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Community care services – provision- closure of building
It was not unlawful for an NHS Trust having given reasonable notice to the local authority and other interested parties to close and sell a building, used by local community care service users, for economic reasons. It was not unlawful for the sale to proceed, even though the trust, the health authority and the local authority had failed to ensure that the service users would receive continuity of service elsewhere. | 394,144 |
Magnify His Name
Why did John write his account of the good news of Jesus?
...that by believing you may have life in his name.
Finding Life in Magnifying Jesus
Fulfillment is found in Magnifying Jesus!
Meaning is found in Magnifying Jesus
The friend of the bridegroom, the shoshben, had a unique place at a Jewish wedding. He acted as the liaison between the bride and the bridegroom; he arranged the wedding; he sent.
Joy in Life comes from Magnifying Jesus
Cf. the rabbinical saying attributed to R. Acha (c. 320), “The Holy Spirit, who rests on the prophets, rests (on them) only by weight לקשׁמב (= by measure)” (SBk, II, p. 431). If this view is as early as the Fourth Gospel there may be in mind a deliberate contrast between Jesus and the prophets. | 163,884 |
Reading has been very scattered the past two weeks. Too much else happening that’s been taking up too much time, and there have been days when I just haven’t had the energy to read late into the night.
So here’s a quick list of what I have managed to get through:
First 160-odd pages (or ‘Part 1’) of Shantaram. As compelling as I’d expected, and a fast read, but so much energy goes in just holding the thing that I haven’t yet managed more than 35-40 pages a night.
Quite a lot of non-fiction, which is unusual for me, but it’s an area I need to brush up on. Finished Thomas Friedman’s The World is Flat (often overweeningly silly, blinkered and patchily written but not without its points of interest). Also Bernard-Henri Levy’s Who Killed Daniel Pearl?, which was especially fascinating for its portrait of the monstrous Omar Sheikh, who masterminded the Pearl kidnapping and murder. Levy’s intense, claustrophobia-inducing book took me back to those dark days in early 2002 when I was working on graveyard shifts for TheNewspaperToday.com and we were all waiting with a mixture of fascination and revulsion for the videos of the killing to be released to the press. The six months post-9/11...what a time that was for a journalist working overnight on a 24-hour website while a new story was breaking in the US every half-hour or so.
Still reading Steven Pinker’s excellent The Blank Slate, about the nature-nurture debate, given to me by Amit Varma a couple of months ago. Riveting though this book is, it requires such concentrated reading that I haven’t been able to get through it once and for all, what with the reviewing obligations that crop up intermittently.
Have started on Orhan Pamuk’s Istanbul; am a big fan of the man’s fiction, especially Snow, and am keen to see what he does with this memoir. Love the early passages about his memories of a childhood spent in his familys’s five-storey house with its unused pianos and untouched Chinese porcelain figures.
A Bunch of Old Letters - a new print of a selection of letters edited by Jawaharlal Nehru, most written by or to him. I’ve been opening this book at random, reading whatever catches my fancy but I did go through the complete correspondence between Nehru and Subhas Chandra Bose between October 1938 and April 1939, when the decisive parting of ways between the two men occurred. A very interesting look at a clash of ideologies that still casts a shadow over the country. Isn’t it such fun when great men scrap (even when they continue signing off their letters “Yours affectionately” until the very end).
Jabberwock,
You make me want to tell the boss to go take a hike so that I can go back home at 6.30 everyday and read books. So many books discussed in this space... many I haven't even heard of before.
I seem to have used the easiest and stupidest excuse ever - lack of time, to stop doing somthing which had been my mainstay before I started working.
Still, its never too late, I guess. Thanks.
If you want toothsome non-fiction, by a legendary globe-trotting journo, coincidentally a great writer, read Ryszard Kapuscinski. Books: start with The Shadow of the Sun and The Soccer War. And if you think this is just my dubious pro-non-fiction op, you have seen the review of his Angola book in Imaginary Homelands.
Let us know what you think of Istanbul...I just finished My Name is Red and plan to start Snow soon.
From various extracts of Levi's book he comes across as the ultimate anti-islamist and anti-pak using plenty of lurid comments to support his beliefs. I plan to read it soon. | 48,104 |
\begin{document}
\begin{abstract}
The main purpose of this paper is to study \emph{$e$-separable spaces}, originally introduced by Kurepa as $K_0'$ spaces; we call a space $X$ $e$-separable iff $X$ has a dense set which is the union of countably many closed discrete sets. We primarily focus on the behaviour of $e$-separable spaces under products and the cardinal invariants that are naturally related to $e$-separable spaces. Our main results show that the statement ``there is a product of at most $\mathfrak c$ many $e$-separable spaces that fails to be $e$-separable'' is equiconsistent with the existence of a weakly compact cardinal.
\end{abstract}
\maketitle
\newtheorem{defin}{Definition}[section]
\newtheorem{prop}[defin]{Proposition}
\newtheorem{question}[defin]{Question}
\newtheorem{obs}[defin]{Observation}
\newtheorem{prob}[defin]{Problem}
\newtheorem{lemma}[defin]{Lemma}
\newtheorem{corol}[defin]{Corollary}
\newtheorem{fact}[defin]{Fact}
\newtheorem{example}[defin]{Example}
\newtheorem{thm}[defin]{Theorem}
\newtheorem{clm}[defin]{Claim}
\newtheorem{subclaim}[defin]{Subclaim}
\newtheorem{rem}[defin]{Remark}
\makeatletter
\newtheorem*{rep@theorem}{\rep@title}
\newcommand{\newreptheorem}[2]{
\newenvironment{rep#1}[1]{
\def\rep@title{#2 \ref{##1}}
\begin{rep@theorem}}
{\end{rep@theorem}}}
\makeatother
\newreptheorem{corollary}{Corollary}
\newreptheorem{theorem}{Theorem}
\section{Introduction}
The goal of this paper is to study a natural generalization of separability: let us call a space $X$ \emph{$e$-separable} iff $X$ has a dense set which is the union of countably many closed discrete sets. The definition is due to Kurepa \cite{kurepa}, who introduced this notion as property $K_0'$ in his study of Suslin's problem. Later, $e$-separable spaces appear in multiple papers related to the study of linearly ordered and GO-spaces \cite{faber, qiao,qiao2, watson}. In particular, Faber \cite{faber} showed that $e$-separable GO-spaces are perfect; however, the converse is famously open: is there, in ZFC, a perfect GO-space (or even just a perfect $T_3$ space) which is not $e$-separable? Let us refer the interested reader to a paper of Benett and Lutzer \cite{lutzer} for more details and results on this topic.
Now, our interest lies mainly in studying $e$-separability with regards to powers and products. Recall that the famous Hewitt--Marczewski--Pondiczery theorem \cite{engelking} states that the product of at most $\mathfrak c$ many separable spaces is again separable. What can we say about $e$-separable spaces in this matter? Historically, another generalization of separable spaces received more attention: \emph{$d$-separable spaces} i.e. spaces with $\sigma$-discrete dense sets. In Kurepa's old notation, $d$-separable spaces were called $K_0$. $d$-separable spaces were investigated in great detail (see \cite{arh81, dD, juhi, moore2, weaksep, tkac}) and they show very interesting behaviour in many aspects, in particular, regarding products. A. Arhangel'ski\u\i\mbox{} proved in \cite{arh81} that any product of $d$-separable spaces is $d$-separable; in \cite{juhi}, the authors show that for every space $X$ there is a cardinal $\kappa$ so that $X^\kappa$ is $d$-separable. Motivated by these results, one of our main objectives is to understand, as much as possible, the behaviour of $e$-separable spaces under products.
Our paper splits into three main parts. First, we make initial observations on $e$-separable spaces in Section \ref{prelim}. Then, in Section \ref{d_and_e}, we investigate if the existence of many large closed discrete sets suffices for a space to be $e$-separable. In particular, we prove that once an infinite power $X^\kappa$ has a closed discrete set of size $d(X^\kappa)$ (the density of $X^\kappa$) then $X^\kappa$ is $e$-separable. As a corollary, we show that certain large powers of non-countably-compact spaces are $e$-separable. Now an interesting open question is whether a countably compact, non-separable space can have an $e$-separable square.
Next, in Section \ref{sizes}, we compare two natural cardinal functions: $d(X)$, the size of the smallest dense set in $X$, with $d_e(X)$, the size of the smallest $\sigma$-closed-discrete dense set. In Theorem \ref{examp}, we show that there is a 0-dimensional space $X$ which satisfies $d(X)<d_e(X)$. We show that a similar example can be constructed for $d$-separable spaces, at least under $\aleph_1<\mathfrak c=2^{\aleph_0}$; we do not know how to remove this assumption. The section ends with a few interesting open problems.
Our main results are finally presented in Section \ref{pressec}: we describe those cardinals $\kappa$ such that the product of $\kappa$ many $e$-separable spaces is $e$-separable again, and hence present the analogue of the Hewitt--Marczewski--Pondiczery theorem for $e$-separable spaces. First, note that $2^{\mathfrak c^+}$ is not $e$-separable (as a compact, non-separable space) and so the question of preserving $e$-separability comes down to products of at most $\mathfrak c$ terms. How could it be possible that $e$-separability is not preserved by small products? The reason must be that there are some large cardinals lurking in the background:
\begin{repcorollary}{main1cor} If the existence of a weakly compact cardinal is consistent with ZFC then so is the statement that there are less than $\mathfrak{c}$ many discrete spaces with non-$e$-separable product.
\end{repcorollary}
\begin{repcorollary}{main2cor}
If there is a non-$e$-separable product of at most $\mathfrak c$ many $e$-separable spaces then there is a weakly compact cardinal in $L$.
\end{repcorollary}
As we shall see, the proof of these results nicely combines various ideas from topology, set theory and logic.
\medskip
Throughout this paper, all spaces are assumed to be $T_1$. Given a product of discrete spaces $X=\prod \{X_\alpha:\alpha<\lambda\}$
and a function $\varepsilon$ satisfying $\dom(\varepsilon)\in[\lambda]^{<\aleph_0}$ and $\varepsilon(\alpha)\in X_\alpha$ for each $\alpha\in\dom(\varepsilon)$, we write $$[\varepsilon]=\{x\in X: \varepsilon\subseteq x\}.$$ Thus, if $x\in X$ is such that $x\uhp \dom(\varepsilon)=\varepsilon$, then $[\varepsilon]$ is a basic open neighbourhood of $x$ in $X$. We let $D(\kappa)$ denote the discrete space on a cardinal $\kappa$.
In general, we use standard notation and terminology consistent with Engelking \cite{engelking}.
\section{Preliminaries}\label{prelim}
The main concept we study in this paper is the following:
\begin{defin}
\label{e-sep}
A topological space $X$ is \emph{$e$-separable} if there is a
sequence $(D_n)_{n\in\omega}$ of closed discrete subspaces of $X$ such
that $\bigcup_{n\in\omega}D_n$ is dense in $X$.
\end{defin}
In this section, we will prove a few general facts about $e$-separable spaces and state some results for later reference. Let us start with simple observations:
\begin{comment}
\begin{defin}
A topological space $X$ is \emph{$d$-separable} if there is a
sequence $(D_n)_{n\in\omega}$ of discrete subspaces of $X$ such
that $\bigcup_{n\in\omega}D_n$ is dense in $X$.
\end{defin}
\end{comment}
\begin{obs} Every separable space is $e$-separable and every $e$-separable space is $d$-separable.
\end{obs}
Recall the following two well known cardinal functions: the density of $X$, denoted by $d(X)$, is the smallest possible size of a dense set in $X$. The extent of a space $X$, denoted by $e(X)$, is the supremum of all cardinalities $|E|$ where $E$ is a closed discrete subset of $X$.
\begin{obs}\label{cardobs} Every $e$-separable space $X$ satisfies $d(X)\leq e(X)$; moreover, if $cf(d(X))>\omega$ then there is a closed discrete set of size $d(X)$ in $X$. In particular, a countably compact space is $e$-separable iff it is separable.
\end{obs}
\begin{example}\label{2^c+} $2^{\mathfrak{c}^+}$ is a compact, $d$-separable but non-$e$-separable space.
\end{example}
\begin{proof} By Arhangel'ski\u\i{}'s \cite{arh81}, $d$-separability is preserved by products. Also, $2^{\mathfrak{c}^+}$ is not $e$-separable as $d(2^{\mathfrak{c}^+})>e(2^{\mathfrak{c}^+})=\oo$, hence Observation \ref{cardobs} can be applied.
\end{proof}
What can we say about metric spaces?
\begin{obs} Every space with a $\sigma$-discrete $\pi$-base is $e$-separable. Hence, every metrizable space is $e$-separable.
\end{obs}
The following result shows that actually a large class of generalized metric spaces are $e$-separable:
\begin{prop}
\label{developD}
Every developable space is $e$-separable.
\end{prop}
Recall that a space $X$ is \emph{developable} iff there is a developement of $X$, i.e. a sequence $(\mathscr{G}_n)_{n\in\omega}$ of open covers of $X$ such that for every $x\in X$ and open $V$ containing $x$ there is an $n\in \omega$ so that $st(x,\mathscr{G}_n)=\bigcup\{U\in \mathscr G_n:x\in U\}\subseteq V$.
\begin{proof}
Let $(\mathscr{G}_n)_{n\in\omega}$ be a development for a topological
space $X$. For each $x\in X$ and $n\in\omega$, let
$V^x_n=st(x,\mathscr{G}_n)$. By Proposition 1.3 of \cite{leandromD},
there is a closed discrete $D_n\subseteq X$ such
that $X=\bigcup_{x\in D_n}V^x_n$ for each $n\in\omega$. We claim that $\bigcup_{n\in\omega}D_n$
is dense in $X$.
Suppose, to the contrary, that there is $p\in
X\setminus\overline{\bigcup_{n\in\omega}D_n}$, and let $m\in\omega$ be
such that $V^p_m\cap\bigcup_{n\in\omega}D_n=\emptyset$. By the choice
of $D_m$, there is $x\in D_m$ such that $p\in V^x_m$; but then
$\{p,x\}\subseteq U$ for some $U\in\mathscr{G}_m$, which implies that
$x\in V^p_m$, thus contradicting the fact that $V^p_m\cap
D_m=\emptyset$.
\end{proof}
It is worth comparing the above result with Proposition 2.3 of
\cite{dD}, which states that every quasi-developable space is
$d$-separable. Note also that the Michael line is a quasi-developable space
(see \cite{bennett}) that is not $e$-separable.
The proof of Proposition \ref{developD} suggests that there might be a connection between $D$-spaces and $e$-separable spaces; recall that a space $X$ is a $D$-space iff for every open neighbourhood assignment $N:X\to \tau$ there is a closed discrete $D\subseteq X$ so that $N''D$ covers $X$. However, we note that the Alexandrov double circle is hereditarily $D$ (see e.g. \cite[Proposition 2.5]{gruen}) but not $e$-separable.\\
For later reference, we would like to state two results on the existence of closed discrete sets in products.
\begin{thm}[\L o\'s \cite{los}, Gorelic \cite{gorelic}]\label{gorelic} $D(\omega)^{2^\kappa}$ contains a closed discrete set of size $\kappa$ for every $\kappa$ less than the first measurable cardinal.
\end{thm}
The above result was first proved by \L o\'s \cite{los} but the reference \cite{gorelic} is more accessible.
\begin{thm}[Mycielski \cite{myc}]\label{myc} $D(\omega)^{\kappa}$ contains a closed discrete set of size $\kappa$ for every $\kappa$ less than the first weakly inaccessible cardinal.
\end{thm}
\section{Density and extent for $e$-separable spaces}\label{d_and_e}
Our goal now is to elaborate further on the observation that if $X$ is $e$-separable then $d(X)\leq e(X)$. In particular, in what context is the implication reversible?
First, note that $d(X)\leq e(X)$ does not imply that there are closed discrete sets of size $d(X)$:
\begin{example} There is a $\sigma$-closed-discrete (hence $e$-separable) space $X$ which contains no closed discrete sets of size $d(X)$.
\end{example}
\begin{proof} Let $X=\oo_\oo+1$ and declare all points in $\oo_\oo$ isolated and let $\{\{\oo_\oo\}\cup A:A\in [\oo_\oo]^{<\aleph_\oo}\}$ form a neighbourhood base at $\oo_\oo$.
\end{proof}
Next, we show that even a significant strengthening of $d(X)\leq e(X)$ fails to imply $e$-separability in general:
\begin{example} There is a 0-dimensional space $X$ such that $|X|=\omega_1$, every somewhere dense subset of $X$ contains a closed discrete subset of size $\omega_1$, while $X$ is not $e$-separable.
\end{example}
\begin{proof} Let $X=\omega_1\mbox{}^{<\omega}$ and declare $U\subseteq X$ to be open iff $x\in U$ implies that $\{\alpha<\omega_1:x\smf (\alpha)\in U\}$ contains a club. Now, $X$ is a Hausdorff, 0-dimensional and dense-in-itself space.
\begin{obs}
A set $E\subseteq X$ is closed discrete iff $\{\alpha<\omega_1:x\smf \alpha\in E\}$ is non-stationary for every $x\in X$.
\end{obs}
This observation immediately implies that the $\sigma$-closed-discrete sets are closed discrete and hence $X$ cannot be $e$-separable.
Suppose that $Y\subseteq X$ is dense in a non-empty open set $V$; $I_x=\{\alpha\in \omg: x\smf \alpha\in Y\}$ must be stationary for any $x\in V$ and so we can select an uncountable but non-stationary $I\subseteq I_x$. Hence $\{x\smf \alpha:\alpha \in I\}$ is an uncountable closed discrete subset of $Y$.
\end{proof}
Now, let us turn to powers of a fixed space $X$. Could it be that $d(X^\kappa)\leq e(X^\kappa)$ implies that $X^\kappa$ is $e$-separable whenever $\kappa$ is an infinite cardinal? The answer is negative, at least under the assumption that there are measurable cardinals:
\begin{example}\label{ex:mble} If $\kappa$ is the first measurable cardinal, then $d(\omega^\kappa)= e(\omega^\kappa)$; however, $\omega^\kappa$ is not $e$-separable.
\end{example}
\begin{proof} It is clear that $d(\omega^\kappa)=\kappa$; also, $2^\lambda<\kappa$ whenever $\lambda<\kappa$, and so Theorem \ref{gorelic} implies that $e(\omega^\kappa)=\kappa$ as well.
If we show that $\omega^\kappa$ has no closed discrete sets of size $\kappa$ then $\omega^\kappa$ cannot be $e$-separable. Suppose that $A=\{x_\alpha:\alpha<\kappa\}\subseteq \omega^\kappa$ and that $\mc U$ is a $\sigma$-complete non-principal ultrafilter on $\kappa$. Note that $$\kappa=\bigcup_{n\in \omega}\{\alpha<\kappa:x_\alpha(\xi)=n\}$$ for each $\xi<\kappa$. So there is a unique $n\in \omega$ such that $\{\alpha<\kappa:x_\alpha(\xi)=n\}\in \mc U$. In turn, we can define $y\in \omega^\kappa$ by $y(\xi)=n$ iff $\{\alpha<\kappa:x_\alpha(\xi)=n\}\in \mc U$. It is easy to see that $\{\alpha<\kappa:x_\alpha\in V\}\in \mc U$ for every open neighbourhood $V$ of $y$, and so $V\cap A$ has size $\kappa$. Hence, $y$ is an accumulation point of $A$.
\end{proof}
However, if we suppose a bit more than $d(X^\kappa)\leq e(X^\kappa)$ then we get
\begin{thm} \label{prodthm}Let $X$ be any space and $\kappa$ an infinite cardinal. If $X^\kappa$ contains a closed discrete set of size $d(X^\kappa)$ then $X^\kappa$ is $e$-separable.
\end{thm}
The above theorem is an analogue of \cite[Theorem 1]{juhi}: if $X^\kappa$ has a discrete subspace of size $d(X)$ then $X^\kappa$ is $d$-separable. Example \ref{ex:mble} shows that assuming ``$X^\kappa$ contains a closed discrete set of size $d(X)$'' does not imply that $X$ is $e$-separable.
We will prove a somewhat technical lemma now which immediately implies Theorem \ref{prodthm} and will be of use later as well:
\begin{lemma}\label{power0} Let $X$ be any space and $\kappa$ an infinite cardinal. Suppose that $D\subseteq X^\kappa$ is dense in $X^\kappa$ and $X^\kappa$ contains a closed discrete set of size $|D|$. Then there is a dense set $E$ in $X^\kappa$ such that
\begin{enumerate}
\item $|D|=|E|$, $d(D)=d(E)$, and
\item $E$ is $\sigma$-closed-discrete.
\end{enumerate}
\end{lemma}
\begin{proof}
Pick a countable increasing sequence $(I_n)_{n\in\omega}$ of subsets of $\kappa$ such that $\kappa=|I_n|=|\kappa\setminus I_n|$ for each $n\in\omega$ and $\kappa=\bigcup_{n\in\omega}I_n$. Fix closed discrete sets $E_n$ of size $|D|$ in $X^{\kappa\setminus I_n}$ and bijections $\varphi_n:D\rightarrow E_n$ for each $n\in \omega$.
We define maps $\psi_n:D\to X^\kappa$ by
\[
\psi_n(d)(\xi) = \begin{cases}
d(\xi), & \text{for } \xi\in I_n, and\\
\varphi_n(d)(\xi), & \text{for } \xi\in\kappa\setm I_n.
\end{cases}
\]
Let $E=\bigcup_{n\in\omega}\ran(\psi_n)$.
Clearly $|D|=|E|$ holds.
It is easy to see that $E$ is dense in $X^\kappa$: if $[\varepsilon]$ is a basic open set in $X^\kappa$ then there is an $n\in \omega$ such that $\dom(\varepsilon)\subseteq I_n$, hence $\ran(\psi_n)\cap [\varepsilon]\neq\emptyset$. Next we show (2) by proving that $\ran(\psi_n)$ is closed discrete as well for each $n\in\omega$. Pick any $x\in X^\kappa$. There is a basic open set $[\varepsilon]$ in $X^{\kappa\setminus I_n}$ such that $x\upharpoonright _{\kappa\setminus I_n} \in [\varepsilon]$ and $|[\varepsilon] \cap E_n|\leq 1$.
Thus the basic open set $\{y\in X^\kappa:\varepsilon\subseteq y\}$ of $X^\kappa$, which we (by abuse of notation) also denote by $[\varepsilon]$, satisfies $x\in [\varepsilon]$ and $|[\varepsilon] \cap \ran(\psi_n)|\leq 1$.
Finally we prove $d(D)=d(E)$. Note that if $D_0$ is dense in $D$ then $\bigcup_{n\in\omega}\psi_n\mbox{}''D_0$ is dense in $E$, hence $d(E)\leq d(D)$. Suppose that $A\in [E]^{<d(D)}$; we want to prove that $A$ is not dense in $E$. If $A$ is finite, there is nothing to prove. If $A$ is infinite, let $$D_A=\bigcup_{n\in\omega}\{d\in D: \psi_n(d)\in A\};$$ then $D_A$ cannot be dense in $D$ as $|D_A|\leq |A|<d(D)$.
Fix a basic open set $U=[\varepsilon]$ such that $[\varepsilon]\cap D_A =\emptyset$. There is an $n^*\in\omega$ such that $\dom(\varepsilon) \subseteq I_{n^*}$.
\begin{clm} If $m\geq n^*$ then $[\varepsilon]\cap \{\psi_m(d):d\in D_A\}=\emptyset$.
\end{clm}
\begin{proof} Suppose that $m\geq n^*$ and $d\in D_A$. Then $d\upharpoonright I_m=\psi_m(d) \upharpoonright I_m$, $d\notin [\varepsilon]$ and $\dom(\varepsilon)\subseteq I_m$, thus $\psi_m(d)\notin [\varepsilon]$.
\end{proof}
Hence $$U\cap \bigcup_{n\in\omega}{\psi_n(D_A)}\subseteq \bigcup_{n<n^*}{\psi_n(D_A)},$$ that is, $U\cap \bigcup_{n\in\omega}{\psi_n(D_A)}$ is closed discrete as each ${\psi_n(D_A)}$ is closed discrete. However, $A\subseteq \bigcup_{n\in\omega}{\psi_n(D_A)}$ which shows that $A\cap U$ cannot be dense in $U$.
\end{proof}
Let us present two corollaries. The aforementioned \cite[Theorem 1]{juhi} implies that $X^{\kappa}$ is always $d$-separable for any $\kappa\ge d(X)$. We know that, say, $[0,1]^\kappa$ is not $e$-separable when $\kappa\ge \mathfrak c^+$ because of Observation \ref{cardobs}; indeed, $[0,1]^{\kappa}$ is compact so $\sigma$-closed-discrete sets are countable, but $[0,1]^{\kappa}$ is not separable. However, the following holds:
\begin{corol}\label{prodcor}
Suppose that $X$ is not countably compact. Then $X^{d(X)}$ is $e$-separable if $d(X)$ is less than the first weakly inaccessible cardinal.
\end{corol}
\begin{proof}
Let $\kappa=d(X)$ and note that it suffices to find a closed discrete subspace of $X^\kappa$ of size $d(X^\kappa)$ by Theorem \ref{prodthm}. First, note that $d(X^\kappa)=\kappa$. Second, $X$ contains an infinite closed discrete subspace $Y$ since $X$ is not countably compact. So $Y^\kappa$ is a closed copy of $D(\omega)^\kappa$ in $X^\kappa$. Finally, $D(\omega)^\kappa$ does contain a closed discrete set of size $\kappa$ by Theorem \ref{myc}.
\end{proof}
\begin{corol}
Suppose that $X$ is not countably compact. Then $X^{2^{d(X)}}$ is $e$-separable if $d(X)$ is less than the first measurable cardinal.
\end{corol}
\begin{proof} The proof is the same as for Corollary \ref{prodcor} but now applying Theorem \ref{gorelic}.
\end{proof}
Interestingly, if $X$ is compact Hausdorff then $X^\omega$ is $d$-separable already (see \cite[Corollary 5]{juhi}). Furthermore, Moore \cite{moore2} showed that there is an $L$-space $X$ such that $X^2$ is $d$-separable. Note that $X$ itself is not $d$-separable since each discrete subspace of $X$ is countable but $X$ has uncountable density. Moore's example was improved by Peng \cite{peng}: there is an $L$-space $X$ such that $X^2$ is $e$-separable. We wonder if the following related question is true:
\begin{prob}
Is there a non-separable, countably compact $X$ so that $X^2$ is $e$-separable?
\end{prob}
\section{The sizes of $\sigma$-discrete dense sets}\label{sizes}
Next, we investigate the size of the smallest $\sigma$-discrete dense set in $e$-separable spaces.
\begin{defin}
For an $e$-separable space $X$, we define $$d_e(X)=\min \{|E|:\text{ E is a dense } \sigma \text{-closed-discrete subset of } X\}.$$
\end{defin}
Clearly $d(X)\leq d_e(X)\leq e(X)$ for any $e$-separable space $X$ and next we show that $d(X)= d_e(X)$ fails to hold in general:
\begin{thm}\label{examp} There is a 0-dimensional $e$-separable space $X$ such that $$\mathfrak{c}=d(X)< d_e(X)=e(X)=w(X)=2^\mathfrak{c}.$$
\end{thm}
\begin{proof} First note the following:
\begin{clm} \label{4.3} Suppose that a space $X$ can be written as $D\cup E$ so that
\begin{enumerate}
\item $D$ is dense in $X$,
\item $E$ is dense and $\sigma$-closed-discrete in $X$,
\item $d(D)<d(E)$, and
\item every $A\in [D]^{\leq e(D)}$ is nowhere dense in $X$ (or equivalently, in $D$).
\end{enumerate}
Then $X$ is $e$-separable and $d(X)< d_e(X)$.
\end{clm}
\begin{proof}
X is $e$-separable by (2) and $d(X)\leq d(D)$ by (1). We prove that if $F\in [X]^{\leq d(X)}$ and $F$ is $\sigma$-closed-discrete then $F$ is not dense in $X$; this proves the claim. Take $F\subseteq X$ as above and note that by (3) there is a non-empty open set $U\subseteq X$ such that $U\cap E \cap F = \emptyset$. As $|F\cap D|\leq e(D)$, $F\cap D$ must be nowhere dense in $X$. Thus there is a non-empty open $V\subseteq U$ such that $V \cap F \cap D =\emptyset$. Thus $V \cap F =\emptyset$ showing that $F$ is not dense.
\end{proof}
Now, it suffices to construct a 0-dimensional space $X=D\cup E$ satisfying (1)-(4). Let us construct $X=D\cup E\subseteq \omega ^{2^\mathfrak{c}}$ such that
\begin{enumerate}[(i)]
\item $D$ is dense in $\omega ^{2^\mathfrak{c}}$,
\item $E$ is dense and $\sigma$-closed-discrete in $\omega ^{2^\mathfrak{c}}$,
\item $|D|=\mathfrak{c}$ and $d(E)=2^{\mathfrak{c}}$, and
\item $e(D)=\omega$.
\end{enumerate}
It is trivial to see that (i)-(iii) implies (1)-(3), respectively, while (iv) implies (4) using the fact that $d(\omega ^{2^\mathfrak{c}})=\mathfrak{c}$.
First we construct $D$. Construct dense subsets $D_n\subseteq n^{2^\mathfrak{c}}$ of size $\mathfrak{c}$ which are countably compact, for each $n\in\omega$; this can be done by choosing a dense subset $D^0_n\subseteq n^{2^\mathfrak{c}}$ of size $\mathfrak{c}$ and adding accumulation points recursively ($\omega_1$ many times) for all countable subsets. Define $D=\bigcup_{n\in\omega}D_n$. Then $D$ is dense in $\omega^{2^\mathfrak{c}}$ as $\bigcup_{n\in\omega} n^{2^\mathfrak{c}}$ is dense in $\omega^{2^\mathfrak{c}}$ and $e(D)=\omega$ as $e(D_n)=\omega$ for all $n\in\omega$; thus $D$ satisfies (i), (iv) and the first part of (iii).
Now, we construct $E$ satisfying (ii) and (iii) which finishes the proof. Let $S=\sigma(\omega^{2^\mathfrak{c}})=\{x\in\omega^{2^\mathfrak{c}}:|\{\alpha\in 2^\mathfrak{c}:x(\alpha)\neq 0\}|<\aleph_0\}$; then $d(S)=2^\mathfrak{c}$ and $S$ is dense in $\omega^{2^\mathfrak{c}}$. Recall that $\omega^{2^\mathfrak{c}}$ contains a closed discrete set of size $2^\mathfrak{c}$ by Theorem \ref{myc}. Now, by applying Lemma \ref{power0}, we find a $\sigma$-closed-discrete $E$ which is dense in $\omega^{2^\mathfrak{c}}$ and satisfies $d(E)=d(S)=2^\mathfrak{c}$.
\end{proof}
Naturally, one can consider the same problem for $d$-separable spaces. Let us present an example along the same lines under the assumption $\aleph_1<\mathfrak c$:
\begin{prop}\label{prop:dsep} Suppose that $\aleph_1<\mathfrak c$. Then there is a $d$-separable space $X$ with $d(X)=\aleph_1$ that contains no dense $\sigma$-discrete sets of size $\aleph_1$.
\end{prop}
\begin{proof} J. Moore \cite[Theorem 5.4]{moore} proved that there is a colouring $c:[\oo_1]^2\to \oo$ such that for every $n\in \oo$, uncountable pairwise disjoint $A\subseteq [\oo_1]^n$, uncountable $B\subseteq \oo_1$ and $h:n\to \oo$ there exist $a\in A$ and $\beta\in B\setm \max(a)$ such that $c(a(i),\beta)=h(i)$ for every $i<n$, where $a=\{a(i):i<n\}$.
Suppose that $D=\{d_n:n\in\omega\}$ is any countable space.
\begin{clm} \label{jm} There is a dense and hereditarily Lindel\"of subspace $Y\subseteq D^{\oo_1}$ that is not separable.
\end{clm}
\begin{proof}
For each $\beta<\omg$, define $y_{\beta}\in D^\omg$ as follows:
\begin{equation}\label{eq:ya}
y_{\beta}({\alpha})=\left \{
\begin{array}{ll}
d_{c(\alpha,\beta)}&\text{if ${\alpha}<{\beta}$,}\\
d_0&\text{if ${\alpha}\geq {\beta}$.}
\end{array}
\right .
\end{equation}
Now let $Y=\{y_{\beta}:\beta<\oo_1\}$.
We claim that there is an $\alpha<\omg$ so that $Y\uhp (\omg\setm \alpha)$ is dense in $D^{\omg\setm \alpha}\simeq D^\omg$. Suppose otherwise: then we can find basic open sets $[\varepsilon_\alpha]$ in $D^{\omg\setm \alpha}$ so that $Y\cap [\varepsilon_\alpha]=\emptyset$. By standard $\Delta$-system arguments, we find $I\in [\omg]^{\aleph_1}$, $n\in \omega$ and $h:n\to \omega$ so that $\dom(\varepsilon_\alpha)=\{a_\alpha(i):i<n\}$ are pairwise disjoint for $\alpha\in I$ and $d_{h(i)}\in \varepsilon_\alpha(a_\alpha(i))$ for each $i<n$. Now, there exist $\alpha\in I$ and $\beta\in \omg\setm \max(\dom(\varepsilon_\alpha))$ so that $c(a_\alpha(i),\beta)=h(i)$ for all $i<n$. This means that $d_{c(a_\alpha(i),\beta)}\in \varepsilon_\alpha(a_\alpha(i))$ for $i<n$ and so $y_\beta\in [\varepsilon_\alpha]$. This contradicts our assumption.
It is clear that $d(Y\uhp (\omg\setm \alpha))=\aleph_1$. It remains to prove that $Y\uhp (\omg\setm \alpha)$ is hereditarily Lindel\"of.
Fix $W\in[\omega_1]^{\aleph_1}$ and, for each $\gamma\in W$, let $[\varepsilon_\gamma]$ be a basic open subset of $D^{\omega_1\setminus\alpha}$ with $y_\gamma\uhp(\omg\setminus\alpha)\in[\varepsilon_\gamma]$; we may assume that $\max(\dom(\varepsilon_\gamma))>\gamma$.
Suppose, by way of contradiction, that for each $\eta<\omg$
we have $\{y_\gamma\uhp(\omg\setminus\alpha):\gamma\in W\}\nsubseteq\bigcup\{[\varepsilon_\gamma]\colon\gamma\in W\cap\eta\}$.
We can then recursively define, for $\zeta<\omg$,
$\cdot$ $\delta_0$ as the least element of $W$;
$\cdot$ $\delta_{\zeta+1}$ as the least $\delta\in W$ satisfying $\delta>\sup_{\eta\le\zeta}\max(\dom(\varepsilon_{\delta_\eta}))$ and $y_\delta\uhp(\omg\setminus\alpha)\notin\bigcup\{[\varepsilon_\gamma]\colon\gamma\in W\cap(\delta_\zeta+1)\}$;
$\cdot$ $\delta_\zeta$ as the least $\delta\in W$ satisfying $\delta>\sup_{\eta<\zeta}\max(\dom(\varepsilon_{\delta_\eta}))$ and $y_\delta\uhp(\omg\setminus\alpha)\notin\bigcup\{[\varepsilon_\gamma]\colon\gamma\in W\cap\sup_{\eta<\zeta}\delta_\eta\}$
if $\zeta$ is a limit ordinal.
Again by $\Delta$-system arguments, there exist
$r\in[\omg\setminus\alpha]^{<\aleph_0}$,
$p:r\to D$,
$Z\in[\omg]^{\aleph_1}$,
$n\in\omega$
and
$h:n\to\omega$
satisfying
\begin{itemize}
\item[$(i)$]
$r\subseteq\dom(\varepsilon_{\delta_\zeta})$ for all $\zeta\in Z$;
\item[$(ii)$]
$\dom(\varepsilon_{\delta_\zeta})\setminus r=\{a_\zeta(i):i<n\}$ are pairwise disjoint for $\zeta\in Z$;
\item[$(iii)$]
$d_{h(i)}\in \varepsilon_{\delta_\zeta}(a_\zeta(i))$ for each $i<n$; and
\item[$(iv)$]
$p\subseteq y_{\delta_\zeta}$ for all $\zeta\in Z$.
\end{itemize}
Now, there are $\zeta,\zeta'\in Z$ such that $\delta_{\zeta'}\ge\max(\dom(\varepsilon_{\delta_\zeta}))$ and $c(a_\zeta(i),\delta_{\zeta'})=h(i)$ for all $i<n$.
Thus $d_{c(a_\zeta(i),\delta_{\zeta'})}\in \varepsilon_{\delta_\zeta}(a_\zeta(i))$ for $i<n$, whence
$y_{\delta_{\zeta'}}\uhp(\omg\setminus\alpha) \in [\varepsilon_{\delta_\zeta}]$
-- which is a contradiction since the fact that $\delta_{\zeta'}\ge\max(\dom(\varepsilon_{\delta_\zeta}))>\delta_\zeta$
implies $\zeta<\zeta'$ by construction.
\end{proof}
Now, by $\mathfrak c \geq \aleph_2$, we can pick a countable dense $D\subseteq\omega^{\omega_2}$. Then $D^{\omega_1}$ is dense in $(\omega^{\omega_2})^{\omega_1} \simeq \omega^{\omega_2}$. By Claim \ref{jm}, there is a dense $Y \subseteq D^{\omega_1}$ such that every discrete subset of $Y$ is countable and hence nowhere dense (as $Y$ is non-separable). Now, by Lemma \ref{power0}, there is a dense $\sigma$-closed-discrete $E \subseteq \omega^{\omega_2}$ satisfying $d(E) = \aleph_2$ -- in view of Theorem \ref{myc} and the fact that e.g. $\sigma(\omega^{\omega_2})=\{x\in\omega^{\omega_2}:|\{\alpha\in\omega_2:x(\alpha)\neq 0\}|<\aleph_0\}$ is a dense subset of $\omega^{\omega_2}$ with density $\aleph_2$.
Let $X=Y\cup E$. An argument strictly analogue to what is done in Claim \ref{4.3} finishes the proof.
\end{proof}
The assumption $\aleph_1<\mathfrak c$ is somewhat unnatural in Proposition \ref{prop:dsep} but we do not know how to remove it:
\begin{prob}
Is there a ZFC example of a $d$-separable space $X$ with the property that every $\sigma$-discrete dense subset of $X$ has cardinality greater than $d(X)$?
\end{prob}
In particular, we cannot answer the following:
\begin{prob}
Is there, in ZFC, a dense $Y\subseteq 2^{\oo_2}$ of size $\aleph_1$ all of whose $\sigma$-discrete subsets are nowhere dense?
\end{prob}
Finally, recall that any compact, $e$-separable space satisfies $d(X)=d_e(X)$. We do not know if the analogue holds for $d$-separable spaces:
\begin{prob}Is there a $\sigma$-discrete dense subset of size $d(X)$ in any compact, $d$-separable space $X$?
\end{prob}
\section{Preservation under products}\label{pressec}
As mentioned in the introduction, the behaviour of separable and $d$-separable spaces under products and powers is very well described: separability is preserved by products of size $\leq \mathfrak c$ but not bigger; on the other hand, the product of $d$-separable spaces is always $d$-separable. Hence our goal in this section is answering the following natural question: for which cardinals $\kappa$ is it true that every product of $\kappa$ many $e$-separable spaces is $e$-separable? As noted earlier in Example \ref{2^c+}, any such $\kappa$ is at most the continuum.
Let us start with powers of a single $e$-separable space.
We would like to thank Ofelia T. Alas for pointing out the following
to us:
\begin{prop}[Alas]
\label{cpower}
Let $X$ be an $e$-separable space and $\kappa\le\mathfrak{c}$. Then
the space $X^\kappa$ is $e$-separable.
\end{prop}
\begin{proof}
Let $(D_n)_{n\in\omega}$ be a sequence of closed discrete subsets of
$X$ with $\bigcup_{n\in\omega}D_n$ dense in $X$.
Fix a subspace $Y\subseteq\mathbb{R}$ with $|Y|=\kappa$, and let
$\mathcal{B}$ be a countable base for $Y$. Now consider
$T=\bigcup_{n\in\omega}(S_n\times\mbox{}^n\omega)$, where
$S_n=\{(B_0,\dots,B_{n-1})\in\mbox{}^n\mathcal{B}:\forall
i,j<n\;(i\neq j\Rightarrow B_i\cap B_j\neq\emptyset)\}$ for every
$n\in\omega$.
Fix an arbitrary $p\in X$.
For each $t=((B_0,\dots,B_{n-1}),(k_0,\dots,k_{n-1}))\in T$, we
define $E_t$ to be the set of those $x\in X^Y$ so that there is an $(a_i)_{i<n}\in\prod\{D_{k_i}:i<n\}$ with
\[
x(\alpha) = \begin{cases}
a_i, & \text{for } \alpha\in B_i \text { and }i<n, \text{ and}\\
p, & \text{for } \alpha\in Y\setminus\bigcup_{i=0}^{n-1}B_i.
\end{cases}
\]
It is routine to verify that each $E_t$ is a closed discrete subspace
of $X^Y$ and that $\bigcup_{t\in T}E_t$ is dense in $X^Y$. Since $T$
is countable and $|Y|=\kappa$, it follows that $X^\kappa$ is
$e$-separable.
\end{proof}
Now, we turn to arbitrary products of $e$-separable spaces. We will see that the heart of the matter is whether we can find \emph{large closed discrete sets in the product of small discrete spaces}.
In \cite{Mrow}, Mr\'owka introduced a class of cardinals denoted by $\mc M^*$: we write $\lambda\in \mc M^*$ iff there is a product of $\lambda$ many discrete spaces $X=\prod \{X_\alpha:\alpha<\lambda\}$ each of size $<\lambda$ so that $X$ has a closed discrete set of size $\lambda$. Equivalently, the product $\prod\{D(\nu)^\lambda:\nu\in \lambda\cap \card\}$ contains a closed discrete set of size $\lambda$.
If a cardinal $\lambda$ is in $\mathcal M^*$ then some degree of compactess fails for $\lambda$. Let us make this statement precise: recall that $\mc L_{\lambda,\omega}$ is the infinitary language which allows conjunctions and disjunctions of $<\lambda$ formulas and universal or existential quantification over finitely many variables. The language $\mc L_{\lambda,\oo}$ is \emph{weakly compact} by definition if every set of at most $\lambda$ sentences $\Sigma$ from $\mc L_{\lambda,\oo}$ has a model provided that every $S\in [\Sigma]^{<\lambda}$ has a model (see \cite{Jech}, p. 382).
\begin{thm}[Mr\'owka \cite{Mrow}, Chudnovsky \cite{Cud}] $\lambda \notin \mc M^*$ if and only if $\mc L_{\lambda,\oo}$ is weakly compact.
\end{thm}
Now, as expected, $\lambda \notin \mc M^*$ -- or, equivalently, the statement ``$\mc L_{\lambda,\oo}$ is weakly compact'' -- has some large cardinal strength. First, we mention two classical results:
\begin{lemma}\cite[Exercises 17.17 and 17.18]{Jech} \label{jechlemma} If $\mc L_{\lambda,\oo}$ is weakly compact then $\lambda$ is weakly inaccessible.
\end{lemma}
\begin{lemma}\cite[Theorem 17.13]{Jech} $\lambda$ is a weakly compact cardinal iff it is strongly inaccessible and $\mc L_{\lambda,\oo}$ is weakly compact.
\end{lemma}
For our current purposes, one can consider the above lemma the definition of weakly compact cardinals. Now, given a weakly compact cardinal $\lambda$, one can enlarge the continuum while the language $\mc L_{\lambda,\oo}$ remains weakly compact:
\begin{thm}[Chudnovsky \cite{Cud}, Boos \cite{Boos}]\label{cud} If $\lambda$ is a weakly compact cardinal and $\mb C_{\lambda^+}$ is the poset for adding $\lambda^+$ many Cohen-reals then $V^{\mb C_{\lambda^+}}\models$ ``$\mc L_{\lambda,\omega}$ is weakly compact hence $\mathfrak{c}\setm \mc M^*\neq \emptyset$''.
\end{thm}
Finally, recently B. Cody, S. Cox, J. D. Hamkins and T. Johnstone \cite{joel, joel2} showed that a weakly compact cardinal can be recovered from $\mc L_{\lambda,\oo}$ being weakly compact:
\begin{thm}\label{joel}
If $\mc L_{\lambda,\oo}$ is weakly compact then $\lambda$ is weakly compact in $L$.
\end{thm}
Now, it is easy to derive our first main result about non-preservation:
\begin{lemma}\label{main1}
If $\lambda\leq \mathfrak c$ and $\lambda\notin \mc M^*$ then there is a non-$e$-separable product of $\lambda$ many discrete spaces.
\end{lemma}
\begin{proof}
$\lambda\notin \mc M^*$ implies that $\mc L_{\lambda,\omega}$ is weakly compact and hence $\lambda$ is a regular limit cardinal. Now take discrete spaces $X_\alpha$ of size $<\lambda$ such that $\sup \{|X_\alpha|:\alpha<\lambda\}=\lambda$. The product $X=\prod \{X_\alpha:\alpha<\lambda\}$ contains no closed discrete subsets of size $\lambda$ as $\lambda\notin \mc M^*$. We claim that $d(X)=\lambda$, which follows from the following more general observation:
\begin{obs}\label{dens} Suppose that $\kappa \leq \mathfrak{c}$ and $X_\alpha$ is discrete for $\alpha<\kappa$. Then $d(\prod \{X_\alpha:\alpha<\kappa\})=\sup \{|X_\alpha|:\alpha<\kappa\}$.
\end{obs}
To prove this observation, simply apply the usual trick appearing in the proof of Proposition \ref{cpower}.
Now, we claim that $X$ cannot be $e$-separable. Indeed, if $X$ is $e$-separable then Observation \ref{cardobs} implies that $X$ has a closed discrete subset of size $d(X)=\lambda=cf(\lambda)>\omega$; however, this is not the case.
\end{proof}
Hence, we immediately get the following:
\begin{corol}\label{main1cor} If the existence of a weakly compact cardinal is consistent with ZFC then so is the statement that there is a non-$e$-separable product of less than $\mathfrak{c}$ many discrete spaces.
\end{corol}
\begin{proof}
Apply Lemma \ref{main1} and Theorem \ref{cud}.
\end{proof}
Now, we will obtain that it is also consistent with ZFC that every product of at most $\mathfrak c$ many $e$-separable spaces is $e$-separable; we will do so by showing that this last statement is implied by the non-existence of weakly compact cardinals in $L$. It will suffice to prove
\begin{thm}\label{main2} Suppose that $\lambda\leq \mathfrak{c}$ is minimal so that there is a family of $\lambda$ many $e$-separable spaces with non-$e$-separable product. Then $\lambda\notin \mc M^*$ and so $\mc L_{\lambda,\omega}$ is weakly compact.
\end{thm}
Let us mention that $\mc L_{\mathfrak c,\omega}$ is not weakly compact \cite{joel} and so $\lambda<\mathfrak c$ in the previous theorem. In any case, if $\mc L_{\lambda,\omega}$ is weakly compact then $\lambda$ is weakly compact in $L$ by Theorem \ref{joel}. In turn, we have the following result:
\begin{corol}\label{main2cor} If there is a non-$e$-separable product of at most $\mathfrak c$ many $e$-separable spaces then there is a weakly compact cardinal in $L$.
\end{corol}
By combining Corollaries \ref{main1cor} and \ref{main2cor}, we obtain:
\begin{corol}\label{maincor}
The following statements are equiconsistent relative to ZFC:
\begin{itemize}
\item[$(a)$]
there is a product of at most $\mathfrak c$ many $e$-separable spaces that fails to be $e$-separable;
\item[$(b)$]
there is a weakly compact cardinal.
\end{itemize}
\end{corol}
Let us now turn to proving Theorem \ref{main2}. First, we start by reducing the problem to products of discrete spaces again:
\begin{lemma}
\label{lemmahmp}
Suppose that $\kappa\le\mathfrak{c}$. Then the following are equivalent:
\begin{itemize}
\item[$(a)$]
every product of at most $\kappa$ many $e$-separable spaces is
$e$-separable;
\item[$(b)$]
every product of at most $\kappa$ many discrete spaces is
$e$-separable.
\end{itemize}
\end{lemma}
\begin{proof}
The implication $(a)\Rightarrow(b)$ holds trivially. We prove
$(b)\Rightarrow(a)$.
Let $X=\prod\{X_\alpha:\alpha\in Y\}$, where $Y\subseteq\mathbb{R}$ has
cardinality at most $\kappa$ and each $X_\alpha$ is $e$-separable. For
each $\alpha\in Y$, fix a point $p_\alpha\in X_\alpha$ and a sequence
$(E^\alpha_k)_{k\in\omega}$ of closed discrete subsets of $X_\alpha$
with $\overline{\bigcup_{k\in\omega}E^\alpha_k}=X_\alpha$.
Fix a countable base $\mathcal{B}$ for $Y$ and, for each
$n\in\omega$, consider
$$
S_n=\{(B_i)_{i<n}\in\mbox{}^n\mathcal{B}:\forall
i,j<n\;(i\neq j\Rightarrow B_i\cap B_j=\emptyset)\};
$$
now, for each $t=((B_i)_{i<n},(k_0,\dots,k_{n-1}))\in S_n\times\mbox{}^n\omega$, define
$Y_t$ to be the set of those $x\in X$ so that
\[
x(\alpha) = \begin{cases}
x'_\alpha & \text{ for some } x'_\alpha\in E^\alpha_{k_i} \text{ for } \alpha\in B_i \text { and }i<n, \text{ and}\\
p_\alpha, & \text{for } \alpha\in Y\setminus\bigcup_{i=0}^{n-1}B_i.
\end{cases}
\]
Note that each $Y_t$ is homeomorphic to the product
$\prod_{i<n}\prod_{\alpha\in B_i}E^\alpha_{k_i}$. Hence $Y_t$ is is
$e$-separable by $(b)$. Let $(D^t_k)_{k\in\omega}$ be a sequence
of closed discrete subsets of $Y_t$ with
$\overline{\bigcup_{k\in\omega}D^t_k}=Y_t$.
Since each $Y_t$ is closed in $X$, we have that each $D^t_k$ is a
closed discrete subset of $X$. Finally, as
$\bigcup_{n\in\omega}\bigcup_{r\in S_n\times\mbox{}^n\omega}Y_t$ is
dense in $X$, it follows that
$$
\overline{\bigcup_{n\in\omega}\bigcup_{t\in
S_n\times\mbox{}^n\omega}\bigcup_{k\in\omega}D^t_k}=X,
$$
thus showing that $X$ is $e$-separable.
\end{proof}
Note that we immediately get the following easy:
\begin{corol}\label{finiteprod}
The product of finitely many $e$-separable spaces is $e$-separable.
\end{corol}
Second, we show that as long as we take the product of large discrete sets relative to the number of terms, we end up with an $e$-separable product:
\begin{lemma}
\label{reg}
Let $\kappa$ be an infinite cardinal. Then the product of
at most
$\kappa$ many discrete spaces of cardinality at least
$\kappa$ is $e$-separable.
\end{lemma}
\begin{proof}
Let $X=\prod\{X_\alpha:\alpha\in\lambda\}$, where $\lambda\le\kappa$
and each $X_\alpha$ is a discrete space with cardinality at least
$\kappa$. We can assume that $\lambda$ is infinite and that
$X_\alpha=|X_\alpha|$ for all $\alpha\in\lambda$.
Define
$$
P^i_j=\{(F,p)\in[\lambda]^i\times
Fn(\lambda,\kappa):|p|=j\textrm{ and }F\cap\mathrm{dom}(p)=\emptyset\}
$$
for each $i,j\in\omega$ where $Fn(\lambda,\kappa)$ denotes the set of finite partial functions from $\lambda$ to $\kappa$. Fix an injective function
$\varphi:\bigcup_{i,j\in\omega}P^i_j\rightarrow\kappa$
such that $\varphi(F,p)>\max(\mathrm{ran}(p))$ for every
$(F,p)\in\bigcup_{i,j\in\omega}P^i_j$.
Now, for every $i,j\in\omega$, let $E^i_j$ be the set of all $x\in X$
for which there is $(F,p)\in P^i_j$ satisfying
\begin{enumerate}
\item $x(\xi)\ge\kappa$ for all $\xi\in F$,
\item $x\in [p]$, and
\item $x(\xi)=\varphi(F,p)$ for all $\xi\in\lambda\setminus(F\cup\mathrm{dom}(p))$.
\end{enumerate}
It is straightforward to verify that $\bigcup_{i,j\in\omega}E^i_j$ is
dense in $X$. We claim that each $E^i_j$ is a closed discrete subset
of $X$, which will conclude our proof.
From this point on, let $i,j\in\omega$ be fixed.
To see that $E^i_j$ is discrete, pick an arbitrary $x\in E^i_j$, and
let this be witnessed by the pair $(F,p)\in P^i_j$. Note that the choice of $\varphi$ ensures that this $(F,p)$ is unique. Pick any
$\eta\in\lambda\setminus(F\cup\mathrm{dom}(p))$ and let $$V=[x\uhp (\dom(p)\cup F\cup \{\eta\})].$$ Then $V$ is an open
neighbourhood of $x$ in $X$ satisfying $E^i_j\cap V=\{x\}$.
It remains to show that $E^i_j$ is closed in $X$. Let then $y\in
X\setminus E^i_j$; we must find an open neighbourhood $V$ of $y$ in
$X$ such that $V\cap E^i_j=\emptyset$. We shall do so by considering
several cases.\\
\emph{$\cdot$ Case 1.}
$G=\{\xi\in\lambda:y(\xi)\ge\kappa\}$ has more than $i$ elements.
Then we may take any $H\in[G]^{i+1}$ and define $V=[y\uhp H]$.\\
\emph{$\cdot$ Case 2.}
$G=\{\xi\in\lambda:y(\xi)\ge\kappa\}$ has cardinality at most $i$.
We will split this case in two:\\
\emph{$\cdot$ Case 2.1.}
$\mathrm{ran}(y)\cap\kappa$ is infinite.
Then we can take $A\in[\kappa]^{j+2}$ such that
$y''A\in[\kappa]^{j+2}$ and define $V=[y\uhp A]$.\\
\emph{$\cdot$ Case 2.2.}
$\mathrm{ran}(y)\cap\kappa$ is finite.
Let $\mu=\max(\mathrm{ran}(y)\cap\kappa)$ and
$H=\{\xi\in\lambda:y(\xi)<\mu\}$.
We divide this case into three subcases:\\
\emph{$\cdot$ Case 2.2.1.}
$|H|>j$.
Pick $H'\in[H]^{j+1}$ and $\beta\in\lambda$ such that
$y(\beta)=\mu$. Now take $V=[g\uhp (H'\cup\{\beta\})]$.\\
\emph{$\cdot$ Case 2.2.2.}
$|H|\le j$ and $\mu\notin\mathrm{ran}(\varphi)$.
Let $B\in[\lambda]^{j+1-|H|}$ be such that
$y''B=\{\mu\}$ and consider $V=[g\uhp (H\cup B)]$.\\
\emph{$\cdot$ Case 2.2.3.}
$|H|\le j$ and $\mu\in\mathrm{ran}(\varphi)$.
Let $(F,p)\in P^i_j$ be such that $\varphi(F,p)=\mu$ and, as in the
previous case, take $B\in[\lambda]^{j+1-|H|}$ satisfying
$y''B=\{\mu\}$. Now define
$$V=[g\uhp (G\cup H\cup B\cup F\cup\mathrm{dom}(p))].$$
Suppose, in order to get a contradiction, that there is $x\in V\cap
E^i_j$ and let $(F',p')\in P^i_j$ witness that $x\in E^i_j$. Since
$|H\cup B|=j+1$ and $x''(H\cup B)=y''(H\cup B)\subseteq\kappa$, we have
that $\varphi(F',p')=\max(x''(H\cup B))=\max(y''(H\cup B))=\mu$. Hence
$(F',p')=(F,p)$ by injectivity of $\varphi$. Now, since
$F=\{\xi\in\lambda:x(\xi)\ge\kappa\}$ and
$G=\{\xi\in\lambda:y(\xi)\ge\kappa\}$, it follows from
$x\upharpoonright(F\cup G)=y\upharpoonright(F\cup G)$ that
$F=G$. Similarly, as $H=\{\xi\in\lambda:y(\xi)<\mu\}$ and
$\mathrm{dom}(p)=\{\xi\in\lambda:x(\xi)<\mu\}$, it follows from
$x\upharpoonright(H\cup\mathrm{dom}(p))=y\upharpoonright(H\cup\mathrm{dom}(p))$
that $H=\mathrm{dom}(p)$. Thus the pair
$(G,y\!\upharpoonright\!H)=(F,p)\in P^i_j$ witnesses that $y\in
E^i_j$, a contradiction.
\end{proof}
Finally, we are ready to present
\begin{proof}[Proof of Theorem \ref{main2}]
Suppose that $\lambda\leq \mathfrak{c}$ is minimal so that there are $e$-separable spaces $X_\alpha$ such that $X=\prod \{X_\alpha:\alpha<\lambda\}$ is not $e$-separable. By Lemma \ref{lemmahmp}, we can suppose that each $X_\alpha$ is discrete.
Note that $$X\simeq\prod \{X_\alpha: \alpha<\lambda, |X_\alpha|<\lambda\}\times \prod \{X_\alpha: \alpha<\lambda, |X_\alpha|\geq \lambda\}.$$ We know that the second term on the right-hand side is $e$-separable by Lemma \ref{reg}. So if $X$ is not $e$-separable then $\prod \{X_\alpha: \alpha<\lambda, |X_\alpha|<\lambda\}$ is not $e$-separable either by Corollary \ref{finiteprod}.
Now, we define $Y_\nu=\prod\{X_\alpha:\alpha<\lambda, |X_\alpha|=\nu\}$ for $\nu\in \lambda\cap \card$. Note that $Y_\nu$ is $e$-separable by Theorem \ref{cpower}. Hence, the minimality of $\lambda$ implies that $I=\{\nu\in \lambda\cap \card:Y_\nu\neq\emptyset\}$ has size $\lambda$; otherwise $X\simeq\prod\{Y_\nu:\nu\in I\}$ is a smaller non-$e$-separable product of $e$-separable spaces. Note that this already shows that $\lambda=\oo_\lambda$.
Let us suppose that $\lambda \in \mc M^*$; we will arrive at a contradiction shortly. Take a decreasing sequence $(I_n)_{n\in\oo}$ of subsets of $I$ so that $\bigcap\{I_n:n\in\oo\}=\emptyset$ and $\lambda=|I_n|=|I\setm I_n|$ for each $n\in\oo$. Note that $d(\prod\{Y_\nu:\nu\in I\setm I_n\})=\lambda$ by Observation \ref{dens}.
\begin{clm} $\prod\{Y_\nu:\nu\in I_n\}$ contains a closed discrete set of size $\lambda$.
\end{clm}
\begin{proof} $\lambda \in \mc M^*$ implies that $Z=\prod \{D(\nu)^\lambda : \nu\in \lambda\cap \card\}$ contains a closed discrete subset of size $\lambda$. Hence, it suffices to show that $Z$ embeds into $\prod\{Y_\nu:\nu\in I_n\}$ as a closed subspace. In order to do that, note that the set $\{\nu\in I_n: \nu>\nu_0\}$ has size $\lambda$ for every $\nu_0\in \lambda\cap \card$. Now it is routine to construct the embedding of $Z$.
\end{proof}
Finally, we can apply Lemma \ref{glue} to see that the product $X=\prod\{Y_\nu:\nu\in I\}$ must be $e$-separable. This contradicts our initial assumption on $X$.
\end{proof}
\section{Final remarks and further questions}
First, referring back to Section \ref{d_and_e}, it is natural to ask if we can say something similar to Theorem \ref{prodthm} about products. Let us present a result in this direction:
\begin{lemma}\label{glue} Suppose that $\kappa$ is an infinite cardinal and there is a decreasing sequence $(I_n)_{n\in\oo}$ of non-empty subsets of $\kappa$ with empty intersection such that $\prod\{X_\alpha:\alpha\in I_n\}$ contains a closed discrete subset of size $\delta_n=d(\prod\{X_\alpha:\alpha\in \kappa\setm I_n\})$ for every $n\in\oo$. Then $X=\prod\{X_\alpha:\alpha<\kappa\}$ is $e$-separable.
\end{lemma}
\begin{proof} Let $X(J)$ denote $\prod\{X_\alpha:\alpha\in J\}$ for $J\subseteq \kappa$. Pick $D_n=\{d^n_\xi:\xi<\delta_n\}\subseteq X(\kappa\setm I_n)$ dense and let $F_n=\{f^n_\xi:\xi<\delta_n\}\subseteq X(I_n)$ be closed discrete.
Now, for each $n\in\omega$, we define $e^n_\xi\in X$ for $\xi<\delta_n$ by
\[
e^n_\xi(\alpha) = \begin{cases}
d^n_\xi(\alpha), & \text{for } \alpha\in \kappa\setm I_n, \text{ and}\\
f^n_\xi(\alpha), & \text{for } \alpha\in I_n.
\end{cases}
\]
We claim that the set $E_n=\{e^n_\xi:\xi<\delta_n\}$ is closed discrete. This follows from
\begin{obs}\label{proj} Suppose that $E\subseteq \prod\{ X_\alpha: \alpha <\kappa\}$ and there is $I\subseteq \kappa$ such that $\pi_I$ is 1-1 on $E$ and the image $\pi_I\mbox{}''E$ is closed discrete in $\prod\{ X_\alpha: \alpha \in I\}$. Then $E$ is closed discrete.
\end{obs}
Now, it is clear that $\bigcup \{E_n:n\in\oo\}$ is a dense and $\sigma$-closed-discrete subset of $X$.
\end{proof}
\begin{comment}
\begin{rem} Consider the product $X=\prod \{D(\alpha):\alpha<\mu\}$ where $\mu$ is a measurable cardinal. Then $d(X)=e(X)=\kappa$ however it can be shown that $X$ does not contain a closed discrete set of size $\kappa$.
\end{rem}
\begin{prob} Suppose that $d(X^\kappa)\leq e(X^\kappa)$ where $\kappa$ is infinite. Is $X^\kappa$ $e$-separable?
\end{prob}
\begin{prob} Do we have something similar to Lemma \ref{power0} for products?
\end{prob}
\begin{defin}$$\mrow=\{\kappa: \omega^\kappa \text{ contains a closed discrete set of size } \kappa\}$$
$$\ulam=\{\kappa:\kappa \text{ is Ulam non-measurable}\}$$
$$\esep=\{\kappa:\omega^\kappa \text{ is $e$-separable}\}$$
\end{defin}
$\mrow$ was introduced by Mr\'owka in \cite{mrow2}. $\mrow \subseteq \ulam$ and by Lemma \ref{power0} we have $\mrow\subseteq \esep$.
\begin{prob} What is the connection between $\mrow$ and $\esep$? Is $\mrow=\esep$? Let $\delta_\kappa=d(\omega^\kappa)$. What is the precise connection between these sets: $$\{\kappa\in\esep:cf(\delta_\kappa)>\omega\}\subseteq \mrow$$ $$\subseteq \{\kappa: \omega^\kappa \text{ contains a closed discrete set of size } \delta_\kappa\}\subseteq \esep \subseteq \{\kappa: \delta_\kappa \leq e(\omega^\kappa)\}$$
\end{prob}
\end{comment}
Second, recall that if $D(\lambda)$ is the discrete space of size $\lambda\geq \kappa$ then $D(\lambda)^\kappa$ is $e$-separable by Lemma \ref{reg}. Actually, we can say a bit more in this case:
\begin{lemma}
\label{discrprod}
Let $(\kappa_i)_{i\in I}$ be a sequence of cardinals and consider the
product space $X=\prod\{D(\kappa_i):i\in I\}$.
Suppose that the set $\{i\in I:\kappa_i=\kappa\}$ is infinite, where
$\kappa=\sum_{i\in I}\kappa_i$.
Then $X$ has a $\sigma$-discrete $\pi$-base.
\end{lemma}
\begin{proof}
Let $J$ be a countable infinite subset of $\{i\in
I:\kappa_i=\kappa\}$. Note that
$\kappa_j=\sum_{i\in I\setminus\{j\}}\kappa_i$ for all $j\in J$. Now let $\{p^j_n(\alpha):\alpha\in\kappa_j\}$ be an
enumeration of the set $$\{p\subseteq\bigcup_{i\in
I\setminus\{j\}}(\{i\}\times\kappa_i):p\textrm{ is a function and
}|p|=n\}$$ for every $j\in
J$ and $n\in\omega$.
Consider
$$A^j_n=\{p^j_n(\alpha)\cup\{(j,\alpha)\}:\alpha\in\kappa_j\};$$
finally, define
$\mathcal{V}^j_n=\{[q]:q\in A^j_n\}$.
Note that each $\mathcal{V}^j_n$ is a discrete family: if
$a=(a_i)_{i\in I}$ is any point of $X$ then $$U=\{(x_i)_{i\in I}\in
X:x_j=a_j\}$$ is an open neighbourhood of $a$ in $X$ such that
$$\{V\in\mathcal{V}^j_n:V\cap
U\neq\emptyset\}=\{V_{p^j_n(a_j)\cup\{(j,a_j)\}}\}.$$
Moreover, $\mathcal{V}=\bigcup_{j\in
J}\bigcup_{n\in\omega}\mathcal{V}^j_n$ is a $\pi$-base for $X$,
since any non-empty open subset of $X$ is determined by a finite number
of coordinates which constitutes a finite subset of
$I\setminus\{j\}$ for some $j\in J$.
\end{proof}
\begin{corol}
\label{discrpower}
If $\lambda\geq \kappa$ then $D(\lambda)^\kappa$ has a $\sigma$-discrete $\pi$-base.
\end{corol}
Finally, selection principles (see e.g. \cite{scheep}) and selective versions of separability and $d$-separability (see e.g. \cite{weaksep}) were proved to be fascinating notions to study. So let us introduce the selective version of $e$-separability:
\begin{defin}
\label{E-sep}
A topological space $X$ is \emph{$E$-separable} if for every sequence of dense sets
$(D_n)_{n\in\omega}$ of $X$ we can select $E_n\subseteq D_n$ so that $E_n$ is closed discrete in $X$ and
$\bigcup_{n\in\omega}E_n$ is dense in $X$.
\end{defin}
Note that every space with a $\sigma$-discrete $\pi$-base is $E$-separable as well. Let us point out that the example of Theorem \ref{examp} is an $e$-separable space which is not $E$-separable.
We ask the following questions:
\begin{prob}
Suppose that $X$ is an $e$-separable space which is the product of discrete spaces. Is $X$ $E$-separable as well?
\end{prob}
\begin{prob}
How does $E$-separability behave under powers and products?
\end{prob}
\section{Acknowledgements}
A significant portion of the above presented research was done at the University of Toronto.
The authors would like to thank Franklin Tall and William Weiss for their helpful comments.
We thank the Set Theory and Topology group of the Alfr\'ed R\'enyi Institute of Mathematics (Hungarian Academy of Sciences) for further help in preparing this paper. We deeply thank Victoria Gitman for bringing to our attention Theorem \ref{joel}, which was imperative for our equiconsistency result.
The first named author was partially supported by Capes (BEX-1088-11-4). The second named author was supported in part by the Ontario Trillium Scholarship and the FWF Grant I1921. | 45,144 |
Hrithik Roshan to host Just Dance on Star Plus
Hrithik Roshan, who is known for his dancing skills, has been signed up by Star Plus TV channel to host their forthcoming dance reality show Just Dance. This is the actor’s debut on the small screen and he is being paid Rs 2 crore per episode.
Hrithik recently participated in the shooting of Just Dance and the promos of the show will be aired soon. This show is said to be totally different from the other dance shows as the format and structure are innovative and more appealing. The highlight of the show would be Hrithik’s dancing at the beginning and conclusion of the show.
Hrithik now joins on TV the other Bollywood bigwigs – Amitabh Bachchan, Shah Rukh Khan, Salman Khan and Akshay Kumar as TV host.
Hrithik’s film Zindagi Na Milegi Dobara directed by Zoya Akhtar is due for release soon. He will also star in Karan Johar’s remake of Agneepath where he will play the character of Vijay Dinanath Chauhan which was played by Amitabh Bachchan in the original. | 144,327 |
\subsection{Proofs for training the output layer (Proof of Theorem \ref{thm sense})}\label{pfoutput}
To begin note that
\begin{align*}
\mtx{\Phi}\mtx{\Phi}^T=\phi\left(\mtx{X}\mtx{W}^T\right)\phi\left(\W\mtx{X}^T\right)=\sum_{\ell=1}^k \phi\left(\mtx{X}\vct{w}_\ell\right)\phi\left(\mtx{X}\vct{w}_\ell\right)^T\succeq \sum_{\ell=1}^k \phi\left(\mtx{X}\vct{w}_\ell\right)\phi\left(\mtx{X}\vct{w}_\ell\right)^T\mathbb{1}_{\{\twonorm{\phi(\X\w_\ell)}\le T_n\}}.
\end{align*}
Here $T_n$ a function of $n$ whose value shall be determined later in the proofs. To continue we need the matrix Chernoff result stated below.
\begin{theorem}[Matrix Chernoff] Consider a finite sequence $\mtx{A}_\ell\in\R^{n\times n}$ of independent, random, Hermitian matrices with common dimension $n$. Assume that $\mtx{0}\preceq \mtx{A}_\ell\preceq R\mtx{I}$ for $\ell=1,2,\ldots,k$. Then
\begin{align*}
\mathbb{P}\Bigg\{\lambda_{\min}\left(\sum_{\ell=1}^k \mtx{A}_\ell\right)\le (1-\delta)\lambda_{\min}\left(\sum_{\ell=1}^k\E[\mtx{A}_\ell]\right)\Bigg\}\le n\left(\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}\right)^{\frac{\lambda_{\min}\left(\sum_{\ell=1}^k\E[\mtx{A}_\ell]\right)}{R}}
\end{align*}
for $\delta\in[0,1)$.
\end{theorem}
Applying this theorem with $\mtx{A}_\ell=\phi\left(\mtx{X}\vct{w}_\ell\right)\phi\left(\mtx{X}\vct{w}_\ell\right)^T\mathbb{1}_{\{\twonorm{\phi(\X\w_\ell)}\le T_n\}}$, $R=T_n^2$ and $\widetilde{\mtx{A}}(\w):=\phi\left(\mtx{X}\vct{w}\right)\phi\left(\mtx{X}\vct{w}\right)^T\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}\le T_n\big\}}$
\begin{align}
\label{maintrunin}
\lambda_{\min}\left(\mtx{\Phi}\mtx{\Phi}^T\right)\ge (1-\delta)k\lambda_{\min}\left(\E[\widetilde{\mtx{A}}(\w)]\right),
\end{align}
holds with probability at least $1-n\left(\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}\right)^{\frac{k\lambda_{\min}\left(\E[\widetilde{\mtx{A}}(\w)]\right)}{T_n^2}}$.
Next we shall connect the the expected value of the truncated matrix $\widetilde{\mtx{A}}(\w)$ to one that is not truncated defined as $\mtx{A}(\w)=\phi(\X\w)\phi(\X\w)^T$. To do this note that
\begin{align}
\label{intertrunc}
\opnorm{\E[\widetilde{\mtx{A}}(\w)-\mtx{A}(\w)]}=&\opnorm{\E\Big[\phi(\X\w)\phi(\X\w)^T\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}\Big]}\nn\\
\overset{(a)}{\le}&\E\Big[\opnorm{\phi(\X\w)\phi(\X\w)^T\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}}\Big]\nn\\
\le&\E\Big[\twonorm{\phi(\X\w)}^2\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}\Big]\\
\overset{(b)}{\le}&2\E\Big[\twonorm{\phi(\X\w)-\phi(\vct{0})}^2\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}\Big]+2\E\Big[\twonorm{\phi(\vct{0})}^2\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}\Big]\nn\\
\overset{(c)}{\le}&2B^2\E\Big[\twonorm{\X\w}^2\mathbb{1}_{\big\{\twonorm{\phi(\X\w)}> T_n\big\}}\Big]+2nB^2\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}\nn\\
\overset{(d)}{\le}&2B^2\sqrt{\E\big[\twonorm{\X\w}^4\big]\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}}+2nB^2\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}\nn\\
\overset{(e)}{\le}&2\sqrt{n}B^2\sqrt{\left(\sum_{i=1}^n \E\big[\abs{\x_i^T\w}^4\big]\right)\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}}+2nB^2\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}\nn\\
\overset{(f)}{\le}&2\sqrt{3}nB^2\sqrt{\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}}+2nB^2\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}\nn\\
\le&6nB^2\sqrt{\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}}.
\end{align}
Here, (a) follows from Jensen's inequality, (b) from the simple identity $(a+b)^2\le 2(a^2+b^2)$, (c) from $\abs{\phi'(z)}\le B$, (d) from the Cauchy-Schwarz inequality, (e) from Jensen's inequality, and (f) from the fact that for a standard moment random variable $X$ we have $\E[X^4]=3$.
To continue we need to show that $\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}$ is small. To this aim note that for any activation $\phi$ with $\abs{\phi'(z)}\le B$ we have
\begin{align*}
\twonorm{\phi\left(\mtx{X}\vct{w}_2\right)-\phi\left(\mtx{X}\vct{w}_1\right)}\le B\opnorm{\mtx{X}}\twonorm{\vct{w}_2-\vct{w}_1}
\end{align*}
Thus by Lipschitz concentration of Gaussian functions for a random vector $\vct{w}\sim\mathcal{N}(\vct{0},\mtx{I}_d)$ we have
\begin{align*}
\twonorm{\phi\left(\mtx{X}\vct{w}\right)}\le& \E[\twonorm{\phi\left(\mtx{X}\vct{w}\right)}]+t,\\
\le&\sqrt{\E[\twonorm{\phi\left(\mtx{X}\vct{w}\right)}^2]}+t,\\
=&\sqrt{n}\sqrt{\E_{g\sim\mathcal{N}(0,1)}[\phi^2(g)]}+t,\\
\le& B\sqrt{2n}+t,
\end{align*}
holds with probability at least $1-e^{-\frac{t^2}{2B^2\opnorm{\mtx{X}}^2}}$. Thus using $t=\Delta B\sqrt{n}$ we conclude that
\begin{align*}
\twonorm{\phi\left(\mtx{X}\vct{w}\right)}\le (\Delta+\sqrt{2})B\sqrt{n},
\end{align*}
holds with probability at least $1-e^{-\frac{\Delta^2}{2}\frac{n}{\opnorm{\mtx{X}}^2}}$. Thus using $\Delta=c\sqrt{\log n}$ and $T_n=CB\sqrt{n\log n}$ we can conclude that
\begin{align*}
\mathbb{P}\{\twonorm{\phi(\X\w)}> T_n\}\le \frac{1}{n^{202}}.
\end{align*}
Thus, using \eqref{intertrunc} we can conclude that
\begin{align*}
\opnorm{\E[\widetilde{\mtx{A}}(\w)-\mtx{A}(\w)]}\le \frac{6B}{n^{100}}.
\end{align*}
Combining this with \eqref{maintrunin} with $\delta=1/2$ we conclude that
\begin{align*}
\lambda_{\min}\left(\mtx{\Phi}\mtx{\Phi}^T\right)\ge \frac{1}{2}k\left(\lambda_{\min}\left(\E[\mtx{A}(\w)]\right)-\frac{6B}{n^{100}}\right)=\frac{1}{2}k\left(\widetilde{\lambda}(\X)-\frac{6B}{n^{100}}\right),
\end{align*}
holds with probability at least $1-ne^{-\gamma \frac{k\widetilde{\lambda}(\X)}{T_n^2}}$. The latter probability is larger than $1-\frac{1}{n^{100}}$ as long as
\begin{align*}
k\ge C \log^2(n)\frac{n}{\widetilde{\lambda}(\X)},
\end{align*}
concluding the proof. | 82,561 |
Professione: Event Planning Business Owner/Certified TESOL/TEFL Tutor
Esperienza di insegnamento: I have over 100 hours of online teaching experience teaching children from ages 3 to 13 years old in China. I worked as an administrator in a private school where I interacted with parents and children from many different cultural backgrounds. I have tutored students in the following areas. ESL beginner to advanced. IELTS,Business and Everyday English.
Interessi: Travel, Reading, Music, Meeting People from other countries. I enjoy watching K-Dramas.
Istruzione:
Lingue parlate: inglese (Madrelingua), coreano (Base)
Profilo video: | 221,197 |
11 comments:
I love this style,... but it looks like you came across with some serious issues. I'm sorry it didn't workout perfectly but I share your opinion when you say it's a very nice top!
Oy. I like the style on you. I'm sorry it had so many issues. I HATE when I can't figure out the sleeves on garments. But.... if you like the top, why not try again? Not that I should talk. I've repeated maybe three things ever.
Sorry this didn't work out, but this is really helpful info nonetheless.
Oh Goody, Nancy, I can't wait to get my pattern!! LOL
I know how you were looking forward to getting this blouse done...and now the disappointment of the outcome, so sorry..
I think when and if I make it up, I will either shorten or lengthen the sleeves and shorten the hemline. Those two changes will look better on my figure type.
I agree with Cidell, maybe try it again and in a fabric with more drape, and see if it changes any of these issues.
The style is nice on you,BTW.
Enid
I think I am going to try it again but I'm going to try to use a sleeve from a different pattern to see if I can get it to fit better and not be shaped funny. I have some Ana Sui charmeuse that I thought I'd try.
Sounds good to me!
New to your blog (via Cidell) but I love the blouse on you. I hear you about the pleat, but with a little tweaking on the next version it is really cute. In fact, I'd wear it just as is. *grin* I hate sleeves and practice raglan just for the safe aspect!
This has been a very informative series of posts. We learned as much as you, which is always one good part of "sewing gone bad!"
You are doing a good job of salvaging what you can of the project. It's looking cute on you.
Thanks for your detailed posts about this style. I've often pondered that one---the sketch is just so intriguing. The sleeve thing would infuriate me; I can handle just about anything going wonky, but sleeve ambiguity---that's the worst!
~Sarah
Thank you thank you. I too keep looking at HotPatterns, because the illustrations for their styles are so compelling, then I get a dose of reality when I see reviews like this one.
Sorry you had the problems you did, thank you for sharing.
BTW, have you read any more of The Collection yet?
Although I really like many HP designs, something has always kept me from trying one. For every item I read in praise of Hot Patterns, I read another item about problems. I'm still a little afraid of them. | 334,216 |
TITLE: Presentation of the fundamental group of an octagon with identified borders
QUESTION [0 upvotes]: I need a little bit of clarification about the presentation of the fundamental group.
My exercise is to consider all the possible identification of the opposite sides of an octagon and calculate their fundamental group, so I begun with
$$G=\langle a,b,c,d|abcdabcd\rangle\;\;\;\;\;\operatorname{and}\;\;\;\;\;G'=\langle a,b,c,d|abcda^{-1}b^{-1}c^{-1}d^{-1}\rangle$$
The second one is equivalent to $\langle x,y|xyx^{-1}y^{-1}\rangle$ so the same as a torus, while the first one I thought that it was equivalent to the projective space and I thought that its representations should be $\langle x,y|xyxy\rangle$.
Here I found out that the correct representation of the fundamental group is $\langle z|z^2\rangle$.
So my question here is why the representation of the torus can't be $\langle z|zz^{-1}\rangle$?
REPLY [2 votes]: Regarding the question in the final paragraph of your post, the group presented by $\langle z \mid z z^{-1} \rangle$ is infinite cyclic, and the fundamental group of the torus is not infinite cyclic, so that cannot be a presentation of the fundamental group of the torus.
Regarding the part of your post asserting the presentations $\langle a,b,c,d \mid abcda^{-1}b^{-1}c^{-1}d^{-1} \rangle$ and $\langle xy \mid xyx^{-1}y^{-1} \rangle$ are the same group, the first has abelianization $\mathbb Z^4$ and the second has abelianization $\mathbb Z^2$, and so those cannot be isomorphic either.
Based on your comment, it looks like you are making mistakes in carrying out Tietze transformations. | 52,906 |
Chad Ochocinco live tweets his wedding, invites a follower
(CBS News) Chad Ochocinco wanted everyone to know that he got married on the Fourth of July, so much so that he live tweeted his wedding.
The avid Twitter user also invited one of his followers, 66-year-old grieving widow Cheryl Minton, to his St. Martin wedding.
Pictures: Chad Ochocinco
Pictures: Celebrity weddings and engagements
Trending News
The Miami Dolphins player and former "Dancing with the Stars" contestant married fellow reality TV star Evelyn Lozada on Wednesday.
"Live tweeting from my wedding...should be a first I'm assuming, music is playing, can't see my guests right now but they're here," he wrote on Twitter early in the evening.
Later he continued on, describing his pre-nuptial jitters. "I don't recall sending butterflies in my stomach an invite...and why am I shaking like I'm in Alaska," he wrote on Twitter.
Before his wedding, Ochocino invited Minton to his wedding after she initiated a conversation with him on Twitter about recently losing her husband. At the wedding the two posed for a picture that was later posted on his Twitter page.
In addition to Ochocinco's tweets, TV cameras also documented the event for the upcoming VH1 reality show "Ev and Ocho," reports People. | 222,563 |
…read more
Via:: Security Week | 193,060 |
Why we like it
- Relais & Chateaux seafront resort with breathtaking views of the Aegean Sea
- Rooms and suites are spread between the Main building with sea view rooms and suites with private balconies or terraces, Bungalows with private pools and Minoan Royal Suites with private pools
-
- A&K guests can enjoy a complimentary canoe, pedalo or stand up paddle board excursion per person for every stay of 7 nights or more (blackout dates apply)
About Elounda Mare
Originally built as a Cretan mansion, Elounda Mare Hotel offers traditional architecture and design combined with antique furniture and artwork. The bungalows are scattered throughout the gardens, spacious yet intimate with private seawater pools and enjoy panoramic views of the Bay of Elounda's calm turquoise waters. It has an unparalleled reputation for its attentive and personalised service. With award-winning restaurants, comprehensive leisure facilities and exquisite design, it is the ideal choice for those seeking a relaxing holiday in a serene and intimate setting.
The relaxing atmosphere of the property means that even at the peak of the season, swimming in the large seawater pool, or the children's paddling pool next to it, are usually a private affair. The extended lawns and sundecks along the shoreline are replete with secluded corners. The private sandy Blue Flag beach is lapped by crystal-clear waters. A Sea-sports facility includes waterskiing and SCUBA diving tuition as well as boats for charter. Don't miss the St. Marina Chapel's square for its art gallery. At it's sister property the Porto Elounda Golf & Spa Resort there is a golf academy, Six Senses Spa and crèche and Children's World with waterpark and a host of activities on offer. | 134,240 |
TITLE: Bases of a covector space
QUESTION [1 upvotes]: I want to prove this proposition.
Let $ V $ be a finite dimensional vector space. Given any basis $ E_1,E_2,...,E_n $ for V, let $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n \in V^* $ be the covectors defined by
$$
\varepsilon^i(E_j)= \delta_j^i
$$
Then $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ is a basis for $ V^* $ and $ dim V^* = dim V$
I have started the proof as follows.
First to prove that the covectors $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ are linearly independent.
If $ \sum_{i=0}^n r_i\varepsilon^i = 0 $, then
$ r_1\varepsilon^1 + r_2\varepsilon^2 + ... + r_n\varepsilon^n (E_1) = 0 (E_1)= 0 $
$r_1\varepsilon^1 (E_1)=0 $
$ r_1=0 $
Similarly each $r_i =0 $ which implies the covectors $ \varepsilon^1,\varepsilon^2,...,\varepsilon^n $ are linearly independent .
Now i am struck at proving that $\varepsilon^1,\varepsilon^2,...,\varepsilon^n $ span $ V^* $
The fact that $ dim V = dim V^* $ is not yet known.
REPLY [0 votes]: For the last part do
\begin{eqnarray*}
F(x)&=&(f-\sum_{i=1}^nf(E_i)\varepsilon^i)(x),\\
&=&f(x)-\sum_{i=1}^nf(E_i)\varepsilon^i(x),
\end{eqnarray*}
then
for $x=E_k$ you get
\begin{eqnarray*}
F(E_k)&=&f(E_k)-\sum_{i=1}^nf(E_i)\varepsilon^i(E_k),\\
&=&f(E_k)-\sum_{i=1}^nf(E_i)\delta^i_k,\\
&=&f(E_k)-f(E_k),\\
&=&0.
\end{eqnarray*}
So $F\equiv0$
and then
$$f=\sum_{i=1}^nf(E_i)\varepsilon^i.$$ | 130,569 |
Symphony No. 4 in A major “Italian”, op. 90
Composed by
Felix Mendelssohn
1809-1847
Orchestration
2 flutes, 2 oboes, 2 clarinets, 2 bassoons, 2 horns, 2 trumpets, timpani and strings
I. Allegro vivace
II. Andante con moto
III. Con moto moderato
IV. Presto & Finale: Saltarello
Composed 1833.
First performance: May 13, 1833, London. London Philharmonic Society. Felix Mendelssohn, conductor.
With his Third and Fourth Symphonies, as well as the concert overture The Hebrides, Mendelssohn managed to musicalize the literary travelogue. In 1829 and 1830, he undertook a “Grand Tour” of England, Scotland, Wales, and Italy. Educated young men had been traveling through Europe like this since the late eighteenth century in order to refine their character and cultural sense (what the Germans called Bildung). Along the way, Mendelssohn conducted the London premiere of his Midsummer Night’s Dream overture and visited the sites in Scotland, like Holyrood Abbey and Fingal’s Cave on the island of Staffa, which would inspire the Hebrides Overture and his Third Symphony. While still working on these pieces, he traveled to Italy, which he managed to portray in music much faster, producing the Symphony No. 4 Italian by 1833, while the “Scottish” Third Symphony would have to wait until 1842.
Mendelssohn came from an era in which some philosophers thought climate profoundly affected human character, language, and culture, which is perhaps why critics at the time heard his Fourth Symphony as “warm” and definitively “southern.” The first movement conveys the breathless excitement of travel and the impression of a pastoral landscape, with its galloping string figures and horn calls, though perhaps not any countryside in particular. In the Andante con moto, we are meant to hear a religious procession (Italian Catholicism being strange to someone from the Protestant north), and the initial melody has something of the placid, meandering character of plainchant. The steady march-like tempo and rhythm has also been compared to the “Marche des pèlerins” in Berlioz’s Harold in Italy (1834), a reminder that Mendelssohn was not the only northern composer fascinated with all things Italian or inspired by literature (in this case Byron). Yet another literary figure, Goethe, may have inspired the third movement of Mendelssohn’s symphony with Lilis Park, a love poem that is at once humorous and sensual. Mendelssohn may have felt the mood suited then-current stereotypes of Italians as openly amorous, even libidinous.
The finale of the symphony comes the closest to a musical postcard in its energetic saltarello rhythm, which Mendelssohn heard when visiting Rome and Naples. With its “little hop” in a quick triple meter, this dance that originated in Tuscany provided the perfect material for a finale. The same is true of the tarantella (already popularized throughout Europe), which Mendelssohn introduces in the development, before combining the two dances to propel the finale into its whirlwind finish. In the Italian Symphony, Mendelssohn created a fine-tuned symphonic drama—an orchestral landscape across which audiences can imagine their own characters and stories. | 259,655 |
Teen sentenced for role in murder
Hiding the knife his brother used to kill a classmate has earned Zachary Butler-Martinez a 180-day stayed sentence and 10 years of probation.
Dakota County judge Richard Spicer sentenced Butler-Martinez, who was 18 at the time of the offense, on Dec. 29. He also ordered Butler-Martinez to stay away from the family of Cody Casey and to write a letter of apology.
Butler-Martinez was at his Rosemount home last March when the 17-year-old Casey came to the door. Butler-Martinez and younger brother Dakota Butler asked Casey several times to leave, but Casey, reportedly looking for another Butler brother for a dispute over a girl, refused. Casey had brought with him a sock filled with rocks.
The confrontation ended with Butler stabbing Casey several times with a kitchen knife. Casey died a short time later.
Butler-Martinez at first told police he didn't know where the knife was but later admitted he'd dumped it in a nearby sewer because he was afraid his brother would get in trouble. | 334,185 |
TITLE: Finding the inverse of a $4×4$ matrix containing only letter values.
QUESTION [0 upvotes]: The question: Find the inverse (if it exists) of the following $4 × 4$ matrix $M$:$$
\begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a\\
\end{bmatrix}$$
In attempting to answer this question, I have receieved a tip from my professor to use the transformation of $M$ in order to find the inverse. Multiplying that, the matrix falls into another $4 × 4$ matrix with a form of the identity matrix but modified with polynomials (of course, since it's only letter values.) I'm wondering how to carry forward with this? I tried zero-ing out rows and columns in order to use cofactor expansion but I'm not finding myself getting anywhere. I appreciate any help.
REPLY [1 votes]: can you complete the following equation?
$$
MM^T= ?
$$ | 7,377 |
\begin{document}
\title[Birationality of moduli spaces of Higgs bundles]{Birationality of moduli spaces of twisted $\U(p,q)$-Higgs bundles}
\author{Peter B. Gothen}
\author{Azizeh Nozad}
\address{Centro de
Matem\'atica da Universidade do Porto \\
Faculdade de Ci\^encias da Universidade do Porto \\
Rua do Campo Alegre, s/n \\ 4169-007 Porto \\ Portugal }
\email{[email protected]}
\email{[email protected]}
\subjclass[2010]{14D20 (Primary) 14H60, 53C07 (Secondary)}
\keywords{Higgs bundles, quiver bundles, indefinite unitary group,
birationality of moduli}
\thanks{
Partially supported by CMUP (UID/MAT/00144/2013), the projects
PTDC/MAT-GEO/0675/2012 and PTDC/MAT-GEO/2823/2014 (first author) and
grant SFRH/BD/51166/2010 (second author), funded by FCT (Portugal)
with national and where applicable European structural funds through
the programme FEDER, under the partnership agreement PT2020.
The authors acknow\-ledge support from U.S. National Science
Foundation grants DMS 1107452, 1107263, 1107367 "RNMS: GEometric
structures And Representation varieties" (the GEAR Network)}
\begin{abstract}
A $\U(p,q)$-Higgs bundle on a Riemann surface (twisted by a line
bundle) consists of a pair of holomorphic vector bundles, together
with a pair of (twisted) maps between them. Their moduli spaces
depend on a real parameter $\alpha$. In this paper we study wall
crossing for the moduli spaces of $\alpha$-polystable twisted
$\U(p,q)$-Higgs bundles. Our main result is that the moduli spaces
are birational for a certain range of the parameter and we deduce
irreducibility results using known results on Higgs bundles. Quiver
bundles and the Hitchin--Kobayashi correspondence play an essential
role.
\end{abstract}
\maketitle
\section{Introduction}
Holomorphic vector bundles with extra structure on a Riemann surface
$X$ have been intensively studied over the last decades. Higgs bundles
constitute an important example, not least due to the non-abelian
Hodge Theorem
\cite{corlette:1988,donaldson:1987,Hitchin:1987,simpson:1988,simpson:1992},
which identifies the moduli space of Higgs bundles with the character
variety for representations of the fundamental group. Another
important example is that of quiver bundles. A quiver $Q$ is a
directed graph and a $Q$-bundle on $X$ is a collection of vector
bundles, indexed by the vertices of $Q$, and morphisms, indexed by the
arrows of $Q$. The natural stability condition for quiver bundles
depends on real parameters and hence so do the corresponding moduli
spaces. The stability condition stays the same in chambers but
wall-crossing phenomena arise and can be used in the study of the
moduli spaces. An early spectacular success for this approach is
Thaddeus' proof of the rank two Verlinde formula \cite{thaddeus:1994},
using Bradlow pairs \cite{bradlow:1991}, which are examples of
\emph{triples}. Triples are $Q$-bundles for a quiver with two vertices
and a single arrow connecting them. Moduli spaces of triples have been
studied extensively, using wall-crossing techniques. Without being
exhaustive, we mention
\cite{bradlow-garcia-prada-gothen:2004}, where connectedness
and irreducibility results for triples were studied, and the work of
Mu\~noz \cite{munoz:2008,munoz:2009,munoz:2010} and
Mu\~noz--Ortega--V{\'a}zquez-Gallo
\cite{munoz-ortega-vazques:2007,munoz-ortega-vazques:2009} on finer
topological invariants, such as Hodge numbers. More generally, chains (introduced by \'Alvarez-C\'onsul--Garc{\'\i}a-Prada in \cite{alvarez-garcia-prada:2001})
are $Q$-bundles for a quiver of type $A_n$. Chains have also been
studied using wall crossing techniques; we mention here the work of
Alvar\'ez-Consul--Garc\'\i{}a-Prada--Schmitt \cite{Schmitt:2006},
Garc\'\i{}a-Prada--Heinloth--Schmitt
\cite{garcia-heinloth-schmitt:2014} and Garc\'\i{}a-Prada--Heinloth
\cite{garcia-heinloth:2013}.
A natural question to ask is to what extent wall crossing techniques
can be extended to moduli of $Q$-bundles for more general quivers. Our
aim in this paper is to investigate the situation when $Q$ has
oriented cycles, as opposed to the case of chains.
Since the number of
effective stability parameters is one less than the number of vertices
of the quiver, in order to encounter wall crossing phenomena, we
are led to considering the following quiver as the simplest
non-trivial case:
\begin{equation}
\label{eq:quiver-Q}
\xymatrix{
\bullet\ar@{<-}@/_1.2pc/[r]&\bullet\ar@{<-}@/_1.2pc/[l].
}
\medskip
\end{equation}
For such quivers it becomes relevant to consider \emph{twisted}
$Q$-bundles, meaning that to each arrow one associates a fixed line bundle
twisting the corresponding morphism.
Quiver bundles for the quiver (\ref{eq:quiver-Q}) are closely related
to Higgs bundles through the notion of $G$-Higgs bundles. These are
the appropriate objects for extending the non-abelian Hodge Theorem to
representations of the fundamental group in a \emph{real} reductive Lie group
$G$ (see, e.g., \cite{garcia-gothen-mundet:2009b,gothen:2014}). The relevant case here is that of $G=\U(p,q)$. Indeed, a
{$\U(p,q)$-Higgs bundle} is a {twisted} $Q$-bundle for the quiver
(\ref{eq:quiver-Q}), twisted by the canonical bundle $K$ of $X$.
Allowing for twisting by an arbitrary line bundle $L$ on $X$, an
\emph{$L$-twisted $\U(p,q)$-Higgs bundle} is a quadruple
$E=(V,W,\beta,\gamma)$, where $V$ and $W$ are vector bundles of rank
$p$ and $q$, respectively, and the morphisms are $\beta\colon W\to
V\otimes L$ and $\gamma\colon V\to W\otimes L$. The stability notion
for $Q$-bundles for the quiver \eqref{eq:quiver-Q} depends on a real
parameter $\alpha$ and the value which is relevant for the non-abelian
Hodge Theorem is $\alpha=0$.
We denote by $\mathcal{M}_\alpha(t)$ the moduli space of
$\alpha$-semistable $L$-twisted $\U(p,q)$-Higgs bundles of type
$t=(p,q,a,b)=(\rk(V),\rk(W),\deg(V),\deg(W))$ and by
$\mathcal{M}^s_\alpha(t) \subset \mathcal{M}_\alpha(t)$ the subspace
of $\alpha$-stable $L$-twisted $\U(p,q)$-Higgs bundles. We show that
the parameter $\alpha$ is constrained to lie in an interval $\alpha_m
\leq \alpha \leq \alpha_M$ (with $\alpha_m=-\infty$ and
$\alpha_M=\infty$ if $p=q$) and the stability condition changes at a
discrete set of critical values $\alpha_c$ for $\alpha$.
Our main result is the following theorem (see
Theorem~\ref{birational} below).
\begin{thma}
Fix a type $t=(p,q,a,b)$. Let $\alpha_c$ be a critical value. If
either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $a/p-b/q>-\deg(L)$, $q\leq p$ and $0\leq\alpha_c^\pm<\frac{2pq}{pq-q^2+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(2)$] $a/p-b/q<\deg(L)$, $p\leq q$ and $\frac{2pq}{pq-p^2+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha_c^\pm\leq 0$.
\end{itemize}
Then the moduli spaces $\mathcal{M}^s_{\alpha _c^-}(t)$ and
$\mathcal{M}^s_{\alpha _c^+}(t)$ are birationally equivalent.
\end{thma}
Under suitable co-primality conditions on the topological invariants
$(p,q,a,b)$ we also have results for the full moduli spaces
$\mathcal{M}_\alpha(t)$; we refer to Theorem~\ref{birational}
below for the precise result.
A systematic study of $\U(p,q)$-Higgs bundles was carried out in
\cite{bradlow-garcia-prada-gothen:2003}, based on results for
holomorphic triples from \cite{Bradlow,bradlow-garcia-prada-gothen:2004}. In
particular, it was shown that the moduli space of $\U(p,q)$-Higgs
bundles is irreducible (again under suitable co-primality
conditions). Using these results, we deduce the following corollary to
our main theorem (see Theorem~\ref{irreducible} below).
\begin{thmb}
Let $L=K$ and fix a type $t=(p,q,a,b)$. Suppose that $(p+q,a+b)=1$ and
that $\tau=\frac{2pq}{p+q}(a/p-b/q)$ satisfies $\abs{\tau}\leq\min\{p,q\}(2g-2)$. Suppose that either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $a/p-b/q>-(2g-2) $, $q\leq p$ and $0\leq\alpha<\frac{2pq}{pq-q^2+p+q}\big(b/q-a/p-(2g-2)\big)+2g-2$,
\item[$(2)$] $a/p-b/q<2g-2$, $p\leq q$ and $\frac{2pq}{pq-p^2+p+q}(b/q-a/p+2g-2)-(2g-2)<\alpha\leq 0$.
\end{itemize}
Then the moduli space $\mathcal{M}_\alpha(t)$ is irreducible.
\end{thmb}
A related work is the recent preprint by Biquard--Garc\'\i{}a-Prada--R\'ubio
\cite{biquard-garcia-rubio:2015}, which studies $G$-Higgs bundles for
\emph{any} non-compact $G$ of hermitian type. Their focus is different from
ours, in that they adopt a general Lie theoretic approach and study
special properties such as rigidity for maximal $G$-Higgs
bundles, whereas wall crossing phenomena are not studied. On the other
hand it
is similar in spirit in allowing for arbitrary values of the stability
parameter and, indeed, our Proposition~\ref{prop-MW1} for twisted
$\U(p,q)$-Higgs bundles is a special
case of the Milnor--Wood inequality for general $G$ proved by these authors (when $L=K$). A different generalization, namely to \emph{parabolic}
$\U(p,q)$-Higgs bundles, has appeared in the work of
Garc\'\i{}a-Prada--Logares--Mu\~noz \cite{garcia-logares-munoz:2009}.
This paper is organized as follows. In Section~\ref{sec:basic-results}
we give some definitions and basic results on quiver bundles. In
Section~\ref{sec:constraints} we analyze how the $\alpha$-stability
condition constrains the parameter range for fixed type
$t=(p,q,a,b)$, prove the Milnor--Wood type inequality for
$\alpha$-semistable twisted $\U(p,q)$-Higgs bundles mentioned
above, and study vanishing of the second hypercohomology group of
the deformation complex and deduce smoothness results for the moduli
space. These results provide essential input for the
analysis in Section~\ref{sec:crossing} of the loci where the moduli
space changes when crossing a critical value. Finally, in
Section~\ref{bi}, we put our results together and prove our main theorems.
This paper is, in part, based on the second author's Ph.D. thesis \cite{nozad:2016}.
\section{Definitions and basic results}
\label{sec:basic-results}
In this section we recall definitions and relevant facts on quiver bundles, from
\cite{Gothen:2005} and \cite{Prada:2003}, that will be needed in the
sequel. We give the results for general $Q$-bundles. This generality is needed
since more general $Q$-bundles naturally appear in the study of
twisted $\U(p,q)$-Higgs bundles (see Section~\ref{sec:bound-chi}).
\subsection{Quivers} A \emph{quiver} $Q$ is a directed graph specified by a set of vertices $Q_0$, a set of arrows $Q_1$ and head and tail maps $h,t : Q_1\to Q_0$. We shall assume that $Q$ is finite.
\subsection{Twisted quiver sheaves and bundles}
Let $X$ be a compact Riemann surface, let $Q$ be a quiver and let $M=\{M_a\}_{a\in Q_1}$ be a collection of finite rank locally free sheaves of $\mathcal{O}_X$-modules.
\begin{definition}
An \emph{$M$-twisted $Q$-sheaf} on $X$ is a pair $E = (V, \varphi)$, where $V$ is a collection of coherent sheaves $V_i$ on $X$, for each $i \in Q_0$, and $\varphi$ is a collection of morphisms $\varphi_a: V_{ta}\otimes M_a\to V_{ha}$, for each $a\in Q_1$, such that $V_i =0$ for all but finitely many $i \in Q_0$, and $\varphi_a=0$ for all but finitely many $a\in Q_1$.
\par A \emph{holomorphic $M$-twisted $Q$-bundle} is an $M$-twisted $Q$-sheaf $E = (V, \varphi)$ such that the sheaf $V_i$ is a holomorphic vector bundle, for each $i \in Q_0$.
\par We shall not distinguish vector bundles and locally free finite rank sheaves.
\end{definition}
\par A morphism between twisted $Q$-sheaves $(V, \varphi)$ and $(W, \psi)$ on $X$ is given by a collection of morphisms $f_i:V_i\to W_i$, for each $i\in Q_0$, such that the diagrams
\[
\xymatrix{
V_{ta}\otimes M_a\ar[r]^{\varphi_a}\ar[d]^{f_{ta}\otimes 1}&V_{ha}\ar[d]^{f_{ha}}\\
W_{ta}\otimes M_a\ar[r]^{\psi_a}&W_{ha}
}
\]
commute for every $a\in Q_1$.
\par In this way $M$-twisted $Q$-sheaves form an abelian category. The
notions of $Q$-subbundles and quotient $Q$-bundles, as well as simple
$Q$-bundles are defined in the obvious way. The subobjects $(0,0)$ and
$E$ itself are called the \emph{trivial subobjects}. The \emph{type}
of a $Q$-bundle $E= (V, \varphi)$ is given by $$t(E)=(\rk(V_i);
\deg(V_i))_{ i\in Q_0},$$ where $\rk(V_i)$ and $\deg(V_i))$ are the
rank and degree of $V_i$, respectively. We sometimes write $\rk(E)=
\rk(\bigoplus V_i)$ and call it the \emph{rank of $E$}. Note that the type is
independent of $\varphi$.
\subsection{Stability}
Fix a tuple $\boldsymbol{\alpha}=(\alpha_i)\in\mathbb{R}^{|Q_0|}$ of
real numbers. For a non-zero $Q$-bundle ${E}=(V,\varphi)$, the
associated \emph{$\boldsymbol{\alpha}$-slope} is defined as
$$\mu_{\boldsymbol{\alpha}}({E})=\frac{\underset{i\in Q_0}{\sum}\big(\alpha_i \rk(V_i)+\deg(V_i)\big)}{\underset{i\in Q_0}{\sum}\rk(V_i)}.$$
\begin{definition}
A $Q$-bundle ${E}=(V,\varphi)$ is said to be
\emph{$\boldsymbol{\alpha}$-(semi)stable} if, for all non-trivial
subobjects ${F}$ of ${E}$,
$\mu_{\boldsymbol{\alpha}}({F})
<(\leq)\mu_{\boldsymbol{\alpha}}({E})$. An
\emph{$\boldsymbol{\alpha}$-polystable} $Q$-bundle is a finite
direct sum of $\boldsymbol{\alpha}$-stable $Q$-bundles, all of them
with the same $\boldsymbol{\alpha}$-slope.
A $Q$-bundle ${E}$ is \emph{strictly
$\boldsymbol{\alpha}$-semistable} if and only if there is a
non-trivial subobject ${F}\subset {E}$ such that
$\mu_{\boldsymbol{\alpha}}({F})=\mu_{\boldsymbol{\alpha}}({E})$.
\end{definition}
\begin{remark}\label{zero}
If we translate the parameter vector
$\boldsymbol{\alpha}=(\alpha_i)_{i\in Q_0}$ by a global constant $c
\in\mathbb{ R}$, obtaining $\boldsymbol{\alpha}'=(\alpha_i')_{i\in
Q_0}$, with $\alpha_i'=\alpha_i+c$, then
$\mu_{\boldsymbol{\alpha}'}({E}) =
\mu_{\boldsymbol{\alpha}}({E})-c$. Hence the stability
condition does not change under global translations. So we may
assume that $\alpha_{0}=0$.
\end{remark}
The following is a well-known fact (see, e.g., \cite[Exercise
$2.5.6.6$]{Schmitt:2008}). Consider a strictly
$\boldsymbol{\alpha}$-semistable $Q$-bundle ${E}=(V,\varphi)$. As it
is not $\boldsymbol{\alpha}$-stable, ${E}$ admits a subobject
${F}\subset {E}$ of the same $\boldsymbol{\alpha}$-slope. If ${F}$ is
a non-zero subobject of ${E}$ of minimal rank and the same
$\boldsymbol{\alpha}$-slope, it follows that ${F}$ is
$\boldsymbol{\alpha}$-stable. Then, by induction, one obtains a flag of
subobjects
\begin{displaymath}
{F}_0=0\subset {F}_1\subset\cdots\subset {F}_m={E}
\end{displaymath}
where
$\mu_{\boldsymbol{\alpha}}({F}_i/{F}_{i-1})=\mu_{\boldsymbol{\alpha}}({E})$
for $1\leq i\leq m$, and where the subquotients
$F_i/F_{i-1}$ are $\boldsymbol{\alpha}$-stable $Q$-bundles. This is
the \emph{Jordan-H\"{o}lder filtration} of $\mathcal{E}$, and it is
not unique. However, the \emph{associated graded
object}
$$\mathrm{Gr}({E}):=\oplus_{i=1}^m{F}_i/{F}_{i-1}$$
is unique up to isomorphism.
\begin{definition}
Two semi-stable $Q$-bundles ${E}$ and
${E}'$ are said to be \emph{$S$-equivalent} if
$\mathrm{Gr}(E)\cong \mathrm{Gr}(E').$
\end{definition}
\begin{remark}
It is a standard fact that each $S$-equivalence class contains a
unique polystable representative. Moreover, if a $Q$-bundle ${E}$
is stable, then the induced Jordan-H\"{o}lder filtration is trivial,
and so the $S$-equivalence class of $E$ coincides with its
isomorphism class.
\end{remark}
\subsection{The gauge theory equations}
Throughout this paper, given a smooth bundle $M$ on $X$, $\Omega^k(M)$
(resp.\ $\Omega^{i,j}(M)$) is the space of smooth $M$-valued $k$-forms
(resp.\ $(i,j)$-forms) on $X$, $\omega$ is a fixed K\"{a}hler form on
$X$, and $ \Lambda: \Omega^{i,j}(M)\to \Omega^{i-1,j-1}(M)$ is
contraction with $\omega$. The gauge equations will also depend on a
fixed collection $q$ of Hermitian metrics $q_a$ on $M_a$, for each
$a\in Q_1$, which we fix once and for all. Let
$\mathcal{E}=(V,\varphi)$ be a $M$-twisted $Q$-bundle on $X$. A
Hermitian metric on $\mathcal{E}$ is a collection $H$ of Hermitian
metrics $H_i$ on $V_i$, for each $i\in Q_0$ with $V_i\neq 0$. To
define the gauge equations on $\mathcal{E}$, we note that
$\varphi_a:V_{ta}\otimes M_a\to V_{ha}$ has a smooth adjoint morphism
$\varphi_a^\ast:V_{ha}\to V_{ta}\otimes M_a$ with respect to the
Hermitian metrics $H_{ta}\otimes q_a$ on $V_{ta}\otimes M_a$ and
$H_{ha}$ on $V_{ha}$, for each $a\in Q_1$, so it makes sense to
consider the compositions $\varphi_a\circ\varphi_a^\ast$ and
$\varphi_a^\ast\circ\varphi_a$. The following definitions are found in
\cite{Prada:2003}. Let $\boldsymbol{\alpha}$ be the stability
parameter.\par Define $\boldsymbol{\tau}$ to be collections of real
numbers $\tau_i$, for which
\begin{equation}
\tau_i=\mu_{\boldsymbol{\alpha}}(\mathcal{E})-\alpha_i, \mbox{ }i\in Q_0.\label{relation}
\end{equation}
Then, by using Remark $\ref{zero}$, $\boldsymbol{\alpha}$ can be recovered from $\boldsymbol{\tau}$ as follows
\begin{align*}
\alpha_i=\tau_{0}-\tau_i, \mbox{ } i\in Q_0.
\end{align*}
\begin{definition}
A Hermitian metric $H$ satisfies the \emph{quiver $\boldsymbol{\tau}$-vortex equations} if
\begin{equation}
\sqrt{-1}\Lambda F(V_i)+\underset{i=ha}{\sum}\varphi_a\varphi_a^{\ast}-\underset{i=ta}{\sum}\varphi_a^{\ast}\varphi_a=\tau_i \mathrm{Id}_{V_i}\label{vortex equations1}
\end{equation}
for each $i\in Q_0$ such that $V_i\neq 0$, where $F(V_i)$ is the
curvature of the Chern connection associated to the metric $H_i$ on
the holomorphic vector bundle $V_i$.
\end{definition}
The following is the \emph{Hitchin-Kobayashi correspondence} between the twisted quiver vortex equations and the stability condition for holomorphic twisted quiver bundles, given in \cite[Theorem 3.1]{Prada:2003}:
\begin{theorem}\label{Hitchin-Kobayashi}
A holomorphic $Q$-bundle ${E}$ is $\boldsymbol{\alpha}$-polystable if and only if it admits a Hermitian metric $H$ satisfying the quiver $\boldsymbol{\tau}$-vortex equations $(\ref{vortex equations1})$, where $\boldsymbol{\alpha}$ and $\boldsymbol{\tau}$ are related by $(\ref{relation})$.
\end{theorem}
\subsection{Twisted $\U(p,q)$-Higgs bundles}
An important example of twisted $Q$-bundles, which is our main object study in this paper, is that of twisted $\U(p,q)$-Higgs bundles on $X$ given in the following. It is to be noted that twisted $\U(p,q)$-Higgs bundles in our study is twisted with the same line bundle for each arrow.
\begin{definition}
Let $L$ be a line bundle on $X$. A \emph{$L$-twisted $\U(p,q)$-Higgs bundle} is a twisted $Q$-bundle for the quiver
\[ \xymatrix{
V\ar@{<-}@/_1.2pc/[r]&W\ar@{<-}@/_1.2pc/[l].
}
\]
\\
where each arrow is twisted by $L$, and such that $\rk(V)=p$ and
$\rk(W)=q$. Thus a $L$-twisted $\U(p,q)$-Higgs bundle is a quadruple $E=(V, W,\beta, \gamma)$, where $V$ and $W$ are holomorphic vector bundles on $X$ of ranks $p$ and $q$ respectively, and$$\beta:W\longrightarrow V\otimes L,$$ $$ \gamma :V\longrightarrow W\otimes L,$$ are holomorphic maps. The \emph{type} of a twisted $\mathrm{U}(p,q)$-Higgs bundle $E=(V,W,\beta,\gamma)$ is defined by a tuple of integers $t(E):=(p,q,a,b)$ determined by ranks and degrees of $V$ and $W$, respectively.
\end{definition}
Note that $K$-twisted $\U(p,q)$-Higgs bundles can be seen as a
special case of $G$-Higgs bundles (\cite{hitchin:1992}, see also \cite{bradlow-garcia-prada-gothen:2003,bradlow-garcia-prada-gothen:hss-higgs,garcia-gothen-mundet:2009b,gothen:2014}), where $G$ is a real form of a
complex reductive Lie group and $K$ is the canonical bundle of the
Riemann surface $X$.
\subsection{Gauge equations}
For this $L$-twisted quiver bundle one can consider the general quiver
equations as defined in (\ref{vortex equations1}).
\par Let $\boldsymbol{\tau}=(\tau_1,\tau_2)$ be a pair of real numbers. A Hermitian metric $H$ satisfies the \emph{$L$-twisted quiver $\boldsymbol{\tau}$-vortex equations} on twisted $\U(p,q)$-Higgs bundle $E$ if
\begin{equation}\label{vortex equations}
\begin{aligned}
\sqrt{-1}\Lambda F_{H_V}+\beta\beta^\ast-\gamma^\ast\gamma&=\tau_1\Id_{V},\\
\sqrt{-1}\Lambda F_{H_W}+\gamma\gamma^\ast-\beta^\ast\beta&=\tau_2\Id_{W}.
\end{aligned}
\end{equation}
where $F_{H_V}$ and $F_{H_W}$ are the curvatures of the Chern connections
associated to the metrics $H_V$ and $H_W$, respectively.
\begin{remark}
\begin{itemize}
\item[(i)] If a holomorphic twisted $\U(p,q)$-bundle $E$ admits a Hermitian metric satisfying the $\boldsymbol{\tau}$-vortex equations, then taking traces in $(\ref{vortex equations})$, summing for $V$ and $W$, and integrating over $X$, we see that the parameters $\tau_1$ and $\tau_2$ are constrained by $p\tau_1+q\tau_2=\deg(V)+\deg(W)$.
\item[(ii)] If $L=K$ the equations are conformally invariant and so depend only on the Riemann surface structure on $X$. In this case they are \emph{the Hitchin equations} for the $\U(p,q)$-Higgs bundle.
\end{itemize}
\end{remark}
\subsection{Stability}
Let $E=(V,W,\beta,\gamma)$ be a twisted $\U(p,q)$-Higgs bundle, and $\alpha$ be a real number; $\alpha$ is called the $\emph{stability parameter}$. The definitions of the previous section specialize as follows.
The \emph{$\alpha$-slope} of $E$ is defined to be
\begin{align*}
\mu_{\alpha}(E)&=\mu(E)+\alpha\frac{p}{p+q},
\end{align*}
where $\mu(E):=\mu(V\oplus W)$. A twisted $\U(p,q)$-bundle $E$ is \emph{$\alpha$-semistable} if, for every proper (non-trivial) subobject $F\subset E$,
$$\mu_{\alpha}(F)\leq\mu_{\alpha}(E).$$
Further, $E$ is \emph{$\alpha$-stable} if this inequality is always strict. A twisted $\U(p,q)$-bundle is called \emph{$\alpha$-polystable} if it is the direct sum of $\alpha$-stable twisted $\U(p,q)$-Higgs bundles of the same $\alpha$-slope.
\begin{remark}
The stability can be defined using quotients as for vector bundles. Note that for any subobject $E'=(V',W')$ we obtain an induced quotient bundle $E/E'=(V/V',W/W',\overline{\beta},\overline{\gamma})$ and $E$ is $\alpha$-(semi)stable if $\mu_\alpha(E/E')(\geq)>\mu_\alpha(E)$.
\end{remark}
The following is a special case of the Hitchin-Kobayashi correspondence between the twisted quiver vortex equations and the stability condition for holomorphic twisted quiver bundles, stated in Proposition $\ref{Hitchin-Kobayashi}$.
\begin{theorem}\label{solution}
A solution to $(\ref{vortex equations})$ exists if and only if $E$ is $\alpha$-polystable for $\alpha=\tau_2-\tau_1$.
\end{theorem}
\subsubsection{Critical values}
\par A twisted $\U(p,q)$-Higgs bundle $E$ is strictly $\alpha$-semistable if and only if there is a proper subobject $F=(V',W')$ such that $\mu_\alpha(F)=\mu_\alpha(E)$, i.e.,
$$\mu(V'\oplus W')+\alpha\frac{p'}{p'+q'}=\mu(V\oplus W)+\alpha\frac{p}{p+q}.$$
The case in which the terms containing $\alpha$ drop from the above equality and $E$ is strictly $\alpha$-semistable for all values of $\alpha$, i.e.,
$$\frac{p}{p+q}=\frac{p'}{p'+q'},\mbox{ and}$$
$$\mu(V\oplus W)=\mu(V'\oplus W')$$
is called \emph{$\alpha$-independent strict semistability}.
\begin{definition}
For a fixed type $(p,q,a,b)$ a value of $\alpha$ is called a
\emph{critical value} if there exist integers $p', q', a'$ and $b'$
such that $\frac{p'}{p'+q'}\neq\frac{p}{p+q}$ and
$\frac{a'+b'}{p'+q'}+\alpha\frac{p'}{p'+q'}=\frac{a+b}{p+q}+\alpha
\frac{p}{p+q}$, with $0\leq p'\leq p$, $0\leq q'\leq q$
and $(p',q')\neq (0,0)$. We say that
$\alpha$ is \emph{generic} if it is not critical.
\end{definition}
\begin{lemma}
\label{lem:no-alpha-independent-semistability}
In the following situations $\alpha$-independent semistability cannot occur:
\begin{itemize}
\item[(i)]\cite[Lemma~2.7]{bradlow-garcia-prada-gothen:2004} There is an integer $m$ such that $\GCD(p+q, d_1+d_2-mp)=1$.
\item[(ii)] If $\GCD(p,q)=1$, for $p\neq q$.
\end{itemize}
\end{lemma}
\begin{proof}
To prove $(ii)$, on the contrary assume that $E=(V,W,\beta,\gamma)$ is a $\alpha$-semistable twisted $\U(p,q)$-Higgs bundle with a proper subobject $E'=(V',W',\beta',\gamma')$ such that
$$
\mu(V'\oplus W')+\alpha\frac{p'}{p'+q'}=
\mu(V\oplus W)+\alpha\frac{p}{p+q}
$$
and
\begin{equation*}
\dfrac{p'}{p'+q'}=\dfrac{p}{p+q},
\end{equation*}
where $p'$ and $q'$ are the ranks of $V'$ and $W'$ respectively. Since $E'$ is proper, either $p'<p$ or $q'<q$ and then the equality $\dfrac{p'}{p'+q'}=\dfrac{p}{p+q}$ contradicts that $p$ and $q$ are co-prime.
\end{proof}
Fix a type $t=(p,q,a,b)$. We denote the moduli space of
$\alpha$-polystable twisted $\U(p,q)$-Higgs bundles
$E=(V,W,\beta,\gamma)$ which have the type $t(E)=(p,q,a,b)$, where
$a=\deg(V)$ and $b=\deg(W)$,
by $$\mathcal{M}_\alpha(t)=\mathcal{M}_\alpha(p,q,a,b),$$ and the
moduli space of $\alpha$-stable twisted $\U(p,q)$-Higgs bundles by
$\mathcal{M}_\alpha^s(t)$. A Geometric Invariant Theory construction
of the moduli space
follows from the general constructions of Schmitt \cite[Theorem
$2.5.6.13$]{Schmitt:2008}, thinking of twisted $\U(p,q)$-Higgs
bundles in terms of quivers.
\subsection{Deformation theory of twisted $\U(p,q)$-Higgs bundles}
\begin{definition}\label{deformation complex of U(p,q)}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs bundle. We introduce the following notation:
\begin{align*}
\mathpzc{Hom}^0&\mathrm{=}\Hom(V',V)\oplus \Hom(W',W), \\
\mathpzc{Hom}^1&\mathrm{=}\Hom(V',W\otimes L)\oplus \Hom(W',V\otimes L).
\end{align*}
With this notation we consider the complex of sheaves
\begin{equation}
\mathpzc{Hom}^\bullet(E',E):\mbox{ }\mathpzc{Hom}^0\overset{a_0}{\longrightarrow} \mathpzc{Hom}^1\label{deformation complex}
\end{equation} defined by
\begin{align*}
a_0(f_1,f_2)&=\big(\phi_a(f_1,f_2),\phi_b(f_1,f_2)\big),\mbox{ for } (f_1,f_2)\in \mathpzc{Hom}^0
\end{align*}
where $$\phi_a: \mathpzc{Hom}^0\to\Hom(V',W\otimes L)\hookrightarrow \mathpzc{Hom}^1\mbox{ and }\phi_b: \mathpzc{Hom}^0\to\Hom(W',V\otimes L)\hookrightarrow \mathpzc{Hom}^1$$
are given by
\begin{align*}
\phi_a(f_1,f_2)&=(f_2\otimes Id_L)\circ \gamma'-\gamma\circ f_1),\\
\phi_b(f_1,f_2)&=(f_1\otimes Id_L)\circ\beta'-\beta\circ f_2).
\end{align*}
The complex $\mathpzc{Hom}^\bullet(E',E)$ is called the
\emph{$\Hom$-complex}. This is a special case of the $\Hom$-complex
for $Q$-bundles defined in \cite{Gothen:2005}, and also for $G$-Higgs
bundles (see, e.g., \cite{Ramanan}). We shall write
$\mathpzc{End}^\bullet(E)$ for $\mathpzc{Hom}^\bullet(E,E)$.
\end{definition}
The following proposition follows from \cite[Theorem 4.1 and Theorem 5.1]{Gothen:2005}.
\begin{proposition}\label{Hom and hyper}
Let $E$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Then
there are natural isomorphisms
\begin{align*}
\Hom(E',E)&\cong \mathbb{H}^0(\mathpzc{Hom}^\bullet (E',E))\\
\Ext^1(E',E)&\cong \mathbb{H}^1(\mathpzc{Hom}^\bullet (E',E))
\end{align*}
and a long exact sequence associated to the complex $\mathpzc{Hom}^{\bullet}(E',E)$:
\begin{multline}\label{hyperH}
0\longrightarrow\mathbb{H}^0(\mathpzc{Hom}^\bullet(E',E))\longrightarrow H^0(\mathpzc{Hom}^0)\longrightarrow H^0(\mathpzc{Hom}^1)\longrightarrow\mathbb{H}^1(\mathpzc{Hom}^\bullet(E',E))\\
\longrightarrow H^1(\mathpzc{Hom}^0)\longrightarrow H^1(\mathpzc{Hom}^1)\longrightarrow\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E))\longrightarrow 0.
\end{multline}
\end{proposition}
When $E=E'$, we have $\End(E)=\Hom(E,E)\cong \mathbb{H}^0(\mathpzc{Hom}^\bullet (E,E))$.
\begin{definition}
We denote by $\chi(E',E)$ the hypercohomology Euler characteristic for the complex $\mathpzc{Hom}^{\bullet}(E',E)$, i.e.
$$\chi(E',E)=\dim\mathbb{H}^0(\mathpzc{Hom}^\bullet(E',E))-\dim\mathbb{H}^1(\mathpzc{Hom}^\bullet(E',E))+\dim\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E)).$$
\end{definition}
As an immediate consequence of the long exact sequence $(\ref{hyperH})$ and the Riemann-Roch formula we can obtain the following.
\begin{proposition}\label{chi}
For any twisted $\U(p,q)$-Higgs bundle $E$ and twisted $\U(p',q')$-Higgs
bundle $E'$ we have
\begin{equation*}
\begin{aligned}[t]
\chi(E',E)&=\chi(\mathpzc{Hom}^0)-\chi(\mathpzc{Hom}^1)\\
&=(1-g)(\rk(\mathpzc{Hom}^0)-\rk(\mathpzc{Hom}^1))
+\deg(\mathpzc{Hom}^0)-\deg(\mathpzc{Hom}^1)\\
&=(1-g)\big(p'p+q'q-p'q-q'p\big)+(q'-p')(\deg(W)-\deg(V))+\\
&\hspace{1.5cm}(q-p)(\deg(V')-\deg(W'))-(pq'+p'q)\deg(L)
\end{aligned}
\end{equation*}
\end{proposition}
Recall that the type of a $\U(p,q)$-Higgs bundle
$E=(V,W,\beta,\gamma)$ is $t(E)=(p,q,a,b)$, where $a=\deg(V)$,
$b=\deg(W)$. The previous proposition shows that $\chi(E',E)$ only
depends on the types $t'=t(E')$ and $t=t(E)$ of $E'$ and $E$,
respectively, so we may use the notation
$$\chi(t',t):=\chi(E',E).$$
\begin{remark}
\label{rem:direct-sum-deform}
Suppose that $E=E'\oplus E''$. Then it is clear that the
$\Hom$-complexes satisfy:
\begin{equation*}
\mathpzc{Hom}^\bullet(E,E)=\mathpzc{Hom}^\bullet(E',E')\oplus \mathpzc{Hom}^\bullet(E'',E'')\oplus \mathpzc{Hom}^\bullet(E'', E')\oplus \mathpzc{Hom}^\bullet(E', E''),
\end{equation*}
and so the hypercohomology groups have an analogous direct sum
decomposition.
\end{remark}
\begin{lemma}\label{extension}
For any extension $0 \rightarrow E' \rightarrow E \rightarrow E'' \rightarrow 0$ of twisted $\U(p,q)$-Higgs bundles,
$$\chi(E,E) = \chi(E', E') + \chi( E'', E'') + \chi( E'', E') + \chi(E', E'').$$
\end{lemma}
\begin{proof}
Since the Euler characteristic is topological, we may assume that
$E=E'\oplus E''$. Now the result is immediate in view of
Remark~\ref{rem:direct-sum-deform}.
\end{proof}
Given the identification of $\mathbb{H}^0(\mathpzc{Hom}^\bullet(E',E))$ with $\Hom(E',E)$, by Proposition $\ref{Hom and hyper}$, the following is the direct analogue of the corresponding result for semistable vector bundles.
\begin{proposition}\label{vanishing}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. If $E$ and $E'$ are both $\alpha$-semistable, then the
following holds:
\begin{itemize}
\item[(1)] If $\mu_\alpha(E)<\mu_\alpha(E')$, then $\mathbb{H}^0(\mathpzc{Hom}^\bullet(E',E))=0$.
\item[(2)] If $\mu_\alpha(E')=\mu_\alpha(E)$, and $E'$ is $\alpha$-stable, then
$$\mathbb{H}^0(\mathpzc{Hom}^\bullet(E',E))\cong\begin{cases}
0 &\mbox{ if } E\ncong E'\\
\mathbb{C}&\mbox{ if } E\cong E'.\\
\end{cases}$$
\end{itemize}
\end{proposition}
\begin{definition}
A twisted $\U(p,q)$-Higgs bundle $E=(V,W,\varphi=\beta+\gamma)$ is \emph{infinitesimally simple} if $\End(E) \cong \C$ and it is \emph{simple} if $\Aut(E) \cong \C^\ast$, where $\Aut(E)$ denotes the automorphism group of $E$.
\end{definition}
Since $L$-twisted $\U(p,q)$-Higgs bundles form an abelian category,
any automorphism is also an endomorphism. Hence, if
$(V, W, \beta, \gamma)$ is infinitesimally simple then it is
simple. Thus Proposition~\ref{vanishing} implies the following lemma.
\begin{lemma}\label{stable implies simple}
Let $(V, W, \beta, \gamma)$ be a twisted $\U(p,q)$-Higgs bundle. If $(V,W,\beta,\gamma)$ is $\alpha$-stable
then it is simple.
\end{lemma}
\begin{proposition}\label{smooth0}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-stable twisted $\U(p,q)$-Higgs bundle of type $t=(p,q,a,b)$.
\begin{itemize}
\item[(1)] The space of infinitesimal deformations of $E$ is isomorphic to the first hypercohomology group $\mathbb{H}^1(\mathpzc{End}^\bullet(E))$.\\
\item[(2)] If $\mathbb{H}^2(\mathpzc{End}^\bullet(E))=0$, then the moduli space $\mathcal{M}^s_\alpha(t)$ is smooth in a
neighborhood of the point defined by $E$ and
\begin{align*}
\dim\mathcal{M}^s_\alpha(t)&=\dim\mathbb{H}^1(\mathpzc{End}^\bullet(E))\\
&=1-\chi(E,E)=(g-1)(q-p)^2+2pq\deg(L)+1.
\end{align*}
\end{itemize}
\end{proposition}
\begin{proof}
Statement $(1)$ follows from \cite[Theorem $2.3$]{Ramanan}. Statement
$(2)$ is analogous to Proposition $2.14$ of \cite{Gothen:2013}.
\end{proof}
\section{Consequences of stability and properties of the moduli space}
\label{sec:constraints}
\subsection{Bounds on the topological invariants and Milnor--Wood inequality}
In this section we explore the constraints imposed by stability on the
topological invariants of $\U(p,q)$-Higgs bundles and on the stability
parameter $\alpha$.
\begin{proposition}\label{bound on beta and gamma}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted
$\U(p,q)$-Higgs bundle. Then the following inequalities hold.
\begin{align}
\label{bound on gamma}
\frac{2pq}{p+q}\big(\mu(V)-\mu(W)\big) &\leq
\rk(\gamma)\deg(L)+\alpha\big(\rk(\gamma)-\frac{2pq}{p+q}\big), \\
\label{bound on beta}
\frac{2pq}{p+q}\big(\mu(W)-\mu(V)\big)&\leq
\rk(\beta)\deg(L)+\alpha\big(\frac{2pq}{p+q}-\rk(\beta)\big).
\end{align}
Moreover, if $\deg(L)+\alpha>0$ and equality holds in \eqref{bound on gamma}
then either $E$ is strictly semistable or $p = q$ and
$\gamma$ is an isomorphism $\gamma\colon V \xra{\cong}W\otimes L$.
Similarly, if $\deg(L)-\alpha>0$ and equality holds in \eqref{bound on
beta} then either $E$ is strictly semistable or $p = q$ and
$\beta$ is an isomorphism $\beta\colon W\xra{\cong}V\otimes L$.
\end{proposition}
\begin{proof}
An argument similar to that given in \cite[Lemma
3.24]{bradlow-garcia-prada-gothen:2003} shows that
\begin{equation*}
2p\big(\mu(V)-\mu_\alpha(E)\big)\leq
\rk(\gamma)\deg(L)+\alpha(\rk(\gamma)-2p);
\end{equation*}
Similarly,
\begin{equation*}
2q\big(\mu(W)-\mu_\alpha(E)\big)
\leq \rk(\beta)\deg(L)-\rk(\beta)\alpha,
\end{equation*}
Using this, the result follows immediately using the following
identities:
\begin{align*}
\mu(V)-\mu_\alpha(E)
&=\frac{q}{p+q}\big(\mu(V)-\mu(W)\big)-\alpha\frac{p}{p+q},\\
\mu(W)-\mu_\alpha(E)
&=\frac{p}{p+q}\big(\mu(W)-\mu(V)\big)-\alpha\frac{p}{p+q}.
\end{align*}
The statement about equality for $\deg(L)-\alpha>0$ also follows as in
loc.\ cit.
\end{proof}
By analogy with the case of $\U(p,q)$-Higgs bundles (cf.\ \cite{bradlow-garcia-prada-gothen:2003}) we make the
following definition.
\begin{definition}
The \emph{Toledo invariant} of a twisted $\U(p,q)$-Higgs bundle
$E=(V,W,\beta,\gamma)$ is
\begin{displaymath}
\tau(E) = 2\frac{q\deg(V)-p\deg(W)}{p+q} =
\frac{2pq}{p+q}\big(\mu(V)-\mu(W)\big).
\end{displaymath}
\end{definition}
The following is the analogue of the Milnor--Wood inequality for
$\U(p,q)$-Higgs bundles
(\cite[Corollary~3.27]{bradlow-garcia-prada-gothen:2003}). When $L=K$,
it is a special case of a general result of
Biquard--Garc\'\i{}a-Prada--Rubio
\cite[Theorem~4.5]{biquard-garcia-rubio:2015}, which is valid for
$G$-Higgs bundles for any semisimple $G$ of Hermitian type.
\begin{proposition}
\label{prop-MW1}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted
$\U(p,q)$-Higgs bundle. Then the following inequality holds:
\begin{displaymath}
-\rk(\beta)\deg(L)+\alpha\big(\rk(\beta)-\frac{2pq}{p+q}\big)
\leq \tau(E)\leq
\rk(\gamma)\deg(L)+\alpha\big(\rk(\gamma)-\frac{2pq}{p+q}\big).
\end{displaymath}
\end{proposition}
\begin{proof}
In view of the definition of $\tau(E)$, we can
write \eqref{bound on gamma} and \eqref{bound on beta} as
\begin{align}\label{bound on gamma2}
\tau(E)&\leq
\rk(\gamma)\deg(L)+\alpha\big(\rk(\gamma)-\frac{2pq}{p+q}\big),\\
-\tau(E)&\leq
\rk(\beta)\deg(L)+\alpha\big(\frac{2pq}{p+q}-\rk(\beta)\big)
\label{bound on beta2}
\end{align}
from which the result is immediate.
\end{proof}
When equality holds in the Milnor--Wood inequality, more information
on the maps $\beta$ and $\gamma$ can be obtained from
Proposition~\ref{bound on beta and gamma}. In this respect we have the
following result.
\begin{proposition}
\label{prop:MW2}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted
$\U(p,q)$-Higgs bundle.
\begin{enumerate}
\item Assume that $\alpha > -\deg(L)$. Then
\begin{displaymath}
\tau(E) \leq \min\{p,q\}\bigl(\deg(L)-\alpha\frac{\abs{p-q}}{p+q}\bigr).
\end{displaymath}
and if equality holds then $p\leq q$ and $\gamma$ is an
isomorphism onto its image.
\item Assume that $\alpha\leq -\deg(L)$. Then
\begin{displaymath}
\tau(E) \leq -\alpha\frac{2pq}{p+q}
\end{displaymath}
and if equality holds and $\alpha< -\deg(L)$ then $\gamma=0$.
\item Assume that $\alpha<\deg(L)$. Then
\begin{displaymath}
\tau(E) \geq \min\{p,q\}\bigl(-\alpha\frac{\abs{p-q}}{p+q}-\deg(L)\bigr)
\end{displaymath}
and if equality holds then $q\leq p$ and $\beta$ is an
isomorphism onto its image.
\item Assume that $\alpha\geq\deg(L)$. Then
\begin{displaymath}
\tau(E) \geq -\alpha\frac{2pq}{p+q}
\end{displaymath}
and if equality holds and $\alpha > \deg(L)$ then $\beta=0$.
\end{enumerate}
\end{proposition}
\begin{proof}
We rewrite \eqref{bound on gamma2} as
\begin{math}
\tau(E) \leq \rk(\gamma)(\deg(L)+\alpha) - \alpha\frac{2pq}{p+q}.
\end{math}
Then (1) and (2) are immediate from
Proposition~\ref{bound on beta and gamma}.
Similarly, (3) and (4) follow rewriting \eqref{bound on beta2} as
\begin{math}
\tau(E) \geq \rk(\beta)(\alpha-\deg(L)) - \alpha\frac{2pq}{p+q}.
\end{math}
\end{proof}
In the case when $\abs{\alpha}<\deg(L)$ we can write the inequality of
the preceding proposition in a more suggestive manner as follows.
\begin{corollary}
\label{cor:MW}
Assume that $\abs{\alpha}<\deg(L)$ and let $E$ be an $\alpha$-semistable
twisted $\U(p,q)$-Higgs bundle. Then
\begin{displaymath}
\abs{\tau(E)} \leq
\min\{p,q\}\bigl(\deg(L)-\alpha\frac{\abs{p-q}}{p+q}\bigr).
\end{displaymath}
\end{corollary}
\begin{remark}
In the cases of Proposition~\ref{prop:MW2} when one of the Higgs
fields $\beta$ and $\gamma$ is an isomorphism onto its image, it is
natural to explore rigidity phenomena for twisted $\U(p,q)$-Hitchin
pairs, along the lines of \cite{bradlow-garcia-prada-gothen:2003}
(for $\U(p,q)$-Higgs bundles) and Biquard--Garc\'\i{}a-Prada--Rubio
\cite{biquard-garcia-rubio:2015} (for parameter dependent $G$-Higgs
bundles when $G$ is Hermitian of tube type). This line of inquiry
will be pursued elsewhere.
\end{remark}
\subsection{Range for the stability parameter}
In the following we determine a range for the stability parameter whenever $p\neq q$. We denote the minimum and the maximum value for $\alpha$ by $\alpha_m$ and $\alpha_M$, respectively.
\begin{proposition}\label{bound for alpha}
Assume that $p\neq q$ and let $E$ be a $\alpha$-semistable twisted
$\U(p,q)$-Higgs bundle. Then $\alpha_m\leq \alpha\leq \alpha_M$,
where
\begin{align*}
\alpha_m&=
\begin{cases}
-\dfrac{2\mathrm{max}\{p,q\}}{|q-p|}\bigl(\mu(V)-\mu(W)\bigr) - \dfrac{p+q}{|q-p|}\deg(L)
&\text{if}\ \mu(V)-\mu(W)>-\deg(L),\\
-\bigl(\mu(V)-\mu(W)\bigr) &\text{if}\ \mu(V)-\mu(W)\leq -\deg(L),
\end{cases}\\
\intertext{and}
\alpha_M&=\begin{cases}
-\dfrac{2\mathrm{max}\{p,q\}}{|q-p|}\bigl(\mu(V)-\mu(W)\bigr) +\dfrac{p+q}{|q-p|}\deg(L)&\text{if}\ \mu(V)-\mu(W)<\deg(L),\\
-\bigl(\mu(V)-\mu(W)\bigr) &\text{if}\ \mu(V)-\mu(W)\geq \deg(L).
\end{cases}
\end{align*}
\end{proposition}
\begin{proof}
First we determine $\alpha_M$. Using \eqref{bound on gamma2} we get
$$\alpha(\dfrac{2pq}{p+q}-\rk(\gamma))\leq\rk(\gamma)\deg(L)-\tau(E)$$
since $p\neq q$ therefore $\dfrac{2pq}{p+q}-\rk(\gamma)> 0$. Hence the above inequality yields
$$\alpha\leq \dfrac{p+q}{2pq-(p+q)\rk(\gamma)}(\rk(\gamma)\deg(L)-\tau(E)).$$
In order to find an upper bound for $\alpha$ we maximize the right
hand side of this inequality as a function of $\rk(\gamma)$. Thus we
study monotonicity of the function $f(r)=\dfrac{rd-\tau}{c-r}$, where
$c=\dfrac{2pq}{p+q}$, $d=\deg(L)$ and $r\in
[0,\mathrm{min}\{p,q\}]$. We obtain the following:
\begin{itemize}
\item[(a)] If $\deg(L)=\mu(V)-\mu(W)$ then $f$ is constant and $$\alpha\leq\mu(W)-\mu(V).$$
\item[(b)] If $\deg(L)>\mu(V)-\mu(W)$ then $f$ is increasing so
\begin{align*}
\alpha&\leq\dfrac{p+q}{|q-p|}
\big(\deg(L)-\dfrac{\tau(E)}{\mathrm{min}\{p,q\}}\big)=\dfrac{p+q}{|q-p|}\deg(L)-\dfrac{2\mathrm{max}\{p,q\}}{|q-p|}(\mu(V)-\mu(W))
\end{align*}
and, if equality holds then $\rk(\gamma)=\mathrm{min}\{p,q\}$.
\item[(c)] If $\deg(L)<\mu(V)-\mu(W)$ then $f$ is decreasing so
$$\alpha\leq \mu(W)-\mu(V)$$
and, if equality holds then $\gamma=0$.
\end{itemize}
Now we determine the lower bound $\alpha_m$. The inequality
\eqref{bound on beta2} yields
$$\alpha\geq \dfrac{\rk(\beta)\deg(L)+\tau(E)}{\rk(\beta)-\dfrac{2pq}{p+q}}.$$
Similarly to the above, by studying the monotonicity of
$g(r)=\dfrac{rd+\tau}{r-c}$, we obtain the following:
\begin{itemize}
\item[(a)$'$] If $\mu(V)-\mu(W)=-\deg(L)$ then $g$ is constant and $$\alpha\geq \mu(W)-\mu(V).$$
\item[(b)$'$] If $\mu(V)-\mu(W)<-\deg(L)$ then $g$ is increasing, so $$\alpha\geq \mu(W)-\mu(V),$$
and, if equality holds then $\beta=0$.
\item[(c)$'$] If $\mu(V)-\mu(W)> -\deg(L)$ then $g$ is decreasing, so $$\alpha\geq-\dfrac{p+q}{|q-p|}(\deg(L)+\dfrac{\tau(E)}{\mathrm{min}\{p,q\}})=-\dfrac{p+q}{|q-p|}\deg(L)-\dfrac{2\mathrm{max}\{p,q\}}{|q-p|}(\mu(V)-\mu(W)),$$
and, if equality holds then $\rk(\beta)=\mathrm{min}\{p,q\}$.
\end{itemize}
Note that if $\mu(V)-\mu(W)\geq 0$ then $\mu(V)-\mu(W)\geq -\deg(L)$, and if $\mu(V)-\mu(W)\leq 0$ then $\mu(V)-\mu(W)<\deg(L)$. Hence the result follows.
\end{proof}
\begin{remark}
\label{rem:extreme-alpha}
The preceding proof gives the following additional information when
$\alpha$ equals one of the extreme values $\alpha_m$ and $\alpha_M$:
\begin{itemize}
\item if $\mu(V)-\mu(W)<\deg(L)$ and $\alpha=\alpha_M$ then
$\rk(\gamma)=\min\{p,q\}$;
\item if $\mu(V)-\mu(W)>\deg(L)$ and $\alpha=\alpha_M$ then
$\gamma=0$;
\item if $\mu(V)-\mu(W)>-\deg(L)$ and $\alpha=\alpha_m$ then
$\rk(\beta)=\min\{p,q\}$, and
\item if $\mu(V)-\mu(W)<-\deg(L)$ and $\alpha=\alpha_m$ then
$\beta=0$.
\end{itemize}
\end{remark}
The following corollary is relevant because $\alpha=0$ is the value of
stability parameter for which the Non-abelian Hodge Theorem gives the
correspondence between $\U(p,q)$-Higgs bundles and representations of
the fundamental group of $X$.
\begin{corollary}
With the notation of Proposition~\ref{bound for alpha}, the inequality $\alpha_M\geq 0$ holds if and only if $\tau(E)\leq
\min\{p,q\}\deg(L)$
and the inequality $\alpha_m\leq 0$ holds if and only if $\tau(E)\geq
-\min\{p,q\}\deg(L)$.
Thus $0\in[\alpha_m,\alpha_M]$ if and only if $\abs{\tau(E)}\leq
\min\{p,q\}\deg(L)$.
\end{corollary}
\begin{proof}
Immediate from Proposition~\ref{bound for alpha}.
\end{proof}
\begin{remark}
Note that the condition $\abs{\tau(E)}\leq \min\{p,q\}\deg(L)$ is
stronger than the condition $\abs{\mu(V)-\mu(W)}\leq \deg(L)$.
\end{remark}
\subsection{Parameters forcing special properties of the Higgs fields}
\label{se:critical-values}
In this section we use a variation on the preceding arguments to find
a parameter range where $\beta$ and $\gamma$ have special
properties. Assume that the twisted $\mathrm{U}(p,q)$-Higgs bundle $E=(V,W,\beta,\gamma)$ has type $(p,q,a,b)$.
For the following proposition it is convenient to introduce the
following notation. For $0\leq i < q \leq p$, let
\begin{displaymath}
\alpha_i=\dfrac{2pq}{q(p-q)+(i+1)(p+q)}
\big(\mu(W)-\mu(V)-\deg(L)\big)+\deg(L),
\end{displaymath}
and for $0\leq j< p\leq q$, let
\begin{displaymath}
\alpha_j'=\dfrac{2pq}{p(q-p)+(j+1)(p+q)}
\big(\mu(W)-\mu(V)+\deg(L)\big)-\deg(L).
\end{displaymath}
\begin{proposition}\label{surjective}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted $\U(p,q)$-Higgs bundle. Then we have the following:
\begin{itemize}
\item[$(i)$] Assume that $p\geq q$ and $\mu(V)-\mu(W)>-\deg(L)$. If
$\alpha<\alpha_{i-1}$ then $\rk(\ker(\beta))<i$. In particular
$\beta$ is injective whenever
\begin{displaymath}
\alpha<\alpha_0 = \frac{2pq}{pq-q^2+p+q}
\big(\mu(W) - \mu(V) - \deg(L)\big) + \deg(L).
\end{displaymath}
\item[$(ii)$] Assume that $p\geq q$ and $\mu(V)-\mu(W)<-\deg(L)$. If
$\alpha<\alpha_{i-1}$ then $\rk(\ker(\beta))>i$. In particular
$\beta$ is zero whenever
\begin{displaymath}
\alpha<\alpha_{q-2} = \frac{2pq}{2pq-p-q}
\big(\mu(W) - \mu(V) - \deg(L)\big) + \deg(L).
\end{displaymath}
\item[$(iii)$] Assume that $p\leq q$ and $\mu(V)-\mu(W)<\deg(L)$. If
$\alpha>\alpha'_j$ then $\rk(\ker(\gamma))<j$. In particular
$\gamma$ is injective whenever
\begin{displaymath}
\alpha>\alpha_0' = \frac{2pq}{pq-p^2+p+q}
\big(\mu(W) - \mu(V) + \deg(L)\big) - \deg(L).
\end{displaymath}
\item[$(iv)$] Assume that $p\leq q$ and $\mu(V)-\mu(W)>\deg(L)$. If
$\alpha>\alpha'_j$ then $\rk(\ker(\gamma))>j$. In particular
$\gamma$ is zero whenever
\begin{displaymath}
\alpha>\alpha_{p-2}' = \frac{2pq}{2pq-p-q}
\big(\mu(W) - \mu(V) + \deg(L)\big) - \deg(L).
\end{displaymath}
\end{itemize}
\end{proposition}
\begin{proof}
We shall only prove parts $(i)$ and $(ii)$. One can deduce the other parts in a similar way. Suppose that $\rk(\ker(\beta))=n>0$. The inequality $(\ref{bound on beta})$ yields
\begin{displaymath}
\alpha \geq \frac{2pq}{n(p+q)+q(p-q)}
\big(\mu(W)-\mu(V)-\deg(L)\big)+\deg(L) = \alpha_{n-1}.
\end{displaymath}
Now suppose $\mu(W)-\mu(V)-\deg(L)<0$, then $\alpha_i$ increases
with $i$ and so, if $n\geq i$ then $\alpha \geq \alpha_{i-1}$. Hence, if
$\alpha < i-1$ then $n<i$. In particular, if $\alpha<\alpha_0$ then $\beta$ is injective, which gives part $(i)$.
On the other hand, if $\mu(W)-\mu(V)-\deg(L)>0$, then $\alpha_i$ decreases
with $i$ and so, if $n\leq i$ then $\alpha \geq \alpha_{i-1}$. Hence, if
$\alpha < \alpha_{i-1}$ then $n>i$. In particular, if $\alpha < \alpha_{q-2}$ then $\beta$ is zero, proving part $(ii)$.
\end{proof}
\begin{remark}
\label{rem:alpha-0-positive}
Note that the signs of $\alpha_0$ and $\alpha'_0$ given in the
preceding proposition are related to the Toledo invariant as follows:
\begin{itemize}
\item $\alpha_0>0$ if and only if $\tau(E)<-(q-1)\deg(L)$.
\item $\alpha_0'<0$ if and only if $\tau(E)>(p-1)\deg(L)$.
\end{itemize}
\end{remark}
\begin{remark}
\label{rem:upq-duality}
Associated to $E=(V,W,\beta,\gamma)$ there is a dual $L$-twisted
$\U(p,q)$-Higgs bundle $E^\ast=(V^*,W^*,\gamma^*,\beta^*)$. Clearly
there is a one-to-one correspondence between subobjects of $E$ and
quotients of $E^*$, and
$\mu_{-\alpha}(E)=-\mu_\alpha(E^*)$. Therefore $\alpha$-stability of
$E^*$ is equivalent to $-\alpha$-stability of $E$.
\end{remark}
\begin{corollary}\label{Cor:Surjective}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted $\U(p,q)$-Higgs bundle. Then we have the following:
\begin{itemize}
\item[(i)] If $p\geq q$ and $\mu(W)-\mu(V)>-\deg(L)$ then $\gamma$ is surjective whenever $$\alpha>\alpha_t:=\frac{2pq}{pq-q^2+p+q}(\mu(W)-\mu(V)+\deg(L))-\deg(L).$$
\item[(ii)] If $p\leq q$ and $\mu(W)-\mu(V)<\deg(L)$ then $\beta$ is surjective whenever $$\alpha<\alpha_t':=\frac{2pq}{pq-p^2+p+q}(\mu(W)-\mu(V)-\deg(L))+\deg(L).$$
\end{itemize}
\end{corollary}
\begin{proof}
Using Proposition~\ref{surjective} we can find a range for the
stability parameter of $E^*$ where $\beta^*$ and $\gamma^*$ are
injective. Hence the result follows by using
Remark~\ref{rem:upq-duality} to relate the stability parameters of
$E$ and $E^*$.
\end{proof}
The following results shows that the bounds in
Proposition~\ref{surjective} are meaningful in view of the bounds for
$\alpha$ of Proposition~\ref{bound for alpha}.
\begin{proposition}
Let $\alpha_0$ and $\alpha_0'$ be given in
Proposition~\ref{surjective}. Then the following holds.
\begin{itemize}
\item[$(i)$] Assume that $p>q$. If $\mu(V)-\mu(W) > -\deg(L)$ then
$\alpha_0>\alpha_m$, and if $\mu(V)-\mu(W) < -\deg(L)$ then $\alpha_{q-2}>\alpha_m$.
\item[$(ii)$] Assume that $p<q$. If $\mu(V)-\mu(W) < \deg(L)$ then
$\alpha_0'<\alpha_M$, and if $\mu(V)-\mu(W) > \deg(L)$ then
$\alpha'_{p-2}<\alpha_M$.
\end{itemize}
\end{proposition}
\begin{proof}
For $(i)$, using $\mu(V)-\mu(W)> -\deg(L)$ we get
\begin{align*}
\alpha_0-\alpha_m&=\bigl(\mu(V)-\mu(W)\bigr)
\Bigl(\dfrac{-2pq}{q(p-q)+p+q}+\dfrac{2p}{p-q}\Bigr)\\
&\phantom{==}+\deg(L)
\Bigl(\dfrac{-2pq}{q(p-q)+p+q}+1+\dfrac{p+q}{p-q}\Bigr)\\
&>\deg(L)\Bigl(-\dfrac{2p}{p-q}+1+\dfrac{p+q}{p-q}\Bigr)=0,
\end{align*}
where we have used that $p>q$ makes the term which multiplies
$\mu(V)-\mu(W)$ positive. Thus $\alpha_0>\alpha_m$. Moreover, when $\mu(V)-\mu(W) < -\deg(L)$
and $p>q$, we have $\alpha_m=\alpha_{q-1}<\alpha_{q-2}$ (cf.\ the
proof of Proposition~\ref{surjective}). This finishes
the proof of $(i)$.
For $(ii)$, using $\mu(V)-\mu(W)<\deg(L)$ we obtain the following
\begin{align*}
\alpha_M-\alpha_0'&=\bigl(\mu(V)-\mu(W)\bigr)
\Bigl(\dfrac{-2q}{q-p}+\dfrac{2pq}{p(q-p)+p+q}\Bigr)\\
&\phantom{==}+\deg(L)
\Bigl(\dfrac{p+q}{q-p}-\dfrac{2pq}{p(q-p)+p+q}+1\Bigr)\\
&>\deg(L)\Bigl(-\dfrac{2q}{q-p}+1+\dfrac{p+q}{q-p}\Bigr)=0,
\end{align*}
where we have used that $p<q$ makes the term which multiplies
$\mu(V)-\mu(W)$ negative.
Hence $\alpha_0'<\alpha_M$. Moreover, when $\mu(V)-\mu(W) > \deg(L)$
and $p<q$, we have $\alpha_M=\alpha'_{p-1}>\alpha'_{p-2}$ (again, cf.\ the
proof of Proposition~\ref{surjective}). This finishes
the proof of $(ii)$.
\end{proof}
\subsection{The comparison between $\U(p, q)$-Higgs bundles and $\GL(p + q, \C)$-Higgs bundles}
Any $\U(p,q)$-Higgs bundle gives rise to a $\GL(p+q,\C)$-Higgs
bundle. In this section we compare the respective stability
conditions. We shall not need these results in the remainder of the
paper but for completeness we have chosen to include them, since the
question is a natural one to consider.
We recall the following about $\GL(n, \C)$-Higgs bundles.
A $\GL(n, \C)$-Higgs bundle on $X$ is a pair $(E, \phi)$, where $E$ is
a rank $n$ holomorphic vector bundle over $X$ and
$\phi\in H^0(\End(E)\otimes K)$ is a holomorphic endomorphism of $E$
twisted by the canonical bundle $K$ of $X$. More generally, replacing
$K$ by an arbitrary line bundle on $X$, we obtain the notion of a
$L$-twisted $\GL(n,\C)$-Higgs bundle on $X$.
The $\GL(n,\C)$-Higgs bundle $(E,\phi)$ is stable if the slope
stability condition
\begin{displaymath}
\mu(E') <\mu(E)
\end{displaymath}
holds for all non-zero proper $\phi$-invariant subbundles $E'$ of
$E$. Semistability is defined by replacing the strict inequality with
a weak inequality. A twisted Higgs bundle is called polystable if it is the
direct sum of stable twisted Higgs bundles with the same slope.
\begin{remark}
Nitsure \cite{nitsure:1991} was the first to study twisted Higgs
bundles in a systematic way. For some of his results he needs to
make the assumption $\deg(L)\geq 2g-2$ (similarly, for example, to
our Proposition~\ref{vanishing main} below). However, the
comparison of stability conditions which we carry out here is valid
for any $L$.
\end{remark}
For any twisted $\U(p,q)$-Higgs bundle $E=(V,W,\beta,\gamma)$ we can
associate a twisted $\GL(p+q,\C)$-Higgs bundle defined by taking
$\widetilde{E}=V\oplus W$ and
$\phi=\left( \begin{array}{cc}0 &\beta \\ \gamma& 0\end{array}
\right)$.
The following result is reminiscent of Theorem~3.26 of
\cite{Gothen:2012}, which is a result for $\mathrm{Sp}(2n, \R)$-Higgs
bundles. The corresponding result for $0$-semistable $\U(p,q)$-Higgs
bundles can be found in the appendix to the first preprint version of
\cite{Gothen:2013} and the proof given there easily adapts to the
present situation. We include it here for the convenience of the
reader.
Recall from Proposition~\ref{surjective} that for $p=q$,
\begin{align}
\label{eq:3}
\alpha_0&=p\big(\mu(W)-\mu(V)-\deg(L)\big)+\deg(L),\\
\label{eq:4}
\alpha_0'&=p\big(\mu(W) - \mu(V) + \deg(L)\big) - \deg(L).
\end{align}
\begin{proposition}
\label{prop:upp-alpha-gl-stability}
Let $E=(V,W,\beta,\gamma)$ be an $\alpha$-semistable twisted
$\U(p,q)$-Higgs bundle such that $p=q$ and let $\alpha_0$ and
let $\alpha_0'$ be given by \eqref{eq:3} and \eqref{eq:4},
respectively. Suppose that one of the following conditions holds:
\begin{enumerate}
\item $\mu(V)-\mu(W)>-\deg(L)$ and $0\leq\alpha<\alpha_0$.
\item $\mu(V)-\mu(W)<\deg(L)$ and $\alpha_0'<\alpha\leq 0$.
\end{enumerate}
Then the associated $\GL(2p,\C)$-Higgs bundle $\widetilde{E}$ is
semistable. Moreover $\alpha$-stability of $E$ implies stability of
$\widetilde{E}$ unless there is an isomorphism $f:V\to W$ such that
$\beta f=f^{-1}\gamma$. In this case $(\widetilde{E},\phi)$ is
polystable and decomposes as
$$
(\widetilde{E},\phi)=(\widetilde{E}_1,\phi_1)\oplus
(\widetilde{E}_2,\phi_2)
$$
where each summand is a stable $\GL(p,\C)$-Higgs bundle isomorphic to $(V,\beta f)$.
\end{proposition}
\begin{proof}
Let $\widetilde{E}'$ be an invariant subbundle of
$\widetilde{E}$. By projecting onto $V$ and $W$ and taking the
kernels and images, we get the following short exact sequences:
\begin{align}\label{eq:short-exact-sequance}
&0\to W''\to \widetilde{E}'\to V'\to 0,\nonumber\\
&0\to V''\to \widetilde{E}'\to W'\to 0.
\end{align}
We can then deduce that
\begin{align}\label{eq:s1}
\deg W'' +\deg V' &=\deg\widetilde{E}' = \deg V'' +\deg W'\nonumber\\
q''+p' &=\rk\widetilde{E}' =p''+q'
\end{align}
where $q''$, $q'$, $p''$ and $p'$ denote the rank of $W''$, $W'$, $V''$ and $V'$, respectively. Note that $(V',W')$ and $(V'',W'')$ define subobjects of $E$. The $\alpha$-semistability conditions applied to these subobjects imply
\begin{align}
\deg V' +\deg W'&\leq \mu(E)(p'+q')+\frac{q'-p'}{2}\alpha\label{eq:slope-condition-E'}\\
\deg V'' +\deg W''&\leq \mu(E)(p''+q'')+\frac{q''-p''}{2}\alpha\label{eq:slope-condition-E''}
\end{align}
Adding these two inequalities and using \eqref{eq:s1}, we get
\begin{equation}
\label{eq:slope-tilde-E-prime}
\mu(\widetilde{E}')\leq
\mu(\widetilde{E})+\frac{q'-p'+q''-p''}{2(p'+p''+q'+q'')}\alpha
= \mu(\widetilde{E})+\frac{q'-p'}{p'+p''+q'+q''}\alpha
\end{equation}
From Proposition~\ref{surjective} we obtain the injectivity of $\beta$
and $\gamma$ by using the hypotheses $(1)$ and $(2)$,
respectively. Injectivity of $\beta$ and $\gamma$ yield $q'\leq p'$
and $q'\geq p'$, respectively. Hence, in either case $(q'-p')\alpha$
is negative. Therefore \eqref{eq:slope-tilde-E-prime} proves that
$\widetilde{E}$ is semistable. \par
Suppose now that $E$ is $\alpha$-stable. Then, by the above argument,
$\widetilde{E}$ is semistable and it is stable if
\eqref{eq:slope-tilde-E-prime} is strict for all non-trivial
subbundles $\widetilde{E}'\subset \widetilde{E}$. The equality holds
in \eqref{eq:slope-tilde-E-prime} if it holds in both
\eqref{eq:slope-condition-E'} and
\eqref{eq:slope-condition-E''}. Since $E$ is $\alpha$-stable the only
way in which a non-trivial subbundle
$\widetilde{E}'\subset \widetilde{E}$ can yield equality in
\eqref{eq:slope-tilde-E-prime} is that
$$V'\oplus W'=V\oplus W\mbox{ and }V''\oplus W''.$$
In this case from \eqref{eq:short-exact-sequance} we obtain
isomorphisms $E'\to V$ and $E'\to W$. Therefore, combining these, we
get an isomorphism $f:V\to W$ such that $\beta f=f^{-1}\gamma$. Hence,
if there is no such isomorphism between $V$ and $W$ then $(\widetilde{E},\phi)$ is $\alpha$-stable.
Now suppose that there exists such an isomorphism $f:V\to W$, define
\begin{align*}
(\widetilde{E}_1,\phi_1)&=(\{(v,f(v))\in\widetilde{E}|v\in V\}, \phi|_{\widetilde{E}_1}),\\
(\widetilde{E}_2,\phi_2)&=(\{(v,-f(v))\in\widetilde{E}|v\in V\}, \phi|_{\widetilde{E}_2}).
\end{align*}
The fact that $\beta f=f^{-1}\gamma$ implies that $(E_i,\phi_i)$, $i=1, 2$, define $\GL(n,\C)$-Higgs bundles isomorphic to $(V,\beta f)$. We have $$(\widetilde{E},\phi)=(\widetilde{E}_1,\phi_1)\oplus(\widetilde{E}_2,\phi_2),$$
with $$\mu(\widetilde{E}_1)=\mu(\widetilde{E})=\mu(\widetilde{E}_2).$$
To show that each summand is a stable $\GL(n,\C)$-Higgs bundle, note that any non-trivial subbundle $\widetilde{E}'$ of $\widetilde{E}_i$ is a subbundle of $\widetilde{E}$ and hence $\mu(\widetilde{E}')<\mu(\widetilde{E})=\mu(\widetilde{E}_i)$.
\end{proof}
\begin{remark}
\label{rem:comparision}
We can also conclude from the proof of the above proposition that a
twisted $\U(p,q)$-Higgs bundle is $\alpha$-semistable for $\alpha=0$
if and only if the associated $\GL(p+q,\C)$-Higgs bundle is
semistable. Equivalence also holds for stability, unless there is
an isomorphism $f:V\to W$ such that $\beta f=f^{-1}\gamma$.
\end{remark}
\subsection{Vanishing of hypercohomology in degree two}
In order to study smoothness of the moduli space we investigate
vanishing of the second hypercohomology group of the deformation
complex (cf.\ Proposition~\ref{smooth0}). This vanishing will also
play an important role in the analysis of the flip loci in
Section~\ref{sec:crossing}. We note that vanishing is not guaranteed by
$\alpha$-stability for $\alpha\neq 0$, in contrast to the case of
triples (and chains), where vanishing is guaranteed for $\alpha>0$.
By using the obvious symmetry of the quiver interchanging the vertices
we can associate to a $\U(p,q)$-Higgs bundle a $\U(q,p)$-Higgs
bundle. The following proposition is immediate.
\begin{proposition}
\label{prop:sigma-alpha-stability}
Let $E=(V,W,\beta,\gamma)$ be a $\U(p,q)$-Higgs bundle and let
$\sigma(E) = (W,V,\gamma,\beta)$ be the associated $\U(q,p)$-Higgs
bundle. Then $E$ is $\alpha$-stable if and only if $\sigma(E)$
is $-\alpha$-stable, and similarly for poly- and
semi-stability. \qed
\end{proposition}
The next result uses this construction and Serre duality to identify the second
hypercohomology of the Hom-complex with the dual of a zeroth
hypercohomology group.
\begin{lemma}
\label{lem:H2-is-H0}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Let
$E''=\sigma(E')\otimes L^{-1}K=(W'\otimes L^{-1}K,V'\otimes
L^{-1}K,\gamma\otimes 1,\beta\otimes 1)$. Then
\begin{displaymath}
\HH^2(\mathpzc{Hom}^\bullet(E',E))
\cong \HH^0(\mathpzc{Hom}^\bullet(E,E''))^*.
\end{displaymath}
\end{lemma}
\begin{proof}
By Serre duality for hypercohomology
$$
\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E))
\cong
\mathbb{H}^0(\mathpzc{Hom}^{\bullet^\vee}(E',E)\otimes K)^*
$$
where the dual complex twisted by $K$ is
\begin{multline*}
\mathpzc{Hom}^{\bullet^\vee}(E',E)\otimes K:\big(\Hom(V,W'\otimes
L^{-1})\oplus\Hom(W,V'\otimes L^{-1})\big)\otimes K \\ \to\big(\Hom(V,V')\oplus\Hom(W,W')\big)\otimes K.
\end{multline*}
One easily checks that the differentials correspond, so that
\begin{displaymath}
\mathpzc{Hom}^{\bullet^\vee}(E',E)\otimes K \cong
\mathpzc{Hom}^\bullet(E,E'').
\end{displaymath}
This completes the proof.
\end{proof}
\begin{lemma}
\label{lem:lambda-f}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. As above
let
$E''=\sigma(E')\otimes L^{-1}K = (W'\otimes L^{-1}K,V'\otimes
L^{-1}K,\gamma'\otimes 1,\beta'\otimes 1)$.
Let $f\in \HH^0(\mathpzc{Hom}^\bullet(E,E''))$ viewed as
morphism $f\colon E\to E''$ and write
$\lambda(f)=\frac{\rk(f(V))}{\rk(f(V))+\rk(f(W))}$. Then, if
$f\neq 0$, the inequality
\begin{equation}
\label{eq:h2-vanishing-ineq}
\alpha(2\lambda(f)-1)+2g-2-\deg(L) \geq 0
\end{equation}
holds. Moreover, if $E$ and $E''$ are $\alpha$-stable, then strict
inequality holds unless $f\colon E\xra{\cong}E''$ is an isomorphism.
\end{lemma}
\begin{proof}
Write $N=\ker(f) \subset
E$ and $I=\im(f) \subset E''$. Then $\alpha$-semistability of $E$
implies that $\mu_\alpha(N)\leq\mu_\alpha(E)$, which is equivalent
to
\begin{equation}
\label{eq:1}
\mu_\alpha(I) \geq \mu_\alpha(E);
\end{equation}
note that this also holds if $N=0$, since then $I\cong E$.
Moreover, by Proposition~\ref{prop:sigma-alpha-stability}, $E''$ is
$-\alpha$-semistable and so
\begin{math}
\mu_{-\alpha}(I)\leq \mu_{-\alpha}(E'').
\end{math}
This, using that $\mu_{-\alpha}(I) = \mu_\alpha(I)
-2\alpha\lambda(f)$ and $\mu_{-\alpha}(E'') =
\mu_\alpha(E)-\alpha+(2g-2-\deg(L))$, is equivalent to
\begin{equation}
\label{eq:2}
\mu_\alpha(I) \leq \mu_\alpha(E) + 2\alpha\lambda(f) - \alpha +
2g-2 - \deg(L).
\end{equation}
Combining \eqref{eq:1} and \eqref{eq:2} gives the result. The
statement about strict inequality is easy.
\end{proof}
The following is our first main result on vanishing of $\HH^2$. It
should be compared with
\cite[Proposition~3.6]{bradlow-garcia-prada-gothen:2004}. The reason
why extra conditions are required for the vanishing is essentially
that the ``total Higgs field'' $\beta+\gamma\in H^0(\End(V\oplus
W)\otimes L)$ is
not nilpotent, contrary to the case of triples.
\begin{proposition}
\label{vanishing main}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle and
$E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Assume that $E$ and $E'$ are $\alpha$-semistable with
$\mu_\alpha(E)=\mu_\alpha(E')$. Let $E''=\sigma(E')\otimes
L^{-1}K$. Assume that one of the following hypotheses hold:
\begin{itemize}
\item[(A)] $\deg(L)>2g-2$;
\item[(B)] $\deg(L)=2g-2$, both $E$ and $E'$ are $\alpha$-stable and
there is no isomorphism $f\colon E \xra{\cong} E''$.
\end{itemize}
Then $\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E))=0$ if one of the
following additional conditions holds:
\begin{itemize}
\item[(1)] $\alpha=0$;
\item[(2)] $\alpha > 0$ and either $\beta'$ is
injective or $\beta$ is surjective;
\item[(3)] $\alpha < 0$ and either $\gamma'$ is
injective or $\gamma$ is surjective.
\end{itemize}
\end{proposition}
\begin{proof}
Suppose first that $\alpha=0$. Then either of the conditions (A) and
(B) guarantee that strict inequality holds in
\eqref{eq:h2-vanishing-ineq}. Hence Lemmas~\ref{lem:H2-is-H0} and
\ref{lem:lambda-f} imply the stated vanishing of $\HH^2$.
Now suppose that $\beta'\colon W'\to V'\otimes L$ is injective. If
$f\colon E\to E''$ is non-zero then, since $f$ is a morphism of
twisted $\U(p,q)$-Higgs bundles, we have $\rk(f(W)) \geq \rk(f(V))$.
Hence $\lambda(f) = \frac{\rk(f(V))}{\rk(f(V))+\rk(f(W))}$ satisfies
$\lambda(f)\leq 1/2$. If additionally $\alpha>0$, it follows that
$\alpha(2\lambda(f)-1) \leq 0$ which contradicts
Lemma~\ref{lem:lambda-f} under either of the conditions (A) and
(B). Therefore there are no non-zero morphisms $f\colon E\to E''$ and so Lemma~\ref{lem:H2-is-H0} implies vanishing of $\HH^2(\mathpzc{Hom}^\bullet(E',E))$.
We have deduced vanishing of $\HH^2$ under the conditions $\alpha>0$
and $\beta'$ injective. The remaining conditions in (2) and (3) for
vanishing of $\HH^2$ can now be deduced by using symmetry arguments as
follows.
Suppose first that $\alpha<0$ and $\gamma'$ is injective.
Then, using
Proposition~\ref{prop:sigma-alpha-stability}, $\sigma(E)$
is an $-\alpha$-semistable $\U(p,q)$-Higgs bundle and similarly for
$\sigma(E')$. Moreover, the $\beta$-map (which is
$\sigma(\gamma')$) of $\sigma(E')$ is injective.
Observe that
\begin{displaymath}
\mathpzc{Hom}^\bullet(\sigma(E'),\sigma(E)) \cong
\mathpzc{Hom}^\bullet(E',E).
\end{displaymath}
Hence, noting that $-\alpha>0$, the conclusion
follows from the previous case applied to the pair $(\sigma(E'),\sigma(E))$.
Next suppose that $\alpha<0$ and $\gamma$ is surjective.
Then the dual $\U(p,q)$-Higgs bundle $E^*$ is $-\alpha$-semistable,
and similarly for $E'^*$. Moreover, the $\beta$-map (which is
$\gamma^*$) of $E^*$ is injective. Observe that
\begin{displaymath}
\mathpzc{Hom}^\bullet(E^*,E'^*) \cong
\mathpzc{Hom}^\bullet(E',E).
\end{displaymath}
Hence again the conclusion follows from the previous case, applied to
the pair $(E^*,E'^*)$.
The final case, $\alpha>0$ and $\beta$ surjective, follows in a
similar way, combining the two previous constructions.
\end{proof}
In the case when $q=1$ we can improve on
Proposition~\ref{vanishing main}, as follows.
\begin{proposition}\label{q=1}
Let $E$ be an $\alpha$-semistable $L$-twisted $\U(p,1)$-Higgs bundle
with $p\geq 2$. Assume that $\deg(L)>2g-2$. Then
$\mathbb{H}^2(\mathpzc{End}^\bullet(E))=0$ for all $\alpha$ in the
range
\begin{displaymath}
p(\mu(V)-\mu(W)) - (p+1)(\deg(L)-2g+2) < \alpha < p(\mu(V)-\mu(W)) + (p+1)(\deg(L)-2g+2).
\end{displaymath}
\end{proposition}
\begin{proof}
Assume first that $\alpha\geq 0$. Note that an isomorphism as in (B) of the
hypothesis of Proposition~\ref{vanishing main} cannot exist when
$p\neq q$. Hence the proposition immediate gives the result if
$\alpha=0$. Moreover, if $\beta\neq 0$, then it is injective, and
hence $\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E))=0$ by (2) of the
proposition. We may thus assume that $\beta=0$ and consider the
$L$-twisted triple $E_T\colon \gamma\colon V\to W\otimes L$. We have
that
\begin{displaymath}
\HH^2(\mathpzc{End}^\bullet(E)) =
\HH^2(\mathpzc{End}^\bullet(E_T)) \oplus H^1(\Hom(W,V)\otimes L),
\end{displaymath}
where $\mathpzc{End}^\bullet(E_T)$ is the deformation complex of the
triple. The vanishing of $\HH^2(\mathpzc{End}^\bullet(E_T))$ for an
$\alpha$-semistable triple when $\alpha>0$ is well
known\footnote{Note that the stability parameter for the
corresponding untwisted triple as considered in
\cite{bradlow-garcia-prada-gothen:2004} is
$\alpha+\deg(L)$.}
(cf.~\cite{bradlow-garcia-prada-gothen:2004}). Hence it remains to
show that $H^1(\Hom(W,V)\otimes L) = 0$ which, by Serre duality, is
equivalent to the vanishing
\begin{displaymath}
H^0(\Hom(V,W)\otimes L^{-1}K) = 0.
\end{displaymath}
So assume we have a non-zero $f\colon V \to W\otimes L^{-1}K$. Then
$f$ induces as non-zero map of line bundles $f\colon V/\ker(f) \to
W\otimes L^{-1}K$ and hence
\begin{equation}
\label{eq:7}
\deg(W) - \deg(L) + 2g-2 \geq \deg(V) - \deg(\ker(f)).
\end{equation}
On the other hand, since $\beta=0$ we can consider the subobject
$(\ker(f),W,0,\gamma)$ of $E$ and hence, by $\alpha$-semistability,
\begin{align}
\mu_\alpha(\ker(f)\oplus W) &\leq \mu_\alpha(V \oplus W) \notag \\
\iff (p+1)\deg(\ker(f)) + \deg(W) &\leq p\deg(V)+\alpha,
\label{eq:6}
\end{align}
where we have used that $\rk(\ker(f)) = p-1$ and $\rk(W)=1$.
Now combining \eqref{eq:7} and \eqref{eq:6} we obtain
\begin{displaymath}
\alpha \geq p(\mu(V)-\mu(W)) + (p+1)(\deg(L)-2g+2).
\end{displaymath}
This establishes the vanishing of $\HH^2$ for $\alpha$ in the range
\begin{displaymath}
0\leq \alpha < p(\mu(V)-\mu(W)) + (p+1)(\deg(L)-2g+2).
\end{displaymath}
On the other hand, if $\alpha\leq 0$, applying the preceding result
to the dual twisted $\U(p,q)$-Higgs bundle
$(V^*,W^*,\gamma^*,\beta^*)$ gives vanishing of $\HH^2$ for $\alpha$
in the range
\begin{displaymath}
0\geq \alpha > p(\mu(V)-\mu(W)) - (p+1)(\deg(L)-2g+2).
\end{displaymath}
This finishes the proof.
\end{proof}
In general the preceding proposition does not guarantee vanishing of
$\HH^2$ for all values of the parameter $\alpha$. But for some values
of the topological invariants, the upper bound of the preceding
proposition is actually larger than the maximal value for the
parameter $\alpha$. More precisely, we have the following result.
\begin{proposition}
\label{prop:q=1-range}
Let $E$ be an $\alpha$-semistable $L$-twisted $\U(p,1)$-Higgs bundle
with $p\geq 2$. Assume that $\deg(L)>2g-2$. We have the following:
\begin{itemize}
\item[(1)]
If $p\bigl(\mu(V)-\mu(W)\bigr) > 2g-2-(p-2)\bigl(\deg(L)-(2g-2)\bigr)$ then $\mathbb{H}^2(\mathpzc{End}^\bullet(E))=0$ for all $\alpha\geq 0$\\
\item[(2)] If $p\bigl(\mu(V)-\mu(W)\bigr) < -2g+2+(p-2)\bigl(\deg(L)-(2g-2)\bigr)$ then $\mathbb{H}^2(\mathpzc{End}^\bullet(E))=0$ for all $\alpha\leq 0$
\end{itemize}
\end{proposition}
\begin{proof}
The upper and lower bound for $\alpha$ given in Proposition~\ref{bound for
alpha} is, in this case
\begin{displaymath}
\alpha_M = -\frac{2p}{p-1}\bigl(\mu(V)-\mu(W)\bigr)
+\frac{p+1}{p-1}\deg(L),
\end{displaymath}
\begin{displaymath}
\alpha_m = -\frac{2p}{p-1}\bigl(\mu(V)-\mu(W)\bigr)
-\frac{p+1}{p-1}\deg(L).
\end{displaymath}
It is simple to check that the inequalities of the statements are
equivalent to $\alpha_M$ being less than the upper bound and $\alpha_m$ being bigger than the lower bound for
$\alpha$ of Proposition~\ref{q=1}.
\end{proof}
The following trivial observation is sometimes useful.
\begin{proposition}
\label{prop:end-hom-vanishing}
Let $E$ and $E'$ be $L$-twisted $\U(p,q)$-Higgs bundles such that
$\HH^2(\mathpzc{End}^\bullet(E\oplus E'))=0$. Then
\begin{displaymath}
\mathbb{H}^2(\mathpzc{Hom}^\bullet(E',E)) =
\mathbb{H}^2(\mathpzc{Hom}^\bullet(E,E')) = 0.
\end{displaymath}
\end{proposition}
\begin{proof}
Immediate in view of Remark~\ref{rem:direct-sum-deform}.
\end{proof}
We can summarize our main results on vanishing of $\HH^2$ as follows.
\begin{lemma}
\label{lem:h2-vanishing-final}
Fix a type $t=(p,q,a,b)$ and let $E$ be an $\alpha$-semistable $L$-twisted
$\U(p,q)$-Higgs bundle of type $t$ with $\deg(L)\geq 2g-2$. If
$\deg(L)=2g-2$ assume moreover that $E$ is $\alpha$-stable.
If either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $q=1$, $p\geq 2$ and $p(a/p-b/q) - \deg(L)(p+1) < \alpha < p(a/p-b/q) + \deg(L)(p+1)$,
\item[$(2)$] $a/p-b/q>-\deg(L)$ and $0\leq\alpha<\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(3)$] $a/p-b/q<\deg(L)$ and $\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha\leq 0$.
\end{itemize}
Then $\mathbb{H}^2(\mathpzc{End}^\bullet(E))$ vanishes.
\end{lemma}
\begin{proof}
For part $(1)$, use Proposition~\ref{q=1}. The other parts follow from
Proposition~\ref{surjective}, Corollary~\ref{Cor:Surjective}, and Proposition~\ref{vanishing main}.
\end{proof}
\subsection{Moduli space of twisted $\U(p,q)$-Higgs bundles}
\label{sec:moduli-space-twisted}
Finally, we are in a position to make statements about smoothness of
the moduli space. Recall that we denote
the moduli space of $\alpha$-polystable twisted $\U(p,q)$-Higgs
bundles with type $t=(p,q,a,b)$ by
$$
\mathcal{M}_\alpha(t)=\mathcal{M}_\alpha(p,q,a,b),
$$
and the moduli space of $\alpha$-stable twisted $\U(p,q)$-Higgs bundle
by $\mathcal{M}_\alpha^s(t)\subset\mathcal{M}_\alpha(t)$.
\begin{proposition}\label{smooth}
Fix a type $t=(p,q,a,b)$.
If either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $q=1$, $p\geq 2$ and $p(a/p-b/q) - \deg(L)(p+1) < \alpha < p(a/p-b/q) + \deg(L)(p+1)$,
\item[$(2)$] $a/p-b/q>-\deg(L)$ and $0\leq\alpha<\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(3)$] $a/p-b/q<\deg(L)$ and $\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha\leq 0$.
\end{itemize}
Then the moduli space $\mathcal{M}_\alpha^s(t)$ is smooth.
\end{proposition}
\begin{proof}
Combine Lemma~\ref{lem:h2-vanishing-final} and Proposition~\ref{smooth0}.
\end{proof}
\section{Crossing critical values}
\label{sec:crossing}
\subsection{Flip loci}\label{Flip loci}
In this section we study the variation with $\alpha$ of the moduli
spaces $\mathcal{M}_\alpha^s(t)$ for fixed type $t=(p,q,a,b)$. We are
using a method similar to the one for chains given in
\cite{Schmitt:2006}, which in turn is based on
\cite{bradlow-garcia-prada-gothen:2004}.
Let $\alpha_c$ be a critical value. We adopt the following
notation:
$$
\alpha_c^+=\alpha_c+\epsilon,\mbox{ }
\alpha_c^-=\alpha_c-\epsilon,
$$
where $\epsilon>0$ is small enough so that $\alpha_c$ is the only
critical value in the interval $(\alpha_c^-, \alpha_c^+)$. We begin
with a set theoretic description of the differences between two spaces
$\mathcal{M}_{\alpha_c^+}$ and $\mathcal{M}_{\alpha_c^-}$.
\begin{definition}
We define \emph{flip loci} $\mathcal{S}_{\alpha_c^{\pm}}
\subset\mathcal{M}^s_{\alpha_c^{\pm}}$ by the condition that the
points in $\mathcal{S}_{\alpha_c^+}$ represent twisted $\U(p,q)$-Higgs
bundles which are $\alpha_c^+$-stable but $\alpha_c^-$-unstable, and
analogously for $\mathcal{S}_{\alpha_c^-}$.
\end{definition}
A twisted $\U(p,q)$-Higgs bundle $E\in\mathcal{S}_{\alpha_c^{\pm}}$ is
strictly $\alpha_c$-semistable and so we can use the Jordan-H\"{o}lder filtrations of $E$ in order to estimate the codimension of $\mathcal{S}_{\alpha_c^{\pm}}$ in $\mathcal{M}_{\alpha_c^{\pm}}$.
The following is an analogue for twisted $\U(p,q)$-Higgs bundles of
\cite[Proposition~4.3]{Schmitt:2006}, which is a result for chains.
\begin{proposition}\label{Prop:bound-for-dim}
Fix a type $t=(p,q,a,b)$. Let $\alpha_c$ be a critical value and let
$\mathcal{S}$ be a family of $\alpha_c$-semistable twisted
$\U(p,q)$-Higgs bundles $E$ of type $t$, all of them pairwise
non-isomorphic, and whose Jordan-H\"{o}lder filtrations has an
associated graded of the form $\mathrm{Gr}(E)=\bigoplus_{i=1}^mQ_i$, with $Q_i$ twisted $\U(p,q)$-Higgs bundle of type $t_i$.
If either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $q=1$, $p\geq 2$ and $p(a/p-b/q) - \deg(L)(p+1) < \alpha_c < p(a/p-b/q) + \deg(L)(p+1)$,
\item[$(2)$] $a/p-b/q>-\deg(L)$ and $0\leq\alpha_c<\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(3)$] $a/p-b/q<\deg(L)$ and $\frac{2pq}{\mathrm{min}\{p,q\}|p-q|+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha_c\leq 0$.
\end{itemize}
Then
\begin{equation}
\label{bund}
\dim \mathcal{S}\leq -\sum_{i\leq j}\chi(t_j,t_i)-\frac{m(m-3)}{2}.
\end{equation}
\end{proposition}
\begin{proof}
Once appropriate vanishing of $\HH^2$ is ensured, the proof is similar
to the proof of \cite[Proposition~4.3]{Schmitt:2006};
we indicate the idea for $m=2$. In view of the definition of $\mathcal{S}$, there exists an injective canonical map $$i: \mathcal{S}\to\mathcal{M}^s_{\alpha_c}(t_1)\times \mathcal{M}^s_{\alpha_c}(t_2)$$
with $i^{-1}(Q_1,Q_2)\cong \mathbb{P}(\Ext^1(Q_2,Q_1))$, where $\mathbb{P}(\Ext^1(Q_2,Q_1))$ parametrizes equivalence classes of extensions
$$0\rightarrow Q_1\rightarrow E \rightarrow Q_2\rightarrow 0.$$ Notice
that $Q_1$ and $Q_2$ satisfy the hypothesis of Proposition
$\ref{vanishing main}$ (or, in case $q=1$, Proposition~\ref{q=1}; cf.\ Proposition~\ref{prop:end-hom-vanishing}) and therefore, cf.\ Proposition $\ref{chi}$, $\dim(\mathbb{P}\Ext^1(Q_2,Q_1))$ is constant as $Q_1$ and $Q_2$ vary in their corresponding moduli spaces.
Hence, we
obtain $$\dim\mathcal{S}\leq\dim\mathcal{M}_{\alpha_c}^s(t_1)+\dim\mathcal{M}_{\alpha_c}^s(t_2)+\dim\mathbb{P}(\Ext^1(Q_2,Q_1)).$$
The general case follows by induction on $m$ as in loc.~cit.
\end{proof}
In order to show that the flip loci $\mathcal{S}_{\alpha_c^{\pm}}$ has
positive codimension we need to bound the values of $\chi(t_i,t_j)$ in
$(\ref{bund})$. This is what we do next.
\subsection{Bound for $\chi$}
\label{sec:bound-chi}
Here we consider a $Q$-bundle associated to the complex $\mathpzc{Hom}^\bullet(E',E)$ and construct a solution to the vortex equations on this $Q$-bundle from solutions on $E'$ and $E$. The quiver $Q$ is the following:
\[
\xymatrix{
\bullet\ar@{<-}@/_1pc/[r]&\bullet\ar@{<-}@/_1pc/[l]\ar@{>}@/_1pc/[r]&\bullet\ar@{>}@/_1pc/[l]\\
}
\]
\\
The construction generalizes the one of \cite{bradlow-garcia-prada-gothen:2004} Lemma $4.2$.
\subsubsection{The $Q$-bundle associated to $\mathpzc{Hom}^\bullet(E',E)$}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs bundle.
Let us consider the following twisted $Q$-bundle $\widetilde{E}$ (the morphisms are twisted by $L$ for each arrow):
\begin{equation}\label{associated}
\xymatrix{
\Hom(W',V)\ar@{<-}@/_2pc/[r]^{\phi_b}&\Hom(V',V)\oplus \Hom(W',W)\ar@{<-}@/_2pc/[l]_{\phi_d} \ar@{>}@/_2pc/[r]^{\phi_a}&\Hom(V',W)\ar@{>}@/_2pc/[l]_{\phi_c}\\
}
\end{equation}
where
\begin{eqnarray*}
\phi_a(f_1,f_2)&=&(f_2\otimes 1_L)\circ \gamma'-\gamma\circ f_1\mbox{ },\\
\phi_b(f_1,f_2)&=&(f_1\otimes 1_L)\circ \beta' -\beta\circ f_2\mbox{ },\\
\phi_c(g)&=&(\beta\circ g, (g\otimes 1_L)\circ \beta')\mbox{ },\\
\phi_d(h)&=&((h\otimes 1_L)\circ\gamma',\gamma\circ h).\\
\end{eqnarray*}
We will write briefly as $\widetilde{E}$
\[
\xymatrix{
\mathpzc{Hom}^{12}\ar@{<-}@/_1pc/[r]^{\phi_b}_L&\mathpzc{Hom}^0\ar@{<-}@/_1pc/[l]_{\phi_d}^L \ar@{>}@/_1pc/[r]^{\phi_a}_L&\mathpzc{Hom}^{11}\ar@{>}@/_1pc/[l]_{\phi_c}^L\\
}.
\]
Note that $\mathpzc{Hom}^1=\mathpzc{Hom}^{11}\oplus
\mathpzc{Hom}^{12}$ and $a_0=(\phi_a,\phi_b)$, where $a_0:
\mathpzc{Hom}^0\to \mathpzc{Hom}^1$ is the Hom-complex
$(\ref{deformation complex})$.
In this section, by using Proposition $\ref{solution}$, we prove that
if $E'$ and $E$ are $\alpha$-polystable then $\widetilde{E}$ is
$\boldsymbol{\alpha}$-polystable for a suitable choice of
$\boldsymbol{\alpha}$.
\begin{lemma}\label{vortex}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Suppose, moreover, we have solutions to the $(\tau_1,\tau_2)$-vortex equations on $E$ and the $(\tau_1',\tau_2')$-vortex equations on $E'$ such that $\tau_1-\tau'_1=\tau_2-\tau'_2$.
Then the induced Hermitian metric on the $Q$-bundle $\widetilde{E}$ satisfies the vortex equations
\begin{eqnarray*}
\sqrt{-1}\Lambda F(\mathpzc{Hom}^{12})+\phi_b\phi_b^{\ast}-\phi_d^{\ast}\phi_d&=&\widetilde{\tau_2} \mathrm{Id}_{\mathpzc{Hom}^{12}},\\
\sqrt{-1}\Lambda F(\mathpzc{Hom}^0)+\phi_c\phi_c^{\ast}+\phi_d\phi_d^{\ast}-\phi_a^{\ast}\phi_a-\phi_b^{\ast}\phi_b&=&\widetilde{\tau_1} \mathrm{Id}_{\mathpzc{Hom}^0},\\
\sqrt{-1}\Lambda F(\mathpzc{Hom}^{11})+\phi_a\phi_a^{\ast}-\phi_c^{\ast}\phi_c&=&\widetilde{\tau_0} \mathrm{Id}_{\mathpzc{Hom}^{11}}.
\end{eqnarray*}
For $\boldsymbol{\tau}=\big(\widetilde{\tau_0},\widetilde{\tau_1},\widetilde{\tau_2}\big)$ given by
\begin{eqnarray*}
\widetilde{\tau_0}&=&\tau_2-\tau'_1,\\
\widetilde{\tau_1}&=&\tau_1-\tau_1'=\tau_2-\tau'_2, \\
\widetilde{\tau_2}&=&\tau_1-\tau'_2.
\end{eqnarray*}
\end{lemma}
\begin{proof}
The vortex equations for $E$ and $E'$ are
\begin{align*}
\sqrt{-1}\Lambda F(V)+\beta\beta^\ast-\gamma^\ast\gamma
&=\tau_1 \mathrm{Id}_{V},\\
\sqrt{-1}\Lambda F(W)+\gamma\gamma^\ast-\beta^\ast\beta
&=\tau_2 \mathrm{Id}_{W},\\
\sqrt{-1}\Lambda F(V')+\beta'\beta'^\ast-\gamma'^\ast\gamma'
&=\tau_1' \mathrm{Id}_{V'},\\
\sqrt{-1}\Lambda F(W')+\gamma'\gamma'^\ast-\beta'^\ast\beta'
&=\tau_2'\mathrm{Id}_{W'}.
\end{align*}
We have
\begin{eqnarray*}
F(\mathpzc{Hom}^0)(\psi,\eta)&=&(F(V)\circ\psi-\psi\circ F(V'),F(W)\circ\eta-\eta\circ F(W')).
\end{eqnarray*}
Now we calculate $\phi_a^\ast$ and $\phi_b^\ast$: for $(f_1,f_2)\in \mathpzc{Hom}^0$, $g\in \mathpzc{Hom}^{11}$ and $h\in \mathpzc{Hom}^{12}$ we have,
\begin{eqnarray*}
\lefteqn{\left\langle \phi_a^\ast(g),(f_1, f_2) \right\rangle _{\mathpzc{Hom}^0}}\\
&=&\left\langle g, \phi_a\big((f_1, f_2)\big) \right\rangle_{\mathpzc{Hom}^{11}}\\
&=& \left\langle g,(f_2\otimes 1_L)\circ\gamma'-\gamma\circ f_1\right\rangle_{\mathpzc{Hom}^{11}}\\
&=&\left\langle g,(f_2\otimes 1_L)\circ\gamma'\right\rangle_{C_{11}} - \left\langle g,\gamma\circ f_1\right\rangle_{\mathpzc{Hom}^{11}}\\
&=&\left\langle (g\circ \gamma'^\ast)\otimes 1_{L^\ast},f_2\right\rangle_{\Hom(W',W)} + \left\langle -\gamma^\ast\circ g,f_1\right\rangle_{\Hom(V',V)}\\
&=&\left\langle \big(-\gamma^\ast\circ g, (g\circ\gamma'^\ast)\otimes1_{L^\ast}\big),(f_1, f_2) \right\rangle _{\mathpzc{Hom}^0}\\
\end{eqnarray*}
and
\begin{eqnarray*}
\left\langle \phi_b^\ast(h),(f_1, f_2) \right\rangle _{\mathpzc{Hom}^0}&=&\left\langle h, \phi_b\big((f_1, f_2)\big) \right\rangle_{\mathpzc{Hom}^{12}}\\
&=& \left\langle h,(f_1\otimes 1_L)\circ\beta'-\beta\circ f_2\right\rangle_{\mathpzc{Hom}^{12}}\\
&=&\left\langle (h\circ\beta'^\ast)\otimes 1_{L^\ast},f_1\right\rangle_{\Hom(V',V)}- \left\langle \beta^\ast\circ h,f_2\right\rangle_{\Hom(W',W)}\\
&=&\left\langle \big((h\circ\beta'^\ast)\otimes 1_{L^\ast},-\beta^\ast\circ h\big),\big(f_1,f_2\big)\right\rangle_{\mathpzc{Hom}^0}\\
\end{eqnarray*}
Hence,
\begin{eqnarray*}
\phi_a^\ast(g)&=&(-\gamma^\ast\circ g,(g\circ\gamma'^\ast)\otimes1_{L^\ast}),\\
\phi_b^\ast(h)&=&((h\circ\beta'^\ast)\otimes 1_{L^\ast},-\beta^\ast\circ h).
\end{eqnarray*}
By a similar calculation as above, we have
\begin{eqnarray*}
\phi_c^\ast(f_1, f_2)&=&(f_2\circ\beta'^\ast)\otimes 1_{L^\ast}-\beta^\ast\circ f_1,\\
\phi_d^\ast(f_1, f_2)&=& (f_1\circ\gamma'^\ast)\otimes1_{L^\ast}-\gamma^\ast\circ f_2.
\end{eqnarray*}
Let $g\in \mathpzc{Hom}^{11} $ and $h\in \mathpzc{Hom}^{12}$, then we have:
\begin{eqnarray*}
\phi_c^\ast\phi_c(g)&=&\phi_c^\ast(\beta\circ g,(g\otimes1_L)\circ\beta')\\
&=&\beta^\ast\beta\circ g+g\circ\beta'\beta'^\ast.
\end{eqnarray*}
\begin{eqnarray*}
\phi_d^\ast \phi_d(h)&=&\phi_d^\ast\big((h\otimes 1_L)\circ\gamma',\gamma\circ h\big)\\
&=&h\circ\gamma'\gamma'^\ast-\gamma^\ast\gamma\circ h.
\end{eqnarray*}
and
\begin{eqnarray*}
\phi_b\phi_b^\ast(h)&=&\phi_b(h\circ\beta'^\ast\otimes 1_{L^\ast},\beta^\ast\circ h)\\
&=&h\circ\beta'^\ast\beta'-\beta\beta^\ast.
\end{eqnarray*}
\begin{eqnarray*}
\phi_a\phi_a^\ast(g)&=&\phi_a(g\circ\gamma'^\ast\otimes1_{L^\ast},-\gamma^\ast\circ g)\\
&=&g\circ\gamma'^\ast\gamma'+\gamma\gamma^\ast\circ g.
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\phi_b\phi_b^\ast-\phi_d^\ast \phi_d(h)&=&h\circ\beta'^\ast\beta'-\beta\beta^\ast\circ h-h\circ\gamma'\gamma'^\ast+\gamma^\ast\gamma\circ h\\
\phi_a\phi_a^\ast-\phi_c^\ast \phi_c(g)&=&g\circ\gamma'^\ast\gamma'+\gamma\gamma^\ast\circ g-\beta^\ast\beta\circ g-g\circ\beta'\beta'^\ast
\end{eqnarray*}
Hence for $g\in \mathpzc{Hom}^{11}$ and $h\in \mathpzc{Hom}^{12}$ we have,
\begin{eqnarray*}
\lefteqn{(\sqrt{-1}\Lambda F(\mathpzc{Hom}^{11})+\phi_a\phi_a^\ast-\phi_c^\ast \phi_c)(g)}\\
&=&\sqrt{-1}\Lambda\big(F(W)\circ g-g\circ F(V')\big)+\phi_a\phi_a^\ast-\phi_c^\ast \phi_c(g)\\
&=&\big(\sqrt{-1}\Lambda F(W)+\gamma\gamma^\ast-\beta^\ast\beta\big)\circ g+g\circ\big(-\sqrt{-1}\Lambda F(V')+\gamma'^\ast\gamma'-\beta'\beta'^\ast\big)\\
&=&\tau_2\mathrm{Id}_{W}\circ g-g\circ\tau'_1\mathrm{Id}_{V'}\\
&=&(\tau_2-\tau'_1)g
\end{eqnarray*}
\begin{eqnarray*}
\lefteqn{(\sqrt{-1}\Lambda F(\mathpzc{Hom}^{12})+\phi_b\phi_b^\ast-\phi_d^\ast \phi_d)(h)}\\
&=&\sqrt{-1}\Lambda\big( \otimes F(V)\circ h-h\circ F(W')\big)+\phi_b\phi_b^\ast-\phi_d^\ast \phi_d(h)\\
& = &\big(\sqrt{-1}\Lambda F(V)+\gamma^\ast\gamma-\beta\beta^\ast\big)\circ h+h\circ\big( -\sqrt{-1}\Lambda F(W')+\beta'^\ast\beta'-\gamma'\gamma'^\ast\big)\\
&=&\tau_1\mathrm{Id}_{V}\circ h-h\circ\tau'_2\mathrm{Id}_{W'}\\
&=&(\tau_1-\tau'_2)h.
\end{eqnarray*}
Similarly for $(f_1,f_2)\in \mathpzc{Hom}^0$ we have,
\begin{eqnarray*}
\phi_c\phi_c^\ast (f_1,f_2)&=&\phi_c((f_2\circ\beta'^\ast)\otimes1_{L^\ast}-\beta^\ast\circ f_1)\\
&=&\big(\beta\beta^\ast\circ f_1-\beta\circ(f_2\circ\beta'^\ast\otimes1_{L^\ast}),f_2\circ\beta'^\ast\beta'-(\beta^\ast\circ f_1\otimes 1_{L})\otimes \beta'\big)
\end{eqnarray*}
\begin{eqnarray*}
\phi_d\phi_d^\ast (f_1,f_2)&=&\phi_d\big((f_1\circ\gamma'^\ast)\otimes1_{L^\ast}-\gamma^\ast\circ f_2\big)\\
&=&\left( f_1\circ\gamma'^\ast\gamma'-\gamma^\ast\circ f_2\otimes1_L\circ\gamma',\gamma\circ(f_1\circ\gamma'^\ast\otimes1_{L^\ast})-\gamma\gamma^\ast\circ f_2\right)
\end{eqnarray*}
and
\begin{eqnarray*}
\phi_a^\ast\phi_a (f_1,f_2)&=&\phi_a^\ast\big(f_2\otimes1_L\circ\gamma'-\gamma\circ f_1\big)\\
&=&\left(-\gamma^\ast\circ f_2\otimes1_L\circ\gamma'+\gamma^\ast\gamma\circ f_1,(f_2\circ\gamma'\gamma'^\ast-\gamma\circ f_1\circ\gamma'^\ast\otimes1_{L^\ast}\right)
\end{eqnarray*}
\begin{eqnarray*}
\phi_b^\ast \phi_b(f_1,f_2)&=&\phi_b^\ast\big(f_1\otimes1_L\circ\beta'-\beta\circ f_2\big)\\
&=&\left(f_1\circ\beta'\beta'^\ast-\beta\circ f_2\circ\beta'^\ast\otimes1_{L^\ast},\beta^\ast\circ f_1\otimes1_L\circ\beta'-\beta^\ast\beta\circ f_2\right)
\end{eqnarray*}
So,
\begin{eqnarray*}
\lefteqn{(\phi_c\phi_c^\ast+\phi_d\phi_d^\ast-\phi_a^\ast\phi_a-\phi_b^\ast\phi_b)(f_1,f_2)}\\
& =&\left(\beta\beta^\ast\circ f_1+f_1\circ\gamma'^\ast\gamma'-\gamma^\ast\gamma\circ f_1-f_1\circ\beta'\beta'^\ast,f_2\circ\beta'^\ast\beta'-\gamma\gamma^\ast\circ f_2-f_2\circ\gamma'\gamma'^\ast+\beta^\ast\beta\circ f_2\right)
\end{eqnarray*}
Hence we have,
\begin{eqnarray*}
\lefteqn{(\sqrt{-1}\Lambda F(\mathpzc{Hom}^0)+\phi_c\phi_c^\ast+\phi_d\phi_d^\ast-\phi_a^\ast\phi_a-\phi_b^\ast\phi_b)(f_1,f_2)}\\
&=&\Big(\sqrt{-1}\Lambda (F(V)\circ f_1-f_1\circ F(V')),\sqrt{-1}\Lambda( F(W)\circ f_2-f_2\circ F(W'))\Big)+\\
& &\Big(\beta\beta^\ast\circ f_1+f_1\circ\gamma'^\ast\gamma'-\gamma^\ast\gamma\circ f_1-f_1\circ\beta'\beta'^\ast,f_2\circ\beta'^\ast\beta'-\gamma\gamma^\ast\circ f_2-f_2\circ\gamma'\gamma'^\ast+\beta^\ast\beta\circ f_2\Big)\\
&=&\Big((\sqrt{-1}\Lambda F(V)+\beta\beta^\ast-\gamma^\ast\gamma)\circ f_1+f_1\circ(-\sqrt{-1}\Lambda F(V')+\gamma'^\ast\gamma'-\beta'\beta'^\ast),\\
& &(\sqrt{-1}\Lambda F(W)+\gamma\gamma^\ast-\beta^\ast\beta)\circ f_2+f_2\circ(-\sqrt{-1}\Lambda F(W')+\beta'^\ast\beta'-\gamma'\gamma'^\ast)\Big)\\
&=&\Big((\tau_1-\tau'_1)f_1,(\tau_2-\tau'_2)f_2\Big).
\end{eqnarray*}
The proof is completed, since by assumption $\tau_1-\tau'_1=\tau_2-\tau'_2$.
\end{proof}
\begin{theorem}\label{Polystable}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Then the $Q$-bundle $\widetilde{E}$ is $\boldsymbol{\alpha}$-polystable for $\boldsymbol{\alpha}=(\alpha,2\alpha)$ .
\end{theorem}
\begin{proof}
Since $E$ and $E'$ are $\alpha$-polystable, from Theorem $\ref{solution}$ follows that they support solutions to the $(\tau_1,\tau_2)$- and $(\tau_1',\tau_2')$-vortex equations where $\alpha=\tau_2-\tau_1=\tau_2'-\tau_1'$. Using Lemma $\ref{vortex}$ it follows that the $Q$-bundle $\widetilde{E}$ admits a Hermitian metric such that vortex equations are satisfied for $\boldsymbol{\tau}=(\tau_2-\tau_1',\tau_2-\tau'_2,\tau_1-\tau'_2)$. Now from Theorem $\ref{Hitchin-Kobayashi}$ we get that $\widetilde{E}$ is $\boldsymbol{\alpha}$-polystable for
\begin{eqnarray*}
\alpha_1=\tau_2-\tau'_1-\tau_2+\tau'_2=\alpha,\\
\alpha_2=\tau_2-\tau'_1-\tau_1+\tau'_2=2\alpha.
\end{eqnarray*}
\end{proof}
\subsubsection{Bound for $\chi(E',E)$}
We are using the method in \cite{bradlow-garcia-prada-gothen:2004} and we start with some lemmas needed to estimate $\chi(E',E)$.
\begin{lemma}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Let $\mathpzc{Hom}^\bullet(E',E)$ be the deformation complex of $E$ and $E'$, as in $(\ref{deformation complex of U(p,q)})$. Then the following inequalities hold.
\begin{align}
\deg(\ker(a_0)&\leq \rk(\ker(a_0))\big(\mu_\alpha(E')-\mu_\alpha(E)\big)\label{kernelH}\\
\deg(\im(a_0)&\leq\big(\rk(\mathpzc{Hom}^{1})-\rk(\im(a_0))\big)\big(\mu_\alpha(E)-\mu_\alpha(E')-\deg(L)\big)-\label{cokernelH}\\
& \alpha\big(\rk(\mathpzc{Hom}^1)-\rk(\im(a_0))
-2\rk(\coker(\phi_b))\big)+\deg(\mathpzc{Hom}^{1}).\nonumber
\end{align}
\end{lemma}
\begin{proof}
Assume that $\rk(\ker(a_0))>0$ as if it is zero then $(\ref{kernelH})$ is obvious. It follows from Proposition $\ref{Polystable}$ that the $Q$-bundle $\widetilde{E}$ is $\boldsymbol{\alpha}=(\alpha,2\alpha)$-polystable. We can define a subobject of $\widetilde{E}$ by \\\\
\[
\xymatrix{
\mathcal{K}:\mbox{ } 0\ar@{<-}@/_2pc/[r]^{}&\ker(a_0)\ar@{<-}@/_2pc/[l]_{} \ar@{>}@/_2pc/[r]^{}&0\ar@{>}@/_2pc/[l]_{}.\\
}
\]
It follows from the $\boldsymbol{\alpha}$-polystability that
\begin{eqnarray*}
\mu_\alpha(\mathcal{K})=\mu(\ker(a_0))+\alpha&\leq& \mu_{\boldsymbol{\alpha}}(\widetilde{E})=\mu_\alpha(E')-\mu_\alpha(E)+\alpha.
\end{eqnarray*}
Thus we have
$$\mu(\ker(a_0))\leq\mu_\alpha(E')-\mu_\alpha(E),$$
which is equivalent to $(\ref{kernelH})$. The second inequality is obvious when $\rk(\im(a_0))=\rk(\mathpzc{Hom}^{1})$. We thus assume $\rk(\im(a_0))<\rk(\mathpzc{Hom}^1)$. We define a quotient of the bundle $\widetilde{E}$ by
\[
\xymatrix{
\mathcal{Q:}\mbox{ }\coker(\phi_b)\otimes L^{-1}\ar@{<-}@/_1pc/[r]^{}&0\ar@{<-}@/_1pc/[l]_{} \ar@{>}@/_1pc/[r]^{}&\coker(\phi_a)\otimes L^{-1}\ar@{>}@/_1pc/[l]_{}\\
}
\]
(we take the saturation if cokernels are not torsion free). By the $\boldsymbol{\alpha}$-polystability of $\widetilde{E}$ we have
\begin{equation}\label{Q}
\mu_{\boldsymbol{\alpha}}(\mathcal{Q})=\mu(\mathcal{Q})+2\alpha\frac{ \rk(\coker(\phi_b))}{\rk(\coker(\phi_a))+\rk(\coker(\phi_b))}\geq \mu_{\boldsymbol{\alpha}}(\widetilde{E})=\mu_\alpha(E')-\mu_\alpha(E)+\alpha.
\end{equation}
Note that $\mu(\mathcal{Q})=\mu(\coker(a_0))-\deg(L)$. This and $(\ref{Q})$, together with the fact that $$\mu(\coker(a_0))\leq\frac{\deg(\mathpzc{Hom}^{1})-\deg(\im(a_0))}{\rk(\mathpzc{Hom}^{1})-\rk(\im(a_0))},$$
lead us to $(\ref{cokernelH})$.
\end{proof}
\begin{lemma}\label{bund for Euler}
Let $E=(V,W,\beta,\gamma)$ be a $L$-twisted $\U(p,q)$-Higgs bundle
and $E'=(V',W',\beta',\gamma')$ a $L$-twisted $\U(p',q')$-Higgs
bundle. Assume that $p-q$ and $p'-q'$ have the same sign, and suppose that the following conditions hold:
\begin{itemize}
\item $-\deg(L)\leq\alpha\leq\deg(L)$ and $\deg(L)\geq 2g-2$,
\item$E$ and $E'$ are $\alpha$-polystable with $\mu_\alpha(E)=\mu_\alpha(E')$,
\item the map $a_0$ is not an isomorphism.
\end{itemize}
Then $$\chi(E',E)\leq1-g,$$
if the map $a_0$ is not generically an isomorphism, otherwise $\chi(E',E)<0$.
\end{lemma}
\begin{proof}
By the estimates $(\ref{kernelH})$ and $(\ref{cokernelH})$, we
obtain
\begin{align*}
\deg(\ker(a_0))+\deg(\im(a_0))&
\leq\big(\mu_\alpha(E')-\mu_\alpha(E)\big)
\Big(\rk(\ker(a_0))+\rk(\im(a_0))-\rk(\mathpzc{Hom}^{1})\Big)\\
&\phantom{\leq}-\alpha\big(\rk(\coker(\phi_a))-\rk(\coker(\phi_a))\big)\\
&\phantom{\leq}-\deg(L)\big(\rk(\mathpzc{Hom}^1)-\rk(\im(a_0))\big)+\deg(\mathpzc{Hom}^{1}).
\end{align*}
As $\mu_\alpha(E)=\mu_\alpha(E')$ we deduce
\begin{multline*}
\deg(\mathpzc{Hom}^0)-\deg(\mathpzc{Hom}^{1})\\
\leq-\alpha\big(\rk(\coker(\phi_a))-\rk(\coker(\phi_a))\big)-\deg(L)\big(\rk\coker(\phi_a)+\rk\coker(\phi_a)\big)
\end{multline*}
and so
\begin{equation}\label{deg}
\deg(\mathpzc{Hom}^0)-\deg(\mathpzc{Hom}^{1})\leq
\begin{cases}
-\deg(L)\rk\coker(\phi_b) &\mbox{ if }-\deg(L)\leq \alpha\leq 0\\
-\deg(L)\rk\coker(\phi_a)&\mbox{ if } 0\leq\alpha\leq \deg(L).\\
\end{cases}
\end{equation}
On the other hand we have
$$\chi(E',E)=(1-g)\big(\rk(\mathpzc{Hom}^0)-\rk(\mathpzc{Hom}^1)\big)+\deg(\mathpzc{Hom}^0)-\deg(\mathpzc{Hom}^1).$$
Combining $(\ref{deg})$ with the above equality, we get
\begin{equation*}
\chi(E',E)\leq\begin{cases}
(1-g)\big(\rk(\mathpzc{Hom}^0)-\rk(\mathpzc{Hom}^1)+2\rk\coker(\phi_b)\big)&\mbox{ if }-\deg(L)\leq \alpha\leq 0\\
(1-g)\big(\rk(\mathpzc{Hom}^0)-\rk(\mathpzc{Hom}^1)+2\rk\coker(\phi_a)\big) &\mbox{ if } 0\leq\alpha\leq \deg(L).\\
\end{cases}
\end{equation*}
From hypothesis we have $\rk(\mathpzc{Hom}^0)\geq\rk(\mathpzc{Hom}^1)$. If $a_0$ is not generically an isomorphism then either cases of the above inequality implies $\chi(E',E)\leq(1-g)$. Otherwise, $$\chi(E',E)=\deg(\mathpzc{Hom}^0)-\deg(\mathpzc{Hom}^1)<0$$since equality happens only if $a_0$ is an isomorphism.
\end{proof}
\section{Birationality of moduli spaces}\label{bi}
Let $\alpha_c$, $\alpha_c^+$ and $\alpha_c^-$ be defined as in Section $\ref{Flip loci}$,where $\epsilon>0$ is small enough so that $\alpha_c$ is the only critical value in the interval $(\alpha_c^-, \alpha_c^+)$. Fix a type $t=(p,q,a,b)$.
\begin{proposition}\label{codimension}
Let $\alpha_c$ be a critical value for twisted $\U(p,q)$-Higgs bundles
of type $t=(a,b,p,q)$.
If either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $a/p-b/q>-\deg(L)$, $q\leq p$ and $0\leq\alpha_c^\pm<\frac{2pq}{pq-q^2+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(2)$] $a/p-b/q<\deg(L)$, $p\leq q$ and $\frac{2pq}{pq-p^2+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha_c^\pm\leq 0$.
\end{itemize}
Then the codimension of the flip loci
$\mathcal{S}_{\alpha_c^\pm}\subset \mathcal{M}^s_{\alpha_c^\pm}(t)$ is strictly positive.
\end{proposition}
\begin{proof}
{}From
Propositions~\ref{smooth} and \ref{smooth0},
$\mathcal{M}^s_{\alpha_c^\pm}$ is smooth of dimension $1-\chi(t,t)$.
Hence, using that by Lemma~\ref{extension}
$\chi(t,t)=\underset{1\leq i,j \leq m}{\sum}\chi(t_i,t_i)$, we have
\begin{align*}
\codim \mathcal{S}_{\alpha _c\frac{+}{}}&=\dim \mathcal{M}_{\alpha _c\frac{+}{}}^s(t)-\dim \mathcal{S}_{\alpha _c\frac{+}{}}\\
&=1-\chi(t,t)-\dim \mathcal{S}_{\alpha _c\frac{+}{}}\\
&=1-\sum_{i,j}\chi(t_i,t_j)-\dim \mathcal{S}_{\alpha _c\frac{+}{}},
\end{align*}
where $t_i$, $t_i$ and $m$ occur in
$\mathrm{Gr}(E)=\bigoplus_{i=1}^mQ_i$ coming from a $\alpha_c$-Jordan-H\"{o}lder filtration of $E$. Now using the inequality $(\ref{bund})$ we get that the codimension of the strictly semistable locus is at least
\begin{align*}
&\min\{1-\sum_{i,j}\chi(t_i,t_j)+\sum_{i\leq
j}\chi(t_j,t_i)+\frac{m(m-3)}{2}\}\\
=&\min\{-\sum_{j<i}\chi(t_j,t_i)+\frac{m(m-3)+2}{2}\},
\end{align*}
where the minimum is taken over all $t_i$ and $m$. Now we show that
$Q_i$ and $Q_j$ satisfy the hypotheses of
Lemma~\ref{bund for Euler}.
Using Proposition~\ref{surjective}, the hypotheses $(1)$ and $(2)$ imply that $\beta$ and $\gamma$
are injective, respectively. Therefore in both cases $p_j-q_j$ and
$p_i-q_i$ have the same sign, for all $i, j$. Note that there are some
$i$ and $j$ such that the map $a_0$ of the $\mathrm{Hom}$-complex
$\mathpzc{Hom}^\bullet(Q_j,Q_i)$ is not an isomorphism, since
otherwise $\mathpzc{End}^\bullet(E)$ will be an isomorphism which is
not possible. This is because for $p\neq q$ we have
$\rk(\mathpzc{End}^0)>\rk(\mathpzc{End}^1)$ which implies that the map
$a_0$ can not be an isomorphism, and for $p = q$ it can be an
isomorphism only if $\beta$ and $\gamma$ both are isomorphisms but
this is not possible since these maps are twisted with a degree
positive line bundle.
\par Hence we have that $-\chi(t_j,t_i)> 0$ and therefore
$$\codim \mathcal{S}_{\alpha _c\frac{+}{}}>\mbox{min}\{\frac{m(m-3)+2}{2}\}.$$
Clearly, the minimum is attained when $m=2$ giving the result.
\end{proof}
\begin{remark}
For $q=1$, one might have hoped to obtain a stronger result in
Proposition~\ref{codimension}, based on (1) of
Proposition~\ref{Prop:bound-for-dim}. The problem is that we also
need to satisfy the hypotheses of Lemma~\ref{bund for Euler} and
this requires injectivity of $\beta$ or $\gamma$.
\end{remark}
From Proposition~\ref{codimension} we immediately obtain the following.
\begin{theorem}\label{birational}
Fix a type $t=(p,q,a,b)$. Let $\alpha_c$ be a critical value. Suppose that either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $a/p-b/q>-\deg(L) $, $q\leq p$ and $0\leq\alpha_c^\pm<\frac{2pq}{pq-q^2+p+q}\big(b/q-a/p-\deg(L)\big)+\deg(L)$,
\item[$(2)$] $a/p-b/q<\deg(L)$, $p\leq q$ and $\frac{2pq}{pq-p^2+p+q}(b/q-a/p+\deg(L))-\deg(L)<\alpha_c^\pm\leq 0$.
\end{itemize}
Then the moduli spaces $\mathcal{M}^s_{\alpha _c^-}(t)$ and
$\mathcal{M}^s_{\alpha _c^+}(t)$ are birationally equivalent. In
particular, if either of the conditions of
Lemma~\ref{lem:no-alpha-independent-semistability} holds then the moduli spaces $\mathcal{M}_{\alpha _c^-}(t)$ and
$\mathcal{M}_{\alpha _c^+}(t)$ are birationally equivalent.
\end{theorem}
\begin{remark}
\label{rem:main-thm-range}
In view of Remark~\ref{rem:alpha-0-positive}, non-emptiness of the
intervals for $\alpha_c^\pm$ in the preceding
theorem bounds the Toledo invariant. Thus the ranges for
the Toledo invariant $\tau=\frac{2pq}{p+q}(a/p-b/q)$ for which the statement of the theorem is
meaningful are:
\begin{itemize}
\item[(1)] $-\frac{2pq}{p+q}\deg(L)<\tau<-(q-1)\deg(L))$;
\item[(2)] $(p-1)\deg(L)<\tau<\frac{2pq}{p+q}\deg(L)$.
\end{itemize}
Note that in case (1) we have $q\leq p$ and hence
$-\frac{2pq}{p+q}\deg(L)\leq -q\deg(L)$, while in case (2) we have
$p\leq q$ and hence $p\deg(L)\leq \frac{2pq}{p+q}\deg(L)$.
\end{remark}
Finally we have the following corollary.
\begin{theorem}\label{irreducible}
Let $L=K$ and fix a type $t=(p,q,a,b)$. Suppose that $(p+q,a+b)=1$ and
that $\tau=\frac{2pq}{p+q}(a/p-b/q)$ satisfies $\abs{\tau}\leq\min\{p,q\}(2g-2)$. Suppose that either one of the following conditions holds:
\begin{itemize}
\item[$(1)$] $a/p-b/q>-(2g-2) $, $q\leq p$ and $0\leq\alpha<\frac{2pq}{pq-q^2+p+q}\big(b/q-a/p-(2g-2)\big)+2g-2$,
\item[$(2)$] $a/p-b/q<2g-2$, $p\leq q$ and $\frac{2pq}{pq-p^2+p+q}(b/q-a/p+2g-2)-(2g-2)<\alpha\leq 0$.
\end{itemize}
Then the moduli space $\mathcal{M}_\alpha(t)$ is irreducible.
\end{theorem}
\begin{proof}
Recall that the value of the parameter for which the non-abelian Hodge
Theorem applies is $\alpha=0$. Thus, using \cite[Theorem~6.5]{bradlow-garcia-prada-gothen:2003}, the moduli
space $\mathcal{M}_0(t)$ is irreducible and non-empty (both the
co-primality condition and the bound on the Toledo invariant are
needed for this). Hence the result follows from Theorem~\ref{birational}.
\end{proof}
\begin{remark}
Note that unless $p=q$, the conditions on $a/b-b/q$ in the preceding theorem are
guaranteed by the hypothesis $\abs{\tau}\leq\min\{p,q\}(2g-2)$
(cf.~Remark~\ref{rem:main-thm-range}).
\end{remark}
\begin{remark}
In the non-coprime case it is known from
\cite{bradlow-garcia-prada-gothen:2003} that the
closure of the stable locus in $\mathcal{M}_0(t)$ is connected
(however, irreducibility is still an open question). Thus, in the
non-coprime case, the closure of the stable locus of
$\mathcal{M}_\alpha(t)$ is connected under the remaining
hypotheses of the preceding theorem.
\end{remark} | 46,862 |
\begin{document}
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\vskip 20pt
\begin{center}
\textcolor{red}{{\large\bf Alternating Permutations and Symmetric
Functions}}
\vskip 15pt
{\textcolor{blue}{\bf Richard P. Stanley}}\\
{\it Department of Mathematics, Massachusetts Institute of
Technology}\\
{\it Cambridge, MA 02139, USA}\\
{\texttt{[email protected]}}\\[.2in]
{\bf\small version of 21 June 2006}\\
\end{center}
\begin{abstract}
We use the theory of symmetric functions to enumerate various classes
of alternating permutations $w$ of $\{1,2,\dots,n\}$. These classes
include the following: (1) both $w$ and $w^{-1}$ are alternating, (2)
$w$ has certain special shapes, such as $(m-1,m-2,\dots,1)$, under the
RSK algorithm, (3) $w$ has a specified cycle type, and (4) $w$ has a
specified number of fixed points. We also
enumerate alternating permutations of a multiset. Most of our
formulas are umbral expressions where after expanding the expression
in powers of a variable $E$, $E^k$ is interpreted as the Euler
number $E_k$. As a small corollary, we obtain a combinatorial
interpretation of the coefficients of an asymptotic expansion
appearing in Ramanujan's Lost Notebook.
\end{abstract}
\section{Introduction.} \label{sec1}
\indent This paper can be regarded as a sequel to the classic paper
\cite{foulkes} of H. O. Foulkes in which he relates the enumeration of
alternating permutations to the representation theory of the symmetric
group and the theory of symmetric functions. We assume familiarity
with symmetric functions as presented in \cite[Ch.~7]{ec2}.
Let $\sn$ denote the symmetric group of all permutations of
$1,2,\dots,n$. A permutation $w=a_1 a_2\cdots a_n\in\sn$ is
\emph{alternating} if $a_1>a_2<a_3>a_4<\cdots$. Equivalently, write
$[m]=\{1,2,\dots,m\}$ and define the \emph{descent set} $D(w)$ of
$w\in\sn$ by
$$ D(w) = \{ i\in[n-1]\st a_i>a_{i+1}\}. $$
Then $w$ is alternating if $D(w)=\{1,3,5,\dots\}\cap [n-1]$.
Similarly, define $w$ to be \emph{reverse alternating} if
$a_1<a_2>a_3<a_4>\cdots$. Thus $w$ is reverse alternating if
$D(w)=\{2,4,6,\dots\}\cap [n-1]$. Also define the \emph{descent
composition} $\co(w)$ by
\beq \co(w) = (\alpha_1,\alpha_2,\dots,\alpha_k),
\label{eq:cow} \eeq
where $D(w)=\{\alpha_1,\alpha_1+\alpha_2,\dots,\alpha_1+\cdots+
\alpha_{k-1}\}$ and $\sum \alpha_i=n$. Thus $\alpha\in\cp(n)$, where
$\cp(n)$ denotes the set of compositions of $n$.
Let $E_n$ denote the number of alternating permutations in $\sn$. Then
$E_n$ is called an \emph{Euler number} and was shown by D. Andr\'e
\cite{andre} to satisfy
\beq \sum_{n\geq 0} E_n\frac{x^n}{n!} = \sec x + \tan x.
\label{eq:eudef} \eeq
(Sometimes one defines $\sum (-1)^nE_nx^{2n}/(2n)! = \sec x$, but we
will adhere to (\ref{eq:eudef}).) Thus $E_{2m}$ is also called a
\emph{secant number} and $E_{2m+1}$ a \emph{tangent number}. The
bijection $w\mapsto w'$ on $\sn$ defined by $w'(i)=n+1-w(i)$ shows
that $E_n$ is also the number of reverse alternating permutations in
$\sn$. However, for some of the classes of permutations considered
below, alternating and reverse alternating permutations are not
equinumerous.
Foulkes defines a certain (reducible) representation of $\sn$ whose
dimension is $E_n$. He shows how this result can be used to compute
$E_n$ and other numbers related to alternating permutations, notably
the number of $w\in\sn$ such that both $w$ and $w^{-1}$ are
alternating. Foulkes' formulas do not give a ``useful'' computational
method since they involve sums over partitions whose terms involve
Littlewood-Richardson coefficients. We show how Foulkes' results can
actually be converted into useful generating functions for computing
such numbers as (a) the number of alternating permutations $w\in\sn$
with conditions on their cycle type (or conjugacy class). The special
case of enumerating alternating involutions was first raised by
Ehrenborg and Readdy and discussed further by Zeilberger
\cite{zeil}. Another special case is that of alternating permutations
with a specified number of fixed points. Our proofs use, in addition
to Foulkes' representation, a result of Gessel and Reutenauer
\cite{g-r} on permutations with given descent set and cycle type. (b)
The number of $w\in\sn$ such that both $w$ and $w^{-1}$ are
alternating, or such that $w$ is alternating and $w^{-1}$ is reverse
alternating. (c) The number of alternating permutations of certain
shapes (under the RSK algorithm). (d) The number of alternating
permutations of a multiset of integers, under various interpretations
of the term ``alternating.''
\textsc{Acknowledgment.} I am grateful to Ira Gessel for providing
some useful background information and references and to an anonymous
referee for several helpful comments, in particular, pointing out a
gap in the proof of Corollary~\ref{cor:asy}.
\section{The work of Foulkes.} \label{sec2}
We now review the results of Foulkes that will be the basis for our
work. Given a composition $\alpha$ of $n$, let $B_\alpha$ denote the
corresponding border strip (or ribbon or skew hook) shape as defined
e.g.\ in \cite{foulkes2}\cite[p.~383]{ec2}. Let $s_{B_\alpha}$ denote
the skew Schur function of shape $B_\alpha$. The following result of
Foulkes \cite[Thm.~6.2]{foulkes2} also appears in
\cite[Cor.~7.23.8]{ec2}.
\begin{theorem} \label{thm:f1}
Let $\alpha$ and $\beta$ be compositions of $n$. Then
$$ \langle s_{B_\alpha},s_{B_\beta}\rangle = \#\{w\in\sn\st
\co(w)=\beta,\ \co(w^{-1})=\alpha\}. $$
\end{theorem}
We let $\tau_n=B_\alpha$ where $\alpha=(1,2,2,\dots,2,j)\in\cp(n)$,
where $j=1$ if $n$ is even and $j=2$ if $n$ is odd. Thus if $'$
indicates conjugation (reflection of the shape about the main
diagonal), then $\tau'_{2k+1}=\tau_{2k+1}$, while $\tau'_{2k} =
(2,2,\dots, 2)$. We want to expand the skew Schur functions
$s_{\tau_n}$ and $s_{\tau'_n}$ in terms of power sum symmetric
functions. For any skew shape $\lambda/\mu$ with $n$ squares, let
$\chi^{\lambda/\mu}$ denote the character of $\sn$ satisfying
ch$(\chi^{\lambda/\mu})=s_{\lambda/\mu}$. Thus by the definition
\cite[p.~351]{ec2} of ch we have
$$ s_{\lambda/\mu} = \sum_{\rho\vdash n} z_\rho^{-1}
\chi^{\lambda/\mu}(\rho)p_\rho, $$
where $\chi^{\lambda/\mu}(\rho)$ denotes the value of
$\chi^{\lambda/\rho}$ at any permutation $w\in\sn$ of cycle type
$\rho$.
The main result \cite[Thm.~6.1]{foulkes}\cite[Exer.~7.64]{ec2}
of Foulkes on the connection between alternating permutations and
representation theory is the following.
\begin{theorem} \label{thm:foulkes}
\emph{(a)} Let $\mu\vdash n$, where $n=2k+1$. Then
$$ \chi^{\tau_n}(\mu) = \chi^{\tau'_n}(\mu) = \left\{
\begin{array}{rl} 0, & \mbox{if $\mu$ has an even part}\\[.05in]
(-1)^{k+r}E_{2r+1}, & \mbox{if $\mu$ has $2r+1$ odd parts
and}\\ & \mbox{\ \ no even parts}. \end{array} \right. $$
\emph{(b)} Let $\mu\vdash n$, where $n=2k$. Suppose that $\mu$ has
$2r$ odd parts and $e$ even parts. Then
\beas \chi^{\tau_n}(\mu) & = & (-1)^{k+r+e}E_{2r}\\[.05in]
\chi^{\tau'_n}(\mu) & = & (-1)^{k+r}E_{2r}. \eeas
\end{theorem}
\textsc{Note.} Foulkes obtains his result from the Murnaghan-Nakayama
rule. It can also be also be obtained from the formula
$$ \sum_{n\geq 0} s_{\tau_n}t^n = \frac{1}{\sum_{n\geq 0}(-1)^n
h_{2n}t^{2n}}+\frac{\sum_{n\geq 0}h_{2n+1}t^{2n+1}}
{\sum_{n\geq 0}(-1)^mh_{2n}t^{2n}}, $$
where $s_{\tau_n}$ denotes a skew Schur function. This formula is due
to Carlitz \cite{carlitz} and is also stated at the bottom of
page~520 of \cite{ec2}.
Foulkes' result leads immediately to our main tool in what follows.
Throughout this paper we will use \emph{umbral notation} \cite{r-t}
for Euler numbers. In other words, any polynomial in $E$ is to be
expanded in terms of powers of $E$, and then $E^k$ is replaced by
$E_k$. The replacement of $E^k$ by $E_k$ is always the \emph{last} step in
the evaluation of an umbral expression. For instance,
$$ (E^2-1)^2 = E^4 -2E^2+1=E_4-2E_2+1=5-2\cdot 1+1=4. $$
Similarly,
\beas (1+t)^E & = & 1+Et+{E\choose 2}t^2+{E\choose 3}t^3+\cdots\\
& = & 1+Et+\frac 12(E^2-E)t^2+\frac 16(E^3-3E^2+2E)t^3+\cdots\\
& = & 1+Et+\frac 12(E_2-E_1)t^2+\frac 16(E_3-3E_2+2E_1)t^3+
\cdots\\ & = &
1+1\cdot t+\frac 12(1-1)t^2+\frac 16(2-3\cdot 1+2\cdot 1)t^3
+ \cdots\\ & = & 1+t+\frac 16 t^3+\cdots. \eeas
If $f=f(x_1,x_2,\dots)$ is a
symmetric function then we use the notation $f[p_1,p_2,\dots]$ for $f$
regarded as a polynomial in the power sums. For instance, if
$f=e_2=\sum_{i<j}x_i x_j=\frac 12(p_1^2-p_2)$ then
$$ e_2[E,-E,\dots]= \frac 12(E^2+E)=1. $$
\begin{theorem} \label{thm:main}
Let $f$ be a homogenous symmetric function of degree $n$. If $n$ is
odd then
\beq \langle f,s_{\tau_n}\rangle =\langle f,s_{\tau'_n}\rangle =
f[E,0,-E,0,E,0,-E,\dots] \label{eq:feo} \eeq
If $n$ is even then
\beas \langle f,s_{\tau_n}\rangle & = &
f[E,-1,-E,1,E,-1,-E,1,\dots]\\
\langle f,s_{\tau'_n}\rangle & = &
f[E,1,-E,-1,E,1,-E,-1,\dots]. \eeas
\end{theorem}
\textbf{Proof.} Suppose that $n=2k+1$. Let $\op(n)$ denote the set
of all partitions of $n$ into odd parts. If $\mu\in\op(n)$ and $\mu$
has $\ell(\mu)=2r+1$ (odd) parts, then write $r=r(\mu)$. Let
$f=\sum_{\lambda\vdash n} c_\lambda p_\lambda$. Then by
Theorem~\ref{thm:foulkes} we have
\beas \langle f,s_{\tau_n}\rangle & = & \left\langle
\sum_\lambda c_\lambda p_\lambda, \sum_{\mu\in\op(n)}z_\mu^{-1}
(-1)^{k+r(\mu)} E_{\ell(\mu)}p_\mu\right\rangle\\ & = &
\sum_{\mu\in\op(n)} c_\mu (-1)^{k+r(\mu)}E_{\ell(\mu)}. \eeas
If $\mu\in \op(n)$ and we substitute $(-1)^jE$ for $p_{2j+1}$ in
$p_\mu$ then we obtain
\beas \prod_{i=1}^{\ell(\mu)}(-1)^{\frac 12(\mu_i-1)}E & = &
(-1)^{\frac 12(2k+1-(2r(\mu)+1))}E^{\ell(\mu)}\\
& = & (-1)^{k+r(\mu)}E^{\ell(\mu)}, \eeas
and equation~(\ref{eq:feo}) follows. The case of $n$ even is
analogous. \qed
\section{Inverses of alternating permutations.}
In this section we derive generating functions for the number of
alternating permutations in $\sn$ whose inverses are alternating or
reverse alternating. This problem was considered by Foulkes
\cite[{\S}5]{foulkes}, but his answer does not lend itself to easy
computation. Such ``doubly alternating'' permutations were also
considered by Ouchterlony \cite{ouch} in the setting of pattern
avoidance. A special class of doubly alternating permutations, viz.,
those that are Baxter permutations, were enumerated by Guibert and
Linusson \cite{g-l}.
\begin{theorem} \label{thm:doubalt}
Let $f(n)$ denote the number of permutations $w\in\sn$ such that both
$w$ and $w^{-1}$ are alternating, and let $f^\ast(n)$ denote the
number of $w\in\sn$ such that $w$ is alternating and $w^{-1}$ is
reverse alternating. Let
\beas L(t) & = & \frac 12\log\frac{1+t}{1-t}\\ & = &
t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots. \eeas
Then
\bea \sum_{k\geq 0}f(2k+1)t^{2k+1} & = & \sum_{r\geq 0}E_{2r+1}^2
\frac{L(t)^{2r+1}}{(2r+1)!} \label{eq:aa1}\\[.1in]
f^\ast(2k+1) & = & f(2k+1) \label{eq:aa2}\\[.1in]
\sum_{k\geq 0}f(2k)t^{2k} & = & \frac{1}{\sqrt{1-t^2}} \sum_{r\geq
0}E_{2r}^2 \frac{L(t)^{2r}}{(2r)!} \label{eq:aa3}\\[.1in]
f^\ast(2k) & = & f(2k)-f(2k-2). \label{eq:aa4} \eea
\end{theorem}
\textbf{Proof.} By Theorem~\ref{thm:f1} we have $f(n) =\langle
s_{\tau_n}, s_{\tau_n}\rangle$. Let $n=2k+1$. Then it follows from
Theorems~\ref{thm:foulkes} and \ref{thm:main} that (writing
$r=r(\mu)$)
\bea f(n) & = & \sum_{\mu\in\op(n)}
z_\mu^{-1}(-1)^{k+r}E_{2r+1}(-1)^{k+r} E^{2r+1}
\nonumber \\ & = &
\sum_{\mu\in\op(n)} z_\mu^{-1}E_{2r+1}^2. \label{eq:fnodd} \eea
Now by standard properties of exponential generating functions
\cite[{\S}5.1]{ec2} or by specializing the basic identity
$$ \sum_\lambda z_\lambda^{-1}p_\lambda=\exp \sum_{n\geq 1}
\frac 1np_n, $$
we have
\beas \sum_{k\geq 0}\sum_{\mu\in\op(2k+1)}
z_\mu^{-1}y^{\ell(\mu)}t^{2k+1} & = &
\exp\left(y\left(t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots\right)\right)
\\ & = & \exp( yL(t)). \eeas
The coefficient of $y^{2r+1}$ in the above generating function is
therefore $L(t)^{2r+1}/(2r+1)!$, and the proof of (\ref{eq:aa1})
follows.
Since $\tau_n=\tau'_n$ for $n$ odd we have
$$ f^\ast(n) =\langle s_{\tau_n},s_{\tau'_n}\rangle =
\langle s_{\tau_n},s_{\tau_n}\rangle = f(n), $$
so (\ref{eq:aa2}) follows.
The argument for $n=2k$ is similar. For $\mu\vdash n$ let $e=e(\mu)$
denote the number of even parts of $\mu$ and $2r=2r(\mu)$ the number
of odd parts. Now the relevant formulas for computing $f(n)$ are
\beas f(n) & = & \sum_{\mu\vdash n}
z_\mu^{-1}(-1)^{k+r+e}E_{2r}(-1)^{k+r+e} E^{2r}\\ & = &
\sum_{\mu\vdash n} z_\mu^{-1}E_{2r}^2 \eeas
and
\beas \sum_{k\geq 0}\sum_{\mu\vdash n}
z_\mu^{-1}y^{2r(\mu)}t^n & = &
\exp\left(y\left(t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots\right)+
\left( \frac{t^2}{2}+\frac{t^4}{4}+\cdots\right)\right)\\
& = & (1-t^2)^{-1/2}(\exp( yL(t)), \eeas
from which (\ref{eq:aa3}) follows.
For the case $f^\ast(n)$ when $n$ is even we have
\beas f^\ast(n) & = & \sum_{\mu\vdash n}
z_\mu^{-1}(-1)^{k+r+e}E_{2r}(-1)^{k+r} E^{2r}\\ & = &
\sum_{\mu\vdash n} z_\mu^{-1}(-1)^eE_{2r}^2 \eeas
and
\beas \sum_{k\geq 0}\sum_{\mu\vdash 2k}
z_\mu^{-1}(-1)^{e(\mu)}y^{2r(\mu)}t^{2k} & = &
\exp\left(y\left(t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots
\right)\right.\\ & & \quad
-\left.\left( \frac{t^2}{2}+\frac{t^4}{4}+
\cdots\right)\right)\\[.1in]
& = & \sqrt{1-t^2}\exp yL(t). \eeas
Hence
$$ \sum_{k\geq 0}f^\ast(2k)t^{2k} =
(1-t^2)\sum_{k\geq 0}f(2k)t^{2k}, $$
from which (\ref{eq:aa4}) follows. \qed
Whenever we have explicit formulas or generating functions for
combinatorial objects we can ask for combinatorial proofs of
them. Bruce Sagan has pointed out that equation (\ref{eq:aa2}) follows
from reversing the permutation, i.e., changing $a_1 a_2 \cdots a_n$ to
$a_n \cdots a_2 a_1$. We do not know combinatorial proofs of
equations (\ref{eq:aa1}), (\ref{eq:aa3}) and (\ref{eq:aa4}). To prove
equations (\ref{eq:aa1}) and (\ref{eq:aa3}) combinatorially, we
probably need to interpret them as exponential generating functions,
e.g., write the left-hand side of (\ref{eq:aa1}) as $\sum_{k\geq 0}
(2k+1)!f(2k+1)t^{2k+1}/(2k+1)!$. Let us also note that if $g(n)$
denotes the number of reverse alternating $w\in\sn$ such that $w^{-1}$
is also reverse alternating, then $f(n)=g(n)$ for all $n$. This fact
can be easily shown using the proof method above, and it is also a
consequence of the RSK algorithm. For suppose that $w$ and $w^{-1}$
are alternating, $w\rsk (P,Q)$ and $w'\rsk (P^t,Q^t)$ (where $^t$
denotes transpose). Then by \cite[Lemma~7.23.1]{ec2} the map $w\mapsto
w'$ is a bijection between permutations $w\in\sn$ such that both $w$
and $w^{-1}$ are alternating, and permutations $w'\in\sn$ such that
both $w$ and $(w')^{-1}$ are reverse alternating. Is there a simpler
proof that $f(n)=g(n)$ avoiding RSK?
\section{Alternating tableaux of fixed shape.}
Let $T$ be a standard Young tableau (SYT). The \emph{descent set}
$D(T)$ is defined by \cite[p.~351]{ec2}
$$ D(T) = \{i\st i+1\ \mbox{is in a lower row than}\ i\}. $$
For instance, if
$$ T = \begin{array}{l}1\,2\,5\\ 3\,4\\ 6, \end{array} $$
then $D(T)=\{2,5\}$. We also define the \emph{descent composition}
$\co(T)$ in analogy with equation~(\ref{eq:cow}). A basic property of
the RSK algorithm asserts that $D(w)=D(Q)$ if $w\rsk (P,Q)$.
An SYT $T$ of size $n$ is called \emph{alternating} if
$D(T)=\{1,3,5,\dots\}\cap[n-1]$ and \emph{reverse alternating} if
$D(T)=\{2,4,6,\dots\}\cap[n-1]$. The following result is an immediate
consequence of Theorem~7.19.7 and Corollary~7.23.6 of \cite{ec2}.
\begin{theorem} \label{thm:dco}
Let $\lambda\vdash n$ and $\alpha\in\con$. Then $\langle
s_\lambda,s_{B_\alpha}\rangle$ is equal to the number of SYT of shape
$\lambda$ and descent composition $\alpha$.
\end{theorem}
Let $\alt(\lambda)$ (respectively, $\ralt(\lambda)$) denote the number
of alternating (respectively, reverse alternating) SYT of shape
$\lambda$. The following result then follows from
Theorems~\ref{thm:main} and \ref{thm:dco}.
\begin{theorem} \label{thm:shape}
Let $\lambda\vdash n$ and $\alpha\in\con$. If $n$ is odd, then
$$ \alt(\lambda) = \ralt(\lambda)
=s_\lambda[E,0,-E,0,E,0,-E,\dots]. $$
If $n$ is even then
\beas \alt(\lambda) & = & s_\lambda[E,-1,-E,1,E,-1,-E,1,\dots]\\
\ralt(\lambda) & = & s_\lambda[E,1,-E,-1,E,1,-E,-1,\dots].
\eeas
\end{theorem}
Theorem~\ref{thm:shape} ``determines'' the number of alternating SYT
of any shape $\lambda$, but the formula is not very enlightening. We
can ask whether there are special cases for which the formula can be
made more explicit. The simplest such case occurs when $\lambda$ is
the ``staircase'' $\delta_m=(m-1,m-2,\dots,1)$. For any partition
$\lambda$ write $H_\lambda$ for the product of the hook lengths of
$\lambda$ \cite[p.~373]{ec2}. For instance,
$$ H_{\delta_m}= 1^{m-1}\,3^{m-2}\,5^{m-3} \cdots (2m-3). $$
\begin{theorem} \label{thm:stair}
If $m=2k$ then
$$ \alt(\delta_m)=\ralt(\delta_m) =
E^k\prod_{j=1}^{m-2} (E^2+j^2)^{k-\lceil j/2 \rceil}. $$
If $m=2k+1$ then
$$ \alt(\delta_m)=\ralt(\delta_m) =
E^k\prod_{j=1}^{m-2} (E^2+j^2)^{k-\lfloor j/2 \rfloor}. $$
\end{theorem}
\textbf{Proof.} By the Murnaghan-Nakayama rule, $s_{\delta_m}$ is a
polynomial in the odd power sums $p_1,p_3,\dots$
\cite[Prop.~7.17.7]{ec2}. Assume that $m$ is odd. Then by the
hook-content formula \cite[Cor.~7.21.4]{ec2} we have
\bea s_{\delta_m}[E,0,E,0,\dots] & = & s_{\delta_m}[E,E,E,\dots]
\nonumber \\ & = &
\frac{E^k\prod_{j=1}^{m-2} (E^2-j^2)^{k-\lfloor j/2 \rfloor}}
{H_{\delta_m}}.
\label{eq:seee} \eea
Let $n={m\choose 2}$, and suppose that $n$ is odd, say $n=2r+1$.
Let $\lambda\in\op_n$ and $2j+1=\ell(\lambda)$. Thus
$$ p_\lambda[E,0,E,0,E,0,\dots]=E^{2j+1}. $$
A simple parity argument shows that
$$ p_\lambda[E,0,-E,0,E,0,-E,0,\dots] = (-1)^{r-j}E^{2j+1}. $$
It follows that we obtain $s_{\delta_m}[E,0,-E,0,E,0,-E,0,\dots]$ from
the polynomial expansion of $s_{\delta_m}[E,0,E,0,E,0,\dots]$ by
replacing each power $E^{2j+1}$ with $(-1)^{r-j}E^{2j+1}$. The proof
for $m$ odd and $n$ odd now follows from equation~(\ref{eq:seee}).
The argument for the remaining cases, viz., (a) $m$ odd, $n$ even, (b)
$m$ even, $n$ odd, and (c) $m$ even, $n$ even, is completely
analogous. \qed
There are some additional partitions $\lambda$ for which
$\alt(\lambda)$ and $\ralt(\lambda)$ factor nicely as polynomials in
$E$. One such case is the following.
\begin{theorem} \label{thm:opsdet}
Let $p$ be odd, and let $p\times p$ denote the partition of $p^2$
whose shape is a $p\times p$ square. Then
\beas \alt(p\times p) & = & \ralt(p\times p)\\ & = &
\frac{E^p(E^2+2^2)^{p-1}(E^2+4^2)^{p-2}\cdots (E^2+(2(p-1))^2)}
{H_{p\times p}}. \eeas
\end{theorem}
\emph{Proof} (sketch). Let $h_n$ denote the complete symmetric
function of degree $n$. From the identity
$$ \sum_{n\geq 0} h_nt^n =\exp\sum_{n\geq 1}\frac{p_nt^n}{n} $$
we obtain
\beas \sum_{n\geq 0}h_n[E,0,E,0,E,0,\dots]t^n & = &
\exp \sum_{n\odd}\frac{Et^n}{n}\\ & = &
\left(\frac{1+t}{1-t}\right)^{E/2}. \eeas
Write
$$ \left(\frac{1+t}{1-t}\right)^{E/2}= \sum_{n\geq 0} a_n(E)t^n. $$
The Jacobi-Trudi identity \cite[{\S}7.16]{ec2} implies that
$s_{p\times p} = \det(h_{p-i+j})_{i,j=1}^p$. Hence
\beq s_{p\times p}[E,0,E,0,E,0,\dots] =
\det(a_{p-i+j}(E))_{i,j=1}^p. \label{eq:oddps} \eeq
I am grateful to Christian Krattenthaler and Dennis Stanton for
evaluating the above determinant. Krattenthaler's argument is as
follows. Write
$$ \left(\frac{1+t}{1-t}\right)^{E/2} =\left( 1+\frac{2t}{1-t}
\right)^{E/2} =1+\sum_{n\geq 1}t^n\sum_{k=1}^n{n-1\choose k-1}
{E/2\choose k}2^k. $$
After substituting $k+1$ for $k$, we see that we want to compute the
Hankel determinant
$$ \det_{0\leq i,j\leq n} \left( \sum_{k=0}^{i+j}
{i+j\choose k}{E/2\choose k+1}2^{k+1}\right). $$
Now by a folklore result \cite[Lemma~15]{kratt} we conclude that this
determinant is the same as
$$ \det_{0\leq i,j\leq n} \left( {E/2\choose i+j+1}2^{i+j+1}
\right). $$
When this determinant is expanded all powers of 2 are the same, so we
are left with evaluating
$$ \det_{0\leq i,j\leq n} \left( {E/2\choose i+j+1}
\right). $$
This last determinant is well-known; see e.g.\ \cite[(3.12)]{kratt}.
\qed
Stanton has pointed out that the determinant of (\ref{eq:oddps}) is a
special case of a Hankel determinant of Meixner polynomials
$M_n(x;b,c)$, viz., $a_p(E)=2EM_{p-1}(E-1;2,-1)$. Since the Meixner
polynomials are moments of a Jacobi polynomial measure
\cite[Thm.~524]{i-s} the determinant will explicitly factor.
Neither of these two proofs of factorization is very enlightening. Is
there a more conceptual proof based on the theory of symmetric
functions?
\textsc{Note.} Permutations whose shape is a $p\times p$ square have
an alternative description as a consequence of a basic property of the
RSK algorithm \cite[Cor.~7.23.11, Thm.~7.23.17]{ec2}, viz., they are
the permutations in $\fs_{p^2}$ whose longest increasing subsequence
and longest decreasing subsequence both have length $p$.
There are some other ``special factorizations'' of $\alt(\lambda)$ and
$\ralt(\lambda)$ that appear to hold, which undoubtedly can be proved
in a manner similar to the proof of Theorem~\ref{thm:opsdet}. Some of
these cases are the following, together with those arising from the
identity $\alt(\lambda)= \alt(\lambda')$ when $|\lambda|$ is odd, and
$\alt(\lambda)=\ralt(\lambda')$ when $|\lambda|$ is even. We write
$\lambda=\langle 1^{m_1}2^{m_2}\cdots\rangle$ to indicate that
$\lambda$ has $m_i$ parts equal to $i$.
\begin{itemize}
\item $\ralt(\langle p^{p-1}\rangle)$
\item $\alt(\langle 1,p^p\rangle)$, $p$ odd
\item certain values of $\alt(b,b-1,b-2,\dots,a)$ or
$\ralt(b,b-1,b-2,\dots,a)$.
\end{itemize}
There are numerous other values of $\lambda$ for which $\alt(\lambda)$
or $\ralt(\lambda)$ ``partially factors.'' Moreover, there are similar
specializations of $s_\lambda$ which factor nicely, although they
don't correspond to values of $\alt(\lambda)$ or $\ralt(\lambda)$,
e.g., $s_{\langle p^p\rangle}[E,0,-E,0,E,0,-E,0,\dots]$ for $p$ even.
\section{Cycle type.}
A permutation $w\in\sn$ has \emph{cycle type} $\rho(w)=(\rho_1,\rho_2,
\dots)\vdash n$ if the cycle lengths of $w$ are $\rho_1,
\rho_2,\dots$. For instance, the identity permutation has cycle type
$\langle 1^n\rangle$. In this section we give an umbral formula for
the number of alternating and reverse alternating permutations
$w\in\sn$ of a fixed cycle type.
Our results are based on a theorem of Gessel-Reutenauer \cite{g-r},
which we now explain. Define a symmetric function
\beq L_n = \frac 1n\sum_{d\mid n}\mu(d)p_d^{n/d}, \label{eq:lndef}
\eeq
where $\mu$ is the number-theoretic M\"obius function.
Next define $L_{\langle m^r\rangle}=h_r[L_m]$
(plethysm). Equivalently, if $f(x)=f(x_1,x_2,\cdots)$ then write
$f(x^r) =f(x_1^r,x_2^r,\cdots)$. Then for fixed $m$ we have
\beq \sum_{r\geq 0} L_{\langle m^r\rangle}(x)t^r
= \exp \sum_{r\geq 1} \frac 1r L_m(x^r)t^r.
\label{eq:lmr} \eeq
Finally, for any partition $\lambda=\langle 1^{m_1}2^{m_2}\cdots
\rangle$ set
\beq L_\lambda=L_{\langle 1^{m_1}\rangle}
L_{\langle 2^{m_2}\rangle}\cdots. \label{eq:mult} \eeq
For some properties of the symmetric functions $L_\lambda$ see
\cite[Exer.~7.89]{ec2}.
\begin{theorem}[Gessel-Reutenauer] \label{thm:gr}
Let $\rho\vdash n$ and $\alpha\in\con$. Let $f(\rho,\alpha)$ denote
the number of permutations $w\in\sn$ satisfying $\rho=\rho(w)$ and
$\alpha= \co(w)$. Then
$$ f(\rho,\alpha) = \langle L_\rho, s_{B_\alpha}\rangle. $$
\end{theorem}
Now for $\rho\vdash n$ let $b(\rho)$ (respectively, $b^\ast(\rho))$
denote the number of alternating (respectively, reverse alternating)
permutations $w\in\sn$ of cycle type $\rho$. The following corollary
is then the special cases $B_\alpha=\tau_n$ and $B_\alpha=\tau'_n$ of
Theorem~\ref{thm:gr}.
\begin{corollary} \label{cor:brho}
We have $b(\rho) =\langle L_\rho, s_{\tau_n}\rangle$ and $b^\ast(\rho)
=\langle L_\rho, s_{\tau'_n}\rangle$.
\end{corollary}
We first consider the case when $\rho=(n)$, i.e., $w$ is an
$n$-cycle. Write $b(n)$ and $b^\ast(n)$ as short for $b((n))$ and
$b^\ast((n))$. Theorem~\ref{thm:bn} below is actually subsumed by
subsequent results (Theorems~\ref{thm:moddfm} and \ref{thm:mevenfm}),
but it seems worthwhile to state it separately.
\begin{theorem} \label{thm:bn}
\emph{(a)} If $n$ is odd then
$$ b(n) = b^\ast(n) = \frac 1n \sum_{d\mid
n}\mu(d)(-1)^{(d-1)/2}E_{n/d}. $$
\emph{(b)} If $n=2^km$ where $k\geq 1$, $m$ is odd, and $m\geq 3$,
then
$$ b(n) = b^*(n) = \frac 1n \sum_{d\mid m}\mu(d)E_{n/d}. $$
\emph{(c)} If $n=2^k$ and $k\geq 2$ then
\beq b(n) = b^\ast(n) = \frac 1n(E_n-1). \label{eq:bnc} \eeq
\emph{(d)} Finally, $b(2)=1$, $b^\ast(2)=0$.
\end{theorem}
\proof (a) By Theorem~\ref{thm:main} and Corollary~\ref{cor:brho} we
have for odd $n$ that
\beas b(n) = b^\ast(n) & = & L_n[E,0,-E,0,E,0,-E,0,\cdots]\\
& = & \frac 1n\sum_{d\mid n} \mu(d)((-1)^{(d-1)/2}E)^{n/d}\\
& = & \frac 1n \sum_{d\mid n}\mu(d)(-1)^{(d-1)/2}E_{n/d}, \eeas
since $n/d$ is odd for each $d\mid n$.
(b) Split the sum (\ref{eq:lndef}) into two parts: $d$ odd and $d$
even. Since $\mu(2d)=-\mu(d)$ when $d$ is odd and since $\mu(4d)=0$
for any $d$, we obtain
\beas b(n) & = & L_n[E,-1,-E,1,E,-1,-E,1,\cdots]\\ & = &
\frac 1n\left(\sum_{d\mid m} \mu(d)((-1)^{(d-1)/2}E)^{n/d} -
\sum_{d\mid m}\mu(d)((-1)^d)^{n/2d}\right)\\ & = &
\frac 1n\left(\sum_{d\mid m} \mu(d)E_{n/d} -
(-1)^{n/2}\sum_{d\mid m}\mu(d)\right). \eeas
The latter sum is 0 since $m>1$, and we obtain the desired formula for
$b(n)$. The argument for $b^\ast(n)$ is completely analogous; the factor
$(-1)^{n/2}$ now becomes $(-1)^{1+\frac n2}$.
(c) When $n=2^k$, $k\geq 2 $, we have
$$ L_n =\frac 1n\left( p_1^n-p_2^{n/2}\right). $$
Substituting $p_1=E$ and $p_2=\pm 1$, and using that $n/2$ is even,
yields (\ref{eq:bnc}).
(d) Trivial. It is curious that only for $n=2$ do we have $b(n)\neq
b^\ast(n)$. \qed
Note the special case of Theorem~\ref{thm:bn}(a) when $m=p^k$, where
$p$ is an odd prime and $k\geq 1$:
$$ b(p^k) =\frac{1}{p^k}\left(E_{p^k}-(-1)^{(p-1)/2}
E_{p^{k-1}}\right). $$
Is there a simple combinatorial proof, at least when $k=1$? The same
can be asked of equation~(\ref{eq:bnc}).
We next turn to the case $\lambda=\langle m^r\rangle$, i.e., all
cycles of $w$ have length $m$. Write $b(m^r)$ as short for $b(\langle
m^r\rangle)$, and similarly for $b^\ast(m^r)$. Set
\beas F_m(t) & = & \sum_{r\geq 0}b(m^r)t^r\\
F^\ast_m(t) & = & \sum_{r\geq 0}b^\ast(m^r)t^r. \eeas
First we consider the case when $m$ is odd.
\begin{theorem} \label{thm:moddfm}
\emph{(a)} Let $m$ be odd and $m\geq 3$. Then
$$ F_m(t) = F^\ast_m(t) = \exp\left[ \frac 1m
\left(\sum_{d\mid m}\mu(d)(-1)^{(d-1)/2}E^{m/d}\right)
(\tan^{-1}t)\right]. $$
\emph{(b)} We have
\beas F_1(t) & = & \sinh(E\tan^{-1} t) +
\frac{1}{\sqrt{1+t^2}}\cosh(E\tan^{-1}t)\\[.05in]
F_1^\ast(t) & = & \sinh(E\tan^{-1} t) +
\sqrt{1+t^2}\cosh(E\tan^{-1}t).
\eeas
\end{theorem}
\proof (a) By equations~(\ref{eq:feo}) and (\ref{eq:lmr}) we have that
the terms of $F_m(t)$ and $F^\ast_m(t)$ of odd degree (in $t$) are
given by
\bea \frac 12(F_m(t) -F_m(-t)) & = & \frac 12(F_m^\ast(t)-
F_m^\ast(-t))\nonumber\\ & = &
\left(\sinh \sum_{r\odd} \frac 1r
L(x^r)t^r\right)\nonumber \\
& & \ \ \left( \exp\sum_{r\even} \frac 1r
L(x^r)t^r\right)[E,0,-E,0,\dots]\nonumber\\ & = &
\left( \sinh \sum_{r\odd}\frac{t^r}{mr}\sum_{d\mid m}
\mu(d)p_{rd}^{m/d}\right)[E,0,-E,0,\dots]\\ & = &
\sinh \sum_{r\odd} \frac{t^r}{mr}\sum_{d\mid m}
\mu(d)(-1)^{(rd-1)/2}E^{m/d}\nonumber\\ & = &
\sinh \frac 1m\sum_{d\mid m} \mu(d)(-1)^{(d-1)/2}E^{m/d}
\left(t-\frac{t^3}{3}+\frac{t^5}{5}-\cdots\right)
\nonumber\\ & = &
\sinh \frac 1m\left(\sum_{d\mid m} \mu(d)(-1)^{(d-1)/2}
E^{m/d}\right) (\tan^{-1} t). \label{eq:sinh} \eea
Similarly the terms of $F_m(t)$ of even degree are given by
\beas \frac 12(F_m(t) +F_m(-t)) & = &
\left(\cosh \sum_{r\odd} \frac 1r
L(x^r)t^r\right)\\
& & \ \ \cdot\left( \exp\sum_{r\even} \frac 1r
L(x^r)t^r\right)[E,-1,-E,1,\dots]\nonumber\\ & = &
\left( \cosh \sum_{r\odd}\frac{t^r}{mr}\sum_{d\mid m}
\mu(d)p_{rd}^{m/d}\right)\\
& & \ \ \cdot\left( \exp \sum_{r\even}
\frac{t^r}{mr}\sum_{d\mid m}\mu(d)p_{rd}^{m/d}\right)
[E,-1,-E,1,\dots]\\ & = &
\left(\cosh\sum_{r\,\mathrm{odd}} \frac{t^r}{mr}\sum_{d\mid m}
\mu(d)((-1)^{(rd-1)/2})^{m/d}E^{m/d}\right) \\
& & \ \ \cdot\left(\exp\sum_{r\,\mathrm{even}}\frac{t^r}
{mr}\sum_{d\mid m}\mu(d)((-1)^{rd/2})^{m/d}\right)
\\ & = &
\left(\cosh \frac 1m\sum_{d\mid m} \mu(d)(-1)^{(d-1)/2}E^{m/d}
\tan^{-1}t\right) \\ & & \ \
\cdot\left(\exp\sum_{r\,\mathrm{even}}\frac{t^r}
{mr}(-1)^{r/2}\sum_{d\mid m}\mu(d) \right) \eeas
\beq \quad = \cosh \frac 1m\left(\sum_{d\mid m}
\mu(d)(-1)^{(d-1)/2} E^{m/d}\right)(\tan^{-1} t). \label{eq:cosh}
\eeq
Adding equations (\ref{eq:sinh}) and (\ref{eq:cosh}) yields (a) for
$F_m(t)$.
The computation for $F^\ast_m(x)$ is identical, except that the factor
$(-1)^{rd/2}$ is replaced by $(-1)^{1+rd/2}$. This alteration does not
affect the final answer.
(b) The computation of the odd part of $F_1(t)$ and $F^\ast_1(t)$ is
the same as in (a), yielding
\beas \frac 12(F_1(t) -F_1(-t)) & = & \frac 12(F_1^\ast(t)-
F_1^\ast(-t))\\ & = & \sinh (E\tan^{-1}t). \eeas
On the other hand,
\beas \frac 12(F_1(t) +F_1(-t)) & = & \cosh( E\tan^{-1}t)
\cdot\left( \exp\sum_{r\even}\frac{t^r}{r}\mu(1)(-1)^{r/2}
\right)\\ & = & \frac{\cosh(E\tan^{-1}t)}
{\sqrt{1+t^2}}, \eeas
and the proof for $F_1(t)$ follows. For $F_1^\ast(t)$ the factor
$(-1)^{r/2}$ becomes $(-1)^{1+r/2}$, so the factor $\sqrt{1+t^2}$
moves from the denominator to the numerator.
\qed
Clearly the only alternating permutation of cycle type $\langle
1^r\rangle$ is $1$ (when $r=1$). Hence from
Theorem~\ref{thm:moddfm}(b) we obtain the umbral identity
\beq \sinh(E\tan^{-1}t)+\frac{1}{\sqrt{1+t^2}}\cosh(E\tan^{-1}t)
= 1+t. \label{eq:umbralid} \eeq
One may wonder what is the point of Theorem~\ref{thm:moddfm}(b) since
$b(1^r)$ is trivial to compute directly. Its usefulness will be seen
below (Theorem~\ref{thm:cycind}), when we consider ``mixed'' cycle
types, i.e., not all cycle lengths are equal.
Theorem~\ref{thm:moddfm} can be restated ``non-umbrally'' analogously
to Theorem~\ref{thm:doubalt}. For instance, if $m=p^k$ where $p$ is
prime and $p\equiv 3\,(\mathrm{mod}\,4)$, then
$$ F_m(t) = F^\ast_m(t) = \sum_{i,j\geq 0} E_{(m/p)i+mj}
\frac{\left( \frac 1p\tan^{-1}t\right)^{i+j}}{i!\,j!}, $$
while if $p\equiv 1\,(\mathrm{mod}\,4)$, then
$$ F_m(t) = F^\ast_m(t) = \sum_{i,j\geq 0} (-1)^iE_{(m/p)i+mj}
\frac{\left( \frac 1p\tan^{-1}t\right)^{i+j}}{i!\,j!}. $$
For general odd $m$, $F_m(t)$ will be expressed as a $2^{\nu(m)}$-fold
sum, where $\nu(m)$ is the number of distinct prime divisors of $m$.
\begin{theorem} \label{thm:mevenfm}
\emph{(a)} Let $m=2^kh$, where $k\geq 1$, $h\geq 3$, and $h$ is
odd. Then
$$ F_m(t)=F_m^\ast(t) =\left( \frac{1+t}{1-t}\right)
^{\frac{1}{2m}\sum_{d\mid h}\mu(d)E^{m/d}}. $$
\emph{(b)} Let $m=2^k$ where $k\geq 2$. Then
$$ F_m(t) = F_m^\ast(t) = \left(\frac{1+t}{1-t}\right)
^{\frac{1}{2m}(E^m-1)}. $$
\emph{(c)} Let $m=2$. Then
\beas F_2(t) & = & \left(\frac{1+t}{1-t}
\right)^{(E^2+1)/4}\\[.05in] F^\ast_2(t) & = & \displaystyle
\frac{F_2(t)}{1+t}\ \ (\mathrm{compare}\ (\ref{eq:aa4})). \eeas
\end{theorem}
\proof (a) The argument is analogous to the proof of
Theorem~\ref{thm:moddfm}. We have
\beas F_m(t) & = & \left(\exp \sum_{r\geq 1} \frac 1r
L(x^r)t^r\right)[E,-1,-E,1,\dots]\\ & = &
\left( \exp \sum_{r\geq 1}\frac{t^r}{mr}\sum_{d\mid m}
\mu(d)p_{rd}^{m/d}\right)[E,-1,-E,1,\dots]\\ & = &
\exp\left( \sum_{r\,\mathrm{odd}}\frac{t^r}{rm}
\sum_{d\mid h} ((-1)^{(rd-1)/2})^{m/d}\mu(d)E^{m/d}\right.\\
& & \ + \left.\sum_{r\,\mathrm{even}}\frac{t^r}{rm}
\sum_{d\mid h} ((-1)^{rd/2})^{m/d} -\sum_r\frac{t^r}{rm}
\sum_{d\mid h} ((-1)^{rd/2})^{m/d}\right)\\ & = &
\exp \sum_{r\,\mathrm{odd}}\frac{t^r}{rm}\sum_{d\mid h}
\mu(d)E^{m/d}\\ & = & \exp \left(\frac 1m\sum_{d\mid h}
\mu(d)E^{m/d}\right)\frac 12\log\frac{1+t}{1-t}, \eeas
and the proof follows for $F_m(t)$. The same argument holds for
$F^\ast_m(t)$ since $-1$ was always raised to an even power or was
multiplied by a factor $\sum_{d\mid h}\mu(d)=0$ in the proof.
(b) We now have
\beas F_m(t) & = & \exp \sum_{r\geq 1}\frac{t^r}{rm}\left(
p_r^m-p_{2r}^{m/2}\right)[E,-1,-E,1,\dots]\\ & = &
\exp \frac 1m\left(\sum_{r\,\mathrm{odd}}\frac{t^r}{r}
\left( ((-1)^{(r-1)/2}E)^m-(-1)^{rm/2}\right)\right.\\ & & \ +
\left.\sum_{r\,\mathrm{even}}\frac{t^r}{r}\left((-1)^{rm/2}
-(-1)^{rm/2}\right)\right)\\ & = &
\left(\frac{1-t}{1+t}\right)^{1/2m}\exp \frac{E^m}{2m}
\log\frac{1+t}{1-t}, \eeas
etc. Again the computation for $F^\ast_m(t)$ is the same.
(c) We have
\beas F_2(t) & = & \exp \frac 12\sum_{r\geq 1}\left(p_r^2-p_{2r}
\right)\frac{t^r}{r}[E,-1,-E,1,\dots]\\ & = &
\exp \frac 12\left[ \sum_{r\,\mathrm{odd}}\left(
\left( (-1)^{(r-1)/2}E\right)^2-(-1)^r\right)
\frac{t^r}{r}\right.\\ & & \ +\left. \sum_{r\,\mathrm{even}}
\left((-1)^r-(-1)^r\right)\right]\\
& = & \exp \frac 12\sum_{r\,\mathrm{odd}}(E^2+1)\frac{t^r}{r}.
\eeas
etc. We leave the case $F_2^\ast(t)$ to the reader. \qed
The expansion of $F_2(t)$ begins
\beq F_2(t) = 1+t+t^2+2t^3+5t^4+17t^5+72t^6+367t^7+2179t^8+\cdots.
\label{eq:ram} \eeq
Ramanujan asserts in Entry 16 of his second notebook (see
\cite[p.~545]{berndt}) that as $t$ tends to $0+$,
\beq 2\sum_{n\geq 0}(-1)^n\left( \frac{1-t}{1+t}\right)^{n(n+1)}
\sim 1+t+t^2+2t^3+5t^4+17t^5+\cdots. \label{eq:berndt} \eeq
Berndt \cite[(16.6)]{berndt} obtains a formula for the complete
asymptotic expansion of $2\sum_{n\geq 0}(-1)^n\left(
\frac{1-t}{1+t}\right)^{n(n+1)}$ as $t\rightarrow 0+$. It is easy to
see that Berndt's formula can be written as $\left(\frac{1+t}{1-t}
\right)^{(E^2+1)/4}$ and is thus equal to
$F_2(t)$. Theorem~\ref{thm:mevenfm}(c) therefore answers a question of
Galway \cite[p.~111]{galway}, who asks for a combinatorial
interpretation of the coefficients in Ramanujan's asymptotic
expansion.
\textsc{Note.} The following formula for $F_2(t)$ follows from
equation (\ref{eq:berndt}) and an identity of Ramanujan proved by
Andrews \cite[(6.3)$_\mathrm{R}$]{andrews}:
$$ F_2(t) = 2\sum_{n\geq 0}q^n\frac{\prod_{j=1}^n(1-q^{2j-1})}
{\prod_{j=1}^{2n+1}(1+q^j)}, $$
where $q=\left(\frac{1-t}{1+t}\right)^{2/3}$. It is not hard to see
that this is a \emph{formal} identity, unlike the asymptotic identity
(\ref{eq:berndt}).
\textsc{Note.} We can put Theorems~\ref{thm:moddfm}(a) into a form
more similar to Theorem~\ref{thm:mevenfm}(a) by noting the identity
$$ \exp (\tan^{-1} t) = \left( \frac{1-it}{1+it}\right)^{i/2}. $$
Hence when $m$ is odd and $m\geq 3$ we have
$$ F_m(t)=F^\ast_m(t) =\left( \frac{1-it}{1+it}\right)^
{\frac{i}{2m}
\sum_{d\mid m}\mu(d)(-1)^{(d-1)/2}E^{m/d}}. $$
The multiplicativity property (\ref{eq:mult}) of $L_\lambda$ allows us
write down a generating function for the number $b(\lambda)$
(respectively, $b^\ast(\lambda)$) of alternating (respectively,
reverse alternating) permutations of any cycle type $\lambda$. For
this purpose, let $t_1,t_2,\dots$ and $t$ be indeterminates and set
$\deg(t_i)=i$, $\deg(t)=1$. If $F(t_1,t_2,\dots)$ is a power series in
$t_1,t_2,\dots$ or $F(t)$ is a power series in $t$, then write
${\cal O}F$ (respectively, ${\cal E}F$) for those terms of $F$ whose
total degree is odd (respectively, even). For instance,
$$ {\cal O}F(t_1,t_2,\dots) = \frac 12\left(F(t_1,t_2,t_3,t_4\dots)-
F(-t_1,t_2,-t_3,t_4,\dots)\right). $$
Define the ``cycle indicators''
\beas Z(t_1,t_2,\dots) & = & \sum_{\lambda=\langle 1^{m_1}2^{m_2}
\cdots\rangle} b(\lambda) t_1^{m_1} t_2^{m_2}\cdots\\
Z^\ast(t_1,t_2,\dots) & = & \sum_{\lambda=\langle 1^{m_1}2^{m_2}
\cdots\rangle} b^\ast(\lambda) t_1^{m_1} t_2^{m_2}\cdots, \eeas
where both sums range over all partitions $\lambda$ of all integers
$n\geq 0$.
\begin{theorem} \label{thm:cycind}
We have
$$ \hspace{-2in}{\cal O}Z(t_1,t_2,\dots) =
{\cal O}Z^\ast(t_1,t_2,\dots) $$
\beq =
{\cal O}\exp
(E\tan^{-1}t_1)\cdot\left(\frac{1+t_2}{1-t_2}\right)^{E^2/4}
F_3(t_3)F_4(t_4)\cdots \label{eq:zodd} \eeq
\beas {\cal E}Z(t_1,t_2,\dots) & = & {\cal E}
\frac{\exp(E\tan^{-1}t_1)}{\sqrt{1+t_1^2}}
\left(\frac{1+t_2}{1-t_2}\right)^{(E^2+1)/4}
F_3(t_3)F_4(t_4)\cdots \\[.1in]
{\cal E}Z^\ast(t_1,t_2,\dots) & = & {\cal E}
\sqrt{1+t_1^2}\,\exp(E\tan^{-1}t_1)\cdot
\frac{1}{1+t_2}\left(\frac{1+t_2}{1-t_2}\right)^{(E^2+1)/4}\\
& & \ \ \cdot F_3(t_3)F_4(t_4)\cdots. \eeas
It is understood that in these formulas $F_j(t_j)$ is to be written
in the umbral form given by Theorems~\ref{thm:moddfm} and
\ref{thm:mevenfm}.
\end{theorem}
\proof Let
\beq G_m(t) = \exp \sum_{r\geq 1} \frac 1r L_m(x^r)t^r.
\label{eq:gmt} \eeq
It follows from equations (\ref{eq:lmr}) and (\ref{eq:mult})
that
$$ {\cal O}Z(t_1,t_2,\dots) = {\cal O}\prod_{m\geq 1}
G_m(t_m)[E,0,-E,0,\dots]. $$
The proofs of Theorems~\ref{thm:moddfm}(a) and \ref{thm:mevenfm}(a,b)
show that for $m\geq 3$,
$$ G_m(t)[E,0,-E,0,\dots] = G_m(t)[E,-1,-E,1,\dots]. $$
Hence we obtain the factors $F_3(t_3)F_4(t_4)\cdots$ in
equation~(\ref{eq:zodd}). It is straightforward to compute
$G_m(t_m)[E,0,-E,0,\dots]$ (and is implicit in
the proofs of Theorems~\ref{thm:moddfm}(b) and \ref{thm:mevenfm}(c))
for $m=1,2$. For instance,
\bea G_1(t_1)[E,0,-E,0\dots] & = & \exp\left(\sum_{r\geq 1} \frac 1r
p_rt_1^r\right)[E,0,-E,0,\dots] \nonumber\\ & = &
\exp \sum_{r\odd}\frac 1r(t_1-\frac 13t_1^3+\cdots) \nonumber\\
& = & \exp(\tan^{-1} t_1). \label{eq:eatt} \eea
Thus we obtain the remaining factors in equation~(\ref{eq:zodd}). The
remaining formulas are proved analogously. \qed
We mentioned in Section~\ref{sec1} that Ehrenborg and Readdy raised
the question of counting alternating involutions $w\in S_n$. An answer
to this question is a simple consequence of Theorem~\ref{thm:cycind}.
\begin{corollary}
Let $c(n)$ (repectively, $c^\ast(n)$) denote the number of alternating
(respectively, reverse alternating) involutions $w\in\sn$. Then
\beas \sum_{n\geq 0} c(2n+1)t^{2n+1} & = & \sinh(E\tan^{-1}t)\cdot
\left(\frac{1+t^2}{1-t^2}\right)^{E^2/4}\\[.05in]
\sum_{n\geq 0} c(2n)t^{2n} & = & \frac{1}{\sqrt[4]{1-t^4}}\,
\cosh(E\tan^{-1}t)\cdot
\left(\frac{1+t^2}{1-t^2}\right)^{E^2/4}\\[.05in]
c^\ast(n) & = & c(n). \eeas
Equivalently,
\beas \sum_{n\geq 0} c(2n+1)t^{2n+1} & = &
\sum_{i,j\geq 0} \frac{E_{2i+2j+1}}{(2i+1)!\,j!\,4^j}
\tan^{-1}(t)^{2i+1} \left(\log \frac{1+t^2}
{1-t^2}\right)^j \\[.05in]
\sum_{n\geq 0} c(2n)t^{2n} & = & \frac{1}{\sqrt[4]{1-t^4}}\,
\sum_{i,j\geq 0} \frac{E_{2i+2j}}{(2i)!\,j!\,4^j}
\tan^{-1}(t)^{2i} \left(\log \frac{1+t^2}
{1-t^2}\right)^j. \eeas
\end{corollary}
\proof
We have
\beas \sum_{n\geq 0}c(2n+1)t^{2n+1} & = & {\cal O}
Z(t,t^2,0,0,\dots) \\
\sum_{n\geq 0}c(2n)t^{2n} & = & {\cal E}
Z(t,t^2,0,0,\dots), \eeas
and similarly for $c^\ast(n)$. The result is thus a special case of
Theorem~\ref{thm:cycind}. \qed
The identity $c(n)=c^\ast(n)$ does not seem obvious. It can also be
obtained using properties of the RSK algorithm, analogous to the
argument after the proof of Theorem~\ref{thm:doubalt} Namely,
$w\in\sn$ is an alternating (respectively, reverse alternating)
involution if and only if $w\rsk(P,P)$, where $P$ is an alternating
(respectively, reverse alternating) SYT. Hence if $w'\rsk (P^t,P^t)$,
then the map $w\mapsto w'$ interchanges alternating involutions
$w\in\sn$ with reverse alternating involutions $w'\in\sn$.
\section{Fixed points.}
P. Diaconis (private communication) raised the question of enumerating
alternating permutations by their number of fixed points. It is easy
to answer this question using Theorem~\ref{thm:cycind}. Write $d_k(n)$
(respectively, $d^*_k(n)$) for the number of alternating
(respectively, reverse alternating) permutations in $\sn$ with $k$
fixed points. Write ${\cal O}_t$ and ${\cal E}_t$ for the odd and
even part of a power series with respect to $t$ (ignoring other
variables), i.e., ${\cal O}_tF(t)=\frac 12(F(t)-F(-t))$ and ${\cal
E}_tF(t)=\frac 12 (F(t)+F(-t))$.
\begin{proposition} \label{prop:fixed}
We have
\bea \sum_{k,n\geq 0} d_k(2n+1)q^k t^{2n+1} & = & {\cal O}_t
\frac{\exp(E(\tan^{-1}qt-\tan^{-1}t))}{1-Et} \label{eq:dkodd}\\
d^*_k(2n+1) & = & d_k(2n+1) \label{eq:dksodd} \\
\sum_{k,n\geq 0} d_k(2n)q^k t^{2n} & = & {\cal E}_t
\sqrt{\frac{1+t^2}{1+q^2t^2}}
\frac{\exp(E(\tan^{-1}qt-\tan^{-1}t))}{1-Et} \nonumber \\
\sum_{k,n\geq 0} d_k^*(2n)q^k t^{2n} & = & {\cal E}_t
\sqrt{\frac{1+q^2t^2}{1+t^2}}
\frac{\exp(E(\tan^{-1}qt-\tan^{-1}t))}{1-Et} \nonumber.
\eea
Equivalently, we have the nonumbral formulas
\beas \sum_{k,n\geq 0} d_k(2n+1)q^k t^{2n+1} & = &
\sum_{\twoline{i,j\geq 0}{i\not\equiv j\,(\mathrm{mod}\,2)}}
\frac{E_{i+j}}{j!}t^i(\tan^{-1}qt-\tan^{-1}t)^j\\
\sum_{k,n\geq 0} d_k(2n)q^k t^{2n} & = &
\sqrt{\frac{1+t^2}{1+q^2t^2}}
\sum_{\twoline{i,j\geq 0}{i\equiv j\,(\mathrm{mod}\,2)}}
\frac{E_{i+j}}{j!}t^i(\tan^{-1}qt-\tan^{-1}t)^j\\
\sum_{k,n\geq 0} d_k^*(2n)q^k t^{2n} & = &
\sqrt{\frac{1+q^2t^2}{1+t^2}}
\sum_{\twoline{i,j\geq 0}{i\equiv j\,(\mathrm{mod}\,2)}}
\frac{E_{i+j}}{j!}t^i(\tan^{-1}qt-\tan^{-1}t)^j.
\eeas
\end{proposition}
\proof
It is not hard to see (e.g., \cite[(1)]{schocker}) that
$$ \sum_{\lambda\vdash n} L_\lambda = p_1^n, $$
where $p_1=x_1+x_2+\cdots$. It follows from equations ({\ref{eq:lmr}),
(\ref{eq:mult}) and (\ref{eq:gmt}) that
$$ G_1(t)G_2(t)\cdots = \sum_{n\geq 0} p_1^nt^n =
\frac{1}{1-p_1t}. $$
Hence by equation (\ref{eq:zodd}) we have
\beas \sum_{k,n\geq 0}d_k(2n+1)q^kt^{2n+1} & = &
\sum_{k,n\geq 0}d_k^*(2n+1)q^kt^{2n+1}\\ & = &
{\cal O}_t \exp(E \at qt)\left(\frac{1+t}{1-t}\right)^{E^2/4}
F_3(t)F_4(t)\cdots\\ & = &
{\cal O}_t \frac{\exp(E \at qt)}{\exp(E\at t)\cdot(1-Et)}, \eeas
proving (\ref{eq:dkodd}) and
(\ref{eq:dksodd}). The proof for $n$ even is analogous.
\qed
\begin{corollary}
For $n>1$ we have $d_0(n)=d_1(n)$ and $d_0^*(n)=d_1^*(n)$.
\end{corollary}
\proof Let
$$ M(q,t) = {\cal O}_t \frac{\exp
E(\tan^{-1}qt-\tan^{-1}t)}{1-Et}. $$
By equation~(\ref{eq:dksodd}) it follows that
\beas \sum_{n\,\mathrm{odd}} d_0(n)t^n & = &
M(0,t)\\
\sum_{n\,\mathrm{odd}} d_1(n)t^n & = &
\left.\frac{\partial}{\partial q} M(q,t)\right|_{q=0}. \eeas
It is straightforward to compute that
$$ \left.\frac{\partial}{\partial q}
M(q,t)\right|_{q=0}-M(0,t) = \sinh(E\tan^{-1} t). $$
By equation~(\ref{eq:umbralid}) we have $\sinh(E\tan^{-1} t)=t$, and
the proof follows for $n$ odd. The proof for $n$ even is completely
analogous.
\qed
We have a conjecture about certain values of $d_k(n)$ and
$d_k^*(n)$. It is not hard to see that
\beas \max\{k\st d_k(n)\neq 0\} & = & \lceil n/2\rceil,\quad n\geq
4\\
\max\{k\st d_k^*(n)\neq 0\} & = & \lceil (n+1)/2\rceil,\quad
n\geq 5.
\eeas
\begin{conjecture}
Let $D_n$ denote the number of derangements (permutations
without fixed points) in $\sn$. Then
\beas d_{\lceil n/2\rceil}(n) & = & D_{\lfloor n/2\rfloor},
\quad n\geq 4\\
d_{\lceil (n+1)/2\rceil}^*(n) & = & D_{\lfloor (n-1)/2\rfloor},
\quad n\geq 5. \eeas
\end{conjecture}
It is also possible to obtain asymptotic information from
Proposition~\ref{prop:fixed}. The next result considers alternating
or reverse alternating derangements (permutations without fixed
points).
\begin{corollary} \label{cor:asy}
\rm{(a)} We have for $n$ odd the asymptotic expansion
\bea d_0(n) & \sim & \frac 1e\left( E_n + a_1 E_{n-2} + a_2
E_{n-4}+\cdots\right) \label{eq:asymodd}\\ & = & \frac 1e\left(
E_n+\frac 13 E_{n-2}-
\frac{13}{90}E_{n-4}+\frac{467}{5760}E_{n-6}+\cdots\right),
\nonumber \eea
where
$$ \sum_{k\geq 0}a_k x^{2k} = \exp\left( 1-\frac 1x \tan^{-1}x
\right). $$
\rm{(b)} We have for $n$ even the asymptotic expansion
\bea d_0(n) & \sim & \frac 1e\left( E_n + b_1 E_{n-2} + b_2
E_{n-4}+\cdots\right) \label{eq:asymeven}\\ & = & \frac 1e\left(
E_n+\frac 56 E_{n-2}-
\frac{37}{360}E_{n-4}+\frac{281}{9072}E_{n-6}+\cdots\right),
\nonumber \eea
where
$$ \sum_{k\geq 0}b_k x^{2k} = \sqrt{1+x^2}\exp\left( 1-\frac 1x
\tan^{-1}x \right). $$
\rm{(c)} We have for $n$ even the asymptotic expansion
\bea d_0^*(n) & \sim & \frac 1e\left( E_n + c_1 E_{n-2} + c_2
E_{n-4}+\cdots\right) \label{eq:asymevens}\\ & = & \frac 1e\left(
E_n-\frac 16 E_{n-2}+
\frac{23}{360}E_{n-4}-\frac{1493}{45360}E_{n-6}+\cdots\right),
\nonumber \eea
where
$$ \sum_{k\geq 0}c_k x^{2k+1} = \frac{1}{\sqrt{1+x^2}}\exp\left(
1-\frac 1x \tan^{-1}x \right). $$
\end{corollary}
\textsc{Note.} Equations~(\ref{eq:asymodd}), (\ref{eq:asymeven}), and
(\ref{eq:asymevens}) are genuine asymptotic expansions since $E_m \sim
2(2/\pi)^{m+1}m!$,
so for fixed $k$,
$$ E_{n-k} \sim 2\left( \frac{\pi}{2}\right)^k\frac{1}{n^k}E_n $$
as $n\rightarrow\infty$. In fact, since
$$ E_m = 2\left(\frac{2}{\pi}\right)^{m+1}m!(1+O(3^{-m})), $$
we can rewrite (\ref{eq:asymodd}) (and similarly (\ref{eq:asymeven}) and
(\ref{eq:asymevens})) as
$$ d_0(n) \sim \frac{E_n}{e}\left( 1+a_1
\left(\frac{\pi}{2}\right)^2 \frac{1}{(n)_2} +a_2
\left(\frac{\pi}{2}\right)^4 \frac{1}{(n)_4}+\cdots\right), $$
where $(n)_j=n(n-1)\cdots(n-j+1)$.
\emph{Proof of Corollary~\ref{cor:asy}.} (a) It follows from
equation~(\ref{eq:dkodd}) that
$$ \sum_{n\,\mathrm{odd}} d_0(n)t^n = {\cal O}_t
\frac{\exp(-E\tan^{-1}t)}{1-Et}. $$
This series has the form
$$ \sum_{n\,\mathrm{odd}} t^n(a_{n0}E^n+a_{n1} E^{n-2} + a_{n2}
E^{n-4}+\cdots). $$
If we replace $t$ with $Et$ and $E$ with $1/E$ we therefore obtain
\beq {\cal O}_t \frac{\exp(-E^{-1}\tan^{-1}tE))}{1-t} =
\sum_{n\,\mathrm{odd}} t^n(a_{n0}+a_{n1} E^2 + a_{n2}
E^4+\cdots). \label{eq:asymp1} \eeq
We claim that for fixed $j$ the coefficients $a_{nj}$ rapidly approach
(finite) limits as $n\rightarrow \infty$. If we expand the left-hand
side of (\ref{eq:asymp1}) as a power series in $E$, it is not hard to
see that the coefficient of $E^{2j}$ has the form $Q_j(t)/(1-t^2)$,
where $Q_j(t)$ is a polynomial in $t,e^t$ and $e^{-t}$. Hence
the coefficient of $t^{2n+1}$ in $Q_j(t)$ has the form
$p_j(n)/(2n+1)!$ for some polynomial $p_j(n)$. It follows that
$$ a_{nj} = Q_j(1)+o(n^{-r}) $$
for all $r>0$. Now
$$ {\cal O}_t \frac{\exp(-E^{-1}\tan^{-1}tE))}{1-t} =
\frac{(1+t)e^{-\frac 1E\tan^{-1}tE}-(1-t)e^{\frac 1E
\tan^{-1} tE}}{2(1-t^2)}. $$
Multiplying by $1-t^2$ and setting $t=1$ gives $e^{-\frac 1E \tan^{-1}
E}$, and the proof follows. The argument for (b) and (c) is
analogous.
\qed
\section{Multisets.}
In this section we give simple umbral formulas for the number of
alternating and reverse alternating permutations of a multiset of
positive integers, with various interpretations of the meaning of
``alternating.'' There has been some previous work on alternating
multiset permutations. Goulden and Jackson \cite[Exer.~4.2.2(b);
solution, pp.~459--460]{g-j} obtain a formula for the number of
alternating permutations of the multiset with one occurrence of $i$
for $1\leq i\leq m$ and two occurrences of $i$ for $m+1\leq i\leq
m+n$. Gessel \cite[pp.~265--266]{gessel} extends this result to
multisets with one, two, or three multiplicities of each part, or with
one or four multiplicities of each part. Upon being told about the
results in this section, Gessel (private communication) was able to
extend his argument to arbitrary multisets, obtaining a result
equivalent to the case $A=\emptyset$ of Theorem~\ref{thm:multi}. Zeng
\cite{zeng} obtains an even more general result concerning the case
$A=\emptyset$.
Our basic tool, in addition to Theorem~\ref{thm:main}, is the
following extension of Theorem~\ref{thm:dco} to skew shapes
$\lambda/\mu$. We define the descent composition of an SYT $T$ of
shape $\lambda/\mu$ exactly as for ordinary shapes, viz., $T$ has
descent composition $\alpha=(\alpha_1,\dots,\alpha_k)$ if $\{\alpha_1,
\alpha_1+\alpha_2, \dots, \alpha_1+\cdots+\alpha_{k-1}\}$ is the set
of those $i$ for which $i+1$ appears in $T$ in a lower row than $i$.
\begin{lemma} \label{lemma:jdt}
Let $\lambda/\mu$ be a skew partition of size $n$, with corresponding
skew Schur function $s_{\lambda/\mu}$ \cite[Def.~7.10.1]{ec2}, and let
$\alpha\in\con$. Then $\langle s_{\lambda/\mu} ,s_{B_\alpha}\rangle$
is equal to the number of SYT of shape $\lambda/\mu$ and descent
composition $\alpha$.
\end{lemma}
\proof Let $s_{\lambda/\mu} =\sum_\nu c^\lambda_{\mu\nu} s_\nu$.
Let $T$ be an SYT of shape $\lambda/\mu$, and apply jeu
de taquin \cite[{\S}A1.2]{ec2} to $T$ to obtain an SYT $T'$ of some
ordinary shape $\nu$. Two fundamental properties of jeu de
taquin assert the following:
\begin{itemize}
\item As $T$ runs over all SYT of shape $\lambda/\mu$,
we obtain by jeu de taquin each SYT $T'$ of shape $\nu$
exactly $c^\lambda_{\mu\nu}$ times.
\item We have $\co(T)=\co(T')$.
\end{itemize}
The first item above appears e.g.\ in \cite[Thm.~A1.3.1]{ec2}, while
the second item is easily proved by showing that the descent
composition is preserved by a single jeu de taquin slide. The proof
of the lemma follows immediately from the two items above. \qed
We can define $\alt(\lm)$ and $\ralt(\lm)$ for skew shapes $\lm$
exactly as we did for ordinary shapes $\lambda$. The following
corollary is then immediate from Theorem~\ref{thm:main} and
Lemma~\ref{lemma:jdt}.
\begin{corollary} \label{cor:skewalt}
Let $\lm$ be a skew shape of odd size $|\lm|$. Then
$$ \alt(\lm) = \ralt(\lm) = s_{\lm}[E,0,-E,0,E,0,-E,0,\dots]. $$
If $|\lm|$ is even then
\beas \alt(\lm) & = & s_{\lm}[E,-1,-E,1,E,-1,-E,1,\dots]\\
\ralt(\lm) & = & s_{\lm}[E,1,-E,-1,E,1,-E,-1,\dots]. \eeas
\end{corollary}
We are now ready to enumerate alternating permutations of a
multiset. If two equal elements $i$ in a permutation appear
consecutively, then we need to decide whether they form an ascent or a
descent. We can make this decision separately for each $i$. Let $k\geq
1$, and let $A,B$ be complementary subsets of $[k]$, i.e.,
$A\cup B=[k]$, $A\cap B=\emptyset$. Let
$\alpha=(\alpha_1,\dots,\alpha_k)$ be a composition of some $n\geq 1$
into $k$ parts. An $\alpha$-\emph{permutation} of $[k]$ is a
permutation of the multiset
$M=\{1^{\alpha_1},\dots,k^{\alpha_k}\}$, i.e., a sequence $a_1 a_2
\cdots a_n$ with $\alpha_i$ occurrences of $i$, for $1\leq i\leq
k$. An $\alpha$-permutation is said to be $(A,B)$-\emph{alternating}
if
$$ a_1>a_2<a_3>a_4<\cdots a_n, $$
where we define $j>j$ if $j\in A$ and $j<j$ if $j\in B$. For instance,
if $A=\{1,3\}$, $B=\{2,4\}$, and $\alpha=(3,2,2,3)$, then the
$\alpha$-permutation $w=1142214343$ is $(A,B)$-alternating since
$$ 1>1<4>2<2>1<4>3<4>3 $$
according to our definition. Similarly we define \emph{reverse
$(A,B)$-alternating}. For example, 2213341414 is a reverse $(A,B)$
$\alpha$-permutation (with $\alpha,A,B$ as before), since
$$ 2<2>1<3>3<4>1<4>1<4. $$
Let $N(\alpha,A,B)$ (respectively, $N^*(\alpha,A,B)$) denote the
number of $(A,B)$-alternating (respectively, reverse $(A,B)$-alternating)
$\alpha$-permutations. Write $e_i$ and $h_i$ for the elementary
and complete symmetric functions of degree $i$.
\begin{theorem} \label{thm:multi}
Let $\alpha=(\alpha_1,\dots,\alpha_k)\in\con$, and let $A,B$ be
complementary subsets of $[k]$. \\
\indent \emph{(a)} If $n$ is odd, then
\beas N(\alpha,A,B) & = & N^*(\alpha,A,B)\\
& = & \prod_{i\in A}e_{\alpha_i}\cdot\prod_{j\in B}h_{\alpha_j}
[E,0,-E,0,E,0,-E,0,\dots]. \eeas
\indent \emph{(b)} If $n$ is even, then
\beas N(\alpha,A,B) & = &
\prod_{i\in A}e_{\alpha_i}\cdot\prod_{j\in B}h_{\alpha_j}
[E,-1,-E,1,E,-1,-E,1,\dots]\\[.1in]
N^*(\alpha,A,B) & = &
\prod_{i\in A}e_{\alpha_i}\cdot\prod_{j\in B}h_{\alpha_j}
[E,1,-E,-1,E,1,-E,-1,\dots]\\[.1in]
\eeas
\end{theorem}
\proof Let $\sigma=\sigma(\alpha,A,B)$ be the skew
shape consisting of a disjoint union of single rows and columns, as
follows. There are $k$ connected components, of sizes
$\alpha_1,\dots,\alpha_k$ from top to bottom. If $i\in A$ then the
$i$th component is a single row, and otherwise a single column. For
instance, $\sigma((3,1,2,2),\{2,4\},\{1,3\})$ and
$\sigma((3,1,2,2),\{4\},\{1,2,3\})$ both have the following diagram:
\vspace{.5em}
\centerline{\psfig{figure=skew.eps}}
\vspace{.5em}
Suppose that $n$ is odd. By Corollary~\ref{cor:skewalt} we have
$$ \alt(\sigma)=\ralt(\sigma) = s_\sigma[E,0,-E,0,\dots]. $$
Given an alternating or reverse alternationg SYT $T$ of shape
$\sigma$, define an $\alpha$-permutation $w=a_1\cdots a_n$ by the
condition that $a_i=j$ if $a_i$ appears in the $j$th component of
$\sigma$. For instance, if
$$ \sigma = \begin{array}{cccc} & & & \!\!\!4\,8\,10\,12\\ & & 2\!\!\!\\
& & 3\!\!\!\\ & & \!11\\ & \!5\,6\,9\\ \!1\\ \!7 \end{array}, $$
then $w=422133413121$. This construction sets up a bijection between
alternating (respectively, reverse alternating) SYT of shape $\sigma$
and $(A,B)$-alternating (respectively, reverse alternating)
$\alpha$-permutations, so the proof follows for $n$ odd. Exactly the
same argument works for $n$ even. \qed
Some values of the relevant specializations of $e_i$ and $h_i$ are as
follows:
$$ \begin{array}{rclcl} e_1[E,0,-E,0,\dots] & = &
h_1[E,0,-E,0,\dots] & = & E\\[.05in]
e_2[E,0,-E,0,\dots] & = & h_2[E,0,-E,0,\dots] &
= & \frac 12 E^2\\[.05in]
e_3[E,0,-E,0,\dots] & = & h_3[E,0,-E,0,\dots] &
= & \frac 16(E^3-2E)\\[.05in]
e_4[E,0,-E,0,\dots] & = & h_4[E,0,-E,0,\dots] &
= & \frac{1}{24}(E^4-8E^2)\\[.05in]
e_5[E,0,-E,0,\dots] & = & h_5[E,0,-E,0,\dots] &
= & \frac{1}{120}(E^5-20E^3+24E) \end{array} $$
$$ \hspace{-.3in}e_1[E,-1,-E,1,\dots]=e_1[E,1,-E,-1,\dots] $$
\vspace{-.2in}
$$ \qquad\qquad =h_1[E,-1,-E,1,\dots]=h_1[E,1,-E,-1,\dots]=E
$$
$$ \begin{array}{rclcl} e_2[E,-1,-E,1,\dots] & = &
h_2[E,1,-E,-1,\dots] & = & \frac 12(E^2+1)\\[.05in]
e_2[E,1,-E,-1,\dots] & = &
h_2[E,-1,-E,1,\dots] & = & \frac 12(E^2-1)\\[.05in]
e_3[E,-1,-E,1,\dots] & = &
h_3[E,1,-E,-1,\dots] & = & \frac 16(E^3+E)\\[.05in]
e_3[E,1,-E,-1,\dots] & = &
h_3[E,-1,-E,1,\dots] & = & \frac 16(E^3-5E)\\[.05in]
e_4[E,-1,-E,1,\dots] & = &
h_4[E,1,-E,-1,\dots] & = & \frac{1}{24}(E^4-2E^2-3)\\[.05in]
e_4[E,1,-E,-1,\dots] & = &
h_4[E,-1,-E,-1,\dots] & = & \frac{1}{24}(E^4-7E^2+9)\\[.05in]
e_5[E,-1,-E,1,\dots] & = &
h_5[E,1,-E,-1,\dots] & = & \frac{1}{120}(E^5-10E^3-11)\\[.05in]
e_5[E,1,-E,-1,\dots] & = &
h_5[E,-1,-E,1,\dots] & = & \frac{1}{120}(E^5-30E^4+89).
\end{array} $$
\indent
It is easy to see (see equations~(\ref{eq:enhn1}), (\ref{eq:enhn2}),
(\ref{eq:enhn3}) below) that for all $i$ we have
\beas e_i[E,0,-E,0,\dots] & = & h_i[E,0,-E,0,\dots]\\
e_i[E,-1,-E,1,\dots] & = & h_i[E,1,-E,-1,\dots]\\
e_i[E,1,-E,-1,\dots] & = & h_i[E,-1,-E,1,\dots]
\eeas
These formulas, together with Theorem~\ref{thm:multi} and the
commutativity of the ring of symmetric functions, yield some results
about the equality of certain values of $N(\alpha,A,B)$. For instance,
if $n$ is odd, then $N(\alpha,A,B)$ depends only on the multiset of
parts of $\alpha$, not on their order, and also not on $A$ and $B$. If
$n$ is even, then $N(\alpha,A,B)$ depends only on the multiset of
parts of $\alpha$ and on which submultiset of these parts index the
elements of $A$ and $B$.
The specialization of $e_i$ and $h_i$ for small $i$ lead to some
nonumbral formulas for certain values of $N(\alpha,A,B)$. For
instance, let $k$ be odd, $\alpha=(3^k)$ (i.e., $k$ parts equal to 3),
$A=\emptyset$, so that $N((3^k),\emptyset,[k])$ is the number of
alternating permutations $a_1>a_2\leq a_3>a_4\leq a_5>\cdots\leq
a_{3k}$ (where $>$ and $\leq$ have their usual meaning) of the
multiset $\{1^3,2^3,\dots,k^3\}$. Then
\beas N((3^k),\emptyset,[k]) & = & h_3^k[E,0,-E,0,\dots]\\
& = & \frac{1}{6^k}E^k(E^2-2)^k\\ & = &
\frac{1}{6^k}\sum_{j=0}^k{k\choose j}(-2)^{k-j}E_{2j+k}. \eeas
In the same way we obtain the formulas in \cite[pp.~265--266]{gessel}.
It is easy to find generating functions for the specializations of
$e_n$ and $h_n$ that we are considering, using the identities
\beas \sum_{n\geq 0} e_nt^n & = & \exp \sum_{j\geq 1} (-1)^{j-1}
\frac{p_j}{j}\\
\sum_{n\geq 0} h_nt^n & = & \exp \sum_{j\geq 1}
\frac{p_j}{j}. \eeas
Namely,
\bea \sum_{n\geq 0}e_n[E,0,-E,0,\dots]t^n & = & \sum_{n\geq
0}h_n[E,0,-E,0,\dots]t^n\nonumber\\ & = &
\exp E\tan^{-1}t \label{eq:enhn1}\\[.1in]
\sum_{n\geq 0}e_n[E,1,-E,-1,\dots]t^n & = &
\sum_{n\geq
0}h_n[E,-1,-E,1,\dots]t^n \nonumber \\ & = &
\displaystyle\frac{1}{\sqrt{1+t^2}}
\exp E\tan^{-1}t \label{eq:enhn2}\\[.1in]
\sum_{n\geq 0}e_n[E,-1,-E,1,\dots]t^n & = & \sum_{n\geq
0}h_n[E,1,-E,-1,\dots] t^n\nonumber\\ & = & \sqrt{1+t^2}
\exp E\tan^{-1}t \label{eq:enhn3} \eea
Equation~(\ref{eq:enhn1}) in fact is a restatement of (\ref{eq:eatt}).
\pagebreak | 141,303 |
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A shop manager, who was awarded £1,000 after six years of voluntarily delivering thousands of hot meals to elderly and disabled people, has donated his prize money to help the charity continue its work.
Sérgio Marcelino, who manages the St. Mary and St. Peter Co-Op En Route stores, has been a dedicated volunteer for 'Meals on Wheels' since 2013.
The charity has been operating in Jersey since 1962 and delivers between 80 and 90 meals a day, four days a week.
Pictured: Sérgio has been volunteering for Meals On Wheels for six years.
They buy all the food, which is cooked at Jersey General Hospital by hospital staff, at cost.
Last year, Sérgio delivered 435 meals in his own time, often covering 17 miles around the island on each ‘round’.
He uses his own car to make his deliveries, covering the cost of his travel as well as offering to undertake extra shifts to cover volunteer shortages and holidays.
Pictured: Sérgio with one of the Meals on Wheels recipients, Michael Horman.
Mr Marcelino said: "A lot of the people who rely on Meals on Wheels are elderly and similar ages to those I serve in En Route. I lived with my grandparents for a long time and know how important it is to look after your elders.
"People write it on their calendars when they know I am coming. They see someone who is younger than them and for many it’s something they look forward to."
Sérgio's commitment was recognised earlier this year at the annual Co-op Training and Recognition Awards, which celebrates employees' contributions to the local community. He was presented with £1,000 by the Society after winning the Society’s Ambassador of the Year Award.
Pictured: Sérgio at the annual Co-op Training and Recognition Awards.
But rather than spend the money on himself, Sérgio has decided to give it to the charity. "I was overwhelmed to be able to present Meals on Wheels with £1,000," Sérgio said. "I wasn’t expecting that from the Co-op and I am totally surprised."
Hilary Grant, chairman of Meals on Wheels, said the money will be put towards the general running costs of the charity. Mrs Grant, who has been involved with the charity for 35 years, added: "I can remember taking Sérgio out on his first round and I knew from that day he was going to be fine.
"Most of us who volunteer either have a part time job or have retired from work. I admire people who can juggle delivering for us with their full-time job."
Pictured: Sérgio was described as an "unsung hero" by Colin Macleod.
Mark Cox, the Co-op CEO, said: "It is an honour to both recognise Sérgio for bringing the Society’s values to life by serving the community and awarding Meals on Wheels with £1,000. At the Co-op we are always eager to support charities but what makes this extra special is that it’s a charity so close to one of our colleague’s hearts.
"Sérgio not only dedicates his own time to Meals on Wheels, but he knows all those he delivers to by name and takes a genuine interest in their lives."
Picture Credit: Callum Thorne. | 292,258 |
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Found 13220 Items Tagged "Karim Rashid"
On View: Dance Headdress (Ci-wara Kun)
These headdresses, called ci-wara, represent antelopes, important animals in Bamana philosophy. The antelope’s power is a metaphor for...
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Hiroshige's One Hundred Famous Views of Edo
Hiroshige's 118 woodblock landscape and genre scenes of mid-nineteenth-century Tokyo, is one of the greatest achievements of Japanese art.
On View: Vase with Three Handles
Some of the finest works of New Kingdom glass were made under Akhenaten, perhaps under the inspiration of Asiatic glassmakers living | 358,067 |
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TITLE: Showing every even number can be factored as a product of E-primes
QUESTION [1 upvotes]: Note this problem is about the investigation of the E-Zone.
Part A: Describe all E-primes.
Answer: E-primes are greater than and equal to 1 and its only E-zone factors are 1 and itself (similar to any primes). Also it is not divisible by 4, which is the same as saying 2 times an odd number.
Part B: Show that every even number can be factored as a product of E-primes.
So my book, which is Joseph Silverman's A Friendly Introduction to Number Theory, states that this proof is just a mimic of the proof of this fact for ordinary numbers, however I was wondering if there was another way of proving this statement?
REPLY [3 votes]: Each even number $n$ is of the form $n=2^ku$ where $u$ is odd and $k\ge 1$. If $k=1$ then $n$ is already an E-prime. Otherwise let the first E-prime be $2u$ and all others be $2$ as many as needed.
Note this method relies on a small part of unique factorization, and it is not the same as the way suggested to solve it in the Silverman text.
Note: For part A you may want to be more explicit and say an E-prime is any number of the form $2u$ with $u$ odd. Also it may or may not need a sign, depending on whether the E-zone includes negative integers.
Elaboration of part B argument: First, the statement that
Every even number $n$ is of the form $2^ku$ with $u$ odd and $k\ge 1$
can be shown by induction. Base case $n=2=2\cdot 1$ is of the desired form with $k=1,u=1$. Now suppose $n>2$ is even so that $n=2k$ for some $k>1$. If $k$ is odd then we have the desired form $n=2^1u$ where $u=k$. If $k$ is even, then $k=2k'$ and then $n=2\cdot(2k').$ At this point $2k'$ is an even number less than $n$ so by the inductive hypotheses $2k'=2^tu$ with $t\ge 1$ and $u$ odd. Putting this back into $n=2 \cdot 2k'$ gives $n=2^{t+1}u$ which is of the desired form, finishing the inductive step.
Now that's done, and you want to show every even number $n$ is a product of E-primes, where these are all numbers of the form $2u$ with $u$ odd (from part A). Note first that $2=2\cdot 1$ qualifies as an E-prime using $u=1$. Since $n$ is even we can use the above shown fact and write $n=2^ku$ with $k\ge 1$ and $u$ odd. There are two cases:
case 1: $k=1$. In this case $n=2^1u=2u$ with $u$ odd, so $n$ is itself an E-prime. It is thus the product of a single E-prime and the result holds in this case.
case 2: $k>1$. Here we can write $k=1+r$ and then
$$n=2^ku=(2u)\cdot 2 \cdot 2 \cdots \cdot 2,$$
where there are $r$ copies of the E-prime $2$ being multiplied together after the initial E-prime $(2u)$. So in case 2 also, we have $n$ as a product of E-primes.
I hope this helps explain part B... | 55,841 |
(*
Title: Code_Matrix.thy
Author: Jose Divasón <jose.divasonm at unirioja.es>
Author: Jesús Aransay <jesus-maria.aransay at unirioja.es>
*)
section\<open>Code generation for vectors and matrices\<close>
theory Code_Matrix
imports
Rank_Nullity_Theorem.Miscellaneous
Code_Set
begin
text\<open>In this file the code generator is set up properly to allow the execution of matrices
represented as funcions over finite types.\<close>
lemmas vec.vec_nth_inverse[code abstype]
lemma [code abstract]: "vec_nth 0 = (%x. 0)" by (metis zero_index)
lemma [code abstract]: "vec_nth 1 = (%x. 1)" by (metis one_index)
lemma [code abstract]: "vec_nth (a + b) = (%i. a$i + b$i)" by (metis vector_add_component)
lemma [code abstract]: "vec_nth (a - b) = (%i. a$i - b$i)" by (metis vector_minus_component)
lemma [code abstract]: "vec_nth (vec n) = (\<lambda>i. n)" unfolding vec_def by fastforce
lemma [code abstract]: "vec_nth (a * b) = (%i. a$i * b$i)" unfolding vector_mult_component by auto
lemma [code abstract]: "vec_nth (c *s x) = (\<lambda>i. c * (x$i))" unfolding vector_scalar_mult_def by auto
lemma [code abstract]: "vec_nth (a - b) = (%i. a$i - b$i)" by (metis vector_minus_component)
definition mat_mult_row
where "mat_mult_row m m' f = vec_lambda (%c. sum (%k. ((m$f)$k) * ((m'$k)$c)) (UNIV :: 'n::finite set))"
lemma mat_mult_row_code [code abstract]:
"vec_nth (mat_mult_row m m' f) = (%c. sum (%k. ((m$f)$k) * ((m'$k)$c)) (UNIV :: 'n::finite set))"
by(simp add: mat_mult_row_def fun_eq_iff)
lemma mat_mult [code abstract]: "vec_nth (m ** m') = mat_mult_row m m'"
unfolding matrix_matrix_mult_def mat_mult_row_def[abs_def]
using vec_lambda_beta by auto
lemma matrix_vector_mult_code [code abstract]:
"vec_nth (A *v x) = (%i. (\<Sum>j\<in>UNIV. A $ i $ j * x $ j))" unfolding matrix_vector_mult_def by fastforce
lemma vector_matrix_mult_code [code abstract]:
"vec_nth (x v* A) = (%j. (\<Sum>i\<in>UNIV. A $ i $ j * x $ i))" unfolding vector_matrix_mult_def by fastforce
definition mat_row
where "mat_row k i = vec_lambda (%j. if i = j then k else 0)"
lemma mat_row_code [code abstract]:
"vec_nth (mat_row k i) = (%j. if i = j then k else 0)" unfolding mat_row_def by auto
lemma [code abstract]: "vec_nth (mat k) = mat_row k"
unfolding mat_def unfolding mat_row_def[abs_def] by auto
definition transpose_row
where "transpose_row A i = vec_lambda (%j. A $ j $ i)"
lemma transpose_row_code [code abstract]:
"vec_nth (transpose_row A i) = (%j. A $ j $ i)" by (metis transpose_row_def vec_lambda_beta)
lemma transpose_code[code abstract]:
"vec_nth (transpose A) = transpose_row A"
by (auto simp: transpose_def transpose_row_def)
lemma [code abstract]: "vec_nth (row i A) = (($) (A $ i))" unfolding row_def by fastforce
lemma [code abstract]: "vec_nth (column j A) = (%i. A $ i $ j)" unfolding column_def by fastforce
definition "rowvector_row v i = vec_lambda (%j. (v$j))"
lemma rowvector_row_code [code abstract]:
"vec_nth (rowvector_row v i) = (%j. (v$j))" unfolding rowvector_row_def by auto
lemma [code abstract]: "vec_nth (rowvector v) = rowvector_row v"
unfolding rowvector_def unfolding rowvector_row_def[abs_def] by auto
definition "columnvector_row v i = vec_lambda (%j. (v$i))"
lemma columnvector_row_code [code abstract]:
"vec_nth (columnvector_row v i) = (%j. (v$i))" unfolding columnvector_row_def by auto
lemma [code abstract]: "vec_nth (columnvector v) = columnvector_row v"
unfolding columnvector_def unfolding columnvector_row_def[abs_def] by auto
end
| 33,114 |
By Tutor and Life Coach Muriel McClymont
I’ve just watched a most interesting TED talk, ‘The Art of Stress Free Productivity’. TED stands for Technology, Entertainment and Design and is a not for profit organisation that aims to spread 'Ideas Worth Sharing'. If you have not yet explored this resource, dip in a toe.
(One of the most amazing TED talks I ever watched is Jill Bolte Taylor talking about how it feels to have a stroke - fascinating insights into our creative brains.)
Stress at this time of year comes in all shapes and sizes. Some people feel stressed just now because they are not going to be with their family for Christmas others because they are going to be with their family for Christmas!
Freelances can be stressed because many of them will have very little work from about now until mid January, which can leave a substantial dent in the coffers. Some will have more work than they feel they can handle.
Whatever the causes, how can we deal with it?
David Allen in ‘The Art of Stress Free Productivity’ offers a deceptively simple three step plan.
- carry a notebook with you and write everything down as it comes into your head
- decide outcomes and then a first action step for every area of your life
- create an overall map of all areas of your life on one sheet of paper
At first I thought this was too simple, then I tried it out. I got a bit carried away with the writing down all the things I was holding in my head and what became my brain dump list got a bit scary in itself. The funny thing was though that I did feel lighter from putting it all down on a piece of paper.
I then grouped my list into specific key areas - work, family, commitments etc. and considered what I wanted to happen short, medium and long term in each area. Then I created a new to do list with specific first step actions to achieve all this.
What was interesting was that at this point many things just dropped off my list because, once I started to think in terms of outcomes, things either became lumped in with others or simply got dropped as no longer relevant.
Then I produced my overall map. I did this by producing a mind map with me in the middle and lines coming off for each area of my life, with sub branches coming off that, which I colour coded it to make it even clearer to read. If you prefer lists, you could make as many headings as you need, and do lists underneath - again you could colour code this to make it more striking visually. If you like drawing, you could make a picture of your life using symbols or pictures that represent all the areas that are important for you. How you create this map doesn’t matter as long as it is meaningful to you.
I have to say this was the most powerful step because I now have that map in my head and this has helped me make better decisions. When I am feeling particularly pressured in one area, I can see at a glance of my map, how any action I take will impact everything else, so, I can avoid being purely reactive and can manage issues in a more proactive, constructive and productive way. All without any angst ridden worry or stress!
I love to find new things that work and before I use it on clients, I like to try it out and work out why it works. This is my take on it:
Writing it Down
The reason this is effective is that once we write things down, not only do we give ourselves space to think, without trying to keep 30 plates spinning in our heads, but we give ourselves an alternative perspective.
By which I mean, when we hold all the information in our head, some of it we will see as vivid pictures, some will be words we hear, many things will have feelings attached. Putting it all down on paper lets you take a step back from any emotions, puts everything on the same footing and gives you some distance to think and a new viewpoint to think from.
Decide Outcomes and First Action Steps
Deciding on the outcome and identifying actions works because it gives us a positive direction, something that no amount of stressing or mulling over a problem, or trying to decide what to do next, can. Nothing is more stress inducing than sitting reviewing a list of problems and getting increasingly overwhelmed by their volume or complexity. Focusing exclusively on the problem gets you nowhere new.
I particularly liked the point made in the video that lack of time is not the problem but a symptom. Einstein, Mozart, and every other genius you might like to mention, only ever had 24 hours in a day!
Creating an Overall Map
Finally, making an overall map of everything in your life that you need to pay attention to allows you to put your whole life and what is important to you in one place which you then have a visual representation for, so you can see each component in context and prioritise effectively.
More information and questions
I highly recommend that you take 20 minutes to watch The Art of Stress Free Productivity and try out these steps for yourself. You may discover something amazing. I’d love to know how you get on. So, let me know or ask questions at our ‘Freelance Challenges’ forum.
For additional tips David Allen provides access to a great range of free articles, see this link.
Wishing you a very merry and stress free Christmas. | 178,581 |
\begin{document}
\date{\today \ {\em AEI ref: AEI-2013-161}}
\begin{abstract}
In this paper we analyze Hertz potentials for free massless spin-$s$ fields
on the Minkowski spacetime, with data in weighted Sobolev spaces.
We prove existence and pointwise estimates for the Hertz potentials using a
weighted estimate for the wave
equation. This is then applied to give
weighted estimates for the
solutions of the spin-$s$ field equations, for arbitrary half-integer $s$. In
particular, the peeling properties of the free massless spin-$s$ fields are
analyzed for initial data in weighted
Sobolev spaces with arbitrary, non-integer weights.
\end{abstract}
\maketitle
\section{Introduction}
The analysis by Christodoulou and Klainerman of the decay of massless fields
of spins 1 and 2 on Minkowski space \cite{Christodoulou:1990dd} served as an
important preliminary for their proof of the non-linear stability of
Minkowski space \cite{Christodoulou:1993vma}. The method used in
\cite{Christodoulou:1990dd} was based on energy estimates using the vector
fields method, see \cite{Klainerman:1985wn}.
This approach was extended to fields of arbitrary spin by Shu
\cite{Shu:1991ta}. The approach of \cite{Christodoulou:1993vma} to the problem of
nonlinear stability of Minkowski space was later extended by Klainerman and
Nicolo \cite{2003CQGra..20.3215K}
to
give the full peeling behavior for the Weyl tensor
at null infinity.
The vector fields method makes use of the conformal symmetries of Minkowski
space to derive conservation laws for higher order energies, which then
via the Klainerman Sobolev inequality \cite{Klainerman:1987hz}
give pointwise estimates for the solution of the wave equation. An analogous
procedure is used for the higher spin fields in the papers cited above.
This procedure gives pointwise
decay estimates for the solution of the Cauchy problem of the wave equation and
the spin-$s$ equation, for
initial data of one particular falloff at spatial infinity. The conditions on the initial data
originate in the
growth properties of the conformal Killing vector fields on Minkowski space, which are used in the energy estimates.
Let $H^j_\delta$ be the weighted $L^2$ Sobolev spaces on $\R^3$, that is to say the space of functions $\phi$ for which
$$
\sum_{k=0}^j\Vert \lAngle r\rAngle^{-\left(3/2 + \delta\right) +k}D^k \phi \Vert_{L^2(\mathbb{R}^3)}^2<\infty, \quad \text{ where }\lAngle r\rAngle = (1+r^2)^{1/2}.
$$ We use the
conventions\footnote{The spaces $H^j_\delta$ are in \cite{Bartnik:1986dq}
denoted by
$W^{j,2}_{\delta}$.}
of Bartnik \cite{Bartnik:1986dq}.
Since we shall use the 2-spinor formalism, as defined in Section \ref{sec:conventions},
we work here and throughout the paper on Minkowski space with signature ${+}\,{-}\,{-}\,{-}$.
Consider the Cauchy problem for
the wave equation
\begin{align}
\square \phi &= 0, \label{eq:wave-intro} \\
\phi \big{|}_{t=0} = f\in H^j_{-3/2}, \quad & \partial_t \phi \big{|}_{t=0} = g\in H^{j-1}_{-5/2}.\nonumber
\end{align}
Then, for $j \geq 2$,
one has the
estimate
\cite{Klainerman:1985wn}
\begin{equation}\label{eq:CKwave}
|\phi(x,t)|
\leq C\lAngle u \rAngle^{-1/2} \lAngle v \rAngle^{-1}
(\Vert f\Vert_{j,-3/2} + \Vert g\Vert_{j-1,-5/2}),
\end{equation}
where $\lAngle u\rAngle = (1 + u^2)^{1/2}$, $u=t-r$ and $v=t+r$.
On the other hand, if one considers the wave equation \eqref{eq:wave-intro}
on the flat 3+1 dimensional Minkowski spacetime
as a special case of the conformally covariant form of the
wave equation
$$
(\nabla^a \nabla_a + R/6) \phi = 0,
$$
the condition on the initial data which is compatible with regular conformal
compactification is
\begin{equation}\label{eq:confdata}
\partial^\ell f = \mathcal{O}(r^{-2-\ell}), \quad \partial^{\ell} g = \mathcal{O}(r^{-3-\ell}).
\end{equation}
Making use of standard energy estimates in the conformal compactification
of Minkowski space one arrives, after undoing the conformal compactification,
at
\begin{equation}\label{eq:confdecay}
|\phi(x,t)| = \mathcal{O}\left (\lAngle u \rAngle^{-1} \lAngle v \rAngle^{-1} \right )
\end{equation}
see the discussion in \cite[\S 6.7]{MR1466700}.
In particular, there is an extra $r^{-1/2}$ falloff in the condition
\eqref{eq:confdata} on the initial data compared to \eqref{eq:wave-intro} as well as an
additional factor $\lAngle u \rAngle^{-1/2}$ decay
in the retarded time coordinate $u$ in \eqref{eq:confdecay}
compared to \eqref{eq:CKwave}.
Let us now consider the case of higher spin fields. Let $2s$ be a positive
integer and let $\phi_{A\dots F}$ be a totally symmetric spinor field of
spin-$s$, i.e with $2s$ indices.
The Cauchy problem for a massless
spin-$s$ field is
\begin{align*}
\nabla_{A'}{}^A \phi_{A \dots F} &= 0 , \\
\phi_{A \dots F} \big{|}_{t=0} &= \varphi_{A \dots F} .
\end{align*}
For $s \geq 1$, the Cauchy datum $\varphi_{A\dots F}$ must satisfy the constraint equation
$$
D^{AB} \varphi_{AB\dots F} = 0,
$$
where $D_{AB}$ is the intrinsic space spinor derivative on $\Sigma$, see
Section~\ref{sec:conventions}. The spin $1/2$ case does not have constraints.
One of the main differences in the asymptotic behavior between a massless
scalar field satisfying a wave equation and a massless higher spin field is
the existence of a hierarchy of decay rates for the different null components
of the field along the outgoing null directions. This property, known as peeling, was first pointed out
by Sachs in 1961 \cite{Sachs:1961tw}.
Let $o_A, \iota_A$ be a spin dyad aligned with the outgoing and ingoing null
directions $\partial_v, \partial_u$, and let $\phi_{\bm{i}}$ be the scalars
of $\phi_{A\dots F}$ defined by
$$
\phi_{\bm{i}} = \phi_{A_1 \dots A_{\bm{i}} B_{\bm{i}+1} \dots B_{2s}} \iota^{A_1} \cdots
\iota^{A_{\bm{i}}}
o^{B_{\bm{i} +1}} \cdots o^{B_{2s}}.
$$
One says that $\phi_{A\dots F}$ satisfies the peeling property if the
components $\phi_{\bm{i}}$ satisfy
$$
\phi_{\bm{i}} = {\mathcal O} ( r^{\bm{i} - 2s -1} ) ,
$$ along
the outgoing null geodesics with affine parameter $r$.
In \cite{MR0175590}, Penrose gave two arguments for peeling of
massless fields on Minkowski space. The first, cf.
\cite[\S 4]{MR0175590}, makes use of a
representation of the field in terms of a Hertz
potential of order $2s$, \emph{i.e.} the field is written as a
derivative of order $2s$ of a potential satisfying a wave equation.
Penrose assumed that the Hertz potential
decays at a specific rate along outgoing
null rays.
He then infered the peeling property from this decay assumption.
The second approach presented by Penrose, cf. \cite[\S 13]{MR0175590},
is based on the just mentioned fact together with the conformal invariance of
the spin-$s$ field equation. Solving the Cauchy problem in the conformally
compactified picture, as was discussed for the wave equation in \cite[\S
6.7]{MR1466700}, and taking into account the effect of the conformal
rescaling, one recovers the peeling property for the solution of the massless
spin-$s$ equation on Minkowski space.
Based on this analysis, Penrose conjectured that the peeling of massless fields at null infinity should be a generic property of asymptotically simple space-times.
The estimate proved in \cite{Christodoulou:1990dd} for the spin-1 or Maxwell
field can be stated in the present notation as
$$
|\phi_{\bm{i}}(t,x)| \leq C \lAngle u \rAngle^{1/2 - \bm{i}} \lAngle v
\rAngle^{\bm{i}-3} \Vert \varphi_{AB} \Vert_{j,-5/2}, \quad
\text{ for } \bm{i}= 1,2,\quad j\geq 2,
$$
while for the component $\phi_0$ one has
$$
|\phi_0(t,x)| \leq C r^{-5/2} \Vert \varphi_{AB} \Vert_{j,-5/2},
$$
along outgoing null rays. Thus, this result does not give the peeling
property for all components of $\phi_{AB}$,
which is due to the fact that the norm $\Vert \varphi_{AB}
\Vert_{j,-5/2}$ is not compatible with the conformal compactification of
Minkowski space. Similarly for the spin-2 case, the result in
\cite{Christodoulou:1990dd} gives peeling for $\phi_{\bm{i}}$, $\bm{i}
= 2,3,4$ for initial data in $H^j_{-7/2}$,
while peeling fails to hold for $\phi_{\bm{i}}$, $\bm{i} =
0,1$.
On the other hand, the condition on the initial datum which is compatible with
a regular conformal compactification, and also gives peeling, is for a spin-$s$ field
\begin{equation}\label{eq:confcond}
\partial^\ell \varphi_{A\dots F} = \mathcal{O}(r^{-2s-2-\ell}).
\end{equation}
In this paper we shall follow an approach outlined by Penrose in
\cite[\S 6]{MR0175590}
to give a weighted decay estimate for spin-$s$ fields of arbitrary, half-integer
spin-$s$. The result proved here admits conditions on the initial data which
include the ones
considered in \cite{Christodoulou:1990dd,Shu:1991ta}, as well as
conditions which
are compatible with peeling, but also general weights.
The results of this paper clarify
the relation between the condition on the initial
datum and the peeling property of the solution of the spin-$s$ field equation.
In this paper we shall make use of some estimates for elliptic equations in
weighted Sobolev spaces, and for technical reasons these are not compatible
with the integer
powers or $r$ as in \eqref{eq:confcond}.
The method we shall use is based on the notion of Hertz potentials.
The reader can refer for background to Stewart \cite{stewart:1979}, Fayos \emph{et al.}
\cite{fayos:llanta:llosa:1985} Benn \emph{et al.} \cite{benn:charlton:kress:1997} and references therein. Since Minkowski space is
topologically trivial, there is no obstruction to representing
a Maxwell field on Minkowski space in terms of a Hertz
potential. However, this general fact does not provide estimates for the
potential. In this
paper we prove the necessary estimates not only for the Maxwell field but for
fields with general half-integer spins.
To introduce the method we here consider the spin-1 case, \emph{i.e.} the
Maxwell field on 3+1 dimensional Minkowski space.
With our choice of signature, the metric on the spatial slices is negative definite.
The Maxwell field, in the absence of sources, is a real differential 2-form $F_{ab}$
which is closed and
divergence free.
For convenience we consider the complex anti-self-dual form
$$
\FF_{ab} = F_{ab} + i * F_{ab},
$$
which corresponds to a symmetric 2-spinor $\phi_{AB}$ via
\begin{equation} \label{eq:F=phi}
\FF_{ab} = \phi_{AB} \eps_{A'B'} .
\end{equation}
In terms of $\FF_{ab}$, the Maxwell equation
is simply
\begin{equation}\label{eq:maxwellinforms}
(\ud \FF)_{abc} = 0.
\end{equation}
\newcommand{\ii}{\bm{i}}
\newcommand{\norm}{\xi}
Let $\norm^a = (\partial_t)^a$
be the unit normal to the Cauchy surface $\Sigma = \{ t = 0\}$.
Given a complex 1-form $\EE_a$ on $\Sigma$,
satisfying the Maxwell constraint equation
\begin{equation}\label{eq:maxwellconstr}
\ud^*\EE=0,
\end{equation}
there is a unique solution of the Maxwell equation such that
$$
(\FF_{ab} \norm^b)\big{|}_{\Sigma} = \EE_a.
$$
Now, let $\HH_{ab}$ be a self-dual
2-form which solves the wave equation
\begin{equation}\label{eq:H-wave}
\square
\HH_{ab} = 0,
\end{equation}
where
$\square = \ud \coder + \coder \ud$
is the Hodge wave operator, and
$\coder = * \ud *$ is the exterior co-derivative in dimension 4.
Defining the form $\FF_{ab}$ by
\begin{equation}\label{eq:F=H}
\FF_{ab} = \ud \coder \HH_{ab} ,
\end{equation}
we have using \eqref{eq:H-wave} that $\FF_{ab}$ is anti-self-dual, and solves the
Maxwell equation. The form $\HH_{ab}$
is called a Hertz-potential for $\FF_{ab}$. Since
we are working on Minkowski space, the wave equation \eqref{eq:H-wave} is
just a collection of scalar wave equations for the components of $\HH_{ab}$, and
hence the solution to \eqref{eq:H-wave} for given Cauchy data can be analyzed
using results for the scalar wave equation. Thus, if we are able to relate
the Cauchy data for the Maxwell field $\FF_{ab}$
to the Cauchy data for $\HH_{ab}$, we
may use the Hertz potential construction to prove estimates for the solution
of the Maxwell field equation, starting from estimates for the wave
equation.
Let the complex
1-form $\KK_a$ be the ``electric field'' corresponding to $\HH_{ab}$,
$$
\KK_a = \HH_{ab} \norm^b.
$$
A calculation shows that if $\FF_{ab}$ is defined in terms of $\HH_{ab}$ by
\eqref{eq:F=H}, the Cauchy data for \eqref{eq:H-wave}
is related to the
Cauchy data for $\FF_{ab}$ by
\begin{equation}\label{eq:E=K}
\EE_a = - *\ud* \ud \KK_a - i *\ud \partial_t\KK_a,
\end{equation}
where in the right hand side we restrict $\KK_a$ and $\partial_t \KK_a$ to
$\Sigma$, and
$\ud, *$
act on objects on $\Sigma$\footnote{Recall that, in dimension 3, for $p$-forms, the exterior co-derivative is given by $\ud^\ast = (-1)^p \ast\ud \ast{}$. }.
The constraint equation $\coder \EE_a = 0$ holds automatically for $\EE_a$
given by \eqref{eq:E=K}.
Now, in order to prove estimates for the Maxwell equation with data
$\EE_a \in H^j_\delta$, satisfying $\coder \EE = 0$,
it is sufficient to show that for any such $\EE_a$, there exists a 1-form
$\LL_a \in H^{j+1}_{\delta+1}$ such that
\begin{equation}\label{eq:E=L}
\EE_a = - i *\ud \LL_a .
\end{equation}
Then taking $\HH_{ab}$ to be a solution of \eqref{eq:H-wave} with Cauchy data
$$
\HH_{ab} \big{|}_{t=0} = 0, \quad
\left ( \partial_t \HH_{ab} \norm^b \right ) \big{|}_{t=0}=\LL_a ,
$$
gives a solution to the Maxwell
equation via \eqref{eq:F=H}. Estimates for the wave equation can thus be
applied to give estimates for the solution of the Maxwell field equation.
The operator $*\ud$ acting on 1-forms, which appears in Eqs.~\eqref{eq:E=K} and
\eqref{eq:E=L} is simply the curl operator. The first important thing to notice about Eq.~\eqref{eq:E=L} is the fact this is an overdetermined system of partial differential equations: the electric field has to satisfy constraints in order to ensure the existence of a solution. The integrability of these equations is well-known to be described by the elliptic complex
\begin{equation}\label{eq:derham}
C^\infty(\R^3,\R)\stackrel{\ud}{\longrightarrow} \Lambda^1\stackrel{\ast \ud}{\longrightarrow} \Lambda^1 \stackrel{\coder}{\longrightarrow}C^\infty(\R^3,\R).
\end{equation}
This complex, which is derived from the standard de Rham complex plays a crucial role in the analysis of Hertz potentials. The geometric constraint $\coder \EE_a =0$ guarantees that Eq.~\eqref{eq:E=L} can be solved, at least formally. This is described in Section~\ref{sec:Integrability}.
In Proposition~\ref{proprepspin1} below, we shall prove
that for non-integer weights $\delta $,
$*\ud : H^{j+1}_{\delta+1} \to \ker \coder \cap H^j_{\delta} $
is a surjection, and the estimate
\begin{equation}\label{estimateLE}
\Vert \LL_a\Vert_{j+1,\delta+1} \leq C \Vert\EE_a\Vert_{j,\delta}
\end{equation}
holds, for some constant $C$.
It is instructive to consider two special weights in the spin-1 case.
First let $\EE_a\in H^j_{-5/2}$ be a solution to the Maxwell constraint equation \eqref{eq:maxwellconstr}. This corresponds to the case considered in
\cite{Christodoulou:1990dd}.
We shall now construct a solution $\LL_a\in H^{j+1}_{-3/2}$ to \eqref{eq:E=L}, which will give us Cauchy data for the Herz potential.
Recall that the Laplacian
$\Delta = \ud \coder + \coder \ud$ is a surjection $H^{j+2}_{-1/2} \to H^j_{-5/2}$, cf. Proposition~\ref{prop:elliptictoolbox}. Hence, we can find $\mathcal{Q}_a\in H^{j+2}_{-1/2}$ such that
$$
\EE_a=\Delta \mathcal{Q}_a.
$$
Making use of the constraint equation \eqref{eq:maxwellconstr}, and the fact that $\Delta$ and $\ud^*$ commute, we find
$$
0=\ud^* \EE=\ud^* \Delta \mathcal{Q}= \Delta \ud^*\mathcal{Q}.
$$
Hence, $\ud^*\mathcal{Q}\in\ker \Delta \cap H^{j+1}_{-3/2}$. Due to injectivity of $\Delta:H^{j+1}_{-3/2} \to H^{j-1}_{-7/2}$, we have $\ud^*\mathcal{Q}=0$, and hence,
$$\EE_a=\ud \coder\mathcal{Q}_a + \coder \ud\mathcal{Q}_a=\coder \ud\mathcal{Q}_a.$$
Therefore we can take $\LL_a=i (*\ud \mathcal{Q})_a$. In this situation, full peeling does not hold, and the Cauchy data for the Hertz potential is
in $H^{j+2}_{-1/2} \times H^{j+1}_{-3/2}$. Secondly, we consider the case $\EE_a \in \ker \coder \cap H^j_{-7/2}$ where
full peeling holds. In this case, the Cauchy data for the Hertz potential is in $H^{j+2}_{-3/2} \times H^{j+1}_{-5/2}$. The relevant fact about the Laplacian is now that $\Delta : H^{j+2}_{-3/2} \to H^j_{-7/2}$ is Fredholm with cokernel consisting of constant forms. Since a constant form $\eta_a$
is automatically closed, using the integrability of the complex \eqref{eq:derham}, cf. Remark~\ref{integrability4}, it is also exact, $\eta_a = (\ud f)_a$ for some $f$. This fact is usually referred to as the Poincar\'e Lemma and can as we will see below, be generalized to higher spin. It follows that the cokernel of $\Delta$ is automatically $L^2$-orthogonal to $\ker \coder \cap H^j_{-7/2}$. Hence,
$$\Delta:H^{j+2}_{-3/2} \to \ker \ud^*\cap H^j_{-7/2}$$
is surjective. Using this fact, by the argument above we can find a solution $\LL_a\in H^{j+1}_{-5/2}$ to equation \eqref{eq:E=L}. For the case of general weights, see Proposition~\ref{proprepspin1}.
It is important to notice that, for weights $\delta < -4$, none of the elements in the cokernel have a preimage under the Laplacian.
However, in Lemma \ref{lemma:orthogonalpreimage}, we prove that one can add an element $\zeta_a$ in the image of $\ast \ud$, so that $E_a-i(\ast \ud\zeta)_a$ is orthogonal to the cokernel of $\Delta$, so that a preimage can be found. This way we can always find a preimage under $\ast \ud$, even if we can not find a preimage under $\Delta$.
In the spin-$s$ case, the argument follows the same outline. However, this requires an extension of
the complex \eqref{eq:derham} to arbitrary spin, and relating the corresponding operators to an elliptic operator, which in this case will be a power of the Laplacian $\Delta^{\lfloor s \rfloor}$.
Nonetheless, two important preliminary results have to be proved to this end: an extension of the complex
\eqref{eq:derham} to arbitrary spin and a decay result for the solution of the scalar wave equation with initial
data in Sobolev spaces of arbitrary non-integer weights and its derivatives.
As far as the authors know, not much work has been performed to extend the Hodge-de Rham theory to arbitrary weighted Sobolev spaces on the one hand and to arbitrary spin on the other hand. Cantor \cite{Cantor:1981bs} proved the first steps of a Hodge decomposition for tensors in weighted Sobolev spaces and Weck and Witsch \cite{Weck:1994ei} gave a Hodge-Helmholtz decomposition for forms in $L^p_{\delta}$ spaces. These two results are nonetheless not sufficient for our purposes. As far as the extension to arbitrary spin is concerned, for analytic solutions of the massless field equation of arbitrary spin on the Euclidean 4-sphere, Woodhouse \cite[Section 10]{Woodhouse:1985wx}, describes the procedure to construct the intermediate potentials, up to the penultimate, in the form language. Penrose \cite{Penrose:te, Penrose:tk} extended this construction for the spin $3/2$ in a wider context. Using this formalism, a topological condition on the domain of validity of the representation on the sphere is given.
In order to apply standard elliptic theory, one has to perform a $3+1$ splitting of the equation relating the field and its Hertz potential
\begin{equation}\label{eq:hertz}
\phi_{A\dots F} = \nabla_{AA'}\dots \nabla_{FF'}\chi^{A'\dots F'}.
\end{equation}
Assuming that the initial data for the potential are
$$
\chi^{A'\dots F'}|_{t=0} = 0 \text{ and } \partial_t\chi^{A'\dots F'}|_{t=0} = \sqrt{2}\tau^{AA'} \dots\tau^{FF'} \zeta_{A\dots F},
$$
the $3+1$ splitting (performed in detail in Section~\ref{sec:spacespinorsplit}) of Eq.~\eqref{eq:hertz} gives
\begin{equation}\label{eq:varphi=zeta}
\varphi_{A\dots F} = \left(\mathcal{G}_{2s}\zeta\right)_{A\dots F},
\end{equation}
where $\mathcal{G}_{2s}$ is a differential operator of order $2s-1$ (see Definition
\ref{def:mathcalg}). In the spin-$1$ case, if one translates the spin formalism into the form language, one recovers Eq.~\eqref{eq:E=L}.
Finding a generalization of the complex~\eqref{eq:derham} to arbitrary spin, which encompasses equation~\eqref{eq:varphi=zeta}, and especially the operator $\mathcal{G}_{2s}$, is then a necessary step in the construction of the initial data of the Hertz potential. A generalization of the de Rham complexes have been introduced in the context of the deformation of conformally flat structures by Gasqui-Goldschmidt \cite{Gasqui:1984vu} and, later generalized by Beig \cite{Beig:1997wo} (see Section~\ref{sec:Integrability}). They obtained differential complexes corresponding to the spin-2 case. We generalize, on $\R^3$, their results to arbitrary spin. One introduces the fundamental operators
\begin{align*}
(\sdiv_{2s}\phi)_{A_1\dots A_{2s-2}}\equiv{}&D^{A_{2s-1}A_{2s}}\phi_{A_1\dots A_{2s}},\\
(\stwist_{2s}\phi)_{A_1\dots A_{2s+2}}\equiv{}&D_{(A_1 A_2}\phi_{A_3\dots A_{2s+2})}.
\end{align*}
We prove that, if $\mathcal{S}_{2s}$ is the space of symmetric space spinor fields on $\R^3$, the sequence
\begin{equation*}
\mathcal{S}_{2s-2}\stackrel{\stwist_{2s-2}}{\longrightarrow} \mathcal{S}_{2s} \stackrel{\mathcal{G}_{2s}}{\longrightarrow} \mathcal{S}_{2s}\stackrel{\sdiv_{2s}}{\longrightarrow} \mathcal{S}_{2s-2},
\end{equation*}
is an elliptic complex. For $s=1$, one recovers the complex~\eqref{eq:derham}. In the situation considered by Gasqui-Goldschmidt and Beig for the spin-2 case, the operator $\mathcal{G}_{4}$ is the linearized Cotton-York tensor. See Section~\ref{sec:Integrability} for details.
The existence of a solution to \eqref{eq:E=L} with the estimate \eqref{estimateLE}
is then used together with a weighted
estimate for the solution of the wave equation with initial $(f,g) \in
H^j_\delta \times H^{j-1}_{\delta-1}$.
As we have not found a
sufficiently general result in the literature, in particular which covers the
range of weights $\delta > -1$ which we need
for the applications to the Hertz
potential in the range where full peeling fails to hold (including the
situation considered in \cite{Christodoulou:1990dd}), we prove the required
result in Section~\ref{sec:estsol}. This result consists in a direct estimate for
the solution of the wave equation, using the representation formula.
For $\delta < -1$, we have in the exterior region
$$
|\phi(t,x)|\leq C \lAngle v\rAngle^{-1}\lAngle u\rAngle^{1+\delta}
\left ( \Vert f\Vert_{3,\delta}+\Vert g\Vert_{2,\delta-1} \right ) ,
$$
see Proposition~\ref{prop:decaywave}.
The main result of the paper, stated in Theorem \ref{maintheorem}, combines the
analysis of the Hertz potential Cauchy data in weighted Sobolev spaces with
the weighted estimate for the solution of the wave equation to provide a
weighted estimate for the solution to the massless spin-$s$ field equation.
The peeling properties of the spin-$s$ field with initial data in weighted
Sobolev spaces are analyzed in detail.
Here it is important to note that the detailed decay estimates for the
components of the \emph{massless spin-$s$ field} $\phi_{A\dots F}$ relies on the
decay of the Hertz potential $\chi^{A' \dots F'}$, for which all components decay as
solutions to the scalar \emph{wave equation}.
The decay properties of the components of $\phi_{A\dots F}$ are
due to their relation to the derivatives of $\chi^{A'\dots F'}$ in terms of a
null tetrad. In particular, the falloff condition on the initial datum of the massless
fields which ensures that the peeling property holds is given explicitly. All the intermediate
states, where the peeling fails because the initial datum does not falloff sufficiently, such
as the decay result obtained by Christodoulou-Klainerman \cite{Christodoulou:1990dd}, are
clearly explained in terms of a decay result for the scalar wave equation.
\subsection*{Overview of this paper}
In Section~\ref{sec:prelim} we state our conventions and recall some basic
facts about elliptic operators on weighted Sobolev spaces.
In particular we introduce the Stein-Weiss operators
divergence $\sdiv$, curl $\scurl$ and the twistor
operator $\stwist$ for higher spin fields, as well as the fundamental higher order
operator $\mathcal G$ originating in the 3+1 splitting of the Hertz potential equation.
In Section~\ref{sec:Integrability} we use these to introduce a
generalization of the de Rham complex for spinor fields.
The problem of constructing initial data
for the Hertz potential is solved in Section~\ref{Section:ConstructPotentials}.
The weighted estimate for the wave equation is given in Section~\ref{sec:estsol},
and the resulting estimates for spin-$s$ fields generated by potentials is given in
Section~\ref{sec:estspin}.
Everything is then tied together in Section~\ref{sec:mainres}, where the 3+1 splitting
of the potential equation is considered and the potential is constructed. The section is
then concluded with the estimates for spin-$s$ fields.
Appendix~\ref{sec:algprop} contains some results on the operator $\mathcal G$ used in the
analysis of the elliptic complex introduced in Section~\ref{sec:Integrability} as
well as for the construction of the initial data for the Hertz potential in
Section~\ref{Section:ConstructPotentials}.
\section{Preliminaries} \label{sec:prelim}
\subsection{Conventions}\label{sec:conventions}
In this paper, we will only work on Minkowski space time. We will use Cartesian coordinates $(t, x^1, x^2, x^3)$ as well as the corresponding spherical coordinates $(t,r,\theta,\phi)$. The spinor formalism with the conventions of \cite{Penrose:1986fk} is extensively used. For important parts of the paper, 3+1 splittings of spinor expressions are performed. The space spinor formalism as introduced in \cite{Sommers:1980dd} is used for this purpose. In this case, the conventions of \cite{Backdahl:2010ig} are adopted. We will always consider the space spinors on the $\{t=\text{const.}\}$ slices of Minkowski space with normal $\tau_{AA'}=\sqrt{2}\nabla_{AA'}t$. Observe that a \emph{negative} definite metric on these slices is used.
The Minkowski space-time $(\mathbb{R}^4, \eta_{\alpha\beta})$ is endowed with its standard connection $\nabla_a = \nabla_{AA'}$. The time slice $\{t=0\}$ is endowed with the connection $D_a = D_{AB}$ defined by
$$
D_{AB} = \tau_{(A}{}^{A'} \nabla_{B)A'}
$$
where $\tau_{AA'}$ is the timelike vector field defined above. Its relation to the connection of the ambient space-time is given by
$$
\nabla_{AA'}=\tfrac{1}{\sqrt{2}}\tau_{AA'}\partial_t-\tau^{B}{}_{A'}D_{AB}.
$$
Let $S_k$ denote the vector bundle of symmetric valence $k$ spinors on $\mathbb{R}^3$. Furthermore, let $\mathcal{S}_k$ denote the space of smooth ($C^\infty$) sections of $S_k$.
\begin{definition}
Let $\mathcal{P}^{<\delta}(S_k)$ denote the finite dimensional subspace of $\mathcal{S}_k$ spanned by constant spinors with polynomial coefficients of degree $<\delta$.
\end{definition}
Observe that with $\delta\leq 0$, $\mathcal{P}^{<\delta}(S_k)$ is just the trivial space $\{0\}$.
\subsection{Analytic framework}
We introduce in this section the analytic framework which is necessary to understand the propagation of the field as well as the geometric constraints. We will use the conventions of Bartnik \cite{Bartnik:1986dq}. Even though Bartnik's paper only gives statements for functions, we can easily extend this to space spinors on Euclidean space.
We recall first the standard norms, coming from the Hermitian space spinor product.
\begin{definition} The Hermitian space spinor product is given by
\begin{equation*}
\langle \zeta_{A\dots F},\phi_{A\dots F}\rangle = \zeta_{A\dots F} \widehat{\phi}^{A\dots F},
\end{equation*}
where
$
\widehat{\phi}^{A\dots F} = \tau^{AA'}\dots \tau^{FF'} \overline{\phi}_{A'\dots F'}
$ and $\tau_{AA'}=\sqrt{2}\nabla_{AA'}t$.
The pointwise norm of a smooth $\phi_{A\dots F}$ is defined via
$$
\vert\phi_{A\dots F} \vert^2 = \phi_{A\dots F} \widehat{\phi}^{A\dots F}.
$$
The pointwise norm of the derivatives of the smooth spinor $\phi_{A\dots F}$ on $\R^3$ is given by
$$
\vert D_a \phi_{A\dots F}\vert^2 = \delta^{ab} D_{a}\phi_{A\dots F} \widehat{D_b\phi}{}^{A\dots F},$$
where $\delta_{ab}$ is the standard Euclidean metric on $\R^3$. The norm of higher order derivatives is defined similarly.
\end{definition}
\begin{remark}\label{rem:DReal} The identity
$$
D_{AB} \widehat{\phi}_{A\dots F} = - \widehat{D_{AB} \phi}_{A\dots F},
$$
holds, due to the fact that the operator $D_{AB}$ is real.
\end{remark}
\begin{definition}
The $L^2$-norm of a smooth spinor field in $\R^3$ is defined by
$$
\Vert\phi_{A\dots F}\Vert_{2} = \left(\int_{\R^3}\vert\phi_{A\dots F} \vert^2 \ud \mu_{\R^3}\right)^{1/2},
$$
where $\ud \mu_{\R^3}$ is the standard volume form on $\R^3$.
The $L^2$-norm is also defined for derivatives in the same way using the pointwise definition above.
\end{definition}
If $u$ is a real scalar, its Japanese bracket
is defined by
$$
\lAngle u\rAngle = (1+u^2)^{1/2}.
$$
We next define the weighted Sobolev norms, which will be used to describe the asymptotic behavior of initial data at space-like infinity.
\begin{definition}[Weighted Sobolev spaces] Let $\delta$ be a real number and $j$ a nonnegative integer.\\
The completion of the space of smooth spinor fields in $\mathcal{S}_{2s}$ with compact support in $\R^3$ endowed with the norm
$$
\Vert \phi_{A\dots F}\Vert^2_{j, \delta} = \sum_{n=0}^j \left\Vert \lAngle r \rAngle^{-(\delta+\frac32) + n} D^n\phi_{A\dots F}\right\Vert^2_{2},
$$
is denoted by $H^j_{\delta}(S_{2s})$.
\end{definition}
For $\delta = - 3/2$, the weighted spaces $H^0_{-3/2}(S_{2s})$
are the standard Sobolev spaces $L^2(S_{2s})$.
Many well-known properties can be proved about these spaces -- see for instance \cite[Theorem~1.2]{Bartnik:1986dq} for more details. The only property, crucial to obtain the pointwise estimates, is the following Sobolev embedding (cf. \cite[Theorem~1.2, (iv)]{Bartnik:1986dq}, specialized to dimension 3).
\begin{proposition}\label{prop:decaysob} Let $\delta$ be a real number and $j\geq 2$ an integer. Then, any spinor field in $H^j_\delta(S_{2s})$ is in fact continuous and there exists a constant $C$ such that, for any $\phi_{A\dots F}$ in $H^j_\delta(S_{2s})$
$$
\vert\phi_{A\dots F}(x) \vert \leq C \lAngle r \rAngle^\delta \Vert\phi_{A\dots F}\Vert_{2, \delta},
$$
and, in fact,
$$
\vert\phi_{A\dots F}(x) \vert = o\bigl( r^{\delta}\bigr) \text{ as } r \rightarrow \infty.
$$
\end{proposition}
We finally recall the following properties of elliptic operators, restricting ourself to powers of the Laplacian. The result stated is a combination of the standard results, see e.g., \cite{Bartnik:1986dq,McOwen:1980gz,Cantor:1981bs,Kat80}.
\begin{proposition}\label{prop:elliptictoolbox} Let $j,l$ be non-negative integers such that $j\geq 2l$, $s$ be in $\tfrac{1}{2}\mathbb{N}_{0}$ and $\delta$ be in $\R\setminus\mathbb{Z}$. The formally self-adjoint elliptic operator of order $2l$
$$
\Delta^l_{2s}: H^{j}_{\delta}(S_{2s}) \longrightarrow H^{j-2l}_{\delta-2l}(S_{2s})
$$
is Fredholm and satisfies
\begin{itemize}
\item its kernel is a subspace of $\mathcal{P}^{<\delta}(S_{2s})$; in particular, $\Delta^l_{2s}$ is injective when $\delta<0$;
\item its co-kernel is a subspace of $\mathcal{P}^{<-3-\delta+2l}(S_{2s})$; in particular, $\Delta^l_{2s}$ is surjective when $\delta>2l-3$.
\end{itemize}
Furthermore,
there exists a constant $C$ such that, for all spinor fields in $H^{j}_{\delta}(S_{2s})$,
$$
\inf_{\psi_{A\dots F} \in \ker(\Delta_{2s}^l)\cap H^{j}_{\delta}(S_{2s}) } \Vert \phi_{A\dots F} + \psi_{A\dots F} \Vert_{j,\delta} \leq C \Vert\Delta^{l}_{2s}\phi_{A\dots F}\Vert_{j-2s,\delta-2s}.
$$
In fact the infimum is attained, and there exists a $\psi_{A\dots F}$ in $\ker(\Delta_{2s}^l)\cap H^{j}_{\delta}(S_{2s})$ such that $\theta_{A\dots F}=\phi_{A\dots F} + \psi_{A\dots F}$ satisfies
$$
\Vert \theta_{A\dots F} \Vert_{j,\delta} \leq C \Vert\Delta^{l}_{2s}\theta_{A\dots F}\Vert_{j-2s,\delta-2s}.
$$
\end{proposition}
\begin{remark}
\begin{enumerate}
\item In this paper, the set $\mathbb{N}$ is the set of positive integers and $\mathbb{N}_0$ the set of non-negative integers.
\item In the range of weights $-1\leq \delta\leq 0$, the operator $\Delta$ is bijective.
\item The inequality comes from the closed range property (see \cite[Theorem~5.2]{Kat80}). Since this infimum corresponds to the distance to the kernel $\ker(\Delta_{2s}^l)$, which is closed, the infimum is in fact attained.
\item We recall here that
the co-kernel of $\Delta^l_{2s}$ in
$H^{j-2l}_{\delta-2l}(S_{2s})$ is $L^2$-orthogonal to the kernel of
$\Delta^{l}_{2s}$ in the dual space $H^{-j+2l}_{-3-\delta+2l}(S_{2s})$.
\item The dimension of the spaces can be computed explicitly -- see for instance \cite{Lockhart:1984kt}. However, we will not make explicit use of this.
\end{enumerate}
\end{remark}
\subsection{Fundamental operators} \label{sec:fundop}
\begin{definition} \label{def:fundop}
Let
$\phi_{A_1\dots A_k}\in \mathcal{S}_k$, that is $\phi_{A_1\dots A_k}=\phi_{(A_1\dots A_k)}$.
Let $D_{AB}$ be the intrinsic Levi-Civita connection. Define the operators
$\sdiv_k: \mathcal{S}_k\rightarrow \mathcal{S}_{k-2}$,
$\scurl_k: \mathcal{S}_k\rightarrow \mathcal{S}_{k}$ and
$\stwist_k: \mathcal{S}_k\rightarrow \mathcal{S}_{k+2}$ via
\begin{align*}
(\sdiv_k\phi)_{A_1\dots A_{k-2}}\equiv{}&D^{A_{k-1}A_{k}}\phi_{A_1\dots A_k},\\
(\scurl_k\phi)_{A_1\dots A_{k}}\equiv{}&D_{(A_1}{}^{B}\phi_{A_2\dots A_{k})B},\\
(\stwist_k\phi)_{A_1\dots A_{k+2}}\equiv{}&D_{(A_1 A_2}\phi_{A_3\dots A_{k+2})}.
\end{align*}
These operators will be called \emph{divergence}, \emph{curl} and \emph{twistor operator} respectively.
\end{definition}
We suppress the indices of $\phi$ in the left hand sides. The label $k$ indicates its valence.
The importance of these operators comes from the following irreducible decomposition which is valid for any $k\geq 1$,
\begin{align*}
D_{A_1 A_2}\phi_{A_3\dots A_{k+2}}={}&(\stwist_k\phi)_{A_1\dots A_{k+2}}
-\tfrac{k}{k+2} \epsilon_{A_1(A_3}(\scurl_k\phi)_{A_4\dots A_{k+2})A_2}\nonumber\\
&-\tfrac{k}{k+2} \epsilon_{A_2(A_3}(\scurl_k\phi)_{A_4\dots A_{k+2})A_1}
+\tfrac{1-k}{1+k} \epsilon_{A_1(A_3}(\sdiv_k\phi)_{A_4\dots A_{k+1}}\epsilon_{A_{k+2})A_2}.
\end{align*}
This irreducible decomposition follows from \cite[Proposition~3.3.54]{Penrose:1986fk}. Contraction with the spin metric $\epsilon^{AB}$ and partial expansion of the symmetries give the coefficients.
We consider now the symbols of these operators
\begin{equation*}
\sigma(\scurl_k) : T^\star M \to L(S_k, S_k),\qquad
\sigma(\sdiv_k) : T^\star M \to L(S_k, S_{k-2}),\qquad
\sigma(\stwist_k) : T^\star M \to L(S_k, S_{k+2}),
\end{equation*}
where $L(S_k, S_j)$ is the space of bundle maps from $S_k$ into $S_j$.
When applied to a 1-form $\xi_{AB}$, one denotes the symbols $\sigma_{\xi}$.
\begin{lemma}\label{symbolcurlreal}
When applied to a 1-form $\xi_{AB}$, the symbol $\sigma_{\xi}(\scurl_k) : S_k \rightarrow S_k$ of $\scurl_k$ is Hermitian and has only real eigenvalues.
\end{lemma}
\begin{proof}
By definition we have
\begin{equation*}
(\sigma_\xi(\scurl_k)\phi)_{A_1\dots A_{k}}\equiv \xi_{(A_1}{}^{B}\phi_{A_2\dots A_{k})B},
\end{equation*}
where $\xi_{AB}$, is real, \emph{i.e.} $\widehat \xi_{AB}=-\xi_{AB}$. For arbitrary $\eta_{A_1\dots A_{k}}$ and $\zeta_{A_1\dots A_{k}}$, we have
\begin{align*}
\langle (\sigma_\xi(\scurl_k)\eta)_{A_1\dots A_{k}},
\zeta_{A_1\dots A_k}\rangle
={}& \xi_{A_1}{}^{B}\eta_{A_2\dots A_{k}B}\hat\zeta^{A_1\dots A_k}
= \eta_{A_2\dots A_{k}B}\widehat \xi^{C_1B}\hat\zeta_{C_1}{}^{A_2\dots A_k}\nonumber\\
={}&\langle \eta_{A_1\dots A_k}, (\sigma_\xi(\scurl_k)\zeta)_{A_1\dots A_{k}}\rangle.
\end{align*}
Hence, the symbol $\sigma_\xi(\scurl_k)$ is for each point Hermitian and, by the spectral theorem, has only real eigenvalues.
\end{proof}
The operators $\sdiv_k$, $\scurl_k$ and $\stwist_k$ are special cases of Stein-Weiss operators. We refer to \cite{Branson1997334} and references therein for general properties of this class of operators.
We will now consider some operators of general order,
that will play an important role in this paper.
The most important second order operator is clearly the Laplacian $\Delta\equiv D_{AB}D^{AB}$. Note that here we are using a \emph{negative} definite metric on $\mathbb{R}^3$. When the Laplacian acts on a spinor field $\phi_{A\dots F}$ in $\mathcal{S}_k$, we will often use the notation $(\Delta_k\phi)_{A\dots F}$, where $k$ indicates the valence of $\phi_{A\dots F}$.
\begin{definition}\label{def:mathcalg}
Define the order $k-1$ operators $\mathcal{G}_k:\mathcal{S}_k\rightarrow \mathcal{S}_{k}$ as
\begin{align*}
(\mathcal{G}_k\phi)_{A_1\dots A_k} &\equiv \sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}\binom{k}{2n+1}(-2)^{-n} \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1} (\Delta^n_{k}\phi)_{A_{k-2n}\dots A_k)B_1\dots B_{k-2n-1}}.
\end{align*}
\end{definition}
The first operators are
\begin{align*}
(\mathcal{G}_1\phi)_{A} &\equiv \phi_{A},\\
(\mathcal{G}_2\phi)_{AB} &\equiv 2 D_{(A}{}^{C}\phi_{B)C}
=2(\scurl_2\phi)_{AB},\\
(\mathcal{G}_3\phi)_{ABC} &\equiv 3 D_{(A}{}^{D}D_{B}{}^{F}\phi_{C)DF}
- \tfrac{1}{2} (\Delta_{3}\phi)_{ABC} \nonumber\\
&=\tfrac{1}{3}(\stwist_1\sdiv_3\phi)_{ABCD}+4(\scurl_3\scurl_3\phi)_{ABC},\\
(\mathcal{G}_4\phi)_{ABCD} &\equiv4 D_{(A}{}^{F}D_{B}{}^{H}D_{C}{}^{L}\phi_{D)FHL}
- 2 D_{(A}{}^{F}(\Delta_{4}\phi)_{BCD)F}\nonumber\\ &=2(\stwist_2\sdiv_4\scurl_4\phi)_{ABCD}+8(\scurl_4\scurl_4\scurl_4\phi)_{ABCD}.
\end{align*}
These operators appear naturally in Proposition~\ref{prop:splittingpotential} below. The
most important properties of these operators are
\begin{equation}\label{eq:divGproperty}
\sdiv_k\mathcal{G}_k=0\quad \text{ and }\quad \mathcal{G}_k\stwist_{k-2}=0,
\end{equation}
which is valid for any $k\geq 2$. The main
idea to prove this is to use that $\sdiv_k\mathcal{G}_k$ and
$\mathcal{G}_k\stwist_{k-2}$ contains derivatives of the kind
$D_{A}{}^{C}D_{BC}=\frac{1}{2}\epsilon_{AB}\Delta$. For a complete proof see
Proposition~\ref{divGproperty}. The operators $\mathcal{G}_k$ also commute
with $\scurl_k$; this is proven in Proposition~\ref{Gcurlexpansion}.
To connect with the standard elliptic theory, we express appropriate powers of the Laplacian in terms of the operators $\mathcal{G}_k$ as
\begin{subequations}
\begin{align}
(\Delta^{k}_{2k}\phi)_{A_1\dots A_{2k}}={}&(\stwist_{2k-2}\mathcal{F}_{2k-2}\sdiv_{2k}\phi)_{A_1\dots A_{2k}}-(-2)^{1-k}(\mathcal{G}_{2k}\scurl_{2k}\phi)_{A_1\dots A_{2k}},\label{LaplacianAsGeven}\\
(\Delta^{k}_{2k+1}\phi)_{A_{1}\dots A_{2k+1}}={}&
(\stwist_{2k-1}\mathcal{F}_{2k-1}\sdiv_{2k+1}\phi)_{A_1\dots A_{2k+1}}
+(-2)^{-k}(\mathcal{G}_{2k+1}\phi)_{A_1\dots A_{2k+1}} \label{LaplacianAsGodd},
\end{align}
\end{subequations}
where the operators $\mathcal{F}_{2s}$ for $s\in \tfrac{1}{2}\mathbb{N}_0$ are defined via
\begin{align*}
(\mathcal{F}_{2s}\phi)_{A_1\dots A_{2s}}={}&
2^{-2s}\sum_{n=0}^{\lfloor s\rfloor}\sum_{m=0}^{\lfloor s\rfloor-n}\binom{2s+2}{2n+2m+2}(-2)^{n}\nonumber\\
&\quad\times \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2n}}{}^{B_{2n}}}_{2n} (\Delta^{\lfloor s\rfloor-n}_{2s}\phi)_{A_{2n+1}\dots A_{2s})B_1\dots B_{2n}}.
\end{align*}
The first operators are
\begin{align*}
(\mathcal{F}_0\phi)={}&\phi,\nonumber\\
(\mathcal{F}_1\phi)_{A}\equiv{}& \tfrac{3}{2}\phi_A,\\
(\mathcal{F}_2\phi)_{AB}={}& \tfrac{7}{4} (\Delta_{2} \phi)_{AB}
-\tfrac{1}{2}D_{(A}{}^{C}D_{B)}{}^{D}\phi_{CD}.
\end{align*}
See Lemma~\ref{lapspinor} in the appendix for the proof of \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd}.
The operator $\scurl_2$ is the spinor equivalent to the operator $\ast \ud$ acting on 1-forms. The tensor equivalent of the operator $\mathcal{G}_4$ is the linearized Cotton-York tensor acting on symmetric trace-free 2-tensors. In the following section, a more detailed description of these relations is given.
\section{Integrability properties of spinor fields}\label{sec:Integrability}
A crucial part of this work relies on integrability properties for
spinors, that is to say proving that a spinor belongs to the image of a certain differential operator. In the case of spin-1, the operator under consideration is the curl operator; its integrability properties are well known
since it is described by the de Rham complex. For
spin-2, one has to resort to a generalization of the de Rham theory to trace free 2-tensors, which happened to have been studied in the context of
conformal deformations of the flat metric by Gasqui and Goldschmidt
\cite{Gasqui:1984vu}, whose
results were
extended by Beig \cite{Beig:1997wo}.
On $\R^3$, we consider a generalization
of these elliptic complexes for arbitrary spin.
We present here the general picture of this integrability result for smooth
spinors. It is well known that for 1-forms the integrability conditions are
given by the following elliptic complex
\begin{equation*}
C^\infty(\R^3,\R)\stackrel{\ud}{\longrightarrow} \Lambda^1\stackrel{\star \ud}{\longrightarrow} \Lambda^1 \stackrel{\coder}{\longrightarrow}C^\infty(\R^3,\R),
\end{equation*}
whose spinorial equivalent is
\begin{equation}\label{exactseq1}
\mathcal{S}_{0}\stackrel{\stwist_0}{\longrightarrow} \mathcal{S}_{2} \stackrel{\scurl_2}{\longrightarrow} \mathcal{S}_{2}\stackrel{\sdiv_2}{\longrightarrow} \mathcal{S}_{0}.
\end{equation}
Gasqui and Goldschmidt were interested in the conformal deformation of a metric on a 3-manifold $M$. A deformation $g_t$ of a metric $g_0$ is said to be conformally rigid if there exist a family of diffeomorphisms $\phi_t^\star$ and of functions $u_t$ such that
$$
\phi^\star_t g_0 = e^{u_t}g_t.
$$
The infinitesimal equation corresponding to this deformation is given by the conformal Killing equation
\begin{equation}\label{confkill}
\mathcal{L}_X g_0-\frac13\text{Tr}_{g_0}(\mathcal{L}_X g_0)g_0=h
\end{equation}
where $X$ is a vector field on $M$ and $h$ is a trace free 2-tensor. The spinor equivalent of this equation is given by
$$
2D_{(AB}X_{CD)} = h_{ABCD}.
$$
Solving \eqref{confkill} requires that the 2-tensor $h$ satisfies the constraint equation. This is stated in \cite[Theorem 6.1, (2.24)]{Gasqui:1984vu} and in \cite{Beig:1997wo}.
\begin{theorem}[Gasqui-Goldschmidt]\label{thm:integrability2}
If $(M,g)$ is a conformally flat 3-dimensional manifold, then the following is an elliptic complex
$$
\Lambda^1(M) \stackrel{L}{\longrightarrow} S_0^2(M, g) \stackrel{\mathcal{R}}{\longrightarrow} S_0^2(M, g) \stackrel{\diverg} {\longrightarrow}\Lambda^1(M),
$$
where $\Lambda^1(M)$ is the space of 1-forms over $M$, $S_0^2(M, g)$ is the space of symmetric trace free 2-tensors and
\begin{eqnarray*}
(LW)_{ab}&=&D_{(a}W_{b)}-\frac{1}{3}g_{ab}D^cW_c\\
(\diverg t)_a&=&2g^{bc}D_ct_{ab}
\end{eqnarray*}
and \begin{equation*}
\begin{array}{c}
\mathcal{R}(\psi)_{ab} = \eps^{cd}{}_a D_{[c}\sigma_{d]b} \text{ where}\\
\sigma_{ab}=D_{(a} D^c \psi_{b)c} - \frac{1}{2} \Delta \psi_{ab} - \frac{1}{4} g_{ab}D^c D^d \psi_{cd}.
\end{array}
\end{equation*}
\end{theorem}
\begin{remark}
\begin{enumerate}
\item A consequence of Theorem \ref{thm:integrability2} is that equation \eqref{confkill} is integrable provided that
$$
\mathcal{R}(h)_{ab}=0.
$$
\item In terms of spinors, the operator $\mathcal{R}_{ab}$ reads
$$
\mathcal{R}_{ab} = \mathcal{R}_{ABCD} = -\frac{i}{2\sqrt{2}} (\mathcal{G}_4)_{ABCD}.
$$
\end{enumerate}
\end{remark}
The spinorial equivalent of this sequence is the following elliptic complex
\begin{equation}\label{exactseq2}
\mathcal{S}_2 \stackrel{\stwist_2}{\longrightarrow} \mathcal{S}_4 \stackrel{\mathcal{G}_4}{\longrightarrow} \mathcal{S}_4 \stackrel{\sdiv_4} {\longrightarrow}\mathcal{S}_2.
\end{equation}
We now state, using the fundamental operators $\stwist_{2s-2}$, $\mathcal{G}_{2s}$ and $\sdiv_{2s}$, a generalization of the elliptic complexes \eqref{exactseq1} and \eqref{exactseq2} for arbitrary spin.
\begin{lemma}\label{lem:ellipticcomplex}
The sequence
\begin{equation*}
\mathcal{S}_{2s-2}\stackrel{\stwist_{2s-2}}{\longrightarrow} \mathcal{S}_{2s} \stackrel{\mathcal{G}_{2s}}{\longrightarrow} \mathcal{S}_{2s}\stackrel{\sdiv_{2s}}{\longrightarrow} \mathcal{S}_{2s-2},
\end{equation*}
is an elliptic complex.
\end{lemma}
\begin{proof}
In view of \eqref{eq:divGproperty},
the sequence is a differential complex. It is therefore enough to
check that the symbol sequence is exact, \emph{i.e.} for a non-zero $\xi$ in $T^\star M$
$$
S_{2s-2} \stackrel{\sigma_\xi(\stwist_{2s-2})}{\longrightarrow} S_{2s} \stackrel{\sigma_\xi(\mathcal{G}_{2s})}{\longrightarrow}
S_{2s}\stackrel{\sigma_\xi(\sdiv_{2s})}{\longrightarrow} S_{2s-2}.
$$
This follows from the vanishing properties \eqref{eq:divGproperty}
and the expression of powers of the Laplacian in terms of these operators, \emph{i.e.} \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd}.
Let $\xi$ be a fixed non-zero element of $T^\star M $. We first notice that
\eqref{eq:divGproperty} implies
$$
\text{im}(\sigma_\xi(\mathcal{G}_{2s})) \subset \ker (\sigma_\xi(\sdiv_{2s}))
\quad \text{ and } \quad
\text{im}(\sigma_\xi(\stwist_{{2s}-2}))\subset \ker(\sigma_\xi(\mathcal{G}_{2s})).
$$
We then notice that the symbol of the Laplacian $\Delta^k_{2s}$ is an invertible symbol which is in the center of the algebra of symbols since its expression is
$$
\sigma_\xi(\Delta_{2s}^k) = |\xi_{AB}|^{2k} I.
$$
Furthermore, using the relations stated in Lemma~\ref{lapspinor}, we have\begin{align}
(\Delta^{s}_{2s}\phi)_{A_1\dots A_{2s}}&=(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\phi)_{A_1\dots A_{2s}}-(-2)^{1-s}(\mathcal{G}_{2s}\scurl_{2s}\phi)_{A_1\dots A_{2s}}\text{ for } s\in \mathbb{N}_0,\label{lapeven}\\
(\Delta^{s-1/2}_{2s}\phi)_{A_{1}\dots A_{2s+1}}&=
(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\phi)_{A_1\dots A_{2s}}
+(-2)^{-1/2-s}(\mathcal{G}_{2s}\phi)_{A_1\dots A_{2s}} \text{ for } s\in \tfrac12+ \mathbb{N}_0\nonumber.
\end{align}
Assume now that the spin is an integer. The proof in the case when the spin is a half integer is left to the reader (the proof is almost identical).
Let $Y$ be an element of $\ker(\sigma_\xi(\sdiv_{2s}))$. Using formula~\eqref{lapeven}, we get
\begin{eqnarray*}
Y &=& \sigma_\xi(\Delta_{2s}^s)^{-1}\sigma_\xi(\Delta_{2s}^s) Y\\
&=&\sigma_\xi(\Delta_{2s}^s)^{-1}\left(\sigma_\xi(\stwist_{2s-2})\sigma_\xi(\mathcal{F}_{2s-2})\sigma_\xi(\sdiv_{2s})-(-2)^{1-s}\sigma_\xi(\mathcal{G}_{2s})\sigma(\scurl_{2s})\right)Y\\
&=&-(-2)^{1-s}\sigma_\xi(\Delta_{2s}^s)^{-1}\sigma_\xi(\mathcal{G}_{2s})\sigma_\xi(\scurl_{2s})Y.
\end{eqnarray*}
Since the symbol of the Laplacian commutes with all other symbols, we consequently get
$$
Y = -(-2)^{1-s}\sigma_\xi(\mathcal{G}_{2s})\sigma_\xi(\Delta_{2s}^s)^{-1}\sigma_\xi(\scurl_{2s}) Y,
$$
that is to say that $Y$ belongs to the image of $\sigma_\xi(\mathcal{G}_{2s})$.
If we now assume that $Y$ is in $\ker(\sigma_\xi(\mathcal{G}_{2s}))$. Using formula~\eqref{lapeven}, we get
\begin{eqnarray*}
Y &=& \sigma_\xi(\Delta_{2s}^s)^{-1}\sigma_\xi(\Delta_{2s}^s) Y\\
&=&\sigma_\xi(\Delta_{2s}^s)^{-1}\left(\sigma_\xi(\stwist_{2s-2})\sigma_\xi(\mathcal{F}_{2s-2})\sigma_\xi(\sdiv_{2s})-(-2)^{1-s}\sigma_\xi(\mathcal{G}_{2s})\sigma_\xi(\scurl_{2s})\right)Y.
\end{eqnarray*}
Since $\mathcal{G}_{2s}$ and $\scurl_{2s}$ commute (Lemma~\ref{Gcurlexpansion}) and since the symbol of the Laplacian commutes with all other symbols, we get
$$
Y=\sigma_\xi(\stwist_{2s-2})\sigma_\xi(\Delta_{2s}^s)^{-1}\sigma_\xi(\mathcal{F}_{2s-2})\sigma_\xi(\sdiv_{2s}),
$$
that is to say that $Y$ belongs to the image of $\sigma_\xi(\stwist_{2s-2})$.
\end{proof}
Using the ellipticity of the sequence, it is finally possible to prove the existence of solutions of equations involving $\mathcal{G}_{2s}$ and $\stwist_{2s}$. This theorem is a direct consequence of \cite[Theorem~1.4]{Anonymous:1996ce}.
\begin{proposition}\label{prop:integrability3} For $x$ in $\R^3$, there exists an open neighborhood $U$ of $x$ such that the sequence
\begin{equation*}
\mathfrak{A}(U,S_{2s-2})\stackrel{\stwist_{2s-2}}{\longrightarrow} \mathfrak{A}(U,S_{2s}) \stackrel{\mathcal{G}_{2s}}{\longrightarrow} \mathfrak{A}(U,S_{2s})\stackrel{\sdiv_{2s}}{\longrightarrow} \mathfrak{A}(U,S_{2s-2}),
\end{equation*}
is exact, where $\mathfrak{A}(U,E)$ denotes the space of real analytic sections of $E$.
\end{proposition}
\begin{remark}\label{integrability4} We in fact only need this result in the context of polynomials. The problem will be to solve, for any real number $\delta$, the equations
$$
\stwist_{2s-2} \phi = \psi, \quad \text{ when }\psi\in \mathcal{P}^{<\delta}(S_{2s}) ,
$$
and
$$
\mathcal{G}_{2s} \xi = \zeta, \quad \text{ when }\zeta\in \mathcal{P}^{<\delta}(S_{2s}).
$$
Proposition~\ref{prop:integrability3} ensures the local existence of solutions to these equations provided that
$$
\mathcal{G}_{2s}\psi =0 \text{ and }\sdiv_{2s} \zeta = 0.
$$
By integration, these solutions are necessarily polynomials.
\end{remark}
\begin{proof} The proof of Proposition~\ref{prop:integrability3} is a direct
consequence of the fact that the fundamental operators $\stwist_{2s-2}$,
$\mathcal{G}_{2s}$ and $\sdiv_{2s}$ are operators with constant
coefficients, which consist only of higher order homogeneous terms. As a
consequence, these operators are all \emph{sufficiently regular}
in the terminology of \cite{Anonymous:1996ce}
(since they have constant coefficients, cf.
\cite[Remark~1.16]{Anonymous:1996ce})
and \emph{formally integrable}
(since they have only homogeneous terms of the highest possible order,
cf. \cite[Remark~1.21]{Anonymous:1996ce}).
Proposition~\ref{prop:integrability3} is then
a direct consequence of \cite[Theorem~1.4]{Anonymous:1996ce}.
\end{proof}
\section{Construction of initial data for the potential} \label{Section:ConstructPotentials}
As seen in the introduction, one of the key points when constructing the initial data for the potential for the massless free field is the ability to solve, at the level of the initial data, the equation
$$
\varphi_{A\dots F} = (\mathcal{G}_{2s} \zeta)_{A\dots F}.
$$
This requires a partial generalization of Proposition \ref{prop:integrability3} to weighted Sobolev spaces.
The main difficulty in the construction of $\zeta_{A\dots F}$ is to obtain the estimate
\begin{equation*}
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\leq C \Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{equation*}
This inequality is similar to those in standard elliptic theory. The main idea of this section is to construct a solution using the relations between the operator $\mathcal{G}_{2s}$ and powers of the Laplacian $\Delta$ as stated in Equations \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd}. The strategy of the proof is as follows:
\begin{enumerate}
\item using the elliptic properties of the Laplacian and its powers, a preimage $\theta_{A\dots F}$
of $\varphi_{A\dots F}$ is constructed;
\item using Equations \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd}, the constraint equation satisfied by the initial datum, and the differential complex stated in Lemma \eqref{lem:ellipticcomplex}, we prove that the only non-vanishing term of Equations \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd} is the one containing $\mathcal{G}_{2s}$.
\end{enumerate}
However, this schematic procedure works only for a certain range of weights. Outside this range, a more thorough discussion has to be performed. One of the key facts which is used is that the polynomial nature of the elements of the kernel of the Laplacian (and its powers) makes it possible to use Proposition \ref{prop:integrability3} (and Remark \ref{integrability4}).
This section deals strictly with the initial data both for the field and the potential. The corresponding Cauchy problems for higher spin fields and for the potential are described in Section \ref{sec:mainres}. More precisely, the details regarding the relation between the Hertz potential and the field are given in Section \ref{sec:spacespinorsplit}, which is devoted to the 3+1 splitting of the potential equation. Furthermore, the representation Theorem \ref{th:repthm} for a massless field in terms of a Hertz potential, based on the uniqueness of solutions of the Cauchy problem, is given in Section \ref{sec:representation}.
For $s=1/2$, we immediately get the desired solution by setting $\zeta_A=\varphi_A$. For higher spin, a more careful analysis is required. To simplify the presentation, the spin-1 case is discussed first in detail, followed by the general spin case.
The following lemma is a technical result describing the orthogonality properties of the range of $\stwist$ and the kernel of $\sdiv$.
\begin{lemma}\label{orthogonalitylemma}
Assume that $\varphi_{A\dots F}\in H^1_\delta(S_{2s})$
satisfies the constraint $(\sdiv_{2s}\varphi)_{C\dots F}=0$, and $\eta_{C\dots F}\in H^1_{-2-\delta}(S_{2s-2})$.
Then $\varphi_{A\dots F}$ is $L^2$ orthogonal to $(\stwist_{2s-2}\eta)_{A\dots F}$.
\end{lemma}
\begin{proof}
Let $\{\varphi^i_{A\dots F}\}_{i=0}^\infty\subset C_0^\infty(S_{2s})$ such that $\Vert \varphi^i_{A\dots F}-\varphi_{A\dots F}\Vert_{1,\delta}\rightarrow 0$ as $i\rightarrow \infty$.
An integration by parts and Remark \ref{rem:DReal} give
\begin{equation*}
\int_{\R^3}\varphi^i_{A\dots F} \widehat{D^{(AB}\eta^{C\dots F)}}\ud\mu_{\R^3}
= \int_{\R^3}D^{AB}\varphi^i_{A\dots F}\widehat\eta^{C\dots F} \ud \mu_{\R^3},
\end{equation*}
that is
\begin{equation*}
\langle \varphi^i_{A\dots F}, (\stwist_{2s-2}\eta)_{A\dots F}\rangle_{L^2} =
\langle (\sdiv_{2s}\varphi^i)_{C\dots F},\eta_{C\dots F}\rangle_{L^2}.
\end{equation*}
Taking the limit as $i\rightarrow \infty$ on both sides gives
$\langle \varphi_{A\dots F}, (\stwist_{2s-2}\eta)_{A\dots F}\rangle_{L^2}=0$.
\end{proof}
\begin{definition}\label{def:Esdelta}
Let $\delta$ be in $\mathbb{R}\setminus\mathbb{Z}$ and $s\in \tfrac{1}{2}\mathbb{N}$.
Furthermore, let
\begin{equation*}
\mathbb{F}_{s,\delta}\equiv\ker \Delta^{\lfloor s\rfloor}_{2s}\cap L^2_{-3-\delta}(S_{2s}).
\end{equation*}
Define the space $\mathbb{E}_{s,\delta}$ to be the $L^2_{-3-\delta}$-orthogonal complement of $\mathbb{F}_{s,\delta}\cap \ker\mathcal{G}_{2s}$ in $\mathbb{F}_{s,\delta}$.
\end{definition}
\begin{remark}\label{rem:Etrivial}
If $\delta\geq -2s-2$, the space $\mathbb{E}_{s,\delta}$ is trivial. This follows from the fact that $\mathbb{F}_{s,\delta}\subset \mathcal{P}^{<-3-\delta}(S_{2s})$ and that $\mathcal{G}_{2s}$ is a homogeneous order $2s-1$ operator. With $-3-\delta \leq 2s-1$ or equivalently $\delta\geq -2s-2$ we have $\mathbb{F}_{s,\delta}\subset \mathcal{P}^{<-3-\delta}(S_{2s})\subset \ker\mathcal{G}_{2s}$ and consequently $\mathbb{E}_{s,\delta}=\{0\}$.
\end{remark}
Before we prove existence of preimages under $\mathcal{G}_{2s}$, we prove a technical lemma that allows us to reduce the problem of finding a preimage of $\ker \sdiv_{2s} \cap H^j_\delta(S_{2s})$ under $\mathcal{G}_{2s}$ to that of finding a preimage of $\ker \sdiv_{2s} \cap H^j_\delta(S_{2s})$ orthogonal to $\mathbb{E}_{s,\delta}$.
\begin{lemma}\label{lemma:orthogonalpreimage}
Let $\delta$ be in $\mathbb{R}\setminus\mathbb{Z}$, $j>0$ integer, $\varphi_{A\dots F}$ in $H^j_\delta(S_{2s})$. If $\delta < -2s-2$ we can find a $\tilde\zeta_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s})$ and a constant $C$ depending only on $s$, $j$ and $\delta$ such that $\varphi_{A\dots F}+(\mathcal{G}_{2s}\tilde\zeta)_{A\dots F}$ is orthogonal to $\mathbb{E}_{s,\delta}$ and
\begin{align*}
\Vert\tilde\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}&\leq C \Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{align*}
\end{lemma}
\begin{proof}
Let $\{\mu^i_{A\dots F}\}_{i}$ be an $L^2_{-\delta-2s-2}(S_{2s})$ orthonormal basis for the finite dimensional space $\mathcal{G}_{2s}(\mathbb{E}_{s,\delta})\subset H^{j+2s-2}_{-\delta-2s-2}(S_{2s})$.
Due to the splitting $\mathbb{F}_{s,\delta}=\mathbb{E}_{s,\delta}\oplus (\mathbb{F}_{s,\delta}\cap \ker\mathcal{G}_{2s})$, we have for each $\mu^i_{A\dots F}$ a unique $\xi^i_{A\dots F}\in \mathbb{E}_{s,\delta}$ such that $(\mathcal{G}_{2s}\xi^i)_{A\dots F}=\mu^i_{A\dots F}$, and $\{\xi^i_{A\dots F}\}_{i}$ span $\mathbb{E}_{s,\delta}$.
Let $\eta^i_{A\dots F}\equiv\lAngle r\rAngle^{2\delta+4s+1}\mu^i_{A\dots F}$.
By \cite[Lemma 5.2]{Cantor:1981bs} we have that $\eta^i_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s})$.
We can now approximate $\xi^i_{A\dots F}\in H^{j+4s-3}_{-\delta-3}$ by choosing a sequence
$\{ \xi^{i,k}_{A\dots F} \}_{k \in \mathbb{N}} \subset C^\infty_0(S_{2s})$ such that $\Vert\xi^i_{A\dots F}-\xi^{i,k}_{A\dots F}\Vert_{j+4s-3,-\delta-3}\rightarrow 0$ for $k \to \infty$.
Repeated integration by parts gives
\begin{align*}
\lAngle(\mathcal{G}_{2s}\eta^i)_{A\dots F},\xi^{j,k}_{A\dots F}\rAngle_{L^2}
={}&-\lAngle\eta^i_{A\dots F},(\mathcal{G}_{2s}\xi^{j,k})_{A\dots F}\rAngle_{L^2},
\end{align*}
where the boundary terms vanish due to the compact support of $\xi^{i,k}_{A\dots F}$.
In the limit $k\rightarrow \infty$, this gives, making use of the definition of the weighted spaces,
\begin{align*}
\lAngle(\mathcal{G}_{2s}\eta^i)_{A\dots F},\xi^j_{A\dots F}\rAngle_{L^2}
={}&-\lAngle\eta^i_{A\dots F},(\mathcal{G}_{2s}\xi^j)_{A\dots F}\rAngle_{L^2} \\
={}& -\lAngle\mu^i_{A\dots F},\mu^j_{A\dots F}\rAngle_{L^2_{-\delta-2s-2}}=-\delta^{ij}.
\end{align*}
Now, let
\begin{align*}
\tilde \zeta_{A\dots F}\equiv{}&\sum_{i}\lAngle \varphi_{A\dots F},\xi^i_{A\dots F}\rAngle_{L^2}\eta^i_{A\dots F}.
\end{align*}
This gives that $\lAngle\varphi_{A\dots F}+(\mathcal{G}_{2s}\tilde\zeta)_{A\dots F},\xi^i_{A\dots F}\rAngle_{L^2}=0$ for all $i$. Hence, $\varphi_{A\dots F}+(\mathcal{G}_{2s}\tilde\zeta)_{A\dots F}$ is orthogonal to $\mathbb{E}_{s,\delta}$.
We also get the estimates
\begin{align}
\Vert\tilde \zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\leq{}&\sum_{i}|\lAngle \varphi_{A\dots F},\xi^i_{A\dots F}\rAngle_{L^2}|\thinspace\Vert\eta^i_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\nonumber\\
\leq{}&\sum_{i}\Vert\varphi_{A\dots F}\Vert_{L^2_\delta}\Vert\xi^i_{A\dots F}\Vert_{L^2_{-3-\delta}}\Vert\eta^i_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\nonumber\\
\leq{}&C \Vert\varphi_{A\dots F}\Vert_{j,\delta},
\end{align}
where $C$ only depends on $s$, $j$ and $\delta$.
\end{proof}
\subsection{The spin-1 case}
To show how we can solve the equation $\varphi_{A\dots F}
=(\mathcal{G}_{2s}\zeta)_{A\dots F}$, we begin with the spin-1 case to illustrate the idea. We then use the same ideas for general spin in Proposition~\ref{proprepspingeneral}.
\begin{proposition}\label{proprepspin1}
Let $\delta$ be in $\mathbb{R}\setminus\mathbb{Z}$, $j$ a positive integer, $\varphi_{AB}$ in $H^j_\delta(S_2)$ such that $D^{AB}\varphi_{AB}=0$.
Then there exist a spinor field $\zeta_{AB}\in H^{j+1}_{\delta+1}(S_2)$ and a constant $C$ depending only on $\delta$ and $j$ such that
\begin{align*}
\varphi_{AB}&=(\mathcal{G}_2\zeta)_{AB},\\
\Vert\zeta_{AB}\Vert_{j+1,\delta+1}&\leq C \Vert\varphi_{AB}\Vert_{j,\delta}.
\end{align*}
\end{proposition}
\begin{proof}
By Remark~\ref{rem:Etrivial} we see that $\varphi_{AB}$ is automatically orthogonal to $\mathbb{E}_{1,\delta}$ if $\delta\geq -4$. If $\delta< -4$, we can use Lemma~\ref{lemma:orthogonalpreimage} to
construct $\tilde\zeta_{AB}\in H^{j+1}_{\delta+1}(S_{2})$ such that $\varphi_{AB}+(\mathcal{G}_{2}\tilde\zeta)_{AB}\in \ker\sdiv_2\cap H^j_\delta(S_2)$ is orthogonal to $\mathbb{E}_{1,\delta}$. The estimates in Lemma~\ref{lemma:orthogonalpreimage} are of the right type, so if we can prove the proposition for $\varphi_{AB}+(\mathcal{G}_{2}\tilde\zeta)_{AB}$ instead of $\varphi_{AB}$, we are done. We can therefore in the rest of the proof without loss of generality assume that $\varphi_{AB}$ is orthogonal to $\mathbb{E}_{1,\delta}$. From now on, to make the link with the language of forms clear, we replace the operator $\mathcal G_2 = 2\scurl_2$ with the curl operator $\scurl_2$.
Let $\theta_{AB}$ be in $\mathbb{F}_{1,\delta}\cap \ker\scurl_2$. The field $\theta_{AB}$ is then in $\mathcal{P}^{<-3-\delta}(S_2)$ and therefore real analytic. Furthermore it is curl-free (\emph{i.e.} in $\ker \scurl_2$). Using Proposition \ref{prop:integrability3}, the sequence
$$
\mathcal{P}^{<-4-\delta}(S_0)\stackrel{\stwist_0}{\longrightarrow} \mathcal{P}^{<-3-\delta}(S_2) \stackrel{\scurl_2}{\longrightarrow} \mathcal{P}^{<-2-\delta}(S_2),
$$
is exact and, therefore, $\theta_{AB}$ can be written as a gradient $D_{AB}\eta = (\stwist_0\eta)_{AB} =\theta_{AB}$, where $\eta\in \mathcal{P}^{<-2-\delta}(S_2)\subset H^1_{-2-\delta}(S_2)$. Then, by Lemma~\ref{orthogonalitylemma}, $\varphi_{AB}$ is $L_2$-orthogonal to $\theta_{AB}$. As $\theta_{AB}$ was arbitrary in $\mathbb{F}_{1,\delta}\cap \ker\scurl_2$ and $\varphi_{AB}$ was by assumption orthogonal to $\mathbb{E}_{1,\delta}$, we have that $\varphi_{AB}$ is orthogonal to all of $\mathbb{F}_{1,\delta}$.
The Laplacian $\Delta_2: H^{j+2}_{\delta+2}(S_2)\rightarrow H^j_\delta(S_2)$ is formally self-adjoint and has closed range and finite dimensional kernel -- see \cite{Cantor:1981bs,McOwen:1980gz,Lockhart:1983ht} for details. By Fredholm's alternative, there exists a $\theta_{AB}\in H^{j+2}_{\delta+2}(S_2)$ such that $\varphi_{AB}=(\Delta_2\theta)_{AB}$. Using Proposition~\ref{prop:elliptictoolbox}, we can modify $\theta_{AB}$ within the class $\ker\Delta_2\cap H^{j+2}_{\delta+2}$ to obtain the estimate
\begin{equation*}
\Vert\theta_{AB}\Vert_{j+2,\delta+2}\leq C \Vert \varphi_{AB}\Vert_{j,\delta},
\end{equation*}
where $C$ only depends on $j$ and $\delta$.
Now, we can re-express the Laplacian $\Delta_2$ as
\begin{equation*}
\varphi_{AB}=(\Delta_2\theta)_{AB}=
-2 (\scurl_2 \scurl_2 \theta)_{AB} + (\stwist_0 \sdiv_2 \theta)_{AB}.
\end{equation*}
We now want to show that $(\stwist_0 \sdiv_2 \theta)_{AB}$ vanishes (for $\delta <0$) or is in the image of $\scurl_2$ (for $\delta>0$).
By the constraint equation and commutations of the divergence and the Laplace operator, we have
$$
0=(\sdiv_2\varphi)=D^{AB}\varphi_{AB} = D^{AB}\left(\Delta \theta_{AB}\right) = \Delta \left(D^{AB}\varphi_{AB}\right) = (\Delta_0\sdiv_2\theta).
$$
Hence, $(\sdiv_2\theta)\in \ker \Delta_0\cap L^2_{\delta+1}(S_0)$.
If $\delta < 0$, we know that $\ker\Delta_0\cap L^2_{\delta+1}(S_0)$ only contains polynomials with degree $<1$, \emph{i.e.} constants, which means that they are in the kernel of the gradient operator $\stwist_0$. Hence,
\begin{equation*}
\varphi_{AB}= -2 (\scurl_2 \scurl_2 \theta)_{AB}= -(\mathcal{G}_2 \scurl_2 \theta)_{AB}
\end{equation*}
and we can therefore choose $\zeta_{AB}=-(\scurl_2\theta)_{AB}$, and we get
\begin{equation*}
\Vert\zeta_{AB}\Vert_{j+1,\delta+1}\leq \Vert\theta_{AB}\Vert_{j+2,\delta+2}\leq C\Vert\varphi_{AB}\Vert_{j,\delta}.
\end{equation*}
If $\delta > 0$, we need to be more careful. Let $\Omega\equiv \ker\Delta_0\cap L^2_{\delta+1}(S_0)$, \emph{i.e.} the set of harmonic polynomials with degree strictly smaller than $\delta+1$. Then $\stwist_0(\Omega)\subset L^2_{\delta}(S_2)$ is also a finite dimensional space of smooth fields. Since $\sdiv_2 \stwist_0 = \Delta_0$, we have the following diagram
\begin{equation*}
\xymatrix{
\Omega\subset\ker\left(\Delta_0\right)\subset \mathcal{S}_0\ar[r]^{\stwist_0}&\stwist_0\left(\Omega\right)\subset\mathcal{S}_2\ar[r]^{\sdiv_2}&\{0\}\subset\mathcal{S}_2\\
&(\scurl_2){}^{-1}\left(\stwist_0\left(\Omega\right)\right)\subset \mathcal{S}_2 \ar[u]^{\scurl_{2}}
}.
\end{equation*}
Using the integrability condition stated in Proposition \ref{prop:integrability3}, and Remark~\ref{integrability4}, and more specifically by
$$
\mathcal{P}^{<\delta+1}(S_2)\stackrel{\scurl_2}{\longrightarrow} \mathcal{P}^{<\delta}(S_2) \stackrel{\sdiv_2}{\longrightarrow} \mathcal{P}^{<\delta-1}(S_0),
$$
we can define an a priori non unique linear mapping $\mathcal{T}: \stwist_0(\Omega)\rightarrow \mathcal{S}_2$, such that $\scurl_2\mathcal{T}$ acts as the identity on $\stwist_0\left(\Omega\right)$. As a linear operator from the finite dimensional space $\stwist_0(\Omega)\subset H^j_\delta(S_2)$ into $ H^{j+2}_{\delta+2}(S_2)$ (endowed with their respective induced Sobolev norms), $\mathcal{T}$ is bounded and, therefore, there exists a constant $C$, depending on the choice of the mapping $\mathcal{T}$, such that
\begin{align*}
\Vert(\mathcal{T}\stwist_0\sdiv_2\theta)_{AB}\Vert_{j+1,\delta+1}
&\leq C \Vert (\stwist_0\sdiv_2\theta)_{AB}\Vert_{j,\delta},\\
(\scurl_2\mathcal{T}\stwist_0\sdiv_2\theta)_{AB}
&=(\stwist_0\sdiv_2\theta)_{AB}.
\end{align*}
Now, let
\begin{equation*}
\zeta_{AB}=-(\scurl_2\theta)_{AB}+(\mathcal{T}\stwist_0\sdiv_2\theta)_{AB}.
\end{equation*}
This gives the desired relations
\begin{align*}
(\mathcal{G}_2\zeta)_{AB}&=
2(\scurl_2\zeta)_{AB}=
-2(\scurl_2\scurl_2\theta)_{AB}+
2(\stwist_0\sdiv_4\theta)_{AB}=\varphi_{AB},\\
\Vert\zeta_{AB}\Vert_{j+1,\delta+1}&\leq
\Vert\theta_{AB}\Vert_{j+2,\delta+2}
+C \Vert (\stwist_0\sdiv_2\theta)_{AB}\Vert_{j,\delta}
\leq C\Vert\varphi_{AB}\Vert_{j,\delta}.
\end{align*}
\end{proof}
\subsection{The spin-$s$ case}
\begin{proposition}\label{proprepspingeneral}
Let $\delta$ be in $\mathbb{R}\setminus\mathbb{Z}$, $j>0$ integer, $\varphi_{A\dots F}$ in $\ker \sdiv_{2s} \cap H^j_\delta(S_{2s})$. Let $m=\lfloor s\rfloor$, \emph{i.e.} the largest integer such that $m\leq s$.
Then there exist a spinor field $\zeta_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s})$ and a constant $C$ depending only on $\delta$ and $j$ such that
\begin{align*}
\varphi_{A\dots F}&=(\mathcal{G}_{2s}\zeta)_{A\dots F},\\
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}&\leq C \Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{align*}
\end{proposition}
\begin{proof}
By Remark~\ref{rem:Etrivial} we see that $\varphi_{A\dots F}$ is automatically orthogonal to $\mathbb{E}_{s,\delta}$ if $\delta\geq -2s-2$. If $\delta< -2s-2$, we can use Lemma~\ref{lemma:orthogonalpreimage} to
construct $\tilde\zeta_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s})$ such that $\varphi_{A\dots F}+(\mathcal{G}_{2s}\tilde\zeta)_{A\dots F}\in \ker\sdiv_{2s}\cap H^j_\delta(S_{2s})$ is orthogonal to $\mathbb{E}_{s,\delta}$. The estimates in Lemma~\ref{lemma:orthogonalpreimage} are of the right type, so if we can prove the proposition for $\varphi_{A\dots F}+(\mathcal{G}_{2s}\tilde\zeta)_{A\dots F}$ instead of $\varphi_{A\dots F}$, we are done. Without loss of generality, we can therefore for the rest of the proof assume that $\varphi_{A\dots F}$ is orthogonal to $\mathbb{E}_{s,\delta}$.
Now, we will establish that $\varphi_{A\dots F}$ is orthogonal to $\mathbb{F}_{s,\delta}$ by using the constraint equation and the orthogonality to $\mathbb{E}_{s,\delta}$. The spinors in $\mathbb{F}_{s,\delta}\subset \mathcal{P}^{<-3-\delta}(S_{2s})$ are polynomial, so we can use Proposition~\ref{prop:integrability3} to conclude that $\mathbb{F}_{s,\delta}\cap \ker\mathcal{G}_{2s}=\mathbb{F}_{s,\delta}\cap \stwist_{2s-2}(\mathcal{P}^{<-2-\delta}(S_{2s-2}))$. But $\mathcal{P}^{<-2-\delta}(S_{2s-2})\subset H^1_{-2-\delta}(S_{2s-2})$, so Lemma~\ref{orthogonalitylemma} gives that $\lAngle \varphi_{A\dots F}, \zeta_{A\dots F}\rAngle_{L^2}=0$ for all $\zeta_{A\dots F}\in \mathbb{F}_{s,\delta}\cap \stwist_{2s-2}(\mathcal{P}^{<-2-\delta}(S_{2s-2}))=\mathbb{F}_{s,\delta}\cap \ker\mathcal{G}_{2s}$. By assumption $\varphi_{A\dots F}$ is orthogonal to $\mathbb{E}_{s,\delta}$ and therefore orthogonal to all of $\mathbb{F}_{s,\delta}$.
The operator $\Delta^m_{2s}: H^{j+2m}_{\delta+2m}(S_{2s})\rightarrow H^j_\delta(S_{2s})$ is formally self-adjoint and has closed range and finite dimensional kernel -- see \cite{Cantor:1981bs,McOwen:1980gz,Lockhart:1983ht} for details. By the Fredholm alternative, and the orthogonality, there exists a $\theta_{A\dots F}\in H^{j+2m}_{\delta+2m}(S_{2s})$ such that $\varphi_{A\dots F}=(\Delta^m_{2s}\theta)_{A\dots F}$.
Using Proposition~\ref{prop:elliptictoolbox} we can modify $\theta_{A\dots F}$ within the class
$\ker\Delta^m_{2s}\cap H^{j+2m}_{\delta+2m}(S_{2s})$ to obtain the estimate
\begin{equation*}
\Vert\theta_{A\dots F}\Vert_{j+2m,\delta+2m}\leq C \Vert \varphi_{A\dots F}\Vert_{j,\delta},
\end{equation*}
where $C$ only depends on $j$ and $\delta$.
For integer spin we can express the $\Delta^m_{2s}$ operator as
\begin{equation*}
(\Delta^{m}_{2s}\theta)_{A\dots F}=
(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}-(-2)^{1-m}(\mathcal{G}_{2s}\scurl_{2s}\theta)_{A\dots F}.
\end{equation*}
For half integer spin we can express the $\Delta^m_{2s}$ operator as
$2s=2m+1$
\begin{equation*}
(\Delta^{m}_{2s}\theta)_{A\dots F}=
(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
+(-2)^{-m}(\mathcal{G}_{2s}\theta)_{A\dots F} .
\end{equation*}
We now want to show that $(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}$ vanishes
(for $\delta<0$) or is in the image of $\mathcal{G}_{2s}$ (for $\delta>0)$.
By the constraint equation and the fact that the divergence and the Laplace operator commute, we have
$$
0=(\sdiv_{2s}\varphi)_{C\dots F}=D^{AB}\varphi_{A\dots F} = D^{AB}\left(\Delta_{2s}^m
\theta_{A\dots F}\right) = \Delta^m_{2s-2} \left(D^{AB}\varphi_{A\dots F}\right) =
(\Delta^m_{2s-2} \sdiv_{2s}\theta)_{C\dots F}.
$$
Hence, $(\sdiv_{2s}\theta)_{C\dots F}$ is in $\ker \Delta^m_{2s-2}\cap
L^2_{\delta+2m-1}(S_{2s-2})$.
If $\delta < 0$, we know that fields in $\ker \Delta^m_{2s-2}\cap L^2_{\delta+2m-1}(S_{2s-2})$ are
in $\mathcal{P}^{<2m-1}(S_{2s-2})$, i. e. they are spanned by constant spinors times polynomials with maximal
degree $2m-2$. They therefore belong to the kernel of the homogeneous order $2m-1$ operator
$\stwist_{2s-2}\mathcal{F}_{2s-2}$. Hence, $(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)
_{A\dots F}=0$ and we get
\begin{equation*}
\varphi_{A\dots F}=
-(-2)^{1-m}(\mathcal{G}_{2s}\scurl_{2s}\theta)_{A\dots F},
\end{equation*}
for integer spin, and
\begin{equation*}
\varphi_{A\dots F}=(-2)^{-m}(\mathcal{G}_{2s}\theta)_{A\dots F},
\end{equation*}
for half integer spin.
For integer spin we can therefore choose $\zeta_{A\dots F}=-(-2)^{1-m}(\scurl_{2s}\theta)_{A\dots F}$, and we get
\begin{equation*}
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\leq C \Vert\theta_{A\dots F}\Vert_{j+2m,\delta+2m}\leq C\Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{equation*}
For half integer spin we can choose $\zeta_{A\dots F}=(-2)^{-m}\theta_{A\dots F}$, and we get
\begin{equation*}
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}=(-2)^{-m} \Vert\theta_{A\dots F}\Vert_{j+2m,\delta+2m}\leq C\Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{equation*}
If $\delta > 0$, we need to be more careful. Let $\Omega\equiv \ker(\Delta^m_{2s-2})\cap\text{im}
(\sdiv_{2s})\cap L^2_{\delta+2m-1}(S_{2s-2})$. We know that it is a finite dimensional space of
polynomial fields. $\stwist_{2s-2}\mathcal{F}_{2s-2}(\Omega)\subset L^2_{\delta}(S_{2s})$ is
therefore also a finite dimensional space in $\mathcal{P}^{<\delta}(S_{2s})$.
Using the relations \eqref{eq:divGproperty}, \eqref{LaplacianAsGeven} and \eqref{LaplacianAsGodd}
we get
\begin{equation*}
\Delta^m_{2s-2} \sdiv_{2s}=\sdiv_{2s}\Delta^m_{2s}
=\sdiv_{2s}\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}.
\end{equation*}
Consequently, on $\Omega\subset\text{im}(\sdiv_{2s})$, the following relation holds
$$
\sdiv_{2s}\stwist_{2s-2}\mathcal{F}_{2s-2}\bigr |_\Omega=\Delta^m_{2s}\bigr |_\Omega=0.
$$
The relations between the considered operators can be summarized by
\begin{equation*}
\xymatrix{
\Omega\subset\ker\left(\Delta^m_{2s-2}\right)\subset \mathcal{S}_{2s-2}\ar[r]^{\stwist_{2s-2}\mathcal{F}_{2s-2}}&\stwist_{2s-2}\mathcal{F}_{2s-2}\left(\Omega\right)\subset\mathcal{S}_{2s}\ar[r]^{\sdiv_{2s}}&\{0\}\subset\mathcal{S}_{2s-2}\\
&(\mathcal{G}_{2s}){}^{-1}\left(\stwist_{2s-2}\mathcal{F}_{2s-2}\left(\Omega\right)\right)\subset
\mathcal{S}_{2s}\ar[u]^{\mathcal{G}_{2s}}
}.
\end{equation*}
Using the integrability condition stated by the exact sequence in Proposition~\ref{prop:integrability3} applied
to polynomials (cf. Remark~\ref{integrability4}), and more specifically
$$
\mathcal{P}^{<\delta + 2s-1}(S_{2s})\stackrel{\mathcal{G}_{2s}}{\longrightarrow} \mathcal{P}^{<\delta }(S_{2s}) \stackrel{\sdiv_{2s}}
{\longrightarrow}\mathcal{P}^{<\delta -1}(S_{2s-2}),
$$
we can define an a priori non unique linear mapping $\mathcal{T}:
\stwist_{2s-2}\mathcal{F}_{2s-2}\left(\Omega\right) \rightarrow \mathcal{S}_{2s} $ such that
$\mathcal{G}_{2s}\mathcal{T}$ is the identity operator on $\stwist_{2s-2}\mathcal{F}_{2s-2}(\Omega)$. As
a linear operator on the finite dimensional space $\stwist_{2s-2}\mathcal{F}_{2s-2}(\Omega)\subset
H^j_\delta(S_{2s})$ into $H^{j+2s-1}_{\delta+2s-1}(S_{2s})$ (endowed with their
respective induced Sobolev norms), $\mathcal{T}$ is bounded and, therefore, there exists a constant $C$,
depending on the choice of the operator $\mathcal{T}$, such that
\begin{align*}
\Vert(\mathcal{T}\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}
&\leq C \Vert (\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}\Vert_{j,\delta},\\
(\mathcal{G}_{2s}\mathcal{T}\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
&=(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}.
\end{align*}
Now, for integer spin we can therefore choose
\begin{equation*}
\zeta_{A\dots F}=
(\mathcal{T}\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
-(-2)^{1-m}(\scurl_{2s}\theta)_{A\dots F}.
\end{equation*}
This gives the desired relations
\begin{align*}
(\mathcal{G}_{2s}\zeta)_{A\dots F}&=
(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
-(-2)^{1-m}(\mathcal{G}_{2s}\scurl_{2s}\theta)_{A\dots F}\nonumber\\
&=(\Delta^m_{2s}\theta)_{A\dots F}=\varphi_{A\dots F},\\
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}&\leq
C(\Vert\theta_{A\dots F}\Vert_{j+2m,\delta+2m}
+ \Vert (\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}\Vert_{j,\delta})\nonumber\\
&\leq C\Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{align*}
For half integer spin, we can choose
\begin{equation*}
\zeta_{A\dots F}=
(\mathcal{T}\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
+(-2)^{-m}\theta_{A\dots F}.
\end{equation*}
This gives the desired relations
\begin{align*}
(\mathcal{G}_{2s}\zeta)_{A\dots F}&=
(\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}
(-2)^{-m}(\mathcal{G}_{2s}\theta)_{A\dots F}\nonumber\\
&=(\Delta^m_{2s}\theta)_{A\dots F}=\varphi_{A\dots F},\\
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}&\leq
C(\Vert\theta_{A\dots F}\Vert_{j+2m,\delta+2m}
+ \Vert (\stwist_{2s-2}\mathcal{F}_{2s-2}\sdiv_{2s}\theta)_{A\dots F}\Vert_{j,\delta})\nonumber\\
&\leq C\Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{align*}
\end{proof}
\section{Estimates for solutions of the scalar wave equation with initial data with arbitrary weight} \label{sec:estsol}
This section contains complementary results for the study of the decay of the solution of the wave equation for the Cauchy problem such as the one stated in \cite{Klainerman:1985wn,Klainerman:1986wra} (using the vector field method), in \cite{MR862696} (using the integral representation), and \cite{DAncona:2001kr} (using Strichartz estimates). The purpose is to link precisely the asymptotic behavior of the initial data at $i^0$ and the asymptotic behavior in the future region $t\geq 0$.
It is important to remark that these different decay results for the wave equation hold for different regularities of the initial data $(f,g)$. In all the aforementioned results, the limiting factor is the use of the Sobolev embedding from $H^2$ into $L^\infty$, and more precisely the way it is used. If the Sobolev embedding is used at the level of both $f$ and $g$, we need $(f, g) \in H^j_\delta \times H^{j-1}_{\delta-1}$ for $j \geq 3$. This phenomenon occurs for instance in the work of Asakura \cite{MR862696}, who is relies on the integral representation of solutions. Energy methods can be used to give a result with weaker regularity assumtions. The Klainerman-Sobolev inequality \cite[Theorem 1]{Klainerman:1987hz} yields a decay estimate with weight $\delta=-3/2$ for $j=2$. Further, using the conformal compactification of Minkowski space to a subset of the Einstein cylinder, and the conformal transformation properties of the wave equation, gives using standard estimates for the wave equation in the Einstein cylinder the decay result for $\delta = -2$ and $j=2$, cf. \cite[Section 6.7]{MR1466700}. It appears to be an open problem to prove the corresponding decay result with $j=2$ for general weight $\delta$.
In this section we shall prove estimates for general weights $\delta$. Although, for $\delta > 0$, these are not actually \emph{decay} estimates, it will be convenient to refer to them using this term. It goes without saying that the most important applications are those with $\delta < 0$. In the following, we shall consider the Cauchy problem
\begin{equation} \label{cauchywave} \left\{
\begin{array}{l}
\square \phi =0 , \\
\phi|_{t=0}=f \in H^j_{\delta}(\R^3, \C) , \\
\partial_t \phi|_{t=0}=g \in H^{j-1}_{\delta-1}(\R^3,\C).
\end{array}\right.
\end{equation}
The following representation formula then holds (\cite{Evans:2010wj} on flat space-time or \cite[Theorem 5.3.3]{Friedlander:1975ub} for arbitrary curved background).
\begin{lemma}\label{integralwave}
The solution of the Cauchy problem \eqref{cauchywave} is given by the representation formula
$$
\phi(t,x) = \frac{1}{4\pi} \left(\int_{\mathbb{S}^2}t\left(g(x+t\omega) + \partial_\omega f (x+t\omega)\right) + f(x+t\omega) \ud \mu_{\mathbb{S}^2}\right),
$$
where $\mathbb{S}^2$ is the unit 2-sphere and $\partial_\omega$ is the derivative in the unit outer normal direction $\omega$ to $\mathbb{S}^2$.
\end{lemma}
Making use of the representation formula stated in Lemma~\ref{integralwave}, and of Proposition~\ref{prop:decaysob}, gives the following result.
\begin{proposition}\label{prop:decaywave} Let $j\geq 3$ and $\delta$ in $\R$, and let $\phi$ be a solution to the Cauchy problem \eqref{cauchywave}. The following inequality holds in the region $t > 0$.
\begin{equation*}
|\phi(t,x)|\leq C \left(\Vert f\Vert_{3,\delta}+\Vert g\Vert_{2,\delta-1}\right)
\begin{cases}
\lAngle v\rAngle^{-1}\lAngle u\rAngle^{1+\delta} &\text{ if } \delta <-1 , \\
\dfrac{\log\lAngle v \rAngle-\log\lAngle u \rAngle}{\lAngle v \rAngle -\lAngle u \rAngle} &\text{ if } \delta =-1 , \\
\lAngle v\rAngle^{\delta} &\text{ if } \delta >-1 .
\end{cases}
\end{equation*}
If, furthermore, $(k,l,m)$ is a triple of non-negative integers, $j\geq 3+k+l+m$, the following pointwise inequality holds, for all $t>0$ and $r>1$,
\begin{equation*}
| \partial_v^k\partial_u^l\snabla^m\phi |\leq C \left( \Vert f\Vert_{3+k+l+m, \delta} + \Vert g\Vert_{2+k+l+m, \delta-1}\right)
\begin{cases}
\lAngle u\rAngle^{1+\delta-l} \lAngle v\rAngle^{-1-k-m} &\text{ if } \delta<l-1\\
\dfrac{\log\lAngle v \rAngle-\log\lAngle u \rAngle}{\lAngle v \rAngle^{l+m}\left(\lAngle v \rAngle -\lAngle u \rAngle\right)} &\text{ if } \delta = l-1\\
\lAngle v\rAngle^{\delta -l-m-k} &\text{ if } \delta>l-1,
\end{cases}
\end{equation*}
where $\partial_u = \frac{1}{2} (\partial_t-\partial_r)$ and $ \partial_v = \frac{1}{2} (\partial_t+\partial_r)$ are respectively the outgoing and ingoing null directions.
\end{proposition}
\begin{remark}\label{rem:decaywave}
\begin{enumerate}
\item For $\delta=-1$, it is important to notice that the following inequalities hold
$$
\dfrac{\log\lAngle v \rAngle-\log\lAngle u \rAngle}{\lAngle v \rAngle -\lAngle u \rAngle} \leq \dfrac{1}{\lAngle u\rAngle } \text{ and } \dfrac{\log\lAngle v \rAngle-\log\lAngle u \rAngle}{\lAngle v \rAngle -\lAngle u \rAngle} \leq C \dfrac{\log\lAngle v \rAngle}{\lAngle v \rAngle}.
$$
As a consequence, in the interior region $t>3r$, the decay result for $\phi$ is
$$
\left|\phi(t,x)\right| \leq \dfrac{\widetilde C}{\lAngle v \rAngle },
$$
and, in the exterior region $3r>t >\frac{r}{3}$,
$$
\left|\phi(t,x)\right| \leq \widetilde C\dfrac{\log\lAngle v \rAngle}{\lAngle v \rAngle }.
$$
The constant $\widetilde C$ depends on the norms of $f$ and $g$ as above.
\item It is important to note that, in the interior region, $\lAngle u \rAngle$ and $\lAngle v \rAngle$ are equivalent, and, as a consequence, the general estimate holds, for all weight $\delta$:
\begin{equation*}
| \partial_v^k\partial_u^l\snabla^m\phi|\leq \widetilde C \lAngle v\rAngle^{\delta -l-m-k}.
\end{equation*}
\end{enumerate}
\end{remark}
For the proof we will need some integral estimates.
\begin{lemma}\label{integralextimateshypergeom}
For any $\delta$ in $\mathbb{R}$, we have the following integral estimates, for all $t>0$ and $x$ in $\R^3$,
$$
\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta} \ud\mu_{\mathbb{S}^2}\leq 8\pi \max\left(1, \dfrac{1}{|2+\delta|}\right)\dfrac{\max\left(\lAngle u\rAngle^{\delta+2}, \lAngle v\rAngle^{\delta+2} \right)}{\lAngle v\rAngle \left(\lAngle u\rAngle + \lAngle v\rAngle\right)} \text{ for }\delta\neq -2
$$
and
$$
\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta} \ud\mu_{\mathbb{S}^2} = 8\pi\dfrac{\log \left(\dfrac{\lAngle u\rAngle}{\lAngle v\rAngle} \right)}{(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} \text{ for } \delta=-2
$$
\end{lemma}
\begin{remark}
\begin{enumerate}
\item For $\delta \neq -2$, the upper bounds become lower bounds if one replaces $\max\left(1, \dfrac{1}{|2+\delta|}\right)$ by $\min\left(1, \dfrac{1}{|2+\delta|}\right)$.
\item It is important to notice that, for $\delta \in (-2,-1]$, the estimate
$$
\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta} \ud \mu_{\mathbb{S}^2} \leq C \lAngle v\rAngle ^\delta
$$
is \emph{stronger} than
$$
\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta} \ud \mu_{\mathbb{S}^2} \leq C \dfrac{\lAngle u\rAngle ^{1+\delta}}{\lAngle v\rAngle}
$$
since
$$
\lAngle v \rAngle^\delta = \lAngle u \rAngle ^{1+\delta} \lAngle v \rAngle^{-1} \lAngle v \rAngle^{\delta+1}\lAngle u \rAngle ^{-(1+\delta)}\leq \lAngle u \rAngle ^{1+\delta} \lAngle v \rAngle^{-1}
$$
(since $\delta+1<0$). It should also be noted that this result agrees with the estimate stated by Asakura in \cite{MR862696}. His assumptions on the initial data
$$
f = \mathcal{O}(\lAngle r\rAngle^\delta), g = \mathcal{O}(\lAngle r\rAngle^\delta),
$$
as well as their derivatives, implies that the integral $\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta}\ud\mu_{\mathbb{S}^2}$ gives the asymptotic behavior of the solutions of the linear wave equation.
\end{enumerate}
\end{remark}
\begin{proof}
Let $(t,x)$ be fixed and consider the sphere $S(x,t)$ with center $x$ and radius $t$. Let $q$ be a point on the sphere $S(x,t)$. The coordinates of $q$ are then given by $(\theta,\phi)$ defined by:
\begin{itemize}
\item in the 2-plane containing the origin $o$, the point $x$ and $q$, $\theta$ is the oriented angle
$$
\theta = (\vec{xo}, \vec{xq}) \in (0,\pi);
$$
\item in the plane orthogonal to $\vec{ox}$ and passing through $x$, one chooses a direction of origin. The direction of the orthogonal projection of $\vec{xq}$ on this plane is labeled by an angle $\phi$ belonging to $(0,2\pi)$.
\end{itemize}
The integral can now be rewritten as
\begin{align*}
\int_{\mathbb{S}^2}\lAngle |x+t\omega|\rAngle^{\delta} \ud \mu_{\mathbb{S}^2} ={}& \int_0^{2\pi}\int_0^{\pi}(1+r^2 +t^2-2tr\cos\theta)^{\delta/2}\sin\theta\ud \theta\ud \phi \\
={}&
\begin{cases}
8\pi\dfrac{\lAngle u\rAngle^{2+\delta}-\lAngle v\rAngle^{2+\delta}}{(2+\delta)(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} &\text{ if } \delta \neq -2 \text{ and } \lAngle u\rAngle\neq \lAngle v\rAngle,\\
8\pi\dfrac{\log \left(\frac{\lAngle u\rAngle}{\lAngle v\rAngle} \right)}{(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} &\text{ if } \delta=-2 \text{ and } \lAngle u\rAngle\neq \lAngle v\rAngle,\\
4\pi \lAngle v\rAngle^{\delta} &\text{ if } \lAngle u\rAngle = \lAngle v\rAngle.\\
\end{cases}
\end{align*}
We note that $\lAngle u\rAngle$ and $\lAngle v \rAngle$ always satisfy
\begin{itemize}
\item $\lAngle u\rAngle\geq 1$ and $\lAngle v\rAngle\geq 1$;
\item for $t\geq 0$, $\lAngle u\rAngle\leq \lAngle v\rAngle$.
\end{itemize}
Let $F_\kappa$ denote the function defined, for $\kappa\geq 0$ and $z$ in $(0,1]$,
$$F_\kappa(z)=
\begin{cases}
\dfrac{1-z^{\kappa}}{\kappa(1-z)} & \text{ if } \kappa > 0 \text{ and } z\neq 1,\\
\dfrac{-\log z}{1-z} &\text{ if } \kappa = 0 \text{ and } z\neq 1,\\
1 &\text{ if } z=1.
\end{cases}
$$
For $\kappa>0$, $F_\kappa$ is a continuous monotonic non-negative function on $(0,1]$. It is bounded from above by $C_\kappa = \max(1, \kappa^{-1})$ and from below by $c_\kappa = \min(1, \kappa^{-1})$ on $(0,1]$. For all $\delta$, the following identities hold,
$$
\dfrac{\lAngle u\rAngle^{2+\delta}-\lAngle v\rAngle^{2+\delta}}{(2+\delta)(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} =
\begin{cases}
\dfrac{\lAngle u\rAngle^{\delta+2}}{\lAngle v\rAngle \left( \lAngle v\rAngle+\lAngle u\rAngle\right)}F_{-\delta-2}\left( \dfrac{\lAngle u\rAngle}{\lAngle v\rAngle}\right) & \text{ if } \delta+2<0.\\
\dfrac{\lAngle v\rAngle^{\delta+1}}{\lAngle v\rAngle+\lAngle u\rAngle}F_{\delta+2} \left(\dfrac{\lAngle u\rAngle}{\lAngle v\rAngle}\right) & \text{ if } \delta+2\geq 0.
\end{cases}
$$
Using the bounds on $F_\kappa$, one gets
\begin{align*}
\dfrac{\lAngle u\rAngle^{2+\delta}-\lAngle v\rAngle^{2+\delta}}{(2+\delta)(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} &\leq
\begin{cases}
C_{-\delta-2}\dfrac{\lAngle u\rAngle^{\delta+2}}{\lAngle v\rAngle \left( \lAngle v\rAngle+\lAngle u\rAngle\right)}& \text{ if } \delta<-2\\
C_{\delta+2}\dfrac{\lAngle v\rAngle^{\delta+1}}{\lAngle v\rAngle+\lAngle u\rAngle} & \text{ if } \delta>-2,
\end{cases}\\
\intertext{ and }\dfrac{\lAngle u\rAngle^{2+\delta}-\lAngle v\rAngle^{2+\delta}}{(2+\delta)(\lAngle u\rAngle^2 -\lAngle v\rAngle^2)} &\geq
\begin{cases}
c_{-\delta-2}\dfrac{\lAngle u\rAngle^{\delta+2}}{\lAngle v\rAngle \left( \lAngle v\rAngle+\lAngle u\rAngle\right)}& \text{ if } \delta<-2\\
c_{\delta+2}\dfrac{\lAngle v\rAngle^{\delta+1}}{\lAngle v\rAngle+\lAngle u\rAngle} & \text{ if } \delta>-2.
\end{cases}
\end{align*}
\end{proof}
\begin{remark} These inequalities actually provide us with \emph{global} estimates for the solution of the wave equation, that is to say inequalities valid both on the exterior and the interior regions.
\end{remark}
\begin{proof}[Proof of Proposition~\ref{prop:decaywave}] Using Proposition~\ref{prop:decaysob}, one knows that if $f\in H^j_{\delta}$, $g\in H^{j-1}_{\delta-1}$, $j\geq 2+n$ and $j\geq 3+m$, there is constant $C$ such that
\begin{eqnarray*}
|D^n f(y)| & \leq & C\lAngle y\rAngle^{\delta-n}\Vert f\Vert_{2+n,\delta}\\
|D^m g(y)| & \leq & C\lAngle y\rAngle^{\delta-1-m}\Vert g\Vert_{2+m,\delta-1}.
\end{eqnarray*}
Using the representation formula stated in Lemma \ref{integralwave}, one gets immediately
\begin{gather*}
|\phi(t,x)|\leq C\left(\Vert f\Vert_{3,\delta}+\Vert g\Vert_{2,\delta-1}\right)\int_{\mathbb{S}^2}\left(\lAngle|x+t\omega|\rAngle^{\delta} +t\lAngle|x+t\omega|\rAngle^{\delta-1}\right)\ud \mu_{\mathbb{S}^2}.
\end{gather*}
We can use the estimate $t\leq (\lAngle u\rAngle^2+\lAngle v\rAngle^2)^{1/2}$ and Lemma~\ref{integralextimateshypergeom} to obtain global estimates for solutions of the wave equation. We compare the contributions from $f$ in $H^j_{\delta}$ and $g$ in $H^{j-1}_{\delta-1}$ in the following table. Its first column is the range of weights $\delta$ considered. The second column contains the asymptotic behavior of
$$
\int_{\mathbb{S}^2}\lAngle|x+t\omega|\rAngle^{\delta}\ud \mu_{\mathbb{S}^2}
$$
coming from $f$ in $H^j_{\delta}$. The third column gives the behaviour of
$$
\int_{\mathbb{S}^2} t\lAngle|x+t\omega|\rAngle^{\delta-1} \ud \mu_{\mathbb{S}^2}
$$
coming from $\partial f$ and $g$ in $H^{j-1}_{\delta-1}$. The estimates stated in these columns are written in such a way that the first factor gives the estimate which is multiplied by bounded quantities, except when considering logarithmic terms. Finally, the last column gives the estimate for the full solution of the wave equation obtained by summing the previous integrals. We use the function $F_0$ defined by
$$
F_{0}(z)=\frac{-\log(z)}{1-z} \text{ which satisfies } F_0(z)\leq z^{-1} \text{ for all } 0\leq z\leq 1.
$$
\begin{equation*}
\begin{array}{|c|c|c|c|}
\hline
&f & t(\partial_\omega f+g) & f+t\partial_\omega f+tg\\
\hline
\delta<-2 & \dfrac{\lAngle u \rAngle ^{\delta+1}}{\lAngle v\rAngle}\cdot \dfrac{\lAngle u \rAngle}{( \lAngle v\rAngle+\lAngle u\rAngle)} & \dfrac{\lAngle u \rAngle ^{\delta+1}}{\lAngle v\rAngle }\cdot \dfrac{t}{\lAngle v\rAngle+\lAngle u\rAngle} & \dfrac{\lAngle u \rAngle^{\delta+1} }{\lAngle v\rAngle} \\
\hline
\delta=-2 & \dfrac{1}{{\lAngle u\rAngle+\lAngle v\rAngle}}\cdot \dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle} & \dfrac{1}{\lAngle u \rAngle\lAngle v\rAngle}\cdot\dfrac{t}{(\lAngle v\rAngle+\lAngle u\rAngle)} & \dfrac{\lAngle u \rAngle^{\delta+1} }{\lAngle v\rAngle}\\
\hline
\delta=-1 & \lAngle v\rAngle^\delta \cdot \dfrac{\lAngle v\rAngle}{\lAngle v\rAngle+\lAngle u\rAngle} & \dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle} \cdot \dfrac{t}{{\lAngle u\rAngle+\lAngle v\rAngle}} & \dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle}
\\
\hline
-2<\delta<-1 & \lAngle v\rAngle^{\delta}\cdot \dfrac{\lAngle v\rAngle}{\lAngle v\rAngle+\lAngle u\rAngle} & \dfrac{\lAngle u \rAngle^{\delta+1} }{\lAngle v\rAngle}\cdot \dfrac{t}{(\lAngle v\rAngle+\lAngle u\rAngle)} & \dfrac{\lAngle u \rAngle^{\delta+1} }{\lAngle v\rAngle}\\
\hline
\hline
\delta>-1 & \lAngle v\rAngle^{\delta}\cdot \dfrac{\lAngle v\rAngle}{\lAngle v\rAngle+\lAngle u\rAngle} & \lAngle v\rAngle^{\delta}\cdot \dfrac{t}{\lAngle v\rAngle+\lAngle u\rAngle} & \lAngle v \rAngle ^\delta\\
\hline
\end{array}
\end{equation*}
As a consequence, the following pointwise estimate holds for $\phi$: for all $t\geq 0$ and $x$ in $\R^3$,
\begin{equation}\label{ineqwave}
|\phi(t,x)|\leq C\left(\Vert f\Vert_{3,\delta}+\Vert g\Vert_{2,\delta-1}\right)\begin{cases}
\lAngle v\rAngle^{-1}\lAngle u\rAngle^{1+\delta}&\text{ for }\delta <-1\\
\dfrac{\log\lAngle v\rAngle- \log\lAngle u\rAngle}{\lAngle v\rAngle-\lAngle u\rAngle} &\text{ for }\delta =-1\\
\lAngle v\rAngle^\delta&\text{ for }\delta >-1
\end{cases}.
\end{equation}
The same process can be applied to the derivatives of $\phi$ in the direction of $u$ and $v$. The integral representations of the derivatives are then
\begin{eqnarray*}
\partial_v \phi (t,x) &=&\frac{1}{8\pi}\int_{\mathbb{S}^2}\big(t\left(\partial_r g(x+t\omega)+\partial_\omega g(x+t\omega)\right)+ g(x+t\omega) \\
&& + t\left(\partial_r\partial_\omega f(x+t\omega)+\partial^2_\omega f(x+t\omega)\right)+\partial_rf(x+t\omega)+2\partial_\omega f(x+t\omega)\big)\ud \mu_{\mathbb{S}^2},\\
\partial_u \phi (t,x) &=&\frac{1}{8\pi}\int_{\mathbb{S}^2}\big(t\left(-\partial_r g(x+t\omega)+\partial_\omega g(x+t\omega)\right)+ g(x+t\omega) \\
&& + t\left(-\partial_r\partial_\omega f(x+t\omega)+\partial^2_\omega f(x+t\omega)\right)-\partial_rf(x+t\omega)+2\partial_\omega f(x+t\omega)\big)\ud \mu_{\mathbb{S}^2}.
\end{eqnarray*}
Using Sobolev embeddings, one gets immediately
\begin{eqnarray*}
|\partial_v \phi (t,x)| &\leq & C\left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)\int_{\mathbb{S}^2}\left(\lAngle |x+t\omega|\rAngle^{\delta-1} +t\lAngle |x+t\omega|\rAngle^{\delta-2}\right)\ud \mu_{\mathbb{S}^2},\\
|\partial_u \phi (t,x)| &\leq &C\left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)\int_{\mathbb{S}^2}\left(\lAngle |x+t\omega|\rAngle^{\delta-1} +t\lAngle |x+t\omega|\rAngle^{\delta-2}\right)\ud \mu_{\mathbb{S}^2}.
\end{eqnarray*}
Again using Lemma~\ref{integralextimateshypergeom} and the same comparison procedure as in the previous table, one gets that, for all $t>0$ and $x$ in $\R^3$,
\begin{equation}\label{ineqwave0}
\max\left(|\partial_v \phi (t,x)|, |\partial_u \phi (t,x)|\right) \leq C\left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)
\begin{cases}
\lAngle v\rAngle^{-1}\lAngle u\rAngle^{\delta}&\text{ for }\delta <0\\
\dfrac{\log\lAngle v\rAngle- \log\lAngle u\rAngle}{\lAngle v\rAngle-\lAngle u\rAngle} &\text{ for }\delta =0\\
\lAngle v\rAngle^{\delta-1} &\text{ for }\delta >0
\end{cases}
\end{equation}
Using these results, one can now refine the estimates for the derivatives of the function $\phi$, using the commutators properties of the wave equation with the vector fields generating the symmetries of the metric. Introducing
$$
S=u\partial_u+v\partial_v \text{ which satisfies } [S,\square]=-2\square,
$$
the function $S\phi$ satisfies the Cauchy problem for the linear wave equation
\begin{equation*} \left\{
\begin{array}{l}
\square\left( S \phi\right) =0 \\
S\phi|_{t=0} \in H^{j-1}_{\delta}(\R^3)\\
\partial_t S\phi|_{t=0} \in H^{j-2}_{\delta-1}(\R^3).
\end{array}\right.
\end{equation*}
As a consequence, one can apply the decay results \eqref{ineqwave} and\eqref{ineqwave0} to $S\phi$. This gives
\begin{equation*}
|S\phi(t,x)|\leq
C \left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)
\begin{cases}
\lAngle v\rAngle^{-1}\lAngle u\rAngle^{1+\delta}&\text{ for }\delta <-1\\
(\lAngle v\rAngle-\lAngle u\rAngle)^{-1}\left(\log\lAngle v\rAngle- \log\lAngle u\rAngle\right) &\text{ for }\delta =-1\\
\lAngle v\rAngle^\delta&\text{ for }\delta >-1
\end{cases}.
\end{equation*}
To get the full decay result, we need to compare carefully the decay for $S\phi$, and the decay for $\partial_u \phi$ to get the full decay result for $\partial_v \phi$. We follow the same procedure as above to compare the decay of these terms in a table using
$$
|u| \leq \lAngle u \rAngle, \quad v \leq \lAngle v \rAngle, \quad \text{ and, for all }r>1, \quad
\left |\dfrac{u}{v}\right | \leq C \dfrac{\lAngle u\rAngle}{\lAngle v \rAngle}\leq C.
$$
\begin{equation*}
\begin{array}{|c|c|c|c|}
\hline
&\frac{ S\phi}{v} & \frac{u}{v}\partial_u\phi & \partial_v \phi \\
\hline
\delta<-1 & \dfrac{\lAngle u \rAngle ^{\delta+1}}{\lAngle v\rAngle}\cdot \dfrac{1}{v} & \dfrac{\lAngle u \rAngle^{\delta} }{\lAngle v\rAngle}\cdot \dfrac{u}{ v }& \dfrac{\lAngle u \rAngle^{\delta+1} }{\lAngle v\rAngle^2} \\
\hline
\delta=-1 & \dfrac{1}{v}\dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle} & \dfrac{1}{\lAngle u \rAngle\lAngle v\rAngle}\cdot\dfrac{u}{v} & \dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle^2}\\
\hline
-1<\delta<0 & \dfrac{\lAngle v\rAngle^{\delta}}{v} & \dfrac{\lAngle u \rAngle^{\delta}}{\lAngle v\rAngle}\cdot \dfrac{u}{ v } & \lAngle v \rAngle ^{\delta-1}\\
\hline
\delta= 0 & \dfrac{\lAngle v\rAngle^{\delta}}{v} & \dfrac{u}{v}\cdot\dfrac{F_0\left(\tfrac{\lAngle u\rAngle }{\lAngle v\rAngle}\right) }{\lAngle v\rAngle} & \lAngle v \rAngle ^{\delta-1}\\
\hline
\delta>0 & \dfrac{\lAngle v\rAngle^{\delta}}{v} & \dfrac{u\lAngle v\rAngle^{\delta-1}}{v} & \lAngle v \rAngle ^{\delta-1}\\
\hline
\end{array}
\end{equation*}
This consequently gives the following
\begin{equation*}
|\partial_v\phi(t,x)|\leq
C \left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)
\begin{cases}
\lAngle v\rAngle^{-2} \lAngle u\rAngle^{\delta+1} &\text{ if } \delta<-1 , \\
\dfrac{1}{\lAngle v\rAngle}\dfrac{\log\lAngle v\rAngle- \log\lAngle u\rAngle}{\lAngle v\rAngle-\lAngle u\rAngle} & \text{ if } \delta=-1, \\
\lAngle v\rAngle^{\delta-1} &\text{ if } \delta>-1.
\end{cases}
\end{equation*}
Finally, the fact that $\square$ commutes with the generators of $SO(3)$,
$$
x^i\partial_j-x^j\partial_i ,
$$
can be used to obtain, for $t>0$ and $r>1$,
$$
|\snabla\phi |\leq C \left(\Vert f\Vert_{4,\delta}+\Vert g\Vert_{3,\delta-1}\right)
\begin{cases}
\lAngle v\rAngle^{-2} \lAngle u\rAngle^{\delta+1} &\text{ if } \delta<-1 , \\
\dfrac{1}{\lAngle v\rAngle}\dfrac{\log\lAngle v\rAngle- \log\lAngle u\rAngle}{\lAngle v\rAngle-\lAngle u\rAngle} & \text{ if } \delta=-1 , \\
\lAngle v\rAngle^{\delta-1} &\text{ if } \delta>-1.
\end{cases}
$$
The proof is completed by a recursion over the number of derivatives, which is not written here in detail. \end{proof}
\begin{remark}
\begin{enumerate}
\item The derivatives $\partial_u$ and $\partial_v$, $\snabla$ play different roles when considering the full scale of weights. This difference is at the origin of the failure of the peeling property for higher spin fields when the rate of decay at $i^0$ of the initial data is too low.
\item This difference between derivatives can be explained by considering the derivatives of the fundamental solution of the wave equation.
\end{enumerate}
\end{remark}
\section{Estimates for spinor fields represented by potentials} \label{sec:estspin}
Penrose, in his original paper on zero-rest mass fields \cite{MR0175590}, proved to the following two results:
\begin{itemize}
\item the existence for analytic massless fields of arbitrary spin of representation of the form
$$
\phi_{A\dots F}= \xi_1^{A'}\dots \xi_{2s}^{F'}\nabla_{AA'}\dots\nabla_{FF'} \chi,
$$
where the $\xi^{A'}$ are constant spinors and $\chi$ is a complex function satisfying the wave equation
$$
\square \chi=0.
$$
\item from a decay ansatz for $\chi$ along outgoing null light rays, he deduced the full peeling result for the considered field.
\end{itemize}
The purpose of this section is to give a similar result for massless field admitting a potential of the form considered by Penrose. The decay result for the solution of the wave equation which is used in this section is given by Proposition \ref{prop:decaywave}.
\subsection{Geometric background and preliminary lemmata}\label{geometricbackground}
The geometric framework and notations are introduced in this section. The geometric background is the Minkowski space-time. We consider on this space time the normalized null tetrad defined by
\begin{align*}
l^a &= \sqrt{2}\partial_v = \frac{1}{\sqrt{2}}\left( \frac{\partial}{\partial t} + \frac{\partial }{\partial r}\right), &
m^a &= \frac{1}{r\sqrt{2}}\left( \frac{\partial}{\partial \theta} + \frac{i}{\sin\theta}\frac{\partial}{\partial \varphi} \right) , \\
n^a&=\sqrt{2}\partial_u = \frac{1}{\sqrt{2}}\left( \frac{\partial}{\partial t} - \frac{\partial}{\partial r}\right), & \overline{m}^a & = \frac{1}{r\sqrt{2}}\left( \frac{\partial}{\partial \theta} - \frac{i}{\sin\theta}\frac{\partial}{\partial \varphi} \right),
\end{align*}
so that $l_a n^a=1$ and $m_a\overline{m}^a=-1$, and
the plane spanned by $l^a, n^a$ is orthogonal to that spanned by $m^a, \overline{m}^a$.
The derivatives in the directions $l^a, n^a, m^a, \overline{m}^a$ are denoted by $D, D', \delta, \delta'$ respectively. Consider finally a spin basis $(o^A,\iota^A)$ arising from this tetrad, i.e.,
$$
\begin{array}{cc}
l^a = o^A\overline{o}^{A'},&\quad m^a= o^A\overline{\iota}^{A'} ,\\
n^a= \iota ^A \overline{\iota}^{A'}, &\quad \overline{m}^a= \iota^A\overline{o}^{A'} .
\end{array}
$$
This basis satisfies
\begin{subequations}\label{basis}
\begin{align}
D o^A&=0 , & D \iota^A&=0 ,\\
D' o^A &= 0 , & D' \iota^A&=0 , \\
\delta o^A&=\frac{\cot{\theta}}{2 r \sqrt{2}} o^A, &\delta \iota^A&=-\frac{\cot{\theta}}{2 r \sqrt{2}} \iota^A-\frac{1}{r\sqrt{2}} o^A , \\
\delta' o^A&= - \frac{\cot{\theta}}{2 r \sqrt{2}}o^A+\frac{\iota^A}{r\sqrt{2}},
& \delta' \iota^A&= \frac{\cot{\theta}}{2r \sqrt{2}}\iota^A .
\end{align}
\end{subequations}
\begin{lemma}[Commutators]\label{commutator}
The following commutator relations hold:
\begin{itemize}
\item $D$ and $D'$ commute.
\item consider the gradient $\snabla$ on the sphere of radius $r$,
$$
\snabla = -\overline{m} \delta -m\delta' ,
$$
then, for any positive integer $k$,
\begin{align*}
\snabla^kD&= D\snabla^k + \frac{k}{r\sqrt{2}} \snabla^k , \\
\snabla^kD ' &= D'\snabla^k - \frac{k}{r\sqrt{2}} \snabla^k,
\end{align*}
where $\snabla^k$ is the $k$-th power of the operator $\snabla$.
\end{itemize}
\end{lemma}
\begin{remark}\label{rem:commute} One can directly infer from this lemma that the operator $r\snabla$ commutes with $D$ and $D'$.
\end{remark}
\begin{proof}
The proof follows directly from the commutator relations
\begin{equation*}
\delta D = D\delta + \frac{1}{r\sqrt{2}}\delta, \quad \delta D' = D\delta - \frac{1}{r\sqrt{2}}\delta,
\end{equation*}
and their complex conjugates, and on
$$
D m^a = D'm^a = 0,
$$
and their complex conjugates.
\end{proof}
\begin{lemma}[Asymptotic behavior of the decomposition of a constant spinor]
Let $\xi^A$ be a constant spinor over $\mathbb{M}$ and consider its decomposition over the basis $(o^A, \iota ^A)$:
$$
\xi^A= \alpha o^A+\beta \iota^A.
$$
Then, for any integer $n$, $\nabla^n \alpha$ and $\nabla^n\beta$ are smooth bounded functions on $\mathbb{M}\backslash \{ \R \times B(0,1)\}$. Furthermore, considering the derivatives in the null directions, the following estimates hold for $\theta \in [c, \pi - c]$ ($c>0$):
\begin{align*}
D\alpha&= D'\alpha=0,&D\beta= D'\beta=0\\
|\delta^n \alpha |&\leq \frac{C}{r^n},&|\delta^n \beta |\leq \frac{C}{r^n},\\
|\delta'^n \alpha |&\leq \frac{C}{r^n},&|\delta'^n \beta |\leq \frac{C}{r^n}.
\end{align*}
\end{lemma}
\begin{proof}
To prove that $\alpha$ and $\beta$ are bounded functions, it suffices to consider the decomposition of the real vector field $\xi{}^A\overline{\xi}{}^{A'}$ in Cartesian coordinates. The time component of the vector field is $|\alpha^2|+|\beta|^2$ and it is constant. As a consequence, $\alpha$ and $\beta$ are smooth bounded functions.
The second step consists in calculating the derivatives of the components in $\xi^{A}$. Since $o^A$ and $\iota^A$ are constant along outgoing and ingoing null rays, the following identities hold:
$$
D\alpha=D'\beta=D\alpha=D'\beta=0.
$$
For the angular derivatives, we have:
\begin{eqnarray*}
\left(\delta \alpha+\alpha \frac{\cot\theta}{2 r \sqrt{2}}-\frac{\beta}{r\sqrt{2}}\right)o^A+\left(\delta \beta - \beta \frac{\cot\theta}{2 r \sqrt{2}}\right)\iota^A=0,\\
\left(\delta'\alpha -\alpha\frac{\cot\theta}{2 r \sqrt{2}}\right)o^A+\left( \delta' \beta+\beta \frac{\cot\theta}{2 r \sqrt{2}}+\frac{\alpha}{r\sqrt{2}}\right)\iota^A=0.
\end{eqnarray*}
An induction using these recursive relations gives the desired results.
\end{proof}
\subsection{Proof of the decay result}
We consider in this section a spin-$s$ field represented as
\begin{equation}\label{representation1}
\phi_{A\dots F}= \xi_1^{A'}\dots \xi_{2s}^{F'}\nabla_{AA'}\dots\nabla_{FF'} \chi,
\end{equation}
where $\chi$ is a complex scalar Hertz potential satisfying the wave equation
$$
\square \chi=0,
$$
and $\xi_1^{A'}, \dots, \xi_{2s}^{F'}$ are constants spinors.
The purpose of this section is to give a result which is similar to the one obtained for the wave (or spin-0) equation in order to retrieve similar decay estimates as in the pioneering work of Christodoulou-Klainerman \cite{Christodoulou:1990dd}.
\begin{proposition}[Decay estimates for arbitrary spin]\label{prop:decay1}
Let $(k,l,m)$ be a triple of non-negative integers and denote by $n$ their sum. We assume that the Hertz potential is a solution of the Cauchy problem, for $j> 2+2s+n$ and $\delta \notin \mathbb{Z}$
\begin{equation*}\left\{
\begin{array}{l}
\square \chi =0 , \\
\chi|_{t=0}\in H^{j}_\delta(\R^3, \mathbb{C}) , \\
\partial_t \chi|_{t=0}\in H^{j-1}_{\delta-1}(\R^3, \mathbb{C}) .
\end{array}\right.
\end{equation*}
The norm of the initial data is denoted by $I_{j,\delta}$,
$$
I_{j,\delta} = \Vert\chi|_{t=0}\Vert_{j, \delta} + \Vert\partial_t\chi|_{t=0}\Vert_{j-1, \delta-1}
$$
Then, the following inequalities hold, for all $\bm{i}$ in $\{0,2s\}$:
\begin{enumerate}
\item for any $t\geq 0$, $x \in \R^3$, such that $t>3r$, that is to say in the interior region,
$$
|\nabla^n \phi_{A\dots F}|\leq c \lAngle t\rAngle^{\delta-2s-n} I_{n+2s+3, \delta} ,
$$
\item for $\bm{i}$ such that $1+\delta-l-\bm{i}<0$, for any $t\geq 0$, $x \in \R^3$, such that $3r>t>\tfrac{r}{3}$, that is to say in the exterior region,
$$
|D^kD'^l\snabla^m\phi_{\bm{i}}|\leq\frac{c \lAngle u\rAngle ^{1+\delta-\bm{i}-l}}{\lAngle v\rAngle^{1+2s-\bm{i}+k+m}}I_{n+2s+3,\delta} ,
$$
\item for $\bm{i}$ such that $1+\delta-l-\bm{i}>0$, for any $t\geq 0$, $x \in \R^3$, such that $3r>t>\tfrac{r}{3}$, that is to say in the exterior region,
$$
|D^kD'^l\snabla^m\phi_{\bm{i}}|\leq c \lAngle v\rAngle^{\delta -2s-l-k-m} I_{n+2s+3,\delta}.
$$
\end{enumerate}
\end{proposition}
\begin{remark}
\begin{enumerate}
\item For the spin-$1$ case with $\delta=-1/2$, that is to say for initial data for the Maxwell fields lying in $H_{-5/2}$, which is the case considered in \cite{Christodoulou:1990dd}, one recovers the decay result stated in that paper. For the spin-$2$ case with $\delta=1/2$, that is to say for initial data in $H_{-7/2}$, which is the case considered by Christodoulou-Klainerman, their results are recovered. We need slightly higher regularity though due to our estimates for the wave equation.
\item It should be noted that in the case when the potential does not decay enough (for $\delta>-2s-2$), the decay rates of some components of the field cannot distinguished, and hence the peeling property fails to hold.
\item The estimates stated in Proposition \ref{prop:decaywave} hold for all weight $\delta$ in $\mathbb{R}$. As a consequence, this should allow us to handle all weights for the initial data of the Hertz potential. Nonetheless, the case of integer weights has been put aside: firstly, the representation of massless free fields, as stated in Proposition~\ref{proprepspingeneral}, does not handle these weights; secondly, the appearance of such weights will lead to an unnecessary lengthy discussion in the proof.
\item The peeling result obtained by Penrose \cite{MR0175590} depends on the assumption that the Hertz potential decays as
$\chi \sim 1/r $
where $r$ is a parameter along the outgoing null rays. For such a decay result to hold, the initial data for the potential have to lie in $H_{\delta}$ with $\delta<-1$. The peeling result by Mason-Nicolas \cite{Mason:2012jq}, which holds for the spins $1/2$ and $1$ on the Schwarzschild space-time, is for initial data lying in a Sobolev space whose weights are not equally distributed on the components.
\end{enumerate}
\end{remark}
\begin{proof}
The proof is made by induction on the spin. The result for the spin-0 case is exactly the one obtained for the wave equation and is the base step of the induction.
We now make the following induction hypothesis for spin-$s$, with $s\in \tfrac{1}{2}\mathbb{N}_0$: for any triple $(k,l,m)$, and for any spin-$s$ field represented by
$$
\psi\underbrace{{}_{A\dots F}}_{\mathclap{2s \text{ indices}}}= \xi_1^{A'}\dots \xi_{2s}^{F'}\nabla_{AA'}\dots\nabla_{FF'} \chi,
$$
where $\chi$ is a potential whose initial data lie in $H^j_{\delta}(\R^3,\mathbb{C})\times H^{j-1}_{\delta-1}(\R^3,\mathbb{C})$, for $j>2+2s+n$ with $n=k+l+m$, the estimates stated in the theorem hold.
Let now $(k,l,m)$ be a triple of non-negative integers and consider a spin-$\left(s+1/2\right)$ field written
$$
\phi\underbrace{{}_{A\dots FG}}_{\mathclap{2s +1\text{ indices}}}= \xi_1^{A'}\dots \xi_{2s+1}^{G'}\nabla_{AA'}\dots\nabla_{GG'} \chi,
$$
where $\chi$ is a potential whose initial data are in $H^j_{\delta}(\R^3,\mathbb{C})\times H^{j-1}_{\delta-1}(\R^3,\mathbb{C})$ ($j>3+2s+n$). Consequently, the induction hypothesis is satisfied for the spin-$s$ field
$$
\psi\underbrace{{}_{B\dots G}}_{\mathclap{2s \text{ indices}}}= \xi_2^{B'}\dots \xi_{2s+1}^{G'}\nabla_{BB'}\dots\nabla_{GG'} \chi,
$$
with the same $\chi$ whose initial data also lies in $H^p_{\delta}(\R^3)\times H^{p-1}_{\delta-1}(\R^3)$ ($p=j-1>2+2s+n$). It remains then to prove that
$$
\phi_{A\dots G}= \xi^{A'}\nabla_{AA'}\psi_{B\dots G}
$$
satisfies the appropriate decay result.
We first consider the interior decay. The result trivially follows from the interior decay result for the wave equation stated in Proposition \ref{prop:decaywave} and Remark \ref{rem:decaywave}. As a consequence, the following relation holds:
\begin{equation*}
|\nabla^n\phi_{A\dots G}|=|\xi^{A'}\nabla_{AA'}\left(\nabla^n\psi _{B\dots G}\right)|,
\end{equation*}
which is a derivative of order $n+1$ of a spinor field of valence $2s$ which satisfies the induction hypothesis. As a consequence, the following decay result is immediate, in the interior region $3t \leq r$:
$$
|\nabla^{n}\phi_{A\dots F}|\leq C\frac{I_{n+2s+3,\delta}}{\lAngle t\rAngle^{-\delta+2s+n+1}},
$$
where $C$ is a constant depending on $n$ and $s$. This closes the induction for the part concerning the interior decay.
Now we consider the problem of the exterior decay, that is to say the decay in the neighborhood of an outgoing light cone:
\begin{equation}\label{descriptionexterior}
\frac{r}{3} \leq t\leq 3 r \Leftrightarrow |t-r|\leq \frac12 \left|t+r\right|.
\end{equation}
Recall that the components of the spinor $\psi_{B\dots F}$ are defined by
$$
\psi_{\bm{i}}= \underbrace{\iota^B\dots \iota^C}_{\bm{i}}\underbrace {o ^D \dots o^G}_{2s -\bm{i}}\psi_{B\dots G}.
$$
The proof in this region is done by induction as in the first part of the proof. Let $(k,l,m)$ be a given triple of non negative integers and denote by $n$ their sum. The induction hypothesis is written as follows for spin-$s$:
\begin{quote}
For any spinor field of valence $2s$ $\psi_{B\dots G}$ satisfying:
$$
\psi_{B\dots G}= \xi_2^{B'}\dots \xi_{2s+1}^{G'}\nabla_{BB'}\dots \nabla_{GG'} \chi
$$
where $\chi$ is a complex scalar solution of the massless wave equation whose initial data lies in $H^j_{\delta}(\R^3, \mathbb{C})\times H^{j-1}_{\delta-1}(\R^3, \mathbb{C})$ ($j>3+2s+n$), the following decay results holds, in the exterior region $\frac{t}{3}\leq r\leq 3t$, for all integer $k,l,m$:
\begin{itemize}
\item for $\bm{i}$ such that $1+\delta-l-\bm{i}<0$:
$$
|D^kD'^l\snabla^m\psi_{\bm{i}}|\leq\frac{c \lAngle u\rAngle ^{\delta+1-\bm{i}-l}}{\lAngle v\rAngle^{1+2s-\bm{i}+k+m}}I_{n+2s+3,\delta}.
$$
\item for $\bm{i}$ such that $1+\delta-l-\bm{i}>0$:
$$
|D^kD'^l\snabla^m\psi_{\bm{i}}|\leq C \lAngle v\rAngle^{\delta -2s-l-k-m} I_{n+2s+3,\delta},
$$
where the constant $C$ depends on the bounds of the exterior domain and the integers $k,l,m$.
\end{itemize}
\end{quote}
There is no need to prove the initial step since it is exactly the result for the standard wave equation. Assume that the induction hypothesis holds for spin-$s$ in $\frac12 \mathbb{N}_0$ and consider the field $\phi_{A\dots G}$ of spin-$\left(s+1/2\right)$ written as:
$$
\phi_{A\dots G}= \xi_1^{A'}\dots \xi_{2s+1}^{G'}\nabla_{AA'}\dots \nabla_{GG'} \chi
$$
where $\chi$ is a complex scalar solution of the massless wave equation whose initial data lies in $H^j_{\delta}(\R^3,\mathbb{C})\times H^{j-1}_{\delta+1}(\R^3,\mathbb{C})$ ($j>2+2s+n$). As a consequence, the spinor:
$$
\psi_{B\dots G}=\xi_2^{B'}\dots \xi_{2s+1}^{G'}\nabla_{BB'}\dots \nabla_{GG'} \chi
$$
is a spinor field of valence $s$ satisfying the requirements of the induction assumption.
To insure the proof of the induction assumption, a relation between the components of $\phi_{A\dots F}$ and the components of the field $\psi_{B\dots G}$ have to established. The components of these fields are related by the following Lemma.
\begin{lemma}\label{relationscoef} Let $s \in \tfrac{1}{2} \mathbb{N}_0$.
The components of $\phi_{A\dots G}$ of spin-$\left(s+1/2\right)$ and $\psi_{B \dots G}$ of spin-$s$, related by
$$
\phi_{A\dots G} = \xi^{A'}\nabla_{AA'} \psi_{B\dots G} \text{ where } \xi^{A'} = \alpha o^{A'} + \beta i^{A'}
$$
are given by the following relations:
\begin{eqnarray}
\phi_0&= &\alpha D\psi_0 +\beta \delta \psi_0-s\beta\frac{\cot \theta}{r\sqrt{2}}\psi_0 ,\\
\phi_{\bm{i}}&= &\alpha\delta'\psi_{\bm{i}-1} +\beta D' \psi_{\bm{i}-1}
+\frac{\alpha}{r\sqrt{2}}\big((s+1-\bm{i}) \cot\theta\psi_{\bm{i}-1}-(2s+1-\bm{i})\psi_{\bm{i}}\big) , \label{relation2}
\end{eqnarray}
for $\bm{i}>0$.
\end{lemma}
\begin{proof} The proof is carried out by using relations \eqref{basis} and is a basic calculation. We have
\begin{eqnarray*}
\phi_0&=& o^A \dots o ^G \phi_{A\dots G} , \\
&=& o^A \dots o ^G\xi^{A'} \nabla_{AA'} \psi_{B\dots G} , \\
&=& \alpha o^B \dots o^G D \psi_{B\dots G} + \beta o^B \dots o^G \delta \psi_{B\dots G} .
\end{eqnarray*}
Since $D o ^A=0$ and $ \delta o^A=\frac{\cot{\theta}}{2 r \sqrt{2}} o^A$, we have
$$
o^B \dots o^G \delta \psi_{B\dots G}= \delta \psi_{0} -2s \frac{\cot\theta}{2r \sqrt{2}}\psi_0 ,
$$
and hence
$$
\phi_0= \alpha D \psi_0 +\beta \delta \psi_0 - s \beta \frac{\cot\theta}{r \sqrt{2}}\psi_0.
$$
Consider now $\bm{i}>0$ fixed; we have
\begin{eqnarray*}
\phi_{\bm{i}}&=& \underbrace{\iota^A \dots \iota ^C}_{\bm{i} \text{ times}} \underbrace{o^D \dots o^G}_{2s + 1-\bm{i} }\phi_{A \dots G}\\
&=& \alpha \underbrace{\iota^B \dots \iota ^C}_{\bm{i}-1 } \underbrace{o^D \dots o^G}_{2s + 1-\bm{i} }\delta ' \psi_{B \dots G}+\beta \underbrace{\iota^B \dots \iota ^C}_{\bm{i}-1 } \underbrace{o^D \dots o^G}_{2s + 1-\bm{i} }D ' \psi_{B \dots G}.
\end{eqnarray*}
Since $\delta' o^A= - \frac{\cot{\theta}}{2 r \sqrt{2}}o^A+\frac{\iota^A}{r\sqrt{2}}$
and $\delta' \iota^A= \frac{\cot{\theta}}{2r \sqrt{2}}\iota^A$ we have
\begin{align*}
\underbrace{\iota^B \dots \iota ^C}_{\bm{i}-1 }& \underbrace{o^D \dots o^G}_{2s + 1-\bm{i} }\delta ' \psi_{B \dots G}
=
\delta' \psi_{\bm{i}-1}
-(\bm{i}-1)\left(\frac{\cot\theta}{2 r \sqrt{2}} \underbrace{\iota^B \dots \iota ^C}_{\bm{i}-1 } \underbrace{o^D \dots o^G}_{2s + 1-\bm{i} }\psi_{B \dots G}\right)\nonumber\\
& - (2s+1-\bm{i})\left(- \frac{\cot\theta}{2 r \sqrt{2}} \underbrace{\iota^B \dots \iota ^C}_{\bm{i}-1 } \underbrace{o^D \dots o^G}_{2s + 1-\bm{i}} \psi_{B\dots G} + \frac{1}{ r \sqrt{2}} \underbrace{\iota^B \dots \iota ^C}_{\bm{i} } \underbrace{o^D \dots o^G}_{2s-\bm{i}} \psi_{B\dots G} \right).
\end{align*}
Consequently, by the relations $D'\iota^A=D'o^A=0$, we get \eqref{relation2}.
\end{proof}
Using Lemma~\ref{relationscoef}, the proof of Proposition \ref{prop:decay1} can be continued. The two cases ($\bm{i}=0$ and $\bm{i}>0$) are treated separately although the method is the same. Here we present the case $\bm{i}=0$, the other case follows similarly.
The expression of the derivative $D^k D'^l\snabla^m\phi_0$ is calculated explicitly, one derivative at a time, using the Leibniz rule:
\begin{gather*}
\snabla^m \phi_0= \sum_{a=0}^m\binom{a}{m}\snabla^{a}\alpha\snabla^{m-a}D\psi_0
+ \sum_{a=0}^m\binom{a}{m}\snabla^{a}\beta\snabla^{m-a}\delta\psi_0\\
-2s\sum_{a+b+c=m}\frac{m!}{a!b!c!}\left(\frac{\partial_{\theta}^b\left(\cot\theta\right)}{2\sqrt{2}r^{b+1}}\right)\snabla^a\beta \partial_{\theta}^b \snabla^c\psi_0
\end{gather*}
since
$$
\snabla^b\cot \theta = \frac{1}{r^b}\frac{\partial^b \cot \theta}{\partial \theta^b}\partial_{\theta}^b,
$$
the power on the vector field have to be understood as a symmetric tensor product.
We then apply simultaneously the derivatives $D$ and $D'$, using the Leibniz rule again. Notice first that $\snabla^a\alpha$ and $\snabla^a\beta$ depend on $r$ but $r^a\snabla^a \alpha$ and $r^a\snabla^a \beta$ do not, since both $\alpha$ and $\beta$ are independent both of time and radius, and $r\snabla$ commutes with $D$ and $D'$ (cf. Remark \ref{rem:commute}). We have
\begin{gather*}
D^k D'^l \snabla^m \phi_0 =\\
\sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m\left[\binom{d}{k}\binom{e}{l}\binom{a}{m}\left(r^a\snabla^{a}\alpha \right)\left((-1)^dA_{a+d+e}^{d+e} \right)\right]\left\{\frac{D^{k-d}D'^{l-e}\snabla^{m-a}D\psi_0}{r^{a+d+e}}\right\}\\
+
\sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m\left[\binom{d}{k}\binom{e}{l}\binom{a}{m}\left(r^a\snabla^{a}\beta \right)\left((-1)^d A_{a+d+e}^{d+e} \right)\right]\left\{\frac{D^{k-d}D'^{l-e}\snabla^{m-a}\delta\psi_0}{r^{a+d+e}}\right\}\\
+2s \sum_{d=0}^k \sum_{e=0}^l \sum_{a+b+c=m}\left[\binom{d}{k}\binom{e}{l}
\frac{m!\partial_{\theta}^b\left(\cot\theta\right)}{2\sqrt{2}a!b!c!} \left(r^a \snabla^a\beta\right)\left((-1)^dA_{1+a+d+e}^{d+e} \right)\right]\\
\times\left\{\frac{D^{k-d}D'^{l-e}\snabla^c\psi_0}{r^{1+a+b+d+e}}\right\},
\end{gather*}
where
$$
A_{m}^{n} = \frac{n!}{ (m-1)! }.
$$
The factors in the square brackets are clearly bounded provided that $\theta$ lies in $[c, \pi- c]$ for a given (arbitrarily small) positive constant $c$. The singularity of the tetrad at the axis $\theta = \pm\pi$ prevents from covering the entire interval $[0,\pi]$. The problem can be easily solved by considering another tetrad associated to another spherical coordinate system. There exists consequently a constant $C$ depending on the spin, the $L^\infty$-bounds on the coefficients of the spinor field $\xi^{A'}$ and their derivatives, such that
\begin{gather}
|D^k D'^l \snabla^m \phi_0|\leq
C \left (\sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m\left|\frac{D^{k-d}D'^{l-e}\snabla^{m-a}D\psi_0}{r^{a+d+e}}\right|\right .\nonumber\\
+ \left . \sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m\left|\frac{D^{k-d}D'^{l-e}\snabla^{m-a}\delta\psi_0}{r^{a+d+e}}\right|
+ \sum_{d=0}^k \sum_{e=0}^l \sum_{a+b+c=m} \left|\frac{D^{k-d}D'^{l-e}\snabla^c\psi_0}{r^{1+a+b+d+e}}\right| \right).\label{Derivativesphi0Estimate1}
\end{gather}
Each of these terms is treated separately.
The first term can be transformed to fit the induction hypothesis using Lemma \ref{commutator}:
\begin{align*}
D^{k-d}D'^{l-e}\snabla^{m-a}D\psi_0={}& D^{k-d+1}D'^{l-e}\snabla^{m-a} \psi_0+D^{k-d}D'^{l-e}\left(\frac{m-a}{r\sqrt{2}}\snabla^{m-a} \psi_0\right)\\
={}&D^{k-d+1}D'^{l-e}\snabla^{m-a} \psi_0\\
+\sum_{f=0}^{k-d}\sum_{g=0}^{l-e}\binom{f}{k-d}\binom{g}{l-e}&\frac{(-1)^f(m-a)(k-d+l-e)!}{r^{1+f+g}\sqrt{2}}D^{k-d-f}D'^{l-e-g}\snabla^{m-a}\psi_0.
\end{align*}
In order to use the decay result stated in the induction hypothesis, the number of derivatives in the ingoing direction has to be taken into account:
\begin{enumerate}
\item[a)]{if $1+\delta -l<0$, then $$ |D^kD'^l \snabla^m \psi_0| \leq C \lAngle u\rAngle^{1+\delta - l}\lAngle v\rAngle^{-1-2s-k-m} I_{n+2s+3,\delta};$$}
\item[b)]{if $1+\delta -l>0$, then $$ |D^kD'^l \snabla^m \psi_0| \leq C \lAngle v\rAngle^{\delta-2s-k-l-m} I_{n+2s+3,\delta}. $$}
\end{enumerate}
In order to simplify the presentation of the proof, we deal specifically with the sum
$$
A = \sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m\left|\frac{D^{k-d+1}D'^{l-e}\snabla^{m-a}\psi_0}{r^{a+d+e}}\right|.
$$
Assume first that $1+\delta-l<0$.
For this case the sum a priori contains terms of both type {\it a} and type {\it b}. We therefore split up the sum over $e$ into two parts corresponding to the different types. Let $e'$ be the largest integer such that $e'\leq l$ and $\delta-l+e'<0$. This means that, if $0\leq e\leq e'-1$, we have $1+\delta-(l-e)<0$ and if $e'\leq e\leq l$ we have $1+\delta-(l-e)>0$.
Using the induction hypothesis, the sum $A$ can then be bounded as follows: there exists a constant $C$ such that
\begin{align*}
A \leq{}& C \sum_{d=0}^k \sum_{a=0}^m \frac{I_{n+2s+4, \delta}}{r^{a+d}}\cdot\Big(
\sum_{e=0}^{e'-1}\frac{\lAngle u\rAngle^{1+\delta-(l-e)}}{\lAngle v\rAngle^{2+2s+k-d+m-a}r^{e}}
+\sum_{e=e'}^{l}\frac{\lAngle v\rAngle^{\delta-2s-k+d-1-l+e-m+a}}{r^{e}}\Big )
\\
\leq{}& C \frac{\lAngle u\rAngle^{1+\delta-l}I_{n+2s+4, \delta}}{\lAngle v\rAngle^{2s+2+k+m}} \left(
\sum_{d=0}^k \sum_{a=0}^m \left(\frac{\lAngle v\rAngle}{r}\right)^{a+d}\right)\\
&\times \left(
\sum_{e=0}^{e'-1}\left(\frac{\lAngle u\rAngle}{\lAngle v\rAngle}\right)^{e}\left(\frac{\lAngle v\rAngle}{r}\right)^{e}
+\sum_{e=e'}^{l}\left(\frac{\lAngle u\rAngle}{\lAngle v\rAngle}\right)^{-(1+\delta-l)}\left(\frac{\lAngle v\rAngle}{r}\right)^{e}
\right).
\end{align*}
Since we are considering the exterior region, that is to say the region defined by
$$
\frac{t}{3}\leq r \leq 3t,
$$
the following inequalities hold (assuming also $r>1$, which is not restrictive, when studying the asymptotic behavior),
$$
\frac{\lAngle v\rAngle}{r}\leq \sqrt{17} \text{ and } \frac{\lAngle u\rAngle}{\lAngle v\rAngle} \leq 1.
$$
As a consequence, there exists a constant $C$ depending only on the considered region and of the number of derivatives such that
$$
A \leq C \lAngle u\rAngle^{1+\delta-l}\lAngle v\rAngle^{-1-2s-1-k-m}I_{n+2s+4,\delta}.
$$
In the case when $1+\delta-l>0$, all the indices $1+\delta-l+e$ are a fortiori positive and, as a consequence, the induction hypothesis gives immediately: there exists a constant $C$ depending on the number of derivatives and on the bounds of the derivatives of $\alpha$ and $\beta$ such that
$$
A\leq C \lAngle v\rAngle^{\delta -2s-1-k-l-m} I_{n+2s+4, \delta} \sum_{d=0}^k \sum_{e=0}^l \sum_{a=0}^m \left(\frac{\lAngle v\rAngle}{r}\right)^{a+e+d}.
$$
There exists consequently, as previously, a constant $C$ depending on the number of derivatives such that
$$
A\leq C \lAngle v\rAngle ^{\delta-2s-1-k-l-m}I_{n+2s+4, \delta}.
$$
The other terms in \eqref{Derivativesphi0Estimate1} can be studied in a similar way and details are left to the reader. Collecting all the inequalities obtained for these derivatives, one gets that there exists a constant $C$, depending only on the Sobolev embeddings and the number of derivatives such that
$$
|D^kD'^l\snabla^m \phi_0 |\leq CI_{n+2s+4,\delta}
\left\{\begin{array}{ll}
\lAngle u\rAngle^{1+\delta-l}\lAngle v\rAngle^{-1-2s-1-k-m}&\text{ if }1+\delta-l<0\\
\lAngle v\rAngle^{-2s-1-k-l-m}&\text{ if }1+\delta-l>0.\end{array}
\right.
$$
The other components $\psi_{\bm{i}}$ of the field can be studied in a similar way. The discussion will this time occur on the sign of $1+\delta-l-\bm{i}$. These complementary computations are left to the reader.
We have now proved that the induction hypothesis holds also for $s+1/2$. We can therefore conclude that it holds for all $s\in \tfrac{1}{2}\mathbb{N}_0$.
\end{proof}
\section{Asymptotic behavior of higher spin fields using Hertz potential} \label{sec:mainres}
This section contains the main result of the paper, which consists, for arbitrary spin, in a decay result for solutions of the Cauchy problem with initial data in weighted Sobolev spaces. This extends the result contained in \cite{Christodoulou:1990dd} for the fixed weight $\delta = -s-3/2$ (for spin-$s$ fields with $s=1,2$) and clarifies the fact that peeling fails for the rapidly decaying components of the field. Furthermore, through Theorem~\ref{th:repthm}, we establish a full correspondence between the decay result of the wave equation and the peeling result for the higher spin fields.
Consider a free massless spin-$s$ field $\phi_{A \dots F}$, \emph{\emph{i.e.}} a symmetric valence $2s$ spinor field on Minkowski space, which solves
\begin{equation}\label{cauchyprobmassless}
\left\{
\begin{array}{l}
\nabla^{AA'}\phi_{A\dots F}=0, \\
\phi_{A\dots F}|_{t=0}=\varphi_{A\dots F} \in H^{j}_{\delta}(S_{2s}).
\end{array}\right.
\end{equation}
For $s\geq 1$, this Cauchy problem is consistent only when the geometric constraint
\begin{equation}\label{geometricconstraint}
D^{AB} \varphi_{ABC\dots F}= (\sdiv_{2s} \varphi)_{C\dots F} = 0
\end{equation}
is satisfied.
In this section, we first investigate which spin-$s$ fields can be represented by a potential of the form
\begin{equation}\label{representation4D}
\phi_{A \dots F} = \nabla_{AA'}\cdots\nabla_{FF'} \widetilde\chi^{A'\dots F'},
\end{equation}
where the Hertz potential satisfies a Cauchy problem
\begin{equation} \label{cauchywavechieq1} \left\{
\begin{array}{l}
\square\chi_{A\dots F}=0, \\
\chi_{A\dots F}|_{t=0}=\xi_{A\dots F}\in H^{j+2s}_{\delta+2s}(S_{2s}),\\
\partial_t\chi_{A\dots F}|_{t=0}=\sqrt{2}\zeta_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s}),\\
\end{array}\right.
\end{equation}
with
\begin{equation*}
\chi_{A\dots F}\equiv \tau_{AA'}\cdots \tau_{FF'}\widetilde\chi^{A'\dots F'}.
\end{equation*}
To achieve this, a 3+1 splitting of Eq.~\ref{representation4D} with respect to the Cauchy surfaces $\{t=\text{const.}\}$ is performed in Section~\ref{sec:spacespinorsplit} so that the initial data for the field $\varphi_{A\dots F}$ and for the potential $(\xi_{A\dots F},\zeta_{A\dots F})$ are related through the operator $\mathcal{G}_{2s}$. Theorem \ref{proprepspingeneral} is then used to construct initial data for the Hertz potential and control their Sobolev norms. In Section~\ref{sec:representation}, the uniqueness of the Cauchy problem for higher spin fields ensures that this field is represented by a Hertz potential satisfying the Cauchy problem~\eqref{cauchywavechieq1} with the constructed data. Finally, in Section~\ref{sec:maintheorem}, the asymptotic behavior of the field is derived from the decay result for the scalar wave equation stated in Proposition~\ref{prop:decaywave} through the technical Proposition~\ref{prop:decay1}.
\subsection{Space spinor splitting}\label{sec:spacespinorsplit}
A 3+1 splitting of the potential equation \eqref{representation4D} is now performed. Let $\tau_{AA'}=\sqrt{2}\nabla_{AA'}t$, which is covariantly constant. The operator $D_{AB}=\tau_{(A}{}^{A'}\nabla_{B)A'}$ is valid everywhere and it coincides with the intrinsic derivative on the slices $\{t=\text{const.}\}$. We therefore can consider it as an operator acting both on space-time spinors and on spatial spinors on a time slice. All other operators defined for fields on $\mathbb{R}^3$ also extend in this way to operators on fields on Minkowski space. With this view, we have the decomposition
$\tau_{B}{}^{A'}\nabla_{AA'}=D_{AB}+\tfrac{1}{\sqrt{2}}\epsilon_{AB}\partial_t$.
The evolution equations of the Cauchy problems \eqref{cauchyprobmassless} and \eqref{cauchywavechieq1} can be re-expressed as
\begin{align}
\partial_t \phi_{A\cdots F} &= \sqrt{2} (\scurl_{2s} \phi)_{A\cdots F},\label{evolutionphieq1}\\
\partial_t \partial_t \chi_{A\cdots F} &= - \Delta_{2s}\chi_{A\cdots F}.\label{wavechieq2}
\end{align}
For the spin-$1/2$ case, we immediately get
\begin{align*}
\phi_{A} &= (\scurl_1 \chi)_{A} + \tfrac{1}{\sqrt{2}}\partial_t \chi_{A}
=(\mathcal{G}_1\scurl_1\chi)_{A}+\tfrac{1}{\sqrt{2}}(\mathcal{G}_1\partial_t\chi)_{A}.
\end{align*}
This simple pattern in fact generalizes to arbitrary spin:
\begin{proposition}\label{prop:splittingpotential}
The equation \eqref{representation4D} together with \eqref{wavechieq2} implies
\begin{equation*}
\phi_{A_1\dots A_{2s}}=(\mathcal{G}_{2s}\scurl_{2s}\chi)_{A_1\dots A_{2s}}
+\tfrac{1}{\sqrt{2}}(\mathcal{G}_{2s}\partial_t\chi)_{A_1\dots A_{2s}}.
\end{equation*}
\end{proposition}
\begin{remark}
The property $\sdiv_{2s}\mathcal{G}_{2s}=0$ of the operators directly gives that the constraint $(\sdiv_{2s} \phi)_{C\cdots F} = 0$ is automatically satisfied for $s\geq 1$.
\end{remark}
\begin{proof}
See Proposition~\ref{splittingpotentialAppendix} in the appendix for a proof.
\end{proof}
\subsection{Representation by a Hertz potential}\label{sec:representation}
We will investigate under which conditions on the initial datum $\varphi_{A\dots F}$ we can construct initial data for the potential.
We immediately see from Proposition~\ref{prop:splittingpotential} that if a potential exists, then
$\varphi_{A\dots F}$ has to be in the image of $\mathcal{G}_{2s}$.
Therefore, we can without loss of generality choose $\xi_{A\dots F}=0$.
This means that we have to solve the equation $\varphi_{A\dots F}
=(\mathcal{G}_{2s}\zeta)_{A\dots F}$. Since the integrability condition \eqref{geometricconstraint} is satisfied, Proposition~\ref{proprepspingeneral} can be used to construct $\zeta_{A\dots F}$.
A key point which will be used later to prove that the field can be represented by a potential
is the uniqueness of the Cauchy problem for first order hyperbolic systems. For such a result, the reader can refer either to \cite{MR0080849} or \cite[Appendix 4]{MR2473363}. That the massless spin-$s$ field equation has a first order symmetric hyperbolic formulation follows immediately from Equation~\eqref{evolutionphieq1} and Lemma~\ref{symbolcurlreal}.
\begin{lemma}\label{uniqueness}
Consider a spinor field $\varphi_{A\dots F}$ in $L^2_{\text{loc}}(S_{2s})$. Then the Cauchy problem \eqref{cauchyprobmassless} admits at most one solution in $C^0(\mathbb{R}, L^2_{\text{loc}}(S_{2s}))$.
\end{lemma}
\begin{proof} This lemma is a direct consequence of the energy estimate.
\end{proof}
\begin{remark}
This lemma does not state existence of solutions to the Cauchy problem for the massless free
fields with initial datum in weighted Sobolev spaces. However, one can use Theorem \ref{th:repthm} to obtain existence of solutions of this Cauchy problem from standard existence theorems
for solutions of the wave equation with initial data in weighted Sobolev spaces.
\end{remark}
We can now use Lemma~\ref{uniqueness} and Proposition~\ref{prop:splittingpotential} to reduce the
problem of constructing a Hertz potential to the level of initial data.
\begin{lemma}\label{lem:potentialconstrainteq1}Let $j\geq 2$ be an integer and $\varphi_{A\dots F}$ be a spinor field in $H^j_\delta(S_{2s})$ satisfying the constraint equation $(\sdiv_{2s}\varphi)_{C\dots F} = 0$.
Assume that there exist spinor fields $\xi_{A\dots F}\in H^{j+2s}_{\delta+2s}(S_{2s})$ and $\zeta_{A\dots F}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s})$ satisfying
\begin{equation*}
\varphi_{A\dots F}
=(\mathcal{G}_{2s}\scurl_{2s}\xi)_{A\dots F}
+(\mathcal{G}_{2s}\zeta)_{A\dots F}.
\end{equation*}
Then the only solution to the Cauchy problem \eqref{cauchyprobmassless} for massless free fields is given by
\begin{equation*}
\phi_{A \dots F} = \nabla_{AA'}\cdots\nabla_{FF'} \widetilde\chi^{A'\dots F'},
\end{equation*}
where $\widetilde \chi^{A'\dots F'}$ is obtained through the Cauchy problem \eqref{cauchywavechieq1} for $\chi_{A\dots F}$ with the initial data $(\xi_{A\dots F}, \sqrt{2}\zeta_{A\dots F})$.
\end{lemma}
\begin{proof} Let
\begin{equation*}
\widetilde\phi_{A \dots F} = \nabla_{AA'}\cdots\nabla_{FF'} \widetilde\chi^{A'\dots F'}.
\end{equation*}
It is a simple calculation to check that $\widetilde \phi_{A \dots F}$ satisfies the massless field equation of spin-$s$ (see \cite{MR0175590}, for instance). Furthermore, the restriction of $\phi_{A\dots F}$ and $\widetilde\phi_{A\dots F}$ agree on $\{t=0\}$ and are equal to $\varphi_{A\dots F}$ which lies in $H^j_\delta(S_{2s})$ and consequently in $L^2_{\delta}(S_{2s})$. Using the uniqueness stated in Lemma \ref{uniqueness}, we can conclude that both agree.
\end{proof}
\begin{theorem}\label{th:repthm} Let $s$ be in $\tfrac{1}{2}\mathbb{N}$, $\delta$ be in $\R\setminus\mathbb{Z}$ and $j\geq2$ an integer. We consider $\varphi_{A\dots F}$ in $H^j_\delta(S_{2s})$ satisfying the constraint equation $D^{AB}\varphi_{A\dots F} =0$.
Then there exists a spinor field $\zeta_{A\dots F}$, solving the equation
$$
\varphi_{A\dots F} = (\mathcal{G}_{2s}\zeta)_{A\dots F}
$$
and satisfying the estimates
\begin{equation*}
\Vert\zeta_{A\dots F}\Vert_{j+2s-1,\delta+2s-1}\leq C \Vert\varphi_{A\dots F}\Vert_{j,\delta}.
\end{equation*}
Furthermore, the unique solution of the Cauchy problem for massless fields \eqref{cauchyprobmassless} with the initial datum $\varphi_{A\dots F}$ is given by
$$
\phi_{A\dots F} = \nabla_{AA'}\dots \nabla_{FF'}\widetilde{\chi}^{A'\dots F'},
$$
where the spinor field $\chi_{A\dots F}$, defined by
$$
\chi_{A\dots F}= \tau_{AA'}\cdots \tau_{FF'}\widetilde\chi^{A'\dots F'},
$$ satisfies the Cauchy problem \eqref{cauchywavechieq1} for the wave equation with initial data $(0,\sqrt{2}\zeta_{A\dots F})$.
\end{theorem}
\begin{proof} This result is a direct consequence of
Lemma~\ref{lem:potentialconstrainteq1}, and of Propositions~\ref{proprepspin1} and \ref{proprepspingeneral}.
\end{proof}
\subsection{Decay result for higher spin fields}\label{sec:maintheorem}
The notations adopted in the formulation of the main theorem is consistent with the ones which are adopted in Section~\ref{geometricbackground}.
The following decay result for higher spin fields recovers the decay result obtained by Christo\-doulou and Klainerman in \cite{Christodoulou:1990dd} for the spins 1 (corresponding to the weight $\delta = -5/2$) and 2 (corresponding to the weight $\delta = -7/2$). The main difference is the regularity of the initial data: Christodoulou and Klainerman rely on weighted Sobolev embedding to obtain their decay result, so that only two derivatives of the initial data are required. The result which is presented here is based on the decay result of solutions of the wave equation stated in Section \ref{sec:estsol} and consequently requires at least three derivatives of the initial data. This restriction can be removed as soon as a decay result for the wave equation for initial data with arbitrary decay at spatial infinity, and regularity $H^2\times H^1$, is established. The following theorem will then also hold with one derivative less in the norms.
\begin{theorem}\label{maintheorem}
Let $s$ be in $\frac12\mathbb{N}$, $\delta$ in $\mathbb{R}\backslash \mathbb{Z}$, $j> 2$ an integer and consider the Cauchy problem for the massless free spin-$s$ fields
$$\left\{
\begin{array}{l}
\nabla^{AA'}\phi_{A\dots F}=0\\
\phi_{A\dots F}|_{t=0} = \varphi_{A\dots F} \in H^j_{\delta}(S_{2s})\\
D^{AB} \varphi_{A\dots F}=0.
\end{array}\right.
$$
We finally consider three nonnegative integers $k,l,m$ whose sum is denoted by $n\leq j-3$.
The following inequalities hold, for all $\bm{i}$ in $\{0,\dots,2s\}$: there exists a constant $C$ depending only on a choice of a constant dyad and of $k,l,m$ such that:
\begin{enumerate}
\item for any $t\geq 0$, $x \in \R^3$ such that $t>3r$, that is to say, in the interior region,
$$
|\nabla^n \phi_{A\dots F}|\leq C \lAngle t\rAngle^{\delta-n} \Vert\varphi_{A\dots F}\Vert_{3+n, \delta};
$$
\item for $\bm{i}$ such that $1+2s+\delta-l-\bm{i}<0$, for any $t\geq 0$, $x \in \R^3$, such that $3r>t>\tfrac{r}{3}$, that is to say in the exterior region,
$$
|D^kD'^l\snabla^m\phi_{\bm{i}}|\leq\frac{C \lAngle u\rAngle ^{1+\delta+2s-l-\bm{i}}}{\lAngle v\rAngle^{1+2s-\bm{i}+k+m}}\Vert\varphi_{A\dots F}\Vert_{3+n, \delta};
$$
\item for $\bm{i}$ such that $1+2s+\delta-l-\bm{i}>0$, for any $t\geq 0$, $x \in \R^3$, such that $3r>t>\tfrac{r}{3}$, that is to say in the exterior region,
$$
|D^kD'^l\snabla^m\phi_{\bm{i}}|\leq C \lAngle v\rAngle^{\delta-n} \Vert\varphi_{A\dots F}\Vert_{3+n, \delta}.
$$
\end{enumerate}
\end{theorem}
\begin{proof} Let $s$ and $\delta$ be such as in the theorem and consider $\varphi_{A\dots F}$ a initial datum in $H^j_{\delta}(S_{2s})$ satisfying the constraints equation
$$
D^{AB}\varphi_{A\dots F}=0.
$$
The initial datum $\varphi_{A\dots F}$ satisfies the assumptions stated in Theorem~\ref{th:repthm}, so that there exists a potential $\tilde{\chi}^{A'\dots F'}$ of order $2s$ such that the solution of the Cauchy problem with initial datum $\varphi_{A\dots F}$ is given by:
$$
\phi_{A\dots F} = \nabla_{AA'}\dots \nabla_{FF'} \tilde{\chi}^{A'\dots F'}
$$
and $\chi_{A\dots F} = \tau_{AA'}\dots \tau_{FF'} \tilde \chi^{A'\dots F'}$ satisfies the Cauchy problem
$$
\left\{
\begin{array}{l}
\square \chi_{A\dots F}=0\\
\chi_{A\dots F}|_{t=0}\in H^{j+2s}_{\delta+2s}(S_{2s})\\
\partial_t \chi_{A\dots F}|_{t=0}\in H^{j+2s-1}_{\delta+2s-1}(S_{2s}).
\end{array}
\right.
$$
Furthermore, the norm of the potential is controlled by the norm of the initial data
$$
\Vert\chi_{A\dots F} \Vert_{3+n+2s,\delta+2s} \leq C\Vert\varphi_{A\dots F}\Vert_{3+n,\delta}.
$$
A constant dyad $(e^A_0, e^A_1)$ on the Minkowski space is chosen. The components of the field $\chi_{A\dots F}$ are then of the form $\chi \xi^1_{A}\dots \xi^{2s}_{F}$,
where the constant spinor $\xi_{A}^{i}$ (for $i\in\{1,\dots, 2s\}$) belongs to $\{e^0_{A}, e^{1}_A\}$ and $\chi$ is a complex function satisfying a Cauchy problem of the form
$$
\left\{
\begin{array}{l}
\square \chi=0 , \\
\chi|_{t=0}\in H^{j+2s}_{\delta+2s}(\R^3, \mathbb{C}) , \\
\partial_t \chi|_{t=0} \in H^{j+2s-1}_{\delta+2s-1}(\R^3, \mathbb{C}).
\end{array}
\right.
$$
Proposition~\ref{prop:decay1} can then be used, on each of the components of the field. All these components decay exactly in the same way and, consequently, the field $\phi_{A\dots F}$ decays exactly as the field under consideration in Proposition~\ref{prop:decay1}.
\end{proof}
\subsection*{Acknowledgments}
We thank Dietrich H\"afner, Jean-Philippe Nicolas and Lionel Mason for
helpful discussions.
\appendix
\section{Algebraic properties of the fundamental operators} \label{sec:algprop}
To prove Proposition~\ref{prop:splittingpotential}, we need the following relation
\begin{lemma}\label{Gcurlexpansion}
The operators $\mathcal{G}_k$ and $\scurl_k$ commute and we have
\begin{eqnarray*}
(\mathcal{G}_k\scurl_k\phi)_{A_1\dots A_k}&=&(\scurl_k\mathcal{G}_k\phi)_{A_1\dots A_k}\\
&=&
\sum_{n=0}^{\lfloor \tfrac{k}{2}\rfloor}\binom{k}{2n}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^n_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}.
\end{eqnarray*}
\end{lemma}
\begin{proof}
We begin by proving that $(\mathcal{G}_k\scurl_k\phi)_{A_1\dots A_k}$ has the desired form.
By partially expanding the symmetry of the $\scurl_k$ operator we get
\begin{align*}
&\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k}\scurl_k\phi)_{A_{k-2n}\dots A_{k})B_1\dots B_{k-2n-1}}\nonumber\\
={}&\frac{2n+1}{k}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^n_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}\nonumber\\
&+\frac{k-2n-1}{k}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}D_{|B_{k-2n-1}|}{}^{B_{k-2n}}(\Delta^n_{k}\phi)_{A_{k-2n}\dots A_{k})B_1\dots B_{k-2n}}\nonumber\\
={}&\frac{2n+1}{k}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^n_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}\nonumber\\
&-\frac{k-2n-1}{2k}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-2}}{}^{B_{k-2n-2}}}_{k-2n-2}(\Delta^{n+1}_{k}\phi)_{A_{k-2n-1}\dots A_{k})B_1\dots B_{k-2n-2}}.
\end{align*}
Where we used $D_{A}{}^{C}D_{BC}=-\tfrac{1}{2}\epsilon_{AB}\Delta$ in the last step. We therefore get
\begin{align}
(\mathcal{G}_k&\scurl_k\phi)_{A_1\dots A_k}\nonumber\\
={}&
\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^n_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}\nonumber\\
&+\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n+1}(-2)^{1-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-2}}{}^{B_{k-2n-2}}}_{k-2n-2}(\Delta^{n+1}_{k}\phi)_{A_{k-2n-1}\dots A_{k})B_1\dots B_{k-2n-2}}\nonumber\\
={}&
\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^n_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}\nonumber\\
&+\sum_{n=1}^{\lfloor \tfrac{k+1}{2}\rfloor}\binom{k-1}{2n-1}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n}(\Delta^{n}_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}.\label{Gkckexpr1}
\end{align}
Where we just changed $n\rightarrow n-1$ in the last sum. The Pascal triangle gives the algebraic identity
\begin{align*}
\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n}A^{k-n}B^{n}
+\sum_{n=1}^{\lfloor \tfrac{k+1}{2}\rfloor}\binom{k-1}{2n-1}A^{k-n}B^{n}
=\sum_{n=0}^{\lfloor \tfrac{k}{2}\rfloor}\binom{k}{2n}A^{k-n}B^{n},
\end{align*}
which in turn gives the desired form for $(\mathcal{G}_k\scurl_k\phi)_{A_1\dots A_k}$. To handle $(\scurl_k\mathcal{G}_k\phi)_{A_1\dots A_k}$ we partially expand the symmetry in the following expression
\begin{align*}
&D_{A_1}{}^{C}\underbrace{D_{(A_2}{}^{B_2}\cdots D_{A_k-2n}{}^{B_{k-2n}}}_{k-2n-1}(\Delta_k^n\phi)_{A_{k-2n+1}\dots A_k C)B_2\dots B_{k-2n}}\nonumber\\
={}&\frac{k-2n-1}{k}D_{A_1}{}^{C}D_{C}{}^{B_2}\underbrace{D_{(A_2}{}^{B_3}\cdots D_{A_k-2n-1}{}^{B_{k-2n}}}_{k-2n-2}(\Delta_k^n\phi)_{A_{k-2n}\dots A_k)B_2\dots B_{k-2n}}\nonumber\\
&+\frac{2n+1}{k}D_{A_1}{}^{C}\underbrace{D_{(A_2}{}^{B_2}\cdots D_{A_k-2n}{}^{B_{k-2n}}}_{k-2n-1}(\Delta_k^n\phi)_{A_{k-2n+1}\dots A_k)C B_2\dots B_{k-2n}}\nonumber\\
={}&-\frac{k-2n-1}{2k}\underbrace{D_{(A_2}{}^{B_3}\cdots D_{A_k-2n-1}{}^{B_{k-2n}}}_{k-2n-2}(\Delta_k^{n+1}\phi)_{A_{k-2n}\dots A_k)A_1 B_3\dots B_{k-2n}}\nonumber\\
&+\frac{2n+1}{k}D_{A_1}{}^{B_1}\underbrace{D_{(A_2}{}^{B_2}\cdots D_{A_k-2n}{}^{B_{k-2n}}}_{k-2n-1}(\Delta_k^n\phi)_{A_{k-2n+1}\dots A_k)B_1\dots B_{k-2n}}.
\end{align*}
Where we in the last step again used $D_{A}{}^{C}D_{BC}=-\tfrac{1}{2}\epsilon_{AB}\Delta$. Using this in the definition of $\mathcal{G}_k$ yields
\begin{align*}
D_{A_1}{}^C&(\mathcal{G}_k\phi)_{CA_2\dots A_k}\nonumber\\
={}&\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n+1}(-2)^{1-n}\underbrace{D_{(A_2}{}^{B_3}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-2}}}_{k-2n-2}(\Delta^{n+1}_{k}\phi)_{A_{k-2n}\dots A_{k})A_1B_3\dots B_{k-2n}}\nonumber\\
&+\sum_{n=0}^{\lfloor \tfrac{k-1}{2}\rfloor}\binom{k-1}{2n}(-2)^{1-n}D_{A_1}{}^{B_1}\underbrace{D_{(A_2}{}^{B_2}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n-1}(\Delta^{n}_{k}\phi)_{A_{k-2n+1}\dots A_{k})B_1\dots B_{k-2n}}.
\end{align*}
After symmetrization we get that $(\scurl_k\mathcal{G}_k\phi)_{A_1\dots A_k}$ has an expansion identical to the one in the first equation in \eqref{Gkckexpr1}. This gives the desired result.
\end{proof}
\begin{proposition}\label{splittingpotentialAppendix}
The equation \eqref{representation4D} together with \eqref{wavechieq2} implies
\begin{equation*}
\phi_{A_1\dots A_{2s}}=(\mathcal{G}_{2s}\scurl_{2s}\chi)_{A_1\dots A_{2s}}
+\tfrac{1}{\sqrt{2}}(\mathcal{G}_{2s}\partial_t\chi)_{A_1\dots A_{2s}}.
\end{equation*}
\end{proposition}
\begin{proof}
Using $\tau_{B}{}^{A'}\nabla_{AA'}=D_{AB}+\tfrac{1}{\sqrt{2}}\epsilon_{AB}\partial_t$ we can write the potential equation in terms of $D_{AB}$ and $\partial_t$. We have
\begin{align*}
\phi_{A_1\dots A_{2s}}
={}&\underbrace{\nabla_{A_1A'_1}\cdots \nabla_{A_{2s}A'_{2s}}}_{2s}\widetilde\chi^{A'_1\dots A'_{2s}}
=\underbrace{\tau^{B_1A'_1}\nabla_{A_1A'_1}\cdots \tau^{B_{2s}A'_{2s}}\nabla_{A_{2s}A'_{2s}}}_{2s}\chi_{B_1\dots B_{2s}}\nonumber\\
={}&\underbrace{(D_{A_1}{}^{B_1}+\tfrac{1}{\sqrt{2}}\epsilon_{A_1}{}^{B_1}\partial_t)\cdots (D_{A_{2s}}{}^{B_{2s}}+\tfrac{1}{\sqrt{2}}\epsilon_{A_{2s}}{}^{B_{2s}}\partial_t)}_{2s}\chi_{B_1\dots B_{2s}}\nonumber\\
={}&\sum_{n=0}^{2s}\binom{2s}{n}2^{-n/2}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2s-n}}{}^{B_{2s-n}}}_{2s-n}\partial_t^n\chi_{A_{2s-n+1}\dots A_{2s})B_1\dots B_{2s-n}}.
\end{align*}
We can now use \eqref{wavechieq2} to eliminate all higher order time derivatives. This gives
\begin{align*}
&\negthickspace\negthickspace\negthickspace\phi_{A_1\dots A_{2s}}\nonumber\\
={}&\sum_{n=0}^{\lfloor s\rfloor}\binom{2s}{2n}2^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2s-2n}}{}^{B_{2s-2n}}}_{2s-2n}\partial_t^{2n}\chi_{A_{2s-2n+1}\dots A_{2s})B_1\dots B_{2s-2n}}\nonumber\\
&+\sum_{n=0}^{\lfloor s-\tfrac{1}{2}\rfloor}\binom{2s}{2n+1}2^{-n-1/2}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2s-2n-1}}{}^{B_{2s-2n-1}}}_{2s-2n-1}\partial_t^{2n+1}\chi_{A_{2s-2n}\dots A_{2s})B_1\dots B_{2s-2n-1}}\nonumber\\
={}&\sum_{n=0}^{\lfloor s\rfloor}\binom{2s}{2n}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2s-2n}}{}^{B_{2s-2n}}}_{2s-2n}(\Delta^n_{2s}\chi)_{A_{2s-2n+1}\dots A_{2s})B_1\dots B_{2s-2n}}\nonumber\\
&+\tfrac{1}{\sqrt{2}}\sum_{n=0}^{\lfloor s-\tfrac{1}{2}\rfloor}\binom{2s}{2n+1}(-2)^{-n}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2s-2n-1}}{}^{B_{2s-2n-1}}}_{2s-2n-1}(\Delta^n_{2s}\partial_t\chi)_{A_{2s-2n}\dots A_{2s})B_1\dots B_{2s-2n-1}}\nonumber\\
={}&(\mathcal{G}_{2s}\scurl_{2s}\chi)_{A_1\dots A_{2s}}
+\tfrac{1}{\sqrt{2}}(\mathcal{G}_{2s}\partial_t\chi)_{A_1\dots A_{2s}}.
\end{align*}
In the last step we used the definition of $\mathcal{G}_{2s}$ and Lemma~\ref{Gcurlexpansion}. In fact we have defined $\mathcal{G}_k$ to match the $\partial_t$ part of this expression.
\end{proof}
\begin{proposition}\label{divGproperty}
For $k\geq 2$, the operators $\mathcal{G}_k$ have the properties $\sdiv_k\mathcal{G}_k=0$ and $\mathcal{G}_k\stwist_{k-2}=0$.
\end{proposition}
\begin{proof}
First we prove that $\sdiv_k\mathcal{G}_k=0$.
By partially expanding the symmetrization in the definition of $\mathcal{G}_k$ and restricting the summation to non-vanishing terms we get
\begin{align*}(\mathcal{G}_k&\phi)_{A_1\dots A_k}\nonumber\\
={}& \sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-3}{2}\right\rfloor}}
\binom{k-2}{2n+1}(-2)^{-n} D_{A_1}{}^{B_1}D_{A_2}{}^{B_2}\underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-3} (\Delta^n_k\phi)_{A_{k-2n}\dots A_k)B_1\dots B_{k-2n-1}}\nonumber\\
&+ \sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}
\binom{k-2}{2n}(-2)^{-n} D_{A_1}{}^{B_2}\underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n-2} (\Delta^n_k\phi)_{A_{k-2n+1}\dots A_k)A_2B_2\dots B_{k-2n}}\nonumber\\
&+ \sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}
\binom{k-2}{2n}(-2)^{-n} D_{A_2}{}^{B_2}\underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n}}{}^{B_{k-2n}}}_{k-2n-2} (\Delta^n_k\phi)_{A_{k-2n+1}\dots A_k)A_1B_2\dots B_{k-2n}}\nonumber\\
&+ \sum_{n=1}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}
\binom{k-2}{2n-1}(-2)^{-n} \underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n+1}}{}^{B_{k-2n+1}}}_{k-2n-1} (\Delta^n_k\phi)_{A_{k-2n+2}\dots A_k)A_1A_2B_3\dots B_{k-2n+1}}.
\end{align*}
Using $D_{A}{}^{C}D_{BC}=-\tfrac{1}{2}\epsilon_{AB}\Delta$ we get $D^{A_1A_2}D_{A_1}{}^{B_1}D_{A_2}{}^{B_2}=\tfrac{1}{2}D^{B_1B_2}\Delta$, $D^{A_1(A_2}D_{A_1}{}^{B_2)}=0$ and
\begin{align*}
(\sdiv_k&\mathcal{G}_k\phi)_{A_1\dots A_k}\nonumber\\
={}& -\sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-3}{2}\right\rfloor}}
\binom{k-2}{2n+1}(-2)^{-n-1} D^{B_1B_2}\underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-3} (\Delta^{n+1}_k\phi)_{A_{k-2n}\dots A_k)B_1\dots B_{k-2n-1}}\nonumber\\
&+ \sum_{n=1}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}
\binom{k-2}{2n-1}(-2)^{-n}D^{B_1B_2} \underbrace{D_{(A_3}{}^{B_3}\cdots D_{A_{k-2n+1}}{}^{B_{k-2n+1}}}_{k-2n-1} (\Delta^n_k\phi)_{A_{k-2n+2}\dots A_k)B_1\dots B_{k-2n+1}}.
\end{align*}
The first sum is identical to the second sum after a variable change $n\rightarrow n-1$, hence $\sdiv_k\mathcal{G}_k=0$.
Now, we turn to the proof of $\mathcal{G}_k\stwist_{k-2}=0$. Partial expansion of the symmetrization in the definition of $\stwist_{k-2}$ gives
\begin{align*}
&\underbrace{D_{(A_1}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_k\stwist_{k-2}\phi)_{A_{k-2n}\dots A_{k})B_1\dots B_{k-2n-1}}\nonumber\\
={}&\frac{(k-2n-1)(k-2n-2)}{k(k-1)}D_{B_1B_2}\underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k-2}\phi)_{A_{k-2n}\dots A_{k})B_3\dots B_{k-2n-1}}\nonumber\\
&+\frac{2(k-2n-1)(2n+1)}{k(k-1)}D_{B_1(A_k}\underbrace{D_{A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k-2}\phi)_{A_{k-2n}\dots A_{k-1})B_2\dots B_{k-2n-1}}\nonumber\\
&+\frac{2n(2n+1)}{k(k-1)}D_{(A_{k-1}A_k}\underbrace{D_{A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k-2}\phi)_{A_{k-2n}\dots A_{k-2})B_1\dots B_{k-2n-1}}\nonumber\\
={}&\frac{(k-2n-1)(k-2n-2)}{2k(k-1)}D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-3}(\Delta^{n+1}_{k-2}\phi)_{A_{k-2n}\dots A_{k})B_3\dots B_{k-2n-1}}\nonumber\\
&+\frac{2n(2n+1)}{k(k-1)}D_{(A_{k-1}A_k}\underbrace{D_{A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k-2}\phi)_{A_{k-2n}\dots A_{k-2})B_1\dots B_{k-2n-1}}.
\end{align*}
Where we again used $D^{A_1A_2}D_{A_1}{}^{B_1}D_{A_2}{}^{B_2}=\tfrac{1}{2}D^{B_1B_2}\Delta$ and $D^{A_1(A_2}D_{A_1}{}^{B_2)}=0$.
We therefore get
\begin{align*}
(\mathcal{G}_k&\stwist_{k-2}\phi)_{A_1\dots A_k}\nonumber\\
={}&-\sum_{n=0}^{\mathclap{\left\lfloor\tfrac{k-3}{2}\right\rfloor}}\binom{k-2}{2n+1}(-2)^{1-n} D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-3}(\Delta^{n+1}_{k-2}\phi)_{A_{k-2n}\dots A_{k})B_3\dots B_{k-2n-1}}\nonumber\\
&\negthickspace\negmedspace{}+\sum_{n=1}^{\mathclap{\left\lfloor\tfrac{k-1}{2}\right\rfloor}}\binom{k-2}{2n-1}(-2)^{-n}D_{(A_{k-1}A_k}\underbrace{D_{A_1}{}^{B_1}\cdots D_{A_{k-2n-1}}{}^{B_{k-2n-1}}}_{k-2n-1}(\Delta^n_{k-2}\phi)_{A_{k-2n}\dots A_{k-2})B_1\dots B_{k-2n-1}}.
\end{align*}
The first sum is identical to the second sum after a variable change $n\rightarrow n-1$, hence $\mathcal{G}_k\stwist_{k-2}=0$.
\end{proof}
To use elliptic theory, we need well behaved elliptic operators. $\mathcal{G}_k$ is in general not elliptic but, through the following lemma, it can related to some power of the Laplacian -- which of course is elliptic.
\begin{lemma}\label{lapspinor}
The formulae \eqref{LaplacianAsGodd} and \eqref{LaplacianAsGeven} hold, that is to say
\begin{align*}
(\Delta^{k}_{2k}\phi)_{A_1\dots A_{2k}}={}&(\stwist_{2k-2}\mathcal{F}_{2k-2}\sdiv_{2k}\phi)_{A_1\dots A_{2k}}-(-2)^{1-k}(\mathcal{G}_{2k}\scurl_{2k}\phi)_{A_1\dots A_{2k}},\\
(\Delta^{k}_{2k+1}\phi)_{A_{1}\dots A_{2k+1}}={}&
(\stwist_{2k-1}\mathcal{F}_{2k-1}\sdiv_{2k+1}\phi)_{A_1\dots A_{2k+1}}
+(-2)^{-k}(\mathcal{G}_{2k+1}\phi)_{A_1\dots A_{2k+1}} .
\end{align*}
\end{lemma}
\begin{proof}
For both formulae, we will use the following help quantity for the spin-$(k+j/2)$ case
\begin{align*}
I^{j,k}_m\equiv{}&\negmedspace\sum_{n=m}^{k-1}\binom{2k+j}{2n+j-2m}(-2)^{-n} \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n} (\Delta^{n}_{2k+j}\phi)_{A_{2k-2n+1}\dots A_{2k+j})B_1\dots B_{2k-2n}}
\end{align*}
Multiplying $D_{A_1}{}^{B_1}D_{C_1}{}^{B_2}\phi_{A_3\dots A_{k}C_2B_2}$ with
$\epsilon_{A_2}{}^{C_1}\epsilon_{B_1}{}^{C_2}=\epsilon_{A_2B_1}\epsilon^{C_1C_2}+\epsilon_{A_2}{}^{C_2}\epsilon_{B_2}{}^{C_1}$ and using $D_{A}{}^{C}D_{BC}=-\tfrac{1}{2}\epsilon_{AB}\Delta$, we get
\begin{align*}
D_{A_1}{}^{B_1}D_{A_2}{}^{B_2}\phi_{A_3\dots A_{k}B_1B_2}
={}&
-\tfrac{1}{2}(\Delta_k\phi)_{A_1\dots A_k}+
D_{A_1 A_2}(\sdiv_k\phi)_{A_3\dots A_{k}}.
\end{align*}
Using this in the definition of $I^{j,k}_{m}$ gives
\begin{align}
I^{j,k}_m
={}&\sum_{n=m}^{k-1}\binom{2k+j}{2n+j-2m}(-2)^{-n-1}\nonumber\\
&\quad\times \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2k-2n-2}}{}^{B_{2k-2n-2}}}_{2k-2n-2} (\Delta^{n+1}_{2k+j}\phi)_{A_{2k-2n-1}\dots A_{2k+j})B_1\dots B_{2k-2n-2}}\nonumber\\
&+\sum_{n=m}^{k-1}\binom{2k+j}{2n+j-2m}(-2)^{-n} \nonumber\\
&\quad\times D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n-2} (\sdiv_{2k+j}\Delta^{n}_{2k+j}\phi)_{A_{2k-2n+1}\dots A_{2k+j})B_3\dots B_{2k-2n}}\nonumber\\
={}&I^{j,k}_{m+1}+\binom{2k+j}{2k+j-2m}(-2)^{-k}(\Delta^{k}_{2k+j}\phi)_{A_1\dots A_{2k+j}}
+\sum_{n=m}^{k-1}\binom{2k+j}{2n+j-2m}(-2)^{-n} \nonumber\\
&\quad\times D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n-2} (\sdiv_{2k+j}\Delta^{n}_{2k+j}\phi)_{A_{2k-2n+1}\dots A_{2k+j})B_3\dots B_{2k-2n}}.\label{IRecursion}
\end{align}
Here, we have changed $n\rightarrow n-1$ in the first sum, and identified that as $I^{j,k}_{m+1}$ plus the term where $n=k$, which gives us the $\Delta^{k}$-term. We can easily solve the recursion \eqref{IRecursion} and get
\begin{align*}
I^{j,k}_0
={}&\sum_{m=0}^{k-1}\binom{2k+j}{2k+j-2m}(-2)^{-k}(\Delta^{k}_{2k+j}\phi)_{A_1\dots A_{2k+j}}
+\sum_{m=0}^{k-1}\sum_{n=m}^{k-1}\binom{2k+j}{2n+j-2m}(-2)^{-n}\nonumber\\
&\quad \times D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n-2} (\sdiv_{2k+j}\Delta^{n}_{2k+j}\phi)_{A_{2k-2n+1}\dots A_{2k+j})B_3\dots B_{2k-2n}}\nonumber\\
={}&\sum_{m=0}^{k-1}\binom{2k+j}{2m}(-2)^{-k}(\Delta^{k}_{2k+j}\phi)_{A_1\dots A_{2k+j}}
+\sum_{n=0}^{k-1}\sum_{m=0}^{k-1-n}\binom{2k+j}{2n+2m+2}(-2)^{n+1-k}\nonumber\\
&\quad \times D_{(A_1A_2}\underbrace{D_{A_3}{}^{B_3}\cdots D_{A_{2n+2}}{}^{B_{2n+2}}}_{2n} (\Delta^{k-1-n}_{2k+j-2}\sdiv_{2k+j}\phi)_{A_{2n+3}\dots A_{2k+j})B_3\dots B_{2n+2}}\nonumber\\
={}&\sum_{m=0}^{k-1}\binom{2k+j}{2m}(-2)^{-k}(\Delta^{k}_{2k+j}\phi)_{A_1\dots A_{2k+j}}
-2^{j-1}(-2)^{k}(\stwist_{2k+j-2}\mathcal{F}_{2k+j-2}\sdiv_{2k+j}\phi)_{A_1\dots A_{2k+j}}.
\end{align*}
In the second sum we have changed the order of summation followed by the change $n\rightarrow k-n-1$.
For the operators acting on an odd number of indices we have
\begin{align*}
(\mathcal{G}_{2k+1}&\phi)_{A_1\dots A_{2k+1}}\nonumber\\
={}&
\sum_{n=0}^{k}\binom{2k+1}{2n+1}(-2)^{-n} \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n} (\Delta^n_{2k+1}\phi)_{A_{2k-2n+1}\dots A_{2k+1})B_1\dots B_{2k-2n}}\nonumber\\
={}&I^{1,k}_0+\binom{2k+1}{2k+1}(-2)^{-k} (\Delta^k_{2k+1}\phi)_{A_1\dots A_{2k+1}}\nonumber\\
={}&\sum_{m=0}^{k}\binom{2k+1}{2k+1-2m}(-2)^{-k}(\Delta^{k}_{2k+1}\phi)_{A_1\dots A_{2k+1}}
-(-2)^k(\stwist_{2k-1}\mathcal{F}_{2k-1}\sdiv_{2k+1}\phi)_{A_1\dots A_{2k+1}}\nonumber\\
={}&(-2)^k (\Delta^{k}_{2k+1}\phi)_{A_1\dots A_{2k+1}}
-(-2)^k(\stwist_{2k-1}\mathcal{F}_{2k-1}\sdiv_{2k+1}\phi)_{A_1\dots A_{2k+1}}.
\end{align*}
Hence, we have \eqref{LaplacianAsGodd}.
For the operators acting on an even number of indices we can use \eqref{Gcurlexpansion} to obtain
\begin{align*}
(\mathcal{G}_{2k}&\scurl_{2k}\phi)_{A_1\dots A_{2k}} \nonumber\\
={}&
\sum_{n=0}^{k}\binom{2k}{2n}(-2)^{-n} \underbrace{D_{(A_1}{}^{B_1}\cdots D_{A_{2k-2n}}{}^{B_{2k-2n}}}_{2k-2n} (\Delta^n_{2k+1}\phi)_{A_{2k-2n+1}\dots A_{2k})B_1\dots B_{2k-2n}}\nonumber\\
={}& I^{0,k}_0+\binom{2k}{2k}(-2)^{-k} (\Delta^k_{2k}\phi)_{A_1\dots A_{2k}}\nonumber\\
={}&\sum_{m=0}^{k}\binom{2k}{2k-2m}(-2)^{-k}(\Delta^{k}_{2k}\phi)_{A_1\dots A_{2k}}
+(-2)^k(\stwist_{2k-2}\mathcal{F}_{2k-2}\sdiv_{2k}\phi)_{A_1\dots A_{2k}}\nonumber\\
={}&-(-2)^{k-1}(\Delta^{k}_{2k}\phi)_{A_1\dots A_{2k}}
+(-2)^{k-1}(\stwist_{2k-2}\mathcal{F}_{2k-2}\sdiv_{2k}\phi)_{A_1\dots A_{2k}}.
\end{align*}
Hence, we have \eqref{LaplacianAsGeven}.
\end{proof}
\bibliographystyle{abbrv}
\bibliography{biblio1}
\end{document} | 107,255 |
Allergy sufferers are used to being unable to do things the same way that non-allergy sufferers do. They have to ensure they don’t come in contact with something they’re allergic to and always be prepared to handle a reaction if they accidentally do. Unfortunately, those allergy concerns don’t go away even when it comes to medical needs. Getting a flu vaccine is something that everyone should do each flu season to reduce the number of cases and protect themselves, as well as the people they’re around, but are allergy sufferers able to safely get vaccinated?
Recombinant Flu Vaccine
The traditional flu vaccine is made using the flu virus along with multiple other ingredients. One of those ingredients that is a common allergy for people is chicken eggs. Over the years, the flu vaccine has been able to be altered to meet the needs of almost anyone, including allergy sufferers! The recombinant flu vaccine is a vaccine that helps prevent the flu but does not contain the actual flu vaccine or chicken eggs, making it safe for allergy sufferers. The recombinant flu vaccine is approved for use in anyone over 18 years of age.
Getting Vaccinated
Getting vaccinated as an allergy sufferer is just as easy as getting vaccinated without allergies! Visit your local urgent care center, inform them that you suffer from allergies, and they’ll give you the recombinant vaccine instead of the traditional egg-based vaccine. The vaccination comes in the form of a shot and the entire process will be over very quickly. Little to no side effects are reported with the recombinant vaccine. The most common side effects include tenderness at the injection site, headache, and muscle aches. All side effects are temporary and won’t impair your day-to-day life.
Making the choice to get vaccinated not only protects you against the flu, but it lowers your risk of giving the flu to a friend, family member, or coworker. A flu vaccine at an urgent care center is covered in full by most major insurance companies. Contact your local urgent care today to learn more about vaccine availability and current procedures amidst the covid-19 pandemic. There’s no reason to remain vulnerable to the flu just because you suffer from allergies! Encourage your friends and family members to visit the urgent care to receive their flu vaccine too. Anyone over 6 months of age is eligible for one version of the vaccine or another, so what are you waiting for?! | 39,709 |
TITLE: On uniform Structure induced by pseudo metric
QUESTION [0 upvotes]: If $ U^{'} $ induced by pseudo metric d, then the induced uniform structure on $ X $ induced by the pseudo metric $ d(f\times f) $?
REPLY [1 votes]: Suppose $f: X \to (Y, \mathcal{U}_d)$ is given for some pseudometric $d$ on $Y$.
So a base of the induced uniformity on $X$ will be all sets of the form
$$(f \times f)^{-1}[U_d(r)] \text{, where } U_d(r)=\{(y,y') \in Y^2: d(y,y') < r\}$$
with $r>0$. Instead, we can define a psueodmetric on $X$ by $d_X(x,x')=d(f(x), f(x'))$ and we have $d_X = d \circ (f \times f)$ as maps and
$$(f \times f)^{-1}[U_d(r)] = U_{d_X}(r)$$ by definition and so we can see this induced uniformity also as the one induced by the pseudometric $d_X$. | 111,137 |
And here we go again.
Not back to Texas... back to the world of "Gotta lie down and close your eyes, and maybe go to sleep. Now. Now. NOW!!!!!"
Ah, it was such a nice dream, to get up and start writing music again. Well, I did manage to get up and put a couple of things on SoundCloud. But that was it. And then, back to bed.
I made it to the vaguely-local dim sum place for lunch, and drove to the oculist to get my glasses adjusted. It was nice being in sun, warmth, and air, but it was a little tricky.
We'll try the "leaving the house" thing again tomorrow, because I have to visit a couple of banks, and go to the oculist again (unfortunately, because the person who said she was going to adjust my glasses got the lenses right but made the right earpiece dig in kinda nastily).
Somehow I made it through the Texas adventure without having to flee back to my hotel room. Barely, sometimes. But now, that I'm back home... I'm trapped in "gotta go to bed."
Is this an attack?
Who knows. And even if it was, or if I knew, would it matter?
Prob'ly not.
Wonder how long it's gonna be before I gotta go back to bed?
Who knows? Which is, of course, par for the course for the Neurological Highway. And, y'know... life.
Well, at least I had the strength to do a little blog-tweaking. And, because I've gotten some requests for it, to make one last visit to TMEA, here's a recording of their amazing performance of my music by the All-State Choir:
Wednesday, February 20, 2013
3 comments:
"...now, that I'm back home... I'm trapped in "gotta go to bed."
Robert, is it possible that something in your home is making you react this way? My neurological performance is very, very sensitive to environmental factors such as building materials, inhalants, molds, cleaning substances, etc. etc. I have been tested for everything under the sun and have altered my living space, cleaning supplies (personal and household), etc. accordingly. I simply find it curious that you got relief from this very oppressive issue while you were away and, now that you are back, the problem has resurfaced. If there is an environmental health practice anywhere near you, it might be worth checking them out. My doctors are in Dallas. I believe they have been a major reason why I have been able to moderate the severity of my MS, through minimizing or avoiding the things that are known to be my antagonists in the environment. For example, we know that beef ingestion makes me wobbly. Adhering to this type of environmental discipline is an exacting life practice but, given the alternative outcome, well worth the effort.
Robert, regarding your music, I am in awe of your ability to create such a wonderful work. When I write something, it never has as many moving parts outside my control as I imagine a performance piece does. Congratulations and well done!
Judy: A very interesting and well-considered point. I'm especially blessed that one of my Medical Team is unbelievably aware of, sensitive to, and right-on-top-of the tragic welath of toxic bogosities that infest today's environment. He read to me some of the cruft that one of his other patients was sensitive to; the list included "drywall dust" and "drywall cement." just try to avoid drywall in this day and age. He has even tested me for sensitivity to the odd stuff I use in my studio for sound abatement--which, fortunately, is something on the short list of "stuff I'm NOT sensitive to." Plus, my wife, now a voice actor, used to work in toxic waste abatement, and she is wicked aware of keeping crap out of the house.
But, still a very well taken point; we most often don't see what's right in front of us... Anyway, the good news is that my Medical Team are specifically chasing the "inescapable fatigue" issue, and who knows what they'll find?
Something much more insidious than "Maybe you're just tired," I'm guessing, will be at the bottom of it. | 213,876 |
TITLE: Is there a subfield $K$ such that $\mathbb{Q} \subset K \subset \mathbb{R}$ with the following property?
QUESTION [1 upvotes]: Is there a subfield $K$ such that $\mathbb{Q} \subset K \subset \mathbb{R}$ (proper subset) as follows:
$\mathbb{R}$ is a vector space over $K$ and has no finite generating set and $K$ is a vector space over $\mathbb{Q}$ and has no finite generating set.
Sum and multiplication are both defined as usual in $\mathbb{R}$.
My guess is that there is one: $K$ should not contain continuum as it is field so in this case it would be equal to $\mathbb{R}$, so $K$ is countable, and $\mathbb{R}$ has no finite generating set over countable sets. On the other hand, $K$ can contain countable but infinite number of elements from $\mathbb{R} - \mathbb{Q}$ such that $\forall a \in \mathbb{R} - \mathbb{Q} \,\,\,\,a \in K \Rightarrow \exists b \in K, q \in \mathbb{Q}:a\neq q\cdot b$, i.e. contains infinite number of irrational numbers so that pairwise their division is not rational. In this case number of elements in $K$ basis is countable, but infinite (each irrational number is in basis, if its product with some rational number isn't in it). $K$ seems to be field and not equal to $\mathbb{R}$, but I'm not sure.
Thanks in advance!
REPLY [5 votes]: Take $K$ to be the field of real algebraic numbers over $\mathbb {Q}$. It is countable so $\mathbb {R}$ is not finitely generated over $K$. Further, using Eisenstein's Criterion, it is simple to show that $[K:\mathbb {Q}]=\infty$ and so $K$ is not finitely generated over $\mathbb{Q}$. | 129,258 |
17.06.2021
Scaldis heavy lift vessels joining forces
The most spectacular part of this trial was the a Double Duo Lift, a simultaneous and synchronised 5000t lift with Rambiz and Gulliver.
For this lift we created a copy of the actual lifts to be performed, by placing a test load pontoon on the transport barge, combined with a unique and safe rigging design.
Scaldis PM, W. Zuidscherwoude says: ‘This test lift has been in the making for over 1,5years. To see it all come together with a perfectly executed lift is a statement of the competence and capabilities of Rambiz, Gulliver and their respective crews’
Scaldis would like to thank everyone who made this possible and especially the Client for the good cooperation, which is essential for this successful project.
« Back to overview | 269,321 |
TITLE: Nonsingular is rational singular
QUESTION [1 upvotes]: I know the following result is well known to the experts, but what are the good references or proofs:
Let $X,Y$ be a smooth variety with projective, birational morphism $f: X \to Y$. Let $L$ be a line bundle on $Y$, then we have the property that the higher direct image $R^if_*f^* L = 0$ when $i > 0$.
This is equivalent to the statement that nonsingular variety is a rational singular variety.
REPLY [4 votes]: There is another way to view this (and prove the second part) that is probably worth explaining (and mentioning some characteristic $p > 0$ connections). As abx already pointed out, this problem reduces to computing $R^i f_* O_X$. You don't need resolution of singularities to prove that vanishing though.
Characteristic zero
In characteristic zero, by Grauert-Riemenschneider vanishing, we have that $R^i f_* \omega_X = 0$ for all $i > 0$ (even if $Y$ is singular). On the other hand, since the relative canonical divisor of a projective birational map between regular schemes $f : X \to Y$ is always effective (this is true whenever the relative canonical divisor makes sense, including mixed characteristic, see the work of Lipman on pseudo-rational singularities), we have that $f_* \omega_X = \omega_Y$.
In other words, $R f_* \omega_X \simeq \omega_Y$ in the derived category (they are quasi-isomorphic). Hence by Grothendieck duality for $f$ we see that $O_Y \simeq R f_* O_X$ in the derived category as well. This implies that $R^i f_* O_X = 0$ for all $i > 0$.
Characteristic $p > 0$
IF $Y$ is smooth, then the same results also holds in characteristic $p > 0$.
See
https://arxiv.org/abs/0911.3599
However, Grauert-Riemenschneider is known to fail in characteristic $p > 0$ for singular $Y$ so you really need $Y$ to at least have mild singularities. See for instance example 3.11 in
http://arxiv.org/pdf/1212.5105.pdf
It is reasonable to perhaps conjecture that if $Y$ has $F$-regular singularities then Grauert-Riemenschnedier vanishing is true and hence $R^i f_* O_X = 0$ for all $i > 0$ and all resolutions of singularities $f : X \to Y$. | 163,820 |
TITLE: Introduction to Representation Theory Problem 1.26 part (b), (c)
QUESTION [2 upvotes]: The Problem: Let $A$ be a Weyl Algebra, generated by two elements $x, y$ with the relation
$$xy - yx - 1 = 0$$
Suppose that char$k = p$.
(b) What is the center of $A$?
Hint: Show that $x^p$ and $y^p$ are central elements.
(c) Find all irreducible finite dimensional representations of $A$.
Hint. Let $V$ be a irreducible finite dimensional representation of $A$, and $v$ be an eigenvector of $y$ in $V$. Show that $\{v, xv, x^2v,...x^{p-1}v\}$ is a basis of $V$.
For part (b) the proposition of this section part (i) tells us that A basis for the Weyl algebra $A$ is $\{x^iy^j, i,j \geq 0\}$. Therefore, to use the hint, we just need to show that $x^px^iy^j = x^iy^jx^p$. In the proof of the proposition they use the $yx - xy = 1$ to interchange $x$ and $y$ at the cost of error terms that have smaller number of letters $x,y$ then the original word. This seemed to be useful for this part until I realized that it didn't allow me to use that char$k = p$. When I see char$k = p$, the first thing I think of is $(x+y)^p = x^p + y^p$ but that's about all that pops to my mind.
For part (c), taking the hint we know that $v$ satisfies $\rho(y)v = k_0v$ for some $k_0 \in k$. I'm confused because $x \in A$ while $v \in V$ so now sure what $xv$ would mean. Are we viewing $V$ as a vector space over $A$ or over $k$?
REPLY [3 votes]: Part a: It would be helpful to have proven as a warm-up exercise that the Weyl algebra is simple in characteristic $0$. I will take the liberty of slightly rewriting the Weyl algebra as $k[x, \partial]/(\partial x - x \partial - 1)$ because I like thinking about its action on $k[x]$ by differential operators. If you prove simplicity using a method like the one in this math.SE answer a key step is the following.
Lemma 1: In any associative algebra, the commutator $[a, -]$ is a derivation: it satisfies the Leibniz rule $[a, bc] = [a, b] c + b [a, c]$.
In the Weyl algebra we have by definition that $[\partial, x] = 1$, so $[\partial, -]$, as a derivation, behaves exactly like the partial derivative with respect to $x$ on the subalgebra generated by $x$ (that's why I like calling it $\partial$); proving this amounts to using the lemma to prove by induction that
$$[\partial, x^n] = nx^{n-1}.$$
Lemma 2: An element $z$ is central iff the commutator $[z, -]$ vanishes iff it vanishes on a set of generators.
Proof. The first half is just a restatement of the definition of centrality; for the second half we use Lemma 1 to show that if $[z, a] = 0$ and $[z, b] = 0$ then $[z, ab] = 0$, so $\text{ker}([z, -])$ is a subalgebra. $\Box$
So to find central elements it's necessary and sufficient to find elements $f$ such that $[f, x] = [f, \partial] = 0$.
Now in characteristic $p$ a funny thing can happen which is that $n$ can be equal to zero if $p \mid n$, and in particular we have that
$$[\partial, x^p] = 0.$$
Switching the roles of $\partial$ and $x^p$ and writing this identity as $[x^p, \partial] = 0$, this tells us that the derivation $[x^p, -]$ vanishes on $\partial$, and since it clearly vanishes on $x$ then by Lemma 2 $x^p$ is central. Exactly the same argument (we could even use an automorphism of the Weyl algebra to formalize this) gives $[\partial^p, x] = 0$ and so $\partial^p$ is central.
So the center contains $k[x^p, \partial^p]$. We now show that this is exactly the center. Write an arbitrary element of the center as
$$f = \sum f_{ij} x^i \partial^j.$$
This is central iff $[f, x] = [f, \partial] = 0$ as we mentioned above. We were thinking of this condition in terms of the derivation $[f, -]$ but we can switch roles again and think of it as $[x, f] = [\partial, f] = 0$: this condition is equivalent to the derivations $[x, -]$ and $[\partial, -]$ both vanishing on $f$, which is a useful observation because it suggests that to do the computations we apply Leibniz some more. This gives
$$[x, f] = \sum f_{ij} x^i [x, \partial^j] = \sum f_{ij} x^i (-j \partial^{j-1})$$
and using that we know $\{ x^i \partial^j \}$ is a basis for the Weyl algebra we conclude that this is equal to zero iff each coefficient $-j f_{ij}$ is equal to zero iff for any $j$ either $p \mid j$ or $f_{ij} = 0$. Similarly
$$[\partial, f] = \sum f_{ij} [\partial, x^i] \partial^j = \sum f_{ij} (ix^{i-1}) \partial^j$$
we conclude that this is equal to zero iff each coefficient $i f_{ij}$ is equal to zero iff for any $i$ either $p \mid i$ or $f_{ij} = 0$. Hence if $f_{ij} \neq 0$ then $p | i, j$ which says exactly that $f \in k[x^p, \partial^p]$ as desired.
The idea to use $(x + y)^p = x^p + y^p$ is a sensible thing to try but unfortunately this identity only holds if $x$ and $y$ commute and so in the noncommutative world it's necessary to do something else.
Part b: $xv$ is shorthand for the action of $x$ on $v$ coming from the module structure. Very formally, a finite-dimensional representation $V$ of $A$ is a finite-dimensional vector space equipped with an algebra homomorphism $\rho : A \to \text{End}(V)$, and then $xv$ is shorthand for $\rho(x)(v)$.
The hint to use eigenvectors implies that we're working over an algebraically closed field $k$ so we'll assume this. Taking the hint, we'll let $v \in V$ be an eigenvector of $x$ (I have switched the variables, it doesn't matter though) and try to say something about the vectors $v, \partial v, \partial^2 v, \dots$. By definition we have
$$xv = \lambda v$$
which gives
$$x(\partial v) = (\partial x - 1)v = \partial (\lambda v) - v = \lambda (\partial v) - v.$$
What this says is that the image of $\partial v$ in the quotient $V / \text{span}(v)$ is an eigenvector of $x$ with the same eigenvalue $\lambda$ again. This is equivalent to a slightly more annoying statement about $x$ acting on the invariant subspace $\text{span}(v, \partial v)$ by the Jordan block $\left[ \begin{array}{cc} \lambda & -1 \\ 0 & \lambda \end{array} \right]$ but I prefer my reformulation which I think is a bit cleaner and more conceptual.
Note that it follows that $v$ and $\partial v$ are linearly independent. Otherwise one would be a multiple of the other, say $\partial v = cv$, and then we would have $x(\partial v) = xcv = \lambda cv = \lambda (\partial v)$ which would give $v = 0$.
Generally we have $[x, \partial^n] = -n \partial^{n-1}$ as explained previously which gives
$$x(\partial^n v) = (\partial^n xv - n \partial^{n-1})v = \lambda (\partial^n v) - n \partial^{n-1} v$$
which gives that the image of $\partial^n v$ in the quotient $V / \text{span}(v, \dots \partial^{n-1} v)$ is an eigenvector with the same eigenvalue $\lambda$ again, and we again have linear independence as long as $n \le p-1$ so that $n \partial^{n-1} \neq 0$. Once we hit $n = p$ the above becomes
$$x(\partial^p v) = \lambda (\partial^p v)$$
so we can no longer rule out the possibility that $\partial^p v$ is a scalar multiple of $v$, and in fact by Schur's lemma $\partial^p$ must act by a scalar $\lambda'^p$ (written this way because $\partial$ must have an eigenvalue $\lambda'$ and so $\partial^p$ must act by $\lambda'^p$). At this point we're done: $\text{span}(v, \dots \partial^{p-1} v)$ is invariant under the action of both $x$ and $\partial$, so by irreducibility it must be all of $V$, and since the vectors are linearly independent they must be a basis.
The scalar $\lambda'$ (equivalently, $\lambda'^p$, since the Frobenius map $x \mapsto x^p$ is invertible over an algebraically closed field of characteristic $p$) can be chosen freely; we can check explicitly that the result is an irreducible module $V_{\lambda, \lambda'}$ of the Weyl algebra by checking that the single relation $[\partial, x] = 1$ is satisfied, which we can check on the basis we just very conveniently computed.
So we've shown that the irreducible representations of the Weyl algebra are $p$-dimensional and parameterized by their central character (the homomorphism $k[x^p, \partial^p] \to k$ from the center to $k$ describing the scalars by which every element acts, by Schur's lemma): every possible central character occurs, and there's a unique irrep $V_{\lambda, \lambda'}$ with a given central character. (This reflects that in characteristic $p$ the Weyl algebra is an Azumaya algebra over its center.)
Note that we could have anticipated that the dimensions of the irreps needed to at least be divisible by $p$. There's a famous argument that shows that the Weyl algebra has no nonzero finite-dimensional representations in characteristic $0$, because if $XD - DX = 1$ where $X, D$ are two $n \times n$ matrices then taking traces of both sides gives $\text{tr}(XD - DX) = 0 = \text{tr}(1) = n$. So in characteristic $0$ this gives $n = 0$ and in characteristic $p$ this gives $p \mid n$. The above argument involving eigenvectors gives a different proof that there are no finite-dimensional representations in characteristic $0$ because it exhibits an arbitrarily long sequence of linearly independent vectors (which then give an infinite-dimensional representation). | 110,156 |
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I got invited by the charming Simone to visit her parish for Stations and a parish Lenten dinner last night, 27 February 2004. Only when I got there did I truly recall how different my Catholic life has been these past few years. I’m registered at Calvert House, which has a pretty fixed clientele. A little under half the people are undergraduates, maybe half graduate students, with a few young professors, babies, professionals usually associated with the University, and the occasional older person mixed in. By older person, here I mean someone over 30. Maybe 35. Then I enter the Stations at St Gregory on the North Side, and things are so radically different. Children between 2 and 17! Families! People older than me! People much older than me! Very very different. Perhaps some other time I’ll talk about the gender gap of American Christianity, or the age composition, or that the parish lists the preferred donation for marriages and funerals on their Web site. This is a personal night.
St Gregory’s was a foreign territory. I’ve never had a real parish as an adult. Between Harvard and Chicago, I attended church routinely at St Andrew the Apostle and St Timothy in the Arlington Diocese, but you could hardly call me a community member. I think I attended one function in four years, and less than a dozen Masses outside Sundays and holy days. The parishes contributed to my minimalism; large poorly designed spaces, bad to nonexistant greeters, defensive focus, little quality liturgy, and inability to plan anything appealing to a single adult were all quite present. I do not absolve myself, though; work devotion, travel, introversion, and an intellectual focus all pulled me to the books and readings and away from activity.
Thus, foreign it was to sit in a school hall and have soup and bread. I still don’t know how to talk to families. Not really knowing anyone didn’t help much, though people were very polite. Several people asked me if I worked at the Medieval Institute. (Simone pursues a doctorate in Medieval Studies.) That made me smile, given that I’m currently trying to program something very far from the time of Aquinas.
The most interesting part was trying to explain just exactly what I do, particularly to the nice lady Nancy. When she went through school, statistics didn’t exist as a discipline. This is a small exaggeration, since departments of statistics did exist in the 1950s. None were large, though, and would not have been mentioned in a secondary school curriculum. Standard deviation is a pretty big step to even my parents’ generation, even at the college level.
By now, I know enough not to give any mathematical words. Instead, I say I have a seven year old’s dream job. Seven was the age of Tinkertoys and dollhouses and sandcastles, worlds from things. They’re toys, sure, but seven is when a young boy or girl creates. Then the discipline of school takes that away, off to a segmented world of rote and memorization, and most people never go back. Except for me. I still build models, now with numbers instead of Lego blocks or dolls, but I get to create, make order, staunch chaos, get lost in a world not quite real.
That explanation appeals to people – though not Nancy, for which I went with the practical result approach. Typically, though, if I look closely, I can see the eyes of my conversant; they flash briefly, back to that general joy. Just for a second, there’s a wish, a wistfulness, that instead of packing vegetable bins or writing reports, they could do that. They often say that I’m very lucky to do something so fun and I love, and that part, at least as much as I get to here, is true. It’s nice to be reminded of the difference in perception between my Gothic tower and the high rises of nonacademic life. | 274,702 |
TITLE: If $p\equiv 3\pmod{4}$ and $p\mid x^2+y^2$, prove $p\mid x,y$.
QUESTION [2 upvotes]: I have to prove that if $p$ is a prime number of the form $p = 4n - 1$, $n\in N$ and $x^2+y^2\equiv 0\pmod{p}$, then $x\equiv 0\pmod{p}$ and $y\equiv 0\pmod{p}$.
I have gone about this as follows and I hope you will correct me if I made any errors:
First of $x^2 + y^2$ is divisible by p so we can write it as being equal to some multiple $k$ of $p$ $+$ a remainder of $0$.
Then let's assume the opposite that neither $x$ or $y$ are congruent to $p$ ergo they aren't divisible by $p$ so we can represent them as being equal to some multiples $l$ and $m$ of $p$ (respectively) and with remainders (nonzero of course) $a$ and $b$ (respectively).
So we get:
$$ x^2 + y^2 = kp + 0$$
$$ x = lp + a $$
$$ y = mp + b$$
If we take the second and third expression and square them we get:
$$ x^2 = l^2p^2 + 2lpa + a^2 $$
$$ y^2 = m^2p^2 + 2mpb + b^2$$
Then we add them up:
$$ x^2 + y^2 = l^2p^2 + 2lpa + a^2 + m^2p^2 + 2mpb + b^2$$
Factorize:
$$ x^2 + y^2 = \left(l^2p + 2la + m^2p + 2mb\right)p + a^2 + b^2$$
We see that $k = l^2p + 2la + m^2p + 2mb$, which gives us:
$$ x^2 + y^2 = kp + a^2 + b^2$$
At the beginning it is stated that the remainder $a$ is equal to zero since $x^2 + y^2$ is divisible by $p$ so get to the conclusion that the sum of two nonzero, positive numbers, more precisely the sum of two squares $a^2 + b^2$ has to be $0$. This is a contradiction.
The contradiction was brought on by the assumption that $x$ and $y$ aren't divisible by $p$ therefore we conclude that $x$ and $y$ must be divisible by $p$.
REPLY [2 votes]: I.e. you want to prove:
$p\equiv 3\pmod{4},\ p\mid x^2+y^2\implies p\mid x,y$.
If $p\mid x$, then $p\mid y$, and vice versa.
So for contradiction assume $\gcd(p,x)=\gcd(p,y)=1$. See Modular Inverse.
$x^2\equiv -y^2\pmod{p}\iff (xy^{-1})^2\equiv -1\pmod{p}$
This can give a contradiction in several ways:
$1)\ $ Just remember Quadratic Reciprocity.
$2)\ $ Raise both sides by $\frac{p-1}{2}$ (which is odd):
$\implies \left(xy^{-1}\right)^{p-1}\equiv (-1)^{\frac{p-1}{2}}\equiv -1\pmod{p}$,
which contradicts Fermat's Little Theorem.
$3)\ $ Square both sides: $\left(xy^{-1}\right)^4\equiv 1\pmod{p}$.
Then $\text{ord}_p\left(xy^{-1}\right)\mid 4$. But it cannot be $1$ or $2$, because $\left(xy^{-1}\right)^2\equiv -1\pmod{p}$, therefore $\text{ord}_p\left(xy^{-1}\right)=4$, so $4\mid p-1$ by Fermat's Little Theorem (see below).
Theorem: If $\text{ord}_m(a)=d$ and $a^k\equiv 1\pmod{m}$, then $d\mid k$.
Proof: For contradiction, assume $k=dl+r$ for some $l,r\in\Bbb Z^+,\, 0<r<d$.
$a^{k}\equiv \left(a^d\right)^la^r\equiv 1^la^r\equiv a^r\equiv 1\pmod{m}$, contradiction. | 132,554 |
TITLE: Combinatorics riddle: keys and a safe
QUESTION [3 upvotes]: There are 8 crew members, The leading member wants that only a crew of 5 people or more could open a safe he bought, To each member he gave equal amount of keys, and locked the safe with several locks.
What's the minimal number of locks he should put:
at least 5 crew members would be required to open the safe?
Any team of 5 crew members could open the safe, but none of the 4 team crew members.
1,2 stands individually of course.
REPLY [6 votes]: Consider the general case that $n$ people know certain secrets such that no subset of $k$ of them know all secrets, but any subset of $k+1$ of them do know all secrets. (Here $n=8$, $k=4$, secrets are locks, and knowing a secret is having a key.)
This can be done using $\binom nk$ secrets, each labelled by a different subset of $k$ among the $n$ people, which is the set of people that does not get to know that secret (everyone else does). Then for any subset of $k$ people the corresponding secret will not be known by them, but for any set of $k+1$ people and any secret there is at least one of them who knowns the secret.
To show one cannot do with less secrets, consider any solution to this problem, and the relation between on one hand the $\binom nk$ subsets of $k$ people and on the other hand the secrets, defined by none of those people knowing the secret. This relation is "one to at least one": it is required that any set of $k$ people not know at least one secret, and if two different subsets of $k$ people do not know the same secret, then the union of those subsets, which contains at least $k+1$ people would not know that secret, and hence not all secrets, which is against the requirement. The relation is in fact a surjective map from secrets to $k$-subsets, so there are al least $\binom nk$ secrets. | 43,015 |
\begin{document}
\maketitle
\begin{abstract}
Sampling of sharp posteriors in high dimensions is a challenging problem, especially when gradients of the likelihood are unavailable. In low to moderate dimensions, affine-invariant methods, a class of ensemble-based gradient-free methods, have found success in sampling concentrated posteriors. However, the number of ensemble members must exceed the dimension of the unknown state in order for the correct distribution to be targeted. Conversely, the preconditioned Crank-Nicolson (pCN) algorithm succeeds at sampling in high dimensions, but samples become highly correlated when the posterior differs significantly from the prior. In this article we combine the above methods in two different ways as an attempt to find a compromise. The first method involves inflating the proposal covariance in pCN with that of the current ensemble, whilst the second performs approximately affine-invariant steps on a continually adapting low-dimensional subspace, while using pCN on its orthogonal complement.
\end{abstract}
\begin{keywords}
Markov chain Monte Carlo, ensemble sampling, Bayesian inference, dimension-robust, affine invariance, gradient-free
\end{keywords}
\begin{AMS}
65N21,
62F15,
65C05,
65N75,
90C56
\end{AMS}
\section{Introduction}
Over the last decade, solving Bayesian inverse problems with high-dimensional parameters has become increasingly feasible due to growing computational resources and the development of methods that scale effectively with the dimension of the state space. To characterize the solution of a Bayesian inference problem, which is the posterior distribution of the parameters, one typically relies on Markov chain Monte Carlo (MCMC) sampling methods.
Abstractly, one is interested in sampling a probability measure $\mu$ that is absolutely continuous with respect to a simpler measure $\mu_0$,
\[
\mu(\dee u) \propto \exp(-\Phi(u))\,\mu_0(\dee u),
\]
where $\Phi(\cdot)$ is a the negative log likelihood, a typically costly-to-evaluate function. Ensemble sampling methods use a set of particles to estimate properties of $\mu$ that can inform the proposal step in MCMC. A particular class of ensemble methods are affine-invariant ensemble methods, whose behavior is invariant under affine transformations.
The nature of affine-invariant sampling methods means that the number of particles required must exceed the dimension of the state $u$, or else they will only sample the distribution restricted to the span of the initial ensemble. \Cref{fig:3d_example} illustrates this effect for a toy example of sampling a standard normal distribution in three dimensions using three ensemble members: the initial ensemble state defines a plane, and the MCMC chains are unable to leave this plane. Additionally, even if a sufficient number of particles are used, the correct distribution may not be targeted; the paper \cite{huijser2015properties} investigates this in the case that the target distribution is a high-dimensional Gaussian. Generally, the performance of ensemble samplers is known to degrade in higher dimensions.
Dimension-robust sampling methods are methods whose performance does not degrade for increasing dimension. However, they are are known to converge slowly when $\mu$ differs substantially from the reference measure $\mu_0$. The aim of this paper is to develop a hybrid version of these two opposite-end sampling approaches, i.e., benefit from the convergence properties of ensemble samplers for concentrated distributions $\mu$ while avoiding degeneration of sampling performance in high dimensions.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{3d_example}\vspace{-3ex}
\caption{An example of the affine-invariant and hybrid samplers for the distribution $N(0,I)$ in $\R^3$, with 5000 samples shown for each sampler and an ensemble consisting of three particles. The planes indicate the span of the initial ensemble of particles; the affine-invariant sampler cannot leave this plane while the particles of the hybrid samplers are not restricted to the plane.}
\label{fig:3d_example}
\end{figure}
\subsection{Related work}
Sampling of high-dimensional posterior distributions via MCMC has received much attention in the past decade, utilizing the formulation of the Metropolis-Hastings algorithm on general state spaces introduced in \cite{tierney1994markov}. The preconditioned Crank-Nicolson (pCN) method \cite{cotter2013mcmc} is a simple gradient-free sampler, in the case of Gaussian priors, with the property that its convergence rate is bounded independently of the dimension of the state space \cite{hairer2014spectral}. Modifications of pCN are considered in \cite{pinski2015algorithms, rudolf2018generalization} wherein proposals may be more adapted to the posterior distribution, typically using derivative information of the likelihood. The paper \cite{vollmer2015dimension} provides a framework for constructing analogous samplers when the prior is non-Gaussian. Other samplers with dimension-independent convergence properties, utilizing derivative information, include $\infty$-MALA, $\infty$-HMC and their manifold variants \cite{BeskosGirolamiLanEtAl17} and DILI \cite{CuiLawMarzouk16}. The papers \cite{KimVillaParnoEtAl21, BeskosGirolamiLanEtAl17} provide a systematic comparison of a number of the above algorithms applied to high-dimensional Bayesian inverse problems. Outside of MCMC, \cite{agapiou2017importance} considers the performance of importance sampling on general state spaces, and its dependence on the discretization dimension and effective dimension of the problem. Variations of the ensemble Kalman Filter (EnKF) \cite{evensen2003ensemble} have also been considered in the context of Bayesian inversion on general state spaces \cite{garbuno2020interacting,pavliotis2022derivative}, allowing for derivative-free exploration of the posterior via approximate natural Langevin dynamics \cite{garbuno2020interacting}.
When the dimension of the state space is finite and relatively low, affine-invariant ensemble samplers (AIES) \cite{GoodmanWeare10,foreman2013emcee} can allow for efficient derivative-free exploration of complicated posterior distributions; in particular those that are highly concentrated due to particularly informative observations. The continuous time limit of one such algorithm has been studied \cite{Garbuno-InigoNuskenReich20}, resulting in certain Langevin dynamics. The dependence of affine-invariant samplers on dimension has also been studied \cite{huijser2015properties}, showing certain degeneration as the dimension increases. To help overcome the dimensional limitations of affine-invariant samplers, \cite{coullon2021ensemble} introduces a sampler that uses AIES on a subspace defined using the prior distribution, and pCN on its complement. In this article we take a similar approach wherein the subspace is not as strongly constrained by the prior, which can be more appropriate in the setting of concentrated posteriors.
\subsection{Contributions and limitations}
Our main contributions are as follows:
(1) We propose two gradient-free ensemble sampling algorithms that are well-defined in infinite dimensions. In these methods, the
subspace spanned by the ensemble is not fixed, and thus there is no minimally required ensemble size.
(2) We numerically study the new methods' performance for different ensemble size, and compare their performance to existing methods for linear, nonlinear and non-smooth infinite-dimensional Bayesian inverse problems.
The proposed methods also have limitations:
(1) Since the ensemble is used to compute a Gaussian proposal distribution in the subspace spanned by the particles, the method loses efficiency for strongly non-Gaussian densities.
(2) Our algorithms require some parameter choices, e.g., the ensemble size, a jump parameter in MCMC, and the dimension of a subspace in one of the methods. However, we will show numerically that the algorithms' performance is rather insensitive to these choices.
\section{Bayesian inverse problems}
In this section we provide an overview of the Bayesian approach to inverse problems, combining the observation model with the prior measure to construct the posterior measure on general state spaces. We then discuss the problem of producing samples from the posterior numerically and various issues that may arise.
\subsection{The prior, likelihood and posterior}
Suppose that we have data $y \in Y$ arising from some nonlinear noisy observations of a state $u \in X$, and our goal is to estimate $u$ from $y$. We write
\[
y = F(\G(u),\eta)
\]
for some forward map $\G:X\to Y_1$, random noise $\eta \in Y_2$ and state-to-observation map $F:Y_1\times Y_2\to Y$. A common setup is that of additive Gaussian noise: $Y_1 = Y_2 = Y$, $F(z,\eta) = z+\eta$ and $\eta \sim N(0,\Gamma)$, so that
\[
y = \G(u) + \eta,\quad \eta\sim N(0,\Gamma).
\]
Such problems are typically ill-posed from a classical perspective: there may exist no solution, the solution may not be unique, or the solution may be highly sensitive to the realization of the noise $\eta$. In this article, we consider the underdetermined case wherein that $X$ is high- or infinite-dimensional Hilbert space and $Y = \R^J$ is finite-dimensional.
We consider the Bayesian approach to the inversion wherein rather than a single state $u \in X$ as a solution, we seek a probability distribution $\mu$ on $X$. If we quantify our prior beliefs about unknown $u$ by a measure $\mu_0(\dee u) = \P(\dee u)$ on $X$ and provide a probability distribution for the noise $\eta$, this induces a likelihood function $\P(y|u)$. For example in case of additive Gaussian noise above, the likelihood is given by
\[
\P(y|u) \propto \exp\left(-\frac{1}{2}\|\cG(u)-y\|_\Gamma^2\right),
\]
where $\|\cdot\|_\Gamma = \|\Gamma^{-\frac{1}{2}}\cdot\|$. We define the solution to the Bayesian inverse problem as the measure $\mu(\dee u) = \P(\dee u|y)$, where by Bayes' theorem
\[
\P(\dee u|y) = \frac{\P(y|u)\P(\dee u)}{\P(y)}.
\]
To be more explicit, suppose that the likelihood takes the form
\[
\P(y|u) \propto \exp(-\Phi(u;y)),
\]
where $\Phi$, referred to as the negative log-likelihood, is sufficiently regular \cite{DashtiStuart16}. Then the Bayesian posterior $\mu$ is absolutely continuous with respect to the prior $\mu_0$, and its Radon-Nikodym derivative takes the form\footnote{In what follows we drop the dependence of $\Phi$ on the data $y$ as we assume it fixed.}
\begin{align}
\label{eq:bayes}
\frac{\dee \mu}{\dee \mu_0}(u) = \frac{1}{Z}\exp(-\Phi(u)),\quad Z = \int_X \exp(-\Phi(u))\,\mu_0(\dee u).
\end{align}
Though our motivation is Bayesian inversion, the methodology introduced in this article may be used to sample general measures with the form \cref{eq:bayes}, for example Gibbs measures. We will still however refer to $\Phi$ as the negative log-likelihood and $\mu_0$ as the prior.
In this article, we focus on the case where the prior measure $\mu_0 = N(0,C_0)$ is a centered Gaussian. In this case, when the dimension of $X$ is finite, the posterior admits a Lebesgue density $\mu(\dee u) = \pi(u)\,\dee u$,
\[
\pi(u) \propto \exp\left(-\Phi(u) - \frac{1}{2}\|u\|_{C_0}^2\right),
\]
which will be useful to consider to provide a formal interpretation of the infinite-dimensional algorithms introduced in the following section.
\begin{remark}
\label{rem:noncenter}
The assumption that the prior measure $\mu_0$ is a centred Gaussian is not as strong as it first appears. For example, suppose that the prior takes the form
\[
\mu_0(\dee u) = \frac{1}{Z_0}\left(-\Phi_0(u)\right)\,\nu_0(\dee u),\quad Z_0 = \int_X \exp(-\Phi_0(u))\,\nu_0(\dee u),
\]
where $\nu_0$ is the pushforward of a Gaussian measure $\nu_r$ via a possibly non-linear map $T:\Xi\to X$, i.e., $\nu_0 = T^\sharp \nu_r$. Then $\mu = T^\sharp \nu$, where
\[
\frac{\dee \nu}{\dee \nu_r}(\xi) = \frac{1}{Z_r}\exp\left(-\Psi(\xi)\right),\quad Z_r = \int_\Xi \exp(-\Psi(\xi))\,\nu_r(\dee \xi)
\]
and $\Psi(\xi) = \Phi(T(\xi)) + \Phi_0(T(\xi))$. The measure $\nu$ is precisely of the form \cref{eq:bayes} with a Gaussian dominating measure and so if we can sample $\nu$, we can sample $\mu$ by transforming the samples with $T$. The simple case $T(\xi) = \xi+m_0$ and $\Phi_0 = 0$ illustrates why it is sufficient to assume the dominating Gaussian is centred, for example. In the algorithms we consider, it can be useful to work with a white noise measure $\nu_r = N(0,I)$, $\Phi_0 = 0$ and define $T(\xi) = C_0^{{1}/{2}}\xi$ to map to the prior Gaussian measure $\mu_0 = \nu_0$.
\end{remark}
\subsection{Probing the posterior}
Though the solution $\mu$ exists abstractly as a measure under relatively mild assumptions on the prior and negative log-likelihood, one is often interested in getting information from this measure numerically. For example, one may desire the mode, the mean or estimates and confidence bounds on quantities of interest. The latter typically require samples from the posterior to estimate via Monte Carlo, since the integrals involved are often high-dimensional. Producing these samples can be challenging in many setups, for example,
\begin{enumerate}[(i)]
\item when the data is particularly informative, the posterior distribution can be concentrated on a lower dimensional submanifold of $X$ which needs to be discovered;
\item effective sampling methods often make use of derivatives of the posterior density, but these may not exist, may be unknown, or may be computationally prohibitive to evaluate;
\item the posterior may have multiple distinct modes, which many sampling methods may struggle to explore; and
\item when the dimension of the space $X$ is infinite, the posterior cannot admit a Lebesgue density $\pi$, however many sampling algorithms are defined in terms of a Lebesgue density. Similarly in high but finite dimensions, the posterior is often almost singular with respect to the Lebesgue measure, leading to statistical issues with said algorithms.
\end{enumerate}
In the remainder of this article we consider certain Markov chain Monte Carlo (MCMC) methods with the aim of partly resolving the above points.
\section{MCMC sampling for Bayesian inverse problems}
\label{sec:mcmc}
In this section we first give an overview of Metropolis-Hastings MCMC algorithms, outline two classes of such algorithms (affine-invariant and dimension-robust) and describe some of their respective advantages and disadvantages. We then introduce two hybrid methods that interpolate between the two classes as an approach to ameliorating some of their disadvantages.
\subsection{Metropolis-Hastings MCMC sampling}
\label{ssec:metropolis}
MCMC methods aim to sample a given probability distribution by constructing a Markov chain for which it is the stationary distribution. A common construction of such a chain is via a Metropolis-Hastings propose-accept-reject mechanism. Given a target probability distribution $\pi$ and a state $u_k$, a new state $\hat{u}_{k}$ is proposed according to a proposal distribution $\hat{u}_{k} \sim q(u_k,u)\,\dee u$. One then sets $u_{k+1} = \hat{u}_{k}$ with probability
\[
\min\left\{1,\frac{\pi(\hat{u}_k)q(u_k,\hat{u}_k))}{\pi(u_k)q(\hat{u}_k,u_k))}\right\},
\]
or else sets $u_{k+1} = u_k$. This choice of acceptance probability ensures that the resulting Markov chain satisfies detailed balance and hence has the desired stationary distribution. A simple choice of proposal distribution is a symmetric random walk proposal, $q(u_k,\cdot) = N(u_k,C)$ for some jump covariance $C$, in which case the algorithm is referred to as Random Walk Metropolis (RWM). However, depending on the structure of the target measure $\pi$, a more complex proposal distribution is typically more efficient computationally. Once the Markov chain has reached stationarity (after a period referred to as burn-in), the samples $\{u_k\}$ may be used to approximate quantities of interest, such as mean, variance, or marginal probability distributions. The samples $\{u_k\}$ are typically correlated -- ideally one wishes to produce a Markov chain whose samples are as least correlated as possible in order to estimate these quantities of interest efficiently.
\subsection{Affine-invariant MCMC sampling}
\label{ssec:affine}
Instead of targeting the posterior density $\pi$ on $X$ directly, an ensemble of particles is used to target the product measure
\begin{align}
\label{eq:pistar}
\pi^*(\dee u^{(1)},\ldots,\dee u^{(N)}) = \prod_{j=1}^N \pi(\dee u^{(j)})
\end{align}
on $X^N$. This immediately provides two advantages over a single chain targeting $\pi$:
\begin{enumerate}[(i)]
\item if the posterior is multimodal, different particles can explore distinct modes without the need to move between them; and
\item the empirical distribution of the ensemble at a given step provides a coarse estimate for the posterior distribution, which can be used to adapt the proposal distribution.
\end{enumerate}
A class of ensemble methods, called affine-invariant methods, were introduced in \cite{GoodmanWeare10}. Suppose that the MCMC update for a particular particle takes the form
\[
u^{(j+1)} = R(u^{(j)},\xi^{(n)}),
\]
where $\xi^{(n)}$ is a random variable. We say that the update is affine-invariant if for any $A \in \mathcal{L}(X,X)$ and $b \in X$,
\[
R(Au+b,\xi) = AR(u,\xi) + b.
\]
If a method has this property then as a consequence, distributions which are concentrated around a hyperplane are as easy to sample as those which are more dispersed; see \cite{GoodmanWeare10} for more details.
We provide an overview of an example of an affine-invariant method introduced in \cite{GoodmanWeare10}, referred to as the \emph{walk move}. Given a particle $u^{(j)}$ we denote $u^{(-j)}$ the complementary ensemble
\[
u^{(-j)} = \{u^{(1)},\ldots,u^{(j-1)},u^{(j+1)},\ldots,u^{(N)}\}
\]
Then given a subcollection $S\subseteq u^{(-j)}$, after the chain has reached stationarity, the sample covariance of $S$ should provide an approximation to posterior covariance. Thus, one can perform RWM updates where the proposal covariance is proportional to this sample covariance. The algorithm is given explicitly in \cref{alg:affine}, and referred to as the Affine Invariant Ensemble Sampler (AIES).
\begin{algorithm}
\begin{algorithmic}
\caption{Affine-invariant MCMC sampling}
\label{alg:affine}
\State Choose initial ensemble of particles $\{u_1^{(n)}\}_{n=1}^N \subseteq X$ and jump parameter $\lambda > 0$.
\For{$k=1:K$}
\For{$n=1:N$}
\State Choose $S \subseteq u_k^{(-n)}$ and propose
\[
\hat{u}_k^{(n)} = u_k^{(n)} + \lambda\cdot\frac{1}{\sqrt{|S|}}\sum_{u_k^{(j)} \in S} z_j(u_k^{(j)} - \overline{u}_S),\quad z_j \iid N(0,1).
\]
\State Set $u_{k+1}^{(n)} = \hat{u}_k^{(n)}$ with probability
\[
\min\left\{1,\exp\left(\Phi(u_k^{(n)}) - \Phi(\hat{u}_k^{(n)}) + \frac{1}{2}\|u_k^{(n)}\|_{C_0}^2 - \frac{1}{2}\|\hat{u}_k^{(n)}\|_{C_0}^2\right)\right\}
\]
\State or else set $u_{k+1}^{(n)} = u_k^{(n)}$.
\EndFor
\EndFor
\State\Return $\{u_k^{(n)}\}_{k,n=1}^{K,N}$.
\end{algorithmic}
\end{algorithm}
Other affine invariant proposals are available, such as the stretch move \cite{GoodmanWeare10}. However, these proposals often have a strong dimensional dependence, and in particular are not well-defined in infinite dimensions. Recently the ALDI method has been introduced \cite{Garbuno-InigoNuskenReich20}, which involves simulating an appropriate affine-invariant Langevin diffusion targeting $\pi^*$ with an Euler-Maruyama scheme; this is a modification of the Ensemble Kalman Sampler \cite{garbuno2020interacting} such that the correct distribution is targeted.
A drawback of the above methods is that in order for them to sample the correct distribution, the number of particles $N$ must be larger than the dimension of the state space $X$. For example, when using \cref{alg:affine} the particles cannot move out of the lowest dimension hyperplane passing through the initial ensemble; see \cref{fig:3d_example} for a simple illustration in three dimensions. When the dimension of the state space is high or infinite this requirement can make the algorithm impractical.
\subsection{Dimension-robust MCMC sampling}
\label{ssec:robustmcmc}
As we are interested in the case when $X$ is high- or infinite-dimensional, we ideally desire a sampling method that is well-defined in infinite dimensions to bypass dimension-dependent issues. Such methods have received much attention recently, though the general Metropolis-Hastings algorithm was formulated on Hilbert space in 1994 \cite{tierney1994markov}. Key to the construction of these algorithms is that the posterior is absolutely continuous with respect to a dominating measure -- in our setup we assume this to be Gaussian, rather than the Lebesgue measure as is typically the case in finite dimensions. The notion of dimension-robustness informally refers to the algorithm being well-defined and ergodic on Hilbert space, and more rigorously defined as the geometric rate of convergence to stationarity with respect to some metric on measures being bounded below by some positive constant independently of dimension.
An example of a dimension-robust MCMC method, assuming a Gaussian prior, is the preconditioned Crank-Nicolson (pCN) method \cite{cotter2013mcmc}. This is a modification of the random walk Metropolis algorithm such that for the proposal, the current state is rescaled and then perturbed by a Gaussian random variable with covariance proportional to the prior covariance:
\[
\hat{u}_k = \sqrt{1-\beta^2}u_k + \beta\xi,\quad \xi\sim N(0,C_0),
\]
for some $\beta \in (0,1]$. The acceptance probability is then simply a likelihood ratio -- the prior information is fully contained in the proposal. When a more general prior mean is $m_0 \in X$ is assumed, the proposal
\[
\hat{u}_k = m_0 + \sqrt{1-\beta^2}(u_k-m_0) + \beta\xi,\quad \xi\sim N(0,C_0),
\]
is instead used, with the same acceptance probability. Whilst this algorithm works in arbitrarily high dimensions when the posterior is absolutely continuous with respect to the prior, its performance in terms of mixing can be poor when the posterior is far from the prior, i.e., when the data is particularly informative and the likelihood is very skewed: in order to maintain a reasonable acceptance rate, the parameter $\beta$ must be chosen extremely small, and so samples are highly correlated. One approach is to, instead of using jumps based on the prior, use jumps from some other Gaussian distribution that has been informed by the likelihood. As long as the jump distribution is equivalent to the prior, the modification to the acceptance probability to ensure detailed balance holds is well-defined in infinite dimensions. Specifically, suppose that the jump distribution is taken to be $N(m,C)$, then the proposal distribution is given by
\begin{align*}
Q(u,dv) &= N\left(m + \sqrt{1-\beta^2}(u-m),\beta^2 C\right).
\end{align*}
Defining the measures $\omega$, $\omega^\top$ on the product space $X\times X$ by
\[
\omega(\dee u,\dee v) = \mu(\dee u)Q(u,\dee v),\quad \omega^\top(\dee u,\dee v) = \omega(\dee v,\dee u),
\]
following \cite{tierney1994markov} the acceptance probability is then given by
\[
\alpha(u,\hat{u}) = \min\left\{1,\frac{\dee \omega}{\dee \omega^\top}(\hat{u},u)\right\},
\]
which is well-defined by the absolute continuity $\mu\ll\mu_0$ and assumed equivalence of the prior and jump distributions. The algorithm, referred to as generalized pCN (gpCN), is given in \cref{alg:gpcn} after calculating this Radon-Nikodym derivative\footnote{A related algorithm, introduced in \cite{rudolf2018generalization}, is also referred to as gpCN. This algorithm also modifies the jump distribution of standard pCN away from the prior, except its mean is modified in such a way that the expression for the acceptance probability remains the same.}. To the authors' knowledge, this algorithm first appeared in \cite{pinski2015algorithms}. If derivatives of the likelihood are available, a typical example of jump distribution one can use is the Laplace approximation to the posterior or some approximation thereof \cite{bui2013computational,pinski2015algorithms}. In the following subsection we consider a different class of jump distributions which do not necessarily require derivatives. The case $m=0$ and $C = C_0$ provides the original pCN algorithm, in which case $I_C \equiv 0$. The parameters $\beta$, $m$ and $C$ may be chosen to depend on the time step $k$, for example if one were to use an adaptive variant of the above; however conditions on the dependence on $k$ are required in order to preserve ergodicity.
\begin{algorithm}
\begin{algorithmic}
\caption{Generalized pCN MCMC sampling}
\label{alg:gpcn}
\State Choose initial state $u_1 \in X$.
\For{$k=1:K$}
\State Propose
\[
\hat{u}_k = m + \sqrt{1-\beta^2}(u_k-m) + \beta \xi,\quad \xi \sim N(0,C)
\]
\State Set $u_{k+1} = \hat{u}_k$ with probability
\[
\min\left\{1,\exp\left(\Phi(u_k) - \Phi(\hat{u}_k) + I_C(u_k) - I_C(\hat{u}_k)\right)\right\}
\]
\State where
\begin{align}
\label{eq:IC}
I_C(u) = \frac{1}{2}\Big\langle u,\big(I-C_0^{\frac{1}{2}}C^{-1}C_0^{\frac{1}{2}}\big)u\Big\rangle_{C_0} - \langle u,m\rangle_C + \frac{1}{2}\|m\|_C^2,
\end{align}
\State or else set $u_{k+1} = u_k$.
\EndFor
\State\Return $\{u_k\}_{k=1}^K$.
\end{algorithmic}
\end{algorithm}
Note that in finite dimensions, $I_C(u)$ can be defined more directly as
\[
I_C(u) = \frac{1}{2}\|u\|_{C_0}^2 - \frac{1}{2}\|u-m\|_C^2.
\]
In infinite dimensions, however, each of these terms is infinite almost surely. In the definition of $I_C$ in \cref{eq:IC} each term is finite, see \cref{appendix}.
\subsection{A hybrid algorithm}
\label{ssec:hybrid}
The affine-invariant MCMC has the advantage of adapting well to skewed posterior distributions. However, it targets the incorrect distribution if there are fewer particles than dimensions, making it impractical for high-dimensional problems. Conversely the gpCN algorithm is well-defined in infinite dimensions, but performs poorly if the posterior is far from the prior (or chosen jump distribution). We balance these issues by interpolating between the two algorithms.
We return to the setup of the affine-invariant MCMC algorithm, and target the product measure $\pi^*$ on $X^N$ given by \cref{eq:pistar}. Let $\R^D$ represent a discretization of $X$. Given a set of particles $S$, of size $|S|$, we define the normalized centred data matrix $V_S \in \R^{D\times |S|}$ by
\[
(V_S)_{:, j} := \frac{1}{\sqrt{|S|-1}}\left(u^{(j)} - \overline{u}_S\right),
\]
where $\overline{u}_S = \frac{1}{|S|}\sum_{u^{(j)} \in S} u^{(j)}$ is the sample mean of the particles in $S$.
The matrix $V_SV_S^\top$ then provides the sample covariance of the particles $S$.
In the inner loop of the affine-invariant MCMC algorithm, we use gpCN with the choice of jump distribution $N(m,C)$ with $C = C_* + \gamma^2V_SV_S^\top$ for some $\gamma > 0$, where the measures $N(0,C_0)$ and $N(0,C_*)$ are assumed equivalent. This jump distribution possesses the requisite prior equivalence due to the following simple proposition.
\begin{proposition}
\label{prop:hybrid_equiv}
Let $C_* \in \mathcal{L}(X;X)$ be positive and trace-class, and let $W \in \mathcal{L}(X;X)$ have finite rank. Then the measures $N(0,C_*)$ and $N(0,C_*+W)$ are equivalent.
\end{proposition}
\begin{proof}
By the Feldman-Hajek theorem \cite{da2002second}, it is sufficient to show that the operator $C_*^{-\frac{1}{2}}(C_*+W)C_*^{-\frac{1}{2}}-I = C_*^{-\frac{1}{2}}WC_*^{-\frac{1}{2}}$ is Hilbert-Schmidt; this follows immediately since $W$ has finite rank.
\end{proof}
The sample covariance $V_SV_S^\top$ has rank at most $|S|-1$; indeed its range is the lowest dimensional hyperplane passing through the elements of $S$, shifted to intersect the origin. By the assumed equivalence of $N(0,C_0)$ and $N(0,C_*)$ the equivalence of the prior and jump distributions follows.
Given a gpCN jump parameter $\beta \in (0,1]$ and RWM jump parameter $\lambda > 0$, we make the choice $\gamma = \lambda/\beta$ so that the proposal distribution is given by
\[
Q(u_k^{(j)},\cdot) = N\left(m + \sqrt{1-\beta^2}(u_k^{(j)}-m), \beta^2\left(C_* + \frac{\lambda^2}{\beta^2}V_SV_S^\top\right)\right).
\]
Hence, as $\beta \to 0$ we recover the affine-invariant proposal as in \cref{alg:affine}, and as $\lambda\to 0$ we recover the gpCN algorithm for each particle . Note that, for $\beta > 0$, the proposals are not restricted to a hyperplane dictated by the initial ensemble. The full algorithm is given in \cref{alg:hybrid}. We refer to this algorithm as \hybrid (\hybridname).
\begin{algorithm}
\begin{algorithmic}
\caption{\hybrid MCMC sampling}
\label{alg:hybrid}
\State Choose initial ensemble of particles $\{u_1^{(n)}\}_{n=1}^N \subseteq X$ and jump parameters $\beta \in (0,1]$, $\lambda > 0$. Set $\gamma = \lambda/\beta$.
\For{$k=1:K$}
\For{$n=1:N$}
\State Choose $S \subseteq u_k^{(-n)}$ and propose
\begin{align*}
\hat{u}_k^{(n)} &= m + \sqrt{1-\beta^2}(u_k^{(n)}-m) + \beta \xi + \lambda\sum_{j \in S} (V_S)_{:, j}z_j,\\
&\quad z_j\iid N(0,1),\quad\xi \sim N(0,C_*).
\end{align*}
\State Set $u^{(n)}_{k+1} = \hat{u}_k^{(n)}$ with probability
\[
\min\left\{1,\exp\left(\Phi(u_k^{(n)}) - \Phi(\hat{u}_k^{(n)}) + I_{C}(u_k^{(n)}) - I_{C}(\hat{u}_k^{(n)})\right)\right\}
\]
\State where $C = C_*+\gamma^2 V_SV_S^\top$, or else set $u_{k+1}^{(n)} = u_k^{(n)}$.
\EndFor
\EndFor
\State\Return $\{u^{(n)}_k\}_{k,n=1}^{K,N}$.
\end{algorithmic}
\end{algorithm}
\begin{remark}
\begin{enumerate}[(i)]
\item A special case of \cref{alg:hybrid} is $m = 0$ and $C_* = C_0$. Denote $P_0 = C_0^{-1}$ the prior precision operator, noting that this is often a local operator and hence sparse when implemented numerically. In this case, $I_{C}$ reduces to
\begin{align}
I_{C}(u) = \frac{1}{2}\left\langle V_S^\top P_0 u, (\gamma^{-2}I + V_S^\top P_0 V_S)^{-1} V_S^\top P_0 u \right\rangle_X,
\end{align}
where the matrix being inverted is small $(N\times N)$.
\item For $\beta > 0$ this algorithm is \emph{not} affine-invariant. However, it is approximately affine-invariant for small $\beta$. Writing $\hat{u} = R(u,\xi)$ for the proposal, as in \cref{ssec:affine}, we have
\begin{align*}
R(Au+b,\xi) &= AR(u,\xi) + b + \beta(I-A)\xi + (\sqrt{1-\beta^2}-1)b\\
&= AR(u,\xi) + b + \mathcal{O}(\beta).
\end{align*}
\item There are two jump parameters that may be tuned in the \hybrid algorithm: $\beta$ corresponding to the pCN jump size and $\lambda$ corresponding to the size of the prior perturbation. Jointly finding the optimal values of these parameters can be difficult in practice; in all numerical examples in this article we simply fix $\lambda = 0.2$ and adapt $\beta$ so that the acceptance rate lies in the interval $(0.15,0.3)$, which appears to be effective empirically. In practice we find starting $\beta$ at a large value is beneficial, allowing for the initial ensemble to adapt to find the effective support of the posterior, before it is reduced to allow for the neighbourhood of the corresponding hyperplane to be explored without too many rejections.
\item In order to provide shift invariance of the proposal it could be tempting to make the choice $m = \overline{u}_S$, the sample mean. However, for a finite number of particles, $\overline{u}_S$ does not lie in the Cameron-Martin space of the prior, and so the acceptance probability is not be well-defined due to measure singularity. One could however consider $m = P\overline{u}_S$ for some projection $P$ onto the Cameron-Martin space.
\item Though one has free choice over the subset $S\subseteq u^{(-n)}$ used to estimate the covariance, we found the choice $S = u^{(-n)}$ to be effective in practice. However, choosing $S$ to be a proper subset of $u^{(-n)}$ may be beneficial when the posterior is multimodal, as well as providing robustness with respect to outliers during burn-in. In the case of multiple separated modes, our sampler would have to additionally be combined with a method that allows samples to switching between modes to accurately measure the relative importance of individual modes \cite{LindseyWeareZhang22,GabrieRotskoffVandeneijnden21}.
\end{enumerate}
\end{remark}
We note that although the dimension $D$ of an inverse problem may be large, often the effective dimension of the problem is much smaller -- the posterior may be concentrated on some low-dimension submanifold of $X$, relative to the prior. It is for this reason that we expect the above algorithm to remain effective when $D$ is large for a finite number of particles. The paper \cite{agapiou2017importance} introduces a quantitative notion of effective dimension for linear Gaussian Bayesian inverse problems, defined in terms of the prior-weighted Gauss-Newton Hessian $Q$. The operator $Q$ may be used to estimate the dimension of the subspace that is informed by the data, relative to the prior. This dimension then gives a rough indication for the order of magnitude of number of particles that should be used in order to achieve good mixing with the \hybrid algorithm. In the notation of \cref{ssec:linear_numerics}, the effective dimension $\mathsf{efd} \leq D$ is defined as\footnote{The paper \cite{agapiou2017importance} also considers an alternative definition of effective dimension simply given by $\tr(Q)$, however this is not bounded above by the dimension of the state space $X$.}
\[
\mathsf{efd} = \tr(Q(I+Q)^{-1}),\quad Q = C_0^{\frac{1}{2}}A^*\Gamma^{-1}AC_0^{-\frac{1}{2}}.
\]
\subsection{An alternative hybrid algorithm}
The paper \cite{coullon2021ensemble} introduces an algorithm that combines affine-invariant sampling with pCN, the Functional Ensemble Sampler (FES), wherein an affine-invariant method is applied on a subspace defined via the prior distribution and pCN is applied on the complementary subspace. Specifically, the affine-invariant method is applied on the subspace spanned by the first $M$ modes of the Karhunen-Lo\'eve expansion of the prior. Thus, if the prior Gaussian distribution on $u$ has Karhune-Lo\'eve expansion
\[
u = \sum_{j=1}^\infty \sqrt{\lambda_j}\xi_j\varphi_j,\quad \xi_j \iid N(0,1),
\]
then a Gibbs-type MCMC algorithm is used to perform affine-invariant updates on the components $\xi_1,\ldots,\xi_M$, and pCN on the remaining components $\xi_{M+1},\ldots$. This is effective when the posterior is relatively close to the prior, however when the data is particularly informative, these prior modes do not represent the posterior well and performance is closer to plain pCN. We consider an adjustment of this algorithm, in the spirit of the hybrid algorithm introduced above, which adapts the subspace based upon the current ensemble. That is, given a subspace dimension $M$ and an ensemble $S$ we diagonalize the sample covariance $V_S V_S^\top$, and truncate this expansion after the first $M$ singular vectors. We then effectively perform an (approximately) affine-invariant update on the span of the first $M$ singular vectors of the sample covariance, and pCN on the orthogonal complement. The specific algorithm is given in \cref{alg:hybrid_proj}, and we refer to this as the \hybridp (\hybridpname) algorithm; for convenience we assume the prior is white as discussed in \cref{rem:noncenter} to avoid the requirement for simultaneous diagonalization of the prior and sample covariances. This method almost agrees with the \hybrid method introduced above, except the contributions to the proposal covariance arising from the sample covariance and the prior are performed on orthogonal subspaces -- again this corresponds to a low-rank update of the prior, and so we may use \cref{prop:hybrid_equiv} to see that the algorithm is well-defined. We compare the \hybrid and \hybridp algorithms with pCN and FES in the following section.
\begin{algorithm}
\begin{algorithmic}
\caption{\hybridp MCMC sampling}
\label{alg:hybrid_proj}
\State Choose initial ensemble of particles $\{u_1^{(n)}\}_{n=1}^N \subseteq X$, jump parameters $\beta \in (0,1]$, $\lambda > 0$, and subspace dimension $M$. Set $\gamma = \lambda/\beta$.
\For{$k=1:K$}
\For{$n=1:N$}
\State Choose $S \subseteq u^{(-n)}$ and let $U\Sigma U^\top = V_SV_S^\top$ be the SVD of the covariance of $S$.
\State Denote $\Sigma_M$ the restriction of $\Sigma$ to the $M$ largest singular values, and $U_M$ the
\State corresponding columns of $U$. Propose
\begin{align*}
\hat{u}_k^{(n)} &= m + \sqrt{1-\beta^2}(u_k^{(n)}-m) + \beta \left[U_M \Big(\gamma\Sigma_M^{\frac{1}{2}} - I\Big)U_M^\top\xi + \xi\right],\quad \xi \sim N(0,I).
\end{align*}
\State Set $u_{k+1}^{(n)} = \hat{u}_k^{(n)}$ with probability
\[
\min\left\{1,\exp\left(\Phi(C_0^{\frac{1}{2}}u_k^{(n)}) - \Phi(C_0^{\frac{1}{2}}\hat{u}_k^{(n)}) + J(U_M^\top u_k^{(n)}) - J(U_M^\top\hat{u}_k^{(n)})\right)\right\}
\]
\State where $J(z) = \frac{1}{2}\|z\|_{\R^M}^2 - \frac{1}{2\gamma^2}\left\langle z,\Sigma_M^{-1}z\right\rangle_{\R^M}$, or else set $u_{k+1}^{(n)} = u_k^{(n)}$.
\EndFor
\EndFor
\State\Return $\{C_0^{\frac{1}{2}}u^{(n)}_k\}_{k,n=1}^{K,N}$.
\end{algorithmic}
\end{algorithm}
\section{Numerical illustrations}
\label{sec:numerics}
We numerically compare the behaviour of the pCN, FES and hybrid algorithms for three different inverse problems. We first consider a linear inverse problem, to investigate the effect of the number of particles and the dimension of the problem. We then consider a nonlinear problem, based on the setup of \cite{Garbuno-InigoNuskenReich20} as well as a generalization, to compare the effect of the sharpness of the posterior distribution relative to the prior on the behavior of the algorithms. Finally we consider a high-dimensional problem with a level-set prior, where no gradients of the likelihood are available.
\subsection{A linear regression problem}
\label{ssec:linear_numerics}
We consider first the case where the forward map is linear, the noise is additive Gaussian, and the prior is Gaussian. In this setup the posterior is Gaussian with a known closed form, and so we can directly compare the result of the sampling with the true posterior in order to assess the accuracy. Let $\Omega = (0,2\pi)$ and define the observation operator $A:C^0(\Omega)\to \R^J$ by $(Av)_j = v(d_j)$. We assume we have data $y \in\ \R^J$ arising from the model
\[
y = Au + \eta,\quad\eta \sim N(0,\gamma^2 I)
\]
for some $\gamma > 0$. The true state $u^\dagger$ generating the data is taken to be $u^\dagger(x) = \sin(x)/2$. We observe the solution at $J = 25$ points, $d_j = 2\pi j/J$, and fix $\gamma = 10^{-3}$ so that the relative $\ell^2$ error on the data is $0.0316\%$. The problem is discretized on a uniform grid of $D$ points.
The prior is taken to be of Mat\'ern type $\mu_0 = N(0,C_0)$, $C_0 = (I-\Delta)^{-1}$, where $\Delta$ is the Laplacian with homogeneous Neumann boundary conditions. The resulting effective dimension of the problem is then approximately $25$. The posterior has closed Gaussian form $\mu = N(m_{\text{p}},C_{\text{p}})$, where
\[
C_{\text{p}}^{-1} = A\Gamma^{-1}A^* + C_0^{-1},\quad C_{\text{p}}^{-1}m_{\text{p}} = A^*\Gamma^{-1}y,
\]
which may be used to assess the accuracy of sampling methods.
\subsubsection{Comparison of algorithms}
We compare the performance of both hybrid algorithms introduced in this paper with the FES algorithm and the pCN algorithm on the above problem. We fix dimension $D=100$ and $N=40$ particles. We do not consider the AIES algorithm, nor the ALDI algorithm of \cite{Garbuno-InigoNuskenReich20}, since we know that these only provide subspace sampling when $N\leq D$. We generate $10^5$ samples per particle so that $4\times 10^6$ samples are generated in total for each method, with the same number of likelihood evaluation required in each case; the first $25\%$ of samples for each particle are discarded as burn-in. For the pCN method, independent chains are run for each particle.
We estimate the autocorrelations for each particle chain and average these over the particles. That is, given samples $\{u_k^{(n)}\}$ and a scalar-valued function $g:X\to\R$, we estimate the function $c:\N\to\R$,
\[
c(j) = \frac{1}{N}\sum_{n=1}^N\frac{c_j^{(n)}}{c_0^{(n)}},\quad c_j^{(n)} = \frac{1}{K}\sum_{k=1}^{K-j}\big(g(u_k^{(n)})-\overline{g}^{(n)}\big)\big(g(u_{k+j}^{(n)})-\overline{g}^{(n)}\big),\quad \overline{g}^{(n)} = \frac{1}{N}\sum_{k=1}^K g(u_k^{(n)}),
\]
where $c_0$ is the sample variance. The area under the graph of $c$ is inversely proportional to the effective number of statistically independent samples in the chain, and so rapid decay of $c$ is desired for an effective sampling algorithm. Throughout this section we will take $g(u) = \|u\|_{L^2}^2$. In \cref{fig:linear_ac_alg} we show the autocorrelations for the four different algorithms. In all cases the jump parameter $\beta$ is adapted so that the acceptance rate lies in the interval $(0.15,0.3)$, and $\lambda = 0.2$ is fixed. In the case of the FES algorithm, the dimension of the subspace AIES is performed upon is chosen as $K=10$, and stretch moves with parameter $a=2$ are performed as suggested in \cite{coullon2021ensemble}; the pCN jump parameter is adapted as above. We see that the autocorrelations decay significantly faster for the algorithms introduced in this paper. The FES method performs similarly to the pCN algorithm as the posterior eigenbasis differs significantly from the prior eigenbasis due to the sharpness of the likelihood. Note that although the autocorrelation for the pCN algorithm decays fast initially due to small scale oscillations, the asymptotic decay of the autocorrelations appears to be faster for the other algorithms. The behavior of the autocorrelations can be further understood from \cref{fig:linear_traces}, which shows the traces of the squared norm of an individual particle for three of the algorithms\footnote{The \hybridp chain is omitted here for brevity; it has the same qualitative behavior as the \hybrid chain.} -- here the long-term correlations for the pCN and FES chains can be observed, and contrasted with the \hybrid chain.
\begin{figure}[bt]
\centering
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={},
xlabel={\large Lag},
ylabel={\large Autocorrelation},
xmin=0, xmax=5000,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
grid=major,
legend style={draw=none,
at={(1.3, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[pcn,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/linear_acs_alg.txt};
\addlegendentry{pCN}
\addplot[fes,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/linear_acs_alg.txt};
\addlegendentry{FES}
\addplot[hybrid,mark=none,style=very thick] table [x index=0, y index=3, col sep=tab] {data/linear_acs_alg.txt};
\addlegendentry{\hybrid}
\addplot[hybrid-proj,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/linear_acs_alg.txt};
\addlegendentry{\hybridp}
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (5000,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:5000,-0.1) -- (axis cs:5000,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\caption{Autocorrelations for the linear regression problem for quantity $\|u\|_{L^2}^2$ for the 4 different sampling algorithms, averaged over particles.}
\label{fig:linear_ac_alg}
\end{figure}
In \cref{fig:linear_densities} we show kernel density estimates for marginals corresponding to two point evaluations, at points $x_1 = 0.01$, $x_2 = 0.63$, for the four different algorithms compared with the true posterior densities, illustrating the accuracy of the methods. We also show (thinned) scatter plots of the point evaluation samples for all particles, illustrating how the different chains mix: the hybrid chains can be seen to be mixing significantly better than the FES chain, which in turn mixes significantly better than the pCN chain. \Cref{tab:linear_errors} compares the sample mean and covariance from the different chains with the true posterior values, along with the multivariate potential scale reduction factor (MPSRF) \cite{brooks1998general}; the latter is computed using inter- and intra-chain correlations, with a value closer to 1 indicating better mixing/convergence. These further illustrate the the mixing properties of the algorithms considered.
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.7]
\begin{axis}[
width=8cm,
height=5cm,
title={pCN},
xlabel={\large Sample Number},
xmin=0, xmax=75000,
ymin=0.74,ymax=0.83,
ytick={0.74,0.77,0.8,0.83},
y tick label style={
/pgf/number format/.cd,
fixed,
fixed zerofill,
precision=2,
/tikz/.cd
},
grid=major,
scaled x ticks = false,
legend style={draw=none,
at={(1.5, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[pcn,mark=none,style=thick] table [x index=0, y expr=\thisrowno{1}*2*pi, col sep=tab, each nth point={5}] {data/linear_traces.txt};
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.7]
\begin{axis}[
width=8cm,
height=5cm,
title={FES},
xlabel={\large Sample Number},
xmin=0, xmax=75000,
ymajorticks=false,
ymin=0.74,ymax=0.83,
ytick={0.74,0.77,0.8,0.83},
grid=major,
legend style={draw=none,
at={(1.5, 0.95)},
anchor=north},
legend style={font=\large},
scaled x ticks = false,
legend cell align=left]
\addplot[fes,mark=none,style=thick] table [x index=0, y expr=\thisrowno{2}*2*pi, col sep=tab, each nth point={5}] {data/linear_traces.txt};
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.7]
\begin{axis}[
width=8cm,
height=5cm,
title={\hybrid},
xlabel={\large Sample Number},
xmin=0, xmax=75000,
ymajorticks=false,
ymin=0.74,ymax=0.83,
ytick={0.74,0.77,0.8,0.83},
grid=major,
scaled x ticks = false,
legend style={draw=none,
at={(1.5, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[hybrid,mark=none,style=thick] table [x index=0, y expr=\thisrowno{3}*2*pi, col sep=tab, each nth point={5}] {data/linear_traces.txt};
\end{axis}
\end{tikzpicture}
\caption{Traces of $\|u\|_{L^2}^2$ for the linear regression problem for 3 of the different sampling algorithms.}
\label{fig:linear_traces}
\end{figure}
\begin{figure}\centering
\begin{tikzpicture}
\node at (-1.5,5) {\includegraphics[width=0.41\textwidth,trim=0cm 0cm 1cm 0cm,clip]{linear_pcn}};
\node at (5,5) {\includegraphics[width=0.41\textwidth,trim=0cm 0cm 1cm 0cm,clip]{linear_kl}};
\node at (-1.5,-1.5) {\includegraphics[width=0.41\textwidth,trim=0cm 0cm 1cm 0cm,clip]{linear_hybrid}};
\node at (5,-1.5) {\includegraphics[width=0.41\textwidth,trim=0cm 0cm 1cm 0cm,clip]{linear_proj}};
\node at (-1.2,8) {\small pCN};
\node at (5.3,8) {\small FES};
\node at (-1.2,1.5) {\small \hybrid};
\node at (5.4,1.5) {\small \hybridp};
\end{tikzpicture}
\caption{Density estimates and scatter plots for the linear regression problem for marginals corresponding to 2 point evaluations, for the 4 different sampling algorithms. The dotted curves represent the true posterior density on the diagonal, and 95\% credible regions below the diagonal.}
\label{fig:linear_densities}
\end{figure}
\begin{table}
\centering
\caption{The relative $\ell^2$ errors in the sample mean $\mathbb{E}(u)$ and sample covariance $\mathrm{Cov}(u)$ for the linear problem, and the MPSRF the 4 different sampling algorithms.}
\label{tab:linear_errors}
\begin{tabular}{r|l|l|l}
& Mean error & Covariance Error & MPSRF\\
\hline
pCN & 0.00834 & 0.964 & 17.4\\
FES & 0.0207 & 0.759 & 4.24\\
\hybrid & 0.00645 & 0.404 & 1.074\\
\hybridp & 0.00784 & 0.390 & 1.075\\
\end{tabular}
\end{table}
\subsubsection{Dependence on number of particles}
As it has been observed that the number of particles $N$ must exceed the dimension of the problem $D$ in order to sample the full posterior when using an affine-invariant sampling method, we study the behavior of the \hybrid algorithm for various numbers of particles $N$. We fix $D = 100$ and vary $N$ between 5 and 40. The number of samples $S$ is varied so that $S\times N = 4\times 10^6$, i.e., the total number of likelihood evaluation remains the same in all cases. The resulting autocorrelations are shown in \cref{fig:linear_ac_npart}. It can be observed that mixing is improved when additional particles are used, however each successive addition of particles yields less of an improvement: the autocorrelation curves accumulate. This is likely related to the effective dimension of the problem being relatively small, as well as the covariance being more accurately estimated.
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={},
xlabel={\large Lag},
ylabel={\large Autocorrelation},
xmin=0, xmax=5000,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
grid=major,
legend style={draw=none,
at={(1.3, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[blue!20,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/linear_acs_npart.txt};
\addlegendentry{$N=5$}
\addplot[blue!40,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/linear_acs_npart.txt};
\addlegendentry{$N=10$}
\addplot[blue!60,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/linear_acs_npart.txt};
\addlegendentry{$N=20$}
\addplot[blue!80,mark=none,style=very thick] table [x index=0, y index=6, col sep=tab] {data/linear_acs_npart.txt};
\addlegendentry{$N=30$}
\addplot[blue!100,mark=none,style=very thick] table [x index=0, y index=8, col sep=tab] {data/linear_acs_npart.txt};
\addlegendentry{$N=40$}
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (5000,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:5000,-0.1) -- (axis cs:5000,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\caption{Autocorrelations of the quantity $\|u\|_{L^2}^2$ for the \hybrid algorithm applied to the linear regression problem, for numbers of particles $N$, averaged over particles.}
\label{fig:linear_ac_npart}
\end{figure}
\subsubsection{Dependence on dimension}
We now consider the effect of the discretization dimension $D$ on the performance of the \hybrid algorithm. We fix $N = 40$, $K = 10^5$ and vary $D=50\times 2^j$, $j=0,\ldots,4$. The resulting autocorrelations are shown in \cref{fig:linear_ac_dim}. The areas under the curves do not increase with discretization level, suggesting that the statistical performance of the algorithm is dimension-robust. This is in contrast to, for example, the AIES algorithm using the stretch move, which fails to be dimension-robust even when sufficient particles are used to ensure the correct distribution is targeted \cite{huijser2015properties}.
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={},
xlabel={\large Lag},
ylabel={\large Autocorrelation},
xmin=0, xmax=5000,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
grid=major,
legend style={draw=none,
at={(1.3, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[blue!20,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/linear_acs_dim.txt};
\addlegendentry{$D=50$}
\addplot[blue!40,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/linear_acs_dim.txt};
\addlegendentry{$D=100$}
\addplot[blue!60,mark=none,style=very thick] table [x index=0, y index=3, col sep=tab] {data/linear_acs_dim.txt};
\addlegendentry{$D=200$}
\addplot[blue!80,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/linear_acs_dim.txt};
\addlegendentry{$D=400$}
\addplot[blue!100,mark=none,style=very thick] table [x index=0, y index=5, col sep=tab] {data/linear_acs_dim.txt};
\addlegendentry{$D=800$}
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (5000,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:5000,-0.1) -- (axis cs:5000,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\caption{Autocorrelations of the quantity $\|u\|_{L^2}^2$ for the \hybrid algorithm applied to the linear problem, for different discretization levels $D$, averaged over particles.}
\label{fig:linear_ac_dim}
\end{figure}
\subsection{A nonlinear inverse problem: Darcy flow}
We now consider a case when the forward map is nonlinear. We consider an example from \cite{Garbuno-InigoNuskenReich20} for reference, and then consider a modified version with smaller observational noise and a slower decaying prior so that the effective dimension of the problem is increased. Specifically, we consider a one-dimensional Darcy flow problem on spatial domain $\Omega = (0,2\pi)$, defining $\mathcal{S}:L^\infty(\Omega)\to C^0(\Omega)$ as the mapping from $u$ to $p$, where
\[
-\frac{\dee}{\dee x}\left(e^{u(x)}\frac{\dee p}{\dee x}(x)\right) = f(x),\quad x \in \Omega
\]
subject to periodic boundary conditions and $\int_\Omega p(x)\,\dee x = 0$. We make the choice
\[
f(x) = \exp\left(-\frac{(x-\pi)^2}{10}\right) - c_f,\quad c_f = \int_\Omega \exp\left(-\frac{(x-\pi)^2}{10}\right)\,\dee x
\]
and define the observation operator $\mathcal{O}:C^0(\Omega)\to\R^J$ as in \cref{ssec:linear_numerics}. The nonlinear forward map is then defined by $\G = \mathcal{O}\circ\mathcal{S}$. We assume we have data $y \in \R^J$ arising from the model
\[
y = \G(u) + \eta,\quad \eta \sim N(0,\gamma^2 I)
\]
for some $\gamma > 0$. As in \cite{Garbuno-InigoNuskenReich20}, we take the true state $u^\dagger$ to be $u^\dagger(x) = \sin(x)/2$. A centered Gaussian prior $N(0,C_0)$ is used, and we consider two problems based on this setup:
\begin{enumerate}[(i)]
\item $C_0^{-1} = 4\left(\mu T - \Delta\right)^2$ where $T:L^2 (\Omega)\to L^2(\Omega)$ is given by $Tv = \frac{1}{|\Omega|}\int_\Omega v$, $\Delta$ is the Laplacian with periodic boundary conditions and $\mu = 100$. Moreover, $\gamma = 10^{-2}$ giving a relative error on the data of $3.81\%$.
\item $C_0^{-1} = I-\Delta$ and $\Delta$ is the Laplacian with homogenous Neumann boundary conditions. Moreover, $\gamma = 10^{-4}$ giving a relative error on the data of $0.0381\%$
\end{enumerate}
We observe the solution at $J = 10$ points, $d_j=2\pi j/J$, so that the first problem (i) is identical to the example considered in \cite{Garbuno-InigoNuskenReich20}. Problem (ii) is a modification with a more concentrated posterior, which is more difficult to sample with methods that heavily rely on the prior, such as pCN.
Note that \cite{Garbuno-InigoNuskenReich20} also proposed a gradient-free method for affine-invariant sampling via simulation of a Langevin-type equation; however, this method suffers the same issue as other affine-invariant methods in that the number of particles must exceed the dimension of the problem.
\subsubsection{Comparison of algorithms}
As for the linear case, we compare the four different algorithms on these problems. We fix dimension $D = 100$ and $N=40$ particles, and generate $10^5$ samples per particles, discarding the first $25\%$ as burn-in. Again, for the pCN method independent chains are run for each particle. We first consider problem (i): the autocorrelations for the quantity $\|u\|_{L^2}^2$ are shown in \cref{fig:nonlinear_ac}, kernel density estimates and scatter plots for marginals corresponding to point evaluations at $x_1 = 0.01$, $x_2 = 0.63$ are shown in the top row in \cref{fig:nonlinear_densities}, and MSPRFs are shown in \cref{tab:psrf}. Note that now there is no analytic form for the true posterior densities to compare with as in the linear case. We first note that, since the likelihood is relatively flat, the posterior is not too far from the prior and so a large step size may be used with pCN leading to fast autocorrelation decay. The \hybrid and \hybridp algorithms achieve similar autocorrelation decay, however that for FES is slower. This is potentially due to FES only using knowledge of the prior eigenmodes but not the decay of its eigenvalues in the space where the AEIS is used, and the prior dominates in this problem. By decreasing the number of modes $M$, better performance could likely be achieved, noting that FES reduces to pCN in the case $M=0$. Nonetheless, the MPSRFs for all algorithms are all close to 1, and the density estimates are similar to one another, since a large number of samples are taken relative to the autocorrelation time.
\begin{table}
\centering
\caption{The MPSRFs for the different nonlinear problems for the 4 different sampling algorithms.}
\label{tab:psrf}
\begin{tabular}{r|l|l|l}
Algorithm & Nonlinear (i) & Nonlinear (ii) & Level Set\\
\hline
pCN & 1.004 & 12.1 & 34.4\\
FES & 1.002 & 1.13 & 68.5\\
\hybrid & 1.005 & 1.03 & 1.50\\
\hybridp & 1.008& 1.03& 1.40\\
\end{tabular}
\end{table}
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={Nonlinear (i)},
xlabel={\large Lag},
ylabel={\large Autocorrelation},
xmin=0, xmax=400,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
grid=major,
legend style={draw=none,
at={(1.5, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[pcn,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/nonlinear1_acs_alg.txt};
\addplot[fes,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/nonlinear1_acs_alg.txt};
\addplot[hybrid,mark=none,style=very thick] table [x index=0, y index=3, col sep=tab] {data/nonlinear1_acs_alg.txt};
\addplot[hybrid-proj,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/nonlinear1_acs_alg.txt};
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (400,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:400,-0.1) -- (axis cs:400,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={Nonlinear (ii)},
xlabel={\large Lag},
xmin=0, xmax=5000,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
ymajorticks=false,
grid=major,
legend style={draw=none,
at={(1.3, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[pcn,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/nonlinear2_acs_alg.txt};
\addlegendentry{pCN}
\addplot[fes,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/nonlinear2_acs_alg.txt};
\addlegendentry{FES}
\addplot[hybrid,mark=none,style=very thick] table [x index=0, y index=3, col sep=tab] {data/nonlinear2_acs_alg.txt};
\addlegendentry{\hybrid}
\addplot[hybrid-proj,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/nonlinear2_acs_alg.txt};
\addlegendentry{\hybridp}
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (5000,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:5000,-0.1) -- (axis cs:5000,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\caption{Autocorrelations of the quantity $\|u\|_{L^2}^2$ for the nonlinear problems (i), (ii) for the 4 different sampling algorithms, averaged over particles.}
\label{fig:nonlinear_ac}
\end{figure}
For problem (ii) the corresponding autocorrelations and density estimates are shown in \cref{fig:nonlinear_ac} and the bottom row in \cref{fig:nonlinear_densities}. The autocorrelation behavior is similar to the linear case, with \hybrid and \hybridp performing similarly to each other and outperforming both pCN and FES. Again the pCN autocorrelation decays faster than FES initially, but FES is faster asymptotically. The density estimates and scatter plots illustrate the poor mixing of pCN compared to the other algorithms. Note that even though the autocorrelation for FES decays only slightly faster than for pCN, the mixing appears much better than pCN and close to that for the hybrid algorithms. The MPSRFs in \cref{tab:psrf} mirror this, with all algorithms significantly outperforming pCN and the hybrid algorithms outperforming FES.
\begin{figure}[bt]\centering
\begin{tikzpicture}
\node at (-5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear1_pcn}};
\node at (0,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear1_kl}};
\node at (5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear1_hybrid}};
\node at (-4.7,7.3) {\small pCN for (i)};
\node at (0.3,7.3) {\small FES for (i)};
\node at (5.3,7.3) {\small \hybrid for (i)};
\end{tikzpicture}\\
\begin{tikzpicture}
\node at (-5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear2_pcn}};
\node at (0,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear2_kl}};
\node at (5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{nonlinear2_hybrid}};
\node at (-4.7,7.3) {\small pCN for (ii)};
\node at (0.3,7.3) {\small FES for (ii)};
\node at (5.3,7.3) {\small \hybrid for (ii)};
\end{tikzpicture}
\caption{Density estimates and scatter plots for marginals corresponding to 2 different point evaluations for the nonlinear problem, for 3 different sampling algorithms. The upper row corresponds to setup (i), and the lower row to setup (ii).}
\label{fig:nonlinear_densities}
\end{figure}
\subsection{A non-differentiable problem: level set prior} We finally consider an example where derivatives do not exist and so gradient-based methods are unavailable. Specifically, we consider a linear inverse problem with a level set prior \cite{iglesias2016bayesian,dunlop2017hierarchical}, with the intention of recovering a piecewise constant field. Whilst the forward map is linear, the level set mapping included in the likelihood ensures that the posterior distribution is non-Gaussian. Specifically, let $\Omega = (0,1)^2$ and define the map $\mathcal{S}:L^2(\Omega)\to C^0(\Omega)$, $u\mapsto p$,
\[
-\Delta p(x) = \mathrm{sgn}(u(x)),\quad x \in \Omega
\]
subject to homogeneous Dirichlet boundary conditions. Define the observation operator $\mathcal{O}:C^0(\Omega)\to \R^J$ as point evaluations on a uniform grid of $J=9$ points, and the nonlinear forward map $\mathcal{G} = \mathcal{O}\circ \mathcal{S}$. The data $y \in \R^J$ is assumed to arise from the model
\[
y = \mathcal{G}(u) + \eta,\quad \eta \sim N(0,\gamma^2 I)
\]
with $\gamma = 10^{-3}$. A continuous Gaussian prior $\mu_0 = N(0,C_0)$ is placed on $u$, $C_0 = (I-\Delta)^2$, with the intention of recovering the binary field $\mathrm{sgn}(u)$. The true binary field is the indicator function of a circle, with the domain discretized on a uniform mesh of $D = 32^2$ points. We fix $N = 80$ particles; for the FES method we choose the number of modes $M = 10$ and for the \hybridp method we choose $M = 20$. For both \hybrid and \hybridp methods we fix $\lambda = 0.2$ as previously. \Cref{fig:levelset_ac} shows the resulting autocorrelations for the quantity $\|u\|_{L^2}^2$, \cref{fig:levelset_densities} shows kernel density estimates for marginals corresponding to two point observations of the field $u$, at points $x_1 = (0.03,0.03)$, $x_2 = (0.5,0.56)$, and \cref{tab:psrf} shows the resulting MSPRFs. The same trends as for the nonlinear problem (ii) are observed, though in this case the FES mixes significantly more slowly.
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
width=8cm,
height=5cm,
title={},
xlabel={\large Lag},
ylabel={\large Autocorrelation},
xmin=0, xmax=8000,
ymax=1,
ymin=-0.1,
ytick={0,0.2,0.4,0.6,0.8,1},
grid=major,
legend style={draw=none,
at={(1.3, 0.95)},
anchor=north},
legend style={font=\large},
legend cell align=left]
\addplot[pcn,mark=none,style=very thick] table [x index=0, y index=1, col sep=tab] {data/level_acs_alg.txt};
\addlegendentry{pCN}
\addplot[fes,mark=none,style=very thick] table [x index=0, y index=2, col sep=tab] {data/level_acs_alg.txt};
\addlegendentry{FES}
\addplot[hybrid,mark=none,style=very thick] table [x index=0, y index=3, col sep=tab] {data/level_acs_alg.txt};
\addlegendentry{\hybrid}
\addplot[hybrid-proj,mark=none,style=very thick] table [x index=0, y index=4, col sep=tab] {data/level_acs_alg.txt};
\addlegendentry{\hybridp}
\addplot[black!60!white,mark=none,style=thick,dashed] coordinates {(0,0) (8000,0)};
\draw[fill=black!60!white,opacity=.2,draw=none] (axis cs:0,-0.1) -- (axis cs:8000,-0.1) -- (axis cs:8000,0.1) -- (axis cs:0,0.1) -- cycle;
\end{axis}
\end{tikzpicture}
\caption{Autocorrelations for the level set problem for 3 different point evaluations, for the 4 different sampling algorithms, averaged over particles.}
\label{fig:levelset_ac}
\end{figure}
\begin{figure}\centering
\begin{tikzpicture}
\node at (-5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{levelset_pcn}};
\node at (0,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{levelset_kl}};
\node at (5,5) {\includegraphics[width=0.32\textwidth,trim=0cm 0.5cm 1.5cm 0cm,clip]{levelset_hybrid}};
\node at (-4.7,7.3) {\small pCN};
\node at (0.3,7.3) {\small FES};
\node at (5.3,7.3) {\small \hybrid};
\end{tikzpicture}
\caption{Density estimates and scatter plots for marginals corresponding to 2 different point evaluations for the level set problem, for 3 different sampling algorithms. }
\label{fig:levelset_densities}
\end{figure}
\section{Conclusions}
By combining affine-invariant with dimension-robust sampling methods, one can find a compromise between the advantages and disadvantages of both. Specifically, in the context of Bayesian inverse problems, when the data is particularly informative and the unknown state is high-dimensional, one can obtain a viable method of sampling the posterior distribution without the need for derivatives of the likelihood.
\appendix
\section{Proofs}\label{appendix}
\begin{proposition}
Define $I_C:X\to\R$ by \cref{eq:IC}. The each term in this expression is finite almost-surely under the posterior.
\end{proposition}
\begin{proof}
As the posterior is absolutely continuous with respect to the prior, it suffices to show that the terms are finite almost-surely under any measure equivalent to the prior. The finiteness of the final two terms follows from the Cameron-Martin theorem applied to the measures $N(m,C)$ and $N(0,C)$, as this is simply the logarithm of the Radon-Nikodym derivative between them. For the first term, note that we have for any Hilbert-Schmidt operator $Z:X\to X$ and $u \sim N(0,C_0)$,
\begin{align*}
\mathbb{E}\langle u,Zu\rangle_{C_0}^2 &= \mathbb{E}\langle \xi,Z\xi\rangle_X^2,\quad \xi \sim N(0,I)\\
&= \mathbb{E}\sum_{i,j,k,l} \xi_i\xi_j\xi_k\xi_l \langle \varphi_i,Z\varphi_j\rangle_X \langle \varphi_k,Z\varphi_l\rangle_X,\quad \xi_j \iid N(0,1)\\
&= 3 \sum_j \langle \varphi_j,Z\varphi_j\rangle_X^2\\
&= 3\|Z\|_{HS}^2,
\end{align*}
where $\{\varphi_j\}$ is any orthonormal basis for $X$. The operator $I-C_0^{\frac{1}{2}}C^{-1}C_0^{\frac{1}{2}}$ is Hilbert-Schmidt by the assumed equivalence of $N(0,C_0)$ and $N(0,C)$ and the Feldman-Hajek theorem, so the result follows.
\end{proof} | 12,810 |
Nuxeo Platform 6.0 Release Notes
For Developers
Elasticsearch in the Platform
Elasticsearch engine and wrapping java service is embedded in Nuxeo CAP distribution, with a dedicated tab in the Admincenter for monitoring the activity. An embedded server is started by default. In production, Nuxeo Plaform is connected to an Elasticsearch cluster. Elasticsearch offers an amazing ability to scale the repository queries while being very easy to set up. The graph below compares number of requests served per second with the number of concurrent users increasing on a given same hardware. In blue dabase queries versus in red Elasticsearch queries.
MongoDB Backend
A deep refactoring of low layers happened in Nuxeo Platform so as to be able to support a non-transactionnal JSON-based storage engines for the repository, such as MongoDB. This refactoring was made so as to be able to implement other NoSQL backends in the future. MongoDB is a perfect candidate when expectations in concurrent writes and high availability are high. Any application existing on top of Nuxeo Platform can benefit form this new persistence option as it is completely transparant to the Content Management model exposed by tree_children Nuxeo Platform. More information on Mongo DB storage and how to set it up .
REST API
Rest API playground
We built a Playground so as to be able to test the REST API and learn how it works.
Content Enricher
Former Rest Providers have been renamed "Content Enricher" for more clarity. The corresponding necessary contribution was also changed. Furthermore, we added the possibility to use parameters when contributing a new content enricher.
Various content enrichers were added to the platform:
- ACL: This adds ACLs of the document to the response. It was previously doable using the @ACL web adaptor. Having a Content Enricher for this allows to get in one same call the document and its ACLs.
- Permission: This one returns the list of permissions the user has on each documents. Useful when building a client side screen, to know if an update button should be displayed.
- Thumbnail: This one adds url of the Thumbnail Picture to the response. You can use it when doing client slide applications.
- Preview: This one adds url of the preview.
- l10nsubjects, l10ncoverage: Exemples of the use of the parameterized enricher based on the VocabularyEnricher class, that returns all the labels of values of the documents that are linked to directories. Useful when you want to make client-side listings of documents and to resolve server-side the values to display (including internationaliation).
The Content Enricher documentation has been updated accordingly.
Query Endpoint
A Query endpoint has been added. It allows to execute NXQL queries (with ability to control if it hits Elasticsearch or the database) and to fetch page providers. More information on the documentation page .
Endpoint for Types and Schemas
We added a new REST API endpoint for document types and schemas. It can be useful when implementing something like a connector. See NXP-14114 for details.
ParentId Ref
parentRef id has been added to the json response.
Improved File Download with the REST API
We've worked on the Blob property JSON representation. See NXP-13616 for details.
Before:
Now:
New Versioning Header
You can now control whether a minor or major version is created when creating or updating a document via the REST API, using the new "X-Versioning-Option" header, with "MAJOR" or "MINOR" as a value.
iOS SDK
We started to provide a SDK to start building your iOS application connected to the Nuxeo Platform through the REST API. We already have some very cool features like a blob offline cache.
JavaScript SDK
The JavaScript Client SDK is fully part of the Nuxeo Platform. Two implementations are provided. One uses node.js primitives, one is based on JQuery XHR. The client exposes:
- Wrappers for resources oriented REST API (at least documents)
- Utilities for automation calls
- Utilities for other resources oriented endpoints REST calls
- Batch upload helper
- Basic and token based authentications.
Client is packaged using bower for the JQuery implementation, and NPM for the node implementation. Clients are under continuous integration using phantom.js.
Box, Mule, CMIS: great new integration capabilities
CMIS Implementation Evolutions
- ACLs are now implemented
- Renditions from the RenditionService are now exposed as CMIS renditions of kind "nuxeo:rendition", with a name that's based on Rendition name. For example "nuxeo:rendition:pdf" for the "pdf" rendition.
- ecm:pos is exposed in CMISQL, so as to handle correctly ordered folders
- Nuxeo CMIS content streams support HTTP cache and last modified headers.
Box API Implementation
The Nuxeo addon nuxeo-box-api is an implementation of the Box API on top of the Nuxeo Platform repository. It transforms the Nuxeo content repository into a Box compliant storage backend. Use cases of such an approach are:
- Light integration on continuous integration chain for your Box development
- On-premise setup of Box stored content
Don't hesitate to read our CTO Thierry Delprat's interview to get a better idea.
Here are some examples:
Folders:
looks like
Files:
looks like
Mule Connector
We wrote a connector for Mule that is based on HTTP (Automation) API.
Mule ESB is a lightweight Java-based enterprise service bus (ESB) and integration platform that allows developers to connect applications together quickly and easily, enabling them to exchange data.
Using this connector, you can build Mule Flows that will use services exposed by Nuxeo Platform.
This connector exposes:
- A predefined set of Operations (getDocument, createDocument, updateDocument, query, ...)
- A generic runOperation to allow call to any operation or chain defined on the Nuxeo server
- Some converters from Nuxeo objects (Document, Documents, Blob) to raw types (Maps, List of Maps, File ...)
We invite you to read the sample and documentation to get a good grasp on the possibilities offered by this module.
Support of Facets and Other Kind of Aggregates on Page Providers and Content Views
Page Provider now integrates aggregates support and content view leveraging such kind of page providers can be configured via Nuxeo Studio easily. That way you can build search interfaces with terms, ranges and histograms filters. Many use cases have been taken into account: support of dates, specific widgets for users, directories, documents. More information about supported aggregates in the platform and how to configure new search screen with aggregates.
JSF2, New User Actions Placeholders, Result Layouts and Other Evolutions of the UI Framework and CAP Web Application
JSF2
Content Application Platform (CAP) and all the plugins have been migrated from JSF 1.2 to JSF 2 and the latest stable release of Richfaces: 4.5. A migration tool has been implemented for helping you migrate your custom code base. That will be necessary when you plan to upgrade your Nuxeo Platform Instance to 6.0. More information can be found on JSF code migration to JSF2 .
Improvement of Content Views Configurability
You can now control the display of the slideshow, the spreadsheet and the edit columns actions at the result layout level. Even better, this area of user actions is now a category of actions. It means that it is easy to contribute new actions there. To evaluate the visibility of such an action, the current content view is made available in the filter context.
Typed Layout
Layouts now have a type, which allows to define properties that can be used to define on a per instance basis if some given features are supported. We currently use it for result layouts, with two types: listingTable and listingThumbnail
Widget Definition Alias
It is possible to define aliases for a widget definition, so as to override former widget definitions with a newer one without having to duplicate that definition. Ex: - document_listing_ajax -> table_listing - search_listing_ajax -> table_listing
Document mode with Fallback on Toggleable layouts widget
Toggleable document layout widget now handles a document mode, since modes for documents are not restricted to "create" or "edit" or "view" and can gather layouts for different rendering use cases (like the drive edit screen), and still need to use "edit" or "view" mode for render. Furthermore, a logic of fallback has been added when the given mode has not been contributed.
Dev Mode: Display of Page Structure
In Development mode, you can display the structure of your page : layouts, widgets, actions, just select the object and display its properties. This makes it a great tool for getting inspiration from the default application for your own project, or for knowing which id should be used for removing or hide conditionnaly a link, a button or any other UI object .
Multi-navigation Category of Actions
The placeholder for adding virtual navigations is now part of Nuxeo CAP. You don’t need to depend of Virtual navigation for adding elements there.
Custom Headers in the Webapp
A custom header has been added: "X-UA-Compatible" with the value "IE=10; IE=11" so as to better control behaviour on Internet Explorer on supported versions. The way this header has been added benefits from a new extension point that allows to contribute additionnal headers.
Select2
Select2 widgets have been improved. There are two noticable new features.
- Select2 widgets can be configured to let the user add entries to its binded directory. Take a look at this example .
- You can now provide a JavaScript function that will be in charge of choosing the id (from serialized entry's fields) that will be submitted by the widget.
Changes in Features Layout, Migration to Maven 3
The build now relies on Maven 3 (Maven 2 previously). There were some big changes in the way features are distributed. Nuxeo DM is a deprecated assembly. You shoud now start systematically from Nuxeo CAP and add addition packages. Faceted Search has been deprecated and exists as a separated package. Virtual Navigation is in a separated package. Tags support has been integrated to Nuxeo CAP. Open Social is deprecated and exits now in a separate marektplace pacakge. Social Collab addon is also deprecated. DAM package contains all the imaging-* addons, and brings Picture, Video and Sound types.
Content Automation
Parameterised Automation Chains and Other Productivity Improvements on Automation
Users can design automation chains with parameters, saving a great number of chains whose structure is repeated with only a few changes in operations parameters.
The readability of an automation chain is improved with a better visibility of arguments values.
Finally, we provide a text mode for automation chain edition, that will facilitate some refactoring, and also make it easy to transfer to support a chain definition, or to build new documentation samples.
Automation: Execution Flow Operations
The list of operations allowing to launch other chains for loops (loop on blobs, loop on documents, loop on arbitrary lists) has been cleaned up. Idea is to have on all those operations support for transaction management, so as to be able to easily design scalable and robusts automation chains.
YAML Converter for Automation Chains
YAML representation of chains is now visible in the platform. Here is an example for the validateDocument chain:
New Operations for Collections Management
We have added operations dedicated to collections management. You can use AddToCollection, RemoveFromCollection, CreateACollection and GetAllDocumentsFromCollection. Their name speak for themselves.
Authentication and Other User Management Evolutions
Token Based Authentication Management & OAuth 2.0 Support
Exploding demand of API use makes it necessary to provide easy means for developers to manage authentication of the applications they develop, let it be on JavaScript based app, iOS, Python (with Drive)… The token based authentication service, implemented initially for the requirements of Nuxeo Drive is extracted and generalized, made available to the Nuxeo Platform developers. In the same time, OAuth 2.0 support is added, so as to provide as an option a very secured way of handling the authentication process.
SCIM 1.1 ImplementationA new scim-server plugin provides a RESTFUL implementation of SCIM 1.1 interface, that allows to provision users and groups using a commonly accepted API. This API is used for instance by third party IDMs like Okta.
Okta SSOA new login-okta plugin allows to use Okta IDP for managing users and authentication.
Duo Web Two Factors authentication pluginA new duoweb-authentication plugin allows to set up a two factors authentication for Nuxeo using Duo Web online sercices.
Audit Service Based on Elasticsearch
A new implementation of the audit service has been made on top of an elastic search index. This provides greater performances which was necessary when audit has millions of entries, and more power in terms of defining what fields are stored.
Various
New Download Servlet
A new download servlet is available for documents that are not bound to a specific document. This new servlet is used by the Download File operation so as to get this operation working even with big files.
UserWorkspace Improvements
It is now possible to override the behaviour of user workspace creation.
Incremental Update for Integer fields
Some document fields can now be updated at the storage level using atomic increment/decrement operations. This is needed for instance to allow concurrent delta updates of fields without losing information. A typical use case is the quotas. See for more details.
Directory Low Level Access Restriction
When contributing a directory declaration, it is now possible to add READ and WRITE restrictions to specific groups.
Java 8 Support
This new release has been tested with Java 8. You can see the list of changes on NXP-13570 .
NXQL Query and Fetch
Using the queryAndFetch method you can now use COUNT, AVG, SUM, MIN and MAX in your NXQL queries. Again, this only work when using queryAndFetch . See NXP-13708 for reference.
Facilitators for Escalation Rules Configuration
Simplify frequent escalation rules use cases by providing some functions usable in the conditional expression:
- timeSinceWorkflowWasStarted()
- timeSinceTaskWasStarted()
- timeSinceDueDateIsOver()
- timeSinceRuleHasBeenFalse()
That way, the following current condition
would be replaced by:
without having to deal with adding a custom node variable etc...
Calendar Facet
We have added a calendar facet that allows to easily set Calendar objects in the repository. They themselves display events in a calendar style, with ability to bind documents to calendars and events. This feature is very useful on Case Management projects, where one wants to review cases on specific events. Thanks to this roadmap item, it is a matter of a few minutes to setup some calendars in your application! The feature is made available in the already existing nuxeo-agenda module.
Giving a Comment When following a Transition
This is something we should have done a long time ago but it's finally here. You can now pass a comment when following a transition. To do this you simply add a comment in the document's context map like this:
Business Days Management
A plugin implemented at first for a customer project has been generically re-branded as it can be useful in many situations: "Business Days Management" module allows to compute a date given a delay and the list of holidays for the year. The delay will be added to the start date without counting the week ends and those holidays. See the nuxeo-business-day-management GitHub repository , we will come up soon with a post around this topic.
Generic Video Class Converter
Thanks to this new Video converter, you can add new video conversion formats without having to write custom Java classes. A sample contribution of a new video converter would be: | 91,247 |
A few recent governmental flip-flops on policies enshrining gender (the social stereotypes based on sex that ritualize male dominance and female subordination) illustrate the politics involved in passing such broad-sweeping yet ill-considered measures.
The first odd policy reversal was the decision in Nova Scotia, widely reported by CBC news and other sources, that the Health Ministry had decided not to fund radical cosmetic surgical procedures based on gender roles. From the CBC:
CBC reported:
“Wilson said there’s a lack of high-quality research about the effectiveness and long-term outcomes of sex reassignment surgery.
He said the decision to deny coverage came after a careful policy review and is declining interviews until after he meets with NSRAP [Nova Scotia Rainbow Action Project] on Wednesday.”
The very NEXT DAY the policy was reversed. From CTV Atlantic News:
“The Nova Scotia government says the province will soon fund gender reassignment surgery .
Health and Wellness Minister David Wilson said he came to the decision after reviewing the medical evidence and the policies of other provinces that fund this procedure.
“Based on the values and direction this government has taken on the issue, I am happy today to confirm funding for gender reassignment surgery,” said Wilson.
“Last November, the province enshrined transgender rights in legislation. We don’t permit others to discriminate against transgender people, and the funding of this surgery is an issue of dignity, and equality for transgender Nova Scotians.”
Wilson said he met with members of the LBGTI community about the decision on Wednesday.”
The other big recent flip-flop involved the sweeping change to school athletic programming in Nebraska. Following widely reported news that sex-based protections for female athletes were being eliminated state-wide and replaced with “sex-role” or “gender” guidelines, it turns out that no such policy was ever approved after all. From Nebraska Watchdog reporter Deena Winter:
.
2 thoughts on “The Politics of Sex-roles: Government Gender Policy flip-flops”
OK, I’m in SE Nebraska and I didn’t hear about this. It also seems that this woman isn’t up to speed on the difference between ‘sexual orientation’ and ‘transgender’. It’s going to be fun watching the shit hit the fan over this one–Nebraskans really, really don’t like others pushing this crap into their lives.
Not sure where to put this. Watch them hem and haw | 254,843 |
TITLE: $p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer
QUESTION [6 upvotes]: Does anyone know if the following problem has ever been studied?
Let $a$ and $b$ be two real numbers and consider the polynomial: $$p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$$
where $n$ is a positive integer.
Does there exist a value of $k$ such that if $p_n(a,b)$ is an integer for $k$ consecutive values of $n$ then $p_n(a,b)$ is an integer for every $n$? If so what is the minimum value of $k$?
What happens if we change the 'integer' condition to 'rational' values at the previous question?
It's not difficult to establish some recurrence relation among the values of $p_n$ but none of them seem to be promising.
I would like to know any reference for this problem or how it could be solved.
Any help would be appreciated.
REPLY [11 votes]: I will do the rational case and assume $a,b\neq 0$ otherwise the problem is trivial. You just need four consecutive values. Note that $p_n(a,b)=\cfrac{a^n-b^n}{a-b}$.
Say you have $p_k$, $p_{k+1}$, $p_{k+2}$, $p_{k+3}$ are all rational.
Note that $p_{k+1}^2-p_kp_{k+2}=(ab)^k$ and $p_{k+2}^2-p_{k+1}p_{k+3}=(ab)^{k+1}$.
Thus $ab$ is rational.Now $p_{k+1}(a+b)=p_{k+2}+abp_{k}$ so it follows that $a+b$ is also rational or $p_{k+1}=0$. But similarly $p_{k+2}(a+b)=p_{k+3}+abp_{k+1}$ so if $p_{k+2}=0$ then $a=b$ and again the problem is trivial.
Thus we have $a+b,ab \in \mathbb{Q}$ and now note that $p_n(a,b)$ is a symmetric polynomial so it can be expressed as a polynomial with rational coefficients in $a+b,ab$ so it is always rational. | 100,880 |
Wednesday, April 6th - - It was a beautiful morning! Cool temperatures required a sweatshirt but it was quite comfortable. There was barely even a breeze!
Yesterday, I mentioned that the traffic couldn't be heard on the Interstate here in the campground. Today was a different day. You could clearly hear the hum/buzz/roar of the cars and trucks passing by. From the top of the sandhill that I was on you could even see the highway. Nevertheless, it was still quite pleasant.
I may be “off the grid” for a few days and no posts have been scheduled so things will be quiet here for a while...
Thursday, April 07, 2011
Morning at Monahans Sandhills
Posted by Becky Wiseman at 5:00 AM
Labels: Journey-2011, Tex.
Wonderful photos, Becky - a good time of the day to be out and about!
These photos are ready for National Geographic! The last one is outstanding!
Have a good few days off the grid! Can't wait to read what you will be up to.
Amazing pictures! Looks like a beautiful morning!
Wow!! Beautiful photos! | 80,114 |
The Federal Government (FG) has requested media and public commentators’ restraint in coverage and comments on the fallouts of what it described as recent farmers and herders clash in Benue State, so as not to inflame the situation, reports ITRealms.
The Minister of Information and Culture, Alhaji Lai Mohammed, made this call at the weekend, told ITRealms in a press statement that Federal Government is working hard to address the situation with a view to preventing a recurrence, not just in Benue State, but nationwide.
“It’s incumbent on the media and public commentators not to engage in actions that can aggravate the crisis,” he said.
Mohammed explained that for avoidance of doubt and fear of being misunderstood that “we are neither saying the media should not cover the crisis nor asking public commentators to desist from commenting on it. All we are saying is that both must be circumspect.”
He pointed out that the current poisoned atmosphere of “incendiary comments, unrestrained use of pictures and footages that offend human sensibilities as well as fingerpointing can only exacerbate the situation and complicate ongoing efforts to end the crisis.”
Further, insisted that FG has mapped out short and long-term solutions to the incessant clashes between farmers and herders, including a planned conference of stakeholders, and assured of the government's strong determination to find a lasting solution to the farmers/herders' clashes.
Ayo Midele/GEE
Linked: | 223,672 |
According to the National Health Care Anti-Fraud Association (NHCAA) and the Blue Cross and Blue Shield Association, the estimated annual costs of health care fraud in the U.S. are more than $70 billion, or about 3% of total national health care expenditures. This is an amount large enough to wipe out the savings from many well intentioned programs. For example, Accountable Care Organizations (ACOs)—one of the key organizational components of the Affordable Care Act (ACA)—are estimated to save only $5.3 billion in the first 10 years of operation.
Efforts to fight health care fraud were reinforced in the Health Insurance Portability and Accountability Act of 1996 (HIPAA), which created the national Health Care Fraud and Abuse Control Program (HCFAC). The HCFAC program is designed to coordinate Federal, State and local law enforcement activities with respect to health care fraud and abuse. Under this program, the government has recovered approximately $28 billion from 2000 to 2012 (see Figure). In the most recent year for data available (2012), the government collected $4.2 billion in criminal fines, civil penalties, asset forfeiture, and other penalties.
The HCFAC coordinated an enforcement effort in May 2012 that involved the highest number of false Medicare billings in the history of the program. The enforcement involved 107 individuals (mostly doctors and nurses) in seven cities charged for alleged participation in false Medicare billings totaling more than $450 million. In FY 2012, 251 guilty pleas and 13 jury trials were litigated, with guilty verdicts against 29 defendants. The average prison sentence in these cases was more than 48 months.
According to the NHCAA, the most common forms of medical care fraud are:
These types of fraudulent activity can be detected a number of ways, ranging from traditional law enforcement investigative techniques to complex statistical analysis of administrative data, such as claims data from government programs like Medicare and Medicaid and private commercial insurance carriers. Most of the fraudulent schemes in the list above can be identified to some extent by performing complex statistical analyses of administrative claims data.
The overall aim of statistical analysis in the identification of fraudulent activity is to separate the “expected” from the “unexpected.” Statistical analysis can also identify “outliers”—those cases that look substantially different from other cases. Prior to the advent of computers and information systems able to process massive databases in minutes, the investigation of fraud typically focused on a “manual” examination of outliers. For example, analysts would “dump” all claims or records that had a certain characteristics, and then manually examine each of those records.
Statistical analysis of expected versus unexpected events, however, is quite different. Using complex regression analyses which make use of all of the data available (e.g., medical claims data merged with patient characteristics), statistical models can be used to predict medical care utilization. Statistical models can make use of data on market demand (e.g., patient characteristics, such as age, gender, income, etc.) and supply characteristics (e.g., number of physicians per capita, number of specialists per capita, number of hospital beds per capita, etc.). For these types of models, medical care utilization (Q) can be expressed as a function of market demand (D) and supply (S), such that Q = f (D, S).
If we estimate this model statistically at the national level, we can generate estimates (“coefficients”) of the role of the variables that make up D and S. We can then apply these estimates to models restricted to a given geographic area (or to a given provider or type of provider) and calculate the expected utilization (Q’). These kinds of predictive models can be designed to address the following types of inquiries:
There are many other types of inquiries that be supported by this basic statistical modeling framework; these are just a few examples. Note that in all of these cases, the actual observed level of utilization (Q) is compared to the expected level of utilization (Q’). In any statistical model, there is “error” associated with the estimates. Thus, we generally do not simply look to see whether Q > Q’. There is a margin of error around Q’, so what we are really interested in is whether the magnitude of the difference between Q and Q’ is “statistically significant.” The calculation of statistical significance takes into account the margin of error of Q’, and provides information as to whether the difference between Q and Q’ is statistically different from zero. If Q is significantly greater than Q’, we are observing levels of utilization that are higher than what we would expect.
Unfortunately, this does not signal the end of the investigation. The statistical analysis is only the first step. Although the statistical analysis points to a significant difference in utilization rates, it is necessary to probe deeper. Does the market area or the provider have characteristics that are not sufficiently captured in the prediction model? If we were to replicate the prediction model on a similar provider or in a similar geographic area, would we observe similar differences? Finally, and perhaps most importantly, is the magnitude of the difference large enough to warrant further investigation? The latter is an interesting question because with limited investigative resources, fraud investigators must look not only for significant differences but for differences that are likely having the biggest financial impact—this may be driven by the overall magnitude of the difference between Q and Q’, but also may be driven by the type of medical procedure in question. Small differences in less expensive procedures have less impact on payers than small differences in more expensive procedures.
At Avalon Health Economics () we have experts with extensive experience in the kind of predictive modeling discussed in this blog. Our consultants have examined administrative claims data from the Medicare and Medicaid programs and have also worked extensively with private insurance claims data and billing data. We have advised national data providers on methodological issues, and have published numerous peer-revised papers wherein the aforementioned statistical methods have been employed.
-John Schneider, PhD, CEO | 79,924 |
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Medford teachers and parents are paying for school supply items that were previously covered by the city or state.
Just prior to the start of the new school year, each public school in Medford received five boxes and one large bag full of donated school supplies.
The donations were made possible by two parents, Cheryl and Alex Rodriguez, who organized a citywide school supply drive – the first event of its kind, at least in recent years. Hundreds of residents dropped off unwanted items and supplies at City Hall in July and August.
“It brought in a ton of supplies,” Cheryl Rodriguez said. “[The] PTOs and principals were very happy.”
Cheryl also started a GoFundMe page over the summer to replace classroom rugs at the Roberts Elementary School, which one of her children attends. The GoFundMe and the Rockin' Roberts PTO together raised enough money to purchase eight rugs for kindergarten and first grade classrooms at the Roberts. The old rugs, according to Rodriguez, had not been replaced since the school was built 15 years ago.
While teachers and parents benefit from these efforts to support the schools, the increased reliance on donations and fundraising to pay for classroom supplies has led some to question whether the district should be covering more of these costs.
“Once you start relying on PTOs to fund those items, you’re getting into a slippery slope of having this GoFundMe educational economy,” said Kristina Gasson, a parent and member of the Rockin' Roberts PTO. “We were happy to buy those [rugs] for our kindergarten and first grade teachers because they get the most wear and tear, but it’s kind of unacceptable that they have 15-year-old rugs that are getting daily use in our district.”
According to Luisa Spirito, a 33-year veteran of the Medford Public Schools, this isn’t the first time the Roberts PTO has stepped up to pay for much-needed items. The PTO also recently purchased chart paper and Chromebooks for teachers at the school.
“We’re very grateful to everything they do for us,” said Spirito, a first-grade teacher at the Roberts.
Years ago, many of these costs were covered by the district, Spirito recalls.
“In past years, we were given a stipend of $250 to purchase supplies, chart paper, whatever you might need,” she said. “You had the discretion of what you wanted to purchase [and use] for the classroom.”
That practice was done away with at least 10 years ago, according to Spirito. Now, she typically spends nearly $300 every fall for a variety of items needed for her first-grade classroom at the Roberts. And she spends another $100-200 as the school year goes on, all without reimbursement from the district.
That’s because the budget for supplies at each school has shrunk, according to School Committee member Mea Quinn Mustone.
“There’s not enough money in the budget to give teachers $200 to buy their own classroom supplies,” Mustone said. “That’s the problem: Each school budget for supplies has gone down.”
Teachers, parents pick up the slack
Unsurprisingly, teachers, parents and PTOs often cover the costs of supplies and other items not included in the school budget – a reality in other school districts in the Boston area as well.
A recent survey conducted by the Medford Transcript asked parents of children at Medford Public Schools how much they spent this fall on school supplies. The survey received 56 responses indicating a wide range of supply costs.
For example, while eight respondents said they spent under $25 for all their children, nine said they spent over $200. The rest fell somewhere in between.
Meanwhile, anonymous comments submitted through the survey suggest that many parents feel obligated to purchase non-mandatory supplies every fall to help the teachers.
“What will be most costly are the tissues, paper towels, hand sanitizer, Band Aids, reams of paper, etc, which the teachers are not allowed to ask for, but are in need of,” one respondent said. “If we don't help supply them, then the teacher has to pay for them out of their own pocket."
Others said they wished the system for purchasing school supplies was more efficient.
“Every year the supply list is posted, and then the kids go to school and teaches provide a different list of school supply requests,” one respondent said. “If you could provide teacher assignments early and have the teachers provide their specific request so that we can buy the material ahead of time, it would save parents a lot of time and money. Or, have all teachers work together to create a more streamlined supply list.”
Gasson and some respondents said they believe the district and teachers have made efforts to reduce the mandatory supply list for parents. For her first grader, Gasson only paid about $20 for school supplies this year.
But $20 per child can still be a challenge for some parents, especially as the demographics of the city continue to change. During the 2017-18 school year, approximately 31 percent of students district-wide were considered economically disadvantaged, according to data from the Department of Elementary and Secondary Education.
“I find that there are more children coming in that have less – children coming in who may not have the right sized backpack, or may not even have one,” Spirito said.
School Committee looks for solutions
At the Sept. 10 School Committee meeting, several committee members proposed ways to decrease costs of school supplies for teachers and parents. Mustone suggested the district request funding for supplies from the state budget, and School Committee member Paul Ruseau mentioned the possibility of purchasing items in bulk.
Spirito is in favor of bulk purchasing and hopes teachers, schools or parents will coordinate with one another to purchase more items in bulk in the future. For example, parents could contribute $5 or $10 prior to the new school year, and teachers could then order notebooks and binders for everyone.
“I’m assuming it would cost much less,” Spirito said.
Though this year’s budget has already been finalized, Ruseau hopes next year's budget will have more money available for supplies. Looking at the fiscal year 2019 budget for mathematics, for example, Ruseau noted that the district spent $10,000 for math-related supplies in all schools.
He questioned whether that amount truly reflected the needs of all mathematics classes in the district.
“That’s just a round number somebody put in the budget,” Ruseau said. “What should the number be? The answer is whatever it costs to provide the supplies to do the math curriculum.”
Outside of the mandatory supply lists, there are also items that parents must pay for if they want their children to participate in certain activities, Ruseau added.
“If you don’t have money to rent instruments, you don’t get to play in band,” he said. “Band is being offered as part of the school day, but it’s not actually being offered to everyone.”
To truly determine the scope of the “GoFundMe educational economy,” Ruseau believes the School Committee needs a little more data: How much are the PTOs spending on school supplies each year, and how many of those expenses should actually be covered by the district?
“Things that belong in our budget ... are being paid for in other ways,” Ruseau said. “So my focus is figuring out what the numbers are. And then, in my mind, we can start identifying how big the problem is and working backwards from there.” | 22,248 |
TITLE: A reverse characterization of „property P can be checked on its finite subsets“
QUESTION [2 upvotes]: (motivation: $\mathscr L$ = sets of n-ary formulas under signature $τ$, $P$ = satisfiability wrt a theory $T$)
Let $\mathscr L = (0,1,\wedge,\vee)$ be a bounded atomic lattice, whose elements will be called finite if they are the join of finitely many atoms.
Let's say a property $P$ on elements in $L$ can be checked finitely if
it is downward closed and
for all $M \in \mathscr L$, $P(M)$ holds if and only if $P(F)$ holds for every finite $F \leq M$.
So given any $M$ satisfying property $P$, we can associate to it the set $\mathscr F_M := \{F\mid F\text{ finite, }F\leq M\}$, every element of which satisfies $P$. Of course, $M = \bigvee \mathscr F$.
Question: If we instead start from an arbitrary set of finite elements $\mathscr F$, all of which satisfy $P$, what does $ℱ$ have to fulfill in addition such that $\bigvee \mathscr F$ satisfies $P$?
REPLY [1 votes]: I unfortunately do not think much can be said in general. If $\mathscr{L}$ is infinitely distributive, then we have the following criterion: $\bigvee\mathscr{F}$ satisfies $P$ if and only if $\bigvee\mathscr{F}_0$ satisfies $P$ for every finite subset $\mathscr{F}_0\subseteq\mathscr{F}$. Since $P$ is downward closed, the "only if" direction is clear. For the "if" direction, suppose that $\bigvee\mathscr{F}$ fails to satisfy $P$. Then, since $P$ can be checked finitely, there exists a finite $F\leqslant\bigvee\mathscr{F}$ not satisfying $P$. Let $a_1,\dots,a_n\in\mathscr{L}$ be atoms such that $F=\bigvee_{i=1}^n a_i$. Since $F\leqslant\bigvee\mathscr{F}$, or in other words $F\wedge\bigvee\mathscr{F}=F$, by distributivity and the fact that the $a_i$ are atoms, there must exist some $b_1,\dots,b_n\in\mathscr{F}$ such that $a_i\leqslant b_i$ for each $i$. Now $F\leqslant \bigvee_{i=1}^n b_i$, so by downward closure $\bigvee_{i=1}^n b_i$ fails to satisfy $P$, whence taking $\mathscr{F}_0=\{b_1,\dots,b_n\}$ gives a finite subset of $\mathscr{F}$ whose join does not satisfy $P$, as desired.
So, for example, $\bigvee\mathscr{F}$ will satisfy $P$ if $\mathscr{F}$ is closed under $\vee$. This is precisely the situation with the set $\mathscr{F}_M$ that you describe, which is closed under $\vee$ for any $M\in\mathscr{L}$. But we unfortunately cannot do much better than that; for instance, it is not in general even enough to ask that $\bigvee\mathscr{F}_0$ satisfies $P$ for every proper subset $\mathscr{F}_0\subset\mathscr{F}$! Indeed, let $X$ be any finite set of size $>1$, and let $\mathscr{L}$ be the power set of $X$. Then the property $P$ of being a proper subset of $X$ can be checked finitely. However, if $\mathscr{F}\subset\mathscr{L}$ is the set of singleton subsets of $X$, then $\bigvee\mathscr{F}_0$ satisfies $P$ for every proper subset $\mathscr{F}_0\subset\mathscr{F}$, even though $\bigvee\mathscr{F}$ does not satisfy $P$.
If $\mathscr{L}$ is not distributive, then things can get even worse; it is possible to have a case where $\mathscr{F}$ is infinite and $\bigvee\mathscr{F}_0$ satisfies $P$ for every proper subset $\mathscr{F}_0\subset\mathscr{F}$, but $\bigvee\mathscr{F}$ nonetheless fails to satisfy $P$. For instance, let $\mathscr{L}$ be the union of the power set of an infinite set $X$ with a single new element $\star$, where $\star\wedge S=\varnothing$ for each $S\neq X$ and $\star\wedge X=\star$. Then, taking $P$ to hold precisely on $L\setminus\{X,\star\}$, $P$ can be checked finitely. However, if $\mathscr{F}$ is the set of singleton subsets of $X$, then $\bigvee\mathscr{F}_0$ satisfies $P$ for every proper subset $\mathscr{F}_0\subset\mathscr{F}$, even though $\bigvee\mathscr{F}=X$ does not. | 81,025 |
\begin{document}
\title{Some new parameterizations for the Diophantine bi-orthogonal monoclinic piped}
\author{Randall L. Rathbun}
\email{[email protected]}
\subjclass[2010]{11D09, 14G05, 11G35}
\keywords{Diophantine piped, monoclinic, asymptotic sequence}
\begin{abstract}
The bi-orthogonal monoclinic Diophantine parallelepiped is introduced, then the $s$-parameters and their
governing equation for the bi-orthogonal monoclinic Diophantine parallelepiped are discussed.
Previous discoveries and parameterizations of the monoclinic piped are noted.
Then two parameterizations $P\left[\frac{1}{2},s_2,s_3,s_4\right], s_i \in \mathbb{Q},\mathbb{Z}$ are given
for a specific type of Diophantine bi-orthogonal monoclinic parallelepiped.
Next, a parameterization $P\left[s_1,s_2,s_3,s_4\right], s_i \in \mathbb{Q},\mathbb{Z}$ is presented
which covers 99.5\% of solutions found by raw computer searches. Several asymptotic sequences approaching
the {\it perfect cuboid} are listed, and some final comments made.
\end{abstract}
\maketitle
\setcounter{section}{0}
\section*{{\bf The bi-orthogonal monoclinic Diophantine parallelepiped}}
Let Figure \ref{fig:mono} be a general bi-orthogonal monoclinic Diophantine parallelepiped, which is a
cuboid with two right angles at any vertice(bi-orthogonal), but the third is not a right angle,
but either acute or obtuse(monoclinic). This figure is composed of two congruent parallelograms (not
rectangles) joined by 4 orthogonal rectangles, all of whose edges are rational or integer.
\begin{figure}[!ht]
\centering
\includegraphics[scale=1.0]{monoclinic}
\vspace{-6pt}
\caption{The Bi-orthogonal Monoclinic Cuboid.}
\label{fig:mono}
\end{figure}
Let $x$, $y$, $z$, denote the three different edges, with $x,y,z \in Z$.
The face rectangle $(x,y)$ has diagonal $a$, the face rectangle $(x,z)$ has diagonal $b$
and the face parallelogram $(y,z)$ has diagonals $c_1, c_2$.
The two different body diagonals are denoted by $d_1, d_2$.
These lengths satisfy the equations
\begin{align}
x^2 + y^2 & = a^2 \\
x^2 + z^2 & = b^2 \\
x^2 + c_1^2 & = d_1^2 \\
x^2 + c_2^2 & = d_2^2 \\
2y^2 + 2z^2 & = c_1^2 + c_2^2 \\
2y^2 + 2b^2 & = d_1^2 + d_2^2 \\
2a^2 + 2z^2 & = d_1^2 + d_2^2
\end{align}
These equations are from Wyss \cite{wyss3}.
The bi-orthogonal monoclinic Diophantine parallelepiped belongs to the family of {\it perfect parallelepipeds}. These
perfect parallelepipeds have all twelve edges rational, (there are only 3 distinct values), all six face diagonals rational,
and all four body diagonals rational. The family contains triclinic, biclinic, and monoclinic pipeds. The {\it perfect cuboid},
if it exists, would belong to this family. It is still unknown if this cuboid exists or not, see Guy \cite{guy}.
\section*{{\bf The $s$-parameters and the governing equation}}
We divide the integer or rational lengths of the piped by $x$ to obtain:
\begin{align}
1 + \left(\frac{y}{x}\right)^2 & = \left(\frac{a}{x}\right)^2
& \text{ let } \frac{y}{x} = u_1 \text{ \& } \frac{a}{x} = v_1 \text{ then } 1 + u_1^2 = v_1^2 \\
1 + \left(\frac{z}{x}\right)^2 & = \left(\frac{b}{x}\right)^2
& \text{ let } \frac{z}{x} = u_2 \text{ \& } \frac{b}{x} = v_2 \text{ then } 1 + u_2^2 = v_2^2 \\
1 + \left(\frac{c_1}{x}\right)^2 & = \left(\frac{d_1}{x}\right)^2
& \text{ let } \frac{c_1}{x} = u_3 \text{ \& } \frac{d_1}{x} = v_3 \text{ then } 1 + u_3^2 = v_3^2 \\
1 + \left(\frac{c_2}{x}\right)^2 & = \left(\frac{d_2}{x}\right)^2
& \text{ let } \frac{c_2}{x} = u_4 \text{ \& } \frac{d_2}{x} = v_4 \text{ then } 1 + u_4^2 = v_4^2
\end{align}
We have four rational equations for Pythagorean triangles.
\begin{equation}
1 + u_i^2 = v_i^2 , \quad i=1\dots 4
\end{equation}
which can be solved in parametric form:
\begin{equation}
u_k = \frac{1-s_k^2}{2s_k}, \quad v_k = \frac{1+s_k^2}{2s_k} \quad \text{ for } k=1 \dots 4
\end{equation}
for $s\in \mathbb{Q}$. We call the quadruple $s_{i=1\dots 4}$, the $s$-parameters $ = \left[s_1,s_2,s_3,s_4\right]$.
Wyss provides the derivation for equations (8-13) in his paper \cite{wyss3}.
If we substitute the $s$-parameters into equations (5-7), we obtain:
\begin{align}
2u_1^2 + 2u_2^2 & = u_3^2 + u_4^2 \\
2u_1^2 + 2v_2^2 & = v_3^2 + v_4^2 \\
2v_1^2 + 2u_2^2 & = v_3^2 + v_4^2
\end{align}
We next substitute $s_{i=1\dots 4}$ for the $u_{i=1\dots 4}$ parameters into equation (14) to derive the governing equation:
\begin{equation}
2 \left(\frac{1-s_1^2}{2s_1}\right)^2 + 2\left(\frac{1-s_2^2}{2s_2}\right)^2 = \left(\frac{1-s_3^2}{2s_3}\right)^2 + \left(\frac{1-s_4^2}{2s^4}\right)^2
\end{equation}
and simplify equation (17) to obtain:
\begin{align}
\label{eq:govern}
\begin{split}
& s_1^2s_2^2s_3^4s_4^2 + s_1^2s_2^2s_3^2s_4^4 - 2s_1^4s_2^2s_3^2s_4^2 - 2s_1^2s_2^4s_3^2s_4^2 + \\
& + 4s_1^2s_2^2s_3^2s_4^2 - 2s_1^2s_3^2s_4^2 - 2s_2^2s_3^2s_4^2 + s_1^2s_2^2s_3^2 + s_1^2s_2^2s_4^2 = 0
\end{split}
\end{align}
which is the governing equation for the $s$-parameters. This equation (\ref{eq:govern}) is given by Sharipov in \cite{shar}.
Ruslan Sharipov \cite{shar} points out that these parameters have to satisfy the following ten polynomial inequalities:
\begin{align}
\label{eq:satisfy}
\begin{split}
& s_1s_2^2s^3 + s_1^2s_2s_3 - s_1s_2s_3^2 + s_1s_2 - s_2s_3 - s_1s_3 < 0, \\
& s_1s_2^2s^4 + s_1^2s_2s_4 - s_1s_2s_4^2 + s_1s_2 - s_2s_4 - s_1s_4 < 0. \\
& 0 < s_i < 1 \; \text{ for } i=1,2,3,4
\end{split}
\end{align}
for rational Diophantine monoclinic pipeds to exist.
Thus we conclude: {\it Every rational Diophantine monoclinic piped corresponds to some
$\left[s_1,s_2,s_3,s_4\right]$ s-parameter satisfying equations (\ref{eq:govern},\ref{eq:satisfy})}.
\section*{{\bf Previous parameterizations of the perfect parallelepiped}}
About 2009, Jorge Sawyer and Clifford Reiter discovered a perfect parallelepiped \cite{saw}. Clifford Reiter and Jordan Tirrell
then released an additional series of papers on parameterizing them \cite{reit1,reit2}. A more recent paper by Sokolowshy,
VanHooft, Volkert, and Reiter has also appeared, which provides an infinite family of monoclinic Diophantine
pipeds \cite{soko}.
Walter Wyss was able to successfully parameterize the perfect parallelogram which has rational sides and two
rational diagonals \cite{wyss1,wyss2}, and he extended this to parameterize a family of bi-orthogonal monoclinic
pipeds \cite{wyss3}.
Ruslan Sharipov recast Wyss' parametrizations, as equations (6.24-6.27), and extended them to three additional mappings for
the bi-orthogonal monoclinic piped, in his equation (6.31), see Sharipov \cite{shar}.
Based upon Wyss's and Sharipov's works, the author recently discovered four parametrizations of the monoclinic piped \cite{rath}.
\section*{{\bf Two parameterizations of the specific $s$-parameter $\left[\frac{1}{2},s_2,s_3,s_4\right]$ }}
This author began a computer search for $s$-parameter solutions to equation (\ref{eq:govern}), which originally was
an $O^8$ type of search, very slow and inefficient. By studying the equation, some ways were found to drop the search space
to an $0^6$ or even an $O^4$ type of search which was much more efficient in discovering solutions.
In particular, it was decided to set $s_1=\frac{1}{2}$ and look for solutions.
After finding about a hundred or so solutions to $s=\left[\frac{1}{2},s_2,s_3,s_4\right]$, it was noticed that the solutions
seemed to occur in two patterns.
{\bf First pattern type $s=\left[\frac{1}{2},\frac{c}{a},\frac{c}{b},\frac{a}{b}\right]$ }
Substituting these $s$-parameters into equation (\ref{eq:govern}), we obtain:
\begin{align}
\frac{a^2 - 2b^2}{4b^6a^2} c^6 + \frac{2a^4 - 9b^2a^2 + 2b^4}{8b^6a^2}c^4 + \left(\frac{1}{4b^2} - \frac{1}{2b^4} a^2 \right)c^2 & = 0 \notag \\
\text{or} \qquad \qquad \qquad \qquad \qquad \notag \\
\label{eq:pat1}
\frac{c^2\left(2c^2-b^2+2a^2\right)\left(2b^2c^2-a^2c^2+2a^2b^2\right)}{8a^2b^6} & = 0
\end{align}
which has 3 solutions for $c$, found by setting each of the 3 factors of the numerator of equation(\ref{eq:pat1}) $=0$.
First solution for the first factor, $c^2 = 0 \;$:
\begin{equation}
c = 0
\end{equation}
This solution is trivial, and thus ignored.
Second solution for the second factor, $2c^2-b^2 + 2a^2 = 0 \;$:
\begin{align}
2c^2 - b^2 + 2a^2 & = 0 \notag \\
2c^2 & = b^2 - 2a^2 \notag \\
4c^2 & = 2b^2 - 4a^2 \notag \\
c & = \pm \frac{1}{2}\sqrt{2b^2-4a^2}
\end{align}
We want $2b^2 - 4a^2 = \square$ which has a general parametric solution in $m,n$.
\begin{align*}
\alpha^2 + \beta^2 & = 2\gamma^2 \\
\alpha & = 2mn+m^2-n^2 \\
\beta & = 2mn+n^2-m^2 \\
\gamma & = m^2 + n^2
\end{align*}
Adapting that parametrization, we obtain the following expression for the second factor
$\left(2c^2-b^2+2a^2\right) = 0$ as:
\begin{align}
& a = mn+\frac{1}{2}\left(m^2-n^2\right) \qquad
b = m^2 + n^2 \qquad
c = mn+\frac{1}{2}\left(n^2-m^2\right) \notag \\
& \text{\hspace{2.0in} or} \label{eq:second} \\
& a = mn+\frac{1}{2}\left(n^2-m^2\right) \qquad
b = m^2 + n^2 \qquad
c = mn+\frac{1}{2}\left(m^2-n^2\right) \notag
\end{align}
Notice the swap between $m,n$ for $a,c$.
Third solution for the third factor, $2b^2c^2 - a^2c^2 + 2a^2b^2 = 0 \;$:
\begin{align}
2b^2c^2 - a^2c^2 + 2a^2b^2 & = 0 \notag \\
c^2(2b^2-a^2) & = -2a^2b^2 \notag \\
c^2(a^2-2b^2) & = 2a^2b^2 \notag \\
c^2 & = \frac{2a^2+b^2}{a^2-2b^2} = \frac{4a^2b^2}{2a^2-4b^2} \notag \\
c & = \pm \frac{2ab}{\sqrt{2a^2-4b^2}}
\end{align}
Notice the similarity in the denominator to the second solution for $c$, except $a,b$ are swapped.
Using the same parametric solution for $2a^2-4b^2 = \square$, we find for the third factor, the
expression:
\begin{align}
& a = m^2 + n^2 \qquad
b = mn+\frac{1}{2}\left(m^2-n^2\right) \qquad
c = \frac{m^4+2nm^3+2n^3m-n^4}{-m^2+2nm+n^2} \notag \\
& \text{\hspace{2.0in} or} \label{eq:third} \\
& a = m^2 + n^2 \qquad
b = mn+\frac{1}{2}\left(n^2-m^2\right) \qquad
c = \frac{-m^4+2nm^3+2n^3m+n^4}{m^2+2nm-n^2} \notag
\end{align}
Collecting the four parametrized solutions, we obtain the $s$-parameter sets as:
\begin{align}
a_1 & = 2mn+m^2-n^2 \notag \\
b_1 & = 2(m^2+n^2) \notag \\
c_1 & = 2mn+n^2-m^2 \notag \\
s_1(m,n) & = \left[\frac{1}{2},\frac{2mn+n^2-m^2}{2mn+m^2-n^2},\frac{2mn+n^2-m^2}{2(m^2+n^2)},\frac{2mn+m^2-n^2}{2(m^2+n^2)}\right]
\end{align}
\begin{align}
a_2 & = 2mn+n^2-m^2 \notag \\
b_2 & = 2(m^2+n^2) \notag \\
c_2 & = 2mn+m^2-n^2 \notag \\
s_2(m,n) & = \left[\frac{1}{2},\frac{2mn+m^2-n^2}{2mn+n^2-m^2},\frac{2mn+m^2-n^2}{2(m^2+n^2)},\frac{2mn+n^2-m^2}{2(m^2+n^2)}\right]
\end{align}
\begin{align}
a_3 & = 2(n^2+2nm-m^2)(m^2+n^2) \notag \\
b_3 & = (n^2+2nm-m^2)(2mn+m^2-n^2) \notag \\
c_3 & = 2(m^4+2nm^3+2n^3m-n^4) \notag \\
s_3(m,n) & = \left[\frac{1}{2},\frac{m^2+2mn-n^2}{n^2+2nm-m^2},\frac{2(m^2+n^2)}{n^2+2nm-m^2},\frac{2(m^2+n^2)}{m^2+2mn-n^2}\right]
\end{align}
\begin{align}
a_4 & = 2(m^2+2nm-n^2)(m^2+n^2) \notag \\
b_4 & = (m^2+2nm-n^2)(2mn+n^2-m^2) \notag \\
c_4 & = 2(-m^4+2nm^3+2n^3m+n^4) \notag \\
s_4(m,n) & = \left[\frac{1}{2},\frac{m^2-2mn-n^2}{n^2-2nm-m^2},\frac{2(m^2+n^2)}{m^2+2mn-n^2},\frac{2(m^2+n^2)}{n^2+2nm-m^2}\right]
\end{align}
The parametric solutions are permutations of $s$-parameter sets, so only 1 is needed.
For example, using $(m,n)=(2,1)$:
\begin{align*}
s_1(2,1) & = [1/2,1/7,1/10,7/10] \\
s_2(2,1) & = [1/2,7,7/10,1/10] \\
s_3(2,1) & = [1/2,7,10,10/7] \\
s_4(2,1) & = [1/2,1/7,10/7,10]
\end{align*}
We choose $s_1(m,n)$ as the best parametrization for our purposes.
In summary, for the first pattern type, $s=\left[\frac{1}{2},\frac{c}{a},\frac{c}{b},\frac{a}{b}\right]$,
we can choose for integers $m,n \in \mathbb{Z}$, the following rational solution for the $s$-parameter:
\begin{equation}
s(m,n) = \left[\frac{1}{2},\frac{2mn+n^2-m^2}{2mn+m^2-n^2},\frac{2mn+n^2-m^2}{2(m^2+n^2)},\frac{2mn+m^2-n^2}{2(m^2+n^2)}\right]
\label{eq:sintg}
\end{equation}
For a rational parameter $q \in \mathbb{Q}$, we substitute into equation (\ref{eq:sintg}), $m=q$, $n=1$ and obtain:
\begin{equation}
s(q) = \left[\frac{1}{2}, \frac{-q^2 + 2q + 1}{q^2 + 2q - 1}, \frac{-q^2 + 2q + 1}{2q^2 + 2}, \frac{q^2 + 2q - 1}{2q^2 + 2}\right]
\label{eq:srat}
\end{equation}
The results from $s(q)$ match the $s$-parameters recoverable from the parameterization of equations (8-11) in this author's paper \cite{rath}, after using the $x$, $y$, $z$, $c_1$, and $c_2$ values to recover the corresponding $s_i$ term.
{\bf Second pattern type $s=\left[\frac{1}{2},\frac{d}{b},\frac{d}{a},\frac{d}{c}\right]$ }
A laborious computer search found the following solutions for $s_1=\frac{1}{2}$, in table \ref{table:search}. which match the second pattern.
\begin{longtable}{lllc}
\hspace{0.5cm} s-series & $b,a,c$ & $d$ & $q$ \\
\toprule
$\left[\frac{1}{2},\frac{16}{7},\frac{16}{5},\frac{16}{35}\right]$ & 7,5,35 & 16 & $\frac{1}{3}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{80}{119},\frac{80}{91},\frac{80}{221}\right]$ & 119,91,221 & 80 & $\frac{1}{5}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{160}{161},\frac{160}{119},\frac{160}{391}\right]$ & 161,119,391 & 160 & $\frac{1}{4}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{224}{527},\frac{224}{425},\frac{224}{775}\right]$ & 527,425,775 & 224 & $\frac{1}{7}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{480}{1519},\frac{480}{1271},\frac{480}{2009}\right]$ & 1519,1271,2009 & 480 & $\frac{1}{9}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{560}{41},\frac{560}{29},\frac{560}{1189}\right]$ & 41,29,1189 & 560 & $\frac{2}{5}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{560}{1081},\frac{560}{851},\frac{560}{1739}\right]$ & 1081,851,1739 & 560 & $\frac{1}{6}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{880}{3479},\frac{880}{2989},\frac{880}{4331}\right]$ & 3479,2989,4331 & 880 & $\frac{1}{11}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{1344}{3713},\frac{1344}{3055},\frac{1344}{5135}\right]$ & 3713,3055,5135 & 1344 & $\frac{1}{8}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{1456}{6887},\frac{1456}{6035},\frac{1456}{8245}\right]$ & 6887,6035,8245 & 1456 & $\frac{1}{13}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{1680}{1241},\frac{1680}{901},\frac{1680}{3869}\right]$ & 1241,901,3869 & 1680 & $\frac{2}{7}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{2240}{12319},\frac{2240}{10961},\frac{2240}{14351}\right]$ & 12319,10961,14351 & 2240 & $\frac{1}{15}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{2464}{2047},\frac{2464}{1495},\frac{2464}{5785}\right]$ & 2047,1495,5785 & 2464 & $\frac{3}{11}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{2640}{9401},\frac{2640}{7979},\frac{2640}{12019}\right]$ & 9401,7979,12019 & 2640 & $\frac{1}{10}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{3264}{20447},\frac{3264}{18415},\frac{3264}{23345}\right]$ & 20447,18415,23345 & 3264 & $\frac{1}{17}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{3520}{721},\frac{3520}{511},\frac{3520}{7519}\right]$ & 721,511,7519 & 3520 & $\frac{3}{8}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{3696}{4633},\frac{3696}{3485},\frac{3696}{9605}\right]$ & 4633,3485,9605 & 3696 & $\frac{2}{9}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{4160}{4879},\frac{4160}{3649},\frac{4160}{10591}\right]$ & 4879,3649,10591 & 4160 & $\frac{3}{13}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{6240}{959},\frac{6240}{679},\frac{6240}{13289}\right]$ & 959,679,13289 & 6240 & $\frac{5}{13}$ \Tstrut\Bstrut \\
$\left[\frac{1}{2},\frac{7280}{4681},\frac{7280}{3379},\frac{7280}{16459}\right]$ & 4681,3379,16459 & 7280 & $\frac{3}{10}$ \Tstrut\Bstrut \\
\bottomrule
\\[-0.5em]
\caption{Computer search results matching $s=\left[\frac{1}{2},\frac{d}{b},\frac{d}{a},\frac{d}{c}\right]$ }
\label{table:search}
\end{longtable}
\vspace{-2.5em}
In the table, $q$ is the variable in equation (\ref{eq:srat}). We also notice from the table that d is divisible by 16.
If we substitute $\left[\frac{1}{2},\frac{d}{b},\frac{d}{a},\frac{d}{c}\right]$ into equation (\ref{eq:govern}), we obtain
\begin{equation*}
\frac{d^4(2b^2c^2d^4-4a^2c^2d^4+2a^2b^2d^4-9a^2b^2c^2d^2+2a^2b^2c^4-4a^2b^4c^2+2a^4b^2c^2)}{8a^4b^4c^4} = 0
\end{equation*}
which is the relationship between $a,b,c,d$. We can solve for $d$ in terms of $a$,$b$,$c$, but an homogeneous sextic equation in $a$,$b$,$c$ results.
Without providing the derivation, the requirements for equation (\ref{eq:govern}) are met by the following two solution sets, except
that the reciprocals, $\frac{b}{d}$, $\frac{a}{d}$, $\frac{c}{d}$, are used instead, but the results are equivalent:
For $m,n \in \mathbb{Z}$:
\begin{equation}
s_{\mathbb{Z}} (m,n) = \left[\frac{1}{2}, \frac{-3m^4 + 18n^2m^2 - 3n^4}{8nm^3 - 8n^3m}, \frac{3m^4 + 6nm^3 + 6n^3m - 3n^4}{8nm^3 - 8n^3m}, \frac{3m^4 - 6nm^3 - 6n^3m - 3n^4}{8nm^3 - 8n^3m}\right]
\end{equation}
For $q \in \mathbb{Q}$:
\begin{equation}
s_{\mathbb{Q}} (q) = \left[\frac{1}{2},\frac{-3q^4+18q^2-3}{8q^3-8q},\frac{3q^4+6q^3+6q-3}{8q^3-8q},\frac{3q^4-6q^3-6q-3}{8q^3-8q}\right]
\end{equation}
Please note that from the first set, that $d = 8nm^3-8n^3m = 8mn(m-n)(m+n)$ is divisible by 16 as previously noted, because $mn(m-n)(m+n)$ is always divisible by 2,
for any $m,n \in \mathbb{Z}$.
\section*{{\bf A new parameterization for the $s$-parameter $\left[s_1,s_2,s_3,s_4\right]$ }}
After carefully examining over 2,000 $s$-parameters found by computer search, it was noticed that a particular $s$-parameter solution $\left[s_1,s_2,1,s_4\right]$
occurred from time to time. A parametric solution was found, which satisfied equation (\ref{eq:govern}), however none of these sets created a rational
Diophantine monoclinic piped, due to the fact that $s_3 = 1$ always leads to a degenerate solution.
After experimenting with these types of sets, a new solution for $s_1 = \frac{r}{s}$ was discovered.
A parametric solution in $\mathbb{Z}$:
\begin{align}
S_{\mathbb{Z}} (r,s,m,n) & = \left[s_1,s_2,s_3,s_4\right] \text{ for } r,s,m,n \in \mathbb{Z} \text{ where} \label{eq:sZ} \\
s_1 & = \frac{r}{s} \notag \\
s_2 & = \frac{(r^2-s^2)m^4+(-6r^2+6s^2)n^2m^2+(r^2-s^2)n^4}{4srnm^3-4srn^3m} \notag \\
s_3 & = \frac{(-r^2+s^2)m^4+(-2r^2+2s^2)nm^3+(-2r^2+2s^2)n^3m+(r^2-s^2)n^4}{4srnm^3-4srn^3m} \notag \\
s_4 & = \frac{(r^2-s^2)m^4+(-2r^2+2s^2)nm^3+(-2r^2+2s^2)n^3m+(-r^2+s^2)n^4}{4srn^3m-4srnm^3} \notag \\
T_{\mathbb{Z}} (r,s,m,n) & = \frac{(-r^2+s^2)m^2+(-r^2+s^2)n^2}{srm^2-srn^2} = s_3 - s_4
\end{align}
where $T_{\mathbb{Z}} (r,s,m,n) = s_3-s_4$ which is a measure of how close the monoclinic piped approaches a rectangular cuboid where $s_3 = s_4$.
A parametric solution in $\mathbb{Q}$:
\begin{align}
S_{\mathbb{Q}} (s,r) & = \left[s_1,s_2,s_3,s_4\right] \text{ for } s,r \in \mathbb{Q} \text{ where} \label{eq:sQ} \\
s_1 & = s \notag \\
s_2 & = \frac{(s^2-1)r^4+(-6s^2+6)r^2+(s^2-1)}{4sr^3-4sr} \notag \\
s_3 & = \frac{(-s^2+1)r^4+(-2s^2+2)r^3+(-2s^2+2)r+(s^2-1)}{4sr^3-4sr} \notag \\
s_4 & = \frac{(s^2-1)r^4+(-2s^2+2)r^3+(-2s^2+2)r+(-s^2+1)}{4sr-4sr^3} \notag \\
T_{\mathbb{Q}} (s,r) & = \frac{(-s^2+1)r^2+(-s^2+1)}{sr^2-s} = s_3 - s_4
\end{align}
and similarly, $T_{\mathbb{Q}} (s,r) = s_3 - s_4$, which is a measure of how close the monoclinic piped approaches
the rectangular cuboid.
\section*{{\bf Asymptotic Sequences for pipeds approaching the perfect cuboid }}
After discovering the parameterization, equation (\ref{eq:sQ}), which satisfies the second pattern type, the computer
was programmed to create 100's of millions of $s$-parameters sets for $S_{\mathbb{Q}}(s,r) = \left[s_1,s_2,s_3,s_4\right]$
for $s=\frac{m}{n}$ a proper fraction, $m<n<=500$, and 202,861 unique values of $r$.
A batch file was created, reduced, and sorted. 121,251,195 $s$-parameter sets were obtained.
It was decided to plot these $s$-parameter points, using the $x = s_3-s_4$ as the coordinate and $y = \text{numerator}(s_2)$
as the ordinate. The idea was to discover if perhaps the points had some type of pattern.
Interestingly, after setting both the x-axis and y-axis to a log scale, the points do follow lines, strongly hinting that
they are found on rational polynomial fraction expressions. See Figure \ref{fig:oneeighthraw}.
\begin{figure}[!ht]
\centering
\includegraphics[width=\textwidth]{plot8raw.png}
\caption{Raw points from the parametric solution for $s_1 = \frac{1}{8}$}
\label{fig:oneeighthraw}
\end{figure}
This is indeed the case, as uncovering the solutions for $r=\frac{1}{8}$ shows in Figure \ref{fig:oneeighthparm}.
\begin{figure}[!ht]
\centering
\includegraphics[width=\textwidth]{plot8param.png}
\caption{Parameterized points from the parametric solution for $s_1 = \frac{1}{8}$}
\label{fig:oneeighthparm}
\end{figure}
As another example, for $s_1 = \frac{1}{18}$, in Figure \ref{fig:oneeighteenth}, we can recognize what is happening.
\begin{figure}[!ht]
\centering
\includegraphics[width=\textwidth]{plot18.png}
\caption{Raw and parameterized points from the parametric solution for $s_1 = \frac{1}{18}$}
\label{fig:oneeighteenth}
\end{figure}
These plots contain rational polynomial fraction expressions for the points on a line. Careful point fitting revealed
that the coordinates of the points were from rational polynomial fractions with a 3rd degree numerator and 4th degree denominator.
The plots enable the author to identify asymptotic sequences, for the monoclinic piped, where $s_3 \approxeq s_4$.
This means that the piped is approaching the {\it perfect cuboid}, either from an obtuse angle $> 90^\degree$ or
an acute angle $< 90^\degree$ as the angle $\to 90^\degree$.
\newpage
{\bf Obtuse angle asymptotic sequence}
The first discovered, because it was the easiest to recognize, was the obtuse angle asymptotic sequence where the
obtuse angle $\to 90^\degree$. In these equations, $n \in \mathbb{Z^+}$:
\begin{align}
s_{ob}(n) & = \left[\frac{n-1}{n},\frac{4n^4-8n^3+4n-1}{4n^4-8n^3+4n^2},\frac{4n^4-12n^3+12n^2-6n+1}{4n^4-8n^3+4n^2},\frac{4n^4-8n^3+4n^2}{4n^4-4n^3+2n-1}\right] \\
t_{ob}(n) & = \frac{-(2n-1)^4}{4(n-1)^2n^2(2n^2-1)(2n^2-2n+1)}
\end{align}
{\bf Acute angle asymptotic sequences}
The acute angle asymptotic sequence initially defied recognition, and it became necessary to gradually build up the bounding
sequences depending upon $s1 = \{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots \}$ in hopes of obtaining
the sequence.
These asymptotic sequences were eventually identified:
\begin{align}
s_{\frac{1}{2}}(n) & = \left[\frac{1}{2},\frac{8n(n+1)(n+2)}{3(n^2-2)(n^2+4n+2)},\frac{8n(n+1)(n+2)}{3(n^2-2)(n^2+2n+2)},\frac{8n(n+1)(n+2)}{3(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{2}}(n) & = \frac{32n(n+1)^2(n+2)}{3(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{3}}(n) & = \left[\frac{1}{3},\frac{3n(n+1)(n+2)}{2(n^2-2)(n^2+4n+2)},\frac{3n(n+1)(n+2)}{2(n^2-2)(n^2+2n+2)},\frac{3n(n+1)(n+2)}{2(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{3}}(n) & = \frac{6n(n+1)^2(n+2)}{(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{4}}(n) & = \left[\frac{1}{4},\frac{16n(n+1)(n+2)}{15(n^2-2)(n^2+4n+2)},\frac{16n(n+1)(n+2)}{15(n^2-2)(n^2+2n+2)},\frac{16n(n+1)(n+2)}{15(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{4}}(n) & = \frac{64n(n+1)^2(n+2)}{15(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{5}}(n) & = \left[\frac{1}{5},\frac{5n(n+1)(n+2)}{6(n^2-2)(n^2+4n+2)},\frac{5n(n+1)(n+2)}{6(n^2-2)(n^2+2n+2)},\frac{5n(n+1)(n+2)}{6(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{5}}(n) & = \frac{10n(n+1)^2(n+2)}{3(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{6}}(n) & = \left[\frac{1}{6},\frac{24n(n+1)(n+2)}{35(n^2-2)(n^2+4n+2)},\frac{24n(n+1)(n+2)}{35(n^2-2)(n^2+2n+2)},\frac{24n(n+1)(n+2)}{35(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{6}}(n) & = \frac{96n(n+1)^2(n+2)}{35(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{7}}(n) & = \left[\frac{1}{7},\frac{7n(n+1)(n+2)}{12(n^2-2)(n^2+4n+2)},\frac{7n(n+1)(n+2)}{12(n^2-2)(n^2+2n+2)},\frac{7n(n+1)(n+2)}{12(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{7}}(n) & = \frac{7n(n+1)^2(n+2)}{3(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{8}}(n) & = \left[\frac{1}{8},\frac{32n(n+1)(n+2)}{63(n^2-2)(n^2+4n+2)},\frac{32n(n+1)(n+2)}{63(n^2-2)(n^2+2n+2)},\frac{32n(n+1)(n+2)}{63(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{8}}(n) & = \frac{128n(n+1)^2(n+2)}{63(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{9}}(n) & = \left[\frac{1}{9},\frac{9n(n+1)(n+2)}{20(n^2-2)(n^2+4n+2)},\frac{9n(n+1)(n+2)}{20(n^2-2)(n^2+2n+2)},\frac{9n(n+1)(n+2)}{20(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{9}}(n) & = \frac{9n(n+1)^2(n+2)}{5(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{10}}(n) & = \left[\frac{1}{10},\frac{40n(n+1)(n+2)}{99(n^2-2)(n^2+4n+2)},\frac{40n(n+1)(n+2)}{99(n^2-2)(n^2+2n+2)},\frac{40n(n+1)(n+2)}{99(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{10}}(n) & = \frac{160n(n+1)^2(n+2)}{99(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{11}}(n) & = \left[\frac{1}{11},\frac{11n(n+1)(n+2)}{30(n^2-2)(n^2+4n+2)},\frac{11n(n+1)(n+2)}{30(n^2-2)(n^2+2n+2)},\frac{11n(n+1)(n+2)}{30(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{11}}(n) & = \frac{22n(n+1)^2(n+2)}{15(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{12}}(n) & = \left[\frac{1}{12},\frac{48n(n+1)(n+2)}{143(n^2-2)(n^2+4n+2)},\frac{48n(n+1)(n+2)}{143(n^2-2)(n^2+2n+2)},\frac{48n(n+1)(n+2)}{143(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{12}}(n) & = \frac{192n(n+1)^2(n+2)}{143(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{13}}(n) & = \left[\frac{1}{13},\frac{13n(n+1)(n+2)}{42(n^2-2)(n^2+4n+2)},\frac{13n(n+1)(n+2)}{42(n^2-2)(n^2+2n+2)},\frac{13n(n+1)(n+2)}{42(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{13}}(n) & = \frac{26n(n+1)^2(n+2)}{21(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{14}}(n) & = \left[\frac{1}{14},\frac{56n(n+1)(n+2)}{195(n^2-2)(n^2+4n+2)},\frac{56n(n+1)(n+2)}{195(n^2-2)(n^2+2n+2)},\frac{56n(n+1)(n+2)}{195(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{14}}(n) & = \frac{224n(n+1)^2(n+2)}{195(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{15}}(n) & = \left[\frac{1}{15},\frac{15n(n+1)(n+2)}{56(n^2-2)(n^2+4n+2)},\frac{15n(n+1)(n+2)}{56(n^2-2)(n^2+2n+2)},\frac{15n(n+1)(n+2)}{56(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{15}}(n) & = \frac{15n(n+1)^2(n+2)}{14(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{16}}(n) & = \left[\frac{1}{16},\frac{64n(n+1)(n+2)}{255(n^2-2)(n^2+4n+2)},\frac{64n(n+1)(n+2)}{255(n^2-2)(n^2+2n+2)},\frac{64n(n+1)(n+2)}{255(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{16}}(n) & = \frac{256n(n+1)^2(n+2)}{255(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{17}}(n) & = \left[\frac{1}{17},\frac{17n(n+1)(n+2)}{72(n^2-2)(n^2+4n+2)},\frac{17n(n+1)(n+2)}{72(n^2-2)(n^2+2n+2)},\frac{17n(n+1)(n+2)}{72(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{17}}(n) & = \frac{17n(n+1)^2(n+2)}{18(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{18}}(n) & = \left[\frac{1}{18},\frac{72n(n+1)(n+2)}{323(n^2-2)(n^2+4n+2)},\frac{72n(n+1)(n+2)}{323(n^2-2)(n^2+2n+2)},\frac{72n(n+1)(n+2)}{323(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{18}}(n) & = \frac{288n(n+1)^2(n+2)}{323(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
\begin{align}
s_{\frac{1}{19}}(n) & = \left[\frac{1}{19},\frac{19n(n+1)(n+2)}{90(n^2-2)(n^2+4n+2)},\frac{19n(n+1)(n+2)}{90(n^2-2)(n^2+2n+2)},\frac{19n(n+1)(n+2)}{90(n^2+2n+2)(n^2+4n+2)}\right] \\
t_{\frac{1}{19}}(n) & = \frac{38n(n+1)^2(n+2)}{45(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
After obtaining these 18 sequences, Maxima was used to derive the coefficients where $d\to\infty$.
{\bf A two parameter acute angle asymptotic sequence}
A solution for the asymptotic sequence where the acute angle $\to 90^\degree$ contains two
parameters for the input, $d,n \in \mathbb{Z^+}$, where $d$ is the value of the denominator in the $s_1 = \frac{1}{d}$ term,
and $n$ is the sequence number in that particular $s_1$ sequence.
This is not the same as the sequence for the obtuse angle $\to 90^\degree$ as it only takes 1 parameter for input.
The solution for the two parameter acute angle asymptotic sequence is:
\begin{align}
\begin{split}
s_{\frac{1}{d}}(d,n) & = \bigg [ \frac{1}{d}, \frac{4dn(n+1)(n+2)}{(d-1)(d+1)(n^2-2)(n^2+4n+2)}, \\
& \qquad \quad \frac{4dn(n+1)(n+2)}{(d-1)(d+1)(n^2-2)(n^2+2n+2)}, \frac{4dn(n+1)(n+2)}{(d-1)(d+1)(n^2+2n+2)(n^2+4n+2)} \bigg ]
\end{split} \\
t_{\frac{1}{d}}(d,n) & = \frac{16dn(n+1)^2(n+2)}{(d-1)(d+1)(n^2-2)(n^2+2n+2)(n^2+4n+2)}
\end{align}
where $d,n \in \mathbb{Z^+}$.
\section*{{\bf Some concluding comments on the $s$-parameters and their pipeds }}
Initially the computer was programmed to do raw brute force searches for $s$-parameter sets for equation (\ref{eq:govern})
and found 3,280 solutions.
After discovering the parameter solution, equation (\ref{eq:sZ}) or equation(\ref{eq:sQ}), the computer created
a large batch of solutions to see how many of the raw solutions were included in this set.
It was discovered that only 16 solutions do not appear in the parameterization, thus the solution covered 99.5\% of
the raw solutions, which is quite remarkable.
These 16 $s$-parameter solutions are
\begin{longtable}{l}
$\left[ s_1, s_2, s_3, s_4 \right]$ \\
\midrule
$\left[\frac{3}{10},\frac{4}{15},\frac{4}{5},\frac{3}{20}\right]$ \Tstrut\Bstrut \\
$\left[\frac{42}{55},\frac{35}{132},\frac{20}{77},\frac{35}{132}\right]$ \Tstrut\Bstrut \\
$\left[\frac{41}{65},\frac{455}{943},\frac{13}{35},\frac{23}{41}\right]$ \Tstrut\Bstrut \\
$\left[\frac{143}{217},\frac{403}{616},\frac{8}{11},\frac{104}{217}\right]$ \Tstrut\Bstrut \\
$\left[\frac{68}{401},\frac{401}{721},\frac{2807}{4964},\frac{4964}{41303}\right]$ \Tstrut\Bstrut \\
$\left[\frac{348}{401},\frac{401}{527},\frac{6817}{8700},\frac{8700}{12431}\right]$ \Tstrut\Bstrut \\
$\left[\frac{314}{415},\frac{415}{527},\frac{1411}{1570},\frac{1570}{2573}\right]$ \Tstrut\Bstrut \\
$\left[\frac{188}{433},\frac{433}{623},\frac{964}{3031},\frac{38537}{45308}\right]$ \Tstrut\Bstrut \\
$\left[\frac{253}{439},\frac{439}{527},\frac{6325}{7463},\frac{6325}{13609}\right]$ \Tstrut\Bstrut \\
$\left[\frac{205}{457},\frac{457}{527},\frac{5125}{7769},\frac{5125}{14167}\right]$ \Tstrut\Bstrut \\
$\left[\frac{294}{473},\frac{473}{697},\frac{3738}{8041},\frac{19393}{26166}\right]$ \Tstrut\Bstrut \\
$\left[\frac{55}{479},\frac{479}{721},\frac{3353}{4015},\frac{4015}{49337}\right]$ \Tstrut\Bstrut \\
$\left[\frac{138}{481},\frac{481}{527},\frac{3450}{8177},\frac{3450}{14911}\right]$ \Tstrut\Bstrut \\
$\left[\frac{341}{661},\frac{661}{791},\frac{1903}{4627},\frac{58993}{74693}\right]$ \Tstrut\Bstrut \\
$\left[\frac{38}{751},\frac{751}{959},\frac{3686}{5257},\frac{3686}{102887}\right]$ \Tstrut\Bstrut \\
$\left[\frac{62}{769},\frac{769}{791},\frac{346}{5383},\frac{10726}{86897}\right]$ \Tstrut\Bstrut \\
\bottomrule
\\[-0.5em]
\caption{Raw search results not matching any parametrization.}
\label{table:anomalous}
\end{longtable}
\vspace{-1.0cm}
It would be interesting to come up with some new rational polynomial fraction expressions for these $s$-parameters,
as the 5th and following terms listed above seem to follow the same type of pattern, $s1,s2 = \frac{a}{b},\frac{b}{c}$.
As a general note, none of the rational polynomial fractions in this paper, for $s_3$, and $s_4$ can be equated in
an attempt to find the {\it perfect cuboid} solution, as doing so caused the parameter $\frac{r}{s}$ or $r$ to
go to 1, giving a degenerate solution. Thus no perfect cuboids exist as a solution of any of these expressions.
This author also checked all the monoclinic pipeds for rational area or volume, from the parameterizations, none had
any face parallelograms with rational area, nor did any have a rational volume. None of the Diophantine monoclinic
piped parametrizations have the vertices embedded in the rational lattice. This is in regard to two questions from
Sawyer and Reiter \cite{saw}.
\noindent \rule[0pt]{3.0in}{0.4pt} | 207,364 |
Kevin Costley
Turkey Talk (Digital Download)D-W9349
Regular price $2.95! Instant download and print. | 359,981 |
Umgullian Racing CommissionTalk0
112,150pages on
this wiki
this wiki
The Umgullian Racing Commission monitored the sport of blob racing. In 22 BBY, they announced the findings of a tainted blob and addressed the authorities. The cheater was subsequently convicted and killed two hours later.
AppearancesEdit
-)}"> Blob Racing Cheater Killed—HoloNet News Vol. 531 48 (content now unavailable; backup links 1 2 on Archive.org)
- Jedi Search (First mentioned) | 85,830 |
February has started and with it comes one of the favorite holidays for many, Valentine’s day. After so many parties, sometimes it’s hard to find the perfect gift for your partner.
But this year will be different. We have selected some products to surprise your soul mate.
Let’s go for it!.
Phone cases for couples Aire – Yeyei Gómez
A proposal that also works very well, the phone cases for couples. This is a special design, designed for you to share with your partner. Yeyei illustrated four different options for you to choose the one you like:
You can see them all here
Phone case for couple Anatomía del Corazón – Yeyei Gómez
If you want an audacious design, will not fins a better than this. Illustrated by Yeyei, you and your partener will be the coolest of the year.
Phone cases for couple Abrazo Pájaro – Kira Diez
We have showed you before, but this design is also available in phone cases. What better way to share the love with your partner?
We hope you you have become inspired with our gift ideas! :) we wish you a very special Valentine’s day! | 152,921 |
Arsenal to move for Bayer Leverkusen striker Kevin Volland?
Arsenal have reportedly “explored” the possibility of signing Bayer Leverkusen striker Kevin Volland in the January transfer window, according to Sky Sports News.
The Gunners are considering a swoop to sign the 27-year-old striker following Mikel Arteta’s appointment as Arsenal manager.
The north London outfit have been scouting the Leverkusen striker over the past couple of months as a potential option at the beginning of 2020.
The report goes on to claim that the German attacker has 18 months left to run on his contract to fuel speculation surrounding his future.
He has contributed five goals and six assists in 17 appearances in the German top flight this season to highlight his credentials.
Arteta is all set to take charge of his first Arsenal game in their next Premier League fixture against Bournemouth on Boxing Day. | 382,289 |
TITLE: Is the Galois group of a given polynomial always a subgroup of the Klein-$4$ group?
QUESTION [2 upvotes]: Let $f(x) = (x^2-ax+b)(x^2-cx+d)$ be a separable polynomial with rational coefficients.
Is it true, that its Galois group over the rationals is always a subgroup of the Klein four group $C_2 \times C_2$?
REPLY [2 votes]: Let $K_1,K_2$ be Galois extension of given factors of $f(x)$ over $\mathbb{Q}$. They are at most degree $2$. (See what happens if one of them is of degree $1$; i.t. $\mathbb{Q}$).
If both are degree $2$ extensions, consider their join in some algebraic closure (it is degree $\leq 4$ extension). Thus the Galois extension of $f(x)$ is inside this extension $K_1K_2$.
If $K_1=K_2$ then $K_1K_2=K_1$ is the Galois extension of $f(x)$ of degree $2$.
If $K_1\neq K_2$, then the Galois extension $K_1K_2$ of $f(x)$ has (at least) two subfields; correspondingly, Galois group (of order $4$) should contain two (at least) subgroups; what it should be? | 44,793 |
\begin{document}
\begin{frontmatter}
\title{Compactly Supported Shearlets are\\ Optimally Sparse}
\author[GK]{Gitta Kutyniok\corref{cor1}\fnref{fn1}}
\author[GK]{Wang-Q Lim\fnref{fn1}}
\address[GK]{Institute of Mathematics, University of Osnabr\"uck, 49069 Osnabr\"uck, Germany}
\cortext[cor1]{Corresponding author}
\fntext[fn1]{G.K. and W.-Q L. would like to thank Wolfgang Dahmen,
David Donoho, Chunyan Huang, Demetrio Labate, Christoph Schwab, and
Gerrit Welper for various discussions on related topics. G.K. and
W.-Q L. acknowledge support from DFG Grant SPP-1324, KU 1446/13. G.K. was
also partially supported by DFG Grant, KU 1446/14.}
\begin{abstract}
Cartoon-like images, i.e., $C^2$ functions which are smooth apart from
a $C^2$ discontinuity curve, have by now become a standard model for measuring sparse (non-linear)
approximation properties of directional representation systems. It was already shown that
curvelets, contourlets, as well as shearlets do exhibit (almost) optimally sparse approximations
within this model. However, all those results are only applicable to band-limited
generators, whereas, in particular, spatially compactly supported generators are of
uttermost importance for applications.
In this paper, we now present the first complete proof of (almost) optimally sparse approximations
of cartoon-like images by using a particular class of directional representation systems, which
indeed consists of compactly supported elements. This class will be chosen as a subset of
shearlet frames -- not necessarily required to be tight -- with shearlet generators having compact
support and satisfying some weak moment conditions.
\end{abstract}
\begin{keyword}
Curvilinear discontinuities, edges, nonlinear approximation, optimal sparsity, shearlets, thresholding, wavelets
\end{keyword}
\end{frontmatter}
\section{Introduction}
{In computer vision, edges were detected as those features} governing an image
while separating smooth regions in between. About 10 years ago, mathematicians
started to design models of images incorporating those findings aiming at
designing representation systems which -- in such a model -- are capable of resolving
edges in an optimally sparse way. However, customarily, at that time an image
was viewed as an element of a compact subset of $L_p$ characterized by a given
Besov regularity with the Kolmogorov entropy of such sets identifying lower
bounds for the distortion rates of encoding-decoding pairs in this model.
Although wavelets could be shown to behave optimally \cite{CDDD} as an
encoding methodology, Besov models are clearly deficient since edges are not
adequately captured. This initiated the introduction of a different model,
called cartoon-like model (see \cite{Don99,WRHB02,CD04}), which revealed the
suboptimal treatment of edges by wavelets.
The introduction of tight curvelet frames in 2004 by Cand\'{e}s and Donoho
\cite{CD04}, which provably provide (almost) optimally sparse approximations
within such a cartoon-like model can be considered a milestone in applied harmonic
analysis. One year later, contourlets were introduced by Do and Vetterli \cite{DV05}
which similarly derived (almost) optimal approximation rates. In the same year,
{\em shearlets} were developed by Labate, Weiss, and the authors in \cite{LLKW05} as
the first directional representation system with allows a unified treatment
of the continuum and digital world similar to wavelets, while also providing
(almost) optimally sparse approximations within such a cartoon-like model \cite{GL07}.
In most applications, spatial localization of the analyzing elements of an
encoding system is of
uttermost importance both for a precise detection of geometric features as well as for a fast
decomposition algorithm. However, none of the previously mentioned results
cover this situation. In fact, the proofs which were provided do by no means
extend to this crucial setting.
\medskip
In this paper, we now present the first complete proof of (almost) optimally sparse approximations
of cartoon-like images by using a particular class of directional representation systems, which
indeed consist of compactly supported elements. This class will be chosen as a subset of
shearlet frames -- not necessarily required to be tight -- with shearlet generators having compact
support and satisfying some weak moment conditions. Interestingly, our proof is very different from
all previous ones caused by the extensive exploration of the compact support of
the shearlet generators and the lack of directional vanishing moments.
\subsection{A Suitable Model for Images: Cartoon-Like Images}
Intuitively, cartoons are smooth image parts separated from other areas by an edge.
After a series of initial models \cite{Don99,WRHB02}, the first complete model of
cartoons has been introduced in \cite{CD04}, and this is what we intend to use also
here. The basic idea is to choose a closed boundary curve and then fill the interior
and exterior part with $C^2$ functions (see Figure \ref{fig:Cartoon}).
\begin{figure}[h]
\begin{center}
\vspace*{0.3cm}
\includegraphics[height=1.25in]{./figs/Cartoon}
\vspace*{-0.3cm}
\end{center}
\caption{Example of a cartoon-like image.}
\label{figure}
\label{fig:Cartoon}
\end{figure}
Let us now be more precise, and introduce $STAR^2(\nu)$, a class of indicator functions of
sets $B$ with $C^2$ boundaries $\partial B$ and curvature bounded by $\nu$, as well as
$\cE^2(\nu)$, a class of cartoon-like images. For this, in polar coordinates, we let $\rho(\theta)
\rightarrow [0,1]$ be a radius function and define the set $B$ by
\[
B = \{x \in \RR^2 : |x| \le \rho(\theta), x = (|x|,\theta) \mbox{ in polar coordinates}\}.
\]
In particular, the boundary $\partial B$ of $B$ is given by the curve
\begin{equation}\label{eq:curve}
\beta(\theta) = \begin{pmatrix} \rho(\theta)\cos(\theta) \\
\rho(\theta)\sin(\theta)\end{pmatrix},
\end{equation}
and the class of boundaries of interest to us are defined by
\begin{equation}\label{eq:curvebound}
\sup|\rho^{''}(\theta)| \leq \nu, \quad \rho \leq \rho_0 < 1.
\end{equation}
The following definition now introduces the notions $STAR^2(\nu)$ and $\cE^2(\nu)$ from \cite{CD04}.
\begin{definition}
For $\nu > 0$, the set $STAR^2(\nu)$ is defined to be the set of all $B \subset [0,1]^2$ such that $B$
is a translate of a set obeying \eqref{eq:curve} and \eqref{eq:curvebound}. Further, $\cE^2(\nu)$
denotes the set of functions $f$ on $\RR^2$ with compact support in $[0,1]^2$ of the form
\[
f = f_0 + f_1 \chi_{B},
\]
where $f_0,f_1 \in C^2(\RR^2)$ with compact support in $[0,1]^2$, $B \in STAR^2(\nu)$, and $\|f\|_{C^2} = \sum_{|\alpha| \leq 2}
\|D^{\alpha}f\|_{\infty} \leq 1.$
\end{definition}
\subsection{Optimal Sparsity of a Directional Representation System}
\label{subsec:optsparse}
The `quality' of the performance of a (directional) representation system with respect to
cartoon-like images is typically measured by taking a non-linear approximation viewpoint.
More precisely, given a cartoon-like image $f \in \cE^2(\nu)$ and a (directional)
representation system $(\sigma_i)_{i \in I}$ which forms an orthonormal basis, the chosen measure
is the asymptotic behavior of the best $N$-term (non-linear) approximation error in $L^2$ norm
in the number of terms $N$, i.e.,
\[
\norm{f-f_N}_2^2 = \Big\|f - \sum_{i \in I_N} \ip{f}{\sigma_i}\sigma_i\Big\|_2^2 \quad \mbox{as } N \to \infty,
\]
where $(\ip{f}{\sigma_i})_{i \in I_N}$ are the $N$ largest coefficients $\ip{f}{\sigma_i}$
in magnitude. Wavelet bases exhibit the approximation rate
\[
\norm{f-f_N}_2^2 \le C \cdot N^{-1} \quad \mbox{as } N \to \infty.
\]
However, Donoho proved in \cite{Don01} that the optimal rate which can be achieved under some
restrictions on the representation system as well as on the selection procedure of
the approximating coefficients is
\[
\norm{f-f_N}_2^2 \le C \cdot N^{-2} \quad \mbox{as } N \to \infty.
\]
It was a breakthrough in 2004, when Cand\'{e}s and Donoho introduced the tight curvelet
frame in \cite{CD04} and proved that this system indeed does satisfy
\[
\norm{f-f_N}_2^2 \le C \cdot N^{-2} \cdot (\log N)^{3} \quad \mbox{as } N \to \infty,
\]
where again the approximation $f_N$ was generated by the $N$ largest coefficients in magnitude.
Although the optimal rate is not completely achieved, the $\log$-factor is typically considered negligible
compared to the $N^{-2}$-factor, wherefore the term `almost optimal' has been adopted into the
language. This result is even more surprising taking into account that in case of a tight frame
the approximation by the $N$ largest coefficients in magnitude does not even always yield the
{\em best} $N$-term approximation.
\subsection{(Compactly Supported) Shearlet Systems}
The directional representation system of {\em shearlets} has recently emerged and rapidly gained
attention due to the fact that -- in contrast to other proposed directional representation systems --
shearlets provide a unified treatment of the continuum and digital world similar to wavelets.
We refer to, e.g., \cite{GKL06,KL09} for the continuum theory, \cite{DKS08,ELL08,Lim09} for the digital
theory, and \cite{GLL09,DK10} for recent applications.
Shearlets are scaled according to a parabolic scaling law encoded in the {\em parabolic scaling matrices}
$A_{2^j}$ or $\tilde{A}_{2^j}$, $j \in \ZZ$, and exhibit directionality by parameterizing slope encoded
in the {\em shear matrices} $S_k$, $k \in \ZZ$, defined by
\[
A_{2^j} =
\begin{pmatrix}
2^j & 0\\ 0 & 2^{j/2}
\end{pmatrix}
\qquad\mbox{or}\qquad
\tilde{A}_{2^j} =
\begin{pmatrix}
2^{j/2} & 0\\ 0 & 2^j
\end{pmatrix}
\]
and
\[
S_k = \begin{pmatrix}
1 & \wql{k} \\ 0 & 1
\end{pmatrix},
\]
respectively.
To ensure an (almost) equal treatment of the different slopes, which is
evidently of significant importance for practical applications, we partition the frequency
plane into the following four cones $\cC_1$ -- $\cC_4$:
\[
\cC_\iota = \left\{ \begin{array}{rcl}
\{(\xi_1,\xi_2) \in \bR^2 : \xi_1 \ge 1,\, |\xi_2/\xi_1| \le 1\} & : & \iota = 1,\\
\{(\xi_1,\xi_2) \in \bR^2 : \xi_2 \ge 1,\, |\xi_1/\xi_2| \le 1\} & : & \iota = 2,\\
\{(\xi_1,\xi_2) \in \bR^2 : \xi_1 \le -1,\, |\xi_2/\xi_1| \le 1\} & : & \iota = 3,\\
\{(\xi_1,\xi_2) \in \bR^2 : \xi_2 \le -1,\, |\xi_1/\xi_2| \le 1\} & : & \iota = 4,
\end{array}
\right.
\]
and a centered rectangle
\[
\cR = \{(\xi_1,\xi_2) \in \bR^2 : \|(\xi_1,\xi_2)\|_\infty < 1\}.
\]
For an illustration, we refer to Figure \ref{fig:shearlet}(a).
\begin{figure}[ht]
\begin{center}
\includegraphics[height=1.4in]{./figs/cones.eps}
\put(-33,58){\footnotesize{$\cC=\cC_1$}}
\put(-70,80){\footnotesize{$\cC_2$}}
\put(-88,30){\footnotesize{$\cC_3$}}
\put(-50,52){\footnotesize{$\cR$}}
\put(-45,15){\footnotesize{$\cC_4$}}
\put(-60,-17){(a)}
\hspace*{4cm}
\includegraphics[height=1.4in]{./figs/shearlettiling.eps}
\put(-60,-17){(b)}
\end{center}
\caption{(a) The cones $\cC_1$ -- $\cC_4$ and the centered rectangle $\cR$ in frequency domain.
(b) The tiling of the frequency domain induced by a (cone-adapted) shearlet system.}
\label{fig:shearlet}
\end{figure}
The rectangle $\cR$ corresponds to the low frequency content of a signal and is customarily
represented by translations of some scaling function. Anisotropy comes into play when
encoding the high frequency content of a signal which corresponds to the cones $\cC_1$ -- $\cC_4$,
where the cones $\cC_1$ and $\cC_3$ as well as $\cC_2$ and $\cC_4$ are treated separately as can be seen
in the following
\begin{definition}
\label{defi:discreteshearlets}
For some sampling constant $c > 0$, the {\em (cone-adapted) shearlet system} $\cSH(c;\phi,\psi,\tilde{\psi})$ generated by
a {\em scaling function} $\phi \in L^2(\mathbb{R}^2)$ and {\em shearlets} $\psi, \tilde{\psi} \in L^2(\mathbb{R}^2)$ is
defined by
\[
\cSH(c;\phi,\psi,\tilde{\psi}) = \Phi(c;\phi) \cup \Psi(c;\psi) \cup \tilde{\Psi}(c;\tilde{\psi}),
\]
where
\[
\Phi(c;\phi) = \{\phi_m = \phi(\cdot-cm) : m \in \bZ^2\},
\]
\[
\Psi(c;\psi) = \{\psi_{j,k,m} = 2^{3j/4} {\psi}({S}_{k} {A}_{2^j}\cdot-cm) :
j \ge 0, |k| \le \lceil 2^{j/2} \rceil, m \in \bZ^2 \},
\]
and
\[
\tilde{\Psi}(c;\tilde{\psi}) = \{\tilde{\psi}_{j,k,m} = 2^{3j/4} \tilde{\psi}(S^T_{k} \tilde{A}_{2^j}\cdot-cm) :
j \ge 0, |k| \le \lceil 2^{j/2} \rceil, m \in \bZ^2 \}.
\]
\end{definition}
The reader should keep in mind that although not indicated by the notation, the functions $\phi_m$, $\psi_{j,k,m}$, and
$\tilde{\psi}_{j,k,m}$ all depend on the sampling constant $c$. For the sake of brevity, we will often write $\psi_\lambda$ and $\tilde{\psi}_{\lambda}$,
where $\lambda = (j,k,m)$ index scale, shear, and position. For later use, we further let $\Lambda_j$ be the indexing sets
of shearlets in $\Psi(c;\psi)$ and $\tilde{\Psi}(c;\tilde{\psi})$ at scale $j$,
respectively, i.e.,
\[
\Psi(c;\psi) = \{\psi_{\lambda} : \lambda \in \Lambda_j, j=0, \ldots, \infty\}
\]
and
\[
\tilde{\Psi}(c;\tilde{\psi}) = \{\tilde{\psi}_{\lambda} : \lambda \in \Lambda_j, j=0, \ldots, \infty\}.
\]
Finally, we define
\[
\Lambda = \bigcup_{j=0}^\infty \Lambda_j.
\]
The tiling of frequency domain induced by $\cSH(c;\phi,\psi,\tilde{\psi})$ is illustrated in Figure \ref{fig:shearlet}(b).
From this illustration, the anisotropic footprints of shearlets contained in $\Psi(c;\psi)$ and $\tilde{\Psi}(c;\tilde{\psi})$
can clearly be seen. However, the reader should notice that the tiling indicated here is based on the {\em essential support} and not
the exact support of the analyzing elements, since our focus will be on shearlet systems associated with spatially compactly
supported generators.
The corresponding anisotropic footprints of shearlets {\em in spatial domain} are of size $2^{-j/2}$ times
$2^{-j}$. A beautiful intuitive extensive explanation of why it is conceivable that such a system -- based on parabolic scaling --
exhibits optimal sparse approximation of cartoon-like images, is provided in \cite{CD04}, and we would like
to refer the reader to this paper. The main idea is to count the number of shearlets intersecting the
discontinuity curves, which is `small' compared to the number of such wavelets, due to their anisotropic footprints.
Certainly, we naturally ask the question when $\cSH(c;\phi,\psi,\tilde{\psi})$ does form a frame for $L^2(\RR^2)$.
The wavelet literature provides various necessary and sufficient conditions for $\Phi(c;\phi)$ to
form a frame for $L^2(\{f \in L^2(\RR^2) : \Sp(\wq{\hat{f}}) \subseteq \cR\})$, also when $\phi$ is compactly
supported in spatial domain. Although not that well-studied as wavelets yet, several answers are also
known for the question when $\Psi(c;\psi)$ forms a frame for
\[
L^2(\{f \in L^2(\RR^2) : \Sp(\wq{\hat f}) \subseteq \cC_1 \cup \cC_3\}),
\]
and we refer to results in \cite{GKL06,KL07,DKST09,KKL10b}.
Since $\Psi(c;\phi)$ and $\tilde{\Psi}(c;\tilde{\psi})$ are linked by a simple rotation of $90^o$, these results
immediately provide conditions for $\tilde{\Psi}(c;\tilde{\psi})$ to constitute a frame for
\[
L^2(\{f \in L^2(\RR^2) : \Sp(\wq{\hat f}) \subseteq \cC_2 \cup \cC_4\}).
\]
Very recent results in \cite{KKL10a}
even focus specifically on the case of spatially compactly supported shearlets -- of uttermost
importance for applications due to their superior localization. For instance, in \cite{KKL10a},
the following special class of compactly supported shearlet frames for $L^2(\bR^2)$ was constructed:
The generating shearlets $\psi$ and $\tilde{\psi}$ were chosen separable, i.e., of the form
$\psi_1(x_1)\cdot\psi_2(x_2)$ and $\psi_1(x_2)\cdot\psi_2(x_1)$, respectively, where $\psi_1$ is a wavelet and $\psi_2$ is
a scaling function both associated with some carefully chosen low pass filter. Intriguingly,
our main result in this paper (Theorem \ref{theo:main}) proves as a corollary that this
class of compactly supported shearlet frames provides (almost) optimally sparse approximations
of cartoon-like images. We refer to \cite{KKL10a} for the precise statement.
Combining those thoughts, we can attest that frame properties of the system $\cSH(c;\phi,\psi,\tilde{\psi})$
including spatially compactly supported generators are already quite well studied.
We however wish to mention that there is a trade-off between {\em compact support} of the shearlet generators, {\em tightness} of the associated frame,
and {\em separability} of the shearlet generators. The known constructions of tight shearlet frames do not use separable generators, and
these constructions can be shown to {\em not} be applicable to compactly supported generators. Tightness is difficult to obtain while
allowing for compactly supported generators, but we can gain separability, hence fast algorithmic realizations. On the other hand, when allowing
non-compactly supported generators, tightness is possible, but separability seems to be out of reach, which
makes fast algorithmic realizations very difficult.
\subsection{Optimally Sparse Approximation of Cartoon-Like Images by Shearlets}
The concept of optimally sparse approximation of cartoon-like images of general (directional)
representation systems was already discussed in Section \ref{subsec:optsparse}.
However, the attentive reader will have realized that only the situation of
tight frames was studied whereas here we need to consider sparse
approximations by arbitrary frames. Hence this situation deserves a careful
commenting.
Let $\cSH(c;\phi,\psi,\tilde{\psi})$ be a shearlet frame for $L^2(\RR^2)$, which for
illustrative purposes for a moment we denote by $\cSH(c;\phi,\psi,\tilde{\psi}) = (\sigma_i)_{i \in I}$,
say. Is it well-known that a frame is associated with a canonical dual frame, which in this
case we want to call $(\tilde{\sigma}_i)_{i \in I}$. Then we define the $N$-term approximation $f_N$
of a cartoon-like image $f \in \cE^2(\nu)$ by the frame $\cSH(c;\phi,\psi,\tilde{\psi})$ to be
\[
f_N = \sum_{i \in I_N} \ip{f}{\sigma_i}\tilde{\sigma}_i,
\]
where $(\ip{f}{\sigma_i})_{i \in I_N}$ are the $N$ largest coefficients $\ip{f}{\sigma_i}$
in magnitude. As in the tight frame case, this procedure does not always yield the {\em best}
$N$-term approximation, but surprisingly even with this `crude' selection procedure -- in
the situation of spatially compactly supported generators -- we can prove an (almost) optimally
sparse approximation rate as our main result shows.
\begin{theorem}\label{theo:main}
Let $c > 0$, and let $\phi, \psi, \tilde{\psi} \in L^2(\RR^2)$ be compactly supported. Suppose
that, in addition, for all $\xi = (\xi_1,\xi_2) \in \RR^2$, the shearlet $\psi$ satisfies\\[-1.25ex]
\bitem
\item[(i)] $|\hat\psi(\xi)| \le C_1 \cdot \min(1,|\xi_1|^{\alpha}) \cdot \min(1,|\xi_1|^{-\gamma}) \cdot \min(1,|\xi_2|^{-\gamma})$ and\\[-0.1ex]
\item[(ii)] $\left|\frac{\partial}{\partial \xi_2}\hat \psi(\xi)\right|
\le \wqqq{|h(\xi_1)|} \cdot \left(1+\frac{|\xi_2|}{|\xi_1|}\right)^{-\gamma}$,\\[-0.2ex]
\eitem
where $\alpha > 5$, $\gamma \ge 4$, $h \in L^1(\bR)$, and $C_1$ is a constant, and suppose that
the shearlet $\tilde{\psi}$ satisfies (i) and (ii) with the roles of $\xi_1$ and $\xi_2$ reversed.
Further, suppose that $\cSH(c;\phi,\psi,\tilde{\psi})$ forms a
frame for $L^2(\RR^2)$.
Then, for any $\nu > 0$, the shearlet frame $\cSH(c;\phi,\psi,\tilde{\psi})$ provides (almost)
optimally sparse approximations of functions $f \in \cE^2(\nu)$, i.e., there exists some $C > 0$ such
that
\[
\|f-f_N\|_2^2 \leq C\cdot N^{-2} \cdot {(\log{N})}^3 \qquad \text{as } N \rightarrow \infty,
\]
where $f_N$ is the nonlinear N-term approximation obtained by choosing the N largest shearlet coefficients
of $f$.
\end{theorem}
Condition (i) can be interpreted as both a condition ensuring (almost) separable behavior
as well as a first order moment condition along the horizontal axis, hence enforcing directional selectivity.
This condition ensures that the support of shearlets in frequency domain is essentially of
the form indicated in Figure \ref{fig:shearlet}(b). Condition~(ii) (together with (i)) is a weak version of a directional vanishing
moment condition\footnote{For the precise definition of directional
vanishing moments, we refer to \cite{DV05}. }, which is crucial for
having fast decay of the shearlet coefficients when the corresponding
shearlet intersects the discontinuity curve. Conditions~(i) and (ii)
are rather mild conditions on the generators. To compare with the
optimality result for band-limited generators we wish to point out
that conditions~(i) and (ii) are obviously satisfied for band-limited
generators.
Notice also that, intriguingly, the -- the `true' optimality destroying -- $\log$-factor has the
{\em same} exponent as in the curvelet-, contourlet-, and shearlet-result on (almost) optimally
sparse approximation.
\subsection{Prior Work and Our Contribution}
In 2004, Cand\'{e}s and Donoho \cite{CD04} achieved a breakthrough when introducing
tight curvelet frames, which provide (almost) optimally sparse approximations of cartoon-like
images (functions in $\cE^2(\nu)$). The main outline of their proof is to break $[0,1]^2$ into smaller
cubes and then separately analyze the
curvelet coefficients essentially centered in the smooth part of the model and those essentially
centered on the discontinuity curve. For both sets of coefficients their weak-$\ell_{2/3}$ norm
is estimated; the estimate for the `non-smooth part' also requiring the usage of the Radon transform.
A year later, Do and Vetterli \cite{DV05} introduced contourlets and proved similar sparsity results
for those. However, although their work includes contourlets with compact support, their construction
is fully based on discrete filter banks so that directional selectivity is problematic. Because of
this fact, infinite directional vanishing moments had to be artificially imposed in order to achieve
(almost) optimal sparsity. However, this is impossible for any function with compact support to
satisfy. Hence, similar to curvelets, optimal sparsity is only proven for {\em band-limited}
contourlets.
In 2005, shearlets were introduced as the first directional representation system ensuring a unified
treatment of the continuum and digital world by Labate, Weiss, and the authors in \cite{LLKW05}.
One year later, Labate and Guo proved (almost) optimally sparse approximations of cartoon-like images
for the at that time customarily utilized shearlet frames \cite{GL07}, which are band-limited
such as curvelets. The proof the authors provided follows the proof in \cite{CD04} very closely
step by step.
Concluding, although those pioneering studies deserve all our credit, these results are far from
including the important class of directional representation systems consisting of compactly
supported functions.
\medskip
The main contribution of this paper is to provide the first complete proof of (almost)
optimally sparse approximations of cartoon-like images using a directional representation system
consisting of compactly supported functions. Our proof is indeed very different from all previous
ones caused by the necessary extensive exploration of the compact support of the shearlet generators,
the only similarity being the breaking of $[0,1]^2$ into smaller cubes and the separate
consideration of shearlet coefficients now being {\em exactly contained} -- in contrast to
being {\em essentially contained} for all
other systems -- in the smooth part and those which intersect the discontinuity curve.
Previous results all require moment conditions along the direction of the discontinuity
curve -- thereby requiring vanishing moments along infinitely many directions asymptotically
in scale --, which
is trivially satisfied for band-limited generators. Intriguingly, a weaker version of
directional vanishing moments, even only in one direction and the shearing taking care of the
remaining directions, is sufficient for our analysis.
\subsection{Outline}
In Section \ref{sec:architecture}, we present the overall structure of the proof, the
results of the analysis of shearlet coefficients being contained in the smooth part and
those which intersect the discontinuity curve, and -- based on these results -- state the proof
of Theorem \ref{theo:main}. The proofs of the results on the behavior of shearlet coefficients
in the smooth and non-smooth part are then carried out in Sections \ref{sec:smoothpart} and
\ref{sec:discontinuity}, respectively.
\section{Architecture of the Proof of Theorem \ref{theo:main}}
\label{sec:architecture}
We now detail the general structure of the proof of Theorem \ref{theo:main}, starting by
introducing useful notions and explaining the blocking into smaller boxes and splitting
into the smooth and non-smooth part. Then the main results concerning the analysis of
shearlet coefficients being entirely contained in the smooth part and those intersecting
the discontinuity curve will be presented followed by the proof of Theorem \ref{theo:main}
based on those.
\subsection{General Organization}
Let now $\cSH(c;\phi,\psi,\tilde{\psi})$ satisfy the hypotheses of Theorem \ref{theo:main}, and
let $f \in \cE^2(\nu)$. Further, we let $A$ denote the lower frame bound of $\cSH(c;\phi,\psi,\tilde{\psi})$.
We first observe that, without loss of generality, we might assume
the scaling index $j$ to be sufficiently large, since $f$ as well as all frame elements in the shearlet
frame $\cSH(c;\phi,\psi,\tilde{\psi})$ are compactly supported in spatial domain, hence a finite
number does not contribute to the asymptotic estimate we aim for. In particular, this means that
we do not need to consider frame elements from $\Phi(c;\phi)$. Also, we are allowed to restrict our
analysis to shearlets $\psi_{j,k,m}$, since the frame elements $\widetilde{\psi}_{j,k,m}$ can be
handled in a similar way.
Our main concern will be to derive appropriate estimates for the shearlet coefficients
$\{\ip{f}{\psi_{\lambda}} : \lambda \in \Lambda\}$ of $f$. Letting $|\theta(f)|_n$ denote
the $n$th largest shearlet coefficient $\ip{f}{\psi_{\lambda}}$ in absolute value and exploring
the frame property of $\cSH(c;\phi,\psi,\tilde{\psi})$, we conclude that
\[
\|f-f_N\|_2^2 \leq \frac{1}{A}\sum_{n>N} |\theta(f)|_n^2,
\]
for any positive integer $N$. Thus, for the proof of Theorem \ref{theo:main}, it suffices to show that
\begin{equation}\label{eq:upper}
\sum_{n>N} |\theta(f)|_n^2 \leq C \cdot N^{-2} \cdot {(\log{N})}^3 \qquad \text{as } N \rightarrow \infty.
\end{equation}
To derive the anticipated estimate in \eqref{eq:upper}, for any shearlet
$\psi_{\lambda}$, we will study two separate cases:
\bitem
\item {\em Case 1}. The compact support of the shearlet $\psi_{\lambda}$ does not intersect the boundary of the set $B$, i.e.,
$\wang{\intt(\Sp(\psi_{\lambda}))} \cap \partial B = \emptyset.$
\item {\em Case 2}. The compact support of the shearlet $\psi_{\lambda}$ does intersect the boundary of the set $B$, i.e.,
$\wang{\intt(\Sp(\psi_{\lambda}))} \cap \partial B \ne \emptyset.$
\eitem
Notice that this exact distinction is only possible due to the spatial compact support of all shearlets
in the shearlet frame.
In the sequel -- since we are concerned with an asymptotic estimate -- for simplicity we will often simply use
$C$ as a constant although it might differ for each estimate. Also all the results in the sequel are independent
on the sampling constant $c>0$, wherefore we now fix it once and for all.
\subsection{The Smooth and the Non-Smooth Part of a Cartoon-Like Image}
To illustrate which conditions on $\psi$ required by Theorem \ref{theo:main} are utilized
for the decay estimates of the different cases, in this section we do not make any initial
assumptions on $\psi$.
Let us start with the smooth part, which is the easier one to handle. Dealing with this part
allows us to consider some $g \in C^2(\RR^2)$ with compact support in $[0,1]^2$ and estimate its shearlet coefficients. This
is done in the following proposition. Notice that the hypothesis on $\psi$ of the following result
is implied by condition (i) in Theorem \ref{theo:main}.
\begin{proposition}
\label{prop:main1}
Let $g \in C^2(\RR^2)$ with compact support in $[0,1]^2$, and let $\psi \in L^2(\RR^2)$ be compactly supported and satisfy
\[
|\hat\psi(\xi)| \le C_1 \cdot \min(1,|\xi_1|^{\alpha}) \cdot \min(1,|\xi_1|^{-\gamma}) \cdot \min(1,|\xi_2|^{-\gamma})
\mbox{ for all }\xi = (\xi_1,\xi_2) \in \RR^2 \hspace*{-0.17cm},
\]
where $\gamma>3$, $\alpha > \gamma+2$, and $C_1$ is a constant. Then, there exists
some $C>0$ such that
\[
\sum_{n>N} |\theta(g)|_n^2 \le C \cdot N^{-2} \qquad \text{as } N \rightarrow \infty.
\]
\end{proposition}
Thus, in this case, optimal sparsity is achieved. The proof of this proposition is given in
Section \ref{sec:smoothpart}.
Next, we turn our attention to the non-smooth part, in particular, to estimating those shearlet
coefficients whose spatial support intersects the discontinuity curve. For this, we first
need to introduce some new notations. For any scale $j \ge 0$ and any grid point $p \in \ZZ^2$, we
let $\cQ_{j,p}$ denote the dyadic cube defined by
\[
\cQ_{j,p} = [-2^{-j/2},2^{-j/2}]^2+2^{-j/2}p.
\]
Further, let $\cQ_j$ be the collection of those dyadic cubes $\cQ_{j,p}$ whose interior, in the following denoted by
$\intt(\cQ_{j,p})$, intersects $\partial B$, i.e.,
\[
\cQ_j = \{\cQ_{j,p}: \intt(\cQ_{j,p}) \cap \partial B \ne \emptyset, p \in \ZZ^2\}.
\]
Of interest to us is also the set of shearlet indices, which are associated with shearlets intersecting
the discontinuity curve inside some $\cQ_{j,p} \in \cQ_{j}$, i.e., for $j\ge0$ and $p \in \bZ^2$ with $\cQ_{j,p} \in \cQ_{j}$,
we will consider the index set
\[
{\Lambda_{j,p}} = \{\lambda \in \Lambda_j : \wqq{\intt(\Sp(\psi_{\lambda})) \cap \intt(\cQ_{j,p})} \cap \partial B \ne \emptyset\}.
\]
Finally, for $j \ge 0$, $p \in \ZZ^2$, and $0 < \eps < 1$, we define $\Lambda_{j,p}(\eps)$ to be \wq{the index set of} shearlets
$\psi_\lambda$, $\lambda \in {\Lambda_{j,p}}$, such that the magnitude of the corresponding shearlet
coefficient $\langle f,\psi_\lambda \rangle$ is larger than $\eps$ and the support of $\psi_\lambda$
intersects $\cQ_{j,p}$ at the $j$th scale, i.e.,
\[
{\Lambda_{j,p}(\eps)} = \{\lambda \in {\Lambda_{j,p}} : |\langle f,\psi_{\lambda}\rangle| > \eps\},
\]
and we define {$\Lambda(\eps)$} to be the \wq{index set for} shearlets so that $|\langle f,\psi_{\lambda} \rangle| > \eps$
across all scales $j$, i.e.,
\[
\Lambda(\eps) = \bigcup_{j,p} \Lambda_{j,p}(\eps).
\]
The expert reader will have noticed that in contrast to the proofs in \cite{CD04} and \cite{GL07},
which also split the domain into smaller scale boxes, we do not apply a weight function to
obtain a smooth partition of unity. In our case, this is not necessary due to the spatial compact
support of the frame elements.
As mentioned at the beginning of this section, we may assume that $j$ is sufficiently large. Given
some scale $j \ge 0$ and position $p \in \ZZ^2$ for which the associated cube $\cQ_{j,p}$
satisfies $\cQ_{j,p} \in \cQ_j$. Then the set
\[
\cS_{j,p}= \bigcup_{\lambda \in \Lambda_{j,p}} \Sp(\psi_\lambda)
\]
is contained in a cubic window of size $C\cdot 2^{-j/2}$ by $C\cdot 2^{-j/2}$, hence is of asymptotically
the same size as $\cQ_{j,p}$. By smoothness assumption on the discontinuity curve $\partial B$, the edge curve can be parameterized by either
$(x_1,E(x_1))$ or $(x_2,E(x_2))$ with $E \in C^2$ in the interior of $\cS_{j,p}$ for sufficiently large $j$.\footnote{In other words, a part of the
edge curve $\partial B$ contained in $\cS_{j,p}$ can be described as a $C^2$ function $x_1 = E(x_2)$ ( or $x_2 = E(x_1)$).}
Thus, we are facing the following two cases (see also Figure \ref{fig:cases2ab}):
\begin{itemize}
\item {\em Case 2a}. The edge curve $\partial B$ can be parameterized \wql{by either $(E(x_2),x_2)$ or $(x_1,E(x_1))$ with $E \in C^2$
in the interior of $\cS_{j,p}$ such that, for any $\lambda \in \Lambda_{j,p}$, there exists some $\hat{x} = (\hat{x}_1,\hat{x}_2)
\in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$ satisfying either $|E'(\hat{x}_2)| \leq 2 $ or $|E'(\hat{x}_1)|^{-1} \leq 2$.}
\item {\em Case 2b}. The edge curve $\partial B$ can be parameterized \wql{by either $(E(x_2),x_2)$ or $(x_1,E(x_1))$ with $E \in C^2$
in the interior of $\cS_{j,p}$ such that, for any $\lambda \in \Lambda_{j,p}$, there exists some $\hat{x} = (\hat{x}_1,\hat{x}_2)
\in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$ satisfying either $|E'(\hat{x}_2)| > 2 $ or $|E'(\hat{x}_1)|^{-1} > 2$.}
Here, we identify $E'(\hat x_1) = 0$ with $|E'(\hat x_1)|^{-1} = \infty > 2$.
\end{itemize}
\begin{figure}[ht]
\begin{center}
\includegraphics[height = 1in]{./figs/case2a.eps}
\put(-40,-15){\footnotesize{(a)}}
\hspace*{4cm}
\includegraphics[height = 1in]{./figs/case2b.eps}
\put(-40,-15){\footnotesize{(b)}}
\end{center}
\caption{(a) A part of the curve $\partial B$ satisfying Case 2a.
(b) A part of the curve $\partial B$ satisfying Case 2b.}
\label{fig:cases2ab}
\end{figure}
For both cases, we will derive the below stated upper estimates \eqref{eq:estimate1} and \eqref{eq:estimate2} for the absolute value of
the associated shearlet coefficients. The proofs of these estimates are contained in Section \ref{sec:discontinuity}.
\begin{proposition}\label{prop:main2}
Let $\psi \in L^2(\RR^2)$ be compactly supported, and assume that $\psi$ satisfies conditions (i) and (ii) of Theorem \ref{theo:main}.
Further, let $j \ge 0$ and $p \in \ZZ^2$, and let $\lambda \in \Lambda_{j,p}$. For fixed $\hat{x}=(\hat{x}_1,\hat{x}_2) \in
\wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$,
let $s$ be the slope\footnote{Notice that here we regard the slope of the tangent to a curve
$(E(x_2),x_2)$, i.e., we consider $s$ of a curve indexed by the $x_2$-axis, for instance, by $x_1 = sx_2+b$. For analyzing shearlets
$\tilde{\psi}_{j,k,m}$, the roles of $x_1$ and $x_2$ would need to be reversed.} of the tangent to the edge curve $\partial B$
at $(\hat{x}_1,\hat{x}_2)$, more precisely,
\begin{enumerate}
\item[{\rm (i)}] if $\partial B$ is parameterized by $(E(x_2),x_2)$ with $E \in C^2$ in the interior of $\cS_{j,p}$, then $s = E'(\hat{x}_2)$,
\item[{\rm (ii)}] if $\partial B$ is parameterized by $(x_1,E(x_1))$ with $E \in C^2$ and $E'(\hat{x}_1) \neq 0$ in the interior of $\cS_{j,p}$,
then $s = ({E'(\hat{x}_1)})^{-1}$, and
\item[{\rm (iii)}] if $\partial B$ is parameterized by $(x_1,E(x_1))$ with $E \in C^2$ and $E'(\hat{x}_1) = 0$ in the interior of $\cS_{j,p}$, then
$s = \infty$.
\end{enumerate}
Then there exists some $C > 0$ such that
\begin{equation}\label{eq:estimate1}
|\langle f,\psi_{\lambda}\rangle| \leq C \cdot \frac{2^{-\frac{3}{4}j}}{|\wq{k+}2^{j/2}s|^3}, \qquad \text{if }|s| \leq \wq{3}.
\end{equation}
and
\begin{equation}\label{eq:estimate2}
|\langle f,\psi_{\lambda}\rangle| \leq C \cdot 2^{-\frac{9}{4}j}, \qquad \text{if }|s| > \wq{\frac{3}{2}} \text{ or } s = \infty,
\end{equation}
\end{proposition}
Notice that in Case 2a, condition (i) or (ii) can occur, whereas in Case 2b, all three conditions can occur.
\subsection{Proof of Theorem \ref{theo:main}}
Let $f \in \cE^2(\nu)$. We first observe that, by Proposition \ref{prop:main1}, we can neglect those shearlet coefficients
whose spatial support of the associated shearlet does not intersect the discontinuity curve.
To estimate the remaining shearlet coefficients, we need to analyze their decay properties.
For this, let $j \ge 0$ be sufficiently large and let $p \in \ZZ^2$, be such that the associated cube
satisfies $\cQ_{j,p} \in \cQ_j$. We note that all sets $\wqq{\intt(\Sp{(\psi_{\lambda})})}$ with $\lambda \in \Lambda_{j,p}$
are contained in the interior of ${\mathcal{S}}_{j,p}$. Therefore weights as in \cite{CD04} and \cite{GL07} are
not required here.
Letting $\eps > 0$, our goal will now be to estimate first $|\Lambda_{j,p}(\eps)|$ and then $|\Lambda(\eps)|$.
WLOG we might assume $\|\psi\|_1 \le 1$, which implies
\[
|\langle f,\psi_{\lambda}\rangle| \le 2^{-3j/4}.
\]
Hence, for estimating $|\Lambda_{j,p}(\eps)|$, it is sufficient to restrict our attention to scales
\beq \label{eq:upperbdonj}
j \le \frac{4}{3}\log_2(\eps^{-1}).
\eeq
We will now deal with Case 2a and Case 2b separately.
\vspace*{0.2cm}
{\em Case 2a}. First, we let $s$ be the slope of the tangent to the edge curve $\partial B$
at $(\hat{x}_1,\hat{x}_2)$ as defined in Proposition \ref{prop:main2}, i.e.,
if $\partial B$ is parameterized by $(E(x_2),x_2)$ in the interior of $\cS_{j,p}$, then $s = E'(\hat x_2)$,
and if $\partial B$ is parameterized by $(x_1,E(x_1))$ in the interior of $\cS_{j,p}$, then $s = E'(\hat x_1)^{-1}$.
By assumption of Case 2a, we have $s \in [-2,2]$. Now observe that, for each shear index $k$,
\begin{equation}\label{eq:number}
|\{\lambda=(j,k,m) : \lambda \in \Lambda_{j,p}\}| \le C \cdot (|k+2^{j/2}s|+1).
\end{equation}
Interestingly, this estimate is independent of the choice of the point $\hat x \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$, which
can be seen as follows: Let $\hat x \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B, \wang{\hat x' \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda'}))} \cap \partial B}$, and let $s$ and $s'$ be the
associated slopes of the tangents to the edge curve $E$ in $\hat x$ and $\hat x'$, respectively. Since $E \in C^2$,
\beq \label{eq:ss'}
|s-s'| \le C_1 \cdot 2^{-j/2},
\eeq
and hence
\begin{equation*}
|k+2^{j/2}s'| \leq C \cdot (|k+2^{j/2}s|+1).
\end{equation*}
This proves that the estimate \eqref{eq:number} remains the same asymptotically, independent of the values of $s$ and $s'$,
and hence of $\hat x$ and $\hat x'$.
We further require the following even stronger observation: Fix some $\hat x \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$,
and let $s$ be the associated slope of the tangent to the edge curve $E$ in $\hat x$. Now consider another $\hat x' \in \wqq{\intt(\cQ_{j,p})
\cap \wang{\intt(\Sp(\psi_{\lambda'}))}} \cap \partial B$, and again let $s'$ be the associated slope of the tangent to the edge curve $E$ in $\hat x'$.
Then, for sufficiently \wq{large} scaling index $j$, by \eqref{eq:ss'}, $|s-s'|$ is sufficiently small \wq{and} we may assume $s' \in [-3,3]$.
Hence the estimate \eqref{eq:estimate1} from Proposition \ref{prop:main2} holds not only for $\hat x$ (with $s$), but also for $\hat x'$
(with $s'$). In fact, it can even be checked that by substituting \wq{$s'$} by \wq{$s$} in \eqref{eq:estimate1} the
asymptotic behavior of the estimate for $|\Lambda_{j,p}(\epsilon)|$ does not change. Let us briefly outline the reasoning. First, we observe that
WLOG we can assume that $|k+2^{j/2}s| \ge 2 \cdot C_1$, where $C_1$ is the constant appearing in \eqref{eq:ss'}, since
$|\{k \in \Z : |k+2^{j/2}s| < 2 \cdot C_1\}| \leq C$ with $C$ being independent on $j$ and hence it can be deduced that the hypothesis $|k+2^{j/2}s| \ge 2 \cdot C_1$
for each $j \ge 0$ does not affect our asymptotic estimate of $|\Lambda_{j,p}(\epsilon)|$. From \eqref{eq:ss'}, it then follows that
\[
|k+2^{j/2}s| \leq 2 \cdot |k+2^{j/2}s'|,
\]
which in turn implies
\[
\frac{2^{-\frac{3}{4}j}}{|k+2^{j/2}s'|^3} \leq 8\frac{2^{-\frac{3}{4}j}}{|k+2^{j/2}s|^3}.
\]
Hence, by substituting \wq{$s'$} by \wq{$s$} in \eqref{eq:estimate1} the asymptotic behavior of the estimate for $|\Lambda_{j,p}(\epsilon)|$ does not change.
Thus, it suffices to consider just one fixed $\hat x \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$ with associated slope $s$
in each $\cQ_{j,p}$. We now turn to estimating $|\Lambda_{j,p}(\eps)|$ in this case.
For this, by estimate \eqref{eq:estimate1} from Proposition \ref{prop:main2}, $|\langle f,\psi_{\lambda}\rangle|>\eps$ implies
\begin{equation}\label{eq:number2}
|k+2^{j/2}s| \le C\cdot \eps^{-1/3}\cdot 2^{-j/4}.
\end{equation}
From \eqref{eq:number} and \eqref{eq:number2}, we then conclude
\begin{equation}\label{eq:case1}
|\Lambda_{j,p}(\eps)| \le C \cdot \sum_{k \in K_j(\eps)}(|k+2^{j/2}s|+1) \le C \cdot (\eps^{-1/3}\cdot 2^{-j/4}+1)^2,
\end{equation}
where $K_j(\eps) = \{k\in\bZ:|k+2^{j/2}s|\le C \cdot \eps^{-1/3} \cdot 2^{-j/4}\}$.
\vspace*{0.2cm}
{\em Case 2b}. Exploiting similar arguments as in Case 2a, it also suffices to consider just one fixed $\hat x \in \wqq{\intt(\cQ_{j,p}) \cap
\intt(\Sp(\psi_{\lambda}))} \cap \partial B$ with associated slope $s$ in each \wqq{$\intt(\cQ_{j,p})$}. Again, our goal is now to estimate $|\Lambda_{j,p}(\eps)|$.
By estimate \eqref{eq:estimate2} from Proposition \ref{prop:main2}, $|\langle f,\psi_{\lambda}\rangle|>\eps$ implies
\[
C \cdot 2^{-\frac 94 j} \ge \eps,
\]
hence
\begin{equation}\label{eq:case2}
\quad \text{and} \quad j \leq \frac 49 \log_2{(\eps^{-1})}+C.
\end{equation}
Since there exists some $C$ with
\[
|\Lambda_{j,p}| \le C \cdot 2^j,
\]
it then follows that
\beq\label{eq:estimateLambda_jp}
|\Lambda_{j,p}(\eps)| \le C \cdot 2^{j}.
\eeq
Notice that this last estimate is exceptionally crude, but will be sufficient for our purposes.
\vspace*{0.2cm}
We now combine the estimates for $|\Lambda_{j,p}(\eps)|$ derived in Case 2a and Case 2b. Since
\[
\#(\cQ_j) \leq C\cdot 2^{j/2}.
\]
by \eqref{eq:case1} (and \eqref{eq:upperbdonj}) and by \eqref{eq:case2} (and \eqref{eq:estimateLambda_jp}), we have
\begin{eqnarray}\label{eq:totalnumber} \nonumber
|\Lambda(\eps)| &\leq& C \cdot \Bigl[ \sum_{j=0}^{\frac 43 \log_2(\eps^{-1})} 2^{j/2}\Bigl(\eps^{-1/3}\cdot 2^{-1/4j}+1\Bigr)^2
+ \hspace*{-0.2cm}\sum_{j=0}^{\frac 49 \log_2(\eps^{-1})+C}2^{\frac 32j}\Bigr]\\
&\leq& C \cdot \eps^{-2/3} \cdot \log_2(\eps^{-1}).
\end{eqnarray}
Having estimated $|\Lambda(\eps)|$, we are now ready to prove our main claim. For this, set $N = |\Lambda(\eps)|$, i.e., the total
number of shearlets $\psi_{\lambda}$ such that the magnitude of the corresponding shearlet coefficient
$\langle f,\psi_{\lambda} \rangle$ is larger than $\eps$. By \eqref{eq:totalnumber}, the value $\eps$ can be written as a function
of the total number of coefficients $N$ in the following way:
\[
\eps(N) \le C \cdot N^{-3/2}\cdot (\log N)^{3/2}, \quad \text{for sufficiently large} \,\,N>0.
\]
This implies that
\[
|\theta(f)|_{N} \le C \cdot N^{-3/2}\cdot (\log N)^{3/2}.
\]
Hence,
\[
\sum_{n > N}|\theta(f)|_{n}^2 \le C\cdot N^{-2} \cdot (\log N)^3 \quad \text{for sufficiently large }\,\,N>0,
\]
which proves \eqref{eq:upper}.
The proof of Theorem \ref{theo:main} is finished.
\section{Analysis of Shearlet Coefficients associated with the Smooth Part of a Cartoon-Like Image}
\label{sec:smoothpart}
In this section, we will prove Proposition \ref{prop:main1}. For this, we first prove a
result which shows that, provided that the shearlet $\psi$ satisfies certain decay conditions
even with strong weights such as $(2^{4j})_j$, the system $\Psi(c;\psi)$ forms a Bessel-like
sequence for $C^2(\RR^2)$ with compact support in $[0,1]^2$.
In the following we will use the notation $r_j \sim s_j$ for $r_j, s_j \in \RR$,
if $C_1 \cdot r_j \le s_j \le C_2 \cdot r_j$ with constants $C_1$ and $C_2$ independent on the
scale $j$.
\begin{lemma}\label{lemma:smoothpart}
Let $\psi \in L^2(\RR^2)$ satisfy
\[
|\hat\psi(\xi)| \le C_1 \cdot \min(1,|\xi_1|^{\alpha}) \cdot \min(1,|\xi_1|^{-\gamma}) \cdot \min(1,|\xi_2|^{-\gamma})
\mbox{ for all }\xi = (\xi_1,\xi_2) \in \RR^2 \hspace*{-0.17cm},
\]
where $\gamma>3$, $\alpha > \gamma+2$, and $C_1$ is some constant. Then, there exists some $C>0$ such that,
for all $g \in C^2(\RR^2)$ with compact support in $[0,1]^2$,
\[
\sum_{j=0}^{\infty}\sum_{|k| \le \lceil 2^{j/2} \rceil}\sum_{m \in \bZ^2} 2^{4j}|\langle \wqqq{g},\psi_{j,k,m}\rangle|^2 \leq C\cdot
\wqqq{\left\|\frac{\partial^2}{\partial x_1^2} g\right\|_2^2}.
\]
\end{lemma}
The proof of this lemma will explore the following result from \cite{KKL10a}, which we state here
for the convenience of the reader.
\begin{proposition}\cite{KKL10a}\label{proposition:upperbd}
Let $\psi \in L^2(\RR^2)$ satisfy
\[
|\hat\psi(\xi)| \le C_1 \cdot \min(1,|\xi_1|^{\alpha}) \cdot \min(1,|\xi_1|^{-\gamma}) \cdot \min(1,|\xi_2|^{-\gamma})
\mbox{ for all }\xi = (\xi_1,\xi_2) \in \RR^2 \hspace*{-0.17cm},
\]
where $\alpha>\gamma>3$ and $C_1$ is some constant. Then, there exists some $C>0$ such that, for all $\eta \in L^2(\RR^2)$,
\[
\sum_{j=0}^{\infty}\sum_{|k| \le \lceil 2^{j/2}\rceil}\sum_{m \in \bZ^2} |\langle \eta,\psi_{j,k,m}\rangle|^2 \leq C\cdot \|\eta\|_2^2.
\]
\end{proposition}
\begin{proof}{\em \hspace*{-0.2cm} (Proof of Lemma \ref{lemma:smoothpart})}.
By the assumption on $\psi$, the parameters $\alpha$ and $\gamma$ can be chosen
such that
\[
|\hat\psi(\xi)| \le C_1 \cdot \min(1,|\xi_1|^{\alpha}) \cdot \min(1,|\xi_1|^{-\gamma}) \cdot \min(1,|\xi_2|^{-\gamma})
\mbox{ for all }\xi = (\xi_1,\xi_2) \in \RR^2 \hspace*{-0.17cm},
\]
where $\alpha > \gamma+2, \gamma > 3$. Now, let $\eta \in L^2(\RR^2)$ be chosen to satisfy
\[
\frac{\partial^2}{\partial x_1^2} \eta = \psi.
\]
Then a straightforward computation shows that $\eta$ satisfies the hypotheses of Proposition \ref{proposition:upperbd}.
Using integration by parts,
\wqqq{
\[
\left|\left\langle \frac{\partial^2}{\partial x_1^2}g,\eta_{j,k,m}\right\rangle\right|^2 = 2^{4j}|\langle g,\psi_{j,k,m}\rangle|^2,
\]
}
hence, by Proposition \ref{proposition:upperbd},
\begin{eqnarray*}
\sum_{j=0}^{\infty}\sum_{|k| \le \lceil 2^{j/2} \rceil}\sum_{m \in \bZ^2} 2^{4j} |\langle g,\psi_{j,k,m}\rangle|^2
&=& \sum_{j=0}^{\infty}\sum_{|k| \le \lceil 2^{j/2} \rceil}\sum_{m \in \bZ^2} \left|\left\langle \frac{\partial^2}{\partial x_1^2}g,
\eta_{j,k,m}\right\rangle\right|^2
\\
&<& C \cdot\left \| \frac{\partial^2}{\partial x_1^2} g \right \|_2^2.
\end{eqnarray*}
The proof is complete.
\end{proof}
This now enables us to derive Proposition \ref{prop:main1} as a corollary.
\begin{proof}{\em \hspace*{-0.2cm} (Proof of Proposition \ref{prop:main1})}.
Set
\[
\tilde{\Lambda}_j = \{\lambda \in \Lambda_j : \Sp(\psi_{\lambda}) \cap \Sp(g) \neq \emptyset \},\quad j > 0,
\]
i.e., $\tilde{\Lambda}_j$ is the set of indices in $\Lambda_j$ associated with shearlets
whose support intersects the support of $g$. Then, for each $J > 0$, we have
\beq \label{eq:NJ}
N_{J} = \Big|\bigcup_{j=0}^{J-1} \tilde{\Lambda}_j \Big| \sim 2^{2J}.
\eeq
Now, first observe that there exists some $C>0$ such that
\begin{eqnarray*}
\sum_{j = 1}^{\infty}2^{4j} \sum_{n>N_{j}} |\theta(g)|_n^2
& \leq & C \cdot \sum_{j=1}^{\infty}\sum_{j'=j}^{\infty}\sum_{k,m} 2^{4j} |\langle \wqqq{g},\psi_{j',k,m}\rangle|^2\\
& = & C\cdot\sum_{j'=1}^{\infty}\sum_{k,m}|\langle \wqqq{g},\psi_{j',k,m}\rangle|^2\left( \sum_{j=1}^{j'}2^{4j}\right).
\end{eqnarray*}
By Lemma \ref{lemma:smoothpart}, this implies
\[
\sum_{j = 1}^{\infty}2^{4j} \sum_{n>N_{j}} |\theta(g)|_n^2
\leq C \cdot \sum_{j'=1}^{\infty}\sum_{k,m}2^{4j'} |\langle \wqqq{g},\psi_{j',k,m}\rangle|^2 < \infty
\]
and hence, also by \eqref{eq:NJ},
\[
\sum_{n>N_{j}} |\theta(g)|_n^2 \leq C \cdot (2^{2j})^{-2} \leq \wq{C} \cdot N_{j}^{-2}.
\]
Finally, let $N>0$. Then there exists a positive integer $j_0>0$ satisfying
\[
N \sim N_{j_0} \sim 2^{2j_0},
\]
and the claim is proved.
\end{proof}
\section{Analysis of Shearlet Coefficients associated with the Discontinuity Curve}
\label{sec:discontinuity}
\subsection{Proof of Proposition \ref{prop:main2}}
Let $(j,k,m) \in \Lambda_{j,p}$, and fix $\hat{x}=(\hat{x}_1,\hat{x}_2) \in \wqq{\intt(\cQ_{j,p}) \cap \intt(\Sp(\psi_{\lambda}))} \cap \partial B$.
Let $s$ be the slope of the tangent to the edge curve $\partial B$
at $(\hat{x}_1,\hat{x}_2)$ as defined in Proposition \ref{prop:main2}, i.e.,
if $\partial B$ is parameterized by $(E(x_2),x_2)$ in the interior of $\cS_{j,p}$, then $s = E'(\hat x_2)$,
and if $\partial B$ is parameterized by $(x_1,E(x_1))$ in the interior of $\cS_{j,p}$, then $s = E'(\hat x_1)^{-1}$,
where we now assume that $E'(\hat x_1) \neq 0$ and consider the case $E'(\hat x_1) = 0$ later.
By translation symmetry, WLOG we can assume
that the edge curve satisfies $E(0)=0$ with $(\hat{x}_1,\hat{x}_2) = (0,0)$.
Further, since the conditions (i) and (ii) in Theorem \ref{theo:main}
are independent on the translation parameter $m$, it does not play a role in our analysis. Hence, WLOG we choose $m = 0$.
Also, since $\psi$ is compactly supported, there exists some $L > 0$ such that $\Sp{(\psi)} \subset [-L/2,L/2]^2$. By a rescaling
argument, we can might \wq{assume} $L=1$. Even more, WLOG we can assume that $\Sp{(\psi)} = [-1/2,1/2]^2$, which implies
\begin{eqnarray} \nonumber
\Sp(\psi_{j,k,0})
& = & \{ x \in \RR^2 : -2^{-j/2} k x_2 -2^{-j-1}\leq x_1 \leq -2^{-j/2} k x_2+2^{-j-1},\\ \label{eq:supp}
& & -2^{-\frac{j}{2}} \leq 2x_2 \leq 2^{-\frac{j}{2}}\},
\end{eqnarray}
since this will not change our asymptotic estimate for $|\Lambda_{j,p}(\epsilon)|$.
Let now $f \in \cE^2(\nu)$, and select ${\mathcal{P}}$ to be the smallest parallelogram which entirely contains the discontinuity curve
parameterized by $(E(x_2),x_2)$ or $(x_1,E(x_1))$ in the interior of $\Sp(\psi_{j,k,0})$ and \wq{whose two sides are parallel to
the tangent $x_1 = sx_2$ to the discontinuity curve at $(\hat{x}_1,\hat{x}_2) = (0,0)$.} For an illustration, we refer to Figure \ref{fig:edgecurve1}.
We now split the coefficients $|\langle f,\psi_{j,k,0}\rangle|$ into the part `inside the parallelogram' and `outside' of it
exploiting the shearing property of shearlets for the second part, and obtain
\begin{equation}\label{eq:shear_coeff}
|\langle f,\psi_{j,k,0}\rangle| = \wq{|\langle \chi_{\mathcal{P}} f,\psi_{j,k,0}\rangle|}
+ |\langle \chi_{{\mathcal{P}}^c}f(S_s \cdot),\psi_{j,\hat k,0}\rangle|
\end{equation}
where $\hat k = k+2^{j/2}s$. From now on, we assume that $\hat k < 0$ with $\hat k = k+2^{j/2}s$. The case $k+2^{j/2}s \ge 0$ can be handled similarly.
Let us start by estimating the first term \wq{$|\langle \chi_{\mathcal{P}}f,\psi_{j,k,0}\rangle|$} in \eqref{eq:shear_coeff}
stated as
{\bf Claim 1.}
\beq \label{eq:term1}
\wq{|\langle \chi_{\mathcal{P}}f,\psi_{j,k,0}\rangle|} \leq C \cdot \frac{(1+|s|^2)^{3/2}}{2^{3j/4} \cdot |\hat{k}|^{3}}.
\eeq
\begin{figure}[h]
\begin{center}
\hspace{-50pt}
\includegraphics[width=13cm]{./figs/Proof22_I_new.eps}
\put(-85,310){\footnotesize{$x_1=E(x_2)$ or $x_2 = E(x_1)$}}
\put(-297,205){\footnotesize{$ 2^{-j/2}$}}
\put(3,275){\footnotesize{$\mathcal{P}$}}
\put(-165,287){\footnotesize{$2^{-j}$}}
\put(-210,300){\footnotesize{$x_1 = sx_2$}}
\put(-70,210){\footnotesize{$d$}}
\put(-60,165){\footnotesize{$\tilde{d}$}}
\end{center}
\vspace{-130pt}
\caption{A shearlet $\psi_{j,{k},0}$ intersecting the edge curve $x_1 = E(x_2)$ or \wq{$x_2=E(x_1)$.} The right hand side shows
a magnification of the parallelogram \wql{ $\mathcal{P}$}.}
\label{fig:edgecurve1}
\end{figure}
First, notice that the tangent to the edge curve $(E(x_2),x_2)$ (or $(x_1,E(x_1))$) is given by $x_1 = sx_2$. We now assume that the edge curve is
contained in a set $\{x \in \R^2 : x_1 \ge sx_2 \}$, and just remark that the general case can be handled similarly.
Let now $d$ be the length of that side of $\mathcal{P}$, which is a part of the tangent $x_1 = sx_2$. We observe that $d$ is in fact the
distance between two points, in which the tangent $x_1=sx_2$ intersects the boundary of $\Sp (\psi_{j,k,0})$. For an illustration we
wish to refer to Figure \ref{fig:edgecurve1}. From this observation, it follows that
\[
d = \frac{2^{-j/2}\sqrt{1+|s|^2}}{|\hat{k}|}.
\]
We now let $\tilde{d}$ be the height of $\mathcal{P}$. Since the edge curve can be parameterized by a $C^2$ function $E$ with bounded curvature,
\[
\tilde{d} \leq C \cdot \Bigl(\frac{2^{-j/2}\sqrt{1+|s|^2}}{|\hat{k}|}\Bigr)^2.
\]
Summarizing, the volume of \wq{$\mathcal{P}$} can be estimated as
\[
|\mathcal{P}| \le C \cdot \frac{(1+|s|^2)^{3/2}}{2^{3j/2} \cdot {|\hat{k}|^3 }}.
\]
This implies
\[
\wq{|\langle f \chi_{\mathcal{P}},\psi_{j,k,0}\rangle|}
\le C \cdot 2^{3j/4} \cdot \|f\|_{\infty}\cdot \|\psi\|_{\infty} \cdot \frac{(1+|s|^2)^{3/2}}{2^{3j/2} \cdot {|\hat{k}|^3}}
\leq C \cdot \frac{(1+|s|^2)^{3/2}}{2^{3j/4} \cdot |\hat{k}|^{3}},
\]
and Claim 1, i.e., estimate \eqref{eq:term1}, is proved.
Next, we estimate the second term, i.e., $|\langle \chi_{\mathcal{P}^c}f(S_s \cdot),\psi_{j,\hat k,0}\rangle|$, in \eqref{eq:shear_coeff} stated as
{\bf Claim 2.}
\beq \label{eq:term2}
|\langle \chi_{\mathcal{P}^c}f(S_s \cdot),\psi_{j,\hat k,0}\rangle|
\leq C \cdot (1+|s|)^2 \cdot \left( \frac{1}{2^{3j/4} \cdot |\hat{k}|^{3}}+\frac{1}{2^{7j/4} \cdot |\hat{k}|^2}\right).
\eeq
Notice that $S_s^{-1}\mathcal{P}$ entirely contains the edge curve of $f(S_s \cdot)$ in the interior of $\Sp(\psi_{j,\hat k,0})$ and that the boundary of
the parallelogram $S_s^{-1}\mathcal{P}$ consists of two vertical line segments in the interior of $\Sp (\psi_{j,\hat k,0})$ (see Figure \ref{fig:edgecurve1_1}).
By translation symmetry, this implies that for proving Claim 2, it suffices to estimate
\[
\langle f_0(S_s\cdot)\chi_{\Omega},\psi_{j,\hat{k},0}\rangle,
\]
where $\Omega = \{(x_1,x_2) \in \bR^2:x_1>0\}$, $f_0 \in C^2(\RR^2)$ compactly supported in $[0,1]^2$
and $\sum_{|\alpha| \le 2}\|D^{\alpha}f_0\|_{\infty} \le 1$. We wish to mention that the consideration of the case $x_1>0$ -- compare the definition
of the set $\Omega$ -- is by no means restrictive, since the case $x_1 \leq 0$ can be handled in a similar way.
\begin{figure}[h]
\begin{center}
\hspace{-50pt}
\includegraphics[width=13cm]{./figs/Proof22_I.eps}
\put(-277,184){\footnotesize{$ 2^{-j/2}$}}
\put(-5,270){\footnotesize{$S^{-1}_s\mathcal{P}$}}
\put(-195,277){\footnotesize{$2^{-j}$}}
\put(-85,200){\footnotesize{$\frac{2^{-j/2}}{|\hat{k}|}$}}
\put(-65,145){\footnotesize{$\sim \frac{(1+|s|^2)^{3/2}}{2^j|\hat k|^2}$}}
\end{center}
\vspace{-130pt}
\caption{A shearlet $\psi_{j,\hat{k},0}$ intersecting \wq{the parallelogram \wql{ $S_s^{-1}\mathcal{P}$}}. The right hand side shows
a magnification of the parallelogram \wql{ $S_s^{-1}\mathcal{P}$}. }
\label{fig:edgecurve1_1}
\end{figure}
Now, again by translation symmetry, we may assume that $\partial (\supp \psi_{j,\hat{k},0})$ intersects the origin.
In particular, we may translate $\Sp (\psi_{j,\hat k,0})$ defined in \eqref{eq:supp} so that we might now consider
\[
\Sp (\psi_{j,\hat k,0}) + (-2^{j-1},0)\]
as the support of $\psi_{j,\hat k, 0}$. We refer the reader to Figure \ref{fig:edgecurve2} for an illustration.
This implies that there is one side of the boundary $\partial (\supp \psi_{j,\hat{k},0})$, which is a part of the line
\[
\mathcal{L} = \{(x_1,x_2) \in \RR^2 : x_2 = (-2^{j/2}/\hat{k}) \cdot x_1\}
\]
with slope $-2^{j/2}/\hat{k}$, as described in Figure \ref{fig:edgecurve2}. Applying the Taylor expansion for \wq{$f_0(S_s \cdot)$} at
each point $x = (x_1,x_2) \in \mathcal{L}$, we obtain
\[
\wq{ f_0(S_s x) }= a(x_1)+b(x_1)\left(x_2+\frac{2^{j/2}}{\hat{k}} \cdot x_1\right)+c(x_1,x_2)\left(x_2+\frac{2^{j/2}}{\hat{k}}\cdot x_1\right)^2,
\]
where $a(x_1),b(x_1)$ and $c(x_1,x_2)$ are all bounded in absolute value by $C(1+|s|)^2$. This implies (compare also an illustration
of the area of integration in Figure \ref{fig:edgecurve2})
{\allowdisplaybreaks
\begin{eqnarray}\label{eq:splitting0}
|\langle \wql{ \wq{f_0(S_s \cdot)\chi_{\Omega}}},\psi_{j,\hat{k},0}\rangle|
&=& \left| \int_{0}^{-\frac{\hat{k}}{2^{j}}}
\int_{-\frac{2^{j/2}}{\hat{k}}\cdot x_1}^{-\frac{2^{j/2}}{\hat{k}}\cdot x_1- \frac{2^{-j/2}}{\hat{k}}}
\wq{f_0(S_s x)} \psi_{j,\hat{k},0}(x) \, dx_2dx_1\right| \nonumber \\
&\le& C\cdot \wql{ (1+|s|)^2} \cdot \left| \int_{0}^{-\frac{\hat{k}}{2^{j}}}\wq{\sum_{\ell=1}^{3}I_{\ell}(x_1)} \, dx_1 \right|,
\end{eqnarray}
}
where
{\allowdisplaybreaks
\begin{eqnarray*}
I_1(x_1) &=& \left| \int_{0}^{- \frac{2^{-j/2}}{\hat{k}}}T_{\beta}\left(\psi_{j,\hat{k},0}(x_1,x_2)\right)dx_2\right|, \\
I_2(x_2) &=& \left| \int_{0}^{- \frac{2^{-j/2}}{\hat{k}}}x_2 \cdot T_{\beta}\left(\psi_{j,\hat{k},0}(x_1,x_2)\right)dx_2\right|, \\
I_3(x_2) &=& \left| \int_{0}^{- \frac{2^{-j/2}}{\hat{k}}}x_2^2 \cdot T_{\beta}\left(\psi_{j,\hat{k},0}(x_1,x_2)\right)dx_2\right|,
\end{eqnarray*}
}
with $T_{\beta}$ being the translation operator defined by $T_{\beta}(f) = f(\cdot -\beta)$ and $\beta \in \RR$ being chosen
to be $\beta = (0,(2^{j/2}/\hat{k}) \cdot x_1).$
\begin{figure}[ht]
\begin{center}
\includegraphics[width=5.5cm]{./figs/Proof22_IIa.eps}
\put(-142,78){\footnotesize{$\frac{2^{-j/2}}{|\hat{k}|}$}}
\put(-70,42){\footnotesize{$\frac{|\hat{k}|}{2^j}$}}
\put(-18,105){\footnotesize{$ 2^{-j/2}$}}
\put(-110,0){\footnotesize{$\mathcal{L}$}}
\put(-46,160){\footnotesize{$ 2^{-j}$}}
\end{center}
\caption{\wang{A shearlet $\psi_{j,\hat{k},0}$ intersecting the edge curve $x_1 = 0$ such
that $\supp(\psi_{j,\hat{k},0})$ intersects the positive $x_2$ axis and
$\partial (\supp \psi_{j,\hat{k},0})$ intersects the origin. The illustration also highlights the integration area for \eqref{eq:splitting0}. }}
\label{fig:edgecurve2}
\end{figure}
We first estimate $I_1(x_1)$. We observe that, since
\[
\{(x_1,x_2) \in \bR^2 : \psi_{j,\hat{k},0}(x_1,x_2) \ne 0\} \subset [0,2^{-j/2}/|\hat{k}|] \quad \text{for a fixed } x_1>0,
\]
WLOG, for any $x_1 > 0$, the interval $[0, 2^{-j/2}/\wqq{|\hat{k}|}]$
for the range of the integration in $I_1(x_1)$ can be replaced by $\bR$ (see also Figure \ref{fig:edgecurve2}).
Therefore, we have
\beq\label{eq:estimateI}
I_1(x_1) = \left|\int_{\bR} \psi_{j,\hat{k},0}(x_1,x_2)dx_2\right|
= \left| \int_{\bR} \hat{\psi}_{j,\hat{k},0}(\xi_1,0) \cdot e^{2\pi i \wql{ x_1}\xi_1}d\xi_1\right|.
\eeq
Now
\[
|\hat{\psi}_{j,\hat{k},0}(\xi_1,\xi_2)| = 2^{-3j/4} \cdot |\hat\psi(2^{-j}\xi_1,2^{-j/2}\xi_2-2^{-j}\hat{k}\xi_1)|.
\]
and hence, by hypothesis (i) from Theorem \ref{theo:main},
\begin{equation}\label{eq:fourier1}
|\hat{\psi}_{j,\hat{k},0}(\xi_1,0)| \leq 2^{j/4} \cdot |2^{-j}h(2^{-j}\xi_1)| \cdot |\hat{k}|^{-\gamma}.
\end{equation}
By \eqref{eq:estimateI} and \eqref{eq:fourier1}, it follows that
\begin{equation}\label{eq:mainestimate1}
I_1(x_1) \le C \cdot \frac{2^{j/4}}{|\hat{k}|^{\gamma}} \quad \mbox{for some } C > 0.
\end{equation}
Next, we estimate $I_2(x_1)$. We have
\begin{eqnarray*}
I_2(x_1)
& \leq & \left|\int_{\bR} x_2 \cdot \psi_{j,\hat{k},0}(x_1,x_2)dx_2\right|
+ \frac{2^{j/2}}{|\hat{k}|} \cdot |x_1| \cdot \left|\int_{\bR}\psi_{j,\hat{k},0}(x_1,x_2)dx_2 \right|\\
& = & S_1 + S_2.
\end{eqnarray*}
To estimate $S_1$, observe that, by hypothesis (ii) from Theorem \ref{theo:main},
\begin{eqnarray}\nonumber
S_1 & = & \wqqq{\frac{1}{2\pi}}\left| \int_{\bR}
\left(\frac{\partial}{\partial \xi_2} \hat{\psi}_{j,\hat{k},0}\right)(\xi_1,0) e^{2\pi i \xi_1 x_1}d\xi_1\right|\\ \label{eq:estimateS1}
& \le & \frac{1}{2\pi}\int_{\bR} (2^{-\frac{j}{4}} \cdot h(2^{-j}\xi_1)) \cdot 2^{-j} \cdot |\hat{k}|^{-\gamma} d\xi_1.
\end{eqnarray}
By \eqref{eq:mainestimate1} and the fact that $0 \le x_1 \le \frac{|\hat{k}|}{2^j}$, the second
term $S_2$ can be estimated as
\beq \label{eq:estimateS2}
S_2 \le C \cdot \left(\frac{2^{j/2}}{|\hat{k}|}|x_1|\right) \cdot \frac{2^{j/4}}{|\hat{k}|^{\gamma}}
\le C \cdot \left( 2^{-j/2}\right) \cdot \frac{2^{j/4}}{|\hat{k}|^{\gamma}}
\le \frac{C}{2^{j/4} \cdot |\hat{k}|^{\gamma}}.
\eeq
Concluding from \eqref{eq:estimateS1} and \eqref{eq:estimateS2}, we obtain
\begin{equation}\label{eq:mainestimate2}
I_2(x_1) \le S_1 + S_2 \le \frac{C}{2^{j/4} \cdot |\hat{k}|^{\gamma}}.
\end{equation}
Finally, we estimate $I_3(x_1)$. For this, notice that $2^{-3j/4}T_{\beta}(\psi_{j,\hat{k},0}(x_1,x_2))$ is bounded, hence
\begin{equation}\label{eq:mainestimate3}
I_3(x_1) \le 2^{\frac{3}{4}j} \cdot C \cdot \left|\int_{0}^{\frac{-1}{2^{j/2}\hat{k}}}x_2^2 \, dx_2\right| \leq \frac{C}{2^{\frac{3}{4}j} \cdot |\hat{k}|^{3}}.
\end{equation}
Summarizing, by \eqref{eq:splitting0}, \eqref{eq:mainestimate1}, \eqref{eq:mainestimate2}, and \eqref{eq:mainestimate3},
\begin{eqnarray*}
|\langle \wq{f_0(S_s \cdot)\chi_{\Omega}},\psi_{j,\hat{k},0}\rangle|
& \le & C \cdot (1+|s|)^2 \cdot \int_{0}^{\frac{|\hat{k}|}{2^j}}\left(\frac{2^{\frac{1}{4}j}}{|\hat{k}|^{\gamma}} \nonumber
+\frac{1}{2^{\frac{3}{4}j} \cdot |\hat{k}|^3}\right)dx_1\\
& \le & C \cdot (1+|s|)^2 \cdot \left( \frac{1}{2^{3j/4} \cdot |\hat{k}|^{3}}+\frac{1}{2^{\frac{7}{4}j} \cdot |\hat{k}|^2}\right),
\end{eqnarray*}
and Claim 2, i.e., estimate \eqref{eq:term2}, is proved.
From Claim 1 and 2, i.e., from \eqref{eq:term1} and \eqref{eq:term2}, we conclude that
\[
|\langle f,\psi_{j,\wq{{k}},0}\rangle| \leq C\Bigl[ (1+|s|)^2\Bigl( \frac{1}{2^{\frac{3}{4}j} \cdot |\hat k|^3}+\frac{1}{2^{\frac{7}{4}j} \cdot |\hat k|^2}\Bigr)
+(1+|s|^2)^{3/2}\frac{1}{2^{\frac{3}{4}j} \cdot |\hat k|^3}\Bigr].
\]
This implies \eqref{eq:estimate1} and \eqref{eq:estimate2} except for the case $s = \infty$, which we will study now.
Finally, we consider the case $s = \infty$, i.e., we assume that the edge curve is parameterized by $(x_1,E(x_1))$ in the interior of $\cS_{j,p}$
such that $E \in C^2$ and $E'(\hat x_1) = 0$. As before, let $f \in \cE^2(\nu)$, and select ${\mathcal{P}}$ to be the smallest parallelogram which
entirely contains the discontinuity curve in the interior of $\Sp(\psi_{j,k,0})$ \wq{and whose two sides are parallel to $x_2 = 0$}. Similarly as before, we consider
\begin{equation} \label{eq:shear_coeff2}
|\langle f,\psi_{j,k,0}\rangle| = |\langle \chi_{\mathcal{P}}f,\psi_{j,k,0}\rangle| + |\langle \chi_{{\mathcal{P}}^c}f,\psi_{j,k,0}\rangle|.
\end{equation}
and estimate both terms on the RHS separately.
We first consider the term $|\langle \chi_{\mathcal{P}}f,\psi_{j,k,0}\rangle|$. Using similar arguments as before, which we decided not to
include in detail to avoid repetitions, one can prove that
\begin{equation}\label{eq:t1}
|\langle \chi_{\mathcal{P}}f,\psi_{j,k,0}\rangle| \leq C \cdot 2^{-\frac{9}{4}j}.
\end{equation}
\wq{Turning our attention} to the second term $|\langle \chi_{{\mathcal{P}}^c}f,\psi_{j,k,0}\rangle|$, we first observe that
\wq{$\mathcal{P}$ entirely contains the edge curve of $f$ in the interior of $\Sp(\psi_{j, k,0})$ and that the boundary of
the parallelogram $\mathcal{P}$ consists of two horizontal line segments in the interior of $\Sp (\psi_{j,k,0})$ (see Figure \ref{fig:edgecurve3}).
By translation symmetry, this implies that for the second term in \eqref{eq:shear_coeff2}, i.e., for $|\langle \chi_{{\mathcal{P}}^c}f,\psi_{j,k,0}\rangle|$,
it suffices to estimate} $|\langle f_0\chi_{\tilde{\Omega}},\psi_{j,k,0}\rangle|$, where $\tilde{\Omega} = \{(x_1,x_2) \in \bR^2:x_2>0\}$ and
$f_0 \in C^2(\RR^2)$ compactly supported in $[0,1]^2$ with $\sum_{|\alpha| \le 2}\|D^{\alpha}f_0\|_{\infty} \le 1$.
As before, the consideration of the case $x_2>0$ -- compare the definition of the set $\tilde{\Omega}$ -- is by no means restrictive, since the case
$x_2 \leq 0$ can be handled in a similar way. Observe that \wq{the (horizontal) vanishing moment condition}
follows from condition (i) in Theorem \ref{theo:main}. Let us briefly think about this: Letting $\xi_2$ be fixed, condition (i)
immediately implies $\hat \psi(0,\xi_2) = 0$. Then, by applying Taylor expansion, it follows that
$\hat \psi(\xi_1,\xi_2) = \frac{\partial\hat \psi}{\partial \xi_1}(0,\xi_2) \cdot \xi_1 + \mathcal{O}(\xi_1^2)$, and,
again by condition (i), we have $\frac{\hat \psi(\xi_1,\xi_2)}{\xi_1} \to 0$ as $\xi_1 \rightarrow 0$, hence
$\frac{\partial\hat \psi}{\partial \xi_1}(0,\xi_2) = 0$. This procedure can now be continued.
Especially, condition (i) from Theorem \ref{theo:main} implies
\begin{equation}\label{eq:moment}
\int_{\bR} x_1^{\ell}\cdot\psi(x_1,x_2)dx_1 = 0 \quad \mbox{for all } x_2 \in \bR \mbox{ and } \ell=0,1.
\end{equation}
Further, we utilize that the shearing operation $S_k$ preserves vanishing moments along the $x_1$ axis.
That is,
\wq{
\[
\int_{\bR}x_1^{\ell} \cdot \psi(S_k (x_1,x_2)^T)dx_1 = 0
\quad \mbox{for all }k, x_2 \in \bR \,\, \mbox{and}\,\, \ell=0,1.
\]
}
This can be seen as follows. For $\ell=0, 1$ and for a fixed $x_2 \in \bR$, the function
$x_1 \mapsto (x_1-kx_2)^{\ell}$ is a polynomial
of degree less than or equal to $\ell$, hence, by condition \eqref{eq:moment} on the number of
vanishing moments on $\psi$, we have
\[
\int_{\bR}x_1^{\ell} \cdot \psi(S_k (x_1,x_2)^T)dx_1 = \int_{\bR}(x_1-kx_2)^{\ell} \cdot \psi(x_1,x_2)dx_1 = 0
\quad \mbox{for all }k \in \bR.
\]
\begin{figure}[ht]
\begin{center}
\includegraphics[width=5.5cm]{./figs/Proof22_IIb.eps}
\put(-38,155){\footnotesize{$2^{-j}$}}
\put(-9,105){\footnotesize{$2^{-j/2}$}}
\end{center}
\caption{\wang{A shearlet $\psi_{j,\hat{k},0}$ intersecting the edge curve $x_2 = 0$ such
that $\supp(\psi_{j,\hat{k},0})$ intersects the positive $x_1$ axis and
$\partial (\supp \psi_{j,\hat{k},0})$ intersects the origin. The illustration also highlights the integration area for \eqref{eq:t2}. }}
\label{fig:edgecurve3}
\end{figure}
Employing Taylor expansion and integration (compare Figure \ref{fig:edgecurve3}) similar to the
proof of the previous case, we finally obtain
\begin{equation}\label{eq:t2}
|\langle f_0\chi_{\tilde{\Omega}},\psi_{j,\wq{{k}},0} \rangle|
\le C \cdot 2^{3j/4} \cdot \int_{0}^{\cdot 2^{-j/2}}\int_{-2^{-j}}^{0} \, x_1^2 dx_1dx_2
\le C \cdot 2^{-11j/4}
\end{equation}
\wql{
By \eqref{eq:t1} and \eqref{eq:t2}, we obtain \eqref{eq:estimate2} when $s = \infty$. This completes the proof. } | 173,970 |
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Our class on Reform Responsa has been going great, and we’ve got one more session – Thursday at 7:30 p.m. Even if you haven’t joined us, yet, you’re welcome to do so – we look at different cases each time, so you can jump in whenever you want to.
The three topics for this week (subject to change, of course) will be:
- What does Reform Judaism say about in vitro fertilization, and about the status of an embryo so created?
- Would Reform Judaism support compulsory immunization of children?
- Does Reform Judaism endorse, allow or forbid same-sex marriages?
As a prelude to the class, I want to open up a discussion on one of these. I really want to do the last one, but I’m going to avoid it, because it’s often hard to have a controlled, measured discussion about topics like that in a forum like this. So, let’s stick with immunization.
Speaking from your own perspective – is it right to compel parents to immunize their children? If you have any knowledge about this, what do you think Reform Judaism would/should say about that? And, what are the reasons behind your opinions? That’s at least as important as the opinions themselves.
Let the games begin…
2 comments:
First off, as you well know, I have no knowledge of Jewish law, so I can only comment on my feelings on this matter.
I have to say, as a parent, I would absolutely despise being forced to do anything to my children, especially something that can affect their health and well-being. While I completely understand the need for universal vaccination for the benefit of everyone, it should still be up to the parents to make that decision. I can identify with those who choose not to vaccinate, since I have made the same choice with my animals, the difference being that we are given the opportunity to titer our animals to see if the vaccinations are needed rather than just blindly giving them shots at suggested intervals. I would much prefer this approach with my kids, too, since I have yet to see studies that have been done showing that kids really need to be vaccinated as often as they are.
As for those parents who choose not to vaccinate at all, I feel that they should be afforded that choice, even though in my mind it's foolish not to protect your children from preventable diseases. That said, I'm not sure I will vaccinate the girls for cervical cancer simply because the vaccine has so little history and some of the side-effects have been quite severe. And really, I'm hoping they will make smart and safe choices in that regard as they get older. I even hesitated about the H1N1 shots this winter for the same reason, but Mike was absolutely insistent about them and there were no immediate reasons to not give them. Hopefully that will continue to be true for the future.
So, no, I don't think anyone should be forced to vaccinate, but I do think education about the importance to the individual's health and the societal benefits should be emphasized to allow parents to make the correct decision for their own children.
So, Missy, being a bit of a Devil's Advocate, you say, "as a parent, I would absolutely despise being forced to do anything to my children, especially something that can affect their health and well-being." How about laws involving car-seats? Isn't that another example of the government telling us how to parent, no matter what we think? | 169,880 |
Apologetics: Pre-Evangelism or Evangelism?
by Jacob Allee
Professional apologists have disagreed on the matter but it is still a worthy question, “What is the relationship of apologetics to evangelism?” Some have, of course, argued that apologetics has no relationship to evangelism and would rather relegate apologetics to be used only for defending and strengthening the faith of believers. Apologetics in their mind is only defensive and has no offensive value. In this view one must simply preach the gospel and then once people believe they can use apologetics to further strengthen their faith.
Others see apologetics having no place in Christian ministry at all because they see reason and evidence as being counter to the gospel that is to be received and held by faith alone (as if having good reasons to believe nullifies faith). Those in this category are they who see Christian faith as necessarily a blind leap. Theologians and philosophers like Karl Barth and Soren Kierkegaard are representative of this ideology which sees reason and evidence as actually destructive to faith since they interpret faith to mean believing against or without reason.
I would argue against both of these view (while finding the first view more acceptable than the second) because I think they misunderstand or don’t take seriously enough what Scripture itself says as it relates to faith and apologetics. In the first case it is clear that, for instance, Paul uses apologetics in an offensive way (offensive as in proactive rather than rude) in Acts 17 at the Areopagus. Paul reasons with the philosophers at Mars Hill who did not know the God of the Bible and did not accept the Old Testament as authoritative and he engages them with an apologetic argument using things from their own culture. Paul makes use of the idol
dedicated to “an unknown god” and he even makes use of secular literature that was known to his audience to show that the ideas he was presenting are no completely foreign to them. So without going to a lot more detail at this point, it’s safe to say that apologetics can be used offensively and not merely defensively.
To the group who see faith and reason as polar opposites I would say that they simply need to look again at Scripture and see that the Bible never uses the term faith to indicate a “blind leap.” Paul is said to have reasoned with the Jews (Acts 17:2; 17:17; 18:4 and 18:19) he also reasoned with the gentiles as we just said in Acts 17 at Mars Hill, and in Acts 24:25 he even reasoned with the governor Felix. Likewise Peters sermon at Pentecost in Acts 2 is not devoid of reason nor evidence as he appeals to the Jews that they need to believe in Jesus as the Messiah. 1 Peter 3:15 tells believers that they need to be prepared to give an answer for the hope that they have when unbelievers ask them about their faith, in other words, what reasons do you have for what you believe? Faith in Scripture communicates the idea of trust and it is not opposed to reason. In fact one could easily argue that the more reasons you have to trust someone or something, the more faith you have in them.
So clearly apologetics can be used offensively and the Bible doesn’t present it as contrary to faith but, rather, supplementary and useful both for sharing with unbelievers and strengthening the faith of believers. So with those two categories out of the way the question still remains “What is apologetics proper relationship to evangelism?” Historically it has often been seen as pre-evangelism. Many have argued that the purpose of apologetics is to clear away objections and to till the ground making it fertile for seed of the gospel to be planted…
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Apologetics: Pre-Evangelism or Evangelism? –: Pre-Evangelism or Evangelism? –
RECOMMENDED APOLOGETICS RESOURCES FOR FURTHER READING:
When God Goes to Starbucks: A Guide to Everyday Apologetics
Mere Apologetics: How to Help Seekers and Skeptics Find Faith
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I’m just an average pew sitter. From my perspective your discussion is just academic grist. Testing it all with my Bible I find no definition of apologetics, pre-evangelism or evangelism. Christ just wants us to share the message without getting hung up on how we define what we are doing. I’ve seen “pre-evangelism” used in the most horrid of ways – the author was attempting to distinguish it with evangelism such that evangelism was said to only be the province of ordained clergy…..what a put down to everyone else! Let us just focus on getting the message out to the people and then let the Holy Spirit move people’s hearts. | 17,759 |
TITLE: Relations between p norms
QUESTION [86 upvotes]: The $p$-norm on $\mathbb R^n$ is given by $\|x\|_{p}=\big(\sum_{k=1}^n
|x_{k}|^p\big)^{1/p}$. For $0 < p < q$ it can be shown that $\|x\|_p\geq\|x\|_q$ (1, 2). It appears that in $\mathbb{R}^n$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$(3), $\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $\mathbb R^n$. For instance, for $n=2$ and $n=3$ one can see that for $0 < p < q$, the spheres with radius $\sqrt n$ with $\|\cdot\|_p$ inscribe spheres with radius $1$ with $\|\cdot\|_q$.
It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $n=2$ it is easily proven (see below), but not for $n>2$. So my questions are:
How can relation (3) be proven for arbitrary $n\,$?
Can this be generalized into something of the form $\|x\|_{p} \leq C \|x\|_{q}$ for arbitrary $0<p < q\,$?
Do any of the relations also hold for infinite-dimensional spaces, i.e. in $l^p$ spaces?
Notes:
$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$, hence $=2\|x\|_{2}^{2}$
$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$.
This works because $|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$, but only because $(|x_{1}| - |x_{2}|)^2 \geq 0$, while for more than two terms $\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$ gives an inequality that never gives the right signs for the cross terms.
REPLY [122 votes]: Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$
$$
\Vert x\Vert_1=
\sum\limits_{i=1}^n|x_i|=
\sum\limits_{i=1}^n|x_i|\cdot 1\leq
\left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}=
\sqrt{n}\Vert x\Vert_2
$$
Such a bound does exist. Recall Hölder's inequality
$$
\sum\limits_{i=1}^n |a_i||b_i|\leq
\left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}
$$
Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$
$$
\sum\limits_{i=1}^n |x_i|^p=
\sum\limits_{i=1}^n |x_i|^p\cdot 1\leq
\left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}}
\left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}
$$
Then
$$
\Vert x\Vert_p=
\left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq
\left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\=
n^{1/p-1/q}\Vert x\Vert_q
$$
In fact $C=n^{1/p-1/q}$ is the best possible constant.
For infinite dimensional case such inequality doesn't hold. For explanation see this answer. | 131,228 |
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Hello everyone,
We switched to a next-gen project beginning of the year. At the end of the year I'm working on a report of all the things we have done, however, I just noticed the sprint reports don't go back all the way until the start of the year. Is there a limit of how far back you can look? Is there a way to see the older data in next-gen some other way?
I remember in the old projects I was able to go back and few reports from more than 4 years ago.
Kind regards,
Bas
Welcome to Atlassian Community!
The only report on a next-gen project that has the option to filter based on a date it's the Cumulative flow diagram.
On my local site, I tested the report on a next-gen project where the older issue is from November 2018, but it's also possible to select another date and it shows on the report as well:
I searched here and didn't find limitations regarding the date range in this report.
Are you facing any issue when selecting the date? Does it show any error?
If possible, share a screenshot with us, just make sure to hide private information.. | 304,551 |
\begin{document}
\maketitle
\begin{abstract}
\paragraph{Abstract.}
Let $G(n,\, M)$ be the uniform random graph with $n$ vertices and $M$ edges.
Let $\circ$ be the maximum block-size of $G(n,\, M)$ or
the maximum size of its maximal $2$-connected induced subgraphs. We determine the expectation of $\circ$
near the critical point $M=n/2$. As $n-2M \gg n^{2/3}$, we find a constant $c_1$ such that
\[
c_1 = \lim_{n \rightarrow \infty} \left( 1 - \frac{2M}{n} \right) \, \qE{(\circ)} \, .
\]
Inside the window of transition of $G(n,\, M)$ with $M=\frac{n}{2}(1+\lambda n^{-1/3})$,
where $\lambda$ is any real number, we find an exact analytic expression for
\[
c_2(\lambda) = \lim_{n \rightarrow \infty}
\frac{\qE{\l(\circarg{\frac{n}{2}(1+\lambda n^{-1/3})}\r)}}
{n^{1/3}} \, .
\]
This study relies on the symbolic method and analytic tools coming from generating function theory which enable us
to describe the evolution of $n^{-1/3} \, \qE{\l(\circarg{\frac{n}{2}(1+\lambda n^{-1/3})}\r)}$ as a function of $\lambda$.
\end{abstract}
\section{Introduction}
Random graph theory~\cite{atrg,Bollobas,JLR2000} is an active area of research
that combines algorithmics, combinatorics, probability theory and graph theory.
The uniform random graph model $G(n, \, M)$ studied in~\cite{ER60} consists in $n$ vertices
with $M$ edges drawn uniformly at random
from the set of ${n \choose 2}$ possible edges. Erd\H os and R\'enyi showed that for many
properties of random graphs, graphs with a number of edges slightly less than a given threshold
are unlikely to have a certain property, whereas graphs with slightly more edges are almost
guaranteed to satisfy the same property, showing paramount changes inside their structures (refer
to as \textit{phase transition}). As shown in their seminal paper~\cite{ER60}, when
$M=\frac{cn}{2}$ for constant $c$ the largest component of $G(n, \, M)$ has a.a.s.
$O(\log{n}), \, \Theta(n^{2/3})$ or $\Theta(n)$ vertices according to
whether $c<1$, $c=1$ or $c>1$. This \textit{double-jump} phenomenon
about the structures of $G(n, \, M)$ was one of the most spectacular results in~\cite{ER60}
which later became a cornerstone of the random graph theory.
Due to such a dramatic change, researchers worked around the critical value $\frac{n}{2}$ and
one can distinguish three different phases: \textit{sub-critical} when
$(M-n/2) n^{-2/3} \rightarrow -\infty$, \textit{critical} $M=n/2 + O(n^{2/3})$ and
\textit{supercritical} as $(M-n/2)n^{-2/3} \rightarrow \infty$.
We refer to Bollob\'{a}s~\cite{Bollobas} and Janson, \L uczak and Ruci\'nski~\cite{JLR2000} for
books devoted to the random graphs $G(n,\, M)$ and $G(n,\, p)$.
If the $G(n,\, p)$ model is the one more commonly used today,
partly due to the independence of the edges, the $G(n,M)$ model
has more enumerative flavors allowing generating functions based approaches.
By setting $p=\frac{1}{n} + \frac{\lambda}{n^{4/3}}$, the stated results of this
paper can be extended to the $G(n,\,p)$ model.\\
\noindent
\textbf{Previous works.}
In graph theory, a block is a maximal 2-connected subgraph (formal definitions are given in Section~\ref{SEC_ENUMERATIVE}).
The problem of estimating the maximum block size has been well
studied for some class of graphs. For a graph drawn uniformly from the class
of simple labeled planar graphs with
$n$ vertices,
the expectation of the number of vertices
in the largest block is $\alpha n$ asymptotically almost surely (a.a.s)
where $\alpha \approx 0.95982$ \cite{PS10,GNR13}.
They found that the largest block in random planar graphs is related to a distribution
of the exponential-cubic type, corresponding to distributions that involve
the Airy function~\cite{Banderier}.
For the labeled connected class, these authors proved also independently
that a connected random planar graph has a unique block of linear size.
When we restrict to sub-critical graph (graph that the block-decomposition looks tree-like),
Drmota and Noy~\cite{DN13} proved that the maximum block size of a random connected graph in an
aperiodic\footnote{In the periodic case, $n \equiv1 \mod d$ for
some $d > 1$ (see ~\cite{DN13} for more details)} sub-critical graph class is $O(\log n)$.
For random maps (a map is a planar graph embedded in the plane),
Gao and Wormald~\cite{GaoW99} proved that a random map with $n$ edges has almost surely $n/3$ edges.
That is, the probability that the size of the largest block is about $n/3$
tends to $1$ as $n$ goes to infinity. This result is improved by Banderier \emph{et al.}~\cite{Banderier}
by finding the density Airy distribution of the map type.
Panagiotou~\cite{P09} obtained more general results for any graph class $\mathcal{C}$.
He showed that the size of largest block of a random graph from $\mathcal{C}$ with $n$ vertices and $m$ edges
belongs to one of the two previous categories ($\Theta(n)$ and $O(\log n)$).
In particular, the author pointed out that random planar graphs with $cn$ edges belong to the
first category, while random outerplanar and series-parallel graphs with fixed average degree belong to
the second category.
For the Erd\H os-R\'enyi $G(n,M)$ model, the maximum block-size is
implicitly a well-studied graph property when $M=\frac{cn}{2}$ for fixed $c<1$.
For this range, $G(n,M)$ contains only trees and unicyclic components a.a.s.~\cite{ER60}.
So, studying maximum block-size and the largest cycle
are the same in this case. Denote by $\circ$ the maximum block-size of $G(n,\, M)$.
It is shown in~\cite[Corollary 5.8]{Bollobas} that as $M=\frac{cn}{2}$ for fixed $c<1$
then $\circ$ is a.a.s
at most $\omega$ for any function $\omega = \omega(n) \rightarrow \infty$.
Pittel~\cite{Pittel88} then obtained the limiting distribution (amongst other results) for
$\circ$ for $c<1$. Note that the results of Pittel are extremely precise and include other
parameters of random graphs with $c$ satisfying $c<1-\varepsilon$ for fixed $\varepsilon>0$.
\noindent \textbf{Our results.}
In this paper, we study the fine nature of the Erd\H os and R\'enyi phase transition,
with emphasis on what happens as the number of edges is close to $\frac{n}{2}$~:
within the window of the phase transition and near to it, we quantify the
maximum block-size of $G(n,\,M)$.
\noindent
For sub-critical random graphs, our finding can be stated precisely as follows~:
\begin{theorem}\label{MAIN_SUBCRITICAL}
If $n-2M \gg n^{2/3}$, the maximum block-size $\circ$ of $G(n,\, M)$ satisfies
\beq\label{eq:MAIN_SUBCRITICAL}
\qE{(\circ)} \sim c_1 \left( \frac{n}{n-2M} \right) ,
\eeq
where $c_1 \approx 0.378\, 911$ is the constant given by
\beq\label{eq:c1}
c_1 = \int_0^{\infty} \left( 1 - e^{-E_1(v)}\right) dv \mbox{ with } E_1(x) = \frac{1}{2} \int_x^{\infty} e^{-t} \frac{dt}{t} \, .
\eeq
\end{theorem}
\noindent
For critical random graphs, we have the following~:
\begin{theorem}\label{MAIN_CRITICAL}
Let $\lambda$ be any real constant and $M=\frac{n}{2}(1+\lambda n^{-1/3})$. The maximum block-size $\circ$ of $G(n,\, M)$
verifies~:
\beq \label{eq:MAIN_CRITICAL}
\qE{(\circ)} \sim c_2(\lambda) \, {n^{1/3}} ,
\eeq
where
\beq \label{eq:c2}
c_2(\lambda) = \frac{1}{\alpha} \int_{0}^{\infty} \l( 1- \sqrt{2\pi} \sum_{r\geq 0} \sum_{d \geq 0}
A\l(3r +\frac{1}{2},\lambda\r) \, e^{-E_1(u)} \, \hardcoeffbis{r}{d}\l(e^{-u}\r) \r) du
\eeq
$E_1(x)$ is defined in (\ref{eq:c1}), $\alpha$ is the positive solution of
\beq \label{eq:alpha}
\lambda \;=\;\alpha^{-1}\,-\,\alpha \, ,
\eeq
the function $A$ is defined by
\ben \label{28INTRO}
A(y,\lambda)={e^{-\lambda^3\!/6}\over3^{(y+1)/3}}\sum_{k\ge0}
{\bigl(\half 3^{2/3}\lambda\bigr)^k\over k!\,\Gamma\bigl((y+1-2k)/3\bigr)} \,,
\een
and the $(\hardcoeffbis{r}{d}(z))$ are polynomials with rational coefficients defined recursively by (\ref{def_erk_bis}).
\end{theorem}
The accuracy of our results is of the same vein as the one on the probability of
planarity of the Erd\H{o}s-R\'enyi critical random graphs~\cite{pams15} or on the finite
size scaling for the core of large random hypergraphs~\cite{DEMBO} which have been
also expressed in terms of the Airy function. This function has been encountered in the physics of
random graphs \cite{JKLP93} and is shown in~\cite{FKP89} related to $A(y,\, \lambda)$
defined by (\ref{28INTRO}) and appearing in our formula (\ref{eq:c2}).
It is important to note that there is \textit{no discontinuity} between Theorems~\ref{MAIN_SUBCRITICAL} and
\ref{MAIN_CRITICAL}. First, observe that as $M=\frac{n}{2} - \frac{\lambda(n) n^{2/3}}{2}$ with
$1 \ll \lambda(n) \ll n^{1/3}$, equation (\ref{eq:MAIN_SUBCRITICAL}) states that
$\qE{(\circ)}$ is about $c_1 \frac{n^{1/3}}{\lambda(n)}$. Next, to see that this value matches
the one from (\ref{eq:MAIN_CRITICAL}), we argue briefly as follows.
In (\ref{eq:alpha}), as $\lambda(n) \rightarrow -\infty$
we have $\alpha \sim |\lambda(n)|$ and (see~\cite[equation (10.3)]{JKLP93})
\[
A\l(3r +\frac{1}{2},\lambda\r) \sim \frac{1}{\sqrt{2\pi} |\lambda(n)|^{3r} } \, .
\]
Thus, all the terms in the inner double summation 'vanish' except
the one corresponding to $r=0$ and $d=0$ (this term is the coefficient for
graphs without multicyclic components $\hardcoeff{0}{0}=1$). It is then remarkable that as $\lambda(n) \rightarrow -\infty$,
$c_2(\lambda(n))$ behaves as $\frac{c_1}{|\lambda(n)|}$.\\
\noindent
\textbf{Outline of the proofs and organization of the paper.}
In~\cite[Section 4]{FO-mappings}, Flajolet and Odlyzko described generating functions based
methods to study extremal statistics on random mappings. Random
graphs are obviously harder structures
but as shown in the masterful work of Janson {\it et al.}~\cite{JKLP93}, analytic combinatorics
can be used to study in depth the development of the connected components of $G(n,\, M)$.
As in~\cite{FO-mappings}, we will characterize the expectation of $\circ$
by means of truncated generating functions.
Given a family $\mathcal{F}$ of graphs, denote by $(F_n)$ the number of graphs of $\mathcal F$ with $n$ vertices.
The \textit{exponential generating function} (EGF for short) associated to the sequence $(F_n)$ (or family $\mathcal{F}$)
is $F(z) = \sum_{n\geq 0} F_n {z^n \over n!}$.
Let $F^{\left[k\right]}(z)$ be the EGF of the graphs in $\mathcal{F}$
but \textit{with all blocks of size at most} $k$.
From the formula for the mean value of a discrete random variable $X$,
$$
\qE(X) = \sum_{k\geq 0} k \qP\left[X=k\right] = \sum_{k\geq 0} \l( 1-\qP\left[X \leq k\right] \r),
$$
we get a generating function version to obtain
$$
\Xi(z) = \sum_{k\geq 0} \left[ F(z) - F^{\left[k\right]}(z)\right] \,
$$
and the expectation of the maximum block-size
of graphs of $\mathcal{F}$ is\footnote{For any power series $A(z) = \sum a_n z^n$, $[z^n]A(z)$
denotes the $n$-th coefficient of $A(z)$, viz. $[z^n]A(z) = a_n$.}
$\frac{n![z^n] \Xi(z)}{F_n}$.
Turning back to $G(n,\,M)$,
realizations of random graphs when $M$ is close to $\frac{n}{2}$ contain a set of trees, some
components with one cycle and complex components with $3$-regular $3$-cores a.a.s. In this paper,
our plan is to apply this scheme above by counting realizations of $G(n,\, M)$ with
all blocks of size less than a certain value. Once we get the forms of their generating functions,
we will use complex analysis techniques to get our results.
This extended abstract is organized as follows. Section~\ref{SEC_ENUMERATIVE} starts with the enumeration of trees of given degree specification.
We then show how to enumerate $2$-connected graphs with $3$-regular $3$-cores. Combining the trees and the blocks graphs
lead to the forms of the generating functions of connected graphs under certain conditions.
Section~\ref{SEC_ENUMERATIVE} ends with the enumeration of complex connected components
with all blocks of size less than a parameter $k$.
Based on the previous results and by means of analytic methods,
Section~\ref{SEC_PROOF_TH1} (resp. ~\ref{SEC_PROOF_TH2}) offers the proof of Theorem~\ref{MAIN_SUBCRITICAL} (resp. ~\ref{MAIN_CRITICAL}).
\section{Enumerative tools}\label{SEC_ENUMERATIVE}
\noindent
\textbf{Trees of given degree specification.}
Let $U(z)$ be the exponential generating function of labelled unrooted trees
and $T(z)$ be the EGF of rooted labelled trees, it is well-known that\footnote{
We refer for instance to Goulden and Jackson \cite{GJ83} for combinatorial operators,
to Harary and Palmer \cite{HP73} for graphical enumeration and to
Flajolet and Sedgewick \cite{FS+} for the symbolic method of generating functions.}:
\begin{equation}\label{eq:cayley-trees}
U(z) = \sum_{n=1}^{\infty} n^{n-2}\frac{z^n}{n!} = T(z)-\frac{T(z)^2}{2} \quad \mbox{and}\quad
T(z) = \sum_{n=1}^{\infty}n^{n-1}\frac{z^n}{n!} = z e^{T(z)}.
\end{equation}
For a tree with exactly $m_i$ vertices of degree $i$,
define its \textit{degree specification} as the $(n-1)$-tuple
$(m_1,\, m_2, \cdots, \,m_{n-1})$. We have the following.
\begin{lemma}\label{tree_spec}
The number of labeled trees with $n$ vertices and
degree specification $(m_1,\, m_2, \cdots m_{n-1})$ with $\sum_{i=1}^{n} m_i = n$ and $\sum_{i=1}^{n} i m_i = 2n-2$ is
\[
a_n(m_1,m_2,\ldots,m_{n-1}) =
\frac{ (n-2)! }{ \prod_{i=1}^{n-1} \left( (i-1)! \right)^{m_i}} {n \choose m_1, \, m_2, \cdots, \, m_{n-1}} \, .
\]
\end{lemma}
\noindent
\SKETCHED~
Using Pr\"{u}fer code, the number of trees with degree sequence $d_1, d_2, \cdots, d_n$
(that is with node numbered $i$ of degree $d_i$) is $\frac{ (n-2)!}{\prod_{i=1}^{n} (d_i-1)!}$.
The result is obtained by regrouping nodes of the same degree.
\ENDPROOF
\noindent
Define the associated EGF to $a_n(m_1,m_2,\ldots,m_{n-1})$ with
\begin{equation}\label{eq:multivariate-unrooted-trees}
U(\delta_1, \, \delta_2, \, \cdots ; \, z) = \sum_{n=2}^{\infty}
\sum a_n(m_1,m_2,\ldots,m_{n-1}) \delta_1^{m_1}\delta_2^{m_2}\cdots \delta_{n-1}^{m_{n-1}}\frac{z^n}{n!}
\end{equation}
where the inner summation is taken other all $i$ such that $\sum i m_i = 2n-2$ and $\sum m_i = n$.
Define $U_n(\delta_1,\delta_2,\ldots,\delta_{n-1})$ as
\begin{equation} \label{eq:sg-tree-given-size}
U_n(\delta_1,\delta_2,\ldots,\delta_{n-1}) = [z^n]U(\delta_1,\delta_2,\ldots,\delta_{n-1}; \,z) \, .
\end{equation}
The following result allows us to compute recursively $U_n(\delta_1, \, \cdots,\, \delta_{n-1})$.
\begin{lemma}{\label{TH:TREE}}
The generating functions $U_n$ defined in \eqref{eq:sg-tree-given-size} satisfy
$U_2(\delta_1)=\frac{\delta_1^2}{2}$ and
for any $n\geq 3$,
\be
U_n(\delta_1,\ldots,\delta_{n-1}) &=& \delta_2 U_{n-1}(\delta_1,\ldots,\delta_{n-2}) \cr
&+& \sum_{i=2}^{n-2} \delta_{i+1} \int_0^{\delta_1} \tiny \frac{\partial}{\partial\delta_i} \normalsize U_{n-1}(x,\delta_2,\ldots,\delta_{n-2})
dx \, .
\ee
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:TH_TREE}.
\noindent
\textbf{Enumerating $2$-connected graphs whose kernels are 3-regular.}
A \textit{bridge} or \emph{cut-edge} of a graph is an
edge whose removal increases its number of connected components.
Especially, the deletion of such an edge disconnects a connected graph.
Similarly an \emph{articulation point} or \emph{cut-vertex} of a connected
graph is a vertex whose removal disconnects a graph. A connected graph
without an articulation point is called a \emph{block} or a $2$-\emph{connected} graph.
Following the terminology of~\cite{JKLP93}, a connected graph has \textit{excess} $r$
if it has $r$ edges more than vertices.
Trees (resp. \textit{unicycles} or \textit{unicyclic components}) are connected
components with excess $r=-1$ (resp. $r=0$).
Connected components with excess $r>0$ are called \textit{complex connected components}.
A graph (not necessarily connected)
is called {\it complex} when all its components are complex.
The \textit{total excess}
of a graph is the number of edges plus the number of acyclic components, minus the number of vertices.
Given a graph, its \emph{$2$-core} is obtained by deleting recursively
all nodes of degree $1$. A \emph{smooth} graph \label{Wr78} is a
graph without vertices of degree one.
The \emph{$3$-core} (also called \emph{kernel}) of a complex graph is the graph
obtained from its $2$-core by repeating the following process
on any vertex of degree two~:
for a vertex of degree two, we can remove it and splice
together the two edges that it formerly touched.
A graph is said \textit{cubic} or \textit{$3$-regular} if all of its vertices are of
degree $3$.
Denote by $\mathcal{B}_r$ the family of
$2$-connected smooth graphs of excess $r$ with $3$-regular $3$-cores
and let
\beq\label{eq:all_block}
\mathcal{B} = \bigcup_{r=1}^{\infty} \mathcal{B}_{r} \, .
\eeq
In this paragraph, we aim to enumerate asymptotically the graphs of $\mathcal{B}_r$.
In~\cite{Chae}, the authors established recurrence relations for the numbers
of labeled cubic multigraphs with given connectivity, number of double edges
and number of loops. For instance, they were able to rederive Wormald's result about
the numbers of labeled connected simple cubic graphs with $3n$ simple edges and
$2n$ vertices \cite[equation (24)]{Chae}. They proved that the number of such
objects is given by
\beq \label{connected_simple_cubic}
\frac{(2n)!}{3n 2^n} \, (t_n - 2t_{n-1}) , \, n \geq 2
\eeq
with
\beq \label{connected_simple_cubic_recurrence}
t_1=0, t_2 =1 \, \text{ and } t_n = 3n t_{n-1} + 2 t_{n-2} + (3n-1) \sum_{i=2}^{n-3} t_i t_{n-1-i}, \, n \geq 2\,.
\eeq
From the sequence $(t_n)$, they found the number of $2$-connected multigraphs.
\begin{lemma}[Chae, Palmer, Robinson]\label{TH_CHAE}
Let $g(s,d)$ be the number
of cubic block ($2$-connected labelled) multigraphs with $s$ single edges and $d$ double-edges.
Then, the numbers $g(s,d)$ satisfy
\[
g(s,\,d)=0 \text{ if } \, s<2 \; , g(s,\,s)= (2s-1)! \text{ and }
g(3s,\,0) = \frac{(2s)!}{3s 2^s} \, (t_s - 2t_{s-1})
\]
with $t_s$ defined as in (\ref{connected_simple_cubic_recurrence}).
In all other cases,
\[
g(s,\,d) = 2n(2n-1) \left( \frac{(s-1)}{d} g(s-1, d-1) + g(s-3,d) \right) \, .
\]
\end{lemma}
We are now ready to enumerate asymptotically the
family $\mathcal{B}_r$. Throughout the rest of this paper
if $A(z)$ and $B(z)$ are two EGFs we write
\[
A(z) \asymp B(z) \text{ if and only if } [z^n]A(z) \sim [z^n] B(z) \; \text{ as } \; n \rightarrow +\infty\, .
\]
\begin{lemma}\label{B_R:equivalence}
For $r \geq 1$, let $B_r(z)$ be the EGF of smooth graphs of excess $r$ whose kernels are $3$-regular
and $2$-connected. $B_r(z)$satisfies
$B_r(z) \asymp \frac{b_r}{\left( 1- z\right)^{3r}}$
where $b_1=\frac{1}{12}$ and for $r\geq 2$
\beq\label{def_B_R}
b_r = \sum_{s+2d=3r} \frac{g(s, \, d)}{2^d (2r)!} \,
\eeq
with the $g(s,\,d)$ defined as in lemma~\ref{TH_CHAE}.
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:B_R:equivalence}.
\noindent
We need to count graphs of excess $r$ with at most $k$ vertices
so that all the blocks of such structures are of size at most $k$.
We begin our task with the graphs with cubic and $2$-connected kernels.
\begin{lemma}\label{lemma:BRK}
Let $\mathcal{B}_r^{\left[k\right]}$ be the family of $2$-connected graphs
of excess $r$, with at most $k-2r$ vertices of degree two in their $2$-cores
and whose $3$-cores are cubic. For any fixed $r \geq 1$, we have
\[
B_r^{\left[k\right]}(z) \asymp b_r \, \frac{ 1- z^{k} }{(1-z)^{3r}}\, .
\]
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:lemma:BRK}.
\noindent
Let ${\mathcal{B}}_r^{\bullet s}$ be the set of graphs of ${\mathcal{B}}_r$
such that $s$ vertices of degree two of their $2$-cores
are distinguished amongst the others.
In other words, an element of ${\mathcal{B}}_r^{\bullet s}$ can be
obtained from an element of ${\mathcal{B}}_r$
by marking (or pointing) $s$ unordered vertices of its $2$-core. In terms of
generating functions, we simply get (see~\cite{HP73,GJ83,FS+})~:
\beq\label{lem:pointed-bloc}
B_r^{\bullet s}(z) = \frac{z^s}{s!} \frac{\partial ^s}{\partial z^s} B_r(z,\,t)\Bigg|_{t=z}
= \frac{z^s}{s!} \frac{\partial ^s}{\partial z^s}\left( b_r \frac{t^{2r}}{(1-z)^{3r}} \right)\Bigg|_{t=z} \, ,
\eeq
where $B_r(z,t)$ is the bivariate EGF of ${\mathcal{B}}_r$
with $t$ the variable
for the vertices of degree $3$.
(The substitution $t=z$ is made after the derivations.)
\noindent
Define
\[
b_r^{\bullet s} = \frac{1}{s!} b_r \prod_{i=1}^{s} [3r + (s-i)] \,
\]
so that $B_r^{\bullet s}(z) \asymp \frac{b_r^{\bullet s}}{\left(1 - z\right)^{3r+s}}$.
Now if we switch to the class of graphs with blocks of size
at most $k$ then by similar arguments, the asymptotic number of
graphs of $\mathcal{B}_r^{\bullet s}$ with $s$ distinguished vertices
and at most $k$ vertices on their $2$-cores behaves as
\[
\brsk(z) \asymp b_r^{\bullet s} \frac{1-z^{k}}{\left(1 - z\right)^{3r+s}} \, .
\]
\noindent
\textbf{Counting $2$-cores with cubic kernels by number of bridges.}
In this paragraph,
we aim to enumerate connected smooth graphs whose $3$-cores
are $3$-regular according to their number of bridges (or cut-edges) and their excess.
To that purpose, let $\mathcal{C}_{r}$ be the family of such graphs with excess $r\geq 0$,
and for any $d\geq 0$ let
$$
\mathcal{C}_{r,d} \eqdef \{G\in \mathcal{C}_r : G \mbox{ is a cycle or its $3$-core is
$3$-regular and has $d$ bridges} \}\, .
$$
Clearly, we have $\mathcal{C}_{r,0} = \mathcal{B}_r$. If we want to mark the
excess of these graphs by the variable $w$, we simply have
\[
C_{r,d}(w,z) = w^r C_{r,d}(z) \, .
\]
\begin{lemma}\label{lemma:CRD}
For any $r\geq 1$ and $d\geq 1$,
\be\label{eq:Crd_from_trees}
C_{r,d}(z) &=& [w^{r}]U_{d+1}\Bigg( B^{\bullet 1}(w,z),\, 2!B^{\bullet 2}(w,z),\,
3!B^{\bullet 3}(w,z)+w^{-1}z,\cr & & 4!B^{\bullet 4}(w,z),\, \ldots ,d!B^{\bullet d}(w,z) \Bigg) \, \frac{w^d}{(1-z)^{d}},
\ee
where
$U_{d+1}$ are the EGF given by lemma~\ref{TH:TREE},
$B_0(w,z)= -\frac{1}{2} \log{(1-z)} - z/2 - z^2/4$,
$B_0^{\bullet s}(w, z) = \frac{1}{s!} \frac{\partial ^s}{\partial z^s} \, B_0(w,z)$
and $B^{\bullet s}(w,z)= \sum_{r \geq 0} w^r B_r^{\bullet s}(z)$.
\end{lemma}
\noindent
\PROOF~
Postponed in the Appendix -- \ref{Proof:lemma:CRD}
\begin{lemma}\label{lemma:behaviourCRD}
For $r\geq 1$ and $d \geq 1$, we have
$$
C_{r,\, d}(z) \asymp \frac{c_{r,\, d}}{(1-z)^{3r}}
$$
where the coefficients $c_{r,\,d}$ are defined by
$$
c_{r,\, d} = [w^{r}]U_{d+1}\l( \beta_1(w), \beta_2(w),\beta_3 (w)+w^{-1},
\beta_4 (w),\ldots,\beta_d (w)\r)w^{d},
$$
with $b_{\ell}$ given by (\ref{def_B_R}) and
$$
\beta_s(w) = \frac{(s-1)!}{2} + \sum_{\ell=1}^{r-1}
w^{\ell} b_\ell \prod_{i=1}^{s} [3\ell + (s-i)] \quad \mbox{with $s\geq 1$}.
$$
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:lemma:behaviourCRD}.
\noindent
Let us restrict our attention to elements of $\mathcal{C}_{r,\,d}$
with blocks of size at most $k$. Denote by $\mathcal{C}_{r,\,d}^{[k]}$ this set of graphs.
Since they can be obtained from a tree with $d+1$ vertices by replacing
each vertex of degree $s$ by a $s$-marked block (block
with a distinguished degree of degree two) of the family $\bigcup_{r=0}^{\infty}\mathcal{B}^{\bullet s,\,[k]}$, we infer the following~:
\begin{lemma}\label{MAIN_LEMMA_CONNECTED}
For fixed values of $r$, the EGF of graphs of $\mathcal{C}_{r,d}^{[k]}$
verifies
$$ {C}_{r,\,d}^{[k]} \asymp c_{r,\,d} \frac{(1-z^k)^{d+1}}{(1-z)^{3r}} \, . $$
\end{lemma}
\noindent
\textbf{From connected components to complex components.}
Denote by $\Erk$ the family of complex graphs (not necessarily connected) of total excess $r$
with all blocks of sized $\leq k$. Let $\erk$ be the EGF of $\Erk$. Using the symbolic method
and sprouting the rooted trees from the smooth graphs counted by ${C}_{r,\,d}^{[k]}(z)$, we get
\[
\sum_{r=0}^\infty w^r\erk(z) = \exp{\left(\sum_{r=1}^{\infty}w^r \sum_{d\geq 0}^{2r-1} {C}_{r,\,d}^{[k]}\l( \, T(z) \,\r) \right)}\, .
\]
We now use a general scheme which relates behavior of connected components and complex components
(see for instance~\cite[Section 8]{JKLP93}).
If $E(w,z)=1+ \sum_{r\geq 1} w^r E_r(z)$ with $E_r(z) \asymp \frac{e_r}{(1-T(z))^{3r}}$ and
$C_r(z) \asymp \frac{c_r}{(1-T(z))^{3r}}$ are EGFs satisfying
\[
1+\sum_{r\geq 1} w^r E_r(z) = \exp{\left(\sum_{r=1}^{\infty} w^r C_r(z) \right)}.
\]
then the coefficients $(e_r)$ and $(c_r)$ are related by
\[
e_0 = 1 \, \text{ and } \, e_r = c_r + \frac{1}{r} \sum_{j=1}^{r-1} j c_j e_{r-j} \text{ as } r \geq 1 \, .
\]
Similarly, after some algebra we get
\begin{lemma}
For fixed $r\geq 1$,
\[
\erk(z) \asymp \sum_{d=0}^{2r-1} \frac{e_{r,d}^{[k]}\l(\, T(z)\,\r)}{(1-T(z))^{3r}}
\]
where the functions $(e_{r,d}^{[k]})$ are defined recursively by
$e_{0,0}^{[k]}(z) = 1$, $e_{r,d}^{[k]}(z)=0$ if $d>2r-1$
and
\beq \label{def_erk}
e_{r,d}^{[k]}(z) = c_{r,d} \left( 1-z^k \right)^{d+1}
+ \frac{1}{r} \sum_{j=1}^{r-1} j c_{j,d} \, e_{r-j,d}^{[k]}(z) \, \l(1-z^k \r)^{d+1} \, .
\eeq
\end{lemma}
\textbf{Remark.} Note that the maximal range $2r-1$ of $d$
appears when the $2$-core is a cacti graph (each edge lies on a path or on a unique cycle),
each cycle have exactly one vertex of degree three and its $3$-core is $3$-regular.
\section{Proof of Theorem~\ref{MAIN_SUBCRITICAL}}\label{SEC_PROOF_TH1}
Following the work of Flajolet and Odlyzko~\cite{FO-mappings} on extremal statistics of
random mappings, let us introduce the relevant EGF for the expectation of the
maximum block-size in $G(n,M)$.
On the one hand, if there are $n$ vertices, $M$ edges and with a total excess $r$
there must be exactly $n-M+r$ acyclic components.
Thus, the number of $(n,M)$-graphs\footnote{Graph with $n$ vertices and $M$ edges}
of total excess $r$ without blocks of size larger than $k$ is
$$
n![z^n]\frac{U(z)^{n-M+r}}{(n-M+r)!} \l(e^{W_0(z)- \sum_{i=k+1}^\infty \frac{T(z)^i}{2i}} \r) E_r^{[k]}(z) \, .
$$
where $W_0(z) = -\frac{1}{2}\log(1-T(z))-\frac{T(z)}{2}-\frac{T(z)^2}{4}$ is the EGF
of connected graphs of excess $r=0$ (see~\cite[equation~(3.5)]{JKLP93}).
On the other hand, the EGF of all $(n,M)$-graphs is
$$ G_{n,M}(z) = \sum_{n\geq 0} \binom{\binom{n}{2}}{M} \frac{z^n}{n!} \, .$$
Define
\beq \label{SG_EXPECTATION_SUB}
\Xi(z) = \sum_{k \geq 0}
\l[ G_{n,M}(z) - \sum_{n\geq 0}\left(n![z^n]\frac{U(z)^{n-M+r}}{(n-M+r)!}
\l(e^{W_0(z)- \sum_{i=k+1}^\infty \frac{T(z)^i}{2i}} \r) E_r^{[k]}(z) \right) \frac{z^n}{n!}\r] \, ,
\eeq
so that
\begin{equation}\label{eq:EXPECTATION_DEF}
\frac{n![z^n]}{\binom{\binom{n}{2}}{M}} \Xi(z) = \sum_{k \geq 0} \l[ 1 -
\frac{n!}{\binom{\binom{n}{2}}{M}}[z^n]\frac{U(z)^{n-M+r}}{(n-M+r)!} \l(e^{W_0(z)- \sum_{i=k+1}^\infty \frac{T(z)^i}{2i}} \r) E_r^{[k]}(z) \r],
\end{equation} is the expectation of $\circ$. \\
We know from the theory of random graphs that in the sub-critical phase when
$n-2M \gg n^{2/3}$ $G(n,M)$ has no complex components with probability $1-O\l( \frac{n^2}{(n-2M)^3} \r)$ (cf~\cite[Theorem~3.2]{DaRa}).
In this abstract, we restrict our attention to the typical random graphs. Otherwise, we will obtain
the same result as stated by bounds on the $E_r^{[k]}(z)$ in (\ref{SG_EXPECTATION_SUB})
since
\[
1 \leq E_r^{[k]}(z) \leq E_r(z) \leq \frac{e_r T(z)}{(1-T(z))^{3r}} \,
\]
(where inequality between the EGFs means that the coefficients of every power of $z$ obeys the same relation and the last
inequality is ~\cite[equation (15.2)]{JKLP93} with $e_{r}={(6r)!\over 2^{5r}3^{2r}(3r)!\,(2r)!}$).
Assuming that the graphs are typical (i.e. without complex components), $\Xi(z)$ behaves as
\begin{align}
\Xi(z) &\asymp \sum_{k\geq 0} \l[G_{n,M}(z) - \sum_{n\geq 0} \l(n![z^n]\frac{U(z)^{n-M}}{(n-M)!} \frac{e^{-\frac{T(z)}{2}-\frac{T(z)^2}{4}}}{(1-T(z))^{1/2}}
\exp\l( - \sum_{j\geq k+1} \frac{T(z)^j}{2j} \r) \r) \frac{z^n}{n!} \r] \label{SUMMATION}
\end{align}
\noindent We need the following Lemma to quantify large coefficients of \eqref{SUMMATION}.
\begin{lemma}\label{lemma_subcritical}
Let $a$ and $b$ be any fixed rational numbers. For any sequence of integers $M(n)$ such that
$\delta n < M$ for some $\delta \in \left[0,\half\right]$ but $n-2M \gg n^{2/3}$, define
$$ f_{a,b}(n,M) = \frac{n!}{ { {n \choose 2} \choose M} } \, \coeff{z^n} \frac{U(z)^{n-M}}{(n-M)!} \,
\frac{ U(z)^b \, e^{-T(z)/2 - T(z)^2/4} }{(1-T(z))^a}\,. $$
We have
$$ f_{a,b}(n,M) \sim 2^b\, \l(\frac{M}{n}\r)^b \, \l( 1 - \frac{M}{n}\r)^b \l(1-\frac{2M}{n}\r)^{1/2 - a} \, .$$
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:lemma_subcritical}.
\noindent
Using Lemma~\ref{lemma_subcritical} with $a=1/2$ and $b=0$,
after a bit of algebra (change of variable $u=T(z)$ and approximating the sum by an integral), we first obtain
\[
\qE{(\circ)} \sim \sum_{k\geq 0} \l( 1 - \exp\l({-\frac{1}{2}\int_{(k+1)(1-\frac{2M}{n})}^{\infty} e^{-v}\frac{dv}{v}}\r)\r).
\]
Then by Euler-Maclaurin summation formula and after a change of variable ($(k+1) (1-\frac{2M}{n}) = u$ so $dk = (1-\frac{2M}{n})^{-1}du$), we get
the result.
\section{Proof of Theorem~\ref{MAIN_CRITICAL}}\label{SEC_PROOF_TH2}
The following technical result is essentially ~\cite[Lemma 3]{JKLP93}. We give it here
in a modified version tailored to our needs (namely involving truncated series).
We refer also to the proof of~\cite[Theorem 5]{FKP89} and~\cite{Banderier} for integrals related to
the Airy function.
\begin{lemma}\label{lemma:Airy}
Let $M=\frac{n}{2}\l(1 + \lambda n^{-1/3}\r)$. Then for any natural integers $a, k$ and $r$ we have
\ben
& & \frac{n!}{{ {n \choose 2} \choose M}} \,
\coeff{z^n} \, \frac{U(z)^{n-M+r}}{(n-M+r)!} \frac{T(z)^{a} \l( 1- T(z)^k \r) }{(1-T(z))^{3r}} \,
\exp\l({W_0(z) - \sum_{i=k}^{\infty} \frac{T(z)^i}{2 i}}\r) \cr
& = & \sqrt{2\, \pi} \, \exp{\l(- \sum_{j=k}^{\infty} \frac{e^{-j \alpha n^{-1/3}}}{2 j}\r)}
\, \left( 1 - e^{-k \alpha n^{-1/3}} \right)
A\left(3r+\frac{1}{2},\, \lambda\right)
\, \left( 1 + O\left(\frac{\lambda^4}{n^{1/3}}\right)\right) \, , \cr
& \, & \,
\een
uniformly for $|\lambda|\le n^{1/12}$ where $A(y,\mu)$ is defined by (\ref{28INTRO})
and $\alpha$ is given by (\ref{eq:alpha}).
\end{lemma}
\noindent
\PROOF~ Postponed in the Appendix -- \ref{Proof:lemma:Airy}.\\
Using this lemma, equation~\eqref{eq:EXPECTATION_DEF} and next approximating
a sum by an integral using Euler-Maclaurin summation, the expectation of $\circ$ is about
\footnotesize
\begin{align}
&\sum_{k=0}^n \l( 1- \sum_r \sum_d
\sqrt{2\pi} \exp \l( -\sum_{j=k}^\infty \frac{e^{-j\alpha n^{-1/3}}}{2j}\r)
\hardcoeff{r}{d}\l(e^{-k\alpha n^{-1/3}}\r) A\l(3r +\frac{1}{2},\lambda\r)\r) \label{integrale-1}\\
&=\alpha^{-1}n^{1/3}\int_{0}^{\alpha n^{2/3}} \l( 1- \sum_r \sum_d
\sqrt{2\pi} \exp \l( -\int_{u}^\infty \frac{e^{-v}}{2v}dv \r) \hardcoeffbis{r}{d}\l(e^{-u}\r)
A\l(3r +\frac{1}{2},\lambda\r)\r) du
\end{align}
\normalsize
where
\beq \label{def_erk_bis}
e_{r,d}(z) = c_{r,d} \left( 1-z \right)^{d+1}
+ \frac{1}{r} \sum_{j=1}^{r-1} j c_{j,d} \, e_{r-j,d}(z) \, \l(1-z \r)^{d+1} \, .
\eeq
\section{Conclusion}
We have shown that the generating function
approach is well suited to make precise the
expectation of maximum block-size of random graphs.
Our analysis is a first step towards a fine description of the
various graph parameters inside the window of transition of random graphs.
\bigskip
\noindent{\textbf{Acknowledgements:}}
The authors thank the reviewers for their thorough reviews and highly appreciate the comments, remarks and
suggestions, which significantly improve the quality of the paper.
The authors express their gratitude to the support of the project ANR 2010 BLAN 0204 -- MAGNUM and
the project PEPS FASCIDO INSMI-INS2I-2015.
\nocite{*}
\bibliographystyle{abbrvnat}
\bibliography{maximum_block_size}
\input{appendix.tex}
\end{document} | 162,365 |
July 11, 2013
Professional article by flexis published in U.S.-magazine „Manufacturing Business Technology”
A professional article by flexis was published in the U.S. magazine "Manufacturing Business Technology" (mbtmag.com). „Creating A Parts Order Processing-oriented supply chain," focuses on the appropriate response to the changed circumstances with regard to the supply chain in order processing. The publication presents the perception flexis has concerning the current challenges and outlines appropriate strategies to handle them.
Here you can read part 1 of the article online.
Here you can read part 2 of the article online. | 216,114 |
TITLE: Pole set of rational function on $V(WZ-XY)$
QUESTION [4 upvotes]: Let $V = V(WZ - XY)\subset \mathbb{A}(k)^4$ (k is algebraically closed). This is an irreducible algebraic set so the coordinate ring is an integral domain which allows us to form a field of fractions, $k(V).$ Let $\overline{W}, \overline{X}$ denote the image of $W$ and $X$ in the coordinate ring. Let $f=\dfrac{\overline{W}}{\overline{X}}\in k(V).$ I want to find the pole set of $f.$ I have a feeling I am over complicating this. Essentially I should just view $f$ as the function $W/X$ restricted to $V.$ I know that points on $V$ where $X=0$ and $W\neq 0$ are in the pole set, and this occurs when $Z=0.$ But what happens if both $W$ and $X$ are $0$ ?
REPLY [7 votes]: Indeed, the pole set of $f=\frac{\overline{W}}{\overline{X}}$ is precisely the set
$$
S(f)=\{(w,0,y,0) \in \Bbb{A}^4 \; \vert \; w,y \in k\}=V(X,Z)
$$
It is clear that $f$ is defined at every point not in $S(f)$, because
$$
f=\frac{\overline{W}}{\overline{X}}=\frac{\overline{Y}}{\overline{Z}}
$$
and from this you see that $f$ is defined at $(w,x,y,z)$ if either $x \neq 0$ or $z \neq 0$.
In order to see that $f$ is not defined anywhere else, let $P=(a,0,b,0) \in \Bbb{A}^4$, and assume you could write
$$
\frac{\overline{W}}{\overline{X}}=\frac{\overline{F}}{\overline{G}}
$$
with $F,G \in k[X,Y,Z,W]$ and $G(P) \neq 0$.
This implies that $G \cdot W-F \cdot X \in (WZ-XY)$, which means that $G \cdot W-F \cdot X=H \cdot (WZ-XY)$ for some $H \in k[X,Y,Z,W]$. Now you get
$$
X \cdot (F-HY)=W \cdot (G-HZ)
$$
from which it follows that $X$ must divide $G-HZ$, i.e. $G-HZ \in (X)$. But then evaluating at $P=(a,0,b,0)$ yields
$$
G(P)=0
$$
which contradicts our choice of $G$. | 61,308 |
TITLE: Proof: Tricky limit going to 0
QUESTION [2 upvotes]: I'm working on a proof and to complete it I need to find a way to choose an $n$ such that $(1-a)^n < \epsilon$ for a fixed $a$ such that $\frac12 < a < 1$ and any small $\epsilon$. I'm trying to prove that a discrete probability space cannot contain an event $\mathcal A$ with probability at most $(1-a)^n$ since this clearly must go to 0. I'm just having a hard time finding an appropriate formula for $n$ to prove this goes to 0.
REPLY [0 votes]: Note that $1-a\leqslant\mathrm e^{-a}$ for every $a$ hence $(1-a)^n\leqslant\mathrm e^{-na}$ for every $a\leqslant1$. Assume that $a\leqslant1$. If $n\geqslant-(\log\varepsilon)/a$, then $\mathrm e^{-na}\leqslant\varepsilon$, which, in turn, ensures that $(1-a)^n\leqslant\varepsilon$.
With the addiditional hypothesis that $a\geqslant1/2$, the condition that $n\geqslant-2\log\varepsilon$ suffices. | 43,867 |
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The Post-Baccalaureate Certificate Program in Communication Sciences and Disorders (CSD), an Extended Education program, is offered through Chapman University's Crean College of Health and Behavioral Sciences. This Speech Language Pathology, Communication Disorders, or similar. | 120,970 |
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TITLE: Two's complement binary arithmetic
QUESTION [1 upvotes]: Suppose binary values are signed 8-bit values, representing twos-complement format with a decimal range from -128 to 127. Which of the following statements are true/false?
1) 00001010 > 00000111
2) 11111111 > 01111111
3) (11111111 + 11111111) > (00000001 - 00000010)
4) (00000100 * 00000100) == 00010000
5) (11111010 * 00000011) == (11101110)
6) (10000000 / 00000100) == 11100000
7) 11110000 - 00000001 == 10001111
I'm studying for my exam and I'm struggling with this problem. Here's what I've got for each of them.
1) In decimal, the LHS is 10 and the RHS is 7. So it is true.
2) In decimal, the LHS is -1, and the RHS is 127. So it's false.
3) I'm really not sure about this one. I know that overflown digits get dropped, but I don't know if there's an overflow here. Here's my guess: The LHS is (-1 + -1) in decimal and the RHS is -1. So my answer is false
4) I think this is true. I don't think anything special is used here.
5) The LHS is (-6 * 3) and the RHS is -18. So it's true?
6) The LHS is -128 / 4 and the RHS is -32, so it's true
7) The LHS is -16 - 1 and the RHS i got -113. So I think it's false?
Can someone please let me know if these are ok?
Thanks
REPLY [1 votes]: Your answers are correct, but I am concerned that you keep having to translate to decimal to reach your conclusions. Here I will demonstrate how I would reason about these questions directly in binary, which you may find useful:
1) Leftmost bits both zero, hence both positive. Leftmost 1 in left number is bit 4, in right number is bit 3; hence left is greater than right. Conclusion: TRUE.
2) Leftmost bit in left number is 1, hence negative; in right number is 0, hence positive. Negative is -not- greater than positive. Conclusion: FALSE.
3) Subtraction in right expression calculated by taking bit complement and adding 1, then adding terms. Result is [sequence of 7 0's followed by 1] plus [sequence of 7 1's followed by 0]. Sum calculable without carry, result is sequence of 8 1's, a negative number. This matches leftmost term in left expression, to which another negative number is added, hence left expression must be smaller. Conclusion: FALSE.
4) Leftmost bits both zero, hence both positive. Leftmost 1 bits in terms of left expression followed by two 0's, leftmost 1 bit in right term followed by four 0's. In product of two terms -carried out in usual on-paper procedure-, result will be eight staggered 8-bit values, of which the third from the top is the first term shifted left by two and the rest are all 0's, hence total sum will be first term shifted by two, which 'inserts two 0's at right', for total of 4. Conclusion: TRUE.
5) [here very tempting to use decimal, but...] Negate the two negative values [note that this leaves the left-right relationship unchanged]; the product on the left is then much simpler, and easy to compare to the right value. Conclusion: TRUE.
6) Multiply both sides by the second term on the left (thereby 'moving the term to the other side'). Second term has a single 1 in bit position 3; hence multiplication 'shift' value on the right two bits left. Cutting out the rightmost 8 bits of the result yields the first term on the left. Conclusion: TRUE.
7) Second term on left consists of a single 1 in bit position 1. Subtraction of 1, quick rule: Flip all 0's to 1's moving right-to-left, when a 1 reached flip it to 0. Doing this with the first term on the left yields three 1's followed by 0 followed by four 1's, which does NOT match the right side. Conclusion: FALSE. | 135,011 |
- Authors: Felipe Campos de Melo Iani1,4, Sérgio Caldas2, Myrian Morato Duarte1, Ana Luisa Furtado Cury1, Alzira Batista Cecílio2, Pedro Augusto Carvalho Costa3, Lis R. Antonelli3, Kenneth J. Gollob4,5,6
- View Affiliations Hide AffiliationsAffiliations: 1 Divisão de Epidemiologia e Controle de Doenças/Instituto Octávio Magalhães, Fundação Ezequiel Dias, Belo Horizonte, Minas Gerais, Brazil. 2 Divisão de Plataformas Tecnológicas/Diretoria de Pesquisa e Desenvolvimento, Fundação Ezequiel Dias, Belo Horizonte, Minas Gerais, Brazil. 3 Centro de Pesquisas René Rachou, Fundação Oswaldo Cruz, Belo Horizonte, Minas Gerais, Brazil. 4 Programa de Pós-graduação em Medicina e Biomedicina, Instituto de Ensino e Pesquisa, Hospital Santa Casa, Belo Horizonte, Minas Gerais, Brazil. 5 Institutos Nacionais de Ciências e Tecnologia-Doenças Tropicais, Belo Horizonte, Minas Gerais, Brazil. 6 Núcleo de Ensino e Pesquisa, Instituto Mario Penna, Belo Horizonte, Minas Gerais, Brazil.
- Publisher: The American Society of Tropical Medicine and Hygiene
- Source: The American Journal of Tropical Medicine and Hygiene, Volume 95, Issue 1, Jul 2016, p. 193 - 200
- DOI:
f Dengue Patients with Early Hemorrhagic Manifestations Lose Coordinate Expression of the Anti-Inflammatory Cytokine IL-10 with the Inflammatory Cytokines IL-6 and IL-8
Abstract
Abstract.
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- Jarman RG, Nisalak A, Anderson KB, Klungthong C, Thaisomboonsuk B, Kaneechit W, Kalayanarooj S, Gibbons RV, , 2011. Factors influencing dengue virus isolation by C6/36 cell culture and mosquito inoculation of nested PCR-positive clinical samples. Am J Trop Med Hyg 84: 218–223.[Crossref]
- Gurukumar KR, Priyadarshini D, Patil JA, Bhagat A, Singh A, Shah PS, Cecilia D, , 2009. Development of real time PCR for detection and quantitation of dengue viruses. Virol J 6: 10.[Crossref]
- World Health Organization (WHO), 2009. CDC protocol of realtime RTPCR for Swine Influenza A(H1N1). Version 2009. World Health Organization, Atlanta, GA. Available at:. Accessed April 15, 2016.
- Caldas S, Caldas IS, Diniz LF, Lima WG, Oliveira RP, Cecílio AB, Ribeiro I, Talvania A, Bahia MT, , 2012. Real-time PCR strategy for parasite quantification in blood and tissue samples of experimental Trypanosoma cruzi infection. Acta Trop 123: 170–177.[Crossref]
- Cecílio AB, Caldas S, Oliveira RA, Santos ASB, Richardson M, Naumann GB, Schneider FS, Alvarenga VG, Estevão-Costa MI, Fuly AL, Eble JA, Sanchez EF, , 2013. Molecular characterization of Lys49 and Asp49 phospholipases A2 from snake venom and their antiviral activities against dengue virus. Toxins 5: 1780–1798.[Crossref]
- Endy TP, Nisalak A, Chunsuttitwat S, Vaughn DW, Green S, Ennis FA, Rothman AL, Libraty DH, , 2004. Relationship of preexisting dengue virus (DV) neutralizing antibody levels to viremia and severity of disease in a prospective cohort study of DV infection in Thailand. J Infect Dis 189: 990–1000.[Crossref]
- Wang WK, Chen HL, Yang CF, Hsieh SC, Juan CC, Chang SM, Yu CC, Lin LH, Huang JH, King CC, , 2006. Slower rates of clearance of viral load and virus-containing immune complexes in patients with dengue hemorrhagic fever. Clin Infect Dis 43: 1023–1030.[Crossref]
- Libraty DH, Acosta LP, Tallo V, Segubre-Mercado E, Bautista A, Potts JA, Jarman RG, Yoon IK, Gibbons RV, Brion JD, Capeding RZ, , 2009. A prospective nested case-control study of dengue in infants: rethinking and refining the antibody-dependent enhancement dengue hemorrhagic fever model. PLoS Med 6: e1000171.[Crossref]
- Guilarde AO, Turchi MD, Siqueira JB, Feres VC, Rocha B, Levi JE, Souza VA, Boas LS, Pannuti CS, Martelli CM, , 2008. Dengue and dengue hemorrhagic fever among adults: clinical outcomes related to viremia, serotypes, and antibody response. J Infect Dis 197: 817–824.[Crossref]
- Akdis CA, Akdis M, , 2009. Mechanisms and treatment of allergic disease in the big picture of regulatory T cells. J Allergy Clin Immunol 123: 735–746.[Crossref]
- Heinrich PC, Behrmann I, Haan S, Hermanns HM, Müller-Newen G, Schaper F, , 2003. Principles of interleukin (IL)-6-type cytokine signalling and its regulation. Biochem J 374: 1–20.[Crossref]
- Matsushima K, Morishita K, Yoshimura T, Lavu S, Kobayashi Y, Lew W, Appella E, Kung HF, Leonard EJ, Oppenheim JJ, , 1988. Molecular cloning of a human monocyte-derived neutrophil chemotactic factor (MDNCF) and the induction of MDNCF mRNA by interleukin 1 and tumor necrosis factor. J Exp Med 167: 1883–1893.[Crossref]
- Burke SM, Issekutz TB, Mohan K, Lee PW, Shmulevitz M, Marshall JS, , 2008. Human mast cell activation with virus-associated stimuli leads to the selective chemotaxis of natural killer cells by a CXCL8-dependent mechanism. Blood 111: 5467–5476.[Crossref]
- Levy A, Valero N, Espina LM, Añez G, Arias J, Mosquera J, , 2010. Increment of interleukin 6, tumour necrosis factor alpha, nitric oxide, C-reactive protein and apoptosis in dengue. Trans R Soc Trop Med Hyg 104: 16–23.[Crossref]
- Lanciotti RS, Gubler DJ, Trent DW, , 1997. Molecular evolution and phylogeny of dengue-4 viruses. J Gen Virol 78: 2279–2284.[Crossref]
- Costa VV, Fagundes CT, Souza DG, Teixeira MM, , 2013. Inflammatory and innate immune responses in dengue infection: protection versus disease induction. Am J Pathol 182: 1950–1961.[Crossref]
- Chaturvedi UC, , 2006. The curse of dengue. Indian J Med Res 124: 467–470.
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- Received : 22 Jul 2015
- Accepted : 07 Feb 2016 | 5,476 |
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\begin{document}
\maketitle
\begin{abstract}
A pro-Lie group $G$ is a topological group such that
$G$ is isomorphic to the projective limit of all
quotient groups $G/N$ (modulo closed normal subgroups $N$)
such that $G/N$ is a finite dimensional real Lie group.
A topological group is almost connected if the totally
disconnected factor group $G_t\defi G/G_0$ of $G$ modulo the
identity component $G_0$ is compact.
In this case it is straightforward that each Lie group
quotient $G/N$ of $G$ has finitely many components.
However, in spite of
a comprehensive literature on pro-Lie groups, the following
theorem, proved here, was not available until now:
\hfill\break
{\sc Theorem}. {\it A pro-Lie group $G$ is almost connected if
each of its Lie group quotients $G/N$ has finitely
many connected components.}
\hfill\break
\nin The difficulty of the proof is the verification of
the completeness of $G_t$.
\end{abstract}
\bsk
\section*{Projective Limits of Almost Connected Lie Groups}
A notorious problem in the structure theory of { pro-Lie
groups} is
the completeness
of quotient groups, notably that of the group $G/G_0$ of connected components.
In one of the sources on pro-Lie groups,
\cite{probook}, the { section} following
Definition 4.24 on pp.195ff.
exhibits some of the characteristic difficulties involving the
completeness of quotients of a pro-Lie group $G$ in general and the
quotient $G_t=G/G_0$ in particular. In their
entirety, these difficulties remain unresolved today. We shall settle
the completeness issue of $G_t$ here for any pro-Lie group whose
Lie group quotients have finitely many components.
\msk
{
Existing literature (see \cite{almost}, Corollary 8.4)
provides the following conclusion, which reinforces the
independent interest in the result of this note:
\msk
{\it An almost connected pro-Lie group $G$ contains a
maximal compact subgroup $C$
and a closed subspace $V$
homeomorphic to $\R^J$ for a set $J$
such that
\cen{$(c,x)\mapsto cx\colon C\times V\to G$}
\vskip-2pt\nin is a homeomorphism.}}
\bsk
So, let $G$ denote a topological group
and ${\cal N}(G)$ the set of all closed normal
subgroups of $G$ for which $G/N$ is a Lie group.
With these conventions we formulate a theorem,
to be proved subsequently.
The proof requires some effort. It is based
on information from \cite{probook}.
\msk
\begin{Theorem} \label{th:main} For a pro-Lie group $G$, the following
statements are equivalent:
\begin{enumerate}[\rm(1)]
\item $G_t$ is compact,
\item There is a compact totally disconnected
subspace $D\subseteq G$
being mapped homeo\-morphically onto $G_t$ by the
quotient map $q_t\colon G\to G_t$.
\item $(G/N)_t$ is finite for all $N\in{\cal N}(G)$.
\end{enumerate}
\end{Theorem}
\nin The proof of the theorem will require the proof of some new lemmas
and some references to existing literature.
The first one is { cited from \cite{almost},
Main Theorem 8.1, Corollary 8.3.}
\begin{Lemma} \label{l:one} Let $G$ be an almost connected
pro-Lie group. Then the following conclusions hold:
\begin{enumerate}[\rm(i)]
\item $G$ contains a maximal compact subgroup $C$,
and any compact subgroup of $G$ has a conjugate inside $C$.
\item $G=G_0C$.
\item $G$ contains a profinite subgroup $D$ such that
$G=G_0D$.
\end{enumerate}
\end{Lemma}
\msk
\nin
For every compact group $K$ there is a compact
totally disconnected subspace $D\subseteq K$ such that
$(k,d)\mapsto kd:K_0\times D\to K$ is a homeomorphism
(see \cite{book3}, Corollary 10.38, p.~573). From
Lemma \ref{l:one}
we know that $G$ is almost connected iff
there is a compact subgroup $K\subseteq G$
such that $G=G_0K$. Write $K=K_0D$ with the topological
direct factor $D\subseteq K$ as we just pointed out. Then
$G=G_0K_0D=G_0D$ and so $(g,d)\mapsto gd:G_0\times D\to G$
is readily seen to be a homeomorphism. Thus, in
Theorem \ref{th:main}, Condition (1) implies (2),
and for (2)\implies(1) there is nothing to prove.
\msk
Let us establish that (1)\implies (3):
\nin
Assume $G/N$ to be a Lie group quotient of $G$.
Then $(G/N)_0=G_0N/N\cong G_0/(G_0\cap N)$ (cf.\ \cite{probook},
Lemma 3.29, p.152). Let $K$ be a compact subgroup of $G$
such that $G=G_0K$, and let $L=G/N$. Then $L=(G_0N/N)(KN/N)
=L_0C$ for the compact Lie group $C=KN/N$. Thus
$L_t=L_0C/L_0\cong C/(C\cap L_0)$ is a compact totally disconnected
Lie group and is therefore finite. This proves (3).
\bsk
There remains a proof of the implication (3)\implies (1):
\nin
For the moment let us assume that the
following hypothesis is satisfied
\msk
\cen{
{\bf (H)} $G_t$ \it is a complete topological group.}
\msk
\nin
By \cite{probook}, Corollary 3.31, hypothesis (H) implies that
$G_t$ is prodiscrete, that is,
$G_t=\lim_{N\in{\cal N}(G_t)} G_t/N$ where $G_t/N$ is discrete.
Now $G_t/N$ is a Lie group quotient of $G$ and so is finite
by (3). Hence $G_t$ is profinite and thus compact. This
proves Condition (1).
\msk
It therefore remains to prove (H).
For this purpose we shall invoke results from \cite{probook}, pp.195ff.
\msk
Firstly, we define ${\cal M}(G)$ to be the subset of
all $M\in{\cal N}(G)$ with the additional property that
each open subgroup $N\subseteq M$ from ${\cal N}(G)$ has
finite index in $M$. We shall then use
\begin{Lemma} \label{l:two} If $G$ is a pro-Lie group such that
\nin{\bf(*)} each Lie group quotient $G/N$, ${ N}
\in{\cal N}(G)$ is almost connected,
\nin
then ${\cal M}(G)$ is cofinal in
${\cal N}({G})$ and thus is a filter basis.
Moreover, $G$ is the strict projective limit
of the $G/M$, $M\in{\cal M}(G)$.
\end{Lemma}
\begin{Proof}
For the proof see Lemma
4.25 in \cite{probook}, pp.195 and 196.
\end{Proof}
\msk
\nin We note that Lemma 4.25 in \cite{probook} states as hypothesis
that $G$ is almost connected which implies {\bf(*)}. But
the hypothesis {\bf(*)} is all that is used in the proof
of Lemma 4.25.
\bsk
\nin
Any set $\cal Z$ of subsets of a set
$G$ may be considered as a subbasis { of closed sets for}
a topology. If $G$ is { a topological group}, and
$\cal Z$ is the set of all cosets $gM$ with $g\in G$ and
$M\in{\cal M}(G)$,
then $\cal Z$ generates { the set of closed sets of}
a topology on $G${,} called the Z-{\it topology}.
\msk
\begin{Lemma} \label{l:three} The {\rm Z}-topology on a
pro-Lie group $G$ satisfying {\rm Condition} {\bf(*)}
of {\rm Lemma \ref{l:two}} is a compact $T_1$-topology.
\end{Lemma}
\begin{Proof} See Proposition 4.27 of \cite{probook}, pp.~197--201.
\end{Proof}
\msk
\nin Again we note that Proposition 4.27 { in \cite{probook}}
assumes the hypothesis that
$G$ is almost connected, but the proof of the conclusion of
Lemma \ref{l:three} only uses Hypothesis {\bf(*)} of
Lemma \ref{l:two}.
\msk
\nin
We now adjust the proof of Theorem 4.28 on p.~202 of \cite{probook}
for our purposes.
\msk
\begin{Lemma} \label{l:four} Let $G$ be a pro-Lie group satisfying
hypothesis {\bf(*)} of {\rm Lemma \ref{l:two}}. Then $G_t$ is complete.
\end{Lemma}
\msk
\nin
We note right away that Lemma \ref{l:four}
will prove hypothesis (H) and therefore
complete the proof of Theorem \ref{th:main}.
\msk
\nin
{\bf Proof} of Lemma \ref{l:four}.
We let $f\colon G\to G_t=G/G_0$ be the
quotient morphism and consider a Cauchy filter $\cal C$ on $G_t$.
We have to show that $\cal C$ converges. By Lemma \ref{l:two},
${\cal M}(G)$
is cofinal in ${\cal N}(G)$. For each $N\in{\cal M}(G)$ let
$N^*=\overline{f(N)}$ and let
$p_{N^*}\colon G\to G_t{/N^*}$ be the
quotient morphism.
Then the image $p_{N^*}({\cal C})$ is a
Cauchy filter in the Lie group $G_t/N^*$ and thus has a limit $g_N$.
Then $(g_N)_{N\in{\cal M}(G)}\in\prod_{N\in{\cal M}(G)}G_t/N^*$
is an element of $\lim_{N\in{\cal M}(G)}G_t/N^*$; indeed $\cal C$
has to converge to a point in the completion of $G_t$.
Now let $F_N=(p_{N^*}\circ f)^{-1}({g}_N)$.
Then $\{F_N:N\in{\cal M}(G)\}$
is a filter basis consisting of cosets
modulo $N_*\defi\overline{G_0N}$ of $G$.
We claim that $N_*\in{\cal M}(G)$. Indeed we have
$N_*\in{\cal N}(G)$. Now we let $M$ be an open subgroup
of $N_*\supseteq G_0$.
Then $G_0\subseteq M$, so $G_0N\subseteq MN$ and $MN$
is open-closed in $N_*=\overline{G_0N}$.
Thus $N_*=MN$. So $MN/M \cong N/(N\cap M)$ is discrete
and then $M\cap N$ is open in $N$. But $N\in{\cal M}(G)$ then implies
that $N_*/M\cong N/(M\cap N)$ is finite. This shows that
$N_*\in{\cal M}(G)$
as claimed. Since $G$ is Z-compact by Lemma { \ref{l:three}},
we find an element
$g\in\bigcap_{N\in{\cal M}(G)}F_N$. But then
$p_{N^*}(f(g))={g}_N$ for
all $N\in{\cal M}(G)$ which implies that $f(g)=\lim{\cal C}$.
Thus every Cauchy filter in $G_t$ converges showing
that $G_t$ is complete.
\qed
\msk
\nin
This completes the proof of Theorem \ref{th:main}.
\bsk
\quad An inspection of \cite{probook} shows that
the following questions appear to be unsettled:
\msk
\nin
{\bf Question 1.}\quad For which pro-Lie groups $G$ is
${\cal M}(G)$ cofinal in ${\cal N}(G)$?
\msk
For each of these groups $G$ we would know that $G$ is (isomorphic to)
the strict projective limit $\lim_{M\in{\cal M}(G)}G/M$.
In \cite{probook} this is proved of all almost
connected pro-Lie groups.
\ssk
\nin
Test examples are the nondiscrete pro-discrete groups
$\Z^{(\N)}$ and $\Z^\N$
(see e.g. \cite{probook}, Example 4.4ff., Proposition 5.2).
\msk
\nin
{\bf Question 2.}\quad For which pro-Lie groups is the Z-topology
compact?
\msk
In \cite{probook}, Proposition 4.27, this is shown for almost connected
pro-Lie groups, and here we have proved it for
those pro-Lie groups $G$ all of whose Lie group quotients are known to be
almost connected.
\msk
Theorem \ref{th:main} suggests the following rather general question:
\msk
\nin {\bf Question 3.}\quad When is the projective limit of a projective system
of almost connected topological groups almost connected?
\msk
Theorem \ref{th:main} says that within the category of pro-Lie groups
we have an affirmative answer
for the projective system of all Lie group quotients.
See also some background information in \cite{probook} in and around
Theorem 1.27, p.~88.
\bsk
\nin
One remark is in order in the context of the Z-topology discussed in
\cite{probook} on pp.~197--203:
In Exercise E4.2(i), {p.~202,} it is pointed out that
{ ${\cal M}(\Z)=\big\{\{0\},\Z\big\}$ and that therefore} the
Z-topology on $\Z$, being the cofinite topology, is compact. {
Whereas the topology generated by the set of cosets $z+N$,
$N\in{\cal N}(\Z)$, the set}
of all subgroups of $\Z$,
fails to be compact.
\bsk
Theorem \ref{th:main} will play a significant role
in the authors' study \cite{dah} of weakly complete real or complex topological
algebras with identity, which will explore in detail their relation
to pro-Lie theory and aims for a systematic treatment of weakly
complete group algebras of topological groups and their representation
and duality theories.
\msk
{
\nin {\bf Acknowledgment.}\quad The authors thank the referee for his
swift, yet thorough contributions to the final form of this note.} | 32,803 |
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\begin{document}
\vspace*{7mm}
\LARGE
\addtocounter{page}{-1}
\thispagestyle{empty}
\begin{center}
\textbf{Monogenic functions in
finite-dimensional\\
commutative associative algebras}
\end{center}
\vskip4mm
\large
\begin{center}\textbf{V.\,S.\,Shpakivskyi}\end{center}
\vspace{7mm}
\small
Let $\mathbb{A}_n^m$ be an arbitrary $n$-dimensional
commutative associative algebra
over the field of complex numbers with $m$ idempotents. Let
$e_1=1,e_2,\ldots,e_k$ with $2\leq k\leq 2n$ be elements of $\mathbb{A}_n^m$ which are linearly
independent over the field of real numbers. We consider monogenic
(i.~e. continuous and differentiable in the sense of Gateaux)
functions of the variable $\sum_{j=1}^k x_j\,e_j$\,, where $x_1,x_2,\ldots,x_k$ are
real, and obtain a constructive description of all mentioned
functions by means of holomorphic functions of complex variables.
It follows from this description that monogenic functions have
Gateaux derivatives of all orders. The present article is generalized of the author's paper \cite{Shpakivskyi-2014}, where mentioned results are obtained for $k=3$.
\vspace{7mm}
\large
\section{Introduction}
Apparently, W.~Hamilton (1843) made the first attempts to
construct an algebra associated with the three-dimensional Laplace
equation\medskip
\begin{equation} \label{Lap3}
\Delta_3 u(x,y,z):=
\left(\frac{{\partial}^2}{{\partial x}^2}+
\frac{{\partial}^2}{{\partial y}^2}+
\frac{{\partial}^2}{{\partial z}^2}\right)u(x,y,z)
=0\,
\end{equation}\medskip
meaning that components of hypercomplex functions satisfy the
equation
(\ref{Lap3}).
He constructed an algebra of noncommutative quaternions over the
field of real numbers $\mathbb{R}$ and made a base for developing
the hypercomplex analysis.
C.~Segre \cite{Segre} constructed an algebra of commutative
quaternions over the field $\mathbb{R}$ that can be considered as
a two-dimensional commutative semi-simple algebra of bicomplex
numbers over the field of complex numbers $\mathbb{C}$.
M.~Futagawa \cite{Futugawa} and J.~Riley \cite{Riley} obtained a constructive
description of analytic function of a bicomplex variable, namely,
they proved that such an analytic function can be constructed with
an use of two holomorphic functions of complex variables.
F.~Ringleb \cite{Ringleb} and S.~N.~Volovel'skaya \cite{Volovel'skaya-1939, Volovel'skaya-1948} succeeded in developing a function
theory for noncommutative algebras with unit over the
real or complex fields, by pursuing a definition of the differential of
a function on such an algebra suggested by Hausdorff in \cite{Hausdorff}. These definitions
make the \textit{a priori} severe requirement that the coordinates of the
function have continuous first derivatives with respect to the coordinates
of the argument element. Namely, F.~Ringleb \cite{Ringleb} considered an arbitrary
finite-dimensional associative (commutative or not) \textit{semi-simple}
algebra over the field
$\mathbb{R}$. For given class of functions which maps the mentioned
algebra onto itself, he obtained a constructive description by
means of real and complex analytic functions.
S.~N.~Volovel'skaya developed the Hausdorff's idea defining the \textit{monogenic} functions on
\textit{non-semisimple} associative algebras and she generalized the Ringleb's results for such algebras.
In the paper \cite{Volovel'skaya-1939} was obtained a
constructive description of monogenic functions in a special three-dimensional non-commutative
algebra over the field $\mathbb{R}$. The results of paper \cite{Volovel'skaya-1939} were generalized in the paper \cite{Volovel'skaya-1948} where Volovel'skaya obtained a
constructive description of monogenic functions in non-semisimple associative algebras of the first category over $\mathbb{R}$.
A relation between spatial potential fields and analytic functions
given in commutative algebras was established by P.~W.~Ketchum
\cite{Ketchum-28} who shown that every analytic function
$\Phi(\zeta)$ of the variable $\zeta=xe_1+ye_2+ze_3$ satisfies the
equation (\ref{Lap3}) in the case where the elements $e_1, e_2,
e_3$ of a commutative algebra satisfy the condition
\begin{equation}\label{garmonichnyj_bazys-ogljad}
e_1^2+e_2^2+e_3^2=0\,,
\end{equation}
because
\begin{equation}\label{garm}
\frac{{\partial}^{2}\Phi}{{\partial x}^{2}}+
\frac{{\partial}^{2}\Phi}{{\partial y}^{2}}+
\frac{{\partial}^{2}\Phi}{{\partial z}^{2}}\equiv{\Phi}''(\zeta) \
(e_1^2+e_2^2+e_3^2)=0\,,\medskip
\end{equation}
where $\Phi'':=(\Phi')'$ and $\Phi'(\zeta)$ is defined by the
equality $d\Phi=\Phi'(\zeta)d\zeta$.
We say that a commutative associative algebra $\mathbb A$ is {\it
harmonic\/} (cf.
\cite{Ketchum-28,Mel'nichenko75,Plaksa}) if in
$\mathbb A$ there exists a triad of linearly independent vectors
$\{e_1,e_2,e_3\}$ satisfying the equality
(\ref{garmonichnyj_bazys-ogljad}) with $e_k^2\ne 0$ for
$k=1,2,3$. We say also that such a triad $\{e_1,e_2,e_3\}$ is {\it
harmonic}.
P.~W.~Ketchum \cite{Ketchum-28} considered the C.~Segre algebra
of quaternions \cite{Segre} as an example of harmonic algebra.
Further M.~N.~Ro\c{s}cule\c{t} establishes a relation between monogenic functions in commutative algebras and partial differential equations. He defined \textit{monogenic} functions $f$ of the variable $w$ by the equality $df(w)\,dw=0$.
So, in the paper \cite{Rosculet} M.~N.~Ro\c{s}cule\c{t} proposed a procedure for constructing an infinite-dimensional topological vector space with commutative multiplication such that monogenic functions in it are the \textit{all} solutions of the equation
\begin{equation}\label{intr}
\sum\limits_{\alpha_0+\alpha_1+\ldots+\alpha_p=N}C_{\alpha_0,\alpha_1,\ldots,\alpha_p}\,
\frac{\partial^N \Phi}{\partial x_0^{\alpha_0}\,\partial x_1^{\alpha_1}\,\ldots\partial x_p^{\alpha_p}}=0,
\end{equation}
with $C_{\alpha_0,\alpha_1,\ldots,\alpha_p}\in\mathbb{R}$.
In particular, such infinite-dimensional topological vector space are constructed for the Laplace
equation (\ref{garm}). In the paper \cite{Rosculet-56} Ro\c{s}cule\c{t} finds a certain connection between monogenic functions in commutative algebras and systems of partial differential equations.
I.~P.~Mel'nichenko proposed for describing solutions of the equation (\ref{intr})
to use hypercomplex functions differentiable in the sense of Gateaux, since in this case the conditions of monogenic are the least restrictive. He started to implement this approach with respect to the thee-dimensional Laplace equation (\ref{garm}) (see \cite{Mel'nichenko75}).
Mel'nichenko proved that there exist exactly $3$
three-dimensional harmonic algebras with unit over the field
$\mathbb C$ (see
\cite{Mel'nichenko75,Melnichenko03,Plaksa}).
In the paper \cite{Pogorui-Rod-Shap}, the authors develop the Melnichenko's idea for the equation (\ref{intr}), and considered several examples.
The investigation of partial differential equations using the hypercomplex methods is effective if hypercomplex monogenic (in any sense) functions can be constructed explicitly. On this way the following results are obtained.
Constructive descriptions of monogenic (i.~e. continuous and
differentiable in the sense of Gateaux) functions taking values in
the mentioned three-dimensional harmonic algebras by means three
corresponding holomorphic functions of the complex variable are
obtained in the papers \cite{Pl-Shp1,Pl-Pukh,Pukh}. Such
descriptions make it possible to prove the infinite
differentiability in the sense of Gateaux of monogenic functions
and integral theorems for these functions that are analogous to
classical theorems of the complex analysis (see, e.~g.,
\cite{Pl-Shp3,Plaksa12}).
Furthermore, constructive descriptions of monogenic functions
taking values in special $n$-dimensional
commutative algebras by means $n$ holomorphic functions of complex
variables are obtained in the papers \cite{Pl-Shp-Algeria,
Pl-Pukh-Analele}.
In the paper \cite{Shpakivskyi-2014}, by author is obtained a constructive description of \textit{all} monogenic
functions of the variable $x_1e_1+x_2e_2+x_3e_3$ taking values in an arbitrary $n$-dimensional
commutative associative algebra with unit by means of holomorphic
functions of complex variables.
It follows from this description
that monogenic functions have Gateaux derivatives of all orders.
In this paper we extend the results of the paper \cite{Shpakivskyi-2014} to
monogenic functions of the variable $\sum\limits_{r=1}^k x_re_r$, where $2\leq k \leq 2n$.
\section{The algebra $\mathbb{A}_n^m$}
Let $\mathbb{N}$ be the set of natural numbers.
We fix the numbers $m,n\in\mathbb{N}$ such that $m\leq n$.
Let $\mathbb{A}_n^m$ be an arbitrary commutative associative algebra with
unit over the field of complex number $\mathbb{C}$.
E.~Cartan \cite[p.~33]{Cartan}
proved that there exist a basis $\{I_r\}_{r=1}^{n}$ in $\mathbb{A}_n^m$
satisfying the following multiplication rules:
\vskip3mm
1. \,\, $\forall\,r,s\in[1,m]\cap\mathbb{N}\,:$ \qquad
$I_rI_s=\left\{
\begin{array}{rcl}
0 &\mbox{if} & r\neq s,\vspace*{2mm} \\
I_r &\mbox{if} & r=s;\\
\end{array}
\right.$
\vskip5mm
2. \,\, $\forall\,r,s\in[m+1,n]\cap\mathbb{N}$\,: \qquad $I_rI_s=
\sum\limits_{p=\max\{r,s\}+1}^n\Upsilon_{r,p}^{s}I_p$\,;
\vskip5mm
3.\,\, $\forall\,s\in[m+1,n]\cap\mathbb{N}$\,\,
$\exists!\;
u_s\in[1,m]\cap\mathbb{N}$ \,$\forall\,
r\in[1,m]\cap\mathbb{N}$\,:\,\,
\begin{equation}\label{mult_rule_3}
I_rI_s=\left\{
\begin{array}{ccl}
0 \;\;\mbox{if}\;\; r\neq u_s\,,\vspace*{2mm}\\
I_s\;\;\mbox{if}\;\; r= u_s\,. \\
\end{array}
\right.\medskip
\end{equation}
Moreover, the structure constants
$\Upsilon_{r,p}^{s}\in\mathbb{C}$
satisfy the associativity conditions:
\vskip2mm (A\,1).\,\, $(I_rI_s)I_p=I_r(I_sI_p)$ \,\,
$\forall\,r,s,p\in[m+1,n]\cap\mathbb{N}$; \vskip2mm (A\,2).\,\,
$(I_uI_s)I_p=I_u(I_sI_p)$ \,\, $\forall\,u\in[1,m]\cap\mathbb{N}$\,\,
$\forall\,s,p\in[m+1,n]\cap\mathbb{N}$.
\vskip2mm
Obviously, the first $m$ basic vectors $\{I_u\}_{u=1}^m$
are idempotents and
form a semi-simple subalgebra of the algebra $\mathbb{A}_n^m$. The
vectors $\{I_r\}_{r=m+1}^n$ form a nilpotent subalgebra of the
algebra $\mathbb{A}_n^m$. The element $1=\sum_{u=1}^mI_u$ is the
unit of $\mathbb{A}_n^m$.
In the cases where $\mathbb{A}_n^m$ has some specific properties,
the following propositions are true.\vskip2mm
\textbf{Proposition 1 \cite{Shpakivskyi-2014}.} \textit{If there exists the unique
$u_0\in[1,m]\cap\mathbb{N}$ such that $I_{u_0}I_s=I_s$ for all
$s=m+1,\ldots,n$, then the associativity condition \em (A\,2) \em
is satisfied.}\vskip2mm
Thus, under the conditions of Proposition 1,
the associativity condition (A\,1) is only required.
It means that the nilpotent subalgebra of $\mathbb{A}_n^m$ with
the basis $\{I_r\}_{r=m+1}^n$ can be an arbitrary commutative
associative nilpotent algebra of dimension $n-m$. We note that such
nilpotent algebras are fully described for the dimensions
$1,2,3$ in the paper \cite{Burde_de_Graaf}, and some four-dimensional nilpotent algebras
can be found in the papers \cite{Burde_Fialowski}, \cite{Martin}.\vskip2mm
\textbf{Proposition 2 \cite{Shpakivskyi-2014}.} \textit{If all $u_r$ are different in the
multiplication rule \em 3\em ,
then $I_sI_p=0$ for all $s,p=m+1,\ldots, n$.}\vskip2mm
Thus, under the conditions of Proposition 2,
the multiplication table of the
nilpotent subalgebra of $\mathbb{A}_n^m$ with the basis
$\{I_r\}_{r=m+1}^n$ consists only of zeros, and all associativity
conditions are satisfied.
The algebra $\mathbb{A}_n^m$ contains $m$ maximal ideals
$$\mathcal{I}_u:=\Biggr\{\sum\limits_{r=1,\,r\neq u}^n\lambda_rI_r:\lambda_r\in
\mathbb{C}\Biggr\}, \quad u=1,2,\ldots,m,
$$
and their intersection is the radical $$\mathcal{R}:=
\Bigr\{\sum\limits_{r=m+1}^n\lambda_rI_r:\lambda_r\in
\mathbb{C}\Bigr\}.$$
Consider $m$ linear functionals
$f_u:\mathbb{A}_n^m\rightarrow\mathbb{C}$ satisfying the
equalities
$$f_u(I_u)=1,\quad f_u(\omega)=0\quad\forall\,\omega\in\mathcal{I}_u\,,
\quad u=1,2,\ldots,m.
$$
Inasmuch as the kernel of functional $f_u$ is the maximal ideal
$\mathcal{I}_u$, this functional is also continuous and
multiplicative (see \cite[p. 147]{Hil_Filips}).
\section{Monogenic functions}
Let us consider the vectors $e_1=1,e_2,\ldots,e_k$ in $\mathbb{A}_n^m$, where $2\leq k\leq 2n$, and these vectors are linearly independent over the field of real numbers
$\mathbb{R}$ (see \cite{Pl-Pukh-Analele}). It means that the equality
$$\sum\limits_{j=1}^k\alpha_je_j=0,\qquad \alpha_j\in\mathbb{R},$$
holds if and only if $\alpha_j=0$ for all $j=1,2,\ldots,k$.
Let the vectors $e_1=1,e_2,\ldots,e_k$ have the following decompositions with respect to
the basis $\{I_r\}_{r=1}^n$:
\begin{equation}\label{e_1_e_2_e_3-k}
e_1=\sum\limits_{r=1}^mI_r\,,
\quad e_j=\sum\limits_{r=1}^na_{jr}\,I_r\,,\quad a_{jr}\in\mathbb{C},\quad j=2,3,\ldots,k.
\end{equation}
Let $\zeta:=\sum\limits_{j=1}^kx_j\,e_j$, where $x_j\in\mathbb{R}$. It is
obvious that
$$\xi_u:=f_u(\zeta)=x_1+\sum\limits_{j=2}^kx_j\,a_{ju},\quad u=1,2,\ldots,m.$$
Let
$E_k:=\{\zeta=\sum\limits_{j=1}^kx_je_j:\,\, x_j\in\mathbb{R}\}$ be the
linear span of vectors $e_1=1,e_2,\ldots,e_k$ over the field
$\mathbb{R}$.
Let $\Omega$ be a domain in $E_k$. With a domain $\Omega\subset E_k$ we associate the domain $$\Omega_{\mathbb{R}}:=\Big\{(x_1,x_2,\ldots,x_k)\in\mathbb{R}^k:\,\zeta=\sum\limits_{j=1}^kx_j\,e_j\in\Omega\Big\}$$ in $\mathbb{R}^k$.
We say that a continuous function
$\Phi:\Omega\rightarrow\mathbb{A}_n^m$ is \textit{monogenic}
in $\Omega$ if $\Phi$ is differentiable in the sense of
Gateaux in every point of $\Omega$, i.~e. if for every
$\zeta\in\Omega$ there exists an element
$\Phi'(\zeta)\in\mathbb{A}_n^m$ such that
\begin{equation}\label{monogennaOZNA}\medskip
\lim\limits_{\varepsilon\rightarrow 0+0}
\left(\Phi(\zeta+\varepsilon
h)-\Phi(\zeta)\right)\varepsilon^{-1}= h\Phi'(\zeta)\quad\forall\,
h\in E_k.\medskip
\end{equation}
$\Phi'(\zeta)$ is the \textit{Gateaux derivative} of the function
$\Phi$ in the point $\zeta$.
Consider the decomposition of a function
$\Phi:\Omega\rightarrow\mathbb{A}_n^m$ with respect to the
basis $\{I_r\}_{r=1}^n$:
\begin{equation}\label{rozklad-Phi-v-bazysi-k}
\Phi(\zeta)=\sum_{r=1}^n U_r(x_1,x_2,\ldots,x_k)\,I_r\,.
\end{equation}
In the case where the functions $U_r:\Omega_{\mathbb{R}}\rightarrow\mathbb{C}$ are
$\mathbb{R}$-differentiable in $\Omega_{\mathbb{R}}$, i.~e. for every $(x_1,x_2,\ldots,x_k)\in\Omega_{\mathbb{R}}$,
$$U_r\left(x_1+\Delta x_1,x_2+\Delta x_2,\ldots,x_k+\Delta x_k\right)-U_r(x_1,x_2,\ldots,x_k)=
$$
$$=\sum\limits_{j=1}^k\frac{\partial U_r}{\partial x_j}\,\Delta x_j+
\,o\left(\sqrt{\sum\limits_{j=1}^k(\Delta x_j)^2}\,\right), \qquad \sum\limits_{j=1}^k(\Delta x_j)^2\to 0\,,$$
the function $\Phi$ is monogenic in the domain $\Omega$ if
and only if the following Cauchy~-- Riemann conditions are
satisfied in $\Omega$:
\begin{equation}\label{Umovy_K-R-k}
\frac{\partial \Phi}{\partial x_j}=\frac{\partial \Phi}{\partial
x_1}\,e_j \qquad \text{for all}\quad j=2,3,\ldots,k.
\end{equation}
\section{An expansion of the resolvent}
Let $b:=\sum\limits_{r=1}^nb_r\,I_r\in\mathbb{A}_n^m$, where $b_r\in\mathbb{C}$, and we note that $f_u(b)=b_u$, \, $u=1,2,\ldots,m$.
It follows form the Lemmas 1, 3 of \cite{Shpakivskyi-2014} that
\begin{equation}\label{obern-el}
b^{-1}=\sum\limits_{u=1}^m\frac{1}{b_u}\,I_u+
\sum\limits_{s=m+1}^{n}\sum\limits_{r=2}^{s-m+1}\frac{\widetilde{Q}_{r,s}}
{b_{u_{s}}^r}\,I_{s}\,.
\end{equation}
where
$\widetilde{Q}_{r,s}$ are determined by the following recurrence
relations:
\begin{equation}\label{Q-hv}
\begin{array}{c}
\displaystyle
\widetilde{Q}_{2,s}:=b_s\,,\qquad
\widetilde{Q}_{r,s}=\sum\limits_{q=r+m-2}^{s-1}\widetilde{Q}_{r-1,q}\,\widetilde{B}_{q,\,s}\,,\; \;\;r=3,4,\ldots,s-m+1,\\
\end{array}
\end{equation}
\begin{equation}\label{B-hv}
\widetilde{B}_{q,s}:=\sum\limits_{p=m+1}^{s-1}b_p \Upsilon_{q,s}^p\,,
\;\;p=m+2,m+3,\ldots,n,
\end{equation}
and the natural numbers $u_s$ are defined in the
rule 3 of the multiplication table of algebra
$\mathbb{A}_n^m$.
In the next lemma we find an expansion of the resolvent $(te_1-\zeta)^{-1}$.
\vskip2mm
\lem\label{lem_1_rezolv_A_n_m-k} \textit{An expansion of the
resolvent is of the form
\begin{equation}\label{lem-rez}
(te_1-\zeta)^{-1}=\sum\limits_{u=1}^m\frac{1}{t-\xi_u}\,I_u+
\sum\limits_{s=m+1}^{n}\sum\limits_{r=2}^{s-m+1}\frac{Q_{r,s}}
{\left(t-\xi_{u_{s}}\right)^r}\,I_{s}\,
\end{equation}
$$\forall\,t\in\mathbb{C}:\,
t\neq \xi_u,\quad u=1,2,\ldots,m,$$
where the coefficients\ $Q_{r,s}$ are determined by the following recurrence
relations:
\begin{equation}\label{Q}
\begin{array}{c}
\displaystyle
Q_{2,s}=T_s\,,\quad
Q_{r,s}=\sum\limits_{q=r+m-2}^{s-1}Q_{r-1,q}\,B_{q,\,s}\,,\; \;\;r=3,4,\ldots,s-m+1,\\
\end{array}
\end{equation}
with
\begin{equation}\label{B}
T_s:=\sum\limits_{j=2}^kx_ja_{js}\,, \quad
B_{q,s}:=\sum\limits_{p=m+1}^{s-1}T_p \Upsilon_{q,s}^p\,,
\;\;p=m+2,m+3,\ldots,n,
\end{equation}
and the natural numbers $u_s$ are defined in the
rule \em 3 \em of the multiplication table of algebra}
$\mathbb{A}_n^m$.\vskip2mm
\textbf{Proof.} Taking into account the decomposition
$$te_1-\zeta=\sum\limits_{u=1}^m(t-\xi_u)I_u-\sum\limits_{r=m+1}^n\sum\limits_{j=2}^kx_ja_{js}\,I_r\,,
$$
we conclude that the relation (\ref{lem-rez}) follows directly from the
equality (\ref{obern-el}) in which instead of $b_u$,\, $u=1,2,\ldots,m$ it should be used the
expansion $t-\xi_u$\,, and instead of $b_s$\,, $s=m+1,m+2,\ldots,n$ it should be used the
expansion $\sum\limits_{j=2}^kx_ja_{js}$.
The lemma is proved.
It follows from Lemma \ref{lem_1_rezolv_A_n_m-k} that the points
$(x_1,x_2,\ldots,x_k)\in\mathbb{R}^k$
corresponding to the noninvertible elements
$\zeta=\sum\limits_{j=1}^kx_j\,e_j$
form the set
\[M_u^{\mathbb{R}}:\quad\left\{
\begin{array}{r}x_1+\sum\limits_{j=2}^kx_j\,{\rm Re}\,a_{ju}=0,\vspace*{3mm} \\
\sum\limits_{j=2}^kx_j\,{\rm Im}\,a_{ju}=0, \\ \medskip
\end{array} \right. \qquad u=1,2,\ldots,m\]
in the $k$-dimensional space $\mathbb{R}^k$.
Also we consider the set
$M_u:=\{\zeta\in E_k: f_u(\zeta)=0\}$ for $u=1,2,\ldots,m$.
It is obvious that the set $M_u^{\mathbb{R}}\subset\mathbb{R}^k$ is congruent with the set $M_u\subset E_k$.
\vskip3mm
\section{A constructive description of monogenic functions}
\vskip3mm
We say that a domain $\Omega\subset E_k$ is \textit{convex with respect to the set of directions}
$M_u$ if $\Omega$ contains the segment $\{\zeta_1+\alpha(\zeta_2-\zeta_1):\alpha\in[0,1]\}$ for all $\zeta_1,\zeta_2\in \Omega$ such that $\zeta_2-\zeta_1\in M_u$.
Denote $f_u(E_k):=\{f_u(\zeta) : \zeta\in E_k\}$. In what follows,
we make the following essential assumption:
$f_u(E_k)=\mathbb{C}$ for all\, $u=1,2,\ldots,m$.
Obviously, it holds if and only if for every fixed $u=1,2, \ldots, m$
at least one of the numbers $a_{2u}$, $a_{3u},\ldots,a_{ku}$ belongs to
$\mathbb{C}\setminus\mathbb{R}$.
\vskip2mm
\lem\label{lem_1_konstruct_opys_A_n_m-k} \textit{Suppose that a
domain $\Omega\subset E_k$ is convex with respect to the set of directions
$M_u$ and $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots, m$. Suppose also that a
function $\Phi:\Omega\rightarrow\mathbb{A}_n^m$ is
monogenic in the domain
$\Omega$. If points
$\zeta_{1},\zeta_{2}\in\Omega$ such that
$\zeta_{2}-\zeta_{1}\in M_u$,
then}
\begin{equation}\label{Fi(dz1')-Fi(dz2')}
\Phi(\zeta_2)-\Phi(\zeta_1)\in\mathcal{I}_u\,.
\end{equation}
\textbf{Proof.} Inasmuch as $f_u(E_k)=\mathbb{C}$, then there exists an element $e_2^*\in E_k$
such that $f_u(e_2^*)=i$. Consider the lineal span $E^*:=\{\zeta=xe_1^*+ye_2^*+ze_3^*:x,y,z\in\mathbb{R}\}$ of the vectors $e_1^*:=1, e_2^*,e_3^*:=\zeta_2-\zeta_1$ and denote $\Omega^*:=\Omega\cap E^*$.
Now, the relations $(\ref{Fi(dz1')-Fi(dz2')})$ can be proved in such a way as Lemma 2.1 \cite{Pl-Shp1}, in the proof of which one must take $\Omega^*,f_u,\{\alpha e_3^*:\alpha\in\mathbb{R}\}$ instead of $\Omega_\zeta, f,L$, respectively. Lemma \ref{lem_1_konstruct_opys_A_n_m-k} is proved.
Let a domain $\Omega\subset E_k$ be convex with respect to the set of directions
$M_u$\,, $u=1,2,\ldots, m$. By
$D_u$ we denote that domain in $\mathbb{C}$
onto which the domain $\Omega$ is mapped by the functional $f_u$.
We introduce the linear operators $A_u$\,, $u=1,2,\ldots,m$, which
assign holomorphic functions $F_u:\,D_u\rightarrow\mathbb{C}$ to
every monogenic function
$\Phi:\Omega\rightarrow\mathbb{A}_n^m$ by the formula
\begin{equation}\label{def_op_A-k}
F_u(\xi_u)=f_u(\Phi(\zeta)),
\end{equation}
where $\xi_u=f_u(\zeta)\equiv x_1+\sum\limits_{j=2}^kx_j\,a_{ju}$ and
$\zeta\in\Omega$. It follows from Lemma
\ref{lem_1_konstruct_opys_A_n_m-k} that the value $F_u(\xi_u)$ does
not depend on a choice of a point $\zeta$ for which
$f_u(\zeta)=\xi_u$.
Now, similar to proof of Lemma 5 \cite{Shpakivskyi-2014} can be proved the following statement.
\vskip2mm
\lem\label{lem_2_konstruct_opys_A_n_m-k} \textit{Suppose
that a domain $\Omega\subset E_k$ is convex with respect to the set of directions
$M_u$ and $f_u(E_k)=\mathbb{C}$
for all $u=1,2,\ldots, m$. Suppose also that for any fixed
$u=1,2,\ldots,m$, a function $F_u:D_u\rightarrow \mathbb{C}$ is
holomorphic in a domain $D_u$ and $\Gamma_u$ is a closed Jordan
rectifiable curve in $D_u$ which surrounds the point $\xi_u$ and
contains no points $\xi_q$, $q=1,2,\ldots, m$,\, $q\neq u$. Then
the function
\begin{equation}\label{lem_5-k}
\Psi_u(\zeta):=I_u\int\limits_{\Gamma_u}F_u(t)(te_1-\zeta)^{-1}\,dt
\end{equation}
is monogenic in the domain
$\Omega$. }
\vskip2mm
\lem\label{Lem-6-osn-const-op-k} \textit{Suppose that a
domain $\Omega\subset E_k$ is convex with respect to the set of directions
$M_u$ and $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots, m$. Suppose also that a function $V:\Omega_{\mathbb{R}}\rightarrow\mathbb{C}$
satisfies the equalities
\begin{equation}\label{lem-6-1-k}
\frac{\partial V}{\partial x_2}=\frac{\partial V}{\partial x_1}\,
a_{2u}\,,\quad\frac{\partial V}{\partial x_3}=\frac{\partial V}{\partial x_1}\,
a_{3u}\,,\quad\ldots, \quad \frac{\partial V}{\partial x_k}=\frac{\partial
V}{\partial x_1}\, a_{ku}
\end{equation}
in $\Omega_{\mathbb{R}}$. Then $V$ is a holomorphic function of the variable
$\xi_u=f_u(\zeta)=x_1+\sum\limits_{j=2}^kx_j\,a_{ju}$ in the domain $D_u$.}
\vskip2mm
\textbf{Proof.} We first separate the real and the imaginary part
of the expression
\begin{equation}\label{lem-6-2-k}
\xi_u=x_1+\sum\limits_{j=2}^kx_j\,{\rm Re}\,a_{ju}+i\,
\sum\limits_{j=2}^kx_j\,{\rm Im}\,a_{ju}=:\tau_u+i\eta_u
\end{equation}
and note that the equalities (\ref{lem-6-1-k}) yield
\begin{equation}\label{lem-6-3-k}
\frac{\partial V}{\partial \eta_u}\,{\rm Im}\,a_{2u} =i\,\frac{\partial
V}{\partial \tau_u}\,{\rm Im}\,a_{2u}\,, \quad\ldots,\quad \frac{\partial
V}{\partial \eta_u}\,{\rm Im}\,a_{ku}=i\,\frac{\partial V}{\partial
\tau_u}\, {\rm Im}\,a_{ku}\,.
\end{equation}
It follows from the condition $f_u(E_k)=\mathbb{C}$ that
at least one of the numbers ${\rm Im}\,a_{2u}$\,, ${\rm Im}\,a_{3u}\,,\ldots,{\rm Im}\,b_u$ is not
equal to zero. Therefore, using (\ref{lem-6-3-k}), we get
\begin{equation}\label{lem-6-4-k}
\frac{\partial V}{\partial \eta_u}=i\,\frac{\partial V}{\partial
\tau_u}\,.
\end{equation}
Now we prove that $V(x'_1,x'_2,\ldots,x'_k)=V(x''_1,x''_2,\ldots,x''_k)$ for points
$(x'_1,x'_2,\ldots,x'_k),(x''_1,x''_2,\ldots,x''_k)\in\Omega$ such that the segment that
connects these points is parallel to a straight line $L_u\subset M_u^{\mathbb{R}}$\,.
To this end we use considerations with the proof of Lemma \ref{lem_1_konstruct_opys_A_n_m-k}.
Since $f_u(E_k)=\mathbb{C}$, then there exists an element $e_2^*\in E_k$
such that $f_u(e_2^*)=i$. Consider the lineal span $E^*:=\{\zeta=xe_1^*+ye_2^*+ze_3^*:x,y,z\in\mathbb{R}\}$ of the vectors $e_1^*:=1$, $e_2^*$, $e_3^*:=\zeta'-\zeta''$, where $\zeta':=\sum\limits_{j=1}^kx'_j\,e_j$\,, \, $\zeta'':=\sum\limits_{j=1}^kx''_j\,e_j$\,, and introduce the denotation $\Omega^*:=\Omega\cap E^*$.
Now, the relation $V(x'_1,x'_2,\ldots,x'_k)=V(x''_1,x''_2,\ldots,x''_k)$ can be proved in such a way as Lemma 6 \cite{Shpakivskyi-2014}, in the proof of which one must take $\Omega^*,\{\alpha e_3^*:\alpha\in\mathbb{R}\}$ instead of $\Omega_\zeta\,, L$, respectively. The lemma is proved.
Thus, a function $V:\Omega_{\mathbb{R}}\rightarrow\mathbb{C}$ of the form
$V(x_1,x_2,\ldots,x_k):=F(\xi_u)$,
where $F(\xi_u)$ is an arbitrary function holomorphic in
the domain $D_u$\,, is a general solution of the system
(\ref{lem-6-1-k}). The lemma is proved.
\vskip2mm
\theor\label{teo_1_konstruct_opys_A_n_m-k} \textit{Suppose that a
domain $\Omega\subset E_k$ is convex with respect to the set of directions
$M_u$ and $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots, m$. Then every
monogenic function $\Phi:\Omega\rightarrow\mathbb{A}_n^m$
can be expressed in the form
\begin{equation}\label{Teor--1-k}
\Phi(\zeta)=\sum\limits_{u=1}^mI_u\,\frac{1}{2\pi
i}\int\limits_{\Gamma_u} F_u(t)(te_1-\zeta)^{-1}\,dt+
\sum\limits_{s=m+1}^nI_s\,\frac{1}{2\pi i}\int\limits_
{\Gamma_{u_s}}G_s(t)(te_1-\zeta)^{-1}\,dt,
\end{equation}\vskip1mm
\noindent where $F_u$ and $G_s$ are certain holomorphic functions in the
domains $D_u$ and $D_{u_s}$, respectively, and $\Gamma_q$ is a
closed Jordan rectifiable curve in $D_q$ which surrounds the point
$\xi_q$ and contains no points $\xi_{\ell}$, $\ell,q=1,2,\ldots,
m$,\,$\ell\neq q$.}\vskip2mm
\textbf{Proof.} We set
\begin{equation}\label{teor__1-k}
F_u:=A_u\Phi,\;\;\;u=1,2,\ldots,m.
\end{equation}
Let us show that the values of monogenic function
\begin{equation}\label{teor__2-k}
\Phi_0(\zeta):=\Phi(\zeta)-\sum\limits_{u=1}^mI_u\,\frac{1}{2\pi
i} \int\limits_{\Gamma_u}F_u(t)(te_1-\zeta)^{-1}\,dt
\end{equation}
belong to the radical $\mathcal{R}$,
i.~e. $\Phi_0(\zeta)\in\mathcal{R}$ for all
$\zeta\in\Omega$.
As a consequence of the equality (\ref{lem-rez}), we have the
equality
$$I_u\,\frac{1}{2\pi i}
\int\limits_{\Gamma_u}F_u(t)(te_1-\zeta)^{-1}\,dt=I_u\,\frac{1}{2\pi
i} \int\limits_{\Gamma_u}\frac{F_u(t)}{t-\xi_u}\,dt+$$
$$
+\frac{1}{2\pi i}\sum\limits_{s=m+1}^{n}
\sum\limits_{r=2}^{s-m+1}\int\limits_{\Gamma_u}\frac{F_u(t)Q_{r,s}}
{\left(t-\xi_{u_{s}}\right)^r}\,dt
\,I_{s}\,I_u\,,
$$
from which we obtain the equality
\begin{equation}\label{teor__3-k}
f_u\left(\sum\limits_{u=1}^mI_u\,\frac{1}{2\pi i}
\int\limits_{\Gamma_u}F_u(t)(te_1-\zeta)^{-1}\,dt\right)=F_u(\xi_u).
\end{equation}
Operating onto the equality (\ref{teor__2-k}) by the functional
$f_u$ and taking into account the relations (\ref{def_op_A-k}), (\ref{teor__1-k}), (\ref{teor__3-k}), we get the equality
$$f_u(\Phi_0(\zeta))=F_u(\xi_u)-F_u(\xi_u)=0
$$
for all $u=1,2,\ldots,m$, i.~e. $\Phi_0(\zeta)\in\mathcal{R}$.
Therefore, the function $\Phi_0$ is of the form
\begin{equation}\label{Fi_0--1-k}
\Phi_{0}(\zeta)=\sum\limits_{s=m+1}^{n} V_{s}(x_1,x_2,\ldots,x_k)\,I_s\,,
\end{equation}
where $V_{s}:\Omega_\mathbb{R}\rightarrow\mathbb{C}$\,, and the Cauchy~--
Riemann conditions (\ref{Umovy_K-R-k}) are satisfied with
$\Phi=\Phi_0$.
Substituting the expressions (\ref{e_1_e_2_e_3-k}), (\ref{Fi_0--1-k}) into the equality
(\ref{Umovy_K-R-k}), we obtain
\begin{equation}\label{teor__5-1-k}
\begin{array}{c}
\displaystyle
\sum\limits_{s=m+1}^{n} \frac{\partial V_{s}}{\partial x_2}\,I_s=
\sum\limits_{s=m+1}^{n} \frac{\partial V_{s}}{\partial x_1}\,I_s
\sum\limits_{r=1}^n a_{2r}\,I_r\,,
\vspace*{1mm}\\ \vdots \vspace*{3mm}\\
\displaystyle
\sum\limits_{s=m+1}^{n} \frac{\partial V_{s}}{\partial x_k}\,I_s=
\sum\limits_{s=m+1}^{n} \frac{\partial V_{s}}{\partial x_1}\,I_s
\sum\limits_{r=1}^n a_{kr}\,I_r\,.\\
\end{array}
\end{equation}
Equating the coefficients of $I_{m+1}$ in these equalities,
we obtain the following system of equations for determining the function
$V_{m+1}(x_1,x_2,\ldots,x_k)$:
$$\frac{\partial V_{m+1}}{\partial x_2}=\frac{\partial V_{m+1}}{\partial x_1}\,
a_{2\, u_{m+1}}\,,
\quad\ldots,\quad
\frac{\partial V_{m+1}}{\partial x_k}=\frac{\partial V_{m+1}}{\partial x_1}\,
a_{k\,u_{m+1}}\,.
$$
It follows from Lemma \ref{Lem-6-osn-const-op-k} that
$V_{m+1}(x_1,x_2,\ldots,x_k)\equiv G_{m+1}(\xi_{u_{m+1}})$, where $G_{m+1}$ is
a function holomorphic in the domain $D_{u_{m+1}}$\,. Therefore,
\begin{equation}\label{teor__4}
\Phi_0(\zeta)=G_{m+1}(\xi_{u_{m+1}})\,I_{m+1}+
\sum\limits_{s=m+2}^{n} V_{s}(x_1,x_2,\ldots,x_k)\,I_s\,.
\end{equation}
Due to the expansion (\ref{lem-rez}), we have the representation
\begin{equation}\label{teor__5}
I_{m+1}\,\frac{1}{2\pi
i}\int\limits_{\Gamma_{u_{m+1}}}G_{m+1}(t)(te_1-\zeta)^{-1}\,dt=
G_{m+1}(\xi_{u_{m+1}})\,I_{m+1}+\Psi(\zeta),
\end{equation}
where $\Psi(\zeta)$ is a function with values in the set
$\big\{\sum_{s=m+2}^n\alpha_s\,I_s:\alpha_s\in\mathbb{C}\big\}$.
Now, consider the function
$$\Phi_1(\zeta):=\Phi_0(\zeta)-I_{m+1}\,\frac{1}{2\pi
i}\int\limits_{\Gamma_{u_{m+1}}}G_{m+1}(t)(te_1-\zeta)^{-1}\,dt.$$
In view of the relations (\ref{teor__4}), (\ref{teor__5}),
$\Phi_1$ can be represented in the form
$$\Phi_{1}(\zeta)=\sum\limits_{s=m+2}^n \widetilde
V_{s}(x_1,x_2,\ldots,x_k)\,I_s\,,$$
where $\widetilde V_{s}:\Omega_\mathbb{R}\rightarrow\mathbb{C}$\,.
Inasmuch as $\Phi_1$ is a monogenic function in $\Omega$,
the functions $\widetilde V_{m+2},\widetilde
V_{m+3},\dots,\widetilde V_{n}$ satisfy the system
\eqref{teor__5-1-k}, where $V_{m+1}\equiv 0$, $V_s=\widetilde V_{s}$
for $s=m+2,m+3,\ldots,n$. Therefore, similarly to the function
$V_{m+1}(x_1,x_2,\ldots,x_k)\equiv G_{m+1}(\xi_{u_{m+1}})$, the function
$\widetilde V_{m+2}$ satisfies the equations
$$
\frac{\partial \widetilde V_{m+2}}{\partial x_2}=\frac{\partial \widetilde V_{m+2}}
{\partial x_1}\,a_{2\,u_{m+2}}\,,\quad\ldots,\quad
\frac{\partial \widetilde V_{m+2}}{\partial x_k}=\frac{\partial
\widetilde V_{m+2}}{\partial x_1}\,a_{k\,u_{m+2}}
$$
and is of the
form $\widetilde V_{m+2}(x_1,x_2,\ldots,x_k)\equiv G_{m+2}(\xi_{u_{m+2}})$,
where $G_{m+2}$ is a function holomorphic in the domain
$D_{u_{m+2}}$\,.
In such a way, step by step, considering the functions
$$\Phi_j(\zeta):=\Phi_{j-1}(\zeta)-I_{m+j}\,\frac{1}{2\pi
i}\int\limits_{\Gamma_{u_{m+j}}}G_{m+j}(t)(te_1-\zeta)^{-1}\,dt$$
for $j=2,3,\dots,n-m-1$, we get the representation
(\ref{Teor--1-k}) of the function $\Phi$. The theorem is proved.
Taking into account the expansion (\ref{lem-rez}), one can rewrite
the equality (\ref{Teor--1-k}) in the following equivalent form:
$$\Phi(\zeta)=\sum\limits_{u=1}^mF_u(\xi_u)I_u+
\sum\limits_{s=m+1}^{n}\sum\limits_{r=2}^{s-m+1}\frac{1}{(r-1)!}\,
Q_{r,s}\,F_{u_s}^{(r-1)}(\xi_{u_s})\,I_{s}+$$
\begin{equation}\label{dopolnenije-1-1-9-k}+\sum\limits_{q=m+1}^nG_q(\xi_{u_q})I_q+
\sum\limits_{q=m+1}^n\sum\limits_{s=m+1}^{n}
\sum\limits_{r=2}^{s-m+1}\frac{1}{(r-1)!}\,Q_{r,s}\,G_q^{(r-1)}(\xi_{u_q})\,I_{q}
\,I_s\,.
\end{equation}\vskip4mm
Thus, the equalities (\ref{Teor--1-k}) and (\ref{dopolnenije-1-1-9-k})
specify methods to construct explicitly any monogenic functions
$\Phi:\Omega\rightarrow \mathbb{A}_n^m$ using $n$
corresponding holomorphic functions of complex variables.
The following statement follows immediately from the equality
(\ref{dopolnenije-1-1-9-k}) in which the right-hand side is a
monogenic function in the domain $\Pi:=\{\zeta\in
E_k:f_u(\zeta)=D_u,\,u=1,2,\ldots,m\}$.\vskip2mm
\theor\label{teo_pro_naslidky-k} {\it Let a domain $\Omega\subset E_k$ is convex with respect to the set of directions $M_u$ and $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots, m$. Then every monogenic function
$\Phi:\Omega\rightarrow \mathbb{A}_n^m$ can be continued
to a function monogenic in the domain $\Pi$.}
The next statement is a fundamental consequence of the equality
(\ref{dopolnenije-1-1-9-k}), and it is true for an arbitrary domain
$\Omega$.\vskip2mm
\theor\label{teo_pro_naslidky2-k} {\it Let $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots,m$. Then for every monogenic function
$\Phi:\Omega\rightarrow \mathbb{A}_n^m$ in an arbitrary
domain $\Omega$, the Gateaux $r$-th derivatives
$\Phi^{(r)}$ are monogenic functions in $\Omega$ for all\,
$r$.}
The proof is completely analogous to the proof of Theorem 4
\cite{Pl-Shp1}.
Using the integral expression
(\ref{Teor--1-k}) of monogenic
function $\Phi:\Omega\rightarrow \mathbb{A}_n^m$ in the case
where a domain $\Omega$ is convex with respect to the set of directions $M_u$\,,\;$u=1,2,\ldots,m$, we obtain the following
expression for the Gateaux $r$-th derivative $\Phi^{(r)}$:
$$\Phi^{(r)}(\zeta)=\sum\limits_{u=1}^mI_u\,\frac{r!}{2\pi i}\int\limits_{\Gamma_u}
F_u(t)\Big((te_1-\zeta)^{-1}\Big)^{r+1}\,dt+$$
$$
+\sum\limits_{s=m+1}^nI_s\,\frac{r!}{2\pi i}\int\limits_
{\Gamma_{u_s}}G_s(t)\Big((te_1-\zeta)^{-1}\Big)^{r+1}\,dt\qquad
\forall\;\zeta\in\Omega\,.\medskip
$$
\section{Remarks}
We note that in the cases where the algebra $\mathbb{A}_n^m$ has some
specific properties (for instance, properties described in
Propositions 1 and 2),
it is easy to simplify the form of the equality
(\ref{dopolnenije-1-1-9-k}). \vskip2mm
\textbf{1.} In the case considered in Proposition 1,
the following equalities hold:
$$u_{m+1}=u_{m+2}=\ldots=u_n=:\eta\,.$$
In this case the representation (\ref{dopolnenije-1-1-9-k}) takes
the form
$$\Phi(\zeta)=\sum\limits_{u=1}^mF_u(\xi_u)I_u+
\sum\limits_{s=m+1}^{n}\sum\limits_{r=2}^{s-m+1}\frac{1}{(r-1)!}\,
Q_{r,s}\,F_\eta^{(r-1)}(\xi_\eta)\,I_{s}+$$
\begin{equation}\label{dopolnenije-1-1-k}+\sum\limits_{s=m+1}^nG_s(\xi_\eta)I_s+
\sum\limits_{q=m+1}^n\sum\limits_{s=m+1}^{n}
\sum\limits_{r=2}^{s-m+1}\frac{1}{(r-1)!}\,Q_{r,s}\,G_q^{(r-1)}(\xi_\eta)\,I_{s}
\,I_q\,.
\end{equation}\vskip2mm
The formula (\ref{dopolnenije-1-1-k}) generalizes representations of
monogenic functions in both three-dimensional harmonic algebras
(see \cite{Pl-Shp1,Pl-Pukh,Pukh}) and specific $n$-dimensional
algebras (see \cite{Pl-Shp-Algeria,Pl-Pukh-Analele}) to the case
of algebras more general form and to a variable of more general form.
\vskip2mm \textbf{2.} In the case considered in Proposition 2, the representation (\ref{Teor--1-k}) takes the form
\begin{equation}\label{dopolnenije-3-k}
\Phi(\zeta)=\sum\limits_{u=1}^mF_u(\xi_u)I_u+\sum\limits_{s=m+1}^nG_s(\xi_{u_s})I_s+
\sum\limits_{s=m+1}^nT_sF_{u_s}^{\,'}(\xi_{u_s})I_s\,.
\end{equation}
The formula (\ref{dopolnenije-3-k}) generalizes representations of
monogenic functions in both a three-dimensional harmonic algebra
with one-dimensional radical (see \cite{Pl-Pukh}) and semi-simple
algebras (see \cite{Pukh,Pl-Pukh-Analele}) to the case of algebras
more general form and to a variable of more general form.
\vskip2mm \textbf{3.} In the case where $n=m$, the algebra
$\mathbb{A}_n^n$ is semi-simple and contains no nilpotent
subalgebra. Then the formulae (\ref{dopolnenije-1-1-k}),
(\ref{dopolnenije-3-k}) take the form
$$
\Phi(\zeta)=\sum\limits_{u=1}^nF_u(\xi_u)I_u\,,
$$
because there are no vectors $\{I_k\}_{k=m+1}^n$.
This formula was obtained in the paper \cite{Pl-Pukh-Analele}.
\section{The relations between monogenic functions and partial differential equations}
Consider the following linear partial differential equation with
constant coefficients:
\begin{equation}\label{dopolnenije----1-k}
\mathcal{L}_NU(x_1,x_2,\ldots,x_k):=\sum\limits_{\alpha_1+\alpha_2\ldots+\alpha_k=N}
C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\,
\frac{\partial^N \Phi}{\partial x_1^{\alpha_1}\,\partial x_2^{\alpha_2}\,\ldots\partial x_k^{\alpha_k}}=0,
\end{equation}\vskip2mm
If a function $\Phi(\zeta)$ is $N$-times
differentiable in the sense of Gateaux in every point of
$\Omega$, then
$$\frac{\partial^{\alpha_1+\alpha_2+\ldots+\alpha_k}\Phi}
{\partial x_1^{\alpha_1}\,\partial x_2^{\alpha_2}\ldots\partial x_k^{\alpha_k}}=$$
$$=
e_1^{\alpha_1}\, e_2^{\alpha_2}\ldots e_k^{\alpha_k}\,\Phi^{(\alpha_1+\alpha_2+\ldots+\alpha_k)}(\zeta)=
e_2^{\alpha_2}e_3^{\alpha_3}\ldots e_k^{\alpha_k}
\,\Phi^{(N)}(\zeta).
$$
Therefore, due to the equality
\begin{equation}\label{dopolnenije----2-1-k}
\mathcal{L}_N\Phi(\zeta)=\Phi^{(N)}(\zeta)
\sum\limits_{\alpha_1+\alpha_2+\ldots+\alpha_k=N}C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\,e_2^{\alpha_2}e_3^{\alpha_3}\ldots e_k^{\alpha_k}\,,
\end{equation}
every $N$-times differentiable in the sense of Gateaux in
$\Omega$ function $\Phi$
satisfies the equation $\mathcal{L}_N\Phi(\zeta)=0$ everywhere in
$\Omega$ if and only if
\begin{equation}\label{dopolnenije----2-k}
\sum\limits_{\alpha_1+\alpha_2+\ldots+\alpha_k=N}C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\,e_2^{\alpha_2}e_3^{\alpha_3}\ldots e_k^{\alpha_k}=0.
\end{equation}
Accordingly, if the condition (\ref{dopolnenije----2-k}) is
satisfied, then the real-valued components ${\rm Re}\,U_k(x_1,x_2,\ldots,x_k)$
and ${\rm Im}\,U_k(x_1,x_2,\ldots,x_k)$ of the decomposition
(\ref{rozklad-Phi-v-bazysi-k})
are solutions of the equation (\ref{dopolnenije----1-k}).
In the case where $f_u(E_k)=\mathbb{C}$ for all $u=1,2, \ldots,
m$, it follows from Theorem \ref{teo_pro_naslidky2-k} that the
equality (\ref{dopolnenije----2-1-k}) holds for every monogenic
function $\Phi:\Omega\rightarrow \mathbb{A}_n^m$.
Thus,
to construct solutions of the equation (\ref{dopolnenije----1-k}) in
the form of components of monogenic
functions,
we must to find $k$ linearly independent over the field $\mathbb{R}$
vectors $(\ref{e_1_e_2_e_3-k})$ satisfying the characteristic
equation (\ref{dopolnenije----2-k}) and to verify the condition:
$f_u(E_k)=\mathbb{C}$ for all $u=1,2,\ldots,m$. Then, the formula
(\ref{Teor--1-k}) gives a constructive description of all mentioned
monogenic functions.
In the next theorem, we assign a special class of equations
(\ref{dopolnenije----1-k}) for which $f_u(E_k)=\mathbb{C}$ for all
$u=1,2,\ldots,m$. Let us introduce the polynomial
\begin{equation}\label{dopolnenije----51-k}
P(b_2,b_3,\ldots,b_k):=\sum\limits_{\alpha_1+\alpha_2+\ldots+\alpha_k=N}C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\,
b_2^{\alpha_2}\,b_3^{\alpha_3}\ldots b_k^{\alpha_k}.
\end{equation}
\vskip1mm
\theor\label{teo_dopolnenije-dlja-uravn-k} {\it Suppose that
there exist linearly independent over the field $\mathbb{R}$
vectors $e_1=1,e_2,\ldots,e_k$ in $\mathbb{A}_n^m$ of the form
$(\ref{e_1_e_2_e_3-k})$ that satisfy the equality
$(\ref{dopolnenije----2-k})$. If $P(b_2,b_3,\ldots,b_k)\neq0$ for all real $b_2,b_3,\ldots,b_k$, then $f_u(E_k)=\mathbb{C}$ for all $u=1,2,\ldots,m$.}\vskip2mm
\textbf{Proof.} Using the multiplication table of
$\mathbb{A}_n^m$, we obtain the equalities
$$e_2^{\alpha_2}=\sum\limits_{u=1}^ma_{2u}^{\alpha_2}\,I_u+\Psi_\mathcal{R}\,,\quad\ldots,\quad
e_k^{\alpha_k}=\sum\limits_{u=1}^ma_{ku}^{\alpha_k}\,I_u+\Theta_\mathcal{R}\,,
$$
where $\Psi_\mathcal{R}\,,\ldots,\Theta_\mathcal{R}\in\mathcal{R}$. Now
the equality
(\ref{dopolnenije----2-k}) takes the form
\begin{equation}\label{dopolnenije----4-k}
\sum\limits_{\alpha_1+\alpha_2+\ldots+\alpha_k=N}C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\Biggr(
\sum\limits_{u=1}^ma_{2u}^{\alpha_2}\ldots a_{ku}^{\alpha_k}\,I_u+\widetilde{\Psi}_\mathcal{R}\Biggr)=0,
\end{equation}
where $\widetilde{\Psi}_\mathcal{R}\in\mathcal{R}$. Moreover, due
to the assumption that the vectors $e_1,e_2,\ldots,e_k$ of the form
$(\ref{e_1_e_2_e_3-k})$ satisfy the equality
$(\ref{dopolnenije----2-k})$, there exist complex coefficients
$a_{jr}$ for $j=1,2,\ldots,k,$ $r=1,2,\ldots,n$ that satisfy the equality
(\ref{dopolnenije----4-k}).
It follows from the equality (\ref{dopolnenije----4-k}) that
\begin{equation}\label{dopolnenije----5-k}
\sum\limits_{\alpha_1+\alpha_2+\ldots+\alpha_k=N}C_{\alpha_1,\alpha_2,\ldots,\alpha_k}\,a_{2u}^{\alpha_2}\ldots a_{ku}^{\alpha_k}=0,
\qquad u=1,2,\ldots,m.
\end{equation}
Since $P(b_2,b_3,\ldots,b_k)\neq0$ for all $\{b_2,b_3,\ldots,b_k\}\subset\mathbb{R}$, the equalities
(\ref{dopolnenije----5-k}) can be satisfied only if for each
$u=1,2,\ldots,m$ at least one of the numbers $a_{2u}$, $a_{3u},\ldots,a_{ku}$ belongs to
$\mathbb{C}\setminus\mathbb{R}$
that implies the relation $f_u(E_k)=\mathbb{C}$ for all\, $u=1,2,\ldots,m$.
The theorem is proved.
We note that
if $P(b_2,b_3,\ldots,b_k)\neq0$ for all $\{b_2,b_3,\ldots,b_k\}\subset\mathbb{R}$, then
$C_{N,0,0,\ldots,0}\neq0$ because otherwise
$P(b_2,b_3,\ldots,b_k)=0$ for $b_2=b_3=\ldots=b_k=0$.
Since the function $P(b_2,b_3,\ldots,b_k)$ is continuous on $\mathbb{R}^k$, the
condition $P(b_2,b_3,\ldots,b_k)\neq0$ means either $P(b_2,b_3,\ldots,b_k)>0$ or $P(b_2,b_3,\ldots,b_k)<0$ for
all real $b_2,b_3,\ldots,b_k$. Therefore, it is obvious that for any
equation (\ref{dopolnenije----1-k}) of elliptic type, the condition
$P(b_2,b_3,\ldots,b_k)\neq0$ is always satisfied for all $\{b_2,b_3,\ldots,b_k\}\subset\mathbb{R}$.
At the same time, there are equations
(\ref{dopolnenije----1-k}) for which $P(b_2,b_3,\ldots,b_k)>0$ for all $\{b_2,b_3,\ldots,b_k\}\subset\mathbb{R}$, but
which are not elliptic. For example, such is the equation
$$\frac{\partial^3 u}{\partial x_1^3}+\frac{\partial^3 u}{\partial x_1\partial x_2^2}+
\frac{\partial^3 u}{\partial x_1\partial x_3^2}+\frac{\partial^3 u}{\partial x_1\partial x_4^2}=0
$$
considered in $\mathbb{R}^4$.
\vskip3mm | 158,022 |
TITLE: What is the manifold structure of U(n)?
QUESTION [8 upvotes]: A Lie group is simultaneously a differentiable manifold. As I understand it, the Lie group is generated via exponentiation of the generators of the Lie algebra.
It is intuitively clear to me that the Lie group U(1) is isomorphic to the manifold of the circle. Are SU(2) and SO(3) isomorphic to the two-sphere?
Furthermore, what is the manifold structure of U(n) in general? Furthermore, can anything be said about the manifold structure of the infinite-dimensional Unitary group? (This group occurs all the time in applications in quantum physics when the Hilbert space is infinite-dimensional).
REPLY [14 votes]: $\text{SU}(2)$ is in fact diffeomorphic to the $3$-sphere $S^3$, rather than the $2$-sphere (which has the wrong dimension; recall that $\mathfrak{su}(2)$ is $3$-dimensional); this comes from its identification with the unit quaternions. $\text{SO}(3)$ is diffeomorphic to the real projective space $\mathbb{RP}^3$, which is the quotient of $S^3$ obtained by identifying antipodes.
In general, $\text{U}(n)$ can't be identified as a more familiar-looking manifold. It is an "iterated extension" of the odd-dimensional spheres $S^1, S^3, ..., S^{2n-1}$, and in fact is rationally homotopy equivalent to the product $S^1 \times S^3 \times ... \times S^{2n-1}$. This means in particular that it has the same rational cohomology and rational homotopy groups as this product; however, it is in general not not homeomorphic or diffeomorphic to this product. "Iterated extension" means that the unitary groups fit into fiber sequences which are ultimately built from odd spheres, starting with
$$\text{SU}(n) \to \text{U}(n) \xrightarrow{\det} S^1$$
and continuing with
$$\text{SU}(n-1) \to \text{SU}(n) \to S^{2n-1}.$$
The first sequence is even a short exact sequence of Lie groups, and it splits smoothly, so $\text{U}(n)$ is diffeomorphic to $\text{SU}(n) \times S^1$; in particular $\text{U}(2)$ is diffeomorphic to $S^3 \times S^1$. But this is not an isomorphism of groups ($\text{U}(n)$ is a semidirect product rather than a direct product), and the corresponding statement for the other fiber sequences should be false, although I haven't verified this. I do know that $\text{SU}(3)$ is not diffeomorphic, and in fact is not even homotopy equivalent, to $S^3 \times S^5$; see this MO question.
The infinite-dimensional unitary group is not a manifold in the usual sense, although it is a Hilbert manifold. Kuiper's theorem implies that it is weakly contractible, so from the perspective of homotopy theory it looks like a point. But of course what actually appears in physics is not the unitary group but the projective unitary group; the infinite-dimensional projective unitary group is an Eilenberg-MacLane space $K(\mathbb{Z}, 2)$. This comes from its identification as a quotient of a weakly contractible space, namely the infinite-dimensional unitary group, by a free action of $\text{U}(1)$; hence it is a model for the classifying space $B \text{U}(1)$, and $\text{U}(1) \cong S^1$ itself is a $K(\mathbb{Z}, 1)$.
A closely related space called the stable unitary group is also very interesting from the perspective of homotopy theory; its homotopy groups are $2$-periodic, which is one way of stating complex Bott periodicity. | 72,600 |
TITLE: What primes divide the discriminant of a polynomial?
QUESTION [11 upvotes]: Given a monic polynomial $p(t) = t^n + ... + c_1 t + c_0$ with integer (or rational) coefficients and with roots $a_1, \dots a_n$, we can compute its discriminant, which is defined to be $\prod_{i< j}(a_i - a_j)^2$.
In my case, I have a polynomial which is the characteristic polynomial of some invertible matrix $T$. It is palindromic -- i.e., $c_{n-i} = c_i$ for all $0 \leq i \leq n$ -- so the roots come in inverse pairs $a$ and $\frac{1}{a}$. There are no repeated roots, so the discriminant is non-zero.
My question is: is there any way of knowing which primes divide this discriminant, i.e. from the coefficients of the polynomial or from the matrix $T$?
REPLY [5 votes]: I disagree with the definition of the discriminant as the resultant of $P$ and $P'$.
When $P$ is a polynomial with integer coefficients, then a prime $q$ should divide the discriminant of $P$ if and only if the reduction of $P$ modulo $q$ has a multiple root (possibly at infinity, when the degree decreases by at least 2 under reduction). But now consider $P=2X^2+ 3X+1$. The resultant of $P$ and $P'$ is $-2$, and the reduction of $P$ modulo 2 has no multiple root. In this case, the well known discriminant $b^2-4ac$ is actually 1. The correct relation between the discriminant and the resultant for a polynomial $P(t)=a_nt^n+\cdots+a_1t+a_0$ is $\mathrm{disc}(P)= (-1)^{n(n-1)/2}\mathrm{res}(P,P')/a_n$. | 19,604 |
Fresno State Men Win Three Straight in MWC
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For the first time since joining the Mountain West Conference, the Fresno State men's basketball team is on a three-game conference win streak. Playing at home Saturday against the last place team in the MWC, the Bulldogs got out to a flying start and rolled to an 82-56 win.
The 'Dogs were led by 23 from sophomore point guard Cezar Guerrero, and also got 16 points from Marvelle Harris and a double double from Tyler Johnson.
Fresno State is back in action at Nevada on Wednesday. | 351,851 |
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TITLE: Log formula proof and explanation
QUESTION [0 upvotes]: How can we prove this identity?
$$a^{\log_b n}=n^{\log_b a}$$
Thanks.
REPLY [0 votes]: The definition of $\log_m w$ where $m>0; m\ne 1; w> 0$ is that there is a unique real $k$ where $m^k = w$ and $\log_m w =k$ is defined to be that value.
Therefore for any $w\ge 0$ and $m\ne 1; m> 0$ we will always have $w = m^{\log_m w}$ because $\log_m w$ is defined to be the unique $k$ so that $m^k = w$ and so $m^{\log_m w} = m^k = w$. Always.
So $a^{\log_b n} = (b^{\log_b a})^{\log_b n} = b^{\log_b a \times \log_b n}$
But $n^{\log_b a} = (b^{\log_b n})^{\log_b a} = b^{\log_b n\times \log_b a}$.
And so both $a^{\log_b n}$ and $n^{\log_b a}$ are both equal to the same thing. | 144,392 |
\begin{document}
\title[Extended Affine Lie Superalgebras]{Extended Affine Lie Superalgebras}
\maketitle
\centerline{\bf Malihe Yousofzadeh\footnote{
Department of Mathematics, University of Isfahan, Isfahan, Iran,
and School of Mathematics, Institute for Research in
Fundamental
Sciences (IPM), Tehran, Iran.\\\\
{\it {\small Key Words:}} {\small Extended affine Lie superalgebra, Extended affine root supersystem.}\\
{\it {\small 2010 Mathematics Subject Classification}.} 17B67}}
\bigskip
\parbox{5in}{
Abstract. We introduce the notion of extended affine Lie superalgebras and investigate the properties of their root systems. Extended affine Lie algebras, invariant affine reflection algebras, finite dimensional basic classical simple Lie superalgebras and affine Lie superalgebras are examples of extended affine Lie superalgebras.}
\section{Introduction} Given an arbitrary $n\times n$-matrix $A$ and a subset $\tau\sub \{1,\ldots,n\},$ one can define the contragredient Lie superalgebra $\gg(A,\tau)$ which is presented by a finite set of generators subject to specific relations. Contragredient Lie superalgebras associated with so-called generalized Cartan matrices are known as Kac-Moody Lie superalgebras. These Lie superalgebras are of great importance among contragredient Lie superalgebras; in particular, affine Lie superalgebras, i.e., those Kac-Moody Lie superalgebras which are of finite growth, but not of finite dimension and equipped with a nondegenerate invariant even supersymmetric bilinear form, play a significant role in the theory of Lie superalgebras.
In the past 40 years, researchers in many areas of mathematics and mathematical physics have been attracted to Kac-Moody Lie superalgebras $\gg(A,\emptyset)$ known as Kac-Moody Lie algebras. These Lie algebras are a natural generalization of finite dimensional simple Lie algebras. One of the differences between affine Lie superalgebras and affine Lie algebras is the existence of nonsingular roots i.e., roots which are orthogonal to themselves but not to all other roots. In 1990, R. H\o egh-Krohn and B. Torresani \cite{HT} introduced irreducible quasi simple Lie algebras as a generalization of both affine Lie algebras and finite dimensional simple Lie algebras over complex numbers. In 1997, the authors in \cite{AABGP} systematically studied irreducible quasi simple Lie algebras under the name extended affine Lie algebras.
The existence of isotropic roots, i.e., roots which are orthogonal to all other roots, is one of the phenomena which occurs in extended affine Lie algebras but not in finite dimensional simple Lie algebras.
Since 1997, different generalizations of extended affine Lie algebras have been studied; toral type extended affine Lie algebras \cite{AKY}, locally extended affine Lie algebras \cite{MY} and invariant affine reflection algebras \cite{N1}, as a generalization of the last two stated classes, are examples of these generalizations.
Basic classical simple Lie superalgebras, orthosymplectic Lie superalgebras of arbitrary dimension as well as specific extensions of particular root graded Lie superalgebras satisfy certain properties which are in fact the super version of the axioms defining invariant affine reflection algebras. In the present work, we study the class of Lie superalgebras satisfying these certain properties; we introduce the notion of extended affine Lie superalgebras. Roughly speaking, an extended affine Lie superalgebra is a Lie superalgebra having a weight space decomposition with respect to a nontrivial abelian subalgebra of the even part and equipped with a nondegenerate invariant even supersymmetric bilinear form such that the weight vectors associated with so-called real roots are ad-nilpotent.
We prove that the even part of an extended affine Lie superalgebra is an invariant affine reflection algebra. We show that corresponding to each nonisotropic root $\a$ of an extended affine Lie superalgebra $\LL,$ there exists a triple of elements of $\LL$ generating a subsuperalgebra $\gg(\a)$ isomorphic to either $\frak{sl}_2$ or $\frak{osp}(1,2)$ depending on whether $\a$ is even or not. Considering $\LL$ as a $\gg(\a)$-module, we can derive some properties of the corresponding root system of $\LL$ which are in fact the features defining extended affine root supersystems \cite{you6}. As $\frak{osp}(1,2)$-modules are important in the theory of extended affine Lie superalgebras, we devote a section to study the module theory of $\frak{osp}(1,2).$ Although, it is an old well-known fact that finite dimensional $\frak{osp}(1,2n)$-modules are completely reducible, using the generic features of $\frak{osp}(1,2),$ we prove that finite dimensional $\frak{osp}(1,2)$-modules are completely reducible in a different approach from the one in the literature. We conclude the paper with some examples showing that starting form an extended affine Lie superalgebra, one can get a new one using an affinization process.
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\section{Finite Dimensional Modules of $\frak{osp}(1,2)$}
Throughout this work, $\bbbf$ is a field of characteristic zero. Unless otherwise mentioned, all vector spaces are considered over $\bbbf.$ We denote the dual space of a vector space $V$ by $V^*.$ If $V$ is a vector space graded by an abelian group, we denote the degree of a homogeneous element $x\in V$ by $|x|;$ we also make a convention that if $|x|$ is appeared in an expression, for an element $x$ of $V,$ by default, we assume that $x$ is homogeneous. If $X$ is a subset of a group $A,$ by $\la X\ra,$ we mean the subgroup of $A$ generated by $X.$ Also we denote the cardinal number of a set $S$ by $|S|;$ and for two symbols $i,j,$ by $\d_{i,j},$ we mean the Kronecker delta. For a map $f:A\longrightarrow B$ and $C\sub A,$ by $f\mid_{_C},$ we mean the restriction of $f$ to $C.$ Also we use $\uplus$ to indicate the disjoint union.
In the present paper, by a module of a Lie superalgebra $\fg,$ we mean a superspace $\v=\v_{\bar 0}\op \v_{\bar 1}$ and a bilinear map $\cdot: \fg\times \v\longrightarrow \v$ satisfying $\fg_{\bar i}\cdot \v_{\bar j}\sub \v_{\bar i+\bar j}$ for $i,j\in\{0,1\}$ and $[x,y]\cdot v=x\cdot (y\cdot v)-(-1)^{|x||y|}y\cdot (x\cdot v)$ for all $x,y\in\fg,$ $v\in\v.$ Also by a $\fg$-module homomorphism from a $\fg$-module $\v$ to a $\fg$-module $\w,$ we mean a linear homomorphism $ \phi$ of parity $\bar i$ ($i\in\{0,1\}$) with $\phi(x\cdot v)=(-1)^{|x||\phi|}x\cdot \phi(v)$ for $x\in \fg,v\in \v.$
Also by a {\it symmetric form} on an additive abelian group $A,$ we mean a map $\fm: A\times A\longrightarrow \bbbf$ satisfying
\begin{itemize}
\item $(a,b)=(b,a)$ for all $a,b\in A,$
\item $(a+b,c)=(a,c)+(b,c)$ and $(a,b+c)=(a,b)+(a,c)$ for all $a,b,c\in A.$
\end{itemize}
In this case, we set $A^0:=\{a\in A\mid(a,A)=\{0\}\}$ and call it the {\it radical} of the form $\fm.$ The form is called {\it nondegenerate} if $A^0=\{0\}.$
We note that if the form is nondegenerate, $A$ is torsion free and we can identify $A$ as a subset of $\bbbq\ot_\bbbz A.$ If $A$ is a vector space over $\bbbf,$ bilinear forms are used in the usual sense.
\
We recall that $\mathfrak{osp}(1,2)$ is a subsuperalgebra of $\frak{sl}(1,2)$ for which
{\small$$
F_+:=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
1 &0 & 0 \\
\end{array}
\right),
F_-:=\left(
\begin{array}{ccc}
0 & 0 &2\\
-2 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right), $$} of parity one, together with {\small $$H:=\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0& -2 & 0 \\
0& 0 &2 \\
\end{array}
\right),E_+=\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 2 & 0 \\
\end{array}
\right), E_-=\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0&-8 \\
0 & 0 & 0 \\
\end{array}
\right),$$}
of parity zero, form a basis. The triple $(F_+,F_-,H)$ is an $\mathfrak{osp}$-triple for $\mathfrak{osp}(1,2)$ in the following sense:
\begin{definition}{\rm
Suppose that $\fg=\fg_{\bar 0}\op\fg_{\bar 1}$ is a Lie superalgebra. We call a triple $(x,y,h)$ of nonzero elements of $\fg$ an {\it $\mathfrak{sl}_2$-super triple} for $\fg$ if \begin{itemize}
\item $\{x,y,h\}$ generates the Lie superalgebra $\fg,$
\item $x,y$ are homogenous of the same degree,
\item $[h,x]=2x,\; [h,y]=-2y,\; [x,y]=h.$
\end{itemize}}If $x,y\in \fg_{\bar 1},$ we refer to $(x,y,h)$ as an {\it $\frak{osp}$-triple} and note that if $x,y\in\fg_{\bar 0},$ $(x,y,h)$ is an $\mathfrak{sl}_2$-triple.
\end{definition}
\begin{lem}\label{iso} Suppose that $(x,y,h)$ is an $\mathfrak{osp}$-triple for a Lie superalgebra $\fg=\fg_{\bar 0}\op\fg_{\bar 1},$ then $(\frac{1}{4}[x,x],-\frac{1}{4}[y,y],\frac{1}{2}h)$ is an $\mathfrak{sl}_2$-triple for $\fg_{\bar 0}$ and $\fg\simeq\frak{osp}(1,2).$
\end{lem}
\pf We have
$$\begin{array}{l}\;[[x,x],[y,y]]=-8h,\;\;[h,[x,x]]=4[x,x],\;\;[h,[y,y]]=-4[y,y],\\\\
\;[[x,x],x]=0,\;\;[[y,y],y]=0.
\end{array}
$$
Therefore, we get that $(\frac{1}{4}[x,x],-\frac{1}{4}[y,y],\frac{1}{2}h)$ is an $\frak{sl}_2$-triple; in particular $[x,x]\neq0$ as well as $[y,y]\neq0$ and we have $\fg=\bbbf [x,x]\op\bbbf [y,y]\op\bbbf h\op\bbbf y\op\bbbf x.$ Now it follows that $\fg$ is isomorphic to $\mathfrak{osp}(1,2).$
\qed
\begin{lem}\label{gen}
Suppose that $(e,f,h)$ is an $\frak{osp}$-triple for a Lie superalgebra $\fg.$ Assume $(\v,\cdot)$ is a $\fg$-module with corresponding representation $\pi.$ If $\lam\in\bbbf\setminus\{-2\}$ and $u\in \v_{\bar i}$ $(i\in\{0,1\})$ are such that $h\cdot u=\lam u$ and $[e,e]\cdot u=0,$ then $\fg_{\bar 0}$-submodule of $\v$ generated by $f\cdot u$ equals to $$T:=\sum_{k\in \bbbz^{\geq0}}\bbbf f^{2k}\cdot (e\cdot u)+ \sum_{k\in \bbbz^{\geq0}}\bbbf f^{2k+1}\cdot(f\cdot(e\cdot u)-(\lam+2)u)$$ in which by the action of $f^k,$ we mean $\pi(f)^k$ for all $k\in\bbbz^{\geq 0}.$
\end{lem}
\pf
Since $h\cdot u=\lam u$ and $e\cdot (e\cdot u)=\frac{1}{2}[e,e]\cdot u=0,$ we have $h\cdot(e\cdot u)=(\lam+2)e\cdot u,$ $h\cdot(f\cdot(e\cdot u))=\lam f\cdot(e\cdot u)$ and that
\begin{eqnarray*}
e\cdot(f\cdot(e\cdot u))=-f\cdot(e\cdot(e\cdot u))+h\cdot (e\cdot u)=(\lam+2)e\cdot u.
\end{eqnarray*}
Now for $x\in\{f\cdot(e\cdot u)-(\lam+2)u,e\cdot u\},$ we have $e\cdot x=0$ and for $$\lam_x:=\left\{\begin{array}{ll}\lam& \hbox{if $x=f\cdot(e\cdot u)-(\lam+2)u$}\\
\lam+2& \hbox{if $x=e\cdot u,$}\end{array}\right.$$ we have
$$\begin{array}{l}\;h\cdot (f^{k}\cdot x)=(\lam_x-2k)f^{k}\cdot x,\\
\;f\cdot (f^{k}\cdot x)=f^{k+1}\cdot x,\\\\
\;e\cdot (f^{k}\cdot x)=
\left\{\begin{array}{ll}
-k f^{k-1} \cdot x& \hbox{$k$ is even}\\
(\lam_x-(k-1))f^{k-1} \cdot x& \hbox{$k$ is odd}
\end{array}\right.
\end{array}$$ for $k\in\bbbz^{\geq 0},$ where $f^{-1}\cdot x$ is defined to be zero. Now it follows that $T$ is invariant under the action of $[f,f],[e,e]$ and $h,$ i.e. it is a $\fg_{\bar 0}$-submodule of $\v.$ On the other hand, $$f\cdot u=\frac{1}{\lam+2}(f^2\cdot( e\cdot u)-f\cdot (f\cdot (e\cdot u-(\lam+2) u)))\in T.$$ Also if $S$ is a $\fg_{\bar 0}$-submodule of $\v$ containing $f\cdot u,$ then \begin{eqnarray*}
-2e\cdot u=-[h,e]\cdot u&=&-h\cdot (e\cdot u)+e\cdot (h\cdot u)\\
&=&-f\cdot (e\cdot (e\cdot u))-e\cdot (f\cdot (e\cdot u))+e\cdot (h\cdot u)\\
&=&-e\cdot (f\cdot (e\cdot u))+e\cdot (h\cdot u)\\
&=& e\cdot (e\cdot (f\cdot u))-e\cdot (h\cdot u)+e\cdot (h\cdot u)\\
&=& \frac{1}{2}[e,e]\cdot (f\cdot u)\in S
\end{eqnarray*}So $T\sub S.$ This completes the proof.
\qed
\begin{lem}\label{int eigen}
Suppose that $(\v,\cdot)$ is a finite dimensional module for a Lie superalgebra $\fg\simeq\frak{osp}(1,2)$ with corresponding representation $\pi.$ Take $(e,f,h)$ to be an $\mathfrak{osp}$-triple for $\fg.$ Then we have the following:
(i) $\pi(h)$ is a diagonalizable endomorphism of $\v$ with even integer eigenvalues each of which occurs with its opposite.
(ii) Suppose that $\v$ is irreducible and $\Lam$ is the set of eigenvalues of $\pi(h).$ Then the corresponding eigenspaces are one-dimensional and there is a nonnegative even integer $\lam$ with $\Lam=\{-\lam,-\lam+2,\ldots,\lam-2,\lam\}.$ Moreover, if $\lam\neq 0,$ $\v_{\bar 0}$ and $\v_{\bar 1}$ are irreducible $\fg_{\bar 0}$-submodules of $\v$ and there is $i\in\{0,1\}$ such that $\{-\frac{\lam}{2},-\frac{\lam}{2}+2,\ldots,\frac{\lam}{2}-2,\frac{\lam}{2}\}$ and $\{-\frac{\lam}{2}+1,-\frac{\lam}{2}+3,\ldots,\frac{\lam}{2}-3,\frac{\lam}{2}-1\}$ are the set of eigenvalues of $\frac{1}{2}\pi(h)|_{\v_{\bar i}}$ and $\frac{1}{2}\pi(h)|_{\v_{\bar i+\bar 1}}$ respectively.
\end{lem}
\pf
$(i)$ We know that $\fg_{\bar 0}\simeq \frak{sl}_2(\bbbf)$ and that $(\frac{1}{4}[e,e],-\frac{1}{4}[f,f],\frac{1}{2}h)$ is an $\mathfrak{sl}_2$-triple for $\fg_{\bar 0}.$ Considering $\v$ as a $\fg_{\bar 0}$-module and using the $\frak{sl}_2$-module theory \cite[\S III.8]{J}, we get $\pi(\frac{1}{2}h)$ acts diagonally on $\v$ with integer eigenvalues each of which occurs with its opposite. This completes the proof.
\begin{comment}
$\bar \v:=\bar \bbbf\ot_\bbbf \v$ is a $\bar \bbbf\ot_\bbbf \fg_{\bar 0}$-module in which by $\bar \bbbf,$ we mean the algebraic closure of $\bbbf.$ Since $\bar \bbbf\ot_\bbbf\fg_{\bar 0}\simeq \mathfrak{sl}_2(\bar \bbbf),$ we know from the $\frak{sl}_2$-module theory on algebraically closed fields that $1\ot\frac{1}{2} h$ acts diagonally on $\bar \v$ with integer eigenvalues, so $1\ot h$ acts diagonally on $\bar \v$ with even integer eigenvalues; in particular, the minimal polynomial of this action is of the form $p(t):=(t-\lam_1)\cdots(t-\lam_n)$ in which $\lam_1,\ldots,\lam_n$ are distinct even integers. Identifying $\v$ as a subset of $\bar \v,$ we can consider $\pi(h)$ as the restriction of the action of $1\ot h$ to $\v.$ So the minimal polynomial $\pi(h)$ is divided by $p(t)$ which in turn implies that $\pi(h)$ is diagonalizable with even integer eigenvalues. Let us return to the action of $\pi(\frac{1}{2}h)$ on $\v.$
\end{comment}
$(ii)$ Suppose that $\v$ is irreducible. Take $\lam$ to be the largest eigenvalue of $\pi(h)$ and fix a homogeneous eigenvector $v_0$ for this eigenvalue. Set $v_{-1}:=0$ and $v_i:=\pi(f)^i(v_0),$ for $i\in\bbbz^{\geq 0}.$
For $i\in\bbbz^{\geq 0},$ we have $$h\cdot v_i=(\lam-2i)v_i,$$ $$f\cdot v_i=v_{i+1}$$ and $$e\cdot v_i=\left\{\begin{array}{ll}
-iv_{i-1} & \hbox{$i$ is even}\\
(\lam-(i-1))v_{i-1}& \hbox{$i$ is odd}.
\end{array}\right.$$ This together with the fact that $\v$ is irreducible shows that $\v=\sum_{k\in\bbbz^{\geq0}}\bbbf f^k\cdot v_0$ and so each eigenspace is one-dimensional.
Since $\lam+2$ is not an eigenvalue, by part $(i),$ $-\lam-2$ is not an eigenvalue; in particular, $v_{\lam+1}=0.$ So $\v=\sum_{k=0}^{\lam}\bbbf f^k\cdot v_0.$
This completes the proof if $\lam=0.$ Now suppose $\lam\neq 0.$ For $i\in\{0,\ldots,\lam\},$ we have $v_i\in \v^{m_i}$ where $m_i:=\lam-2i$ and $$\v^{m_i}:=\{v\in \v\mid h\cdot v=m_iv\}=\{v\in \v\mid \frac{1}{2}h\cdot v=\frac{1}{2}m_iv\}.$$
Set $$U:=\hbox{span}_\bbbf\{v_{2i}\mid i\in\{0,\ldots, \frac{\lam}{2}\} \}\andd W:=\hbox{span}_\bbbf\{v_{2i+1}\mid i\in\{0,\ldots, \frac{\lam}{2}-1\} \}.$$
Both $U$ and $W$ are invariant under the actions of $\frac{1}{4}[e,e],-\frac{1}{4}[f,f],\frac{1}{2}h$ and so they are $\fg_{\bar 0}$-submodules of $\v.$ Since $\lam$ is the largest eigenvalue for $\pi(h)$ and $h\cdot( e\cdot v_0)=(\lam+2)v_0,$ we have $e\cdot v_0=0$ and so we get $0\neq \lam v_0=h\cdot v_0=e\cdot f\cdot v_0+f\cdot e\cdot v_0=e\cdot f\cdot v_0.$ This implies that $f\cdot v_0\neq 0.$ So $U$ and $W$ are nonzero $\fg_{\bar 0}$-submodules.
For $i\in\{0,\ldots,\lam\},$ take $U^{\frac{m_i}{2}}:=\v^{m_i}\cap U$ and $W^{\frac{m_i}{2}}:=\v^{m_i}\cap W.$ Then we have $$U=\sum_{i=0}^{\lam/2}\bbbf v_{2i}=U^{-\frac{\lam}{2}}\op U^{-\frac{\lam}{2}+2}\op\cdots\op U^{\frac{\lam}{2}-2}\op U^{\frac{\lam}{2}}$$ and $$W=\sum_{i=0}^{\lam/2-1}\bbbf v_{2i+1}=W^{-\frac{\lam}{2}+1}\op W^{-\frac{\lam}{2}+3}\op\cdots\op W^{\frac{\lam}{2}-3}\op W^{\frac{\lam}{2}-1}.$$ Using the standard $\mathfrak{sl}_2$-module theory, we get that $v_i\neq 0$ for $0\leq i\leq \lam$ and that both $U$ and $W$ are irreducible $\fg_{\bar 0}$-modules. If the homogeneous element $v_0$ is of degree $\bar i$ $(i\in \{0,1\}),$ we have $\v_{\bar i}=U$ and $\v_{\bar i+\bar 1}=W.$ This completes the proof.
\qed
\begin{cor}\label{cor1} Each (nonzero) finite dimensional irreducible $osp(1,2)$-module is of odd dimension. Moreover, suppose that $\lam$ is a nonnegative even integer and $\v$ is a superspace with a basis $\{v_i\mid 0\leq i\leq \lam\}$ of homogeneous elements such that $\{v_{2i}\mid 0\leq i\leq \frac{1}{2}\lam\}$ is a basis for either of $\v_{\bar 0}$ or $\v_{\bar 1}.$ Take $(e,f,h)$ to be an $\frak{osp}(1,2)$-triple for a superalgebra $\fg.$ Set $v_{\lam+1}=v_{-1}:=0$ and define $\cdot:\fg\times \v\longrightarrow \v$ by
$$\begin{array}{ll}
\begin{array}{l}
f\cdot v_i:=v_{i+1},\\
e\cdot v_i:=\left\{\begin{array}{ll}
-iv_{i-1} & \hbox{$i$ is even}\\
(\lam-(i-1))v_{i-1}& \hbox{$i$ is odd,}
\end{array}\right.\\
h\cdot v_i=(\lam-2i)v_i,
\end{array}& \begin{array}{l}
\;[f,f]\cdot v_i:=2f\cdot(f\cdot v_i),\\
\;[e,e]\cdot v_i:=2e\cdot(e\cdot v_i),\\
\end{array}
\end{array} $$
for $0\leq i\leq \lam.$ Then up to isomorphism, $(\v,\cdot)$ is the unique finite dimensional irreducible $\fg$-module of dimension $\lam+1.$
\end{cor}
\begin{lem}\label{com red}
Each finite dimensional $\frak{osp}(1,2)$-module is completely reducible.
\end{lem}
\pf Fix an $\frak{osp}$-triple $(e,f,h)$ for $\fg:=\frak{osp}(1,2)$ and consider the $\frak{sl}_2$-triple $(\frac{1}{4}[e,e],-\frac{1}{4}[f,f],\frac{1}{2}h)$ for $\fg_{\bar 0}$ as in
Lemma \ref{iso}. Suppose that $\v$ is a finite dimensional $\frak{osp}(1,2)$-module with corresponding representation $\pi$. We know from Lemma \ref{int eigen} that $\pi(h)$ is diagonalizable with even integer eigenvalues. We also know that $\v_{\bar 0}$ and $\v_{\bar 1}$ are both finite dimensional $\fg_{\bar 0}$-submodules of $\v.$ Suppose that $\v_{\bar 0}=\op_{j=1}^n W^j$ is a decomposition of $\v$ into irreducible $\fg_{\bar 0}$-modules. For each $1\leq j\leq n,$ take $w_j$ to be an eigenvector of the largest eigenvalue $\lam_j$ of $\pi(\frac{1}{2}h)$ on $W^j.$ We have \begin{equation}\label{final}
W^j=\op_{k=0}^{\lam_j} \bbbf [f,f]^k\cdot w_j=\op_{k=0}^{\lam_j} \bbbf f^{2k}\cdot w_j
\end{equation}
by the $\frak{sl}_2$-module theory. For $j\in\{1,\ldots,n\},$ set $$\displaystyle{ T^j:=\sum_{k=0}^{\infty}\bbbf f^{2k}\cdot (e\cdot w_j)}\andd S^j:=\displaystyle{\sum_{k=0}^{\infty}\bbbf f^{2k+1}\cdot(f\cdot (e\cdot w_j)-(2\lam_j+2)w_j)}.$$ We carry out the proof in the following steps:
\textbf{Step 1}. If $y$ is an eigenvector of $\pi(h)$ of eigenvalue $2\lam$
such that $e\cdot y=0,$ then $ f^{2\lam+1}\cdot y=0:$ As for $k\in\bbbz^{\geq 0},$ $h\cdot(f^k\cdot y)=(\lam-2k)f^{k}\cdot y$ and $\v$ is finite dimensional, there is $k\in\bbbz^{\geq0}$ such that $f^k\cdot y\neq0$ but $f^{k+1}\cdot y=0.$ Therefore, we have $$0=e\cdot (f^{k+1}\cdot y)=\left\{\begin{array}{ll}
-(k+1) f^k\cdot y& \hbox{$k$ is odd}\\
(2\lam-k)f^{k}\cdot y& \hbox{$k$ is even.}
\end{array}\right.$$ This implies that $k=2\lam$ and so we are done.
\textbf{Step 2}. For each $j\in\{1,\ldots,n\},$ $$\displaystyle{T^j=\sum_{k=0}^{\lam_j+1}\bbbf f^{2k}\cdot (e\cdot w_j)}\andd \displaystyle{S^j=\sum_{k=0}^{\lam_j-1}\bbbf f^{2k+1}\cdot(f\cdot (e\cdot w_j)-(2\lam_j+2)w_j):}
$$
Since $w_j$ is an eigenvector of $\pi(h)$ with eigenvalue $2\lam_j,$ we have $h\cdot(e\cdot w_j)=(2\lam_j+2)(e\cdot w_j)$ and $h\cdot(f\cdot(e\cdot w_j))=2\lam_j f\cdot(e\cdot w_j).$ Also since $[e,e]\cdot w_j=0,$ as before, we have $$e\cdot(e\cdot w_j)=0\andd e\cdot (f\cdot (e\cdot w_j)-(2\lam_j+2)w_j)=0 $$ and so we are done using Step 1.
\textbf{Step 3}. {\small $\displaystyle{\v_{\bar 0}=\sum_{j=1}^n\sum_{k=0}^{\lam_j}\bbbf f^{2k+1}\cdot (e\cdot w_j)+\sum_{j=1}^n\sum_{k=0}^{\lam_j}\bbbf f^{2k}\cdot(f\cdot (e\cdot w_j)-(2\lam_j+2)w_j)}:$} Take $X$ to be the right hand side of this equality.
For each $j\in\{1,\ldots,n\},$ $$w_j=\frac{1}{2\lam_j+2}(f\cdot (e\cdot w_j)-(f\cdot (e\cdot w_j)-(2\lam_j+2)w_j)).$$ Therefore $w_j\in \displaystyle{\sum_{k=0}^{\lam_j}\bbbf f^{2k+1}\cdot (e\cdot w_j)+\sum_{k=0}^{\lam_j}\bbbf f^{2k}\cdot(f\cdot (e\cdot w_j)-(2\lam_j+2)w_j)\sub X}$
for all $j\in\{1,\ldots,n\}.$ This completes the proof as $X\sub \v_{\bar 0}$ and $\{w_j\mid 1\leq j\leq n\}$ is a set of generators for the $\fg_{\bar 0}$-module $\v_{\bar 0}.$
\textbf{Step 4}.
For each $j\in\{1,\ldots,n\},$ set
$U^j$ to be the $\fg_{\bar 0}$-submodule of $\v$ generated by $f\cdot w_j,$ then $\v_{\bar 1}=P+ \sum_{j=1}^n U^j$ in which $P:=\{0\}$ if $\v_{\bar 1}$ has no one-dimensional irreducible $\fg_{\bar 0}$-submodule and otherwise, we take it to be the summation of all one-dimensional irreducible $\fg_{\bar 0}$-submodules of $\v_{\bar 1}:$
Take $U:=P+ \sum_{j=1}^n U^j.$ Since $\v_{\bar 1}$ is a completely reducible $\fg_{\bar 0}$-module, there is a $\fg_{\bar 0}$-submodule $K$ of $\v_{\bar 1}$ such that $\v_{\bar 1}=U\op K.$ If $K\neq \{0\},$ we pick an irreducible $\fg_{\bar 0}$-submodule $S$ of $K$ and suppose $u$ is an eigenvector for the largest eigenvalue $\lam$ of the action $\frac{1}{2}h$ on $S.$ Since $S\cap U\sub K\cap U=\{0\},$ $S$ is not one-dimensional, so $\lam$ is positive and $2f\cdot (f\cdot u)=[f,f]\cdot u\neq 0.$ But $f\cdot u\in \v_{\bar 0}=\sum_{j=1}^n\sum_{k=0}^{\lam_j}\bbbf f^{2k}\cdot w_j,$ so $f\cdot(f\cdot u)\in \sum_{j=1}^n\sum_{k=0}^{\lam_j}\bbbf f^{2k}\cdot f\cdot w_j\in U.$ This means that $0\neq f\cdot(f\cdot u)\in S\cap U\sub K\cap U=\{0\}, $ a contradiction. Therefore, $K=\{0\}$ and we are done.
\textbf{Step 5}. $\v$ is a summation of irreducible $\fg$-modules: Consider $U^j$ and $P$ as in Step 4.
Fix a basis $\{v_1,\ldots,v_m\}$ of $P$ if $P$ is not zero and set $x_i:=f\cdot(e\cdot v_i)-2v_i$ for $i\in\{1,\ldots,m\}.$ Then for $i\in\{1,\ldots,m\},$ $\bbbf x_i$ is a trivial one-dimensional $\fg$-submodule of $\v$ and as $e\cdot v_i\in \v_{\bar0},$ (\ref{final}) implies that $v_i=\frac{1}{2}(f\cdot(e\cdot v_i)-x_i)\in \sum_{j=1}^n U^j+\bbbf x_i.$ Moreover, by Lemma \ref{gen}, for $j\in\{1,\ldots,n\},$ $U^j=T^j+S^j $ and by Steps 2,3,4, we have $$\v=\sum_{j=1}^n\sum_{k=0}^{2\lam_j+2}\bbbf f^{k}\cdot (e\cdot w_j)+\sum_{j=1}^n\sum_{k=0}^{2\lam_j}\bbbf f^{k}\cdot(f\cdot (e\cdot w_j)-(\lam_j+2)w_j)+\sum_{i=1}^m\bbbf x_i$$ in which the last part is disappeared if $P=\{0\}.$ This together with Step 1 and Corollary \ref{cor1} completes the proof.
\qed
\section{Extended Affine Lie Superalgebras}
We call a triple $(\LL,\hh,\fm),$ consisting of a nonzero Lie superalgebra $\LL=\LL_{\bar 0}\op\LL_{\bar1},$ a nontrivial subalgebra $\hh$ of $\LL_{\bar 0}$ and a nondegenerate invariant even supersymmetric bilinear form $\fm$ on $\LL,$ a {\it super-toral triple} if
\begin{itemize}
\item
{\rm $\LL$ has a weight space decomposition $\LL=\op_{\a\in\hh^*}\LL^\a$ with respect to $\hh$ via the adjoint representation. We note that in this case $\hh$ is abelian; also as $\LL_{\bar 0}$ as well as $\LL_{\bar 1}$ are $\hh$-submodules of $\LL,$ we have $\LL_{\bar 0}=\op_{\a\in \hh^*}\LL_{\bar 0}^\a$ and $\LL_{\bar 1}=\op_{\a\in\hh^*}\LL_{\bar 1}^\a$ with $\LL_{\bar i}^\a:=\LL_{\bar i}\cap \LL^\a,$ $ i=0,1$ \cite[Pro. 2.1.1]{MP}},
\item {\rm the restriction of the form $\fm$ to $\hh$ is nondegenerate.}
\end{itemize}
We call $R:=\{\a\in \hh^*\mid \LL^\a\neq\{0\}\},$ the {\it root system} of $\LL$ (with respect to $\hh$). Each element of $R$ is called a {\it root.} We refer to elements of $R_0:=\{\a\in \hh^*\mid \LL_{\bar 0}^\a\neq\{0\}\}$ (resp. $R_1:=\{\a\in \hh^*\mid \LL_{\bar1}^\a\neq\{0\}\}$) as {\it even roots} (resp. {\it odd roots}). We note that $R=R_0\cup R_1.$
Suppose that $(\LL,\hh,\fm)$ is a super-toral triple and $\mathfrak{p}:\hh\longrightarrow \hh^*$ is the function mapping $h\in\hh$ to $(h,\cdot).$ Since the form is nondegenerate on $\hh,$ this map is one to one. So for each element $\a$ of the image $\hh^\mathfrak{p}$ of $\hh$ under $\mathfrak{p},$ there is a unique $t_\a\in\hh$ representing $\a$ through the form $\fm.$ Now we can transfer the form on $\hh$ to a form on $\hh^\mathfrak{p},$ denoted again by $\fm,$ and defined by $$(\a,\b):=(t_\a,t_\b)\;\;\;(\a,\b\in \hh^\fp).$$
\begin{lem}
\label{symm} Suppose that $(\LL,\hh,\fm)$ is a super-toral triple with corresponding root system $R=R_0\cup R_1.$ Then we have the following:
(i) For $\a,\b\in \hh^*,$ $[\LL^\a,\LL^\b]\sub\LL^{\a+\b}.$ Also for $i=0,1$ and $\a,\b\in R_i,$ we have $(\LL_{\bar i}^\a,\LL_{\bar i}^\b)=\{0\}$ unless $\a+\b=0;$ in particular, $R_0=-R_0$ and $R_1=-R_1.$
(ii) Suppose that $\a\in \hh^\fp$ and $x_{\pm\a}\in\LL^{\pm\a}$ with $[x_\a,x_{-\a}]\in\hh,$ then we have $[x_\a,x_{-\a}]=(x_\a,x_{-\a})t_\a.$
(iii) Suppose that $\a\in R_i\setminus\{0\}$ $(i\in\{0,1\}),$ $x_\a\in\LL_{\bar i}^\a$ and $x_{-\a}\in\LL_{\bar i}^{-\a}$ with $[x_\a,x_{-\a}]\in\hh\setminus\{0\},$ then we have $(x_\a,x_{-\a})\neq0$ and that $\a\in\hh^\fp.$
\end{lem}
\pf $(i)$ It is easy to see.
\begin{comment}
\pf Suppose that $\a,\b\in\hh^*.$ For $h\in\hh,$ $x\in\LL_\a$ and $y\in\LL_\b,$ then we have
\begin{eqnarray*}
[h,[x,y]]=-[[x,y],h]&=&-([x,[y,h]]-(-1)^{|x||y|}[y,[x,h]])\\&=&-([x,-[h,y]]-(-1)^{|x||y|}[y,-[h,x]])\\
&=&[x,[h,y]]-(-1)^{|x||y|}[y,[h,x]]\\
&=&[x,\b(h)y]-(-1)^{|x||y|}[y,\a(h)x]\\
&=&\b(h)[x,y]-\a(h)(-1)^{|x||y|}[y,x]\\
&=&\b(h)[x,y]+\a(h)(-1)^{|x||y|}(-1)^{|x||y|}[x,y]\\
&=&(\a+\b)(h)[x,y].
\end{eqnarray*}
Next suppose $\a+\b\neq0.$ Fix $h\in\hh$ with $(\a+\b)(h)\neq 0,$ then for $x\in\LL_{\bar i}^\a$ and $y\in \LL_{\bar i}^\b,$ we have \begin{eqnarray*}(\a+\b)(h)(x,y)=\a(h)(x,y)+\b(h)(x,y)&=&([h,x],y)+(x,[h,y])\\&=&-([x,h],y)+(x,[h,y])=0.\end{eqnarray*} So we have $(x,y)=0.$ Now as the form is positive and nondegenerate, we get that the form restriction to both $\LL_{\bar 0}$ and $\LL_{\bar1}$ are nondegenerate and so $$(\LL_{\bar i}^\a,\LL_{\bar i}^\b)=\{0\} \hbox{ unless } \a+\b=0;\;\; (i\in\{1,2\},\;\a,\b\in R_i).$$ This in particular implies that $$R_0=-R_0\andd R_1=-R_1.$$
This completes the proof.\qed
\end{comment}
$(ii)$ For $h\in\hh,$ we have \begin{equation}\label{p}(h,[x_\a,x_{-\a}])=([h,x_\a],x_{-\a})=\a(h)(x_\a,x_{-\a}).\end{equation} Therefore we have $$(h,[x_\a,x_{-\a}])=\a(h)(x_\a,x_{-\a})=(t_\a(x_\a,x_{-\a}),h).$$ This together with the fact that the form on $\hh$ is symmetric and nondegenerate completes the proof.
$(iii)$ Suppose to the contrary that $(x_\a,x_{-\a})=0,$ then (\ref{p}) implies that for all $h\in\hh,$ $(h,[x_\a,x_{-\a}])=0$ but the form on $\hh$ is nondegenerate, so $[x_\a,x_{-\a}]=0,$ a contradiction. Again using (\ref{p}), we get that $(h,\frac{1}{(x_\a,x_{-\a})}[x_\a,x_{-\a}])=\a(h)$ for all $h\in\hh$ and so $\a=\mathfrak{p}(\frac{1}{(x_\a,x_{-\a})}[x_\a,x_{-\a}])\in\hh^\fp.$
\qed
\begin{definition}
{\rm A super-toral triple $(\LL=\LL_{\bar 0}\op\LL_{\bar1},\hh,\fm)$ (or $\LL$ if there is no confusion), with root system $R=R_0\cup R_1,$ is called an {\it extended affine Lie superalgebra} if}
\begin{itemize}{\rm
\item{(1)} for $\a\in R_i\setminus\{0\}$ ($i\in\{0,1\}$), there are $x_\a\in\LL_{\bar i}^\a$ and $x_{-\a}\in\LL_{\bar i}^{-\a}$ such that $0\neq[x_\a,x_{-\a}]\in\hh,$
\item{(2)} for $\a\in R$ with $(\a,\a)\neq 0$ and $x\in \LL^\a,$ $ad_x:\LL\longrightarrow\LL,$ mapping $y\in\LL$ to $[x,y],$ is a locally nilpotent linear transformation.}
\end{itemize}
{\rm The extended affine Lie superalgebra $(\LL,\hh,\fm)$ is called an {\it invariant affine reflection algebra} \cite{N1} if $\LL_{\bar1}=\{0\}$ and it is called a {\it locally extended affine Lie algebra} \cite{MY} if $\LL_{\bar1}=\{0\}$ and
$\LL^0=\hh.$ Finally a locally extended affine Lie algebra $(\LL,\hh,\fm)$ is called an {\it extended
affine Lie algebra} \cite{AABGP} if $\LL^0=\hh$ is a finite dimension subalgebra of $\LL.$}
\end{definition} We immediately have the following lemma:
\begin{prop}\label{aff-ref}
If $(\LL,\hh,\fm)$ is an extended affine Lie superalgebra, then the triple $(\LL_{\bar 0},\hh,\fm|_{\LL_{\bar 0}\times\LL_{\bar 0}})$ is an invariant affine reflection algebra.
\end{prop}
\begin{Example}
{\rm Finite dimensional basic classical simple Lie superalgebras \cite{K1} and affine Lie superalgebras \cite{van-de} are examples of extended affine Lie superalgebras.}\hfill $\diamondsuit$
\end{Example}
In the sequel, we shall prove that the root system of an extended affine Lie superalgebra $(\LL,\hh,\fm)$ is an extended affine root supersystem in the following sense:
\begin{definition}\label{iarr}
{\rm Suppose that $A$ is a nontrivial additive abelian group, $\fm:A\times A\longrightarrow \bbbf$ is a symmetric form and $R$ is a subset of $A.$ Set
$$\begin{array}{l}
R^0:=R\cap A^0,\;\;\;\;\;
\rcross:=R\setminus R^0,\\\\
\rcross_{re}:=\{\a\in R\mid (\a,\a)\neq0\},\;\;\;\rre:=\rcross_{re}\cup\{0\},\\\\
\rcross_{ns}:=\{\a\in R\setminus R^0\mid (\a,\a)=0 \},\;\;\; \rim:=\rcross_{ns}\cup\{0\}.
\end{array}$$
We say $(A,\fm,R)$ is an {\it extended affine root supersystem} if the following hold:
$$\begin{array}{ll}
(S1)& \hbox{$0\in R,$ and $\la R\ra= A,$}\\\\
(S2)& \hbox{$R=-R,$}\\\\
(S3)&\hbox{for $\a\in \rre^\times$ and $\b\in R,$ $2(\a,\b)/(\a,\a)\in\bbbz,$}\\\\
(S4)&\parbox{4.5in}{ {\it root string property} holds in $R$ in the sense that for $\a\in \rre^\times$ and $\b\in R,$ there are nonnegative integers $p,q$ with $2(\b,\a)/(\a,\a)=p-q$ such that \begin{center}$\{\b+k\a\mid k\in\bbbz\}\cap R=\{\b-p\a,\ldots,\b+q\a\},$\end{center}
} \\\\
(S5)&\parbox{4.5in}{for $\a\in \rim$ and $\b\in R$ with $(\a,\b)\neq 0,$
$\{\b-\a,\b+\a\}\cap R\neq \emptyset.$ }
\end{array}$$
If there is no confusion, for the sake of simplicity, we say {\it $R$ is an extended affine root supersystem in $A.$}}
\end{definition}
\begin{lem}\label{sl2}
Suppose that $(\LL,\hh,\fm)$ is an extended affine Lie superalgebra with root system $R=R_0\cup R_1.$ Suppose that $\a\in R_i$ $(i\in\{0,1\})$ with $(\a,\a)\neq 0.$ Recall that $t_\a$ is the unique element of $\hh$ representing $\a$ through the form $\fm$ restricted to $\hh$ and set $h_\a:=2t_\a/(\a,\a).$ Then there are $y_{\pm\a}\in\LL_{\bar i}^{\pm \a}$ such that $(y_\a,y_{-\a},h_\a)$ is an $\mathfrak{sl}_2$-super triple for the subsuperalgebra $\gg(\a)$ generated by $\{y_\a,y_{-\a},h_\a\};$ in particular, if $\a\in R_1\cap \rre^\times,$ then $2\a\in R_0.$
\end{lem}
\pf Suppose that $i\in\{0,1\}$ and $\a\in R_i$ with $(\a,\a)\neq0.$ Fix $x_{\pm\a}\in\LL_{\bar i}^{\pm\a}$ with $0\neq[x_\a,x_{-\a}]\in\hh.$
Considering Lemma \ref{symm} and setting $e_\a:=x_\a$ and $e_{-\a}:=x_{-\a}/(x_\a,x_{-\a}),$ we have $[e_\a,e_{-\a}]=t_\a.$ Now we get that $(y_\a:=2e_\a/(t_\a,t_\a),y_{-\a}:=e_{-\a},h_\a)$ is an $\mathfrak{sl}_2$-super triple for the subsuperalgebra $\gg(\a).$ Next suppose $\a\in R_1\cap\rre^\times.$ Using Lemma \ref{iso}, we get that $0\neq[y_\a,y_\a]\sub\LL^{2\a}_{\bar 0}$ and so $2\a\in R_0.$
\qed
\begin{lem}\label{final3}
If $i,j\in\{0,1\},$ $\a\in R_i$ and $\b\in R_j$ with $(\a,\b)\neq0,$ then either $\b-\a\in R$ or $\b+\a\in R.$
\end{lem}
\pf Fix $0\neq z\in\LL_{\bar j}^\b$ and $x\in\LL_{\bar i}^\a,y\in\LL_{\bar i}^{-\a}$ with $[x,y]\in \hh\setminus\{0\}.$ Using Lemma \ref{symm}, we have $[x,y]=(x,y)t_\a.$
Therefore we have
\begin{eqnarray*}
0\neq(x,y)(\a,\b)z=(x,y)[t_\a,z]=[[x,y],z]=[x,[y,z]]-(-1)^{|x||y|}[y,[x,z]].
\end{eqnarray*}
This in turn implies that either $[y,z]\neq 0$
or $[x,z]\neq 0.$ Therefore either $\b-\a\in R$ or $\b+\a\in R.$
\qed
\begin{prop}\label{no hole}Suppose that $(\LL,\hh,\fm)$ is an extended affine Lie superalgebra with root system $R=R_0\cup R_1.$
For $\a,\b\in R$ with $(\a,\a)\neq0,$ we have the following:
(i) $\frac{2(\b,\a)}{(\a,\a)}\in\bbbz,$ in particular if $k\in\bbbf$ and $k\a\in R,$ then $k\in\{0,\pm1,\pm2,\pm1/2\}.$
(ii) $r_\a(\b):=\b-\frac{2(\a,\b)}{(\a,\a)}\a\in R.$
(iii) There are nonnegative integers $p,q$ such that $p-q=2(\b,\a)/(\a,\a)$ and $\{k\in\bbbz\mid \b+k\a\in R\}=\{-p,\ldots,q\}.$
\end{prop}
\pf
Suppose that $\a,\b\in R$ with $(\a,\a)\neq0.$ Assume $\a\in R_i$ for some $i\in\{0,1\}.$ Using Lemma \ref{sl2}, there are $y_{\a}\in\LL_{\bar i}^{\a}, y_{-\a}\in\LL_{\bar i}^{- \a}$ such that $[y_\a,y_{-\a}]=h_\a=\frac{2t_\a}{(\a,\a)}$ and $(y_\a,y_{-\a},h_\a)$ is an $\mathfrak{sl}_2$-super triple for the subsuperalgebra $\gg(\a)$ of $\LL$ generated by $\{y_\a,y_{-\a},h_\a\}.$ Consider $\LL$ as a $\gg(\a)$-module via the adjoint representation, then $\m:=\sum_{k\in\bbbz}\LL^{\b+k\a}$ is a $\gg(\a)$-submodule of $\LL.$ For $k\in\bbbz$ and $x\in\LL^{\b+k\a},$ set $$\m(x):=\hbox{span}_\bbbf\{\hbox{ad}_{y_{-\a}}^n\hbox{ad}_{y_\a}^mx\mid m,n\in\bbbz^{\geq 0}\}.$$ We claim that $\m(x)$ is the $\fg(\a)$-submodule of $\m$ generated by $x.$ Indeed, as $\gg(\a)$ is generated by $\{y_\a,y_{-\a},h_\a\},$ it is enough to show that $\m(x)$ is invariant under $\hbox{ad}_{h_\a},\hbox{ad}_{y_\a},\hbox{ad}_{y_{-\a}}.$ By Lemma \ref{symm}($i$), for $m,n\in\bbbz^{\geq0},$ we have $\hbox{ad}_{y_{-\a}}^n\hbox{ad}_{y_\a}^mx\in \LL^{\b+k\a+m\a-n\a},$ so $\m(x)$ is invariant under the action of $h_\a.$ Also it is trivial that $\m(x)$ is invariant under the action of $y_{-\a}.$ We finally show that it is invariant under $\hbox{ad}_{y_\a}.$ We use induction on $n$ to prove that $[y_\a,\hbox{ad}_{y_{-\a}}^n\hbox{ad}_{y_\a}^mx]\in\m(x)$ for $n,m\in\bbbz^{\geq0}.$
If $n=0,$ there is nothing to prove, so we assume $n\in\bbbz^{\geq1}$ and that $[y_\a,\hbox{ad}_{y_{-\a}}^{n-1}\hbox{ad}_{y_\a}^mx]\in\m(x)$ for all $m\in\bbbz^{\geq0}.$ Now for $m\in\bbbz^{\geq0},$ we have
\begin{eqnarray*}
[y_\a,\hbox{ad}_{y_{-\a}}^n\hbox{ad}_{y_\a}^mx]&=&[y_\a,[y_{-\a},\hbox{ad}_{y_{-\a}}^{n-1}\hbox{ad}_{y_\a}^mx]]\\
&=&(-1)^{|y_\a|}[y_{-\a},[y_{\a},\hbox{ad}_{y_{-\a}}^{n-1}\hbox{ad}_{y_\a}^mx]]+[h_\a,\hbox{ad}_{y_{-\a}}^{n-1}\hbox{ad}_{y_\a}^mx].
\end{eqnarray*}
This together with the induction hypothesis and the fact that $\m(x)$ is invariant under $\hbox{ad}_{h_\a}$ and $\hbox{ad}_{y_{-\a}},$ completes the induction process. Now we are ready to prove the proposition. Keep the notations as above.
$(i)$ Since $\hbox{ad}_{y_{-\a}}$ and $\hbox{ad}_{y_\a}$ are locally nilpotent linear transformations, for $x\in\LL^{\b+k\a}$ $(k\in\bbbz),$ $\m(x)$ is finite dimensional, so $\m$ is a summation of the finite dimensional $\gg(\a)$-submodules $\m(x)$ $(x\in\LL^{\b+k\a};\; k\in\bbbz).$ We know that
\begin{equation}\label{eigen}
\parbox{3.8in}{
$h_\a$ acts diagonally on $\m$ with the set of eigenvalues $\{\b(h_\a)+2k\mid k\in\bbbz,\;\b+k\a\in R\}.$ Moreover, this set of eigenvalues is the union of the set of eigenvalues of the action of ${h_\a}$ on the finite dimensional $\gg(\a)$-submodules $\m(x)$ ($x\in\LL^{\b+k\a};\;k\in\bbbz$) of $\m.$}
\end{equation}
Since $\LL^\b\neq \{0\},$ each nonzero element of $\LL^\b$ is an eigenvector of $\hbox{ad}_{h_\a}$ restricted to $\m$ corresponding to the eigenvalue $\b(h_\a).$ Therefore $\b(h_\a)$ is an eigenvalue of $\hbox{ad}_{h_\a}$ restricted to a finite dimensional $\gg(\a)$-submodule $\m(x)$ for some $x\in\LL^{\b+k\a}$ $(k\in\bbbz)$ and so using $\frak{sl}_2$-module theory together with Lemma \ref{int eigen}, we get that $2(\b,\a)/(\a,\a)=\b(h_\a)\in\bbbz.$ This completes the proof.
$(ii)$ As in the previous case, $\b(h_\a)$ is an eigenvalue of $\hbox{ad}_{h_\a}$ restricted to a finite dimensional $\gg(\a)$-submodule $\m(x)$ of $\m,$ for some $x\in\LL^{\b+k\a}$ ($k\in\bbbz$). From Lemma \ref{int eigen} and $\frak{sl}_2$-module theory, we know that $-\b(h_\a)$ is also an eigenvalue for $\hbox{ad}_{h_\a}$ on $\m(x).$ So there is an integer $k$ such that $\b+k\a\in R$ and $-\b(h_\a)=\b(h_\a)+2k.$ This implies that $k=-\b(h_\a).$ In particular, $\b-2\frac{(\b,\a)}{(\a,\a)}\a=\b-\b(h_\a)\a=\b+k\a\in R.$
$(iii)$
We first prove that $\{k\in\bbbz\mid \b+k\a\in R\}$ is an interval.
To this end, we take $r,s\in\bbbz$ with $\b+r\a,\b+s\a\in R$ and show that $\b+t\a\in R$ for all $t$ between $r,s.$ Without loss of generality, we may assume $|\b(h_\a)+2r|\geq|\b(h_\a)+2s|.$ Suppose that $t$ is an integer between $r,s.$
Since $\b+t\a\in R$ if and only if $-\b-t\a\in R,$ we simultaneously replace $\b$ with $-\b$ and $(r,s)$ with $(-r,-s)$ if it is necessary and assume $\b(h_\a)+2r\geq0.$ So $-\b(h_\a)-2r\leq \b(h_\a)+2s\leq \b(h_\a)+2r.$ But using (\ref{eigen}), we get that $\b(h_\a)+2r$ is an eigenvalue of the action of $h_\a$ on $\m(x)$ for some $x\in\LL^{\b+k\a}$ $(k\in\bbbz).$ So by Lemmas \ref{int eigen} and \ref{com red}, $\b(h_\a)+2t$ is an eigenvalue for the action of $h_\a$ on $\m(x).$ Therefore, we have $\b+t\a\in R$ by (\ref{eigen}).
We next show that this interval is bounded.
For $k\in\bbbz,$ we have $(\b+k\a,\b+k\a)=(\b,\b)+2k(\b,\a)+k^2(\a,\a).$ So there are at most two integer numbers such that $\b+k\a\not\in \rre^\times.$ Now to the contrary assume that the mentioned interval is not bounded. Without loss of generality, we may assume there is a positive integer $k_0$ such that for $k\in\bbbz^{\geq k_0},$ $\b+k\a\in \rre^\times$ and $(\a,\b+k\a)\neq 0.$ For $k\in\bbbz^{\geq k_0},$ we have
\begin{eqnarray*}
\frac{\frac{2(\a,\b)}{(\a,\a)}+2k}{\frac{(\b,\b)}{(\a,\a)}+\frac{2k(\a,\b)}{(\a,\a)}+k^2}&=&\frac{2(\a,\b)+2k(\a,\a)}{(\b,\b)+2k(\a,\b)+k^2(\a,\a)}\\
&=&\frac{2(\a,\b+k\a)}{(\b+k\a,\b+k\a)}\in\bbbz.\end{eqnarray*}
Now as $k,\frac{2(\a,\b)}{(\a,\a)}\in\bbbz,$ we get that $\frac{(\b,\b)}{(\a,\a)}\in\bbbq,$ so $\lim_{k\rightarrow \infty}\frac{\frac{2(\a,\b)}{(\a,\a)}+2k}{\frac{(\b,\b)}{(\a,\a)}+\frac{2k(\a,\b)}{(\a,\a)}+k^2}=0.$ This is a contradiction as it is a sequence of nonzero integer numbers. Therefore we have a bounded interval. Suppose $p,q$ are the largest nonnegative integers with $\b-p\a,\b+q\a\in R.$ Since $r_\a(\b-p\a)=\b+q\a,$ we get that $p-q=\frac{2(\b,\a)}{(\a,\a)}.$ This completes the proof.
\begin{comment}
Take $\gamma:=\b-p\a,\eta:=\b+q\a.$ Since $\gamma-\a\not\in R$ and $r_\a(\gamma-\frac{2(\gamma,\a)}{(\a,\a)}\a+\a)=\gamma-\a,$ we get that $q=-\frac{2(\gamma,\a)}{(\a,\a)}=-\frac{2(\b,\a)}{(\a,\a)}+p.$
\end{comment}
\qed
\begin{cor}
\label{cor2}
Suppose that $(\LL,\fm,\hh)$ is an extended affine Lie superalgebra with root system $R,$ then $R$ is an extended affine root supersystem in its $\bbbz$-span.
\end{cor}
\begin{proof}
It is immediate using Lemma \ref{final3} together with Proposition \ref{no hole}.
\end{proof}
\begin{prop}
\label{symm2}
Suppose that $(\LL,\fm,\hh)$ is an extended affine Lie superalgebra with root system $R.$
(i) For $\a\in \rre^\times,$ we have $2\a\not \in R_1;$ also if $\a\in \rre$ and $2\a\not\in R,$ we have $\a\in R_0.$
(ii) If $\a\in R_0$ with $(\a,\a)=0,$ then $(\a, R_0)=\{0\};$ moreover, $R_0\cap \rim=\{0\}.$
(iii) If $\LL^0\sub \LL_{\bar 0},$ then we have $ R^\times\cap R_0\cap R_1=\emptyset.$
\end{prop}
\pf
\begin{comment}
Let $\a\in R$ be such that $2\a\not\in R.$ Fix $e_\a\in \LL^\a$ and $f_\a\in\LL^{-\a}$ with $[e_\a,f_\a]=t_\a,$ Now the map $ad_{f_\a}$ restricted to $\LL_\a$ is an injection with image $\bbbf t_\a.$ This implies that $\dim(\LL_\a)=1.$
\end{comment}
$(i)$ We know from Proposition \ref{no hole}($i$) that $4\a\not\in R.$ Now the result is immediate using Lemma \ref{sl2}($i$).
$(ii)$ Although using a modified argument as in \cite[Pro. 3.4]{MY} and \cite[Pro. I.2.1]{AABGP}, one can get the first assertion, for the convenience of readers, we give its proof. To the contrary, suppose $\a,\b\in R_0$ with $(\a,\a)= 0$ and $(\a,\b)\neq 0.$
If there are infinitely many consecutive integers $n$ with $\b+n\a\in R_0\cap \rre^\times,$ for such integer numbers $n,$ we have $\frac{2(\a,\b+n\a)}{(\b+n\a,\b+n\a)}=\frac{2(\a,\b)}{(\b,\b)+2n(\a,\b)}\in\bbbz.$ Therefore, we get $(\b,\b)\neq 0$ and $\frac{\frac{2(\a,\b)}{(\b,\b)}}{1+2n\frac{(\a,\b)}{(\b,\b)}}\in\bbbz$ which is absurd as it is a sequens of nonzero integer numbers converging to $0.$ Therefore, there is $p\in \bbbz$ with $\gamma:=\b+p\a\in R_0$ and $\gamma-\a\not \in R_0.$
Fix $0\neq x\in\LL_{\bar 0}^\gamma$ and $y_{\pm\a}\in \LL_{\bar 0}^{\pm\a}$ with $[y_{-\a},y_\a]=t_\a.$ Setting $x_0:=x, x_n:=(ad_{y_\a})^nx,$ for $n\in\bbbz^{\geq 1},$ we have $ad_{y_{-\a}}(x_n)=n(\a,\gamma)x_{n-1}.$ So $x_n\neq0$ for all $n\in\bbbz^{\geq 0}.$ This means that for consecutive integer numbers $k_n:=p+n$ $(n\in\bbbz^{\geq0}),$ $\b+k_n\a\in R_0,$ a contradiction.
For the last assertion, suppose $0\neq \a\in R_0\cap \rim.$ Since $\a\not\in R^0,$ there is $\b\in R$ with $(\a,\b)\neq 0.$ If $\b\in \rim,$ there is $r\in\{\pm 1\}$ with $\a+r\b\in \rre;$ see Lemma \ref{final3}. Since $(\a+r\b,\a)\neq0,$ we always may assume there is a real root $\gamma$ with $(\a,\gamma)\neq0.$ But by the first assertion, $\gamma\in R_1\cap \rre^\times$ and so $2\gamma\in R_0$ by Lemma \ref{sl2}. This implies that $(\a,2\gamma)=0$ which is a contradiction. This completes the proof.
$(iii)$ To the contrary, suppose that $\a\in R^\times\cap R_0\cap R_1.$ By part ($ii$), we get that $\a\in\rre^\times.$ Consider Lemma \ref{symm} and fix $x_\a\in\LL_{\bar 0}^\a,y_\a\in \LL_{\bar 0}^{-\a}$ such that $[x_\a,y_\a]=t_\a;$ also fix $e_\a\in\LL_{\bar1}^\a,$ $f_\a\in\LL_{\bar1}^{-\a}$ with $[e_\a,f_\a]=t_\a.$ We have $[y_\a,e_\a]\in\LL_{\bar1}^0=\{0\}$ and by part ($i$), $[x_\a,e_\a]\in\LL_{\bar1}^{2\a}=\{0\}$, so we get
\begin{eqnarray*}
0\neq (\a,\a)e_\a=[t_\a,e_\a]=[[x_\a,y_{\a}],e_\a]=[x_\a,[y_\a,e_\a]]-[y_\a,[x_\a,e_\a]]=0
\end{eqnarray*}
which is a contradiction.
\qed
\begin{Example}{\rm
Suppose that $\Lam$ is a torsion free additive abelian group and $\gg$ is a locally finite basic classical simple Lie superalgebra i.e., a direct union of finite dimensional basic classical simple Lie superalgebras (see \cite{you7} for details) with a Cartan subalgebra $\hh$ and a fixed nondegenerate invariant even bilinear form $f\fm.$ One knows $(\gg,\hh,f\fm)$ is an extended affine Lie superalgebra with $\gg^0=\hh.$
Suppose that $\theta:\Lam\times\Lam\longrightarrow \bbbf\setminus\{0\}$ is a commutative 2-cocycle, that is, $\theta$ satisfies the following properties:
$$\theta(\zeta,\xi)=\theta(\xi,\zeta)\andd \theta(\zeta,\xi)\theta(\zeta+\xi,\eta)=\theta(\xi,\eta)\theta(\zeta,\xi+\eta)$$
for all $\zeta,\xi,\eta\in \Lam$.
Suppose that $\theta(0,0)=1$ and note that this in turn implies that $\theta(0,\lam)=1$ for all $\lam\in\Lam.$ Consider the $\bbbf$-vector space $\aa:=\sum_{\lam\in\Lam}\bbbf t^\lam$ with a basis $\{t^\lam\mid \lam\in\Lam\}.$ Now $\aa$ together with the product defined by $$t^\zeta\cdot t^\xi:=\theta(\zeta,\xi)t^{\zeta+\xi}\;\;\;(\xi,\zeta\in\Lam)$$ is a $\Lam$-graded unital commutative associative algebra with $\aa^\lam:=\bbbf t^\lam$ (for $\lam\in\Lam$). We refer to $\aa$ as the {\it commutative associative torus} corresponding to $(\Lam,\theta).$
Set $$\hat \gg:=\gg\ot\aa$$ and define $$|x\ot a|:=|x|;\;\; x\in\gg,a\in\aa.$$ Then $\hat\gg$ together with
$$[x\ot a,y\ot b]_{_{\hat \gg}}:=[x,y]\ot ab$$ for $x,y\in\gg$ and $a,b\in\aa,$ is a Lie superalgebra.
\begin{comment}
\begin{eqnarray*}
[x\ot a,y\ot b]&=&(-1)^{|a||y|}[x,y]\ot ab\\
&=&-(-1)^{|a||y|}(-1)^{|x||y|}(-1)^{|a||b|}[y,x]\ot ba\\
&=&-(-1)^{|a||y|}(-1)^{|x||y|}(-1)^{|a||b|}(-1)^{|x||b|}[y\ot b,x\ot a]\\
&=&-(-1)^{|x\ot a||y\ot b|}[y\ot b,x\ot a].
\end{eqnarray*}
\begin{eqnarray*}
[[x\ot a,y\ot b],z\ot c]&=&[(-1)^{|a||y|}[x,y]\ot ab,z\ot c]\\
&=&(-1)^{|a||y|}(-1)^{|a||z|+|b||z|}[[x,y],z]\ot abc\\
&=&(-1)^{|a||y|}(-1)^{|a||z|+|b||z|}([x,[y,z]]-(-1)^{|x||y|}[y,[x,z]])\ot abc\\
&=&(-1)^{|b||z|}([x\ot a,[y,z]\ot bc]\\
&-&(-1)^{|a||y|}(-1)^{|a||z|+|b||z|}(-1)^{|x||y|}[y,[x,z]])\ot abc\\
&=&(-1)^{|b||z|}([x\ot a,[y,z]\ot bc]\\
&-&(-1)^{|a||y|}(-1)^{|a||z|+|b||z|}(-1)^{|x||y|}(-1)^{|a||b|}[y,[x,z]])\ot bac\\
&=&(-1)^{|b||z|}([x\ot a,[y,z]\ot bc]\\
&-&(-1)^{|a||y|}(-1)^{|a||z|+|b||z|}(-1)^{|x||y|}(-1)^{|a||b|}(-1)^{|b||x|+|b||z|}[y\ot b,[x,z]\ot ac])\\
&=&(-1)^{|b||z|}([x\ot a,[y,z]\ot bc]\\
&-&(-1)^{|a||y|}(-1)^{|a||z|}(-1)^{|x||y|}(-1)^{|a||b|}(-1)^{|b||x|}[y\ot b,[x,z]\ot ac])\\
&=&[x\ot a,[y\ot b,z\ot c]]\\
&-&(-1)^{|a||y|}(-1)^{|x||y|}(-1)^{|a||b|}(-1)^{|b||x|}[y\ot b,[x\ot a,z\ot c]])\\
&=&[x\ot a,[y\ot b,z\ot c]]-(-1)^{|x\ot a||y\ot b|}[y\ot b,[x\ot a,z\ot c]])\\
\end{eqnarray*}
\end{comment}
Now define $$(x\ot t^\lam,y\ot t^\mu):=\theta(\lam,\mu)\d_{\lam+\mu,0}f(x,y)$$ for $x,y\in \gg$ and $\lam,\mu\in\Lam.$
This defines a nondegenerate invariant even supersymmetric bilinear form on $\hat\gg.$
\begin{comment}
\begin{eqnarray*}
(x\ot t^\lam,[y\ot t^\mu,z\ot t^\eta])&=&(x\ot t^\lam,(-1)^{|\mu||z|}\theta(\mu,\eta)[y,z]\ot t^{\mu+\eta})\\
&=&(-1)^{|\mu||z|}\theta(\mu,\eta)(-1)^{|\lam||y|+|\lam||z|}\theta(\lam,\mu+\eta)\d_{\lam+\mu+\eta,0}str(x[y,z])\\
&=&(-1)^{|\mu||z|+|\lam||y|+|\lam||z|}\theta(\lam,\mu)\theta(\lam+\mu,\eta)\d_{\lam+\mu+\eta,0}str([x,y],z])\\
&=&(-1)^{|\lam||y|}\theta(\lam,\mu)([x,y]\ot t^{\lam+\mu},z\ot t^\eta)\\
&=&([x\ot t^{\lam},y\ot t^{\mu}],z\ot t^\eta).
\end{eqnarray*}
\end{comment}
Next take $\v:=\bbbf\ot_\bbbz\Lam.$ Identify $\Lam$ with a subset of $\v$ and fix a basis $B:=\{\lam_i\mid i\in I\}\sub \Lam$ of $\v.$ Suppose that $\{d_i\mid i\in I\}$ is the dual basis of $B$ and $\v^\dag$ is the restricted dual of $\v$ with respect to this basis.
Each $d\in \v^\dag$ can be also considered as a derivation on $\hat \gg$ mapping $x\ot t^\lam$ to $d(\lambda) x\ot t^\lam$ for $x\in \gg$ and $\lambda \in \Lambda;$ indeed, for $a,b\in \gg$ and $\lam,\mu\in\Lam,$ we have \begin{eqnarray}
d([a\ot t^\lam,b\ot t^\mu]_{_{\hat\gg}})&=&d(\lam+\mu)[a,b]\ot t^\lam t^\mu\nonumber\\
&=&d(\lam)[a,b]\ot t^\lam t^\mu+d(\mu)[a,b]\ot t^\lam t^\mu\label{e2}\nonumber\\
&=&[d(a\ot t^\lam),b\ot t^\mu]_{_{\hat\gg}}+[a\ot t^\lam,d(b\ot t^\mu)]_{_{\hat\gg}}.\nonumber
\end{eqnarray}
Also for $a,b\in \gg,$ $d,d'\in \v^\dag$ and $\lam,\mu\in \Lam, $ we have
\begin{eqnarray}(d(a\ot t^\lam),b\ot t^\mu)=d(\lam)(a\ot t^\lam,b\ot t^\mu)
&=&d(\lam)\d_{\lam,-\mu}\theta(\lam,\mu)f(a,b)\nonumber\\
&=&-d(\mu)\d_{\lam,-\mu}\theta(\lam,\mu)f(a,b)\nonumber\\
&=&-d(\mu)(a\ot t^\lam,b\ot t^\mu)\nonumber\\
&=&-(a\ot t^\lam,d(b\ot t^\mu))\nonumber
\end{eqnarray}
and
\begin{eqnarray}
(dd'(a\ot t^\lam),b\ot t^\mu)=d(\lam)d'(\lam)(a\ot t^\lam,b\ot t^\mu)&=&d(\lam)d'(\lam)\d_{\lam+\mu,0}\theta(\lam,\mu)f(a,b)\nonumber\\
&=&-d(\lam)d'(\mu)\d_{\lam+\mu,0}\theta(\lam,\mu)f(a,b)\nonumber\\
&=&-d(\lam)d'(\mu)(a\ot t^\lam,b\ot t^\mu)\nonumber\\
&=&-(d(\lam)a\ot t^\lam,d'(\mu)b\ot t^\mu)\nonumber\\
&=&-(d(a\ot t^\lam),d'(b\ot t^\mu)).\nonumber
\end{eqnarray}
Therefore, we have
{\small\begin{equation}\label{e1}
(d(x),y)=-(x,d(y))\andd (dd'(x),y)=-(d(x),d'(y))\;\;\;\;\; (x,y\in \hat\gg,d,d'\in\v^\dag).
\end{equation}}
Set
$$\fl:=\hat \gg\op\v\op\v^\dag=(\gg\ot \aa)\op\v\op\v^\dag$$ and define
\begin{equation}\label{bracketloop2}\begin{array}{l}
\;[d,x]=-[x,d]=d(x), \quad d\in\v^\dag,x\in \hat \gg,\\
\;[\v,\fl]=[\fl,\v]=\{0\},\\
\;[\v^\dag,\v^\dag]=\{0\},\\
\;[x,y]=[x,y]_{_{\hat\gg}}+\sum_{i\in I}(d_i(x),y)\lam_i, \quad
x,y\in \hat\gg. \end{array}\end{equation}
\begin{lem}
Set $\fl_{\bar 0}:=\hat\gg_{\bar 0}\op\v\op\v^\dag$ and $\fl_{\bar 1}:=\hat\gg_{\bar 1},$ then $\fl=\fl_{\bar 0}\op\fl_{\bar 1}$ together with the Lie bracket as in (\ref{bracketloop2}) is a Lie superalgebra.
\end{lem}
\begin{proof} Since the form on $\gg$ is supersymmetric, (\ref{e1}) implies that the Lie bracket defined in $(\ref{bracketloop2})$ is anti-supercommutative. So we just need to verify the Jacobi superidentity.
We recall that the form on $\hat\gg$ is invariant and supersymmetric and that $d\in \v^\dag$ acts as a derivation on $\hat \gg.$ Take $x,y,z\in \hat \gg$ and $d,d'\in\v^\dag.$ Then we have
{\small
\begin{eqnarray*}
(d([x,y]_{\hat \gg}),z)
&\stackrel{(\ref{e1})}{=}&-([x,y]_{\hat \gg},d(z))\\
&=&-(x,[y,d(z)]_{\hat \gg})\\
&=&(-1)^{|y||z|}(x,[d(z),y]_{\hat \gg})\\
&=&(-1)^{|y||z|}(x,d([z,y]_{\hat \gg}))-(-1)^{|y||z|}(x,[z,d(y)]_{\hat \gg})\\
&=&-(x,d([y,z]_{\hat \gg}]))-(-1)^{|y||z|}([x,z]_{\hat \gg},d(y))\\
&=&-(x,d([y,z]_{\hat \gg}))+(-1)^{|y||z|+|z||x|}([z,x]_{\hat \gg},d(y))\\
&\stackrel{(\ref{e1})}{=}&-(-1)^{|x||z|}((-1)^{|x||y|}(d([y,z]_{\hat \gg}),x)+(-1)^{|y||z|}(d([z,x]_{\hat \gg}),y)).
\end{eqnarray*}}
Therefore we have
{\small \begin{eqnarray*}
&&(-1)^{|x||z|}[[x,y],z]+(-1)^{|z||y|}[[z,x],y]+(-1)^{|y||x|}[[y,z],x]\\
&=&(-1)^{|x||z|}[[x,y]_{\hat \gg},z]_{\hat \gg}+(-1)^{|x||z|}\sum_{i\in I}(d_i([x,y]_{\hat \gg}),z)\lam_i\\
&+&(-1)^{|z||y|}[[z,x]_{\hat \gg},y]_{\hat \gg}+(-1)^{|y||z|}\sum_{i\in I}(d_i([z,x]_{\hat \gg}),y)\lam_i\\
&+&(-1)^{|y||x|}[[y,z]_{\hat \gg},x]_{\hat \gg}+(-1)^{|x||y|}\sum_{i\in I}(d_i([y,z]_{\hat \gg}),x)\lam_i\\
&=&0.
\end{eqnarray*}}
Also we have
{\small
\begin{eqnarray*}
[[x,y],d]&=&[[x,y]_{_{\hat\gg}},d]\\
&=&-d([x,y]_{_{\hat\gg}}) \\
&=&-[d(x),y]_{_{\hat\gg}}-[x,d(y)]_{_{\hat\gg}}\\
&=&-[x,d(y)]_{_{\hat\gg}}+\sum_{i\in I}(dd_i(x),y)\lam_i-\sum_{i\in I}(dd_i(x),y)\lam_i-[d(x),y]_{_{\hat\gg}}\\
&=&-[x,d(y)]_{_{\hat\gg}}+\sum_{i\in I}(d_id(x),y)\lam_i-\sum_{i\in I}(dd_i(x),y)\lam_i-[d(x),y]_{_{\hat\gg}}\\
&\stackrel{(\ref{e1})}{=}&-[x,d(y)]_{_{\hat\gg}}-\sum_{i\in I}(d_i(x),d(y))\lam_i+\sum_{i\in I}(d(x),d_i(y))\lam_i-[d(x),y]_{_{\hat\gg}}\\
&=&-[x,d(y)]_{_{\hat\gg}}-\sum_{i\in I}(d_i(x),d(y))\lam_i+(-1)^{|x||y|}(\sum_{i\in I}(d_i(y),d(x))\lam_i+[y,d(x)]_{_{\hat\gg}})\\
&=&-[x,d(y)]+(-1)^{|x||y|}[y,d(x)]\\
&=&[x,[y,d]]-(-1)^{|x||y|}[y,[x,d]]
\end{eqnarray*}}
and
\begin{eqnarray*}
[[d,d'],x]=0=dd'(x)-dd'(x)=dd'(x)-d'd(x)=[d,[d',x]]-[d',[d,x]].
\end{eqnarray*}
Now the result immediately follows.
\end{proof}
\begin{lem}
Extend the form on
$\hat\gg$
to a supersymmetric bilinear form $\fm_{_\fl}$ on $\fl$ by
\begin{equation}\label{formloop}
\begin{array}{l}
(\v, \v)_{_\fl}=(\v^\dag,\v^\dag)_{_\fl}=(\v,\gg\ot\aa)_{_\fl}=(\v^\dag,\gg\ot\aa)_{_\fl}:=\{0\},\\
(v,d)_{_\fl}:=d(v),\quad d\in \v^\dag, v\in \v.
\end{array}
\end{equation}
Then $\fm_{_\fl}$ is a nondegenerate invariant even supersymmetric bilinear form.
\end{lem}
\begin{proof}
It is trivial that this form is nondegenerate, even and supersymmetric, so we just prove that it is invariant.
We first consider the following easy table:
$${\small
\begin{tabular}{|l|l|l|}
\hline
$(x,y,z)\in$&$([x,y],z)\in $&$(x,[y,z])\in$\\
\hline
$(\hat\gg,\hat\gg,\v)$& $(\hat\gg+\v,\v)_{_\fl}=\{0\}$ &$ (\hat\gg,\{0\})_{_\fl}=\{0\}$\\
\hline
$(\v,\hat\gg,\hat\gg)$& $(\{0\},\hat\gg)_{_\fl}=\{0\}$ &$ (\v,\hat\gg+\v)_{_\fl}=\{0\}$\\
\hline
$(\hat\gg,\v, \v\cup\v^\dag\cup \hat\gg)$& $(\{0\},\fl)_{_\fl}=\{0\}$& $(\hat\gg,\{0\})_{_\fl}=\{0\}$\\
\hline
$(\hat \gg,\v^\dag,\v\cup\v^\dag)$&$(\hat\gg,\v\cup\v^\dag)_{_\fl}=\{0\}$& $(\hat\gg,\{0\})_{_\fl}=\{0\}$\\
\hline
$(\v\cup\v^\dag,\hat\gg,\v\cup\v^\dag)$& $(\hat\gg,\v\cup\v^\dag)_{_\fl}=\{0\}$&$(\v\cup\v^\dag,\hat\gg)_{_\fl}=\{0\}$\\
\hline
$ (\v\cup\v^\dag,\v\cup\v^\dag,\hat\gg)$ & $(\{0\},\hat\gg)_{_\fl}=\{0\}$ &$(\v\cup\v^\dag,\hat\gg)_{_\fl}=\{0\}$\\
\hline
$ (\v\cup\v^\dag,\v\cup\v^\dag,\v\cup\v^\dag)$ & $(\{0\},\v\cup\v^\dag)_{_\fl}=\{0\}$ &$(\v\cup\v^\dag,\{0\})_{_\fl}=\{0\}$\\
\hline
\end{tabular}}$$
Then we note that if $x,y,z\in\hat\gg,$ we have {\small$$([x,y],z)_{_\fl}=([x,y]_{_{\hat\gg}},z)_{_\fl}=([x,y]_{_{\hat\gg}},z)=(x,[y,z]_{_{\hat\gg}})=(x,[y,z]_{_{\hat\gg}})_{_\fl}=(x,[y,z])_{_\fl}$$}and for $x,y\in\hat\gg$ and $z=d_j\in\v^\dag$ $(j\in I),$ we get {\small \begin{eqnarray*}
([x,y],z)_{_\fl}=([x,y]_{\hat\gg}+\sum_{i\in I}(d_i(x),y)\lam_i,d_j)_{_\fl}=(d_j(x),y)
&\stackrel{(\ref{e1})}{=}&
-(x,d_j(y))\\
&=&-(x,[d_j,y])_{_\fl}\\
&=&(x,[y,z])_{_\fl}.
\end{eqnarray*}}
Considering the latter equality, as the form is supersymmetric, for $y,z\in\hat\gg$ and $x=d_j\in\v^\dag$ $(j\in I),$ we have
{\small \begin{eqnarray*}
([x,y],z)_{_\fl}=(-1)^{|y||z|}(z,[x,y])=-(-1)^{|y||z|}(z,[y,x])&=&-(-1)^{|y||z|}([z,y],x)\\
&=&-(-1)^{|y||z|}(x,[z,y])\\
&=&(x,[y,z]).
\end{eqnarray*}}
Finally, for $x,z\in\hat\gg$
and $y=d_j\in\v^\dag$ $(j\in I),$ one has {\small $$([x,y],z)_{_\fl}=-([d_j,x],z)_{_\fl}=-(d_j(x),z)\stackrel{(\ref{e1})}{=}(x,d_j(z))=(x,[d_j,z])=(x,[y,z])_{_\fl}.$$}
This completes the proof.
\end{proof}
Now set $\fh:=(\hh\ot\bbbf)\op\v\op\v^\dag$ and take $R$ to be the root system of $\gg$ with respect to $\hh.$ We consider $\a\in R$ as an element of $\fh^*$ by $$\a({\v\op\v^\dag}):=\{0\} \andd \a(h\ot 1):=\a(h)\;\;\; (h\in\hh).$$ We also consider $\lam\in \v$ as an element of $\fh^*$ by $$\lam((\fh\ot\bbbf)\op\v):=\{0\}\andd \lam(d):=d(\lam)\;\;\; (d\in\v^\dag).$$ Then $\fl$ has a weight space decomposition with respect to $\fh$ with the corresponding root system $\mathfrak{R}=\{\a+\lam\mid \a\in R,\lam\in\Lam\};$ moreover, we have
$$\begin{array}{l}
\fl^0=\fh\andd \fl^{\a+\lam}=\gg^\a\ot \bbbf t^\lam \;\;\;(\a\in R,\lam\in\Lam \hbox{ with $\a+\lam\neq0$}).
\end{array}$$
Suppose $\lam\in \Lam$ and $\a\in R_i$ ($i\in\{0,1\}$) with $\a+\lam\neq0.$ Use Lemma \ref{symm}($iii$) together with the fact that $f\fm$ is nondegenerate on $\hh$ to fix $x\in\gg_{\bar i}^\a,y\in \gg_{\bar i}^{-\a}$ with $f(x,y)=1$ and $[x,y]\in \hh.$ Take $t_\a$ to be the unique element of $\hh$ representing $\a$ through $f\fm.$ We have
\begin{eqnarray*}
[x\ot t^\lam,\theta(\lam,-\lam)^{-1}y\ot t^{-\lam}]=(t_\a\ot 1)+\sum_{i\in I}d_i(\lam)\lam_i= (t_\a\ot 1)+\lam\in \fh\setminus\{0\}.
\end{eqnarray*}
It follows that $(\fl,\fh,\fm)$ is an extended affine Lie superalgebra with root system $\mathfrak{R}.$
\hfill$\diamondsuit$}
\end{Example}
For a unital associative algebra $\aa$ and nonempty index supersets $I,J,$ by an $I\times J$-matrix
with entries in $\aa,$ we mean a map $A:I\times J\longrightarrow \aa.$ For $i\in I,j\in J,$ we set
$a_{ij}:=A(i,j)$ and call it {\it $(i,j)$-th entry} of $A.$ By a convention, we denote the matrix
$A$ by $(a_{ij}).$ We also denote the set of all $I\times J$-matrices with entries in $\aa$ by
$\aa^{I\times J}$ and note that it is a vector superspace, under the componentwise summation and scalar product, with $$\aa^{I\times J}_{\bar i}:=\{A\in \aa^{I\times J}\mid A(I_{\bar t}\times J_{\bar s} )=0;\; \hbox{ $t,s\in\{0,1\}$ with $\bar t+\bar s=\bar i+\bar 1$}\},$$ for $i=0,1.$
If $A=(a_{ij})\in\aa^{I\times J}$ and
$B=(b_{jk})\in \aa^{J\times K}$ are such that for all $i\in I$ and $k\in K,$ at most for finitely many $j\in J,$ $a_{ij}b_{jk}$'s are nonzero, we define the product $AB$ of $A$ and $B$ to be the
$I\times K$-matrix $C=(c_{ik})$ with $c_{ik}:=\sum_{j\in J}a_{ij}b_{jk}$ for all $i\in I,k\in K.$
We note that if $A,B,C$ are three matrices such that $AB,$ $(AB)C,$ $BC$ and $A(BC)$ are defined,
then $A(BC)=(AB)C.$
For $i\in I,j\in J$ and $a\in \aa,$ we define $E_{ij}(a)$ to be the matrix in $\aa^{I\times J}$ whose $(i,j)$-th entry is
`` $a$ " and other entries are zero and if $\aa$ is unital, we set $$e_{i,j}:=E_{i,j}(1).$$ Take $M_{I\times J}(\aa)$ to be the subsuperspace of $\aa^{I\times J}$ spanned by
$\{E_{ij}(a)\mid i\in I,j\in J,a\in A\};$ in fact $M_{I\times J}(\aa)$ is a superspace with $M_{I\times J}(\aa)_{\bar i}=\hbox{span}_\bbbf\{E_{r,s}(a)\mid |r|+|s|=\bar i\},$ for $i=0,1.$ Also with respect to the multiplication of matrices, the vector superspace
$M_{I\times I}(\aa)$ is an associative $\bbbf$-superalgebra and so is a Lie superalgebra under the Lie bracket
$[A,B]:=AB-(-1)^{|A||B|}BA$ for all $A,B\in M_{I\times I}(\aa).$ We denote this Lie superalgebra by
$\mathfrak{pl}_I(\mathcal{A}).$
For an element $X\in \mathfrak{pl}_I(\mathcal{A}),$ we set $str(X):=\sum_{i\in I}(-1)^{|i|}x_{i,i}$ and call it the {\it supertrace} of $X.$ We finally make a convention that if $I$ is a disjoint union of nonempty subsets
$I_1,\ldots, I_t$ of $I,$ then for an $I\times I$-matrix $A,$ we write
$$A=\left [\begin{array}{ccc} A_{1,1}&\cdots& A_{1,t}\\
A_{2,1}&\cdots&A_{2,t}\\
\vdots&\vdots&\vdots\\
A_{t,1}&\cdots&A_{t,t}\\
\end{array}\right ]$$
in which for $1\leq r,s\leq t,$ $A_{r,s}$ is an $I_r\times I_s$-matrix whose $(i,j)$-th entry
coincides with $(i,j)$-th entry of $A$ for all $i\in I_r,j\in I_s.$ In this case, we say that $A\in \aa^{I_1\uplus\cdots\uplus I_t}$ and note that the defined matrix product obeys the product of block matrices.
\smallskip
In the next example, we realize a certain extended affine Lie superalgebra using an ``{\it affinization}'' process. To this end, we need to fix some notations. Suppose that $A$ is a unital associative algebra and ``$\;*\;$'' is an involution on $\aa$ that is a self-inverting linear endomorphism of $A$ with $(ab)^*=b^*a^*$ for all $a,b\in A.$ We next assume $\dot I,\dot J,\dot K$ are nonempty index sets with disjoint copies $\bar {\dot I}=\{\bar i\mid i\in \dot I\},$ $\bar {\dot J}=\{\bar j\mid j\in \dot J\}$ and $\bar {\dot K}=\{\bar k\mid k\in \dot K\}$ respectively. Suppose that $0,0',0''$ are three distinct symbols and by a convention, take $\bar0:=0,$ $\bar0':=0'$ and $\bar0'':=0''.$ We take $I$ to be either $\dot I\uplus\bar{\dot I}$ or $\{0\}\uplus \dot I\uplus\bar{\dot I},$ $J$ to be either $\dot J\uplus\bar{\dot J}$ or $\{0'\}\uplus\dot J\uplus\bar{\dot J},$ and $K$ to be either $\dot K\uplus\bar{\dot K}$ or $\{0''\}\uplus \dot K\uplus\bar{\dot K}.$ For a matrix $X=(x_{ij})\in M_{I\times J}(\aa),$ define $X^{\diamond}$ to be the matrix $(y_{ji})$ of $M_{J\times I}(\aa)$ with $y_{ji}:= x_{\bar i\bar j}^*$ ($i\in I,j\in J$) where for an index $t\in I\cup J,$ by $\bar{\bar t},$ we mean $t.$
It is immediate that for a matrix $X=(x_{ij})\in M_{I\times I}(\aa),$ \begin{equation}\label{dia1}tr(X^{\diamond})=(tr(X))^*.\end{equation} Also if $X=(x_{ij})\in M_{I\times J}(\aa)$ and $Y\in M_{J\times K}(\aa),$ then for $i\in I$ and $k\in K,$ we have
\begin{eqnarray*}(XY)^\diamond_{ki}=(\sum_{j\in J}x_{\bar ij}y_{j\bar k})^*=\sum_{j\in J} y_{j\bar k}^* x_{\bar ij}^*=\sum_{j\in J} y_{\bar j\bar k}^* x_{\bar i\bar j}^*=\sum_{j\in J} Y^\diamond_{kj} X^\diamond_{ji}=(Y^\diamond X^\diamond)_{ki}\end{eqnarray*} which implies that \begin{equation}\label{dia2}(XY)^\diamond=Y^\diamond X^\diamond.\end{equation}
\begin{Example}
{\rm
In this example, we assume that the field $\bbbf$ contains a forth primitive root of unity and call it $\zeta.$ Suppose that $G$ is a torsion free additive abelian group and $\lam$ is a commutative $2$-cocycle satisfying $\lam(0,0)=1.$ Take $\aa$ to be the commutative associative torus corresponding to $(G, \lam).$ Take $*$ to be an involution of $\aa$ mapping $\aa^\tau$ to $\aa^\tau$ for all $\tau\in G$ and suppose that $I$ and $J$ are as in the previous paragraph such that $I\cap J=\emptyset$ and that $|I|\neq|J|$ if $I$ and $J$ are both finite.
Consider $I\uplus J$ as a superset with $|i|:=\bar 0$ and $|j|:=\bar 1$ for $i\in I$ and $j\in J$ and take $\LL:=\mathfrak{pl}_{I \uplus J}(\aa).$ One knows that for ${}_{[\tau]}\LL:=\{(x_{ij})\in \LL\mid x_{ij}\in\aa^\tau;\;\;\forall i,j\in I\uplus J\}$ ($\tau\in G$),
${\displaystyle\LL=\bigoplus_{\tau\in G}{}_{[\tau]}\LL}$ is a $G$-graded Lie superalgebra.
Set $$\gg:=\mathfrak{sl}_\aa(I,J):=\{A\in \mathfrak{pl}_{I\uplus J}(\aa)\mid str(A)=0\}\simeq \frak{sl}_\bbbf(I,J)\ot \aa.$$ As $ \gg$ is a subsuperalgebra of $\LL$ generated by $\{E_{i,j}(a)\mid i,j\in I\uplus J, i\neq j, a\in\aa\},$ it follows that $\gg$ is a $G$-graded subsuperalgebra of $\LL.$
\begin{comment}
In fact if $A\in \mathfrak{sl}_\aa(I,J)$ and $i_1,\ldots,i_n\in I,j_1,\ldots,j_m\in J$ are the only indices with $a_{i_1i_1},\ldots,a_{i_ni_n},a_{j_1j_1},\ldots,a_{j_{m}j_{m}}\neq 0,$ we have
\begin{eqnarray*}
\sum_{i\in I\cup J}a_{ii}e_{ii}&=&a_{i_1i_1}e_{i_1i_1}+\cdots +a_{i_ni_n}e_{i_ni_n}+a_{j_1j_1}e_{j_1j_1}+\cdots a_{j_mj_m}e_{j_mj_m}\\
&=&\sum_{t=1}^{n-1}(a_{i_1i_1}+\cdots+a_{i_ti_t})[e_{i_ti_{t+1}},e_{i_{t+1}i_t}]\\
&+&(a_{i_1i_1}+\cdots+a_{i_ni_n})[e_{i_nj_{1}},e_{j_{1}i_n}]\\
&+&\sum_{t=2}^{m}(-a_{i_1i_1}-\cdots-a_{i_ni_n}+a_{j_1j_1}+\cdots+a_{j_{t-1}j_{t-1}})[e_{j_{t-1}j_{t}},e_{j_{t}j_{t-1}}].
\end{eqnarray*}
\end{comment}
Setting $\hh:=\hbox{span}_\bbbf\{e_{i,i}-e_{r,r},e_{j,j}-e_{s,s},e_{i,i}+e_{j,j}\mid i,r\in I, j,s\in J\},$
one knows that $\gg$ has a weight space decomposition $\gg=\op_{\a\in \hh^*}\gg^\a$ with respect to $\hh$ with the corresponding root system $$R=\{\ep_i-\ep_r,\d_j-\d_s,\ep_i-\d_j,\d_j-\ep_i\mid i,r\in I, j,s\in J\},$$ where for $t\in I$ and $p\in J,$
$$\begin{array}{l}\ep_t:\hh\longrightarrow \bbbf;\;\;\; e_{i,i}-e_{r,r}\mapsto \d_{i,t}-\d_{r,t},e_{j,j}-e_{s,s}\mapsto 0,e_{i,i}+e_{s,s}\mapsto \d_{i,t},\\
\d_p:\hh\longrightarrow \bbbf;\;\;\; e_{i,i}-e_{r,r}\mapsto0,e_{j,j}-e_{s,s}\mapsto \d_{j,p}-\d_{p,s},e_{i,i}+e_{j,j}\mapsto \d_{p,j},
\end{array} $$ $(i,r\in I,\;\; j,s\in J)$
and for $ i,r\in I$ and $ j,s\in J$ with $i\neq r$ and $j\neq s,$ we have
$$
\begin{array}{ll}
\gg^{\ep_i-\ep_r}=\aa e_{i,r},\;\; \gg^{\d_j-\d_s}=\aa e_{j,s},\;\;
\gg^{\ep_i-\d_j}=\aa e_{i,j},\;\;\gg^{\d_j-\ep_i}=\aa e_{j,i},\\
\displaystyle{\gg^0=\{A=\sum_{t\in I\uplus J}a_{tt}e_{t,t}\in \frak{pl}_{I\uplus J}(\aa)\mid str(A)=0\}.}
\end{array}
$$ For $\a\in R$ and $\tau\in G,$ setting $${}_{[\tau]}\gg^\a:={}_{[\tau]}\gg\cap\gg^\a,$$ we have
$$
\begin{array}{ll}
{}_{[\tau]}\gg^{\ep_i-\ep_r}=\aa^\tau e_{i,r},\;\; {}_{[\tau]}\gg^{\d_j-\d_s}=\aa^\tau e_{j,s},\;\;
{}_{[\tau]}\gg^{\ep_i-\d_j}=\aa^\tau e_{i,j},\;\;{}_{[\tau]}\gg^{\d_j-\ep_i}=\aa^\tau e_{j,i},\\
\displaystyle{{}_{[\tau]}\gg^0=\{A=\sum_{t\in I\uplus J}a_{tt}e_{t,t}\in \frak{pl}_{I\uplus J}(\aa)\mid a_{tt}\in\aa^\tau\;(t\in I\uplus J), str(A)=0\}}
\end{array}
$$ for $ i,r\in I$ and $ j,s\in J$ with $i\neq r$ and $j\neq s.$
Now take $\ep:\aa\longrightarrow \bbbf$ to be a linear function defined by $$x^\tau\mapsto \left\{\begin{array}{ll}0&\hbox{if $\tau\neq0$}\\
1&\hbox{if $\tau=0$}\end{array}\right.\;\;\;\; (\tau\in G). $$ Define $$\fm:\gg\times\gg\longrightarrow\bbbf;\;\;(x,y)\mapsto \ep(str(xy)) .$$
This defines a nondegenerate invariant even supersymmetric bilinear form on $\gg.$
\begin{comment}
If $a=\sum_{\tau}\sum_{r,t\in I\cup J}a^{t,r}_\tau x^\tau e_{r,t}$ is an element of the radical of the form, for $p,q\in I\cup J$ with $p\neq q$ and $\gamma\in G,$ we have
$$\ep(str((\sum_{\tau}\sum_{r,t\in I\cup J}a^{t,r}_\tau x^\tau e_{r,t})(x^{-\gamma}e_{p,q})))=0.$$
$$\ep(str((\sum_{\tau}\sum_{r\in I\cup J}a^{p,r}_\tau x^\tau x^{-\gamma} e_{r,q}))=\ep(str((\sum_{\tau}\sum_{r,t\in I\cup J}a^{t,r}_\tau x^\tau x^{-\gamma} \d_{t,p}e_{r,q}))=0.$$ So
$$(-1)^{|q|}a^{p,q}_\gamma\lam(\gamma,-\gamma)=\ep(\sum_{\tau}(-1)^{|q|}a^{p,q}_\tau x^\tau x^{-\gamma} )=0$$ so $$a_\gamma^{p,q}=0.$$
Therefore $$a=\sum_{\tau}\sum_{r\in I\cup J}a^{r,r}_\tau x^\tau e_{r,r}$$
$$\ep(str((\sum_{\tau}\sum_{r\in I\cup J}a^{r,r}_\tau x^\tau e_{r,r})((-1)^{|p|}x^{-\gamma}e_{p,p}-(-1)^{|q|}x^{-\gamma}e_{q,q})))=0.$$
So we have $$\ep(str((\sum_{\tau}\sum_{r\in I\cup J}a^{r,r}_\tau x^\tau e_{r,r})((-1)^{|p|}x^{-\gamma}e_{p,p})=\ep(str((\sum_{\tau}\sum_{r\in I\cup J}a^{r,r}_\tau x^\tau e_{r,r})((-1)^{|q|}x^{-\gamma}e_{q,q})))$$
$$\ep(str(\sum_{\tau}\sum_{r\in I\cup J}(-1)^{|p|}a^{r,r}_\tau x^\tau x^{-\gamma} \d_{r,p}e_{p,p}))=\ep(str(\sum_{\tau}\sum_{r\in I\cup J}(-1)^{|q|}a^{r,r}_\tau x^\tau x^{-\gamma} \d_{r,q}e_{q,q}))$$
$$\ep(str(\sum_{\tau}(-1)^{|p|}a^{p,p}_\tau\lam(\tau,-\gamma) x^{\tau-\gamma} e_{p,p}))=\ep(str(\sum_{\tau}(-1)^{|q|}a^{q,q}_\tau \lam(\tau,-\gamma) x^{\tau-\gamma} e_{q,q}))$$
$$\ep(\sum_{\tau}a^{p,p}_\tau\lam(\tau,-\gamma) x^{\tau-\gamma})=\ep(\sum_{\tau}a^{q,q}_\tau \lam(\tau,-\gamma) x^{\tau-\gamma} )$$
$$a^{p,p}_\gamma\lam(\gamma,-\gamma) =a^{q,q}_\gamma \lam(\gamma,-\gamma) $$
Setting $k:=a_\gamma^{p,p}$ for some $p,\gamma,$ we have $a=k\sum_{\tau}\sum_{r\in I\cup J} x^\tau e_{r,r}.$ But $0=str(a)$ which together with the fact that $|I|\neq |J|$ if $I,J$ are both finite, implies that $k=0.$ This completes the proof.
\end{comment}
Next take $\v:=\bbbf\ot_\bbbz G.$ Since $G$ is torsion free, we can identify $G$ with a subset of $\v$ in a usual manner. We next fix a basis $B:=\{\tau_t\mid t\in T\}\sub G$ of $\v.$ Suppose that $\{d_t\mid t\in T\}$ is its dual basis and take $\v^\dag$ to be the restricted dual of $\v$ with respect to this basis. Each $d\in \v^\dag$ can be considered as a derivation of $\gg$ (of degree $0$) by $d(x):=d(\tau)x$ for each $x\in {}_{[\tau]}\gg$ ($\tau\in G$).
Set $$\fl:=\gg\op\v\op\v^\dag.$$
We extend the form on
$\gg$
to a nondegenerate even supersymmetric bilinear form $\fm$ on $\fl$ by
\begin{equation*}\label{formloop2}
\begin{array}{c}
{\small (\v, \v)=(\v^\dag,\v^\dag)=(\v,\gg)=(\v^\dag,\gg):=\{0\}\andd (d,v):=d(v)\quad (d\in \v^\dag, v\in \v).}
\end{array}
\end{equation*}
We also define
\begin{equation*}\label{bracketloop22}\begin{array}{l}
\;[d,x]=-[x,d]=d(x), \quad (d\in\v^\dag,x\in\gg)\\
\;[\v,\fl]=\{0\},\\
\;[\v^\dag,\v^\dag]=\{0\},\\
\;[x,y]=[x,y]_{_{\gg}}+\sum_{t\in T}(d_t(x),y)\tau_t, \quad
(x,y\in \gg) \end{array}\end{equation*}
in which $[\cdot,\cdot]_{_\gg}$ is the bracket on $\gg.$ Setting $\fh:=\hh\op\v\op\v^\dag,$ as in the previous example, one gets that $(\fl,\fh,\fm)$ is an extended affine Lie superalgebra
with root system $$\mathfrak{R}=\{\a+\tau\mid \a\in R,\tau\in G\}$$ in which $\a\in R$ is considered as an element of $\fh^*$ by $$\a({\v\op\v^\dag}):=\{0\},$$ and $\tau\in \v$ is considered as an element of $\fh^*$ by $$\tau(\hh\op\v):=\{0\}\andd \tau(d):=d(\tau)\;\;\; (d\in\v^\dag).$$ We also have
$$\begin{array}{l}
\fl^0=\fh\andd \fl^{\a+\tau}={}_{[\tau]}\gg^\a \;\;\;(\a\in R,\tau\in G \hbox{ with $\a+\tau\neq0$}).
\end{array}$$
Next for $A=\left(\begin{array}{ll}X&Y\\Z&W\end{array}\right)\in\LL=\frak{pl}_{I\uplus J}(\aa),$ define $A^\#:=\left(\begin{array}{ll}-X^\diamond&Z^\diamond\\-Y^\diamond&-W^\diamond\end{array}\right).$ We have $[A,B]^\#=[A^\#,B^\#]$ and so $\#$ is a Lie superalgebra automorphism of $\LL$ of order 4. Since $\#$ maps $\gg$ to $\gg$ (see (\ref{dia1})), we consider $\#$ as a Lie superalgebra automorphism of $\gg$ as well.
Suppose that $M=\left(\begin{array}{ll}X&Y\\Z&W\end{array}\right),N=\left(\begin{array}{ll}A&B\\C&D\end{array}\right)$ are elements of $\gg,$
then we have
{\small
\begin{eqnarray}
(M^\#,N^\#)&=&\ep(str (M^\#N^\#))\nonumber\\&=&\ep(tr (X^\diamond A^\diamond-Z^\diamond B^\diamond)-tr(-Y^\diamond C^\diamond+ W^\diamond D^\diamond))\nonumber\\
&\stackrel{(\ref{dia2})}{=}&\ep(tr ((AX)^\diamond +(CY)^\diamond- (BZ)^\diamond -(DW)^\diamond))\nonumber\\
&=& \ep(tr ((AX)^\diamond) +tr ((CY)^\diamond)- tr ((BZ)^\diamond) -tr ((DW)^\diamond))\nonumber\\
&\stackrel{(\ref{dia1})}{=}&\ep((tr (AX))^*) +(tr (CY))^*)- (tr (BZ))^*) -(tr (DW))^*))\label{equality}\\
&=&\ep(tr (AX) +tr (CY)- tr (BZ) -tr (DW))\nonumber\\
&=&\ep(tr (XA)+tr (YC)- tr (ZB) -tr (WD))\nonumber\\
&=& \ep(str(MN))=(M,N).\nonumber
\end{eqnarray}}
We also have \begin{equation}\label{equality1} d(x^\#)=(d(x))^\#;\;\; d\in \v^\dag,\; x\in \gg.\end{equation}
Now extend $\#$ to $\fl$ by $v^\#:=v$ for $v\in\v\op\v^\dag.$ It follows from (\ref{equality}) and (\ref{equality1}) that $\#$ is an automorphism of $\fl$ of order $4$ mapping $\fh$ to $\fh.$
So we have $\fl=\op_{i=0}^3{}^{[i]}\fl,$ where for $i\in\bbbz,$ $${}^{[i]}\fl:=\{x\in\fl\mid x^\#=\zeta^i x\}$$ in which $[i]$ indicates the congruence of $i\in\bbbz$ modulo $4\bbbz.$ Using (\ref{equality}) together with the fact that the form on $\fl$ is nondegenerate, for $i,j\in\bbbz,$ we have \begin{equation}\label{form}({}^{[i]}\fl,{}^{[j]}\fl)\neq \{0\}\hbox{ if and only of $i+j\in 4\bbbz.$}\end{equation}
Next take $\sg$ to be the restriction of $\#$ to $\fh$ and set $\mathfrak{h}^\sg$ to be the set of fixed points of $\fh$ under $\sg.$ Consider a linear endomorphism of $\fh^*$ mapping $\a\in\fh^*$ to $\a\circ\sg^{-1}$ and denote it again by $\sg.$ The Lie superalgebra $\fl$ has a weight space decomposition
$\fl=\sum_{\{\pi(\a)\mid\a\in \frak{R}\}}\fl^{\pi(\a)}$ with respect to $\fh^\sg$ where $$\pi(\a):=(1/4)(\a+\sg(\a)+\sg^2(\a)+\sg^3(\a))\;\;\; (\a\in \frak{R})$$ (see \cite[(2.11) \& Lem. 3.7]{ahy}). Moreover, we have
\begin{eqnarray*} \pi(\frak{R})&=&\{\pi(\a)\mid \a\in\mathfrak{R}\}\\
&=&\{\tau\mid \tau\in G\}\\
&\cup&\{\frac{1}{2}((\ep_i-\ep_r)+(\ep_{\bar r}-\ep_{\bar i}))+\tau\mid \tau\in G,i,r\in I; i\neq r\}\\
&\cup& \{\frac{1}{2}((\d_j-\d_s)+(\d_{\bar s}-\d_{\bar j}))+\tau\mid \tau\in G, j,s\in J; j\neq s\}\\
&\cup&
\{\frac{1}{2}((\ep_i-\d_j)+(\d_{\bar j}-\ep_{\bar i}))+\tau\mid \tau\in G, i\in I,j\in J\}\\
&\cup& \{\frac{1}{2}((\d_j-\ep_i)+(\ep_{\bar i}-\d_{\bar j})+\tau\mid \tau\in G, i\in I, j\in J\}\\
&=&\{\tau\mid \tau\in G\}\\
&\cup&
\{\frac{1}{2}((\ep_i-\ep_{\bar i})-(\ep_{ r}-\ep_{\bar r}))+\tau\mid \tau\in G, i,r\in I; i\neq r\}\\
&\cup&
\{\frac{1}{2}((\d_j-\d_{\bar j})-(\d_{s}-\d_{\bar s}))+\tau\mid \tau\in G, j,s\in J; j\neq s\}\\
&\cup&
\{\frac{1}{2}((\ep_i-\ep_{\bar i})-(\d_{ j}-\d_{\bar j}))+\tau\mid \tau\in G, i\in I, j\in J\}\\
&\cup&
\{\frac{1}{2}((\d_{ j}-\d_{\bar j})-(\ep_i-\ep_{\bar i}))+\tau\mid \tau\in G, i\in I, j\in J\}
\end{eqnarray*}
which is an extended affine root supersystem of type $BC(I,J)$ if $\{0,0'\}\cap (I\cup J)\neq \emptyset,$ and it is of type $C(I,J),$ otherwise; see \cite{you6} for the notion of type for an extended affine root supersystem.
We next note that for $i\in\{0,1,2,3\},$ as ${}^{[i]}\fl$ is an $\fh^\sg$-submodule of $\fl,$ it inherits the weight space decomposition ${\displaystyle{}^{[i]}\fl=\sum_{\{\pi(\a)\mid\a\in \frak{R}\}}{}^{[i]}\fl^{\pi(\a)}}$ from $\fl$ in which ${}^{[i]}\fl^{\pi(\a)}:={}^{[i]}\fl\cap\fl^{\pi(\a)}.$ We recall (\ref{form}) together with the fact that the form $\fm$ is nondegenerate. So if $\a,\b\in \frak{R}$ and $i,j\in\bbbz,$ we get that \begin{equation}\label{form1}
\begin{array}{l}\hbox{ for $0\neq x\in{}^{[i]}\fl^{\pi(\a)},$} (x,{}^{[j]}\fl^{\pi(\b)})\neq \{0\}\\
\hbox{ if and only if $i+j\in 4\bbbz$ and $\pi(\a)+\pi(\b)=0.$}
\end{array}\end{equation}
Now we set $$\tilde\fl:=\sum_{i\in\bbbz}({}^{[i]}\fl\ot\bbbf t^i)\op\bbbf c\op\bbbf d$$ where $c,d$ are two symbols. Since $\#$ preserves the $\bbbz_2$-grading on $\fl,$ $\tilde\fl$ is a superspace with $$\tilde\fl_{\bar 0}:=\sum_{i\in\bbbz}(({}^{[i]}\fl\cap \fl_{\bar 0})\ot\bbbf t^i)\op\bbbf c\op\bbbf d\andd \tilde\fl_{\bar 1}:=\sum_{i\in\bbbz}(({}^{[i]}\fl\cap \fl_{\bar 1})\ot\bbbf t^i).$$ Moreover, $\tilde\fl$ together with the following bracket \\
$\begin{array}{ll}
[x\ot t^i+rc+sd,y\ot t^j+r'c+s'd]^{\;\tilde{}}&:=[x,y]\ot t^{i+j}+i\d_{i,-j}(x,y)c\\
&+sj y\ot t^j-s'ix\ot t^i
\end{array}$
\noindent is a Lie superalgebra equipped with a weight space decomposition with respect to $(\mathfrak{h}^\sg\ot \bbbf)\op\bbbf c\op\bbbf d.$ More precisely, if we define
$$\d:(\fh^\sg\ot\bbbf)\op\bbbf c\op\bbbf d\longrightarrow \bbbf;\;\;c\mapsto0,d\mapsto1,h\ot1\mapsto 0\;\;(h\in\fh^\sg),$$ then we get that $\tilde{\mathfrak{R}}:=\{\pi(\a)+i\d\mid \a\in \mathfrak{R} ,i\in\bbbz,\;{}^{[i]}\fl\cap \fl^{\pi(\a)}\neq\{0\}\}$ is the corresponding root system of $\tilde\fl$ in which $\pi(\a)$ is considered as an element of the dual space of $(\fh^\sg\ot\bbbf)\op\bbbf c\op\bbbf d$ mapping $h\ot 1\in \fh^\sg\ot\bbbf$ to $\a(h)$ and $c,d$ to $0.$ Furthermore, since for each $\a\in \frak{R}\setminus\{0\},$ $\pi(\a)\neq 0,$ as in \cite[Cor. 3.26]{ABP}, we have $ {}^{[0]}\fl^{\pi(0)}=\fh^\sg.$ So for $\a\in \mathfrak{R}$ and $i\in\bbbz,$ we have
$$\tilde\fl^{\pi(\a)+i\d}=\left\{\begin{array}{ll}
({}^{[i]}\fl\cap \fl^{\pi(\a)})\ot t^i& \hbox{if }(\a,i)\neq(0,0)\\
(\fh^\sg\ot 1)\op \bbbf c\op\bbbf d& \hbox{if } (\a,i)=(0,0).\\
\end{array}
\right.$$
\begin{comment}
Note that for each $\a\in \frak{R}\setminus\{0\},$ $\pi(\a)\neq 0,$ so we have $\tilde\fl^0=(\fh^\sg\ot 1)\op \bbbf c\op\bbbf d$ using the same argument as in [ABP, Cor. 3.26].
\end{comment}
We extend the form on
$\fl$
to a supersymmetric bilinear form $\fm^{\tilde{}}$ on $\tilde\fl$ by
\begin{equation}\label{formloop}
\begin{array}{l}
(c,d)^{\tilde{}}=1,\;(c,c)^{\tilde{}}=(d,d)^{\tilde{}}=(c,{}^{[i]}\fl\ot\bbbf t^i)^{\tilde{}}=(d,{}^{[i]}\fl\ot\bbbf t^i)^{\tilde{}}:=\{0\}\;\;(i\in\bbbz)\\
(x\ot t^i,y\ot t^j)^{\tilde{}}=\d_{i+j,0}(x,y)\;\; (i,j\in\bbbz,x\in{}^{[i]}\fl,y\in {}^{[j]}\fl).
\end{array}
\end{equation}
Since the form on $\fl$ is even and nondegenerate, $\fm^{\tilde{}}$ is also even and nondegenerate; moreover, by (\ref{form1}), if $j\in\{0,1\},$ $\a\in \mathfrak{R}$ and $i\in\bbbz$ with $(i,\a)\neq(0,0)$ such that ${}^{[i]}\fl^{\pi(\a)}\cap \fl_{\bar j} \neq\{0\},$ we have $({}^{[i]}\fl^{\pi(\a)}\cap \fl_{\bar j},{}^{[-i]}\fl^{\pi(-\a)}\cap \fl_{\bar j})\neq \{0\}.$ Using this together with the fact that $ {}^{[0]}\fl^{\pi(0)}=\fh^\sg$ and Lemma \ref{symm}($iii$), one finds $x\in {}^{[i]}\fl^{\pi(\a)}\cap \fl_{\bar j},y\in {}^{[-i]}\fl^{\pi(-\a)}\cap \fl_{\bar j}$ with $0\neq [x,y]\in \fh^\sg.$ So $$[x\ot t^i,y\ot t^{-i}]^{\;\tilde{}}=[x,y]\ot 1+i(x,y)c\in (\fh^\sg\ot 1)\op \bbbf c\op\bbbf d\setminus\{0\}.$$ Also as $\pi({\mathfrak{R}})$ is an extended affine root supersystem, it follows that $ad_x$ is locally nilpotent for all $x\in\tilde\fl^{\tilde\a}$ where $\tilde\a\in {\tilde{\mathfrak{R}}}_{re}^\times.$
So $$(\tilde\fl,(\fh^\sg\ot \bbbf)\op\bbbf c\op\bbbf d,\fm^{\tilde{}})$$ is an extended affine Lie superalgebra with corresponding root system $\tilde{\mathfrak{R}}.$ \hfill{$\diamondsuit$}}
\begin{comment}
Since the form is nondegenerate on $\tilde\LL$ and $\tilde\LL^0=\fh^\sg+\bbbf c+\bbbf d,$ all conditions hold.
\end{comment}
\end{Example}
\centerline{Acknowledgement}
This research was in part
supported by a grant from IPM (No. 92170415) and partially carried out in IPM-Isfahan branch. | 121,275 |
Lovely time at the Ferry Street Growers Market last Saturday – nice weather, a vibrant market, friendly people, a few CDs sold, and the lovely thing about playing at a farmer’s market: on the way out, a number of vendors saying “hey, you’re the musician, right? How about some [insert one or more: bread/cabbage/potatos/beans/summer squash/cherries]?”
Only one regret: I didn’t get any photos. If you took one, let me know!
(And thanks to Jeremy for loan of the tent. I would’ve gotten all red and crispy without it.) | 125,565 |
Code of Conduct for Culture Funds
At the end of 2012 the directors of the culture funds have determined the Code of Conduct for Culture Funds. From 1 May 2013, this new code of conduct replaces the existing Code of Conduct for Culture Funds from 2008.
The code of conduct aims to create more transparency about the way in which the culture funds carry out their public tasks, and in doing so to contribute to the confidence of applicants and citizens in the culture funds. In drawing up the code of conduct, special attention has been paid to the position of the applicant in the grant process. The code of conduct comprises six sections:
1) integrity and impartiality;
2) openness;
3) involvement of stakeholders;
4) quality of service;
5) evaluation;
6) accountability.
The code of conduct is endorsed by the Mondriaan Fund, the Netherlands Film Fund, the Dutch Foundation for Literature, the Performing Arts Fund NL, the Cultural Participation Fund, and the Creative Industries Fund NL.
Apart from the Code of Conduct for Culture Funds, the Mondriaan Fund follows the Governance Code for Culture and takes into account the Code Cultural Diversity. | 142,545 |
JCA Home • Issue Contents
Catalan Numbers and Power Laws in Cellular Automaton Rule 14
Henryk Fuks and Jeff Haroutunian
We discuss example of an elementary cellular automaton for which the density of ones decays toward its limiting value as a power of the number of iterations n. Using the fact that this rule conserves the number of blocks 10 and that preimages of some other blocks exhibit patterns closely related to patterns observed in rule 184, we derive expressions for the number of n-step preimages of all blocks of length 3. These expressions involve Catalan numbers, and together with basic properties of iterated probability measures they allow us to compute the density of ones after n iterations, as well as probabilities of occurrence of an arbitrary block of length smaller or equal to 3. | 316,541 |
By James E. Mitchell, MD and Kristine J. Steffen, PharmD
University of North Dakota, Fargo
Reprinted from Eating Disorders Review
January/February 2008 Volume 19, Number 1
©2008 Gürze Books
In thinking about areas on which to focus this review, the possibilities seemed endless. Therefore, we have chosen a few representative areas but realize we do so at the expense of much very important work.
Two issues of ongoing concern are the health problems of patients with eating disorders and their use of health services. Spoor and colleagues (2007), using community-recruited samples, found that the total number of occurrences of binge eating and compensatory behaviors, or simply compensatory behaviors occurring twice a week, resulted in greater self-reported health-care utilization using items from the Health Survey Utilization Scale that assessed both mental and physical health care utilization.Mond and colleagues (2007), again using a community sample, found that women with bulimic-type eating disorders rarely received treatment for an eating disorder but often received treatment for other mental health problems and/or for weight loss.
Ecological momentary assessment (EMA) has emerged as a helpful tool, as shown in several studies. bulimia nervosa .
Treatment update cognitive behavioral therapy . “Improving self-esteem,” “improving body experience,” and “learning problem-solving skills” were all considered.
Long-term outcome.
Comorbidity
Relative to comorbidity, Claes and Vandereycken (2007) found that patients who engage in self-injurious behaviors were more likely to report a history of physical and/or sexual abuse, as is now noted repeatedly in the literature. Swinbuorne and Touyz (2007) found that although the literature suggests an increase in the prevalence of anxiety disorders in patients with eating disorders relative to the rates of the general population, the research has striking inconsistencies, and much additional research in this area is indicated.
References
Engelberg MJ, Steiger H, Gauvin L, Wonderlich SA. Binge eating antecedents in bulimic syndromes: an examination of dissociation and negative affect. Int J Eat Disord. 2007; 40:531.
Fichter MM, Quadflieg N. Long-term stability of eating disorder diagnoses. Int J Eat Disord. 2007; Nov; 40 Suppl:S61-6.
Grilo CM, Pagano ME, Skodol AE, et al. Natural course of bulimia nervosa and of eating disorder not otherwise specified: 5-year prospective study of remissions, relapses, and the effects of personality disorder. J Clin Psychiatry. 2007; 68:738.
Hilbert A, Tuschen-Caffier B. Maintenance of binge eating through negative mood: a naturalistic comparison of binge eating disorder and bulimia nervosa. Int J Eat Disord. 2007; 40:521.
leGrange D, Crosby RD, Rathous PH, Leventhal BL. A randomized controlled comparison of family-based treatment and supportive psychotherapy for adolescent bulimia nervosa. Arch Gen Psychiatry. 2007; 64:1049.
Mond JM, Hay PJ, Rodgers B, Owen C. Health service utilization for eating disorders: findings from a community-based study. Int J Eat Disord. 2007;40:399.
Shapiro JR, Berkman ND, Brownley KA, et al. Bulimia nervosa treatment: a systematic review of randomized controlled trials. Int J Eat Disord. 2007; 40:321.
Smyth JM, Wonderlich SA, Heron KE, et al. Daily and momentary mood and stress are associated with binge eating and vomiting in bulimia nervosa patients in the natural environment. J Consult Clin Psychol. 2007; 75:629.
Spoor STP, Stice E, Burton E, Bohon C. Relations of bulimic symptom frequency and intensity to psychosocial impairment and health care utilization: results from a community-recruited sample. Int J Eat Disord. 2007;40:505.
Swinbourne JM, Touyz SW. The co-morbidity of eating disorders and anxiety disorders: a review. Eur Eat Disord Rev. 2007; 15:253.
Vanderlinden J, Buis H, Pieters G, Probst M. Which elements in the treatment of eating disorders are necessary “ingredients” in the recovery process?—A comparison between the patient’s and therapist’s view. Eur Eat Disord Rev. 2007;15:537.
Wilson GT. Psychological treatment of eating disorders. Annual Rev Clin Psychol. 2005; 1:439. | 411,247 |
A few people have asked me when I am going to do my next blog recently. And I do admit the regularity of my blogging has dropped off slightly of late. And I guess there are two reasons for this.
Firstly, TIME. It is well documented that the working parent is time poor, splitting themselves between family and work and feeling that they are constantly compromising on one or the other. As the working parent of a child with additional needs, that ramps up a further notch or ten. Marigold now has sooooooo many therapies. What started off as “a bit of physio” has become: physio, occupational therapy, speech and language therapy, hydrotherapy, visits to a paediatric chiropractor, feeding assessments with another therapist to see whether we can get the the root cause of her ongoing chest infections (possible aspiration due to weak muscle tone). Every five or six weeks Marigold will go down with one of the said infections, which leads to more appointments; emergency doctors for medication, the salt caves to try to give her some respiratory relief. There is a constant merry-go-round of the other necessary appointments for her – just in for next month ophthalmology, ENT and a video fluoroscopy. Actually, writing it all down makes me feel slightly panicky. In addition to Marigold’s needs we of course have another little person, our four-year old son Harrison, who has just started school and needs love and attention, someone to play Power Ranger and Super Hero games with him, laugh about toilet humour and help him with his phonics homework. Also: our families (especially devoted grandparents without whom we would simply drown) and friends who are going through their own challenging life events, all people we want to give our time and attention to. Oh yes, and we work of course, 4 days a week for me and full-time for my husband.
So blogging has really slipped down the list. I think having read the above, you will let me off the hook for that.
But there is another reason as well. All of the above eventually just becomes…..the norm. I am now used to trying to squeeze one to two therapies in for Marigold in to an already hectic working week. We have accepted the fact that every six weeks or so, she will have an infection of some sort and we will have a week from hell of no sleep and horrendous anxiety whilst we try to help her get better (those are the hardest times.) This is just life with Marigold.
And it is hard, but rich and beautiful and so blessed. We would not have it any other way.
So I do I find myself wondering what to write about sometimes.
Anyway, you would think on top of all that, it would probably be best not to take anything else on. Yet my husband, he of “the glue that holds us together fame” has decided to run the London Marathon in 2016 for Unique, the Rare Chromosome Disorder Support Group, who have provided us with a huge amount of knowledge and support since Marigold’s diagnosis of 3q11 trisomy mosaicism, a unique genetic condition.
My husband has run many marathons before. Upon completing last one in 2013, his exact words were “I am never, ever doing that again.” He hated it. He was practically sick half way round. All his toenails fell off. Getting up in the freezing cold in the winter to go and do a 12 mile training run is definitely not fun. Even less fun when you have been up all night with small people. But this year he is inspired to run for a cause that means so very much for us, for our beautiful Marigold who tries so very hard to overcome her challenges. I know running it for her and other Unique children everywhere will give him the strength to get through his training and over the finish line on the day, where me, Harrison and Marigold will be cheering him on.
I am so proud that with everything going on in our lives, my husband is going to commit to such a huge task. I had a rare half an hour just now, and rather than sitting down and enjoying a cup of tea and some much-needed head space, I found the inspiration to write this blog. Because if he can do the London Marathon, I can find some time to write a little something, non?
@HBurness
@unique_charity
@LondonMarathon | 292,361 |
Frequently.
The question clients most often ask is:
How can I make the transition during this decision-making process easier?
Here are a few tips about life decisions to keep in mind:
- I quote Mr. Miyagi from Karate Kid, “have balance, and focus“
- Write out your choices… allowing you to empty your mind of emotion and use more logical reasoning when deciding
- Trust your inner voice and stick to your decision. You’ll feel confident you made the decision and stuck to the plan.
These are just a few basic ideas to help, there are usually more variables involved in the process.
But remember, do the things that make yourself proud!
Stay happy and healthy,
Ashley.
Please follow and like us: | 396,260 |
TITLE: Stochastic differential equation with trigonometric functions
QUESTION [0 upvotes]: I heard that the following SDE can be solved analitically by substitution:
$dX(t) = - \left[ \sin (2 X(t) ) + \frac{1}{4} \sin (4 X(t) ) \right] dt + \sqrt{2} \cos^2 X(t) dB(t),$
$X(0) = 1, \; t \in [0, \tau)$
But is isn't as simple as it sounds. Any ideas? Of course $\mathbb{B}$ is a brownian motion.
REPLY [2 votes]: Motivation: Only few stochastic differential equations (SDEs) can be solved explicitely; one of them is a linear SDE with determinstic coefficients, i.e. an SDE of the form
$$dZ_t = (\alpha(t)+\beta(t) Z_t) \, dB_t + (\gamma(t)+\delta(t) Z_t) \, dt. \tag{1}$$
That's why one usually tries to transform SDEs into a linear SDE. For SDEs of the form
$$dX_t = b(X_t) \, dt + \sigma(X_t) \, dB_t, \tag{2}$$
i.e. SDEs where the coefficient do not depend on the time $t$, there are necessary and sufficient criterions for a transformation into a linear SDE (see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Section 19.4, 2nd edition).
Roughly speaking, the only two possibilities two transform an SDE $(2)$ into a linear SDE $(1)$ are the following two substitutions: Set
$$Z_t := f(X_t)$$
where either
$$f(x) :=\gamma \int_0^x \frac{1}{\sigma(y)} \, dy \tag{3} $$
or
$$f(x) := \exp \left( \gamma \cdot \int_0^x \frac{1}{\sigma(y)} \, dy \right) \tag{4}$$
for some suitable constant $\gamma>0$. (In fact, there are criterions in terms of the derivatives when to use which one; but they are quite messy.)
So here we go. The SDE
$$dX_t = - \left(\sin(2X_t) + \frac{1}{4} \sin(4X_t)\right) \, dt + \sqrt{2} \cos^2 X_t \, dB_t \tag{5}$$
is obviously of the form $(5)$ where $\sigma(y) = \sqrt{2} \cos^2(y)$. If we try substitution $(3)$, then
$$Z_t := f(X_t) \stackrel{(3)}{=} \gamma \int_0^{X_t} \frac{1}{\sqrt{2} \cos^2(y)} \, dy = \gamma' \tan X_t$$
where $\gamma' := \frac{\gamma}{\sqrt{2}}$. Applying Itô's formula yields
$$\begin{align*} Z_t-Z_0 &= \gamma' \int_0^t \frac{1}{\cos^2(X_s)} \, dX_s + \gamma \int_0^t \frac{\sin X_s}{\cos^3 X_s} \, d\langle X \rangle_s \\ &\stackrel{(5)}{=} \sqrt{2}\gamma' \int_0^t dB_s - \gamma' \int_0^t \left(\frac{\sin(2X_s)}{\cos^2(X_s)}+ \frac{\sin(4X_s)}{4\cos^2(X_s)} \right) \, ds + 2\gamma' \int_0^t \frac{\sin X_s}{\cos^3 X_s} \cos^4 X_s \, ds. \end{align*}$$
In the last step we used that the quadratic variation $\langle X \rangle_s$ equals, by $(5)$,
$$d \langle X \rangle_t = 2 \cos^4 X_t \, dt.$$
Using the well-known formulas
$$\sin(2x) = 2 \sin x \cdot \cos x \qquad \sin (4x) = 8 \sin x \cdot \cos^3(x) - 4 \sin x \cdot \cos x$$
we get
$$Z_t - Z_0 = \sqrt{2}\gamma' \int_0^t dB_s -\gamma' \int_0^t Z_s \, ds.$$
This means that we have transformed the SDE $(5)$ into a linear SDE of the form $(1)$. This SDE can be solved explicitely, and therefore we find an explicit expression for the solution
$$X_t = \arctan \left(\frac{Z_t}{\gamma'} \right).$$
(Here $\gamma'>0$ is an arbitrary constant; we may choose $\gamma'=1$.) | 140,462 |
Westfield-Mayville Rotary Club Members Deliver Club Donation
Photo Credit to Felicia Lawson
On Aug. 12 two Westfield-Mayville Rotary Club members delivered a club donation of a large supply of paper goods and meal containers to the Westfield Community Kitchen. Pictured (l. to r.) are Rotarians Steve Stratton and Patty Benton with Jean McCausland, Westfield Community Kitchen volunteer. | 270,828 |
So now what?
After the Timberwolves missed out on D’Angelo Russell late Sunday night, it sent a wave of disappointment for those fans who had high hopes while others who may have been more jaded now have the right to say, “I told you so.”
But the Wolves still need to field a roster for next season, and while you can put to rest those dreams of a Russell-Karl-Anthony Towns pick and roll, perhaps Russell’s decision to play for Golden State opens up room for Tyus Jones to come back.
Jones had a meeting with the Wolves via Skype on Sunday and he’s a restricted free agent, meaning the Wolves can match any offer sheet that Jones gets. A lot of guards have already come off the free-agent market, and it’ll be interesting to see where it lands for Jones. It’s hard to pin down where exactly Jones’ value is. He could command an annual salary in the $8 million range, but that could fluctuate up or down.
The Wolves are expected to continue their conversations with Jones and after missing out on Russell, the Wolves are still in need of point guards, especially now that Derrick Rose’s return isn’t an option as he is set to sign a two-year deal with Detroit. Jeff Teague and his expiring $19 million deal are still on the roster for next season.
The Wolves still have a need for power forwards after Taj Gibson signed with the Knicks and the team traded Dario Saric to the Suns in the deal that netted Jarrett Culver with the No. 6 pick in the draft.
As it stands now, Gorgui Dieng is the only natural four on the Wolves roster or else the team will have to play small ball with Robert Covington potentially at that position, something President Gersson Rosas said the team was open to trying.
One name to watch in that market is JaMychal Green, who has played five seasons with the Clippers, Grizzlies and Spurs. He averaged 9.4 points per game in playing 65 games with the Grizzlies and Clippers.
Both Marcus and Markieff Morris are also available.
As of Sunday, there was little contact between Anthony Tolliver and the Wolves about bringing Tolliver back.
><< | 25,197 |
Hi Len,I hope this isn't too late for the .31 merge window. If so, no big deal;this patchset isn't urgent, it's just a nice cleanup that was inspiredfrom investigating the recent bug in acpi_pci_bind() (fixed by dacd254).This patch series eliminates static boot-time binding of ACPIand PCI devices, and introduces an API to perform this lookupduring runtime.This change has the following advantages: - eliminates struct acpi_device vs struct pci_dev lifetime issues - lays groundwork for eliminating .bind/.unbind from acpi_device_ops - lays more groundwork for eliminating .start from acpi_device_ops and thus simplifying ACPI drivers - whacks out a lot of codeThis patchset is based on lenb/test (66c74fa1d4), and has been boottested on ia64 and x86. I also performed physical PCI hotplug testsusing acpiphp on ia64. I do not have an acpiphp-capable x86 platform.I didn't test the changes in the video driver either, as I don't havethe hardware.v1 -> v2 - rearrange series into a more logical order - much simpler acpi_is_root_bridge() implementation - no longer export acpi_pci_find_root() - no longer leak memory in acpi_get_pci_dev() - no longer leak references in acpi_pci_unbind/acpi_pci_bind - convert video driver to use acpi_get_pci_dev() - kill off acpi_get_physical_pci_device() - incorporate Bjorn's other comments. :)---Alex Chiang (11): ACPI: kill acpi_get_physical_pci_device() ACPI: video: convert to acpi_get_pci_dev ACPI: kill acpi_get_pci_id PCI Hotplug: acpiphp: convert to acpi_get_pci_dev ACPI: acpi_pci_unbind should clean up properly after acpi_pci_bind ACPI: simplify acpi_pci_irq_del_prt() API ACPI: simplify acpi_pci_irq_add_prt() API ACPI: eviscerate pci_bind.c ACPI: Introduce acpi_get_pci_dev() ACPI: Introduce acpi_is_root_bridge() ACPI: make acpi_pci_bind() static drivers/acpi/glue.c | 40 ----- drivers/acpi/pci_bind.c | 315 +++++------------------------------- drivers/acpi/pci_irq.c | 17 +- drivers/acpi/pci_root.c | 112 ++++++++++++- drivers/acpi/video.c | 6 - drivers/acpi/video_detect.c | 9 + drivers/pci/hotplug/acpi_pcihp.c | 40 ----- drivers/pci/hotplug/acpiphp_glue.c | 27 +-- include/acpi/acpi_bus.h | 2 include/acpi/acpi_drivers.h | 10 - include/linux/pci_hotplug.h | 1 11 files changed, 179 insertions(+), 400 deletions(-) | 182,525 |
TITLE: The most Efficient Algorithm for Factoring Polynomial Over Finite Field
QUESTION [4 upvotes]: I have a polynomial defined over a finite field $\mathbb{F}_p$, where $p$ is a large prime number (e.g. 256-bit).
The polynomail's degree is big (e.g. at least $10^5$).
My Goal is: To find the roots of the polynomial.
One way to do that is to factorize the polynomial then find the low degree polynomials'roots. My polynomial can be constructed to be a monic polynomial (if this helps to speed up the proceess). Here, how fast I can find the roots matters to me.
Question: What is the most efficient algorithm that factorize the polynomial over a finite field.
I am aware of this paper: http://www.shoup.net/papers/lille.pdf
but I want something faster e.g. $O(n)$
REPLY [5 votes]: Concerning the complexity of factorizations, I believe the best current result is
Kedlaya, Umans:
Fast polynomial factorization and modular composition.
SIAM J. Comput. 40 (2011), no. 6, 1767–1802.
which gives complexity of roughly $(n^{1.5}+n\log q)\log q$ for a field with $q$ elements. | 168,730 |
Divested ventures
Responsible divestments are a key part of transitioning our portfolio to deliver upon our Powering Progress strategy. In 2021, total divestment proceeds were approximately $15 billion. See the business sections in our Annual Report 2021 for details.
We carry out due diligence on potential buyers when divesting parts of our business. We collaborate with both in-house and external experts, where appropriate, to conduct checks and examine key attributes of potential buyers.
Find out more about how we divest at. | 312,703 |
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\chapter{ORTHONORMAL BASIS FUNCTIONS}\label{app:BasisFunctions}
This appendix provides the reader with an elementary understanding of orthogonal basis functions. Any reader interested in this subject may refer to Reference \cite{MultiVarOrthPolyBook} for a more in-depth understanding. In essence, basis functions are for a function space what vectors are for a vector space. In other words, a linear combination of basis functions spans the function space, just as a linear combination of basis vectors spans the vector space. Thus, a linear combination of basis functions is a useful free function choice for optimization problems.
\section{Mathematical Preliminaries}
This section introduces some mathematical preliminaries needed to understand the properties of basis functions, and in particular, the properties of orthogonal basis functions. The content introduced here is designed to give the reader a basic understanding and will only scratch the surface of this field of mathematics. As such, when appropriate, references will be provided so that the reader can delve deeper into these topics if desired. Moreover, this section assumes the reader is familiar with the properties of vector spaces. If the reader is unfamiliar with these topics, then they may consider reading Reference \cite{TriebelFunctionSpacesBook} for function spaces and reviewing the portion of Reference \cite{GilbertStrangLinearAlgebra} dedicated to vector spaces.
This dissertation is primarily concerned with function spaces that can be used to describe continuous, non-infinite functions, as these will be particularly useful for describing the solutions of differential equations: the extended Lebesgue spaces, also known as $L^{pe}$ spaces, are the function spaces that contain these functions. The extended Lebesgue spaces are defined based on a generalization of the $p$-norm used to describe vector spaces. Recall that the $p$-norm for a vector is
\begin{equation*}
||\B{x}||_p = \bigg(\sum_{k=1}^n |x_k|^p\bigg)^{1/p},
\end{equation*}
where $\B{x}\in\mathbb{R}^n$ is an arbitrary vector, $x_k$ are the components of $\B{x}$, and $p \geq 1$. The $p$-norm of functions is,
\begin{equation*}
||f(z)||_p = \bigg( \int_\Omega |f(z)|^p \dd z \bigg)^{1/p},
\end{equation*}
for some arbitrary function $f (z)$ defined on the domain $\Omega$. Note that this $p$-norm can also be defined with a measure $\dd\mu (z)$, in which case the $p$-norm is written as
\begin{equation*}
||f(z)||_p = \bigg(\int_\Omega |f(z)|^p \dd \mu (z) \bigg)^{1/p}.
\end{equation*}
The rigorous mathematical definition of a measure will not be discussed here; the interested reader can refer to Reference \cite{MeasureTheoryBook} for more information. For the material in this dissertation, it is sufficient to note that the measure $\dd\mu(z) = W(z) \dd z$ where $W(z) \geq 0 \, \forall\ z\in\Omega$. The measure for a function is analogous to the weights in a weighted vector norm. An arbitrary function, $f(z)$, defined over the domain $\Omega$ is part of the $L^{pe}(\Omega,\mu)$ space if
\begin{equation*}
||f(z)||_p = \bigg(\int_\Omega |f(z)|^p \dd \mu (z) \bigg)^{1/p} < \infty.
\end{equation*}
This appendix will focus on basis functions in the $L^{2e}$ space, i.e., for $p=2$.
The generalization of the $p$-norm is sufficient for describing which functions are in the $L^{pe}$ space. However, the $p$-norm gives no information about the orthogonality of two functions. For this, an inner product is needed. Fortunately, the $L^{2e}(\Omega,\mu)$ space already comes equipped with an inner product,
\begin{equation*}
\langle f, g \rangle = \int_\Omega f(z) \, g(z) \dd \mu (z),
\end{equation*}
where $f (z)$ and $g (z)$ are arbitrary functions in the $L^{2e}(\Omega,\mu)$ space, and $\langle f,g \rangle$ is used to denote an inner product between these functions. The functions $f$ and $g$ are considered orthogonal if $\langle f, g \rangle = 0$. Just as orthogonal basis vectors can be convenient for describing an arbitrary vector in a vector space, so too are orthogonal basis functions for describing an arbitrary function in a function space\footnote{In addition, choosing orthogonal basis functions can also guarantee certain solution properties.}.
Of course, spanning the entirety of $L^{2e}$ space would require an infinite number of basis functions, as the dimension of the $L^{2e}$ space is infinite. Thus, to make problems computationally tractable, a finite number, $m$, of basis functions is used. In general, as the number $m$ increases, the error between the estimated and actual solution will decrease. Finally, note that the basis set domain need not coincide with the domain of the problem. If a bijective map exists that transforms the basis function domain into the problem domain, then that basis may be used to describe the problem's solution. This notion is used frequently throughout this dissertation.
Based on the description of orthogonal basis function sets thus far, one has two parameters that can be used to describe a basis set for $L^{2e}$:
\begin{enumerate}
\item The domain on which the basis is defined, $\Omega$.
\item The measure used for the basis, $\mu$.
\end{enumerate}
In the following sections, some frequently used orthogonal basis sets will be presented. The presentation will include the domain and measure for each set and recursive generating functions for the set if they exist. The section that follows explains how to extend these basis sets to the multivariate case and concludes with a table that summarizes all the basis functions presented.
\section{Chebyshev Orthogonal Polynomials}
Chebyshev orthogonal polynomials are two sets of basis functions, the first and the second kind. They are usually indicated as $T_k (z)$ and $U_k (z)$, respectively. This section summarizes the main properties of the first kind, $T_k (z)$, only, which are defined on the domain $z\in[-1,+1]$ and with the measure $\dd \mu (z) = \dfrac{1}{\sqrt{1 - z^2}} \dd z$. These polynomials can be generated using the following useful recursive function,\footnote{Note that in this recursive formulation and those that follow, the $z$ argument is dropped for clarity, i.e., $T_k(z)\to T_k$.}
\begin{equation*}
T_{k + 1} = 2 \, z \, T_k - T_{k - 1} \qquad \text{starting from:} \; \begin{cases} T_0 = 1, \\ T_1 = z.\end{cases}
\end{equation*}
Also, all the derivatives of Chebyshev orthogonal polynomials can be computed recursively, starting from
\begin{equation*}
\dfrac{\dd T_0}{\dd z} = 0, \quad \dfrac{\dd T_1}{\dd z} = 1 \qquad \text{or} \qquad \dfrac{\dd^d T_0}{\dd z^d} = \dfrac{\dd^d T_1}{\dd z^d} = 0 \quad \forall \; d > 1,
\end{equation*}
and then using,
\begin{align*}
\dfrac{\dd T_{k+1}}{\dd z} &= 2 \, \left(T_k + z \, \dfrac{\dd T_k}{\dd z}\right) - \dfrac{\dd T_{k-1}}{\dd z} \\
\dfrac{\dd^2 T_{k+1}}{\dd z^2} &= 2 \left(2 \, \dfrac{\dd T_k}{\dd z} + z \, \dfrac{\dd^2 T_k}{\dd z^2}\right) - \dfrac{\dd^2 T_{k-1}}{\dd z^2} \\
&\ \ \vdots \\
\dfrac{\dd^d T_{k+1}}{\dd z^d} &= 2 \left( d \, \dfrac{\dd^{d-1} T_k}{\dd z^{d-1}} + z \, \dfrac{\dd^d T_k}{\dd z^d}\right) - \dfrac{\dd^d T_{k-1}}{\dd z^d} \quad \forall \; d \ge 1.
\end{align*}
for $k \ge 1$.
The integral of $ T_k (z)$ has the following useful property,
\begin{equation*}
\int_{-1}^{+1} T_k (z) \dd z = \left\{\begin{array}{lcl} = 0 & {\rm if} & k = 1 \\ = \dfrac{(-1)^k + 1}{1 - k^2} & {\rm if} & k \ne 1\end{array}\right.
\end{equation*}
while the inner product of two Chebyshev orthogonal polynomials satisfies the orthogonality property,
\begin{equation*}
\langle T_i (z), T_j (z)\rangle = \int_{-1}^{+1} T_i (z) \, T_j (z) \, \dfrac{1}{\sqrt{1 - z^2}} \dd z = \begin{cases} = 0 & {\rm if} \quad i \ne j \\ = \pi & {\rm if} \quad i = j = 0 \\ = \pi/2 & {\rm if} \quad i = j \ne 0\end{cases}.
\end{equation*}
Figure \ref{fig:ICOPfig} shows the first five Chebyshev orthogonal polynomials.
\begin{figure}[ht]
\centering\includegraphics[width=0.9\linewidth]{Figures/BasisFunctions/COP.pdf}
\caption{First five Chebyshev orthogonal polynomials.}
\label{fig:ICOPfig}
\end{figure}
\section{Legendre Orthogonal Polynomials}
The Legendre orthogonal polynomials, $L_k (z)$, are defined on the domain $z\in[-1, +1]$ with measure $\dd \mu (z) = \dd z$. These polynomials can also be generated recursively by,
\begin{equation*}
L_{k+1} = \dfrac{2k+1}{k+1} \, z \, L_k - \dfrac{k}{k+1} \, L_{k-1} \qquad \text{starting with:} \; \begin{cases} L_0 = 1 \\ L_1 = z.\end{cases}
\end{equation*}
All derivatives of Legendre orthogonal polynomials can be computed in a recursive way, starting from,
\begin{equation*}
\dfrac{\dd L_0}{\dd z} = 0, \quad \dfrac{\dd L_1}{\dd z} = 1 \qquad \text{or} \qquad \dfrac{\dd^d L_0}{\dd z^d} = \dfrac{\dd^d L_1}{\dd z^d} = 0 \quad \forall \; d > 1,
\end{equation*}
and continuing with,
\begin{align*}
\dfrac{\dd L_{k+1}}{\dd z} &= \dfrac{2k+1}{k+1} \left(L_k + z \dfrac{\dd L_k}{\dd z}\right) - \dfrac{k}{k+1} \dfrac{\dd L_{k-1}}{\dd z} \\
\dfrac{\dd^2 L_{k+1}}{\dd z^2} &= \dfrac{2k+1}{k+1} \left(2\dfrac{\dd L_k}{\dd z} + z \dfrac{\dd^2 L_k}{\dd z^2}\right) - \dfrac{k}{k+1} \dfrac{\dd^2 L_{k-1}}{\dd z^2} \\
& \ \ \vdots \\
\dfrac{\dd^d L_{k+1}}{\dd z^d} &= \dfrac{2k+1}{k+1} \left(d\dfrac{\dd^{d-1} L_k}{\dd z^{d-1}} + z \dfrac{\dd^d L_k}{\dd z^d}\right) - \dfrac{k}{k+1} \dfrac{\dd^d L_{k-1}}{\dd z^d} \quad \forall \; d \ge 1,
\end{align*}
for $k \ge 1$. In addition, the inner products of the Legendre polynomials highlight their orthogonality,
\begin{equation*}
\langle L_i (z), L_j (z)\rangle = \int_{-1}^{+1} L_i (z) \, L_j (z) \, \dd z = \dfrac{2}{2 i + 1} \, \delta_{ij}.
\end{equation*}
Figure \ref{fig:I_LeP} shows the first five Legendre orthogonal Polynomials.
\begin{figure}[ht]
\centering\includegraphics[width=0.9\linewidth]{Figures/BasisFunctions/LeP.pdf}
\caption{First five Legendre orthogonal polynomials.}
\label{fig:I_LeP}
\end{figure}
\section{Laguerre Orthogonal Polynomials}
Laguerre orthogonal polynomials, $L_k (z)$, are defined on the domain $[0,\infty)$ and by the measure $\dd\mu (z) = e^{-z} \dd z$. They are generated using the recursive function,
\begin{equation*}
L_{k+1} = \dfrac{2k + 1 - z}{k + 1} \, L_k - \dfrac{k}{k + 1} \, L_{k-1} \qquad \text{starting with:} \; \begin{cases} L_0 =& 1, \\ L_1 =& 1 - z.\end{cases}
\end{equation*}
All derivatives of Laguerre orthogonal polynomials can be computed recursively, starting from
\begin{equation*}
\dfrac{\dd L_0}{\dd z} = 0, \quad \dfrac{\dd L_1}{\dd z} =-1 \qquad \text{or} \qquad \dfrac{\dd^d L_0}{\dd z^d} = \dfrac{\dd^d L_1}{\dd z^d} = 0 \quad \forall \; d > 1,
\end{equation*}
then using
\begin{align*}
\dfrac{\dd L_{k+1}}{\dd z} &= \dfrac{2k + 1 - z}{k + 1} \dfrac{\dd L_k}{\dd z} - \dfrac{1}{k + 1} L_k - \dfrac{k}{k + 1} \dfrac{\dd L_{k-1}}{\dd z} \\
\dfrac{\dd^2 L_{k+1}}{\dd z^2} &= \dfrac{2k + 1 - z}{k + 1} \dfrac{\dd^2 L_k}{\dd z^2} - \dfrac{2}{k + 1} \dfrac{\dd L_k}{\dd z} - \dfrac{k}{k + 1} \dfrac{\dd^2 L_{k-1}}{\dd z^2} \\
& \ \ \vdots \\
\dfrac{\dd^d L_{k+1}}{\dd z^d} &= \dfrac{2k + 1 - z}{k + 1} \dfrac{\dd^d L_k}{\dd z^d} - \dfrac{d}{k + 1} \dfrac{\dd^{d-1} L_k}{\dd z^{d-1}} - \dfrac{k}{k + 1} \dfrac{\dd^d L_{k-1}}{\dd z^d} \quad \forall \; d \ge 1,
\end{align*}
for $k \geq 1$.
Figure \ref{fig:I_LaP} shows the first five Laguerre orthogonal Polynomials.
\begin{figure}[ht]
\centering\includegraphics[width=0.9\linewidth]{Figures/BasisFunctions/LaP.pdf}
\caption{First five Laguerre orthogonal polynomials.}
\label{fig:I_LaP}
\end{figure}
\section{Hermite Orthogonal Polynomials}
There are two Hermite orthogonal polynomials, the probabilists, indicated by $E_k (z)$, and the physicists, indicated by $H_k (z)$. The probabilists are defined on the domain $z\in(-\infty,\infty)$ and with the measure $\dd \mu(z) = e^{-(z^2/2)} \dd z$, and the physicists are defined on the domain $z\in(-\infty,\infty)$ and with the measure $\dd\mu(z) = e^{-z^2}\dd z$. They are both generated using recursive functions.
The probabilists' polynomials can be defined recursively by,
\begin{equation*}
E_{k + 1} = z \, E_k - k E_{k - 1} \qquad \text{starting with:} \; \begin{cases} E_0 =& 1 \\ E_1 =& z.\end{cases}
\end{equation*}
All derivatives can be computed recursively, starting from
\begin{equation*}
\dfrac{\dd E_0}{\dd z} = 0, \quad \dfrac{\dd E_1}{\dd z} = 1 \qquad \text{or} \qquad \dfrac{\dd^d E_0}{\dd z^d} = \dfrac{\dd^d E_1}{\dd z^d} = 0 \quad \forall \; d > 1,
\end{equation*}
then using,
\begin{align*}
\dfrac{\dd E_{k+1}}{\dd z} &= E_k + z \dfrac{\dd E_k}{\dd z} - k \dfrac{\dd E_{k-1}}{\dd z} \\
\dfrac{\dd^2 E_{k+1}}{\dd z^2} &= 2 \dfrac{\dd E_k}{\dd z} + z \dfrac{\dd^2 E_k}{\dd z^2} - k \dfrac{\dd^2 E_{k-1}}{\dd z^2} \\
& \ \ \vdots \\
\dfrac{\dd^d E_{k+1}}{\dd z^d} &= d \dfrac{\dd^{d-1} E_k}{\dd z^{d-1}} + z \dfrac{\dd^d E_k}{\dd z^d} - k \dfrac{\dd^d E_{k-1}}{\dd z^d} \quad \forall \; d \ge 1,
\end{align*}
for $k \geq 1$.
The physicists' polynomials can be defined by the recursive relationship,
\begin{equation*}
H_{k + 1} = 2 z \, H_k - 2 k \, H_{k - 1} \qquad \text{starting with:} \; \begin{cases} H_0 =& 1 \\ H_1 =& 2z.\end{cases}
\end{equation*}
All derivatives can be computed recursively, starting from
\begin{equation*}
\dfrac{\dd H_0}{\dd z} = 0, \quad \dfrac{\dd H_1}{\dd z} = 2 \qquad \text{or} \qquad \dfrac{\dd^d H_0}{\dd z^d} = \dfrac{\dd^d H_1}{\dd z^d} = 0 \quad \forall \; d > 1,
\end{equation*}
then using,
\begin{align*}
\dfrac{\dd H_{k+1}}{\dd z} &= 2 H_k + 2 z \dfrac{\dd H_k}{\dd z} - 2k \dfrac{\dd H_{k-1}}{\dd z} \\
\dfrac{\dd^2 H_{k+1}}{\dd z^2} &= 4 \dfrac{\dd H_k}{\dd z} + 2 z \dfrac{\dd^2 H_k}{\dd z^2} - 2k \dfrac{\dd^2 H_{k-1}}{\dd z^2} \\
& \ \ \vdots \\
\dfrac{\dd^d H_{k+1}}{\dd z^d} &= 2 d \dfrac{\dd^{d-1} H_k}{\dd z^{d-1}} + 2 z \dfrac{\dd^d H_k}{\dd z^d} - 2 k \dfrac{\dd^d H_{k-1}}{\dd z^d} \quad \forall \; d \ge 1,
\end{align*}
for $k \geq 1$.
Figure \ref{fig:I_HoP} shows the first five probabilists' and physicists' Hermite orthogonal polynomials.
\begin{figure}[ht]
\centering\includegraphics[width=\linewidth]{Figures/BasisFunctions/HoP.pdf}
\caption{First five Hermite orthogonal polynomials.}
\label{fig:I_HoP}
\end{figure}
\section{Fourier Basis}
The Fourier basis is defined on the domain $z\in[-\pi,\pi]$ and with the measure $\dd \mu (z) = \dd z$. The basis does not have a recursive generating function. Rather, the basis can be mathematically written as,
\begin{equation*}
g_k(z) = \begin{cases}1, &k=0\\\cos(\ceil{k/2}z), &\text{$k$ is even}\\\sin(\ceil{k/2}z), &\text{$k$ is odd}\end{cases}
\end{equation*}
where $\ceil{x}$ rounds $x$ to the next largest integer and $k=0,\dots,m$. There is no recursive relationship to compute the subsequent derivatives of Fourier bases. However, the $n$-th derivative can be computed using,
\begin{equation*}
\dfrac{\dd^d g (z)}{\dd z^d} = \begin{cases} \begin{cases}0, &k=0\\\ceil{k/2}^d\cos(\ceil{k/2}z), &\text{$k$ is even}\\\ceil{k/2}^d\sin(\ceil{k/2}z), &\text{$k$ is odd}\end{cases} &\mod(d,4) = 0 \\ \begin{cases}0, &k=0\\-\ceil{k/2}^d\sin(\ceil{k/2}z), &\text{$k$ is even}\\\ceil{k/2}^d\cos(\ceil{k/2}z), &\text{$k$ is odd}\end{cases} &\mod(d,4) = 1 \\ \begin{cases}0, &k=0\\-\ceil{k/2}^d\cos(\ceil{k/2}z), &\text{$k$ is even}\\-\ceil{k/2}^d\sin(\ceil{k/2}z), &\text{$k$ is odd}\end{cases} &\mod(d,4) = 2 \\ \begin{cases}0, &k=0\\\ceil{k/2}^d\sin(\ceil{k/2}z), &\text{$k$ is even}\\-\ceil{k/2}^d\cos(\ceil{k/2}z), &\text{$k$ is odd}\end{cases} &\mod(d,4) = 3\end{cases}
\end{equation*}
whenever $d > 0$. Figure \ref{fig:I_FS} shows the first five Fourier basis functions.
\begin{figure}[ht]
\centering\includegraphics[width=0.9\linewidth]{Figures/BasisFunctions/FS.pdf}
\caption{First five Fourier basis functions.}
\label{fig:I_FS}
\end{figure}
\section{Extension to Multivariate Domains}
In general, multivariate orthogonal basis sets can be created by taking all possible products of functions in the basis sets that make up the individual variables. The measure that makes up this new basis set will be the product of measures of the individual basis sets, and the domain of the multivariate basis set will be the union of the domains that make up the individual basis sets. More details and insights on the 2-dimensional and $n$-dimensional orthogonal basis functions are contained in Reference \cite{Ye} and References \cite{MultiVarOrthPolyBook, Xu}, respectively.
Consider $n$ independent variables in the vector $\B{x} = \{x_1, x_2, \cdots, x_n\}\T$. Moreover, let the orthogonal basis set for each of these independent variables be denoted by $\p{k}{B}_j$, where the subscript $j$ denotes the $j$-th basis function and the pre-superscript $k$ denotes the $k$-th independent variable. For example, the third basis function for $x_2$ would be $\p{2}{B}_3$. The measure of the multivariate basis set will be denoted by $\mu (\B{x}) = \ds\prod_{k=1}^n \p{k}{\mu}(x_k)$ where $\p{k}{\mu}(x_k)$ is the measure for the $k$-th independent variable. The domain of the multivariate basis will be denoted by $\Omega = \p{1}{\Omega} \times \p{2}{\Omega} \times \cdots \times \p{n}{\Omega}$, where the generic $\p{k}{\Omega}$ denotes the domain of the $k$-th basis set. Then, an arbitrary basis function for the multivariate domain can be written as,
\begin{equation}\label{eq:nDbasisAsTensorProduct}
\mathcal{B}_{i_1i_2\dots i_n} = \p{1}{B}_{i_1} \p{2}{B}_{i_2} \cdots \p{n}{B}_{i_n},
\end{equation}
where $i_1,\cdots ,i_n \in \mathbb{Z^+}$. In other words, Equation \eqref{eq:nDbasisAsTensorProduct} generates a multivariate basis via a tensor product of univariate basis functions \cite{nDbasisFunctions}. If one were to use all possible products of the functions in the individual basis sets which span $L^{2e}(\p{k}{\Omega},\p{k}{\mu})$, i.e., use all possible combinations of $i_1,\cdots ,i_n \in \mathbb{Z^+}$, an infinite set, then the resulting multivariate basis would span the multivariate function space $L^{2e}(\Omega,\mu)$. Of course, in practice this is not possible, so a finite number of basis functions from the set is used.
Consider the inner product of two different basis functions $\mathcal{B}_{i_1\dots i_n}$ and $\mathcal{B}_{j_1\dots j_n}$ where at least one $i_k\neq j_k$,
\begin{equation}\label{eq:MultiVarInnerProd}
\langle \mathcal{B}_{i_1\dots i_n}, \mathcal{B}_{j_1\dots j_n} \rangle = \int_\Omega \mathcal{B}_{i_1\dots i_n} \, \mathcal{B}_{j_1\dots j_n} \dd\mu = \ds\prod_{k=1}^n \int_{\Omega_k} \p{k}{B}_{i_k} \, \p{k}{B}_{j_k} \dd\mu_k.
\end{equation}
Since these are different basis functions, there must be some $k=\kappa$ such that $i_{\kappa} \neq j_{\kappa}$. For $k=\kappa$, the integral
\begin{equation*}
\int_{\Omega_\kappa} \p{\kappa}{B}_{i_{\kappa}} \p{k}{B}_{j_{\kappa}} \ \dd \mu_\kappa = 0,
\end{equation*}
and thus, the product of integrals in Equation \eqref{eq:MultiVarInnerProd} is equal to zero. It follows that,
\begin{equation*}
\langle \mathcal{B}_{i_1\dots i_n}, \mathcal{B}_{j_1\dots j_n} \rangle = 0 \quad \text{if} \quad \exists\ \kappa \mid i_\kappa \neq j_\kappa.
\end{equation*}
Hence, the resulting multivariate basis set is orthogonal.
Just as in the univariate case, the problem being solved must be made tractable by choosing basis functions up to some finite degree $m$. All the multivariate basis functions of order $m$ are defined by choosing $i_1, \cdots, i_n$ to be on the set,
\begin{equation*}
\{\B{i} \mid i_k\in\mathbb{Z}^{+}, \sum_{k=1}^n (i_k-1) = m\},
\end{equation*}
where $i_k$ denotes the elements of $\B{i}$.
Table \ref{tab:BasisSummary} summarizes the orthogonal basis sets described in this section.
\begin{table}[!ht]
\centering
\caption{Univariate orthogonal basis functions summary.}
\begin{tabular}{|l|c|c|}
\hline
Basis function name & Domain, $\Omega$ & Measure, $\dd \mu (z)$ \\
\hline\hline
Chebyshev polynomials & $[-1,1]$ & $\dfrac{1}{1 - z^2} \dd z$ \\
Legendre polynomials & $[-1,1]$ & $\dd z$ \\
Laguerre polynomials & $[0,\infty)$ & $e^{-z} \dd z$ \\
Hermite probabilists polynomials & $(-\infty,\infty)$ & $e^{-(z^2/2)} \dd z$ \\
Hermite physicists polynomials & $(-\infty,\infty)$ & $e^{-z^2} \dd z$ \\
Fourier series & $[-\pi,\pi]$ & $\dd z$ \\
\hline
\end{tabular}
\label{tab:BasisSummary}
\end{table} | 35,828 |
- Grammy-Nominated AEMM Almunus Visits
- Experiential Learning: Talent Agency Practicum
- Fall Hokin Exhibition: ¡VIVA LA SOUL POWER!
- Faculty Spotlight: Jason Stephens
- Fall 2013 Haus Season presented by Club Management Class
- Nikki on Faculty Spotlight: Joe Bogdan
- 2013 Season for Haus @ The Quincy Wong Center
Announcing.
Posted on 25.02.2013
Post by aemm
Categories: Arts Management, Columbia College 0 Comments | 143,878 |
\section{Chu-Vandermonde Identity}
Tags: Binomial Coefficients, Chu-Vandermonde Identity
\begin{theorem}
:$\displaystyle \sum_k \binom r k \binom s {n - k} = \binom {r + s} n$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sum_n \binom {r + s} n x^n
| r = \paren {1 + x}^{r + s}
| c = [[Binomial Theorem]]
}}
{{eqn | r = \paren {1 + x}^r \paren {1 + x}^s
| c = [[Exponent Combination Laws]]
}}
{{eqn | r = \sum_k \binom r k x^k \sum_m \binom s m x^m
| c = [[Binomial Theorem]]
}}
{{eqn | r = \sum_k \binom r k x^k \sum_{n - k} \binom s {n - k} x^{n - k}
| c =
}}
{{eqn | r = \sum_n \paren {\sum_k \binom r k \binom s {n - k} } x^n
| c =
}}
{{end-eqn}}
As this has to be true for all $x$, we have that:
:$\displaystyle \binom {r + s} n = \sum_k \binom r k \binom s {n - k}$
{{qed}}
\end{proof}
\begin{proof}
This is a special case of [[Gauss's Hypergeometric Theorem]]:
:${}_2F_1 \left({a, b; c; 1}\right) = \dfrac{\Gamma \left({c}\right) \Gamma \left({c - a - b}\right)} {\Gamma \left({c - a}\right) \Gamma \left({c - b}\right)}$
where:
:${}_2F_1$ is the [[Definition:Hypergeometric Series|hypergeometric series]]
:$\Gamma \left({n + 1}\right) = n!$ is the [[Definition:Gamma Function|Gamma function]].
One regains the [[Chu-Vandermonde Identity]] by taking $a = -n$ and applying [[Negated Upper Index of Binomial Coefficient]]:
:$\dbinom n k = (-1)^k \dbinom {k - n - 1} k$
throughout.
{{qed}}
\end{proof}
\begin{proof}
From [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk|Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$]]:
{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk}}
where $r, s, t \in \R, n \in \Z$.
Setting $t = 0$:
:$\displaystyle \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$
which is the result required.
{{qed}}
\end{proof}
| 191,261 |
Syntax checker for python using inotify
“Because syntax checking should happen on file writes, not VCS commits”
quick (or not so quick) and dirty tool to run syntax checking and doc tests on python code in a separate screen window-bang/*/
Docstring checking imports your modules, if you run your code in a virtualenv with libs not avalible to the system then import errors may occur. to prevent this install and run icheck from your virtualenv
Bug Fixes
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Download the file for your platform. If you're not sure which to choose, learn more about installing packages. | 30,130 |
TITLE: Analogy between the exterior power and the power set
QUESTION [26 upvotes]: The symmetric algebra of an object exists in every cocomplete $\otimes$-category. For the category of sets $\mathrm{Sym}(X)$ is the set of multi-subsets of $X$.
The usual definition of the exterior power works in every cocomplete linear $\otimes$-category in which $2$ is invertible. But what about the non-linear case? Are there also "exterior powers" in $\otimes$-categories which are not linear? Of course the usual definition using alternating maps does not work. But isn't it striking that for the cartesian category of sets there is a quite natural candidate, namely the power set? Here are some analogies (here $P(X)$ denotes the power set of $X$ if $X$ is finite; in general it is the set of all finite subsets of $X$; $P_n(X)$ is the set of all subsets of $X$ with $n$ elements):
$P(X) = \coprod_n P_n(X)$ and $\Lambda(M) = \oplus_n \Lambda^n(M)$
$P(X \sqcup Y) = P(X) \times P(Y)$ and $\Lambda(M \oplus N) = \Lambda(M) \otimes \Lambda(N)$
It follows the "categorified Vandermonde identity":
$P_n(X \sqcup Y) = \coprod_{p+q=n} P_q(X) \times P_q(Y)$ and $\Lambda^n(M \oplus N) = \oplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$
$(P(X),\cup)$ is a commutative monoid and $(\Lambda(M),\wedge)$ is a graded-commutative algebra, i.e. commutative monoid object in the tensor category of graded modules equipped with with twisted symmetry
If $M$ is free with (ordered) basis $X$, then $\Lambda(M)$ is free with basis $P(X)$, and $\Lambda^n(M)$ is free with basis $P_n(X)$. In particular, $\dim \Lambda^n(M)=\dim P_n(X)$.
If $T$ is a commutative monoid, then homomorphisms $P(X) \to T$ correspond to maps $f : X \to T$ with $f(x)^2=f(x)$, and if $A$ is a graded-commutative algebra, then homomorphisms $\Lambda(M) \to A$ correspond to homomorphisms of modules $f : M \to A_1$ with $f(x)^2=0$ or rather $f(x)f(y)+f(y)f(x)=0$ in the context of $\otimes$-categories (so these conditions are not the same, but both use $f(x)^2$).
Therefore I would like to ask: Is there a notion of exterior algebra for certain cocomplete $\otimes$-categories, including categories of modules and the category of sets? In the latter case, do we get the power set?
REPLY [1 votes]: To a set $X$ associate the free vector space $M(X)$ over $X$; conversely, for a vector space $M$ let $X$ be the index set of a basis of $M$. Then the analogy is just how one does exterior algebra in terms of a basis.
This fits into the way how representation theory for $GL(n)$ and for the symmetric group $S(n)$
are related to each other, both using Young projectors in interated tensor products of $\mathbb C^n$. This becomes more striking even if we take the direct limit for $n\to \infty$.
See books and papers by Yuri Neretin (in arXiv).
Edit: For modules $M$ over an algebra $A$, one could consider the corresponding algebra of dual numbers $A\circledS M$ (i.e., $A\oplus M$ with multiplication
$(a,m).(a',m') = (a.a', a.m' + m.a')$ and the Kaehler differentials over this algebra.
See 2.3 of here. | 139,543 |
Do you feel like you have a pit in your stomach? When you turn on the TV do you feel confused, maybe raged, and possibly disgusted. Do you have no clue what to believe anymore? One person did this and another did that, and you don’t have the secret powers to decipher fact from fiction. Maybe you are exhausted to the point that you have given up. Do you feel like the world will be what it is and there is nothing you can do about it?
Maybe you feel like you can’t trust the media. Maybe you just wish that every third status in your social media feed was not negative news related. Wouldn’t it be nice to be able to see through the fog that lies between propaganda and reality?
If you have read this far, I’m guessing the presidential election has crossed your mind. But in reality, this is not just about the election. It’s about the new norm? This new norm has us all waiting on our heels for the next “worst” thing. It’s like we are in the middle of a thriller action movie, or better yet we are stuck in a movie such as Ground Hog’s Day.
Is there a bigger picture that we are missing? Has the shadow (no pun intended from previous paragraph) become so dark that we have lost hope in the light? Has the negative in this world brought you to the acceptance of this new norm?
Let me try for a moment to bring back the light. Don’t allow the pit in your stomach to remain there. You know why? It’s simple…God’s Got This! Bigger than a negative news story and certainly bigger than the election itself is the plans that God has for us. He has our back. He tells us in 1 Peter 5:7 when he says “cast all your anxiety on him because he cares for you.”
Does this mean that you can breathe easy now that you don’t have to carry your anxiety? It’s not that easy is it? The new norm has triple stamped its mark on our livelihood. We must battle this with our faith. We must allow prayer to trump (no pun intended again) politics. We must allow love to trump hate. We must allow positives to illuminate the dark negative shadows. We must say to ourselves…”God’s got this.”
In doing so we need continue to see the good that this world has to offer. We need to share good news with each other. We need to stop putting ourselves down and we certainly need to stop putting others down. We need to stop judging people. God tells us in Ecclesiastes 12:14 he is the one that will bring every act to judgement and he will decide between good and evil.
You will make a judgement in November, but it’s not the final judgement. Let’s have faith in God’s plan and let’s have faith along that journey. If you are struggling with this election or if you are fighting your own personal battles, tell yourself over and over again that God’s got this. Give your fear and your anxiety to him. Rest and be comforted that your daily struggles and the negatives you see on the news is minuscule compared to the master plan that has been given to you.
Vote God today, in November, and every day of your life. Prayer over Politics and Positive over Negative. Good over Evil and Love over Hate. No worries my friends because God’s got this!
I promise you. God’s Got This. No more worries. No more fear. When the world has you down simply yield to God. | 109,430 |
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