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\begin{document}

\title{SOME TYPES AND COVERS FOR QUATERNIONIC HERMITIAN GROUPS}
\author{Thomas Laebel Madsen}
\date{July 2014}

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{ %\large
UNIVERSITY OF OKLAHOMA
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GRADUATE COLLEGE
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SOME TYPES AND COVERS FOR 
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QUATERNIONIC HERMITIAN GROUPS
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A DISSERTATION
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SUBMITTED TO THE GRADUATE FACULTY
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in partial fulfillment of the requirements for the
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Degree of
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DOCTOR OF PHILOSOPHY
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By
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THOMAS LAEBEL MADSEN
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%Department of Mathematics
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Norman, Oklahoma
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%Norman, OK 73019 USA
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2014
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SOME TYPES AND COVERS FOR 
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QUATERNIONIC HERMITIAN GROUPS
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A DISSERTATION APPROVED FOR THE
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DEPARTMENT OF MATHEMATICS
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BY
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\  \  \  \  \  \  \  & Dr.~Alan Roche, Chair\\
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  & Dr.~Ralf Schmidt \\
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 & Dr.~Jonathan Kujawa \\
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 & Dr.~Lucy Lifschitz \\
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   & Dr.~James Hawthorne \\
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\copyright \    Copyright by THOMAS LAEBEL MADSEN \ 2014

All Rights Reserved.
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\chapter*{Acknowledgements}

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First, I would like to thank my advisor Dr. Alan Roche for his never ceasing support and encouragement. I am very grateful to him for not giving up on me. I would also like to thank Dr. Lucy Lifschitz, Dr. Ralf Schmidt, Dr. Jonathan Kujawa, and Dr. James Hawthorne for their willingness to serve on my doctoral committee. 

A special thanks to the late Dr. Edward Cline who initially encouraged me to pursue a Ph.D. in mathematics. I would also like to thank the Department of Mathematics and its staff. Their work and support did not go unnoticed. 

Lastly I thank my wife Kelsey and son James for putting up with me.
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In this dissertation we determine the reducibility of certain induced representations. We do this using Bushnell and Kutzko's method of types and covers. We consider certain quaternionic hermitian groups over a $p$-adic field. While the types and covers can be found in the literature, the computation of the associated Hecke algebras has not been done before. 


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\chapter{Introduction}
The area of interest in this dissertation is the representation theory of reductive $p$-adic groups. Specifically we are interested in how Bushnell-Kutzko's theory of types and covers can be used to obtain explicit results on the reducibility of parabolically induced representations. 

Representation theory is a central area of  modern mathematics. In particular, the representation theory of reductive $p$-adic groups is central to the web of conjectures known as the Langlands program whose influence pervades current research in automorphic forms and number theory. The origins of representation theory go back to Frobenius and others in the 1890s in the study of finite groups. A key tool introduced by Frobenius is the method of induction, a way of building representations of a group from representations of subgroups. To make this an effective means of constructing or classifying representations, one needs to be wise in the choice of both the subgroups one induces from and the representations one induces. 

In the case of reductive real or $p$-adic groups, it is natural to induce from what are called parabolic subgroups. For general linear groups, these are (up to conjugacy) the subgroups of invertible block upper-triangular matrices. 
A proper parabolic subgroup is not reductive but admits a canonical reductive quotient.  For example, in the case of block upper-triangular matrices, 
this reductive quotient is isomorphic to the corresponding group of block diagonal matrices. One takes a representation of this reductive quotient, views it as a representation of the parabolic subgroup
and then induces. The whole process is called parabolic induction. 

In contrast to reductive real groups, a central feature of reductive $p$-adic groups is that there are irreducible representations that  never occur as subrepresentations of parabolically induced representations. These are the \emph{supercuspidal} representations. 
They serve as fundamental building blocks. Indeed, suppose $\Pi$ is an irreducible representation of a reductive $p$-adic group $G$. 
By insights of Harish-Chandra and others it is known that there is a parabolic subgroup of $G$ and an irreducible supercuspidal representation $\pi$ of its reductive quotient such that $\Pi$ occurs in the representation obtained from $\pi$ via parabolic induction. Moreover, only finitely many 
$\Pi$ are related to the supercuspidal representation $\pi$ in this way. 

Thus, a core problem  in  $p$-adic representation theory is  to understand when and how parabolically induced representations decompose, 
especially when the inducing representation is supercuspidal. This is the problem I study in a very special situation. 

In the case of finite groups, Mackey theory provides an efficient way of decomposing induced representations through the action of 
certain intertwining operators.  The same operators can be used to study parabolically induced representations for $p$-adic groups. 
Their construction, however, is considerably more subtle and involves a process of analytic continuation.  
By work of Langlands and Shahidi \cite{S1990}, one knows that properties of these intertwining operators give rise to local $L$-functions and that these $L$-functions are the key to many reducibility questions. In \cite{S1992} Shahidi studies certain reducibility questions in this way for the split classical groups. 

Bushnell-Kutzko's method of types and covers \cite{BK1998} provides another way of studying reducibility questions.  It relies on detailed
knowledge of the internal structure of the inducing representation and certain related constructions. In circumstances where this is available, the method can lead to strikingly explicit results. Indeed, in \cite{KM2006} Kutzko and Morris use the method to reconsider a special case of the situation studied by Shahidi in \cite{S1992} and obtain a sharper form of his results. In this dissertation, I study a situation that is analogous to the one considered by Kutzko and Morris where the underlying group is non-split. Pursuing the same basic strategy, I obtain an explicit reducibility result as described in more detail below. 

With the first chapter being this introduction, in the second chapter we give some background that mainly aims to fix some notation and terminology. In the third chapter we pose the question and walk through the answer.

Our approach is local. We note that Mui{\'c} and Savin in \cite{MS2000} have answered the reducibility question that we consider using global methods. Their work is more general in that it is not restricted to the depth zero case. However, their results are less explicit. Further, their approach only works when the underlying field has characteristic zero. Our only restriction is that the residual characteristic of the underlying field is odd.


We adopt the following conventions and notations.  All rings are assumed to have multiplicative identities.  All representations are complex representations. Given a category $\mathcal{C}$, by $x\in \mathcal{C}$ we mean that $x$ is an object in the category. We often identify representations if they are equivalent as representations. In this same way, we often say that one representation is a subrepresentation of another representation if it is equivalent to a subrepresentation of the other representation.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Background %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Background}
In this chapter we lightly touch on some of the background needed. We will not go into great detail, but only highlight tools and results that we shall use later. The main sources for background are \cite{C1985}, \cite{BH2006}, and \cite{K1998}.

\section{Algebraic groups}

We assume some familiarity with algebraic groups. There are several ways to present the basic notions. We adopt here the view found in \cite{S2009}. For this section only, let $K$ be an algebraically closed field. We recall that an algebraic group is an algebraic variety over $K$ that is a group, such that the multiplication and taking inverse are morphisms of varieties. When the variety is affine we call the group an affine (or linear) algebraic group. It is a fact that every affine algebraic groups is isomorphic to some closed (in the Zariski topology) subgroup of $\GL_n(K)$ for some $n$. We follow standard abuses of notation in often not distinguishing between algebraic groups and their groups of $F$-points. 

\subsection{About finite groups}
We now describe finite groups of Lie type as in \cite{C1985}.  Assume that $K$ has characteristic $p$ and let $q = p^e$ for some positive integer $e$. Let $F_q : \GL_n(K) \to \GL_n(K)$ be the homomorphism given by
\begin{align*}
F_q(a_{ij}) = (a_{ij}^q).
\end{align*}
A map $F: G \to G$ is called a standard Frobenius map if for some embedding $i: G \to \GL_n(K)$ we have
\begin{align*}
i(F(g)) = F_{q}(i(g))
\end{align*}
for all $g\in G$. A map $F: G \to G$ is called a Frobenius map if $F^{m}$ is a standard Frobenius map for some $m$. For a Frobenius map $F: G\to G$ we can talk about the $F$-points of $G$:
\begin{align*}
G^F = \{g\in G : F(g) = g\}.
\end{align*}
Then $G^F$ is a finite subgroup of $G$.  The finite groups we obtain in this way are called the finite groups of Lie-type. Consider the following examples.
\begin{enumerate}
\item Let $G = \GL_n(K)$. Let $i:G \to G$ be the identity map. Let $F: G \to G$ be the map: $(a_{ij}) \mapsto (a_{ij}^q)$. Then $G^F = \GL_n(\F_q)$.
\item Let $G = \GL_n(K)$. Let again $i:G\to G$ be the identity. Let $F: G\to G$ be the map $F: (a_{ij}) \mapsto {}^\top(a_{ij}^q)^{-1}$. Then $G^F = \{(a_{ij}) \in \GL_n(\F_{q^2}): (a_{ij})(a_{ij}^q)^{\top} = 1\}$. This is the unitary group $U_n(\F_{q^2})$.
\item Let $G = \Sp_{n}(K)$. Let $i:G\to \GL_{n}(K)$ be inclusion. Let $F:G\to G$ be the map $F: (a_{ij}) \mapsto (a_{ij}^q)$. In this case, $G^F = \Sp_{n}(\F_q)$. 
\end{enumerate}



\subsection{Locally profinite groups}
The following is mainly based on the treatment in \cite{BH2006}. We say that a topological group $G$ is profinite if it is compact, Hausdorff, and totally disconnected. We can also define a profinite group as the inverse limit of finite groups. Let $(I, \leq)$ be a partially ordered set. Let $\{H_i\}_{i\in I}$ be a collection of finite groups all equipped with the discrete topology. Let $f_{ij} : H_j \to H_i$ for $i \leq j$ be group homomorphisms. We require that $f_{ii}$ is the identity of $H_i$ and that $f_{ki}\circ f_{ij} = f_{kj}$ for $ k\leq i \leq j$. Then we can form the inverse limit
\begin{align*}
 \lim_{\longleftarrow} H_n = \left\{ (h_{i}) \in \prod H_n : f_{ij}(h_j) = h_i\right\}. 
\end{align*} 
This is a group under $(a_i)(b_i) = (a_ib_i)$.

For $p$ a prime, $H_n = \Z / p^n\Z$, and the maps $f_{ij}: \Z / p^j\Z \to \Z / p^i\Z$ being given by $f_{ij}: x + p^j\Z \mapsto x + p^i\Z$, the inverse limit is the $p$-adic integers $\Z_p$. 

We also have the definition of a locally profinite group. We say that a topological group $G$ is locally profinite if it is Hausdorff and if each neighborhood of the identity in $G$ contains a compact open subgroup. 
\begin{lemma}
A topological group is locally profinite if and only if it is locally compact and totally disconnected.
\end{lemma}

A compact locally profinite group is a profinite group. Finite groups are locally profinite.




\section{Facts from representation theory}
We recall some basic facts from representation theory. The following is not in any way a systematic treatment, but merely serves to fix notation and highlight tools that we shall need later. 
\subsection{Smooth representations}

Let $G$ be a locally profinite group. A representation $(\pi, V)$ of $G$ is a complex vector space $V$ together with a group homomorphism $\pi: G \to \GL(V)$. 

Let $K$ be a compact open subgroup of $G$. Then we let $V^K$ be the set of $\pi(K)$-fixed vectors: $V^{K} = \{v\in V \mid \pi(k)v = v$ for all $k\in K \}$. We call $(\pi, V)$ a smooth representation if
\begin{align*}
V = \bigcup_{K} V^K.
\end{align*}
A smooth representation $(\pi, V)$ is called admissible if the spaces $V^{K}$ have finite dimension.

Let 
\begin{align*}
V^{*} = \Hom_{\C}(V, \C)
\end{align*}
be the dual of $V$. Define a representation $(\pi^*, V^*)$ of $G$ by 
\begin{align*}
(\pi^*(g)v')(v) = v'(\pi(g^{-1})v)
\end{align*}
for $v\in V$, $v'\in V^*$ and $g\in G$. This defines a representation of $G$, but it is not necessarily smooth. Therefore we consider the space of smooth vectors
\begin{align*}
V^{\lor} = (V^*)^{\infty} = \bigcup_{K} (V^*)^K
\end{align*} 
where the union is taken over all compact open subgroups of $G$. We then define the representation $(\pi^\lor, V^{\lor})$ by $\pi^\lor(g)(v) = \pi^*(g)v$ for $v\in V^{\lor}$ and $g\in G$. This representation is smooth and we call it the \emph{smooth dual} or the \emph{contragredient} of $(\pi, V)$. 

Given a representation $\pi$ of $H\leq G$ and $g\in G$ we let $\pi^g$ denote the representation $\pi^g$ of $H^{g} = g^{-1}Hg$ given by $\pi^g(h') = \pi(gh'g^{-1})$.


\subsection{Induced representations}

Let $G$ be a locally profinite group and let $H$ be a closed subgroup of $G$. Let $(\sigma, W)$ be a smooth representation of $H$. Let
\begin{align*}
\Ind_{H}^{G}(\sigma, W) =\{f: G \to W \mid &f(hg) = \sigma(h)f(g) \text{ for all } h\in H, g\in G,\\ &
\text{ there is a compact open subgroup } K\leq G  \\ 
& \text{ such that } f(gk) = f(g) \text{ for all } g\in G, k\in K\}.
\end{align*}
Let $G$ act on $\Ind_H^G(\sigma, W)$ by
\begin{align*}
(g.f)(x) = f(xg).
\end{align*}
This defines a smooth representation of $G$. We also consider the space
\begin{align*}
\ind_H^G(\sigma, W) =\{ f\in \Ind_H^G(\sigma, W) \mid f \text{ has compact support modulo } H\}.
\end{align*}
The same action as above defines a smooth representation of $G$. We call this method of induction \emph{compact induction}. That $f$ has compact support modulo $H$ means that $\supp(f)$ is compact in $G / H$. That is, if $p: G \to G / H$ is the projection map, then $p(\supp(f))$ is compact. 

We also define \emph{normalized induction}. Let $P$ be a parabolic subgroup of the $F$-points of a reductive algebraic group defined over a p-adic field $F$. Let
\begin{align*}
\i_P^G(\sigma, W) = \ind_{P}^G(\delta_{P}^{1/2}\tensor \sigma).
\end{align*}
Here $\delta_{P}$ is a certain function: $P \to (0,\infty)$. See section $3$ in \cite{BH2006} for more on this function. As a matter of notation we will write $\i_P^G(\sigma)$ for $\i_P^G(\sigma, V)$. 

One important property of normalized induction is that 
\begin{align*}
\i_P^G(\sigma)^\lor \simeq \i_P^G(\sigma^\lor).
\end{align*}
In particular parabolic induction takes self-dual representations to self-dual representations. 



\subsection{Cuspidal representations}
\begin{definition}
Let $G$ be a finite group of Lie type. Let $\pi$ be an irreducible representation of $G$. We call $\pi$ \emph{cuspidal} if
\begin{align*}
\Hom_{G}(\pi , \Ind_{P}^G \tau) = \{0\}
\end{align*}
for all proper parabolic subgroups $P$ of $G$ and $\tau$ representation of a Levi factor of $P$ inflated to $P$.
\end{definition}

In the following chapter we will, as mentioned earlier, study reducibily of induced representations. A key requirement is that the representation being induced has \emph{depth zero}. The general definition of depth can be found in \cite{MP1996}. In chapter \ref{mainchapter} we will state the definition in the setup that we use there. 

We recall \emph{Frobenius Reciprocity}. Let $G$ be a locally profinite group. Let $H$ be a open subgroup of $G$. Let $(\pi,V)$ be a representation of $G$ and $(\sigma,W)$ a representation of $H$. Then
\begin{align*}
\Hom_{G}(\ind_{H}^G\sigma, \pi) \simeq \Hom_H(\sigma, \pi\rvert_H).
\end{align*}
For $H$ a closed subgroup of $G$, we have
\begin{align*}
\Hom_{G}(\pi, \Ind_H^G \sigma) \simeq \Hom_{H}(\pi\rvert_H, \sigma). 
\end{align*}
We also recall \emph{Mackey's Irreducibility Criterion} which is often used in conjunction with Frobenius Reciprocity. The original reference for this is \cite{M1951}. Let $G$ be a locally profinite group and $H$ an open and compact mod center subgroup of $G$. Let $(\pi,V)$ be a smooth irreducible representation of $H$. Then $\ind_{H}^{G} \pi$ is irreducible exactly when 
\begin{align*}
\Hom_{}(\pi^g\rvert_{H\cap H^g}, \pi\rvert_{H\cap H^g}) = \{0\}
\end{align*}
for $g\not\in H$.



\section{Finite groups of Lie type}\label{sec:finitegroups}

The following highlights certain aspects that we shall make reference to later. A fuller treatment of the theory of finite groups of Lie Type can be found in \cite{C1985} where precise statements and definitions can be found. Another good resource is \cite{DM1991}.

In the following we abuse notation and let $G = G^F$ be a finite group of Lie type with a Frobenius map $F: G \to G$. As described well in \cite{C1985} we have the so-called Deligne-Lusztig (generalized) characters $R_{T,\theta}$ where $T$ is a maximal torus in $G$ and $\theta$ an irreducible character of $G$. We recall that $\theta$ is said to be in general position if the only element of $W(T) = N_G(T) / T$ that fixes $\theta$ is the identity. In the case where $\theta$ is in general position $\pm R_{T,\theta}$ is an irreducible character of $G$. 

For $\chi \in \Irr(G)$ a character there is a maximal torus $T$ and $\theta \in \Irr(T)$ such that $\langle \chi, R_{T,\theta}\rangle \neq 0$. Here $\langle\cdot, \cdot \rangle$ is the usual scalar product. This means that all irreducible characters of $G$ occur as constituents of some $R_{T,\theta}$. We say that two pairs $(T,\theta)$ and $(T',\theta')$ are geometrically conjugate if
\begin{itemize}
\item there is a $g\in G$ such that $T = {}^gT'$ and 
\item if $g\in G^{F^{n}}$ for some $n$, then $\theta \circ N_{F^n/F} = \theta'\circ N_{F^{n}/F}\circ \Ad(g)$.
\end{itemize}
Here $\Ad(g)$ is conjugation by $g$ and $N_{F^n / F}$ is the reduced norm.

If $(T,\theta)$ and $(T',\theta')$ are not geometrically conjugate, then $R_{T,\theta}$ and $R_{T', \theta'}$ have no irreducible components in common.

Each irreducible character of $G$ determines a unique geometric conjugacy class $[T,\theta]$. We say that two characters are (geometrically) conjugate if in this way they give rise to the same class. An irreducible character $\chi$ is called unipotent if $\langle R_{T, 1} , \chi\rangle \neq 0$. Note that all unipotent characters form a single conjugacy class. An irreducible character $\chi$ is called regular if $\langle \Gamma, \chi\rangle \neq 0$ where $\Gamma$ is the Gelfand-Graev character (See chapter 14 in \cite{DM1991} for more on this). An irreducible  character will be called semi-simple if its dual is (up to sign) a regular irreducible character. Note that the only semisimple unipotent character is the trivial character.

If $(T^*, F^*)$ is dual to $(T, F)$, then $\Irr(T^{F}) \simeq (T^{*})^{F^*}$ (see chapter 13 in \cite{DM1991} for more on this). Note that the $\theta$ in the $R_{T,\theta}$'s are exactly characters of $T$, so using this isomorhism (of groups) we get elements of the dual torus $T^*$.

Suppose that $(G,F)$ and $(G^*, F^*)$ are in duality with corresponding dual tori $T$ and $T^*$. Then we have have a bijection between the pairs $\{[T, \theta]\}$ and $F^*$-stable semisimple conjugacy classes in $G^*$. 



\section{Types and covers}
The theory of types and covers is described in detail in \cite{BK1998}. We adopt the approach given there. We will, though, also rely on the presentations given in \cite{K1998} and \cite{BHK2011}. Let $F$ be a non-Archimedean local field. Let $G$ be the $F$-rational points of a connected, reductive, algebraic group defined over $F$. Then $G$ is (isomorphic to) a closed subgroup of $GL_n(F)$ for some $n$. Since $GL_n(F)$ is locally profinite, $G$ is locally profinite. We recall that this means that $G$ is locally compact and totally disconnected, or equivalently, that each neighborhood of the identity in $G$ contains a compact open subgroup.

We note that $G$ is unimodular, so a left Haar measure is a right Haar measure  and vice versa. We fix a Haar measure $\mu$ on $G$.

Let $\R(G)$ be the category of smooth representations of $G$. That is, the objects in $\R(G)$ are the smooth representations and the morphisms are $G$-maps between representations. Given a smooth representation $\pi$ of $G$, there is a parabolic subgroup $P$ of $G$ with Levi decomposition $P = LU$ (with $L$ the Levi factor) and an irreducible supercuspidal representation $\sigma$ of $L$ such that $\pi$ is a composition factor of $\i_P^G(\sigma)$. We note that if $\pi$ is supercuspidal then $L = G$. The pair $(L,\sigma)$ is not unique. We say that two pairs $(L_1, \sigma_1)$ and $(L_2, \sigma_2)$ are inertially equivalent if there is a $g\in G$ such that $L_2 = L_1^g$ and $\sigma_2\tensor \chi \simeq \sigma_1^g$ for some unramified character $\chi$ of $L_2$. We let $[L, \sigma] = [L,\sigma]_G$ be the equivalence class and we let $\B(G)$ be the set of equivalence classes. Given $\pi \in \R(G)$ we then obtain a unique equivalence class $\I(\pi) \in \B(G)$. We call $\I(\pi)$ the inertial support (or just support) of $\pi$. 

For $\s \in \B(G)$ we let $\R^\s(G)$ be the (full) subcategory of $\R(G)$ consisting of all smooth representations $\pi'$ with inertial support $\I(\pi') = \s$. 
We can now state the Bernstein decomposition:
\begin{align*}
\R(G) = \prod_{s\in \B(G)} \R^{\s}(G).
\end{align*}

Let $\s \in \B(G)$. Then fix $L$ and $\sigma$ such that $\s = [L,\sigma]$. Then let $\s_L = [\sigma, L]_L \in \B(L)$. Given this setup with $P$ a parabolic subgroup of $G$ with Levi-factor $L$, we have the normalized induction functor
\begin{align*}
\i_P^G : \R^{\s_L}(L) \longrightarrow \R^{\s}(G).
\end{align*}

\subsection{Types}

Let $K$ be a compact open subgroup of $G$. Let $(\rho, W)$ be an irreducible smooth representation of $K$. For a smooth representation $(\pi, V)$ of $G$ we let $V^{\rho}$ be the $\rho$-isotypic subspace of $V$. That is, $V^{\rho}$ is the sum of all irreducible $K$-subspaces of $V$ which are equivalent to $\rho$:
\begin{align*}
V^{\rho} = \sum_{W'} W'
\end{align*}
where the sum is over all $W'$ such that $(\pi\mid_K, W') \simeq (\rho, W)$.

Let $\H(G)$ be the space of all locally constant compactly supported functions $f: G \to \C$. This is a $\C$-algebra under convolution $\star$. That is, for $f,g\in \H(G)$ we have
\begin{align*}
(f\star g)(x) = \int_G f(y)g(y^{-1}x) \; d\mu(y).
\end{align*}
We have here fixed a Haar measure $\mu$ on $G$. We note that $\H(G)$ only has an identity when $G$ is a discrete group. For $(\pi, V)$ a representation of $G$, $\H(G)$ acts on $V$ via
\begin{align*}
hv = \int_{G} h(x)\pi(x)vd\mu(x)
\end{align*}
for $h\in \H(G)$ and $v\in V$. We identify $\R(G)$ and $\H(G)-$Mod.

Let $e_\rho$ be the element in $\H(G)$ defined with support $K$ such that for each $x\in K$,
\begin{align*}
e_\rho(x)  = \frac{\dim \rho}{\mu(K)}\tr_W(\rho(x^{-1})).
\end{align*}
Then $e_\rho$ is an idempotent in $\H(G)$, that is, $e_\rho \star e_{\rho} = e_{\rho}$.

Let $\R_{\rho}(G)$ be the (full) subcategory of $\R(G)$ consisting of all $(\pi, V)$ where $V$ is generated by $V^\rho = e_\rho V$ (the $\rho$-isotypic vectors). That is, $(\pi, V) \in \R_\rho(G)$ if and only if
\begin{align*}
V = \H(G) \star e_\rho V.
\end{align*}
We can now state the definition of a type.
\begin{definition}
Let $\s\in \B(G)$. We say that $(K, \rho)$ is an $\s$-type in $G$ if $\R_\rho(G) = \R^{\s}(G)$.
\end{definition}



\subsection{Hecke algebras}
We define the Hecke algebra:
\begin{align*}
\H(G, \rho) = \{f: G \to \End_\C(\rho) : &\supp(f)\text{ is compact and  }\\
&f(k_1gk_2) = \rho(k_1)f(g)\rho(k_2)\;\forall k_i \in K, g\in G\}.
\end{align*}
There is a canonical isomorphism
\begin{align*}
\H(G, \rho) \tensor_{\C} \End_\C(W) \stackrel{\simeq}{\longrightarrow} e_\rho \star \H(G) \star e_\rho,
\end{align*}
and a corresponding functor
\begin{align*}
V \mapsto V^{\rho}:\quad \R_\rho(G) \longrightarrow e_\rho\star \H(G)\star e_\rho -\text{Mod}.
\end{align*}
So we get a functor 
\begin{align*}
M_\rho: \R_\rho(G) \longrightarrow \H(G,\rho)-\text{Mod}.
\end{align*}
The importance of types is seen in the following theorem
\begin{theorem}
The following are equivalent:
\begin{enumerate}
\item $(K,\rho)$ is an $\s$-type in $G$.
\item For $\pi\in \Irr(\R^\s(G))$ we have $\I(\pi) = \s$.
\item The functor $M_\rho$ is an equivalence of categories.
\end{enumerate}
\end{theorem}


\subsection{Covers}

\begin{definition}
Let $P = LU$ be a parabolic subgroup with Levi factor $L$ and let $\overline{P} = L\overline{U}$ be its opposite. We will in general denote the $L$-opposite of a parabolic subgroup $P$ by $\overline{P}$. Let $K$ be a compact open subgroup of $G$ and let $(\sigma , W)$ be an irreducible representation of $K$. Then we say that $(K, \sigma)$ is decomposed with respect to $(L, P)$ if the following hold:
\begin{enumerate}
\item $K = (K \cap \overline{U})(K \cap L)(K\cap U)$.
\item $K\cap \overline{U}$, $K\cap U  \leq \ker(\sigma)$. 
\end{enumerate}
\end{definition}

For $(K, \rho)$ decomposed with respect to $(L, P)$ we write $K_L = K \cap L$ and we write $\rho_{L}$ for $\rho\mid_{K_L}$. Let $I_G(\rho) = \{g\in G : \text{there exists } f\in \H(G,\rho) \text{ such that } g\in \supp(f)\}$. Let $\H(G,\rho)_L =\{f\in \H(G, \rho) : \supp(f) \subseteq KLK\}$. 

\begin{proposition}
Let $(K,\rho)$ decompose with respect to $(L, P)$. Then
\begin{enumerate}
\item $\rho_L$ is irreducible.
\item $I_L(\rho_L) = I_G(\rho) \cap L$.
\item There is an embedding
\begin{align*}
T: \H(L, \rho_L) \longrightarrow \H(G, \rho)
\end{align*}
such that if $f\in \H(L, \rho_L)$ has support $K_LzK_L$ for some $z\in L$, then $T(f)$ has support contained in $KzK$.
\item The map $T$ induces an isomorphism of vector spaces
\begin{align*}
\H(L, \rho_L) \stackrel{\simeq}{\longrightarrow} \H(G,\rho). 
\end{align*}
\end{enumerate}
\end{proposition}

\begin{definition}
An element $z\in L$ is called $(K,P)$-positive if 
\begin{enumerate}
\item $z(K\cap U)z^{-1} \subseteq K\cap U$,
\item $z^{-1}(K\cap \overline{U})z \subseteq K\cap \overline{U}$.
\end{enumerate}
\end{definition}


\begin{definition}
An element $z\in L$ is called strongly $(K,P)$-positive if
\begin{enumerate}
\item $z$ is $(K,P)$-positive,
\item $z$ is in the center of $M$: $z\in Z(M)$,
\item for any compact open subgroups $K_1$ and $K_2$ of $U$, there is an integer $m\geq 0$ such that $z^mK_1z^{-m}\subseteq K_2$,
\item for any compact open subgroups $K_1'$ and $K_2'$ of $\overline{U}$, there is an integer $m\geq 0$ such that $z^{-m}K_1'z^m \subseteq K_2$.
\end{enumerate}
\end{definition}
Let
\begin{align*}
\H(L,\rho_L)^+ = \{&f\in \H(L,\rho_L) : \\ &\supp(f) \text{ consists of } (K,P)\text{-positive elements}\}.
\end{align*}
The isomorphism of vector spaces $T$ restricts to an embedding of algebras
\begin{align*}
T: \H(L, \rho_L)^+ \longhookrightarrow \H(G,\rho).
\end{align*}
This embedding extends to $\H(L,\rho_L)$ under the following condition.
\begin{theorem}
The embedding $T$ extends to an embedding of algebras 
\begin{align*}
t: \H(L,\rho_L) \longhookrightarrow \H(G,\rho)
\end{align*}
if and only if $T(\phi_z)$ is invertible for some strongly $(K,P)$-positive element $z$. If this embedding exists, it is unique.
\end{theorem}


\begin{proposition} We have the following.
\begin{enumerate}
\item Strongly $(K,P)$-positive elements exist.
\item Given a strongly $(K,P)$-positive element $z\in L$, there is a unique function $\phi_z \in \H(L, \rho_L)$ with support $K_LzK_L$ such that $\phi_z(z)$ is the identity function on $W$.
\end{enumerate}
\end{proposition}


Let $\H(G,\rho)_L = \{f \in \H(G,\rho) : \supp(f) = KLK\}$. This is a subspace of $\H(G,\rho)$.
%%%%%%%%%%%%%%%% Definition of cover

\begin{definition}
Let $L$ be a proper Levi subgroup of $G$. Let $K_L$ be a compact open subgroup of $L$, and let $\rho_L$ be an irreducible smooth representation of $K_L$. Let $K$ be a compact open subgroup of $G$ and let $\rho$ be an irreducible smooth representation of $K$. Then we say that $(K,\rho)$ is a $G$-cover of $(K_L, \rho_L)$ if
\begin{enumerate}
\item the pair $(K,\rho)$ is decomposed with respect to $(L, P)$ for every parabolic subgroup $P$ of $G$ with Levi factor L;
\item $L\cap L = K_L$, and $\rho\rvert_{L} \simeq \rho_L$;
\item The embedding $T : \H(L, \rho_L)^+ \hookrightarrow \H(G,\rho)$ extends to an embedding of algebras $t': \H(L,\rho_L) \hookrightarrow \H(G,\rho)$.
\end{enumerate}
\end{definition}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Hermitian Quaternionic Groups %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\chapter{Quaternionic Hermitian Groups}\label{mainchapter}
We now turn to the study of quaternionic hermitian groups.
\section{Setup}
Let $F$ be a non-Archimedean local field of residual characteristic not equal to $2$ with uniformizer $\varpi_F$. Write $\nu_F: F \to \Z \cup \{\infty\}$ for the normalized valuation on $F$ so that $\nu_{F}(\varpi_F) = 1$. Thus the ring of integers $\O_F = \{x\in F : \nu(x) \geq 0 \}$ and the unique maximal ideal $\p_F = \{x\in F : \nu(x) > 0\}$. Write $q$ for the order of the residue field $k_F = \O_F / \p_F$. Then $q = p^f$ for some odd prime $p$ and some positive $f$. Let $D = F \oplus iF \oplus jF \oplus kF$ be the unique quaternionic division algebra over $F$. Here $i^2$, $j^2$, $k^2 \in F$ are non-squares in $F$. For $x = x_0 + ix_1 + jx_2 + kx_3$ we denote by $\bar{x}$ the involution
\begin{align*}
\overline{x_0 + ix_1 + jx_2 + kx_3} = x_0 - ix_1 - jx_2 - kx_3.
\end{align*}
For a matrix $(g_{ij}) \in M_n(D)$, let $\bar{g} = (\bar{g}_{ij})$. We have the reduced norm $\Nrd_D: D \to F$. The map
\begin{align*}
\nu_D: D \to \Z \cup \{\infty\}
\end{align*}
given by
\begin{align*}
\nu_D(x) = \nu_F(\Nrd_D(x))
\end{align*}
is a valuation on $D$. As with $F$, we have the ring of integers $\O_D$, the maximal ideal $\p_D$, and the finite residue field $k_D$. Note $[k_D:k_F] = 2$, so $\lvert k_D\rvert = q^2$.
\begin{lemma}
One can choose a uniformizer $\varpi_D$ in $D$ such that $\varpi_D^2 = \varpi_F$.
\end{lemma}
See \cite{R2003} for a proof of this. We fix $\varpi_D$ as in the lemma. 

\begin{definition}
Let $\pi$ be an irreducible supercuspidal representation of $\GL_n(D)$. Let $K = \GL_n(D)$ and $K_1 = 1 + \varpi_DM_n(\O_D)$. We say that $\pi$ has depth zero if $\pi^{K_1} \neq \{0\}$. That is, $\pi$ has depth zero exactly when it has non-trivial fixed vectors under $K_1$. 
\end{definition}

\subsection{The groups}

Let $\epsilon\in \{\pm 1\}$ and let $n > 1$ be an integer. Let $I_n$ be the $n\times n$ identity matrix and let $J_\epsilon$ be the $2n\times 2n$ matrix
\begin{align*}
J_\epsilon = \begin{pmatrix} 0 & I_n \\ \epsilon I_n & 0\end{pmatrix}. 
\end{align*}
Define the group
\begin{align*}
G_\epsilon = \{g\in \GL_{2n}(D) : g^* J_\epsilon g = J_\epsilon\}
\end{align*}
where for $g = (g_{ij})$, $g^{*} = (\overline{g_{ij}})^{\top} = (\overline{g_{ji}})$. Then $G_1$ is an inner form of $\Sp(4n, F)$ and $G_{-1}$ is an inner form of $\SO(4n,F)$. We will sometimes drop the subscript $\epsilon$ and just write $G$. 

Let $P$ be the standard Siegel parabolic subgroup of $G_\epsilon$ with standard Levi decomposition
\begin{align*}
P = LU_\epsilon.
\end{align*}
Thus
\begin{align*}
L = \left\{\begin{pmatrix} a & 0 \\ 0 & {}^\top\bar{a}^{-1}\end{pmatrix} : a\in \GL_n(D)\right\} \simeq \GL_n(D)
\end{align*}
and
\begin{align*}
U_\epsilon =  \left\{ \begin{pmatrix} 1 & X \\ 0 & 1\end{pmatrix} : \epsilon X + \overline{X}^{\top} = 0\right\}.
\end{align*}

\subsection{The question}

Let $z\in \C$ be a complex number. Let $\Nrd$ denote the reduced norm from $M_{n}(D)$ to $F$ (see \cite{R2003}). The characters $\chi_z$ given by $\chi_{z}(g) = z^{-\nu_{F}(\Nrd(g))}$ for $g\in L$ are the unramified characters of $L$. We write $X_{nr}(L)$ for the group of unramified characters of $L$. Thus $\C^{\times} \simeq X_{nr}(L)$ via the map $z\mapsto \chi_z$.

Let $\pi_0$ be an irreducible unitary supercuspidal representation of $L$ of depth zero and let $\chi\in X_{nr}(L)$. The question we wish to answer is
\begin{center}
When is $\i_P^{G_\epsilon}(\pi_0\tensor \chi)$ reducible?
\end{center}
That is, we want to find the characters $\chi$ such that this induced representation is reducible. 

First, by Corollary 2 in \cite{R2002}, if $\i_P^{G_\epsilon}(\pi_0\tensor \chi)$ is reducible for some $\chi$, then 
\begin{align*}
{}^w\pi_0 \simeq \pi_0\tensor \nu
\end{align*}
for some $\nu\in X_{nr}(L)$ where $w$ denotes the non-trivial element of $N_G(L) / L$, represented by $\left(\begin{smallmatrix} 0 & \epsilon \\ 1 & 0\end{smallmatrix}\right)$. We note the effect on an element from $L$:
\begin{align*}
{\vphantom{\Bigg(}}^w\!\!\begin{pmatrix} a & 0 \\ 0 & {}^\top\bar{a}^{-1}\end{pmatrix} = \begin{pmatrix} {}^\top\bar{a}^{-1} & 0 \\ 0 & a\end{pmatrix}.
\end{align*}

Thus ${}^w\pi_0(g) =\pi_0({}^{\top}\bar{g}^{-1})$. By Lemma 1.1 of \cite{MS2000}, ${}^w\pi_0 \simeq \pi_0^{\lor}$. Suppose ${}^w\pi_0 \simeq \pi_0\nu$. We write $\nu = \eta / {}^w\eta$ for some $\eta \in X_{nr}(L)$. Note this is possible since ${}^w\eta = \eta^{-1}$, so we are simply choosing a square root of $\nu$. Then ${}^w\pi_0{}^w\eta \simeq \pi_0\eta$, that is,
\begin{align*}
{}^w(\pi_0\eta) \simeq \pi_0\eta.
\end{align*}

Hence, replacing $\pi_0$ by a suitable unramified twist, we can arrange for $\pi_0$ to be self-dual (and still unitary): $\pi_0^{\lor} \simeq \pi_0$. From now on, we assume that $\pi_0$ is self-dual.

\section{Construction of $\pi_0$}

We want to understand how $\pi_0$ is constructed. Consider the maximal compact open subgroup $K = \GL_{n}(\O_D)$ of $L$. Let $K_1 = 1 + \varpi_D M_n(\O_D)$. Then $K / K_1 \simeq \GL_n(k_D)$. Since $\pi_0$ has depth zero, 
\begin{align*}
\pi_0\rvert_{K} \supseteq \rho.
\end{align*}
where $\rho$ is the inflation to $K$ of an irreducible cuspidal representation $(\sigma, U)$ of $K / K_1 \simeq \GL_n(k_D) = \GL_{n}(\F_{q^2})$. To understand $\pi_0$ we therefore ask the question: What are the irreducible representations of $L \simeq \GL_n(D)$ that contain $\rho$ upon restriction to $K$? First we focus on $\sigma$. 

We know that the character of $\sigma$ is a Deligne-Lusztig character:
\begin{align*}
\text{char}(\sigma) = \pm R_{T, \theta}
\end{align*}
for some minisotropic torus $T \simeq \F_{q^{2n}}^{\times}$ of $\GL_{n}(\F_{q^2})$ and some regular character $\theta: \F_{q^{2n}}^{\times} \to \C^{\times}$. Regularity of $\theta$ means that
\begin{align*}
\theta^{\gamma} \neq \theta
\end{align*}
for all non-trivial $\gamma$ in the Galois group $\Gal(\F_{q^{2n}} / \F_{q^2})$ where ${}^\gamma\theta(x) = \theta(\gamma(x))$. We now fix this pair $(T,\theta)$. Since $\rho$ is inflated from $\sigma$ we write
\begin{align*}
\rho = \rho_{\theta}.
\end{align*}



\subsection{Structure of $\pi_0$}

Let
\begin{align*}
\wt{K} = N_G(\rho) = \{x\in G \mid {}^xK = K\text{ and } {}^x\rho \simeq \rho\}.
\end{align*}

Let $\wt{\rho}$ be any extension of $\rho$ to $\wt{K}$ that contains $\rho$. This extension exists because $\wt{K} / K$ is cyclic. Then we claim that
\begin{align*}
\ind_{\wt{K}}^L \wt{\rho}
\end{align*}
is irreducible. To show this we have to show that the elements intertwining $\wt{\rho}$ are exactly the elements from $\wt{K}$. Say that $g\in L$ intertwines $\wt{\rho}$. So
\begin{align*}
\Hom_{\wt{K}\cap {}^g\wt{K}}(\wt{\rho}\rvert_{\wt{K}\cap {}^g\wt{K}}, {}^g\wt{\rho}\rvert_{\wt{K}\cap {}^g\wt{K}}) \neq \{0\}. 
\end{align*}

Restricting to $K$,
\begin{align*}
\Hom_{K\cap {}^gK}(\rho\rvert_{K\cap {}^gK} , \rho\rvert_{K\cap {}^gK}) \neq \{0\}.
\end{align*}
That is, $g$ intertwines $\rho$. By ``intertwining implies conjugacy'' (see \cite{GSZ2001}), $g\in \wt{K}$. Hence by Mackey's criterion, $\ind_{\wt{K}}^L\wt{\rho}$ is irreducible. 

Now say that $\wt{\rho}_1$ and $\wt{\rho}_2$ irreducible representations of $\wt{K}$ that contain $\rho$ (upon restriction to $K$). Then the induced representations $\ind_{\wt{K}}^L \wt{\rho}_1$ and $\ind_{\wt{K}}^L \wt{\rho}_2$ are irreducible. Suppose that
\begin{align*}
\ind_{\wt{K}}^L \wt{\rho}_1 \simeq \ind_{\wt{K}}^L \wt{\rho}_2.
\end{align*}
Then
\begin{align*}
\{0\} &\neq \Hom_L(\ind_{\wt{K}}^L \wt{\rho}_1, \ind_{\wt{K}}^L \wt{\rho}_2) \\
	&\simeq \Hom_{\wt{K}}(\wt{\rho}_1, \ind_{\wt{K}}^L \wt{\rho}_1\rvert_{\wt{K}})\\
	&\simeq \Hom_{\wt{K}}(\wt{\rho}_1, \oplus_{g\in \wt{K} \backslash L / \wt{K}} \ind_{\wt{K}\cap {}^g\wt{K}}^{\wt{K}} {}^g\wt{\rho}_2\rvert_{\wt{K}\cap {}^g\wt{K}}). 
\end{align*}
Thus there is a $g\in \wt{K} \backslash L / \wt{K}$ such that 
\begin{align*}
\Hom_{\wt{K}\cap{}^g\wt{K}}(\wt{\rho}_1\rvert_{\wt{K}\cap{}^g\wt{K}}, {}^g\wt{\rho}_2\rvert_{\wt{K}\cap{}^g\wt{K}}) \simeq \Hom(\wt{\rho}_1,  \ind_{\wt{K}\cap {}^g\wt{K}}^{\wt{K}} {}^g\wt{\rho}_2\rvert_{\wt{K}\cap {}^g\wt{K}}) \neq \{0\}. 
\end{align*}
That is, $g$ intertwines $\wt{\rho}_1$ with $\wt{\rho}_2$. As before this means that $g$ intertwines $\rho$. So $g\in \wt{K}$ and $\wt{\rho}_1 \simeq \wt{\rho}_2$.
 
In sum, we have a bijection from equivalence classes of irreducible representations of $\wt{K}$ containing $\rho$ to equivalence classes of irreducible representations of $L$ containing $\rho$ given by
\begin{align*}
\wt{\rho} \mapsto \ind_{\wt{K}}^L \wt{\rho}.
\end{align*}

We will show in the next section that $\wt{K} = K\langle \varpi_F\rangle$. This implies
\begin{align*}
\pi_0 = \ind_{\wt{K}}^{L} \wt{\rho}.
\end{align*}
Note $\pi_0$ self-dual implies $\omega_{\pi_0}(\varpi_F) = \pm 1$, where $\omega_{\pi_0}$ denotes the central character of $\pi_0$. As $\wt{\rho}$ is completely determined by its value at $\varpi_F$, it follows that there are exactly two possibilities for $\pi_0$. 

\section{Computing $\wt{K}$}

We have
\begin{align*}
\langle \varpi_F \rangle K = Z(G)K \leq \wt{K} \leq N_{G}(K) =  \langle \varpi_D \rangle K.
\end{align*}
Note the index of $\wt{K}$ in $\langle\varpi_D\rangle K$ is $1$ or $2$ since 
\begin{align*}
\lvert \langle \varpi_D \rangle K / \langle \varpi_F \rangle K \rvert =2. 
\end{align*}
We want to prove the following:
\begin{align*}
\wt{K} = \langle \varpi_F \rangle K.
\end{align*}
That is, we want to prove that $\varpi_D \not\in \wt{K}$. This will take some time. The proof divides naturally into two cases.

\subsection{Two cases}
We have
\begin{align*}
\pi_0 &= \ind_{\wt{K}}^{L}\wt{\rho}, \\
\pi_0^{\lor} &= \left(\ind_{\wt{K}}^L\wt{\rho}\right)^{\lor} \simeq \ind_{\wt{K}}^{L}\wt{\rho}^{\lor}.
\end{align*}
Thus $\wt{\rho}$ and $\wt{\rho}^{\lor}$ intertwine in $L$ which implies that $\rho$ and $\rho^\lor$ intertwine in $L$ and so $\rho$ and $\rho^\lor$ are conjugate in $L$ (see \cite{GSZ2001}).

Hence we are in one of two cases:
\begin{align*}
\text{(A):}\;\; \rho \simeq \rho^{\lor} \quad\quad\text{or}\quad \quad \text{(B):}\;\;{}^{\varpi_D}\rho \simeq \rho^{\lor}. 
\end{align*}
We consider these seperately. We first need some preliminary observations. Let $k = \F_{q^2}$ and $l = \F_{q^{2n}}$, so the extension $l / k$ has degree $n$. 
\begin{lemma}\label{tracelemma}
Let $\psi$ and $\phi$ be regular characters of $l^\times$. If 
\begin{align*}
\tr(\rho_{\psi}) = \tr(\rho_\phi)
\end{align*}
on $l^{\times}$, then
\begin{align*}
\rho_{\psi} \simeq \rho_{\phi}.
\end{align*}

\end{lemma}
\begin{proof}
In chapter 7 of $\cite{C1985}$ one finds the character formula for the $R_{T,\theta}$'s. Using a version from \cite{L1972}, we have for $\alpha \in l^{\times}$
\begin{align*}
\tr_{\rho_{\psi}}(\alpha) = c\sum_{\nu \in \Gal(l / k)} \psi^{\nu}(\alpha) = c\sum_{\nu \in \Gal(l / k)} \psi(\nu(\alpha))
\end{align*}
where the constant $c$ depends only on the field extension $k(\alpha) / k$. Hence, if $\tr \rho_{\psi} = \tr \rho_{\phi}$ on $l^{\times}$, then \begin{align*}
\sum_{\nu \in \Gal(l / k)} \psi^{\nu}(\alpha)  = \sum_{\nu \in \Gal(l/k)} \phi^{\nu}(\alpha),
\end{align*}
for all $\alpha\in l^{\times}$. Now let $\Gamma = l^{\times}\rtimes \Gal(l/k)$ where $\Gal(l/k)$ acts on $l^\times$ in the obvious way. We form the induced representations
\begin{align*}
\tau_{\psi} = \ind_{l^\times}^{\Gamma} \psi \quad\quad\text{and}\quad\quad \tau_{\phi} = \ind_{l^\times}^{\Gamma}\phi.
\end{align*}

Note $\tau_{\psi}$ and $\tau_{\phi}$ are irreducible by Mackey's criterion. The support of $\tr\tau_{\psi}$ is contained in $l^{\times}$. Further, for $\alpha \in l^{\times}$,
\begin{align*}
\tr\tau_{\psi}(x) = \sum_{\nu\in \Gal(l / k)} \psi^{\nu}(x).
\end{align*}
We have the same formula for $\tau_{\phi}$ and so
\begin{align*}
\tr\tau_{\psi} = \tr\tau_{\phi}.
\end{align*}
Hence $\tau_{\psi} \simeq \tau_{\phi}$. That is, $\ind_{l^\times}^\Gamma \rho_\psi$ and $\ind_{l^\times}^\Gamma \rho_\phi$ are isomorphic. Thus, by Frobenius Reciprocity,
\begin{align*}
1 &= \dim\Hom_{\Gamma} (\ind_{l^\times}^\Gamma \rho_\psi , \ind_{l^\times}^\Gamma \rho_\phi) \\
	&=\dim \Hom_{l^\times} ((\ind_{l^\times}^\Gamma \rho_{\psi}) \rvert_{l^{\times}},\rho_{\phi}) \\
	&=\dim  \Hom_{l^\times}\left( \bigoplus_{\nu \in \Gal(l / k)}  \rho_{\psi}^{\nu}       ,\rho_{\phi}  \right).
\end{align*}
So there is a $\nu \in \Gamma$ such that $\psi^{\nu} = \phi$. Hence $\rho_{\psi} \simeq \rho_{\phi}$.
\end{proof}

\begin{proposition}
With notation as above,
\begin{align*}
\rho_{\theta}^{\lor} \simeq \rho_{\theta^{-1}}.
\end{align*}
\end{proposition}
\begin{proof}
For $x\in l^{\times}$, there is a constant $c$ (which depends only on $k(x) = k(x^{-1})$) such that
\begin{align*}
\tr \rho_{\theta}^{\lor}(x) &= \tr \rho_{\theta}(x^{-1}) \\
&= \sum_{\nu \in \Gal(l / k)} \theta^{\nu}(x^{-1}) \\ 
&=  \sum_{\nu \in \Gal(l / k)} (\theta^{-1})^{\nu}(x) \\ 
&= \tr \rho_{\theta^{-1}}(x). 
\end{align*}
So from lemma \ref{tracelemma}, $\rho_{\theta}^{\lor} \simeq \rho_{\theta^{-1}}$. 
\end{proof}

Write $\overline{\rho_\theta}$ for the representation given by $\overline{\rho_{\theta}}(g) = \rho_{\theta}(\bar{g})$. For $g = (a_{ij})\in \GL_n(\F_{q^2})$, $\bar{g} = (a_{ij}^q)$.

\begin{proposition}\label{prop1}
Let $\theta: l^{\times} \to \C^{\times}$ be a regular character. Then
\begin{align*}
\overline{\rho_\theta} \simeq \rho_{\theta^q}.
\end{align*}
\end{proposition}
\begin{proof}
We recall that $l = \F_{q^{2n}}$ and $k = \F_{q^2}$. Fix the $k$-vector space $V = k^n$. Then $M_{n}(k) \simeq \End_{k}(V)$. Let $i: l \hookrightarrow M_n(k)$ be an embedding of $k$-algebras. Write $\phi: l \to l$ for the map $\lambda \mapsto \lambda^q$ and $\Phi: M_n(k) \to M_n(k)$ for the map $(a_{ij}) \mapsto (a_{ij}^q)$. Consider the diagram
\begin{align*}
\xymatrix{ l\ar[r]^{i}\ar[d]_{\phi} & M_n(k)\ar[d]^{\Phi} \\ l\ar[r]_{i} & M_n(k).
}
\end{align*}
Let $f = \Phi\circ i$ and $g = i\circ \phi$. We can make $V$ into an $l$-module in two ways:
\begin{itemize}
\item $\lambda.v = \Phi(i(\lambda))v$,
\item $\lambda_{*}v = i\circ \phi(\lambda)v = i(\lambda^q)v$.
\end{itemize}
Note that for $\lambda \in k$ we have $\lambda.v = \Phi(i(\lambda))v = \Phi(\diag(\lambda))v = \diag (\lambda^q)v = i(\lambda^q)v = \lambda_*v$. Hence the two different actions agree on $k$. These $l$-module structures must be isomorphic as there is only one $l$-module (up to isomorphism) of a given dimension. Thus there is an isomorphism (of $l$-modules) $h: V\to V$ such that
\begin{align*}
h(\lambda.v) = \lambda_*h(v).
\end{align*}
Note that $h$ is $k$-semi-linear since the module structures agree on $k$. Thus, for $v\in V$
\begin{align*}
h(\lambda v) &= \lambda^qh(v).
\end{align*}
Likewise, $h^{-1}(\mu v) = \mu^{q}h^{-1}(v)$ for $v\in V$ and $\mu \in k$. Now, for $a \in \End_k(V)$ we have $hah^{-1} \in \End_k(V)$. We see this from
\begin{align*}
(hah^{-1})(\lambda v) 
	&= ha(\lambda^{q}h^{-1}(v)) \\
	&= h(\lambda^q ah^{-1}(v)) \\
	&= \lambda^{q^2}hah^{-1}(v) \\
	&= \lambda (hah^{-1})(v)
\end{align*}
for $\lambda \in k$. Thus
\begin{align*}
h(\Phi(i(\lambda))v) &= i(\lambda^q)h(v).
\end{align*}
That is
\begin{align*}
h \circ \Phi(i(\lambda)) \circ h^{-1} = i(\lambda^q).
\end{align*}
Hence for $\alpha\in i(l^{\times})$, 
\begin{align*}
\tr \overline{\rho_{\theta}}(\alpha)
	&= \tr \rho_{\theta}(\bar{\alpha})\\
	&= \tr \rho_{\theta}(\Phi(\alpha)) \\
	&= \tr \rho_{\theta}(h^{-1}) \alpha^qh) \\
	&= \tr \rho_{\theta}(\alpha^{q}).
\end{align*}
By Lemma \ref{tracelemma}, $\overline{\rho_{\theta}} \simeq \rho_{\theta^q}$.
\end{proof}


\subsection{Case (A): $\rho\simeq\rho^{\lor}$}

We begin the treatment of Case (A). Recall $\Gal(l/k) = \langle \Fr\rangle$ where $\Fr(\lambda) = \lambda^{q^2}$. In this case we have $\rho_{\theta^{-1}} \simeq \rho_\theta = \rho$. Thus there exists a $\gamma \in \Gal(l / k)$ such that
\begin{align*}
\theta^{\gamma}  = \theta^{-1}.
\end{align*}
Applying $\gamma$ twice we get
\begin{align*}
\theta^{\gamma^2} = \theta.
\end{align*}
Since $\theta$ is regular, $\gamma^2 = 1$. Assume $\gamma = 1$. Then $\theta = \theta^{-1}$, so $\theta^2 = 1$ and the image of $\theta$ is contained in $\{\pm 1\}$. Then
\begin{align*}
\theta^{\Fr}(\lambda) = \theta(\lambda^{q^2}) =  \theta(\lambda)^{q^2} = \theta(\lambda)
\end{align*}
for all $\lambda \in \F_{q^2}$. That is, $\theta^{\Fr} = \theta$, a contradiction. Hence $\gamma$ must have order $2$. Thus $n = 2m$ for some $m$ and $\gamma = \Fr^m$ is the unique element of order $2$ in $\Gal(l / k)$. In all,
\begin{align*}
\theta^{-1} = \theta^\gamma = \theta^{\Fr^m} = \theta^{(q^2)^m} = \theta^{q^{2m}} = \theta^{q^n}.
\end{align*}
We keep this in mind as we study case (B).

\subsection{Case (B): ${}^{\varpi_D}\rho\simeq\rho^{\lor}$}

From \cite{BH2006} we have that for $x\in \O_D$,
\begin{align*}
\varpi_D x \varpi_D^{-1} \equiv x^q \mod \p_D.
\end{align*}
That is, conjugating by $\varpi_D$ induces the non-trivial $\F_{q}$-automorphism $\lambda \mapsto \lambda^q : \F_{q^2} \to \F_{q^2}$. This extends to a map $\GL_{n}(\F_{q^2}) \to \GL_{n}(\F_{q^2})$ via
\begin{align*}
(a_{ij}) \mapsto (a_{ij}^q).
\end{align*}
which we also denote by $\overline{(a_{ij})} = (\overline{a_{ij}})$. Recall that $\overline{\rho}$ denotes the representation
\begin{align*}
\overline{\rho}(g) = \rho(\overline{g}).
\end{align*}
\begin{proposition}\label{prop2}
We have
\begin{align*}
{}^{\varpi_D}\rho_{\theta} \simeq \rho_{\theta^q}.
\end{align*}
\end{proposition}
\begin{proof}
By Proposition \ref{prop1}, $\overline{\rho_{\theta}} \simeq \rho_{\theta^q}$. So we need to prove that $\overline{\rho_{\theta}} \simeq {}^{\varpi_D}\rho_\theta$. Recall that conjugating by $\varpi_D$ is the same as raising to the power $q$ mod $\p_D$. Therefore, for $x\in \F_{q^{2n}}^{\times}$, $\tr \rho_{\theta^q}(x) = \tr \overline{\rho_{\theta}}(x)$ on $\F_{q^{2n}}^{\times}$. So by Lemma \ref{tracelemma}, $\overline{\rho_{\theta}} \simeq \rho_{\theta^q}$.
\end{proof}
Thus
\begin{align*}
\rho_{\theta^{-1}} \simeq \rho_{\theta}^{\lor}  \simeq {}^{\varpi_D}\rho \simeq \overline{\rho_\theta} \simeq \rho_{\theta^q}.
\end{align*}

As before this gives the existence of an element $\gamma \in \Gal(\F_{q^{2n}} / \F_{q^2})$ such that 
\begin{align*}
(\theta^q)^{\gamma} = \theta^{-1}.
\end{align*}
Since the operators $\phi \mapsto \phi^q$ and $\phi \mapsto \phi^{\gamma}$ commute,
\begin{align*}
(((\theta^q)^\gamma)^q)^\gamma =  \theta^{q^2\gamma^2} = \theta.
\end{align*}
Therefore $\Fr \gamma^2 = 1$, or $\gamma^{2} = \Fr^{-1}$. Recall that $\Fr$ is a generator of
\begin{align*}
\Gamma = \Gal(l/k) = \{x\mapsto x^{q^{2k}} : 0\leq k \leq n-1\}.
\end{align*}
Then $\Fr^{-1}$ is a generator, so $\gamma$ is also a generator. The order of $\gamma^2$ is equal to $n / \gcd(2,n)$. Hence $\gcd(2,n) = 1$ and $n$ must be odd. Say that $n = 2m + 1$. Now $\gamma$ is the unique element in $\gamma \in \Gal(\F_{q^{2n}} / \F_{q^2})$ satisfying $\gamma^{2} = \Fr^{-1}$. Clearly,
\begin{align*}
\Fr^{n} = \Fr^{2m + 1} = 1.
\end{align*}
So
\begin{align*}
(\Fr^{m})^2 = \Fr^{-1}.
\end{align*}
Hence
\begin{align*}
\gamma^2 = \Fr^{-1} = \Fr^{2m},
\end{align*}
so
\begin{align*}
\gamma = \Fr^{m}.
\end{align*}

Thus we obtain the same relation as in the other case:
\begin{align*}
\theta^{q^n} = \theta^{q^{2m + 1}} = (\theta^{q})^{q^{2m}} = (\theta^q)^\gamma = \theta^{-1}.
\end{align*}
We can distinguish case (A) and case (B) by noting that case (A) occurs when $n$ is even and case (B) when $n$ is odd.

\subsection{Showing $\wt{K} = \langle \varpi_F\rangle K$}

\begin{lemma}\label{lemmaK}
We have
\begin{align*}
\wt{K} = \langle \varpi_F\rangle K.
\end{align*}
\end{lemma}
\begin{proof}
In both cases above, we concluded that
\begin{align*}
\theta^{q^n} = \theta^{-1}.
\end{align*}
Assume that $\varpi_D\in \wt{K}$. That is ${}^{\varpi_D}\rho_\theta \simeq \rho_{\theta}$. From Proposition \ref{prop2} we recall that
\begin{align*}
{}^{\varpi_D}\rho_\theta \simeq \rho_{\theta^q}. 
\end{align*}
Thus there is a $\gamma \in \Gal(l/k)$ such that $\theta^\gamma = \theta^q$. So $\theta^{\gamma^2} = \theta^{q^2} = \theta^{\Fr}$. That is $\gamma^2 = \Fr$. As before, $\Fr$ is a generator for the Galois group $\Gal(l/k)$, hence $\gamma$ is a generator. The order of $\gamma^2$ is $n / \gcd(2,n)$. Hence $n$ must be odd.

Thus we are in case (B) where $\rho_{\theta}^{\lor} \simeq {}^{\varpi_D}\rho_{\theta}$. By assumption ${}^{\varpi_D}\rho_{\theta} \simeq \rho_{\theta}$, so $\rho_{\theta} \simeq \rho_{\theta}^{\lor}$. This is case (A) where $n$ is even. Hence we have a contradiction. So $\varpi_D \not\in \wt{K}$.
\end{proof}

\section{Affine roots}
We now want to view $\rho$ as a representation of certain other groups. First we relate these groups to a set of affine roots.

Let 
\begin{align*}
S = \{\begin{pmatrix} \text{diag}(s_1, s_2, \dots , s_n) & 0 \\ 0 & \text{diag}(s_1^{-1}, s_{2}^{-1}, \dots, s_n^{-1} ) \end{pmatrix} : s_i\in F^{\times}
\}
\end{align*}
be a maximal split $F$-torus in $G_\epsilon$. 

Write $\Phi = \Phi(S,G)$ for the roots of $S$ in $G$. This root system is of type $C_n$. Concretely, if
\begin{align*}
e_i: \begin{pmatrix} \text{diag}(s_1, s_2, \dots , s_n) & 0 \\ 0 & \text{diag}(s_1^{-1}, s_{2}^{-1}, \dots, s_n^{-1} ) \end{pmatrix} \mapsto s_i,
\end{align*} 
then
\begin{align*}
\Phi = \{\pm e_i \pm e_j, \pm 2e_i\ : 0\leq i< j \leq n\}.
\end{align*}
The positive roots are
\begin{align*}
\Phi^+ = \{e_i + e_j, e_i - e_j, 2e_i\ : 0\leq i < j  \leq n\}
\end{align*}
and we have the basis
\begin{align*}
\Delta = \{\alpha_1 = e_1 - e_2, \dots, \alpha_{n-1} = e_{n-1} - e_n, \alpha_n = 2e_n\}.
\end{align*}
The longest root is $\tilde{\alpha} = 2e_1$. Now we consider the affine roots $a_{\alpha, k} = \alpha + k$ for $\alpha$ a root and $k$ an integer. (Here by $k$ we mean the translation by $k$.) Specifically, let $a_0 = 1 - \tilde{\alpha}$, and for $1\leq i \leq n$ let $a_n = a_{\alpha_i, 0}$. Then
\begin{align*}
\{a_0, a_1, \dots, a_n\}
\end{align*}
is a basis for the affine roots. And we have the affine Weyl group $\wt{W} = \langle s_{\alpha_i} : 0\leq i \leq n \rangle$ where $s_{\alpha_i}$ is a fundamental reflection in a hyperplane. 


\section{The types and covers}

Morris in \cite{M1993} gives us the types and covers in our situation. We now describe them in some detail. We have $\rho$, a representation of $K = \GL_n(\O_D)$. Let $\Theta = \{a_1, \dots , a_{n-1}\}$. Let $I$ be the Iwahori subgroup of $G$, that is, the set of $(2n) \times (2n)$ matrices in $G$ of the form:
\begin{align*}
I = \begin{pmatrix} \O_D & \O_D & \dots & \O_D \\
 \p_D & \O_D & \dots & \O_D \\
 \vdots & & \ddots & \\
 \p_D & \p_D & \dots & \O_D
   \end{pmatrix} \cap G_{\epsilon}.
\end{align*}
This is the set of matrices with elements from $\O_D$ on the diagonal and above and elements from $\p_D$ below the diagonal. Then we get the Siegel parahoric subgroup (compact open subgroup, $2n\times 2n$ matrices)
\begin{align*}
\P &= \P_{\Theta} = IW_{\Theta}I = \begin{pmatrix}\O_D & \O_D \\ \p_D & \O_D\end{pmatrix}\cap G.
\end{align*}
Note that $\P$ is not maximal. We recall that $\rho$ is a representation of $K= \GL_n(\O_D)$, and now we want to view $\rho$ as a representation of $\P$. Then let $\M = \P \cap L \simeq K$. Concretely,
\begin{align*}
\M &= \left\{ \begin{pmatrix} a & 0 \\ 0 & {}^\top\bar{a}^{-1}\end{pmatrix} : a\in \GL_{n}(\O_D) \right\}.
\end{align*}

Using the isomorphism $\M\simeq K$ we view $\rho$ as a representation of $\M$. We often abuse notation and write $\rho$ for this representation. Later we will need to distinguish this representation, and so we use the notation $\rho_{\M}$. Let 
\begin{align*}
\N &= \P \cap U_\epsilon = \{\begin{pmatrix} I_n & X \\ 0 & I_n  \end{pmatrix}: X\in M_{n}(\O_D), \epsilon X + \overline{X}^{\top} = 0 \}, \\
\overline{\N} &= \P \cap \overline{U}_\epsilon = \{\begin{pmatrix} I_n & 0 \\ X & I_n \end{pmatrix} : X\in M_n(\O_D), \epsilon X + \overline{X}^{\top} = 0 \}.
\end{align*}

\begin{lemma}
We have
\begin{align*}
\P = \overline{\N}\cdot \M \cdot \N = (\P\cap \overline{U_{\epsilon}})(\P\cap L)(\P\cap U_{\epsilon})
\end{align*}
with every element of $\P$ being written uniquely as a product $n'\cdot m\cdot n$ where $n' \in \overline{N}, m\in M, n\in N$.
\end{lemma}

We now extend $\rho$ to a representation of $\P$ by $\rho(\bar{n}mn) = \rho(m)$. We must show that this is well-defined. Let $\M_1$ be the image of $K_1 = 1 + \varpi_DM_n(\O_D)$ under the isomorphism $K\simeq \M$. Thus $\rho$ is trivial on $\M_1$. Note that
\begin{align*}
\M \cap \langle \overline{\N}, \N \rangle \subseteq \M_1. 
\end{align*}
Suppose $\bar{n}mn = \bar{n}'m'n' \in \P$. Then $m\M_1 = m'\M_1$. Hence $\rho(\bar{n}mn) = \rho(m) = \rho(m') = \rho(\bar{n}'m'n')$.  

Now let $\s = [L,\pi_0]_{G}$ and $\s_L = [L,\pi_0]_L$.  From Morris \cite{M1999}, we have:

\begin{theorem}
$(K, \rho)$ is an $\s_L$-type in $L$ and $(\P,\rho)$ is a $G$-cover of $(K,\rho)$. In particular, $(\P,\rho)$ is an $\s$-type in $G$.
\end{theorem}
By the theory of types and covers, we have the following commutative diagram
\xymatrixcolsep{0.8in}
\begin{align*}
\xymatrix{ \R^\s(G)\ar[r]^{m_{G}} &\H(G, \rho)-\text{Mod}\\ \R^{\s_L}(L)\ar[r]_{m_L}\ar[u]^{\i_P^{G}} & \H(L,\rho)-\text{Mod}.\ar[u]_{t_{P*}}}
\end{align*}
We note the importance that $\pi_0$ has depth zero in the use of $\cite{M1999}$. We recall that  $t_{P*}$ is the map $U \mapsto \Hom_{t_P(\H(L, \rho_L))}(\H(G,\rho), U)$ induced by the embedding of $\C$-algebras $t_P : \H(L,\rho) \hookrightarrow \H(G,\rho)$.



\section{The Hecke algebra}

We have now reduced the problem to finding the Hecke algebras $\H(L,\rho)$ and $\H(G, \rho)$. Since the intertwining of $\rho$ in $L$ is 
\begin{align*}
\I_{L}(\rho) &= \wt{K} = F^{\times}K = \langle \varpi_F\rangle K \\ 
&= \coprod_{i\in \mathbbm{Z}} K\varpi_F^i K,
\end{align*}
we have
\begin{align*}
\H(L, \rho) \simeq \C[d, d^{-1}].
\end{align*}
Here $d$ is an indeterminate. It corresponds to the element in $\H(L,\rho)$ that has support $\varpi_F K$ and whose value at $\varpi_F$ is $1$. Let $\D = \C[d, d^{-1}]$. 

Our main result is:

\begin{theorem}\label{theoremmain}
\begin{align*}
\H(G, \rho) = \langle f_0, f_1 : f_i^2 = q^{n} + (q^{n} - 1)f_i \rangle.
\end{align*}
\end{theorem}
Note that the relations $\H(G,\rho)$ can also be presented by\label{otherrelations}
\begin{align*}
\H(G, \rho) = \langle h_0, h_1 : h_i^2 = 1 + (q^{n/2} - q^{-n/2})h_i \rangle,\label{usingthes}
\end{align*}
where $f_i = q^{n/2}h_i$, $i=0,1$.

%%%%%%%%%%%%%%%%%%%%% THE PROOF %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The proof will occupy some space and we divide it up. 

\subsection{General preliminaries}
We now fix $\epsilon = 1$ and let $G = G_1$. The case of $\epsilon = -1$ is very similar. We indicate in section \ref{sec:negone} the minor changes needed to treat this case.

Let
\begin{align*}
N_{G}(\rho_\M) &= \{g\in L : {}^g\M = \M, {}^g\rho_\M \simeq \rho_\M\}.
\end{align*}
From 4.15 in \cite{M1993},
\begin{align*}
\I_G(\rho) = \P N_{G}(\rho_\M)\P.
\end{align*}
We have
\begin{align*}
N_{G}(\rho_\M) \subseteq N_{G}(\M) = \langle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} \varpi_D & 0 \\ 0 & -\varpi_D^{-1} \end{pmatrix} \rangle\M.
\end{align*}

As in the proof of Lemma \ref{lemmaK} we now split the proof into two cases depending on whether $n$ is even or $n$ is odd.

\subsection{Case (A)}\label{caseA} Assume first that $n$ is even. That means that $\rho \simeq \rho^{\lor}$. Also recall that ${}^{\varpi_D}\rho\simeq \overline{\rho}$. 

Let 
\begin{align*}
w_0 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \\
w_D &= \begin{pmatrix} \varpi_D & 0 \\ 0 & -\varpi_D^{-1} \end{pmatrix},\\
s_0 &= \begin{pmatrix} 0 & \varpi_D \\ -\varpi_D^{-1} & 0 \end{pmatrix} =  w_Dw_0, \text{ and}\\
s_1 &= \begin{pmatrix} 0 & -\varpi_D^{-1} \\ \varpi_D & 0 \end{pmatrix} = w_0w_D.
\end{align*}
Then
\begin{align*}
{}^{w_0}\rho_{\M} &\simeq \overline{\rho}_{\M}^{\lor}, \\
{}^{w_D}\rho_{\M} &\simeq \overline{\rho}_{\M}.
\end{align*}
Since $\rho_{\M} \simeq \rho_{M}^{\lor}$, we have 
\begin{align*}
{}^{s_i}\rho_M \simeq \rho_M
\end{align*}
for $i=0,1$.
Now 
\begin{align*}
\begin{pmatrix}0 & 1 \\ 1 &  0\end{pmatrix} \not\in N_{G}(\rho_M),
\end{align*}
so
\begin{align*}
N_{G}(\rho_{\M}) &\subsetneq N_{G}(\M).
\end{align*}
Thus
\begin{align*}
\langle s_0, s_1 \rangle \M \subseteq N_{G}(\rho_{\M}) \subsetneq N_{G}(\M).
\end{align*}
Now 
\begin{align*}
s_0s_1 = \begin{pmatrix} \varpi_F & 0 \\ 0 &\varpi_F^{-1} \end{pmatrix},
\end{align*}
so
\begin{align*}
\langle s_0, s_1\rangle = \langle s_0\rangle \ltimes \langle s_0s_1\rangle.
\end{align*}
As $s_0 = w_Dw_0$,
\begin{align*}
N_{G}(\M) &= \langle w_0, w_D\rangle \M =\langle s_0, w_D\rangle\M = (\langle s_0\rangle\ltimes \langle w_D\rangle)\M.
\end{align*}
Thus
\begin{align*}
\lvert N_{G}(\M) / \langle s_0, s_1\rangle\M\rvert = \lvert \langle w_D\rangle / \langle s_0s_1\rangle \rvert = 2.
\end{align*}
We conclude that
\begin{align*}
N_{G}(\rho_{\M}) = \langle s_0, s_1 \rangle \M.
\end{align*}
Hence
\begin{align*}
\I_{G}(\rho) = \P N_{G}(\rho_M)\P = \P \langle s_0, s_1 \rangle \P.
\end{align*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $W(\M) = \langle w_0, s_1 \rangle / \{\pm 1\}\simeq N_{G}(\M) / \M$. That is, $W(\M)$ is an infinite dihedral group generated as a Coxeter group by $w_0$ and $s_1$ where we abuse notation and write $w_0$ and $s_1$ for the corresponding elements in the quotient $W(\M)$. 

We now want to consider $\P \backslash G / \P$ double cosets. First, by general structure theory or direct matrix multiplications as in section 6.9 in \cite{AB2008}, the map 
\begin{align*}
w \mapsto \P w\P  : W(\M) \to \P \backslash G / \P
\end{align*}
is injective. 

Second, let $l$ be the length function on $W(\M)$. We have, for example, $l(s_1) = 1$ and $l(s_1w_0) = 2$. By \cite{AB2008},
\begin{itemize}
\item  $\P w \P w' \P = \P ww' \P$ if $l(ww') = l(w) + l(w')$.
\item $\P w_0 \P w_0 \P = \P \cup \P w_0 \P$.
\end{itemize}
One checks that
\begin{align*}
s_0 = w_0s_1w_0^{-1}.
\end{align*}
With this in mind we see that
\begin{align*}
\I_{G_{\epsilon}}(\rho) = \P\langle s_0, s_1\rangle\P.
\end{align*}
Write $W(\rho_{\M})$ for $\langle s_0, s_1\rangle / \{\pm 1\}$. Then $W(\rho_{\M})$ is a subgroup of $W(\M)$. 

Under the isomorphism $W(\M) \simeq N_{G}(\M) / \M$ this subgroup corresponds to $N_{G}(\rho_{\M}) / \M$:
\begin{align*}
W(\rho_{\M}) \simeq N_{G}(\rho_{\M}) / \M. 
\end{align*}

The group $W(\rho_{\M})$ is again an infinite dihedral group is generated as a Coxeter group by $\{s_0, s_1\}$ (again abusing notation). This Coxeter group has the length function $l_{\rho}$. Furthermore, if $l_{\rho}(ww') = l_{\rho}(w) + l_{\rho}(w')$ then $l(ww') = l(w) + l(w')$. Thus if $l_{\rho}(ww') = l_{\rho}(w) + l_{\rho}(w')$, then $\P w\P w' \P = \P ww'\P$. 

For $i=0,1$, let $f_i$ be any element in $\H(G_{\epsilon}, \rho)$ with support $\P s_i\P$. By Schur's Lemma, $f_i$ is unique up to multiplication by non-zero scalars. Each element of $W(\rho_{\M})$ is a word in $\{s_0, s_1\}$. Hence $\H(G_{\epsilon}, \rho)$ is generated as a $\C$-algebra by $f_0$ and $f_1$. We now want to find the relations that the two generators satisfy.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let
\begin{align*}
K_1 &= \langle \P, s_1\rangle.
\end{align*}
Note that $K_1$ is a compact open subgroup of $G$.

We will show that
\begin{align*}
\H(K_1, \rho) \simeq \End_{\Sp_{2n}(k)}(\ind_{\overline{P}(k)}^{\Sp_{2n}(k)} \overline{\rho}).
\end{align*}
To motivate the following sections, we assume for the moment that we have established this isomorphism. By Proposition 3.2 in \cite{KM2006}, there exists a unique $f_1$ in $\H(G,\rho)$ with $\supp f_1 = \P s_1 \P$ such that 
\begin{align*}
f_1^2 = \lambda 1 + (\lambda - 1)f_1
\end{align*}
where $\lambda$ is given as follows. The representation $\ind_{\overline{P}(k)}^{\Sp_{2n}(k)} \overline{\rho} = \pi_1 \oplus \pi_2$ where $\pi_1$ and $\pi_2$ are distinct irreducible representations of $\Sp_{2n}(k)$. Assume $\dim(\pi_1) \leq \dim(\pi_2)$. We need to find the parameter $\lambda$ in the relation above to determine the Hecke algebra $\H(K_1, \rho)$. By \cite{KM2006}, \label{lambdaref}
\begin{align*}
\lambda = \frac{\dim(\pi_2)}{\dim(\pi_1)}.
\end{align*}

We will compute this quotient in our situation. First we recall that $\theta$ corresponds to a semi-simple element $s^*\in T^*$ in the dual group $G^*$. Then by theorems 8.4.8 and 8.4.9 in \cite{C1985},
\begin{align*}
\lambda = \lvert C_{G^*}(s^*) \rvert_p.
\end{align*}
In section \ref{symplecticcase} below we explain how the finite group $\Sp_{2n}(k)$ appears.



\subsubsection{Quadratic relations I}\label{Sp-quadratic-relations I}

It is easier to find $\H(K_1^g, \rho^g) \simeq \H(K_1, \rho)$ for a suitable $g$. Let $\Psi$ be the composition of $\Ad\left(\begin{smallmatrix}\varpi_D & 0 \\ 0 & 1 \end{smallmatrix}\right)$ with reduction mod $\p_D$. By ``reduction mod $\p_D$'' we mean that we take each entry in the matrix mod $\p_D$.  

We need to find $\Psi(K_1)$. Recall first that 
\begin{align*}
\P = \N\M\overline{\N}.
\end{align*}
We find the image of each of these factors. Clearly,
\begin{align*}
\Psi(\N) = \begin{pmatrix} 1 & \p_D \\ 0 & 1 \end{pmatrix} \mod \p_D = \{1\}.
\end{align*}
Writing $\Ad$ for $\Ad\left(\begin{smallmatrix}\varpi_D & 0 \\ 0 & 1  \end{smallmatrix}\right)$,
\begin{align*}
\Ad(\M) = \left\{\begin{pmatrix} \varpi_Dg\varpi_D^{-1} & 0 \\ 0 & {}^\top\bar{g}^{-1}   \end{pmatrix}  : g\in \GL_n(\O_D)  \right\}.
\end{align*}
Hence
\begin{align*}
\Psi(\M) =\left \{\begin{pmatrix} a & 0 \\ 0 & {}^\top a^{-1} \end{pmatrix} : a\in \GL_n(\F_{q^2})  \right\}.
\end{align*}
Now recall that
\begin{align*}
\overline{\N} = \left\{ \begin{pmatrix} 1 & 0 \\ y & 1  \end{pmatrix} : y + \overline{y}^{-1}, y\in M_{n}(\O_D)\right\}.
\end{align*}
Then 
\begin{align*}
\Ad\begin{pmatrix} \varpi_D & 0 \\ 0 & 1  \end{pmatrix}\begin{pmatrix} 1 & 0 \\ y & 1 \end{pmatrix} &= \begin{pmatrix} 1 & 0 \\ y\varpi_D^{-1} & 1  \end{pmatrix}.
\end{align*}

We write $x = y\varpi_D^{-1} \in M_n(\O_D)$ and find that
\begin{align*}
\overline{x}^{\top} 
   &= \overline{y\varpi_D^{-1}}^{\top} \\
   &= \overline{\varpi_D^{-1}} \overline{y}^{\top} \\
   &= -\varpi_D^{-1}(-y) \\
   &= \varpi_D^{-1}v \\
   &= \varpi_D^{-1}y\varpi_D^{-1}\varpi_D\\
   &= \varpi_D^{-1} x \varpi_D.
\end{align*}
We now reduce mod $\p_D$ and use that $\varpi_D\alpha \varpi_D^{-1} = \bar{\alpha} = \alpha^q$ mod $\p_D$ to get (now in $M_n(\F_{q^{2}}$))
\begin{align*}
x^{\top}  = x.
\end{align*}
Hence
\begin{align*}
\Psi(\overline{\N}) = \{\begin{pmatrix} 1 & 0 \\ x & 1 \end{pmatrix}  : x^{\top} = x, x\in M_n(\F_{q^2})  \}.
\end{align*}
We find that
\begin{align*}
\Psi(\P) = \overline{P}(\F_{q^2}).
\end{align*}
This is the opposite standard Siegel parabolic subgroup of $\Sp_{2n}(\F_{q^2}$) and so
\begin{align*}
\Psi(K_1) &= \langle\overline{P}(\F_{q^2}), \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \rangle \\
&\leq \Sp_{2n}(\F_{q^2}).
\end{align*}
In fact, we have equality since $\overline{P}(\F_{q^2})$ is a maximal proper subgroup of $\Sp_{2n}(\F_{q^2})$ and $\overline{P}(\F_{q^2})$ does not contain $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix}\right)$. So the image is $\Sp_{2n}(\F_{q^2})$:
\begin{align*}
\Psi(K_1) = \Sp_{2n}(\F_{q^2}).
\end{align*}
 Hence
\begin{align*}
\H(K_1, \rho) \simeq \End_{K_1}(\ind_{\P}^{K_1}\rho) \simeq \End_{\Sp_{2n}(\F_{q^2})}(\ind_{\overline{P}(\F_{q^2})}^{\Sp_{2n}(\F_{q^2})} \tilde{\rho}).
\end{align*}
It now follows that there is an $f_1\in \H(G,\rho)$ with support $\P s_1 \P$ such that
\begin{align*}
f_1^2 = q^{n} + (q^n - 1)f_1,
\end{align*}
provided the parameter $\lambda = q^n$. We show that $\lambda$ takes this value in section \ref{symplecticcase}. 




%Here $\tilde{\rho}$ is the twisted $\rho$ we obtain under $\Ad$ (conjugation).
\subsubsection{Quadratic relations II}\label{Sp-quadratic-relations II}

We need a different approach to find the quadratic relations for $f_0$. Let
\begin{align*}
w_0 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \\
s_1 &= \begin{pmatrix} 0 & -\varpi_D^{-1} \\ \varpi_D & 0 \end{pmatrix}. 
\end{align*}
Note that
\begin{align*}
w_0s_1 = \begin{pmatrix} \varpi_D & 0 \\ 0 & -\varpi_D^{-1} \end{pmatrix}.
\end{align*}
Thus
\begin{align*}
N_{G_{\epsilon}}(\M) &= \langle w_0, s_1\rangle \M,\\
\langle w_0, s_1\rangle\cap \M &= \{\pm 1\}. 
\end{align*}

Now we want to show that $f_0$ can be chosen so that it satisfies the same relation as $f_1$. First note that since $s_0 = w_0s_1w_0^{-1}$ ($w_0 = w_0^{-1}$),
\begin{align*}
\P s_0\P = \P w_0\P s_1 \P w_0\P.
\end{align*}
Now $w_0$ intertwines $\rho$ with $\overline{\rho}$. 
\begin{lemma}
There exist elements $\Lambda\in \H(G_{\epsilon}, \rho)$ and $\Upsilon\in \H(G_{\epsilon} ,\overline{\rho})$ such that
\begin{align*}
\Lambda(p_1w_0p_2) &= \rho(p_1)\Lambda(w_0)\rho(p_2), \\
\Upsilon(p_1w_0p_2) &= \overline{\rho}(p_1)\Upsilon(w_0)\overline{\rho}(p_2) 
\end{align*}
for all $p_1, p_2 \in \P$ and 
\begin{align*}
\Lambda\Upsilon = 1_{\rho}\in \H(G_{\epsilon}, \rho), \\
\Upsilon\Lambda = 1_{\overline{\rho}} \in \H(G_{\epsilon}, \overline{\rho}).
\end{align*}
Here $1_{\rho}$ and $1_{\overline{\rho}}$ are the identities in their respective Hecke algebras.
\end{lemma}
\begin{proof}
First we write ${}^\phi g = {}^\top g^{-1}$ for any $g\in \GL_{n}(\F_{q^2})$. Let $\rho^{\phi}$ be the representation defined by
\begin{align*}
\rho^{\phi}(g) = \rho({}^{\phi} g). 
\end{align*}
Then $\rho^{\phi} \simeq \rho^{\lor}$. We see this by considering the characters of the representations. In particular we note that $\tr \rho({}^\top g^{-1}) = \tr \rho(g^{-1})$. Let $V$ be the space of $\rho$. Thus there is an invertible map $t: V \to V$ satisfying that for all $g\in \GL_n(\F_{q^2})$
\begin{align*}
t\circ \rho({}^\phi g) = \rho(g) \circ t.
\end{align*}
Hence
\begin{align*}
t^2 \rho(g) 
	&= t^2 \circ {}^\phi({}^\phi\rho)(g) \\
	&= t \circ {}^\phi (\rho(g))\circ t \\
	&= \rho(g)\circ t^2.
\end{align*}
By Schur's Lemma, $t^2= c1_V$ for some $c\in \C^{\times}$. Adjusting $t$ by a suitable scalar multiple, we may assume $t^2 = 1_V$. Observe that for $k\in K$,
\begin{align*}
t\rho({}^\top k^{-1}) 
	&=  t{}^\phi\rho(k) \\
	&= \rho(k)t.
\end{align*}
Hence, for $m\in \M$,
\begin{align*}
t\rho_{\M}({}^\top m^{-1}) = \rho_{\M}(m) t.
\end{align*}
Now let $\Lambda_1 $ be the map with support $\P w_0\P$ given by
\begin{align*}
\Lambda_1(p_1 w_0 p_2) = \rho(p_1)t\overline{\rho}(p_2)
\end{align*}
for $p_i\in \P$. We need to show that $\Lambda_1$ is well-defined. To see this, suppose
\begin{align*}
p_1 w_0 p_2 = r_1 w_0 r_2
\end{align*}
where $p_i, r_i \in \P$. Then 
\begin{align*}
r_1^{-1}p_1 = w_0r_2p_2^{-1}w_0^{-1} \in \P \cap w_0\P w_0^{-1}.
\end{align*}
Recall the decomposition
\begin{align*}
\P = \overline{\N}\M\N.
\end{align*}
Write
\begin{align*}
r_1^{-1}p_1 &= k^-k^0k^+, \\
r_2p_2^{-1} &= l^+l^0l^-
\end{align*}
for $k^-,l^- \in \overline{\N}$, $k^0, l^0 \in \M$, $k^+, l^+ \in \N$. We have
\begin{align*}
k^-k^0k^+ = (w_0l^+w_0^{-1})(w_0l^0w_0^{-1})(w_0l^-w_0^{-1}) \in \overline{N}MN.
\end{align*}
By uniqueness of expression in $\overline{N}MN$,
\begin{align*}
k^0 = w_0l^0w_0^{-1}. 
\end{align*}
Further,
\begin{align*}
\rho(r_1^{-1}p_1) &= \rho_{\M}(k^0), \\
\overline{\rho}(r_2p_2^{-1}) &= \overline{\rho}_{\M}(l^0).
\end{align*}
We want to prove that
\begin{align*}
\Lambda_1(p_1 w_0 p_2) = \Lambda_1(r_1 w_0 r_2).
\end{align*}
That is, we want
\begin{align*}
 \rho(p_1)t\overline{\rho}(p_2) &=  \rho(r_1)t\overline{\rho}(r_2).
\end{align*}
We can state this as
\begin{align*}
\rho(r_1^{-1}p_1)t &= t\overline{\rho}(r_2p_2^{-1}).
\end{align*}
That is,
\begin{align*}
\rho_\M(k^0)t = t\overline{\rho_\M}(l^0).
\end{align*}
But $l^0 = w_0k^0w_0^{-1}$ ($w_0^{-1} = w_0$), so we need
\begin{align*}
\rho_{\M}(k^0)t = t \overline{\rho_\M}(w_0k^0w_0^{-1}).
\end{align*}
This comes from the definition of $t$. Hence $\Lambda_1$ is well-defined. 

Likewise we obtain a well-defined function $\Upsilon_1$ with support $\P w_0\P$ satisfying
\begin{align*}
\Upsilon_1(p_1w_0p_2) = \overline{\rho}(p_1)t \rho(p_2)
\end{align*}
for $p_i\in \P$.

Now the convolution $\Lambda_1\Upsilon_1\in \H(G, \rho)$ has support of contained in
\begin{align*}
\P w_0\P w_0\P = \P \sqcup \P w_0 \P. 
\end{align*}
If $\Lambda_1\Upsilon_1(w_0) \neq 0$, then there is a function in $\H(G, \rho)$ which is supported on $\P w_0 \P$. This implies ${}^{\varpi_D}\rho_\M \simeq\rho_{\M}$, so ${}^{\varpi_D}\rho \simeq \rho$. As we are in case (A), this is a contradiction. Hence the support of $\Lambda_1\Upsilon_1$ is contained in $\P$ and $\Lambda_1\Upsilon_1 = c1_{\rho} \in \H(G, \rho)$ for some constant $c \in \C^\times$. The identity element $1_{\rho}$ is supported on $\P$ and satisfies
\begin{align*}
1_{\rho}(p) = \frac{1}{\vol(\P)}\rho(p)
\end{align*}
for $p\in \P$.
Hence
\begin{align*}
\Lambda_1\Upsilon_1(1) = \frac{c}{\vol(\P)}1_V. 
\end{align*}
We now determine the value of $c$. 
\begin{align*}
\frac{c}{\vol(\P)}1_V
	&= \Lambda_1\Upsilon_1(1) \\
	&= \int_{G_{\epsilon}} \Lambda_1(g)\Upsilon_1(g^{-1})\; dg \\
	&= \int_{\P w_0 \P} \Lambda_1(g)\Upsilon_1(g^{-1})\; dg.
\end{align*}
We use here that the support of $\Lambda_1$ and $\Upsilon_1$ is $\P w_0\P$. For $g = p_1 w_0 p_2 \in \P w_0 \P$, we have $g^{-1} = p_2^{-1} w_0 p_1^{-1}$, so
\begin{align*}
\Lambda_1(g)\Upsilon_1(g^{-1})
	&= \rho(p_1)t\overline{\rho}(p_2)\overline{\rho}(p_2^{-1})t\rho(p_1^{-1}) \\
	&= \rho(p_1)tt\rho(p_1^{-1})\\
	&= \rho(p_1)\rho(p_1^{-1}) \\
	&= 1_V.
\end{align*}
Thus
\begin{align*}
\frac{c}{\vol(\P)}1_V
	&= \int_{\P w_0\P} 1_V \; dg \\
	&= \vol(\P w_0 \P)1_V.
\end{align*}
Hence
\begin{align*}
c &= \vol(\P)\vol(\P w_0 \P) \\
  &= \vol(\P)\vol(\P) [\P w_0 \P : \P].
\end{align*}
Let $\Lambda = c^{-1/2}\Lambda_1$ and $\Upsilon = c^{-1/2}\Upsilon_1$. In all 
\begin{align*}
\Lambda \Upsilon 
	&= c^{-1}\Lambda_1\Upsilon_1 \\
	&= c^{-1}c 1_{\rho} \\
	&= 1_{\rho}.
\end{align*}
In the same way
\begin{align*}
\Upsilon\Lambda &= 1_{\overline{\rho}}.
\end{align*}
\end{proof}

We can say a bit more about the constant $c$ in the proof above. In fact, we can compute the index $[\P w_0 \P : \P] = [\P : \P\cap w_0 \P w_0^{-1}]$. Here 
\begin{align*}
\P \cap w_0 \P w_0^{-1} = \begin{pmatrix}\O_D & \O_D \\ \p_D & \O_D \end{pmatrix} \cap \begin{pmatrix}\O_D & \p_D \\ \O_D & \O_D \end{pmatrix} = \begin{pmatrix}\O_D & \p_D \\ \p_D & \O_D \end{pmatrix}.
\end{align*}
(This is, of course, all intersected with $G_1$.) Hence we see that the index of $\P\cap w_0 \P w_0^{-1}$ is exactly the size of the set\
\begin{align*}
N(\F_{q^2}) = \left\{\begin{pmatrix}1 & X \\ 0 & 1\end{pmatrix}  :  X \in M_n(\O_D / \p_D), X + \overline{X}^{\top} = 0\right\}. 
\end{align*}
The size of this set is
\begin{align*}
\lvert N(\F_{q^2})\rvert = q^{n^2}.
\end{align*}
Note that $X = (x_{ij})\in N(\F_{q^2})$ means that $x_{ji}= -x_{ij}^q = 0$ for $j> i$. For $i = j$ we get $x_{ii} + x_{ii}^q = 0$. This equation has exactly $q$ solutions. In all, the number of elements in $N(\F_{q^2})$ is
\begin{align*}
(q^2)^{\sum_{i=1}^{n-1} i} q^n 
	&=(q^2)^{\frac{1}{2}(n-1)n} q^n \\
	&=q^{(n-1)n}q^n \\
	&= q^{n^2}.
\end{align*}
Thus
\begin{align*}
c = \vol(\P)^2q^{n^2}.
\end{align*}
Since $n=2m$ is even,
\begin{align*}
\sqrt{c} = \vol(\P)q^{n^2/2} = \vol(\P)q^{2m^2}.
\end{align*}

We can (and do) choose an element $f_1' \in \H(G_{\epsilon}, \overline{\rho})$ such that 
\begin{align*}
(f_1')^2 = q^n + (q^n - 1)f_{1}'.
\end{align*}
Let $f_0 = \Lambda f_{1}'\Upsilon$. Then
\begin{align*}
f_0^2 
	&= (\Lambda f_{1}'\Upsilon)^2 \\
	&= \Lambda f_1'\Upsilon \Lambda f_1'\upsilon \\
	&= \Lambda (f_{1}')^2 \upsilon \\
	&= \Lambda(q^n + (q^n - 1)f_{1}')\Upsilon \\
	&= q^n \Lambda \Upsilon + (q^n - 1)\Lambda f_1'\Upsilon \\
	&= q^n + (q^n-1)f_0.
\end{align*}

\subsection{Case (B)}  We now turn to the case where $n = 2m + 1$ is  odd and ${}^{\varpi_D}\rho \simeq \rho^{\lor}$. The arguments here are broadly similar to those in case (A), so we do not provide all the details, but just point out the differences. The groups $\N, \M$, and $\overline{\N}$ are as before. Again we recall that $k = \F_{q^2}$ and $l = \F_{q^{2n}}$. Let
\begin{align*}
w_0 &= \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}, \\
s_1 &= \begin{pmatrix} 0 & -\varpi_D^{-1} \\ \varpi_D & 0 \end{pmatrix}.
\end{align*}
Now
\begin{align*}
s_0 &= s_1 w_0 s_1^{-1} = \begin{pmatrix} 0 & \varpi_F^{-1} \\ \varpi_F & 0\end{pmatrix}.
\end{align*}
Then
\begin{align*}
N_{G}(\M) = \langle w_0, s_1\rangle\M
\end{align*}
and
\begin{align*}
\I_{G}(\rho) &= \P\langle w_0, s_0 \rangle \P.
\end{align*}
Let
\begin{align*}
K_0 &= \langle \P, w_0\rangle. \\
\end{align*}

\subsubsection{Quadratic relations I}

Note $K_0\subseteq \GL_{2n}(\O_D)$, so we can reduce $K_0$ mod $\p_D$. We write $\Psi$ for this operation.

As in \ref{Sp-quadratic-relations I}, $\Psi(\N)$ is trivial. 
\begin{align*}
\Psi(\M) &= \left\{ \begin{pmatrix} a & 0 \\ 0 & {}^{\top}\bar{a}^{-1} \end{pmatrix} : a\in \GL_{n}(k) \right\}.
\end{align*}
Now $\Psi(\P)$ is a maximal parabolic subgroup of $U_{2n}(\F_{q^2})$, where we realize the unitary group using the form given by
\begin{align*}
J = \begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}.
\end{align*}
That is
\begin{align*}
U_{2n}(\F_{q^2}) &= \{g\in \GL_{2n}(\F_{q^2}) : {}^\top\bar{g}Jg = J\}.
\end{align*}
As $\Psi(w_0)\not\in \Psi(\P)$, it follows that $\Psi(K_0) = U_{2n}(\F_{q^2})$. Hence we need to computer the parameter $\lambda$ from section \ref{caseA} (page \pageref{lambdaref}) and we do this in section \ref{unitarycase}.

\subsubsection{Quadratic relations II}

Everything works out as in section \ref{Sp-quadratic-relations II}. 

\section{The $\epsilon = -1$ case}\label{sec:negone}

Until now we have silently assumed that $\epsilon = 1$. We now touch on the case when $\epsilon = -1$. Much is the same and independent of the value of $\epsilon$. Now, the group $G_{-1}$ is realized with respect to the form corresponding to
\begin{align*}
J = \begin{pmatrix} 0 & I_n \\ -I_n & 0\end{pmatrix}.
\end{align*}
We again get two cases and two sets of quadratic relations for each case.

We have $\P = \overline{\N}\M\N$ with $\M$ as above and 
\begin{align*}
\N = \left\{\begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix} :  \overline{X}^{\top} = X \right\},\quad \overline{\N} = \left\{\begin{pmatrix} I_n & 0 \\ X & I_n\end{pmatrix} :  \overline{X}^{\top} = X \right\}.
\end{align*}
As in section \ref{Sp-quadratic-relations I} we get a situation where we want to reduce mod $\p_D$. The key thing now is that
\begin{align*}
\Psi(\overline{\N}) &= \left\{ \begin{pmatrix} I_n & 0 \\ x & I_n \end{pmatrix} : x^{\top} = -x\right\}.
\end{align*}

For case (A) with the corresponding parahoric $K$, $\Psi(K)$  is the special orthogonal group $\SO_{2n}(\F_{q^2})$. As $\SO_{2n}$ does not have connected center, we cannot directly appeal to Lusztig's work to determine the parameter $\lambda$. Instead, we appeal to \cite{KM2006} to see that $\lambda = q^n$. For case (B) we again get the unitary group.

\section{Unitary case}\label{unitarycase}

It remains to compute the parameter $\lambda$ in the unitary and the symplectic cases. We start with the unitary case.

Let $G = U_{2n}(F_{q^2})$ (not to be confused with $G_{\epsilon}$). Since the dual group $G^* \simeq G$, we will leave out the $*$ in the notation.

We realize $G$ using the form
\begin{align*}
J = \begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}.
\end{align*}
That is,
\begin{align*}
G = \{g\in \GL_{2n}(\F_{q^2}) : {}^{\top}\bar{g} J g = J\}.
\end{align*}
We have 
\begin{align*}
P = M \ltimes N
\end{align*}
where
\begin{align*}
M & = \{\begin{pmatrix} a & 0 \\ 0 & {}^\top\bar{a}^{-1}\end{pmatrix} : a\in \GL_{n}(\F_{q^2}) \}, \\
N &= \{\begin{pmatrix} I_n & X \\ 0 & I_n \end{pmatrix} : X + \overline{X}^{\top} = 0\}.
\end{align*}

We also recall that
\begin{align*}
\theta^{q^{n}} = \theta^{-1}. 
\end{align*}
As mentioned in the background material, $\theta$ corresponds to an element $s$ in $M^*\simeq M$. We write
\begin{align*}
s = \begin{pmatrix}\alpha & 0 \\ 0 & {}^\top\bar{\alpha}^{-1}\end{pmatrix} \in M. 
\end{align*}
Here $\alpha \in \F_{q^{2n}}^{\times}$, or more precisely, $\alpha$ is in the image of $\F_{q^{2n}}^{\times}$ under a fixed embedding $\F_{q^{2n}}^\times \hookrightarrow \GL_{n}(\F_{q^2})$. That $\theta$ is regular implies that $\alpha$ generates $\F_{q^{2n}} / \F_{q^2}$. That is, $\F_{q^{2n}} = \F_{q^2}[\alpha]$. Our goal is to compute
\begin{align*}
\lvert C_G(s)\rvert_p.
\end{align*}
That is, we want the order of a Sylow $p$-subgroup of $C_{G}(s)$. By 3.19 in \cite{DM1991} the Sylow $p$-subgroups of $C_{G}(s)$ are the sets of $\F_{q^2}$-points of the unipotent radicals of the Borel subgroups of $C_{\bf G}(s)$. By 2.2 in \cite{DM1991}, the Borel subgroups of $C_{\bf G}(s)$ have the form ${\bf B}\cap C_{\bf G}(s)$ where ${\bf B}$ is a Borel subgroup of ${\bf G}$. Hence $C_{P}(s) = P \cap C_{G}(s)$ contains a Sylow $p$-subgroup of $C_{G}(s)$. Now suppose that $x\in P$ commutes with $s$ and that $x = mn$ with $m\in M$ and $n\in N$. Then $sx = xs$ and so $sxs^{-1} = x$. Hence
\begin{align*}
sxs^{-1} = x = smns^{-1} sms^{-1} sns^{-1}. 
\end{align*}
Since $P = M \ltimes N$ is a direct product, $s$ commutes with both $m$ and $n$. Therefore
\begin{align*}
C_{P}(s) = C_{M}(s) \ltimes C_N(s). 
\end{align*}
The centralizer of $s$ in $M$ is $T$ and thus has order prime to $p$. Furthermore $C_N(s)$ is a $p$-subgroup. In all, 
\begin{align*}
C_{N}(s) = \{n\in N : sn = ns\}
\end{align*}
is a Sylow $p$-subgroup of $C_{G}(s)$. We now compute $\lvert C_N(s)\rvert$. That is, we must compute the number of elements $n\in N$ that commute with $s$.

\begin{lemma}
We have
\begin{align*}
\lvert C_N(s)\rvert = q^n.
\end{align*}
\end{lemma}
\begin{proof}
The elements of $N$ are have the form
\begin{align*}
\begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix},
\end{align*}
where $X + \overline{X}^{\top} = 0$. If 
\begin{align*}
n = \begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix},
\end{align*}
then $ns = sn$ gives
\begin{align*}
\begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix}\begin{pmatrix} \alpha & 0 \\ 0 & {}^\top\bar{\alpha}^{-1} \end{pmatrix} &= \begin{pmatrix} \alpha & 0 \\ 0 & {}^\top\bar{\alpha}^{-1} \end{pmatrix}\begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix},
\end{align*}
which is equivalent to $\alpha X {}^\top\bar{\alpha} = X$. That is, we need to count the number of solutions to the following equations:
\begin{enumerate}
\item $\alpha X {}^\top\bar{\alpha} = X$.
\item $X + {}^\top \overline{X} = 0$.
\end{enumerate}
Now let $\Xi = M_n(\F_{q^2})$. Let
\begin{align*}
\Xi_\epsilon = \{ X\in \Xi : {}^\top \overline{X} = \epsilon X\}.
\end{align*}
Then 
\begin{align*}
\Xi = \Xi_1 \oplus \Xi_{-1},
\end{align*}
since, for any $X\in \Xi$,
\begin{align*}
X = \frac{1}{2}(X + \overline{X}^{\top}) + \frac{1}{2}(X - \overline{X}^\top). 
\end{align*}
Let
\begin{align*}
\Xi(\alpha) &= \{X\in \Xi : \alpha X \alpha^{\top} = X\}, \\
\Xi_\epsilon(\alpha) &= \{X\in \Xi_{\epsilon} : \alpha X \bar{\alpha}^\top = X\}. 
\end{align*}
Then
\begin{align*}
\Xi(\alpha) = \Xi_{1}(\alpha)\oplus \Xi_{-1}(\alpha).
\end{align*}
Choose $\gamma \in \F_{q^2}$ such that $\gamma \neq 0$ and $\overline{\gamma} = -\gamma$. Then the map $\Xi_{\epsilon}(\alpha) \to \Xi_{-\epsilon}(\alpha)$ given by $X \mapsto \gamma X$ is a bijection. 

We also have the bijection from $\Xi_{-1}$ to $C_{N}(s)$ given by
\begin{align*}
X \mapsto \begin{pmatrix} I_n & X \\ 0 & I_n\end{pmatrix}. 
\end{align*}
Hence
\begin{align*}
\lvert C_{N}(s)\rvert &= \lvert \Xi_{-1}(\alpha)\rvert = \lvert \Xi_1(\alpha)\rvert. 
\end{align*}

If we can show that $\lvert \Xi(\alpha)\rvert = q^{2n}$, then we are done. That is, we want to count the number of elements $X \in \Xi$ satisfying that $\alpha X = X {}^\top\bar{\alpha}^{-1}$ for fixed $\alpha$. Consider an embedding $\phi_1: \F_{q^{2n}} \hookrightarrow M_n(\F_{q^2})$: $\beta\mapsto m_{\beta}$. Let $f(x)$ be the minimal polynomial of $\alpha$ over $\F_{q^2}$. Then
\begin{align*}
\F_{q^{2n}} \simeq \F_{q^2}[x] / \langle f(x) \rangle. 
\end{align*}
Under $\phi_1$, a polynomial $p(\alpha)$ in $\alpha$ is mapped to $p(m_\alpha)$. We get another embedding $\phi_2: \F_{q^{2n}} \simeq \F_{q^2}[x] / \langle f(x) \rangle \hookrightarrow M_n(\F_{q^2})$ satisfying
\begin{align*}
p(\alpha) \mapsto p({}^\top\bar{\alpha}^{-1}).
\end{align*} 
We first must show that this map is well-defined. 

Let $\alpha^*$ denote ${}^\top\bar{\alpha}$. Also, for
\begin{align*}
p(x) = \sum_{j} b_j x^j \in \F_{q^2}[x],
\end{align*}
we let
\begin{align*}
\bar{p}(x) = \sum_{j} \overline{b_j}x^{j}. 
\end{align*}
We need to show that the minimal polynomial evaluated at ${}^\top \overline{m}_{\alpha}^{-1}$ is zero. We note that
\begin{align*}
f({}^\top \overline{m}_{\alpha}^{-1}) = f(m_{\alpha}^{-1})^* = (\bar{f}(m_{\alpha}^{-1}))^*.
\end{align*}
We want to show that $\bar{f}(m_{\alpha}^{-1}) = 0$, equivalently $\bar{f}(\alpha^{-1}) = 0$. We have $\theta^{-1} = \theta^{q^n}$, so the same relation applies to $s$ and also to $\alpha$. Thus $\alpha^{-1} = \alpha^{q^n}$. Let $c \in \Gal(\F_{q^{}2n} / \F_{q})$ be the map $c(\lambda) = \lambda^{q^n}$. We have $f(\alpha) = 0$, so
\begin{align*}
0 = f(\alpha)^{q^n} = \bar{f}(\alpha^{q^n}) = \bar{f}(\alpha^{-1}). 
\end{align*}
The second equality uses the fact that $n$ is odd. Hence the embedding is well-defined. 

We now have two different embeddings $\phi_1$ and $\phi_2$ of $l$ in $M_n(\F_{q^2})$. We want to find the number of $X$ satisfying $\alpha X = X {}^\top\bar{\alpha}^{-1}$. That is, we want to find the number of $X$ satisfying
\begin{align*}
X\phi_2(\lambda) = \phi_1(\lambda)X
\end{align*}
for all $\lambda \in \F_{q^{2n}}$. Let $k = \F_{q^2}$ and $l = \F_{q^{2n}}$. As in the proof of Proposition \ref{prop1}, we make $k^n$ into an $l$-module in two ways:
\begin{align*}
\lambda.v &= \phi_1(\lambda)v, \\
\lambda_* v &= \phi_2(\lambda)v.
\end{align*}
We then obtain two $l$-modules that we denote ${}_1k^n$ and ${}_2 k^n$. That means that $X\phi_2(\lambda) = \phi_1(\lambda)X$ is equivalent to
\begin{align*}
X \in \Hom_l({}_1k^n, {}_2k^{n}) \simeq \Hom_l(l, l) \simeq l,
\end{align*}
so $\lvert\Hom_l({}_1k^n, {}_2k^{n})\rvert = q^{2n}$. 
\end{proof}





\section{Symplectic case}\label{symplecticcase}
We again let $k = \F_{q^2}$ and $l = \F_{q^{2n}}$. Let $G = \Sp_{2n}(\F_{q^2})$. The dual group $G^{*} = \SO_{2n+1}(\F_{q^2})$. We realize $G^*$ using
\begin{align*}
J = \begin{pmatrix}0 & 0 & I_n \\ 0 & 1 & 0 \\ I_n & 0 & 0\end{pmatrix}.
\end{align*}
That is,
\begin{align*}
G^* = \{g\in \SL_{2n+1}(\F_{q^2}) : {}^{\top}g J g = J\}.
\end{align*}
Let
\begin{align*}
M^* &= \left\{ \begin{pmatrix}a & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & {}^\top a^{-1}\end{pmatrix} : a\in \GL_{n}(\F_{q^2})\right\}
\end{align*}
and
\begin{align*}
N^* &= \left\{\begin{pmatrix}I_n & u & X \\ 0 & 1 & -u^{\top} \\ 0 & 0 & I_n\end{pmatrix} : X^{\top} + uu^{\top} + X = 0\right\}.
\end{align*}
Then
\begin{align*}
P^* &= M^* \ltimes N^*
\end{align*}
is the standard Siegel parabolic subgroup. As mentioned in section \ref{sec:finitegroups}, the character $\theta$ corresponds to an element
\begin{align*}
s^* = \begin{pmatrix} \alpha & & \\ & 1 & \\ & & {}^\top\alpha^{-1}\end{pmatrix}
\end{align*}
in $M^*$. Here $\alpha \in \F_{q^{2n}}^\times$. That $\theta$ is regular means that $k[\alpha] = \F_{q^{2n}}$. We want to compute
\begin{align*}
\lvert C_{G^*}(s^*)\rvert_{p}.
\end{align*}
First we observe (as in the unitary case) that
\begin{align*}
\lvert C_{G^*}(s^{*})\rvert = \lvert C_{N^*}(s^{*})\rvert.
\end{align*}
\begin{lemma}
We have
\begin{align*}
\lvert C_{N^*}(s^*)\rvert = q^{n}.
\end{align*} 
\end{lemma}
\begin{proof}
Let 
\begin{align*}
\begin{pmatrix}I_n & u & X \\ 0 & 1 & -u^{\top} \\ 0 & 0 & I_n\end{pmatrix} \in N^*.
\end{align*}
Then $s^*n = ns^*$ means
\begin{enumerate}
\item $\alpha u =u$,
\item ${}^\top u{}^\top a\alpha^{-1} = {}^{\top} u$, and
\item $\alpha X = X{}^\top\alpha^{-1}$.
\end{enumerate}
The first condition gives $u = 0$ since $\alpha \neq 1$ which holds because $k(\alpha) = l$. We are left with the third condition. As before, let $\Xi = M_n(k)$ and $\Xi_\epsilon = \{X\in \Xi : {}^\top X = \epsilon X\}$. Again,
\begin{align*}
\Xi = \Xi_1 \oplus \Xi_{-1}.
\end{align*}
We define $\Xi(\alpha)$, $\Xi_{\epsilon}(\alpha)$ as before, so that
\begin{align*}
\Xi(\alpha) = \Xi_1(\alpha) \oplus \Xi_{-1}(\alpha).
\end{align*}
We again have a bijection from $\Xi_{-1}(\alpha) \to C_{N^*}(s^*)$ now given by
\begin{align*}
X \mapsto \begin{pmatrix} I_n & 0 & X \\ 0 & 1 & 0 \\ 0 & 0 & I_n\end{pmatrix}. 
\end{align*}
As in the unitary case, we define two $l$-module structures on $k^n$. Again let $f(x)$ be the minimal polynomial of $\alpha$ over $k$. As before, $l \simeq k[x] / \langle f(x)\rangle$. For $p(x) \in k[x]$ we define an $l$-action on $k^n$ by
\begin{align*}
p(\alpha).v = p(\alpha)v.
\end{align*}
Note this makes sense since $p(\alpha)\in M_n(k)$. We define the second $l$-module structure via
\begin{align*}
p(\alpha)_* v = p({}^\top\alpha^{-1})v = {}^\top p(\alpha^{-1})v. 
\end{align*}
Again this makes sense since $p(\alpha^{-1})\in M_n(k)$. Again we need to check that this second structure is well-defined. That is, we need to check that $f(\alpha^{-1}) = 0$. Again we use that $\alpha^{q^n} = \alpha^{-1}$, so
\begin{align*}
0 
	&= f(\alpha)\\
	&= f(\alpha)^{q^n} \\
	&= f(\alpha^{q^n}) \\
	&= f(\alpha^{-1}).
\end{align*}
As before, we now have two $l$-module structures ${}_1k^n$ and ${}_2k^n$ on $k^n$. Once again $\Xi(\alpha) \simeq l$, and so
\begin{align*}
\lvert \Xi(\alpha)\rvert = \lvert l \rvert = q^{2n}.
\end{align*}
Since $\Xi(\alpha) = \Xi_1(\alpha)\oplus \Xi_{-1}(\alpha)$,
\begin{align*}
\lvert \Xi(\alpha)\rvert = \lvert \Xi_1(\alpha)\rvert \lvert \Xi_{-1}(\alpha)\rvert.
\end{align*}
In particular, each $\lvert \Xi_{\epsilon}(\alpha)\rvert$ is non-zero. Let $X\in \Xi_{-1}(\alpha)$, $X\neq 0$. Then
\begin{align*}
\Xi(\alpha) = \{p(\alpha)X : p(x) \in k[x]\}. 
\end{align*}
We have $p(\alpha)X \in \Xi_{-1}(\alpha)$ exactly when 
\begin{align*}
(p(\alpha)X)^\top = -p(\alpha)X.
\end{align*}
Here the left hand side is
\begin{align*}
X{}^\top p(\alpha)^{\top} = -Xp(\alpha)^{\top}
\end{align*}
because $X \in \Xi_{-1}(\alpha)$. The right hand side is
\begin{align*}
-Xp(\alpha^{-1})^\top .
\end{align*}
So $p(\alpha)X$ is in $\Xi_{-1}(\alpha)$ exactly when
\begin{align*}
Xp(\alpha)^\top = X p(\alpha^{-1})^\top,
\end{align*}
that is, when
\begin{align*}
p(\alpha) = p(\alpha^{-1}). 
\end{align*}
Now let $\sigma \in \Gal(l / k)$ denote the unique element of order $2$ given by $\sigma(x) = x^{q^{n}}$. Then $\sigma(\alpha) = \alpha^{-1}$. So the condition above is equivalent to 
\begin{align*}
p(\alpha) = \sigma(p(\alpha)). 
\end{align*}
Let $l_0$ be the fixed field of $\sigma$. Then $\langle \sigma\rangle = \Gal(l / l_0)$ and so $[l: l_0] = 2$. In conclusion we get
\begin{align*}
\Xi_{-1}(\alpha) = \{\lambda X : \lambda \in l_0\}
\end{align*}
and this set has order $\lvert l_0 \rvert = q^{n}$. Hence,
\begin{align*}
\lvert \Xi_{1}(\alpha)\rvert = \lvert \Xi_{-1}(\alpha)\rvert = q^n.
\end{align*}
\end{proof}

\subsection{Conclusion}
We have now proved Theorem \ref{theoremmain} on page \pageref{theoremmain}. That is, we have proved that in all cases the Hecke algebra $\H(G_\epsilon, \rho)$ is generated by two elements $f_0$ and $f_1$ subject to the same relation $f_i^2 = q^n + (q^n - 1)f_i$. This will now allow us to give the answer to the question that we posed in the beginning:
\begin{center}
When is $\i_P^{G_{\epsilon}}(\pi_0\tensor \chi)$ reducible?
\end{center}
That is, for what $\chi$ is the induced representation reducible?

\section{Answering the question}
We recall the diagram
\begin{align*}
\xymatrix{ \R^\s(G_\epsilon)\ar[r]^{m_{G_\epsilon}} &\H(G_\epsilon, \rho)-\text{Mod}\\ \R^{\s_L}(L)\ar[r]_{m_L}\ar[u]^{\i_P^{G}} & \H(L,\rho)-\text{Mod}.\ar[u]_{t_{P*}}}
\end{align*}
We are interested in the induced representation, that is, in the functor
\begin{align*}
\i_P^{G_{\epsilon}}: \R^{\s_L}(L) \longrightarrow \R^{\s}(G_{\epsilon}).
\end{align*}
We have now translated the question into a question about the map
\begin{align*}
t_{P*}: \H(L, \rho)-\text{Mod} \longrightarrow \H(G_{\epsilon}, \rho)-\text{Mod}. 
\end{align*}
We have found generators and relations for $\H(G_{\epsilon}, \rho)$. Let 
\begin{align*}
\D &= \C[d,d^{-1}] \\
\H &=  \langle h_0, h_1 : h_i^2 = 1 + (q^{n/2} - q^{-n/2})h_i\rangle
\end{align*}
with $d$ an indeterminate. We recall from page \pageref{otherrelations} the note that with $h_i = q^{-n/2}f_i$, $\H$ is generated by $h_0$ and $h_1$ subject to the relation $h_i^2 = 1 + (q^{n/2} - q^{-n/2})h_i$. We recall also that $\H(L, \rho)\simeq \C[d, d^{-1}]$.

Then 
\begin{align*}
\H(L, \rho)-\text{Mod} &\simeq \D-\text{Mod}\\
\H(G_{\epsilon}, \rho)-\text{Mod} &\simeq \H-\text{Mod}.
\end{align*}
Therefore, we have the map
\begin{align*}
\i_*: \D-\text{Mod} \longrightarrow \H-\text{Mod}.
\end{align*}
This map is given by $\i_{*}(T) = \Hom_{\H}(\H, T)$ where $\H$ acts by right translation. 
\begin{proposition}
We have $\i_{*}(d) = h_0h_1$.
\end{proposition}
The proof of this Proposition is like the proof of Proposition 6.4 in \cite{KM2006}. From Proposition 1.6 in \cite{KM2009}, $\i_{*}(\psi)$ is reducible exactly when $\psi(d)\in \{q^{-n} , -1 , q^n\}$. Recall that $\pi_0$ is self-dual, and so the square of its central character $\omega_{\pi_0}$ is $1$: $\omega_{\pi_0}^2 = 1$. That is, the image of $\omega_{\pi_0}$ is contained in  $\{\pm 1\}$.

For $\chi$ an unramified character of $L$,
\begin{align*}
m_{L}(\pi_{0}\tensor \chi)(d) = \omega_{\pi_0}(\varpi_F)\chi(\varpi_F).
\end{align*}
If $\chi$ is trivial, then we have reducibility if and only if $\omega_{\pi_0}(\varpi_F) = -1$. Assume that $\chi$ is not trivial. We know that $\chi(x) = \lvert \Nrd(x)\rvert_F^s$. So we need
\begin{align*}
\{q^{-n}, -1, q^n\} 
	&\ni \chi(\varpi_F)  \\
	&= \lvert \Nrd(\varpi_F)\rvert_F^s \\
	&= \lvert \varpi_F^{2n} \rvert_F^s \\
	&= q^{-\nu_F(\varpi_F^{2n})s} \\
	&= q^{-2ns}.
\end{align*}
Hence we need $s\in \left\{\pm\frac{1}{2}\right\}$

We state this a bit differently. One can choose $\pi_0$ such that $\omega_{\pi_0}(\varpi_F) = 1$, and so we end with the following theorem.

\begin{theorem}
Let $\pi_0$ be a depth zero irreducible unitary supercuspidal representation of $G_\epsilon$. Choose $\pi_0$ such that it has trivial central character. Let $\chi$ be an unramified character of $L$. Then
\begin{align*}
\i_P^G(\pi_0\tensor \chi)
\end{align*}
is reducible if and only if $\chi(\varpi_F) = -1$ or $\chi(\cdot) = \lvert \Nrd(\cdot)\rvert_F^{\pm 1/2}$.
\end{theorem}
















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