_g[\mathcal{A}(q)(v_q)],
\]
for any $v_q \in T_q Q $ and any $g \in G$, where
$\operatorname{Ad}$ denotes the adjoint representation of $G $ on
$\mathfrak{g}$;
\item $\mathcal{A}(q)(\xi _{Q}(q))=\xi $, for any $\xi \in
\mathfrak{g}$.
\end{itemize}
If $\mu\in \mathfrak{g}^\ast$ is given, we denote by
$\mathcal{A}_{\mu } \in \Omega^1(Q)$ the $\mu$--component of
$\mathcal{A}$, that is, the one--form on
$Q$ defined by $\langle\mathcal{A}_{\mu }(q),v_{q}\rangle
=\langle\mu ,\mathcal{A}(q)(v_{q})\rangle$ for any $v_q \in T_qQ$.
The $G$-invariance of the metric and the relation
\[
(\operatorname{Ad}_{g}\xi
)_{Q}(q)=g\cdot \xi _{Q}(g^{-1}\cdot q),
\]
implies that
\begin{equation}
\label{equivariance of I}
\mathbb{I(}g\cdot q)=\operatorname{Ad}_{g^{-1}}^{\ast }\circ
\mathbb{I(}q)\circ \operatorname{Ad}_{g^{-1}}.
\end{equation}
We shall also need later the infinitesimal version of the above
identity
\begin{equation}
\label{infinitesimal equivariance of I}
T_q\mathbb{I} \left(\xi_{Q}(q)\right)
= -\operatorname{ad}_{\xi }^{\ast} \circ \mathbb{I}(q)
- \mathbb{I}(q) \circ \operatorname{ad}_{\xi},
\end{equation}
which implies
\begin{equation}
\label{useful identity for I}
\left\langle T_q \mathbb {I} ( \zeta_Q(q)) \xi, \eta
\right\rangle
= \mathbf{d}\langle\mathbb{I}(\cdot )\xi ,\eta
\rangle(q)\left(\zeta _{Q}(q)\right)
=\langle\mathbb {I}(q)[\xi,\zeta ],\eta
\rangle+\langle\mathbb{I(}q)\xi ,[\eta ,\zeta ]\rangle.
\end{equation}
for all $q\in Q$ and all $\xi, \eta, \zeta\in \mathfrak{g}$.
\section{Relative equilibria}
\label{Relative equilibria}
This section recalls the basic facts about relative equilibria
that will be needed in this paper. For proofs see \cite{f of m},
\cite{lm}, \cite{marsden 92}, \cite{ims}, \cite{slm}.
\subsection{Basic definitions and concepts}
Let $ \Psi: G \times Q \rightarrow Q$ be a left action of the Lie
group on the manifold
$Q$. A vector field $X:Q\rightarrow TQ$ is said to be $G$-{\bfi equivariant\/} if
\[
T_{q}\Psi _{g}(X(q)) = X(\Psi _{g}(q))\quad
\text{or,~equivalently,} \quad
\Psi_g^\ast X = X
\]
for all $q\in Q$ and $g\in G$. If $X$ is $G$-equivariant, then $G$ is said
to be a {\bfi symmetry group\/} of the dynamical system $\dot{q}=X(q)$.
A {\bfi relative equilibrium\/} of a $G$--equivariant vector field $X$ is a
point $q_{e}\in Q$ at which the value of $X$ coincides with the
infinitesimal generator of some element $\xi \in \mathfrak{g}$,
usually called the {\bfi velocity\/} of $q_e $, i.e.,
\[
X(q_{e})=\xi _{Q}(q_{e}).
\]
A relative equilibrium $q_{e}$ is said to be {\bfi asymmetric\/}
if the isotropy subalgebra $\mathfrak{g}_{q_{e}} : = \{ \eta
\in \mathfrak{g} \mid \eta_Q(q_e) = 0 \} =\{0\}$, and {\bfi
symmetric} otherwise. Note that if
$q_{e}$ is a relative equilibrium with velocity $\xi \in
\mathfrak{g}$, then for any $g\in G$, $g\cdot q_{e}$ is a
relative equilibrium with velocity $\operatorname{Ad}_{g}\xi $.
The flow of an equivariant vector field induces a flow on the
quotient space. Thus, if the $G $--action is free and proper, a
relative equilibrium defines an equilibrium of the induced vector
field on the quotient space and conversely, any element in the
fiber over an equilibrium in the quotient space is a relative
equilibrium of the original system.
\subsection{Relative equilibria in Hamiltonian $G$-systems}
Given is a symplectic manifold $(P, \omega)$, a left Lie group
action of $G $ on $P $ that admits a momentum map $\mathbf{J}: P
\rightarrow \mathfrak{g}^\ast$, that is, $X_{\mathbf{J}^ \xi} =
\xi_P $, for any $\xi\in \mathfrak{g}$, where $\mathbf{J}^ \xi(p):
= \langle \mathbf{J}(p), \xi \rangle $, $p \in P$, is the
$\xi$--component of $\mathbf{J} $. We shall also assume
throughout this paper that the momentum map $\mathbf{J}$ is
equivariant, that is, $\mathbf{J}(g \cdot p ) =
\operatorname{Ad}^\ast_{g^{-1}} \mathbf{J} (p)$, for any $g \in G
$ and any $p \in P $. Given is also a $G $--invariant function
$H: P \rightarrow \mathbb{R}$. Noether's theorem states that the
$\mathbf{J} $ is conserved along the flow $F_t $ of the
Hamiltonian vector field $X_H $. In what follows we shall call
the quadruple $(Q,\omega ,H,\mathbf{J},G)$ a {\bfi Hamiltonian
$G$--system\/}. Consistent with the general definition presented
above, a point
$ p_{e}\in P$ is a {\bfi relative equilibrium\/} if
\[
X_{H}(p_{e})\in T_{p_{e}}(G\cdot p_{e}),
\]
where $G\cdot p_e : \{ g \cdot p_e \mid g \in G \} $ denotes the
$G $--orbit through $p_e $. Relative equilibria are
characterized in the following manner.
\begin{proposition}
\label{characterization of relative equilibria}
(\textbf{Characterization of relative equilibria}). Let $p_{e}\in
P$ and $ p_{e}(t)$ be the integral curve of $X_{H}$ with initial
condition $p_{e}(0)=p_{e}$. Let $\mu:= \mathbf{ J}(p_{e})$.
Then the following are equivalent:
\begin{enumerate}
\item[\textbf{(i)}] $p_{e}$ is a relative equilibrium.
\item[\textbf{(ii)}] There exists $\xi \in \mathfrak{g}$ such that
$p_{e}(t)=\exp (t\xi )\cdot p_{e}$.
\item[\textbf{(iii)}] There exists $\xi \in \mathfrak{g}$ such
that $p_{e}$ is a critical point of the {\bfi augmented
Hamiltonian\/}
\[
H_{\xi }(p):=H(p)-\langle\mathbf{J(}p)-\mu ,\xi \rangle.
\]
\end{enumerate}
\end{proposition}
Once we have a relative equilibrium, its entire $G$-orbit consists
of relative equilibria and the relation between the velocities of
the relative equilibria that are on the same $G$-orbit is given by
the adjoint action of $G$ on $\mathfrak{g}$.
\begin{proposition}
\label{orbita}
With the notations of the previous proposition, let $p_{e}$ be a
relative equilibrium with velocity $\xi$. Then
\begin{enumerate}
\item[\textbf{(i)}] for any
$g\in G$, $g\cdot q_{e}$ is also a relative
equilibrium whose velocity is $\operatorname{Ad}_g \xi$;
\item[\textbf{(ii)}] $\xi(q_{e})\in \mathfrak{g}_\mu : = \{\eta
\in \mathfrak{g}\mid \operatorname{ad}^\ast_\eta \mu = 0 \}$, the
coadjoint isotropy subalgebra at $\mu\in \mathfrak{g}^\ast$, i.e.,
$\operatorname{Ad}_{\exp t\xi}^{\ast}\mu=\mu$ for any $t \in
\mathbb{R}$.
\end{enumerate}
\end{proposition}
\subsection{Relative equilibria in simple mechanical $G$-systems}
In the case of simple mechanical $G$-systems, the
characterization $\mathbf{(iii)}$ in Proposition
\ref{characterization of relative equilibria} can be simplified in
such way that the search of relative equilibria reduces to the
search of critical points of a real valued function on
$Q$.
Depending on whether one \ keeps track of the velocity or
the momentum of a relative equilibrium, this simplification
yields the \textit{augmented} or the \textit{amended} potential
criterion, which we introduce in what follows.
Let $(Q,\langle\!\langle \cdot ,\cdot \rangle\!\rangle _{Q},V,G)$
be a simple mechanical $G$--system.
\begin{itemize}
\item For $\xi \in \mathfrak{g}$, the {\bfi augmented potential\/}
$V_{\xi }:Q\rightarrow \mathbb{R}$ is defined by $V_{\xi
}(q):=V(q)-\frac{1}{2}\langle
\mathbb{I}(q)\xi ,\xi \rangle$.
\item For $\mu \in \mathfrak{g}^{\ast }$, the {\bfi amened
potential\/} $V_{\mu }:Q\rightarrow \mathbb{R}$ is defined by
$V_{\mu }(q):=V(q)+\frac{1}{2}\langle
\mathbb{\mu },\mathbb{I}(q)^{-1}\mu \rangle$.
\end{itemize}
Note that the amended potential is defined at $q\in Q$ only if $q$
in an asymmetric point.
There is an alternate expression for the amended potential,
namely, $V_{\mu }(q)=(H\circ \mathcal{A}_{\mu })(q)$.
\begin{proposition}
(\textbf{Augmented potential criterion}).
\label{augmented potential criterion}
A point
$(q_{e},p_{e})\in T^{\ast}Q$ is a relative equilibrium if and only
if there exists a $\xi \in
\mathfrak{g}$ such that:
\begin{enumerate}
\item[\textbf{(i)}] $p_{e}=\mathbb{F}L(\xi _{Q}(q_{e}))$ and
\item[\textbf{(ii)}] $q_{e}$ is a critical point of $V_{\xi }$.
\end{enumerate}
\end{proposition}
\begin{proposition}
(\textbf{Amended potential criterion}).
\label{amended potential criterion}
A point $(q_{e},p_{e})\in
T^{\ast }Q$ is a relative equilibrium if and only if there exists
a $\mu \in \mathfrak{g} ^{\ast }$ such that:
\begin{enumerate}
\item[\textbf{(i)}] $p_{e}=\mathcal{A}_{\mu }(q_{e})$ and
\item[\textbf{(ii)}] $q_{e}$ is a critical point of $V_{\mu }$.
\end{enumerate}
\end{proposition}
\section{Some basic results from the theory of Lie group actions}
\label{Some basic results from the theory of group actions}
We shall need a few fundamental results form the theory of group
actions which we now review. For proofs and further information
see \cite{br}, \cite{dk}, \cite{kawakubo}, \cite{or}.
\subsection{Maximal tori}
Let $V$ be a representation space of a compact Lie group $G$.
A point $v\in V$ is {\bfi regular\/} if there is no $G$--orbit in
$V$ whose dimension is strictly greater than the dimension of the
$G$--orbit through $v$. The set of regular points, denoted
$V_{reg}$, is open and dense in $V$.
In particular, $\mathfrak{g}_{reg}$ and
$\mathfrak{g}_{reg}^{\ast}$, denote the set of regular points in
$\mathfrak{g}$ and $\mathfrak{g}^{\ast }$ with respect to adjoint
and coadjoint representation, respectively.
A subgroup of a Lie group is said to be a {\bfi torus\/} if it is
isomorphic to $S^{1}\times \cdot \cdot \cdot \times S^{1}$.
Every Abelian subgroup of a compact connected Lie group is
isomorphic to a torus.
A subgroup of a Lie group is said to be a {\bfi maximal torus\/}
if it is a torus that is not properly contained in some other
torus.
Every $\xi \in \mathfrak{g}$ belongs to at least one maximal
Abelian subalgebra and every $\xi \in \mathfrak{g\cap g}_{reg}$
belongs to exactly one such maximal Abelian subalgebra.
Every maximal Abelian subalgebra is the Lie algebra of some
maximal torus in $G$.
Let $\mathfrak{t}$ be the maximal Abelian subalgebra corresponding
to a maximal torus $T$. Then for any $\xi \in \mathfrak{t\cap
g}_{reg}$, we have that $G_{\xi }=T$. The space
$[\mathfrak{g},\mathfrak{t}]$ is the orthogonal complement to
$\mathfrak{t}$ in $\mathfrak{g}$ with respect to any $G$--invariant
inner product on $\mathfrak{g}$. Such an inner product exists by
compactness of $G$ by simply averaging any inner product on
$\mathfrak{g}$. Therefore, we have
$\mathfrak{ g} =\mathfrak{t}\oplus [\mathfrak{g},\mathfrak{t}]$.
Let $[\mathfrak{g},\mathfrak{t}]^{\circ }$ the annihilator of
$[\mathfrak{g},\mathfrak{t}]$. Then $G_{\mu }=T$ for every $\mu
\in [\mathfrak{g},\mathfrak{t} ]^{\circ }\cap
\mathfrak{g}_{reg}^{\ast }$. Since
$[\mathfrak{g},\mathfrak{t}]^{\circ }\cap
\mathfrak{g}_{reg}^{\ast }$ is dense in
$[\mathfrak{g},\mathfrak{t}]^{\circ }$, it follows that
$T\subset G_{\mu }$ for every $\mu \in
[\mathfrak{g},\mathfrak{t}]^{\circ }$.
\subsection{Twisted products}
Let $G$ be a Lie group and $H\subset G$ be a Lie subgroup. Suppose
that $H$ acts on the left on a manifold $A$. The {\bfi twisted
action\/} of $H$ on the product $G\times A$ is defined by
\[
h\cdot (g,a)=(gh,h^{-1}\cdot a), \quad h\in H, \quad g\in G,
\quad a\in A.
\]
Note that this action is free and proper by the freeness and properness of
the action on the $G$--factor. The {\bfi twisted product\/}
$G\times _{H}A$ is defined as the orbit space $(G\times A)/H$
of the twisted action. The elements of $G\times
_{H}A$ will be denoted by $[g,a],$ $g\in G,$
$a\in A$.
The twisted product $G\times _{H}A$ is a $G$--space relative to
the left action defined by $g^{\prime }\cdot \lbrack
g,a]=[g^{\prime }g,a]$. Also, the action of $H$ on $A$ is proper
if and only if the $G$--action on $G\times _{H}A$ is proper.
The isotropy subgroups of the $G$--action on the twisted product
$G\times _{H}A$ satisfy
\[
G_{[g,a]}=gH_{a}g^{-1}, \quad g\in G, \quad a\in A.
\]
\subsection{Slices}
Throughout this paragraph it will be assumed that $\Psi : G
\times Q \rightarrow Q $ is a left proper action of the Lie group
$G$ on the manifold $Q$. This action
will not be assumed to be free, in general. For $q\in Q$ we will
denote by $H:=G_{q} := \{ g \in G \mid g \cdot q = q \}$ the
isotropy subgroup of the action $\Psi$ at $q $.
We shall introduce also the following convenient notation: if $K
\subset G $ is a Lie subgroup of $G $ (possibly equal to $G $),
$\mathfrak{k}$ is its Lie algebra, and $q\in Q$, then
$\mathfrak{k}\cdot q := \{ \eta_Q(q) \mid \eta \in
\mathfrak{k} \}$ is the tangent space to the orbit $K\cdot q $ at
$q $.
A {\bfi tube\/} around the orbit
$G\cdot q$ is a $G$-equivariant diffeomorphism $\varphi :G\times
_{H}A\rightarrow U$, where $U$ is a $G$-invariant neighborhood of
$G\cdot q$ and $A$ is some manifold on which $H$ acts. Note that
the $G$-action on the twisted product $ G\times _{H}A$ is proper
since the isotropy subgroup $H$ is compact and, consequently, its
action on $A$ is proper. Hence the $G$-action on $G\times _{H}A$
is proper.
Let $S$ be a submanifold of $Q$ such that $q\in S$ and $H\cdot
S=S$. We say that $S$ is a {\bfi slice\/} at $q$ if the map
\[
\varphi :G\times _{H}S\rightarrow U
\]
\[
\lbrack g,s]\mapsto g\cdot s
\]
is a tube about $G\cdot q$, for some $G$--invariant open
neighborhood of $G\cdot q$. Notice that if $S$ is a slice at
$q$ then $g\cdot S$ is a slice at the point $g\cdot q$.
The following statements are equivalent:
\begin{enumerate}
\item[\textbf{(i)}] There is a tube $\varphi :G\times
_{H}A\rightarrow U$ about $G\cdot q$ such that $\varphi
([e,A])=S$.
\item[\textbf{(ii)}] $S$ is a slice at $q$.
\item[\textbf{(iii)}] The submanifold $S$ satisfies the following
properties:
\begin{enumerate}
\item[\textbf{(a)}] The set $G\cdot S$ is an open neighborhood of
the orbit $G\cdot q$ and
$S$ is closed in $G\cdot S$.
\item[\textbf{(b)}] For any $s\in S$ we have
$T_{s}Q=\mathfrak{g}\cdot s+T_{s}S$. Moreover, $\mathfrak{g}\cdot
s\cap T_{s}S=\mathfrak{h}\cdot s$, where $\mathfrak{h}$ is the
Lie algebra of $H $. In particular
$T_{q}Q=\mathfrak{g}\cdot q\oplus T_{q}S$.
\item[\textbf{(c)}] $S$ is $H$-invariant. Moreover, if $s\in S$
and $g\in G$ are such that
$g\cdot s\in S$, then $g\in H$.
\item[\textbf{(d)}] Let $\sigma :U\subset G/H\rightarrow G$ be a
local section of the submersion $G\rightarrow G/H$. Then the map
$F:U\times S\rightarrow Q$ given by $F(u,s):=\sigma (u)\cdot s$
is a diffeomorphism onto an open set of $Q$.
\end{enumerate}
\item[\textbf{(iv)}] $G\cdot S$ is an open neighborhood of $G\cdot
q$ and there is an equivariant smooth retraction
\[
r:G\cdot S\rightarrow G\cdot q
\]
of the injection $G\cdot q\hookrightarrow G\cdot S$ such that $r^{-1}(q)=S$.
\end{enumerate}
\begin{theorem}
\textbf{(Slice Theorem)} Let $Q$ be a manifold and $G$ be a Lie
group acting properly on $Q$ at the point $q\in Q$. Then, there
exists a slice for the $G$--action at $q$.
\end{theorem}
\begin{theorem}
\label{tube theorem}
\textbf{(Tube Theorem)} Let $Q$ be a manifold and $G$ be a Lie group acting
properly on $Q$ at the point $q\in Q$, $H:=G_{q}$. There exists a tube $
\varphi :G\times _{H}B\rightarrow U$ about $G\cdot q$ such that $\varphi
([e,0])=q$, $\varphi ([e,B])=:S$ is a slice at $q$; $B$ is an
open $H$--invariant neighborhood of $0$ in the vector space
$T_{q}Q/T_{q}(G\cdot q) $, on which $H$ acts linearly by $h\cdot
(v_{q}+T_{q}(G\cdot q)):=T_{q}\Psi _{h}(v_{q})+T_{q}(G\cdot
q)$.
\end{theorem}
If $Q$ is a Riemannian manifold then $B$ can be chosen to be a
$G_{q}$--invariant neighborhood of $0$ in $(\mathfrak{g}\cdot
q)^{\perp }$, the orthogonal complement to $\mathfrak{g}\cdot q$
in $T_{q}Q$. In this case $ U=G\cdot \operatorname{Exp}_{q}(B)$,
where $\operatorname{Exp}_{q}: T_q Q \rightarrow Q $ is the
Riemannian exponential map.
\subsection{Type submanifolds and fixed point subspaces}
Let $G$ be a Lie group acting on a manifold $Q$. Let $H$ be a closed
subgroup of $G$. We define the following subsets of $Q$ :
\begin{eqnarray*}
Q_{(H)} &=&\{q\in Q\mid G_{q}=gHg^{-1},g\in G\}, \\
Q^{H} &=&\{q\in Q\mid H\subset G_{q}\}, \\
Q_{H} &=&\{q\in Q\mid H=G_{q}\}.
\end{eqnarray*}
All these sets are submanifolds of $Q $.
The set $Q_{(H)}$ is called the $(H)$--{\bfi orbit type
submanifold\/},
$ Q_{H}$ is the $H$--{\bfi isotropy type submanifold\/}, and
$Q^{H}$ is the $H$--{\bfi fixed point submanifold\/}. We
will collectively call these subsets the {\bfi type
submanifolds}. We have:
\begin{itemize}
\item $Q^{H}$ is closed in $Q$;
\item $Q_{(H)}=G\cdot Q_{H}$;
\item $Q_H $ is open in $Q^H $.
\item the tangent space at $q \in Q_H $ to $Q_H $ equals
\[
T_{q}Q_{H}=\{v_{q}\in T_{q}Q\mid T_{q}\Psi_{h}(v_{q})
=v_{q},\, \forall h\in H\}=(T_q Q)^H=T_q Q^H;
\]
\item $T_{q}(G\cdot q)\cap (T_{q}Q)^{H}=T_{q}(N(H)\cdot q)$,
where $N(H) $ is the normalizer of $H $ in $G $;
\item if $H$ is compact then $Q_{H}=Q^{H}\cap Q_{(H)}$ and $Q_H $
is closed in $Q_{(H)}$.
\end{itemize}
If $Q$ is a vector space on which $H$ acts linearly, the set
$Q^{H}$ is found in the physics literature under the names of
{\bfi space of singlets\/} or {\bfi space of invariant vectors\/}.
\begin{theorem}
\textbf{(The stratification theorem)}. Let $Q$ be a smooth manifold and $G$
be a Lie group acting properly on it. The connected components of the orbit
type manifolds $Q_{(H)}$ and their projections onto orbit space $Q_{(H)}/G$
constitute a Whitney stratification of $Q$ and $Q/G$, respectively. This
stratification of $Q/G$ is minimal among all Whitney stratifications of $Q/G$.
\end{theorem}
The proof of this result, that can be found in \cite{dk} or
\cite{pflaum}, is based on the Slice Theorem and on a series
of extremely important properties of the orbit type manifolds
decomposition that we enumerate in what follows. We start by
recalling that the set of conjugacy classes of subgroups of a Lie
group $G$ admits a partial order by defining $ (K)\preceq (H)$ if
and only if $H$ is conjugate to a subgroup of $K$. Also, a point
$q\in Q$ in a proper $G$--space $Q$ (or its corresponding
$G$--orbit, $ G\cdot q$) is called {\bfi principal\/} if its
corresponding local orbit type manifold is open in $Q$. The orbit
$G\cdot q$ is called
{\bfi regular\/} if the dimension of the orbits nearby coincides
with the dimension of $ G\cdot q$. The set of principal and
regular orbits will be denoted by $ Q_{princ}/G$ and $Q_{reg}/G$,
respectively. Using this notation we have:
\begin{itemize}
\item For any $q\in Q$ there exists an neighborhood $U$ of $q$
that intersects only finitely many connected components of
finitely many orbit type manifolds. If $Q$ is compact or a linear
space where $G$ acts linearly, then the $G$--action on $Q$ has
only finitely many distinct connected components of orbit type
manifolds.
\item For any $q\in Q$ there exists an open neighborhood $U$ of
$q$ such that $(G_{q})\preceq (G_{x})$, for all $x\in U$. In
particular, this implies that $\dim G\cdot q\leq \dim G\cdot x$,
for all
$x\in U$.
\item {\bfi Principal Orbit Theorem}: For every connected
component
$ Q^{0}$ of $Q$ the subset $Q_{reg}\cap Q^{0}$ is connected,
open, and dense in $Q^{0}$. Each connected component $(Q/G)^{0}$
of $Q/G$ contains only one principal orbit type, which is
connected open and dense in $(Q/G)^0 $.
\end{itemize}
\section{Regularization of the amended potential criterion}
\label{Regularization of the amended potential criterion}
In this section we shall follow the strategy in \cite{hm} to give
sufficient criteria for finding relative equilibria emanating
from a given one and to find a method that distinguishes
between the distinct branches. The criterion will involve a
certain regularization of the amended potential. The main
difference with \cite{hm} is that all hypotheses but one have been
eliminated and we work with a general torus and not just a
circle. The conventions, notations, and method of proof are those
in \cite{hm}.
\subsection{The bifurcation problem}
Let $(Q,\langle\!\langle \cdot ,\cdot \rangle\!\rangle _{Q},V,G)$
be a simple mechanical
$G$-system, with $G$ a compact Lie group with the Lie algebra
$\mathfrak{g}$. Recall that the left $G $--action $\Psi:G \times
Q \rightarrow Q $ is by isometries and that the potential $V:Q
\rightarrow \mathbb{R}$ is $G$--invariant. Let $ q_{e}\in Q$ be a
symmetric point whose isotropy group $G_{q_{e}}\subset
\mathbb{T}$ is contained in a maximal torus $\mathbb{T}$ of $G$.
Denote by $\mathfrak{t}
\subset \mathfrak{g}$ the Lie algebra of $\mathbb{T}$; thus
$\mathfrak{t}$ is a maximal Abelian Lie subalgebra of
$\mathfrak{g}$. Throughout this section we shall make the
following hypothesis:
\smallskip
\begin{center}
\textbf{(H)} \textit{ every} $v_{q_{e}}\in \mathfrak{t}\cdot
q_{e}$ \textit{is~a~relative~equilibrium}.
\end{center}
\smallskip
The following result was communicated to us by J. Montaldi.
\begin{proposition}
\label{montaldi}
In the context above we have that:
\[
\begin{array}{cc}
(i) & \mathbf{d}V(q_{e})=0 \\
(ii) & \mathbb{I(}q_{e}\mathbb{)}\mathfrak{t}\subseteq
[\mathfrak{g},
\mathfrak{t}]^{\circ }.
\end{array}
\]
\end{proposition}
\begin{proof}
$(i)$ Because all the elements in $\mathfrak{t}\cdot q_{e}$ are relative
equilibria, we have by the augmented potential criterion
$\mathbf{d}V_{\xi }(q_{e})=0$, for any $\xi \in \mathfrak{t}$.
Consequently for $\xi =0$ we will obtain
$0=\mathbf{d}V_{0}(q_{e})=\mathbf{d}V(q_{e})$.
$(ii)$ Substituting in the relation \eqref{useful identity for I}, $q$ by
$q_{e}$ and setting $\eta =\xi
\in \mathfrak{t}$ we obtain:
\[
\mathbf{d}\langle\mathbb{I(}\cdot \mathbb{)}\xi ,\xi \rangle(q_{e})
(\zeta _{Q}(q_{e})) =\langle
\mathbb{I(}q_{e}\mathbb{)[}\xi ,\zeta ],\xi \rangle
+\langle\mathbb{I(}q_{e}\mathbb{)}\xi
,[\xi ,\zeta ]\rangle = 2\langle \mathbb{I(}q_{e}\mathbb{)}\xi
,[\xi ,\zeta ]\rangle
\]
for any $\xi \in \mathfrak{t}$ and $\zeta \in \mathfrak{g}$.
The augmented potential criterion yields
\[
0 = \mathbf{d}
V_{\xi}(q_{e})
=\mathbf{d}V(q_{e})-\frac{1}{2}\mathbf{d}\langle\mathbb{I(}\cdot
\mathbb{)}\xi ,\xi
\rangle(q_{e}).
\]
Since $\mathbf{d}V(q_{e})=0$ by (i), this implies
$\mathbf{d}\langle\mathbb{I(}\cdot
\mathbb{)}\xi ,\xi \rangle(q_{e})=0$ and consequently
$\langle\mathbb{I(}q_{e}\mathbb{)}\xi , [\xi ,\zeta ]\rangle=0$,
for any $\xi \in \mathfrak{t}$ and $\zeta \in \mathfrak{g}$. So we have the inclusion
\[
\mathbb{I(}q_{e}\mathbb{)}\xi \subseteq [\mathfrak{g},\xi
]^{\circ }.
\]
Now we will prove that $[\mathfrak{g},\xi ]^{\circ }=[\mathfrak{g},\mathfrak{
t}]^{\circ }$ for regular elements $\xi \in \mathfrak{t}$. For this it is enough
to prove that $[\xi ,\mathfrak{g}]=[\mathfrak{t},
\mathfrak{g}]$ for regular elements $\xi \in \mathfrak{t}$.
It is obvious that $[\xi ,\mathfrak{g}]\subseteq \lbrack \mathfrak{t},
\mathfrak{g}]$. Equality will follow by showing that both spaces
have the same dimension. To do this, let
$F_{\xi }:\mathfrak{g}\rightarrow
\mathfrak{g}$, $F_{\xi }(\eta):=\operatorname{ad}_{\xi }\eta $,
which is obviously a linear map whose image and kernel are
$\operatorname{Im} (F_{\xi })=[\xi ,\mathfrak{g}]$ and
$\ker (F_{\xi })=\mathfrak{g} _{\xi }$. Because $\xi \in
\mathfrak{t}$ is a regular element we have that $\mathfrak{g}_{\xi
}=\mathfrak{t}$ and so $\ker (F_{\xi })=\mathfrak{t}$. Thus
$\dim (\mathfrak{g})=\dim (
\mathfrak{t})+\dim ([\xi ,\mathfrak{g}])$ and so using the fact that $\dim (
\mathfrak{g})=\dim (\mathfrak{t})+\dim ([\mathfrak{t},\mathfrak{g}])$ (since
$\mathfrak{g}=\mathfrak{t}\oplus [\mathfrak{g},\mathfrak{t}]$,
$\mathfrak{g}$ being a compact Lie algebra), we obtain the equality
$\dim ([\xi ,\mathfrak{g}])=\dim ([\mathfrak{t},\mathfrak{g}])$.
Therefore, $[\xi,\mathfrak{g}]=[\mathfrak{t},\mathfrak{g}]$ for
any regular element $\xi \in \mathfrak{t}$.
Summarizing, we proved
\[
\mathbb{I(}q_{e}\mathbb{)}\xi \subseteq
[\mathfrak{g},\mathfrak{t} ]^{\circ },
\]
for any regular element $\xi \in \mathfrak{t}$.
The continuity of $\mathbb{I(}q_{e}\mathbb{)}$, the closedness of
$[\mathfrak{g},\mathfrak{t}]^{\circ }$, and that fact that the
regular elements $\xi \in \mathfrak{t}$ form a dense subset of
$\mathfrak{t}$, implies that
\[
\mathbb{I(}q_{e}\mathbb{)}\xi \subseteq
[\mathfrak{g},\mathfrak{t} ]^{\circ },
\]
for any $\xi \in \mathfrak{t}$ and hence $\mathbb{I(}q_{e}\mathbb{)}
\mathfrak{t}\subseteq [\mathfrak{g},\mathfrak{t}]^{\circ }$.
\end{proof}
\begin{lemma}
\label{same isotropy}
For each $v_{q_{e}}\in \mathfrak{t}\cdot q_{e}$ we have
$G_{v_{q_{e}}}=G_{_{q_{e}}}$.
\end{lemma}
\begin{proof}
The inclusion $G_{v_{q_{e}}}\subseteq G_{_{q_{e}}}$ is obviously
true, so it will be enough to prove that $G_{v_{q_{e}}}\supseteq
G_{_{q_{e}}}$. To see this, let $g\in G_{_{q_{e}}}$ and
$v_{q_{e}}=\xi _{Q}(q_{e})\in
\mathfrak{t}\cdot q_{e}$, with $\xi\in \mathfrak{t}$. Then, since
$G_{q_e} $ is Abelian, we get
\begin{align*}
T_{q_{e}}\Psi _{g}\left(v_{q_{e}}\right)
&=T_{q_{e}}\Psi _{g}\left(\xi_{Q}(q_{e})\right)
=T_{q_{e}}\Psi _{g}\left(\left.\frac{d}{dt}\right|_{t=0}\Psi
_{\exp (t \xi )}(q_{e})\right) \\
&=\left.\frac{d}{dt}\right|_{t=0}\left(\Psi _{g}\circ \Psi _{\exp
(t\xi )}\right)(q_{e})=
\left.\frac{d}{dt}\right| _{t=0}(\Psi _{\exp (t \xi )}\circ \Psi
_{g})(q_{e}) \\ &=\left.\frac{d}{dt}\right|_{t=0}\Psi _{\exp
(t\xi )}(q_{e})= \xi_Q(q_e) = v_{q_{e}},
\end{align*}
that is, $g\cdot v_{q_{e}} = v_{q_{e}} $, as required.
\end{proof}
The bifurcation problem for relative equilibria on $TQ$ can be
regarded as a bifurcation problem on the space $Q\times\mathfrak{g}
^{\ast}$ as the following shows.
\begin{proposition}
\label{map f}
The map $f:TQ\rightarrow Q\times \mathfrak{g}^{\ast }$ given by $
v_{q}\mapsto (q,\mathbf{J}_{L}(v_{q}))$ restricted to the set
of relative equilibria is one to one and onto its image.
\end{proposition}
\begin{proof}
The only thing to be proved is that the map is injective. To see
this, let $(q_{1},(\xi_{1})_{Q}(q_{1}))$ and
$(q_{2},(\xi_{2})_{Q}(q_{2}))$ be two relative equilibria such
that $f(q_{1},(\xi_{1})_{Q}(q_{1}))=
f(q_{2},(\xi_{2})_{Q}(q_{2}))$. Then $q_{1}=q_{2}=:q$ and
$\mathbf{J}_{L}(q,(\xi _{1}-\xi _{2})_{Q}(q))=
\mathbb{I}(q)(\xi_{1}-\xi_{2})=0$ which shows that $\xi_{1}-\xi_{2}\in \ker
\mathbb{I}(q)=\mathfrak{g}_{q}$ and hence $(\xi_{1})_
{Q}(q)=(\xi_{2})_{Q}(q)$.
\end{proof}
We can thus change the problem: instead of searching for relative
equilibria of the simple mechanical system in $TQ$, we shall set
up a bifurcation problem on $Q \times \mathfrak{g}^\ast$ such
that the image of the relative equilibria by the map $f $ is
precisely the bifurcating set.
To do this, we begin with some geometric considerations. We
construct a $G$-invariant tubular neighborhood of the orbit
$G\cdot q_{e}$ such that the isotropy group of every point in
this neighborhood is a subgroup of $G_{q_{e}}$. This follows from
the Tube Theorem \ref{tube theorem}. Indeed, let
$B\subset (\mathfrak{g}\cdot q_{e})^{\perp }$ be a
$G_{q_{e}}$-invariant open neighborhood of
$0_{q_{e}}\in (\mathfrak{g}\cdot q_{e})^{\perp }$ such that on
the open $G$-invariant neighborhood
$G\cdot \operatorname{Exp}_{q_{e}}(B)$ of $G\cdot q_{e}$, we
have $(G_{q_{e}})\preceq (G_{q})$ for every $q\in G\cdot
\operatorname{Exp}_{q_{e}}(B)$. Moreover
$G$ acts freely on $G\cdot \operatorname{Exp}_ {q_{e}}\left(B\cap
(T_{q_{e}}Q)_{\{e\}}\right)$. It is easy to see that $B\times
\mathfrak{g}^{\ast }$ can be identified with a slice at
$(q_{e},0)$ with respect to the diagonal action of $G$ on $
(G\cdot \operatorname{Exp}_{q_{e}}(B))\times \mathfrak{g}^{\ast
}$.
The strategy to prove the existence of a bifurcating branch of relative
equilibria with no symmetry from the set of relative equilibria
$\mathfrak{t}
\cdot q_{e}$ is the following. Note that we do not know a priori
which relative equilibrium in $\mathfrak{t} \cdot q_{e}$ will
bifurcate. We search for a local bifurcating branch of relative
equilibria in the following manner. Take a vector $v_{q_{e}}\in
B\cap (T_{q_{e}} Q)_{\{e\}}$ and note that
$\operatorname{Exp}_{q_e} (v_{q_{e}})
\in Q$ is a point with no symmetry, that is,
$G_{\operatorname{Exp}_{q_e} (v_{q_{e}})} = \{e\}$. Then $\tau
v_{q_{e}} \in B\cap (T_{q_{e}} Q)_{\{e\}}$, for $\tau\in
I $, where $I $ is an open interval containing $[0,1]$, and
$\operatorname{Exp}_{q_e} (\tau v_{q_{e}})$ is a smooth path
connecting $q_e $, the base point of the relative equilibrium in
$\mathfrak{t}\cdot q_e $ containing the branch of bifurcating
relative equilibria, to
$\operatorname{Exp}_{q_e} (v_{q_{e}})
\in Q$. In addition, we shall impose that the entire path
$\operatorname{Exp}_{q_e} (\tau v_{q_{e}})$ be formed by
base points of relative equilibria. We still need the vector part
of these relative equilibria which we postulate to be of the form
$\zeta(\tau)_Q(\operatorname{Exp}_{q_e} (\tau v_{q_{e}}) )$,
where $\zeta(\tau) \in \mathfrak{g}$ is a smooth path of Lie
algebra elements with $\zeta(0) \in \mathfrak{t}$. Since
$\operatorname{Exp}_{q_e} (\tau v_{q_{e}})$ has no symmetry for
$\tau> 0 $, the locked inertia tensor is invertible at these
points and the path
$\zeta(\tau) $ will be of the form
\[
\zeta(\tau) = \mathbb {I}(\operatorname{Exp}_{q_e} (\tau
v_{q_{e}}))^{-1} (\beta(\tau)),
\]
where $\beta(\tau) $ is a smooth path in $\mathfrak{g}^\ast$ with
$\beta(0) \in \mathbb {I}(q_e) \mathfrak{t}$. Now we shall use
the characterization of relative equilibria involving the amended
potential to require that the path
$\left(\operatorname{Exp}_{q_e}(\tau v_{q_e}), \beta(\tau) \right)
\in (G\cdot \operatorname{Exp}_{q_{e}}(B)) \times
\mathfrak{g}^\ast $ be such that
$f^{-1}(\left(\operatorname{Exp}_{q_e}(\tau v_{q_e}), \beta(\tau)
\right)$ are all relative equilibria.
The amended potential criterion is applicable along the path
$\operatorname{Exp}_{q_e}(\tau v_{q_e})$ for $\tau >0 $, because
these points have no symmetry. As we shall see below, we shall
look for $\beta(\tau)$ of a certain form and
then the characterization of relative equilibria via the amended
potential will impose conditions on both $\beta(\tau) $ and
$v_{q_e}$. We begin by specifying the form of $\beta(\tau)$.
\subsection{Splittings}
We shall need below certain direct sum decompositions of
$\mathfrak{g}$ and $\mathfrak{g}^\ast$. The compactness of
$G$ implies that $\mathfrak{g}$ has an invariant inner
product and that $\mathfrak{g}=\mathfrak{t}\oplus
[\mathfrak{g},\mathfrak{t}]$ is an orthogonal direct sum. Let
$\mathfrak{k}_{1}\subset\mathfrak{t}$ be the orthogonal complement
to $\mathfrak{k}_{0}:=\mathfrak{g}_{q_{e}}$ in
$\mathfrak{t}$. Denoting
$\mathfrak{k}_{2}:=[\mathfrak{g},\mathfrak{t}]$ we obtain the
orthogonal direct sum $\mathfrak{g}=\mathfrak{k}_{0}\oplus
\mathfrak{k} _{1}\oplus \mathfrak{k}_{2}$.
For the dual of the Lie algebra, let
$\mathfrak{m}_{i}:=(\mathfrak{k} _{j}\oplus
\mathfrak{k}_{k})^{\circ }$ where $(i,j,k)$ is a cyclic
permutation of $(0,1,2)$. Then $\mathfrak{g}^{\ast }=\mathfrak{m}
_{0}\oplus \mathfrak{m}_{1}\oplus \mathfrak{m}_{2}$ is also an
orthogonal direct sum relative to the inner product on
$\mathfrak{g}^\ast$ naturally induced by the invariant inner
product on $\mathfrak{g}$.
\begin{lemma}
The subspaces defined by the above splittings have the following
properties:
\begin{enumerate}
\item[\textbf{(i)}] $\mathfrak{k}_{0}$, $\mathfrak{k}_{1}$,
$\mathfrak{k}_{2}$ are $G_{q_{e}}$-invariant and $G_{q_{e}}$ acts
trivially on $\mathfrak{k}_{0}$ and $\mathfrak{k}_{1}$;
\item[\textbf{(ii)}] $\mathfrak{m}_{0}$, $\mathfrak{m}_{1}$,
$\mathfrak{m}_{2}$ are $G_{q_{e}}$-invariant and $G_{q_{e}}$ acts
trivially on $\mathfrak{m}_{0}$ and $\mathfrak{m}_{1}$.
\end{enumerate}
\end{lemma}
\begin{proof}
\textbf{(i)} Because $G_{q_{e}}$ is a subgroup of $ \mathbb{T}$
it is obvious that $G_{q_{e}}$ acts trivially on
$\mathfrak{t}=\mathfrak{k}_{0}\oplus \mathfrak{k}_{1}$ and hence
on each summand. To prove the
$G_{q_{e}}$-invariance of
$\mathfrak{k}_{2}=[\mathfrak{g},\mathfrak{t}]
$, we use the fact that
$\operatorname{Ad}_{g}[\xi _{1},\xi
_{2}]=[\operatorname{Ad}_{g}\xi _{1},\operatorname{Ad}_{g}\xi
_{2}]$, for any
$\xi _{1},\xi _{2}\in
\mathfrak{g}$ and $g\in G$. Indeed, if $\xi _{1}\in
\mathfrak{g}$,
$\xi _{2}\in \mathfrak{t}$, $g\in G_{q_{e}}$ we get
$\operatorname{Ad}_{g}[\xi _{1},\xi _{2}]\in
\lbrack \mathfrak{g},\mathfrak{t}]=\mathfrak{k}_{2}$.
\textbf{(ii)} For $g\in G_{q_{e}}$, $\mu \in \mathfrak{m}_{0}$ we
have to prove that $ \operatorname{Ad}_{g}^{\ast }\mu \in
\mathfrak{m}_{0}$. Indeed, if $\xi =\xi _{1}+\xi _{2}\in
\mathfrak{k}_{1}\oplus \mathfrak{k}_{2}$, we have
\begin{align*}
\langle \operatorname{Ad}_{g}^{\ast }\mu ,\xi \rangle
&=\langle \operatorname{Ad}_{g}^{\ast
}\mu ,\xi _{1}+\xi _{2}\rangle
=\langle \mu, \operatorname{Ad}_{g}(\xi _{1}+\xi _{2})\rangle \\
&=\langle \mu ,\xi _{1}+\operatorname{Ad}_{g}\xi _{2}\rangle
=0
\end{align*}
since $G_{q_{e}}$ acts trivially on $\mathfrak{k}_{1}$,
$\mathfrak{k}_{2}$ is $G_{q_{e}}$--invariant and
$\mathfrak{m}_{0}=(\mathfrak{k}_{1}\oplus
\mathfrak{k}_{2})^{\circ }$.
The same type of proof holds for $\mathfrak{m}_{1}$and
$\mathfrak{m}_{2}$.
For $g\in G_{q_{e}}$, $\mu \in \mathfrak{m}_{0}$ we have to prove that $
\operatorname{Ad}_{g}^{\ast }\mu =\mu $.
Let $\xi =\xi _{0}+\xi _{1}+\xi _{2}\in \mathfrak{g}$, with
$\xi_i \in \mathfrak{k}_i $, $i = 0,1,2 $.
We have
\begin{align*}
\langle \operatorname{Ad}_{g}^{\ast }\mu -\mu ,\xi
\rangle
&=\langle \operatorname{Ad}_{g}^{\ast }\mu ,\xi _{0}+\xi _{1}+\xi
_{2}\rangle -\langle \mu ,\xi _{0}+\xi _{1}+\xi _{2}\rangle \\
&=\langle \mu ,\operatorname{Ad}_{g}(\xi _{0}+\xi _{1}+\xi
_{2})\rangle - \langle \mu ,\xi _{0}+\xi _{1}+\xi _{2}\rangle \\
&=\langle \mu ,\xi _{0}+\xi _{1}+\operatorname{Ad}_{g}\xi
_{2}\rangle - \langle \mu ,\xi _{0}\rangle
=\langle \mu ,\xi_{1}+\operatorname{Ad}_{g}\xi _{2} \rangle
=0
\end{align*}
because $G_{q_{e}}$ acts trivially on $\mathfrak{k}_{0}\oplus \mathfrak{k}
_{1}$, $\mathfrak{k}_{2}$ is $G_{q_{e}}$--invariant, and
$\mathfrak{m}_{0}=(
\mathfrak{k}_{1}\oplus \mathfrak{k}_{2})^{\circ }$.
The same type of proof holds for $\mathfrak{m}_{1}$.
\end{proof}
Recall from \S \ref{Simple mechanical systems with symmetry} that
$\ker \mathbb {I}(q_e) =
\mathfrak{g}_{q_e} = \mathfrak{k}_0 $. In particular, $\mathbb
{I}(q_e) \mathfrak{k}_0 = \{0\}$. The value of $\mathbb {I}(q_e)$
on the other summands in the decomposition $\mathfrak{g}=
\mathfrak{k}_0 \oplus \mathfrak{k}_1 \oplus \mathfrak{k}_2 $ is
given by the following lemma.
\begin{lemma}
\label{moment of inertia isomorphism}
For $i\in \{1,2\}$ we have that $\mathfrak{m}_{i}=\mathbb{I}(q_{e})
\mathfrak{k}_{i}$.
\end{lemma}
\begin{proof}
Let $\kappa_{i}\in\mathfrak{k}_{i}$ with $i\in \{0,1,2\}$ be
arbitrary. Then
\begin{equation*}
\langle \mathbb{I}(q_{e})\kappa_{1},
\kappa_{0}+\kappa_{2}\rangle
=\langle \mathbb{I}(q_{e})\kappa_{1},\kappa_{0}\rangle
+\langle\mathbb{I}(q_{e})\kappa_{1},\kappa_{2}\rangle
=\langle \mathbb{I}(q_{e})\kappa_{0},\kappa_{1}\rangle
+\langle\mathbb{I}(q_{e})\kappa_{1},\kappa_{2}\rangle
=0
\end{equation*}
as $\ker \mathbb{I}(q_{e})=\mathfrak{k}_{0}$ and, by Proposition
\ref{montaldi} (ii),
$\mathbb{I}(q_{e})\mathfrak{t}\subset \mathfrak{k}_{2}^{\circ}$.
This proves that $\mathbb{I}(q_{e})\mathfrak{k}_{1}\subset
\mathfrak{m}_{1}$. Counting dimensions we have that
$\dim\mathbb{I}(q_{e})\mathfrak{k}_{1}= \dim \mathfrak{k}_1 -
\dim \ker \left(\mathbb {I}(q_e)|_{\mathfrak{k}_1} \right) =
\dim \mathfrak{g}-\dim \mathfrak{k}_{0}-\dim \mathfrak{k}_{2}
=\dim \mathfrak{m}_{1}$,
since $\ker \left(\mathbb {I}(q_e)|_{\mathfrak{k}_1} \right) =
\{0\} $. This proves that
$\mathfrak{m}_{1}=\mathbb{I}(q_{e})
\mathfrak{k}_{1}$. In an analogous way we prove the equality for
$i=2$.
\end{proof}
In the next paragraph we shall need the direct sum decomposition
$\mathfrak{g}^\ast = \mathfrak{m}_1 \oplus \mathfrak{m}$, where
$\mathfrak{m}_1 = \mathbb {I}(q_e) \mathfrak{t}$ and
$\mathfrak{m} := \mathfrak{m}_0 \oplus \mathfrak{m}_2 $.
Let $\Pi_1: \mathfrak{g}^\ast\rightarrow \mathbb {I}(q_e)
\mathfrak{t}$ be the projection along $\mathfrak{m} $. Similarly,
denote
$\mathfrak{k}: =
\mathfrak{k}_1
\oplus\mathfrak{k}_2$, and write $\mathfrak{g} =
\mathfrak{g}_{q_e} \oplus\mathfrak{k}$. Thus there is another
decomposition of $\mathfrak{g}^\ast$, namely, $\mathfrak{g}^\ast
= \mathfrak{g}_{q_e} ^\circ \oplus \mathfrak{k}^\circ $. However,
for any $\zeta \in \mathfrak{g}_{q_e} $ and any $\xi\in
\mathfrak{g}$, we have
$\langle \mathbb {I}(q_e) \xi, \zeta \rangle = \langle\!\langle
\xi_Q(q_e), \zeta_Q(q_e) \rangle\!\rangle = 0 $ since
$\zeta_Q(q_e) = 0 $, which shows that $\mathbb {I}(q_e)
\mathfrak{g} \subset \mathfrak{g}_{q_e}^\circ $. Since $ \ker
\mathbb {I}(q_e) = \mathfrak{g}_{q_e} $, it follows that $\dim
\mathbb {I}(q_e) \mathfrak{g} = \dim \mathfrak{g} - \dim \ker
\mathbb {I}(q_e) = \dim \mathfrak{g} - \dim \mathfrak{g}_{q_e} =
\dim \mathfrak{g}_{q_e}^\circ $, which shows that
$\mathfrak{g}_{q_e}^\circ = \mathbb {I}(q_e)\mathfrak{g}$. Thus
we also have the direct sum decomposition $\mathfrak{g}^\ast =
\mathbb {I}(q_e)\mathfrak{g} \oplus \mathfrak{k}^\circ $. Note
that $\mathbb {I}(q_e) \mathfrak{g} = \mathfrak{m}_1 \oplus
\mathfrak{m}_2 $, by Lemma \ref{moment of inertia isomorphism}
and that $\mathfrak{m}_0 = \mathfrak{k}^\circ $. Summarizing we
have:
\[
\mathfrak{g}^\ast = \mathfrak{m}_0 \oplus \mathfrak{m}_1 \oplus
\mathfrak{m}_2 = \mathfrak{k}^\circ \oplus \mathbb {I}(q_e)
\mathfrak{g}, \quad \text{where} \quad \mathbb {I}(q_e) \mathfrak{g} = \mathfrak{m}_1 \oplus
\mathfrak{m}_2 \quad \text{and} \quad \mathfrak{m}_0 =
\mathfrak{k}^\circ.
\]
\subsection{The rescaled equation}
Recall that $B\subset (\mathfrak{g}\cdot q_{e})^{\perp }$ is a
$G_{q_{e}}$-invariant open neighborhood of
$0_{q_{e}}\in (\mathfrak{g}\cdot q_{e})^{\perp }$ such that on
the open $G$-invariant neighborhood
$G\cdot \operatorname{Exp}_{q_{e}}(B)$ of $G\cdot q_{e}$, we
have $(G_{q_{e}})\preceq (G_{q})$ for every $q\in G\cdot
\operatorname{Exp}_{q_{e}}(B)$.
Consider the following rescaling:
\begin{equation*}
v_{q_{e}}\in B\cap (T_{q_{e}}Q)_{\{e\}}\mapsto \tau v_{q_{e}}\in
B\cap (T_{q_{e}}Q)_{\{e\}}
\end{equation*}
\begin{equation*}
\mu \in \mathfrak{g}^{\ast }\mapsto \beta (\tau ,\mu )\in \mathfrak{g}^{\ast }
\end{equation*}
where, $\tau \in I$, $I$ is an open interval containing
$[0,1]$, and
$\beta :I\times \mathfrak{g}^{\ast }\rightarrow
\mathfrak{g}^{\ast }$ is chosen such that $\beta (0,\mu
)=\Pi _{1}\mu $. So, for $(v_{q_{e}},\mu )$ fixed, $(\tau
v_{q_{e}},\beta (\tau ,\mu ))$ converges to $(0_{q_{e}},\Pi
_{1}\mu )$ as $\tau \rightarrow 0$. Define
\begin{equation*}
\beta (\tau ,\mu ):=\Pi _{1}\mu +\tau \beta '(\mu
)+\tau^{2} \beta ''(\mu )
\end{equation*}
for some arbitrary smooth functions $\beta ',\beta '':
\mathfrak{g}^{\ast }\rightarrow \mathfrak{g}^{\ast }$. Since
$\mathbb{I}$ is invertible only for points with no symmetry, we
want to find conditions on
$\beta '$, $\beta ''$ such that the expression
\begin{equation}
\label{32}
\mathbb{I}(\operatorname{Exp} _{q_{e}}(\tau v_{q_{e}}))^{-1}\beta
(\tau ,\mu )
\end{equation}
extends to a smooth function in a neighborhood of $\tau =0$. Note
that $v_{q_{e}}$ is different from $0_{q_{e}}$ since
$G_{v_{q_{e}}} = \{e\}$ by construction and $G_{0_{q_{e}}} =
G_{q_e}
\neq \{e\}$. Define
\begin{equation*}
\Phi :I\times \left(B\cap (T_{q_{e}}Q)_{\{e\}}\right)\times
\mathfrak{g}^{\ast }\times
\mathfrak{g}_{q_{e}}\times \mathfrak{k}\rightarrow \mathfrak{g}^{\ast }
\end{equation*}
\begin{equation}
\label{definition of Phi}
\Phi (\tau ,v_{q_{e}},\mu ,\xi ,\eta )
:=\mathbb{I}(\operatorname{Exp}_{q_{e}}(\tau
v_{q_{e}})) (\xi +\eta )-\beta (\tau ,\mu ).
\end{equation}
Now we search for the velocity $\xi + \eta $ of relative
equilibria among the solutions of
$\Phi (\tau ,v_{q_{e}},\mu ,\xi ,\eta )=0$. We shall prove below
that $\xi $ and $\eta$ are smooth functions of
$\tau $, $v_{q_{e}}$, $\mu$, even at $\tau= 0$.
Then \eqref{32} shows that $\xi+
\eta $ is a smooth function of $\tau $, $v_{q_{e}}$, $\mu$, for
$\tau$ in a small neighborhood of zero.
\subsection{The Lyapunov-Schmidt procedure}
To solve $\Phi =0$ we apply the standard Lyapunov-Schmidt
method. This equation has a unique solution for
$\tau
\neq 0$, because $\tau v_{q_{e}}\in B\cap (T_{q_{e}}Q)_{\{e\}}$ so
$\mathbb{I}(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}}))$ is
invertible. It remains to prove that the equation has a solution
when $\tau =0$.
Denote by $D_{\mathfrak{g}_{q_{e}}\times \mathfrak{k}}$ the
Fr\'echet derivative relative to the last two factors
$\mathfrak{g}_{q_{e}}\times \mathfrak{k}$ in the definition of
$\Phi$. We have
\begin{equation*}
\ker D_{\mathfrak{g}_{q_{e}}\times \mathfrak{k}}\Phi (0,v_{q_{e}},\mu ,\xi
,\eta )=\ker \mathbb{I}(q_{e})=\mathfrak{g}_{q_{e}}.
\end{equation*}
We will solve the equation $\Phi =0$ in two steps. For this, let
\begin{equation*}
\Pi :\mathfrak{g}^{\ast }\rightarrow \mathbb{I}(q_{e})\mathfrak{g}
\end{equation*}
be the projection induced by the splitting $\mathfrak{g}^{\ast }=\mathbb{I}
(q_{e})\mathfrak{g}\oplus \mathfrak{k}^{\circ }$.
\textbf{Step1}. Solve $\Pi \circ \Phi =0$ for $\eta $ in terms of
$\tau $, $v_{q_{e}}$, $\mu $, $\xi $. For this, let
\begin{align*}
\widehat{\mathbb{I}}(\operatorname{Exp} _{q_{e}}(\tau v_{q_{e}}))
&:= (\Pi\circ \mathbb{I})(\operatorname{Exp}_{q_{e}}
(\tau v_{q_{e}}))|_{\mathfrak{k}}: \mathfrak{k}\rightarrow
\mathbb {I}(q_e) \mathfrak{g}\\
\overset{\thicksim}{\mathbb{I}}(\operatorname{Exp} _{q_{e}}
(\tau v_{q_{e}}))
&:=(\Pi \circ \mathbb{I})(\operatorname{Exp}_{q_{e}}(\tau
v_{q_{e}}))|_{\mathfrak{g}_{q_{e}}}: \mathfrak{g}_{q_e}
\rightarrow \mathbb {I}(q_e) \mathfrak{g}
\end{align*}
where $\widehat{\mathbb{I}}(\operatorname{Exp} _{q_{e}}(\tau
v_{q_{e}})) $ is an isomorphism even when $\tau =0$. Then we
obtain
\begin{equation}
\label{pi composed with Phi}
(\Pi \circ \Phi )(0,v_{q_{e}},\mu ,\xi ,\eta )
= \Pi [\mathbb{I} (q_{e})(\xi +\eta )-\beta (0,\mu)]
=\widehat{\mathbb{I}}(q_{e}) \eta -\Pi _{1}\mu .
\end{equation}
Denoting $\eta _{\mu }:=\widehat{\mathbb{I}}
(q_{e})^{-1}(\Pi _{1}\mu )$, we have $(\Pi \circ \Phi)
(0,v_{q_{e}},\mu ,\xi ,\eta _{\mu })\equiv 0$.
Denoting by $D_\eta$ the partial Fr\'echet derivative relative
to the variable $\eta
\in \mathfrak{k}$ we get at any given point $(0, v_{q_e}^0,
\mu^0, \xi^0, \eta^0 )$
\begin{equation}
\label{isomorphism condition for ift}
D_\eta (\Pi \circ \Phi )(0, v_{q_e}^0, \mu^0, \xi^0, \eta^0)
=\widehat{\mathbb{I}}(q_{e})
\end{equation}
which is invertible. Thus the implicit function theorem gives a
unique smooth function
$\eta (\tau ,v_{q_{e}},\mu ,\xi )$ such that $\eta
(0, v_{q_e}^0,
\mu^0, \xi^0)=\eta^0$ and
\begin{equation}
\label{definition of eta}
(\Pi \circ \Phi) (\tau ,v_{q_{e}},\mu ,\xi ,\eta (\tau
,v_{q_{e}},\mu ,\xi ))\equiv 0.
\end{equation}
The function $\eta $ is defined in some open set in $I\times
\left(B\cap (T_{q_{e}}Q)_{\{e\}}\right)\times
\mathfrak{g}^{\ast}\times
\mathfrak{g}_{q_{e}}$ containing $(0, v_{q_e}^0, \mu^0, \xi^0)
\in \{0\}\times
\left(B\cap (T_{q_{e}}Q)_{\{e\}}\right)\times
\mathfrak{g}^{\ast }\times \mathfrak{g}_{q_{e}}$.
If we now choose $\eta^0 = \eta_{\mu^0} = \widehat{\mathbb
{I}}(q_e) ^{-1}( \Pi_1 \mu ^0)$, then uniqueness of the
solution of the implicit function theorem implies that $\eta(0,
v_{q_e}, \mu, \xi) = \eta_\mu$ in the
neighborhood of $(0, v_{q_e}^0, \mu^0, \xi^0)$.
\medskip
Later we will need the following result.
\begin{proposition}
\label{belongs}
We have $\eta _{\mu }:=\widehat{\mathbb{I}}
(q_{e})^{-1}(\Pi _{1}\mu ) \in \mathfrak{k}_1
\subset \mathfrak{t}$.
\end{proposition}
\begin{proof}
Since we can write $\mathfrak{t}=\ker \mathbb{I}(q_{e})\oplus
\mathfrak{k}_1$ we obtain
\begin{equation*}
\widehat{\mathbb{I}}(q_{e})\mathfrak{k}_1
=(\Pi \circ \mathbb{I}(q_{e}))
\mathfrak{k}_1
=\mathbb{I} (q_{e})
\mathfrak{k}_1=\mathbb{I}(q_{e})(\mathfrak{t})
=\operatorname{Im}\Pi _{1}.
\end{equation*}
Now, because $\widehat
{\mathbb{I}}(q_{e})$ is an isomorphism, it follows that
$\widehat{\mathbb{I}}
(q_{e})^{-1}(\Pi _{1}\mu ) \in \mathfrak{k}_1$.
\end{proof}
\textbf{Step2}. Now we solve the equation $(Id-\Pi)\circ\Phi=0$.
For this, let
\begin{equation*}
\varphi :I\times \left(B\cap (T_{q_{e}}Q)_{\{e\}}\right)\times
\mathfrak{g}^{\ast }\times
\mathfrak{g}_{q_{e}}\rightarrow \mathfrak{k}^{\circ }
\end{equation*}
\begin{equation}
\label{definition of phi}
\varphi (\tau ,v_{q_{e}},\mu ,\xi )
:=(Id-\Pi ) \Phi (\tau ,v_{q_{e}},\mu ,\xi ,\eta (\tau
,v_{q_{e}},\mu ,\xi )).
\end{equation}
In particular, $\varphi (0,v_{q_{e}},\mu ,\xi )=(Id-\Pi)
(\mathbb{I} (q_{e})(\xi +\eta _{\mu })-\Pi _{1}\mu )$. Since
$\operatorname{Im}\mathbb{I} (q_{e}) = \operatorname{Im}\Pi$ and
$\operatorname{Im}\Pi _{1} = \mathbb {I}(q_e) \mathfrak{t}
\subset \mathbb {I}(q_e) \mathfrak{g}$, it follows that $\varphi
(0,v_{q_{e}},\mu ,\xi )\equiv 0$. We shall solve for $\xi
\in\mathfrak{g}_{q_e}$, in the
neighborhood of $(0, v_{q_e}^0, \mu^0, \xi^0)$ found in Step 1,
the equation $\varphi (\tau ,v_{q_{e}},\mu ,\xi ) = 0 $. To do
this, we shall need information about the higher derivatives of
$\varphi $ with respect to $\tau $, evaluated at $\tau =0$.
\begin{lemma}
Let $\xi $, $\eta \in \mathfrak{g}$ and $q\in Q$. Suppose that
$\mathbf{d}V_{\eta }(q)=0$, where $V_{\eta}$ is the augmented
potential and suppose that both $\xi $ and $[\xi ,\eta ]$ belong
to $\mathfrak{g}_{q}$. Then
$\mathbf{d}\langle \mathbb{I} (\cdot )\xi ,\eta \rangle (q)=0$.
\end{lemma}
\begin{proof}
Since $\mathbf{d}V_{\eta}(q)=0$, $\eta _{Q}(q)$ is a relative
equilibrium by Proposition \ref{augmented potential criterion},
that is,
$X_{H}(\alpha _{q})=\eta_{T^{\ast }Q}(\alpha _{q})$, where $\alpha
_{q}=\mathbb{F}L(\eta _{Q}(q))$. Now suppose that both $\xi $,
$[\xi ,\eta ]\in \mathfrak{g}_{q}$. Then
\begin{equation*}
\xi _{T^{\ast }Q}(\alpha _{q})=\left. \frac{d}{dt}\right|
_{t=0}\mathbb{F}L(\exp(t\xi )\cdot \eta
_{Q}(q))=\mathbb{F}L([\xi ,\eta ]_{Q}(q))=0,
\end{equation*}
where we have used that $g\cdot \eta
_{Q}(q)=(\operatorname{Ad}_{g}\eta )_{Q}(g\cdot q)$. It follows
that
$(\eta +\xi )_{T^{\ast }Q}(\alpha _{q})=X_{H}(\alpha _{q})$ and
hence, again by Proposition \ref{augmented potential criterion},
that $0=\mathbf{d}V_{\eta +\xi }(q)=\mathbf{d}V_{\eta}(q)
-\mathbf{d}\langle\mathbb{I}(\cdot )\eta ,\xi
\rangle(q)-\frac{1}{2}\mathbf{d}\|
\xi _{Q}(\cdot )\| ^{2}(q)$. However, $\mathbf{d}\| \xi
_{Q}(\cdot )\| ^{2}(q) = 0 $ since $\xi \in \mathfrak{g}_q $, as
an easy coordinate computation shows. Since $\mathbf{d}V_\eta(q) =
0 $ by hypothesis, we have $\mathbf{d}\langle\mathbb{I}(\cdot
)\eta ,\xi \rangle(q) = 0 $. Symmetry of $\mathbb {I}(q)$ proves
the result.
\end{proof}
Let now $\xi \in \mathfrak{g}_{q_{e}}$ and $\eta \in
\mathfrak{t}$. Since $\mathfrak{g}_{q_e} \subset \mathfrak{t}$,
we have $[ \xi, \eta ] = 0 \in \mathfrak{g}_{q_e}$. In
addition, hypothesis \textbf{(H)} and Proposition
\ref{augmented potential criterion}, guarantee that
$\mathbf{d}V_\xi(q_e) = 0 $ which shows that all hypotheses of
the previous lemma are satisfied. Therefore,
\begin{equation}
\label{differential of i}
\mathbf{d}\langle\mathbb{I}(\cdot )\xi,\eta
\rangle(q_{e})=0 \quad \text{for} \quad \xi \in
\mathfrak{g}_{q_{e}}, \; \eta \in
\mathfrak{t}.
\end{equation}
\subsection{The bifurcation equation}
Now we can proceed with the study of equation $
\varphi= (Id-\Pi)\circ\Phi=0$.
We have
\begin{align}
\label{first tau derivative of phi}
\frac{\partial \varphi}{\partial \tau}(\tau, v_{q_e}, \mu, \xi)
= (Id - \Pi) &\left[ T_{\tau v_{q_e}}(\mathbb {I} \circ
\operatorname{Exp}_{q_e})(v_{q_e})(\xi + \eta(\tau, v_{q_e}, \mu,
\xi))
+ \mathbb {I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))
\frac{\partial \eta}{\partial \tau}(\tau, v_{q_e}, \mu, \xi)
\right. \nonumber \\
&\qquad \left. - \frac{\partial \beta}{\partial \tau}(\tau, \mu)
\right].
\end{align}
\begin{proposition}
$\frac{\partial }{\partial \tau }\varphi (0,v_{q_{e}},\mu ,\xi )\equiv
-(Id-\Pi )\beta '(\mu )$.
\end{proposition}
\begin{proof}
Formula \eqref{first tau derivative of phi} gives for $\tau= 0 $
\begin{equation*}
\frac{\partial \varphi }{\partial \tau }(0,v_{q_{e}},\mu ,\xi )
=(Id-\Pi ) \left[\big(T_{q_e}
\mathbb{I}(v_{q_{e}})\big)(\xi +\eta _{\mu })
+\mathbb{I}(q_{e})\frac{\partial \eta }{\partial
\tau }(0,v_{q_{e}},\mu ,\xi )-\frac{\partial \beta }{\partial \tau
}(0,\mu )\right].
\end{equation*}
Now, because $\operatorname{Im}\mathbb{I}(q_{e})=\operatorname{Im}\Pi $
we obtain $(Id-\Pi )
\circ \mathbb{I}(q_{e})=0$ and hence the second summand vanishes.
From
\eqref{differential of i} we have that
$(T_{q_e}\mathbb{I}
(v_{q_{e}}))(\mathfrak{t})\subset \mathfrak{g}_{q_{e}}^{\circ
}=\operatorname{Im}\Pi $. Using Proposition \ref{belongs} and
since $\xi \in \mathfrak{g}_{q_{e}}\subset
\mathfrak{t}$, we obtain that $\xi +\eta_{\mu }\in \mathfrak{t}$. Therefore
$(Id-\Pi )[(T_{q_e}\mathbb{I}(v_{q_{e}}))(\xi +\eta
_{\mu })]=0$. Since $\frac{\partial \beta }{\partial \tau }(0,\mu
)=\beta ^{\prime }(\mu )$, we obtain the desired equality.
\end{proof}
Let us impose the additional condition $\beta '(\mu)
\subset\operatorname{Im}\Pi $. Then it follows that
\begin{equation*}
\varphi (\tau ,v_{q_{e}},\mu ,\xi )=\tau ^{2}\psi (\tau ,v_{q_{e}},\mu ,\xi).
\end{equation*}
for some smooth function $\psi $ where
\begin{equation*}
\psi (0,v_{q_{e}},\mu ,\xi )=\frac{1}{2}\frac{\partial ^{2}\varphi }{
\partial \tau ^{2}}(0,v_{q_{e}},\mu ,\xi )
\end{equation*}
We begin by solving the equation
\begin{equation*}
\psi (0,v_{q_{e}},\mu ,\xi )=0
\end{equation*}
for $\xi $ as a function of $v_{q_{e}}$ and $\mu $. Equivalently,
we have to solve
\begin{equation*}
\frac{1}{2}\frac{\partial ^{2}\varphi }{\partial \tau ^{2}}(0,v_{q_{e}},\mu
,\xi )=0.
\end{equation*}
To compute this second derivative of $\varphi$ we shall use
\eqref{first tau derivative of phi}. We begin by noting that
$\tau\in I \mapsto T_{\tau v_{q_e}}(\mathbb {I} \circ
\operatorname{Exp}_{q_e})(v_{q_e})$ is a smooth path in
$L(\mathfrak{g}, \mathfrak{g}^\ast)$ and so we can define the
linear operator from $\mathfrak{g}$ to $\mathfrak{g}^\ast$
by
\[
A_{v_{q_e}}: = \left.\frac{\partial}{\partial \tau}\right|_{\tau=
0}T_{\tau v_{q_e}}(\mathbb {I} \circ
\operatorname{Exp}_{q_e})(v_{q_e}) \in L(\mathfrak{g},
\mathfrak{g}^\ast).
\]
With this notation, formulas \eqref{first tau derivative of phi},
\eqref{definition of Phi}, \eqref{definition of phi}, and
Proposition \ref{belongs} yield
\begin{align}
\label{second tau derivative of phi at zero}
\frac{\partial ^{2}\varphi }{\partial \tau^{2}} (0,v_{q_{e}},\mu,
\xi )
&=(Id-\Pi)\left[ A_{v_{q_e}}(\xi +\eta _{\mu})
+2T_{q_e}\mathbb{I}(v_{q_{e}})\frac{\partial \eta}{\partial
\tau }(0,v_{q_{e}},\mu ,\xi ) \right.\\
& \qquad \qquad \qquad \left. + \mathbb{I}(q_{e})\frac{\partial
^{2}\eta }{\partial
\tau ^{2}} (0,v_{q_{e}},\mu ,\xi )-2\beta ''(\mu
)\right] \nonumber \\
&= (Id-\Pi)\left[ A_{v_{q_e}}(\xi +\eta _{\mu})
+2T_{q_e}\mathbb{I}(v_{q_{e}})\frac{\partial \eta}{\partial
\tau }(0,v_{q_{e}},\mu ,\xi ) - 2\beta ''(\mu) \right]
\nonumber
\end{align}
since $(Id - \Pi)\mathbb{I}(q_{e})\frac{\partial
^{2}\eta }{\partial \tau ^{2}} (0,v_{q_{e}},\mu ,\xi) = 0$.
Let $\{\xi _{1},...,\xi _{p}\}$ be a basis of
$\mathfrak{g}_{q_{e}}$. Since $\partial^2 \varphi ( \tau, v_{q_e},
\mu, \xi)/\partial \tau ^2 \in \mathfrak{k}^\circ $ and
$\mathfrak{g} = \mathfrak{g}_{q_e} \oplus \mathfrak{k}$ the
equation $\partial ^2 \varphi(0, v_{q_e}, \mu, \xi)/\partial
\tau^2 = 0 $ is equivalent to the following system of $p $
equations
\[
\left\langle\frac{\partial ^{2}\varphi }{\partial \tau ^{2}}(0,v_{q_{e}},\mu
,\xi ) , \xi_b \right \rangle = 0, \quad \text{for~all} \quad b =
1, \dots, p,
\]
which, by \eqref{second tau derivative of phi at zero}, is
\[
\left\langle (Id-\Pi)\left[ A_{v_{q_e}}(\xi +\eta _{\mu})
+2T_{q_e}\mathbb{I}(v_{q_{e}})\frac{\partial \eta}{\partial
\tau }(0,v_{q_{e}},\mu ,\xi ) - 2\beta ''(\mu
)\right] ,\xi_b \right\rangle = 0, \quad \text{for~all}
\quad b = 1, \dots, p.
\]
We shall show that in this expression we can drop the projector
$ Id - \Pi $. Indeed, let $\alpha = \alpha_0 + \alpha_1 +
\alpha_2 \in \mathfrak{g}^\ast = \mathfrak{m}_0 \oplus
\mathfrak{m}_1 \oplus \mathfrak{m}_2 $, where $\alpha_i \in
\mathfrak{m}_i $, for $i = 0,1,2 $. Since $\Pi:
\mathfrak{g}^\ast \rightarrow \mathbb {I}(q_e) \mathfrak{g} =
\mathfrak{m}_1 \oplus \mathfrak{m}_2 $, we have
\[
\langle (Id - \Pi) \alpha, \xi_b \rangle = \langle \alpha, \xi_b
\rangle - \langle \alpha_1, \xi_b \rangle - \langle \alpha_2,
\xi_b \rangle = \langle \alpha, \xi_b \rangle
\]
because $\langle \alpha_1, \xi_b \rangle = 0$, since $\alpha_1 \in
\mathfrak{m}_1 = (\mathfrak{k}_0 \oplus \mathfrak{k}_2)^\circ $,
$\xi_b
\in \mathfrak{g}_{q_e} = \mathfrak{k}_0 $, and
$\langle \alpha_2, \xi_b \rangle = 0 $, since $\alpha_2 \in
\mathfrak{m}_2 = (\mathfrak{k}_0 \oplus \mathfrak{k}_1)^\circ $,
$\xi_b
\in \mathfrak{g}_{q_e} = \mathfrak{k}_0 $. The system to be
solved is hence
\begin{equation}
\label{linear system to be solved}
\left\langle A_{v_{q_e}}(\xi +\eta _{\mu})
+2T_{q_e}\mathbb{I}(v_{q_{e}})\frac{\partial \eta}{\partial
\tau }(0,v_{q_{e}},\mu ,\xi ) - 2\beta ''(\mu
) ,\xi_b \right\rangle = 0, \quad \text{for~all}
\quad b = 1, \dots, p.
\end{equation}
In what follows we need the expression for $\frac{\partial \eta}
{\partial \tau }(0,v_{q_{e}},\mu ,\xi)$. Differentiating
\eqref{definition of eta} relative to $\tau$ at zero and taking
into account \eqref{isomorphism condition for ift} and
\eqref{definition of Phi}, we get
\begin{align}
\label{first tau derivative of eta at zero}
&\frac{\partial \eta }{\partial \tau }(0,v_{q_{e}},\mu ,\xi)
= - \widehat{\mathbb {I}}(q_e)^{-1}\frac{\partial}{\partial
\tau}(\Pi\circ \Phi)(0, v_{q_e}, \mu, \xi, \eta_\mu)\\
&\qquad = - \widehat{\mathbb {I}}(q_e)^{-1} \Pi\left[
T_{q_e}\mathbb {I}(v_{q_e})(\xi+ \eta_\mu) - \beta'(\mu) \right]
\nonumber \\
&\qquad = - \left(
\widehat{\mathbb{I}}(q_e)^{-1} \circ T_{q_e}\overset{\thicksim}
{\mathbb {I}}(v_{q_e})\right) \xi
- \left(\widehat{\mathbb
{I}}(q_e)^{-1} \circ T_{q_e}\widehat{\mathbb
{I}}(v_{q_e}) \circ
\widehat{\mathbb{I}} (q_{e})^{-1}\right)(\Pi _{1}\mu ) +
\widehat{\mathbb{I}} (q_{e})^{-1}(\beta'(\mu)) \nonumber
\end{align}
since $T_{q_e}\overset{\thicksim}{\mathbb {I}} = \Pi \circ
T_{q_e}\mathbb {I}|_{\mathfrak{g}_{q_e}} $ and
$T_{q_e}\widehat{\mathbb {I}} = \Pi \circ
T_{q_e}\mathbb {I}|_{\mathfrak{k}}$.
Expanding
$\xi$ in the basis
$\{\xi_1,
\dots, \xi_p\}$ as $\xi= \alpha^i \xi_i $ and taking into account
the above expression, the system \eqref{linear system to be
solved} is equivalent to the following system of linear
equations in the unknowns $\alpha^1, \dots, \alpha^p$
\[
A_{a b} \alpha^a + B_b = 0, \quad a,b = 1, \dots , p,
\]
where
\begin{align}
A_{a b}&: = \left\langle A_{v_{q_e}} \xi_a, \xi_b \right\rangle -
2 \left\langle \left(T_{q_e}\mathbb {I}(v_{q_e}) \circ
\widehat{\mathbb {I}}(q_e)^{-1} \circ T_{q_e} \overset{\thicksim}
{\mathbb {I}}(v_{q_e})\right) \xi_a, \xi_b
\right\rangle
\label{A}\\
B_b&:=\left\langle \left(A_{v_{q_e}} \circ \widehat{\mathbb{I}}
(q_{e})^{-1}\circ \Pi_1\right) \mu, \xi_b \right\rangle
- 2 \left\langle \left(T_{q_e}\mathbb {I}(v_{q_e}) \circ
\widehat{\mathbb
{I}}(q_e)^{-1} \circ T_{q_e}\widehat{\mathbb
{I}}(v_{q_e}) \circ
\widehat{\mathbb{I}} (q_{e})^{-1} \circ \Pi_1 \right)\mu, \xi_b
\right\rangle
\label{B} \\
&\qquad +2 \left\langle \left(T_{q_e}\mathbb {I}(v_{q_e}) \circ
\widehat{\mathbb{I}}(q_e)^{-1}\right)\beta'(\mu), \xi_b
\right\rangle
- \left\langle \beta''(\mu), \xi_b \right\rangle.
\nonumber
\end{align}
Denote by $A: = [A_{a b}] $ the $p\times p $ matrix with entries
$A_{a b} $. Thus, if $v_{q_{e}}\notin
\mathcal{Z}_{\mu}=:\{v_{q_{e}}\in B\cap (T_{q_{e}}Q)_{\{e\}}\mid
\det A=0\}$ this linear system has a unique solution for
$\alpha^1, \dots , \alpha^p $, that is for $\xi$, as function of $
v_{q_{e}}$, $\mu $. we shall denote this solution by
$\xi_0(v_{q_e}, \mu)$.
\textit{Summarizing, if $v_{q_e} \notin
\mathcal{Z}_{\mu}$, then $\xi_0(v_{q_e}, \mu) $ is the unique
solution of the equation}
\begin{equation}
\label{second step in LS}
\frac{\partial ^{2}\varphi }{\partial
\tau ^{2}}(0,v_{q_{e}},\mu ,\xi )=0.
\end{equation}
\begin{lemma}
\label{properties of Z} The set $\mathcal{Z}_{\mu}$ is closed and
$G_{q_{e}}$--invariant in $B\cap (T_{q_e}Q)_{\{e\}}$.
\end{lemma}
\begin{proof}
The set $\mathcal{Z}_{\mu}$ is obviously closed. Since
$\mathfrak{k}$ is $G_{q_{e}}$--invariant it follows that
$\mathfrak{k}^\circ$ is $G_{q_e}$--invariant. Formula
\eqref{equivariance of I} shows that
$\mathbb{I}(q_{e})\mathfrak{g}$ is also $G_{q_e}$--invariant.
Thus the direct sum $\mathbb{I}(q_{e}) \mathfrak{g}\oplus
\mathfrak{k}^{\circ }$ is a $G_{q_{e}}$--invariant decomposition
of $\mathfrak{g}^{\ast }$ and therefore $\Pi : \mathfrak{g}^\ast
\rightarrow \mathbb {I}(q_e) \mathfrak{g}$ is
$G_{q_{e}}$--equivariant. From the $G_{q_{e}}$--equivariance of
$\operatorname{Exp}_{q_{e}}$ and \eqref{equivariance of I}, it
follows that $\mathbb{I} (\operatorname{Exp} _{q_{e}}(h\cdot
v_{q_{e}}))=\operatorname{Ad}^\ast_{h^{-1}}\circ
\mathbb{I}(\operatorname{Exp} _{q_{e}}(v_{q_{e}})) \circ
\operatorname{Ad}_{h^{-1}} = \operatorname{Ad}^\ast_{h^{-1}}\circ
\mathbb{I}(\operatorname{Exp} _{q_{e}}(v_{q_{e}}))$ for any $h
\in G_{q_e}$ since $G_{q_e} \subset \mathbb{T}$ and is therefore
Abelian. Thus
\begin{align*}
\overset{\thicksim}{\mathbb{I}}(\operatorname{Exp}
_{q_{e}}(h\cdot v_{q_{e}}))
&= \Pi \circ \mathbb{I}(\operatorname{Exp}
_{q_{e}}(T_{q_{e}}\Psi _{h}\cdot v_{q_{e}}))|_{\mathfrak{g}_{q_e}}
= \Pi\circ \operatorname{Ad}^\ast_{h^{-1}}\circ
\mathbb{I}(\operatorname{Exp}
_{q_{e}}(v_{q_{e}}))|_{\mathfrak{g}_{q_e}}\\
&= \operatorname{Ad}^\ast_{h^{-1}} \circ
\Pi\circ \mathbb{I}(\operatorname{Exp}
_{q_{e}}(v_{q_{e}}))|_{\mathfrak{g}_{q_e}}
= \operatorname{Ad}^\ast_{h^{-1}} \circ
\overset{\thicksim}{
\mathbb{I}}(\operatorname{Exp} _{q_{e}}(v_{q_{e}}))
\end{align*}
for all $h\in G_{q_{e}}$ and $v_{q_{e}}\in B$. Replacing here
$v_{q_e} $ by $sv_{q_e} $ and taking the $s $--derivative at
zero, shows that
$T_{q_e}\overset{\thicksim}{\mathbb {I}}(h\cdot v_{q_e}) \xi =
\operatorname{Ad}^\ast_{h^{-1}}\left(
T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \xi
\right)$ for any $h\in G_{q_{e}}$ and $\xi \in
\mathfrak{g}_{q_{e}}$, that is,
$T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \xi $
is $G_{q_{e}}$--equivariant
as a function of $v_{q_{e}}$, for all $\xi \in
\mathfrak{g}_{q_{e}}$. Similarly $ T_{q_e} \mathbb
{I}(h\cdot v_{q_e}) = \operatorname{Ad}^\ast_{h^{-1}} \circ
T_{q_e} \mathbb{I}( v_{q_e}) \circ \operatorname{Ad}_{h^{-1}}$.
From \eqref{equivariance of I} and the definition of
$\widehat{\mathbb {I}}(q_e)^{-1}$, it follows that
$\widehat{\mathbb {I}}(q_e)^{-1} = \operatorname{Ad}_h \circ
\widehat{\mathbb{I}}(q_e)^{-1} \circ \operatorname{Ad}^\ast_h$ for
any $h \in G_{q_e}$. Thus, for $h \in G_{q_e}$, the second summand
in $A_{ab}$ becomes
\begin{align*}
&\left\langle \left(T_{q_e}\mathbb {I}(h\cdot v_{q_e}) \circ
\widehat{\mathbb {I}}(q_e)^{-1} \circ T_{q_e} \overset{\thicksim}
{\mathbb {I}}(h\cdot v_{q_e})\right) \xi_a, \xi_b
\right\rangle \\
&= \left\langle \left(\operatorname{Ad}^\ast_{h^{-1}} \circ
T_{q_e} \mathbb{I}( v_{q_e}) \circ \operatorname{Ad}_{h^{-1}}
\circ \widehat{\mathbb {I}}(q_e)^{-1} \circ
\operatorname{Ad}^\ast_{h^{-1}}\circ
T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \right)
\xi_a,
\xi_b \right\rangle\\
&= \left\langle \left(\operatorname{Ad}^\ast_{h^{-1}} \circ
T_{q_e} \mathbb{I}( v_{q_e})
\circ \widehat{\mathbb {I}}(q_e)^{-1} \circ
T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \right)
\xi_a,
\xi_b \right\rangle\\
&= \left\langle \left(
T_{q_e} \mathbb{I}( v_{q_e})
\circ \widehat{\mathbb {I}}(q_e)^{-1} \circ
T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \right)
\xi_a,
\operatorname{Ad}_{h^{-1}}\xi_b \right\rangle\\
&= \left\langle \left(
T_{q_e} \mathbb{I}( v_{q_e})
\circ \widehat{\mathbb {I}}(q_e)^{-1} \circ
T_{q_e}\overset{\thicksim}{\mathbb {I}}(v_{q_e}) \right)
\xi_a,
\xi_b \right\rangle
\end{align*}
since $\operatorname{Ad}_{h^{-1}}\xi_b = 0 $ because $h \in
G_{q_e}$ and $\xi_b \in \mathfrak{g}_{q_e}$. This shows that the
second summand in $A_{ab}$ is $G_{q_e}$-- invariant.
Next, we show that the first summand in $A_{ab}$ is $G_{q_e}$--
invariant. To see this note that
\[
\left\langle A_{v_{q_e}}\xi_a, \xi_b \right\rangle
= \left.\frac{\partial}{\partial
\tau}\right|_{\tau= 0}\left\langle T_{\tau v_{q_e}}(\mathbb {I}
\circ \operatorname{Exp}_{q_e})(v_{q_e})\xi_a, \xi_b \right\rangle
= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\mathbb{I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))\xi_a, \xi_b
\right\rangle.
\]
Therefore, for any $h \in G_{q_e} $ we get from
\eqref{equivariance of I}
\begin{align*}
\left\langle A_{h\cdot v_{q_e}}\xi_a, \xi_b \right\rangle
&= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\mathbb{I}(\operatorname{Exp}_{q_e}(\tau h \cdot v_{q_e}))\xi_a,
\xi_b
\right\rangle
= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\mathbb{I}(h \cdot \operatorname{Exp}_{q_e}(\tau v_{q_e}))\xi_a,
\xi_b \right\rangle \\
&= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\operatorname{Ad}_{h^{-1}}^\ast
\mathbb{I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))
\operatorname{Ad}_{h^{-1}}\xi_a,
\xi_b \right\rangle\\
&= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\mathbb{I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))
\operatorname{Ad}_{h^{-1}}\xi_a,
\operatorname{Ad}_{h^{-1}}\xi_b \right\rangle\\
&= \left.\frac{\partial^2}{\partial
\tau^2}\right|_{\tau= 0} \left\langle
\mathbb{I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))\xi_a,
\xi_b \right\rangle
= \left\langle A_{v_{q_e}}\xi_a, \xi_b \right\rangle,
\end{align*}
as required.
\end{proof}
\begin{proposition}
\label{determination of xi} The equation $\varphi (\tau
,v_{q_{e}},\mu ,\xi)=0 $ for $(\tau ,v_{q_{e}},\mu ,\xi) \in
I\times \left(B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu}\right) \times \mathfrak{g}^{\ast }\times
\mathfrak{g}_{q_{e}}$ has a unique smooth solution $\xi (\tau
,v_{q_{e}},\mu )\in \mathfrak{g}_{q_{e}}$ for $(\tau, v_{q_e},
\mu) \in I\times (B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu})\times \mathfrak{g}^{\ast }$.
\end{proposition}
\begin{proof} Denote by $D_\xi$ the Fr\'echet derivative relative
to the variable $\xi \in \mathfrak{g}_{q_e}$. Recall that
$\xi_0(v_{q_e}, \mu) \in \mathfrak{g}_{q_e}$ is the unique
solution of the equation $\frac{\partial ^{2} \varphi}{\partial
\tau ^{2}}(0,v_{q_{e}},\mu ,\xi )=0$. Formulas
\eqref{second tau derivative of phi at zero} and \eqref{first
tau derivative of eta at zero} yield
\begin{align}
\label{expanded second tau derivative of phi at zero}
\frac{\partial ^{2}\varphi }{\partial \tau^{2}} (0,v_{q_{e}},\mu,
\xi )
&= (Id-\Pi)\left[ A_{v_{q_e}}(\xi +\eta _{\mu})
- 2\left(T_{q_e}\mathbb{I}(v_{q_{e}}) \circ
\widehat{\mathbb{I}}(q_e)^{-1} \circ T_{q_e}\overset{\thicksim}
{\mathbb {I}}(v_{q_e})\right) \xi \right. \\
& \phantom{T_{q_e}\overset{\thicksim}
{\mathbb{I}}(v_{q_e})}
- 2\left(T_{q_e}\mathbb{I}(v_{q_{e}}) \circ
\widehat{\mathbb {I}}(q_e)^{-1} \circ
T_{q_e}\widehat{\mathbb{I}}(v_{q_e}) \circ
\widehat{\mathbb{I}} (q_{e})^{-1}\right)(\Pi _{1}\mu )
\nonumber \\
&\left. \phantom{T_{q_e}\overset{\thicksim}
{\mathbb{I}}(v_{q_e})}
+ 2\left(T_{q_e}\mathbb{I}(v_{q_{e}}) \circ\widehat{\mathbb{I}}
(q_{e})^{-1}\right)(\beta'(\mu)) - 2\beta ''(\mu) \right]
\nonumber
\end{align}
and hence
\[
D_\xi\frac{\partial ^{2}\varphi }{\partial \tau^{2}} (0,v_{q_{e}},
\mu, \xi_0(v_{q_e}, \mu))
= (Id-\Pi)\left[ A_{v_{q_e}}|_{\mathfrak{g}_{q_e}}
- 2T_{q_e}\mathbb{I}(v_{q_{e}}) \circ
\widehat{\mathbb{I}}(q_e)^{-1} \circ T_{q_e}\overset{\thicksim}
{\mathbb {I}}(v_{q_e})
\right] : \mathfrak{g}_{q_e} \rightarrow \mathfrak{k}^\circ.
\]
We shall prove that this linear map is injective. To see this,
note that relative to the basis $\{\xi_1, \dots , \xi_p\}$ of
$\mathfrak{g}_{q_e}$ this linear operator has matrix $A $ by
\eqref{A}. Thus, if $v_{q_e} \notin \mathcal{Z}_{\mu}$, this
matrix is invertible. In particular, this linear operator is
injective.
Since $\mathfrak{g} = \mathfrak{g}_{q_e} \oplus
\mathfrak{k}$, it follows that $\dim \mathfrak{g}_{q_e} = \dim
\mathfrak{g} - \dim \mathfrak{k} = \dim \mathfrak{k}^\circ $, so
the injectivity of the map $D_\xi\frac{\partial ^{2}\varphi
}{\partial \tau^{2}} (0,v_{q_{e}}^0, \mu^0, \xi_0(v_{q_e}^0,
\mu^0))$ implies that it is an isomorphism. Therefore, if
$v_{q_{e}}\in B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu}$ is near $v_{q_e}^0 $, the implicit function
theorem, guarantees the existence of an open neighborhood
$V_{0}\subset I\times (B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu})\times
\mathfrak{g}^{\ast }$ containing $(0, v_{q_e}^0, \mu^0) \in
\{0\}\times (B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu}) \times
\mathfrak{g}^{\ast }$ and of a unique smooth function $\xi
:V_{0}\rightarrow \mathfrak{g}_{q_{e}}$ satisfying $\varphi (\tau
,v_{q_{e}},\mu , \xi(\tau
,v_{q_{e}},\mu) )=0$ such that $\xi(0,v_{q_{e}}^0,\mu^0) =
\xi_0(v_{q_e}^0,
\mu^0)$.
On the other hand, for $\tau \neq 0$, the equation $\varphi (\tau
,v_{q_{e}},\mu ,\cdot )=0$ has a unique solution for $\xi $,
namely the $\mathfrak{g}_{q_{e}}$-component of
$\mathbb{I}(\operatorname{Exp} _{q_{q}}(\tau
v_{q_{e}}))^{-1}\beta (\tau ,\mu )$, which is a smooth function
of $\tau ,v_{q_{e}},\mu $. This is true since $\xi + \eta =
\mathbb{I}(\operatorname{Exp} _{q_{q}}(\tau v_{q_{e}}))^{-1}\beta
(\tau ,\mu )$ by construction and we determined the two
components $\xi\in \mathfrak{g}_{q_e} $ and $\eta \in
\mathfrak{k}$ in $\mathfrak{g} = \mathfrak{g}_{q_e} \oplus
\mathfrak{k}$ via the Lyapunov-Schmidt method, precisely in order
that this equality be satisfied. Therefore, the solution
$\xi(\tau, v_{q_e}, \mu)$ obtained above by the implicit
function theorem must coincide with the $\mathfrak{g}_{q_{e}}$-component of
$\mathbb{I}(\operatorname{Exp} _{q_{q}}(\tau
v_{q_{e}}))^{-1}\beta (\tau ,\mu )$ for $\tau>0 $.
Since this entire argument involving the Lyapunov-Schmidt
procedure was carried out for any $(v_{q_e}^0, \mu^0)$, it
follows that the equation
$\varphi (\tau,v_{q_{e}},\mu ,\xi)=0$ has a unique smooth
solution $\xi (\tau ,v_{q_{e}},\mu )\in \mathfrak{g}_{q_{e}}$
for $(\tau, v_{q_e},
\mu) \in I\times (B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu})\times \mathfrak{g}^{\ast }$.
\end{proof}
\begin{remark}
\label{remark on smoothness}
\normalfont
The previous proposition says that if we define
\begin{equation*}
\zeta(\tau ,v_{q_{e}},\mu )=\mathbb{I}(\operatorname{Exp}
_{q_{e}}(\tau v_{q_{e}}))^{-1}\beta (\tau ,\mu )
\end{equation*}
on $(I\setminus \{0\})\times (B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu})\times \mathfrak{g}^{\ast }$, then $\zeta(\tau
,v_{q_{e}},\mu )$ can be smoothly extended for $\tau=0$.
We have, in fact, $\zeta(\tau ,v_{q_{e}},\mu )= \xi(\tau
,v_{q_{e}},\mu ) + \eta (\tau ,v_{q_{e}},\mu ,\xi(\tau
,v_{q_{e}},\mu ) )$, where $\eta(\tau ,v_{q_{e}},\mu, \xi )$ was
found in the first step of the Lyapunov-Schmidt procedure and
$\xi(\tau ,v_{q_{e}},\mu )$ in the second step, as given in
Proposition \ref{determination of xi}. Note also that $\zeta(0,
v_{q_e}, \mu) = \xi_0(v_{q_e}, \mu) + \widehat{\mathbb
{I}}(q_e)^{-1} \Pi_1 \mu \in \mathfrak{t}$.
\end{remark}
\subsection{A simplified version of the amended potential
criterion} At this point we have a candidate for a bifurcating
branch from the set of relative equilibria $\mathfrak{t} \cdot
q_e $. This branch will start at $\zeta(0, v_{q_e}, \mu)_Q(q_e)
\in \mathfrak{t}\cdot q_e \subset T_{q_e}Q$. By Lemma \ref{same
isotropy}, the isotropy subgroup of $\zeta(0, v_{q_e}, \mu)_Q(q_e)
$ equals $G_{q_e} $, for any $v_{q_e} \in B\cap
(T_{q_{e}}Q)_{\{e\}}\setminus \mathcal{Z}_{\mu}$ and $\mu\in
\mathfrak{g}^\ast$. The isotropy groups of the points on the
curve $\zeta(\tau, v_{q_e}, \mu)_Q(\operatorname{Exp}_{q_{e}}
(\tau v_{q_e}))$, for $\tau\neq 0 $, are all trivial, by
construction. Hence $\zeta(\tau, v_{q_e},
\mu)_Q(\operatorname{Exp}_{q_{e}}(\tau v_{q_e}))$ is a curve
that has the properties of the bifurcating branch of relative
equilibria with broken symmetry that we are looking for. We do
not know yet that all points on this curve are in fact relative
equilibria. Thus, we shall search for conditions on $v_{q_e}$ and
$\mu$ that guarantee that each point on the curve
$\tau\mapsto\zeta(\tau, v_{q_e},
\mu)_Q(\operatorname{Exp}_{q_{e}}(\tau v_{q_e}))$ is a relative
equilibrium. This will be done by using the amended potential
criterion (see Proposition \ref{amended potential criterion}) which
is applicable because all base points of this curve, namely
$\operatorname{Exp}_{q_e}(\tau v_{q_e})$, have trivial isotropy
for $\tau\neq 0$. To carry this out, we need some additional
geometric information.
\medskip
From standard theory of proper Lie group actions (see e.g.
\cite{dk}, \S 2.3, or \cite{kawakubo}) it follows that
the map
\begin{equation}
\label{corespondenta}
[v_{q_{e}},\mu ]_{G_{q_{e}}} \in (B
\times \mathfrak{g}^{\ast })/G_{q_{e}}
\longmapsto [\operatorname{Exp}_{q_{e}}(v_{q_{e}}),\mu ]_{G} \in
((G \cdot \operatorname{Exp}_{q_e} B) \times \mathfrak{g}^{\ast
})/G
\end{equation}
is a homeomorphism of $(B\times \mathfrak{g}^{\ast })/G_{q_{e}}$
with $((G \cdot \operatorname{Exp}_{q_e} B) \times
\mathfrak{g}^{\ast })/G$ and that its restriction to $ ((B\cap
(T_{q_{e}}Q)_{\{e\}} \setminus \mathcal{Z}_{\mu})\times
\mathfrak{g}^{\ast })/G_{q_{e}}$ is a diffeomorphism onto its
image. We think of a pair $(\operatorname{Exp}_{q_e}(v_{q_e}),
\mu)$ as the base point of a relative equilibrium and its
momentum value. All these relative equilibria come in
$G$-orbits. The homeomorphism \eqref{corespondenta} allows the
identification of $G$-orbits of relative equilibria with
$G_{q_e}$-orbits of certain pairs $(v_{q_e},\mu)$. We shall work
in what follows on both sides of this identification, based on
convenience.
We will need the
following lemma, which is a special case of stability of
the transversality of smooth maps (see e.g. \cite{gp}).
\begin{lemma}
Let $G$ be a Lie group acting on a Riemannian manifold $Q$, $q\in
Q$, and let $\mathfrak{k}\subset \mathfrak{g}$ be a subspace
satisfying $\mathfrak{k}\cap \mathfrak{g}_{q}=\{0\}$. Let
$V\subset T_{q}Q$ be a subspace such that $\mathfrak{k}\cdot
q\oplus V=T_{q}Q$. Then there is an $\epsilon >0$ such that if
$\|v_{q}\| <\epsilon $,
\begin{equation*}
T_{\operatorname{Exp} _{q}(v_{q})}Q=\mathfrak{k}\cdot \operatorname{Exp} _{q}(v_{q})\oplus
(T_{v_{q}}\operatorname{Exp} _{q})V.
\end{equation*}
\end{lemma}
To deal with $G $-orbits of relative equilibria, we need a
different splitting of the same nature. The following result
is modeled on a proposition in \cite{hm}.
\begin{proposition}
\label{decomposition of the tangent space adapted to sigma bar}
Let $v_{q_e} \in B\cap (T_{q_{e}}Q)_{\{e\}} \setminus
\mathcal{Z}_{\mu}$ be given. Consider the principal
$G_{q_e}$-bundle $B\cap (T_{q_{e}}Q)_{\{e\}} \setminus
\mathcal{Z}_{\mu} \rightarrow [B\cap (T_{q_{e}}Q)_{\{e\}}
\setminus \mathcal{Z}_{\mu}]/G_{q_{e}}$ (this is implied by Lemma
\ref {properties of Z}). Let $\widetilde{U}$ be a neighborhood of
$[0_{q_e}] \in (T_{q_e}Q)/G_{q_e}$ and define the open set $U: =
\widetilde{U} \cap [B\cap (T_{q_{e}}Q)_{\{e\}} \setminus
\mathcal{Z}_{\mu}]/G_{q_{e}} $ in $[B\cap (T_{q_{e}}Q)_{\{e\}}
\setminus \mathcal{Z}_{\mu}]/G_{q_{e}} $. Let $\sigma :U\subset
[B\cap (T_{q_{e}}Q)_{\{e\}})\setminus
\mathcal{Z}_{\mu}]/G_{q_{e}}\rightarrow B\cap
(T_{q_{e}}Q)_{\{e\}}\setminus \mathcal{Z}_{\mu}$ be a smooth
section, $[v_{_{q_{e}}}]\in U$, and $\overline{\sigma
}:=\operatorname{Exp} _{q_{e}}\circ \sigma : U \rightarrow Q$.
Then there exists $\epsilon >0$ such that for $ 0<\tau <\epsilon
$ sufficiently small, we have
\begin{equation*}
T_{\overline{\sigma }([\tau v_{q_{e}}])}Q
=\mathfrak{t}\cdot \overline{\sigma }([\tau v_{q_{e}}])
\oplus T_{[\tau v_{q_{e}}]}
\overline{\sigma}(T_{[\tau v_{q_{e}}]}U)
\oplus (T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}_{q_{e}})
(\mathfrak{k} _{2}\cdot q_{e}).
\end{equation*}
\end{proposition}
\begin{proof}
Since $\mathfrak{g}= \mathfrak{k}_0 \oplus \mathfrak{k}_1 \oplus
\mathfrak{k}_2 $ and $\mathfrak{k}_0 = \mathfrak{g}_{q_e}$ we
have
$T_{q_{e}}Q=\mathfrak{k}_{1}\cdot q_{e}\oplus
\mathfrak{k}_{2}\cdot q_{e}\oplus (\mathfrak{g}\cdot
q_{e})^{\perp }$. Apply the above lemma with
$\mathfrak{k}=\mathfrak{k}_{1}$ and $V=\mathfrak{k}_{2}\cdot
q_{e}\oplus (\mathfrak{g}\cdot q_{e})^{\perp }$. For the
$\epsilon >0 $ in the statement choose $\tau $ such that $0<
\tau< \epsilon $ and $\| \sigma([ \tau v_{q_e}]) \| <
\epsilon$. Then
\begin{align}
\label{decomposition of q for the section}
T_{\overline{\sigma }([\tau v_{q_{e}}])}Q
&=\mathfrak{k}_{1}\cdot \overline{
\sigma }([\tau v_{q_{e}}])
\oplus (T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}})(\mathfrak{k}_{2}\cdot q_{e} \oplus (\mathfrak{g}\cdot
q_e) ^\perp)\\
&= \mathfrak{k}_{1}\cdot \overline{\sigma }([\tau v_{q_{e}}])
\oplus (T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}})((\mathfrak{g}\cdot q_e) ^\perp )
\oplus (T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}})(\mathfrak{k}_2 \cdot q_e). \nonumber
\end{align}
since $\operatorname{Exp}_{q_e}$ is a diffeomorphism on $B
\subset (\mathfrak{g}\cdot q_e)^\perp$. Since $(\sigma ,U)$ is a
smooth local section, $\mathcal{Z}_{\mu}$ is closed and
$G_{q_e}$-invariant in $B\cap (T_{q_e}Q)_{\{e\}}$, and
$(T_{q_e}Q)_{\{e\}}$ is open in $T_{q_e} Q $, it follows that
$B\cap (T_{q_e}Q)_{\{e\}}$ is open in $(\mathfrak{g}\cdot
q_e)^\perp$ and thus we get
\begin{equation*}
(\mathfrak{g}\cdot q_e)^\perp = T_{\sigma ([\tau
v_{q_{e}}])}(B\cap (T_{q_{e}}Q)_{\{e\}}\setminus
\mathcal{Z}_{\mu}) = T_{[\tau v_{q_{e}}]}\sigma (T_{[\tau
v_{q_{e}}]}U) \oplus \mathfrak{k}_{0}\cdot \sigma ([\tau
v_{q_{e}}]),
\end{equation*}
where $\mathfrak{k}_{0}\cdot \sigma ([\tau v_{q_{e}}]) =
\{\zeta_{T_{q_e}Q} (\sigma([ \tau v_{q_e}]) \mid \zeta\in
\mathfrak{k}_0 \}$. The
$G_{q_e}$-equivariance of
$\operatorname{Exp}_{q_e}$ implies that
\[
T_{u_{q_{e}}}\operatorname{Exp} _{q_{e}}(\xi
_{T_{q_e}Q}(u_{q_{e}}))=\xi _{Q}(\operatorname{Exp}
_{q_{e}}(u_{q_{e}})) \quad \text{for~all} \quad
\xi \in \mathfrak{k}_{0}, \quad u_{q_{e}}\in T_{q_{e}}Q
\]
and hence
\begin{align}
\label{decomposition of one summand}
&(T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}})((\mathfrak{g}\cdot q_e) ^\perp ) \\
&\qquad \qquad =(T_{\sigma([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}}
\circ T_{[\tau v_{q_e}]}\sigma )(T_{[\tau v_{q_{e}}]}U)\oplus
(T_{\sigma ([\tau v_{q_{e}}])}\operatorname{Exp}
_{q_{e}})(\mathfrak{k}_{0}\cdot
\sigma ([\tau v_{q_{e}}])) \nonumber \\
&\qquad \qquad =T_{[\tau v_{q_{e}}]}\overline{\sigma
}(T_{[\tau v_{q_{e}}]}U)\oplus
\mathfrak{k}_{0}\cdot \overline{\sigma }([\tau v_{q_{e}}]).
\nonumber
\end{align}
Introducing \eqref{decomposition of one summand} in
\eqref{decomposition of q for the section} and taking into
account that $\mathfrak{t}=\mathfrak{k}_{0}\oplus
\mathfrak{k}_{1}$ we get the statement of the proposition.
\end{proof}
We want to find pairs $(v_{q_e}, \mu) $ such that
$\mathbf{d}V_{\beta(\tau, \mu)} (\operatorname{Exp} _{q_{e}}(\tau
v_{q_{e}}))=0$ for $\tau> 0 $. Since $V_{\beta(\tau, \mu)}$ is
$G_{\beta(\tau,
\mu)}$-invariant, this condition will hold if we only verify it
on a subspace of $T_{\operatorname{Exp}_{q_e}(\tau v_{q_e})} Q$
complementary to $\mathfrak{g}_{\beta(\tau,\mu)} \cdot
\operatorname{Exp} _{q_{e}}(\tau v_{q_{e}}) = \mathfrak{t}\cdot
\operatorname{Exp} _{q_{e}}(\tau v_{q_{e}})$. The previous
decomposition of the tangent space immediately yields the
following result.
\begin{corollary}
\label{criteriu}
Suppose that $\mu\in \mathfrak{g}^\ast$ is such
that $\mathfrak{g}_{\beta( \tau, \mu)} = \mathfrak{t}$ for all
$\tau$ in a neighborhood of zero. Let $U$ and $\sigma $ be as in
Proposition \ref{decomposition of the tangent space adapted to
sigma bar}, $[v_{q_{e}}]\in U$, and $\overline{\sigma
}:=\operatorname{Exp} _{q_{e}}\circ \sigma $. Then there is an
$\epsilon >0$ such that $\mathbf{d}V_{\beta (\tau,
\mu)}(\overline{\sigma }([\tau v_{q_{e}}])=0$ if and only if
$\mathbf{d}(V_{\beta (\tau, \mu )}\circ \overline{\sigma })([\tau
v_{q_{e}}])=0$ and $\mathbf{d}(V_{\beta (\tau, \mu )}\circ
\operatorname{Exp} _{q_{e}})( \sigma ([\tau v_{q_{e}}]))|
_{\mathfrak{k}_{2}\cdot q_{e}}=0$ for $ 0<\tau <\epsilon $.
\end{corollary}
\subsection{The study of two auxiliary functions}
Let $I $ be an open interval containing zero. Recall that $p =
\dim \mathfrak{g}_{q_e} = \dim \mathfrak{m}_0 $. Let
$\vartheta_{1}$ be an element of a basis
$\{\vartheta _{1},\vartheta _{2},...,
\vartheta _{p}\}$ for $\mathfrak{m}_{0}$ and define $\beta
:(I\setminus \{0\})\times (\mathfrak{m}_{1}\oplus
\mathfrak{m}_{2})\rightarrow
\mathfrak{g}^{\ast }$ by
\begin{equation*}
\beta (\tau ,\mu)=\Pi _{1}\mu+\tau \Pi_{2}\mu
+\tau ^{2}\vartheta _{1},
\end{equation*}
where $\Pi_1 : \mathfrak{g}^\ast\rightarrow \mathfrak{m}_1 =
\mathbb {I}(q_e) \mathfrak{t}$ and $\Pi_2: \mathfrak{g}^\ast
\rightarrow \mathfrak{m}_2 = \mathfrak{t}^\circ $. Notice
that this function is a particular case of
\begin{equation*}
\beta (\tau ,\mu )=\Pi _{1}\mu +\tau \beta ^{\prime }(\mu )+\tau ^{2}\beta
^{\prime \prime }(\mu ),
\end{equation*}
by choosing $\beta' (\mu) = \Pi_2 \mu$ and $\beta '' (\mu) =
\vartheta_1 $.
Recall that $\mathbb {I}(q_e) = \mathfrak{m}_1
\oplus\mathfrak{m}_2 $ by Lemma \ref{moment of inertia
isomorphism} and that $\mathbf{J}_L(\mathfrak{g}\cdot q_e) =
\mathbb {I}(q_e) \mathfrak{g}$ from the definition of
$\mathbf{J}_L $.
\begin{theorem}
\label{theorem about F}
The smooth function $F_{1}:(I\setminus \{0\})\times U\times
\mathbf{J}_{L}(\mathfrak{g}\cdot q_{e})\rightarrow \mathbb{R}$
defined by
\begin{equation*}
F_{1}(\tau ,[v_{q_{e}}],\mu ):=(V_{\beta (\tau, \mu )}\circ
\overline{\sigma })(\tau [v_{q_{e}}]).
\end{equation*}
can be extended to a smooth function on $I\times U\times
\mathbf{J}_{L}(\mathfrak{g}\cdot q_{e})$, also denoted by $F_1
$. In addition
\begin{equation*}
F_{1}(\tau ,[v_{q_{e}}],\mu )=F_{0}(\mu )+\tau ^{2}F(\tau
,[v_{q_{e}}],\mu ).
\end{equation*}
where $F_{0}$, $F$ are defined on $\mathbf{J}_{L}(
\mathfrak{g}\cdot q_{e})$ and on $I\times U\times \mathbf{J}_{L}(
\mathfrak{g}\cdot q_{e})$ respectively.
\end{theorem}
\begin{proof}
Denote $v_{q_{e}}:=\sigma ([v_{q_{e}}])\in B\cap
(T_{q_{e}}Q)_{\{e\}} \setminus \mathcal{Z}_{\mu}$. One can easily
see that
\begin{equation*}
(V_{\beta (\tau, \mu)}\circ \overline{\sigma })(\tau \lbrack
v_{q_{e}}])=V(\operatorname{Exp} _{q_{e}}(\tau
v_{q_{e}}))+\frac{1}{2}\left\langle\beta (\tau ,\mu
),\mathbb{I}(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}}))^{-1}
\beta(\tau ,\mu )\right\rangle.
\end{equation*}
By Remark \ref{remark on smoothness}, the second term is smooth
even in a a neighborhood of $\tau= 0 $. Since the first term is
obviously smooth, it follows that $V_{\beta (\tau, \mu)}\circ
\overline{\sigma }$ is smooth also in a neighborhood of $\tau= 0
$. This is the smooth extension of $F_1 $ in the statement.
Let $\{\xi _{1},...,\xi _{p}\}$ be a
basis for $\mathfrak{g}_{q_{e}} \subset \mathfrak{t}$. Then,
again by Remark \ref{remark on smoothness}, we have
\begin{equation*}
\mathbb{I}(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}}))^{-1}\beta
(\tau ,\mu )=\overset{p}{\underset{a=1}{
\sum }}\alpha _{a}(\tau ,v_{q_{e}},\mu )\xi _{a}+\eta \left(\tau
,v_{q_{e}},\mu ,
\overset{p}{\underset{a=1}{\sum }}\alpha _{a}(\tau ,v_{q_{e}},\mu
)\xi _{a}\right)
\end{equation*}
where $\alpha _{1},...,\alpha _{p},\eta $ are smooth real
functions of all their arguments. In what follows we will denote
\begin{equation*}
\eta \left(\tau ,v_{q_{e}},\mu ,\overset{p}{\underset{a=1}{\sum
}}\alpha _{a}(\tau ,v_{q_{e}},\mu )\xi _{a}\right)
=\eta (\tau,v_{q_{e}},\mu ,\alpha _{1}(\tau ,v_{q_{e}},\mu
),...,\alpha _{p}(\tau ,v_{q_{e}},\mu )).
\end{equation*}
Let $\mu \in \mathbf{J}_{L}(\mathfrak{g}\cdot q_{e})=
\mathfrak{m}_1 \oplus\mathfrak{m}_2$ and $v_{q_{e}}\in B\cap
(T_{q_{e}}Q)_{\{e\}}\backslash \mathcal{Z}_{\mu}$. Since in the
computations that follow, the arguments $v_{q_e}$ and $\mu$ play
the role of parameters, we shall denote temporarily $\alpha
_{a}(\tau )=\alpha _{a}(\tau, v_{q_{e}},\mu)$, $a\in
\{1,...,p\}$, and $\eta (\tau ,\alpha _{1},...,\alpha _{p})=\eta
(\tau ,v_{q_{e}},\mu ,\alpha _{1}(\tau ,v_{q_{e}},\mu ),...,\alpha
_{p}(\tau ,v_{q_{e}},\mu ))$. Then by \eqref{first tau derivative
of eta at zero} we get
\begin{align*}
\frac{\partial \eta }{\partial \tau }(0,\alpha _{1},...,
\alpha _{p})
= &-\overset{p}{\underset{a=1}{\sum }}\alpha_{a}
\left(\widehat{\mathbb{I}}(q_{e})^{-1}\circ
T_{q_e}\overset{\thicksim}{\mathbb{I}}
(v_{q_{e}})\right)\xi _{a} \\
& -\left(\widehat{\mathbb{I}}(q_{e})^{-1} \circ
T_{q_e}\widehat{\mathbb{I}}(v_{q_{e}}) \circ
\widehat{\mathbb{I}}(q_{e})^{-1} \right)\Pi_{1}\mu
+\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_{2}\mu.
\end{align*}
Formula \eqref{pi composed with Phi} shows that
\[
\frac{\partial \eta }{\partial \alpha_a }(0,\alpha_1, \dots ,
\alpha_p) = 0
\]
Note that
\begin{equation*}
\left.V_{\beta (\tau ,\mu )}(\operatorname{Exp}_{q_{e}}(\tau
v_{q_{e}}))\right|_{\tau =0} = V(q_{e})+\frac{1}{2}
\left\langle \Pi_{1}\mu,\widehat
{\mathbb{I}}(q_{e})^{-1}\Pi_{1}\mu\right\rangle
\end{equation*}
is independent of $v_{q_{e}}$. This shows that $F_1(0, [v_{q_e}],
\mu) = F_0(\mu)$ for some smooth function on $\mathfrak{m}_1
\oplus \mathfrak{m}_2 $.
Using Remark \ref{remark on smoothness}, we get
\begin{align*}
\left.\frac{d}{d\tau }\right|_{\tau =0}&V_{\beta
(\tau ,\mu )}(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}}))
=\mathbf{d}V(q_{e})(v_{q_{e}}) + \frac{1}{2}\left\langle
\Pi_{2}\mu ,\overset{p}{
\underset{a=1}{\sum }}\alpha _{a}(0)\xi _{a}+\eta (0,\alpha _{1},...,\alpha
_{p})\right\rangle\\
&+\frac{1}{2}\left\langle \Pi_1\mu, \sum_{a=1}^p \frac{\partial
\alpha_a}{\partial \tau}(0)\left(\xi_a + \frac{\partial
\eta}{\partial \alpha_a}(0, \alpha_1, \dots , \alpha_p) \right)
+ \frac{\partial \eta}{\partial \tau}(0, \alpha _1, \dots ,
\alpha_p) \right\rangle.
\end{align*}
The first term
$\mathbf{d}V(q_{e})=0$ by
Proposition \ref{montaldi} (i).
Since $\eta(0, v_{q_e}, \mu, \xi) =
\eta_\mu = \widehat{\mathbb {I}}(q_e)^{-1}\Pi_1 \mu \in
\mathfrak{t}$ by Proposition \ref{belongs}, we get
\[
\overset{p}{\underset{ a=1}{\sum
}}\alpha _{a}(0)\xi _{a}+\eta (0,\alpha _{1},...,\alpha _{p})=
\overset{p}{\underset{a=1}{\sum }}\alpha _{a}(0)\xi _{a}
+\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_{1}\mu\in \mathfrak{t}.
\]
Thus the second term vanishes because
$\mathfrak{m}_{2}=\mathfrak{t}^{\circ }$.
As $\frac{\partial \eta }{\partial \alpha _{a}}(0, \alpha_1,
\dots , \alpha_p) =0$ and
$\mathfrak{m}_{1}$ annihilates $\mathfrak{g} _{q_{e}}$, the third
term becomes
\begin{align*}
\left\langle \Pi_{1}\mu,\frac{\partial \eta }{\partial \tau }
(0,\alpha _{1},...,\alpha_{p})\right\rangle
=&-\overset{p}{\underset{a=1}{\sum}}\alpha _{a}
\left\langle \Pi_1\mu,
\left(\widehat{\mathbb{I}}(q_{e})^{-1}\circ
T_{q_e}\overset{\thicksim}{\mathbb{I}}
(v_{q_{e}})\right)\xi _{a}\right\rangle \\
&-\left\langle\Pi_{1}\mu,\left(\widehat{\mathbb{I}}(q_{e})^{-1}
\circ T_{q_e}\widehat{\mathbb{I}}(v_{q_{e}}) \circ
\widehat{\mathbb{I}}(q_{e})^{-1} \right)\Pi_{1}\mu\right\rangle\\
&+\left\langle\Pi_{1}\mu,\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_{2}\mu
\right\rangle.
\end{align*}
We will prove that each summand in this expression vanishes.
$\bullet$ Since
$\langle \mathfrak{m}_0 , \mathfrak{k}_1 \rangle = 0$, we get
\begin{align*}
&\left\langle \Pi_1\mu,
\left(\widehat{\mathbb{I}}(q_{e})^{-1}\circ
T_{q_e}\overset{\thicksim}{\mathbb{I}}
(v_{q_{e}})\right)\xi _{a}\right\rangle
= \left\langle
T_{q_e}\overset{\thicksim}{\mathbb{I}}
(v_{q_{e}})\xi_{a},\;
\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_1\mu \right\rangle \\
&\qquad = \left\langle
T_{q_e}{\mathbb{I}}
(v_{q_{e}})\xi_{a},\;
\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_1\mu \right\rangle
= \mathbf{d}\left\langle \mathbb {I}(\cdot) \xi_a,
\eta_\mu \right\rangle(q_e)(v_{q_e}) =0
\end{align*}
by \eqref{differential of i} because $\xi_a \in
\mathfrak{g}_{q_e}$ and $\eta_\mu \in \mathfrak{t}$. Thus the
first summand vanishes.
$\bullet$ The second summand equals
\begin{equation*}
\left\langle\Pi_{1}\mu,\left(\widehat{\mathbb{I}}(q_{e})^{-1}
\circ T_{q_e}\widehat{\mathbb{I}}(v_{q_{e}}) \circ
\widehat{\mathbb{I}}(q_{e})^{-1} \right)\Pi_{1}\mu\right\rangle
= \left\langle T_{q_e}\widehat{\mathbb{I}}(v_{q_{e}})
\eta_\mu, \eta_\mu \right\rangle
= \left\langle T_{q_e}\mathbb{I}(v_{q_{e}})
\eta_\mu, \eta_\mu \right\rangle
\end{equation*}
because $\langle \mathfrak{m}_0, \mathfrak{k}_1 \rangle = 0 $. We
shall prove that this term vanishes in the following way. Recall
that $\eta_\mu \in \mathfrak{k}_1 \subset \mathfrak{t}$. For any
$\zeta \in \mathfrak{t}$, hypothesis \textbf{(H)} states that
$\zeta_Q(q_e) $ is a relative equilibrium and thus, by the
augmented potential criterion (see Proposition \ref{augmented
potential criterion}), $\mathbf{d}V_\zeta(q_e) = 0 $. Since
\[
\mathbf{d}V_\zeta(q_e) (u_{q_e}) = \mathbf{d}V(q_e)(u_{q_e}) -
\frac{1}{2} \left\langle T_{q_e} \mathbb{I}(u_{q_e}) \zeta,
\zeta \right\rangle
\]
for any $u_{q_e} \in T_{q_e} Q $ and $\mathbf{d}V(q_e) = 0 $ by
Proposition \ref{montaldi} (i), it follows that $\left\langle
T_{q_e}
\mathbb{I}(u_{q_e}) \zeta,\zeta\right\rangle = 0 $. Thus the
second summand vanishes.
$\bullet$ The third summand is
\begin{equation*}
\left\langle\Pi_{1}\mu,\widehat{\mathbb{I}}(q_{e})^{-1}\Pi_{2}\mu
\right\rangle
=\langle \Pi_{2}\mu,\eta_\mu \rangle =0
\end{equation*}
because $\mathfrak{m}_{2}=\mathfrak{t}^{\circ }$.
So, we finally conclude that
\begin{equation*}
\left.\frac{d}{d\tau }\right|_{\tau =0}V_{\beta
(\tau ,\mu )}(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}})) =0
\end{equation*}
and hence, by Taylor's theorem, we have
\begin{equation*}
F_{1}(\tau ,[v_{q_{e}}],\mu )=F_{0}(\mu )+\tau ^{2}F(\tau,
[v_{q_{e}}],\mu )
\end{equation*}
for some smooth function $F $.
\end{proof}
\begin{theorem}
\label{theorem about G}
The smooth function $G_{1}:(I \setminus \{0\}) \times U\times
\mathbf{J}_{L}(\mathfrak{g}\cdot q_{e})\rightarrow
\mathfrak{k}_{2}^{\ast }$ defined by
\begin{equation*}
\left\langle G_{1}(\tau ,[v_{q_{e}}],\mu ),\varsigma
\right\rangle
=\mathbf{d}(V_{\beta (\tau ,\mu )}\circ
\operatorname{Exp}_{q_{e}})(\sigma (\tau \lbrack v_{q_{e}}]))\big(
\varsigma _{Q}(q_{e})\big), \quad \varsigma \in \mathfrak{k}_2,
\end{equation*}
can be smoothly extended to a function on $I\times U\times
\mathbf{J}_{L}(\mathfrak{g}\cdot q_{e})$, also denoted by $G_1$.
In addition,
\begin{equation*}
G_{1}(\tau ,[v_{q_{e}}],\mu )=\tau G(\tau ,[v_{q_{e}}],\mu )
\end{equation*}
where $G:I\times U\times \mathbf{J}_{L}(\mathfrak{g}\cdot
q_{e})\rightarrow \mathfrak{k}_{2}^{\ast }$ is a smooth function.
\end{theorem}
\begin{proof}
We will show that $G_{1}$ is a smooth function at $\tau =0$ and
that $G_{1}(0,[v_{q_{e}}],\mu )=0$.
Let $v_{q_{e}}=\sigma ([v_{q_{e}}])$. Then
\begin{align*}
&\left\langle G_{1}(\tau ,[v_{q_{e}}],\mu ),\varsigma
\right\rangle
= \mathbf{d}V_{\beta(\tau,\mu)}
\left(\operatorname{Exp}_{q_e}(\tau v_{q_e})\right)
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right) \\
&\qquad =\mathbf{d}V(\operatorname{Exp}_{q_{e}}(\tau v_{q_{e}}))
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right) \\
& \qquad \qquad \qquad +\frac{1}{2}\left\langle \beta (\tau
,\mu),\; T_{\operatorname{Exp}_{q_e}( \tau v_{q_e})}
(\mathbb{I}(\cdot)^{-1})
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right)
\beta (\tau ,\mu )
\right\rangle \\
&\qquad = \mathbf{d}V
\left(\operatorname{Exp}_{q_e}(\tau v_{q_e})\right)
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right)
-\frac{1}{2}\left\langle \beta(\tau,\mu),\;
\phantom{T_{\operatorname{Exp}_{q_e}( \tau v_{q_e})}\mathbb{I}}
\right.\\
&\qquad \qquad \qquad \left.
\left[ \mathbb {I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))^{-1}
\circ
T_{\operatorname{Exp}_{q_e}( \tau v_{q_e})}\mathbb{I}
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right)
\circ \mathbb {I}(\operatorname{Exp}_{q_e}(\tau v_{q_e}))^{-1}
\right] \beta( \tau, \mu)
\right\rangle \\
&\qquad =\mathbf{d}V
\left(\operatorname{Exp}_{q_e}(\tau v_{q_e})\right)
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right) \\
& \qquad \qquad \qquad -\frac{1}{2}\left\langle \zeta (\tau,
v_{q_e}, \mu),\; T_{\operatorname{Exp}_{q_e}( \tau v_{q_e})}
\mathbb{I}
\left(T_{\tau v_{q_e}} \operatorname{Exp}_{q_e}
\big(\varsigma_Q(q_e) \big) \right)
\zeta (\tau , v_{q_e}, \mu )
\right\rangle ,
\end{align*}
where $\zeta (\tau ,v_{q_{e}},\mu)
:=\mathbb{I}^{-1}((\operatorname{Exp}_{q_{e}}(\tau
v_{q_{e}}))\beta (\tau ,\mu )$. Since $\zeta (\tau ,v_{q_{e}},\mu
)$ is smooth in all variables also at $\tau =0$ by Remark
\ref{remark on smoothness}, it follows that
$\langle G_{1}(\tau ,[v_{q_{e}}],\mu ),\varsigma \rangle$ is a
smooth function of all its variables.
This expression at $\tau= 0 $ equals
\begin{align*}
&\langle G_{1}(0,[v_{q_{e}}],\mu ),\varsigma \rangle
=\mathbf{d}V(q_{e})
(\varsigma _{Q}(q_{e}))
-\frac{1}{2}\left\langle \zeta (0,
v_{q_e}, \mu),\; T_{q_e} \mathbb{I}
\left( \varsigma_Q(q_e) \right)
\zeta (0 , v_{q_e}, \mu )
\right\rangle \\
&\qquad =\mathbf{d}V(q_{e})
(\varsigma _{Q}(q_{e}))
-\frac{1}{2}\left\langle(\mathbb{I}(q_{e})[\zeta
(0,v_{q_{e}},\mu ),\varsigma ], \zeta (0,v_{q_{e}},\mu)
\right\rangle
- \frac{1}{2} \left\langle \mathbb {I}(q_e)
\zeta(0,v_{q_{e}},\mu), [ \zeta(0,v_{q_{e}},\mu), \varsigma]
\right\rangle\\
&\qquad =\mathbf{d}V(q_{e})
(\varsigma _{Q}(q_{e}))
- \left\langle \mathbb {I}(q_e)
\zeta(0,v_{q_{e}},\mu), [ \zeta(0,v_{q_{e}},\mu), \varsigma]
\right\rangle
\end{align*}
by \eqref{useful identity for I}. Since $V$ is $G$-invariant it
follows that $\mathbf{d}V(q_{e})(\varsigma _{Q}(q_{e}))=0$.
Since $\zeta (0,v_{q_{e}},\mu ) = \xi(0,v_{q_{e}},\mu ) +
\eta_\mu \in \mathfrak{g}_{q_e} \oplus \mathfrak{k}_1 =
\mathfrak{t}$ (see Remark \ref{remark on smoothness}) it follows
that
$[\zeta (0,v_{q_{e}},\mu ),\varsigma ]\in [ \mathfrak{t},
\mathfrak{g}]$. By Proposition \ref{montaldi} (ii), we have
$\mathbb{I}(q_{e})\mathfrak{t} \subset [ \mathfrak{g},
\mathfrak{t}]^{\circ }$ and hence the second term above also
vanishes. Thus we get
$\langle G_{1}(0,[v_{q_{e}}],\mu ),\varsigma \rangle =0$ for any
$\varsigma \in \mathfrak{k}_2 $, that is, $G_{1}(0,[v_{q_{e}}],\mu
) = 0 $ which proves the theorem.
\end{proof}
\subsection{Bifurcating branches of relative equilibria}
Let $(Q,\langle\!\langle \cdot ,\cdot \rangle\!\rangle _{Q},V,G)$
be a simple mechanical $G$-system, with $G$ a compact Lie group
with the Lie algebra $\mathfrak{g}$. Let $ q_{e}\in Q$ be a
symmetric point whose isotropy group $G_{q_{e}}$ is contained in a
maximal torus $\mathbb{T}$ of $G$. Denote by $\mathfrak{t} \subset
\mathfrak{g}$ the Lie algebra of
$\mathbb{T}$.
Let $B\subset (\mathfrak{g}\cdot q_{e})^{\perp }$ be a
$G_{q_{e}}$--invariant open neighborhood of $0_{q_{e}}\in
(\mathfrak{g}\cdot q_{e})^{\perp }$ such that the exponential map
is injective on $B $ and for any $ q \in G\cdot
\operatorname{Exp}_{q_{e}}(B)$ the isotropy subgroup $G_{q} $ is
conjugate to a (not necessarily proper) subgroup of $G_{q_{e}}$.
Define the closed $G_{q_e}$--invariant subset
$\mathcal{Z}_{\mu^{0}}=:\{v_{q_{e}}\in B\cap
(T_{q_{e}}Q)_{\{e\}}\mid \det A=0\}$, where ${\mu^{0}}\in
\mathfrak{m}_{1}\oplus \mathfrak{m}_{2}$ is arbitrarily chosen and
the entries of the matrix $A $ are given in \eqref{A}. Let $U
\subset [B\cap (T_{q_{e}}Q)_{\{e\}} \setminus
\mathcal{Z}_{\mu^{0}}]/G_{q_{e}} $ be open and consider the
functions $F$ and $G $ given in Theorems \ref{theorem about F} and
\ref{theorem about G}.
Define $G^{i}:I\times U\times (\mathfrak{m}_{1}\oplus
\mathfrak{m}_{2})\rightarrow \mathbb{R}$ by
\[
G^{i}(\tau,[v_{q_{e}}],\mu_{1}+\mu _{2}):=\langle
G(\tau,[v_{q_{e}}],\mu_{1}+\mu _{2}),\varsigma_{i}\rangle,
\]
where $\{\varsigma_{i} \mid i=1,... ,\operatorname{dim}
\mathfrak{k_{2}}\}$ is a basis for ${\mathfrak k}_{2}$.
Choose $([v_{q_{e}}],\mu _{1}+\mu _{2})\in U\times
(\mathfrak{m}_{1}\oplus \mathfrak{m}_{2})$ such that
\[
\frac{\partial F}{\partial u}(0,[v_{q_{e}}],\mu _{1}+\mu _{2})=0,
\]
where the partial derivative is taken relative to the variable $u
\in U $. Define the matrix
\begin{equation*}
\Delta_{([v_{q_{e}}],\mu _{1},\mu _{2})} := \left[
\begin{array}{cc}
\frac{\partial ^{2}F}{\partial u^{2}}(0,[v_{q_{e}}],\mu _{1}+\mu
_{2}) & \frac{\partial^{2}F}{\partial \mu_{2}\partial
u}(0,[v_{q_{e}}],\mu _{1}+\mu _{2})\\
\frac{\partial G^{i}}{\partial u}(0,[v_{q_{e}}],\mu _{1}+\mu _{2})
& \frac{\partial G^{i}}{\partial \mu_{2}}(0,[v_{q_{e}}],\mu
_{1}+\mu _{2})
\end{array}
\right],
\end{equation*}
where the partial derivatives are evaluated at $\tau
=0, [v_{q_{e}}],\mu = \mu_1 + \mu_2$. Here $ \frac{\partial}
{\partial \mu _{2}}$ denotes the partial derivative with respect to
the $\mathfrak{m}_{2}$-component $\mu_2 $ of $\mu $.
In the framework and the notations introduced above we will state
and prove the main result of this paper. Let $\pi :TQ\rightarrow
(TQ)/G$ be the canonical projection and $\mathcal{R} _{e}:=\pi
(\mathfrak{t}\cdot q_{e})$.
\begin{theorem}
\label{principala}
Assume the following:
\begin{equation*}
{\bf (H)}\; \text{every } v_{q_{e}}\in \mathfrak{t}\cdot q_{e}
\text{ is a relative equilibrium.}
\end{equation*}
If there is a point $([v_{q_{e}}^{0}],\mu
_{1}^{0}+\mu _{2}^{0})\in U\times (\mathfrak{m}_{1}\oplus
\mathfrak{m}_{2})$ such that
\begin{eqnarray*}
&&
\begin{array}{cc}
1) & \frac{\partial F}{\partial u}(0,[v_{q_{e}}^{0}],\mu
_{1}^{0}+\mu _{2}^{0})=0,
\end{array}
\\
&&
\begin{array}{cc}
2) & G^{i}(0,[v_{q_{e}}^{0}],\mu _{1}^{0}+\mu _{2}^{0})=0
\end{array}
\\
&&
\begin{array}{cccc}
3) & \Delta _{([v_{q_{e}}^{0}],\mu _{1}^{0},\mu _{2}^{0})} & is &
non degenerate,
\end{array}
\end{eqnarray*}
then there exists a family of continuous curves
$\gamma_{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}:[0,1]\rightarrow (TQ)/G$
parameterized by
$\mu_{1}$ in a small neighborhood $\mathcal{V}_{0}$ of $\mu_{1}^{0}$
consisting of classes of relative equilibria with trivial
isotropy on $\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}(0,1)$ satisfying
\begin{equation*}
\operatorname{Im}\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0}, \mu
_{2}^{0})}}^{\mu_{1}}\bigcap \mathcal{R} _{e}=\left\{\gamma
_{_{({[v_{q_{e}}^{0}]}, \mu _{1}^{0}, \mu
_{2}^{0})}}^{\mu_{1}}(0)\right\}
\end{equation*}
and $\gamma _{_{({[v_{q_{e}}^{0}]}, \mu _{1}^{0}, \mu
_{2}^{0})}}^{\mu_{1}}(0) = [ \zeta_Q(q_e)]$, where $\zeta =
\widehat{\mathbb {I}}(q_e)^{-1} \mu_1 \in \mathfrak{t}$.
For $\mu_{1},\mu_{1}^{\prime}\in \mathcal{V}_{0}$ with
$\mu_{1}\neq\mu_{1}^{\prime}$, where $\mathcal{V}_{0}$ is as
above, the above branches do not intersect, that is,
\begin{equation*}
\left\{ \gamma _{_{({[v_{q_{e}}^{0}]}, \mu _{1}^{0}, \mu
_{2}^{0})}}^{\mu_{1}}(\tau) \,\Big|\, {\tau\in [0,1]} \right\}
\bigcap
\left\{\gamma _{_{({[v_{q_{e}}^{0}]}, \mu _{1}^{0}, \mu
_{2}^{0})}}^{\mu_{1}^{\prime}}(\tau) \,\Big|\, {\tau\in
[0,1]}\right\} = \varnothing.
\end{equation*}
Suppose that $ ([v_{q_{e}}^{0}],\mu _{1}^{0},\mu
_{2}^{0}) \neq ([v_{q_{e}}^{1}],\mu _{1}^{1},\mu _{2}^{1})$.
(i) If $\mu_{1}^{0}\neq \mu_{1}^{1}$ then the families of relative
equilibria do not intersect, that is,
\begin{equation*}
\left\{\gamma_{_{({[v_{q_{e}}^{0}]}, \mu _{1}^{0},
\mu_{2}^{0})}} ^{\mu_{1}}(\tau) \,\Big|\, {(\tau,\mu_{1})\in
[0,1]\times \mathcal{V}_{0}} \right\}
\bigcap
\left\{\gamma _{_{({[v_{q_{e}}^{1}]}, \mu_{1}^{1},
\mu_{2}^{1})}}^{\mu_{1}^{\prime}}(\tau) \,\Big|\,
{(\tau,\mu_{1}^{\prime})\in [0,1]\times \mathcal{V}_{1}}\right\}
=\varnothing,
\end{equation*}
where $\mathcal{V}_{0}$ and $\mathcal{V}_{1}$ are
two small neighborhoods of $\mu_{1}^{0}$ and $\mu_{1}^{1}$
respectively such that $\mathcal{V}_{0}\cap
\mathcal{V}_{1}=\varnothing$.
(ii) If $\mu_{1}^{0}=\mu_{1}^{1}=\overline \mu$ and
$[v_{q_{e}}^{0}]\neq[v_{q_{e}}^{1}]$ then $\gamma
_{_{({[v_{q_{e}}^{0}]},\overline\mu , \mu
_{2}^{0})}}^{\overline\mu}(0)=\gamma
_{_{({[v_{q_{e}}^{1}]},\overline\mu , \mu
_{2}^{1})}}^{\overline\mu}(0)$ and for $\tau>0$ we have
\begin{equation*}
\left\{\gamma
_{_{({[v_{q_{e}}^{0}]},\overline\mu , \mu
_{2}^{0})}}^{\overline\mu}(\tau) \,\Big|\, {\tau\in
(0,1]}\right\}
\bigcap \left\{\gamma
_{_{({[v_{q_{e}}^{1}]},\overline\mu , \mu
_{2}^{1})}}^{\overline\mu}(\tau) \,\Big|\, {\tau\in (0,1]}\right\}
=\varnothing.
\end{equation*}
\end{theorem}
\begin{proof}
Let $({[v_{q_{e}}^{0}]},\mu _{1}^{0}+\mu _{2}^{0})\in U\times
(\mathfrak{m}_{1}\oplus \mathfrak{m} _{2}) $ be such that the
conditions 1-3 hold. Because $\Delta _{({[v_{q_{e}}^{0}]},\mu
_{1}^{0}+\mu _{2}^{0})}$ is nondegenerate, we can apply the
implicit function theorem for the system $(\frac {\partial
F}{\partial u},G^{i})(\tau,[v_{q_{e}}],\mu _{1}+\mu _{2})=0$
around the point $(0,{[v_{q_{e}}^{0}]},\mu _{1}^{0}+\mu
_{2}^{0})$ and so we can find an open neighborhood $J\times
\mathcal{V}_{0}$ of the point $(0,\mu_{1}^{0})$ in $I\times
\mathfrak {m}_{1}$ and two functions $u:J\times
\mathcal{V}_{0}\rightarrow U$ and $\mu _{2}:J\times
\mathcal{V}_{0}\rightarrow \mathfrak{m}_{2}$ such that
$u(0,\mu _{1}^{0})={[v_{q_{e}}^{0}]}$, $\mu _{2}(0,\mu
_{1}^{0})=\mu _{2}^{0}$ and
\begin{eqnarray*}
&&
\begin{array}{cc}
i) & \frac{\partial F}{\partial u}(\tau ,u(\tau,\mu _{1} ),\mu
_{1}+\mu _{2}(\tau,\mu _{1}))=0
\end{array}
\\
&&
\begin{array}{cc}
ii) & G^{i}(\tau ,u(\tau,\mu _{1} ),\mu _{1}+\mu _{2}(\tau,\mu
_{1} ))=0.
\end{array}
\end{eqnarray*}
Therefore, from Theorems \ref{theorem about F} and \ref{theorem
about G} it follows that the relative equilibrium conditions of
Corollary \ref{criteriu} are both satisfied. Thus
we obtain the following family of branches of relative equilibria
$[(\overline{\sigma}(\tau\cdot
u(\tau,\mu_{1})),\beta(\tau,\mu_{1}+\mu_{2}(\tau,\mu_{1})))]_{G}$
parameterized by $\mu_{1} \in \mathcal{V}_{0}$. For $\tau>0$ the
isotropy subgroup is trivial and for
$\tau=0$ the corresponding points on the branches are
$[(\overline{\sigma}([0_{q_{e}}]),\mu_{1}]_{G}=[q_{e},\mu_{1}]_{G}$
which have the isotropy subgroup equal to $G_{q_{e}}$. This shows
that there are points in $\mathcal{R}_e $ from which there are
emerging branches of relative equilibria with broken trivial
symmetry.
Using now the correspondence given by Proposition \ref{map f} and
a rescaling of $\tau$ we obtain the desired family of continuous
curves $\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}:[0,1]\rightarrow (TQ)/G$ parameterized by
$\mu_{1}$ in a small neighborhood $\mathcal{V}_{0}$ of $\mu_{1}^{0}$
consisting of classes of relative equilibria with trivial
isotropy on $\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}(0,1)$ and such that
\begin{equation*}
\operatorname{Im}\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}\bigcap \mathcal{R} _{e}=\{\gamma_{
_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu _{2}^{0})}}^ {\mu_1}(0)\}
\end{equation*}
and $\gamma_{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu _{2}^{0})}}
^{\mu_1}(0) = [ \zeta_Q(q_e)]$, where $\zeta =
\widehat{\mathbb {I}}(q_e)^{-1} \mu_1$.
Equivalently, using the identification given by
(\ref{corespondenta}) and by Proposition \ref{map f} we obtain that
the branches of relative equilibria $\gamma
_{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}(\tau)\in (TQ)/G$ are identified with
$[\sigma(\tau\cdot
u(\tau,\mu_{1})),\beta(\tau,\mu_{1}+\mu_{2}(\tau,\mu_{1}))]
_{G_{q_{e}}}$. It is easy to see that for
$\mu_{1}\neq\mu_{1}^{\prime}$ we have that
$\beta(\tau,\mu_{1}+\mu_{2}(\tau,\mu_{1}))\neq\beta(\tau^{\prime},
\mu_{1}^{\prime}+\mu_{2}(\tau,\mu_{1}^{\prime}))$ for every
$\tau,\tau^{\prime}\in [0,1]$. Using now the fact that $G_{q_{e}}$
acts trivially on $\mathfrak{m}_{1}$ we obtain
\begin{equation*}
\left\{ \gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}}(\tau) \,\Big| \, {\tau\in [0,1]} \right\}
\bigcap
\left\{\gamma _{_{({[v_{q_{e}}^{0}]},\mu _{1}^{0},\mu
_{2}^{0})}}^{\mu_{1}^{\prime}}(\tau) \,\Big| \, {\tau\in
[0,1]}\right\} = \varnothing.
\end{equation*}
In an analogous way, using the same argument we can prove $(i)$.
For $(ii)$ we start with two branches of relative equilibria,
$b_{1}(\tau,\overline\mu):=[\sigma(\tau\cdot
u(\tau,\overline\mu)),\beta(\tau,\overline\mu+\mu_{2}
(\tau,\overline\mu))]_{G_{q_{e}}}$ and
$b_{2}(\tau^{\prime},\overline\mu):=[\sigma(\tau^{\prime}\cdot
u^{\prime}(\tau^{\prime},\overline\mu)),\beta(\tau^{\prime},
\overline\mu+\mu_{2}(\tau,\overline\mu))]_{G_{q_{e}}}$. For
$\tau=\tau^{\prime}=0$ we have
$b_{1}(0,\overline\mu)=[0,\overline\mu]_{G_{q_{e}}}
=b_{2}(0,\overline\mu)$. We also have
$u(0,\overline\mu)=[v_{q_{e}}^{0}]\neq
[v_{q_{e}}^{1}]=u^{\prime}(0,\overline\mu)$ and so, from
the implicit function theorem, we obtain
$u(\tau,\overline\mu)\neq u^{\prime}(\tau^{\prime},
\overline\mu)$ for $\tau,\tau^{\prime}>0$ small
enough.
Suppose that there exist $\tau,\tau^{\prime}>0$ such that
$b_{1}(\tau,\overline\mu)=b_{2}(\tau^{\prime},\overline\mu)$. Then
using the triviality of the ${G_{q_{e}}}$-action on
$\mathfrak{m}_{0}$ we obtain that
$\tau^{2}\nu_{0}={\tau^{\prime}}^{2}\nu_{0}$ and consequently
$\tau=\tau^{\prime}$. The conclusion of $(ii)$ follows now by
rescaling.
\end{proof}
\begin{remark}
\normalfont
We can have two particular forms for the rescaling $\beta $
according to special choices of the groups $G$ and $G_{q_{e}}$,
respectively.
(a) If $G$ is a torus, then from the splitting $\mathfrak{g}=
\mathfrak{k}_{0}\oplus \mathfrak{k}_{1}\oplus \mathfrak{k}_{2}$, where $
\mathfrak{k}_{0}=\mathfrak{g}_{q_{e}}$, $\mathfrak{k}_{0}\oplus \mathfrak{k}
_{1}=\mathfrak{t}$, and
$\mathfrak{k}_{2}=[\mathfrak{g},\mathfrak{t}]$, we conclude
that $\mathfrak{k}_{2}=\{0\}$ (since $\mathfrak{g}= \mathfrak{t}$)
and consequently
$\mathfrak{m} _{2}=\{0\}$. In this case we will obtain the special
form for the rescaling $
\beta :I\times \mathfrak{m}_{1}\rightarrow \mathfrak{g}^{\ast }$,
$\beta (\tau ,\mu )=\mu +\tau ^{2}\nu _{0}$.
(b) If is $G_{q_{e}}$ a maximal torus in $G$, so $\mathfrak{g}_{q_e}
= \mathfrak{t}$, then the same splitting implies that
$\mathfrak{k}_{1}=\{0\}$ and consequently $\mathfrak{m}_{1}=\{0\}$.
In this case we will obtain the special form for the rescaling
$\beta :I\times \mathfrak{m} _{2}\rightarrow
\mathfrak{g}^{\ast }$, $\beta (\tau ,\mu )=\tau \mu +\tau ^{2}\nu
_{0}$.
\end{remark}
\section{Stability of the bifurcating branches of relative
equilibria}
\label{stability section}
In this section we shall study the stability of the
branches of relative equilibria found in the previous section.
We will do this by applying a result of Patrick \cite{patrick
thesis} on $G_{\mu}$-stability to our situation. First we shortly
review this result.
\begin{definition}
Let $z_{e}$ be a relative equilibrium with velocity $\xi_{e}$ and
$J(z_{e})=\mu_{e}$. We say that $z_{e}$ is {\bfi formally
stable\/} if $\mathbf{d}^{2}(H-J^{\xi_{e}})(z_{e})|_
{T_{z_{e}}J^{-1}(\mu_{e})}$ is a positive or negative definite
quadratic form on some (and hence any) complement to ${\mathfrak
g}_{\mu_{e}}\cdot z_{e}$ in $T_{z_{e}}J^{-1}(\mu_{e})$.
\end{definition}
We have the following criteria for formal stability.
\begin{theorem}[Patrick, 1995]
\label{patrick}
Let $z_{e}\in T^{\ast }Q$ be a relative equilibrium with momentum
value $\mu _{e} \in \mathfrak{g}^\ast$ and base point $q_e \in
Q$. Assume that $\mathfrak{g}_{q_{e}}=\{0\}$. Then
$z_{e}$ is formally stable if and only if
$\mathbf{d}^{2}V_{\mu }(q_{e})$ is positive definite on one (and
hence any) complement $\mathfrak{g}_{\mu }\cdot q_{e}$ in
$T_{q_{e}}Q$.
\end{theorem}
To apply this theorem to our case in order to obtain the formal
stability of the relative equilibria on a bifurcating branch
we proceed as follows. First notice that if we fix $\mu
\in
\mathfrak{m}_{1}\oplus \mathfrak{ m}_{2}$ and $[v_{q_{e}}]\in U$
as in Theorem \ref{principala}, we obtain locally a branch
of relative equilibria with trivial isotropy bifurcating from
our initial set. More precisely, this branch starts at the
point
\[
\left(\widehat{ \mathbb{I}}(q_{e})^{-1}\Pi_{1}\mu \right)_Q(q_e).
\]
The momentum values along this branch are
$\beta (\tau ,\mu )$, and for $\tau
\neq 0$ the velocities have the expression $\mathbb{I}
(\operatorname{Exp} _{q_{e}}(\sigma(\tau
u(\tau,\mu_{1}))^{-1}\beta (\tau ,\mu )$. The base points of this
branch are
$\operatorname{Exp}_{q_{e}}(\sigma(\tau u(\tau,\mu_{1}))$.
Recall from Corollary \ref{criteriu} that we
introduced the notation $\overline{\sigma} : =
\operatorname{Exp}{q_e} \circ \sigma $ that will be used below.
By the definition of $\beta (\tau ,\mu )$ we have
$\mathfrak{g}_{\beta (\tau ,\mu )}=\mathfrak{t}$ for all $\tau $,
even for $\tau =0$. The base points for the
entire branch have no symmetry for $\tau>0 $ so we can
characterize the formal stability (in our case the $\mathbb{T}$
-stability) of the whole branch (locally) in terms of Theorem
\ref{patrick}. We begin by giving sufficient conditions that
guarantee the $\mathbb{T}$-stability of the branch, since
$G_{\beta(\tau,
\mu) } = \mathbb{T} $. To do this, one needs to find conditions
that insure that for
$\tau
\neq 0 $ (where the amended potential exists)
\[
\mathbf{d}^2 V_{\beta(\tau, \mu)}(
\overline{\sigma}(\tau u(\tau,\mu_{1}))|_{T_{[\tau
u(\tau,\mu_{1})]}
\overline{\sigma}(T_{[\tau u(\tau,\mu_{1})]}U) \oplus
(T_{\sigma ([\tau u(\tau,\mu_{1})])}\operatorname{Exp}_{q_{e}})
(\mathfrak{k} _{2}\cdot q_{e})}
\]
is positive definite. We do not know how to control the cross
terms of this quadratic form. This is why we shall work only with
Abelian groups $G $ since in that case the subspace
$\mathfrak{k}_2 = \{0\}$ and the second summand thus vanishes.
So, let $G $ be a torus $\mathbb{T}$. By Proposition
\ref{decomposition of the tangent space adapted to sigma bar} and
Theorem
\ref{theorem about F}, the second variation
\[
\mathbf{d}^2 V_{\beta(\tau, \mu)}(
\overline{\sigma}(\tau u(\tau,\mu_{1}))|_{T_{[\tau
u(\tau,\mu_{1})]}
\overline{\sigma}(T_{[\tau u(\tau,\mu_{1})]}U)}
\]
coincides for $\tau\neq 0 $, with the second variation
\begin{equation}
\label{hessian of f one}
\mathbf{d}_{U}^2 F_1(\tau,
u(\tau,\mu_{1}), \mu_1 + \mu_2( \tau, \mu_1))|_{T_{[\tau
u(\tau,\mu_{1})]}U}
\end{equation}
of the auxiliary function $F_1 $, where $\mathbf{d}^2_{U}$ denotes the second variation relative to the
second variable in $F_1$. But, unlike
$V_{\beta(\tau,\mu)}$, the function
$F_1$ is is defined even at $\tau = 0$. Recall from Theorem
\ref{theorem about F} that on the bifurcating branch the amended
potential has the expression
\begin{equation*}
F_{1}(\tau ,u(\tau,\mu_{1}),\mu_1 + \mu_2( \tau, \mu_1)
)=F_{0}(\mu_1 + \mu_2( \tau, \mu_1))+\tau ^{2}F(\tau,
u(\tau,\mu_{1}),\mu_1 +
\mu_2(\tau, \mu_1) ),
\end{equation*}
where $F_0$ is smooth on $\mathbf{J}_L (\mathfrak{g}\cdot
q_e) = \mathbb {I}(q_e)\mathfrak{g}$ and $F, F_1 $ are both
smooth functions on $I\times U\times \mathbf{J}
_{L}(\mathfrak{g}\cdot q_{e})$, even around $\tau =0$. So, if the
second variation of $F $ at $(0, [v_{q_e}^0], \mu_1 ^0 + \mu_2
^0)$ is positive definite, then the quadratic form \eqref{hessian
of f one} will remain positive definite along the branch for $\tau
> 0 $ small. So we get the following result.
\begin{theorem}
Let $\mu_1 ^0 + \mu_2 ^0 \in \mathfrak{m}_{1}\oplus
\mathfrak{m}_{2}$ and
$[v_{q_{e}}^0]\in U$ be as in the Theorem \ref{principala} and
assume that
$\mathbf{d}^2_U F(0,[v_{q_{e}}^0],\mu_1^0 + \mu_2 ^0 )$ is
positive definite. Then the branch of relative
equilibria with no symmetry which bifurcate form
$\left(\widehat{ \mathbb{I}}(q_{e})^{-1}\mu_1^0
\right)_Q(q_e)
$ will be $\mathbb{T}$-stable for $\tau>0$ small.
\end{theorem}
\medskip
A direct application of this criterion to the double spherical
pendulum recovers the stability result on the bifurcating branches
proved directly in \cite{ms}.
\medskip
\addcontentsline{toc}{section}{Acknowledgments}
\noindent\textbf{Acknowledgments.} We would like to thank J.
Montaldi for telling us the result of Proposition \ref{montaldi}
and for many discussions that have influenced our presentation
and clarified various points during the writing of this paper.
Conversations with A. Hern\'andez and J. Marsden are also
gratefully acknowledged. The third and fourth authors were
partially supported by the European Commission and the Swiss
Federal Government through funding for the Research Training Network
\emph{Mechanics and Symmetry in Europe} (MASIE). The first and
third author thank the Swiss National Science Foundation for
partial support. | 118,630 |
TITLE: Can a set be closed without its limit points?
QUESTION [0 upvotes]: By definition, a set A is closed if it contains all of its accumulation(limit) points.
St A = A'
Also, a set is closed if it is the closure of the set.
where A = ClosureA = A∪A'
What happens if the accumulation points of the set is empty? is that set automatically open or closed?
Is the closure of a set without an accumulation point closed or open?is the closure of every set always closed?
REPLY [0 votes]: Does Tom Hanks live with all his pet elephants? What if Tom Hanks doesn't have any pets elephants? Then does Tom Hanks live with all his pet elephants or doesn't he?
To a mathematician the the answer to "Does Tom Hanks live with all his pet elephants" is YES. To say, "No, Tom Hanks doesn't live with all of his pet elephants" means that there is at least one pet elephant Tom Hanks doesn't live with. But that is not the case.
So can a set be closed without its limit points. For a set to be "without" its limit points means, there must be at least one limit point of the set that is not in the set. And by definition, this means the set isn't closed because a set being closed means it does contain all it's limit point.
But that doesn't mean the set HAS any limit points. If a set, (say for example the set $\mathbb Z$ as a subset of $\mathbb R$ with the usual metric-- $\mathbb Z$ does not have any limit points-- none) doesn't have any limit points at all, then the set IS closed.
Because the answer to "Does $\mathbb Z$ contain all its limit points" is "YES". $\mathbb Z$ does not have any limit points so there are no limit points that $\mathbb Z$ does not contain, so it is true that $\mathbb Z$ contains all its limit points.
So $\mathbb Z$ is closed. | 82,008 |
TITLE: Prove that if $\phi$ is invertible operator of V space, then $\phi$ and $\phi^{-1}$ has the same invariant subspaces.
QUESTION [0 upvotes]: Prove that if $\phi$ is invertible operator of V space, then $\phi$ and $\phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $\phi$, then $\phi(W)\subset W$. If W is finite and $\phi$ is invertible $\Rightarrow$ $\phi\mid_W : W \rightarrow W$ is bijection, that's why for every $v\in W$ there exist $v'\in W$ such that $\phi(v) = v' \Rightarrow \phi^{-1}(\phi(v)) = \phi^{-1}(v')\Rightarrow v = \phi^{-1}(v') \Rightarrow \phi^{-1}(W)\subset W$.
But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?
REPLY [1 votes]: I suppose that the assertion that you wish to prove is this:
If $V$ is a vector space, $\phi\colon V\longrightarrow V$ is an invertible linear map and $W\subset V$ is a subspace such that $\phi(W)\subset W$, then $\phi^{-1}(W)\subset W$.
If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $\mathbb R$ into itself, let $\bigl(\phi(f)\bigr)(x)=f(x-1)$ and let $W=\{f\colon\mathbb{R}\longrightarrow\mathbb{R}\,|\,x<0\implies f(x)=0\}$. Then $\phi(W)\subset W$. However, if you define$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<0\\1&\text{ otherwise,}\end{cases}\end{array}$$then $f\in W$, but $\phi^{-1}(f)\notin W$, since $\bigl(\phi^{-1}(f)\bigr)\left(-\frac12\right)=1$. | 54,625 |
Great news! I have a new curry recipe to share. This time, originating from India/Pakistan. I haven’t written about many curries lately, and I’ve decided that this needs to change as it is undeniably one of the loves of my life. The new curry discovery mission starts here. This one is wonderful if you have a pack of lamb mince to use up (a far more interesting and challenging option than a Shepherd’s Pie, come on let’s be honest). In fact you could use any type of mince – the dish is not only interchangeable but fairly quick to rustle up (in curry terms – around half an hour).
The recipe features in ‘The Complete Asian Cookbook’ by Charmaine Solomon where you can find the full ingredients list and cooking method. This book comprehensively covers a diverse range of food across Asia and we highly recommend it to anyone that is partial to cuisine from these parts.
To make Keema Kari (serves two generously) you will need:
– Two tablespoons of butter
– One large onion (finely sliced)
– Two cloves of garlic (peeled/crushed)
– One teaspoon of ground turmeric
– Half a teaspoon of freshly grated ginger
– Half a teaspoon of chilli powder
– 500g of minced meat (we used lamb)
– Half a cup of natural yoghurt
– One cup of fresh peas (we used frozen and thawed)
– One teaspoon of garam masala
– One and a half teaspoons of salt
– Two tablespoons of fresh coriander leaves (finely chopped)
– One fresh red chilli (finely sliced)
Method:
– Firstly, heat the butter in a stove top (lidded) pan and fry the onion until soft. Next, add in the ginger and garlic, frying until the garlic is nicely golden brown.
– Add in the tumeric and chilli powder, then fry for a few seconds. Add in the meat and move it around in the pan constantly, until completely browned and making sure all large lumps are broken down.
– Stir in the peas and yoghurt, then cook (covered) for around fifteen minutes. Add in the garam masala and cook further until both the meat and peas are tender.
– Garnish with the fresh coriander leaves and chilli, then serve.
We had ours alongside some basmati rice, a Tarka dal, Kachumber salad (both favourites in the H of G) and a couple of parathas. It certainly needed some accompaniments as the meat itself, although providing a decent amount of heat, is a little on the dry side for our tastes. That said, we do like a fairly saucy curry. However, the peas provide a pleasing bit of extra texture and freshness. It worked perfectly with all of the side dishes and a cheeky spoonful of mango chutney on the side. | 5,565 |
Breivik spared me, then shot me: Witness
Oslo:.. | 378,718 |
\begin{abstract}
We prove the convergence of $ \nN $-particle systems of Brownian particles with logarithmic interaction potentials onto a system described by the infinite-dimensional stochastic differential equation (ISDE).
For this proof we present two general theorems on the finite-particle approximations of interacting Brownian motions.
In the first general theorem, we present a sufficient condition for a kind of tightness of solutions of stochastic differential equations (SDE) describing finite-particle systems, and prove that the limit points solve the corresponding ISDE. This implies, if in addition the limit ISDE enjoy a uniqueness of solutions, then the full sequence converges.
We treat non-reversible case in the first main theorem.
In the second general theorem, we restrict to the case of reversible particle systems and simplify the sufficient condition. We deduce the second theorem from the first.
We apply the second general theorem to Airy$ _{\beta }$ interacting Brownian motion with $ \beta = 1,2,4$, and the Ginibre interacting Brownian motion.
The former appears in the soft-edge limit of Gaussian (orthogonal/unitary/symplectic) ensembles in one spatial dimension, and the latter in the bulk limit of Ginibre ensemble in two spatial dimensions, corresponding to a quantum statistical system for which the eigen-value spectra belong to non-Hermitian Gaussian random matrices. The passage from the finite-particle stochastic differential equation (SDE) to the limit ISDE is a sensitive problem because the logarithmic potentials are long range and unbounded at infinity. Indeed, the limit ISDEs are not easily detectable from those of finite dimensions.
Our general theorems can be applied straightforwardly to the grand canonical Gibbs measures with Ruelle-class potentials such as Lennard-Jones 6-12 potentials and and Riesz potentials.
\end{abstract}
\maketitle
\section{Introduction}\label{s:1}
Interacting Brownian motion in infinite dimensions
is prototypical of diffusion processes of infinitely many particle systems,
initiated by Lang \cite{lang.1,lang.2}, followed by Fritz \cite{Fr}, Tanemura \cite{tane.2}, and others. Typically, interacting Brownian motion $ \mathbf{X}=(X^i)_{i\in\N }$ with Ruelle-class (translation invariant) interaction $ \Psi $ and inverse temperature $ \beta \ge 0 $ is given by
\begin{align}\label{:10a}& \quad
dX_t^i = dB_t^i - \frac{\beta }{2}
\sum_{j;j\not=i} ^{\infty}\nabla \Psi (X_t^i-X_t^j ) dt
\quad (i\in\N )
.\end{align}
Here an interaction $ \Psi $ is called Ruelle-class if $ \Psi $ is super stable in the sense of Ruelle, and integrable at infinity \cite{ruelle.2}.
The system $ \mathbf{X}$ is a diffusion process with state space
$ \mathbf{S}_0 \subset (\Rd )^{\N }$, and has no natural invariant measures. Indeed, such a measure $ \check{\mu }$, if exists, is informally given by
\begin{align}\label{:10b}&
\check{\mu } = \frac{1}{\mathcal{Z}}
e^{- \beta \sum_{(i,j);\, i<j} ^{\infty}\Psi (x_i-x_j)}
\prod_{k\in\N } dx_k
,\end{align}
which cannot be justified as it is because of the presence of an infinite product of Lebesgue measures.
To rigorize the expression \eqref{:10b}, the Dobrushin--Lanford--Ruelle (DLR) framework introduces the notion of a Gibbs measure. A point process $ \mu $ is called a $ \Psi $-canonical Gibbs measure if it satisfies the DLR equation: for each $ m \in \N $ and $ \mu $-a.s.\! $ \xi =\sum_i \delta_{\xi _i}$
\begin{align}\label{:10c}&
\mu _{r,\xi }^m (d\mathsf{s} )
= \frac{1}{\mathcal{Z}_{r,\xi }^m }
e^{- \beta \{\sum_{i<j ,\, s_i , s_j \in \Sr }^m \Psi (s_i-s_j) +
\sum_{ s_i \in \Sr, \xi _j \in \Sr ^c }^m \Psi (s_i-\xi _j) \}
} \prod_{k=1}^m ds_k
,\end{align}
where $ \mathsf{s} = \sum_i \delta_{s_i}$, $ \Sr = \{ |x| \le r \} $,
$ \pi_r (\mathsf{s}) = \mathsf{s} (\cdot \cap \Sr )$, and
$ \xi $ is the outer condition.
Furthermore, $ \mu _{r,\xi }^m$ denotes the regular conditional probability:
\begin{align*}&
\mu _{r,\xi}^m (d\mathsf{s} ) =
\mu (\pi_r (\mathsf{s}) \in d\mathsf{s} | \, \mathsf{s} (\Sr ) = m ,\, \pi_r^c (\mathsf{s})=\pi_r^c (\xi ))
.\end{align*}
Then $ \mu $ is a reversible measure of
the delabeled dynamics $ \mathsf{X}$ such that
$ \mathsf{X}_t = \sum_{i\in \N } \delta _{X_t^i}$.
If the number of particles is finite, $ \nN $ say, then SDE \eqref{:10a} becomes
\begin{align}\label{:10e}&
dX_t^{\nN ,\, i } = dB_t^i
- \frac{\beta }{2} \{ \nabla \Phi ^{\nN }(X_t^{\nN ,\, i })
+
\sum_{j;j\not=i} ^{\nN }\nabla \Psi (X_t^{\nN ,\, i }-X_t^{\nN ,\, j }) \} dt
\quad (1 \le i \le \nN )
,\end{align}
where $ \Phi ^{\nN }$ is a confining free potential vanishing zero as $ \nN $
goes to infinity.
The associated labeled measure is then given by
\begin{align}\label{:10f}&
\muNcheck = \frac{1}{\mathcal{Z}}
e^{- \beta \{ \sum_{i=1}^{\nN } \Phi ^{\nN } (x_i)
+ \sum_{(i,j);\, i < j} ^{\nN }\Psi (x_i-x_j) \}
}
\prod_{k=1 }^{\nN } dx_k
.\end{align}
The relation between \eqref{:10e} and \eqref{:10f} is as follows.
We first consider the diffusion process associated with the Dirichlet form
with domain $ \mathcal{D}^{\muNcheck } $ on $ L^2 ((\Rd )^{\nN } ,\muNcheck )$,
called the distorted Brownian motion, such that
$$ \mathcal{E}^{\muNcheck } (f,g) = \int_{(\Rd )^{\nN } } \half \sum_{i=1}^{\nN }
\nabla _i f \cdot \nabla _i g
\, \muNcheck (d\mathbf{x}_{\nN })
,$$
where $ \nabla _i = (\PD{}{x_{ij}})_{j=1}^d$,
$ \mathbf{x}_{\nN } =(x_1,\ldots,x_{\nN } ) \in (\Rd )^{\nN } $,
and $ \cdot $ denotes the inner product in $ \Rd $.
The generator $ - L^{\muNcheck } $ of $ \mathcal{E}^{\muNcheck }$ is then given by
$$ \mathcal{E}^{\muNcheck } (f,g) =
( - L^{\muNcheck } f, g )_{L^2((\Rd )^{\nN } , \muNcheck )}
.$$
Integration by parts yields the representation of
the generator of the diffusion process such that
\begin{align*}&
L^{\muNcheck } = \half \Delta - \frac{\beta }{2}
\sum_{i=1}^{\nN } \{
\nabla {\Phi ^{\nN }} (x_i)
+ \sum_{j;\, j\not=i } ^{\nN }\nabla \Psi (x_i-x_j)
\}\cdot \nabla _i
,\end{align*}
which together with It$ \hat{\mathrm{o}}$ formula yields SDE \eqref{:10e}.
For a finite or infinite sequence $ \mathbf{x}=( x_i ) $, we set
$ \ulab (\mathbf{x}) = \sum_i \delta_{x_i}$ and call $ \mathfrak{u}$
a delabeling map.
For a point process $ \mu $, we say a measurable map $ \lab = \lab (\mathsf{s})$
defined for $ \mu $-a.s.\! $ \mathsf{s}$
with value $ \S ^\infty \cup \{\bigcup _{\nN =1}^{\infty}\S ^{\nN } \} $
is called a label with respect to $ \mu $ if
$\ulab \circ \lab (\mathsf{s}) = \mathsf{s}$.
Let $ \labN $ be a label with respect to $ \muN $.
We denote by $ \lab _m $ and $ \labNm $ the first $ m $-components of these labels, respectively. We take $ \Phi ^{\nN } $ such that the associated point process $ \muN = \muNcheck \circ \ulab ^{-1}$ converges weakly to $ \mu $:
\begin{align}\label{:10g}&
\limi{\nN } \muN = \mu \quad \text{ weakly}
.\end{align}
The associated delabeling
$ \mathsf{X}^{\nN } = \sum_{i=1}^{\nN }
\delta _{X^{\nN ,\, i}}$ is reversible with respect to $ \muN $. The labeled process $ \mathbf{X}=(X^i)$ and $ \mathbf{X}^{\nN } = (X^{\nN ,\, i})$
can be recovered from $ \mathsf{X}$ and $ \mathsf{X}^{\nN }$ by taking suitable initial labels $ \lab $ and $ \labN $, respectively.
Choosing the labels in such a way that for each $ m \in \N $
\begin{align}\label{:10h}&
\limi{\nN } \muN \circ \labNm ^{-1 } =
\mu \circ \labm ^{-1}\quad \text{ weakly}
,\end{align}
we have the convergence of labeled dynamics $ \mathbf{X}^{\nN }$ to $ \mathbf{X}$ such that for each $ m $
\begin{align}\label{:10i}&
\limi{\nN } (X^{\nN ,1},\ldots,X^{\nN ,m }) =
(X^{1},\ldots,X^{m })
\quad \text{ in law in } C([0,\infty ); (\Rd )^m )
.\end{align}
We expect this convergence because of
the absolute convergence of the drift terms in \eqref{:10a} and
energy in the DLR equation \eqref{:10c} for well-behaved initial distributions
although it still requires some work to justify this rigorously even if $ \Psi \in C_0^3(\Rd )$ \cite{lang.1}.
If we take logarithmic functions as interaction potentials,
then the situation changes drastically. Consider the soft-edge scaling limit of
Gaussian (orthogonal/unitary/symplectic) ensembles.
Then the $ \nN $-labeled density is given by
\begin{align}\label{:10k}&
\check{\mu } _{\mathrm{Airy}, \beta } ^{\nN }(d\mathbf{x}_{\nN })=
\frac{1}{Z}
\{ \prod_{i<j}^{\nN }|x_i-x_j|^\beta \}
\exp\bigg\{-\frac{\beta}{4}\sum_{k=1}^{\nN } |2\sqrt{\nN }+\nN ^{-1/6}x_k|^2 \bigg\}
d\mathbf{x}_{\nN }
\end{align}
and the associated $ \nN $-particle dynamics described by SDE
\begin{align}\label{:10l}&
dX_t^{\n ,i} = dB_t^i +
\frac{\beta }{2} \sum_{j=1,\, j\not= i}^{\n }
\frac{1}{X_t^{\n ,i} - X_t^{\n ,j} } dt
- \frac{\beta }{2 }
\{ \n ^{1/3} + \frac{1}{2\n ^{1/3}}X_t^{\n ,i} \}dt
.\end{align}
The correspondence between \eqref{:10k} and \eqref{:10l} is transparent and
same as above. Indeed, we first consider distorted Brownian motion
(Dirichlet spaces with $ \check{\mu } _{\mathrm{Airy}, \beta } ^{\nN }$
as a common time change and energy measure),
then we obtain the generator of the associated diffusion process by integration by parts. SDE \eqref{:10l} thus follows from the generator immediately.
It is known that the thermodynamic limit $\mu _{\mathrm{Airy},\beta }$
of the associated point process $ \mu _{\mathrm{Airy}, \beta } ^{\nN }$
exists for each $ \beta >0$ \cite{rrv.airy}.
Its $ m $-point correlation function is explicitly given as a determinant of certain kernels if $ \beta = 1,2,4$ \cite{AGZ,mehta}. Indeed, if $ \beta = 2 $, then the $ m $-point correlation function of
the limit point process $ \mu _{\mathrm{Airy},2}$ is
\begin{align} \notag &
\rho _{\Ai ,2}^{m} (\mathbf{x}_{m}) =
\det [K_{\Ai , 2}(x_i,x_j) ]_{i,j=1}^{m}
,\end{align}
where $ K_{{\Ai },2}$ is the continuous kernel such that, for $ x \not= y $,
\begin{align} & \notag
K_{{\Ai },2}(x,y) =
\frac{{\Ai }(x) {\Ai }'(y)-{\Ai }'(x) {\Ai }(y)}{x-y}
.\end{align}
We set here ${\Ai }'(x)=d {\Ai }(x)/dx$ and
denote by ${\Ai }(\cdot)$ the Airy function given by
\begin{align}& \notag
{\Ai }(z) = \frac{1}{2 \pi} \int_{\R} dk \,
e^{i(z k+k^3/3)},
\quad z\in\R
.\end{align}
For $ \beta = 1,4 $ similar expressions in terms of the quaternion determinant
are known \cite{AGZ,mehta}.
From the convergence of equilibrium states, we may expect the convergence of solutions of SDEs \eqref{:10l}.
The divergence of the coefficients in \eqref{:10l} and the very long-range nature of the logarithmic interaction however prove to be problematic.
Even an informal representation of the limit coefficients is nontrivial but
has been obtained in \cite{o-t.airy}. Indeed, the limit ISDEs are given by
\begin{align} \label{:10H} &
dX_t^i=dB_t^i+
\frac{\beta }{2}
\limi{r} \{ \sum_{|X_t^j|<r, j\neq i} \frac{1}{X_t^i-X_t^j} -\int_{|x|<r}\frac{\varrho (x)}{-x}\,dx\}dt \quad (i\in\N)
.\end{align}
Here $ \varrho (x) = 1_{(-\infty , 0)} (x) \sqrt{-x}$, which is the shifted and rescaled semicircle function at the right edge.
As an application of our main theorem (\tref{l:22}),
we prove the convergence \eqref{:10i} of solutions from \eqref{:10l} to \eqref{:10H} for
$ \{ \mu _{\mathrm{Airy}, \beta } ^{\nN } \} $ with $ \beta = 2 $.
We also prove that the limit points of solutions of \eqref{:10l}
satisfy ISDE \eqref{:10H} with $ \beta = 1,2,4 $.
For general $ \beta \not=1,2,4$, the existence and uniqueness of solutions of \eqref{:10H} is still an open problem.
Indeed, the proof in \cite{o-t.airy} relies on a general theory developed in
\cite{o.tp,o.isde,o.rm,o.rm2,o-t.tail}, which reduces the problem to the quasi-Gibbs property and the existence of the logarithmic derivative of the equilibrium
state. These key properties are proved only for $ \beta = 1,2,4 $ at present.
We refer to \cite{o.rm,o.rm2} for the definition of the quasi-Gibbs property and
\dref{d:21} for the logarithmic derivative.
Another typical example is the Ginibre interacting Brownian motion, which is an infinite-particle system in $ \R ^2$ (naturally regarded as $ \mathbb{C}$), whose equilibrium state is the Ginibre point process $ \mu _{\mathrm{gin}} $.
The $ m $-point correlation function $ \rho_{\mathrm{gin}}^m $ with respect to
Gaussian measure $ (1/\pi ) e^{-|x|^2}dx $ on $ \mathbb{C}$ is then given by
\begin{align}& \notag
\rho_{\mathrm{gin}}^m (\mathbf{x}_m)=\det [e^{x_i\bar{x}_j}]_{i,j=1}^m
.\end{align}
The Ginibre point process $ \mu _{\mathrm{gin}} $
is the thermodynamic limit of $ \nN $-particle point process
$ \mu _{\mathrm{gin}}^{\nN } $ whose labeled measure is given by
\begin{align}& \notag
\check{\mu }_{\mathrm{gin}}^{\nN } (d\mathbf{x}_{\nN }) =
\frac{1}{\mathcal{Z} } \prod_{i<j}^{\nN }|x_i-x_j|^2
e^{-\sum_{i=1}^{\nN }|x_i|^2} d\mathbf{x}_{\nN }
.\end{align}
The associated $ \nN $-particle SDE is then given by
\begin{align}\label{:10r}
dX_t^{N,i} = \, & dB_t^i-X_t^{N,i}dt +\sum_{j=1, j\neq i}^{\nN } \frac{X_t^{N,i}-X_t^{N,j}}{|X_t^{N,i}-X_t^{N,j}|^2}dt \quad (1\le i\le \nN )
.\end{align}
We shall prove that the limit ISDEs are
\begin{align}\label{:10s}
dX_t^i= \, &
dB_t^i+\limi{r} \sum_{|X_t^i-X_t^j|<r, j\neq i} \frac{X_t^i-X_t^j}{|X_t^i-X_t^j|^2}dt \quad (i\in\N)
\\\intertext{
{\em and }}
\label{:10t}
dX_t^i= \, &
dB_t^i - X_t^idt +
\limi{r} \sum_{|X_t^j|<r, j\neq i} \frac{X_t^i-X_t^j}{|X_t^i-X_t^j|^2}dt \quad (i\in\N)
.\end{align}
In \cite{o.isde, o-t.tail}, it is proved that these ISDEs have the same pathwise unique strong solution for $ \mu _{\mathrm{gin}}\circ \lab^{-1} $-a.s. $ \mathbf{s}$,
where $ \lab $ is a label and $ \mathbf{s}$ is an initial point.
As an example of applications of our second main theorem (\tref{l:22}),
we prove the convergence of
solutions of \eqref{:10r} to those of \eqref{:10s} and \eqref{:10t}.
This example indicates again the sensitivity of the representation of the limit ISDE. Such varieties of the limit ISDEs are a result of the long-range nature of the logarithmic potential.
The main purpose of the present paper is to develop a general theory for finite-particle convergence applicable to logarithmic potentials, and in particular, the Airy and Ginibre point processes. Our theory is also applicable to essentially all Gibbs measures with Ruelle-class potentials such as the Lennard-Jones 6-12 potential and Riesz potentials.
In the first main theorem (\tref{l:21}), we present a sufficient condition for a kind of tightness of solutions of stochastic differential equations (SDE) describing finite-particle systems, and prove that the limit points solve the corresponding ISDE. This implies, if in addition the limit ISDE enjoy uniqueness of solutions, then the full sequence converges.
We treat non-reversible case in the first main theorem.
In the second main theorem (\tref{l:22}), we restrict to the case of reversible particle systems and simplify the sufficient condition. Because of reversibility, the sufficient condition is reduced to the convergence of logarithmic derivative of $ \mu ^{\nN }$ with marginal assumptions.
We shall deduce \tref{l:22} from \tref{l:21} and
apply \tref{l:22} to all examples in the present paper.
If $ \Psi (x) = -\log |x|$, $ \beta = 2 $ and $ d=1$, there exists an algebraic method to construct the associated stochastic processes \cite{j.02,j.03,KT07b,KT11}, and to prove the convergence of finite-particle systems \cite{o-t.sm,o-t.core}. This method requires that interaction $ \Psi $ is the logarithmic function with $ \beta = 2$ and depends crucially on an explicit calculation of space-time determinantal kernels. It is thus not applicable to $ \beta \not=2$ even if $ d=1$.
As for Sine$_{\beta } $ point processes, Tsai proved the convergence of finite-particle systems for all $ \beta \ge 1$ \cite{tsai.14}. His method relies on a coupling method based on monotonicity of SDEs, which is very
specific to this model.
The organization of the paper is as follows:
In \sref{s:2}, we state the main theorems (\tref{l:21} and \tref{l:22}).
In \sref{s:3}, we prove \tref{l:21}.
In \sref{s:4}, we prove \tref{l:22} using \tref{l:21}.
In \sref{s:5}, we present examples.
\section{Set up and the main theorems}\label{s:2}
\subsection{Configuration spaces and Campbell measures}\label{s:21}
Let $ \S $ be a closed set in $ \Rd $ whose interior
$ \S _{\mathrm{int}}$ is a connected open set
satisfying $ \overline{\S _{\mathrm{int}}} = \S $ and
the boundary $ \partial \S $ having Lebesgue measure zero.
A configuration
$ \mathsf{s} = \sum_i \delta _{s_i}$ on $ \S $
is a Radon measure on $ \S $ consisting of delta masses.
We set $ \Sr = \{ s \in \S \, ;\, |s| \le r \} $.
Let $ \SS $ be the set consisting of all configurations of $ \S $.
By definition, $ \SS $ is given by
\begin{align} & \notag
\SS = \{ \mathsf{s} = \sum_{i} \delta _{s_i}\, ;\,
\text{ $\mathsf{s} ( \Sr ) < \infty $ for each $ r\in\N $} \}
.\end{align}
By convention, we regard the zero measure as an element of $ \SS $.
We endow $ \SS $ with the vague topology, which makes
$ \SS $ a Polish space.
$ \SS $ is called the configuration space over $ \S $ and a probability measure $ \mu $ on $ (\SS , \mathcal{B}(\SS ) )$
is called a point process on $ \S $.
A symmetric and locally integrable function
$ \map{\rho ^n }{\S ^n}{[0,\infty ) } $ is called
the $ n $-point correlation function of a point process $ \mu $
on $ \S $ with respect to the Lebesgue measure if $ \rho ^n $ satisfies
\begin{align} & \notag
\int_{A_1^{k_1}\ts \cdots \ts A_m^{k_m}}
\rho ^n (x_1,\ldots,x_n) dx_1\cdots dx_n
= \int _{\SS } \prod _{i = 1}^{m}
\frac{\mathsf{s} (A_i) ! }
{(\mathsf{s} (A_i) - k_i )!} d\mu
\end{align}
for any sequence of disjoint bounded measurable sets
$ A_1,\ldots,A_m \in \mathcal{B}(\S ) $ and a sequence of natural numbers
$ k_1,\ldots,k_m $ satisfying $ k_1+\cdots + k_m = n $.
When $ \mathsf{s} (A_i) - k_i < 0$, according to our interpretation,
${\mathsf{s} (A_i) ! }/{(\mathsf{s} (A_i) - k_i )!} = 0$ by convention.
Hereafter, we always consider correlation functions with respect to Lebesgue measures.
A point process $ \mu _{x}$ is called the reduced Palm measure of $ \mu $
conditioned at $ x \in \S $ if $ \mu _{x}$ is the regular conditional probability defined as
\begin{align} & \notag
\mu _{x} = \mu (\cdot - \delta_x | \mathsf{s} (\{ x \} ) \ge 1 )
.\end{align}
A Radon measure $ \muone $ on $ \S \times \SS $
is called the 1-Campbell measure of $ \mu $ if $ \muone $ is given by
\begin{align}\label{:21h}&
\muone (dx d\mathsf{s}) = \rho ^1 (x) \mu _{x} (d\mathsf{s}) dx
.\end{align}
\subsection{Finite-particle approximations (general case)}\label{s:22}
Let $ \{\muN \} $ be a sequence of point processes on $\S $
such that $ \muN (\{ \mathsf{s}(\S ) = \nN \} ) = 1 $. We assume: \\
\As{H1} Each $ \muN $ has a correlation function $ \{\rho ^{N,n}\} $
satisfying for each $r \in\N$
\begin{align}\label{:20f} &
\limi{N} \rho ^{N,n} (\mathbf{x})=
\rho ^{n} (\mathbf{x}) \quad \text{ uniformly on $\Sr ^{n}$ for all $n\in\N$}
,\\ \label{:20g}&
\sup_{N\in\N } \sup_{\mathbf{x} \in \Sr ^{n}} \rho ^{N,n} (\mathbf{x}) \le
\cref{;40b} ^{n} n ^{\cref{;40c}n}
,\end{align}
where $ 0 < \Ct{;40b}(r) < \infty $ and
$ 0 < \Ct{;40c}(r)< 1 $ are constants independent of $ n \in \N $.
It is known that \eqref{:20f} and \eqref{:20g} imply weak convergence
\eqref{:10g} \cite[Lemma A.1]{o.rm}. As in \sref{s:1}, let $ \lab $ and $ \labN $
be labels of $ \mu $ and $ \muN $, respectively. We assume:
\medskip
\noindent
\As{H2} For each $m\in\N$, \eqref{:10h} holds. That is,
\begin{align}\tag{\ref{:10h}}& \quad
\limi{N}\mu^{\nN } \circ \labNm ^{-1} =\mu \circ \labm ^{-1}
\quad \text{ weakly in $ \S ^m $}
.\end{align}
\medskip
We shall later take $ \mu^{\nN } \circ \labN ^{-1} $
as an initial distribution of a labeled finite-particle system.
Hence \As{H2} means convergence of the initial distribution of the labeled dynamics.
There exist infinitely many different labels $\lab $, and
we choose a label such that the initial distribution of the labeled dynamics converges. \As{H2} will be used in \tref{l:22} and \tref{l:21}.
\medskip
For $ \mathbf{X}=(X^i)_{i=1}^{\infty}$ and
$ \mathbf{X}^{\nN }=(X^{\nN ,i})_{i=1}^{\nN }$, we set
\begin{align} & \notag
\Xidt = \sum_{j\not=i}^{\infty} \delta_{X_t^{j}}, \quad \text{ and }\quad
\mathsf{X}_t^{N,\diai }= \sum_{j\not=i}^{ N } \delta_{X_t^{N,j}}
,\end{align}
where $X_t^{N,\diai }$ denotes the zero measure for $ \nN = 1 $.
Let $ \map{\sN ,\sigma }{\S \ts \SSS }{\R ^{d^2}}$ and
$ \map{\bbb ^{\nN } ,\bbb }{\S \ts \SSS }{\Rd }$ be measurable functions.
We introduce the finite-dimensional SDE
of $ \mathbf{X}^{\nN } =(X^{\nN ,i})_{i=1}^{\nN } $
with these coefficients such that for $ 1\le i\le N $
\begin{align}\label{:20m}
dX_t^{N,i} &=
\sN (X_t^{N,i},\XNidt )dB_t^i +
\bbb ^{\nN }(X_t^{N,i},\XNidt )dt
\\\label{:20n}
\XX _0^{\nN } & = \mathbf{s}
.\end{align}
We assume:
\medskip
\noindent \As{H3}
SDE \eqref{:20m} and \eqref{:20n}
has a unique solution for $ \muN \circ \labN ^{-1}$-a.s.\! $ \mathbf{s}$
for each $ \nN $: this solution does not explode.
Furthermore, when $ \partial \S $ is non-void, particles never hit the boundary.
\bs
We set $ \aN = \sN {}^{t}\sN $ and assume:
\medskip
\noindent
\As{H4}
$\sN $ are bounded and continuous on $ \S \ts \SS $,
and converge uniformly to $\sigma $ on $ \SrSS $ for each $ r \in \N $.
Furthermore, $\aN $ are uniformly elliptic on $ \Sr \ts \SS $ for each $ r \in \N $ and $ \nablax \aN $ are uniformly bounded on $ \S \ts \SS $.
\medskip
From \As{H4} we see that
$ \aN $ converge uniformly to $ \aaa := \sigma {}^t\sigma $
on each compact set $ \SrSS $, and that
$ \aN $ and $\aaa $ are bounded and continuous on $ \S \ts \SS $.
There thus
exists a positive constant $\Ct{;3} $
such that
\begin{align}\label{:20o}&
||\aaa ||_{\S \ts \SS } ,\
||\nabla_x \aaa ||_{\S \ts \SS } , \
\sup_{\nN \in\N } ||\aN ||_{\S \ts \SS } ,\
\sup_{\nN \in\N } ||\nablax \aN ||_{\S \ts \SS } \le \cref{;3}
.\end{align}
Here $ \| \cdot \|_{\S \ts \SS } $
denotes the uniform norm on $ \S \ts \SS $.
Furthermore, we see that $ \aaa $ is uniformly elliptic on each $ \SrSS $.
From these, we expect that SDEs \eqref{:20m}
have a sub-sequential limit.
\begin{align} \notag
\limi{\nN }\{X_t^{N,i} - X_0^{N,i}\} &=
\limi{\nN }\int_0^t \sN (X_t^{N,i},\XNidt )dB_u^i +
\limi{\nN } \int_0^t \bbb ^{\nN }(X_t^{N,i},\XNidt )du
\\ \notag
&=
\int_0^t \sigma (X_t^{N,i},\XNidt )dB_u^i +
\limi{\nN } \int_0^t \bbb ^{\nN }(X_t^{N,i},\XNidt )du
.\end{align}
To identify the second term on the right-hand side and to justify the convergence, we make further assumptions. As the examples in \sref{s:1} suggest, the identification of the limit is a sensitive problem, which is at the heart of the present paper.
\bs
We set the maximal module variable $ \overline{\mathbf{X}}^{N,m}$
of the first $ m $-particles by
\begin{align}&\notag
\overline{\mathbf{X}}^{N,m}= \max_{i=1}^m
\sup_{t\in [0,T] }|X_t^{N,i}|
.\end{align}
and by $ \mathcal{L}_r^{N}$ the maximal label with which the particle
intersects $ \Sr $; that is,
\begin{align}&\notag
\mathcal{L}_r^{N} =
\max \{ i \in \N \cup \{ \infty \} \, ;\, |X_t^{N,i}| \le r
\text{ for some } 0\le t \le T \}
.\end{align}
We assume the following.
\\
\As{I1} For each $ m \in \N $
\begin{align}\label{:40x}&
\limi{a} \liminfi{\nN }P ^{\muNl }
(\overline{\mathbf{X}}^{N,m} \le a ) = 1
\end{align}
and there exists a constant $\Ct{;41a}=\cref{;41a}(m,a)$ such that
for $ 0 \le t , u \le T $
\begin{align} \label{:40z} &
\supN \sum_{i=1}^m
\mathrm{E} ^{\muNl }
[|X _t^{N,i} - X _u^{N,i}|^{4 };\overline{\mathbf{X}}^{N,m} \le a ]
\le \cref{;41a} |t-u|^{2}
.\end{align}
Furthermore, for each $ r \in \N $
\begin{align}\label{:40p}&
\limi{L} \liminfi{\nN }P ^{\muNl }
(\mathcal{L}_r^{N} \le L ) = 1
.\end{align}
Let $ \muNone $ be the one-Campbell measure of $ \muN $
defined as \eqref{:21h}. Set $ \Ct{;43}(\rrr , N)= \muNone ( \Sr \ts \SSS )$.
Then by \eqref{:20g} $\sup_{N}\cref{;43}(\rrr ,N)<\infty $
for each $\rrr \in \N $. Without loss of generality,
we can assume that $\cref{;43} > 0 $ for all $\rrr , N $.
Let $ \muNone _r = \muNone (\cdot \cap \{\Sr \ts \SSS \}) $.
Let $ \muNonebar $ be the probability measure defined as
$\muNonebar (\cdot )=
\muNone (\cdot \cap \{\Sr \ts \SSS \})/\cref{;43}$.
Let $ \varpi _{r,s}$ be a map from $ \Sr \ts \SS $ to itself such that
$ \varpi _{r,s} (x,\mathsf{s}) = (x,\sum_{|x-s_i|<s} \delta_{s_i})$, where
$ \mathsf{s}= \sum_i \delta_{s_i}$.
Let $ \mathcal{F}_{\rrr ,s} = \sigma [\varpi _{r,s}]$
be the sub-$ \sigma $-field of $ \mathcal{B}(\Sr \ts \mathcal{\SS }) $
generated by $ \varpi _{r,s}$. Because
$ \Sr $ is a subset of $ \S $, we can and do regard
$ \mathcal{F}_{\rrr ,s} $ as a $ \sigma $-field on $ \S \ts \SS $,
which is trivial outside $ \Sr \ts \SS $.
We set a tail-truncated coefficient $ \bNrs $ of $ \bN $ and their tail parts $ \6 $ by
\begin{align}
\label{:40a}&
\bbb _{r,s}^{\nN } =
\mathrm{E}^{\muNonebar }
[ \bbb ^{N}|\mathcal{F}_{\rrr ,s} ] ,\quad
\bN = \bbb _{r,s}^{\nN } + \6
.\end{align}
We can and do take a version of $ \bNrs $ such that
\begin{align}
\label{:40b}&
\bNrs (x,\mathsf{y}) = 0 \quad \text{ for } x \not\in \Sr
,\\
\label{:40c}&
\bNrs (x,\mathsf{y}) = \bbb _{r+1,s}^{N} (x,\mathsf{y})
\quad \text{ for } x \in \S _{r}
.\end{align}
We next introduce a cut-off coefficient $ \bNrsp $ of $ \bNrs $.
Let $ \bNrsp $ be a continuous and $ \mathcal{F}_{\rrr ,s}$-measurable function
on $ \S \ts \SS $ such that
\begin{align} \label{:40ff}
\bNrsp (x,\mathsf{y}) &=0
\quad \text{ for } x \not\in \Sr
\\
\label{:40f}
\bNrsp (x,\mathsf{y}) & = \bbb _{r+1,s,\p }^{N} (x,\mathsf{y}) ,\quad
\quad \text{ for } x \in \S _{r-1}
\end{align}
and that, for $ (\S \ts \SS )_{r,\p } = \{ (x,\mathsf{y} ) \in \S _{r} \ts \SS
;\, |x- y_i|\le 1/2^{\p } \text{ for some } y_i \}$, where
$ \mathsf{y}=\sum_i\delta_{y_i}$,
\begin{align}
\label{:40F}&
\bNrsp (x,\mathsf{y}) = 0
\quad \text{ for $ (x,\mathsf{y}) \in (\S \ts \SS )_{r,\p +1} $}
,\\
\label{:40g}&
\bNrsp (x,\mathsf{y}) = \bbb _{r,s}^{N} (x,\mathsf{y})
\quad \text{ for $ (x,\mathsf{y}) \not\in (\S \ts \SS )_{r,\p } $}
.\end{align}
The main requirements for $ \bN $ and $ \bNrsp $ are the following:
\ms
\noindent
\As{I2} There exists a $ \phat $ such that
$ 1 < \phat $ and that for each $ r \in \N $
\begin{align}\label{:40h}&
\limsupi{\nN } \int _{\SrSS }|\bN | ^{\phat } d\muNone < \infty
.\end{align}
Furthermore, for each $ r , i \in \N $, there exists a constant $ \Ct{;40i}$ such that
\begin{align}\label{:40i}&
\sup_{\p \in\N }
\sup_{\nN \in \N } E ^{\muNl }
[\int_0^T | \bNrsp (X_t^{N,i},\XNidt ) | ^{\phat } dt ] \le
\cref{;40i}
.\end{align}
\ms
We set $ \SS _r^m = \{\mathsf{s}\, ; \mathsf{s}(\Sr ) = m\}$.
Let $ \| \cdot \|_{\S \ts \SS _r^m} $ denote the uniform norm on
$ \S \ts \SS _r^m $ and set
$ L^{\phat }(\muNone _r ) = L^{\phat }(\Sr \ts \SS , \muNone )$.
For a function $ f $ on $\S \ts \SS _r^m $ we denote by
$ \nabla f = (\nablax \check{f}, \nabla_{y_i} \check{f})$,
where $ \check{f} $ is a function on $ \Sr \ts \Sr ^m $ such that
$ \check{f}(x,(y_i)_{i=1}^m)$ is symmetric in $ (y_i)_{i=1}^m $
for each $ x $ and $ f(x,\sum_i\delta_{y_i}) = \check{f}(x,(y_i)_{i=1}^m)$.
We decompose $ \bNrs $ as
\begin{align}\label{:40d}&
\bNrs = \bNrsp + \7
\end{align}
and we assume:
\medskip
\noindent
\As{I3}
For each $ m ,\p , r , s \in \N $ such that $ r < s $, there exists
$ \bbb _{r,s,\p } $ such that
\begin{align}
\label{:41r} &
\limi{\nN } \| \bbb _{r,s,\p }^{\nN } - \bbb _{r,s,\p }
\|_{\S \ts \SS _r^m }
= 0
.\\
\intertext{Moreover, $ \bbb _{r,s,\p }^{\nN } $ are differentiable in $ x $ and satisfying the bounds: }
\label{:41U}&
\supN \| \nabla \bbb _{r,s,\p }^{\nN } \|_{\S \ts \SS _r^m } < \infty
,\\\label{:41s}&
\limi{\p } \supN \| \bNrsp - \bNrs \|_{ L^{\phat }( \muNone _r ) } =0
.\end{align}
Furthermore, we assume for each $ i , r < s \in \N $
\begin{align}\label{:41t}&
\limi{\p } \limsupi{\nN }
E ^{\muNl }
[\int_0^T | \{ \bNrsp - \bNrs \} (X_t^{N,i},\XNidt ) | ^{\phat }
dt ] = 0
,\\\label{:41tt}&
\limi{\p }
E ^{\mul }
[\int_0^T | \{ \brsp - \brs \} (X_t^{i},\Xidt ) | ^{\phat }
dt ] = 0
,\end{align}
where $ \bbb _{r,s} $ is such that
\begin{align}\label{:50l}&
\bbb _{r,s} (x,\mathsf{y}) =
\limi{\nN } \bbb _{r,s}^{\nN } (x,\mathsf{y})
\quad \text{ for each } (x,\mathsf{y}) \in \bigcup_{\p \in \N }
(\S \ts \SS )_{r,\p } ^c
.\end{align}
\begin{rem}\label{r:J3}
We see that $ \bigcup_{\p \in \N }
(\S \ts \SS )_{r,\p } ^c = \{\Sr ^c \ts \SS \} \cup
\{ (x,\mathsf{y}) ; x \not= y_i \text{ for all }i\} $ by definition and
$ \bbb _{r,s} (x,\mathsf{y}) = 0 $ for $ x \not\in \Sr $ by \eqref{:40b}.
The limit in \eqref{:50l} exists because of
\eqref{:40F}, \eqref{:40g}, and \eqref{:41r}.
\end{rem}
\noindent
\As{I4}
There exists a $ \btail \in C (\S ; \Rd )$ independent of $ r \in \N $ and
$ \mathsf{s} \in \SSS $ such that
\begin{align} \label{:41u} &
\limi{s} \limsupi{\nN } \| \bNrstail - \btail
\|_{ L^{\phat }( \muNone _r ) }
=0
.\end{align}
Furthermore, for each $ r , i \in \N $:
\begin{align}\label{:41v}&
\limi{s} \limsupi{\nN }
E ^{\muNl }
[\int_0^T | ( \bNrstail - \btail ) (X_t^{N,i},\XNidt ) | ^{\phat }
dt ] = 0
.\end{align}
We remark that $ \btail $ is automatically independent of $ r $
for consistency \eqref{:40g}.
By assumption, $ \btail = \btail (x)$ is a function of $ x $.
From \eqref{:40a} and \eqref{:40d} we have
\begin{align}\label{:41w}&
\bN = \bNrsp + \btail + \{\7 \} + \{ \6 -\btail \}
.\end{align}
In \As{I3} and \As{I4}, we have assumed that the last two terms
$ \{\7 \} $ and $ \{ \6 -\btail \} $ in \eqref{:41w} are asymptotically negligible.
Under these assumptions, we prove in \lref{l:50} that
there exists $ \bbb $ such that for each $ r \in \N $
\begin{align}
\label{:41a}&
\limi{s} \| \brs - \bbb \|_{ L^{\phat }( \muNone _r ) } = 0
.\end{align}
We assume: \\
\As{I5} For each $ i , r \in \N $
\begin{align}\label{:41z}&
\limi{s }
E ^{\mul }
[\int_0^T | ( \brs - \mathsf{b} ) (X_t^{i},\Xidt ) | ^{\phat }
dt ] = 0
.\end{align}
We say a sequence $ \{ \mathbf{X}^{\nN } \} $ of $ C([0,T];\S ^{\nN })$-valued
random variables is tight if for any subsequence we can choose a subsequence
denoted by the same symbol such that
$ \{ \mathbf{X}^{\nN ,m} \}_{\nN \ge m } $ is convergent in law
in $ C([0,T]; \S ^m )$ for each $ m \in \N $.
With these preparations, we state the main theorem in this section.
\begin{thm} \label{l:21}
Assume \As{H1}--\As{H4} and \0.
Then, $ \{ \mathbf{X}^{\nN } \} _{\nN \in \N } $
is tight in $ C([0,T];\S ^{\N } )$ and, any limit point
$ \mathbf{X} = (X^i)_{i\in\N } $ of $ \{ \mathbf{X}^{\nN } \} _{\nN \in \N } $
is a solution of the ISDE
\begin{align}\label{:41b}&
dX_t^i = \sigma (X_t^i,\mathsf{X}_t^{\diai }) dB_t^i +
\{\bbb (X_t^i,\mathsf{X}_t^{\diai })+
\btail (X_t^i)\} dt
.\end{align}
\end{thm}
\begin{rem}\label{r:49}
If diffusion processes are symmetric, we can dispense with
\eqref{:40z}, \eqref{:40i}, \eqref{:41t}, \eqref{:41tt},
\eqref{:41v}, and \eqref{:41z} as we see in Subsection \ref{s:23}.
Indeed, using the Lyons-Zheng decomposition
we can derive these from {\em static} conditions
\As{H4}, \eqref{:40h}, \eqref{:41r}, \eqref{:41s}, \eqref{:41u}, and \eqref{:41a}.
We remark that we can apply \tref{l:21} to non-symmetric diffusion processes by assuming these {\em dynamical} conditions.
\end{rem}
\subsection{Finite-particle approximations (reversible case)}\label{s:23}
For a subset $ A $, we set $ \map{\pi_{A}}{\SS }{\SS } $ by
$ \pi_{A} (\mathsf{s}) = \mathsf{s} (\cdot \cap A )$.
We say a function $ f $ on $ \SS $ is local if
$ f $ is $ \sigma [\pi_{K}]$-measurable for some compact set $ K $ in $ \S $.
For a local function $ f $ on $ \SS $, we say $ f $ is smooth if $ \check{f}$ is smooth, where
$ \check{f}(x_1,\ldots )$ is a symmetric function such that
$ \check{f}(x_1,\ldots ) = f (\mathsf{x})$ for $ \mathsf{x} = \sum _i \delta _{x_i}$.
Let $ \di $ be the set of all bounded, local smooth functions on $ \SS $.
We write $ f \in L_{\mathrm{loc}}^p(\muone )$ if
$ f \in L^p(\Sr \ts \SS , \muone )$ for all $ r \in\N $.
Let
$ C_{0}^{\infty}(\S )\ot \di = \{ \sum_{i=1}^Nf_i (x) g_i (\mathsf{y})\, ;\, f_i \in C_{0}^{\infty}(\S ),\, g_i \in \di ,\, N \in \N \} $
denote the algebraic tensor product of $ C_{0}^{\infty}(\S ) $ and $\di $.
\begin{dfn}\label{d:21}
A $ \Rd $-valued function $ \dmu \in L_{\mathrm{loc}}^1(\muone )$ is called
{\em the logarithmic derivative} of $\mu $ if, for all
$\varphi \in C_{0}^{\infty}(\S )\ot \di $,
\begin{align}&\notag
\int _{\S \times \SS }
\dmu (x,\mathsf{y})\varphi (x,\mathsf{y})
\muone (dx d\mathsf{y}) =
- \int _{\S \times \SS }
\nablax \varphi (x,\mathsf{y}) \muone (dx d\mathsf{y})
.\end{align}
\end{dfn}
\begin{rem}\label{r:21}
\thetag{1}
The logarithmic derivative $ \dmu $ is determined uniquely (if exists).
\\\thetag{2}
If the boundary $ \partial \S $ is nonempty and particles
hit the boundary, then $ \dmu $ would contain a term arising from the boundary condition.
For example, if the Neumann boundary condition is imposed on the boundary,
then there would be local time-type drifts.
We shall later assume that particles never hit the boundary,
and the above formulation is thus sufficient in the present situation.
\\
\thetag{3}
A sufficient condition for the explicit expression of
the logarithmic derivative of point processes is given in \cite[Theorem 45]{o.isde}.
Using this, one can obtain the logarithmic derivative of point processes appearing in random matrix theory such as sine$_{\beta } $, Airy$_{\beta }, $ ($ \beta=1,2,4$), Bessel$_{2,\alpha } $ ($ 1\le \alpha $),
and the Ginibre point process (see Examples in \sref{s:5}). For canonical Gibbs measures with Ruelle-class interaction potentials, one can easily calculate the logarithmic derivative employing DLR equation \cite[Lemma 10.10]{o-t.tail}.
\end{rem}
We assume:
\bs
\noindent
\As{J1}
Each $ \muN $ has a logarithmic derivative $ \dmuN $,
and the coefficient $ \bbb ^{\nN } $ is given as
\begin{align}\label{:21j}&
\bbb ^{\nN }=\frac{1}{2}\{\nablax \aN + \aN \dmuN \}
.\end{align}
Furthermore, the vector-valued functions $ \{\nablax \aN \}_{\nN }$ are continuous and converge to $ \nablax \aaa $ uniformly on each $ \Sr \ts \SS $, where
$ \nablax \aN $ is the $ d $-dimensional column vector such that
\begin{align}\label{:21k}&
\nablax \aN (x,\mathsf{y}) =
{}^{t}\Big(
\sum_{i=1}^d \PD{}{x_i}\aN _{1i} (x,\mathsf{y}),\ldots,
\sum_{i=1}^d \PD{}{x_i}\aN _{di} (x,\mathsf{y})
\Big)
.\end{align}
\begin{rem}\label{r:26}
From \As{J1} we see that the delabeled dynamics
$ \mathsf{X}^{\nN } = \sum_{i=1}^{\nN } \delta_{X^i}$
of $ \mathbf{X}^{\nN } $ is reversible with respect to $ \muN $.
Thus \As{J1} relates the measure $ \muN $
with the labeled dynamics $ \mathbf{X}^{\nN } $.
For each $ \nN < \infty $, $\XX ^{\nN }$ has a reversible measure.
Indeed, the symmetrization $ (\muN \circ \labN ^{-1})_{\mathrm{sym}}$
of $\muN \circ \labN ^{-1} $ is a reversible measure of $\XX ^{\nN }$
as we see for $ \muNcheck $ in Introduction, where
$ (\muN \circ \labN ^{-1})_{\mathrm{sym}}= \frac{1}{\nN ! }
\sum_{\sigma \in \mathfrak{S}_N}(\muN \circ \labN ^{-1})\circ \sigma ^{-1} $ and
$ \mathfrak{S}_N$ is the symmetric group of order $ N $.
When $ \nN = \infty $, $ \mathbf{X}$ does not have any reversible measure
in general. For example, infinite-dimensional Brownian motion $ \mathbf{B}=(B^i)_{i\in\N }$ on $ (\Rd )^{\N }$ has no reversible measures.
We also remark that the Airy$_{\beta }$ ($ \beta = 1,2,4$) interacting Brownian motion
defined by \eqref{:10H} has a reversible measure given by
$ \mu _{\mathrm{Airy}, \beta }\circ\lab ^{-1}$
with label $ \lab (\mathsf{s}) = (s_1,s_2,\ldots)$ such that $ s_i > s_{i+1}$
for all $ i \in \N $ because $ \lab $ gives a bijection from (a subset of) $ \SS $ to $ \R ^{\N }$ defined for $ \mu _{\mathrm{Airy}, \beta } $-a.s.\ $ \mathsf{s}$, and thus the relation
$ \mathbf{X}_t = \lab (\mathsf{X}_t)$ holds for all $ t $.
\end{rem}
We prove that convergence of the logarithmic derivative
implies weak convergence of the solutions of the associated SDEs.
Each logarithmic derivative $ \dmuN $ belongs to a different $ L^p $-space
$ L^p (\muNone )$, and $ \muNone $ are mutually singular.
Hence we decompose $ \dmuN $ to define a kind of $ L^p $-convergence.
Let $ \map{\uu ,\ \uN ,\ \ww }{\S }{\Rd }$ and
$\map{\g ,\, \gN ,\, \vvv ,\, \vN }{\S ^{2} }{\Rd }$
be measurable functions. We set
\begin{align}
\label{:21l}&
\ggs (x,\mathsf{y})= \vs + \sum_{i} \chi _s (x-y_i) \g (x,y_i)
,\\&\notag
\ggNs (x,\mathsf{y})= \vNs + \sum_{i} \chi _s (x-y_i) \gN (x,y_i)
,\\&\notag
\rrNs (x,\mathsf{y})= \vNsinfty + \sum_{i}(1- \chi _s (x-y_i))\gN (x,y_i)
,\end{align}
where $ \mathsf{y}=\sum_{i}\delta_{y_i}$ and
$ \chi _s \in C_0^{\infty} (\S )$ is a cut-off function such that
$ 0 \le \chi _s \le 1 $, $ \chi _s (x) = 0 $ for $ |x| \ge s + 1 $, and
$ \chi _s (x) = 1$ for $ |x| \le s $.
We assume the following.
\medskip
\noindent
\As{J2} Each $ \muN $ has a logarithmic derivative $ \dmuN $ such that
\begin{align}\label{:21m}&
\dmuN (x,\mathsf{y})= \uN (x) + \ggNs (x,\mathsf{y}) + \rrNs (x,\mathsf{y})
.\end{align}
Furthermore, we assume that
\begin{itemize}
\item[(1)]
$\uN $ are in $ C^1 (\S ) $. Furthermore,
$ \uN $ and $ \nabla \uN $ converge uniformly to $ u $ and $ \nabla u $, respectively, on each compact set in $ \S $.
\item[(2)]
For each $ s \in \N $, $\vNs $ are in $ C^1 (\S ) $.
Furthermore, functions
$\vNs $ and $ \nablax \vNs $ converge uniformly to $\vs $ and
$ \nablax \vs $, respectively, on each compact set in $ \S $.
\item[(3)]
$ g^{\nN }$ are in $ C^1 (\S ^2 \cap \{x \not= y \}) $.
Furthermore,
$ g^{\nN }$ and $\nablax g^{\nN }$ converge uniformly to
$ g $ and $\nablax g $, respectively, on $ \S ^2 \cap \{ |x-y| \ge 2^{-\p } \} $ for each $ \p >0 $.
In addition, for each $ r \in \N $,
\begin{align}\label{:21q}&
\limi{\p } \limsupi{\nN }
\int_{x \in \Sr , |x-y| \le 2^{-\p } }
\chi _s (x-y) | \gN (x,y) | ^{\phat }\, \rho _x^{\nN ,1}(y) dxdy = 0
,\end{align}
where $ \rho _x^{\nN ,1} $ is a one-correlation function of the reduced Palm measure $ \mu _x^{\nN }$.
\item[(4)] There exists a continuous function $ \map{w }{\S }{\R }$ such that
\begin{align}\label{:21r}&
\limi{s}\limsupi{N}
\int_{\Sr \times \SSS } |\rrNs (x,\mathsf{y}) - \www |^{\phat } d\muNone
= 0 , \quad \ww \in L^{\phat }_{\mathrm{loc}} (\S ,dx)
.\end{align}
\end{itemize}
Let $ p $ be such that $ 1 < p < \phat $.
Assume \As{H1} and \As{J2}.
Then from \cite[Theorem 45]{o.isde} we see that
the logarithmic derivative $\dmu $ of $ \mu $ exists in
$ \Lploc (\muone )$ and is given by
\begin{align} \label{:21v}
\dmu (x,\mathsf{y})=
\uu (x) + \mathsf{g} (x,\mathsf{y}) + \www
.\end{align}
Here $ \mathsf{g} (x,\mathsf{y})= \limi{s} \ggs (x,\mathsf{y}) $ and
the convergence of $ \lim \ggs $ takes place in $ \Lploc (\muone )$.
We now introduce the ISDE of $ \mathbf{X}=(X^i)_{i\in\N }$:
\begin{align}\label{:21w}&
dX_t^i = \sigma (X_t^i,\mathsf{X}_t^{\diai }) dB_t^i
+
\frac{1}{2} \{
\nablax \aaa (X_t^i,\mathsf{X}_t^{\diai })
+
\aaa (X_t^i,\mathsf{X}_t^{\diai })
\dmu (X_t^i,\mathsf{X}_t^{\diai })
\} dt
\\\label{:21W}&
\mathbf{X}_0 = \mathbf{s}
.\end{align}
Here $ \nablax \aaa $ is defined similarly as \eqref{:21k}.
If $ \sigma $ is the unit matrix and \As{J2} is satisfied, we have
\begin{align}\label{:21x}
dX_t^i &= dB_t^i
+ \frac{1}{2} \{
u (X_t^i) + w (X_t^i)
+
\mathsf{g} (X_t^i,\mathsf{X}_t^{\diai })
\} dt
.\end{align}
In the sequel, we give a sufficient condition
for solving ISDE \eqref{:21w} (and \eqref{:21x}).
Let $ \mathbb{D}$ be the standard square field on $ \SS $ such that
for any $ f , g \in \di $ and $ \mathsf{s}=\sum_i\delta_{s_i}$
\begin{align*}&
\mathbb{D}[f,g] (\mathsf{s})=
\half \{\sum_i \nabla _i{\check{f}} \cdot \nabla _i{\check{g}}\} \,
(\mathsf{s})
,\end{align*}
where $ \cdot $ is the inner product in $ \Rd $. Since the function
$ \sum_i
\nabla _i{\check{f}} (\mathbf{s})\cdot \nabla _i{\check{g}} (\mathbf{s})$,
where $ \mathbf{s}=(s_i)_i$ and $ \mathsf{s}= \sum_i \delta_{s_i}$,
is symmetric in $ (s_i)_i$, we regard it as a function of $ \mathsf{s}$.
We set $ \Lm = L^2(\SS ,\mu )$ and let
\begin{align*}&
\mathcal{E}^{\mu }(f,g) = \int_{\SS } \mathbb{D}[f,g] (\mathsf{s}) \mu (d\mathsf{s}),
\\ &
\di ^{\mu } =\{ f \in \di \cap \Lm \, ;\,
\mathcal{E}^{\mu }(f,f) < \infty \}
.\end{align*}
We assume:
\ms
\noindent
\As{J3} \quad
$ (\mathcal{E}^{\mu }, \mathcal{D}_{\circ }^{\mu } )$ is closable on $ \Lm $.
\bs
From \As{J3} and the local boundedness of correlation functions given by
\As{H1}, we deduce that
the closure $ (\mathcal{E}^{\mu }, \mathcal{D}^{\mu } )$
of $ (\mathcal{E}^{\mu }, \mathcal{D}_{\circ }^{\mu } )$
becomes a quasi-regular Dirichlet form \cite[Theorem 1]{o.dfa}.
Hence, using a general theory of quasi-regular Dirichlet forms,
we deduce the existence of
the associated $ \SS $-valued diffusion $ (\mathsf{P},\mathsf{X})$ \cite{mr}.
By construction,
$ (\mathsf{P},\mathsf{X})$ is $ \mu $-reversible.
If one takes $ \mu $ as Poisson point process with Lebesgue intensity, then the diffusion $ (\mathsf{P},\mathsf{X})$ thus obtained
is the standard $ \SS $-valued Brownian motion $ \mathsf{B}$ such that
$ \mathsf{B}_t=\sum_{i\in\N } \delta_{B_t^i}$, where
$ \{ B^i \}_{i\in\N } $ are independent copies of the standard Brownian motions on $ \Rd $. This is the reason why we call $ \mathbb{D}$ the standard square field.
\bigskip
Let $ \mathrm{Cap}^{\mu }$ denote the capacity given by the Dirichlet space
$ (\mathcal{E}^{\mu }, \mathcal{D}^{\mu }, \Lm )$ \cite{FOT.2}.
Let
\begin{align}& \notag
\SSSsi =
\{ \sss \in \SSS \, ;\, \, \sss (x)\le 1 \text{ for all }x \in \S ,\, \,
\sss (\S )= \infty \}
\end{align}
and assume:
\medskip
\noindent
\As{J4}
$\mathrm{Cap}^{\mu } (\{\SSSsi \}^c) = 0 $.
\ms
\noindent
Let $\erf (t)=\frac{1}{\sqrt{2\pi }} \int_t^\infty e^{ - {|x|^2}/{2}}\,dx$ be the error function.
Let $ \Sr = \{ |x| < r \} $ as before. We assume:
\medskip
\noindent
\As{J5}
There exists a $ Q >0 $ such that for each $ R >0 $
\begin{align}\label{:21t}
\liminfi{r} \sup_{\nN \in\N }
\Bigl\{ \int_{\S _{r+R}} \rho ^{\nN ,1} (x)\,dx
\Bigr\} \erf \Bigl(\frac{r}{\sqrt{(r+R ) Q }}\Bigr)= 0
.\end{align}
We write $ s_i=\labN (\mathsf{s})_i$ and assume for each $ r \in \N $
\begin{align}\label{:21T}&
\limi{L} \limsupi{\nN } \sum_{i>L}
\int_{\SS } \mathrm{Erf} (\frac{|s_i|-r}{\sqrt{\cref{;3}}T})
\muN (d\mathsf{s}) = 0
.\end{align}
We remark that \eqref{:21T} is easy to check. Indeed,
we prove in \lref{l:67} that, if $ s_i=\labN (\mathsf{s})_i$ is taken
such that
\begin{align}\label{:25g}&
|s_1|\le |s_2|\le \cdots
,\end{align}
then \eqref{:21T} follows from \As{H1} and \eqref{:25f} below.
\begin{align}\label{:25f}& \quad
\limi{q}\limsupi{N} \int_{\S \backslash \S _{q}}
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx =0
.\end{align}
Let $ \lab $ be the label as before.
Let $ \mathbf{X} = (X^i)_{i\in\N }$ be a family of solution of
\eqref{:21w} satisfying
$ \mathbf{X}_0 = \mathbf{s} $ for $ \mu \circ \lab ^{-1}$-a.s.\! $ \mathbf{s}$.
We call $ \mathbf{X}$ satisfies $ \mu $-absolute continuity condition if
\begin{align}\label{:21y}&
\mu _t \prec \mu \quad \text{ for all } t \ge 0
,\end{align}
where $ \mu _t $ is the distribution of $ \mathsf{X}_t $ and
$ \mu _t \prec \mu $ means $ \mu _t $
is absolutely continuous with respect to $ \mu $.
Here $ \mathsf{X}_t = \sum_{i\in\N } \delta_{X_t^i}$,
for $ \mathbf{X}_t = (X_t^i)_{i\in\N }$. By definition
$ \mathsf{X} =\{ \mathsf{X}_t \} $ is the delabeled dynamics of
$ \mathbf{X}$ and by construction $ \mathsf{X}_0 = \mu $ in distribution.
We say ISDE \eqref{:21w} has $ \mu $-uniqueness of solutions in law
if $ \mathbf{X}$ and $ \mathbf{X}'$ are solutions with the same initial distributions
satisfying the $ \mu $-absolute continuity condition, then
they are equivalent in law.
We assume:
\medskip
\noindent
\As{J6}
ISDE \eqref{:21w} has $ \mu $-uniqueness of solutions in law.
\medskip
Let $ \mathbf{X}^{N} $ be a solution of \eqref{:20m}. From \eqref{:21j}
we can rewrite \eqref{:20m} as
\begin{align}\label{:21z}&
dX_t^{N,i} = \sigmaN (X_t^{N,i} ,\mathsf{X}_t^{\nN , \diai }) dB_t^i
+
\frac{1}{2} \{
\nablax \aN
+
\aN
\dmuN
\} (X_t^{N,i} ,\mathsf{X}_t^{\nN , \diai }) dt
.\end{align}
We set $ \mathbf{X}^{\nN , m }= (X^{\nN ,1},X^{\nN ,2},\ldots,X^{\nN ,m}) $
$ 1 \le m \le \nN $ and
$ \mathbf{X}^{m}= (X^{1},X^{2},\ldots,X^{m})$.
We say $ \{ \mathbf{X}^{\nN } \} $ is tight in $ C([0,\infty );\S ^{\N })$
if each subsequence $ \{ \mathbf{X}^{\nN '} \} $ contains a subsequence
$ \{ \mathbf{X}^{\nN ''} \} $ such that
$ \{ \mathbf{X}^{\nN '',m} \} $ is convergent weakly in $ C([0,\infty );\S ^m )$
for each $ m \in \N $.
\begin{thm}\label{l:22}
Assume \As{H1}--\As{H4} and \As{J1}--\As{J5}.
Assume that $ \mathbf{X}_0^{\nN }= \muN \circ \labN ^{-1} $ in distribution.
Then $ \{ \mathbf{X}^{\nN } \} $ is tight in $ C([0,\infty );\S ^{\N })$ and
each limit point $ \mathbf{X}$ of $ \{ \mathbf{X}^{N} \} $ is a solution of
\eqref{:21w} with initial distribution $ \mul $.
Furthermore, if we assume \As{J6} in addition, then for any $m \in \mathbb{N}$
\begin{align}\label{:21a}
\limi{\nN } \mathbf{X}^{\nN , m } = \mathbf{X}^m
\quad\text{ weakly in } C([0,\infty) ,\S ^m)
.\end{align}
\end{thm}
\begin{rem}\label{r:21a}
To prove \eqref{:21a} it is sufficient to prove the convergence in $ C([0,T];\S ^m )$ for each $ T \in \N $. We do this in the following sections.
\end{rem}
\begin{rem}\label{r:k4} \thetag{1}
A sufficient condition for \As{J3} is obtained in \cite{o.rm, o.rm2}.
Indeed, if $ \mu $ is a $ (\Phi ,\Psi )$-quasi-Gibbs measure
with upper semi-continuous potential $ (\Phi ,\Psi )$,
then \As{J3} is satisfied. This condition is mild and is satisfied by
all examples in the present paper.
We refer to \cite{o.rm, o.rm2} for the definition of quasi-Gibbs property.
\\
\thetag{2}
From the general theory of Dirichlet forms,
we see that \As{J4} is equivalent to the non-collision of particles \cite{FOT.2}.
We refer to \cite{inu} for a necessary and sufficient condition of this non-collision property of interacting Brownian motions in finite-dimensions, which gives a sufficient condition of non-collision in infinite dimensions.
We also refer to \cite{o.col} for a sufficient condition for non-collision property of interacting Brownian motions in infinite-dimensions applicable to, in particular, determinantal point processes.
\\\thetag{3}
From \eqref{:21t} of \As{J5}, we deduce that each tagged particle $ X^i $
does not explode \cite{FOT.2,o.tp}. We remark that the delabeled dynamics
$ \mathsf{X}=\sum_i \delta_{X^i}$ are $ \mu $-reversible, and they thus
never explode. Indeed, as for configuration-valued diffusions, explosion occurs if and only if infinitely many particles gather in a compact domain,
so the explosion of tagged particle does not imply that of the configuration-valued process.
\\\thetag{4}
It is known that, if we suppose \As{H1}, \As{J1}--\As{J5}, then ISDE \eqref{:21w} has a solution for $ \mul $-a.s. $ \mathbf{s}$ satisfying
the non-collision and non-explosion property \cite{o.isde}.
Indeed, let $ \mathbf{X}=(X^i)$ be the $ \SN $-valued continuous process
consisting of
tagged particles $ X^i$ of the delabeled diffusion process $ \mathsf{X}=\sum_{i\in\N } \delta_{X^i}$ given by the Dirichlet form of \As{J3}.
Then from \As{J4} and \As{J5} \eqref{:21t} we see
$ \mathbf{X}$ is uniquely determined by its initial starting point.
It was proved that $ \mathbf{X}$ is a solution of
\eqref{:21w} in \cite{o.isde}.
\end{rem}
\begin{rem}\label{r:K5}
Assumption \As{J6} follows from tail triviality of $ \mu $ \cite{o-t.tail}, where tail triviality of $ \mu $ means that the tail $ \sigma $-field
$ \mathcal{T}=\bigcap_{r=1}^{\infty} \sigma [\pi _{\Sr ^c}]$ is $ \mu $-trivial.
Indeed, from tail triviality of $ \mu $ and marginal assumptions (\As{E1}, \As{F1}, and \As{F2} in \cite{o-t.tail}), we obtain \As{J6}.
Tail triviality holds for all determinantal point processes \cite{o-o.tt} and
grand canonical Gibbs measures with sufficiently small inverse temperature
$ \beta > 0 $.
\end{rem}
\section{Proof of \tref{l:21}}\label{s:3}
The purpose of this section is to prove \tref{l:21}.
We assume the same assumptions as \tref{l:21}
throughout this section. We begin by proving \eqref{:41a}.
\begin{lem} \label{l:50}
\eqref{:41a} holds.
\end{lem}
\begin{proof}
From \As{H1} and \eqref{:41r}, we obtain
\begin{align} \label{:50z}&\quad \quad \quad
\limi{N} \bNrsp = \, \brsp
\quad \text{ for $ \muonebar $-a.s.\! and in $ \Lp $. }
\end{align}
We next prove the convergence of $ \{ \brsp \} $ as $ \p \to \infty $.
Note that
\begin{align}\label{:50i}&
\| \brsp - \brsq \|_{ \Lp }
\\\le \, & \notag
\| \brsp - \bNrsp \|_{ \Lp } +
\| \bNrsp - \bNrsq \|_{ \Lp } +
\| \bNrsq - \brsq \|_{ \Lp }
.\end{align}
From \eqref{:41s} for each $ \epsilon $ there exists a $ \p _0$ such that
for all $ \p ,\q \ge \p _0 $
\begin{align}\label{:50p}&
\supN \| \bNrsp -\bNrsq \|_{ L^{\phat }( \muNone _r ) } < \epsilon
\end{align}
By \eqref{:50z} there exists an $ \nN = \nN_{\p ,\q }$ such that
\begin{align}\label{:50q}&
\| \brsp - \bNrsp \|_{ \Lp } < \epsilon ,\quad
\| \brsq - \bNrsq \|_{ \Lp } < \epsilon
.\end{align}
Putting \eqref{:50p} and \eqref{:50q} into \eqref{:50i}, we deduce that
$ \{ \brsp \}_{\p \in \N } $ is a Cauchy sequence in $ \Lp $.
Hence from \eqref{:40g}, \eqref{:41s}, and \eqref{:50l} we see
\begin{align}\label{:50a}
\limi{\p } \brsp = \, &\brs \quad \text{ in } \Lp
.\end{align}
Recall that
$ \bbb _{r,s}^{\nN } =
\mathrm{E}^{\muNonebar } [ \bbb ^{N}|\mathcal{F}_{\rrr ,s} ] $
by \eqref{:40a}. Then, because
$ \mathcal{F}_{\rrr ,s} \subset \mathcal{F}_{\rrr ,s+1}$, we have
\begin{align}\label{:50J}&
\bbb _{r,s}^{\nN } =
\mathrm{E}^{\muNonebar } [ \bbb _{r,s+1}^{N}|\mathcal{F}_{\rrr ,s} ]
.\end{align}
From $ \bbb _{r,s}^{\nN } =
\mathrm{E}^{\muNonebar } [ \bbb ^{N}|\mathcal{F}_{\rrr ,s} ] $
we have
\begin{align}& \notag
\| \bbb _{r,s}^{\nN } \|_{\LNp } \le \| \bN \|_{\LNp }
\end{align}
From this and \eqref{:40h} we obtain
\begin{align}\label{:50X}&
\sup_{r < s } \limsupi{\nN } \| \bbb _{r,s}^{\nN } \|_{\LNp } \le
\limsupi{\nN } \| \bN \|_{\LNp } <
\infty
.\end{align}
Combining \eqref{:50l}, \eqref{:50J} and \eqref{:50X}, we have
\begin{align}\label{:50n}&
\bbb _{r,s} =
\limi{\nN} \bbb _{r,s}^{\nN } = \limi{\nN}
\mathrm{E}^{\muNonebar } [\bbb _{r,s+1}^{N}|\mathcal{F}_{\rrr ,s} ] =
\mathrm{E}^{\muonebar } [ \bbb _{r,s+1}|\mathcal{F}_{\rrr ,s} ]
.\end{align}
From \As{H1}, \eqref{:50l}, \eqref{:50X}, and Fatou's lemma, we see that
\begin{align}\label{:50m}&
\sup_{r < s } \| \bbb _{r,s} \|_{\Lp } \le
\sup_{r < s } \liminfi{\nN } \| \bbb _{r,s}^{\nN } \|_{\LNp }
<
\infty
.\end{align}
From \eqref{:50n} we deduce that
$ \{ \bbb _{r,s} \}_{s=r+1}^{\infty} $ is martingale in $ s $.
Applying the martingale convergence theorem to $ \{\brs \}_{s=r+1}^{\infty}$ and
using \eqref{:50m}, we deduce that there exists a $ \br $ such that
\begin{align}\label{:50o}&
\brs = \mathrm{E}^{\muonebar } [ \br |\mathcal{F}_{\rrr ,s}]
\end{align}
and that
\begin{align}&\notag
\limi{s} \bbb _{r,s} = \br
\quad \text{for $ \muonebar $-a.s.\! and in }
L^{\phat }(\muonebar )
.\end{align}
By the consistency of $ \{ \muonebar \}_{r \in \N } $ in $ r $,
the function $ \br $ in \eqref{:50o} can be taken to be independent of $ r $.
This together with \eqref{:50a} completes the proof of \eqref{:41a}.
\end{proof}
\bs
We proceed with the proof of the latter half of \tref{l:21}.
Recall SDE \eqref{:20m}. Then
\begin{align}\label{:52a} &
X_t^{N,i} - X_0^{N,i} =
\int_0^{t} \sN (X_u^{N,i},\XNidu )dB _u^i +
\int_0^t \bbb ^{N} (X_u^{N,i},\XNidu ) du
.\end{align}
Using the decomposition in \eqref{:41w}, we see from \eqref{:52a} that
\begin{align}\label{:52b}
X_t^{N,i} - X_0^{N,i} =\, &
\int_0^{t} \sN (X_u^{N,i},\XNidu )dB _u^i
+
\int_0^t \{\bNrsp + \btail \} (X_u^{N,i},\XNidu ) du
\\ \notag + &
\int_0^t \Big[ \{\7 \}
+
\{ \6 - \btail \} \Big] (X_u^{N,i},\XNidu ) du
.\end{align}
Let $ \partial _{i,j} = \PD{}{x_{i,j}}$,
$ x_i = (x_{i,j})_{j=1}^d \in \Rd $,
and
$ \mathbf{x}_m = (x_i)_{i=1}^m \in (\Rd )^m $.
Set $ \nabla _i =(\partial _{i,j} )_{j=1}^d$.
Let $ \hhh \in C_0^{\infty} (\S ^m) $ and
$ \aN _i\nabla _i\nabla _i \hhh (\mathbf{x}_m) =
\sum_{k,l=1}^d \aN _{kl} (x_i) \partial_{i,k}\partial_{i,l}
\hhh (\mathbf{x}_m)$.
Applying the It$ \hat{\mathrm{o}}$ formula to $ \hhh $ and \eqref{:52b}, and
putting $ \mathbf{X}_t^{\nN , m } = (X_t^{N,1},\ldots,X_t^{N,m})$, we deduce that
\begin{align}\label{:52d}&
\hhh (\mathbf{X}_t^{N,m} ) - \hhh (\mathbf{X}_0^{N,m} ) =
\sum_{i=1}^m \Big ( \int_0^{t}
\nc \cdot \sN (X_u^{N,i},\XNidu )dB _u^i
\\ \notag &
+
{\int_0^t \half
\aN _i\nabla _i\nabla _i \hhh (\mathbf{X}_u^{\nN , m } ) +
\{\bNrsp + \btail \} (X_u^{N,i},\XNidu ) \cdot \nc du }
\Big)
\\ \notag & +
\sum_{i=1}^m \int_0^t \nc \cdot \{ \7 \} (X_u^{N,i},\XNidu ) du
\\ \notag & +
\sum_{i=1}^m \int_0^t
\nc
\cdot
\{\6 - \btail \} (X_u^{N,i},\XNidu ) du
.\end{align}
We set
\begin{align}&\notag
\QN =
\sum_{i=1}^m \int_0^T \Big| \{ \7 \} (X_u^{N,i},\XNidu )
\Big|
du
,\\ &\notag
\RN =
\sum_{i=1}^m \int_0^T \Big|
\{\6 - \btail \} (X_u^{N,i},\XNidu ) \Big|du
.\end{align}
\begin{lem} \label{l:56}For each $ m , r < s \in \N $
\begin{align}& \notag
\limi{\p } \limsupi{\nN } \mathrm{E} ^{\muNl }
\big[ (\QN )^{\phat }\big]
= 0
,\\& \notag
\limi{s} \limsupi{\nN } \mathrm{E} ^{\muNl }
\big[ (\RN )^{\phat }\big]
= 0
.\end{align}
\end{lem}
\begin{proof}
\lref{l:56} follows from \eqref{:41t} and \eqref{:41v} immediately.
\end{proof}
Let $ \Xi = \S ^m \ts (\R ^{d^2})^m \ts (\Rd )^m $ and $ \hhh \in C_0^{\infty} (\S ^m) $.
Let $ \map{F }{C([0,T];\Xi )}{C([0,T]; \R )}$ such that
\begin{align}\label{:52p}
F (\xi , \eta , \zeta )& (t) = \hhh (\xi (t)) - \hhh (\xi (0) ) -
\int_0^t \sum_{i=1}^m \zeta _i (u) \cdot \nabla _i \hhh (\xi (u)) du
\\ \notag &
-
\int_0^t \sum_{i=1}^m \Big(
\half \eta _{i} (u) \Delta _i \hhh (\xi (u)) +
\btail (\xi _i(u)) \cdot \nabla _i \hhh (\xi (u)) \Big) du
,\end{align}
where $ \xi = (\xi _i)_{i=1}^m $,
$ \eta = (\eta _i)_{i=1}^m $, $ \eta _i = (\eta _{i,kl})_{k,l=1}^d$,
$ \zeta = (\zeta _i)_{i=1}^m $, and
$ \Delta _i = \sum_{j=1}^d \partial_{i,j} ^2 $.
As $ \hhh \in C_0^{\infty} (\S ^m) $ and $ \btail \in C (\S ^m) $ by definition, we see that $ F $ satisfies the following.
\\
\thetag{1} $ F $ is continuous. \\
\thetag{2} $ F (\xi , \eta , \zeta ) $ is bounded in
$ (\xi , \eta ) $ for each $ \zeta $, and linear in $ \zeta $ for each $ (\xi , \eta )$.
\medskip
Let
$ \mathbf{A}^{\nN ,m}= ( \mathsf{A}^{\nN ,i})_{i=1}^m $ and
$ \BNrspm = ( \BNirsp )_{i=1}^m $
such that
\begin{align}\label{:52g}&
\ANirs (t) = \aNrs (X_t^{N,i},\XNidt )
,\quad
\BNirsp (t) = \bNrsp (X_t^{N,i},\XNidt )
.\end{align}
Then we see from \eqref{:52d}--\eqref{:52g} that for each $ m \in \N $
\begin{align}\label{:52h}&
\Big|F (\mathbf{X}^{N,m}, \mathbf{A}^{\nN ,m} , \BNrspm ) -
\sum_{i=1}^m \int_0^{\cdot }
\nc \cdot \sN (X_u^{N,i},\XNidu )dB _u^i \Big|\\ \notag &
\le
\cref{;52h}\{ \QN + \RN \}
,\end{align}
where $ \Ct{;52h} =\cref{;52h}(\psi )$ is the constant such that
$ \cref{;52h}= \max_{i=1}^m \|\nabla _i \psi \|_{\S ^m}$
($ \| \cdot \|_{A}$ is the uniform norm over $ A$ as before).
We take the limit of each term in \eqref{:52h} in the sequel.
\begin{lem} \label{l:51}
$ \{X^{N,i}\}_{\nN \in\N }$, $\{ \ANirs \}_{N\in\N }$ and
$\{ \BNirsp \}_{N \in\N }$
are tight for each $ i , r , s ,\p \in \N $.
\end{lem}
\begin{proof}
The tightness of $ \{X^{N,i}\}_{\nN \in\N }$ is clear from \As{I1}.
We note that $ \{\nablax \aN \}_{\nN }$ is uniformly bounded
on $ \Sr \ts \SS $ for each $ r \in \N $ by \As{H4}.
Hence from this and \As{I1} there exists a constant $ \Ct{;51d}$ independent of $ \nN $ such that for all $ 0 \le u,v \le T $
\begin{align}&\notag
\mathrm{E} ^{\muNl }
[|\ANirs (u) - \ANirs (v) |^{4}; \sup_{t\in [0,T] }|X_t^{N,i}| \le a ]
\le \cref{;51d} |u-v|^{2}
.\end{align}
By \As{I1} we see that $\{\ANirs (0) \}_{\nN \in \N }$
is tight. Combining these deduces the tightness of $\{ \ANirs \}_{N\in\N }$.
Recall that $ \BNirsp (t) = \bNrsp (X_t^{N,i},\XNidt ) $ and that
$ \bNrsp $ is $ \mathcal{F}_{\rrr ,s}$-measurable by assumption.
By construction
\begin{align}& \label{:51b}
P ^{\muNl } ( X_t^{N,j} \in \Sr \text{ for all } 1\le j \le m ,\,
0 \le t \le T
| \ \mathcal{L}_{r+s}^{N} \le m ) = 1
.\end{align}
Let $ \Ct{;51a}=
\supN \| \nabla \bbb _{r,s,\p }^{\nN } \|_{\S \ts \SS _s^{m-1} } $.
From \eqref{:52g}, \eqref{:41U}, \eqref{:51b}, and \eqref{:40z}
we see
\begin{align}& \notag
E ^{\muNl }
[|\BNirsp (u) - \BNirsp (v)| ^{4}; \sup_{t\in [0,T] }|X_t^{N,i}| \le a ,\
\mathcal{L}_{r+s}^{N} \le m
]
\\\notag = \, &
E ^{\muNl }
[|\bNrsp (X_u^{N,i},\XNidu ) - \bNrsp (X_v^{N,i},\XNidv )
| ^{4}; \sup_{t\in [0,T] }|X_t^{N,i}| \le a ,\
\mathcal{L}_{r+s}^{N} \le m
]
\\ \notag \le \, &
E ^{\muNl }
[ \sum_{j=1 }^{m}
\cref{;51a}^4
| X_u^{N,j} - X_v^{N,j}| ^{4} ;
\sup_{t\in [0,T] }|X_t^{N,i}| \le a ,\ \mathcal{L}_{r+s}^{N} \le m ]
\\ \notag
\le \, & \cref{;51a}^4 \cref{;40i}|u-v| ^{2}
\quad \text{ for all $ 0 \le u, v \le T $}
.\end{align}
From this, \eqref{:40x}, and \eqref{:40p}, we deduce
the tightness of $\{ \BNirsp \}_{N\in\N }$.
\end{proof}
\begin{lem} \label{l:53}
$ \{((X^{N,i}, \ANirs , \BNirsp ) )_{i=1}^m\}_{N \in\N } $ is tight in
$C([0,T], \Xi ^m )$ for each $ m , r , s , \p \in \N $.
\end{lem}
\begin{proof}
\lref{l:53} is obvious from \lref{l:51}.
Indeed, the tightness of the probability measures on a countable product space follows from that of the distribution of each component.
\end{proof}
Assumption \As{I1} and \lref{l:53} combined with the diagonal argument
imply that for any subsequence of
$ \{((X^{N,i}, \ANirs , \BNirsp ) )_{i=1}^m\}
_{N, \, \p \in\N ,\, r < s < \infty } $,
there exists a convergent-in-law subsequence, denoted by the same symbol.
That is, for each $ \p, s, r , m\in\N $,
\begin{align}& \label{:54a}
\limi{\nN }
(X^{N,i}, \ANirs , \BNirsp ) _{ i=1}^{m} =
(X^{i}, \Airs , \Birsp ) _{ i=1}^{m}
\quad \text{ in law}
.\end{align}
We thus assume \eqref{:54a} in the rest of this section.
Let $ \mathbf{A}^m = (\Airs )_{i=1}^m $,
$ \BNrspm = ( \BNirsp )_{i=1}^m $, and
$ \mathbf{X}^{m}= (X^i)_{i=1}^m $
for $ \mathbf{X}=(X^i)_{i\in\N }$ in \tref{l:21}.
\begin{lem} \label{l:54}
For each $ m \in \N $
\begin{align}\label{:54b} &
\limi{\nN }
F (\mathbf{X}^{N,m}, \mathbf{A}^{\nN ,m} , \BNrspm ) =
F (\mathbf{X}^{m}, \mathbf{A}^m , \Brspm )
\quad \text{ in law}
.\end{align}
Moreover, $ \Airs $ and $ \Birsp $ are given by
\begin{align}\label{:54u}&
\Airs (t) = \aaa (X_t^{i},\Xidt ) , \quad
\Birsp (t) = \brsp (X_t^{i},\Xidt )
.\end{align}
\end{lem}
\begin{proof}Recall that $ F (\xi , \eta , \zeta )$ is continuous.
Hence \eqref{:54b} follows from \eqref{:54a}.
By \As{H4} we see $ \{\aN \}$ converges to $ \aaa $ uniformly on each
$ \SrSS $. Then, from this, \eqref{:41r}, and \eqref{:52g}
we obtain \eqref{:54u}.
\end{proof}
\begin{lem} \label{l:55} For each $ m \in \N $
\begin{align}\notag
& \limi{\nN } \sum_{i=1}^m \int_0^{\cdot }
\nc \cdot \sN (X_u^{N,i},\XNidu )dB _u^i =
\mart
\quad \text{ in law},\end{align}
where $ (\hat{B}^i)_{i=1 }^m $ is the first $ m $-components of
a $ (\Rd )^{\N } $-valued Brownian motion $ (\hat{B}^i)_{i\in\N } $.
\end{lem}
\begin{proof}
By the calculation of quadratic variation, we see
\begin{align*}&
\bra
\int_{0}^{\cdot}
\partial _{i,k} \hhh (\mathbf{X}_u^{N,m})
\sum_{n=1}^d \sN _{kn}(X_u^{N,i},\XNidu ) dB_{u}^{i,n}
,
\int_{0}^{\cdot}
\partial _{j,l} \hhh (\mathbf{X}_u^{N,m})
\sum_{n=1}^d \sN _{ln}(X_u^{N,j},\XNjdu ) dB_{u}^{j,n}
\ket_u
\\ \notag &
= \delta_{ij}
\int_0^{\cdot } \aN _{kl}(X_u^{N,i},\XNidu )
\partial _{i,k} \hhh (\mathbf{X}_u^{N,m})
\partial _{i,l} \hhh (\mathbf{X}_u^{N,m}) du
.\end{align*}
From \As{H4}, we see that
$\aN $ converges to $ \mathsf{a}$ uniformly on $ \Sr $ for each $ r \in \N $.
Hence we deduce from \As{I1} and $ \hhh \in C_0^{\infty} (\S ^m)$
the convergence in law such that
\begin{align} \notag
\limi{\nN }&\sum_{i=1}^m
\int_0^{\cdot } \aN _{kl}(X_u^{N,i},\XNidu )
\partial _{i,k} \hhh (\mathbf{X}_u^{N,m})
\partial _{i,l} \hhh (\mathbf{X}_u^{N,m}) du
\\ \notag &
= \sum_{i=1}^m \int_0^{\cdot } \aaa _{kl}(X_u^{i},\Xidu )
\partial _{i,k} \hhh (\mathbf{X}_u^{m})
\partial _{i,l} \hhh (\mathbf{X}_u^{m}) du
.\end{align}
Then the right-hand side gives the quadratic variation of
$\Mart $. This completes the proof.
\end{proof}
\bs
We are now ready for the proof of \tref{l:21}.
\noindent
{\em Proof of \tref{l:21}. }
From \lref{l:56} and \eqref{:52h} we deduce that
\begin{align}\notag
\limsupi{\nN } &\mathrm{E} ^{\muNl }\Big[
\sup_{0\le t \le T }
\Big|F (\mathbf{X}^{N,m}, \mathbf{A}^{\nN ,m} , \BNrspm )(t)
\\ \notag & \quad \quad
-
\sum_{i=1}^m \int_0^{t}
\nc \cdot \sN (X_u^{N,i},\XNidu )dB _u^i \Big|^{\phat } \Big]
\\ & \notag
\le \limsupi{\nN }
\mathrm{E} ^{\muNl }\big[ (\QN )^{\phat }+ (\RN )^{\phat }\big]
=: \cref{;41}(s,\p )
,\end{align}
where $ 0\le \Ct{;41}(s,\p )=\cref{;41}(s,\p ,\hhh ) \le \infty $ is a constant
depending on $ s , \p , \hhh $.
Applying \lref{l:54} and \lref{l:55} to \eqref{:52h}, we then deduce that
\begin{align*}&
\mathrm{E} ^{\mul }\big[ \sup_{0\le t \le T } \big|
F (\mathbf{X}^{m}, \mathbf{A}^m , \Brspm )(t)
-
\sum_{i=1}^m
\int_0^{t } \8 \cdot \sigma (X_u^{i},\Xidu )d\hat{B }_u^i
\big|^{\phat } \big]
\le \cref{;41}(s,\p )
.\end{align*}
From this and \eqref{:52p}, we obtain that
\begin{align}\label{:59f}&
\mathrm{E} ^{\mul }\big[\sup_{0\le t \le T } \big|
\hhh (\mathbf{X}_{t }^{m} ) - \hhh (\mathbf{X}_0^{m} ) -
\sum_{i=1}^m
\int_0^{t }\8 \cdot
\sigma (X_u^{i},\Xidu )d\hat{B }_u^i
\\ \notag & \quad
- \sum_{i=1}^m
\9
\\ \notag & \quad
- \sum_{i=1}^m \int_0^{t }
\mathsf{b}_{r,s,\p}(X_u^{i},\Xidu ) \cdot
\8 du \big|^{\phat } \big]
\\ \notag &
\le \cref{;41}(s,\p )
.\end{align}
Take $ \hhh = \hhh _R \in C_0(\S ^m )$
such that $ \hhh (x_1,\ldots,x_m) = x_i $ for $ \{ |x_j|\le R; j=1,\ldots,m \} $ while keeping $ |\nabla_i \hhh |$ bounded in such a way that
\begin{align*}&
\cref{;41}(\p , s )=\sup_R\cref{;41}(\p , s ,R) = o (\p , s )
.\end{align*}
Then we deduce from \eqref{:59f} that
\begin{align}\label{:59h}&
\mathrm{E} ^{\mul }\big[\sup_{0\le t \le T }\big|
X_{t\wedge \tau_R}^i -
X_0^i
- \int_0^{t\wedge \tau_R }
\sigma (X_u^i,\mathsf{X}_u^{\diai })
d\hat{B }_u^i
\\ \notag &
-
\int_0^{t\wedge \tau_R }
\{\brsp (X_u^i,\mathsf{X}_u^{\diai })+
\btail (X_u^i)\} du
\big|^{\phat } \big]
\le \cref{;41}(s,\p )
,\end{align}
where $ \tau_R $ is a stopping time such that, for
$ \mathbf{X}^m=(X^{i},\Xid )_{i=1}^m \in C([0,T];(\S \ts \SS )^m)$,
\begin{align} \notag
&
\tau_R = \inf\{ t>0; |X_t^i| \ge R \text{ for some } i=1,\ldots,m \}
.\end{align}
As $ R > 0 $ is arbitrary, \eqref{:59h} holds for all $ R > 0$.
Taking $ R \to \infty $, we thus obtain
\begin{align}\label{:59i}&
\mathrm{E} ^{\mul }\big[\sup_{0\le t \le T } \big|
X_{t}^i -
X_0^i
- \int_0^{t }
\sigma (X_u^i,\mathsf{X}_u^{\diai })
d\hat{B }_u^i
\\ \notag &\quad \quad \quad \quad \quad \quad \quad
-
\int_0^{t }
\{\brsp (X_u^i,\mathsf{X}_u^{\diai })+
\btail (X_u^i)\} du
\big| \big]
\\ \notag \le \, &
\liminfi{R} \mathrm{E} ^{\mul }\big[\sup_{0\le t \le T }\big|
X_{t\wedge \tau_R}^i -
X_0^i
- \int_0^{t\wedge \tau_R}
\sigma (X_u^i,\mathsf{X}_u^{\diai })
d\hat{B }_u^i
\\ \notag & \quad \quad \quad \quad \quad \quad \quad
-
\int_0^{t\wedge \tau_R}
\{\brsp (X_u^i,\mathsf{X}_u^{\diai })+
\btail (X_u^i)\} du
\big| \big]
\\ \notag
\le \, &\cref{;41}(s,\p )^{1/{\phat }}
.\end{align}
We note here that the integrands in the first and second lines of \eqref{:59i}
are uniformly integrable because of \eqref{:59h}.
Taking $ \p \to \infty $, then $ s \to \infty $ in \eqref{:59i}, and using assumptions \eqref{:41tt} and \eqref{:41z} we thus obtain
\begin{align}& \notag
\mathrm{E} ^{\mul }\big[\sup_{0\le t \le T } \big|
X_{t}^i - X_0^i
- \int_0^{t} \sigma (X_u^i,\mathsf{X}_u^{\diai }) d\hat{B }_u^i
-
\int_0^{t}
\{\bbb (X_u^i,\mathsf{X}_u^{\diai })+
\btail (X_u^i)\} du
\big| \big] = 0
.\end{align}
This implies for all $ 0\le t \le T $
\begin{align}\label{:59l}&
X_{t}^i - X_0^i
- \int_0^{t} \sigma (X_u^i,\mathsf{X}_u^{\diai })
d\hat{B }_u^i
-
\int_0^{t}
\{\bbb (X_u^i,\mathsf{X}_u^{\diai })+
\btail (X_u^i)\} du = 0
.\end{align}
We deduce \eqref{:41b} from \eqref{:59l}, which completes the proof of \tref{l:21}.
\qed
\section{Proof of \tref{l:22} } \label{s:4}
Is this section we prove \tref{l:22} using \tref{l:21}.
\As{H1}--\As{H4} are commonly assumed in \tref{l:22} and \tref{l:21}.
Hence our task is to derive condition \0 from conditions stated in \tref{l:22}.
From \As{J2} we easily deduce that
\begin{align} \label{:60a}&
\limi{N}\uN =\uu \quad \text{ in } L^{\phat }_{\mathrm{loc}}(\S ,dx)
,\\ \label{:60b}&
\limi{N}\ggNs = \ggs \quad \text{ in }L^{\phat }_{\mathrm{loc}}(\muone )
\quad \text{for all }s
.\end{align}
\begin{lem} \label{l:61}
$ \mu $ has a logarithmic derivative $ \dmu $ in $ \Lplochat $,
where $ 1 \le p < \phat $.
\end{lem}
\begin{proof}
We use a general theory developed in \cite{o.isde}.
\As{H1} corresponds to \thetag{4.1} and \thetag{4.2} in \cite{o.isde}.
\eqref{:60a}, \eqref{:60b}, \eqref{:21m}, and \eqref{:21r}
correspond to \thetag{4.15}, \thetag{4.30}, \thetag{4.29},
and \thetag{4.31} in \cite{o.isde}.
Then all the assumptions of \cite[Theorem 45]{o.isde} are satisfied.
We thus deduce \lref{l:61} from \cite[Theorem 45]{o.isde}.
\end{proof}
\bs
Let $ \{ \mathbf{X}^{\nN } \}_{\nN \in \N } $ be a sequence of solutions in \eqref{:20m} and \eqref{:20n}. We set the $ m $-labeling
\begin{align}\label{:62p}&
\mathbf{X}^{\nN , [m] } =
(X^{\nN , 1},\ldots, X^{\nN , m},\sum_{j= 1+m }^{\nN } \delta_{X^{\nN , j}})
.\end{align}
It is known \cite{o.tp, o.isde} that $ \mathbf{X}^{\nN , [m] } $
is a diffusion process associated with the Dirichlet form
$ \mathcal{E}^{\muNm } $ on $ L^2(\S ^{m}\ts \SS , \muNm ) $ such that
\begin{align}\label{:62q}&
\mathcal{E}^{\muNm } (f,g) =
\int_{\S ^{m}\ts \SS }
\half \{ \sum_{i=1}^m \nabla _i{f} \cdot \nabla _i{g} \} +
\mathbb{D}[f,g] d\muNm
,\end{align}
where the domain $ \mathcal{D}^{[m]}$ is taken as the closure of
$ \mathcal{D}_0^{[m]}= C_0^{\infty} (\S ^m)\ot \mathcal{D}_{\circ } $.
Note that the coordinate function $ x_i = x_i\ot 1 $ is locally in $ \mathcal{D}^{[m]}$.
From this we can regard $ \{X _t^{N,i}\}$ as a Dirichlet process of the
$ m $-labeled diffusion $ \mathbf{X}^{\nN } $
associated with the Dirichlet space as above.
In other words, we can write
\begin{align}&\notag
X _t^{N,i} - X _0^{N,i} =
f_i (\mathbf{X}^{\nN } _t) - f_i (\mathbf{X}^{\nN } _0)
=: A _t^{[f_i]}
,\end{align}
where $ f_i (\mathbf{x},\mathsf{s}) = x_i \ot 1 $, $ x_i \in \Rd $, and
$ \mathbf{x} = (x_j)_{j=1}^m \in (\Rd )^m$.
By the Fukushima decomposition of $ X_t^{N,i} $, there exist a unique continuous local martingale additive functional $ \MNi = \{ \MNi _t \} $ and
an additive functional of zero energy
$ \mathsf{N}^{\nN ,i } = \{ \mathsf{N}^{\nN ,i } _t \} $ such that
\begin{align}&\notag
X_t^{N,i} - X_0^{N,i} = \MNi _t + \mathsf{N}^{\nN ,i } _t
.\end{align}
We refer to \cite[Chapter 5]{FOT.2} for the Fukushima decomposition.
Because of \eqref{:20m}, we then have
\begin{align}&\notag
\MNi _t = \int_0^t \sN (X_u^{N,i},\XNidu )dB_u^i
,\quad
\mathsf{N}^{\nN ,i } _t = \int_0^t \bbb ^{\nN }(X_u^{N,i},\XNidu )du
.\end{align}
\begin{lem} \label{l:62}
Let $ \map{r_T}{C([0,T];\S )}{C([0,T];\S )}$ be such that
$ r_T (X )_t = X_{T-t}$.
Suppose that $ \mathbf{X}_0^{\nN , [m] } = \mu ^{N,[m]} $ in law.
Then
\begin{align} \label{:62a}
X_t^{N,i} - X_0^{N,i}
&=\half \MNi _t + \half (\MNi _{T-t} (r_{T}) - \MNi _{T} (r_{T}))
\quad \text{ a.s.}
.\end{align}
\end{lem}
\begin{proof}
Applying the Lyons-Zheng decomposition \cite[Theorem 5.7.1]{FOT.2} to
additive functionals $ A ^{[f_i]} $ for $1 \le i\le m$,
we obtain \eqref{:62a}.
\end{proof}
\begin{lem} \label{l:63}
\As{I1} holds.
\end{lem}
\begin{proof}
Although $ \MNi $ is a $ d $-dimensional martingale by definition,
we assume $ d =1 $ here and prove only this case for simplicity.
The general case $ d \ge 1 $ can be proved in a similar fashion.
Let $ \cref{;3}$ be the constant in \eqref{:20o} (under the assumption $ d=1$). Then we note that for $ u \ge v$
\begin{align}\label{:63z}&
0 \le \langle \MNi \rangle _u - \langle \MNi \rangle _v =
\int_v^u \ANirs (t) dt \le \cref{;3} (u-v)
\end{align}
We begin by proving \eqref{:40z}.
From a standard calculation of martingales and \eqref{:63z}, we obtain
\begin{align}\notag
\mathrm{E} ^{\muNl }[|\MNi _u - \MNi _v |^{4}] &=
\mathrm{E} ^{\muNl }
[|B _{\langle \MNi \rangle _u} - B _{\langle \MNi \rangle _v }|^{4 }]
\\&\notag
= 3\mathrm{E} ^{\muNl }
\bigl[|\langle \MNi \rangle _u - \langle \MNi \rangle _v |^{2 }\bigr]
\\& \notag
\le \cref{;53} |u-v|^2
,\end{align}
where $ \Ct{;53}=3\cref{;3}^2 $ and $ \{B_t \}$
is a one-dimensional Brownian motion.
Applying the same calculation to
$ \MNi _{T-t} (r_{T}) - \MNi _{T} (r_{T}) $, we have
\begin{align}\label{:63b} & \quad \quad \quad
\mathrm{E} ^{\muNl }[|\MNi _{T-t} (r_{T}) - \MNi _{T-u} (r_{T})|^{4 }]
\le \cref{;53} |t-u|^{2}
\quad \text{ for each $ 0 \le t,u \le T $}
.\end{align}
Combining \eqref{:62a} and \eqref{:63b} with the Lyons-Zheng decomposition
\eqref{:62a}, we thus obtain
\begin{align}&\label{:63c}
\quad \quad \quad
\mathrm{E} ^{\muNl }[|X _t^{N,i} - X _u^{N,i}|^{4 }] \le
2\cref{;53} |t-u|^{2}
\quad \text{ for each $ 0 \le t,u \le T $}
.\end{align}
Taking a sum over $ i=1,\ldots,m$ in \eqref{:63c}, we deduce \eqref{:40z}.
We next prove \eqref{:40x}. From \eqref{:62a} we have
\begin{align} \notag
2 |X_t^{N,i} - X_0^{N,i} |
& \le |\MNi _t | + | \MNi _{T-t} (r_{T}) - \MNi _{T} (r_{T}) |
\quad \text{ a.s.}
.\end{align}
From this and a representation theorem of martingales, we obtain
\begin{align}\label{:63e} &
P ^{\muNl } (\sup_{t\in[0,T]} |X _t^{N,i} - X_0^{N,i} |\ge a )
\\ \notag
\le &
P ^{\muNl } (\sup_{t\in[0,T]} |\MNi _t |\ge a ) +
P ^{\muNl }
(\sup_{t\in[0,T]} |\MNi _{T-t} (r_{T}) - \MNi _{T} (r_{T}) |\ge a )
\\ \notag
= &
2 P ^{\muNl } (\sup_{t\in[0,T]} |\MNi _t |\ge a )
\\ \notag
= &
2 P ^{\muNl } (\sup_{t\in[0,T]} |B_{\langle \MNi \rangle _t }|\ge a )
.\end{align}
A direct calculation shows
\begin{align}\label{:63f}&
P ^{\muNl } (\sup_{t\in[0,T]} |B_{\langle \MNi \rangle _t }|\ge a )
\le
P ^{\muNl } (\sup_{t\in[0,\sqrt{\cref{;3}}T]}
|B_{t}|\ge a ) \le
\mathrm{Erf} (\frac{a}{\sqrt{\cref{;3}}T})
\end{align}
From \eqref{:63e}, \eqref{:63f}, and \As{H2}, we obtain \eqref{:40x}.
We proceed with the proof of \eqref{:40p}.
Similarly as \eqref{:63e} and \eqref{:63f}, we deduce
\begin{align}\label{:63h}
P ^{\muNl } (\inf_{t\in[0,T]} |X _t^{N,i} |\le r )
\le &
P ^{\muNl }
(\sup_{t\in[0,T]} |X _t^{N,i} - X_0^{N,i} |\ge |X_0^{N,i} |-r )
\\ \notag \le &
2 P ^{\muNl } (\sup_{t\in[0,T]} |\MNi _t |\ge |X_0^{N,i} |-r )
\\ \notag \le &
2 \int_{\SS } \mathrm{Erf} (\frac{|s_i|-r}{\sqrt{\cref{;3}}T})
\muN (d\mathsf{s})
,\end{align}
where $ s_i = \labN (\mathsf{s})_i$.
We note that $ X_0^{N,i} =s_i $ by construction.
From \eqref{:63h} and \eqref{:21T}, we deduce
\begin{align}\notag
\limsupi{\nN } P ^{\muNl } (\mathcal{L}_r^{N} > L )
\le \, & \limsupi{\nN }
\sum_{i>L} P ^{\muNl } (\inf_{t\in[0,T]} |X _t^{N,i} |\le r )
\\ \notag \le \, & 2
\limsupi{\nN } \sum_{i>L}
\int_{\SS } \mathrm{Erf} (\frac{|s_i|-r}{\sqrt{\cref{;3}}T})
\muN (d\mathsf{s})
\\ \notag \to &\, 0 \quad (L \to \infty )
.\end{align}
This completes the proof.
\end{proof}
\begin{lem} \label{l:64}
\As{I2} holds.
\end{lem}
\begin{proof}
\eqref{:40h} follows from \eqref{:60a}, \eqref{:60b}, and \eqref{:21r}.
For each $ i \in \N $ we deduce that
\begin{align}\label{:64a}
E ^{\muNl }
[\int_0^T | \bNrsp (X_t^{N,i},\XNidt ) | ^{\phat } dt ]
\le & \,
\sum_{i=1}^{\nN } E ^{\muNl }
[\int_0^T | \bNrsp (X_t^{N,i},\XNidt ) | ^{\phat } dt ]
\\ \notag = & \,
E ^{\muNl } [ \sum_{i=1}^{\nN }
\int_0^T | \bNrsp (X_t^{N,i},\XNidt ) | ^{\phat } dt ]
\\ \notag = &\,
E ^{\muNone }
[\int_0^T | \bNrsp (\mathbf{X}_t^{\nN , [1] } ) | ^{\phat } dt
]
.\end{align}
Diffusion process $ \mathbf{X}^{\nN , [1] } $ in \eqref{:62p} with $ m = 1 $
given by the Dirichlet form $ \mathcal{E}^{\muNone } $ in
\eqref{:62q} is $ \muNone $-symmetric.
Hence we see that for all $ 0 \le t \le T $
\begin{align} \notag
E ^{\muNone }
[| \bNrsp (\mathbf{X}_t^{\nN , [1] } ) | ^{\phat }
]
\le \,
\int_{\S \ts \SS } |\bNrsp | ^{\phat } d\muNone
.\end{align}
This yields
\begin{align}\label{:64b}&
\int_0^T dt \, E ^{\muNone }
[| \bNrsp (\mathbf{X}_t^{\nN , [1] } ) | ^{\phat }
]
\le \, T
\int_{\S \ts \SS } |\bNrsp | ^{\phat } d\muNone
.\end{align}
From \eqref{:64a} and \eqref{:64b} we obtain \eqref{:40i}.
\end{proof}
\begin{lem} \label{l:65}
\As{I3}--\As{I5} hold.
\end{lem}
\begin{proof} Conditions \eqref{:41r}, \eqref{:41U}, and \eqref{:41s}
follow from \As{J1}, \As{J2}, \As{I1}, \As{I2}, and \eqref{:21l}.
Similarly, as \lref{l:64}, we obtain for each $ i \in \N $
\begin{align}\label{:65a}&
E ^{\muNl }
[\int_0^T | (\7 ) (X_t^{N,i},\XNidt ) | ^{\phat } dt ]
\le T \int_{\S \ts \SS } |\7 | ^{\phat } d\muNone
.\end{align}
Hence \eqref{:41t} follows from \eqref{:65a} and \eqref{:41s}.
\eqref{:41tt} follows from \eqref{:50a} and an inequality similar to \eqref{:65a}.
We have thus obtained \As{I3}.
Condition
\eqref{:41u} follows from \As{J1} and \As{J2}.
Similarly, as \lref{l:64}, we obtain for each $ i \in \N $
\begin{align}&\notag
E ^{\muNl }
[\int_0^T | ( \bNrstail - \btail ) (X_t^{N,i},\XNidt ) | ^{\phat } dt ]
\le T \int_{\S \ts \SS } | \bNrstail - \btail | ^{\phat } d\muNone
.\end{align}
This together with \eqref{:41u} implies \eqref{:41v}. Hence we have \As{I4}.
Similarly as \lref{l:64}, we obtain \eqref{:41z} from \eqref{:41a}.
We have thus obtained \As{I5}.
\end{proof}
\bs
\noindent {\em Proof of \tref{l:22}. }
\0 follows from \lref{l:63}--\lref{l:65}.
Hence we deduce \tref{l:22} from \tref{l:21}.
\qed
We finally present a sufficient condition of \eqref{:21T}.
\begin{lem} \label{l:67}
Assume \As{H1} and \eqref{:25f} for each $ r \in \N $
as \sref{s:2}. We take the label $ \labN $ as \eqref{:25g}.
Then
\eqref{:21T} holds.
\end{lem}
\begin{proof}
Let
$ \Ct{;67} = \cref{;67}(\nN )$ be such that
\begin{align}\notag
\cref{;67} &=\,
\int_{\S }
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx
.\end{align}
Let $ \Ct{;68}=\limsupi{N}\cref{;67}(\nN )$.
Then from \As{H1} and \eqref{:25f}, we see that for each large
$ r $
\begin{align}\label{:67x}
\cref{;68}& \le \limi{\nN }
\int_{\Sr }
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx
+
\limsupi{\nN }
\int_{\S \backslash \Sr }
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx
\\ \notag &
< \infty
.\end{align}
From \As{H1} we see that $ \{\muN \}_{\nN \in \N }$
converges to $ \mu $ weakly.
Hence $ \{\muN \}_{\nN \in \N }$ is tight.
This implies that there exists a sequence of increasing sequences of
natural numbers $ \mathbf{a}_n = \{ a_n(m) \}_{m=1}^{\infty} $
such that $ \mathbf{a}_n < \mathbf{a}_{n +1}$
and that for each $ m $
\begin{align*}&
\limi{n} \limsupi{\nN } \muN (\mathsf{s} (\S _{m}) \ge a_n(m) ) = 0
. \end{align*}
Without loss of generality, we can take
$ a_n(m) > m $ for all $ m , n \in \N $.
Then from this, we see that there exists a sequence
$ \{\mathsf{p}(L) \}_{L\in\N }$ converging to $ \infty $
such that $ \mathsf{p}(L) < L $ for all $ L \in \N $ and that
\begin{align} \label{:67c}&
\limi{L}\limsupi{N} \muN (\mathsf{s}(\S _{\mathsf{p}(L)}) \ge L ) = 0
.\end{align}
Recall that the label $ \labN (\mathsf{s}) = (s_i)_{i\in\N } $
satisfies $ |s_1|\le |s_2|\le \cdots $.
Using this, we divide the set $ \SS $ as in such a way that
\begin{align*}&
\text{$ \{ s_L \in \S _{\mathsf{p}(L)} \} $ and
$ \{ s_L \not\in \S _{\mathsf{p}(L)} \} $.}
\end{align*}
Then $ \mathsf{s} \in \{ s_L \in \S _{\mathsf{p}(L)} \} $ if and only if
$ \mathsf{s}(\S _{\mathsf{p}(L)}) \ge L $. Hence
we easily see that
\begin{align}\notag &
\sum_{i>L}
\int_{\SS } \mathrm{Erf} (\frac{|s_i|-r}{\sqrt{\cref{;3}}T})
\muN (d\mathsf{s})
\le
\cref{;67}(\nN )
\muN (\{ \mathsf{s} (\S _{\mathsf{p}(L)}) \ge L \} )
+
\int_{\S \backslash \S _{\mathsf{p}(L)}}
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx
.\end{align}
Taking the limits on both sides, we obtain
\begin{align}& \notag
\limi{L}\limsupi{N}
\sum_{i>L}
\int_{\SS } \mathrm{Erf} (\frac{|s_i|-r}{\sqrt{\cref{;3}}T})
\muN (d\mathsf{s})
\le
\\\notag
\, \cref{;68}&
\limi{L}\limsupi{N}
\muN (\{ \mathsf{s} (\S _{\mathsf{p}(L)}) \ge L \} )
+
\limi{L}\limsupi{N}
\int_{\S \backslash \S _{\mathsf{p}(L)}}
\mathrm{Erf} (\frac{|x|-r}{\sqrt{\cref{;3}}T}) \rho^{\nN , 1} (x) dx
.\end{align}
Applying \eqref{:67x} and \eqref{:67c} to the second term, and
\eqref{:25f} to the third, we deduce \eqref{:21T}.
\end{proof}
\section{Examples}\label{s:5}
The finite-particle approximation in \tref{l:22} contains many examples
such as Airy$ _{\beta }$ point processes ($ \beta = 1,2,4$), Bessel$ _{2,\alpha }$ point process, the Ginibre point process, the Lennard--Jones 6-12 potential, and Riesz potentials.The first three examples are related to random matrix theory and the interaction
$ \Psi (x) = - \log |x| $, the logarithmic function.
We present these in this section. For this we shall confirm the assumptions in \tref{l:22},
that is, assumptions \As{H1}--\As{H4} and \As{J1}--\As{J6}.
Assumption \As{H1} is satisfied for the first three examples \cite{mehta,Sos00}.
As for the last two examples, we assume \As{H1}. We also assume \As{H2}.
\As{H3} can be proved in the same way as given in \cite{o-t.tail}.
In all examples, $ \aaa $ is always a unit matrix. Hence it holds that
\As{H4} is satisfied and that
\eqref{:21j} in \As{J1} becomes $ \bbb ^{\nN } = \half \dmuN $.
From this we see that SDEs \eqref{:21z} and \eqref{:21w} become
\begin{align}\label{:90b}
dX_t^{N,i}=\, & dB_t^{N,i}+\half \dmuN (X_t^{N,i}, \XNidt)\,dt\quad (1\le i\le N)
,\\\label{:90a}
dX_t^i=\, & dB_t^i+\half \dmu (X_t^i, \Xidt)\,dt\quad (i\in\N)
,\end{align}
where $ \dmu $ is the logarithmic derivative of $ \mu $ given by \eqref{:21v}.
Assumption \As{J6} for the first three examples with $ \beta = 2 $
can be proved in the same way as \cite{o-t.tail} as we explained in \rref{r:K5}.
Thus, in the rest of this section, our task is to check assumptions \As{J2}--\As{J5}.
\subsection{The Airy$ _{\beta }$ interacting Brownian motion ($ \beta = 1,2,4$)}
\label{s:52}
Let $ \mu _{\mathrm{Airy}, \beta } ^{\nN }$ and
$\mu _{\mathrm{Airy},\beta }$ be as in \sref{s:1}.
Recall SDEs \eqref{:10l} and \eqref{:10H} in \sref{s:1}.
Let $ \mathbf{X}^{\nN } = (X^{\nN ,i})_{i=1}^{\nN }$ and
$ \mathbf{X}=(X^i)_{i\in\N }$ be solutions of
\begin{align} \tag{\ref{:10l}}
dX_t^{\n ,i} = \,& dB_t^i +
\frac{\beta }{2} \sum_{j=1, \, j\not= i}^{\n }
\frac{1}{X_t^{\n ,i} - X_t^{\n ,j} } dt
- \frac{\beta }{2 }
\{ \n ^{1/3} + \frac{1}{2\n ^{1/3}}X_t^{\n ,i} \}dt
,\\%\end{align}
\tag{\ref{:10H}}
dX_t^i= \, & dB_t^i+
\frac{\beta }{2}
\limi{r} \{ \sum_{|X_t^j|<r, j\neq i} \frac{1}{X_t^i-X_t^j} -\int_{|x|<r}\frac{\varrho (x)}{-x}\,dx\}dt \quad (i\in\N)
.\end{align}
\begin{prop}\label{l:32}
If $ \beta = 1,4$, then each sub-sequential limit of solutions $ \mathbf{X}^{\nN } $
of \eqref{:10l}
satisfies \eqref{:10H}.
If $ \beta =2$, then the full sequence converges to \eqref{:10H}.
\end{prop}
\begin{proof}
Conditions \As{J2}--\As{J5} other than \eqref{:21q} can be proved in the same way
as given in \cite{o-t.airy}. In \cite{o-t.airy}, we take $ \chi_s (x) = 1_{\S _s} (x)$; its
adaptation to the present case is easy.
We consider estimates of correlation functions such that
\begin{align}\label{:92aa}&
\inf_{\nN \in \N }\rho_{\mathrm{Airy}, \beta } ^{\nN ,1} (x)
\ge
\cref{;32a} \quad \text{ for all } x \in \Sr
\\\label{:92a}&
\sup_{\nN \in \N }\rho_{\mathrm{Airy}, \beta } ^{\nN ,2} (x,y)
\le
\cref{;32} |x-y| \quad \text{ for all } x,y \in \Sr
,\end{align}
where $ \Ct{;32a}(r)$ and $ \Ct{;32}(r) $ are positive constants.
The first estimate is trivial because $ \rho_{\mathrm{Airy}, \beta } ^{\nN ,1} $
converges to $ \rho_{\mathrm{Airy}, \beta } ^{1} $ uniformly on $ \Sr $ and,
all these correlation functions are continuous and positive.
The second estimate follows
from the determinantal expression of the
correlation functions and bounds on derivative of determinantal kernels.
Estimates needed for the proof can be found in \cite{o-t.airy} and
the detail of the proof of \eqref{:92a} is left to the reader.
Equation \eqref{:21q} follows from \eqref{:92aa} and \eqref{:92a}.
Indeed, the integral in \eqref{:21q} is taken on the bounded domain and
the singularity of integral of $ \gN (x,y)=\beta /(x-y)$ near $ \{x=y\}$ is logarithmic.
Furthermore,
the one-point correlation function $ \rho_{\mathrm{Airy}, \beta, x } ^{\nN ,1}$
of the reduced Palm measure conditioned at $ x$
is controlled by the upper bound of the two-point correlation function and
the lower bound of one-point correlation function because
\begin{align*}&
\rho_{\mathrm{Airy}, \beta, x } ^{\nN ,1} (y) =
\frac{\rho_{\mathrm{Airy}, \beta } ^{\nN ,2} (x,y)}
{ \rho_{\mathrm{Airy}, \beta } ^{\nN ,1} (x)}
.\end{align*}
Using these facts, we see that \eqref{:92aa} and \eqref{:92a} imply \eqref{:21q}.
\end{proof}
\subsection{The Bessel$ _{2,\alpha }$ interacting Brownian motion} \label{s:53}
Let $\S =[0,\infty)$ and $\alpha \in[1,\infty)$.
We consider the Bessel$ _{2,\alpha }$ point process
$ \mu _{\mathrm{bes},2, \alpha }$ and their $ \nN $-particle version.
The Bessel$ _{2,\alpha }$ point process $ \mu _{\mathrm{bes},2,\alpha }$ is
a determinantal point process with kernel
\begin{align}\label{:93p}
\mathsf{K} _{\mathrm{bes},2,\alpha } (x,y) & =
\frac{J_{\alpha } (\sqrt{x}) \sqrt{y} J_{\alpha }' (\sqrt{y}) -
\sqrt{x} J_{\alpha }' (\sqrt{x}) J_{\alpha }(\sqrt{y})
}{2(x-y)}
\\ \notag & =
\frac{\sqrt{x} J_{\alpha +1} (\sqrt{x}) J_{\alpha } (\sqrt{y}) -
J_{\alpha } (\sqrt{x}) \sqrt{y} J_{\alpha +1}(\sqrt{y})
}{2(x-y)}
,\end{align}
where $ J_{\alpha }$ is the Bessel function of order $ \alpha $ \cite{Sos00,o-h.bes}.
The density $ \mathsf{m} _{\alpha }^{\n }(\mathbf{x}) d\mathbf{x}$
of the associated $ \nN $-particle systems $ \mu _{\mathrm{bes},2,\alpha }^{\nN }$
is given by
\begin{align}\label{:13}&
\mathsf{m} _{\alpha }^{\n} (\mathbf{x}) =
\frac{1}{\mathcal{Z} _{\alpha }^{\n }}
e^{-\sum_{i=1}^{\n }x_i/4\n } \prod_{j=1}^{\n }x_j^{\alpha }
\prod_{k<l}^{\n } |x_k-x_l|^{2 }
.\end{align}
It is known that $ \mu _{\mathrm{bes},2,\alpha }^{\nN }$ is also determinantal \cite[945p]{Sos00} and \cite[91p]{forrester}
The Bessel$ _{2,\alpha }$ interacting Brownian motion is
given by the following \cite{o-h.bes}.
\begin{align}\label{:93a}
dX_t^{N,i} = &dB_t^i+ \{-\frac{1}{8N}+
\frac{\alpha }{2X_t^{N,i}} +
\sum_{j=1, j\neq i}^{\nN }
\frac{1}{X_t^{N,i}-X_t^{N,j}}\} dt \quad (1\le i\le N)
,\\
\label{:93b}
dX_t^i= &dB_t^i+
\{ \frac{\alpha }{2X_t^i} +
\sum_{ j\neq i}^\infty \frac{1}{X_t^i-X_t^j} \} dt \quad (i\in\N)
.\end{align}
This appears at the hard edge of one-dimensional systems.
\begin{prop} Assume $ \alpha > 1 $.
Then \eqref{:21a} holds for \eqref{:93a} and \eqref{:93b}.
\end{prop}
\begin{proof}
\As{J2}--\As{J5} except \eqref{:21T} are proved in \cite{o-h.bes}.
We easily see that the assumptions of \lref{l:67} hold and yield \eqref{:21T}.
We thus obtain \As{J5}.
\end{proof}
\begin{rem}\label{r:33}
There exist other natural ISDEs and
$ \nN $-particle systems related to the Bessel point processes.
They are the non-colliding square Bessel processes and their square root.
The non-colliding square Bessel processes are reversible to the Bessel$ _{2,\alpha }$ point processes, but the associated Dirichlet forms are different from the Bessel$ _{2,\alpha }$ interacting Brownian motion.
Indeed, the coefficients
$ \mathsf{a}^{\nN } $ and $ \mathsf{a} $ in \sref{s:2} are taken to be
$ \mathsf{a}^{\nN }(x.\mathsf{y}) = \mathsf{a}(x.\mathsf{y})= 4x $.
On the other hand, each square root of the non-colliding Bessel processes
is not reversible to
the Bessel$ _{2,\alpha }$ point processes, but has the same type of Dirichlet forms as
the Bessel$ _{2,\alpha }$ interacting Brownian motion. In particular, the coefficients
$ \mathsf{a}^{\nN } $ and $ \mathsf{a} $ in \sref{s:2} are taken to be
$ \mathsf{a}^{\nN }(x.\mathsf{y}) = \mathsf{a}(x.\mathsf{y})= 1 $.
That is, they are constant time change of
distorted Brownian motion with the standard square field.
We refer to \cite{KT11,KT11-b,o-t.sm} for these processes.
For reader's convenience we provide an ISDE describing the non-colliding square Bessel processes and their square root. We note that SDE \eqref{:93d} is
a constant time change of that in \cite{KT11-b,o-t.sm}.
Let $ \mathbf{Y}^{N} = (Y^{N,i})_{i=1}^N $ and
$ \mathbf{Y} = (Y^i)_{i\in\N } $ be the non-colliding square Bessel processes.
Then for $ 1\le i\le N$
\begin{align}\label{:93c}
dY_t^{N,i} = & 2 \sqrt{Y_t^{N,i}}dB_t^i+
4 \{-\frac{Y_t^{N,i}} {8N}+ \frac{\alpha +1}{2} +
\sum_{j=1, j\neq i}^{\nN }
\frac{ Y_t^{N,i} }{Y_t^{N,i}-Y_t^{N,j}}\} dt
,\\
\label{:93d}
dY_t^i= & 2\sqrt{Y_t^{i}} dB_t^i+
4 \{ \frac{\alpha +1}{2} +
\sum_{ j\neq i}^\infty \frac{ Y_t^{i}}{Y_t^i-Y_t^j} \} dt \quad (i\in\N)
.\end{align}
Let $ \mathbf{Z}^{N} = (Z^{N,i})_{i=1}^N $ and $ \mathbf{Z} = (Z^i)_{i\in\N } $ be square root of the non-colliding square Bessel processes. Then applying It$ \hat{\mathrm{o}}$ formula we obtain from \eqref{:93c} and \eqref{:93d}
\begin{align}\label{:93e}
dZ_t^{N,i} = & dB_t^i+ \{-\frac{Z_t^{N,i}}{ 4 N}+
\frac{\alpha + \half }{Z_t^{N,i}} +
\sum_{j=1, j\neq i}^{\nN }
\frac{2 Z_t^{N,i}} {(Z_t^{N,i})^2-(Z_t^{N,j})^2}\} dt \ (1\le i\le N)
,\\
\label{:93f}
dZ_t^i= & dB_t^i+
\{ \frac{\alpha + \half }{ Z_t^i} +
\sum_{ j\neq i}^\infty \frac{ 2 Z_t^{N,i}}{(Z_t^i)^2-(Z_t^j)^2} \} dt \quad (i\in\N)
.\end{align}
We remark that \tref{l:22} can be applied to the non-colliding square Bessel processes because the equilibrium states are the same as the Bessel interacting Brownian motion and
coefficients are well-behaved as
$ \mathsf{a}^{\nN }(x.\mathsf{y}) = \mathsf{a}(x.\mathsf{y})= 4x $.
\end{rem}
\subsection{The Ginibre interacting Brownian motion} \label{s:54}
Let $\S =\R ^2$.
Let $ \mu _{\mathrm{gin}}^{\nN } $ and $ \mu _{\mathrm{gin}} $
be as in \sref{s:1}.
Let $ \Phi ^{\nN } = |x|^2$ and $\Psi (x)=-\log |x|$.
Then the $ \nN $-particle systems are given by
\begin{align}\tag{\ref{:10r}}
dX_t^{N,i} = & dB_t^i-X_t^{N,i}dt +
\sum_{j=1, j\neq i}^{\nN }
\frac{X_t^{N,i}-X_t^{N,j}}{|X_t^{N,i}-X_t^{N,j}|^2}dt
\quad (1\le i\le N)
.\end{align}
The limit ISDEs are
\begin{align}\tag{\ref{:10s}}
dX_t^i= &
dB_t^i+\limi{r} \sum_{|X_t^i-X_t^j|<r, j\neq i}
\frac{X_t^i-X_t^j}{|X_t^i-X_t^j|^2}dt \quad (i\in\N)
\\\intertext{and } \tag{\ref{:10t}}
dX_t^i=
&
dB_t^i - X_t^idt +
\limi{r} \sum_{|X_t^j|<r, j\neq i} \frac{X_t^i-X_t^j}{|X_t^i-X_t^j|^2}dt \quad (i\in\N)
.\end{align}
\begin{prop}\label{l:33}
\eqref{:21a} holds for \eqref{:10r} and both \eqref{:10s} and \eqref{:10t}.
\end{prop}
\begin{proof}
\As{J2}--\As{J5} except \eqref{:21T} are proved in \cite{o.rm,o.isde}.
\eqref{:21T} is obvious for a Ginibre point process because
their one-correlation functions with respect to the Lebesgue measure have
a uniform bound such that $ \rho _{\mathrm{gin}}^{\nN , 1} \le 1/\pi $.
This estimate follows from \thetag{6.4} in \cite{o.isde} immediately.
Let $ \mathsf{d}_1$ and $ \mathsf{d}_2$ be the logarithmic derivative associated with ISDEs \eqref{:10s} and \eqref{:10t}.
Then $ \mathsf{d}_1 = \mathsf{d}_2 $ a.s.\ \cite{o.isde}. Hence we conclude \pref{l:33}
\end{proof}
\subsection{Gibbs measures with Ruelle-class potentials} \label{s:55}
Let $ \mu ^{\Psi }$ be Gibbs measures
with Ruelle-class potential $ \Psi (x,y) = \Psi (x-y)$
that are smooth outside the origin.
Let $ \Phi ^N \in C^{\infty}(\S )$ be a confining potential for the $ N $- particle system.
We assume that the correlation functions of $ \mu ^{\Phi ^N ,\Psi }$
satisfy bounds
$ \sup_N \rho ^{\nN , m } \le \cref{;35a} ^m $ for some constants
$ \Ct{;35a}$; see the construction of \cite{ruelle.2}.
Then one can see in the same fashion as \cite{o-t.tail} that
$ \mu ^{\Psi }$ satisfy \As{J2}--\As{J5} except \eqref{:21T}.
Under the condition $ \sup_N \rho ^{\nN , m } \le \cref{;35a} ^m $,
\eqref{:21T} is obvious.
Moreover, if $ \mu ^{\Psi }$ is a grand canonical Gibbs measure with
sufficiently small inverse temperature $ \beta $, then $ \mu ^{\Psi }$
is tail trivial. Hence we can obtain \As{J6} in the same way as \cite{o-t.tail} in this case.
We present two concrete examples below.
\subsubsection{Lennard--Jones 6-12 potentials} \label{s:55a}
Let $\S =\R ^3$ and $\beta >0$.
Let $\Psi _{6-12}(x)=|x|^{-12}-|x|^{-6}$ be the Lennard-Jones potential.
The corresponding ISDEs are given by the following.
\begin{align}\notag
dX_t^{N,i}=
&
dB_t^i+\frac{\beta }{2} \{
\nabla \Phi ^N (X_t^{N,i}) +
\sum_{j=1, \atop j\neq i}^{\nN }\frac{12(X_t^{N,i}-X_t^{N,j})}{|X_t^{N,i}-X_t^{N,j}|^{14}}-
\frac{6(X_t^{N,i}-X_t^{N,j})}{|X_t^{N,i}-X_t^{N,j}|^{8}} \} dt
\ (1\le i\le N)
,\\ \notag
dX_t^i=
&
dB_t^i+\frac{\beta }{2}\sum_{j=1,j\neq i}^\infty
\{ \frac{12(X_t^i-X_t^j)}{|X_t^i-X_t^j|^{14}}- \frac{6(X_t^i-X_t^j)}{|X_t^i-X_t^j|^{8}} \}dt\quad (i\in\N)
.\end{align}
\subsubsection{Riesz potentials} \label{s:56}
Let $d<a\in\N$ and $\beta >0$. Let $\Psi _a(x)=\frac{\beta }{a }|x|^{-a}$ the Riesz potential.
The corresponding SDEs are given by
\begin{align}\notag
dX_t^{N,i}= &
dB_t^i+\frac{\beta }{2}
\{
\nabla \Phi ^N (X_t^{N,i}) +
\sum_{j=1,j\neq i}^{\nN } \frac{X_t^{N,i}-X_t^{N,j}}{|X_t^{N,i}-X_t^{N,j}|^{2+a}}
\}
dt \quad (1\le i\le N)
,\\\notag
dX_t^i= &
dB_t^i+\frac{\beta }{2}\sum_{j=1,j\neq i}^\infty \frac{X_t^i-X_t^j}{|X_t^i-X_t^j|^{2+a}}dt \quad (i\in\N)
.\end{align}
\section{Acknowledgments}
{}\quad
\noindent
Y.K. is supported by Grant-in-Aid for JSPS JSPS Research Fellowships (No.\!\! 15J03091).
\noindent
H.O. is supported in part by a Grant-in-Aid for Scenic Research (KIBAN-A, No.\!\! 24244010; KIBAN-A, No.\!\! 16H02149; KIBAN-S, No.\!\! 16H06338) from the Japan Society for the Promotion of Science. | 158,030 |
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An Usbekistan national, being held on terrorism-related charges since his Boise arrest May 16, has a new attorney.
30-year-old Fazliddin Kurbanov was arrested May 16 when FBI agents raided his Boise apartment, saying the suspect conspired to use a weapon of mass destruction and made regular shopping trips to purchase bomb-making materials.
But two Boise public defenders, Richard Rubin and Thomas Monaghan. who had initially been assigned to represent Kurbanov, told a federal court judge that due to the federal sequestration and its financial squeeze on their resources, they they should be removed from the case. Rubin told the Associated Press that federal budget cuts had reduced the budget of his office by 10 percent.
The AP's John Miller reports that a federal judge has now appointed Charles Peterson to take over Kurbanov's defense. The AP reports that Peterson was part of the defense team that helped win Randy Weaver's acquittal on charges that he murdered a federal marshall during the Ruby Ridge standoff in 1992. Peterson also helped to defend a Saudi Arabian University of Idaho student was was acquitted of using his computer skills to support terrorism following the September 11, 2001, attacks.
Kurbanov has pleaded not guilty to charges including that he helped teach people to build bombs which targeted public transportation.
Join the conversation at facebook.com/boiseweekly
or send letters to [email protected]. | 33,255 |
TITLE: Verification of simple probability excercise
QUESTION [2 upvotes]: Task: We have a 10 floor building, 7 people stand at the botton, waiting for the elevator. What is the chance that all people will exit on different floors?
Proposed Solution: I assume the cardinality of favorable outcomes is $$ \binom{10}{7}, $$ for it is given by the amount of ways you can choose 7 floors out of 10, without repetition.
Then the cardinality of total outcomes is $$ \binom{10+7-1}{7}=\binom{16}{7}, $$ as it is given by the amount of ways you can choose 7 floors out of 10 with possible repetition.
In this case, the probability is given by $$ P=\frac{\binom{10}{7}}{\binom{16}{7}}\approx0,0104895... $$
The problem: I believe my solution is correct, however the end result is ugly, which would not be a problem, but pretty much all excercises I have found in the batch this came from had "pretty" solutions, therefore, I am worried that this might be incorrect.
I would be terribly appreciative if someone could verify this solution.
REPLY [4 votes]: I find it easier to treat each person as a unique (distinct) thing. Thus the total number of ways for 7 people to exit a total of 10 floors is 10^7 (each person can pick one of the 10 floors). Next, the number of ways that these 7 people can exit on different floors is 10!/3! (the first person picks one of 10 floors, the next picks from the remaining 9 and so on). Thus the probability (assuming all outcomes are equally likely) is (10!/3!)/10^7 which comes to 0.06048 or 6.048%. | 216,382 |
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- Targeted etanercept for treatment-refractory pain due to bone metastasis: two case reports. Edward L. Tobinick MD.Clinical Therapeutics. 2003 Aug;25(8):2279-88.
- Perispinal TNF-alpha inhibition for discogenic pain. Edward L. Tobinick MD, Susan Britschgi-Davoodifar. Swiss Med Wkly. 2003 Mar; 133(11-12):170–7.
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- Tumor Necrosis Factor-Alpha Antagonist Reduces Apoptosis of Neurons and Oligodendroglia in Rat Spinal Cord Injury.Chen KB, Uchida K, Nakajima H, Yayama T, Hirai T, Watanabe S, Guerrero AR, Kobayashi S, Ma WY, Liu SY, Baba H. Spine (Phila Pa 1976). 2011 Jan 8. [Epub ahead of print] PMID: 21224756.
- Etanercept restores the antinociceptive effect of morphine and suppresses spinal neuroinflammation in morphine-tolerant rats. Shen CH, Tsai RY, Shih MS, Lin SL, Tai YH, Chien CC, Wong CS. Anesth Analg. 2011 Feb;112(2):454-9.PMID:21081778.
- Direct application of the tumor necrosis factor-alpha inhibitor, etanercept, into a punctured intervertebral disc decreases calcitonin gene-related peptide expression in rat dorsal root ganglion neurons. Horii M, Orita S, Nagata M, Takaso M, Yamauchi K, Yamashita M, Inoue G, Eguchi Y, Ochiai N, Kishida S, Aoki Y, Ishikawa T, Arai G, Miyagi M, Kamoda H, Kuniyoshi K, Suzuki M, Nakamura J, Toyone T, Takahashi K, Ohtori S. Spine (Phila Pa 1976). 2011 Jan 15;36(2):E80-5. PMID: 21057386.
- 2018 IN. | 242,595 |
TITLE: Help on second order linear equations
QUESTION [0 upvotes]: Find the solution of the given initial value problem:
$y'' + 8y' -9y = 0, y(2)=1, y'(2)=0$
What I did was solve the characteristic equation and then I got the solution to the homogeneous equation: $y(x)= {C1e}^x + {C2e}^{-9x}$, but I'm not sure how to solve it with the initial conditions.
Solve the initial value problem $y''-y'-2y=0, y(0)= \alpha, y'(0)=2$. Then find $\alpha$ so that the solution approaches zero as t becomes infinity.
This problem is more challenging for me. What I tried to do here was solve the characteristic equation which gave me $y(t)=C1{e}^{2t} + C2{e}^{-t}$. Then I found
$y'(t)=2{C1e}^{2x} {-C2e}^{-x}$
so then I used the initial conditions and I got:
$C1 + C2 = \alpha$
$2C1 -C2 = 2$
and by inspection of the general solution it seems that only way this will approach $0$ is if $C1 = 0$ since there is no way that $e^{2t}$ will ever be zero, but $e^-t$ will. So does $\alpha = -2$?
Any tips/solution would be appreciated :)
REPLY [4 votes]: For the first one, you found a good solution:
$$y(x)= c_1e^x + c_2e^{-9x}$$
Now, you have two unknowns, so you need two equations in order to solve for those constants. You are given $y(2)=1, y'(2)=0$, so we use this to set up and solve those two equations and determine $c_1$ and $c_2$. Using the first initial condition, , $y(2) = 1$, we have (just substitute):
$$y(2) = c_1 e^2 + c_2 e^{-18} = 1$$
Using the second initial condition, $y'(2) = 0$, we first need to find the derivative and then substitute, so we have
$$y(x)= c_1e^x + c_2e^{-9x} \implies y'(x) = c_1e^x - 9 c_2e^{-9x}$$
We can now substitute $y'(2) = 0$ and have:
$$ y'(2) = c_1e^2 - 9 c_2e^{-18} = 0$$
Now, we have the $2\times 2$ system:
$$c_1 e^2 + c_2 e^{-18} = 1 \\ c_1e^2 - 9 c_2e^{-18} = 0$$
Use your favorite method (Elimination, Cramer's Rule, Gaussian Elimination...) to solve that $2x2$, yielding:
$$c_1 = \dfrac{9}{10e^2}, c_2 = \dfrac{e^{18}}{10}$$
The final result is (you can clean it up a bit, but left this way on purpose):
$$y(x)= \dfrac{9}{10e^2}~e^x + \dfrac{e^{18}}{10}~e^{-9x}$$
We repeat the same process for the second equation:
$$y''-y'-2y=0, y(0)= \alpha, y'(0)=2$$
Again you found the correct solution as:
$$y(t)=c_1e^{2t} + c_2e^{-t}$$
So, we repeat the above steps to get (note: $y'(t) = 2c_1e^{2t} -c_2e^{-t}$):
$$y(0) = c_1 + c_2 = \alpha \\ y'(0) = 2c_1-c_2 = 2$$
Solving this $2x2$, yields:
$$c_1 = \dfrac{\alpha + 2}{3}, c_2 = \dfrac{2(\alpha -1)}{3}$$
Our final result is:
$$y(t) = \dfrac{\alpha + 2}{3}~e^{2t} + \dfrac{2(\alpha -1)}{3}~e^{-t}$$
We now want to find $\alpha$ such that the solution approaches zero as $t \rightarrow \infty$, so you are correct, we need to get rid of the $e^{2t}$ term. We can let $\alpha = -2$, so that leaves:
$$y(t) = -2~e^{-t}$$ | 69,587 |
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TITLE: An urn contains 2 red balls and 3 blue balls.
QUESTION [0 upvotes]: An urn contains 2 red balls and 3 blue balls. Six drawing are made, the ball being placed back into the urn after each drawing. What is the probability, expressed as a rational number,that exactly 2 red balls are draw?
My solution
$\frac{10\times 5^4}{5^{6}}$ (10 from 5C2)
$\frac{2}{5}$
REPLY [1 votes]: This is an example of a problem that can be solved via knowledge of the binomial distribution.
The probability of drawing a red ball is $\frac{2}{5}$. The probability of drawing a blue ball is $\frac{3}{5}$. There are $6$ drawings to be made (with replacement) and we are interested in the probability that exactly two of them will be red.
In other words, $n=6$, $k=2$ and $p=\frac{2}{5}$.
The probability of $k$ "successes" out of $n$ trials where probability of success if $p$ is:
$$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$
Plugging in the appropriate values of $n,k,p$ give you the probability.
$\binom{6}{2}\left(\frac{2}{5}\right)^2\left(\frac{3}{5}\right)^4$
"Took red fixed for first two slots then there are 10 such arrangements with 2 red balls"
"My solution: $\frac{10\times 5^4}{5^6}$"
I do not understand why you thought this would work. If you fix the first two slots as red and then have no restriction on the remaining slots, what is to prevent you from having more than two reds? I further do not see how the expression you wrote has anything to do with your proposed solution. | 54,663 |
The initiation of the Collaborative Influenza Vaccine Innovation Centers (CIVICs) program was announced by the National Institute of Allergy and Infectious Diseases (NIAID), part of the National Institutes of Health (NIH), on September 30, 2019 with press releases through both NIH and NIAID. The CIVICs Network of research centers will work together in a coordinated, multidisciplinary effort to develop more durable, broadly protective and longer-lasting influenza vaccines. Seasonal influenza causes hundreds of thousands of hospitalizations and tens of thousands of deaths in the United States each year, according to the Centers for Disease Control and Prevention. While current seasonal influenza vaccines are widely available and provide important public health benefits, they could be improved. Notably, they do not always protect against all strains of circulating influenza viruses.
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This article excerpts text from these sources: | 377,214 |
45949 Wellington Ave., Chilliwack BC V2P 2C6 Tel: 604-
Dear Friends of Art,
How you are?
Hopefully, the nice days that we are having here in Chilliwack recently indicate that winter is at last passing. The sun is shining, the temperature is rising, the birds are singing, and all in all you can smell spring in the air.
And to celebrate the brightness and lightness of the season ahead, I'm hosting a wonderful exhibition of glass and pottery works at my gallery downtown, Asai's Art Gallery.
So why not venture downtown Chilliwack to take in the show and to show your support of our fine local artists?
The show, Metamorphosis of Glass and Clay, began on Monday, February 18, and continues through Wednesday, April 5. It includes the work of five local female artists, each of whom is an exemplar of the richness of artistic talent to be found in and around the Chilliwack area.
I take great pride in helping to promote the work of local artists and dearly want everyone to become familiar with the artists and their artistry.
I look forward to seeing you soon,
Asai Wu-
Asai's Art Gallery
Tuesday to Saturday, 12:00 p.m. to 5:00 p.m.
Dear Friends of Art,
For the information regarding “A Tribute to Dr. N. Bethune -
Dear Friends of Art,
We had another successful year for the Hands Across the Pacific International art show in Chilliwack.
This year marks the 8th exhibition.
It is wonderful to see, as the year goes by, the steady increase of participants and the quality of art work.
To review the art work, please, click here!!!
Dear Friends of Art,
We have stated a new exhibition, “It’s All About the Light -
The show is held from December 10, 2008 through January 10, 2008.
For the images from the exhibition, please, click on the image below!! | 259,526 |
Product listing: Keencut 18" Laser Trued Safety Straight Edge - Foster (60008) to Crystal Clear 8.5 x 14" Heat Resistant 7mil Legal Size Covers - 100PK (TC7HT8.5X14S)
Keencut 18" Laser Trued Safety Straight Edge - Foster
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Renz Premium 3/4" Transparent Coat 2:1 Twin Loop Ring Wire
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Wilson Jones 2" Black Half Size Basic Opaque Round Ring Binders - 6pk - Specialty Binders
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2-3/8" x 7-1/4" Crystal Clear Adhesive Vinyl Pockets 100pk
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20mm Neon Red 4:1 Pitch Spiral Binding Coil - 100pk
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1.5" Premium White Round Ring Clear View Binders - 12pk
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GBC ECO Friendly Black Recycled Poly Covers 100pk - Specialty Covers
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GBC 5mil HeatSeal Ultra Clear Menu Size Laminating Pouches - 100pk
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3" Aluminum Screw Posts - 100pk
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33mm Clear 4:1 Pitch Spiral Binding Coil - 100pc
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Wilson Jones 2" White Opaque Round Binders with Labels - 12pk - Non View Binders
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- Ring Style: Round-Ring
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- Color: White
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Crystal Clear 8.5 x 14" Heat Resistant 7mil Legal Size Covers - 100PK
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Your Price: $60.69 | 296,260 |
TITLE: isomorphism of extensions by abelian varieties
QUESTION [4 upvotes]: Let $A$ and $G$ be abelian varieties over $\mathbb{C}$. An element $P$ of $\text{Ext}(A, G)$ is an exact sequence
$0 \to G \to P \to A \to 0$,
here one can give $P$ the structure of an abelian variety. And $P$ can be viewed as principal $G$-bundle over $A$. In general, is there a way to determine if two given elements $P$ and $P'$ in $\text{Ext}(A, G)$ are isomorphic?
In particular, assume that $A$ is an elliptic curve. Given two extensions $0 \to G \to P \to A \to 0$ and $0 \to G \to P' \to A \to 0$ with morphisms $g : G \to G$, $f : P \to P'$, and $h : A \to A$ so that the resulting diagram commutes, $g$ is an isomorphism, and $h$ is an isogeny with kernel $(\mathbb{Z}/n\mathbb{Z})^2$. Can we show that $P$ and $P'$ are isomorphic or is there an example otherwise?
REPLY [1 votes]: The answer to your second question is no. Take $E_1,E_2$ non-isogenous elliptic curves without complex endomorphisms. Take $P_1,P_2$ 5-torsion points in $E_1,E_2$ resp. Now let $P=E_1\oplus E_2/(P_1-P_2)$ and $P'=E_1\oplus E_2/(P_1-2P_2)$. Now, there is clearly a map of the type you described with $n=2$ descending from the endomorphism of $E_1\oplus E_2$ which is multiplicatino by 1 on $E_1$ and $2$ on $E_2$.
However, $P$ and $P'$ are not isomorphic: as there are only $2$ injections from each of $E_1,E_2$ to each of $P,P'$, they can each only be written as 5-quotients of $E_1\oplus E_2$ in essentially one way.
In general, in your setup the class of $P'$ is $n$ times the class of $P$ in $\textrm{Ext}(A,G)$. This is because $P$ is the fiber product of $P$ with $A$ over $A$ under the multiplication by $n$ map. Thus, any class of $\textrm{Ext}(A,G)$ which is not $(n-1)$-torsion OR $(n+1)$ - torsion, and there are lots of them, will yield a distinct $P$ from $P'$. You have to include $(n+1)$ since the $-1$ automorphism flips the Ext class.
Of course, if $A,G$ are isogenous and have automorphisms you have to be a little more careful, as you might have distinct elements of $\textrm{Ext}(A,G)$ which are abstractly isomorphic. | 210,873 |
Dog Pens
Why invest in a dog pen?
Containing your dog when you can’t supervise them, a dog pen can be an extremely useful acquisition. Robust and durable, an indoor dog pen or outdoor dog pen will serve you well for as long as you need it and could make all the difference. At Time for Paws, we are delighted to bring you the outstanding Henry Wag 6 Sided Pet Pen With Base which folds flat for easy storage when not in use.
Indoor or outdoor, you decide!
If there are times when you need to contain your pooch, a dog pen could be for you. Your dog will learn to remain calm in their pen and it will enable you to introduce new pets safely before allowing your animals to be left alone together. Young dogs can be confined to a pen at night to prevent them from exploring or destroying your home. Better still, their toilet accidents will be limited to a small area. You can also grab a little peace while you are eating or entertaining guests and keep your pet away from small children.
Choosing the right dog pen
Always ensure that your pen is of a suitable size for your pooch. Your dog should be able to stand up, lie down and turn around inside their pen with ease. Metal pens can sometimes be joined together to form a larger space, if you need to contain more than one dog and you should check your chosen pen for any issues such as sharp edges before using it.
Introducing your dog to its new pen
It’s best to introduce your puppy to its pen from an early age as they will be more accepting of their confinement. Older dogs may require a bribe or several to lure them in and settle them down. Never force your dog into the pen as this will inspire fear and don’t give up if your pet is initially resistant to being confined. If your household is busy and you have several pets, your dog may eventually view their pen as a haven where they can rest undisturbed.
A pen is not a prison
You should limit use of your pen to periods of no more than four hours during the day. It is not a prison and dogs will eventually become bored, leading to behavioural issues. Praise your dog for remaining calm and only confine them to their pen when absolutely necessary.
Want to read more, then Purina has an excellent article about dog/puppy pens | 169,177 |
Atlanta Criminal Defense Attorney Discusses Gun Charges
Recently, the State of Georgia has been featured in national news stories because of new gun regulations, or reductions thereto, that was signed into law this spring and went into effect in July, known as the Safe Carry Protection Act. This new law means that people are able to carry their guns in many locations that previously were restricted, including bars, religious buildings, and school zones in certain circumstances. However, the bar owners or religious leaders do have the right to ban guns from their premises. Individuals even may carry guns into some government buildings that do not have security measures that prohibit this, specifically metal detectors or security personnel. However, despite the ability to carry a firearm in many more situations, there still are regulations that can get a person into serious trouble if he or she violates them.
It is important to remember that even though people now may carry a gun with far fewer restrictions, there are actions that may lead to weapons charges in Georgia, including:
- Being in possession of a firearm when a person has been convicted of a felony;
- Being in possession of a handgun as a minor – there are certain exceptions to this provision, including a minor who is using a gun owned by his parents on land belonging to his parents, with permission;
- Possessing a weapon that is restricted or prohibited – restricted weapons include sawed off rifles and shotguns, silencers, racket launchers, and mortars. The recent gun law does permit hunters to use silencers as long as the property owner of the land on which they are hunting are aware of the use of the silencers or other similar devices;
- Possessing a gun in a school zone without specific authorization to do so – the school board may designate officials to carry firearms, but handguns generally are still prohibited in these locations. The penalty for carrying a handgun without authorization on a college campus has been reduced in accordance with the provisions of the new law to a fine;
- Discharging a firearm while under the influence of alcohol or drugs;
- Pointing a gun at someone or firing a gun in the direction of another person – it is illegal to use a gun to scare someone, even if the person holding the firearm has no intention of firing the gun;
- Using a firearm in the commission of a crime.
The new gun law does remove some restrictions that prevented individuals convicted of specific misdemeanors from obtaining a carry permit. Georgia does require people to have a carry permit, although they do not have to possess a permit to purchase a firearm. Under the provisions of the new law, it is not acceptable for a police officer to stop a person strictly for the purpose of asking to see their carry permit. This new provision is likened to the prohibition of a police officer from pulling over a driver merely to request to see his driver’s license.
Penalties for a weapons charge vary widely depending in the nature of the violation, but a felony conviction can lead to a mandatory two-year sentence and a fine up to $10,000.00. A person may face incarceration of up to 10 years, so it is important to get the right legal assistance when being charged with a gun law violation.
The Abt Law Firm, LLC. Aggressively Fights for its Clients
Despite the fact that there is a new gun law in place that eases many restrictions on carrying a handgun in Georgia, a person who is charged with being in violation of a gun law faces the possibility of serious prison time and large fines. The Atlanta Criminal Defense Firm of the Abt Law Firm, LLC. knows how to fight against these charges in order to minimize the risk to our clients.! | 373,104 |
ISA is a company that operates throughout Latin America in the electricity, transportration, and telecommunications fields. Its mission statement, “Connections that Inspire”, on the underpins its belief that sustainability underpins good business practice.
The company currently leads a corporate sustainability program called Conexión Jaguar, in alliance with South Pole and Panthera, which helps conserve biodiversity, including the rare jaguar, mitigate climate change, and develop rural communities. The program protects and restores forests in the countries where ISA operates.
Aligned with the goals of Initiative 20x20, Conexión Jaguar has set the initial goal of supporting at least 20 projects that conserve or restore 400,000 hectares of land in Latin America and store 9 million tons of CO2 emissions. To achieve their goal, ISA is looking to cooperate with other Initiative 20x20 partners.
Protecting the world's largest wetland for people and the climate in Brazil
Safeguarding biodiversity and farming sustainably in Peru's Amazon
Building sustainable cocoa agroforestry systems in Colombia
Stephanie Cardona Muñoz, [email protected] | 255,871 |
MECHANIC JOBS
OVERVIEW:
Mechanics assemble and build engines and mechanical components, inspect machinery and run diagnostic tests to pinpoint problems. Mechanics are responsible for maintaining equipment and systems by cleaning them, replenishing fluids and replacing worn out parts. They perform safety tests and ensure that engines meet environmental standards.
Sound knowledge of hydraulic and electrical systems is often a requirement for mechanic jobs. When completing tasks, mechanics use a range of hand tools such as screwdrivers, hammers and pliers. They also use computerised equipment to diagnose issues, and are required to keep up with advances in technology.
Types of mechanics:
- Auto mechanics
- Heavy vehicle mechanics
- Small engine mechanics
- Diesel mechanics
- Marine mechanics
- Aircraft mechanics
CURRENT VACANCIES
PROSPECTS FOR MECHANIC JOBS:
Mechanics are employed across multiple industries and their skills are in demand. Job opportunities vary between industries, but employment opportunities are generally steady.
Employment prospects for auto mechanics rose strongly over the past five years and growth is expected to be above average up to 2019. As this is large occupational group, jobs should be available in most areas throughout Australia over this period. Opportunities for air-conditioning and refrigerator mechanics increased significantly over the previous ten years and are forecasted to remain steady up to 2020.
Click here for current job vacancies with Australia’s leading recruiter for mechanic jobs.
EMPLOYMENT PROJECTIONS FOR MECHANIC JOBS:
Projections from the Australian Department of Employment, 2015 to 2020. Note data is '000.
ASSOCIATED OCCUPATIONS:
Motor Mechanic, Auto Mechanic, Heavy Vehicle Mechanic, HD Mechanic, Small Engine Mechanic, Diesel Mechanic, Marine Mechanic, Aircraft Mechanic, Aircraft Maintenance Engineer, Motorbike Mechanic, Brake Mechanic, Light Engine Mechanic, Vehicle Mechanic, Automatic Transmission Mechanic, Automotive Technician, Exhaust Fitter, Air Conditioning Fitter, Diesel Fitter, Heavy Duty Fitter.
DUTIES:
Assemble mechanical equipment
Inspect engines and run a range of diagnostic tests
Conduct repairs on engines and electrical components
Maintain machinery and equipment by cleaning and replacing fluids and parts
Rotate tyres and balance wheels
Interpret and follow work orders
Follow checklists during inspections to ensure all necessary parts are examined
Consult with customers and clients about problems and maintenance
Repair hydraulic systems
Repair cooling systems and hosing
Install, maintain and repair air conditioning systems
Keep detailed logs of completed work
Report on functionality and safety issues
Conduct test drives to confirm repairs and maintenance have been completed
Comply with safety and environmental standards
Monitor supplies and place orders for stock when necessary
Personal attributes required in mechanic jobs:
- Strong Verbal Communication Skills.
- Strong Written Communication Skills.
- Problem Solving Skills.
- Fast Learner.
- Dexterity.
- Physical Fitness.
- Safety Awareness.
PAY FOR MECHANIC JOBS:
Auto mechanics earn between $17 to $29.78 per hour depending on skills and experience. A full-time auto mechanic in Australia earns a pre-tax average of $1000 per week. Full-time weekly earnings for air conditioning and refrigeration mechanics are higher at $1618 per week before tax.
Aircraft mechanics earn a weekly average of $1759 pre tax, while diesel mechanics earn a median rate of $27 per hour. A skilled diesel mechanic can make up up to $93,165 per year.
QUALIFICATIONS FOR MECHANICS:
In Australia, the most common pathway to becoming a mechanic is through an apprenticeship. Apprentices complete four years of on and off-the-job training. This involves hands-on experience in a workshop combined with formal classroom instruction.
Pre-apprenticeship studies are available for high school students who wish to increase their chances of gaining an apprenticeship. Students learn about the care and maintenance of vehicles, and the basics of engine construction. Pre-apprenticeship training may be credited towards later qualifications, shortening the length of an apprenticeship.
After completing their apprenticeship, mechanics often undertake specialised certificate courses in areas such car repairs, diesel engines or aircraft maintenance.
Suitable courses for those seeking to become mechanics or update qualifications include:
Certificate IV in Automotive Management
Certificate IV in Automotive Electrical Technology
Certificate IV in Automotive Mechanical Diagnosis
Certificate III in Motorcycle Mechanical Technology
Certificate III in Light Vehicle Mechanical Technology
Certificate III in Outdoor Power Equipment
Many mechanic jobs specify the following:
Trade qualifications
Proven work history
Own tools
Heavy vehicle licence
Forklift ticket
Driver’s licence and own vehicle
Pre-employment screening for health conditions, drugs and alcohol | 285,435 |
Thanks to WSB for the tip.
Participate in a community forum about the potential location of a new Seattle Municipal Jail in West Seattle:
Saturday, July 26, from 9 a.m. to noon, in the Brockey Conference Center at South Seattle Community College, located at 6000 16th Ave. S.W. – focus: West Marginal Way and Myers Way sites;
Here's the full announcement from the City's website:
NEWS ADVISORY
SUBJECT: Seattle seeks feedback on possible municipal jail sites; Four dates set for public forums
FOR IMMEDIATE RELEASE: 6/4/2008 10:00:00 AM
FOR MORE INFORMATION CONTACT: Katherine Schubert-Knapp (206) 684-0909 Kathy Sugiyama (206) 684-0909
Seattle seeks feedback on possible municipal jail sitesFour dates set for public forums
S
EATTLE - The city of Seattle announced today it will hold public forums around the city to provide information and hear feedback about possible sites for a new Seattle Municipal Jail. While residents are welcome to attend any of the forums, each will focus primarily on a specific potential jail site, as outlined below. The forum dates and locations are:
Thursday, June 26, from 6 to 9 p.m. at the Aerospace Machinists Union Hall A, located at 9125 15th Place S. – focus: West Marginal Way and Myers Way sites;
Saturday, July 12, from 9 a.m. to noon, in the Wellness Center at North Seattle Community College, located at 9600 College Way N. – focus: Aurora site;
Saturday, July 26, from 9 a.m. to noon, in the Brockey Conference Center at South Seattle Community College, located at 6000 16th Ave. S.W. – focus: West Marginal Way and Myers Way sites;
Wednesday, July 30, from 6 to 9 p.m. at the Seattle Center Exhibition Hall, located at 225 Mercer Street – focus: Interbay site..
For more information, visit the city’s municipal jail Web site:
### | 264,911 |
A good example from Saint Louis Martin:
In the last post, we looked at the tremendous affection and intimacy that St. Louis Martin exhibited toward his children. Louis was a man of great tenderness, and at the same time, he was also a man with backbone. In this post, we look at the holy firmness of Saint Louis.
St. Louis Martin was very much active and engaged in the lives of his children. He would even scold them when love required such a method. Celine, the seventh child in the Martin family, gives examples of how St. Louis demonstrated a rightly ordered firmness in his fathering of Therese (the youngest child):
“My father had a very special consideration for his youngest child, and was as attentive to her as a mother. But if it is true that little Therese was, as she says, always ‘surrounded with love,’ it is also true that she was never spoiled. The proof that my father did not spoil her, and that she did not do just what she liked at home, is shown by a fact which made a deep impression on her, and which she relates in her Autobiography: how she was severely reprimanded for not wishing to leave her games at the first call of her father.”
“A little later, at Liseux – she might have been six years old – she took great pleasure in carrying the newspaper to our father every morning. One day I wanted to take it to him; but Therese, quicker than I, had already caught it up and ran to him. This disappointed me and I showed it. Papa reproached little Therese for not having yielded to me, and he scolded her very severely, so much so that I was extremely upset myself.”
Another story from Celine relates a similar firmness toward Marie-Louise (the oldest child):
“One year at the end of the summer holidays Marie was taking a walk with our father in a small family property, which formerly belonged to a person called Roulee. Marie began to gather some flowers saying: ‘I will take these back with me to the Visitation School, as a souvenir of the Roullees.’ Our father, wishing to teach her a lesson, replied: ‘That’s it! And then you can look down on your little friends by showing them the flowers from your estate.’ Poor Marie, seeing that he had guessed her thoughts, threw away her bouquet to show that she was above vainglory.”
It was not only in correction of his children that Louis was firm, but also in the way he trained them to grow in self-discipline. Celine remarks that “without being severe, he raised his children in fidelity to all their duties.” Celine further describes how St. Louis provided a disciplined schedule of prayer for the family:
“At home, our education had piety as its chief lever. There was a complete liturgy of household life: evening prayers all together, Month of May Devotions, Sunday Offices, spiritual reading before Feasts, etc. Our father aided as much as he could the development of the spiritual life.”
How will I apply this? (The Challenge):
Step 1: I will answer the following questions honestly – How well am I providing a well-ordered and disciplined schedule of prayer for my family? What is one decision I can make to incorporate a greater firmness in this regard for the sake of my family?
Step 2: I will write this one resolution on a piece of paper and commit to it by signing my name.
Step 3: I will tape the commitment to my bathroom mirror to help me remember to take action.
Some possible results:
- I may increase the value I place on scheduled prayer – transmitting this highly important value more effectively to my children.
- I may become more attentive to the actions and intentions of my children – increasing my ability to both affirm and correct as needed.
- I may gain a deeper understanding of the vital need to be actively engaged in the lives of my children, especially as a spiritual leader – helping me to follow through in this area more consistently.
Saint Louis Martin, pray for us! | 223,072 |
A conversation with Dr Rob Moodie
October. | 93,872 |
TITLE: If a function is undefined at a point, is it also discontinuous at that point?
QUESTION [5 upvotes]: I posted a solution here with an illustration (see below) and commented that the function was discontinuous at $x=1$, where it is undefined. Someone told me, no, it is undefined but continuous.
Now I'm confused.
I would have thought that the point $(1, -1/2)$ in the graph below should be designated with a hollow point (a point that isn't filled in) to demonstrate that that point is not actually on the graph. Furthermore, I thought that this hole would constitute a discontinuity.
If I need to be set straight here, could someone please help me out?
REPLY [4 votes]: Excerpt from my answer to this question:
The term continuous function is defined with respect to its domain. Therefore it is crucial to specify the domain of a function, if we want to analyse the function with respect to continuity. Outside of the domain of a function this function is not continuous, since it's not even defined there.
Note that when we talk about discontinuities of a one variable function we classify them as either being a removable discontinuity, a jump discontinuity or an essential resp. infinite discontinuity. The key point here is, that each of these discontinuities is defined with respect to the domain of $f$. We conclude, discontinuities are defined solely within the domain of $f$.
Informally: The domain and codomain specify where the function lives and we can't say anything about the function outside of its region of existence. | 87,553 |
TITLE: Real $2\times 2$ matrix $X$ such that $X^2 + 2X= -5I$
QUESTION [1 upvotes]: Find a real $2\times 2$ matrix $X = \left(\begin{matrix} a& b\\ c & d\end{matrix}\right)$ such that $X^2 + 2X = -5I.$
With this question I'm kinda lost with the $2X$ part but a full explanation or a little bit of a lead would be greatly appreciated.
REPLY [3 votes]: A number of answers to this question have been given but I just want to indicate the simple solution and its connection with the general theory. The question is for a matrix with minimal polynomial $x^2+2x+5$. This is irreducible over the ground field ($\mathbb{R}$). Let $V$ be a two dimensional space and $T$ a linear transform with the given polynomial as minimal polynomial. Let $v \in V$ be any vector Then we have also $Tv$ and finally
$T^2v=-5-2Tv$ writing vector coordinates with respect to the basis $\{v,Tv\}$ we see that the matrix for $T$, acts as follows,
$$\begin{pmatrix}
a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0
\end{pmatrix}= \begin{pmatrix}0\\1
\end{pmatrix}$$
and
$$\begin{pmatrix}
a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1
\end{pmatrix}= \begin{pmatrix}-5\\-2
\end{pmatrix}$$
so we see immediately that
$$X=
\begin{pmatrix}
0&-5\\1&-2\end{pmatrix}$$
This is called the companion matrix and we see that any solution will be conjugate to this solution. The advantage here is that we can simply read the solution off of the polynomial.
In general for an irreducible minimal polynomial of the form
$x^n+a_{n-1}x^{n-1}+\cdots +a_0=0$ we have companion matrix
$$\begin{pmatrix}
0&0&\cdots&0&-a_0\\
1&0&\cdots&0 &-a_1\\
0&1&\cdots &0&-a_1\\
\vdots&\vdots&\ddots &\vdots&\vdots\\
0&0&\cdots &1&-a_{n-1}\\
\end{pmatrix}$$
which has minimal polynomial $x^n+a_{n-1}x^{n-1}+\cdots +a_0=0$. | 193,803 |
Music
[Song + Video] Peruzzi x El fuego – Majesty (Remix)
Unarguably the biggest wedding song of 2019 Majesty by Peruzzi gets an equally amazing refix by all standards, they say dont touch a classic, well a classic birthed another classic. Peep and share your thoughts guys
Dont forget to follow Elfuego on all social media platforms @ichie_fuego Produced and mixed by luminary beats | 201,767 |
Versace Bright Crystal 1.7 oz/ 50 mL Eau de Toilette Spray
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Fragrance Family: Floral Scent Type: Fruity Florals Key Notes: Red Fruits, Vanilla About: The playful scent of Angel Eau Sucree wi...ll captive her. Irresistibly sweet, the sparkling fragrance will mesmerize her with its mouthwatering and tangy facets. Notes of red berry sorbet and creamy caramelized meringue give way to a tantalizing oriental accord of warm notes. read more
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Boats for votes?
May 06, 2012, 04:05 AM. | 73,838 |
TITLE: Tips on resolving a Lagrange Multipliers equation system
QUESTION [4 upvotes]: I'm having a very hard time resolving the system of equations after using the Lagrange Multipliers optimization method. For instance:
The plane $ x + y + 2z = 2 $ intersects the paraboloid $ z = x^2 + y^2 $ over an ellipse. Find the ellipse points that are nearer and farther from the origin.
I know that the Lagrange equation is going to be:
$ L(x,y,z) = \sqrt{x^2 + y^2 + z^2} - L_1(x + y + 2z - 2)-L_2(z - x^2 - y^2)$
For simplicity it can be rewritten as:
$ L(x,y,z) = x^2 + y^2 + z^2 - L_1(x + y + 2z - 2)-L_2(z - x^2 - y^2)$
And that the equation system will be:
$ \frac{\partial L}{\partial x} = 2x - L_1 + 2 L_2 x = 0 $
$ \frac{\partial L}{\partial y} = 2y - L_1 + 2 L_2 y = 0 $
$ \frac{\partial L}{\partial z} = 2z - 2L_1 - L_2 = 0 $
$ \frac{\partial L}{\partial L_1} = x + y + 2z - 2 = 0 $
$ \frac{\partial L}{\partial L_2} = z - x^2 - y^2 = 0 $
Solving this system is really giving me a hard time, so instead of asking for the answer (which it could also help to see what are your procedures) I'd like to know if there are any known tips that could lead on a solution for the system.
REPLY [0 votes]: Hint
As commented by David, some of your partial derivatives are wrong and I suppose that you have fixed them.
In any manner, the first four derivatives make a linear system. So, just solve it the manner you want and, as a result, $x,y,z,L_1$ will simply express as function of $L_2$. Now, plug these expressions in the fifth derivative : this will lead to an equation in $b$ you can solve for $b$. Now, go backward and get the value of each parameter.
I am sure that you can take from here. | 42,188 |
There is no doubt that Frankie Peters’ mother hanged herself with a silken cord in her bedroom; or that she had dressed herself like Marie Antoinette, complete with powdered wig, put on Poulenc’s Requiem, and died at the age of 54.
Gossipy rumors, like prejudice, have their basis in fact. Although the facts of Estelle Peters’ death remain unclear, there was no doubt in the community that she was either a strong candidate for suicide or a likely victim of her husband’s humiliating abuse.
Gossip has always been a social lubricant, a lot like the weather but more compelling. We all have our suspicions about people and want to share and corroborate them.
Take The Creep at the gym. He is there every day at 7:15, dressed like Ghost Dog, groaning on the machines, grimacing, cosseted in rubber tubing and flogging himself on the treadmill in some twisted, interior Passion play. If he showed up only occasionally, or changed his outfit; if he smiled, nodded in recognition, or showed even a hit of sociality or civility, the gossip would never have started. But started it did.
The Creep was an ex-con, the most current and popular gossip went. His sweatpants hung down just over his ass-crack, his tats were of avenging angels and Mongol raiders, and his attitude was pure ghetto. He was so intimidating – fearful in fact – that no one had the gumption to talk to him; so horror vacui, gym patrons had to make something up. They created a heady mix of prejudice, DC street stats, and pure fantasy, and made him into one of the Four Horsemen of the Apocalypse and avoided him.
The rumor at my boarding school was that the headmaster had a dog’s jaw. He had been horribly mutilated by a Nazi land mine, survived, and underwent emergency battlefield surgery during which the mandibular joint of a captured German shepherd was transplanted into William Farnsworth. None of us had any doubt of the veracity of the account. He himself referred to his war wounds, the trauma of battle, and the sacrifice he and others of his generation had made for the Republic. He was the spitting image of Helmut, the mascot of the football team, a German Shepherd with bad hips and the long, pointed, and poorly calibrated jaw of the Head of School.
Even today, many decades removed from those adolescent years at Lefferts, most of us still remember Mr. Farnsworth’s dog’s jaw. Poor bastard. He was not a bad headmaster. He led the movement to make the school coed, joining forces with the girls’ school across the Farmington River. He loosened lights-out rules, encouraged ecumenism, and was the first headmaster to invite a rabbi to give the invocation at graduation; but he will always and only be remembered for his dog’s jaw.
My mother had a story for every house along Madison Street. Despite outward appearances – well-heeled, J.Press, Vineyard, and Yale – the Ballards at 483 were drunkards, atheists, and modern day slavers. They had to be, for someone of their Old Line community had blackballed us from the Meadows Country Club, and my mother had to tar them with some thick, gooey slime. They were nothing of the sort, of course. The empty vodka bottles in the trash on Mondays were remnants of lively, gay parties of the West End. Divided by 100 the fifths of Smirnoff amounted to nothing, let alone the Satanic orgy imagined by my spurned mother.
Yet, the more she told the story of the ‘Monday Morning Bottles” to her friends and neighbors, all of whom had been denied entrance to the all-WASP country club, the more the rumor spread. Before you knew it, the whole of Madison Street was a devilish sinkhole of vice and moral corruption worse than Bourbon Street.
Betsy Bosworth was a bull-dyke because she flicked boys’ crotches as she passed them on the down staircase. We all conflated her severe looks, lean frame, and boyish hairdo into Lesbianism, although in 1956 we had no idea what a lesbian was, what they did, or why. Did they rub cunts together? Betsy got all A’s and was headed to Miss Porter’s. Her father was a captain of industry, the CEO of one of New Brighton’s premier industries, so how could she be a dyke?
We all decided that even the daughter of a scion of New Brighton and New England society could eat pussy, even though few of us knew what that was all about either.
Bartley Jennings was a Senior Vice President of Forward International, a non-profit agency whose charter codified the company’s commitment to the world’s poor and underprivileged. Bartley was a harridan, a take-no-prisoners, scorched earth, victory-at-all-costs executive for whom the company’s charitable goals never stood in the way of Simon Legree management. Bartley managed by fear and intimidation. More young minions left under her watch than any SVP before her. She was the champion of Senior Management because of her numbers, but everyone who worked for her looked for The Great Escape.
Not surprisingly, all of her employees tried to imagine what her personal life could be like. Did she pussy-whip her husband of whom she always spoke fondly? Did she run her household like a Soviet gulag, consigning her children to the salt mines for minor infractions of her domestic code?
Male employees wanted to fuck her, despite her severe and uncompromising demeanor. They were convinced that her Amazon demeanor meant nothing at all; that she was a pussycat waiting for the right man and hard cock to release her feminine passions. They imagined her coming home after a hard day’s work, downing three martinis, dismissing her impotent husband, and pleasuring herself to multiple orgasms over imaginary but passionately desired lovers.
Sociologists estimate that at least eighty-five percent of human discourse is spent in gossip. Men and women gossip equally although differently. Women speculate on who was fucked; and men fantasize about who was doing the fucking; but in the end they both dream about sex.
Too much is made of the downside of gossip – catty, frustrated women attributing their own inadequacies and social ineptness to others; horny, frustrated men thinking the most prurient and licentious thoughts about the women they will never have. Gossip is the truest expression of a competitive, hierarchical society. Think the worst of your rivals, and it might come true. Spread vile rumors, vilify, suspect, and impugn the reputation of others and you might come out on top.
What is forgotten in all this psycho-speak is the pure, unadulterated fun of gossipy speculation. Estelle Peters is dead whether by her own hand or by an insidious psychological push by her husband; but the rumor of her going to her own personal gallows in a bungalow on Corbin Avenue dressed like Marie Antoinette is far better than any reality. Headmaster Farnsworth most certainly had no dog’s jaw, but the story of his battlefield rescue, the sacrifice of the loyal German Shepherd, the acumen and heroism of the field MASH unit who braved bombs and dismemberment to save the jaw of Corporal Farnsworth is and should be legend.
Let’s face it. Rumor, innuendo, fantasy, and melodrama are far more interesting than reality. Why are we held to such an unrealistic standard? Truth, veracity, and reality are very overrated.
Anyone who has the least bit of imagination has created an alternate world for everyone. Brad Pitt and Angelina Jolie fight in the bedroom - great vase-throwing, vituperative encounters by two driven, competitive people. Billy Graham must have buggered his valet. The rumors of LBJ pissing over the balustrade of the White House overlooking the South Lawn must have been true. JFK not only bedded Marilyn Monroe, but half the starlets of Hollywood. Nixon tortured rabbits at 3am on the lanes of the Presidential bowling lanes. The neighbor at 4923 holds Satanic rituals.
Rumor, innuendo, gossip, and tall tales are good for you. They satisfy natural urges to feel superior, to invent, to fantasize, and to disparage. They make us feel good.
My mother told me that my cousin’s son had divorced his wife of long standing, married a Gypsy prostitute, and was living like a pasha in Brooklyn Heights. He had turned a blind eye to her continuing sexual libertinage because he was enthralled that after fifteen years with a New England blueblood he had finally freed himself from bourgeois propriety.
A more objective source told me that Randall had indeed married a Romanian woman – Jewish, not Gypsy – and was living happily on the Main Line of Philadelphia. Rumor had it that his wife was more like Mary Magdalene before she repented, more insatiable Jewish whore than saint, but reports of her becoming a Catherine Deneuve Belle de Jour were most certainly exaggerated.
The brave new world of virtuality reality is upon us. The ‘what if’, ‘of course’, and ‘don’t be stupid’ will replace logical, disciplined disaggregation of the facts. When mind and machine are fully and seamlessly linked, we all will wander through worlds of imagined reality. Rumor will no longer be catty speculation but the truth.
My favorite rumor concerns David Halloran, history teacher at Lefferts. He walked with an exaggerated limp, affected bow ties and spoke with a British accent. We all were convinced that he was a true war hero, an RAF pilot shot down by the Germans over hostile territory, incarcerated in brutal POW camps, but who escaped thanks to ingenuity, courage, and good luck. His arrogance was fitting and appropriate. He was our hero.
An obituary in the Hartford Express many years later praised Mr. Halloran for being a dedicated teacher, devout Lutheran, and faithful family man. He had been born with a congenitally-deformed hip which had kept him out of the army, but he had volunteered at re-entry centers for wounded and damaged GIs where he performed selfless, Christian service.
Our adolescent gossipy fantasy was far more appealing, and most of us now still believe that he rained terror down on Germans from his Spitfire. Such is the power of rumor, gossip, and innuendo. Fantasy trumps reality every time.
Note: Only a member of this blog may post a comment. | 114,930 |
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Looking for inspiration for a Gulf Coast trip? Beach destinations in Florida are famous for their crystalline waters, sugary sand, and sea breeze. After a long day of exploring its shores, you may find yourself wondering where to stay in Clearwater Beach.
A stone’s throw away from Tampa Bay, Clearwater Beach is an award-winning vacation destination. The Clearwater Beach area bursts with life. It teems with areas & neighborhoods perfect for families, art lovers, and solo travelers. Whether you’re looking for fresh seafood in luxury restaurants or a secluded cove, it’s all waiting after near one of the best hotels in Clearwater.
Not yet sure where to begin your Clearwater accommodations search? Find the perfect neighborhood and lodgings with this complete guide to the best places to stay in Clearwater, FL.
Looking for more ideas for your trip? Check out our other hotel & accommodations guides and our Florida Travel Guide for recommendations on when to visit, where to go & what to do!
Best places to stay in Clearwater Beach, FL
From the pristine beaches of Sand Key to the art galleries of Downtown, there’s something for everyone around Clearwater Beach. Clearwater’s small-town atmosphere attracts visitors all year long. Travelers flock here to enjoy the white sand adorning the Gulf Coast and the city’s friendly character.
Families, couples, and solo travelers alike are certain to fall in love with the area, from sunset celebrations on Pier 60 to secluded beaches. Different qualities attract, from sun-kissed shores to migratory birds and endangered sea turtles.
While it’s a small area, there are different options for finding accommodations in Clearwater. Some areas are full of luxury resorts; others offer more affordable hotels & apartments.
Whether you want to be in the heart of the town or a quieter area away from the beach, there are lodgings in Clearwater to suit all types of travelers.
The best areas to stay in Clearwater, FL, are:
South Beach
✓ Best area to stay in Clearwater Beach for first-time visitors
Visiting Clearwater for the first time? Set your compass for South Beach. If you’re searching for a sun-kissed stretch of sand and a prime location, South Beach is the perfect place to stay in Clearwater Beach.
Looking over the Gulf of Mexico, South Beach places you right in the heart of the action. The area is full of mid-range and luxury condominiums. All are strategically placed overlooking the lovely turquoise waters. Accommodation options in South Clearwater Beach vary, with oceanfront hotels leading the way. Many are adorned with swimming pools, tennis courts, and other leisure facilities.
Grab your shoes and get ready for an adventure. Pier 60 is just a short walk away. This long pier is one of the top attractions in Clearwater. It offers some of the best sunset views in Clearwater, as well as excellent spots for fishing. A real hit with families, Pier 60 features a smattering of vendors and performers.
There’s the added benefit of plenty of dining options in Clearwater Beach South. Excellent restaurants can be found around every corner, as well as bars serving iced drinks and craft beer. Looking for somewhere with lots of coastal flavors and fun? Grab seafood and drinks at Jimmy’s Fish House & Iguana Bar.
As you’d expect with its central location, accommodation prices in South Beach start in the mid-range and higher. For budget lodgings, you’ll need to look elsewhere.
Recommended hotels in South Beach
Pier House 60 Clearwater Beach Marina Hotel
One of the best hotels for sun-seekers in South Beach, Pier House 60 Clearwater Beach Marina Hotel is perfect for relaxing. From the swimming pool to the short walk to the beach, everything you need is right on your doorstep. Experience fine panoramic views of the area from Jimmy’s Crow’s Nest, the hotel’s popular rooftop bar.
Hyatt Regency Clearwater Beach Resort & Spa
Looking for a slice of luxury? The Hyatt Regency Clearwater Beach Resort & Spa is a 4-star hotel with all the top amenities. From spacious suites to stretch your legs into a rooftop pool on the 8th floor, one of the best spots for watching the ocean is waiting for you.
Wyndham Grand Clearwater Beach
Ready for a high-rise beachfront getaway? The Wyndham Grand Clearwater Beach comes complete with an outdoor swimming pool and incredible views of the Gulf Coast. You can step out onto your furnished balcony to see how beautiful these are. This hotel places you by the main pier and the marina where excursions leave.
Opal Sands
Searching for ocean views? Opal Sands might tick your boxes. This ultra-modern resort is unique. You’ll love peering through floor-to-ceiling windows, offering a perfect view over the ocean. Staying here, you’ll be on Beach Walk, a seaside promenade close to dozens of hotels and bars.
Opal Sands is also a stone’s throw away from Clearwater Marine Aquarium. If you’re looking for top attractions, contemporary surrounds, and enchanting artwork you’ll feel at home here.
North Beach
✓ Best neighborhood to stay in Clearwater Beach for quieter seaside pleasures
Ready to lose yourself to natural beauty? North Beach attracts nature lovers from around the world. The area is much less crowded than Clearwater Beach South and comes with a small-town charm that’s hard to beat.
Clearwater Beach is a huge stretch of sugary sand on this barrier island. The quieter areas of the sandy shore fall around Clearwater Beach North. You can grab your sunscreen and find a spot of your own in minutes.
When you’re done relaxing on the beach and visiting nearby attractions, restaurants nearby are aplenty. The party areas are all closer to South Beach, making North Beach the best option for families and couples who want to relax. In North Clearwater Beach, you can watch the stars with nothing but the sound of the ocean surrounding you.
A short excursion away, you can find Caladesi Island State Park and Honeymoon Island State Park. Caladesi Island is untouched, full of pristine beaches and lush greenery. Honeymoon Island attracts migratory birds and hosts a wide array of native flora and fauna. As its name implies, It’s the perfect place for a romantic getaway.
Recommended hotels in North Beach
Barefoot Bay Resort Hotel
It’s time for a journey back to the past with the heritage of seaside vacations. Barefoot Bay Resort Hotel offers simple surrounds, great value, and an outdoor pool. The location is perfect, too; it’s just a few minutes away from Clearwater Beach, Pier 60, and many of the town’s top shops and restaurants. Don’t forget to rent a jet ski and take to the waves.
Palm Pavilion Inn
Ready to watch the sunset from an observation deck? Palm Pavilion Inn offers that and more. The hotel places you right on the doorstep of Clearwater Beach. This award-winning boutique property features only thirty rooms and a swimming pool. At these ideal lodgings, you will be able to relax and unwind.
Camelot Beach Suites
Ready to stretch your legs? At Camelot Beach Suites, every room is a suite. You’ll be able to destress after a long day at the beach or on the pier with the extra space. Renowned for its old Florida charm, this award-winning boutique hotel draws families and couples for its privacy. With twenty suites, there’s space for everyone to enjoy. The sandy shores are just a short walk away, too.
Sandpearl Resort Private Beach
Dreaming of seclusion? Sandpearl Resort is a favorite among North Beach visitors. You can dive headfirst into luxury with a private beach and a lagoon-style pool. Afterward, tuck into delicious meals at the award-winning restaurant. Families with children are certain to appreciate Camp Ridley. Campfire stories and treasure hunts are the tip of the iceberg. A holistic approach leads the way at Sandpearl Resort, with everything you need right on your doorstep.
Downtown Clearwater
✓ Best place to stay in Clearwater Beach for budget travelers
Traveling to the Floridian coast on a budget? Downtown Clearwater offers cheaper accommodations than the beachfront. The is, however, a more limited selection.
Still, Downtown Clearwater is easy to access. A 30-minute drive across Old Tampa Bay to Tampa International Airport shows just how easy it is to get to.
The Clearwater city center has undergone an amazing beautification process over the years. Cleveland Street District, in particular, offers a breathtaking view. The neighborhood overlooks Clearwater Harbor and Marina. Here, you can find dining, entertainment, and a realm of urban activities. This attractive spot is perfect if you’re looking to relax away from the beach.
When it comes to activities, they’re a-plenty. For festivals and concerts, head over to Coachman Park. Interested in baseball? BayCare Ballpark (Spectrum Field) is the best spot if you want to see the Phillies spring training camp.
Get your camera ready for this next one. Downtown Clearwater is home to Clearwater Aquarium. With a huge focus on “Rescue Rehab Release,” it comes as no surprise that marine life thrives here. The aquarium is also home to Winter the Atlantic bottlenose. With her prosthetic tail, Winter is the star of the heart-warming Dolphin Tale.
Recommended hotels in Downtown Clearwater
La Quinta by Wyndham Clearwater Central
Ready for a day at the park? La Quinta by Wyndham Clearwater Central is one of the closest hotels to Spectrum Field. Start your day with a free continental breakfast before dipping in the outdoor swimming pool or exploring downtown. Many top attractions are a short drive from the hotel.
Residence Inn by Marriott Downtown Clearwater
Want to be right in the heart of the action? Residence Inn by Marriott Downtown Clearwater offers a great home base for exploring. This all-suite hotel is minutes away from the beach and attractions, including SimCenter Tampa Bay. A short drive away from the barrier island and Clearwater Beach, the hotel has everything you’ll need.
Fairfield Inn & Suites by Marriott Clearwater
Happy to be a little further away from the hustle and bustle? Fairfield Inn & Suites is conveniently located between Clearwater Beach and Tampa International Airport. With a car, you can get to the beach in minutes. It’s the exceptional environment here that guests talk about the most, with amazing service and a heated pool available. Not to mention the recent renovation, leaving everything looking pristine.
Sand Key
✓ Top area to stay in Clearwater Beach for a romantic getaway
Looking to escape to a pristine paradise with your partner? Sand Key is one of the best areas for a romantic getaway around Clearwater Beach. The atmosphere here is much more relaxed than on the barrier island or in downtown.
Get your swimsuit ready and head to the beach. Sand Key is surrounded by turquoise waters and pale, white sand. Cabanas, umbrellas, and beach chairs are all available. When you want to relax for the day, picnic grills and tables will lend a welcome hand, too.
From sailing classes to birdwatching, there’s a lot happening in Sand Key Park. Endangered birds and sea turtles are frequent visitors to the barrier island, using the rare salt marsh to nest. The sky stretches forever, making it easy to spot pelicans and herons in flight.
If you want to relax and watch the sunset in peace, this is the place. You’re only a short journey away from Clearwater’s top attractions, but you have the privacy of a secluded haven.
Recommended hotels in Sand Key
Clearwater Beach Marriott Suites on Sand Key
Ready for some of the best ocean views in Florida? Clearwater Beach Marriott Suites on Sand Key offers an enviable location. The all-suite hotel comes complete with private balconies, where you can look out on dolphins jumping in the bay or the ocean. The less-crowded beaches offer the perfect backdrop. After a day of fun, amazing restaurants are right down the road.
Sheraton Sand Key Resort
Vintage vibes fuel the Floridian shore. Sheraton Sand Key Resort offers a luxurious blast into the past, as one of the first hotels to be built here. With a private beach, you’re only moments away from your own secluded haven. Award-winning dining options are available to you here.
Get ready to enjoy yourself if you love animals, too. This hotel is on a sea turtle nesting ground, with workers checking for nests every day. The wildlife conservation here is easy to see and so important.
Nautical Watch
Do you like the privacy an aparthotel offers? Nautical Watch on Sand Keys is full of Belleair Beach apartments, each overlooking the crystalline ocean or pool. With a hot tub, manicured garden, and private beach, you’ll want for nothing here. Expect amazing sunsets, sugar sand, and ultimate privacy.
Indian Rocks Beach
✓ Top district to stay near Clearwater Beach for low-key Old Florida beach vibes
Dreaming of the vacation your grandparents would have loved? Indian Rocks Beach offers that intimate appeal only smaller beach towns pull off. On its three miles of beautiful beaches, you can grab your sunscreen and find a stretch of your own.
View dramatic sunsets from its pristine white-sand shores. Dunes line these beaches. They also rent out cabanas to lay in the sun and watch the day go by. Indian Rocks Beach and Indian Shores are part of a historic hamlet, best showcased in the Indian Rocks Historical Museum.
Go for a walk here and you’ll find the communities your older relatives would have ventured to in decades past. Everything is casual, with small, family-owned eateries. Fun is on the menu, too. Smuggler’s Cove Adventure Golf has an alligator exhibit, while Splash Harbour Water Park is home to a lazy river and water slides.
Accommodation options along Indian Rocks Beach are plentiful. Depending on how you like spending your vacations, rental lodgings might be ideal. There are also luxurious hotels and lush suites. With 25 minutes to St. Pete-Clearwater International Airport, this is one accessible getaway spot.
Recommended hotels in Indian Rocks Beach
New Hotel Collection Harbourside Resort
On a family vacation? New Hotel Collection Harbourside Resort is a short walk away from the white shores of Indian Rocks Beach. A great option for families with children, this hotel offers an on-site water park. When you’re keeping little ones entertained, the slides and lazy river are bound to go down a treat. You can get to the nearest beach in just a few minutes.
Legacy Vacation Resorts Indian Shores
Enjoy the sea breeze? Legacy Vacation Resorts Indian Shores offers the perfect spot to unwind, between the bay and the ocean. This tranquil resort overlooks the Intercoastal Waterway. During dinner, you’ll be able to watch dolphins and manatees in the water. The Indian Rocks Beach attractions are a short walk down the road. There’s also a pool, game room, and shuffleboard court at the hotel for your entertainment.
BeachTrail Lodging
In need of a little extra space? BeachTrail Lodging offers you your own unit, tucked away from the noise of the main road with two parking spaces. It can be found on a secluded spot of beach that you’re certain to enjoy. Nearby, hop on a sightseeing boat tour to explore the area. A street away, you’ll find the Indian Rocks History Museum. | 92,785 |
Ideal for small or medium-sized groups, Laskey.
Square feet: 1461
Ceiling Height: 9ft
Cost Per Day: $275
Accommodates A/V: Yes
In-Room Catering: Yes
Maximum Capacity: 50
Conference Square: 32
Classroom: 40
U-Shape: 24
Banquet Rounds: 48
Theatre/Lecture: 50
Circle of Chairs: 40
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Idle Speculations on Alien Intelligence and the star KIC 8462852.
This is just a personal observation subject to error but as someone far more interested in the Search for Extraterrestrial Intelligence (SETI) than the average person I have noticed something over the last few years. While all the excellent and totally professional science guys and gals still say we are only now reaching the technological level that could allow us to gleam an artificial signal out of all the normal background static I sometimes detect a hint of uncertainty in their voices concerning their research.
Don't misunderstand me, I totally support SETI and do believe that our galaxy does host some number of intelligent, technological advanced alien species. The chief debate to me centers on just how many stars have planets that can spawn and then support for extended periods of time lifeforms that evolve into intelligent creatures. Current research suggests habitable planets are plentiful but whether or not the circumstances that allow intelligence to evolve and then survive are common is still a huge unknown.
For further, and sadly, just as needed clarification. NO, I do not put any stock in all the UFO reports and alien abduction stories that have made it to the popular media over the last fifty to sixty years. On the rare occasions I get to have an reasonable conversation about the possibility of intelligent alien life, someone nearly always chimes in about Roswell, Area 51, or some other tripe about UFO's and the government covering up their existence.
As much as I dream of manned interstellar flight being possible, right now it looks incredibly hard and so expensive it would take a truly prolonged global effort to pay for the project and build. We naked primates are still fighting over religion and whose nation is most special, the idea of us working together to cross the distance between stars is lubricious. Of course, a totally unexpected and out-of-left-field technological breakthrough could change the equation but that is just wishful thinking on my part.
Getting back on point, it seems some of the SETI folks are getting a little worried. For the last twenty years or so we have been scanning huge swaths of the galaxy and millions of frequencies and except for periodic false alarms and the rare and mysterious burp that doesn't repeat, come up with nothing. This has lead some in the biological sciences to suggest that complex life might be quite rare. They point out that while Earth has been around for over 4 billion years it was only 542 million years ago during the Cambrian Explosion that complex life appeared in the fossil record. Before that, all the evidence suggests that life on this planet was limited to single-cell organisms. These “Rare Earth” types argue that there are a complex array of interlocking conditions required before life can move beyond simple organisms and that if you remove a few from the equation further development is stymied.
Even if some complex organisms on an alien world evolves into an intelligent species there is a whole host of different disasters that might drive them extinct before reaching the technological point where they can hope to make their presence known to the wider galaxy. The late, and great, Carl Sagan speculated that all intelligent species might go through a cultural adolescence where their moral and ethical development is outpaced by their technological abilities. In short, they might nuke themselves into oblivion over inconsequential things like religion, nationalistic bullshit, resources, or whose penis is bigger.
(Important side note: Let me go on the record to state I not just talking about us arrogant Americans. At least here in the United States many of us blow off this “Exceptionalism” bullshit. I frankly find Russians more obnoxious with their ethnic based belief that their shit doesn't stink which is made worse by the chip they perpetually carry on their shoulder. It also appears that the Chinese harbor grand ideas of taking charge of the world and get quite upset when everyone doesn't go along with what they consider promotes harmony, which they define as anything that advances their national interests.)
It was the author and astrophysicist David Brin who in his book Existence upped the ante on intelligent species survival by mentioning there are a multitude of cosmic booby traps just waiting to cause an extinction level event for the members of any unwary civilization. These traps ranged from the known threats of a major asteroid or comet impact to being too close to a star going supernova or a passing neutron star. He also included “man-made” disasters such as a genetically created pandemic, the release of nano-sized robots that consume all matter, to climate disaster, which might be the one that already has our name.
Despite the fact that I am a huge fan of Neil deGrasse Tyson and his buddy Bill Nye I highly disagree with their skepticism when it comes to talk about the near-term human colonization of Mars. They tend to think colonization requires a planet with a breathable atmosphere, and while that would be nice such worlds are in short supply in our solar system. I would much rather deal with domed or underground cities on Mars if it meant the survival of the human race in the face of some man-made extinction event.
Long story short, when you take into consideration all the possible disasters-- or filters-- awaiting to befall intelligent species not only do they need to damn near be saints but extremely lucky as well to survive much beyond our current technological level. None of this bodes well for us humans looking to find other intelligent life in our galaxy.
Just speaking for myself, I was quite happy to hear last week that researches have spotted a strange star that does something none of the science types have ever seen before. The star, KIC 8462852, located about 1400 lightyears away dims in way never seen in any of the other stars observed by the famous Kepler orbiting telescope.
The Kepler space observator was designed to monitor the brightness of stars and detect when a body, such as a planet passes in front. Since its launch, the Kepler spacecraft has detected over a thousand confirmed exoplanets with some believed to be close to the size of Earth and orbit in the habitable zone of their parent star where water could exist as a liquid.
In the case of KIC 8462852 what has everyone buzzing is that Kepler detected two massive dips in the brightness of the star roughly every 750 days. One dimming event blocked up to 15 percent of the star's brightness with a later event blocked up to 22 percent. These changes in brightness are consistent with many small masses orbiting the star in tight formation. One thing seems to be clear, that whatever is causing this bizarre event is not a planet. Something the size of our Jupiter would only block 1 percent of our star's light. Whatever is blocking the light of KIC 8462852 is covering half the width of the star.
The overwhelming possibility of what we are seeing is either a cloud of disintegrating comets, essentially a bunch of massive balls of ice and dirty involved in a cosmic-level traffic pileup. Or a recently captured asteroid field kind of like the one Han Solo had to navigate through to escape the clutches of Darth Vader in The Empire Strikes Back.
KIC 8462852 does have a close by stellar companion that might have stirred up the cloud of comets that surround all stars, sending them crashing inward and after 1400 years of its light traveling our way screw with H. sapien astronomers here on Earth desperate to make a first contact with aliens. And while we believe it to be an older star it's possible KIC 8462852 is actually young and still has a lot of primordial junk floating around it. For this one I guess the best analogy would be your average teenager's room.
There is another possibility that is only barely mentioned in passing and only half seriously. What we might be seeing is evidence of an artificial construction called a Dyson Sphere. The idea being that stars put out a lot of free energy and that if an advanced race built a shell around it they could collect all that light and use it for some super advanced purpose.
Needless to say, I hope we have found evidence of an alien civilization that has somehow found a way to survive all the natural and self-inflicted wounds an uncaring universe can throw on an intelligent lifeform. I'm not looking for some event that knocks humanity out of its tried and true combination of ignorance and apathy. Just a flicker of some awareness that we are not alone in the universe could be enough to defer our appointment with extinction.
One final note, as of this writing the Allen Telescope Array is even now scanning KIC 8462852 for signs of artificially generated radio signals. No, I'm not optimistic but when you're playing this type of game patience is a vital element in this research.
For further reading see: Popular Science-Have We Just Detected Megastructures Built By Aliens Around A Distant Star.
4 comments:
Interesting. I do believe there is live beyond our planet but as you say they just start to develop technology to be able to pick up signs and I even think it is by long not enough.
And than why would they ever be interested in us? There is not that much to learn from human species as we don't keep our planet clean and fight each other all the time
...patience is a vital element...yeah, it seems like a few hundred years of patience. I don't think we or our children will know - but maybe our great great grandchildren. However, I do think colonizing Mars is somewhat more possible - my youngest son would totally go for that trip (I told him he should wait until after I'm dead, LOL).
I have enough troubles and wondering if there is intelligent life out there is something that I leave up to other folks, like you! However, I do find that the possibilities makes for great Science Fiction. Have you read Lee Modesitt? Two of his books (Parafaith War and Ethos) deal with a religious society gone bad and then a secular one gone bad. Of course, sci-fi tends to imagine worlds that look a lot like us and instead of nations fighting, it is planets or galaxies (and maybe that is the only way the world would be united is a common enemy from out there.
Marja: What you mentioned might be the reason aliens haven't contacted us. They might think we're on the same level we think of ants.
Pixel: I actually quite surprised Elon Musk is so open about his desire to colonize Mars. I remember back in the 80's a scientist brought up the subject and was laughed out of the room.
Sage: I'll have to check out those books. As far as humanity uniting, yeah it would probably take some outside threat. | 226,811 |
TITLE: What can we say about the differentiability of $f(z)= \Im(z)^n$ at $z=0$ for $n \in \mathbb{Z}$?
QUESTION [0 upvotes]: What can we say about the differentiability of $f(z)= \Im(z)^n$ at $z=0$ for $n \in \mathbb{Z}$?
It is clear to me that $\Im(z)$ is not differentiable at $0$. This follows from the fact that for $\Re(z) \neq 0$, $\dfrac{\Im(z)}{z} = \dfrac{0}{z} = 0$ but for $z \neq 0$ purely imaginary, $\dfrac{\Im(z)}{z} = \dfrac{z}{z} = 1$
It is also clear to me that $\Im(z)^2$ is differentiable at $z=0$. I have come up with two different ways to arrive at this conclusion, one of them is the following:
Notice that $\displaystyle \lim_{z \to 0} \frac{\Im(z)^2}{z} = 0$ if and only if $\displaystyle \left| \lim_{z \to 0} \frac{\Im(z)^2}{z}\right| = \lim_{z \to 0} \left|\frac{\Im(z)^2}{z}\right| =0$
Then, observe that, for $z \neq 0$, $$\left| \frac{\Im(z)^2}{z}\right| = \left|\frac{z^2 - 2z \bar{z} + \bar{z}^2 }{-4z}\right| \leq \frac{1}{4} \left(|z| + 2| - \bar{z} |+ \left|\frac{\bar{z}^2}{z}\right|\right)= \frac{1}{4} \cdot 4|z| = |z| $$
With correct algebraic manipulation we can get from this inequality that $ \displaystyle \lim_{z \to 0} \left|\frac{\Im(z)^2}{z}\right| \leq \lim_{z \to 0} |z|$. Hence $\displaystyle \lim_{z \to 0} \left|\frac{\Im(z)^2}{z}\right|=0$.
Now I would like to find a way to determine the differentiability of $\Im(z)^n$ for arbitrary $n \in \mathbb{Z}$. I am fairly confident that I will be able to show that $\Im(z)^n$ is differentiable for $n \geq 2$. But for $n \leq -1$ I'm not sure what will happen.
REPLY [1 votes]: $|\Im (z)| \leq |z|$.
So $\left|\frac {\Im(z)^{n}-0} z\right| \leq |z|^{n-1} \to 0$ as $z \to 0$ for any $n >1$.
The function is not even defined in any neighborhood of $0$ if $n <0$. | 165 |
Learn how to ‘green’ your office space
The Office of Sustainability and the UF Thompson Earth Systems Institute have collaborated on a guide filled with resources to help you green your office space. The guide provides tips on the types of plants you can incorporate into your space, where you can find them and other sustainable ways to ‘green’ your space.
Visit the Office of Sustainability website to learn more:. | 309,643 |
All Nippon Airways: Ana's Adventures
Agency: Carrot Creative
Client: All Nippon Airways
Role: Lead Animator / 3D / Editing / Sound Design
Art Direction: Adam Lowe
Illustration: Jay Fleckenstein
Animation: Whitney Brown, Krissan Pattugalan
Ana's Adventures is an animated series made monthly for the Japanese airline, All Nippon Airways. Each episode is set in a region voted on by the audience and highlights several attractions which travelers can experience for themselves when visiting Japan. Videos are posted to the North American social media accounts where the audience votes for the next episode's destination.
I have personally only worked on this episode (Kagawa) of Ana's adventures. | 269,348 |
Changing Your Life For The Best Famous Quotes & Sayings
List of top 34 famous quotes and sayings about changing your life for the best to read and share with friends on your Facebook, Twitter, blogs.
Top 34 Quotes About Changing Your Life For The Best
#1. You didn't," she said in a resolute voice. "And you won't in the future, because with each purging of your soul, your anger will subside until the only release you'll need will be in my arms. - Author: Monica Burns
#2. Changing the destructive things you say to yourself when you experience the setbacks that life deals all of us is the central skill of optimism. - Author: Martin Seligman
#3. Big decisions in my life have always come easy and are made without hesitation. It is easier for me to make a life-changing decision than to decide what to get for dessert. - Author: Tony Hawk
#4._3<<
#5.. - Author: Jake Roberts
#6. When you get to a place where you don't go for what you can get, but you go for what you can give, you gonna see your life change tremendously. - Author: Eric Thomas
#7. I'm not in this just to change the law. It's about changing society. I want gay kids to grow up believing that they can get married, that they can join the Scouts, that they can choose the life they want to live. - Author: Evan Wolfson
#8. Life is rich, always changing, always challenging, and we architects have the task of transmitting into wood, concrete, glass and steel, of transforming human aspirations into habitable and meaningful space. - Author: Arthur Erickson
#9. You never know until you give something your best and keep working at it. Follow your dreams no matter what - Author: Santonu Kumar Dhar
#10. And what's the point of changing when I'm happy as I am? - Author: Paul McCartney
#11. To change your language you must change your life. - Author: Derek Walcott
#12. So if you don't like your life, change it. How would you change it? You decide. There's no action without first decision. Decision is the mother or the father of action, and action is what changes your life. - Author: Tony Robbins
#13. We can change our whole life and attitude of the people around us simply by changing ourselves.. - Author: Rudolf Dreikurs
#14. You better start swimming, or you'll sink like a stone. Because the Time's they are a-changing. - Author: Bob Dylan
#15.
#16. So don't be worried if you have the sub-standard version barking out the orders. It will continue to do so, just be aware of it. Than call it for what it is, gremlin for the best you. Then grab that vision of the best you and get up and give it everything you have got. - Author: Tony Curl
#17. Take one minute every day and think hard about your life. You will always change for the best. - Author: Bangambiki Habyarimana
#18.. - Author: A.P. Herbert
#19. One minute of sincere gratitude can wash away a lifetime's disappointments. - Author: Silvia Hartmann
#20. It's nice sometimes to be the river rather than the rock. - Author: Jo Beverley
#21. Because there is liberating power in each and every truth, the one who walks in the truth in all his ways will be set free. A lie, no matter how "little," gives the powers of darkness an opening for attack, but the truth chases them far away. - Author: Johan Oscar Smith
#22. I love the flowers for their beauty and dazzling smile. I love the moon for its soothing light and changing style. - Author: Debasish Mridha
#23._22<<
#24. We suffer because we want life to be different from what it is. We suffer because we try to make pleasurable what is painful, to make solid what is fluid, to make permanent what is always changing. - Author: Sakyong Mipham
#25. Destiny is variable, not fixed; it is forever changing depending upon your free will to make choices for what you want your life to be. - Author: Steven Redhead
#26. If there is no fate and our interactions depend on such a complex system of chance encounters, what potentially important connections do we fail to make? What life changing relationships or passionate and lasting love affairs are lost to chance? - Author: Simon Pegg
#27. It is not the nature of man, as I see it, ever to be quite satisfied with what he has in life ... Contentment tends to breed laxity, but a healthy discontent keeps us alert to the changing needs of our time. - Author: Frances Perkins
#28. You have to know yourself and know what you can carry. If you're not comfortable, no matter how well dressed you are, you're not going to look right. I'd rather be happy and feel comfy. - Author: Iris Apfel
#29. Getting what you want will not change your life at the Being Level, so don't let what you have determine 'who you are', otherwise you will always feel dissatisfied. - Author: Robert Anthony
#30. When it is mid week, pause and ponder! The very single days we disregard are what become the very years we wished to have used effectively and efficiently. If we disregard today, we shall remember our had I know tomorrow. Time changes therefore think of the changing times. - Author: Ernest Agyemang Yeboah
#31. Asking the questions - that's what changes lives. Every cell in your body is awake with inquiry. And you cannot believe the old thoughts again. - Author: Byron Katie
#32. I am really curious about life, about why we are all here. I notice my skin is ageing, things are changing, I've seen people dying, so that's the train we are all on. - Author: Damien Rice
#33. That was the way things worked. When you were looking for the big fight, the moment that you thought would knock everything over, nothing much happened at all. - Author: Ann Patchett
#34. Change your thinking, and you will change your life. - Author: Brian Tracy
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\begin{document}
\maketitle
\begin{abstract}
The anti-forcing number of a perfect matching $M$ of a graph $G$ is
the minimum number of edges of $G$ whose deletion results in a subgraph
with a unique perfect matching $M$, denoted by $af(G,M)$.
When $G$ is a plane bipartite graph, Lei et al. established a minimax result:
For any perfect matching $M$ of $G$,
$af(G,M)$ equals the maximum number of $M$-alternating cycles of $G$
where any two either are disjoint or intersect only at edges in $M$;
For a hexagonal system, the maximum anti-forcing number equals the fries number. In this paper we show that for every perfect matching $M$ of a hexagonal system $H$ with the maximum anti-forcing number or minus one, $af(H,M)$ equals the number of $M$-alternating hexagons of $H$. Further we show that a hexagonal system $H$ has a triphenylene as nice subgraph if and only $af(H,M)$ always equals the number of $M$-alternating hexagons of $H$ for every perfect matching $M$ of $H$.
\vspace{0.3cm} \noindent\textbf{Keywords:}
Hexagonal system; Perfect matching; Anti-forcing number; Fries number; Triphenylene.
\end{abstract}
\section{Introduction}
All graphs considered in this paper are finite and simple connected graphs.
Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$.
A \emph{perfect matching} $M$ of $G$ is a set of edges of $G$
such that each vertex of $G$ is incident with exactly one edge in $M$.
This graph-theoretical concept coincides with a Kekul\'{e} structure of chemical molecules.
In 1987, Randi\'{c} and Klein \cite{Klein-Randic-1987} proposed
the \emph{innate degree of freedom} of a Kekul\'{e} structure,
i.e. the minimum number of double bonds which simultaneously belong to
the given Kekul\'{e} structure and to no other one.
This notion has arisen in the study of finding resonance structures of a given molecule in chemistry.
Later, it is named ``forcing number'' by Harary et al. \cite{Harary-1991}.
A \emph{forcing set} $S$ for a perfect matching $M$ of a graph $G$ is a subset of $M$ which
is not contained in other perfect matchings of $G$.
The \emph{forcing number} of $M$ is the smallest cardinality over all forcing sets of $M$,
denoted by $f(G,M)$.
The \emph{maximum forcing number} of $G$ is the maximum value of forcing numbers
of all perfect matchings of $G$, denoted by $F(G)$.
For further information on this topic,
we refer the reader to a survey \cite{Che-Chen-2011} and other references
\cite{Adams-2004,Afshani-2004,Diwan2019,Pachter-1998,
Riddle-2002,Zhang-Li-1995, WYZ16}.
Let $M$ be a perfect matching of a graph $G$.
A cycle (resp. a path $P$) $C$ of $G$ is \emph{$M$-alternating} if
the edges of $C$ (resp. $P$) appear alternately in $M$ and $E(G)\backslash M$.
It was revealed \cite{Adams-2004,Riddle-2002} that
a subset $S\subseteq M$ is a forcing set of $M$ if and only if
$S$ contains at least one edge of each $M$-alternating cycle of $G$.
This implies a simple inequality $f(G,M)\geq c(G,M)$,
where $c(G,M)$ denotes the maximum number of disjoint $M$-alternating cycles in $G$.
In the case where $G$ is a plane bipartite graph, Pachter and Kim \cite{Pachter-1998} observed that
these two numbers are always equal to each other.
\begin{thm}\label{Forcing-feedback}
\cite{Pachter-1998}
Let $G$ be a plane bipartite graph with a perfect matching $M$. Then $$f(G,M) = c(G,M).$$
\end{thm}
A more general result on bipartite graphs due to B. Guenin and R. Thomas is given as follows; see \cite[Corollary 5.8]{Guenin-2011}.
\begin{thm}\cite{Guenin-2011}
Let $G$ be a bipartite graph, and let $M$ be a perfect matching in $G$. Then
$G$ has no matching minor isomorphic to $K_{3,3}$ or the Heawood graph if and only if
$f(G',M') = c(G',M')$ for every subgraph $G'$ of $G$ such that $M' =
M\cap E(G')$ is a perfect matching in $G'$.
\end{thm}
In 2007, the anti-forcing number of a graph
was introduced by Vuki\v{c}evi\'{c} and Trinajsti\'{c} \cite{Vukicevic-2007} as
the smallest number of edges whose
removal results in a subgraph with a unique perfect matching. In an early paper, Li
\cite{Li-1997} proposed a forcing single edge
(i.e. anti-forcing edge) of a graph, which belongs to all but one perfect matching.
For other researches on this topic, see Refs
\cite{Che-Chen-2011,Deng-2007,Vukicevic-2008,Bian-2011}.
More recently, by an analogous manner as the forcing number, Klein and Rosenfeld \cite{Klein-Rosenfeld-2014} and
Lei et al. \cite{Lei-Zhang} independently defined the anti-forcing number of a single perfect matching in a graph.
Let $M$ be a perfect matching of a graph $G$. A subset $S\subseteq E(G)\backslash M$ is
called an \emph{anti-forcing set} of $M$ if $G-S$ has a unique perfect matching $M$,
where $G-S$ denotes the graph obtained from $G$ by deleting all edges in $S$.
The following lemma shows an equivalent condition.
\begin{lem}\label{Anti-forcing-cycle}
\cite{Lei-Zhang}
Let $G$ be a graph and $M$ be a perfect matching of $G$.
A subset $S$ of $E(G)\backslash M$ is an anti-forcing set of $M$ if and only if
$S$ contains at least one edge of every $M$-alternating cycle.
\end{lem}
The minimum cardinality of anti-forcing sets of $M$ is called the \emph{anti-forcing number} of $M$, denoted by $af(G,M)$. From these concepts, we can see that the anti-forcing number
of a graph $G$ is just the minimum value of anti-forcing numbers of all perfect matchings of $G$.
The \emph{maximum anti-forcing number} of $G$ is the maximum value of anti-forcing numbers
of all perfect matchings of $G$, denoted by $Af(G)$. Two sharp upper bounds on maximum anti-forcing number and anti-forcing spectrum of a graph, we may refer to recent refs. \cite{DengZhang17b, shiZ2017, DK2015}.
Given a graph $G$ with a perfect matching $M$, two $M$-alternating cycles of $G$
are said to be \emph{compatible} if they either are disjoint or intersect only at edges in $M$.
A set $\mathcal{A}$ of pairwise compatible $M$-alternating cycles of $G$ is called
a \emph{compatible $M$-alternating set}.
Let $c'(G,M)$ denote the maximum cardinality of compatible $M$-alternating sets of $G$. We also have $af(G,M)\geq c'(G,M)$.
For plane bipartite graphs $G$, Lei et al. \cite{Lei-Zhang} established the equality.
\begin{thm}\label{Anti-forcing-feedback}
\cite{Lei-Zhang}
Let $G$ be a plane bipartite graph with a perfect matching of $M$. Then $$af(G,M) = c'(G,M).$$
\end{thm}
A \emph{hexagonal system} (or benzenoid) is a 2-connected finite plane graph such that
every interior face is bounded by a regular hexagon of side length one \cite{Sachs-1984}.
It can also be formed by a cycle with its interior in the infinite hexagonal lattice
on the plane (graphene) \cite{Cyvin-book-1988}.
A hexagonal system with a perfect matching is regarded as
a molecular graph (carbon-skeleton) of a benzenoid hydrocarbon.
Hence these kinds of graphs are called
benzenoid systems and have been extensively investigated;
We may refer to a detailed review \cite{Randic-review-2003} due to Randi\'{c}.
Recently it was known \cite{Xu-2013,Lei-Zhang} that maximum forcing number and anti-forcing number of a hexagonal system are equal to the famous Clar number and Fries number respectively, which can measure the stability of polycyclic benzenoid hydrocarbons
\cite{Abeledo-2007,Clar-1972, Hansen-Zheng-1992}. The same results hold for (4,6)-fullerene graphs \cite{shiWZ2017}.
Let $H$ be a hexagonal system with a perfect matching. A subgraph $H_0$ of $H$ is called {\em nice} if $H-V(H_0)$ has a perfect matching.
A set of disjoint hexagons of $H$ is called {\em sextet pattern} if they form a nice subgraph of $H$.
The size of a maximum resonant set of $H$ is the \emph{Clar number} of $H$, denoted by $Cl(H)$.
\begin{thm}\label{HS-F(H)-Cl(H)}
\cite{Xu-2013}
Let $H$ be a hexagonal system with a perfect matching. Then $F(H)=Cl(H)$.
\end{thm}
Xu et al. \cite{Xu-2013} obtained the theorem by using Zheng and Chen's result \cite{Zheng-1985} that if $H- K$
has at least two different perfect matchings for a resonant set $K$ of $H$, then $Cl(H)\geq |K|+1$.
By rising this bound by one, the present authors obtained a stronger result.
\begin{thm}\label{HS-F(H)-Cl(H)-stronger}
\cite{Zhou-Zhang-2015}
Let $H$ be a hexagonal system with a perfect matching.
For every perfect matching $M$ of $H$ with forcing number $F(H)$,
$H$ has $F(H)$ disjoint $M$-alternating hexagons.
\end{thm}
However, Theorem \ref{HS-F(H)-Cl(H)-stronger} does not necessarily hold for perfect matchings of $H$
with the forcing number $F(H)-1$.
For example, the Coronene (see Figure \ref{fig:counterexample-1}(a)) has the maximum forcing number 3.
For the specific perfect matching of Coronene marked by the bold lines, it
has forcing number two, but it has only one alternating hexagon.
\begin{figure}[H]
\begin{center}
\includegraphics[height=3cm]{fig1.pdf}
\caption{{\small (a) Coronene}, and (b) Triphenylene.}
\label{fig:counterexample-1}
\end{center}
\end{figure}
For a hexagonal system $H$ with a perfect matching $M$, let $fr(H,M)$ denote the number of
all $M$-alternating hexagons in $H$. The maximum value of $fr(H,M)$
over all perfect matchings $M$ is just the {\em Fries number} of $H$, denoted by $Fr(H)$.
Since all $M$-alternating hexagons of $H$ are compatible,
Theorem \ref{Anti-forcing-feedback} implies $af(H,M)=c'(H,M)\geq fr(H,M)$.
The second equality does not hold in general.
For example, the bold lines of Triphenylene in Figure \ref{fig:counterexample-1}(b)
constitute a perfect matching with anti-forcing number two,
whereas it has only one alternating hexagon.
However, Lei et al. \cite{Lei-Zhang} obtained the following result by finding a perfect matching $M$
with the equality.
\begin{thm}\label{HS-F(H)-Fries(H)}
\cite{Lei-Zhang}
Let $H$ be a hexagonal system with a perfect matching. Then $$Af(H)=Fr(H).$$
\end{thm}
In this article, we show that $af(H,M) = fr(H,M)$ always holds for every perfect matching $M$
of a hexagonal system $H$ with the maximum anti-forcing number or minus one.
\begin{thm}\label{main-thm}
Let $H$ be a hexagonal system with a perfect matching. Then for every perfect matching $M$ of $H$
with anti-forcing number $Af(H)$ or $Af(H)-1$, we have
\begin{equation}af(H,M)=fr(H,M).\end{equation}
\end{thm}
To prove this main result, in Section 2 we introduce some auxiliary terms relevant to
our studies and give a crucial lemma that states that for a non-crossing compatible $M$-alternating set of $H$ with two members whose interiors have a containment relation, the maximum anti-forcing number of $H$ is larger than the cardinality by at least two.
In Section 3, by using this lemma we obtain a stronger result: for any perfect matching $M$ of $H$ whose anti-forcing number reaches the maximum value or minus one, any two members in any given maximum non-crossing compatible $M$-alternating set of $H$ have disjoint interiors and any member bounds a linear hexagonal chain, then give a proof of Theorem \ref{main-thm}.
In Section 4 we give a complete characterization to hexagonal systems $H$ that always have Eq. (1) for each perfect matching $M$ of $H$ by forbidding a triphenylene as nice subgraph.
\section{A crucial lemma}
In what follows, we assume that all hexagonal systems are embedded in the plane such that
some edges parallel to each other are vertical except for Figure \ref{fig:hexagonal-system-11}.
A \emph{peak} (resp. \emph{valley}) of a hexagonal system is a vertex whose neighbors
are below (resp. above) it.
For convenience, the vertices of a hexagonal system are colored with white and black such that
any two adjacent vertices receive different colors, and the \emph{peaks} are colored black.
Let $G$ be a plane bipartite graph with a perfect matching $M$, and
let $\mathcal{A}$ be a compatible $M$-alternating set of $G$.
Two cycles in $\mathcal{A}$ are said to be \emph{non-crossing} if
their interiors either are disjoint or have a containment relation.
Further, we say $\mathcal{A}$ is \emph{non-crossing} if any two cycles in $\mathcal{A}$ are non-crossing.
The following useful lemma was described in the first claim of the proof of {\cite[Theorem 3.1]{Lei-Zhang}}.
\begin{lem}\cite{Lei-Zhang}\label{Crossing-Noncrossing}
Let $H$ be a hexagonal system with a perfect matching $M$. Then
for any compatible $M$-alternating set $\mathcal{A}$ of $H$,
$H$ has a non-crossing compatible $M$-alternating set $\mathcal{A}'$
with $|\mathcal{A}'|=|\mathcal{A}|$.
\end{lem}
We now state a crucial lemma as follows.
\begin{lem}\label{Crucial-lemma}
For a hexagonal system $H$ with a perfect matching $M_0$, let $\mathcal{A}_0$
be a non-crossing compatible $M_0$-alternating set of $H$. Suppose
$\mathcal{A}_0$ has a pair of members so that their interiors have a containment relation.
Then $Af(H)\geq |\mathcal{A}_0|+2$.
\end{lem}
In order to prove this lemma, we need some further terminology.
Let $M$ be a perfect matching of $H$.
An edge of $H$ is called an $M$-\emph{double} \emph{edge} if
it belongs to $M$, and an $M$-\emph{single} \emph{edge} otherwise. $M$-double edges are often indicated by bold or double edges in figures.
An $M$-alternating cycle $C$ of $H$ is said to be \emph{proper} (resp. \emph{improper}) if
each $M$-double edge in $C$ goes from white end to black end (resp. from black end to white end)
along the clockwise direction of $C$.
The boundary of the infinite face of $H$ is called the \emph{boundary} of $H$,
denoted by $\partial(H)$. An edge on the boundary of $H$ is a \emph{boundary edge}.
A hexagon of $H$ is called an \emph{external hexagon} if
it contains a boundary edge, and an \emph{internal hexagon} otherwise.
A hexagonal system $H$ is \emph{cata-condensed} if all vertices of $H$ lie on $\partial(H)$.
A hexagon of a cata-condensed hexagonal system is a \emph{branch} if
it has three adjacent hexagons.
For example, the graph showed in Figure \ref{fig:counterexample-1}(b)
is a cata-condensed hexagonal system with exactly one branch. A cata-condensed hexagonal system without branch is a \emph{ hexagonal chain}. In particular, it is a \emph{linear chain} if
the centers of all hexagons lie on a straight line. A maximal linear chain of a hexagonal chain is called a {\em segment}.
The symmetric difference of finite sets $A_1$ and $A_2$ is defined as
$A_1\oplus A_2:= (A_1\cup A_2)\backslash (A_1\cap A_2)$.
This operation can be defined among many finite sets in a natural way and
is associative and commutative.
If $C$ is an $M$-alternating cycle of $H$, then $M\oplus C$ is also a perfect matching of $H$
and $C$ is an $(M\oplus C)$-alternating cycle of $H$, where $C$ may be regarded as its edge-set.
For a cycle $C$ of a hexagonal system $H$, let $I[C]$ denote the subgraph of $H$ formed by
$C$ together with its interior, and let $h(C)$ be the number of hexagons in $I[C]$.
\vskip 0.2cm
\noindent\textbf{Proof of Lemma \ref{Crucial-lemma}.}
By the assumption, we can choose a perfect matching $M$ of $H$ and a non-crossing compatible $M$-alternating set $\mathcal{A}$
of $H$ so that the following three conditions hold.\\
(i) $|\mathcal{A}|=|\mathcal{A}_0|$, \\
(ii) $\mathcal{A}$ has a pair of members so that their interiors have a containment relation, \\
(iii) $h(\mathcal{A}):=\sum_{C\in \mathcal{A}}h(C)$ is as small as possible subject to (i) and (ii).
Since $\mathcal{A}$ is non-crossing, we have that for any two cycles in $\mathcal{A}$
their interiors either are disjoint or one contains the other one.
Hence the cycles in $\mathcal{A}$ form a \emph{poset} according to
the containment relation of their interiors.
Since each $M$-alternating cycle has an $M$-alternating hexagon in its interior (cf. [32]),
we immediately have the following claim.
\vskip 0.2cm
\noindent\textbf{Claim 1.} Every minimal member of $\mathcal{A}$ is a hexagon.
\vskip 0.2cm
By the choice of $\mathcal{A}$, we can see that $\mathcal{A}$ has at least one non-hexagon member.
Let $C$ be a minimal non-hexagon member of $\mathcal{A}$.
Then $C$ is an $M$-alternating cycle whose interior contains only minimal members of $\mathcal{A}$.
By Claim 1 $C$ contains at least one hexagon as a member of $\mathcal{A}$ in its interior.
Set $H':=I[C]$. So it follows that $H'$ is not a linear hexagonal chain.
\vskip 0.2cm
\noindent\textbf{Claim 2.} For any $M$-alternating hexagon $h$ in $H'$,
either $h\in \mathcal{A}$ or at least one of the three $M$-double edges of $h$ does not belong to $C$.
\begin{proof}
If $h$ belongs to $\mathcal{A}$, then the claim holds.
If not, suppose to the contrary that the three $M$-double edges of $h$ belong to $C$.
Then $M\oplus h$ is a perfect matching of $H$,
and all (one to three) components of $C\oplus h$ are $(M\oplus h)$-alternating cycles.
We can see that every minimal member of $\mathcal{A}$ in $H'$ is disjoint with $h$.
By the choice of $C$, $C\oplus h$ has a component as a cycle $C'$ which
is not a hexagon and contains a minimal member of $\mathcal{A}$ in its interior.
Since each vertex of $H$ has degree 2 or 3,
each $M$-double edge of $H$ is contained in at most two cycles of $\mathcal{A}$.
This implies that $\mathcal{A}\backslash \{C\}$ has at most one member intersecting $h$.
If such a member exists, denote it by $C''$ and
let $\mathcal{A'}:=(\mathcal{A}\cup \{h, C'\})-\{C, C''\}$;
otherwise, let $\mathcal{A'}:=(\mathcal{A}\cup \{C'\})-\{C\}$.
Then $\mathcal{A'}$ is a compatible $(M\oplus h)$-alternating set of $H$
satisfying Conditions (i) and (ii). But $h(\mathcal{A'})<h(\mathcal{A})$,
contradicting the choice for $\mathcal{A}$. Hence Claim 2 holds.
\end{proof}
Now we focus our attention on hexagonal system $H'$ with the boundary $C$ as some preliminaries.
Without loss of generality, suppose that $C$ is a proper $M$-alternating cycle (for the other case,
analogous arguments are implemented on right-top and right-bottom corners of $H'$).
We apply an approach and notion appeared in Ref. \cite{Zhou-Zhang-2015}. Along the boundary $C$ of $H'$,
we will find two substructures of $H'$ in its left-top corner and
left-bottom corner as Figures \ref{fig:hexagonal-system-01} and
\ref{fig:hexagonal-system-03} respectively as follows.
A \emph{b-chain} of hexagonal system $H'$ is a maximal horizontal linear chain consisting of
the consecutive external hexagons when traversing (counter)clockwise the boundary $\partial (H')$.
A b-chain is called \emph{high} (resp. \emph{low})
if all hexagons adjacent to it are below (resp. above) it.
For example, in Figure \ref{fig:substructure} (taken from \cite{Zhou-Zhang-2015}),
$D_0$, $D_1$, $D_2$, $G_1, G_2,\ldots, G_9, G_1', D_5, D_6$ and $D_7$ are b-chains.
In particular, $D_0$, $D_1$, $D_2$ and $G_1$ are high b-chains,
while $G_1'$, $D_5$ and $D_6$ are low b-chains.
But $G_2, G_3, \ldots, G_9$ and $D_7$ are neither high nor low b-chains.
\begin{figure}[H]
\begin{center}
\includegraphics[height=6.50cm]{fig2.pdf}
\caption{{\small
Various b-chains of a hexagonal system, taken from \cite{Zhou-Zhang-2015}.}}
\label{fig:substructure}
\end{center}
\end{figure}
Given a high b-chain and a low b-chain of $H'$, they are distinct,
otherwise $H'$ itself is a linear chain, contradicting the choice of $C$.
When traversing the b-chains along the boundary $\partial (H')$ counterclockwise from the high b-chain to the low b-chain,
let $G_1$ be the last high b-chain and let $G_1'$ be the first low b-chain after $G_1$.
Then the b-chains between $G_1$ and $G_1'$ descend monotonously.
From high b-chain $G_1$ we have a sequence of consecutive b-chains $G_1, G_2,\ldots, G_m$, $m\geq 1$,
with the following properties:
(1) for each $1\leq i<m$, $G_{i+1}$ is next to $G_{i}$, and
the left end-hexagon of $G_{i+1}$ lies on the lower left side of $G_{i}$,
(2) either $G_m$ is just the low b-chain $G_1'$ or
$G_{m+1}$ is a b-chain next to $G_m$ such that $G_{m+1}$
has no hexagon lying on the lower left side of $G_{m}$.
Let $G$ be the subgraph of $H'$ consisting of b-chains $G_1, G_2,\ldots, G_{m-1}$
and the hexagons of $G_m$ lying on the lower left side of $G_{m-1}$.
Then $G$ is a ladder-shape hexagonal chain.
Similarly, from low b-chain $G_1'$ we have a sequence of consecutive b-chains
$G_1', G_2',\ldots, G_s'$, $s\geq 1$, with the following properties:
(1) for each $1\leq j<s$, $G_{j+1}'$ is next to $G_{j}'$, and
the left end hexagon of $G_{j+1}'$ lies on the upper left side of $G_{j}'$,
(2) either $G_s'$ is just the high b-chain $G_1$ or
$G_{s+1}'$ is next to the b-chain $G_{s}'$ such that $G_{s+1}'$
has no hexagon lying on the upper left side of $G_{s}'$.
Let $G'$ be the subgraph of $H'$ consisting of b-chains $G_1', G_2',\ldots, G_{s-1}'$
and the hexagons of $G_{s}'$ lying on the higher left side of $G_{s-1}'$. So $G'$ is an inverted ladder-shaped hexagonal chain.
For example, given a high b-chain $D_1$ and a low b-chain $D_5$ in Figure \ref{fig:substructure},
we can get two required hexagonal chains $G=G_1\cup G_2\cup G_3\cup G_4$ and $G'=G_9\cup G_1'$.
The following claim is obvious.
\vskip 0.2cm
\noindent\textbf{Claim 3.} Either $G$ and $G'$ are disjoint or
they intersect only in the b-chain $G_m=G'_s$.
\vskip 0.2cm
To analyze the substructure $G$ of $H'$, we label some edges of $G$ as follows
(see Figure \ref{fig:hexagonal-system-01}):
let $e_{1,1}$ be the slant $M$-double edge of the right end hexagon of $G_1$ which
belongs to $C$ and contains a peak of $H'$.
Neither $A$ nor $A'$ is contained in $H'$.
Denote by $e_{i,j}$, $1\leq i\leq m$ and $1\leq j\leq n(i)$, the $j$-th edge of $G_i$ that is parallel to $e_{1,1}$ and on the boundary $C$ of $H'$,
and denote the specific edges in $G_1$ and $G_m$
by $a, a'$ and $e_0, e_0'$ respectively.
\begin{figure}[H]
\begin{center}
\includegraphics[height=6cm]{fig3.pdf}
\caption{{\small
Hexagonal chain $G$ on the left-top corner of $H'$
(bold edges are $M$-double edges, $m=5$,
n(1)=3, n(2)=1, n(3)=3, n(4)=2, n(5)=2 and $A,A'\notin H'$) and the corresponding broken line segment $L_1$.}}
\label{fig:hexagonal-system-01}
\end{center}
\end{figure}
Since the boundary $C$ of $H'$ is a proper $M$-alternating cycle,
all the edges $e_0$, $e_0'$, $e_{i,j}$, $1\leq i\leq m$, $1\leq j\leq n(i)$,
are $M$-double edges. In order to simplify our discussions,
we draw a ladder-shape broken line segment $L_1=P_0P_1\cdots P_{q+1}(q\geq 1)$
(see Figure \ref{fig:hexagonal-system-01}) satisfying:
(1) $L_1$ only passes through hexagons of $G$,
(2) the endpoints $P_0$ and $P_{q+1}$ are the midpoints of the edges $a$ and $a'$ respectively,
(3) $L_1$ passes through the centers of all hexagons of $G$, and
(4) each $P_i$ ($1\leq i\leq q$) is a turning point, which is the center of a hexagon $S_i$ of $G$. Then
each line segment $P_iP_{i+1}$ ($0\leq i\leq q$) is orthogonal to an edge direction, and
$P_{i+1}$ ($0\leq i\leq q$) lies on the lower left side or the left side of $P_i$
according as $i$ is even or odd.
Similarly we treat substructure $G'$ of $H'$ as follows (see Figure \ref{fig:hexagonal-system-03}).
Let $f_{k,\ell}$, $1\leq k\leq s$ and $1\leq \ell\leq t(k)$, and $f_0,f_0', b,b'$
be a series of boundary edges on this structure as indicated in Figure
\ref{fig:hexagonal-system-03}.
Neither hexagon $B$ nor hexagon $B'$ is contained in $H'$.
Since the boundary of $H'$ is a proper $M$-alternating cycle,
we can see that all the edges $f_0$, $f_0'$, $f_{k,\ell}$,
$1\leq k\leq s$ and $1\leq \ell\leq t(k)$,
are $M$-double edges.
\begin{figure}[H]
\begin{center}
\includegraphics[height=4.5cm]{fig4.pdf}
\caption{{\small
Hexagonal chain $G'$ on the left-bottom corner of $H'$
(bold edges are $M$-double edges, $s=4$, t(1)=3, t(2)=1, t(3)=3, t(4)=1
and $B,B'\notin H'$) and the corresponding broken line segment $L_2$.}}
\label{fig:hexagonal-system-03}
\end{center}
\end{figure}
Like $L_1$, we also draw a ladder-shape broken line segment
$L_2=Q_0Q_1\cdots Q_{r+1}(r\geq 1)$ as indicated in Figure
\ref{fig:hexagonal-system-03} so that $L_2$ only passes through hexagons of $G'$ and each turning point $Q_i$ ($1\leq i\leq r$) is the center of a hexagon $T_i$ of $G'$.
It is obvious that both $L_i$, $i=1,2$, have an odd number of turning points.
By Claim 3, we immediately obtain the following claim.
\vskip 0.2cm
\noindent\textbf{Claim 4.} Either the broken line segments $L_1$ and $L_2$ are disjoint
or the last segment $P_qP_{q+1}$ of $L_1$ is identical to the last segment $Q_rQ_{r+1}$ of $L_2$.
\vskip 0.2cm
Since the boundary of $H'$ is a proper $M$-alternating cycle,
we have that all the edges of $H$ intersected by $L_i$, $i=1,2$, are $M$-single edges.
We now have the following claim.
\vskip 0.2cm
\noindent\textbf{Claim 5.}
(a) The boundary of $G$ (resp. $G'$) is a proper $M$-alternating cycle, and\\
(b) $n(1)=$ 1 or 2 (resp. $t(1)=$ 1 or 2), and $m\geq 2$ (resp. $s\geq 2$).
\begin{proof}
We only consider $G$ (the other case is almost the same).
Let $Z_1$ be the path induced by those vertices of $G$ which are just upon $L_1$.
Let $Z_2$ be the path induced by those vertices of $G$ which are just below $L_1$.
Since the boundary $C$ of $H'$ is a proper $M$-alternating cycle,
$Z_1$ is an $M$-alternating path with two end edges in $M$.
To prove statement (a), it suffices to show that
$Z_2$ is also an $M$-alternating path with two end edges in $M$.
Let $w_1(=e_0'), w_2, \ldots, w_{\ell_2}$ be all parallel edges of $G$ below $P_qP_{q+1}$
and let $h_1(=e_0), h_2, \ldots, h_{\ell_1}$ be all vertical edges of $G$
on the right of $P_0P_1$ (see Figure \ref{fig:hexagonal-system-02}).
Note that all the edges intersected by $L_1$ are $M$-single edges. It follows from $\{e_0, e_0'\}\subseteq M$ that
$h_1, h_2, \ldots, h_{\ell_1}$ (resp. $w_1, w_2, \ldots, w_{\ell_2}$)
are forced by $e_0$ (resp. $e_0'$) in turn to belong to $M$.
If $q=1$, $Z_2$ is an $M$-alternating path with two end edges in $M$.
Let $q\geq 3$. For each even $i$, $2\leq i\leq q-1$,
let $e_i''$ be the slant edge of $S_i$ in $Z_2$.
Let $e_i$ and $e_i'$ be the two edges of $Z_2$ adjacent to $e_i''$
(see Figure \ref{fig:hexagonal-system-02}(a)).
Clearly, $e_i$ is parallel to $e_0$, and $e_i'$ is parallel to $e_0'$.
We assert that $e_i''\notin M$. Otherwise, $e_i''\in M$, and $e_i''$ does not lie on the boundary $C$ of $H'$
since $C$ is a proper $M$-alternating cycle.
So $H'$ has a hexagon $S_i'$ containing $e_i, e_i'$ and $e_i''$.
Let $C':=C\oplus S_i$ and let $\mathcal{A}':=(\mathcal{A}\cup \{C'\})-\{C\}$ (see Figure \ref{fig:hexagonal-system-02}(b)).
Then $\mathcal{A}'$ is a compatible $M$-alternating set of $H$ satisfying conditions (i) and (ii).
But $h(\mathcal{A}')=h(\mathcal{A})-1$, contradicting the choice of $\mathcal{A}$.
So the assertion is true.
Note that all the edges intersected by $L_1$ are $M$-single edges.
We have $\{e_0, e_0', e_2, e_2', \ldots, e_{q-1}, e_{q-1}'\}\subseteq M$.
So it follows that $Z_2$ is an $M$-alternating path with two end edges in $M$
(see Figure \ref{fig:hexagonal-system-01}). Hence statement (a) holds.
\begin{figure}[H]
\begin{center}
\includegraphics[height=5cm]{fig5.pdf}
\caption{{\small
Illustration for Claim 5 in the proof of Lemma \ref{Crucial-lemma}.}}
\label{fig:hexagonal-system-02}
\end{center}
\end{figure}
Next we prove statement (b).
Suppose to the contrary that $n(1)\geq 3$. Let $S_{1,1}$ and $S_{1,2}$ be the first and second hexagons of high b-chain $G_1$ from right to left. Then $P_1$ is the center of $S_{1,1}=S_1$.
For $i=1,2$, let $g_i$ be the edge of $S_{1,i}$ parallel to $a$ and below $L_1$
(see Figure \ref{fig:hexagonal-system-01}).
By statement (a), we have $g_1, g_2\in M$.
Therefore, $S_1$ is a proper $M$-alternating hexagon, but not in $\mathcal{A}$.
By Claim 2, $g_1\notin C$. Since the boundary $C$ of $H'$ is an $M$-alternating cycle,
$g_1$ has no end-vertices in $C$. This implies that
$S_1$ has three consecutively adjacent hexagons in $H'$.
We can see that none of members of $\mathcal{A}$ except $C$ intersect $S_1$.
Let $M':=M\oplus S_1$ and $\mathcal{A'}:=(\mathcal{A}\cup \{C\oplus S_1\})-\{C\}$.
Then $M'$ is a perfect matching of $H$, and
$\mathcal{A'}$ is a compatible $M'$-alternating set satisfying conditions (i) and (ii).
But $h(\mathcal{A'})<h(\mathcal{A})$, contradicting the choice for $\mathcal{A}$.
Hence $n(1)=$ 1 or 2.
Suppose to the contrary that $m=1$.
By statement (a), we can see that $g_1\in M$ and $g_1$ is an edge of $S_1$. We have that $S_1$ is a proper $M$-alternating hexagon, but not in $\mathcal{A}$.
By analogous arguments as above, we arrive in
a similar contradiction no matter $n(1)=1$ or 2.
Hence $m\geq 2$ and statement (b) holds.
\end{proof}
Claim 5 implies that for all odd integers $i$ and $j$, $S_i$ ($1\leq i\leq q$) and $T_j$ ($1\leq j\leq r$) are
proper $M$-alternating hexagons, and the other hexagons of $G$ and $G'$ are not $M$-alternating.
For each even $i$, $2\leq i\leq q-1$, let $S_i'$ denote the hexagon (in the hexagonal lattice, but not necessarily contained in $H$)
adjacent to $S_i$ and below $L_1$ (see Figure \ref{fig:hexagonal-system-01}).
Similarly, for each even $j$, $2\leq j\leq r-1$, let $T_j'$ denote the hexagon
adjacent to $T_j$ and above $L_2$ (see Figure \ref{fig:hexagonal-system-03}).
By the above discussions to $H'$, we now go back to the discussion to $H$ and will get our result. We now get a new perfect matching $M'$ of $H$ from $M$ by rotating all $M$-alternating hexagons of $G$ and $G'$ as follows (see Figure \ref{fig:hexagonal-system-04}),
$$M':=M\oplus S_1\oplus S_3\oplus \cdots\oplus S_q\oplus T_1\oplus T_3\oplus\cdots\oplus T_r.$$
Let $\mathcal{B}$ be the set of $M'$-alternating hexagons in $G\cup G'$ and let $\mathcal{B}'=\{S_2',S_4',\dots,S_{q-1}'\}\\ \cup\{T_2',T_4',\dots,T_{r-1}'\}$. Then $\mathcal B\supseteq \{S_1, S_3, \dots, S_q, T_1, T_3,\dots,T_r\}$. We can have the following system of cycles of $H$,
$$\mathcal{A}':=( \mathcal{A}\cup \mathcal{B})\setminus (\mathcal{B}'\cup \{C\}).$$
\begin{figure}[H]
\begin{center}
\includegraphics[height=6.5cm]{fig6.pdf}
\caption{{\small
Illustration for Claim 6: The gray hexagons form $G\cup G'$, and perfect matchings $M$ and $M'$ of $H$ have restrictions on $G\cup G'$ as left and right respectively.}}
\label{fig:hexagonal-system-04}
\end{center}
\end{figure}
\vskip 0.2cm
\noindent\textbf{Claim 6.} $\mathcal A'$ is an $M'$-alternating compatible set and $|\mathcal A'|\geq |\mathcal A|+2.$
\begin{proof}Given any member $C'$ in $\mathcal A'$. Then $C'\in \mathcal A$ or $\cal B$. First we want to show that $C'$ is an $M'$-alternating cycle. If $C'$ does not intersect anyone of $S_1, S_3, \dots, S_q, T_1, T_3,\dots,T_r$, then $C'\in \mathcal A$ and $C'$ is both $M$- and $M'$-alternating cycle. If $C'=S_i$ or $T_j$ for odd $1\leq i\leq q$ and odd $1\leq j\leq r$, then $C'\in \mathcal B$ and $C'$ is both $M$- and $M'$-alternating cycle. The remaining case is that $C'$ intersects some $S_i$ or $T_j$ for odd $1\leq i\leq q$ and odd $1\leq j\leq r$, say the former $S_i$, but $C'\not=S_i$. We assert that $C'\in \mathcal B$, which implies that $C'$ is an $M'$-alternating hexagon. Suppose to the contrary that $C'\in \mathcal A\setminus \mathcal B$. Then $C'$ is an $M$-alternating cycle. If $I[C']\subset H'=I[C]$, then $C'$ is an $M$-alternating hexagon not in $G$ since each member of $\mathcal A$ lying in the interior of $C$ is a hexagon. So $C'=S'_{i-1}$ for $i\geq 3$ or $C'=S'_{i+1}$ for $i\leq q-2$, a contradiction. Otherwise, $C'$ lies outside $C$ since $C$ and $C'$ are non-crossing. Since $C$ and $C'$ are compatible $M$-alternating cycles, $C'$ passes through only either the right vertical edge $e_0$ of $S_1$ for $n(1)\geq 2$ or $e_0'$ of $S_q$ for $n(m)=1$ and $G$ and $G'$ being disjoint. In such either case, three $M$-double edges of $S_1$ or $S_q$ belong to $C$, contradicting Claim 2, so the assertion holds.
Next we show that $\mathcal A'$ is an $M'$-alternating compatible set. For the members of $\mathcal A'$ lying in the interior of $C$, they are $M'$-alternating hexagons and thus compatible. For the members $C'$ of $\mathcal A'$ lying in the exterior of $C$, $C'$ is disjoint with everyone of $S_1, S_3, \dots, S_q, T_1, T_3,\dots,T_r$. Otherwise, $C'\in \mathcal B$ by the above assertion, contradicting that $C'$ lies on the exterior of $C$. So such members $C'$ are $M$-alternating cycles in $\mathcal A$ and compatible. Suppose that $C'$ intersects some member $h$ of $\mathcal A'$ inside $C$.
By the Jordan Curve Theorem we know that $C'\cap h \subset C$. That is, each edge of $C'\cap h$ belong to $C$. Since $M$ and $M'$ have the same restriction on $C'\cap h$ and $C'$ and $C$ are compatible $M$-alternating cycles, each edge of $C'\cap h$ belong to $M$, thus to $M'$, so $C'$ and $h$ are compatible $M'$-alternating cycles.
Finally we show the remaining inequality. For each odd $i$ with $1\leq i\leq q-2$, the hexagons between $S_i$ and $S_{i+2}$ in $G$ are not $M$-alternating, but at least one and at most two of them are $M'$-alternating hexagons, which correspond to $S'_{i+1}$. Similarly for each odd $j$ with $1\leq i\leq r-2$, the hexagons between $T_i$ and $T_{i+2}$ in $G'$ are not $M$-alternating, but at least one and at most two of them are $M'$-alternating hexagons, which correspond to $T'_{i+1}$. Next we consider the end segments of hexagonal chains $G$ and $G'$. If $n(1)\geq 2$, then $S_1$ is the right end-hexagon of $G_1$, and $S_1\notin \mathcal A$ since $S_1$ and $C$ are not compatible $M$-alternating cycles. But, $S_1\in \mathcal B$. Otherwise, $G_1$ is a single hexagon other than $S_1$, the upper end segment of $G$ has a unique $M$-hexagon and two $M'$-alternating hexagons, $S_1$ and its neighbor. Similarly we have that the last row $G_m$ of $G$ has more members of $\mathcal B$ than $\mathcal A$ by at least one. In analogous arguments as above we also have that the first segment and last row of $G'$ each has more members of $\mathcal A$ than $\mathcal B$ by at least one. Note that if both last rows of $G$ and $G'$ are identical, then their extra members together count one, and $C$ is moved out $\mathcal A$. So we have that
$|\mathcal A'|\geq |\mathcal A|+2.$ \end{proof}
By Theorem 1.3 and Claim 6, we have
$Af(H)\geq af(H,M')=c'(H,M')\geq |\mathcal{A}'|\geq |\mathcal{A}|+2=|\mathcal{A}_0|+2$,
that is, $Af(H)\geq |\mathcal{A}_0|+2$.
Now the entire proof of the lemma is complete.
\hfill $\square$
\section{Minimax results for large anti-forcing numbers}
We can describe a minimax result stronger than Theorem \ref{main-thm} as follows.
\begin{thm}\label{containment-relation}
For a hexagonal system $H$, let $M$ be a perfect matching of $H$ with $af(H,M)=Af(H)$ or $Af(H)-1$,
and let $\mathcal{A}$ be a maximum non-crossing compatible $M$-alternating set of $H$.
Then (1) any two members in $\mathcal{A}$ have disjoint interiors, and
(2) for any $C\in \mathcal{A}$, $I[C]$ is a linear chain.
\end{thm}
From Statement (1) of Theorem \ref{containment-relation}, which is implied by Lemma \ref{Crucial-lemma}, we immediately obtain our main result. \\
\noindent\textbf{{Proof of Theorem} \ref{main-thm}.}
Let $\mathcal{A}$ be a maximum non-crossing compatible $M$-alternating set of $H$.
Then by Theorem \ref{Anti-forcing-feedback} and Lemma \ref{Crossing-Noncrossing},
we have $af(H,M)=|\mathcal{A}|$.
By Theorem \ref{containment-relation}(1), we know that for any two cycles in $\mathcal{A}$
their interiors are disjoint.
It was shown in \cite{Zhang-Zhang-2000} that for each $C\in \mathcal{A}$,
$H$ has an $M$-alternating hexagon $h$ in $I[C]$.
All such cycles $C$ in $\mathcal{A}$ are replaced with $M$-alternating hexagons $h$ of $I[C]$
to get a set $K$ of $M$-alternating hexagons with $|K|=|\mathcal{A}|$.
So we have $fr(H,M)\geq |K|=af(H,M)$.
On the other hand, $af(H,M)\geq fr(H,M)$ since $K$ is also a compatible $M$-alternating set. Both inequalities imply the result.
\hfill $\square$
\vskip 0.2cm
In order to prove Theorem \ref{containment-relation}(2), the characterization for hexagonal systems $H$ with $af(H) = 1$ due to Li \cite{Li-1997} are presented here.
It is clear that $H$ has an anti-forcing edge if and only if $af(H) = 1$.
\begin{figure}[H]
\begin{center}
\includegraphics[height=3cm]{fig7.pdf}
\caption{{\small
Truncated parallelograms $H(6,6,5,4)$ and $H(6,6,6,6)$: anti-forcing edges are marked.}}
\label{fig:hexagonal-system-10}
\end{center}
\end{figure}
For integers $n_1 \geq n_2 \geq \ldots \geq n_k$, $k\geq 1$, let $H(n_1, n_2, \ldots, n_k)$
be a hexagonal system with $k$ horizontal rows of $n_1 \geq n_2 \geq \ldots \geq n_k$ hexagons and first
hexagon of each row being immediately below and to the right of the first one in the previous row, and
we call it \emph{truncated parallelogram} \cite{Cyvin-book-1988};
For example, see Figure \ref{fig:hexagonal-system-10}.
In particular, $H(r, r, \ldots, r)$ with $k \geq 2$ is parallelogram,
both $H(1, 1, \ldots, 1)$ with $k \geq 1$ and $H(r)$ with $r \geq 1$ are linear chains.
\begin{thm}\label{anti-forcing-edge}
\cite{Li-1997} Let $H$ be a hexagonal system. Then
$af(H)=1$ if and only if $H$ is a truncated parallelogram.
\end{thm}
\vskip 0.2cm
\noindent\textbf{{Proof of Theorem} \ref{containment-relation}.}
(1) By Theorem \ref{Anti-forcing-feedback} and Lemma \ref{Crossing-Noncrossing}, we have $af(H,M)=|\mathcal{A}|$.
Suppose to the contrary that statement (1) does not hold. Then by Lemma \ref{Crucial-lemma}
we have $Af(H)\geq |\mathcal{A}|+2=af(H,M)+2\geq Af(H)+1$, a contradiction.
So statement (1) holds.
(2) Let $n:=af(H,M)=|\mathcal{A}|$ and let $\mathcal{A}=:\{C_1,C_2,\ldots,C_n\}$.
Choose an anti-forcing set $S$ of $M$ with $|S|=n$. Let $S_i:=S\cap E(C_i)$, $i=1,2,\ldots,n$.
By Lemma \ref{Anti-forcing-cycle}
we have $S_i\neq \emptyset$ for each $i$. Since $\mathcal{A}$ is a compatible $M$-alternating set,
$S_i\cap S_j=\emptyset$ for any $1\leq i< j\leq n$. So we can assume that $S:=\{e_1,e_2,\ldots, e_n\}$ with $e_i\in E(C_i)$ for all $1\leq i\leq n$.
For any $1\leq i \leq n$, since $C_i$ is an $M$-alternating cycle of $H$,
the restriction $M_i$ of $M$ on $I[C_i]$ is a perfect matching of $I[C_i]$.
By Theorem \ref{containment-relation}(1), we can see that only edge $e_i$ of $S$ lies in $I[C_i]$,
$1\leq i \leq n$.
So all $M_i$-alternating cycles in $I[C_i]$ pass through edge $e_i$, $1\leq i \leq n$.
By Lemma \ref{Anti-forcing-cycle}, we have that
$\{e_i\}$ is an anti-forcing set of $M_i$, $1\leq i \leq n$.
That is, $e_i$ is an anti-forcing edge of $I[C_i]$. By Theorem \ref{anti-forcing-edge}, each $I[C_i]$ is a truncated parallelogram.
If some $I[C_i]$
is not a linear chain, i.e. $I[C_i]$ has at least two rows and at least two columns of hexagons,
then $I[C_i]$ has a unique perfect matching $M_i$ not containing edge $e_i$
(see Figure \ref{fig:hexagonal-system-10} (left)). Let $h_1$ be the hexagon of $I[C_i]$ with edge $e_i$, and let $h_2$ and $h_3$ be the two hexagons of $I[C]$ adjacent to $h_1$ in the first column and the first row respectively.
It was pointed out in \cite{Li-1997} that
$h_1$ is only $M_1$-alternating hexagon in $I[C_1]$.
So $M':=M\oplus h_1$ is a perfect matching of $H$, and $h_1,h_2$ and $h_3$ are $M'$-alternating hexagons. We can see that $\mathcal{A}':=\mathcal{A}\cup \{h_1, h_2, h_3\}-\{C\}$ is a compatible $M'$-alternating set of $H$.
Hence
$Af(H)\geq af(H,M')\geq |\mathcal{A}'|= |\mathcal{A}|+2\geq Af(H)+1$,
a contradiction. Hence each $I[C_i]$ is a linear chain and statement (2) holds.
\hfill$\Box$
\section{Minimax result for all perfect matchings}
It is natural to ask whether Theorem \ref{main-thm} holds for all perfect matchings $M$ of a hexagonal system $H$. A counterexample can show that the minimax relation does not necessarily hold for a perfect matching of a hexagonal system $H$
with the third maximum anti-forcing number $Af(H)-2$.
\begin{figure}[H]
\begin{center}
\includegraphics[height=2.5cm]{fig8.pdf}
\caption{{\small
A hexagonal system $R_n$ with a perfect matching $M$.}}
\label{fig:hexagonal-system-11}
\end{center}
\end{figure}
Let $R_n$ be a hexagonal system with $2n+4$ hexagons and a perfect matching $M$ as shown in Figure
\ref{fig:hexagonal-system-11} (the edges in $M$ are indicated by double edges). Then $R_n$ contains one triphenylene whose central hexagon is denoted $h$. Let $M'=M\oplus h$. Then all hexagons of $R_n$ are $M'$-alternating. So
the Fries number of $R_n$ is the number of hexagons in $R_n$ (see also \cite{Harary-1991}).
By Theorem \ref{HS-F(H)-Fries(H)}, we have $Af(R_n)=Af(R_n,M')=Fr(R_n)=2n+4$.
However we can confirm that $af(R_n, M)=Af(H)-2>fr(H,M)$. By counting $M$-alternating hexagons in $R_n$, we have that $fr(H,M)=2n+1$. On the other hand, we can find a compatible $M$-alternating set of size $2n+2$. So $af(R_n, M)\geq 2n+2> fr(H,M)$.
By a direct check or Theorem \ref{main-thm}, we have $af(R_n, M)\leq Af(R_n)-2$.
So $af(R_n, M)=Af(H)-2$.
We can see that the above counterexample contains a triphenylene as nice subgraph. In fact we can give a characterization for a hexagonal system $H$ to have the mini-max relation $af(H,M)=fr(H,M)$ by forbidding triphenylene as nice subgraph (see Theorem \ref{Triphenylene}).
To the end we present some concepts and known results. Let $H$ be a hexagonal system with a perfect matching. Let $r(H)$ and $k(H)$ be the numbers of sextet patterns and Kekul\'e structures of $H$ respectively.
\begin{thm}[\hskip -0.02mm\cite{Zhang-Chen-1986, Shiu-2002}]\label{Coronoid-nice-subgraph}
Let $H$ be a hexagonal system with a perfect matching. Then $r(H)\leq k(H)$, and the following statements are equivalent.\\
(i) $r(H)= k(H)$,\\
(ii) $H$ has a coronene as a nice subgraph, and\\
(iii) $H$ has two disjoint cycles $R$ and $C$ so that $R$ lies in the interior of $C$ and $R\cup C$ is a nice subgraph of $G$.
\end{thm}
\begin{figure}[H]
\begin{center}
\includegraphics[height=5.5cm]{fig9.pdf}
\caption{A hexagonal system with a unique $M$-alternating hexagon for a perfect matching $M$ marked by double lines.}
\label{Chapter-5-12}
\end{center}
\end{figure}
Let $H$ be a hexagonal system with a hexagon $h$. Draw three rays $OA, OB$ and $OC$ from the center $O$ of $h$ so that they pass through the centers of three disjoint edges of $h$ respectively, which divide the plane into three areas $AOB, BOC$ and $COA$. Such three regional coordinate system is denoted by $O-ABC$.¡¡ Zhang et al. \cite{Zhang-Guo-Chen-1988-2} ever gave the following fact.
\begin{lem}\label{Chapter-fact}
A hexagonal system $H$ has a perfect matching $M$ with a unique $M$-alternating hexagon $h$ if and only if it has the coordinate system $O-ABC$ so that $O$¡¡ is the center of $h$,¡¡rays $OA, OB$ and $OC$ do not intersect edges in $M$ and all edges of $M$ in anyone of areas $AOB, BOC$ and $COA$ are parallel to each other.
\end{lem}
\begin{thm}\label{Triphenylene}
Let $H$ be a hexagonal system with a perfect matching. Then $H$ has no triphenylenes as nice subgraph if and only if for each perfect matching $M$ of $H$, $af(H,M)=fr(H,M)$.
\end{thm}
\begin{proof}
We first prove the sufficiency. Suppose to the contrary that $H $ contains a triphenylene as a nice subgraph. Let $M'$ be a perfect matching of the triphenylene as shown in Figure \ref{fig:counterexample-1}(b). Because the triphenylene is a nice subgraph of $H$, $M'$ can be extended to a perfect matching $M $ of $H $. So $M'\subseteq M $. Let $C$ be the boundary of the triphenylene. It is easy to see that $C $ is an $M$-alternating cycle of $H$ and $C$ is compatible with each $M$-alternating hexagon of $H$. By Theorem \ref{Anti-forcing-feedback}, we have $af(H,M)\geq fr(H,M)+1>fr(H,M)$, a contradiction. Hence the sufficiency holds.
We now prove the necessity. Suppose to the contrary that $H $ has a perfect matching $M_0 $ so that $af(H, M_0) > fries (H, M_0)$. Let $\mathcal {A} $ be a maximum non-crossing compatible $M_0 $-alternating set of $H$. By Theorem \ref{Anti-forcing-feedback} and Lemma \ref{Crossing-Noncrossing} we have $af(H, M_0) = |\mathcal {A}|$ and there are two cycles in $\mathcal {A}$ so that their interiors have a containment relation; Otherwise, since there is an $M_0$-alternating hexagon in the interior of each $M_0$-alternating cycle \cite{Zhang-Zhang-2000}, $H$ has at least $af(H, M_0)$ $M _ 0 $-alternating hexagons, that is, $af(H, M_0)\leq fr(H, M_0)$, a contradiction. So we can select two cycles $C_1$ and $C_2$ in $H$ to meet the following conditions:\\
\indent (i) $I[C_1] \subseteq I[C_2] $, \\
\indent(ii) $H$ has a perfect matching $M$ so that $C_1$ and $C_2$ are compatible $M $-alternating cycles, and\\
\indent (iii) $h(C_1) + h(C_2)$ is as small as possible subject to Conditions (i) and (ii) (recall that $h(C_i) $ is the number of hexagons inside $C_i$).
If $I[C_2]$ has an $M$-alternating hexagon $h$ which is disjoint with $C_2$, then
$h\cup C_2$ is a nice subgraph of $H$. By Theorem \ref{Coronoid-nice-subgraph}, $H$ contains a coronene as a nice subgraph. So $H$ also contains triphenylene as a nice subgraph.
From now on suppose that all the $M $-alternating hexagons in $I[C_2]$ intersect $C_2$.
Obviously, if $C_2$ is a proper (resp. improper) $M$-alternating cycle, then each of the $M$-alternating hexagons in $I[C_2]$ is also proper (resp. improper). Without loss of generality, suppose that the $C_2$ and $M$-alternating hexagons in $I[C_2]$ are proper. Take an $M $-alternating hexagon $h$ inside $C_1 $. Since $C_1$ and $C_2$ are compatible $M $-alternating, $h$ is compatible with $C_2$. We can show the following fact.
\vskip 2mm
\noindent\textbf {Claim.} $h$ is the only $M$-alternating hexagon in $I[C_2]$.
\begin{proof}Suppose to the contrary that $h'$ is an $M$-alternating hexagon of $I[C_2]$ different from $h$. Then $h'$ and $h$ are disjoint because any two proper $M $-alternating hexagons do not intersect. Let $M': = M\oplus h'$ and $C'_2: =C_2\oplus h'$. Then $M'$ is a perfect matching of $H$, and each component of $C_2\oplus h'$ is an $M'$-alternating cycle. Take a component $C'_2$ of $C_2\oplus h'$ so that $h$ lies inside $C'_2$. Then $h$ and $C_2'$ are also compatible $M'$-alternating cycles. It is clear that the cycles $C'_2 $ and $h$ satisfy the above conditions (i) and (ii). But $h(C'_2)+1<h(C_2)+h(C_1)$, a contradiction with the minimality of $h(C_2)+h(C_1)$.\end{proof}
Note that the restriction of $M$ on $I[C_2]$ is a perfect matching of $I[C_2]$. From the Claim and Lemma \ref{Chapter-fact}, from center $O $ of $h$ we establish coordinate system $O-ABC$ so that $OA, OB$ and $OC$ do not pass through $M$-edges in $I[C_2]$, and all $M$-double edges of $I[C_2]$ in anyone of areas $AOB, BOC$ and $COA$ are parallel to each other (see Figure \ref {Chapter-5-12}). Because $h$ and $C_2$ are two compatible $M $-alternating cycles, every $M $-single edge of $h $ is not on $C_2$. This shows that $h$ has three adjacent hexagons $h_1, h_2$, and $h_3$ in $I[C_2]$ which intersect $OA, OB$, and $OC $ respectively. Further, hexagons $h, h_1, h_2 $ and $h_3$ form a triphenylene whose boundary is $M$-alternating cycle. So the triphenylene is a nice subgraph of $H$, a contradiction. So $af(H, M_0) =fr(H, M_0)$ and the necessity holds.
\end{proof} | 154,357 |
Valentine’s Day Gifts for Her: Ideas for the Female Traveler
I’ve never been that big on Valentine’s Day. I like the general sentiment of the holiday, but like many people, I get a little repulsed by the over-commercialization of the holiday. Plus, I’m just not that into pink, cheesiness or heart-shaped boxes of chocolate.
That being said, I’m all about showing your significant other your love and affection and I’m always all about travel gear as gifts. Best thing is great travel gifts don’t have to break the bank. They are many affordable items that would make a perfect Valentine’s gift for your lovely travel loving lady and plenty of pricier items if that is your think.
So, for all the outdoorsy, travel-loving ladies in your life, here are some great gifts for the female traveler in your life.
1. Flip and Tumble Bags
I’m a big fan of using a lightweight, reusable bag when I travel—and the Flip and Tumble bags have quickly becoming my favorite. These bags are made from durable ripstop Nylon (similar to parachute material) and fold into a ball with just a 3 inch diameter (smaller than a peach). Plus, they come in a host of very cute bright colors and prints.
Flip & Tumble bags also fold up into a delightfully small little pouch that’s super easy to stuff and makes for ideal storage. These are the perfect travel and daily life companion.
>>Read more about and buy a set of Flip & Tumble bags
2. Travel Dress or Skirt
I know the guys out there may be a little hesitant about buying their ladies clothing for Valentine’s Day, but getting your travel loving gal a great, versatile travel dress or skirt is a great way to go. Patagonia’s Morning Glort Dress will not only look great on many body types, but won’t wrinkle when she packs it.
>>Patagonia Morning Glory Dress or Isis Melbourne Travel Skirt
3. A Personalized Rolling Suitcase
If she’s the rolling suitcase kind of gal, then buying her a new, great carry-on sized rolling suitcase for Valentine’s Day is sure to put a smile on her face. If you really want to go the extra mile, you can get a personalized embroidery to go on the bag.
>>Read more about and buy a Swiss Army personalized suitcase
4. Backpack
If your lady is the rough, outdoorsy backpacker style traveler, than getting her a new women’s specific backpack is a great Valentine’s Day gift. The Deuter Futura Pro is a great option for petite women and the Gregory Jade is a fabulous all-around pack.
>>Read more about women’s backpacks
5. Netbook Case
If your girlfriend or wife has a tiny little traveling netbook, then getting her a sleek, stylish case to keep it in makes for a great, useful Valentine’s Day gift. There are many other great Netbook accessories out there too.
>>Read about and buy more of the Best Netbook Accessories and Best iPad Accessories
6. Kindle
If your special lady is a book worm and a traveler, investing in a portable, electronic reader makes for the perfect gift. This way, you can take dozens of books with you on the road, and it won’t take up any more room than the slender little Kindle.
>>Read more about and buy a Kindle 3G
7. Yoga Travel Bag
Whether your gal is a globe-trotting yoga retreat fanatic or just a wii yoga junkie, there is a yoga bag designed totally for her. Yoga bags are styling with plenty of room to stash your yoga pants, mat, and other yoga accessories.
>>Read more about a buy Yoga Bags and Accessories
8. Cozy Fleece or Down Jacket
My heart could easily be won over with a cozy Patagonia Fleece or a down puffy coat. Nothing says love better than a lightweight, warm, super comfy jacket!
>>Read more about and buy fleece jackets and down puffy coats
9. Travel Bag
I really hope my boyfriend reads this post because I would love one of these cutey Overland Equipment Bayliss Shoulder Bags. With plenty of compartments and organizer pockets and pouches, the Bayliss would be great every day bag for the urban traveler. There are plenty of slots for your credit cards and pockets for your passport or checkbook, a fleece-lined pocket for your cell phone or sunglasses, and an external organizer for all your small goods plus it is cute as can be.
Ladies, what is your dream Valentine gift? | 105,498 |
8 comments:
Laurie's Thoughts and Reviews would be delighted to host Sept 18th
Up 'Til Dawn Book Blog - September 28th
Buried Under Romance
Sept 19 with review
We have the book.
Independent Authors is available any date.
I can host September 20
I can do Sept. 21st.
I can host on September 25th.
Words of Wisdom from The Scarf Princess
I would like to host on September 26 @The Avid Reader | 67,838 |
\begin{document}
\title[Integral Representation of $ax^2+by^2$]{On the Integral Representation of $ax^2+by^2$ and the Artin Condition}
\author[C. Lv]{Chang Lv}
\address{Key Laboratory of Mathematics Mechanization\\
NCMIS, Academy of Mathematics and Systems Science\\
Chinese Academy of Sciences, Beijing 100190, P.R. China}
\email{[email protected]}
\author[J. Shentu]{Junchao Shentu}
\address{Institute of Mathematics\\
Academy of Mathematics and Systems Science\\
Chinese Academy of Sciences, Beijing 100190, P.R. China}
\email{[email protected]}
\subjclass[2000]{Primary 11D09, 11E12, 11D57; Secondary 11D57, 14L30, 11R37}
\keywords{integral points, ring class field}
\date{\today}
\begin{abstract}
Given a number field $F$ with $\o_F$ its ring of integers. For certain $a,b$ and $\alpha$ in $\o_F$, we show that
the Artin condition is the only obstruction to the local-global principle for integral solutions of equation $ax^2+by^2=\alpha$.
Some concrete examples are presented at last.
\end{abstract}
\maketitle
\section{Introduction}\label{sec_intro}
The main theorem of David A. Cox \cite{cox} is a beautiful criterion of the solvability of the diophantine equation $p=x^2+ny^2$. The specific statement is
\thmus{
Let $n$ be an positive integer. Then there is a monic irreducible polynomial $f_n(x)\in\ZZ[x]$ of degree $h(-4n)$ such that if an odd prime $p$ divides neither $n$ nor the discriminant of $f_n(x)$, then $p=x^2+ny^2$ is solvable over $\ZZ$ if and only if $\fracl{-n}{p}{}=1$ and $f_n(x)=0$ is solvable over $\ZZ/p\ZZ$. Here $h(-4n)$ is the class number of primitive positive definite binary forms of discriminant $-4n$. Furthermore, $f_n(x)$ may be taken to be the minimal polynomial of a real algebraic integer $\alpha$ for which $L=K(\alpha)$ is the ring class field of the order $\ZZ[\sqrt{-n}]$ in the imaginary quadratic field $K=\QQ(\sqrt{-n})$.
}
There are some generalizations considering the problem over quadratic fields.
Recently, Harari \cite{bmob} showed that the Brauer-Manin obstruction is the only obstruction for the existence of integral points of a scheme over the ring of integers of a number field, whose generic fiber is a principal homogeneous space of a torus.
After then Dasheng Wei and Fei Xu gave another proof of the result in \cite{multi-norm-tori,multip-type} where the Brouer-Manin obstruction is constructive. This can be used to determine the existence of integral points for the scheme.
Then Dasheng Wei applied the method in \cite{multi-norm-tori} to determine which integers can be written as a sum of two integral squares for some of the quadratic fields $\QQ(\sqrt{\pm p})$ (in \cite{wei1}), $\QQ(\sqrt{-2p})$ (in \cite{wei2}) and so on. That is to determine which integers have integral representations of the form $x^2+y^2$.
In this paper, we consider integral representations of the form $ax^2+by^2$ in some other number fields. Although we are interested in certain specific number fields, our formal criterion is for general number fields.
Given a number field $F$ with $\o_F$ its ring of integers. Suppose $a,b$ and $\alpha$ are in $\o_F$ such that $d=-ab$ is not a square in $F$. Let $\XX=\mathbf{Spec}(\o_F[x,y]/(ax^2+by^2-\alpha))$. The general result is:
\thmus{[Theorem \ref{thm_artin_cond}]
Suppose
one of the following conditions holds:
\enmt{
\it[$(1)$] For every $u$ in $ \o_F^\t$, the equation $x^2-dy^2=u$ is solvable in $\o_F$.
\it[$(2)$] For every $u\neq1$ in $\o_F^\t$ there exists a place $\p$ such that the equation $x^2-dy^2=u$ is not solvable in $F_\p$.
}
Then $\XX(\o_F)\neq\emptyset$ if and only if there exists
\eqn{
\prod_{\p\in\Omega_F}(x_\p, y_\p)\in\prod_{\p\in\Omega_F}\XX(\o_{F_\p})
}
such that
\eq{
\psi_{H_L/E}(\tilde f_E(\prod_\p(x_\p,y_\p)))=1.
}}
For notations one can see Section \ref{sec_nota} and for proofs Section \ref{sec_main}.
Due to Wei \cite{wei1,wei2}, the last condition in the theorem is called \emph{Artin condition}.
The theorem shows that if $(1)$ or $(2)$ holds, the Artin condition is the only obstruction to the local-global principle for integral solutions of equation $ax^2+by^2=\alpha$.
In Section \ref{sec_appl}, we set our focus on some applications of the this theorem.
For a special case ($H_L/F$ is abelian, see Section \ref{sec_abelian}), the Artin condition always holds. Hence the Hasse principal is true.
For the case $F=\QQ$ (Section \ref{sec_q}) and the case $F=\QQ(\sqrt{-p})$ with $a=1,b=q$
where $p$ and $q$ are rational primes satisfying some conditions
(Section \ref{sec_pq}), condition $(1)$ holds and then we prove that
the integral local condition in together with the Artin condition completely describe the global integral solvability.
We also give some examples to show the explicit criteria of the solvability.
\section{Solvability by the Artin Condition}\label{sec_artin_cond}
\subsection{Notations}\label{sec_nota}
We fix some notations. Let $F$ be a number field, $\o_F$ the ring of integers of $F$,
$\Omega_F$ the set of all places in $F$ and $\infty_F$ the set of all infinite places in $F$. We write $\p<\infty_F$ for $\p\in\Omega_F\setminus\infty_F$.
Let $F_\p$ be the completion of $F$ at $\p$ and $\o_{F_\p}$ be the valuation ring of $F_\p$ for each $\p<\infty_F$. We also write $\o_{F_\p}=F_\p$ for $\p\in\infty_F$.
The adele ring (resp. idele group) of $F$ is denoted as $\AA_F$ (resp. $\II_F$).
Let $a,b$ and $\alpha$ be nonzero elements in $\o_F$ and suppose that $d=-ab$ is not a square in $F$. Let $E=F(\sqrt d)$ and $\XX=\mathbf{Spec}(\o_F[x,y]/(ax^2+by^2-\alpha))$ be the affine scheme defined by the equation $ax^2+by^2=\alpha$ over $\o_F$.
The equation $ax^2+by^2=\alpha$ is solvable over $\o_F$ if and only if $\XX(\o_F)\neq\emptyset$.
Denote $R_{E/F}(\GG_m)$ the Weil restriction (see \cite{jmilne-ag}) of $\GG_{m,E}$ to $F$. Let
\eqn{ \varphi: R_{E/F}(\GG_m)\lr\GG_m }
be the homomorphism of algebraic groups which represents
\eqn{ x\mapsto N_{E/F}(x): (E\otimes_FA)^\t\lr A^\t}
for any $F$-algebra $A$. Define the torus $\TT:=\ker\varphi$.
Let $\XX_F$ be the generic fiber of $\XX$.
Then $\XX_F$ is naturally a $\TT$-torsor by the action of $A$-points:
\aln{
\TT(A)\t \XX_F(A) &\lr \XX_F(A)\\
(u+\sqrt{d}v,x+\frac{\sqrt{d}}{a}y) &\mapsto (u+\sqrt{d}v)(x+\frac{\sqrt{d}}{a}y)
}
It can be seen that
\eq{\label{eq_in_stab}
\TT(\o_{F_\p})\subseteq \mathbf{Stab}(\XX(\o_{F_\p})):=\set{g\in \TT(F_\p) | g\XX(\o_{F_\p})=\XX(\o_{F_\p})}.
}
Denote by $\lambda$ the embedding of $\TT$ into $R_{E/F}(\GG_m)$.
Tautologically, $\lambda$ induces a natural injective group homomorphism
\eqn{ \lambda_E: \TT(\AA_F)\lr\II_E. }
Let $L=\o_F+\o_F\sqrt{d}$ in $E$ and $L_\p=L\otimes_{\o_F}\o_{F_\p}$ then
\eqn{ \TT(\o_{F_\p})=\set{\beta\in L_\p^\t|N_{E_\p/F_\p}(\beta)=1}. }
It follows that $\lambda_E(\TT(\o_{F_\p})) \subseteq L_\p^\t.$
Note that $\lambda_E( \TT(F)) \subseteq E^\t$ in $\II_E$. Hence the following map induced by $\lambda_E$ is well-defined:
\eqn{
\tilde\lambda_E: \TT(\AA_F)/\TT(F)\TT(\prod_{\p\in \Omega_F} \o_{F_\p})\lr \II_E/E^\t \prod_{\p\in\Omega_F} L_\p^\t.
}
\lemm{\label{lambda_inj}
The map $\tilde\lambda_E$ is injective if one of the following conditions holds:
\enmt{
\it[$(1)$] For every $u$ in $ \o_F^\t$, the equation $x^2-dy^2=u$ is solvable in $\o_F$.
\it[$(2)$] For every $u\neq1$ in $\o_F^\t$ there exists a place $\p$ such that the equation $x^2-dy^2=u$ is not solvable in $F_\p$.
}}
\pf{
Recall that $\TT=\ker(R_{E/F}(\GG_m)\lr\GG_m)$. Therefore we have
\eqn{ \TT(F)=\set{\beta\in E^\t|N_{E/F}(\beta)=1} }
and
\eqn{ \TT(\o_{F_\p})=\set{\beta\in L_\p^\t|N_{E_\p/F_\p}(\beta)=1}. }
Suppose $t\in \TT(\AA_F)$ such that $\tilde\lambda_E(t)=1$. Write $t=\beta i$ with $\beta\in E^\t$ and $i\in \prod_\p L_\p^\t$.
Since $t\in \TT(\AA_F)$ we have
\eqn{ N_{E/F}(\beta)N_{E/F}(i)=N_{E/F}(\beta i)=1. }
It follows that
\eqn{ N_{E/F}(\beta)=N_{E/F}(i^{-1})\in F^\t\cap \prod_\p\o_{F_\p}^\t=\o_F^\t. }
If $N_{E/F}(\beta)=N_{E/F}(i)=1$ then we have $\beta\in \TT(F)$ and $i\in\prod_\p\TT(\o_{F_\p})$. Hence $u=\beta i\in \TT(F)\prod_\p\TT(\o_{F_\p})$.
Otherwise $N_{E/F}(\beta)=u$ for some $u\neq1$ in $\o_F^\t$. Then we know that
the equation $x^2-dy^2=u$ is solvable in $F_\p$ for every place $\p$ of $F$.
This is to say condition $(2)$ does not holds and therefore condition $(1)$ holds.
It follows that $x^2-dy^2=u$ is solvable in $\o_F$.
Let $(x_0,y_0)\in \o_F^2$ be such a solution and let
\eqn{ \zeta=x_0+y_0\sqrt{-1},\gamma=\beta\zeta^{-1}\text{ and }j=i\zeta. }
Then $N_{E/F}(\gamma)=N_{E/F}(j)=1$. Hence $\gamma\in \TT(F)$ and $j\in\prod_\p\TT(\o_{F_\p})$. It follows that $u=\beta i=\gamma j\in \TT(F)\prod_\p\TT(\o_{F_\p})$.
This finishes the proof.
}
Now we assume that
\eq{\label{eq_nonempty}
\XX(F)\neq\emptyset.
}
Fixing a rational point $P\in \XX(F)$, since $\XX_F$ is a trivial $\TT$ torsor, we have an isomorphism
\aln{
\phi_P: \prod_{\p\in\Omega_F}\XX(\o_{F_\p}) &\cong \prod_{\p\in\Omega_F}\TT(\o_{F_\p})\\
x &\mapsto P^{-1}x
}
induced by $P$.
Since we can view $\XX(\o_{F_\p})$ as an subset of $\XX(F_\p)$, the composition $f_E:=\lambda_E\phi_P: \prod_\p\XX(\o_{F_\p})\lr \II_E$ makes sense, mapping
$x$ to $P^{-1}x$ in $\II_E$. Note that $P$ is in $E^\t\subset \II_E$ since it is a rational point over $F$. It follows that we can
define the map $\tilde f_E$ to be the composition
\eqn{\xymatrix{
\prod_\p\XX(\o_{F_\p}) \ar@{->}[r]^-{ f_E} &\II_E \ar@{->}[r]^-{\t P} &\II_E\\
x \ar@{|->}[r] &P^{-1}x \ar@{|->}[r] &x.
}}
It can be seen that the restriction to $\XX(\o_{F_\p})$ of $\tilde f_E$ is defined by
\eq{\label{eq_tilde_f_E}
\tilde f_E[(x_\p,y_\p)]=
\cs{
(ax_\p+y_\p\sqrt{d},ax_\p-y_\p\sqrt{d})\in E_{\P_1}\otimes E_{\P_2} &\text{if }\p\text{ splits in }E/F,\\
ax_\p+y_\p\sqrt{d} \in E_\P &\text{otherwise},
}}
where $\P_1$ and $\P_2$ (resp. $\P$) are places of $E$ above $\p$.
Recall that $L=\o_F+\o_F\sqrt{d}$ in $E$ and $L_\p=L\otimes_{\o_F}\o_{F_\p}$. Then $\Xi_L:=E^\t\prod_\p L_\p^\t$ is an open subgroup of $\II_E$. By the ring class field corresponding to $L$ we mean the class field $H_L$ corresponding to $\Xi_L$ under the class field theory.
Let $\psi_{H_L/E}: \II_E\lr\Gal(H_L/E)$ be the Artin map.
For any $\prod_{\p\in\Omega_F}(x_\p, y_\p)\in\prod_{\p\in\Omega_F}\XX(\o_{F_\p})$, noting that $P$ is in $E$, we have
$\psi_{H_L/E}(f_E(\prod_\p(x_\p,y_\p)))=1$ if and only if
$\psi_{H_L/E}(\tilde f_E(\prod_\p(x_\p,y_\p)))=1$.
\rk{\label{rk_hasse_min}
If $\prod_{\p\in\Omega_F}\XX(\o_{F_\p})\neq\emptyset$, then assumption \eqref{eq_nonempty} we made before, that is $\XX(F)\neq\emptyset$, holds automatically by Hasse-Minkowski theorem on quadratic equation. Hence We can pick a $F$-point $P$ of $\XX$ and define $\phi_P$.
}
\subsection{The General Theorem}\label{sec_main}
The following is the general theorem we mainly use in this paper, which is motivated by \cite[Corollary 1.4 and 1.6]{multi-norm-tori}.
\thm{\label{thm_artin_cond}
Let symbols be as before. Suppose
one of the following conditions holds:
\enmt{
\it[$(1)$] For every $u$ in $ \o_F^\t$, the equation $x^2-dy^2=u$ is solvable in $\o_F$.
\it[$(2)$] For every $u\neq1$ in $\o_F^\t$ there exists a place $\p$ such that the equation $x^2-dy^2=u$ is not solvable in $F_\p$.
}
Then $\XX(\o_F)\neq\emptyset$ if and only if there exists
\eqn{
\prod_{\p\in\Omega_F}(x_\p, y_\p)\in\prod_{\p\in\Omega_F}\XX(\o_{F_\p})
}
such that
\eq{\label{eq_artin_cond}
\psi_{H_L/E}(\tilde f_E(\prod_\p(x_\p,y_\p)))=1.
}}
\pf{
If $\XX(\o_F)\neq\emptyset$, then
\eqn{
\tilde f_E\left(\prod_\p\XX(\o_{F_\p})\right)\cap E^\t\prod L_\p^\t\supseteq \tilde f_E(\XX(\o_F))\cap E^\t\neq\emptyset
}
Hence there exists $x\in \prod_{\p\in\Omega_F}\XX(\o_{F_\p})$ such that $\psi_{H_L/E}\tilde f_E(x)=1$.
Conversely,
we know from Lemma \ref{lambda_inj} that
\eqn{
\tilde \lambda_E: \TT(\AA_F)/\TT(F)\TT(\prod_\p \o_{F_\p})\lr \II_E/E^{\t}\prod L_\p^\t
}
is injective.
Suppose there exists $x\in \prod_\p\XX(\o_{F_\p})$ such that $\psi_{H_L/E}\tilde f_E(x)=1$
(here $\tilde f_E$ makes sense by Remark \ref{rk_hasse_min} in Section \ref{sec_intro}) , i.e.
$f_E(x)\in E^{\t}\prod L_\p^\t$.
Since $\tilde \lambda_E$ is injective, there are $\tau\in \TT(F)$ and $\sigma\in\TT(\prod_\p \o_{F_\p})$ such that $\tau\sigma=f_E(x)=P^{-1}x$,
i.e. $\tau\sigma(P)=x$. Since $P\in\XX(F)$ and $\TT(\o_{F_\p})\subseteq \mathbf{Stab}(\XX(\o_{F_\p}))$ (see \eqref{eq_in_stab}), it follows that
\eqn{
\tau(P)=\sigma^{-1}(x)\in \XX(F)\cap \prod_\p\XX(\o_{F_\p})=\XX(\o_F).
}
The proof is complete.
}
The condition \eqref{eq_artin_cond} is called \emph{Artin condition} in Wei's \cite{wei1,wei2}. It interprets the fact that the Brauer-Manin obstruction is the only obstruction for existence of the integral points by conditions in terms of class field theory.
Consequently, the integral local condition in together with the Artin condition completely describe the global integral solvability.
As a result, in the cases where the ring class fields are known it is possible to compute the Artin condition, giving a explicit criteria for the solvability.
\section{Applications}\label{sec_appl}
We now apply the result stated in Theorem \ref{thm_artin_cond} to proof the following corollaries.
\subsection{On the Cases $H_L/F$ Abelian}\label{sec_abelian}
Recall that $H_L/F$ is the extension of fields corresponding to $E^\t\prod_\p L_\p^\t$ under the class field theory, where $L=\o_F+\o_F\sqrt{d}$ in $E$ and $L_\p=L\otimes_{\o_F}\o_{F_\p}$.
In the cases that $H_L/F$ is abelian, the Artin condition always holds. Hence the Hasse principal is true.
\thm{\label{thm_trivial_artin_cond}
Suppose $H_L/F$ is abelian, then
$\XX(\o_F)\neq\emptyset$ if and only if $\prod_{\p\in\Omega_F}\XX(\o_{F_\p})\neq \emptyset$.
}
\pf{
Since $H_L/F$ is abelian, the following diagram commutes:
\eqn{\xymatrix{
\II_E \ar@{->}[r]^-{\psi_{H_L/E}} \ar@{->}[d]_{N_{E/F}} &\Gal(H_L/E) \ar@{_(->}[d]\\
\II_F \ar@{->}[r]^-{\psi_{H_L/F}} &\Gal(H_L/F)
}}
It follows that
\eqn{
\psi_{H_L/E}(\tilde f_E(\prod_\p(x_\p,y_\p)))
= \psi_{H_L/F}(N_{E/F}( \tilde f_E(\prod_\p(x_\p,y_\p))))
= \psi_{H_L/F}(\alpha) = 1
}
where the second equation comes from the definition \eqref{eq_tilde_f_E} of $\tilde f_E$ and
\eqn{
(x_\p+ \sqrt d y_\p)(x_\p- \sqrt d y_\p) = \alpha\text{ in }E_\P\text{ with }\P\mid \p,
}
while the last equation is obtained by the assumption that $\alpha\in F$.
Hence the Artin condition always holds and the result follows.
}
\eg{
Let $F=\QQ(\sqrt{-59})$ and $\alpha\in\o_F$. Then the equation $x^2+2y^2=\alpha$ is solvable over $\o_F$ if and only if
it is solvable over $\o_{F_\p}$ for all $p\in\Omega_F$.
}
\pf{
In this example, we have $a=1$, $b=2$ and $d=-2$. Let $E=F(\sqrt{-2})$.
It follows that $L=\o_F+\o_F\sqrt{-2}=\o_E$ and $H_L=H_E$ the Hilbert field of $E$.
It can be shown that $H_E=EH_F$, and hence $H_L/F$ is abelian. Then the result follows from Theorem \ref{thm_trivial_artin_cond}.
}
\subsection{On the Equation $ax^2+by^2=n$ Over $\QQ$}\label{sec_q}
We consider the case where $F=\QQ$.
\thm{\label{thm_q}
Let $a$, $b$ and $n$ be positive integers. Suppose $d = -ab$ is not a square. Set $E=\QQ(\sqrt d)$,
$L=\ZZ+ \ZZ\sqrt d $ and $H_L$ the ring class field corresponding to $L$.
Let $\XX=\mathbf{Spec}(\ZZ[x,y]/(ax^2+by^2-n))$.
Then $\XX(\ZZ)\neq\emptyset$ if and only if there exists
\eqn{
\prod_{p\le \infty}(x_p, y_p)\in\prod_{p\le \infty}\XX(\ZZ_p)
}
such that
\eqn{
\psi_{H_L/E}(\tilde f_E(\prod_p(x_p,y_p)))=1
}
where $\tilde f_E$ is defined the same as in \eqref{eq_tilde_f_E}
except $F=\QQ$.
}
\pf{
Since $d=-ab<0$ it is clear that $x^2-dy^2=-1$ is not solvable over $\RR$, which is to
say the condition $(2)$ of Theorem \ref{thm_artin_cond} holds since the only units of $\ZZ$ are $\{\pm1\}$.
Then the Proposition applies, whence the result follows.
}
We now give an example where explicit criterion is obtained using this result.
\eg{
Let $n$ be a positive integer and $f(x)=x^4-x^3+x+1\in\ZZ[x]$.
Write $n=2^{s_1}\t7^{s_2}\t\prod_{k=1}^g p_k^{e_k}$ and define
$D=\{p_1,p_2,\dots,p_g\}$ and
\aln{
D_1&=\set{p\in D| \fracn{-14}{p}=1\text{ and }f(x)\mod p\text{ irreducible}},\\
D_2&=\set{p\in D| \fracn{-14}{p}=1\text{ and }f(x)\text{ splits into two irreducible factors}}.
}
Then the diophantine equation $2x^2+7y^2=n$ is solvable over $\ZZ$ if and only if
\enmt{
\itm{1} $n\t 2^{-s_1}\equiv\pm1\pmod8$,
\itm{2} $\fracn{n\t 7^{-s_2}}{7}=1$,
\itm{3} for all $p\nmid2\t7$, $\fracn{-14}{p}=1$ for odd $e, v_p(n)=e$,
\itm{4} and
\eqn{\cs{
\sum{p_i\in D_1}e_i\equiv0\pmod2 &\text{ if }D_1\neq\emptyset,\\
1+s_1+s_2+\sum_{p_j\in D_2}e_j\equiv0\pmod2 &\text{ if }D_1=\emptyset.
}}}}
\pf{
In this example, we have $a=2$, $b=7$,$d=-14$ and $d\equiv2\pmod4$. Let $E=\QQ(\sqrt{-14})$.
It follows that $L=\ZZ+\ZZ\sqrt{-14}=\o_E$ and $H_L=H_F=E(\alpha)$ the Hilbert field of $E$
where the minimal polynomial of $\alpha$ is $f(x)$. The Galois group $\Gal(H_L/E)=<\sqrt{-1}>\cong\ZZ/4\ZZ$.
Let $\XX=\mathbf{Spec}(\ZZ[x,y]/(2x^2+7y^2-n))$ and
\eqn{
\tilde f_E[(x_p,y_p)]=
\cs{
(2x_p+y_p\sqrt{-14},ax_p-y_p\sqrt{-14}) &\text{if }p\text{ splits in }E/\QQ,\\
2x_p+y_p\sqrt{-14} &\text{otherwise}.
}}
Then by Theorem \ref {thm_q}, $\XX(\ZZ)\neq\emptyset$ if and only if there exists
\eqn{
\prod_{p\le \infty}(x_p, y_p)\in\prod_{p\le \infty}\XX(\ZZ_p)
}
such that
\eqn{
\psi_{H_L/E}(\tilde f_E(\prod_p(x_p,y_p)))=1.
}
Next we compute these conditions in details.
By a simple computation the local condition \eqn{
\prod_{p\le \infty}\XX(\ZZ_p)\neq\emptyset
} is equivalent to
\eq{\label{eq_eg_q_local}
\cs{
n\t 2^{-s_1}\equiv\pm1\pmod8,\\
\fracn{n\t 7^{-s_2}}{7}=1,\\
\text{for all }p\nmid2\t7, \fracn{-14}{p}=1\text{ for odd }e, v_p(n)=e.
}}
For Artin condition, let $(x_p,y_p)_p\in\prod_p\XX(\ZZ_p)$. Then
\eq{\label{eq_local_decomp}
(x_p+ \sqrt {-14} y_p)(x_p- \sqrt {-14} y_p) = n\text{ in }E_\P\text{ with }\P\mid p.
}
And since $H_L/E$ is unramified, for any $p$ we have
\eq{\label{eq_psi_1}
1=\cs{
\psi_{H_L/E}(p_\P)\psi_{H_L/E}(p_{\bar\P}), &\text{ if }p=\P\bar\P\text{ splits in }E/\QQ,\\
\psi_{H_L/E}(p_\P), &\text{ if }p=\P\text{ inert in }E/\QQ,
}}
where $p_\P$ (resp. $p_{\bar\P}$) is in $\II_E$ such that its $\P$ (resp. $\bar\P$) component is $p$ and the other
components are $1$.
We calculate $\tilde f_E[(x_p,y_p)]$ separately:
\enmt{
\it If $p=2$, $2=\P_2^2$ in $E/\QQ$. Suppose $\P_2=\pi_2\o_{E_{\P_2}}$ for $\pi_2\in\o_{E_{\P_2}}$.
Since $\P_2^2$ is principal in $E$ but $\P_2$ is not, we have $\psi_{H_L/E}((\pi_2)_{\P_2})=-1$.
By \eqref{eq_local_decomp} we have
\eqn{ v_{\P_2}(2x_2+\sqrt{-14}y_2)=v_{\P_2}(2x_2-\sqrt{-14}y_2)=\frac{1}{2}v_{\P_2}(2n)=v_2(2n)=s_1+1.}
It follows that \aln{
\psi_{H_L/E}(\tilde f_E[(x_2,y_2)])&=\psi_{H_L/E}((2x_2+\sqrt{-14}y_2)_{\p_2})\\
&=(-1)^{v_{\P_2}(2x_2+\sqrt{-14}y_2)}=(-1)^{s_1+1},
}where $\tilde f_E[(x_2,y_2)]$ is also regarded as an element in $\II_E$ such that the components above $2$ are given by the value
of $\tilde f_E[(x_2,y_2)]$ and $1$ otherwise.
\it If $p=7$, a similar argument shows that $\psi_{H_L/E}(\tilde f_e[(x_2,y_2)])=(-1)^{s_2}$.
\it If $\fracn{-14}{p}=1$ then by \eqref{eq_psi_1} we can distinguish the following cases:
\enmt{[(i)]
\it $f(x)\mod p$ splits into linear factors.
Then $\psi_{H_L/E}(p_\P)=\psi_{H_L/E}(p_{\bar\P})=1$ and $\psi_{H_L/E}(\tilde f_E[(x_p,y_p)])=1$.
\it $f(x)\mod p$ splits into two irreducible factors.
Then $\psi_{H_L/E}(p_\P)=\psi_{H_L/E}(p_{\bar\P})=-1$.
It follows that \aln{
\psi_{H_L/E}(\tilde f_E[(x_p,y_p)])&=\psi_{H_L/E}((2x_p+\sqrt{-14}y_p)_\P)(2x_p-\sqrt{-14}y_p)_{\bar\P})\\
&=(-1)^{v_\P(2x_p+\sqrt{-14}y_p)+v_{\bar\P}(2x_p-\sqrt{-14}y_p)}=(-1)^e,
}where $v_p(n)=e$ since \aln{
v_\P(2x_p+\sqrt{-14}y_p)&+v_{\bar\P}(2x_p-\sqrt{-14}y_p)\\
&=v_p(2x_p+\sqrt{-14}y_p)+v_p(2x_p-\sqrt{-14}y_p)=v_p(n).
}
\it $f(x)\mod p$ irreducible.
Then $\psi_{H_L/E}(p_\P)=-\psi_{H_L/E}(p_{\bar\P})=\pm\sqrt{-1}$.
It follows that \aln{
\psi_{H_L/E}&(\tilde f_E[(x_p,y_p)])=\psi_{H_L/E}((2x_p+\sqrt{-14}y_p)_\P)(2x_p-\sqrt{-14}y_p)_{\bar\P})\\
&=(\pm\sqrt{-1})^{v_\P(2x_p+\sqrt{-14}y_p)+v_{\bar\P}(2x_p-\sqrt{-14}y_p)}
(-1)^{v_{\bar\P}(2x_p-\sqrt{-14}y_p)}\\
&=(\pm\sqrt{-1})^e(-1)^a
}where $v_p(n)=e$ and $a=v_p(2x_p-\sqrt{-14}y_p)$ (in $\QQ_p$, $0\le a\le e$). By Hensel lemma, we can choose the local solution
$(x_p,y_p)$ suitably, such that $a$ can rich any value between $0$ and $e$.
Hence $ \psi_{H_L/E}(\tilde f_E[(x_p,y_p)]) =\pm(\sqrt{-1})^e$ with the sign chosen freely.
}
\it If $\fracn{-14}{p}=-1$ then $p$ inert in $E/\QQ$. By \eqref{eq_psi_1} we have $\psi_{H_L/E}(\tilde f_E[(x_p,y_p)])=1$.
\it At last if $p=\infty$, since $H_L/E$ is unramified, we have $\psi_{H_L/E}(\tilde f_E[(x_\infty,y_\infty)])=1$.
}
Putting the above argument together, we know the Artin condition is
\eq{\label{eq_eg_q_artin}
\cs{
\sum{p_i\in D_1}e_i\equiv0\pmod2 &\text{ if }D_1\neq\emptyset,\\
1+s_1+s_2+\sum_{p_j\in D_2}e_j\equiv0\pmod2 &\text{ if }D_1=\emptyset.
}}
The proof is done if we put the local condition \eqref{eq_eg_q_local} and the Artin condition \eqref{eq_eg_q_artin} together.
}
\subsection{On the Equation $x^2+qy^2=\alpha$ Over $\QQ(\sqrt{-p})$}\label{sec_pq}
At last we consider the representations of the form $x^2+qy^2$ over a class of imaginary quadratic fields $\QQ(\sqrt{-p})$ where
\eq{\label{eq_pq_cond}
p\text{ and }q\text{ are rational primes of the form }4k+1\text{ satisfying }\fracn{p}{q}=-1\text{ or }\fracl{p}{q}{4}=-1.
}
\thm{\label{thm_pq}
Let $p$, $q$ satisfy \eqref{eq_pq_cond} and $F=\QQ(\sqrt{-p})$ be quadratic field.
Let $\alpha$ be an element in $\o_F$. Set $d = -q$, $E=F(\sqrt d)$,
$L=\o_F+ \o_F\sqrt d $ and $H_L$ the ring class field corresponding to $L$.
Let $\XX=\mathbf{Spec}(\o_F[x,y]/(x^2+qy^2-\alpha))$.
Then $\XX(\o_F)\neq\emptyset$ if and only if there exists
\eqn{
\prod_{\p\in\Omega_F}(x_\p, y_\p)\in\prod_{\p\in\Omega_F}\XX(\o_{F_\p})
}
such that
\eqn{
\psi_{H_L/E}(\tilde f_E(\prod_\p(x_\p,y_\p)))=1
}
where $\tilde f_E$ is defined the same as in \eqref{eq_tilde_f_E}
except $a=1$ and $b=q$.
}
\pf{
Suppose that \eqref{eq_pq_cond} holds. Since $p$ and $q$ are distinct and of the form $4k+1$, we have $\o_F=\ZZ+\ZZ\sqrt{-p}$ and $q$ is not a square in $F$. We claim that the condition $(1)$ of Theorem \ref{thm_artin_cond} holds.
Since $\o_F^\t={\pm 1}$, it's enough to prove that $x^2+qy^2=-1$ has an integral solution over $\o_F$.
Since $p$ and $q$ are distinct and of the form $4k+1$, we have $\o_F=\ZZ+\ZZ\sqrt{-p}$ and $q$ is not a square in $F$.
If $\fracn{p}{q}=-1$, or $\fracn{p}{q}=1$ and $\fracl{p}{q}{4}=\fracl{q}{p}{4}=-1$, then $x^2-pqy^2=-1$ is solvable over $\ZZ$ (\cite[p. 228]{dirichlet}); otherwise $\fracn{p}{q}=1$, $\fracl{p}{q}{4}=-1$ and $\fracl{q}{p}{4}=1$, so $x^2-pqy^2=p$ is solvable over $\ZZ$ (\cite[Corollary 1.4]{wei_diophantine}).
Both cases indicate that there exists $x_0,y_0\in\o_F$ such that $x_0^2+qy_0^2=-1$.
Hence Theorem \ref{thm_artin_cond} yields the result.
}
\newnoindbf{{Acknowledgment}
The authors would like to thank Dasheng Wei and Congjun Liu for many helpful discussions and comments.
}
\bibliography{\thefilename}
\bibliographystyle{amsplain}
\end{document} | 14,717 |
Privacy
This week our star is Lilly, a Labrador Retriever, who lives on a
farm in Pennsylvania with her people and lots of other animals.
Lilly is extremely small for a Labrador -- she was the runt of her
litter (along with her sister Sweetpea, whom we met 2 weeks ago).
Lilly has a lot of medical problems, including with her joints, but
she gets along well and seems to be a happy little dog.
Despite her small size, Lilly thinks she is a fierce protector and
will growl very convincingly. But she's also a pushover -- a family
friend can come over, pick her up (while she's still growling), and
scratch her belly. She loves it!
Previous Star of the Week | 418,648 |
Perhaps money is too good for Apple and AT&T, Steve Job’s official word on iPhone tethering the iPad is a simple, “no”.
Well, the wi-fi version of the iPad is poised to hit store shelves on April 3, and given that a large number of interested purchasers for the device presumably own iPhones, it’s only natural to wonder if AT&T will enable users to tether their iPhones to the iPad. Now seeing as how a wi-fi version of the iPad will soon be accompanied by a 3G equipped iPad with service supplied exclusively by AT&T, there’s really no reason why AT&T would permit iPad tethering when it can make some extra cash with month to month 3G contracts on the 3G enabled iPad.
Of course, let’s not forget AT&T is charging multiple times for the things they should be only charging once.
For example, iPad will have 3G connection, provided users pay “separately” for the service. There will be a WiFi-only version, which I should be able to tether using my Sprint Palm Pre, or even my Nexus One.
And iPhone users can also do it too using PDANet.
The point is, all the consumers who can’t even install an app (or doesn’t know about it) will get screwed, paying 3 times for their 3G service, once for an iPhone, once for their 3G modem, and once more for an iPad when all this could be solved with iPhone tether.
I guess it’s not a big deal since people can use PDANet or use other methods of wifi tethering. But the point remains that AT&T is still clearly being “evil” by charging customers more than they should.
So what?
Let’s start a war, let’s start boycotting all AT&T services. I have, it’s been 3 years since I have purchased any AT&T products. Unless consumers start doing something about it, we are all going to be treated like bunch of guinea pigs that can’t think for ourselves.
Seriously, do you like being “owned” by AT&T as an iPhone user? I bet you don’t, the only way to make AT&T to change is for you to step up and start DOING SOMETHING ABOUT IT.
There, I did something about it by writing this blog post and other great hideous posts outlining what is wrong with AT&T, I just need your help to get this ball rollin’
All I want you to do is this, just one tweet or even tell your friend how much AT&T sucks because they make you pay for every little thing. You can still like Apple, I like Apple too but AT&T? That’s something I get peeved about. I am not even a customer of AT&T but I just hate what they are doing to people, screwing them in the a**. There, I said it, someone has to speak the truth around here.
| 174,624 |
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TITLE: Is there a field extension $E$ containing $S$ which has finite dimension when considered as a vector space over $S?$
QUESTION [0 upvotes]: An algebraic number field (or simply number field) is an extension field
$K$
of the field of rational numbers
$\mathbb {Q}$ such that the field extension
$K/\Bbb Q$ has finite degree (and hence is an algebraic field extension). Thus
$K$ is a field that contains
$\mathbb {Q}$ and has finite dimension when considered as a vector space over
$\mathbb {Q}.$
Consider:
$$ \exp: \Bbb Q \to (S \subset \Bbb R^+) $$
$(S,\times)$ is a group. Furthermore $(S,\times, *)$ is a field by:
$$ m: S \times S \to S $$
with $(e^a,e^b)$ to $e^{ab},$ where $a,b \in \Bbb Q.$
Is there a field extension $E$ containing $S$ which has finite dimension when considered as a vector space over $S?$
REPLY [2 votes]: Yes, such an extension exists. One possibility is to take any extension field $K$ of $\mathbb{Q}$, given you have one at hand: $K$ is, up to isomorphism, a field extension of $S$, because $S$ is, up to isomorphism, just $\mathbb{Q}$.
An easy possibility, if I understand your question correctly, would be $E = S$. This is trivially a field extension of $S$ of dimension $1$. Please give me a hint if I didn't understand your question correctly. | 2,334 |
\section{Correction terms in Heegaard Floer homology}\label{s:prelim}
Heegaard Floer homology is a family of invariants of 3--manifolds introduced by Ozsv\'ath and Szab\'o \cite{OzsvathSzabo-HF}; in this paper we are concerned with the `plus' version, which associates to a rational homology sphere $Y$ equipped with a \spinc structure $\ft$ a $\Q$--graded $\Z[U]$--module $\HFp\Yt$. Recall that the action of $U$ decreases the degree by 2.
The group $\HFp\Yt$ further splits as a direct sum of $\Z[U]$--modules $\Tp\oplus\HFpred\Yt$, where $\Tp = \Z[U,U^{-1}]/U\cdot\Z[U]$. The degree of the element $1\in \Tp$ is called the \emph{correction term} of $\Yt$, and it is denoted by $d\Yt$.
\begin{thm}[\cite{OzsvathSzabo-absolutely}]
The correction term satisfies the following properties:
\begin{itemize}\itemsep -0,5pt
\item $d(Y,\overline\ft) = d\Yt = -d(-Y,\ft)$, where $\overline\ft$ is the conjugate of $\ft$;
\item if $(W,\fs)$ is a negative definite \spinc $4$--manifold with boundary $\Yt$, then
\[
c_1(\fs)^2 + b_2(W) \le 4d\Yt.
\]
\end{itemize}
In particular $d\Yt$ is invariant under \spinc rational homology cobordisms.
\end{thm}
\begin{cor}\label{c:QHDvsd}
If $W$ is a rational homology ball with boundary $Y$, and $\ft$ is a \spinc structure on $Y$ that extends to $W$, then $d\Yt=0$.
\end{cor}
When $Y$ is obtained as rational surgery along a knot $K$ in $S^3$, one can recover the correction terms of $Y$ in terms of a family of invariants of $K$, first introduced by Rasmussen \cite{Rasmussen-Goda} and then further studied by Ni and Wu \cite{NiWu}, and Hom and Wu \cite{HomWu}. We call these invariants $\{V_i(K)\}_{i\ge 0}$, adopting Ni and Wu's notation instead of Rasmussen's --- who used $h_i(K)$ instead --- as this seems to have become more standard.
We denote with $\unknot$ the unknot.
\begin{thm}[\cite{Rasmussen-Goda, NiWu}]
The sequence $\{V_i(K)\}_{i\ge 0}$ takes values in the non-negative integers and is eventually $0$. Moreover, $V_{i}(K)-1 \le V_{i+1}(K) \le V_i(K)$ for every $i\ge 0$.
For every rational number $p/q$ and for an appropriate indexing of \spinc structures on $S^3_{p/q}(\unknot)$ and $S^3_{p/q}(K)$, we have
\begin{equation}\label{e:NiWu}
d(S^3_{p/q}(K),i) = -2\max\{V_{\lfloor i/q\rfloor}(K),V_{\lceil (p-i)/q\rceil}(K)\} + d(S^3_{p/q}(\unknot),i).
\end{equation}
\end{thm}
\begin{defn}[\cite{HomWu}]
The minimal index $i$ such that $V_i(K)=0$ is called $\nu^+(K)$.
\end{defn}
In order to know the correction terms of $S^3_{p/q}(K)$ it suffices to know the values of $V_i(K)$ for each $i$ as well as the values of the correction terms of lens spaces.
\begin{prop}[{\cite[Proposition~4.8]{OzsvathSzabo-absolutely}}]
Let $p,q,i$ be integers with $p>q>0$, $\gcd(p,q) = 1$ and $0\le i<p+q$. Denote with $r$ and $j$ the reductions of $p$ and $i$ mod $q$ respectively. Then, for an appropriate indexing of \spinc structures,
\begin{equation}\label{e:Lrecursion}
d(L(p,q),i) = \frac14 - \frac{(p+q-2i-1)^2}{4pq} - d(L(q,r),j).
\end{equation}
\end{prop}
\begin{rem}
Notice that in \cite{OzsvathSzabo-absolutely, NiWu, HomWu} the lens space $L(p,q)$ is defined as obtained by doing $p/q$--surgery along the unknot $\unknot$. We adopt the convention that $L(p,q)$ is obtained by doing $-p/q$--surgery along $\unknot$, following \cite{Neumann, Lisca-ribbon} (among others): this explains the sign difference in \cref{e:Lrecursion} with respect to the original.
\end{rem} | 187,018 |
It is no surprise then that there was serious excitement in our house when I showed Zoey and her brothers that we had received a Fairy Flowerpot and Fairy Watering Can to review.
My Fairy Garden Fairy Flowerpot
The My Fairy Garden Fairy Flowerpot is perfect for introducing your child to gardening and teaching them how you can grow things from seeds. There are so many beautiful features to this flowerpot including the little ornate designs on the door and the scalloped edging pattern around the top.
The Fairy Flowerpot is far more than just a flowerpot though as it comes with it's very own magical fairy called Blossom who is a Spring Fairy and loves flowers and being surrounded by nature.
The Fairy Flowerpot is far more than just a flowerpot though as it comes with it's very own magical fairy called Blossom who is a Spring Fairy and loves flowers and being surrounded by nature.
Blossom has her own little door which allows her to enter her pretty flowerpot home. She has brought with her some special easy-grow seeds which Zoey has been calling special magic fairy seeds. We have just set them and are eagerly awaiting signs of growth. Zoey has been pretty much checking them every 10 minutes and as her big brother told her singing and music might help speed up their appearance, she has been treating them to her very own little concert complete with harmonica playing.
My Fairy Garden Fairy Watering Can
The Fairy Watering Can is perfect for any little budding gardener. It is so pretty with it's very own little fairy Fenn who swings off the handle.
The Fairy Watering can is fully functional with a detachable sprinkler rose along with a spade and fork which fit either side of the fairy door. Like the Flowerpot, the Watering Can has a special door which opens up and leads to Fenn the Fairy's little home for when she wants a little break from swinging on the handle.
The twins have decided that Fenn uses her fairy magic to add special powers to the water which we put in the Watering Can which will make seeds grow faster and the flowers to be even more beautiful. We can't wait to see some how long the magic will take to work, it's definitely all very exciting.
Our Verdict
I was so excited to give these products to Zoey and her brothers, I knew she would squeal with delight and I wasn't wrong. The products are both made of high quality materials and are well designed to be both functional and meet the demands of play.
The fairies included are so pretty and add that extra special touch of magic, I love the fact that their clothes are clearly inspired by nature with Blossoms dress looking like it is made of flower petals and Fenn's skirt of leaves.
The fairy doors and compartments are such a lovely way to integrate the fairy toy itself into the fully functional flowerpot and watering can as they are more than just a pot or a watering can but also a home to a magic little fairy.
I think the products are a wonderful idea and I love that they are helping to instil an interest in nature in my children. I highly recommend Fairy Garden products, they are kid friendly, educational, aesthetically pleasing and sprinkled with magic. Children will love them and their grown ups will too!
*Disclaimer: I received the My Fairy Garden Fairy Flowerpot and Fairy Watering Can for review. All views and opinions are my own and unbiased.
*Disclaimer: I received the My Fairy Garden Fairy Flowerpot and Fairy Watering Can for review. All views and opinions are my own and unbiased. | 52,023 |
\begin{document}
\begin{abstract}
For a symmetric algebra $A$ over a field $K$ of characteristic $p>0$
K\"ulshammer constructed a descending sequence of ideals of the
centre of $A$. If $K$ is perfect this sequence was shown to be an
invariant under derived equivalence and for algebraically closed
$K$ the dimensions of their image in the stable centre was shown to be invariant
under stable equivalence of Morita type.
Erdmann classified algebras of tame representation type which
may be blocks of group algebras, and Holm classified Erdmann's list
up to derived equivalence. In both classifications certain parameters
occur in the classification, and
it was unclear if different parameters lead to different algebras.
Erdmann's algebras fall into three classes,
namely of dihedral, semidihedral and of quaternion type.
In previous joint work with Holm we used K\"ulshammer ideals to distinguish
classes with respect to these parameters in case of algebras of dihedral and
semidihedral type. In the present paper we determine the K\"ulshammer
ideals for algebras of quaternion type and distinguish again algebras with
respect to certain parameters.
\end{abstract}
\maketitle
\section*{Introduction}
Erdmann gave in \cite{Erdmann} a list of basic symmetric
algebras of tame representation type which include all the algebras which
may be Morita equivalent
to blocks of finite groups of tame representation type. She obtained these algebras
by means of properties of the Auslander-Reiten quiver which are
known to hold for blocks of group rings with tame representation type.
These algebras
are subdivided into three classes, those of dihedral type, of
semidihedral type and of quaternion type, corresponding to the
possible defect group in case of group algebras, and actually defined
by the behaviour of their Auslander-Reiten quiver.
Holm refined in \cite{Holmhabil}
Erdmann's classification of those algebras which may occur as blocks of group algebras to a classification up to derived equivalence.
However, in \cite{Erdmann,Holmhabil} the algebras
are defined by quivers with relations, and the relations involve
certain parameters, corresponding mostly to deformations of the socle of the algebras.
It was unclear in some cases if different parameters lead to
different derived equivalence classes of algebras. The question of non trivial socle deformations
appears to be a very subtle one in this special case, but also in general,
and little progress was made on this question until
very recently.
In \cite{kuelsquest} we showed that a certain sequence of ideals of
the centre of a symmetric algebra defined previously by K\"ulshammer \cite{Ku1}
is actually invariant under derived equivalences if the base field is perfect. We call this
sequence of ideals the K\"ulshammer ideals.
In joint work \cite{LZZ}
with Liu and Zhou we showed that if the base field is algebraically closed, then
the dimension of the
image of this invariant in the stable centre is also
an invariant under stable equivalences of Morita type.
In joint work \cite{hztame} with Holm we observed that the K\"ulshammer ideals
behave in a very subtle manner with respect to the deformation parameters.
Using this observation we showed that some of the
parameters are invariants under derived equivalence for certain families
of algebras of dihedral and of semidihedral type. The present paper is
a continuation and completion of \cite{hztame}.
In order to apply the
theory of K\"ulshammer ideals we need to use the symmetrising form
explicitly, and in the present work we progress in avoiding the ad-hoc arguments used in our
previous work to determine the symmetrising form. In this note we
compute the K\"ulshammer ideals for algebras of quaternion type and distinguish
this way the derived equivalence classes of the
algebras with two simple modules. Over algebraically closed fields of
characteristic different from $2$ we can classify completely the derived equivalence classes of
the algebras of quaternion type occurring in Holm's list, except for a case of very small
parameters. If the field is
algebraically closed of characteristic $2$ then we have an almost complete classification
in case of two simple modules. The result in this case is displayed in Corollary~\ref{Finalcorollarytwosimples} and Theorem~\ref{main}.
We also deal with the case of algebras of quaternion type with
three simple modules, where K\"ulshammer ideals distinguish the isomorphism classes of algebras
in characteristic $2$ with parameter $d$ according to whether or not $d$ is a square in $K$.
The invariance of K\"ulshammer ideals under derived or Morita equivalence is proved only
in case $K$ is perfect, which implies that all elements of $K$ are squares when $K$ is of characteristic $2$.
Hence, we cannot say more about this case, and the derived equivalence classification remains open for this class of $20$-dimensional algebras.
Derived equivalent local algebras are actually Morita equivalent
(cf \cite[Proposition 6.7.4]{reptheobuch}), so that the derived equivalence classification of the class of algebras of quaternion type with one simple module coincides with its classification up to isomorphism. Isomorphic algebras have isomorphic K\"ulshammer ideal structure.
For the reader's convenience we give the somewhat technical result for the
class of algebras with two simples here.
Blocks of quaternion type with two simple modules are derived equivalent to an
algebra $A^{k,s}(a,c)$ for parameters $a\in K^\times$ and $c\in K$ and integers
$s\geq 3$ and $k\geq 1$.
\begin{itemize}
\item
In particular, if $K$ is an algebraically closed field of characteristic different from $2$, then
there is $a'\in K^\times$ such that
$A^{k,s}(a,c)\simeq A^{k,s}(a',0)$, and if $(k,s)\neq(1,3)$, then $A^{k,s}(a,c)\simeq A^{k,s}(1,0)$. Moreover, if $A^{k,s}(1,0)$ and $A^{k',s'}(1,0)$ are derived equivalent, then $(k,s)=(k',s')$ or $(k,s)=(s',k')$.
\item
If $K$ is a perfect field of characteristic $2$, we have the following situation.
The algebra $A^{k,s}(a,c)$ is not derived equivalent to $A^{k,s}(a',0)$ for
any $a,a',c\in K^\times$. If $K$ is algebraically closed of characteristic $2$, and
if $c\neq 0$, then $A^{k,s}(a,c)$ is isomorphic to $A(a'',1)$ for some $a''\in K^\times$, and
if $(k,s)\neq (1,3)$, then $A^{k,s}(a,0)\simeq A^{k,s}(1,0)$.
Further, again for algebraically closed $K$, if
$A^{k,s}(a',c')$ is derived equivalent to $A^{k',s'}(a'',c'')$ then
$(k,s)=(k',s')$ or $(k,s)=(k',s')$.
We do not know for which parameters $a,a'\in K^\times$ we get
$A^{k,s}(a,1)$ is derived equivalent to $A^{k,s}(a',1)$, and we do not know when
$A^{(1,3)}(a,0)$ is derived equivalent to $A^{(1,3)}(a',0)$ for $a,a'\in K^\times$.
\end{itemize}
The K\"ulshammer ideal structure depends in a quite subtle way on the parameters, and
we want to stress the fact that we need to compute the ideals as ideals, and as in \cite{hztame}
it is not sufficient to consider the dimensions only.
The paper is organised as follows. In Section~\ref{Sec-Reynolds} we recall basic facts
about K\"ulshammer ideals and improve the general methods needed to compute the K\"ulshammer
ideal structure for symmetric algebras.
In Section~\ref{quaterniontypesection} we apply the general theory
to algebras of quaternion type,
and we prove our main result Theorem~\ref{main} there.
\subsection*{Acknowledgement}
I wish to thank Oyvind Solberg for giving me during the
Oberwolfach conference ``Hochschild cohomology and applications'' in February 2016
a GAP program to compute the K\"ulshammer ideals. The GAP program \cite{GAP}
uses the package ``qpa'' and encouraged me to study the quaternion type algebras.
I also wish to thank Rachel Taillefer for pointing out the particularity of $(k,s)=(1,3)$
for two simple modules which I forgot to consider in a previous version. I thank the referee for numerous very useful remarks, and in particular for alerting me on some mistake
in the initial version concerning symmetrising forms.
\section{Review on K\"ulshammer ideals and how to compute them}
\label{Sec-Reynolds}
The aim of this section is to briefly give the necessary
background on K\"ulshammer ideals,
as introduced by B. K\"ulshammer
\cite{Ku1}. Morita invariance of K\"ulshammer ideals (then named generalised Reynolds'
ideals) was shown in \cite{BHHKM,HHKM} for perfect fields $K$.
K\"ulshammer ideals were proved to be a derived invariant in \cite{kuelsquest},
were used in \cite{HS,HS2,BSkow} to classify weakly
symmetric algebras of polynomial growth or domestic type up to derived
equivalences, in \cite{hztame} for a derived equivalence classification
of algebras of dihedral or semidihedral type, in \cite{Ln}
for deformed preprojective algebras of type $L$, and in \cite{ST}
for the derived equivalence
classification of certain special biserial algebras. The concept was generalised
to general finite-dimensional algebras in \cite{BHZ}, to an invariant of
Hochschild (co)homology for symmetric algebras \cite{gerstenhaber}
and in \cite{TAHochschild} for general algebras. The image of the K\"ulshammer ideals
in the stable centre were shown to be an
invariant under stable equivalences of Morita type \cite{LZZ,KLZ}.
An overview is given in \cite{Iran,reptheobuch}.
Let $K$ be a field of characteristic $p > 0$.
Any finite-dimensional symmetric $K$-algebra $A$ has
an associative, symmetric, non-degenerate $K$-bilinear form
$\langle -,-\rangle : A
\times A \rightarrow K$. For any $K$-linear subspace $M$ of $A$ we
denote the orthogonal space by $M^{\bot}$ with respect to this form.
Moreover, let $[A,A]$ be the $K$-subspace
of $A$ generated by all commutators $[a,b]:=a b - b a$,
where $a, b \in A$. For any $n \geq 0$ set
$$T_n (A) = \left\{ x \in A \mid x^{p^n} \in [A,A]\right\}.$$
Then, by \cite{Ku1}, for any $n\ge 0$, the orthogonal
space $T_n (A)^{\bot}$ is an ideal of the center
$Z(A)$ of $A$, called {\it $n$-th K\"ulshammer ideal}.
These ideals form a descending sequence
$$Z(A) = [A,A]^{\perp} = T_0(A)^{\perp} \supseteq T_1(A)^{\perp}
\supseteq T_2(A)^{\perp}
\supseteq \ldots \supseteq T_n(A)^{\perp} \supseteq \ldots$$
with intersection of all ideals $T_n(A)^{\perp}$ for $n\in\N$
being the Reynolds' ideal
$R(A)=Z(A)\cap\soc(A)$.
In \cite{HHKM} it has been shown that if $K$ is perfect, then
the sequence of K\"ulshammer ideals is invariant under Morita equivalences.
Later, it was shown that the sequence of K\"ulshammer ideals
is invariant under derived equivalences, and the image of the
sequence of K\"ulshammer ideals in the stable centre
is invariant under stable equivalences of Morita type.
The following theorem recalls part of what is known.
\begin{Theorem}
\label{prop:zimmermann}
\begin{itemize}\item \cite[Theorem 1]{kuelsquest}
Let $A$ and $B$ be finite-dimensional symmetric algebras over a perfect
field $K$ of positive characteristic $p$.
If $A$ and $B$ are derived equivalent, then
there is an isomorphism $\varphi : Z(A) \rightarrow Z(B)$ between
the centers of $A$ and $B$ such that $\varphi(T_n (A)^{\bot}) = T_n
(B)^{\bot}$ for all positive integers $n$.
\item (cf e.g. \cite[Proposition 6.8.9]{reptheobuch})
Let $A$ and $B$ be derived equivalent finite dimensional $K$-algebras
over a field $K$, which is a splitting field for $A$ and for $B$. Then
the elementary divisors of the Cartan matrices of $A$ and of $B$ coincide. In particular,
the determinant of the Cartan matrices coincides.
\item \cite[Corollary 6.5]{LZZ}
If $A$ and $B$ are stably equivalent of Morita type, and if $K$ is an
algebraically closed field, then $\dim_K(T_n(A)/[A,A])=\dim_K(T_n(B)/[B,B])$.
\end{itemize}
\end{Theorem}
We note that in the proof of \cite[Theorem 1]{kuelsquest} the
hypothesis that $K$ is algebraically closed is never used.
The assumption on the field $K$ to be perfect is sufficient.
The aim of the present note is to show how these derived
invariants can be applied to some subtle questions in the
derived equivalence classifications of algebras of quaternion type.
In order to compute the K\"ulshammer ideals we need a symmetrising form.
However, the K\"uls\-hammer ideals do not depend on the choice of the
symmetrising form if $K$ is perfect (cf \cite[Proof of Claim 3]{kuelsquest}).
We showed in \cite{Ln} (see also \cite{reptheobuch})
that every Frobenius form arises as in the following proposition.
\begin{Prop} \cite{hztame,Ln}
\label{prop:form}
Let $A$ be a basic Frobenius algebra such that $K$ is a splitting field for
$A$, and let $\{e_1,\dots,e_n\}$ be a choice of orthogonal primitive idempotents
with $\sum_{i=1}^n e_i=1$. Then there are bases ${\mathcal B}_{i,j}$ of $e_iAe_j$ such that
${\mathcal B}=\bigcup_{i,j=1}^n{\mathcal B}_{i,j}$ is a basis of $A$ containing
a basis of $\soc(A)$
and such that the following statements hold:
\begin{enumerate}
\item[{(1)}] Define an $K$-linear mapping $\psi$ on the basis elements by
$$
\psi(b)=\left\{
\begin{array}{ll} 1 & \mbox{if $b\in\soc(A)$} \\
0 & \mbox{otherwise}
\end{array} \right.
$$
for $b\in {\mathcal B}$.
Then an associative non-degenerate $K$-bilinear
form $\langle-,-\rangle$ for $A$ is given by
$\langle x,y\rangle := \psi(xy).$
\item[{(2)}] Any Frobenius form arises this way for some choice of a basis $\mathcal B$.
\end{enumerate}
\end{Prop}
Note that the hypothesis in \cite{hztame,Ln} is slightly different, however equivalent to the one
given here.
If $A$ is a basic symmetric algebra over an algebraically closed field $K$, then $A=KQ/I$
and we want to determine those bases ${\mathcal B}_s$ of $\soc(A)$ which yield
a symmetric form. This problem is addressed in previous papers
dealing with K\"ulshammer ideals (cf \cite[Remark 2.9]{Ln}, \cite[Remark 3.2]{hztame}).
The following remark indicates a necessary condition for the problem.
\begin{Rem}\label{symmetrisingform}
If $A$ is an indecomposable, basic symmetric algebra over an algebraically closed field $K$ and
let $\{e_1,\dots,e_n\}$ be a choice of orthogonal primitive idempotents
with $\sum_{i=1}^n e_i=1$. Suppose that
${\mathcal B}_s$ is a $K$-basis of $\soc(A)$ and suppose that for each $b\in{\mathcal B}_s$
there is a unique $e_i$ such that $e_ibe_i=b$.
Using \cite[Proposition 2.7.4]{reptheobuch} it is not hard
to see that we can always find a basis
${\mathcal B}_s$ of $\soc(A)$ such that the difference of two elements of ${\mathcal B}_s$
is in the commutator subspace. Moreover, since the elements of ${\mathcal B}_s$
are uniquely determined up to scalars by this property, Proposition~\ref{prop:form} then shows that we can complete the basis ${\mathcal B}_s$ to a basis ${\mathcal B}$ as in the proposition.
If $\psi$ is a $K$-linear map as in
Proposition~\ref{prop:form}, then $\psi([A,A])=0$. In particular, if $b,b'\in \soc(A)$ with
$b-b'\in[A,A]$, then $\psi(b)=\psi(b')$.
\end{Rem}
\section{Algebras of quaternion type}
\label{quaterniontypesection}
\subsection{Two simple modules}
Erdmann gave a classification of algebras which could appear as blocks of
tame representation type. These algebras fall in three classes, the
algebras of dihedral, the algebras of semidihedral and the algebras of
quaternion type. Erdmann's classification was up to Morita equivalence.
Holm \cite[Appendix B]{Holmhabil} gave a classification up to derived
equivalence and obtained for non-local algebras of quaternion type two
families, one containing algebras with two simple modules, one containing algebras
with three simple
modules. The algebras in each family share a common quiver, and the relations depend on a
number of parameters.
The quiver for the algebras with two simples is the following.
\unitlength1cm
\begin{center}
\begin{picture}(10,2)
\put(2,1){$\bullet$}\put(4,1){$\bullet$}
\put(2,1.3){$1$}\put(4,1.3){$2$}
\put(2.2,1.2){\vector(1,0){1.6}}
\put(3.8,.9){\vector(-1,0){1.6}}
\put(1.2,1){\circle{2}}
\put(1.9,.98){\vector(0,1){.1}}
\put(5,1){\circle{2}}
\put(4.3,1.08){\vector(0,-1){.1}}
\put(.6,1){$\alpha$}
\put(3,1.3){$\beta$}
\put(3,.7){$\gamma$}
\put(5.4,1){$\eta$}
\end{picture}
\end{center}
Let $k\geq 1, s\geq 3$ integers and $a\in K^\times, c\in K$. Then we get an
algebra $Q(2{\mathfrak B})_1^{k,s}(a,c)$ by the above quiver with relations
$$\beta\eta=(\alpha\beta\gamma)^{k-1}\alpha\beta,\;\;\;\;
\eta\gamma=(\gamma\alpha\beta)^{k-1}\gamma\alpha,\;\;\;\;
\alpha^2=a\cdot(\beta\gamma\alpha)^{k-1}\beta\gamma+c\cdot(\beta\gamma\alpha)^k,$$
$$\gamma\beta=\eta^{s-1}, \;\;\;\;\alpha^2\beta=0,\;\;\;\;\gamma\alpha^2=0.$$
\begin{Rem}\label{CentreandCartandet} Using \cite{Holmhabil} we see that
the centre of this algebra is of dimension $k+s+2$ and the Cartan matrix is
$\left(\begin{array}{cc}4k&2k\\2k&k+s\end{array}\right)$ with determinant $4ks$.
Hence, using Theorem~\ref{prop:zimmermann}, if
$D^b(Q(2{\mathfrak B})_1^{k,s}(a,c))\simeq D^b(Q(2{\mathfrak B})_1^{k',s'}(a',c'))$,
then $4ks=4k's'$ and $k+s+2=k'+s'+2$. Therefore $(k+s)^2=(k'+s')^2$ and
$(k-s)^2=(k'-s')^2$, which implies $k=k'$ and $s=s'$, or $k=s'$ and $k'=s$.
\end{Rem}
\begin{Lemma}\label{centreofQ2}
Let $K$ be a field, and let $A^{k,s}(a,c):=Q(2{\mathfrak B})_1^{k,s}(a,c)$.
Then, $Z(A^{k,s}(a,c))$ has a $K$-basis formed by the disjoint union
$$\{\eta-(\alpha\beta\gamma)^{k-1}\alpha\}
\stackrel{\cdot}{\cup}\{\eta^t\;|2\leq t\leq s\}
\stackrel{\cdot}{\cup}\{(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u\;|\;
1\leq u\leq k-1\}
\stackrel{\cdot}{\cup}\{1,(\alpha\beta\gamma)^k,\alpha^2\}$$
and is isomorphic, as commutative $K$-algebra, with
$$ K[U,Y,S,T]/(Y^{s+1},U^k-Y^s-2T,S^2,T^2,SY,SU,ST,UY,UT,YT)$$
where
\begin{eqnarray*}
U&:=&(\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\\
Y&:=&\eta-(\alpha\beta\gamma)^{k-1}\alpha\\
S&:=&\alpha^2\\
T&:=&(\alpha\beta\gamma)^{k}
\end{eqnarray*}
\end{Lemma}
Proof. First, $\eta-(\alpha\beta\gamma)^{k-1}\alpha$ commutes trivially with $\eta$, since $\eta\alpha=0=\alpha\eta$.
Now
\begin{eqnarray*}
\alpha(\eta-(\alpha\beta\gamma)^{k-1}\alpha)-(\eta-(\alpha\beta\gamma)^{k-1}\alpha)\alpha
&=&\alpha^2(\beta\gamma\alpha)^{k-1}-(\alpha\beta\gamma)^{k-1}\alpha^2\\
&=&(a(\beta\gamma\alpha)^{k-1}\beta\gamma+c(\beta\gamma\alpha)^k)(\beta\gamma\alpha)^{k-1}\\
&&-(\beta\gamma\alpha)^{k-1}(a(\beta\gamma\alpha)^{k-1}\beta\gamma+c(\beta\gamma\alpha)^k)\\
&=&a\left((\beta\gamma\alpha)^{k-1}\beta\gamma(\beta\gamma\alpha)^{k-1}-
(\beta\gamma\alpha)^{2k-2}\beta\gamma\right)\\
&=&a(\beta\gamma\alpha)^{k-1}\left(\beta\gamma(\beta\gamma\alpha)^{k-1}-
(\beta\gamma\alpha)^{k-1}\beta\gamma\right).
\end{eqnarray*}
This is trivially $0$ if $k=1$, and if $k>1$, then $$(\beta\gamma\alpha)^{2k-2}\beta\gamma=\beta(\gamma\alpha\beta)^{2k-2}\gamma
=\beta(\gamma\alpha\beta)^{k}(\gamma\alpha\beta)^{k-2}\gamma=
\beta\eta^s(\gamma\alpha\beta)^{k-2}\gamma=0.$$
Hence
\begin{eqnarray*}
a(\beta\gamma\alpha)^{k-1}\left(\beta\gamma(\beta\gamma\alpha)^{k-1}-
(\beta\gamma\alpha)^{k-1}\beta\gamma\right)
&=&a(\beta\gamma\alpha)^{k-1}\beta\gamma(\beta\gamma\alpha)(\beta\gamma\alpha)^{k-2}\\
&=&a(\beta\gamma\alpha)^{k-1}\beta\eta^{s-1}(\gamma\alpha)(\beta\gamma\alpha)^{k-2}\\
&=&a(\beta\gamma\alpha)^{k-1}\beta\eta^{s-2}(\gamma\alpha\beta)^{k-1}\gamma\alpha^2(\beta\gamma\alpha)^{k-2}\\
&=&0
\end{eqnarray*}
since $\gamma\alpha^2=0$.
The relations
$\beta\eta=(\alpha\beta\gamma)^{k-1}\alpha\beta$ and
$\eta\gamma=(\gamma\alpha\beta)^{k-1}\gamma\alpha=\gamma(\alpha\beta\gamma)^{k-1}\alpha$
show that $\eta-(\alpha\beta\gamma)^{k-1}\alpha$ commutes with $\beta$ and with $\gamma$.
Now, if $k>1$, then $\left(\eta-(\alpha\beta\gamma)^{k-1}\alpha\right)^2=\eta^2$, and if
$k=1$, then $\left(\eta-(\alpha\beta\gamma)^{k-1}\alpha\right)^2=\eta^2+\alpha^2$. Since $\alpha^2\beta=\gamma\alpha^2=0$, it is clear that $\alpha^2$ is central.
Hence $\eta^t$ is central for each $t\geq 2$.
Now,
$$\alpha\beta\gamma\beta=\alpha\beta\eta^{s-1}=\alpha(\alpha\beta\gamma)^{k-1}\alpha\beta\eta^{s-2}=0$$
since $\alpha^2\beta=0$. Likewise $\gamma\beta\gamma\alpha=0$. Hence
$$
\beta U=\beta\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)=
\beta\gamma\alpha\beta=\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)\beta=U\beta
$$
and
$$
\gamma U=\gamma\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)
=\gamma\alpha\beta\gamma
=\gamma\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)=
U\gamma.
$$
Now,
\begin{eqnarray*}
\eta U&=&\eta\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)\\
&=&\eta(\gamma\alpha\beta)=(\gamma\alpha\beta)^{k-1}\gamma\alpha^2\beta\\
&=&0\\
&=&\gamma\alpha(\alpha\beta\gamma)^{k-1}\alpha\beta\\
&=&\gamma\alpha\beta\eta\\
&=&\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)\eta\\
&=&U\eta
\end{eqnarray*}
and
$$\alpha U=\alpha\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)
=\alpha\beta\gamma\alpha=\left((\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)\right)\alpha=
U\alpha.
$$
Hence $U$ is central, and then we only need to compute $U^u$ to get the result. Finally, socle
elements in basic symmetric algebras over splitting fields
are always central, and $1$ is central of course. We know by \cite{Holmhabil}
that the centre is $(2+k+s)$-dimensional,
and obtain therefore the result.
\dickebox
\begin{Rem} \label{remarkparameters}
Erdmann and Skowro\'nski show in \cite[Lemma 5.7]{ErdSkow}
that if $K$ is an algebraically closed field, then
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(1,c')$ for some $c'\in K$
and if $K$ is of characteristic different from $2$, then
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(a,0)$.
We can examine their computations again to get slightly better results. We assume here $k+s>4$.
Suppose that $K$ admits any $k$-th root, i.e. for all $x\in K$ there is
$y\in K$ with $y^k=x$.
We want to simplify the parameters $a,c$. Replace $\alpha$ by $x_\alpha\alpha$,
$\beta$ by $x_\beta\beta$,
$\gamma$ by $x_\gamma\gamma$ and $\eta$ by $x_\eta\eta$ for non zero scalars
$x_\alpha,x_\beta,x_\gamma,x_\eta$.
Then the relations above are equivalent to
\begin{eqnarray*}
x_\eta\beta\eta&=&x_\alpha^k(x_\beta x_\gamma)^{k-1}(\alpha\beta\gamma)^{k-1}\alpha\beta,\\
x_\eta \eta\gamma&=&x_\alpha^{k}(x_\beta x_\gamma)^{k-1}(\gamma\alpha\beta)^{k-1}\gamma\alpha,\\
x_\alpha^2\alpha^2&=&a\cdot x_\alpha^{k-1}(x_\beta x_\gamma)^{k}
(\beta\gamma\alpha)^{k-1}\beta\gamma+c\cdot x_\alpha^{k}(x_\beta x_\gamma)^{k}(\beta\gamma\alpha)^k,\\
x_\gamma x_\beta\gamma\beta&=&x_\eta^{s-1}\eta^{s-1},\\
\alpha^2\beta&=&0,\\
\gamma\alpha^2&=&0.
\end{eqnarray*}
We first choose $x_\beta$ such that $x_\beta x_\gamma=x_\eta^{s-1}$ to get the system
\begin{eqnarray*}
\beta\eta&=&x_\alpha^kx_\eta^{(k-1)(s-1)-1}(\alpha\beta\gamma)^{k-1}\alpha\beta,\\
\eta\gamma&=&x_\alpha^{k}x_\eta^{(k-1)(s-1)-1}(\gamma\alpha\beta)^{k-1}\gamma\alpha,\\
\alpha^2&=&a\cdot x_\alpha^{k-3}x_\eta^{k(s-1)}
(\beta\gamma\alpha)^{k-1}\beta\gamma+c\cdot x_\alpha^{k-2}x_\eta^{k(s-1)}(\beta\gamma\alpha)^k,\\
\gamma\beta&=&\eta^{s-1},\\
\alpha^2\beta&=&0,\\
\gamma\alpha^2&=&0.
\end{eqnarray*}
Then we put $x_\alpha=x_\eta^{-\frac{(k-1)(s-1)-1}{k}}$ and obtain the system
\begin{eqnarray*}
\beta\eta&=&(\alpha\beta\gamma)^{k-1}\alpha\beta,\\
\eta\gamma&=&(\gamma\alpha\beta)^{k-1}\gamma\alpha,\\
\alpha^2&=&a\cdot x_\eta^{-\frac{(k-1)(s-1)-1}{k}\cdot(k-3)+k(s-1)}
(\beta\gamma\alpha)^{k-1}\beta\gamma+
c\cdot x_\eta^{-\frac{(k-1)(s-1)-1}{k}(k-2)+k(s-1)}(\beta\gamma\alpha)^k,\\
\gamma\beta&=&\eta^{s-1},\\
\alpha^2\beta&=&0,\\
\gamma\alpha^2&=&0.
\end{eqnarray*}
Now, $-\frac{(k-1)(s-1)-1}{k}\cdot(k-3)+k(s-1)=0$ implies $k=3$ and $s=1$, or $k=1$ and $s=3$,
which are excluded parameters, where the first case is already excluded since the algebra is
defined only for $s\geq 3$, and where both cases are excluded by our hypothesis.
This number can be simplified and we therefore define
$u(k,s):=(k-3)+(4k-3)(s-1)>0$ for our admissible parameters.
Hence if there is an element
$y_a\in K$ such that $y_a^{u(k,s)}=a^{-k}$, then we may choose such a
parameter $x_\eta$, such that we may assume $a=1$. This holds in particular if $K$
is algebraically closed.
We obtain this way $A^{k,s}(a,0)\simeq A^{k,s}(1,0)$ if $K$ is sufficiently big, i.e. there is an element $y_a$ satisfying $y_a^{u(k,s)}=a^{-k}$.
Moreover, since $s\geq 3$ and $k\geq 1$ we get
$$-\frac{(k-1)(s-1)-1}{k}(k-2)+k(s-1)= \frac{(k-2)+(3k-2)(s-1)}k\geq \frac{(k-2)+2(3k-2)}k=\frac{7k-6}k>0.$$
Let
$v(k,s):=(k-2)+(3k-2)(s-1)$. If there is $y_c\in K$ such that $y_c^{v(k,s)}=c^{-k}$, then
we can therefore choose $x_\eta$ such that we can assume $c=1$ if $c\neq 0$.
Again, this is trivially true if $K$ is algebraically closed.
As a consequence, combining our computation and the result \cite[Lemma 5.7]{ErdSkow},
if $K$ is algebraically closed of characteristic different from $2$, then
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(1,0)$.
\end{Rem}
\begin{Theorem}\label{main}
Let $A^{k,s}(a,c)$ be the algebra $Q(2{\mathfrak B})_1^{k,s}(a,c)$
over a field $K$ of characteristic $p$. Let $a,c\in K\setminus\{ 0\}$.
We get the following cases.
\begin{enumerate}
\item \label{item1} Suppose $p=2$.
\begin{enumerate}
\item If $k=1$, and
\begin{enumerate}
\item if $s$ is even or if $a$ is not a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0))),$$
\item if $s$ is odd and $a$ is a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0)))-1.$$
\end{enumerate}
\item If $k>1$ is odd, and
\begin{enumerate}
\item if $s$ is even and if $c$ is a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0))),$$
\item if $s$ is odd or if $c$ is not a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0)))-1.$$
\end{enumerate}
\item If $k$ is even, and
\begin{enumerate}
\item if $c$ is a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0))),$$
\item if $c$ is not a square in $K$, then
$$\dim_K(T_1^\perp(A^{k,s}(a,c)))=\dim_K(T_1^\perp(A^{k,s}(a,0)))-1.$$
\end{enumerate}
\end{enumerate}
\item \label{item2} Suppose $K$ is a perfect field of characteristic $p=2$.
\begin{enumerate}
\item Then $A^{k,s}(a,0)\simeq A^{k,s}(1,0)$ and
\begin{enumerate}
\item if $k$ and $s$ are even, then
$$Z(A^{k,s}(a,0))/T_1^\perp(A^{k,s}(a,0))\simeq K[U,Y,S]/(U^{k/2}-Y^{s/2},S^2,UY,US,YS),$$
\item if $k>1$ or $s$ is odd, then
$$Z(A^{k,s}(a,0))/T_1^\perp(A^{k,s}(a,0))\simeq K[U,Y,S]/(U^{\lceil k/2\rceil},Y^{\lceil s/2\rceil},S^2,UY,US,YS).$$
\end{enumerate}
\item
If $c\neq 0$, then
\begin{enumerate}
\item if $k$ and $s$ are even, then
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y]/(U^{k/2}-Y^{s/2},UY),$$
\item if $k>1$ or $s$ is odd, then
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y]/(U^{\lceil (k+1)/2\rceil},Y^{\lceil (s+1)/2\rceil},UY).$$
\end{enumerate}
\item If $k=1$, then
\begin{enumerate}
\item if ($s$ is odd and $c=0$) or ($s$ is even and $c\neq 0$),
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y,S]/(Y^{\lceil (s+1)/2\rceil},S^2,YS),$$
\item if ($s$ is odd and $c\neq 0$) or ($s$ is even and $c= 0$),
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y]/(Y^{\lceil s/2\rceil}).$$
\end{enumerate}
\end{enumerate}
\item \label{item3} If $p\neq 2$, and if $K$ is algebraically closed, then $A^{k,s}(a,c)\simeq A^{k,s}(a,c)$ for some $a'\in K^\times,$ and if $(k,s)\neq (1,3)$, then $A^{k,s}(a,c)\simeq A^{k,s}(1,0)$.
\item \label{item4} If $p>3$ or $n\geq 2$, then the dimension of the K\"ulshammer ideal
$T_n(A^{k,s}(a,c))^\perp$ does not depend on
the parameters $a,c$.
\end{enumerate}
Suppose that $K$ is algebraically closed.
Then $A^{k,s}(a,0)\simeq A^{k,s}(1,0)$ if $(k,s)\neq (1,3)$ and if $c\neq 0$ then
$A^{k,s}(a,c)\simeq A^{k,s}(a',1)\simeq A^{k,s}(1,c')$ for some $a',c'\in K^\times$.
\end{Theorem}
\begin{Rem}
Consider the case $s=3$ and $k=1$. The computations in Remark~\ref{remarkparameters}
show that if $K$ is algebraically closed, and $c\neq 0$, then for
each $a'$ there is $a$ such that
$A^{1,3}(a',c)\simeq A^{1,3}(a,1)$.
This case is quite particular which allows
an alternative argument for distinguishing derived equivalence classes.
If $c=0$, then the relations of $A^{1,3}(a,0)$ are homogeneous. Therefore
the algebra $A^{1,3}(a,0)$ is graded by path lengths with semisimple degree $0$ component.
A theorem of Rouquier \cite[Theorem 6.1]{Rouqgrad} shows that
if $A^{1,3}(a,0)$ is derived equivalent to another algebra $B$, then the induced
stable
equivalence of Morita type induces a grading on $B$. Moreover,
by \cite[Lemma 5.21]{Rouqgrad}
the degree $0$ component of $A^{1,3}(a,0)$ is of finite global dimension if and only if
the degree $0$ component of $B$ is of finite global dimension.
\end{Rem}
\begin{Rem}
The hypothesis that $K$ is algebraically closed is stronger than required. A more precise,
somewhat technical statement is given at the end of Remark~\ref{remarkparameters}.
\end{Rem}
Proof of Theorem~\ref{main}. The isomorphisms
$A^{k,s}(a,0)\simeq A^{k,s}(1,0)$ for $(k,s)\neq (1,3)$ and if $c\neq 0$ then $A^{k,s}(1,c')\simeq A^{k,s}(a,c)\simeq A^{k,s}(a',1)$
for some $a',c'\in K^\times$
follow from Remark~\ref{remarkparameters}.
Define the following subsets of $Q(2{\mathfrak B})_1^{k,s}(a,c)$.
$${\mathcal B}_1:=\{\alpha(\beta\gamma\alpha)^n,(\beta\gamma\alpha)^n\beta\gamma,
(\beta\gamma\alpha)^m, e_1(\alpha\beta\gamma)^\ell\;|\;0\leq\ell\leq k,
1\leq m\leq k-1,0\leq n\leq k-1\},$$
$${\mathcal B}_2:=\{e_2\eta^t,(\gamma\alpha\beta)^m\;|\;0\leq t\leq s,1\leq m\leq k-1\},$$
$${\mathcal B}_3:=\{(\beta\gamma\alpha)^n\beta,\alpha(\beta\gamma\alpha)^n\beta \;|\;
0\leq n\leq k-1\},$$
$${\mathcal B}_4:=\{(\gamma\alpha\beta)^n\gamma,(\gamma\alpha\beta)^n\gamma\alpha \;|\;
0\leq n\leq k-1\}.$$
The (disjoint) union of these sets forms a basis of $Q(2{\mathfrak B})_1^{k,s}(a,c)$, using
the known Cartan matrix of $Q(2{\mathfrak B})_1^{k,s}(a,c)$.
As a next step we need to compute the commutator space. Clearly, non
closed paths are commutators, since if $e_ipe_j\neq 0$ for some path $p$
and $e_i\neq e_j$, then
$p=e_ip-pe_i$. Hence ${\mathcal B}_3\cup {\mathcal B}_4\subseteq [A^{k,s}(a,c),A^{k,s}(a,c)]$.
Moreover, $$\alpha(\beta\gamma\alpha)^n=
\alpha(\beta\gamma\alpha)^n-(\beta\gamma\alpha)^n\alpha\in [A(a,c),A(a,c)]\;\forall n\geq 1,$$
$$(\beta\gamma\alpha)^m=(\alpha\beta\gamma)^m=
(\gamma\alpha\beta)^m\in A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]\;\forall m\geq 0,$$
$$(\beta\gamma\alpha)^m\beta\gamma=\gamma(\beta\gamma\alpha)^m\beta=0
\in A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]\;\forall m\geq 1,$$
and hence $${\mathcal B}_{comm}:=\{\alpha,(\alpha\beta\gamma)^m,
\eta^t,e_1,e_2\;|\;1\leq t\leq s-1,1\leq m\leq k\}$$
is a generating set of $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$. Since the dimension of
the centre of $A^{k,s}(a,c)$ equals the dimension of $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$,
the algebra $A^{k,s}(a,c)$ being symmetric, and both are of dimension $2+k+s$,
by \cite{Holmhabil}, we get that the classes represented by the elements
${\mathcal B}_{comm}$ form actually a basis
of $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$.
We only need to work in $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$, and therefore
we need to consider linear combinations
of paths in ${\mathfrak B}_{comm}$ only. We can omit idempotents,
since computing modulo the radical these idempotents
remain idempotents, and are hence never nilpotent modulo commutators. Hence
we only need to consider linear combinations of elements in the set
$$\{\eta^t,\alpha,(\beta\gamma\alpha)^m\;|\;1\leq t\leq s-1,1\leq m\leq k\}.$$
We deal with the case $p=2$.
Let hence $p=2$.
In the commutator quotient squaring is semilinear
(cf e.g. \cite{Ku1},\cite[Lemma 2.9.3]{reptheobuch}).
{\bf If $k>1$ is odd,} then
\begin{eqnarray*}
0&=&\left(\sum_{t=1}^{s-1}x_t\eta^t+u\alpha+\sum_{m=1}^ky_m(\beta\gamma\alpha)^m\right)^2\\
&=&\sum_{t=1}^{s-1}x_t^2\eta^{2t}+
u^2c(\beta\gamma\alpha)^k+\sum_{m=1}^ky_m^2(\beta\gamma\alpha)^{2m}\\
&=&\sum_{1\leq t\leq s/2}x_t^2\eta^{2t}+
u^2c(\beta\gamma\alpha)^k+\sum_{m=1}^{(k-1)/2}y_m^2(\beta\gamma\alpha)^{2m},
\end{eqnarray*}
which implies $x_t=0$ for all $t\leq \frac{s}2$, $y_m=0$ for all $m\leq \frac{k-1}2$.
If $c=0$ then there is no other constraint.
Suppose $c\neq 0$.
Then, if $s$ is odd we get $u=0$. If $s$ is even, then
$x_{s/2}^2+cu^2=0$ which has a non trivial solution if and only if $c$ is a square in $K$.
Hence, computing in $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$ we get
$$T_1(A^{k,s}(a,c))=\left\{\begin{array}{ll}
\langle \alpha,\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac s2,m>\frac{k-1}2\rangle_K&\mbox{ if $c=0$}\\
\langle \eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac s2,m>\frac{k-1}2\rangle_K&\mbox{ if $s$ is odd and $c\neq 0$}\\
\langle \eta^t,(\beta\gamma\alpha)^m\;|\;
t>\frac s2,m>\frac{k-1}2\rangle_K&\mbox{ if $s$ is even and $0\neq c\not\in K^2$}\\
\langle \eta^t,(\beta\gamma\alpha)^m,\eta^{s/2}+d\alpha\;|\;
t>\frac s2,m>\frac{k-1}2\rangle_K&\mbox{ if $s$ is even and $0\neq c=d^2$}
\end{array}\right.$$
{\bf If $k$ is even,} then
\begin{eqnarray*}
0&=&\left(\sum_{t=1}^{s-1}x_t\eta^t+u\alpha+\sum_{m=1}^ky_m(\beta\gamma\alpha)^m\right)^2\\
&=&\sum_{t=1}^{s-1}x_t^2\eta^{2t}+
u^2c(\beta\gamma\alpha)^k+\sum_{m=1}^ky_m^2(\beta\gamma\alpha)^{2m}\\
&=&\sum_{1\leq t\leq (s-1)/2}x_t^2\eta^{2t}+
u^2c(\beta\gamma\alpha)^k+\sum_{m=1}^{k/2}y_m^2(\beta\gamma\alpha)^{2m},
\end{eqnarray*}
which implies $y_m=0$ for $1\leq m<k/2$ and $x_t=0$
for $t\leq \frac{s-1}2$.
If $c=0$, then $x_{s/2}+y_{k/2}=0$ in case $s$ is even, and $y_{k/2}=0$ in case $s$ is odd.
Suppose $c\neq 0$.
If $s$ is odd, then $y_{k/2}^2+cu^2=0$, and if $s$ is even, then $y_{k/2}^2+x_{s/2}^2+cu^2=0$.
Again $y_{k/2}^2+cu^2=0$ and
$y_{k/2}^2+x_{s/2}^2+cu^2=0$ has non zero solutions if and only if $c$ is a square.
Computing in $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$ we get
$$T_1(A(a,c))=\left\{\begin{array}{ll}
\langle \alpha,\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac{s-1}2,m>\frac{k}2\rangle_K&
\mbox{ if $s$ is odd and $c=0$}\\
\langle \alpha,\eta^{s/2}+(\beta\gamma\alpha)^{k/2},\eta^t,(\beta\gamma\alpha)^m\;|\;
t>\frac{s}2,m>\frac{k}2\rangle_K&\mbox{ if $s$ is even and $c=0$}\\
\langle \eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac{s-1}2,m>\frac{k}2\rangle_K&
\mbox{ if $s$ is odd and $0\neq c\not\in K^2$}\\
\langle \eta^t,(\beta\gamma\alpha)^m,(\beta\gamma\alpha)^{k/2}+d\alpha\;|\;
t>\frac{s-1}2,m>\frac{k}2\rangle_K&\mbox{ if $s$ is odd and $0\neq c=d^2$}\\
\langle \eta^t,(\beta\gamma\alpha)^m,
(\beta\gamma\alpha)^{k/2}+\eta^{s/2}\;|\;t>\frac{s}2,m>\frac{k}2\rangle_K&
\mbox{ if $s$ is even and $0\neq c\not\in K^2$}\\
\langle \eta^t,(\beta\gamma\alpha)^m,(\beta\gamma\alpha)^{k/2}+\eta^{s/2},\eta^{s/2}+d\alpha\;|\;
t>\frac s2,m>\frac{k}2\rangle_K&\mbox{ if $s$ is even and $0\neq c=d^2$}
\end{array}\right.$$
{\bf If $k=1$,} then, since $\beta\gamma=\gamma\beta=\eta^{s-1}$ in
the commutator quotient,
\begin{eqnarray*}
0=\left(\sum_{t=1}^{s-1}x_t\eta^t+u\alpha+y_1(\beta\gamma\alpha)\right)^2
&=&\sum_{t=1}^{s-1}x_t^2\eta^{2t}+u^2a\eta^{s-1}+
u^2c(\beta\gamma\alpha)
\end{eqnarray*}
which implies $x_t=0$ for $1\leq t\leq \frac{s-2}2$.
If $c=0$, then $x_{s/2}=0$ in case $s$ is even, and $x_{(s-1)/2}^2+au^2=0$ in case $s$ is odd.
This last equation has non zero solutions if and only if $a\in K^2$.
Suppose $c\neq 0$. If $s$ is even, then
$x_{s/2}^2+cu^2=0$. This has non zero solutions if and only if $c\in K^2$.
If $s$ is odd, then $cu^2=0$ and $x_{(s-1)/2}^2+au^2=0$. Hence
$s$ odd implies $u=0=x_{(s-1)/2}$. Computing again in $A^{k,s}(a,c)/[A^{k,s}(a,c),A^{k,s}(a,c)]$,
$$T_1(A^{k,s}(a,c))=\left\{\begin{array}{ll}
\langle \eta^{(s-1)/2}+b\alpha,\eta^t,(\beta\gamma\alpha)\;|\;
t>\frac{s-1}2\rangle_K&\mbox{ if $s$ is odd, $a=b^2$ and $c=0$}\\
\langle \eta^t,(\beta\gamma\alpha)\;|\;t>\frac{s-1}2\rangle_K&\mbox{ if $s$ is odd, $a\not\in K^2$ and $c=0$}\\
\langle \eta^t,(\beta\gamma\alpha)\;|\;t>\frac{s}2\rangle_K&\mbox{ if $s$ is even and $c=0$}\\
\langle \eta^t,(\beta\gamma\alpha)\;|\;t>\frac{s-1}2\rangle_K&\mbox{ if $s$ is odd and $c\neq 0$}\\
\langle \eta^t,(\beta\gamma\alpha)\;|\;t>\frac{s}2\rangle_K&\mbox{ if $s$ is even and $0\neq c\not\in K^2$}\\
\langle \eta^t,(\beta\gamma\alpha),\eta^{s/2}+d\alpha\;|\;t>\frac s2\rangle_K&\mbox{ if $s$ is even and $0\neq c=d^2$}
\end{array}\right.$$
It is easy to see that computing $T_n(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]$
for $n\geq 2$ (and any $p\geq 2$ in this case)
yields expressions which are independent of $a,c$.
\bigskip
In order to compute the K\"ulshammer ideal $T_1(A^{k,s}(a,c))^\perp$ we need to give the symmetrising form of $A^{k,s}(a,c)$.
Recall that we have a basis ${\mathcal B}=\bigcup_{i=1}^4{\mathcal B}_i$
of $A^{(k,s)}(a,c)$ given by
$${\mathcal B}_1:=\{\alpha(\beta\gamma\alpha)^n,(\beta\gamma\alpha)^n\beta\gamma,
(\beta\gamma\alpha)^m, e_1(\alpha\beta\gamma)^\ell\;|\;0\leq\ell\leq k,
1\leq m\leq k-1,0\leq n\leq k-1\},$$
$${\mathcal B}_2:=\{e_2\eta^t,(\gamma\alpha\beta)^m\;|\;0\leq t\leq s,1\leq m\leq k-1\},$$
$${\mathcal B}_3:=\{(\beta\gamma\alpha)^n\beta,\alpha(\beta\gamma\alpha)^n\beta \;|\;
0\leq n\leq k-1\},$$
$${\mathcal B}_4:=\{(\gamma\alpha\beta)^n\gamma,(\gamma\alpha\beta)^n\gamma\alpha \;|\;
0\leq n\leq k-1\}.$$
We define a trace map $$A(a,c)\stackrel{\psi}{\lra} K$$ by
$$\psi(\eta^s)=\psi((\alpha\beta\gamma)^k)=1,\mbox{ and }\psi(x)=0\mbox{ if }x \mbox{ is a path in the quiver such that } x\in {\mathcal B}\setminus\soc(A^{k,s}(a,c)).$$
Note that $(\alpha\beta\gamma)^k=\beta\eta\gamma=(\beta\gamma\alpha)^k$.
Remark~\ref{symmetrisingform} indicates that $\psi$ should coincide on these
socle elements for $\psi$ to define a symmetric form. Indeed,
$\eta^s=\eta\gamma\beta=(\gamma\alpha\beta)^k$
and hence
$$\eta^s-(\alpha\beta\gamma)^k=(\gamma\alpha\beta)^k-(\alpha\beta\gamma)^k=
[\gamma,\alpha\beta(\gamma\alpha\beta)^{k-1}]$$
is a commutator.
We need to prove that $\psi(c_1c_2)=\psi(c_2c_1)$ for all elements $c_1,c_2\in{\mathcal B}$.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_1$:}
We obtain
$\alpha(\beta\gamma\alpha)^{n_1}\cdot \alpha(\beta\gamma\alpha)^{n_2}=0$ if $n_1+n_2>0$, and the case $n_1=n_2=0$ is clearly symmetric.
$$\alpha(\beta\gamma\alpha)^{n_1}\cdot (\beta\gamma\alpha)^{n_2}\beta\gamma=
(\alpha\beta\gamma)^{n_1+n_2+1}=
(\beta\gamma\alpha)^{n_2}\beta\gamma\cdot \alpha(\beta\gamma\alpha)^{n_1},$$
$$\alpha(\beta\gamma\alpha)^{n_1}\cdot (\beta\gamma\alpha)^{m_2}=
(\alpha\beta\gamma)^{n_1+m_2}\alpha\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)$$
is mapped to $0$ by $\psi$, and $m_2>0$ implies
$$(\beta\gamma\alpha)^{m_2}\cdot \alpha(\beta\gamma\alpha)^{n_1}=0.$$
Now, if $\ell_2>0$, then
$$\alpha(\beta\gamma\alpha)^{n_1}\cdot (\alpha\beta\gamma)^{\ell_2}=0$$
and
$$(\alpha\beta\gamma)^{\ell_2}\cdot\alpha(\beta\gamma\alpha)^{n_1}=
(\alpha\beta\gamma)^{\ell_2+n_1}\alpha\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)$$
is mapped to $0$ by $\psi$. If $\ell_2=0$, then the two elements commute trivially.
$$(\beta\gamma\alpha)^{n_1}\beta\gamma\cdot (\beta\gamma\alpha)^{n_2}\beta\gamma=0=
(\beta\gamma\alpha)^{n_2}\beta\gamma\cdot (\beta\gamma\alpha)^{n_1}\beta\gamma$$
and
$$(\beta\gamma\alpha)^{n_1}\beta\gamma\cdot (\beta\gamma\alpha)^{m_2}=0$$
whereas
$$(\beta\gamma\alpha)^{m_2}\cdot(\beta\gamma\alpha)^{n_1}\beta\gamma=
(\beta\gamma\alpha)^{m_2+n_1}\beta\gamma\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)$$
is mapped to $0$ by $\psi.$ If $\ell_2>0$, then
$$(\beta\gamma\alpha)^{n_1}\beta\gamma\cdot (\alpha\beta\gamma)^{\ell_2}=
(\beta\gamma\alpha)^{n_1+\ell_2}\beta\gamma\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)$$
is mapped to $0$ by $\psi$, whereas
$$(\alpha\beta\gamma)^{\ell_2}\cdot(\beta\gamma\alpha)^{n_1}\beta\gamma=0.$$
Clearly $e_1$ commutes with $(\beta\gamma\alpha)^{n_1}\beta\gamma$.
Now, trivially
$$(\beta\gamma\alpha)^{m_1}\cdot (\beta\gamma\alpha)^{m_2}=(\beta\gamma\alpha)^{m_2}\cdot (\beta\gamma\alpha)^{m_1}$$
and
$$(\alpha\beta\gamma)^{\ell_1}\cdot (\alpha\beta\gamma)^{\ell_2}=(\alpha\beta\gamma)^{\ell_2}\cdot (\alpha\beta\gamma)^{\ell_1}.$$
Finally, if $\ell_1>0$ then
$$(\alpha\beta\gamma)^{\ell_1}\cdot (\beta\gamma\alpha)^{m_2}=0
= (\beta\gamma\alpha)^{m_2}\cdot(\alpha\beta\gamma)^{\ell_1}.$$
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_2$:}
Since ${\mathcal B}_1\subseteq e_1A^{k,s}(a,c)e_1$, and since
${\mathcal B}_2\subseteq e_2A^{k,s}(a,c)e_2$ we get $\psi(c_1c_2)=\psi(c_2c_1)=0$ for
$c_1\in{\mathcal B_1}$ and $c_2\in{\mathcal B_2}$.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_3$:}
Since ${\mathcal B}_1\subseteq e_1A^{k,s}(a,c)e_1$, and since
${\mathcal B}_3\subseteq e_1A^{k,s}(a,c)e_2$ we get $c_1c_2=0$ and $c_2c_1\in e_1A^{k,s}(a,c)e_3$ for
$c_1\in{\mathcal B_3}$ and $c_2\in{\mathcal B_1}$. Non closed paths are mapped to $0$ by $\psi$.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_4$:}
Since ${\mathcal B}_4\subseteq e_2A^{k,s}(a,c)e_1$ the same arguments as in the case $(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_3$ apply.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_2\times {\mathcal B}_2$:}
Clearly $\eta^{t_1}\cdot\eta^{t_2}=\eta^{t_2}\cdot\eta^{t_1}$ and
$(\gamma\alpha\beta)^{m_1}\cdot(\gamma\alpha\beta)^{m_2}=
(\gamma\alpha\beta)^{m_2}\cdot(\gamma\alpha\beta)^{m_1}.$
Moreover, if $t>0$ then
$$\eta^t\cdot(\gamma\alpha\beta)^m=0=(\gamma\alpha\beta)^m\cdot\eta^t.$$
If $t=0$, then trivially $\eta^t\cdot(\gamma\alpha\beta)^m=(\gamma\alpha\beta)^m\cdot\eta^t.$
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_2\times {\mathcal B}_3$:}
Since then $c_1c_2$ and $c_2c_1$ are non closed paths, the same arguments as in the case
$(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_3$ apply.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_2\times {\mathcal B}_4$:}
Again since then $c_1c_2$ and $c_2c_1$ are non closed paths, the same arguments as in the case
$(c_1,c_2)\in {\mathcal B}_1\times {\mathcal B}_3$ apply.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_3\times {\mathcal B}_3$:}
Then $c_1c_2=0=c_2c_1$.
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_3\times {\mathcal B}_4$:}
$$(\beta\gamma\alpha)^{n_1}\beta\cdot (\gamma\alpha\beta)^{n_2}\gamma=
(\beta\gamma\alpha)^{n_1+n_2}\beta\gamma\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)$$
is mapped to $0$ by $\psi$,
and
$$(\gamma\alpha\beta)^{n_2}\gamma\cdot (\beta\gamma\alpha)^{n_1}\beta=
(\gamma\alpha\beta)^{n_2}(\gamma\beta)(\gamma\alpha\beta)^{n_1}=0.$$
Now,
$$(\beta\gamma\alpha)^{n_1}\beta\cdot (\gamma\alpha\beta)^{n_2}\gamma\alpha-
(\gamma\alpha\beta)^{n_2}\gamma\alpha\cdot (\beta\gamma\alpha)^{n_1}\beta=
(\beta\gamma\alpha)^{n_1+n_2+1}-(\gamma\alpha\beta)^{n_1+n_2+1}.$$
and the value of $\psi$ on each of the summands is equal.
$$\alpha(\beta\gamma\alpha)^{n_1}\beta\cdot(\gamma\alpha\beta)^{n_2}\gamma=
(\alpha\beta\gamma)^{n_1+n_2+1}$$
and
$$(\gamma\alpha\beta)^{n_2}\gamma\cdot\alpha(\beta\gamma\alpha)^{n_1}\beta=
(\gamma\alpha\beta)^{n_2+n_1+1}$$
both have identical values under $\psi.$
Finally
$$\alpha(\beta\gamma\alpha)^{n_1}\beta\cdot(\gamma\alpha\beta)^{n_2}\gamma\alpha=
(\alpha\beta\gamma)^{n_1+n_2+1}\alpha\in({\mathcal B}_1\cup\{0\})\setminus\soc(A^{(k,s)}(a,c)
$$
is mapped to $0$ by $\psi$
and
$$(\gamma\alpha\beta)^{n_2}\gamma\alpha\cdot\alpha(\beta\gamma\alpha)^{n_1}\beta
=0.$$
\noindent
{\bf Case $(c_1,c_2)\in {\mathcal B}_4\times {\mathcal B}_4$:}
Then $c_1c_2=0=c_2c_1$.
Altogether this shows that $\psi$ is symmetric. The fact that $\psi$ defines a non degenerate
form follows as in the proof of Proposition~\ref{prop:form}. For the reader's convenience
we recall the short argument. Suppose that the form defined by $\psi$ is degenerate. Then
there is a $0\neq x\in A^{k,s}(a,c)$ with $\psi(xy)=0$ for all $y$, and since $1=e_1+e_2$ there is a
primitive idempotent $e\in\{e_1,e_2\}$ of $A^{k,s}(a,c)$ such that we may suppose that
$x\in eA^{k,s}(a,c)$. Let $S$ be a simple submodule of
$xA^{k,s}(a,c)$ and there is $y$ such that
$0\neq s=xy\in S$. Since $S\leq eA^{k,s}(a,c)$ is one-dimensional,
and included in the socle, and since $e{\mathcal B}e$ contains a basis of $S$ we get $\psi(xy)\neq 0$.
The form defined by $\psi$ is trivially associative. Hence $\psi$ defines
a symmetrising form.
\medskip
We come to the main body of the proof.
Recall from Lemma~\ref{centreofQ2} that $\dim_K(Z(A^{k,s}(a,c)))=k+s+2$.
We proceed case by case.
\medskip
{\bf $k>1$ odd and $c=0$}: Recall that in this case
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle\alpha,\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac s2; m>\frac{k-1}2
\right\rangle_K.$$
Hence
\begin{eqnarray*}\dim_K(T_1(A^{k,s}(a,0))/[A^{k,s}(a,0),A^{k,s}(a,0)])&=&
\left\{\begin{array}{ll}1+\frac s2+\frac{k+1}2-1&\mbox{ if $s$ is even}\\
1+\frac{s+1}2+\frac{k+1}2-1&\mbox{ if $s$ is odd}
\end{array}\right.\\
&=&\left\{\begin{array}{ll}\frac s2+\frac{k+1}2&\mbox{ if $s$ is even}\\
\frac{s+1}2+\frac{k+1}2&\mbox{ if $s$ is odd}
\end{array}\right.
\end{eqnarray*}
observing that $(\beta\gamma\alpha)^k-\eta^s\in [A^{k,s}(a,c),A^{k,s}(a,c)]$.
Therefore
\begin{eqnarray*}
\dim_K(T_1(A^{k,s}(a,0))^\perp)&=&k+s+2-\left\{\begin{array}{ll}\frac s2+\frac{k+1}2&\mbox{ if $s$ is even}\\
\frac{s+1}2+\frac{k+1}2&\mbox{ if $s$ is odd}
\end{array}\right.\\
&=&\left\{\begin{array}{ll}\frac s2+1+\frac{k+1}2&\mbox{ if $s$ is even}\\
\frac{s+1}2+\frac{k+1}2&\mbox{ if $s$ is odd}
\end{array}\right.
\end{eqnarray*}
But, in case $s$ is even,
$$\left\{\eta^t,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u,(\alpha\beta\gamma)^k\;|\;
u\geq\frac{k+1}2, t\geq \frac s2\right\}\subseteq T_1(A(a,c))^\perp,$$
and in case $s$ is odd,
$$\left\{\eta^t,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u,(\alpha\beta\gamma)^k\;|\;
u\geq\frac{k+1}2, t\geq \frac{s+1}2\right\}\subseteq T_1(A(a,c))^\perp.$$
This is a basis of a subspace of the centre of the dimension as required, and hence
the set above is a basis of $T_1(A^{k,s}(a,c))^\perp$. Hence, with these parameters, if $s$ is even then
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y,S]/(Y^{s/2},U^{(k+1)/2},S^2,YS,US,UY),$$
and if $s$ is odd, then
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y,S]/(Y^{(s+1)/2},U^{(k+1)/2},S^2,YS,US,UY).$$
{\bf $k>1$ odd, $c\neq 0$, and $s$ is odd}:
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac s2; m>\frac{k-1}2
\right\rangle_K.$$
In this case
we get $\alpha^2\in T_1(A^{k,s}(a,c))^\perp$, and using the preceding discussion we get that
$$\left\{\alpha^2, (\beta\gamma\alpha)^k,\eta^t,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u\;|\;
u\geq\frac{k+1}2,t\geq \frac{s+1}2\right\}$$
is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp$.
Hence in this case
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y]/(Y^{(s+1)/2},U^{(k+1)/2},UY).$$
{\bf $k>1$ odd, $d^2=c\neq 0$, and $s$ is even.}
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle d\alpha+\eta^{s/2},\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac s2; m>\frac{k-1}2
\right\rangle_K.$$
Then
$$\left(\eta^{s/2}+d\alpha\right)\cdot\left(\eta^{s/2}+\frac d{ca}\alpha^2\right)=
\eta^s+\frac{d^2}{ca}\alpha^3=
\eta^s+\frac c{ca}\alpha^3=\eta^s+\frac 1a\cdot a(\alpha\beta\gamma)^k
$$
and this is mapped to $0$ by $\psi$. Hence,
$$\left\{\frac{d}{ca}\alpha^2+\eta^{s/2}, (\alpha\beta\gamma)^k,
\eta^t, (\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u\;|\;
u\geq\frac{k-1}2,t\geq \frac s2+1\right\}$$ is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp$.
Therefore in this case
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y]/(Y^{\frac s2+1},U^{(k+1)/2},UY).$$
{\bf $k$ even and $c=0$ and $s$ is odd}.
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle d\alpha+(\beta\gamma\alpha)^{k/2},\eta^t,(\beta\gamma\alpha)^m\;|\;t>\frac {s-1}2;
m>\frac{k}2
\right\rangle_K.$$
Then the discussion of the case $k>1$ odd and $c=0$ shows that
$$\left\{\eta^t,(\alpha\beta\gamma)^k,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u\;|\;
u\geq\frac{k}2,t\geq \frac {(s+1)}2\right\}$$ is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp.$
Hence in this case
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y,S]/(Y^{ (s+1)/2},U^{k/2},S^2,YS,US,UY).$$
{\bf $k$ even and $c=0$ and $s$ is even}.
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle \alpha,\eta^{s/2}+(\beta\gamma\alpha)^{k/2},\eta^t,(\beta\gamma\alpha)^m\;|\;
t>\frac s2; m>\frac{k}2
\right\rangle_K.$$
Then
$$\left\{(\alpha\beta\gamma)^k,\eta^t,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u,\eta^{s/2}+(\beta\gamma\alpha)^{k/2}\;|\;
u\geq\frac{k}2+1,t\geq \frac s2+1\right\}$$ is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp.$
Hence in this case
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y,S]/(Y^{s/2}-U^{k/2},S^2,YS,US,UY).$$
{\bf $k$ even, $0\neq c=d^2$, $s$ odd:}
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle d\alpha+(\beta\gamma\alpha)^{k/2},\eta^t,(\beta\gamma\alpha)^m\;|\;
t>\frac {s-1}2; m>\frac{k}2
\right\rangle_K.$$
Since $\dim(Z(A^{k,s}(a,c)))=k+s+2$, and since
$$\dim(T_1(A^{k,s}(a,c)))/[A^{k,s}(a,c),A^{k,s}(a,c)]=\frac k2+\frac{s+1}2,$$
we get $$\dim(T_1(A^{k,s}(a,c))^\perp)=2+k+s-\frac k2-\frac{s+1}2=\frac k2+\frac{s+1}2+1.$$
Moreover,
$$
\left(\frac{d}{c}\alpha^2+U^{k/2}\right)\cdot\left(d\alpha+(\beta\gamma\alpha)^{k/2}\right)=
\frac{d^2}c\alpha^3+dU^{k/2}\alpha+\frac{d}c(\beta\gamma\alpha)^{k/2}+(\beta\gamma\alpha)^k
$$
and this maps to $0$ by the map $\psi$.
Therefore
$$\left\{(\beta\gamma\alpha)^k,
\frac{d}{c}\alpha^2+(\alpha\beta\gamma)^{k/2}+(\beta\gamma\alpha)^{k/2}+(\gamma\alpha\beta)^{k/2},
\eta^t,(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u\;|\;
u\geq\frac{k}2+1,t\geq \frac {s+1}2\right\}$$ is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp,$
and therefore
$$Z(A^{k,s}(a,c))/T_1(A^{k,s}(a,c))^\perp\simeq K[U,Y]/(U^{k/2},Y^{(s+1)/2},UY).$$
{\bf $k$ even, $c=d^2\neq 0$, and $s$ even:}
Recall
$$T_1(A^{k,s}(a,c))/[A^{k,s}(a,c),A^{k,s}(a,c)]=
\left\langle
d\alpha+\eta^{s/2},(\beta\gamma\alpha)^{k/2}+\eta^{s/2},\eta^t,(\beta\gamma\alpha)^m\;|\;
t>\frac s2; m>\frac{k}2
\right\rangle_K.$$
Then
$$\left\{\alpha^2+d\eta^{s/2},(\beta\gamma\alpha)^k,\eta^t,
(\alpha\beta\gamma)^u+(\beta\gamma\alpha)^u+(\gamma\alpha\beta)^u,\eta^{s/2}+(\beta\gamma\alpha)^{k/2}\;|\;
u\geq\frac{k}2+1,t\geq \frac {s}2+1\right\}$$ is a $K$-basis of $T_1(A^{k,s}(a,c))^\perp.$
Hence in this case
$$Z(A^{k,s}(a,c))/T_1^\perp(A^{k,s}(a,c))\simeq K[U,Y]/(Y^{ s/2}-U^{k/2},UY).$$
{\bf If $k=1$ and $c=0$ and $s$ odd:}
Since $\dim(Z(A^{1,s}(a,c))=3+s$, and since $$\dim(T_1(A^{1,s}(a,c))/[A^{1,s}(a,c),A^{1,s}(a,c)])=3+\frac{s-1}2,$$ we obtain $\dim(T_1(A^{1,s}(a,c))^\perp)=\frac{s+1}2$.
Observe that $\eta^s=(\alpha\beta\gamma)+(\beta\gamma\alpha)+(\gamma\alpha\beta)=U$.
Then we get
$$\left\{\beta\gamma\alpha,\eta^t,\;|\;
t\geq \frac {s+1}2\right\}$$ is a $K$-basis of $T_1(A^{1,s}(a,c))^\perp$.
Therefore
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y,S]/(Y^{(s+1)/2},S^2,YS).$$
{\bf If $k=1$ and $c=0$ and $s$ even:}
Since $\dim(Z(A^{1,s}(a,c))=3+s$, and since $$\dim(T_1(A^{1,s}(a,c))/[A^{1,s}(a,c),A^{1,s}(a,c)])=1+\frac{s}2,$$ we obtain $\dim(T_1(A^{1,s}(a,c))^\perp)=2+\frac{s}2$.
Hence
$$\left\{\alpha^2,\beta\gamma\alpha,\eta^t,\;|\;
t\geq \frac {s}2\right\}$$ is a $K$-basis of $T_1(A^{1,s}(a,c))^\perp$ and
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y]/Y^{s/2}.$$
{\bf If $k=1$ and $c\neq 0$ and $s$ odd:}
Since $\dim(Z(A^{1,s}(a,c))=3+s$, and since $$\dim(T_1(A^{1,s}(a,c))/[A^{1,s}(a,c),A^{1,s}(a,c)])=2+\frac{s-1}2,$$ we obtain $\dim(T_1(A^{1,s}(a,c))^\perp)=1+\frac{s+1}2$.
Hence
$$\left\{\alpha^2,\beta\gamma\alpha,\eta^t,\;|\;
t\geq \frac {s+1}2\right\}$$ is a $K$-basis of $T_1(A^{1,s}(a,c))^\perp$ and
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y]/Y^{(s+1)/2}.$$
{\bf If $k=1$ and $c\neq 0$ and $s$ even:}
Since $\dim(Z(A^{1,s}(a,c))=3+s$, and since $$\dim(T_1(A^{1,s}(a,c))/[A^{1,s}(a,c),A^{1,s}(a,c)])=2+\frac{s}2,$$ we obtain $\dim(T_1(A^{1,s}(a,c))^\perp)=1+\frac{s}2$.
Hence
$$\left\{\beta\gamma\alpha,\eta^t,\;|\;
t\geq 1+\frac {s}2\right\}$$ is a $K$-basis of $T_1(A^{1,s}(a,c))^\perp$ and
$$Z(A^{1,s}(a,c))/T_1^\perp(A^{1,s}(a,c))\simeq K[Y,S]/(Y^{(s+2)/2},S^2,YS).$$
\dickebox
In order to be more concise we summarise the results from Theorem~\ref{main} and Remark~\ref{CentreandCartandet}
in case $K$ is algebraically closed in the following corollary.
\begin{Cor}\label{Finalcorollarytwosimples}
Let $K$ be an algebraically closed field of characteristic $p\in\N\cup\{\infty\}$, and let
$a,a',c$ be non-zero elements in $K$, and let $c',c''\in K$.
\begin{itemize}
\item
If $p\neq 2$, then there is $a'\in K^\times$ such that
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(a',0)$,
and if $(k,s)\neq (1,3)$, then
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(1,0)$.
\item
If $p=2$, then $D^b(Q(2{\mathfrak B})_1^{k,s}(a,c))\not\simeq D^b(Q(2{\mathfrak B})_1^{k,s}(a',0))$.
Moreover, there is $a''\in K^\times$ such that
$Q(2{\mathfrak B})_1^{k,s}(a,c)\simeq Q(2{\mathfrak B})_1^{k,s}(a'',1)$ and if $(k,s)\neq (1,3)$, then $Q(2{\mathfrak B})_1^{k,s}(a',0)\simeq Q(2{\mathfrak B})_1^{k,s}(1,0)$.
\item
For any characteristic of $K$ we get
$$\left(D^b(Q(2{\mathfrak B})_1^{k,s}(a,c''))\simeq D^b(Q(2{\mathfrak B})_1^{k',s'}(a',c'))\right)\Rightarrow
\left(\mbox{$(k=k'$ and $s=s'$) or ($k=s'$ and $s=k')$.}\right)$$
\end{itemize}
\end{Cor}
Proof. The first statement is an immediate consequence of Theorem~\ref{main} item (\ref{item3}) and
\cite[Lemma 5.7]{ErdSkow}.
The second statement follows from Theorem~\ref{main} item (\ref{item2})(a), (\ref{item2})(b), (\ref{item2})(c), and Theorem~\ref{prop:zimmermann}.
Indeed, the isomorphism type of the centre modulo the first K\"ulshammer ideal
differs in case $c=0$ and $c\neq 0$.
More precisely,
the commutative algebras from case (\ref{item2})(a) (i) and (\ref{item2})(a) (ii) are non isomorphic since
the dimensions of the socles of these algebras differ by $1$. Likewise,
the commutative algebras from case (\ref{item2})(b) (i) and (\ref{item2})(b) (ii) are non isomorphic since
the dimensions of the socles of these algebras differ by $1$.
The dimension of the socle of the centre modulo the K\"ulshammer ideal distinguish the algebras also in case (\ref{item2})(c), i.e. $k=1$.
The third statement follows from Remark~\ref{CentreandCartandet}.
\dickebox
\begin{Rem} Let $K$ be an algebraically closed field of characteristic $2$.
We do not know for which pair of parameters $a,a'\in K^\times$ we get that
$Q(2{\mathfrak B})_1^{k,s}(a',1)$ is derived equivalent to $Q(2{\mathfrak B})_1^{k,s}(a,1)$.
We do not know for which parameters $k,s$ the algebras $Q(2{\mathfrak B})_1^{k,s}(a,c)$
and $Q(2{\mathfrak B})_1^{s,k}(a,c)$ are derived equivalent.
\end{Rem}
\begin{Rem}
The case $p=3$ is special if $K$ is not perfect.
Let $p=3$ and use the notations used in the proof of Theorem~\ref{main}. Then $\alpha^3=a(\beta\gamma\alpha)^m$. In
the commutator quotient taking third power is semilinear
(cf e.g. \cite{Ku1},\cite[Lemma 2.9.3]{reptheobuch}), and therefore
\begin{eqnarray*}
0&=&\left(\sum_{t=1}^{s-1}x_t\eta^t+u\alpha+\sum_{m=1}^ky_m(\beta\gamma\alpha)^m\right)^3\\
&=&\sum_{t=1}^{s-1}x_t^3\eta^{3t}+
u^3a(\beta\gamma\alpha)^k+\sum_{m=1}^ky_m^3(\beta\gamma\alpha)^{3m}\\
&=&\sum_{1\leq t\leq (s-1)/3}x_t^3\eta^{3t}+
u^3a(\beta\gamma\alpha)^k+\sum_{1\leq m\leq k/3}y_m^3(\beta\gamma\alpha)^{3m}.
\end{eqnarray*}
We have again various cases. If $3$ does not divide $k$ and $3$ does not divide $s$, then
$x_t=0$ for all $t\leq s/3$ and $y_m=0$ for all $m\leq k/3$ and $u=0$.
If $3$ does not divide $k$ but $3|s$, then $x_t=0$ for all $t< s/3$ and
$y_m=0$ for all $m\leq k/3$ and $x_{s/3}^3+au^3=0$, which has a non zero
solution if and only if $a$ is a cube. If $3$ divides $k$ and $3$ does not divide $s$, then
$x_t=0$ for all $t\leq s/3$ and $y_m=0$ for all $m< k/3$ and $y_{m/3}^3+au^3=0$,
which has a non zero solution if and only if $a$ is a cube.
If $3$ divides $k$ and $3$ divides $s$, then
$x_t=0$ for all $t< s/3$ and $y_m=0$ for all $m< k/3$ and
$x_{s/3}^3+y_{m/3}^3+au^3=0$, which has a non zero solution if and only if $a$ is a cube.
As seen above, the first K\"ulshammer ideal
detects if the parameter $a$ is a third power in case $k$ or $s$ is divisible by $3$.
This shows that
the isomorphism $A^{k,s}(a,0)\simeq A^{k,s}(1,0)$, which we proved for
algebraically closed base fields,
is false if the base field is not perfect.
\end{Rem}
\subsection{Three simple modules}
Holm shows that there are two families of algebras $Q(3{\mathcal K})^{a,b,c}$ and
$Q(3{\mathcal A})^{2,2}_1(d)$ with three simple modules such that any block with
quaternion defect group and three simple modules is derived equivalent to an algebra
in one of these families. According to \cite{Holmhabil}
the derived classification of the case
$Q(3{\mathcal K})^{a,b,c}$ is complete, whereas the classification for the case
$Q(3{\mathcal A})^{2,2}_1(d)$ is complete up to the scalar $d\in K\setminus\{0,1\}$.
The quiver $3{\mathcal A}$ is
\unitlength1cm
\begin{center}
\begin{picture}(10,2)
\put(0,.9){$\bullet$}\put(2,.9){$\bullet$}\put(4,.9){$\bullet$}
\put(0,.5){$1$}\put(2,.5){$2$}\put(4,.5){$3$}
\put(.1,1.1){\vector(1,0){1.7}}
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\put(3.9,.9){\vector(-1,0){1.7}}
\put(1,1.2){$\beta$}
\put(3,1.2){$\delta$}
\put(1,.6){$\gamma$}
\put(3,.6){$\eta$}
\end{picture}
\end{center}
$B(d):=Q(3{\mathcal A})^{2,2}_1(d)$ is the quiver algebra of $3{\mathcal A}$ modulo the relations
$$\beta\delta\eta=\beta\gamma\beta,\;\;\;\;
\delta\eta\gamma=\gamma\beta\gamma,\;\;\;\;
\eta\gamma\beta=d\cdot \eta\delta\eta,\;\;\;\;
\gamma\beta\delta=d\cdot \delta\eta\delta, \;\;\;\;\beta\delta\eta\delta=0,\;\;\;\;\eta\gamma\beta\gamma=0$$
for $d\in K\setminus\{0,1\}$.
Following \cite{Erdmann} the Cartan matrix of $B(d)$ is
$\left(\begin{array}{ccc}4&2&2\\ 2&3&1\\ 2&1&3\end{array}\right)$ and
the centre is $6$-dimensional. The Loewy series of the projective indecomposable modules
are given below.
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We obtain a basis of the socle of $B(d)$ by
$$\{s_1:=\beta\delta\eta\gamma,s_2:=\eta\gamma\beta\delta,s_3:=\gamma\beta\delta\eta\}.$$
The closed paths of the algebra are
$$\{e_0,e_1,e_2,\beta\gamma, \gamma\beta, \delta\eta, \eta\delta,
\beta\delta\eta\gamma, \eta\gamma\beta\delta, \gamma\beta\delta\eta\}.$$
The centre is formed by linear combinations of closed paths and has a basis
$$\{1,\beta\gamma+\gamma\beta+\frac 1d\eta\delta,
\beta\gamma+\delta\eta+\eta\delta,
\beta\delta\eta\gamma,\eta\gamma\beta\delta,\gamma\beta\delta\eta\}$$
as is easily verified.
Non closed paths are clearly commutators. Obviously
$$\beta\delta\eta\gamma\equiv\eta\gamma\beta\delta\equiv\gamma\beta\delta\eta
\text{ mod }[B(d),B(d)].$$
Moreover,
$$\beta\gamma-\gamma\beta\in [B(d),B(d)]\mbox{ and }\delta\eta-\eta\delta\in [B(d),B(d)]. $$
Since the dimension of the centre of $B(d)$ coincides with the dimension of the commutator quotient, we get a basis of $B(d)/[B(d),B(d)]$
by $$\{e_0,e_1,e_2,\beta\gamma,\delta\eta,\beta\delta\eta\gamma\}.$$
We now suppose that the characteristic $p$ of $K$ is $p=2$. If $x$ is a square in $K$, then denote
$y=\sqrt x$ if $y^2=x$. We compute
$$
(\beta\gamma)^2=\beta\gamma\beta\gamma=\beta\delta\eta\gamma\mbox{ and }
(\delta\eta)^2=\delta\eta\delta\eta=\frac{1}{d}\delta\eta\gamma\beta.
$$
If $d$ is a square in $K$, then
$$((\gamma\beta)+\sqrt{d}(\eta\delta))^2=
\gamma\eta\gamma\beta+d\eta\delta\eta\delta=
\delta\eta\gamma\beta+\eta\gamma\beta\delta\in[B(d),B(d)]$$
so that $T_1(B(d))/[B(d),B(d)]$ is $1$-dimensional.
If $d$ is not a square, then $T_1(B(d))=[B(d),B(d)]$.
Let us consider the centre. Denote $\beta\gamma+\gamma\beta+\frac 1d\eta\delta=x$ and
$\beta\gamma+\delta\eta+\eta\delta=y$. Then we get
\begin{eqnarray*}
x^2=(\beta\gamma+\gamma\beta+\frac 1d\eta\delta)^2&=&
\beta\delta\eta\gamma+\frac 1d\delta\eta\gamma\beta+\frac 1d\eta\gamma\beta\delta
=s_1+\frac 1d s_2+\frac 1d s_2,\\
y^2=(\beta\gamma+\delta\eta+\eta\delta)^2&=&
\beta\delta\eta\gamma+\delta\eta\gamma\beta+\frac{1}{d^3}\eta\gamma\beta\delta
=s_1+s_2+\frac{1}{d^3}s_3,\\
xy=(\beta\gamma+\delta\eta+\eta\delta)\cdot(\beta\gamma+\gamma\beta+\frac 1d\eta\delta)&=&
\beta\delta\eta\gamma+\delta\eta\gamma\beta+\frac{1}{d^2}\eta\gamma\beta\delta
=s_1+s_2+\frac{1}{d^2}s_3.
\end{eqnarray*}
The coefficient matrix above has determinant $\frac{(d-1)^2}{d^4}$ and since $d\neq 1$,
the elements $x^2,y^2,xy$ are linearly independent, and hence
$Z(B(d))\simeq K[x,y]/(x^3,y^3,x^2y,xy^2)$. Moreover, choose the Frobenius
form given by
$$\psi(\beta\delta\eta\gamma)=\psi(\delta\eta\gamma\beta)=\psi(\eta\gamma\beta\delta)=1
\mbox{ and $\psi(c)=0$ if $c$ is a path of length at most $3$,}$$ following
Remark~\ref{symmetrisingform}. The relations are homogeneous, which shows that in order to prove symmetry of the form
we only need to consider paths $c_1$ and $c_2$ such that the lengths of $c_1$ and
$c_2$ sum up to $4$. The verification is a trivial and short computation which can be left to the reader.
Suppose now that $K$ is a perfect field. An elementary computation gives that $T_1^\perp(B(d))$
has a basis $\{x,s_1,s_2,s_3\}$, and therefore
$Z(B(d))/T_1^\perp(B(d))\simeq K[y]/y^2$, independently of $d$.
\begin{Theorem} \label{Qthreesimpletheorem}
Let $K$ be a field of characteristic $2$, and let $B(d)$ be the algebra
$Q(3{\mathcal A})^{2,2}_1(d)$. Then $\dim_K(T_1^\perp(B(d))/R(B(d)))=1$ if $d$
is a square in $K$, and $\dim_K(T_1^\perp(B(d))/R(B(d)))=0$ if $d$
is not a square in $K$.
\end{Theorem}
Proof: is done above.
\dickebox
\begin{Rem}
Unlike in case of Theorem~\ref{main} and its Corollary~\ref{Finalcorollarytwosimples},
using K\"ulshammer ideals we cannot distinguish the derived category of
$Q(3{\mathcal A})^{2,2}_1(d)$ from the derived category of
$Q(3{\mathcal A})^{2,2}_1(d')$ for two parameters $d,d'$.
If $K$ is perfect of characteristic $2$, then all elements of $K$ are squares.
Theorem~\ref{prop:zimmermann} needs that $K$ is perfect for
the invariance of K\"ulshammer ideals under derived equivalences and
$K$ is even algebraically closed for the invariance under stable equivalences
of Morita type.
We can only say that the algebra $Q(3{\mathcal A})^{2,2}_1(d)$ is not isomorphic to the
algebra $Q(3{\mathcal A})^{2,2}_1(d')$ if $d$ is a square and $d'$ is not.
\end{Rem} | 81,213 |
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Accounting & Finance sees highest hiring in Jan: Report
The Naukri JobSpeak Index for the month of January 2020 (2,381), marked a 6% rise in hiring activity as compared to January 2019 (2,251).
The major contributors to growth in hiring were industries like Accounting & Finance (37%), Education/Teaching (23%), Insurance (19%) and BPO/ITES (18%) while industries like Pharma (-5%) and Auto (-25%) witnessed a dip in recruitment activity. Demand for job roles in Teaching & Education (20%), BPO/ITES (17%), Accounting (14%) and Sales/Business Development (10%) also led to the overall uptick in hiring.
The job markets in Hyderabad (7%), Pune (7%) and Kolkata (6%) picked up on hiring activity and have consistently grown over the past year. Bangalore and Chennai saw negative growth in hiring at 3% and 4% respectively. Demand for entry-level professionals (0-3 years exp.) and senior-level executives (4-7 years exp.) grew by 10% and 7% respectively in January’20 YoY.
Commenting on the report, Pawan Goyal, Chief Business Officer, Naukri.com said, “The year has started on a cautious note with hiring activity growing by 6% in January’ 20 versus an average of 10% in the last quarter. While the Auto and Hotels & Hospitality industry continues to see a dip in hiring, the demand in Accounting & Finance and Education & Teaching industry maintains a buoyant mood. After a negative hiring growth in most of 2019, the telecom industry sees revival in recruitment activity for the past four months.”
Key Highlights of Naukri JobSpeak for January 2020 vs January 2019
Hiring Trends - Industry
Besides Accounting/Finance, other industries that showed growth in hiring activity during the month of January 2020 were - Education/Teaching(23%), BPO/ITES (18%), Insurance (19%), Telecom (17%), Banking (4%) and Oil & Gas (4%). The IT-Software industry posted a single-digit growth of 9% in January 20 versus an average YoY growth of 29% in 2019. The hiring in industries like Pharma (-5%), Hotels & Hospitality (5%) and Auto (-25%) witnessed a dip in January 20 versus last year same time.
The hiring activity in the Telecom Industry sees revival trends with consistent growth in the past four months when compared year on year. After a negative hiring growth in most of 2019, the growth picked up momentum in September 2019. The recruitment activity sees a growth of 17% in January 20.
Hiring Trends – Functional Area
Recruitment activities for professionals in the Teaching/Education domain witnessed an encouraging growth of 20%. Functional roles in ITES/BPO (17%), Accounting (14%), Sales/ Business Development (10%) and Marketing & Advertising (7%) also witnessed a growth in hiring. However, Hotels & Hospitality, HR/Administration and Legal roles saw a dip of 3%, 7%, and 16% respectively.
Hiring Trends - Experience
Hiring by experience levels was a mixed bag with demand for entry-level executives (0-3 yrs. exp.) leading the way with 10% YoY growth. The demand for senior-level executives (4-7 yrs. exp.) and middle - management roles (8-12 yrs. exp.) grew by 7% and 1% respectively. Hiring for both senior management roles (13-16 yrs. exp.) and leadership roles (16+ yrs. exp.) saw a dip of 8% & 10% respectively.
Hiring Trends - City
Recruitment across cities was positive during the month of January’20 with Hyderabad & Pune witnessing an increase of 7% YoY, followed by Kolkata with an increase of 6%. Some of the hiring trends observed across key cities are as follows -
●Hyderabad: Hiring in Hyderabad increased by 7%. BPO (24%), Accounting (22%) and IT-Software (13%) lead the way, while BPO/ITES roles witnessed an increase of 20%. Hiring for entry-level executives (0-3 years) and mid-level executives (4-7 years) also grew by 15% and 11% respectively.
●Pune: The city saw a 7% rise in recruitment activity. Accounting, BPO and IT-Software industries recorded growth rates of 29%, 28% and 19% respectively. Amongst experience bands, Pune mirrored the trends prevalent nationally, with growth across entry-level executive (10%), mid-management (9%) and senior-level executive (8%) hiring. Hiring activity in IT-Hardware & BPO/ITES roles grew by 31% and 25% respectively.
●Delhi/NCR: Hiring activity in the capital city witnessed an increase of 3% primarily led by hiring in the BPO sector (40%). Other industries that witnessed an uptick in hiring were Accounting (28%) & IT-Hardware (28%). Demand for job roles across Pharma & BPO/ ITES grew by 10% and 8% respectively. Hiring in the entry-level executive band grew by 2%.
●Mumbai: Recruitment in Mumbai increased by 4% in January 2020. Hiring in the BPO sector was up by 41%. This was followed by the IT - Software & IT-Hardware industries with an increase of 28% & 23% respectively. The demand for job roles across BPO/ ITES & IT - Software increased by (29%) and (20%) respectively.
●Bangalore: Overall hiring activity in Bangalore witnessed a drop of 3%. There has been a dip in hiring across experience bands. However, hiring in Accounts and BPO industry witnessed a growth of 56% and 22% respectively. The demand for Accounting roles also grew by 53%.
●Chennai: Recruitment activities in Chennai dropped by 4%. Auto & BFSI Industry showed a de-growth of 34% & 22% respectively. However, demand for Banking (43%) & Accounting (24%) roles recorded growth. Hiring in entry-level professionals witnessed positive growth of 5% whereas growth in senior-management bands was negative.
| 348,395 |
During the entire month of July, your San Antonio Area Chevy Dealers need your help to “Stuff The Silverado.” Soon the school bells will be ringing again, but unfortunately, some South Texas students can’t afford basic school supplies.
We’re asking for your help by bringing new pens, pencils, notebooks, crayons and other school supplies for students of all ages to your San Antonio area Chevy dealers and to help us make sure all San Antonio students are back-to-school ready.
“Stuff The Silverado,” benefitting the Salvation Army, is a KSAT12 Community in Partnership program with your San Antonio Area Chevy Dealers, Energy Transfer and University Health System.
If you wish to donate click on the button below or text EZSTUFF to 474747 to make a donation now. | 56,455 |
GreenWatch Sunday: seiche
Rising Waters, a New Microplastics Study, and New Anti-plastic Supermarket Strategies in this Weeks GreenWatch.
Last week in Buffalo was characterized by wild weather. A huge blunderbuss of a storm came ripping down through the Great Lakes with high winds (gusts of over 75mph were reported in WNY) rain, sleet, cold, snow, and a whole lot of water. On Wednesday, 4 April, a seiche, driven by winds and atmospheric pressure crashed into our Buffalo/Lake Erie/Niagara River and Buffalo River shorelines.
A seiche is a tsunami like movement of water in a lake or a bay. Unlike a tsunami, a seiche is not created by an earthquake. It is a weather and wind driven event. It comes ashore in oscillating waves, rapidly raising the water levels. Buffalo is particularly vulnerable to seiches. Wind drives water from the western basin of Lake Erie into the eastern end where we are.
Scientists call our lake seiches meteotsunamis.
Seiches, like tsunamis, can wreak considerable damage to property as the rising body of water, washes over and floods shorelines, and surges up creeks and rivers.
The wind and pressure push water down the shallow and narrow bathtub that is Lake Erie, piling up a lot of water, waves, and energy in the front. The 4 April seiche raised our shoreline water levels 5-8 feet.
This storm created wind damage throughout our region, and significant flooding in the Outer Harbor. Ice was pushed over the Ice Boom into the Niagara River. You may have seen the tremendous amount of ice in the river this week.
I toured the the Outer Harbor during the event, trying to navigate Furhmann Blvd. The winds were fierce. Smashed light fixtures were in the street. Power lines and the new traffic lights were waving precariously and dangerously. The waters had surged to the edge of the riprap at Gallagher Beach as an angry Lake Erie crashed into the pier. I was very concerned that something (like a truck) would blow off the skyway and down onto where I was. Thankfully the skyway was closed to trucks. Times Beach Nature Preserve was particularly effected. The waters had risen as high as I have ever seen them, and I have 40 years of watching here. They came to the edge of the Fuhrmann/Andrle trail. Waves were crashing over it and into the streetside fence. Much of the vegetation inside the preserve was overtopped. Geese and ducks had been blown against this new shoreline and were struggling to hold their perilous positions. Next door at Wilkeson Pointe Park and at the Michigan Ave Pier, ice had filled the narrow slips, trapping many ducks in thick wind-driven slush. Some sadly had perished. The waters overtopped the slip walls. The seiche effect here was at least 5 feet, probably more. At the Queen City Landing site where developer Gerry Buchheit is trying to build a 23 story glass condo tower, water had overtopped a considerable part of the site. This site is undergoing brownfield remediation at public expense and by the looks of it, the water, waves, and ice were washing an awful lot of contaminated soil into the Lake, and the nearby boat slips and fish habitat. I hope that the state takes appropriate action on this very contaminated and exposed site of deep and soft fill. I would like to believe that the developer, lawyers, and public officials responsible observed this storm. Perhaps we can say that “they know not what they have done” but the truth is that they know exactly what they have done. Protect us from greed and bad decisions. As I stood against the wind and rain all along the Outer Harbor it was impossible for me to imagine anyone wanting, or being able to safely live out here. I feel for those first responders that soon will be called upon to make all weather rescues for those foolish condo residents that do not recognize history or the power of nature. More on that history a little later.
Buffalo is known to have several seiche events a year. The Great Lakes have over 100 recordable seiches each year. According to a report issued last year, Calumet Harbor on Lake Michigan averages the most at 29 yearly seiches. Buffalo comes in second, averaging 19.
Most of the seiches that we experience in Buffalo are minor and not always obvious to the casual observer. But if you are a boater, you known them when you see them. Sometimes they can be astonishingly destructive. The 4 April seiche was created and accompanied by a strong weather system that brought significant pressure changes, high winds, and plenty of precipitation. Everyone of us in the region experienced the power of this set of storms.
Climate science is telling us that we have been experiencing, and should expect, increasingly unstable and extreme weather patterns. Last weeks storms were definitely organized around extreme weather. Climate scientists say that climate change includes a disrupted atmosphere that leads to these kinds of extreme events.
In recent years we have experienced a number of extreme events. The National Climate Assessment for extreme weather shows a significant observed increase in precipitation for the United States in recent decades.
We should plan for more extreme precipitation and wind events. Coupled with winter temperatures, we could have a catastrophic event hard-wired into our future. In every conceivable way, the Great Lakes, Lake Erie, and Buffalo are in for dramatic escalations. Destructive seiches are our future. We have to plan for climate resiliency and our shorelines are ground zero.
While this seiche was modest in lake level rise, we have had many seiches that have been gigantic with a historic topping out reported at 22 feet. The wind and pressure system made this recent seiche come ashore with substantial repercussions. Imagine a big seiche in really bad, i.e winter weather?
On Friday December 1, 2006 a springtime seiche flowed over the lake and river shorelines to the tune of a 9 feet rise in a couple of hours. Thankfully little ice was involved, the winds were less than catastrophic and so most of the damage came from flooding. I was out at midnight documenting the damage for GreenWatch. Lakeshore Commons, the very expensive condominiums surrounding the Buffalo’s Erie Basin Marina, and LaSalle Park were under seige as the water overtopped all walls. At the more or less unprotected LaSalle Park huge waves flooded the lower portion and brought in trash including tree trunks, an unbelievable mountain of plastic debris, and parts of an old car.
I wrote about that event for GreenWatch/Artvoice in December 2006 :
At midnight, I toured the Lakeshore Commons and made my way into and angry waves. the wind was blowing with some velocity. Rain and sleet hitting my face painfully, I was drenched in a matter of seconds and I got the heck out of there. I drove out towards the outer harbor. Ohio Street was bad with wind and rain, so I drove to Tifft Street, and then out along Fuhrmann Boulevard. The whole world was shaking as water dumped out of the sky in a furious rain. The road was submerging as I neared Gallagher Beach. The beach itself was gone and the howling foamy waves crested the roadway there. I had no option but to turn around and quickly get out. My thoughts turned to those at the Coast Guard Station on the other side of the dark intense water. My thoughts turned to those that perished out here during the Blizzard of 77 and those many souls that vanished from the Irish shanty town that spread out along the old seawall. Over the decades of its existence starting in the early days of the Erie Canal storms, seiches, snow and ice hit this historical Buffalo community with a cruel vengeance. Never forget. Many Buffalo families do not.
The Great Seiche of 1844
Buffalo’s most historic seiche occurred in October 1844. Buffalo was then a young but growing powerhouse bolstered by two decades of supergrowth stimulated by the opening of the Erie Canal. At this point the city, its homes, businesses, wharfs, and streets, occupied low-lying lands along the Buffalo Creek/river corridor, up to the “Hydraulics” which is now characterized by the location of Larkinville. The “Hydraulics” area had become the industrial center of the new city and a canal had been built to service this district because what was then called the Little Buffalo Creek could not handle the commercial shipping needs.
At the Lake end of the river, there were no breakwall as we have today. There was a modest wall that lead to the historic lighthouse, but it hardly was a storm surge protector.
On the night of October 18, 1844, a 22 foot seiche swept down the lake and overtopped the sleeping city of Buffalo. It unloaded on the lower districts, icely demolished scores of structures, and caused the deaths of a lot of people, many unknown and unrecorded. The exact number of deaths is estimated from the time range from 30 to over 100. Throughout the region, including those on ships outside the harbor and between Buffalo and Erie, many more are known to have perished.
In a Buffalo News Story from 2016, Mike Vogel of the Buffalo Lighthouse Association, is quoted as saying
“The city lacked an extensive system of protective breakwaters in the lake to blunt the force of the waters,” “Buffalo Harbor was just two stone and timber piers jutting into the lake from the mouth of Buffalo Creek; the city’s still-standing land mark lighthouse marked the end of one of the piers, and a crude stone seawall stretched south from the base of that pier along part of what is now Fuhrmann Boulevard.”
It was Abraham Lincoln that first funded the construction of a more resilient breakwall. Initial Construction was completed in 1866 with subsequent construction and more breakwalls constructed and maintained through the present day. Today we have to continue to think of how to make our coastline resilient. Is building a new community on the Outer Harbor a good idea? many people including this writer, think not. Creating a more natural coastal area that includes recreation, parks, natural restoration and ecological conservation components may be the new coastline direction that we have to think about. Climate change provides an escalating urgency. Only fools will try to fight nature there.
Links
The October Surprise of 1844 Buffalo History Gazette
Great Flood of 1844 Devastates Buffalo Buffalo News
The Great Buffalo Flood of 1844 Maritime History of the Great Lakes
A Seawall Shantytown in Buffalo by Edward J Patton A fine account about the Shantytown -published on the Facebook page of he Buffalo Irish Festival.
In Other News
Microplastics Discovered in 90% of Bottled Water
A recent report led by our friend Dr. Shari Mason from SUNY Fredonia has received international attention. Her group studied major brands of bottled water from around the world and discovered that most contained microscopic particles of plastic, aka microplastics at dangerous levels. The most prominent plastic is polypropylene. Nylon and polyethylene terephthalate is also found in significant concentrations. Some brands contain up to 10,000 individual particles per liter.
Previously Mason’s team had studied tap water from around the world which revealed alarming amounts of dangerous microplastics. The US has the highest contamination rate at 94%. The new study indicates that bottled water, in some instance may have twice as much microplastics as tap water.
Links:
World Health Organization launches review after micro plastics found in 90% of bottled water The Guardian 14 March 2018
PLUS PLASTIC Orb Media
Plastic fibers found in tap water around the world The Guardian, 6 September 2017
Straw Free and EkoPlaza Initiatives
Speaking of microplastics. You may know that microbeads were the target of local legislation sponsored by Erie County Legislator Pat Burke and signed into law by Erie County Executive Mark Poloncarz in the late summer of 2015. Legislative bodies across the U.S. followed suit and eventually the Microbead Free Water Act was signed into law in 2016 by President Obama. Great leadership from Burke and Poloncarz helped put our region on the environmental defenders map. Recently Malibu and Davis California, Seattle, Washington, Miami Beach and Fort Meyers, Florida have initiated a ban regarding the use of plastic straws. If you see beach debris you will note that plastic straws make up a lot of debris. Maybe this is a new step that we can initiate in Buffalo and Erie County. Mark and Pat?
Links
Is Plastic-Free the New Organic? Organic Authority 6 April 2018
Now an important new anti-plastics initiative has been announced by the Dutch supermarket chain Ekoplaza. They are beginning to remove plastic packaging from grocery items in 74 locations. We should watch this closely and perhaps initiate a campaign asking Wegmans to do the same. Who’s in?
More here:
Dutch Supermarket Adds Plastic-Free Grocery Aisle Organic Authority 1 March 2018
A short Video produced by NowThisFuture
Greenwatch has covered water issues since our inception. If you search the Public database you will find dozens of articles about these crucial issues.
Here are some important links:
GreenWatch: A Big Win for the Public Trust and Local Action on Microbeads August 3, 2013
GreenWatch: Poloncarz signs Microbeads Legislation into Law August 15, 2015
GreenWatch Sunday: Our Public Health Crisis October 22, 2017
GreenWatch Sunday: Water March 4, 2018
GreenWatch Sunday Morning Television
April 4 Seiche on Buffalo’s Outer Harbor
I toured the Outer Harbor during the April 4 seiche that I have written about above. here is a short video of observations. | 176,072 |
When you select a shampoo for your baby, you need to strictly avoid ingredients such as parabens, parfum, ethylated surfactants, DEA and TEA, MI etc.. Organic baby shampoos are produced by employing all-natural ingredients, which can be helpful for reducing the number of chemicals. Nothing surpasses the potency and gentleness of organic baby shampoo. It’s made of plant-based and natural ingredients to safeguard your baby stays secure and shield.
Table of Contents
- List of 10 Best Organic Baby Shampoo 2020
- 1. Puracy Natural Baby Shampoo & Body Wash
- 2. Babyganics Baby Shampoo and Body Wash
- 3. Dry Scalp Treatment Shampoo
- 4. Aveeno Baby Gentle Wash & Shampoo
- 5. Baby Mantra 3-in-1 Bubble Bath, Shampoo and Body Wash
- 6. Natures Baby Organics Shampoo and Body Wash
- 7. Fresh Monster Kids Shampoo
- 8. Weleda Calendula Baby Shampoo and Body Wash
- 9. Cetaphil Baby Wash And Shampoo
- 10. CeraVe Baby Wash & Shampoo
- Conclusion
List of 10 Best Organic Baby Shampoo 2020
1. Puracy Natural Baby Shampoo & Body Wash
One of the most well-known choices and safest infant shampoos, which also happens to be among my personal favorites is Puracy Natural Baby Shampoo.
It is an awesome alternative because it is super clean, natural, and secure, but also stays at a relatively low cost that’s affordable by many. They ought to reconsider the tagging of their product since tear-free, but all in all, it’s amazing and incredibly worth-while!
This incredible baby shampoo is the winner of the title of best natural baby shampoo and its impressive features will make it clear why. This can be an insanely clean and natural formulation that’s not just super safe, but also impressively good at cleaning your child’s hair. As an even greater bonus, in addition, it doubles as a baby body scrub!
- Hypoallergenic
- Baby shampoo and body wash
- No harsh chemicals
- Essential oils
- Jojoba seed oil
- Vegan-friendly
- The BUMP 2019 award winner
2. Babyganics Baby Shampoo and Body Wash
Clean your baby from head to toe using the Babyganics Baby Shampoo and Body Wash.. It is formulated using plant-based ingredients that help to support wholesome development and provide antioxidant protection. This shampoo has been analyzed and approved by dermatologists and pediatricians so you can purchase with confidence. The neoNourish all-natural seed oil mix found in this item is non-allergenic, tear-free, and extra gentle on your baby’s skin. What’s more, the shampoo is washed off easily to promote smooth combing of hairfollicles.
- Smells good
- Leaves the hair clean and soft
- The 3 bottles will last a very long period
- Doesn’t dry out the skin
3. Dry Scalp Treatment Shampoo
If you want to promote a healthy scalp and gorgeous hair, it is possible to try Dry Scalp Treatment Shampoo. Pure, gentle, natural, organic ingredients are used for making this shampoo. This shampoo contains Manuka Honey, that has antifungal, and antibacterial properties. Another ingredient, Cehami is useful for reducing itching, redness, inflammation, and stimulating hair growth. Dry Scalp Treatment Shampoo is also beneficial for restoring equilibrium, rebuilding and maintaining acid mantle, which is the natural protector of skin. This shampoo is extremely effective for treating various hair and scalp problems.
- Stimulates hair growth
- Soothe and cleansing damaged, sensitive skin
- Non-allergenic
- ideal for sensitive scalp of babies
- Reduces itching and redness
- All-natural ingredients
4. Aveeno Baby Gentle Wash & Shampoo
This amazing baby shampoo is actually a truly great all-around baby shampoo which can work well for virtually any baby! To actually top it off, it’s super affordable too!
Aveeno labels this to be lightly scented, but it’s still too much for a lot of individuals. Many parents say they aren’t a huge fan of the odor and wish that it was either unscented or at least a different scent. I’ve mixed feelings on it myself. It is kind of difficult to describe the odor — almost kind of like oats. I do not hate it, but I really do want unscented hair shampoo.
- Lightly scented
- Hypoallergenic
- Baby shampoo and body wash
- Tear-free
- Natural oat kernel extract
- Sunflower seed oil
5. Baby Mantra 3-in-1 Bubble Bath, Shampoo and Body Wash
This list is filled with some great baby skincare products, but components such as coconut oil, olive oil, and aloe make this Baby Mantra Shampoo 3-in-1 stand above the rest! This regular shampoo-bubble bath-body wash is made from the finest non-toxic, natural and organic ingredients.
Fits ages, for children and adults with skin allergies, it does not include any dangerous sulfates, parabens, phthalates, dyes or some other nasty chemicals we often mention here.
Last, it’s a beautiful and light scent of lavender oil mixed with all the soothing aloe vera, leaving your infant calm, relaxed, and ready for bed. The item is really cheap since the jar lasts quite long. A capful is enough to see joyful bubbles appear while holding it under the flowing water
- Lasts about 3 months/bottle
- PETA Certified
- NPA Accredited
- Healthy Child Healthy World Approved
- No compounds, dyes, and fragrances
6. Natures Baby Organics Shampoo and Body Wash
Natures Baby Organics Shampoo and Body Wash contain no synthetic chemicals neither does it borrow components from animals. As such, it makes the ideal choice vegans and individuals who favor organic products. The coconut pineapple odor is natural and fabulous to provide you a refreshing feel after a tub.
This shampoo is soft, making it appropriate for newborns, infants, teens as well as the whole family. It doesn’t dehydrate the skin. Instead, it moisturizes and leaves the skin glow while leaving the hair soft and luxurious.
- Ultra-mild and secure for your babies eyes
- Leaves fine plant and flower smell afterwards
7. Fresh Monster Kids Shampoo
Affordability is one of the most important qualities of New Monster Kids Shampoo. Designed specifically for children, this plant-based multi-tasker bath is ideal for cleaning head . This shampoo is completely free from harmful chemicals, artificial preservatives, colors, paraben, etc..
All-natural botanical and fruit extracts are used in this shampoo. It’s also free of toxins, thus the shampoo is safe for your kids. The formulation is analyzed by a dermatologist, so thus it’s safe for kids with sensitive skin.
- Really affordable
- Free from poisonous substances
- Highly effective shampoo and body wash
- Soothing fragrance
- Gentle and effective
- Effective for sensitive skin
8. Weleda Calendula Baby Shampoo and Body Wash
Weleda is Europe’s leading baby care manufacturer, made in Germany, with a vast assortment of baby skincare products that are calendula based. Organic Calendula in Weleda gardens in Germany is used to create these gentle and safe baby products.
Free of nasty compounds, sulfates, parabens, phthalates, and mineral oil, using a tear-free formula, this shampoo and body wash will neutralize the infant and take wonderful care of the delicate skin of our youngest. Additionally, this shampoo is naturally fragranced with pure Calendula essential oils, free from heavy scents and artificial fragrances.
Weleda Calendula shampoo and body wash includes gentle surfactant made from coconut and sugar which lathers and cleanses without drying. You may notice that the baby needs nobody lotion after washing, and that’s something good to know when it comes to purchasing of merchandise which can treat eczema or cradle cap of our infants.
- Organic merchandise
- Calendula is handpicked in their gardens
- Cruelty free
- Organic Calendula odor
9. Cetaphil Baby Wash And Shampoo
If you would like to purchase the most inexpensive and effective shampoo to the infant, Cetaphil Baby Wash and Shampoo is the perfect option. It is formulated on the basis of the tear-free formulation. Thus, it blends to a wealthy and lathering wash. Thus, it can clean the fragile skin and hair of your baby softly.
Calming organic calendula is employed for making shampoo and body scrub. This shampoo is free of colorant, mineral oil, paraben, and tears. Cetaphil Baby Wash and Shampoo is mild to the skin of babies. Fresh fragrance is helpful for getting a calming effect.
- Free from colorant, Mineral oil, paraben, etc
- Includes calming, organic calendula
- Makes hair and skin soft and smooth
- Tear-free formulation
- Highly affordable
10. CeraVe Baby Wash & Shampoo
From the famous brand CeraVe Baby comes the best baby shampoo and clean! This amazing 2-in-1 formulation is perfect for using as both a human body wash for your infant’s body in addition to shampoo for their scalp and hair.
Most baby shampoo choices can double as a body scrub for the infant as well, however, there are a number of things about this CeraVe Baby Wash and Shampoo we think really does make it the best 2-in-1 baby shampoo option!
CeraVe Baby Wash and Shampoo may be a superb selection for a lot of reasons. It’s great for all skin types but can be particularly helpful for babies that suffer from eczema, cradle cap, or other skin irritations. It’s extremely moisturizing but will not leave your child’s hair or skin feeling oily either. It’s safe, powerful, and very helpful for your child’s hair and skin!
- Tear-free
- Fragrance-free
- Paraben-free
- Sulfate-free
Conclusion
What’s so important is your kid’s safety and health, purchasing the best organic baby shampoo won’t only enable you to clean and soften your child’s hair but also will protect and alleviate your baby of a dry and annoying scalp. At the exact same time, the best baby shampoo is multi-purpose so you can use it to clean your baby from head to toe. Thus, take features, ingredients, and price into thoughts as you make the choice for your baby. | 347,800 |
Kyrgyzstan was recognized the unhappiest state in CIS. The Sustainable Development Solutions Network presents such data in the annual report on the level of happiness of countries.
When assessing the level of happiness of the population in 155 countries, such indicators as general
In the overall ranking, Kyrgyzstan took the 98th place in terms of happiness among the countries of the world, and the last place — among the countries of Central Asia.
Kyrgyzstan also takes the last place in CIS. Uzbekistan took the 47th place, Russia — 49th, Turkmenistan — 59th, Kazakhstan — 60th, and Tajikistan — 96th place.
Last year, Kyrgyzstan was ranked 85th. Thus, over the year, Kyrgyzstan dropped by 13 positions.
The top three happiest countries include Norway, Denmark and Iceland. Rwanda, Syria, Tanzania, Burundi and the Central African Republic also became the unhappiest countries. | 318,510 |
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