text:
\begin{document}

\title{SOME IDENTITIES AND INEQUALITIES FOR G-FUSION FRAMES}

\author[R. Zarghami Farfar]{Ramazan Zarghami Farfar}
\address{Dapartement of Geomatic and Mathematical\\Marand Faculty of Technical and Engineering\\University of Tabriz\\, Iran\\}
\email{zarghamir@gmail.com}

\author[V. Sadri]{Vahid Sadri}
\address{Department of Mathematics, Faculty of Tabriz  Branch\\ Technical and Vocational University (TUV), East Azarbaijan
, Iran}
\email{vahidsadri57@gmail.com}

\author[R. Ahmadi]{Reza Ahmadi}
\address{Institute of Fundamental Sciences\\University of Tabriz\\, Iran\\}
\email{rahmadi@tabrizu.ac.ir}

\begin{abstract}
G-fusion frames which are obtained from the combination of g-frames and fusion frames were recently introduced for Hilbert spaces. In this paper, we present a new identity for g-frames which was given by Najati in a special case. Also, by using the idea of this identity and their dual frames, some equlities and inequalities will be presented for g-fusion frames.
\end{abstract}

\subjclass[2010]{Primary 42C15; Secondary 46C99, 41A58}

\keywords{g-frame, dual g-frame, g-fusion frame, dual g-fusion frame.}

\maketitle

\section{Introduction}
Recent developments in frame theory and their applications are the result of some mathematician's efforts in this topic (see \cite{ds}, \cite{ga}, \cite{fe}, \cite{aklr}, \cite{ck} and \cite{ch}). By more than half a century, this theory has a interesting applications in different branches of science such as the filter bank theory, signal and image processing, wireless communications, atomic systems and Kadison-Singer problem. In 2005, Balan, Casazza and others  were able to find some useful identities in the frames and fusion frames by studying the properties of the Parseval frames \cite{balan}.
In \cite{sad}, a special kind of frames is introduced which are called g-fusion frames that are the combination of g-frames and fusion frames. We  present those identities in these frames.

\section{Preliminaries}
Throughout this paper, $H$ and $K$ are separable Hilbert spaces, $\pi_{V}$ is the orthogonal projection from $H$ onto a closed subspace $V\subset H$ and $\mathcal{B}(H,K)$ is the collection of all the bounded linear operators of $H$ into $K$. If $K=H$, then $\mathcal{B}(H,H)$ will be denoted by $\mathcal{B}(H)$.
Also, $\lbrace H_j\rbrace_{j\in\Bbb J}$ is a sequence of Hilbert spaces  and $\Lambda_j\in\mathcal{B}(H,H_j)$ for each $j\in\Bbb J$ where $\Bbb J$ is a subset of $\Bbb Z$.
The following lemmas  from Operator Theory are useful.
\begin{lemma}(\cite{ga})\label{l0}
Let $V\subseteq H$ be a closed subspace, and $T$ be a linear  bounded operator on $H$. Then
$$\pi_{V}T^*=\pi_{V}T^* \pi_{\overline{TV}}.$$
\end{lemma}
\begin{lemma}\label{l1}
Let $u\in\mathcal{B}(H)$ be adjoint and $v:=au^2+bu+c$ where $a,b,c\in\Bbb R$.
\begin{enumerate}
\item[(I)] If $a>0$, then
$$\inf_{\Vert f\Vert=1}\langle vf, f\rangle\geq\frac{4ac-b^2}{4a}.$$
\item[(II)] If $a<0$, then
$$\sup_{\Vert f\Vert=1}\langle vf, f\rangle\leq\frac{4ac-b^2}{4a}.$$
\end{enumerate}
\end{lemma}
\begin{lemma}\label{l2}
If $u,v$ are operators on $H$ satisfying $u+v=id_{H}$, then $u-v=u^2-v^2$.
\end{lemma}
\begin{proof}
We can write
$$u-v=2u-id_{H}=u^2-(id_H-u)^2=u^2-v^2.$$
\end{proof}
 We define the space $\mathscr{H}_2:=(\sum_{j\in\Bbb J}\oplus H_j)_{\ell_2}$ by
\begin{eqnarray*}
\mathscr{H}_2=\big\lbrace \lbrace f_j\rbrace_{j\in\Bbb J} \ : \ f_j\in H_j , \ \sum_{j\in\Bbb J}\Vert f_j\Vert^2<\infty\big\rbrace,
\end{eqnarray*}
with the inner product defined by
$$\langle \lbrace f_j\rbrace, \lbrace g_j\rbrace\rangle=\sum_{j\in\Bbb J}\langle f_j, g_j\rangle.$$
It is clear that $\mathscr{H}_2$ is a Hilbert space with pointwise operations.
\begin{definition}
We call the sequence $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$ a g-frame for $H$ with respect to $\{H_j\}_{j\in\Bbb J}$ if there exist $0<A\leq B<\infty$ such that for each $f\in H$
\begin{eqnarray}\label{g}
A\Vert f\Vert^2\leq\sum_{j\in\Bbb J}\Vert \Lambda_j f\Vert^2\leq B\Vert f\Vert^2.
\end{eqnarray}
\end{definition}
If $A=B=1$, we call $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$ a Parseval g-frame.
The synthesis and analysis operators in  g-frames are defined by
\begin{align*}
T_{\Lambda}&:\mathscr{H}_2\longrightarrow H \ \ \ \ \ \ \ \ \ , \ \ \ \ \ \ T_{\Lambda}^*:H\longrightarrow\mathscr{H}_2\\
T_{\Lambda}(\lbrace f_j\rbrace_{j\in\Bbb J})&=\sum_{j\in\Bbb J} \Lambda_{j}^{*}f_j \ \ \ , \ \ \ \ T_{\Lambda}^*(f)=\lbrace \Lambda_j f\rbrace_{j\in\Bbb J}.
\end{align*}
Therefore, the g-frame operator is defined by
$$Sf=TT^*f=\sum_{j\in\Bbb J}\Lambda^*_j \Lambda_j f.$$
The operator $S$ is a bounded, positive and invertible.
If $\tilde{\Lambda}_j:=\Lambda_j S^{-1}$, then $\{\tilde{\Lambda}_j\}_{j\in\Bbb J}$ is called a (canonical) dual g-frame of $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$ and we can write\begin{eqnarray}\label{g1}
f=\sum_{j\in\Bbb J}\tilde{\Lambda}^{\ast}_{j}\Lambda_j f=\sum_{j\in\Bbb J}\Lambda^{\ast}_j\tilde{\Lambda}_j f.
\end{eqnarray}
If $\{\Lambda_j\}_{j\in\Bbb J}$ is a g-frame for $H$ with bounds $A$ and $B$, respectively, then $\{\tilde{\Lambda}_j\}_{j\in\Bbb J}$ is a g-frame for $H$ too with bounds $B^{-1}$ and $A^{-1}$, respectively (for more details, see \cite{sun}).

\begin{definition}
Let $W=\lbrace W_j\rbrace_{j\in\Bbb J}$ be a family of closed subspaces of $H$, $\lbrace v_j\rbrace_{j\in\Bbb J}$ be a family of weights, i.e. $v_j>0$. We say $\Lambda:=(W_j, \Lambda_j, v_j)$ is a  g-fusion frame for $H$ if there exists $0<A\leq B<\infty$ such that for each $f\in H$
\begin{eqnarray}\label{gf}
A\Vert f\Vert^2\leq\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2\leq B\Vert f\Vert^2.
\end{eqnarray}
\end{definition}
We call $\Lambda$ a Parseval g-fusion frame if $A=B=1$. When the right hand of (\ref{gf}) holds, $\Lambda$ is called a g-fusion Bessel sequence for $H$ with bound $B$. Throughout this paper, $\Lambda$ will be a triple $(W_j, \Lambda_j, v_j)$ with $j\in\Bbb J$.

The synthesis and analysis operators in the g-fusion frames are defined by (for more details, see \cite{sad})
\begin{align*}
T_{\Lambda}&:\mathscr{H}_2\longrightarrow H \ \ \ \ \ \ \ \ \ , \ \ \ \ \ \ T_{\Lambda}^*:H\longrightarrow\mathscr{H}_2\\
T_{\Lambda}(\lbrace f_j\rbrace_{j\in\Bbb J})&=\sum_{j\in\Bbb J}v_j \pi_{W_j}\Lambda_{j}^{*}f_j \ \ \ , \ \ \ \ T_{\Lambda}^*(f)=\lbrace v_j \Lambda_j \pi_{W_j}f\rbrace_{j\in\Bbb J}.
\end{align*}
Thus, the g-fusion frame operator is given by
$$S_{\Lambda}f=T_{\Lambda}T^*_{\Lambda}f=\sum_{j\in\Bbb J}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f.$$
Therefore
$$A\  id_{H}\leq S_{\Lambda}\leq B\  id_H.$$
This means that $S_{\Lambda}$ is a bounded, positive and invertible operator (with adjoint inverse) and we have
$$B^{-1}id_{H}\leq S_{\Lambda}^{-1}\leq A^{-1}id_H.$$
 So, we have the folloeing reconstruction formula for any $f\in H$
\begin{equation}\label{gf1}
f=\sum_{j\in\Bbb J}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}S^{-1}_{\Lambda}f
=\sum_{j\in\Bbb J}v_j^2 S^{-1}_{\Lambda}\pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f.
\end{equation}
Let $\tilde{\Lambda}:=(S^{-1}_{\Lambda}W_j, \Lambda_j \pi_{W_j}S_{\Lambda}^{-1}, v_j)$. Then, $\tilde{\Lambda}$ is called the \textit{(canonical) dual g-fusion frame}  of $\Lambda$. Hence, for each $f\in H$, we get (see \cite{sad})
\begin{align}\label{gfd}
f=\sum_{j\in\Bbb J}v_j^2\pi_{W_j}\Lambda^*_j\tilde{\Lambda_j}\pi_{\tilde{W_j}}f=
\sum_{j\in\Bbb J}v_j^2\pi_{\tilde{W_j}}\tilde{\Lambda_j}^*\Lambda_j\pi_{W_j}f,
\end{align}
where $\tilde{W_j}:=S^{-1}_{\Lambda}W_j  \ , \ \tilde{\Lambda_j}:=\Lambda_j \pi_{W_j}S_{\Lambda}^{-1}.$
Thus, we obtain
\begin{align}\label{inverse}
\langle S^{-1}_{\Lambda}f, f\rangle=\sum_{j\in\Bbb J}v_j^2\Vert\tilde{\Lambda}_j\pi_{\tilde{W}_j}f\Vert^2.
\end{align}

\section{The Main Results}
 Let $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$ be a g-frame for $H$ with respect to $\{H_j\}_{j\in\Bbb J}$ with boinds $A, B$ and $\{\tilde{\Lambda}_j\}_{j\in\Bbb J}$ be a (canonical) dual g-frame of $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$. Suppose that $\Bbb I\subseteq\Bbb J$  and let
\begin{align*}
S_{\Bbb I}&:H\rightarrow H\\
S_{\Bbb I}f&:=\sum_{j\in\Bbb I} \Lambda^{\ast}_j\tilde{\Lambda}_j f.
\end{align*}
We have
\begin{align*}
\Vert S_{\Bbb I}f\Vert^2&=\big(\sup_{\Vert h\Vert=1}\vert\langle S_{\Bbb I}f,h\rangle\vert\big)^2\\
&=\sup_{\Vert h\Vert=1}\big(\sum_{j}\vert\langle\tilde{\Lambda}_j f, \Lambda_j h\rangle\vert\big)^2\\
&\leq\sum_{j}\Vert\tilde{\Lambda}_j f\Vert^2 .\sup_{\Vert h\Vert=1}\sum_{j}\Vert \Lambda_j h\Vert^2\\
&\leq BA^{-1}\Vert f\Vert^2.
\end{align*}
Thus, $S_{\Bbb I}\in\mathcal{B}(H)$ and is positive. From (\ref{g1}), we obtain that $S_{\Bbb I}+S_{\Bbb I^c}=id_{H}$.
\begin{theorem}\label{t1}
For $f\in H$, we have
\begin{eqnarray*}
\sum_{j\in\Bbb I}\langle \tilde{\Lambda}_j f,\Lambda_j f\rangle-\Vert S_{\Bbb I}f\Vert^2 =\sum_{j\in\Bbb I^c}\overline{\langle \tilde{\Lambda}_j f,\Lambda_j f\rangle}-\Vert S_{\Bbb I^c}f\Vert^2
\end{eqnarray*}
where $\Bbb I^c$ is the complement of $\Bbb I$.
\end{theorem}
\begin{proof}
For each $f\in H$, we have
\begin{align*}
\sum_{j\in\Bbb I}\langle \tilde{\Lambda}_j f,\Lambda_j f\rangle-\Vert \sum_{j\in\Bbb I} \Lambda^{\ast}_j \tilde{\Lambda}_j f\Vert^2&=\langle S_{\Bbb I}f,f\rangle-\Vert S_{\Bbb I}f\Vert^2\\
&=\langle S_{\Bbb I}f,f\rangle-\langle S^{\ast}_{\Bbb I} S_{\Bbb I}f,f\rangle\\
&=\langle(id_{H}-S_{\Bbb I})^{\ast}S_{\Bbb I}f,f\rangle\\
&=\langle S^{\ast}_{\Bbb I^c}(id_{H}-S_{\Bbb I^c})f,f\rangle\\
&=\langle S^{\ast}_{\Bbb I^c}f,f\rangle-\langle S^{\ast}_{\Bbb I^c}S_{\Bbb I^c}f,f\rangle\\
&=\langle f,S_{\Bbb I^c}f\rangle-\langle S_{\Bbb I^c}f,S_{\Bbb I^c}f\rangle\\
&=\sum_{j\in\Bbb I^c}\langle \Lambda_j f,\tilde{\Lambda}_j f\rangle-\Vert \sum_{j\in\Bbb I^c} \Lambda^{\ast}_j \tilde{\Lambda}_j f\Vert^2\\
&=\sum_{j\in\Bbb I^c}\overline{\langle \tilde{\Lambda}_j f,\Lambda_j f\rangle}-\Vert \sum_{j\in\Bbb I^c} \Lambda^{\ast}_j \tilde{\Lambda}_j f\Vert^2
\end{align*}
and the proof is completed.
\end{proof}
Now, if $\lbrace \Lambda_j\rbrace_{j\in\Bbb J}$ is a Parseval g-frame, then $\tilde{\Lambda_j}=\Lambda_j$ and we obtain the following famous formula:
\begin{align*}
\sum_{j\in\Bbb I}\Vert\Lambda_jf\Vert^2-\Vert S_{\Bbb I}f\Vert^2=\sum_{j\in\Bbb I^c}\Vert\Lambda_j f\Vert^2-\Vert S_{\Bbb I^c}f\Vert^2,
\end{align*}
where $S_{\Bbb I}f=\sum_{j\in\Bbb I} \Lambda^{\ast}_j\Lambda_j f$.

The same can be presented for g-fusion frames. Let $\Lambda$ be a g-fusion frame for $H$ with (canonical) dual g-fusion frame $\tilde{\Lambda}=(\tilde{W_j}, \tilde{\Lambda_j}, v_j)$ where $\tilde{W_j}=S^{-1}_{\Lambda}W_j$ and $\tilde{\Lambda_j}=\Lambda_j \pi_{W_j}S_{\Lambda}^{-1}$. For simplicity, we show the following operator with the same symbol $S_{\Bbb I}$ in which again $\Bbb I$ is a finite subset of $\Bbb J$:
\begin{align}
S_{\Bbb I}f=\sum_{j\in\Bbb I}v_j^2\pi_{W_j}\Lambda^*_j\tilde{\Lambda_j}\pi_{\tilde{W_j}}f , \ \ \ \ \ (\forall f\in H).
\end{align}
It is easy to check that, $S_{\Bbb I}\in\mathcal{B}(H)$ and positive. Again, we have
$$S_{\Bbb I}+S_{\Bbb I^c}=id_{H}.$$
\begin{remark}
Let $\Lambda$ be a Parseval g-fusion frame for $H$. Since $\mathcal{B}(H)$ is a $C^*$-algebra and $S_{\Bbb I}$ is a positive, so
$r(S_{\Bbb I})=\Vert S_{\Bbb I}\Vert$, where $r$ is the spectral radius. Thus
$$\max_{\lambda\in\sigma(S_{\Bbb I})}\vert\lambda\vert=r(S_{\Bbb I})\leq 1$$
and we conclude that $\sigma(S_{\Bbb I})\in[0, 1]$.
\end{remark}
\begin{theorem}\label{tg1}
Let $f\in H$, then
\begin{eqnarray*}
\sum_{j\in\Bbb I}v_j^2\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}f\rangle-\Vert S_{\Bbb I}f\Vert^2 =\sum_{j\in\Bbb I^c}v_j^2\overline{\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}f\rangle}-\Vert S_{\Bbb I^c}f\Vert^2.
\end{eqnarray*}
\end{theorem}
\begin{proof}
The proof follows by a similar argument to the proof of Theorem \ref{t1}.
\end{proof}
\begin{corollary}\label{cor1}
Let $\Lambda$ be a Parseval g-fusion frame for $H$. Then
\begin{small}
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2-\Vert \sum_{j\in\Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j &\Lambda_j \pi_{W_j}f\Vert^2 =\\
&=\sum_{j\in\Bbb I^c}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2-\Vert \sum_{j\in\Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2.
\end{align*}
\end{small}
Moreover
\begin{eqnarray*}
\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2+\Vert \sum_{j\in\Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2\geq
\frac{3}{4}\Vert f\Vert^2.
\end{eqnarray*}
\end{corollary}
\begin{proof}
If $f\in H$, we obtain
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2+\Vert S_{\Bbb I^c}f\Vert^2&=
\big\langle(S_{\Bbb I}+S^2_{\Bbb I^c})f, f\big\rangle\\
&=\big\langle(S_{\Bbb I}+id_H-2S_{\Bbb I}+S^2_{\Bbb I})f, f\big\rangle\\
&=\langle(id_H-S_{\Bbb I}+S_{\Bbb I}^2)f, f\rangle.
\end{align*}
Now, by Lemma \ref{l1} for $a=1$, $b=-1$ and $c=1$ the inequality holds.
\end{proof}
\begin{corollary}\label{cor2}
Let $\Lambda$ be a Parseval g-fusion frame for $H$. Then
$$0\leq S_{\Bbb I}-S_{\Bbb I}^2\leq\frac{1}{4}id_H$$
\end{corollary}
\begin{proof}
We have $S_{\Bbb I}S_{\Bbb I^c}=S_{\Bbb I^c}S_{\Bbb I}$. Then $0\leq S_{\Bbb I}S_{\Bbb I^c}=S_{\Bbb I}-S_{\Bbb I}^2$. Also, by Lemma \ref{l1}, we get
$$S_{\Bbb I}-S_{\Bbb I}^2\leq\frac{1}{4}id_H.$$
\end{proof}
\begin{theorem}\label{t3.3}
Let $\Lambda$ be a g-fusion frame with  the g-fusion frame operator $S_{\Lambda}$. If $\Bbb I\subseteq\Bbb J$  and $f\in H$, then
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2+\Vert S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I^c}f\Vert^2=\sum_{j\in\Bbb I^c}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2+\Vert S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I}f\Vert^2.
\end{align*}
\end{theorem}
\begin{proof}
Let $\Theta_j:=\Lambda_j\pi_{W_j}S_{\Lambda}^{-\frac{1}{2}}$ and $X_j:=S_{\Lambda}^{-\frac{1}{2}}W_j$. In \cite{sad} has been showed that $(X_j, \Theta_j, v_j)$ is a Parseval g-fusion frame for $H$. Then, by Corollary \ref{cor1}, we have
\begin{small}
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert \Theta_j \pi_{X_j}f\Vert^2-\Vert \sum_{j\in\Bbb I}v_j^2 \pi_{X_j}\Theta^*_j &\Theta_j \pi_{X_j}f\Vert^2 =\\
&=\sum_{j\in\Bbb I^c}v_j^2\Vert \Theta_j \pi_{X_j}f\Vert^2-\Vert \sum_{j\in\Bbb I^c}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}f\Vert^2.
\end{align*}
\end{small}
By replacing $S_{\Lambda}^{\frac{1}{2}}f$ instead of $f$ and this fact that
\begin{align*}
\sum_{j\in\Bbb I}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}f&=\sum_{j\in\Bbb I}v_j^2 (\Theta^*_j\pi_{X_j})^* \Theta_j \pi_{X_j}f\\
&=\sum_{j\in\Bbb I}v_j^2(\Lambda_j \pi_{W_j}S_{\Lambda}^{-\frac{1}{2}}\pi_{X_j})^*\Lambda_j \pi_{W_j}S_{\Lambda}^{-\frac{1}{2}}\pi_{X_j}f\\
&=\sum_{j\in\Bbb I}v_j^2 S_{\Lambda}^{-\frac{1}{2}}\pi_{W_j}\Lambda_j^*\Lambda_j \pi_{W_j}S_{\Lambda}^{-\frac{1}{2}}f\\
&=S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I}S_{\Lambda}^{-\frac{1}{2}}f,
\end{align*}
the proof follows.
\end{proof}
\begin{corollary}\label{cor3}
Let $\Lambda$ be a g-fusion frame with the g-fusion frame operator $S_{\Lambda}$. If $\Bbb I\subseteq\Bbb J$, then
$$0\leq S_{\Bbb I}-S_{\Bbb I}S^{-1}_{\Lambda}S_{\Bbb I}\leq\frac{1}{4}S_{\Lambda}.$$
\end{corollary}
\begin{proof}
In the proof of Theorem \ref{t3.3}, we showed that
$$\sum_{j\in\Bbb I}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}f=S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I}S_{\Lambda}^{-\frac{1}{2}}f.$$
By Corollary \ref{cor2} we get
$$0\leq\sum_{j\in\Bbb I}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}f-\big(\sum_{j\in\Bbb I}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}f\big)^2\leq\frac{1}{4}id_H.$$
Therefore,
$$0\leq S_{\Lambda}^{-\frac{1}{2}}(S_{\Bbb I}-S_{\Bbb I}S^{-1}_{\Lambda}S_{\Bbb I})S_{\Lambda}^{-\frac{1}{2}}\leq\frac{1}{4}id_H$$
and the proof is completed.
\end{proof}
\begin{corollary}
Suppose that $\Lambda$ is a g-fusion frame with the g-fusion frame operator $S_{\Lambda}$. If $\Bbb I\subseteq\Bbb J$  and $f\in H$, then
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2+\Vert S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I^c}f\Vert^2\geq\frac{3}{4}\Vert S_{\Lambda}^{-1}\Vert^{-1} \Vert f\Vert^2.
\end{align*}
\end{corollary}
\begin{proof}
By Theorem \ref{t3.3} and Corollary \ref{cor1}, we can write
\begin{small}
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2+\Vert S_{\Lambda}^{-\frac{1}{2}}S_{\Bbb I^c}f\Vert^2&=\sum_{j\in\Bbb I}v_j^2\Vert \Theta_j \pi_{X_j}S^{\frac{1}{2}}_{\Lambda}f\Vert^2+\Vert \sum_{j\in\Bbb I^c}v_j^2 \pi_{X_j}\Theta^*_j \Theta_j \pi_{X_j}S^{\frac{1}{2}}_{\Lambda}f\Vert^2\\
&\geq\frac{3}{4}\Vert S^{\frac{1}{2}}_{\Lambda}f\Vert^2\\
&=\frac{3}{4}\langle S_{\Lambda}f, f\rangle\\
&\geq\frac{3}{4}\Vert S_{\Lambda}^{-1}\Vert^{-1} \Vert f\Vert^2.
\end{align*}
\end{small}
\end{proof}
\begin{theorem}
Let $\Lambda$ be a Parseval g-fusion frame for $H$ and $\Bbb I\subseteq\Bbb J$. Then
\begin{enumerate}
\item [(I)] $0\leq S_{\Bbb I}-S_{\Bbb I}^2\leq\dfrac{1}{4}id_H$.
\item [(II)] $\dfrac{1}{2}id_H\leq S_{\Bbb I}^2+S_{\Bbb I^c}^2\leq\dfrac{3}{2}id_H$.
\end{enumerate}
\end{theorem}
\begin{proof}
(I). Since $S_{\Bbb I}+S_{\Bbb I^c}=id_H$, then $S_{\Bbb I}S_{\Bbb I^c}+S^2_{\Bbb I^c}=S_{\Bbb I^c}$. Thus,
$$S_{\Bbb I}S_{\Bbb I^c}=S_{\Bbb I^c}-S^2_{\Bbb I^c}=S_{\Bbb I^c}(id_H-S_{\Bbb I^c})=S_{\Bbb I^c}S_{\Bbb I}.$$
But, $\Lambda$ is a Parseval, so $0\leq S_{\Bbb I}S_{\Bbb I^c}=S_{\Bbb I}-S_{\Bbb I}^2$. On the other hand, by Lemma \ref{l2}, we get
$$S_{\Bbb I}-S_{\Bbb I}^2\leq\frac{1}{4}id_H.$$
(II). We saw that $S_{\Bbb I}S_{\Bbb I^c}=S_{\Bbb I^c}S_{\Bbb I}$, then by Lemma \ref{l2},
$$S_{\Bbb I}^2+S_{\Bbb I^c}^2=id_H-2S_{\Bbb I}S_{\Bbb I^c}=2S_{\Bbb I}^2-2S_{\Bbb I}+id_H\geq\frac{1}{2}id_H.$$
On the other hand, we have again by Lemma \ref{l2} and $0\leq S_{\Bbb I}-S_{\Bbb I}^2$,
$$S_{\Bbb I}^2+S_{\Bbb I^c}^2\leq id_H+2S_{\Bbb I}-2S_{\Bbb I}^2\leq\frac{3}{2}id_H.$$
\end{proof}
\begin{corollary}
Let $\Lambda$ be a g-fusion frame with the g-fusion frame operator $S_{\Lambda}$. If $\Bbb I\subseteq\Bbb J$, then
$$\frac{1}{2}S_{\Lambda}\leq S_{\Bbb I}S^{-1}_{\Lambda}S_{\Bbb I}-S_{\Bbb I^c}S^{-1}_{\Lambda}S_{\Bbb I^c}\leq\frac{3}{2}S_{\Lambda}.$$
\end{corollary}
\begin{proof}
The proof is similar to the proof of Corollary \ref{cor3}.
\end{proof}
\begin{theorem}
Let $\Lambda$ be a g-fusion frame with the g-fusion frame operator $S_{\Lambda}$. If $\Bbb I\subseteq\Bbb J$, then for any $f\in H$
\begin{small}
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2-\sum_{j\in\Bbb J}v_j^2\Vert\tilde{\Lambda}_j\pi_{\tilde{W}_j}M_{\Bbb I}f\Vert^2=\sum_{j\in\Bbb I^c}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2-\sum_{j\in\Bbb J}v_j^2\Vert\tilde{\Lambda}_j\pi_{\tilde{W}_j}M_{\Bbb I^c}f\Vert^2
\end{align*}
\end{small}
where
\begin{align*}
M_{\Bbb I}f=\sum_{j\in\Bbb I}v_j^2\pi_{W_j}\Lambda^*_j\Lambda_j\pi_{W_j}f.
\end{align*}
\end{theorem}
\begin{proof}
Via the definition of $S_{\Lambda}$, it is clear that $M_{\Bbb I}+M_{\Bbb I^c}=S_{\Lambda}$. Therefore, $S^{-1}_{\Lambda}M_{\Bbb I}+S^{-1}_{\Lambda}M_{\Bbb I^c}=id_H$. Hence, by Lemma \ref{l2}
$$S^{-1}_{\Lambda}M_{\Bbb I}-S^{-1}_{\Lambda}M_{\Bbb I^c}=(S^{-1}_{\Lambda}M_{\Bbb I})^2-(S^{-1}_{\Lambda}M_{\Bbb I^c})^2.$$
Thus, for each $f,g\in H$ we obtain
\begin{equation*}
\langle S^{-1}_{\Lambda}M_{\Bbb I}f, g\rangle-\langle S^{-1}_{\Lambda}M_{\Bbb I}S^{-1}_{\Lambda}M_{\Bbb I}f, g\rangle=\langle S^{-1}_{\Lambda}M_{\Bbb I^c}f, g\rangle-\langle S^{-1}_{\Lambda}M_{\Bbb I^c}S^{-1}_{\Lambda}M_{\Bbb I^c}f, g\rangle.
\end{equation*}
We choose $g$ to be $g=S_{\Lambda}f$ and we can get
\begin{equation*}
\langle M_{\Bbb I}f, f\rangle-\langle S^{-1}_{\Lambda}M_{\Bbb I}f, M_{\Bbb I}f\rangle=\langle M_{\Bbb I^c}f, f\rangle-\langle S^{-1}_{\Lambda}M_{\Bbb I^c}f, M_{\Bbb I^c}f\rangle
\end{equation*}
and by (\ref{inverse}),  the proof is completed.
\end{proof}


TITLE: Function $f : X \to Y$ continuous on a dense subspace $D \subset X$.
QUESTION [0 upvotes]: Let $X, Y$ be topological spaces and let $D \subset X$ be dense. Suppose $f : D \to Y$ is continuous. Then $f : X \to Y$ is continuous on all of $X$.

The above stated assertion does clearly not hold in general. However, I am curious if the assertion holds under certain conditions on the spaces $X, Y$ or function $f$, or whether there are similar results.
Any comment is appreciated.

REPLY [2 votes]: As a counterexample you can consider $D=(0,1]$, $X=[0,1]$, and $Y=\mathbb{R}$ with the Euclidean topology. Take $f:D\to\mathbb{R}$ as
$$f(x)=\frac 1x.$$
There is no way to extend continuously this function on the whole interval $[0,1]$.
However there is a requirement on $f$ that makes it possible (if for example $X,Y$ are metric spaces and $Y$ is complete), namely uniform continuity of $f$.


TITLE: Amplitude $\langle0|e^{-iHT}|0\rangle$ in A. Zee's QFT In A Nutshell
QUESTION [5 upvotes]: In his Quantum Field Theory In a Nutshell, in page 12, (Second Ed), A Zee says that conventionally, the amplitude $\langle0|e^{-iHT}|0\rangle$ is denoted by $Z$. In the next paragraph, he considers the amplitude in eq. (6), and performs a Wick rotation to Euclidean time, $t\rightarrow -it$, and writes,
$$
Z=\int Dq(t)e^{-\int_0^Tdt[\frac{1}{2}m\dot{q}^2+V(q)]}
$$
which is the Euclidean Path Integral. I am confused because he denotes this quantity by $Z$. Is it the same $Z$ as $\langle0|e^{-iHT}|0\rangle$? If so, then I cannot see how. Can anybody help me?
I understand that
$$
Z=\langle0|e^{-iHT}|0\rangle=\int dq_F\int dq_I\langle F|q_F\rangle\langle q_F|e^{-iHT}|q_I\rangle\langle q_I|I\rangle
$$
with $|F\rangle=|I\rangle=|0\rangle$. How is this similar to the former integral?

REPLY [4 votes]: The step of evaluating your second expression for $Z$ and arriving at the path integral form, which is the first expression you mention, is a nontrivial but standard derivation in the path integral formulation of QFT. It is quite lengthy, but can be found in most books that cover the path integral formulation. The general idea it to split the time interval $T$ into $N$ intervals (with $N$ large), and to insert a full set of eigenstates at each intermediate time, such that
$$
\langle q_F|e^{-iHT}|q_I\rangle= \langle q_F|\left(e^{-iHT/N}\right)^N|q_I\rangle\\
=\int dq_1\ldots dq_{N-1}\langle q_F|e^{-iHT/N}|q_{N-1}\rangle\langle q_{N-1}|e^{-iHT/N}|q_{N-2}\rangle\langle q_{N-2}|\ldots|q_1\rangle\langle q_{1}|e^{-iHT/N}|q_I\rangle
$$
This can be solved in the following way. Since Hamiltonian also contains momentum operators you additionally have to insert a complete set of conjugate momentum eigenstates at each intermediate time. Then, all the operators in the Hamiltonian can be evaluated on these inserted states and you are left with C-number functions only. Since the Hamiltonian is quadratic in the momentum operator the momentum integrals are Gaussian and can be performed explicitly. The $N$ integral over intermediate field configurations enter the definition of the path integral (in essence, the path integral is just defined as such an integral over $n$ intermediate time field configurations in the limit where $N\to\infty$).
For more details, see for example the Advanced QFT lecture notes of Hugh Osborn, which can be found here. The derivation you are looking for is at the very beggining of the script.


text:
\begin{document}
\maketitle

\begin{abstract}
We prove a conjecture of Prasad and Takloo-Bighash for discrete series of inner forms of the general linear group over a non archimedean local field, in the case
of Steinberg representations.
\end{abstract}

 \section*{Introduction}

 Let $F$ be a non archimedean local field of characteristic not $2$ and $E$ a quadratic extension of $F$. Let $D$ be a central division $F$-algebra
 of dimension $d^2$. If $n$ is a positive integer, then $\mathcal{M}_n(D)$ is a simple central $F$-algebra of dimension $n^2d^2$. As $E/F$ is quadratic, $E$ is embedded in $\mathcal{M}_n(D)$ as an $F$-subalgebra
 if and only if $nd$ is even.
 Then $C_{\mathcal{M}_n(D)}(E)$ (the centralizer of $E$ in
 $\mathcal{M}_n(D)$) is an $E$-algebra. We denote by $N_{rd,F}$ (resp. $N_{rd,E}$) the reduced norm of $GL(n,D)$ (resp. $(C_{\mathcal{M}_n(D)}(E))^\times$) as well as its restriction to any subgroup.
 \paragraph{}We consider the following conjecture of Prasad and Takloo-Bighash:
\begin{conje}[\cite{P-TB}, Conjecture 1]
Let $A=\mathcal{M}_n(D)$. Let $\pi$ be an irreducible admissible representation of $A^\times$ such that the corresponding representation (via Jacquet-Langlands) of $GL(nd,F)$ is generic with central
character $\omega_\pi$. Let $\mu$ be a character of $E^\times$ such that $\mu^{\frac{nd}{2}}|_{F^\times}=\omega_\pi$. If the character $\mu\circ N_{rd,E}$ of $(C_{\mathcal{M}_n(D)}(E))^\times$ appears as a quotient in $\pi$ restricted to $(C_{\mathcal{M}_n(D)}(E))^\times$, then :
\begin{enumerate}
\item the Langlands parameter of $\pi$ takes values in $GSp_{nd}(\mathbb{C})$ with similitude factor $\mu_{|F^\times}$.
\item the epsilon factor $\epsilon(\frac{1}{2},\pi\otimes Ind_E^F(\mu^{-1}))=(-1)^n\omega_{E/F}(-1)^{\frac{nd}{2}}\mu(-1)^{\frac{nd}{2}}$ ($\omega_{E/F}$ is the quadratic character of $F^\times$ with kernel the norms of $E^\times$).
\end{enumerate}
If $\pi$ is a discrete series representation of $A^\times$, then these two conditions are necessary and sufficient for the character $\mu\circ N_{rd,E}$ of $(C_{\mathcal{M}_n(D)}(E))^\times$ to appear as a quotient in $\pi$ restricted to $(C_{\mathcal{M}_n(D)}(E))^\times$.
\end{conje}

Let us set $G=GL(n,D)$ and $H=(C_{\mathcal{M}_n(D)}(E))^\times$. We consider this conjecture for the Steinberg representation and we prove it by first
establishing some $H$-distinction results. The main results of this paper are :

\begin{theo}
Let $n$ be a positive integer and let $\mu$ be a character of $E^*$. $E$ is embedded in $\mathcal{M}_n(D)$ if and only if $nd$ is even. We set $St(1)=St(n,1)$ the Steinberg representation of $G$ and
$\tilde{\mu}:=\mu\circ N_{rd,E}$.
\begin{itemize}
\item If $d$ is even, $H=(C_{\mathcal{M}_n(D)}(E))^\times=GL(n,C_D(E))$ and $St(n,1)$ is $\tilde{\mu}$-distinguished under $H$ if and only if
\begin{itemize}
\item $\mu_{|F^*}=1$ and $\mu\neq1$ if $n$ is even.
\item $\mu=1$ if $n$ is odd.
\end{itemize}
\item If $d$ is odd and $n$ is even, $H=(C_{\mathcal{M}_n(D)}(E))^\times=GL(n/2,D\otimes_FE)$ and $St(n,1)$ is $\tilde{\mu}$-distinguished under $H$ if and only if $\mu_{|F^*}=1$ and $\mu\neq1$.
\end{itemize}
\end{theo}

\begin{theo}(Prasad and Takloo-Bighash conjecture, Steinberg case) Let $A=\mathcal{M}_n(D)$ and $\pi=St(n,1)$ which is an irreducible admissible representation of $A^\times=GL(n,D)=G$. Recall that $\pi$ corresponds via Jacquet-Langlands correspondence to
$St(nd,1)$ (the Steinberg representation of $GL(nd,F)$) with central character $\omega_\pi=1$. Let $\mu$ be a character of $E^\times$ such that $\mu^{\frac{nd}{2}}|_{F^\times}=\omega_\pi=1$. Then, the character $\mu\circ N_{rd,E}$ of
$H=(C_{\mathcal{M}_n(D)}(E))^\times$ appears as a quotient in $\pi$ restricted to $H$ if and only if :
\begin{enumerate}
 \item the Langlands parameter of $\pi$ takes values in $GSp_{nd}(\mathbb{C})$ with similitude factor $\mu_{|F^\times}$.
 \item the epsilon factor satisfies $\epsilon(\frac{1}{2},\pi\otimes Ind_E^F(\mu^{-1}))=(-1)^n\omega_{E/F}(-1)^{\frac{nd}{2}}$ (where $\omega_{E/F}$ is the quadratic character of $F^\times$ with kernel the norms of $E^\times$).
\end{enumerate}
\end{theo}

Notice that when the character $\mu\circ N_{rd,E}$ of $H$ extends to $G$, the conjecture can be reformulated in the following more appealing way, and this is the version of the conjecture considered in \cite{FMW}.

\begin{theo}
\begin{sloppypar}
  Let $St(n,\chi)$ be the Steinberg representation of $G=GL(n,D)$. $St(n,\chi)$ is $H$-distinguished if and only if it is symplectic and $\epsilon(\frac{1}{2},BC_E(St(n,\chi)))=(-1)^n$ (where $BC_E$ denotes the base change to $E$).
\end{sloppypar}
\end{theo}

In this latter reference, the authors partially prove the conjecture
for supercuspidal representations of $GL(n,\mathbb{H})$
 where $\mathbb{H}$ is the quaternion division algebra over a $p$-adic field.

 \paragraph{}We recall that, for $\mu$ a character of $H$, a representation $(\pi, V)$ of $G$ is said to be $\mu$-distinguished under $H$ ($H$-distinguished if $\mu=1$ and just distinguished if
 there is no possible confusion) if the space of $H$-homomorphisms $\mathrm{Hom}_H(\pi,\mu)$ is non-zero \textit{i.e.} if there exists a non-zero linear form $L$ on $V$ such that $L(\pi(h).v)=\mu(h)L(v)$
 for all $h\in H$, for all $v\in V$. \\
 Now, any character of $GL(n,D)$ can be written as $\chi\circ N_{rd,F}$ where $\chi$ is a character of $F^*$. For $\chi$ a character of $F^*$,
 the Steinberg representation $St(n,\chi)$ of $GL(n,D)$ (denoted $St(\chi)$ if the context is clear)
 is given by $ind_{P_\emptyset}^G(\chi\circ N_{rd,F})/\underset{P}\sum ind_P^G(\chi\circ N_{rd,F})$ where $P_\emptyset$ denotes the minimal standard parabolic subgroup of $G$ and where the standard parabolic subgroups $P$
 in the sum correspond to a partition of $n$ with all elements equal to $1$ except one of them which is equal to $2$.

 \paragraph{}In the first part of this paper, we recall some useful elementary results to study the distinction, mainly Frobenius reciprocity and Mackey theory.\\

 In Sections 2 and 3, we study the distinction of the Steinberg representation according to the parity of $d$.
 We follow the method used by Matringe in \cite{M.16} which we recall now.
 We start by determining a set of representatives of double cosets $P\backslash G/H$ for $P$ a standard parabolic
 subgroup of $G$. These representatives allow us to apply Mackey theory, which is, with Frobenius reciprocity and modulus characters computations, an
 essential tool to establish a necessary condition for the distinction of the Steinberg representation. \\
First, we show that $\mathrm{Hom}_H(ind_{P_\emptyset}^G(1),\tilde{\mu})$ is at most one dimensional, hence $\mathrm{Hom}_H(St(1),\tilde{\mu})$ as well.
It moreover implies that $\mathrm{dim}(\mathrm{Hom}_H(St(1),\tilde{\mu}))=1$ if and only if there is a nonzero $\tilde{\mu}$-equivariant linear form on $ind_{P_\emptyset}^G(1)$
which vanishes at each term of $\underset{P\text{ of type }(1,\dots,1,2,1,\dots,1)}\sum ind_P^G(1)$. In particular the necessary condition on $\mu$ for
$St(1)$ to be $\tilde{\mu}$-distinguished will come from the fact that $ind_{P_\emptyset}^G(1)$ must also be $\tilde{\mu}$-distinguished in this case.
On the other hand, if $ind_{P_\emptyset}^G(1)$ is $\tilde{\mu}$-distinguished, we have an explicit $\tilde{\mu}$-equivariant linear form on its space given by an integral. To get our sufficient condition, we
will show that this linear form does not vanish on $ind_{P_\emptyset}^G(1)$ for a well chosen standard parabolic subgroup $P$ of type $(1,\dots,1,2,1,\dots,1)$ when
$\mu$ not of the correct form.\\

In the last section, we explain how to reformulate the Prasad and Takloo-Bighash conjecture for the Steinberg representation and we do the epsilon factor calculation in order to prove it.

\paragraph{Notation}
We let $P_\emptyset$ denote the minimal standard parabolic subgroup (of upper triangular matrices), $M_\emptyset$ its standard Levi subgroup and $N_\emptyset$ its unipotent radical.
We set $P_\emptyset^-$ the subgroup of $G$ of lower triangular matrices. We denote by $\Phi$ the roots of the center $Z({M_{\emptyset}})$ of ${M_{\emptyset}}$ acting on the Lie algebra $Lie(G)$, by $\Phi^+$ those corresponding to the restriction
of this action on $Lie(N_\emptyset)$, and by $\Phi^-$ those corresponding to the restriction
of this action on $Lie(N_\emptyset^-)$. In particular $Lie(N_\emptyset)=\oplus_{\alpha \in \Phi^+} Lie(N_\alpha)$ and
$Lie(N_\emptyset^-)=\oplus_{\alpha \in \Phi^-} Lie(N_\alpha)$, with
obvious notation. If $P$ is a parabolic subgroup of $G$ containing $P_\emptyset$, with standard Levi factor $M$, we denote by $\Phi_M$
the roots of $Z({M_{\emptyset}})$ on $Lie(M)$. We define $\Phi_M^+$ and $\Phi_M^-$ in a similar fashion as above.
\paragraph{}
 For $\chi$ a character of a parabolic subgroup $P$, $ind_P^G(\chi)$ is the un-normalized parabolic induction. We let $\Delta_X$ denote the modular character of a locally compact totally disconnected topological
 group $X$ \textit{i.e.} such that $\lambda(xg)=\Delta_X(g)\lambda(x)$ $\forall x,g\in X$, for $\lambda$ a left Haar measure on $X$. We set $\delta_X:=\Delta_X^{-1}$.

 \paragraph{Acknowledgements}
 I am very grateful to my supervisor Nadir Matringe for all his help and his essential explanations. I also thank Paul Broussous, my co-supervisor, for his useful explanations.

 \section{General results}

 We gather together some useful results (written as in \cite{M.16}).

 \paragraph{}We only consider smooth representations on complex vector spaces.
Let $X$ be a locally compact totally disconnected space, and $L$ a locally compact totally disconnected group acting
continuously and properly on $X$. If $\chi$ is a character of $L$, we denote
by $\mathcal{C}_c^\infty(L\backslash X,\chi)$ the space of smooth functions on $X$, with compact support mod $L$, and which transform by $\chi$ under left translation by elements of $L$. If $X$ is a group $Q$ which contains $L$, then we write
$ind_{L}^Q(\chi)$ for $\mathcal{C}_c^\infty(L\backslash Q,\chi)$, which is a representation of $Q$ by right translation.
 We shall often use the following two theorems, which are respectively Frobenius reciprocity and Mackey theory for compactly induced representations. The first one is a consequence of Proposition 2.29 of \cite{BZ.76}

\begin{prop}\label{Frob}
Let $\chi$ be a character of $L$, then the vector space $Hom_Q(ind_{L}^Q(\chi),\mu)$ is isomorphic to
$Hom_L(\Delta\chi,\mu)$, where $\Delta$ is the quotient of the modulus character of $L$ by that of $Q$.
\end{prop}

The next one is a consequence of Theorem 5.2 of \cite{BZ.77}.

\begin{prop}\label{Mackey}
Let $P$ be a parabolic subgroup of $G$, and $\mu$ be a character of $P$. Take a set of representatives $(u_1,\dots,u_r)$
of $P\backslash G/ H$, ordered such that $X_i=\coprod_{k=1}^i Pu_k H$ is open in $G$ for each $i$. Then $ind_P^G(\mu)$ is filtered by
the $H$-submodules $\mathcal{C}_c^\infty(P\backslash X_i,\mu)$, and
$$\mathcal{C}_c^\infty(P\backslash X_i,\mu)/\mathcal{C}_c^\infty(P\backslash X_{i-1},\mu)\simeq
\mathcal{C}_c^\infty(P\backslash Pu_i H,\mu).$$
\end{prop}

Finally, we recall the following result from \cite{H-W.93}, which is Proposition 3.4 in there. It in particular implies that if
$P$ contains a minimal $\tau$-split parabolic subgroup (see below), then $\mathcal{C}_c^\infty(P\backslash P H,\mu)$ is
a subspace of $ind_P^G(\mu)$.

\begin{prop}\label{ouvert}
Let $P$ be a parabolic subgroup of $G$, and $\tau$ be an $F$-rational involution of $G$. The class $PH$ is open if and only
if $P$ contains a minimal parabolic $\tau$-split subgroup $P'$ (which means that $\tau(P')$ and $P'$ are opposite parabolic subgroups).
\end{prop}

\section{Case $d$ even}

\subsection{Preliminaries}
We fix $F$ a non archimedean local field of characteristic not $2$ and $E$ a quadratic extension. We let $|\cdot|_F$ denote the normalized absolute value on $F$ and $N_{E/F}$ the norm map from $E^*$ to $F^*$.
 Let $D$ be a central division $F$-algebra of dimension $d^2$ with $d=2d'$ an even positive integer. As $2$ divides $d^2$, $E$ can be seen as a subfield of $D$. Moreover, as car($F$)$\neq2$, there exists $\delta$ in $E$ with
$\delta^2\in F\setminus F^2$ such that $E=F[\delta]$.\\
As $E$ is a subfield of $D$, we can consider the centralizer of $E$ in $D$ : $C_D(E)$. It is easy to see that $C_D(E)$ is a division $E$-algebra. The double centralizer property says that $[E:F][C_D(E):F]=[D:F]=d^2$
so $[C_D(E):F]=d^2/2$ and that $E=C_D(C_D(E))$ so the center of $C_D(E)$ is $E$. In summary, $D':=C_D(E)$ is a central divsion $E$-algebra of dimension $d^2/4=d'^2$.
We recall that $N_{rd,F}$ denotes the reduced norm on $GL(n,D)$ and $N_{rd,E}$ denotes the reduced norm of $GL(n,D')$ as well as its restriction to any subgroup.
\paragraph{}It is easy to see that $C_D(E)$ is the fixed points set $D^\sigma$ of $D$ under the involution $\sigma=int(\delta)$. As $\sigma$ is an involution different of identity, $-1$ is an eigenvalue of $\sigma$ so we can chose $\iota\in D$
(which is thus not in $D'$) such that $\sigma(\iota)=-\iota$ and $D=D'\bigoplus \iota D'$. We can easily see that $C_{\mathcal{M}_n(D)}(E)=\mathcal{M}_n(C_D(E))=\mathcal{M}_n(D')$.\\
If we denote again by $\sigma$ the involution of $GL(n,D)$ which is given by applying $int(\delta)$ to each entry of a matrix of $GL(n,D)$, then $H=GL(n,D')$ is the fixed points of $G=GL(n,D)$ under $\sigma$~: $H=G^\sigma$.

\subsection{Representatives of $P\backslash G/H$}
Let $P$ be a standard parabolic subgroup of $G=GL(n,D)$, corresponding to a partition $\bar{n}=(n_1,\dots,n_r)$ of $n$. We denote $I(\bar{n})$ the set of symmetric matrices with natural number entries such that the sum of the $i$-th
row equals $n_i$ for all $i$ in $\{1,\dots,r\}$. Let $\mathcal{B}=(e_1,\dots,e_n)$ be the canonical basis of $D^n$. For $s=(n_{i,j})_{1\leq i,j\leq r}\in I(\bar{n})$, we set
$$\mathcal{B}_{i,j}=(e_{(n_1+\dots+n_{i-1}+n_{i,1}+\dots+n_{i,j-1}+1)},\dots,e_{(n_1+\dots+n_{i-1}+n_{i,1}+\dots+n_{i,j-1}+n_{i,j})})$$
We set $u_s\in GL(n,D)$ the matrix in the basis $\mathcal{B}$ of the endomorphism of $D^n$ mapping Vect$(\mathcal{B}_{i,i})_D$ to itself and Vect$(\mathcal{B}_{i,j}\cup\mathcal{B}_{j,i})_D$ to itself for $i\neq j$ such that
$u_s$ restricted to Vect$(\mathcal{B}_{i,i})_D$ is $I_{n_{i,i}}$ and $u_s$ restricted to Vect$(\mathcal{B}_{i,j}\cup\mathcal{B}_{j,i})_D$ is $\begin{pmatrix}I_{n_{i,j}}&-\iota I_{n_{i,j}}\\I_{n_{i,j}}&\iota I_{n_{i,j}}\end{pmatrix}$.

\begin{prop}
 The set of elements $u_s$ for $s$ in $I(\bar{n})$ as described above is a set of representatives of the double cosets $P\backslash G/H$.
\end{prop}
\begin{proof}
 The proof is similar to the odd case of \cite{M.16} and with more precisions in \cite{M.11}.
\end{proof}

\begin{rem}
 If we denote $\mathbb{H}$ the division algebra of quaternions over $F$ (\textit{i.e.} of dimension $4$ over $F$), then $C_\mathbb{H}(E)=E$. Moreover, if we denote $\iota'$ an element of $\mathbb{H}\setminus E$ such that $\mathbb{H}=E\oplus \iota'E$
 and $\sigma(\iota')=-\iota'$ and $u_s'$ the same matrix as $u_s$ where $\iota$ have been replaced by $\iota'$, then the map $u_s\mapsto u_s'$ for $s\in I(\bar{n})$ induces a bijection between $P(D)\backslash GL(n,D)/GL(n,D')$ and
 $P(\mathbb{H})\backslash GL(n,\mathbb{H})/GL(n,E)$.
\end{rem}

\paragraph{}Now, for $s$ in $I(\bar{n})$, we set $w_s=u_s\sigma(u_s^{-1})$. We can write $I=\llbracket 1,n\rrbracket$ as the ordered disjoint union
$$I=I_{1,1}\cup I_{1,2}\cup \dots \cup I_{1,r} \cup I_{2,1}\cup \dots \cup I_{r,1}\cup \dots \cup I_{r,r-1}\cup I_{r,r},$$
with $I_{i,j}$ of length $n_{i,j}$. We can check that $w_s$ is the matrix of the permutation of $n$
sending $I_{i,i}$ to itself identically and $I_{i,j}$ to $I_{j,i}$ when $i\neq j$ such that $w_s(n_{1,1}+\dots+n_{i,j-1}+k)=n_{1,1}+\dots+n_{j,i-1}+k$ for $k\in\{1,\dots,n_{i,j}\}$.\\
We also set
$$\begin{matrix}&&&&\\\sigma_s&:&G=GL(n,D)&\rightarrow&GL(n,D)\\&&x&\mapsto&w_s\sigma(x)w_s^{-1}\end{matrix}.$$
Then, $u_sHu_s^{-1}$ is the group of fixed points of $G$ under the involution $\sigma_s$ : $u_sHu_s^{-1}=:G^{\sigma_s}$.

\paragraph{}Now, for $s\in I(\bar{n})$, we can consider the standard parabolic subgroup of $G$ attached to $s$ and its standard decomposition denoted by : $P_s=M_sN_s$. We notice that as $s$ can be seen as a sub-partition of $(n_1,\dots,n_r)$
(corresponding to the standard parabolic subgroup $P$), $P_s$ is included  in $P$. We need now to study $P_s\cap u_sHu_s^{-1}$ and especially its decomposition.
\paragraph{}The same proof as in Lemma 3.2 and Proposition 3.2 of \cite{M.16} shows the following proposition~:

\begin{prop}
 For $s\in I(\bar{n})$, we have $w_s(\Phi_M^-)\subset\Phi^-$, $w_s(\Phi_M^+)\subset\Phi^+$. Moreover, $P\cap u_sHu_s^{-1}=P_s\cap u_sHu_s^{-1}$ and $P_s\cap u_sHu_s^{-1}$ is the semidirect product of $M_s\cap u_sHu_s^{-1}$ and $N_s\cap u_sHu_s^{-1}$.
\end{prop}

We can make the subgroup $M_s^{\sigma_s}=M_s\cap u_sHu_s^{-1}$ explicit :
$$M_s^{\sigma_s}=\{diag(a_{1,1},a_{1,2},\dots,a_{1,r},a_{2,1},\dots,a_{r,r})\;;\; a_{j,i}=\sigma(a_{i,j})\in GL(n_{i,j},D)\}.$$

\paragraph{}Finally, we have the following equality of characters :
\begin{prop}\label{car2}
 $(\delta_{P_s^{\sigma_s}})_{|M_s^{\sigma_s}}=(\delta_{P_s})_{|M_s^{\sigma_s}}$
\end{prop}
\begin{proof}
As said in Proposition 4.4 of \cite{M.11}, thanks to Lemma 1.10 of \cite{K-T.08}, as $\sigma_s$ is defined over $E$, it is enough to check the equality on the $E$-split component $Z_s^{\sigma_s}$ (the maximal $E$-split
 torus in the center of $M_s$) of $M_s^{\sigma_s}$. Here,
 $$Z_s^{\sigma_s}=\left\{\begin{pmatrix}\lambda_{1,1}I_{n_{1,1}}&&&\\&\lambda_{1,2}I_{n_{1,2}}&&(0)\\(0)&&\ddots&\\&&&\lambda_{r,r}I_{n_{r,r}}\end{pmatrix};\;\lambda_{i,j}=\lambda_{j,i}\in E^*\right\}$$
 For $t\in Z_s^{\sigma_s}$, $\delta_{P_s}(t)=|N_{rd,F}(Ad(t)_{|Lie(N_s)})|_F=\prod_{\alpha\in\Phi^+-\Phi_s^+}|N_{rd,F}(\alpha(t))|_F$.
 \paragraph{}Now, for $\alpha\in\Phi$, let $\mathcal{N}_{\alpha,w_s(\alpha)}=\{x\in Lie(N_\alpha)+Lie(N_{w_s(\alpha)});\;\sigma_s(x)=x\}$; it is a right $D'$-vector space of dimension $|\alpha,w_s(\alpha)|$. Then, for $t\in Z_s^{\sigma_s}$, we have :
 $$\delta_{P_s^{\sigma_s}}(t)=\prod_{\{\alpha,w_s(\alpha)\}\subset\Phi^+-\Phi_s^+}|N_{rd,E}(Ad(t)_{|\mathcal{N}_{\alpha,w_s(\alpha)}})|_E=\prod_{\{\alpha\in\Phi^+-\Phi_s^+;\;w_s(\alpha)\in\Phi^+-\Phi_s^+\}}|N_{rd,E}(\alpha(t))|_E.$$

 Moreover, we have (see Proposition 4.4 of \cite{M.11}) :
\begin{equation}\label{relation2}
\prod_{\{\alpha\in \Phi^+-\Phi_s^+; w_s(\alpha)\notin \Phi^+-\Phi_s^+\}}|N_{rd,E}(\alpha(t))|_E= \prod_{\{\alpha\in \Phi^+-\Phi_s^+; w_s(\alpha)\in \Phi^--\Phi_s^-\}}|N_{rd,E}(\alpha(t))|_E= 1
\end{equation}

 \paragraph{}Thus, $\delta_{P_s^{\sigma_s}}(t)=\underset{\{\alpha\in\Phi^+-\Phi_s^+\}}\prod|N_{rd,E}(\alpha(t))|_E$.
 \paragraph{}Finally, as $\alpha(t)\in D'$, we have
 $$|N_{rd,F}(\alpha(t))|_F=|N_{E/F}\circ N_{rd,E}(\alpha(t))|_F=|N_{E/F}\circ N_{rd,E}(\alpha(t))|_E^{1/2}=|N_{rd,E}(\alpha(t))|_E$$
 which gives the equality of characters.
\end{proof}

\subsection{Distinguished Steinberg representations}
For $\mu$ a character of $E^*$, we set $\tilde{\mu}=\mu\circ N_{rd,E}$. In this section, we will study whether $St(1)$ is $\tilde{\mu}$-distinguished under $H$ or not, according to the character $\mu$ of $E^*$. We denote by $St(1)$ the Steinberg representation
$ind_{P_\emptyset}^G(1)/\sum_P ind_P^G(1)$
where $P$ describes the standard parabolic subgroups of $G$ corresponding to a partition of $n$ of type $(1,\dots,1,2,1,\dots,1)$.

\paragraph{}First, we suppose that $St(1)$ is $\tilde{\mu}$-distinguished under $H$ and we find a necessary condition on $\mu$ in the following proposition :
\begin{prop}\label{necessaire2}
 Suppose that $St(1)$ is $\tilde{\mu}$-distinguished under $H$. Then, $\mu_{|F^*}=1$ if $n$ is even and $\mu=1$ if $n$ is odd. Moreover, only the open orbit $P_\emptyset u_sH$ where $s=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$
 supports a $\tilde{\mu}$-equivariant linear form and dim$(Hom_H(St(1),\tilde{\mu}))=1$.
\end{prop}
\begin{proof}
The idea of the proof is the same as those of Propositions 3.4 and 3.5 of \cite{M.16} so we do not give all details.
Supppose that $St(1)$ is $\tilde{\mu}$-distinguished, then $ind_{P_\emptyset}^G(1)$ is also $\tilde{\mu}$-distinguished so
\begin{eqnarray*}
\exists s\in I(\bar{n})\text{ such that }\mathrm{Hom}_{P_\emptyset^{\sigma_s}}\left(\frac{\Delta_{P_\emptyset^{\sigma_s}}}{\Delta_{G^{\sigma_s}}}\tilde{\mu}_s^{-1},1\right)\neq\{0\}
\end{eqnarray*}
where $\tilde{\mu}_s(x)=\tilde{\mu}(u_s^{-1}xu_s)$ for $x$ in $u_s H u_s^{-1}$.
\paragraph{}\begin{sloppypar}Now, suppose that $w_s$ has at least one fixed point, then if we consider $M=diag(1,\dots,1,a,1,\dots,1)$ with $a\in F^*$ in the $i$-th row, the previous equality of characters and the fact that $\tilde{\mu}_{|F^*}$ is unitary (considering the central character of
$ind_{P_\emptyset}^G(1)$ and its $\tilde{\mu}$-distinction) imply that
 $i=\frac{n+1}{2}$ so $n$ is odd and $w_s$ has only one fixed point.
 \end{sloppypar}

 \paragraph{}Thus, if $n$ is even, $w_s$ has no fixed point. Then we get  $s=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$ and $\tilde{\mu}(u_s^{-1}M_{\emptyset}^{\sigma_s}u_s)=1$.
 Let us prove that $N_{rd,E}(u_s^{-1}M_\emptyset^{\sigma_s}u_s)=F^*$. First, we show how $D$ can be embedded in $\mathcal{M}_2(D')$. $\mathcal{M}_2(D')$ identifies to $\mathrm{End}(D)_{D'}$ via the basis $(1,\iota)$ of the right
 $D'$-vector space $D$. Moreover, we have
 $$\begin{matrix}D\otimes_FE&\tilde{\longrightarrow}&\mathrm{End}(D)_{D'}\\a\otimes e&\mapsto&(d\mapsto ade)\end{matrix}$$
 so $D$ can be embedded in $\mathcal{M}_2(D')$ via
 $$\begin{matrix}f\::\:&D&\longrightarrow&\mathcal{M}_2(D')\\&a&\mapsto&a\otimes 1\in D\otimes_FE\simeq\mathcal{M}_2(D')\end{matrix}$$
 Thus, $N_{rd,E}(f(D^*))=N_{rd,F}(D^*)=F^*$ (a splitting field $L$ for $D$ is also a splitting field for $D\otimes_FE$ and $(D\otimes_FE)\otimes_EL=D\otimes_FL$). We finish by noticing that $u_s^{-1}M_\emptyset^{\sigma_s}u_s=f(D)$.
 Thus, $N_{rd,E}(u_s^{-1}M_\emptyset^{\sigma_s}u_s)=F^*$ so we obtain $\mu_{|F^*}=1$.\\
 If $n$ is odd, we get $s=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$, $\mu_{|F^*}=1$ and, as $w_s$ has one fixed point, we have the extra condition $\tilde{\mu}(D'^*)=1$. As $N_{rd,E}(D'^*)=E^*$, we obtain $\mu=1$.

 \paragraph{}We notice that, as $\sigma(u_s^{-1}P_\emptyset u_s)=u_s^{-1}P_\emptyset^- u_s$, then $u_s^{-1}P_\emptyset u_sH$ is open in $G$ (thanks to Proposition \ref{ouvert}) so $P_\emptyset u_sH$ is open in $G$ too.
\end{proof}

Now, we will exhibit a non-zero $\tilde{\mu}$-equivariant linear form on $ind_{P_\emptyset}^G(1)$. To do that, we will follow the strategy of Sections 4.2 and 4.3 of \cite{M.16}.

\paragraph{}For $z\in\mathbb{C}$, we denote by $\delta_z$ the character $(\delta_{P_\emptyset})^z$. For $f\in ind_{P_\emptyset}^G(1)$, we denote by $f_z$ the only element in $ind_{P_\emptyset}^G(\delta_z)$ such that
$f_{z|K}=f_{|K}$ with $K=GL(n,\mathcal{O}_D)$. The map
$$\phi : \begin{matrix}&&\\ind_{P_\emptyset}^G(1)&\rightarrow&ind_{P_\emptyset}^G(\delta_z)\\f&\mapsto&f_z\end{matrix}$$
is a $K$-isomorphism. We also set $s_0=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}\in I(\bar{n})$,
$u_0=u_{s_0}$ and $\sigma_0=\sigma_{s_0}$.

\begin{prop}
 Suppose that $\mu_{|F^*}=1$ if $n$ is even and $\mu=1$ if $n$ is odd. For $f$ in $ind_{P_\emptyset}^G(1)$, the integral $I_{n,z}(f_z)=\int_{u_0^{-1}P_\emptyset u_0\cap H\backslash H}\tilde{\mu}^{-1}(h)f_z(u_0h)dh$ converges for $Re(z)$ large enough and
 there exists $Q\in\mathbb{C}[X]$ such that $Q(q^{-z})I_{n,z}(f_z)$
 belongs to $\mathbb{C}[q^{\pm z}]$ for all $f$ in $ind_{P_\emptyset}^G(1)$ (with $q=|k_F|$).\\
 Moreover, for $z'$ in $\mathbb{C}$, there exists $l_{z'}$ in $\mathbb{N}$ such that
 $$L_{z'}=\underset{z\rightarrow z'}{lim}(1-q^{z'-z})^{l_{z'}}I_{n,z}$$
 belongs to $\mathrm{Hom}_H(ind_{P_\emptyset}^G(\delta_{z'}),\tilde{\mu})\backslash\{0\}$.
\end{prop}
\begin{proof}
 This is due to the Bernstein principle for meromorphic continuation of equivariant linear forms (see Corollary 2.12 in \cite{M.15} which is stated for $GL(n,F)$ but is also true for $GL(n,D)$). First, notice that
 $$P_\emptyset^{\sigma_0}=P_\emptyset\cap u_0Hu_0^{-1}=\{diag(a_1,\dots,a_{n/2},\sigma(a_{n/2}),\dots,\sigma(a_1)); a_i\in D^*\}\text{ if $n$ is even and }$$
 $$P_\emptyset^{\sigma_0}=\{diag(a_1,\dots,a_{\frac{n-1}{2}},b,\sigma(a_{\frac{n-1}{2}}),\dots,\sigma(a_1)); a_i\in D^*, b\in D^{'*}\}\text{ if $n$ is odd.}\hspace{1.5cm}$$
 Thus, as $N_{rd,F}(a)=N_{rd,F}(\sigma(a))$ for $a\in D^*$, then $\delta_z(u_0^{-1}P_\emptyset u_0\cap H)=1$ and $\tilde{\mu}(u_0^{-1}P_\emptyset u_0\cap H)=1$ so $I_{n,z}(\mathcal{C}_c^\infty(P_\emptyset\backslash P_\emptyset u_0 H,\delta_{z}))$ is
 non-zero and well defined for all $z$ in $\mathbb{C}$. As $\mathcal{C}_c^\infty(P_\emptyset\backslash P_\emptyset u_0 H,\delta_{z})\subset ind_{P_\emptyset}^G(\delta_{z})$ (because $P_\emptyset u_0 H$ is open in $G$), it means that for all $z$ in $\mathbb{C}$,
 there exists $\tilde{f}_{z}$ in $ind_{P_\emptyset}^G(\delta_{z})$ such that $I_{n,z}(\tilde{f}_{z})\neq0$.

\paragraph{}To see that $I_{n,z}(f_z)$ is absolutely convergent for $Re(z)$ large enough, first we notice that $|\tilde{\mu}^{-1}|=|\mu^{-1}|\circ N_{rd,E}$ can be written $|\tilde{\mu}^{-1}|=|\cdot|_E^\alpha\circ N_{rd,E}$ for $\alpha$ a complex number. As $|\cdot|_E=|\cdot|_F\circ N_{E/F}$ and as $N_{rd,F|H}=N_{E/F}\circ N_{rd,E|H}$, then $|\tilde{\mu}^{-1}|=|\cdot|_F^\alpha\circ N_{rd,F|H}$ so $|\tilde{\mu}^{-1}|$ can be extended to a character of $G$ that we denote by $\chi$. Now, we have to prove that   $\int_{u_0^{-1}P_\emptyset u_0\cap H\backslash H}\chi(h)f_z(u_0h)dh$ is absolutely convergent. We can see $f_z\mapsto \int_{u_0^{-1}P_\emptyset u_0\cap H\backslash H}\chi(h)f_z(u_0h)dh$ as a function of $\chi f_z$ which belongs to $\chi\otimes ind_{P_\emptyset}^G(\delta_z)$. Then, the absolute convergence comes from Theorems 2.8 and 2.16 of \cite{B-D.08}.

 \paragraph{}Finally, we can use the Bernstein principle for meromorphic continuation of equivariant linear forms, because the space $\mathrm{Hom}_H(ind_{P_\emptyset}^G(\delta_{z}),\tilde{\mu})$ is of dimension $\leq1$ for all $q^{-s}$. Indeed, we prove it
 as in the beginning of the proof of Proposition \ref{necessaire2}. We suppose that $ind_{P_\emptyset}^G(\delta_{z})$ is $\tilde{\mu}$-distinguished under $H$. It implies that $\tilde{\mu}_{|F^*}$ is unitary (considering the central character). Then, we apply
 Mackey theory and Frobenius reciprocity and we get that there exists a unique $s$ (it is $s_0$) such that $\mathrm{Hom}_{P_\emptyset^{\sigma_s}}(\Delta_{P_\emptyset^{\sigma_s}}\delta_z,\tilde{\mu}_s)$ is non-trivial. Thus, we obtain that
 $\mathrm{Hom}_{P_\emptyset^{\sigma_s}}(\Delta_{P_\emptyset^{\sigma_s}}\delta_z,\tilde{\mu}_s)$ is of dimension $1$ so $\mathrm{Hom}_H(ind_{P_\emptyset}^G(\delta_{z}),\tilde{\mu})$ is $1$-dimensional too.
 \end{proof}

\paragraph{}Now, we can give two results of distinction, according to the parity of $n$ :

\begin{prop}\label{dist}
 If $n$ is odd and if $\mu=1$, then $St(1)$ is $\tilde{\mu}$-distinguished.
\end{prop}
\begin{proof}
 Taking $z'=0$, we get that $L_0$ is a non-zero $\tilde{\mu}$-equivariant linear form on $ind_{P_\emptyset}^G(1)$ which is thus $\tilde{\mu}$-distinguished. We end this proof as in Proposition 3.6 of \cite{M.16}.

\end{proof}

\begin{prop}
 If $n$ is even and if $\mu_{|F^*}=1$ and $\mu\neq1$, then $St(1)$ is $\tilde{\mu}$-distinguished.
\end{prop}
\begin{proof}
The proof is the same as in Proposition \ref{dist}.
\end{proof}

To finish this case, we have the following proposition :

\begin{prop}
 If $n$ is even and if $\mu=1$, then $St(1)$ is not $\tilde{\mu}$-distinguished.
\end{prop}
To prove this theorem, we need first two lemmas about the integral $I_{n,z}$. We denote by $\Phi$ the constant function equal to $1$ in $ind_{P_\emptyset}^G(1)$.  Then, for any $z\in\mathbb{C}$, $f_z=f\Phi_z$. If $n=2$, we denote $\Phi$ by $\Phi_2$.

\begin{lem}
Suppose that $\mu=1$ and that $n=2$, then up to a unit in $\mathbb{C}[q^{\pm z}]$, we have :
$$I_{2,z}(\Phi_{2,z})=L(dz-\frac{d}{2},1_{E^*})L(dz,1_{E^*})/L(2dz,1_{F^*}),$$
where $L$ is the usual Tate $L$-factor. In particular, $I_{2,0}(\Phi_2)\neq0$.
\end{lem}
\begin{proof}
The proof is the same as Proposition 4.5 of \cite{M.16} with $u_0=\begin{pmatrix}1&-\iota\\1&\iota\end{pmatrix}$. We only give the beginning and the end. We set $\nu_F:=|N_{rd,F}(\cdot)|_F$.
For $Re(z)$ large enough, we have :
$$\Phi_{2,z}(g)=\nu_F(g)^{dz}\int_{D^*}\Phi((0,t)g)\nu_F(t)^{2dz}dt/L(2dz,1_{F^*}).$$
\paragraph{}Finally, if we set $\epsilon=(\nu_F(h_\iota^{-1})\nu_F(u_0))^{dz}$ and denote by $\Phi_0$ the characteristic function of $\mathcal{M}(2,\mathcal{O}_{D'})$ (and recalling that $d'=\frac{d}{2}$ is the index of $D'$ over $E$), we obtain :
\begin{eqnarray*}
 I_{2,z}(\Phi_{2,z})&=&\epsilon\int_H\Phi_0(h)\nu_F(h)^{dz}dh/L(2dz,1_{F^*})\\
 &=&\epsilon\int_H\Phi_0(h)\nu_E(h)^{2d'z}dh/L(2dz,1_{F^*})\\
 &=&\epsilon L(2d'z-\frac{1}{2}(2d'-1),1_H)/L(2dz,1_{F^*})\\
 &&\text{ by definition of $Z$ and $L$-functions}\\
 &=&\epsilon L(2d'z-\frac{1}{2}(3d'-1),1_{D'^*})L(2d'z-\frac{1}{2}(d'-1),1_{D'^*})/L(2dz,1_{F^*})\\
 &&\text{by inductivity relation of the Godement-Jacquet $L$-factor }L(z,1_H)\\
 &=&\epsilon L(2d'z-d',1_{E^*})L(2d'z,1_{E^*})/L(2dz,1_{F^*})
\end{eqnarray*}
If $z=0$, $L(2d'z,1_{E^*})$ and $L(2dz,1_{F^*})$ have one simple pole and $L(2d'z-d',1_{E^*})$ has no pole so $I_{2,0}(\Phi_2)\neq0$.
\end{proof}

We recall Proposition 4.6 of \cite{M.16} :
\begin{lem}\label{nonzero}
Suppose that $\mu=1$. For $n=2m$, let $P$ be the standard parabolic subgroup of $G$ corresponding to the partition $\bar{n}=(1,\dots,1,2,1,\dots,1)$ with $n_{m}=2$. Then, there is $f$ in $ind_P^G(1)$ such that $I_{n,z}(f\Phi_z)=I_{2,z}(\Phi_{2,z})$. In particular, taking $z=0$, one has $I_{n,0}(f)=I_{2,0}(\Phi_2)\neq0$.
\end{lem}
\begin{proof}
The proof is exactly the same as in Proposition 4.6 of \cite{M.16}.
\end{proof}
Finally, we come back to the proof of the proposition :
\begin{proof}[Proof of the proposition]
 Suppose that $\mu=1$ and $St(1)$ is $\tilde{\mu}$-distinguished (\textit{i.e.} $H$-distinguished). Then, $ind_{P_\emptyset}^G(1)$ is $H$-distinguished so there exists $L$ a non-zero $H$-invariant linear form on $ind_{P_\emptyset}^G(1)$.
 As $\mathrm{dim}(\mathrm{Hom}_H(ind_{P_\emptyset}^G(1),1))=1$ (thanks to Proposition \ref{necessaire2}), then $L$ equals to $L_0$ up to a non-zero scalar. As $St(1)$ is distinguished, $L_{|ind_P^G(1)}$ must be equal to zero
 for all standard parabolic subgroups $P$ of type $(1,\dots,1,2,1,\dots,1)$. Moreover, as said in Proposition \ref{necessaire2}, $L_0$ restricts non trivially to
 $\mathcal{C}_c^\infty(P_\emptyset\backslash P_\emptyset u_0H,1)\subset ind_{P_\emptyset}^G(1)$.\\
 As $I_{n,0}(\mathcal{C}_c^\infty(P_\emptyset\backslash P_\emptyset u_0H,1))$ is non zero (and is well defined), this implies that $L_0=I_{n,0}$. Now, we take $f$ and $P$ as in Lemma \ref{nonzero}. Then we have
 $L_0(f)=I_{n,0}(f)\neq0$ which contradicts the distinction of $St(1)$.
\end{proof}

\section{Case $d$ odd and $n$ even}

 \subsection{Preliminaries}

 We set $n=2m$ and we suppose that $d$ (the index of $D$ over its center $F$) is odd. Let us consider $D\otimes_F E$ which is a central division $E$-algebra of dimension $d^2$ (thanks
 to Wedderburn structure theorem and Hasse's invariant). We can choose $\delta\in (D\otimes_F E)\backslash D$ such that $\Delta:=\delta^2\in F$ (for example, $\delta=1\otimes x$ with $x$ in $E\backslash F$
 such that $x^2$ is in $F$). Then, $D\otimes_F E$ is of dimension $2$ over $D$ so
 we can write $D\otimes_F E$ as $D\oplus \delta D$ and $D\otimes_FE$ identifies as
 a right $D$-vector space of dimension $2$.\\
 Now, if we let $(e_1,\dots,e_m)$ denote the canonical basis of $(D\otimes_FE)^m$, then the right $D$-vector space $(D\otimes_FE)^m$ identifies with the $D$-vector space $D^{2m}$ via the basis $\mathcal{B}=(e_1,\dots,e_m,\delta e_m,\dots,\delta e_1)$ of $D^{2m}$ so
 $\mathrm{End}((D\otimes_FE)^m)_D \simeq \mathrm{End}(D^{2m})_D$. Now, if $u\in \mathrm{End}((D\otimes_FE)^m)_D$, it is easy to see that $u\in \mathrm{End}((D\otimes_FE)^m)_{D\otimes_FE}$ if and only if $u$ commutes with the multiplication by $\delta$ (denoted $\mu_\delta$). In the basis $\mathcal{B}$, the matrix of
 the endomorphism $\mu_\delta$ is given by
 $$\begin{pmatrix}&&&&&\Delta\\&&&&\iddots&\\&&&\Delta&&\\&&1&&&\\&\iddots&&&&\\1&&&&& \end{pmatrix}=:U_{\Delta_{2m}}$$
 so $\mathcal{M}_m(D\otimes_FE)$ can be viewed as the fixed points of $\mathcal{M}_{2m}(D)$ under the involution $int_{U_{\Delta_{2m}}}$ of $\mathcal{M}_{2m}(D)$ (where $int_{U_{\Delta_{2m}}}(x) = U_{\Delta_{2m}} x {U_{\Delta_{2m}}}^{-1}$). \\
 Now, $E$ can be embedded in $\mathcal{M}_{2m}(D)$ via
 $$x\in E\mapsto \begin{pmatrix}1\otimes x&&\\&\ddots&\\&&1\otimes x\end{pmatrix}\in \mathcal{M}_m(D\otimes_FE)\subset\mathcal{M}_{2m}(D)$$
 and it is easy to check that $g\in C_{\mathcal{M}_{2m}(D)}(E)$ if and only if $g$ commutes with $U_{\Delta_{2m}}$ \textit{i.e.} if and only if $g\in\mathcal{M}_m(D\otimes_FE)$. Thus $C_{\mathcal{M}_{2m}(D)}(E)=\mathcal{M}_m(D\otimes_FE)$
 and we set $G=GL(2m,D)$, $H=GL(m,D\otimes_FE)=G^\sigma$ with $\sigma = int_{U_{\Delta_{2m}}}$.
 We recall that $N_{rd,F}$ denotes the reduced norm on $GL(n,D)$ and $N_{rd,E}$ denotes the reduced norm on $GL(m,D\otimes_FE)$ (as well as its restriction to any subgroup).

 \subsection{Representatives of $P\backslash G/H$}

 Let $P$ be a standard parabolic subgroup of $G=GL(2m,D)$ corresponding to a partition $\bar{n}=(n_1,\dots,n_r)$ of $n=2m$. We define $I(\bar{n})$ to be the set of symmetric matrices $s=(n_{i,j})\in\mathcal{M}_r(\mathbb{N})$ with positive integral
 entries, even on the diagonal, and such that the sum of the $i$-th row is equal to $n_i$ for all $i$ in $\{1,\dots,r\}$.
 \paragraph{}As each $n_{i,i}$ is even for $i\in\mathbb\llbracket 1,r\rrbracket$, we can write $n_{i,i}=2m_{i,i}$ and we can write $n$ as an ordered sum of integers in two different ways :
  \begin{equation*}n=m_{1,1}+n_{1,2}+\dots+n_{1,r}+m_{2,2}+n_{2,3}+\dots+n_{2,r}+m_{3,3}+\dots+m_{r-1,r-1}+n_{r-1,r}+m_{r,r}\end{equation*}
\begin{equation} +m_{r,r}+n_{r,r-1}+m_{r-1,r-1}+\dots+m_{3,3}+n_{r,2}+\dots+n_{3,2}+m_{2,2}+n_{r,1}+\dots+n_{2,1}+m_{1,1}\end{equation}
\begin{flushright}(1\textsuperscript{st} ordering)\end{flushright}

 \begin{equation}n=n_{1,1}+\dots+n_{1,r}+n_{2,1}+\dots+n_{2,r}+\dots+n_{r,1}+\dots+n_{r,r}\hspace{1cm}\end{equation}
 \begin{flushright}(2\textsuperscript{nd} ordering corresponding to the lexicographical ordering)\end{flushright}

We denote by $w_s$ the matrix of the permutation (still denoted $w_s$) defined as follows:\\
If $i\in \llbracket{1,r \rrbracket}$, then for $k\in \llbracket{1,m_{i,i}\rrbracket}$, we set
$$w_s(m_{1,1}+\dots+m_{i-1,i-1}+n_{i-1,i}+\dots+n_{i-1,r}+k)=n_{1,1}+\dots+n_{i-1,1}+\dots+n_{i,i-1}+k,$$ and
$$w_s(m_{1,1}+\dots+m_{i+1,i+1}+n_{r,i}+\dots+n_{i+1,i}+k)=n_{1,1}+\dots+n_{i-1,1}+\dots+n_{i,i-1}+k+m_{i,i}.$$
If $i<j$, for $k\in \llbracket{1,n_{i,j}\rrbracket}$, we set
$$w_s(m_{1,1}+ \dots+ m_{i,i}+ n_{i,i+1}+ \dots +n_{i,j-1}+k)=n_{1,1}+ \dots + n_{1,r}+\dots+ n_{i,1}+ \dots+ \dots +n_{i,j-1}+k $$ and
$$w_s(m_{1,1}+ \dots+ m_{i+1,i+1}+ n_{r,i} +\dots+ n_{j+1,i}+k)= n_{1,1}+ \dots + n_{1,r}+\dots+ n_{j,1}+ \dots+ \dots +n_{j,i-1}+k .$$

In other words, $w_s$ sends an integer of rank $k$ according to the 1\textsuperscript{st} ordering to the integer of rank $k$ corresponding to the 2\textsuperscript{nd} ordering.

\paragraph{}A proof similar to Proposition 3.1 of \cite{M.16} shows the following result :

\begin{prop}
Let $\bar{n}$ be a partition of $n$ and $P$ be a standard parabolic subgroup of $G$ corresponding to this partition, then $G=\bigsqcup_{s\in I(\bar{n})}Pw_sH$.
\end{prop}

\begin{rem}
 \begin{sloppypar}
There is a bijection between $P_n(D)\,\backslash \,GL(n,D)\,/\,GL(m,D\otimes_F E)$ and $P_n(F)\backslash GL(n,F)/GL(m,E)$ via the identity map of $\{w_s|s\in I(\bar{n})\}$.
 \end{sloppypar}
\end{rem}

Now, for $s\in I(\bar{n})$, we set $t_s=w_sU_{\Delta_{2m}}w_s^{-1}$. It is a monomial matrix (so it is in $N_G(M_\emptyset)$) and if we let $\tau_s$ denote the image of $t_s$ in $\mathcal{\sigma}_n=N_G(M_\emptyset)/M_\emptyset$, then
$\tau_s$ is a permutation matrix of order $2$, given by the formula $\tau_s=w_sww_s^{-1}$ (where $w=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$).\\
Then, we can see that $w_sHw_s^{-1}$ is the group of the fixed points of $G$ under the involution :
$$\sigma_s : x\mapsto t_sxt_s^{-1}.$$

\paragraph{}We need to know exactly how acts the permutation $\tau_s$. One checks that $\tau_s$ is the involution of $\llbracket 1,n\rrbracket=I=I_{1,1}\cup I_{1,2}\cup \dots \cup I_{1,r} \cup \dots \cup I_{r,1}\cup \dots \cup I_{r,r-1}\cup I_{r,r}$
(with $I_{i,j}$ of length $n_{i,j}$), which stabilises each $I_{i,i}$, acting on it as the symmetry with respect to its midpoint, and which stabilises $I_{i,j}\cup I_{j,i}$ (for $i<j$) and acts
on this union of intervals as the symmetry with center the midpoint of the interval joining the left end of $I_{i,j}$ and the right end of $I_{j,i}$.

\paragraph{}For $s\in I(\bar{n})$, we denote by $P_s$ the standard parabolic subgroup of $G$ corresponding to the sub-partition $s$ of $\bar{n}$. As usual, we denote $P=MN$ and $P_s=M_sN_s$ the standard Levi decomposition of $P$ and $P_s$.
Then, again as in the even case, we have the following proposition :

\begin{lem}\label{lemme}
For $s\in I(\bar{n})$, one has $\tau_s(\Phi_M^-)\subset \Phi^-$, $\tau_s(\Phi_M^+)\subset \Phi^+$.
\end{lem}

\begin{prop}\label{decomp}
For any $s\in I(\bar{n})$, one has $P\cap w_s H w_s^{-1}=P_s\cap w_s H w_s^{-1}$, and
$P_s\cap w_s H w_s^{-1}$ is the semidirect product of $M_s\cap w_s H w_s^{-1}$ and $N_s\cap w_s H w_s^{-1}$.
\end{prop}

We will now let $P_s^{\sigma_s}$ denote $P_s\cap w_s H w_s^{-1}$ and $M_s^{\sigma_s}$ denote $M_s\cap w_s H w_s^{-1}$. We can explicitly describe the group $M_s^{\sigma_s}$ : an element $a\in M_s^{\sigma_s}$ is of the form :
$$a=diag(a_{1,1},a_{1,2},\dots,a_{1,r},a_{2,1},\dots,a_{r,r-1},a_{r,r})$$

where $a_{i,i}\in GL(n_{i,i},D)$ satisfies $a_{i,i}=U_{\Delta}a_{i,i}U_{\Delta}^{-1}$ ($U_{\Delta}:=U_{\Delta_{n_{i,i}}}$)
\textit{i.e.} $a_{i,i}\in GL(m_{i,i},D\otimes_FE)$ and $a_{i,j}\in GL(n_{i,j},D)$ satisfies $a_{i,j}=wa_{j,i}w^{-1}$ if $i\neq j$ ($w=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$) for all $i$, $j$ in $\{1,\dots,r\}$.

\begin{prop}\label{car}
We have the following equality of characters :
$$(\delta_{P_s^{\sigma_s}})_{|M_s^{\sigma_s}}=(\delta_{P_s}^{1/2})_{|M_s^{\sigma_s}} .$$
\end{prop}

\begin{proof}
Again, the method is the same as in Proposition 4.4 of \cite{M.11}.
Thanks to Lemma 1.10 of \cite{K-T.08}, it is enough to check the equality on the $F$-split component $Z_s^{\sigma_s}$ (the maximal $F$-split torus in the center of $M_s$) of $M_s^{\sigma_s}$ because $\sigma_s$ is defined over $F$.
Thus, we will consider each character on $Z_s^{\sigma_s}$, that is to say, the subgroup of the matrices of the form
$$\begin{pmatrix}\lambda_{1,1}I_{n_{1,1}}&&&\\&\lambda_{1,2}I_{n_{1,2}}&&\\&&\ddots&\\&&&\lambda_{r,r}I_{n_{r,r}}\end{pmatrix}$$
with $\lambda_{i,j}=\lambda_{j,i}\in F^*$ and $n_{i,i}$ even.
\paragraph{}For $\alpha\in\Phi$, we set $\mathcal{N}_{\alpha,\tau_s(\alpha)}=\{x\in Lie(N_\alpha)+Lie(N_{\tau_s(\alpha)}\: ;\:\sigma_s(x)=x\}$. It's a $D$-vector space of dimension $1$.\\
For $t\in Z_s^{\sigma_s}$, we have :

\begin{equation}\label{relation1}
\delta_{P_s^{\sigma_s}}(t) = \prod_{\{\alpha,\tau_s(\alpha)\}\subset \Phi^+-\Phi_s^+}|N_{rd,F}(Ad(t)_{|\mathcal{N}_{\alpha,\tau_s(\alpha)}})|_F = \prod_{\{\alpha\in \Phi^+-\Phi_s^+; \tau_s(\alpha)\in \Phi^+-\Phi_s^+\}}|N_{rd,F}(\alpha(t))|_F^{1/2}
\end{equation}
\begin{sloppypar}
The second equality of \eqref{relation1} comes from the fact that if
$\{\alpha_0,\tau_s(\alpha_0)\}\subset \Phi^+-\Phi_s^+$, then $|N_{rd,F}(Ad(t)_{|\mathcal{N}_{\alpha_0,\tau_s(\alpha_0)}})|_F =|N_{rd,F}(\alpha_0(t))|_F$ and the power $1/2$ comes from the fact that $\tau_s$ has no fixed point
whereas $\mathcal{N}_{\alpha,\tau_s(\alpha)}$ is of dimension $1$.
\end{sloppypar}

As in the even case, $\underset{{\{\alpha\in\Phi^+-\Phi_s^+;\;\tau_s(\alpha)\notin\Phi^+-\Phi_s^+\}}}\prod|N_{rd,F}(\alpha(t))|_F=1$. This implies that
$$\delta_{P_s}^{\sigma_s}(t)=\underset{\{\alpha\in\Phi^+-\Phi_s^+\}}{\prod}|N_{rd,F}(\alpha(t))|_F^{1/2}$$

\paragraph{}Finally, by definition we have :
$$\delta_{P_s}(t)=|N_{rd,F}(Ad(t)_{|Lie(N_s)})|_F = \prod_{\{\alpha\in\Phi^+-\Phi_s^+\}}|N_{rd,F}(\alpha(t))|_F.$$
and we have the characters equality.

\end{proof}

\subsection{Distinguished Steinberg representations}
In this part, we will study whether the Steinberg representation is $\mu\circ N_{rd,E}$-distinguished under $H$ or not according to the character of $E^*$ considered $\mu$.
For $\mu$ a character of $E^*$, we set $\tilde{\mu}:=\mu\circ N_{rd,E}$. We recall that $St(1)$ is the Steinberg representation
$ind_{P_\emptyset}^G(1)/S$ with $S=\sum_P ind_P^G(1)$ where the standard parabolic subgroups $P$ in the sum correspond to a partition $\bar{n}$ of $n$ with all $n_i$'s equal to $1$ except one which
is $2$.
\paragraph{}First, we give a necessary condition on $\mu$ to allow $St(1)$ to be $\tilde{\mu}$-distinguished.

\begin{prop}\label{necessaire}
If $St(1)$ is $\tilde{\mu}$-distinguished under $H$, then $\mu_{|F^*}=1$. Moreover, only the open orbit $P_\emptyset H$ supports a $\tilde{\mu}$-equivariant linear form and $dim\left(Hom_H(St(1),\tilde{\mu})\right)= 1$.
\end{prop}

\begin{proof}
 As the method is the same as in the proof of Proposition \ref{necessaire2}, we will only underline the most important points.\\
 First, we recall that $\tilde{\mu}_{|F^*}$ is unitary. Then, $St(1)$ being $\tilde{\mu}$-distinguished implies, by Frobenius reciprocity, Mackey theory and Proposition \ref{car2}, that there
 exists $s$ in $I(\bar{n})$ such that $(\delta_{P_\emptyset}^{1/2})_{|M_\emptyset^{\sigma_s}}=\tilde{\mu}_{s|M_\emptyset^{\sigma_s}}$ where $\tilde{\mu}_s(x)=\tilde{\mu}(w_s^{-1}xw_s)$ for $x$ in $w_sHw_s^{-1}$.

\paragraph{}Thus, we get $s=\begin{pmatrix}&&1\\&\iddots&\\1&&\end{pmatrix}$ and $\mu\circ N_{rd,E}\left(\begin{pmatrix}a_1&&&&&\\&\ddots&&&&\\&&a_m&&&\\&&&a_m&&\\&&&&\ddots&\\&&&&&a_1\end{pmatrix}\right)=1$ for all $(a_1,\dots,a_m)$
in $(D^*)^m$. As $\begin{pmatrix}a_1&&&&&\\&\ddots&&&&\\&&a_m&&&\\&&&a_m&&\\&&&&\ddots&\\&&&&&a_1\end{pmatrix}$ is the embedding of $\begin{pmatrix}a_1&&\\&\ddots&\\&&a_m\end{pmatrix}$ of $\mathcal{M}_m(D\otimes_FE)$ in
$\mathcal{M}_{2m}(D)$, it implies that $\mu\circ N_{rd,E}(D^*)=1$.\\
Let us show that $N_{rd,E}(D^*)=1$. There exists $L/F$ an extension of dimension $d$ such that $D\otimes_FL\simeq \mathcal{M}_d(L)$. As $d$ is odd, $L\otimes_FE$ is a field and it is a $d$-dimensional extension of $E$.
Thus, we have the following natural commutative diagram :
$$\begin{matrix}D\otimes_FE&\longrightarrow&\mathcal{M}_d(L\otimes_FE)\\\uparrow&\circlearrowleft&\uparrow\\D&\longrightarrow&\mathcal{M}_d(L)\end{matrix}$$
which implies $N_{rd,E}(D^*)=N_{rd,F}(D^*)=F^*$. We deduce that $\mu_{|F^*}=1$.
\paragraph{}We end this proof by noticing that as $\sigma(P_\emptyset)=P_\emptyset^-$, then $P_\emptyset H$ is open in $G$ thanks to Proposition \ref{ouvert}.
\end{proof}

Now, we show that $St(1)$ is $\tilde{\mu}$-distinguished under $H$ if and only if $\mu_{|F^*}=1$ and $\mu\neq 1$ in the following two propositions :
\begin{prop}
If $\mu_{|F^*}=1$ and $\mu\neq 1$, then $St(1)$ is $\tilde{\mu}$-distinguished under $H$.
\end{prop}
\begin{proof}
Suppose $\mu_{|F^*}=1$ and $\mu\neq1$. We have $\sigma(P_\emptyset)=P_\emptyset^-$ and if $a$ is in
$M_\emptyset^\sigma$, then $a$ is of the form
$a=\begin{pmatrix}a_1&&&&&\\&\ddots&&&&\\&&a_m&&&\\&&&a_m&&\\&&&&\ddots&\\&&&&&a_1\end{pmatrix}$ with $a_1,\dots,a_m \in D^*$ and $\delta_{P_\emptyset}^{-1/2}\tilde{\mu}^{-1}(a)=1$. Thus, thanks to Theorem 2.8 of \cite{B-D.08},
$ind_{P_\emptyset}^G(1)$ is $\tilde{\mu}$-distinguished. We end the proof as in Proposition 3.6 of \cite{M.16}.
\end{proof}

Finally, we get the non-distinguished case :

\begin{prop}
If $\mu=1$, then $St(1)$ is not $\tilde{\mu}$-distinguished under $H$.
\end{prop}
\begin{proof}
We do not give the proof because it is similar to the one of Theorem 3.1 in \cite{M.16}.
\end{proof}

\section{Prasad and Takloo-Bighash conjecture}\label{epsilon}

\paragraph{}Let us summarize our results :

\begin{thm}
Let $n$ be a positive integer and let $\mu$ be a character of $E^*$. $E$ is embedded in $\mathcal{M}_n(D)$ if and only if $nd$ is even. We set $G=GL(n,D)$ and $St(1)=St(n,1)$ the Steinberg representation of $G$.
We recall that $\tilde{\mu}$ denotes $\mu\circ N_{rd,E}$.
\begin{itemize}
\item If $d$ is even, $H=(C_{\mathcal{M}_n(D)}(E))^\times=GL(n,C_D(E))$ and $St(n,1)$ is $\tilde{\mu}$-distinguished under $H$ if and only if
\begin{itemize}
\item $\mu_{|F^*}=1$ and $\mu\neq1$ if $n$ is even.
\item $\mu=1$ if $n$ is odd.
\end{itemize}
\item If $d$ is odd and $n$ is even, $H=(C_{\mathcal{M}_n(D)}(E))^\times=GL(n/2,D\otimes_FE)$ and $St(n,1)$ is $\tilde{\mu}$-distinguished under $H$ if and only if $\mu_{|F^*}=1$ and $\mu\neq1$.
\end{itemize}
\end{thm}

\paragraph{}Now, let us write and prove the Conjecture 1 of \cite{P-TB} for the Steinberg representation $St(n,1)$ (which is a discrete series representation) :

\begin{thm}\label{TB}(Prasad and Takloo-Bighash conjecture, Steinberg case) Let $A=\mathcal{M}_n(D)$ and $\pi=St(n,1)$ which is an irreducible admissible representation of $A^\times=GL(n,D)=G$. Recall that $\pi$ corresponds via Jacquet-Langlands correspondence to
$St(nd,1)$ (a representation of $GL(nd,F)$) with central character $\omega_\pi=1$. Let $\mu$ be a character of $E^\times$ such that $\mu^{\frac{nd}{2}}|_{F^\times}=\omega_\pi=1$. Then, the character $\mu\circ N_{rd,E}$ of
$H=(C_{\mathcal{M}_n(D)}(E))^\times$ appears as a quotient in $\pi$ restricted to $H$ if and only if :
\begin{enumerate}
 \item the Langlands parameter of $\pi$ takes values in $GSp_{nd}(\mathbb{C})$ with similitude factor $\mu_{|F^\times}$.
 \item the epsilon factor satisfies $\epsilon(\frac{1}{2},\pi\otimes Ind_E^F(\mu^{-1}))=(-1)^n\omega_{E/F}(-1)^{\frac{nd}{2}}$ (where $\omega_{E/F}$ is the quadratic character of $F^\times$ with kernel the norms of $E^\times$).
\end{enumerate}

\end{thm}

As $N_{rd,F|H}=N_{E/F}\circ N_{rd,E}$, then if $\mu$ is a character of $E^*$, $\mu\circ N_{rd,E}$ can be extended to a character of $G$ if and only if there exists $\chi$ a character of $F^*$ such that $\mu=\chi\circ N_{E/F}$.

\paragraph{}Let us rephrase Theorem \ref{TB} in this case. Let $\pi$ and $\mu$ be as in the conjecture and suppose that there exists $\chi$ a character of $F^*$ such that $\mu=\chi\circ N_{E/F}$. We denote by $W_F$ the Weil group
of $F$ and $BC_E$ denotes the base change to $E$.

\begin{itemize}
 \item As $\mu=\chi\circ N_{E/F}$, the statement ``$\mu\circ N_{rd,E}$ appears as a quotient in $\pi$ restricted to $H$'' is equivalent to saying that $St(n,1)$
 is $\chi\circ N_{rd,F}$-distinguished under $H$. This is again equivalent to $(\chi\circ N_{rd,F})^{-1}\otimes St(n,1)=St(n,\chi^{-1})$ is $H$-distinguished.
 \item Now, let us consider the 1\ts{st} point of the conjecture. The Langlands parameter of $\pi=St(n,1)$ is $Sp(nd)=:\Phi$ (where $Sp(nd)$ denotes the unique irreducible representation of $SL(2,\mathbb{C})$ of dimension $nd$) . The 1\ts{st} assertion in the conjecture means :
 \begin{eqnarray}
\text{there exists }<\cdot,\cdot> \text{ a nondegenerate alternating  bilinear form on } \mathbb{C}^{nd}=:V \text{ such that } \nonumber\\
\forall w\in W_F, \,\forall v,v'\in V, \hspace{0.5cm}<\Phi(w).v,\Phi(w).v'>=\mu_{|F^\times}(w)<v,v'>=\chi^2<v,v'>. \label{forme}
 \end{eqnarray}
As the Langlands parameter of $St(n,\chi^{-1})$ is $Sp(nd)\otimes \chi^{-1}=:\Psi$, statement \eqref{forme} is equivalent to :
\begin{eqnarray*}
 &&\text{there exists } <\cdot,\cdot> \text{ a nondegenerate alternating bilinear form on } V \text{ such that } \\
 &&\forall w\in W_F,\;\forall v, v'\in V,\hspace{.5cm} <\Psi(w).v,\Psi(w).v'>=<v,v'>
\end{eqnarray*}
which is exactly the definition of $St(n,\chi^{-1})$ being symplectic.
\item Finally, we consider the 2\ts{nd} point of the conjecture and we formulate the epsilon factor in another way :
\begin{eqnarray*}
 \epsilon(\frac{1}{2},\pi\otimes Ind_E^F(\mu^{-1}))&=&\epsilon(\frac{1}{2},Sp(nd)\otimes Ind_E^F((\chi\circ N_{E/F})^{-1}))\\
 &=&\epsilon(\frac{1}{2},Ind_E^F(Sp(nd)\otimes \chi_E^{-1}))\hspace{1cm}\text{where }\chi_E=\chi\circ N_{E/F}=\mu\\
 &=&\omega_{E/F}(-1)^{\frac{nd}{2}}\epsilon(\frac{1}{2},Sp(nd)\otimes \chi_E^{-1})\\
 &=&\omega_{E/F}(-1)^{\frac{nd}{2}}\epsilon(\frac{1}{2},BC_E(St(n,\chi^{-1})))
\end{eqnarray*}
so the 2\ts{nd} point of the conjecture is equivalent to : $\epsilon(\frac{1}{2},BC_E(St(n,\chi^{-1})))=(-1)^n$.
\end{itemize}

\paragraph{}To sum up, under the additional hypothesis that the character $\mu\circ N_{rd,E}$ of $H$ can be extended to a character of $G$, Theorem \ref{TB} is equivalent to the following, which is a reformulation similar to
Conjecture 1.4 of \cite{FMW}:

\begin{thm}
 Let $St(n,\chi)$ be the Steinberg representation of $G=GL(n,D)$. $St(n,\chi)$ is $H$-distinguished if and only if it is symplectic and $\epsilon(\frac{1}{2},BC_E(St(n,\chi)))=(-1)^n$.
\end{thm}

\begin{rem}
 In \cite{FMW}, Conjecture 1.4 (a reformulation of Conjecture 1 of Prasad and Takloo-Bighash in \cite{P-TB}) is stated for general representations but only in the quaternionic case. It is checked for supercuspidal representations
 with extra conditions.
\end{rem}

\begin{proof}[Proof of Theorem \ref{TB}]

We use the usual $\epsilon$, $\gamma$ and $L$-factors as defined in Godement-Jacquet (see \cite{G-J.72}); we omit the third parameter in $\epsilon$ and $\gamma$ which is a non-trivial additive character of $E$ but trivial on $F$.

\paragraph{}According to the preceding reformulation (the three points above), we have to prove that $St(n,1)$ is $\tilde{\mu}$-distinguished if and only if $\mu_{|F^*}=1$ and $\epsilon(\frac{1}{2},Sp(nd)\otimes\mu^{-1})=(-1)^n$
that is to say $\epsilon(\frac{1}{2},St(nd,\mu^{-1}))=(-1)^n$ where $St(nd,\mu^{-1})$ is the Steinberg representation of $GL(nd,E)$ with parameter $\mu^{-1}$.

\paragraph{}For $s\in\mathbb{C}$,
$$\gamma(s,St(nd,\mu^{-1}))=(\mu^{-nd}(-1))^{nd-1}\epsilon(s,St(nd,\mu^{-1}))\frac{L(1-s,St(nd,\mu^{-1}))}{L(s,St(nd,\mu^{-1}))}.$$

\begin{sloppypar}On the other hand, $$\gamma(s,St(nd,\mu^{-1}))=\gamma(s+\frac{1-nd}{2},\mu^{-1})\times\gamma(s+\frac{1-nd}{2}+1,\mu^{-1})\times\dots\times\gamma(s+\frac{nd-1}{2},\mu^{-1}).$$\\

If $\mu_{|F^*}=1$, then $\bar{\mu}=\check{\mu}$ (where $\bar{\mu}$ denotes the Galois twist of $\mu$ and $\check{\mu}$ denotes the contragredient of $\mu$) so
$$L(s,\mu^{-1})=L(s,\check{\mu}^{-1})\hspace{0.5cm}\text{and}\hspace{0.5cm}\epsilon(s,\mu^{-1})\epsilon(1-s,\mu^{-1})=1\hspace{0.5cm}\forall s\in \mathbb{C}.$$
Hence, if $\mu_{|F^*}=1$, $\gamma(s,\mu^{-1})=\epsilon(s,\mu^{-1})\frac{L(1-s,\mu^{-1})}{L(s,\mu^{-1})}\:\forall s\in\mathbb{C}$, so $\gamma(s,St(nd,\mu^{-1}))$ equals
\begin{equation*}
\begin{array}{l}
\epsilon(s+\frac{1-nd}{2},\mu^{-1})\times\dots\times\epsilon(s+\frac{nd-1}{2},\mu^{-1})\times\frac{L(1-s+\frac{nd-1}{2},\mu^{-1})\times\dots\times L(1-s+\frac{1-nd}{2},\mu^{-1})}{L(s+\frac{1-nd}{2},\mu^{-1})\times\dots\times L(s+\frac{nd-1}{2},\mu^{-1})}\\
=\frac{L(\frac{1}{2}+\frac{nd-1}{2},\mu^{-1})\times\dots\times L(\frac{1}{2}+\frac{1-nd}{2},\mu^{-1})}{L(\frac{1}{2}+\frac{1-nd}{2},\mu^{-1})\times\dots\times L(\frac{1}{2}+\frac{nd-1}{2},\mu^{-1})}.
\end{array}
\end{equation*}
Now for $\mu$ a character of $E^*$ such that $\mu_{|F^*}=1$ and $s$ a real number, we need to know when $L(s,\mu^{-1})$ has a pole. Let $\varpi_E$ be a uniformizer of $E$ and $q_E$ denote the cardinality of the residue field of $E$.\\
If $\mu$ is non-ramified, $L(s,\mu^{-1})=\frac{1}{1-\mu^{-1}(\varpi_E)q_E^{-s}}$ and $L(s,\check{\mu}^{-1})=L(s,\mu^{-1})=\frac{1}{1-\mu(\varpi_E)q_E^{-s}}$ so $L(s,\mu^{-1})$ has a pole is equivalent to :
$$\mu^{-1}(\varpi_E)=q_E^s=\mu(\varpi_E)\Leftrightarrow s=0\text{ and }\mu(\varpi_E)=1\textit{ i.e. }\mu=1$$
\begin{flushleft}because $\mu$ is non-ramified.\end{flushleft}
If $\mu$ is ramified, $L(s,\mu^{-1})=1$ so $L(s,\mu^{-1})$ has no pole.\\
To conclude, $L(s,\mu^{-1})$ has a pole if and only if $s=0$ and $\mu=1$.

\paragraph{}Finally, if $\mu_{|F^*}=1$, $\gamma(\frac{1}{2},St(nd,\mu^{-1}))=\epsilon(\frac{1}{2},St(nd,\mu^{-1}))$ and also
$\gamma(\frac{1}{2},St(nd,\mu^{-1}))=\frac{L(\frac{nd}{2},\mu^{-1})\times\dots\times L(1-\frac{nd}{2},\mu^{-1})}{L(1-\frac{nd}{2},\mu^{-1})\times\dots\times L(\frac{nd}{2},\mu^{-1})}$ so :
\begin{itemize}
\item ($\mu_{|F^*}=1$ and $\mu\neq1$) $\Leftrightarrow$ ($\epsilon(\frac{1}{2},St(nd,\mu^{-1}))=1$ and $\mu_{|F^*}=1$).
\item ($\mu=1$) $\Leftrightarrow$ ($\epsilon(\frac{1}{2},St(nd,\mu^{-1}))=\underset{s\rightarrow \frac{nd}{2}}{lim}\frac{L(\frac{nd}{2}-s,\mu^{-1})}{L(s-\frac{nd}{2},\mu^{-1})}=\underset{s\rightarrow 0}{lim}\frac{1-q_E^{-s}}{1-q_E^s}=\underset{s\rightarrow 0}{lim}\frac{-1}{q_E^s}=-1$
\\and $\mu_{|F^*}=1$).
\end{itemize}
This is equivalent to $St(n,1)$ is $\tilde{\mu}$-distinguished if and only if $\mu_{|F^*}=1$ and $\epsilon(\frac{1}{2},St(nd,\mu^{-1}))=(-1)^n$.
\end{sloppypar}
\end{proof}

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config: arxiv


TITLE: Sufficient condition for function given as integral to be continuous on $[0,c]$
QUESTION [0 upvotes]: Let $f: \mathbb{R} \times \mathbb{R} \rightarrow  \mathbb{R}$ and suppose that $F(x)=\int_0 ^c   f(x,y)  dy$, where $c$ is some positive constant. Is it sufficient that $f$ is uniformly continuous on $[0,c]\times [0,c]$, so that $F(x) $ is continuous on $[0,c]$?

REPLY [1 votes]: Yes, it is sufficient. Given $\epsilon>0$, there is a $\delta>0$ so that
for $x_i,y  \in [0,c]$,
$$|x_2-x_1|<\delta \Rightarrow |f(x_2,y )-f(x_1,y)|<\epsilon/c  \,,
$$
so for such $x_1,x_2$ we have, by https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals
$$|F(x_2)-F(x_1)| \le \int_0^c |f(x_2,y )-f(x_1,y)| \, dy <\epsilon \,.$$