\chapter{Unitarizability in Mixed Case for Corank \texorpdfstring{$\leq 3$}{Lg}} \label{mixed-case}

In this chapter we shall use notation and terms introduced in sections 8 and 9 of \cite{MR3969882} regarding
Jantzen decomposition of an irreducible representation of a classical $p$-adic group We shall recall
some of the most basic definitions. See section 8 of \cite{MR3969882} for more details.

\section{Jantzen decomposition}

\subsection{Support of representation of classical group}
Let $X \subseteq \Cusp$ and suppose that $X$ is $F'/F$-selfcontragredient, i.e. that
$$
\check X=X,
$$
where $\check X=\{\check\rho;\rho\in\ X\}$, and let $\s\in\Cuspcl$.
We say that $\g\in\Irrcl$ is supported in $X\cup\{\s\}$ if there exist $\rho_1,\dots,\rho_k\in X$
(not necessarily distinct) such that
$$
\g\leq \rho_1\times\dots\times\rho_k\rtimes\s.
$$
We denote by $\Irrcl_{X\cup\{\s\}}$ the set of representations in $\Irrcl$ supported in $X\cup\{\s\}$.
For a not-necessarily irreducible representation $\pi$ of a classical group, one says that it is supported on $X\cup\{\s\}$
if each irreducible subquotient of it is supported on that set.

\subsection{Regular partition} Let
$$
X=X_1\cup X_2
$$
be a partition of an $F'/F$-selfcontragredient $X\subseteq \Cusp$. We shall say that this partition is regular if $X_1$ (and hence also $X_2$)
is $F'/F$-selfcontragredient,
and if among $X_1$ and $X_2$ there is no reducibility, i.e. if $\rho_1\times\rho_2$ is irreducible for all $\rho_1\in X_1$ and $\rho_2\in X_2$.


\subsection{Decomposition}
Let $\pi\in\Irrcl_{X\cup\{\s\}}$, where $X$ is $F'/F$-selfcontragredi\-ent, and let $X=X_1\cup X_2$ be a regular partition of $X$.
Fix $i \in \{1, 2 \}$.
Then, there exists $\beta\in\Irr$ supported in $X_{3-i}$ and $\g\in\Irrcl_{X_i\cup\{\s\}}$ such that
\[
\pi \hookrightarrow \beta\rtimes \g.
\]
Moreover,
$
\g
$
is uniquely determined by the above requirement. It is denoted by
$$
X_i(\pi)
$$
and called the Jantzen component of $\pi$ corresponding to the member $X_i$ in the regular partition $X=X_1\cup X_2$.
Furthermore, let $X\subset \Cusp$ be such that $\Cusp=X\cup (\Cusp\setminus X)$ is a regular partition of $\Cusp$ and let
$\pi\in\Irrcl$. Then,
$
X(\pi)
$
is defined as above for the regular partition $\Cusp=X\cup (\Cusp\setminus X)$
(and it is called again the Jantzen component of $\pi$ corresponding to $X$).


For $\rho\in\Cuspsd$, denote $X_\rho=\{\nu^x\rho:x\in \R\}$.
Let $\pi\in\Irrcl$ be weakly real.
Then, there exist finitely many distinct $\rho_1,\dots,\rho_k\in\Cuspsd$ and $\s\in\Cuspcl$ such that the support of $\pi$ is in
$X_{\rho_1}\cup\dots\cup X_{\rho_k}\cup\{\s\}$. The representations
$$
(X_{\rho_1}(\pi), \dots,X_{\rho_k}(\pi))
$$
determine $\pi$, and this defines a bijection
\[
\Irrcl_{X_{\rho_1}\cup\dots\cup X_{\rho_k}\cup\{\s\}}\rightarrow\prod_{i=1}^k\Irrcl_{X_{\rho_i}\cup\{\s\}}.
\]
The inverse map is denoted by
$$
\Psi_{X_{\rho_1},\dots,X_{\rho_k}}.
$$
The correspondence $\pi \mapsto (X_{\rho_1}(\pi), \dots,X_{\rho_k}(\pi))$ has a number of very nice properties
(see \cite{MR1481814} or section 8 of \cite{MR3969882}).
We shall now prove one additional very simple property which we shall use often.

\begin{lemma} \label{J-ind}
Let $X$ be an $F'/F$-selfcontragredient subset of $\Cusp$, and let $X=X_1\cup X_2$ be a regular partition of $X$.
Let $\theta_i\in\Irr$ supported in $X_i$ and $\pi_i\in\Irrcl_{X_i\cup\{\s\}}$, $i=1,2$.
Suppose that $\theta_i\rtimes\pi_i$, $i=1,2$ are both irreducible.
Then,
\begin{equation} \label{JDI}
\Psi_{X_1,X_2}(\theta_1\rtimes\pi_1,\theta_2\rtimes\pi_2)\cong \theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2).
\end{equation}
\end{lemma}

\begin{proof}
Note that $\theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2)$ is irreducible by (1) of Remark 8.9 of \cite{MR3969882}.

By the definition of $\Psi_{X_1,X_2}(\pi_1,\pi_2)$, we know that $\Psi_{X_1,X_2}(\pi_1,\pi_2)\h \tau \rtimes\pi_1$,
where $\tau $ is irreducible and supported on $X_2$. This implies $\tau\times\theta_1\cong\theta_1\times\tau$. Therefore,
$$
\theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2)\h \theta_1\times \theta_2\rtimes\tau \rtimes\pi_1\cong
\theta_2\times\tau \times\theta_1 \rtimes\pi_1.
$$
If $\theta_2\times\tau$ is not irreducible, we can easily show that it admits an irreducible subquotient $\varphi$ such that
$\theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2)\h \varphi \times\theta_1 \rtimes\pi_1$.
Therefore, $X_1(\theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2))\cong \theta_1 \rtimes\pi_1$.
Analogously $X_2(\theta_1\times \theta_2\rtimes \Psi_{X_1,X_2}(\pi_1,\pi_2))\cong \theta_2 \rtimes\pi_2$. This proves \eqref{JDI}.
\end{proof}

\section{Preservation of unitarizability by decomposition in corank \texorpdfstring{$\leq 3$}{leq3}}

\begin{lemma}
Let $\pi$ be a weakly real irreducible subquotient of $\theta_1\times\dots\times\theta_k\rtimes\s$, where $\theta_i\in\Cusp $ and
$$
k\leq 3.
$$
Suppose that all $X_{\rho_i}(\pi)$ in the Jantzen decomposition of $\pi$ are unitarizable. Then, $\pi$ is unitarizable.
\end{lemma}

\begin{proof}
For $k=1$, there is nothing to prove.
We shall now prove the case $k=3$. (The case $k=2$ is easier, and will be omitted.)


Let $\pi \mapsto (X_{\rho_1}(\pi), \dots,X_{\rho_\ell}(\pi))$ be the Jantzen decomposition of $\pi$.
For the proof, we consider only those $\rho_i$ for which
$X_{\rho_i}(\pi)\ne \s$. We assume this for the rest of the proof.
Denote
$$
\alpha_i=\alpha_{\rho_i,\s}.
$$
If $\ell=1$, the claim obviously holds (since then $\pi=X_{\rho_1}(\pi)$).

Consider first the case $\ell=3$. Then, $\pi$ is a subquotient of $[x_1]^{(\rho_1)}\times[x_2]^{(\rho_2)} \times [x_3]^{(\rho_3)}\rtimes\s$,
where $x_i\geq0$.
Then, $X_{\rho_i}(\pi)$ are irreducible subquotients of $[x_i]^{(\rho_i)}\rtimes\s$.
Since $X_{\rho_i}(\pi)$ are unitarizable, then we know $x_i\leq\alpha_i$, $1\leq i\leq 3$.
But then each irreducible subquotient of $[x_1]^{(\rho_1)}\times[x_2]^{(\rho_2)} \times [x_3]^{(\rho_3)}\rtimes\s$ is unitarizable
(since we are in complementary series or its ends). Therefore, $\pi$ is unitarizable.


Suppose $\ell=2$. We may assume that $\pi$ is a subquotient of $[x_1]^{(\rho_1)}\times[x_2]^{(\rho_1)} \times [x_3]^{(\rho_2)} \rtimes\s$.
Then, $X_{\rho_2}(\pi)$ is an irreducible unitarizable subquotient of $[x_3]^{(\rho_2)} \rtimes\s$.
This implies $x_3\leq \alpha_2$. Furthermore, $\pi$ is an irreducible subquotient of
$[x_3]^{(\rho_2)} \rtimes X_{\rho_1}(\pi)$. Since $X_{\rho_1}(\pi)$ is unitarizable and $x_3\leq\alpha_2$,
this implies that $\pi$ is unitarizable (again we are in complementary series or its ends).
\end{proof}


\begin{lemma} \label{k=2}
Let $\pi$ be a weakly real irreducible representation of a classical group.
Suppose that some $X_{\rho_i}(\pi)$ is a non-unitarizable subquotient of $\theta_1\times\dots\times\theta_k\rtimes\s$,
where $\theta_i\in\Cusp $ and
$$
k\leq 2.
$$
Then, $\pi$ is not unitarizable.
\end{lemma}

\begin{proof}
Suppose on the contrary that $\pi$ is unitarizable. Denote $\rho_i$ simply by $ \rho$ and let
$X_\rho^c=\Cusp\setminus X_\rho.$ We also denote $\alpha=\alpha_{\rho,\s}$. Now $X_\rho(\pi)$ is a subquotient of
$$
[x_1]^{(\rho)}\times\dots\times[x_k]^{(\rho)}\rtimes\s,
$$
where $x_i\geq0 $ and $k\leq2$. Denote $\pi_\rho=X_\rho(\pi)$ and $\pi_\rho^c=X_\rho^c(\pi)$.
Clearly
$$
\pi= \Psi_{X_{\rho},X_{\rho}^c}( \pi_\rho ,\pi_\rho^c ).
$$

If $k=1$, then non-unitarizability of $\pi_\rho $ implies $\pi_\rho \cong [x_1]^{(\rho)}\rtimes\s$ where $x_1>\alpha$.
Now Lemma \ref{J-ind} implies
$$
\pi\cong [x_1]^{(\rho)}\rtimes\pi_\rho^c .
$$
This cannot be unitarizable, since we can deform $x_1$ to the right as far as we want.
We get a contradiction (with the fact that unitarizability can show up only in bounded domains -- see \cite{MR733166} for more details).

Consider now the case $k=2$. We shall suppose as usually $0\leq x_1\leq x_2$. Recall that $\pi_\rho $ is a subquotient of
$$
[x_1]^{(\rho)}\times[x_2]^{(\rho)}\rtimes\s.
$$
We consider several cases.
The first is
$$
\alpha=0.
$$
The non-unitarizability of $\pi_\rho $ implies that $x_1+x_2>1$.
This implies $\pi_\rho \cong [x_1]^{(\rho)}\times[x_2]^{(\rho)}\rtimes\s$. Now Lemma \ref{J-ind} implies
$$
\pi\cong [x_1]^{(\rho)}\times[x_2]^{(\rho)}\rtimes\pi_\rho^c.
$$
Further, we can deform $x_1$ to $x_2$, use the unitary parabolic reduction and get a contradiction with the
unitarizability in the case of general linear groups (more precisely, with the existence of the complementary series there).


It remains to consider the case
$$
\alpha>0.
$$
First recall that Theorem 1.2 of \cite{MR3969882} implies that
\begin{equation} \label{Psi1}
\Psi_{X_{\rho},X_{\rho}^c}( \tau ,\pi_\rho^c ) \text{ is not unitarizable}
\end{equation}
for any non-unitarizable irreducible subquotient $\tau$ of $ [\alpha]^{(\rho)}\times[\alpha+1]^{(\rho)}\rtimes\s $.
Therefore, we assume
$$
(x_1,x_2)\ne (\alpha,\alpha+1)
$$
in the sequel.


Consider first the case
$$
\alpha=\tfrac12.
$$
Since $\pi_\rho $ is not unitarizable, we have $x_1>\tfrac12$ or $x_2>\tfrac12$.

Suppose $x_i\ne \tfrac12$ for $i=1,2$.

Assume first that $x_2\pm x_1\ne1$. Then, $\pi\cong [x_1]^{(\rho)}\times [x_2]^{(\rho)}\rtimes\s$. Now Lemma \ref{J-ind} implies
$$
\pi\cong [x_1]^{(\rho)}\times [x_2]^{(\rho)}\rtimes\pi_\rho^c.
$$
If $x_1>\tfrac12$, then we can deform $x_1$ to $x_2$, switch one $x_2$ to $-x_2$,
use the unitary parabolic reduction and get a contradiction (with existence of complementary series
for general linear group), which implies that $\pi$ cannot be unitarizable.

If $x_1<\tfrac12$, then we can deform $x_2$ to get $x_2+\e x_1=1$ for some $\e\in\{\pm1\}$,
and take there an irreducible subquotient denoted again by $\pi$ (which must be unitarizable).
Now $\pi_\rho \cong \tau \rtimes\s$ for some irreducible subquotient of the reducible representation
$[\e x_1]^{(\rho_1)}\times [x_2]^{(\rho_1)}$. Lemma \ref{J-ind} implies
$$
\pi\cong \tau \rtimes\pi_\rho^c.
$$
Now we can deform $\tau$ to exponents $(\tfrac12,\tfrac32)$.
The properties of the Jantzen decomposition and \eqref{Psi1}
imply that $\pi$ is not unitarizable (since in the limit we have a non-unitarizable subquotient).

Assume now that $x_2+\e x_1=1$ for some $\e\in\{\pm1\}$. Then, in the same way as above we get
$$
\pi\cong \tau \rtimes\pi_\rho^c
$$
for some irreducible subquotient of the reducible representation $[\e x_1]^{(\rho)}\times [x_2]^{(\rho)}$.
Now we finish this case as the previous one.

It remains to consider the case $x_1=\tfrac12$.
Then, $\pi_\rho \cong [x_2]^{(\rho)}\rtimes\theta$ where $\theta$ is an irreducible subquotient of
$[\tfrac12]^{(\rho_2)}\rtimes\s$ (recall $x_2\ne\tfrac32$). Lemma \ref{J-ind} implies
$$
\pi\cong [x_2]^{(\rho)}\rtimes \Psi_{X_{\rho},X_{\rho}^c}( \theta ,\pi_\rho^c ).
$$
We now deform $x_2$ to $\tfrac32$, and in a similar way as before, we get a contradiction.


It remains to consider the case
$$
\alpha\geq1.
$$
First assume
$$
x_2>\alpha.
$$
We know from Proposition 2.2 of \cite{MR3711838} that $x_1\leq \alpha$ and $x_2-x_1\leq1$, which implies $x_2\leq \alpha+1$.
Therefore, if $x_2=\alpha+1$, then $x_1=\alpha$. Then, we know that $\pi$ is not unitarizable by \eqref{Psi1}.
It remains to consider the case
$$
x_2<\alpha+1.
$$

Let
$$
x_2>\alpha.
$$
Consider first the case $x_1= \alpha$. Then, $\pi_\rho =[x_2]^{(\rho)}\rtimes \theta$ where $\theta$ is $
\delta([\alpha]^{(\rho)}; \s)$ or $L([\alpha]^{(\rho)}; \s)$.
One directly gets that
$$
\pi\cong [x_2]^{(\rho)}\rtimes \Psi_{X_{\rho},X_{\rho}^c}( \theta ,\pi_\rho^c ).
$$
We now deform $x_2$ to $\alpha+1$, and in a similar way as before, we get a contradiction.

Therefore, we need to consider the case
$$
x_1<\alpha.
$$

Assume first that $x_2-x_1=1$. Then, a short analysis using Lemma \ref{J-ind} implies that
$\pi\cong \tau\rtimes\pi_\rho^c $, where $\tau$ is an irreducible subquotient of $[x_1]^{(\rho_1)}\times [x_2]^{(\rho_2)}$.
Now deforming $\tau$ to exponents $\alpha,\alpha+1$ and using the properties of the Jantzen decomposition,
we would get a contradiction.


If $x_2-x_1<1$, then similarly we get $\pi\cong [x_1]^{(\rho_1)}\times [x_2]^{(\rho_2)}\rtimes\pi_\rho^c$.
We can now deform (increase) $x_2$ to the previous case, consider a limit, and repeat the above argument.
Therefore, we get again a contradiction.


Consider now the case
$$
x_2=\alpha.
$$
We need to consider the case $\alpha-1<x_1\leq\alpha$.
Then, we know that $\pi_\rho \cong [x_1]^{(\rho)}\rtimes \theta$ for some irreducible subquotient
$\theta$ of $[\alpha]^{(\rho)}\rtimes\s$, which implies by Lemma \ref{J-ind}
$$
\pi\cong [x_1]^{(\rho)}\rtimes \Psi_{X_{\rho},X_{\rho}^c}( \theta ,\pi_\rho^c ).
$$
Now we can deform $x_1$ to $\alpha+1$, and get a contradiction in the same way as in the previous cases
(here we use that $[\alpha]\rtimes L([\alpha];\sigma)$ and $[\alpha]\rtimes \delta([\alpha];\sigma)$ are irreducible, and properties of the Jantzen decomposition).


It remains to consider the case
$$
x_2<\alpha
$$
and the region
$$
1-x_1<x_2<x_1+1.
$$
Then, representations $\pi\cong [x_1]^{(\rho)}\times [x_2]^{(\rho)}\rtimes\pi_\rho^c $, and moreover,
for the exponents satisfying above relations, $[x_1]^{(\rho)}\times [x_2]^{(\rho)}\rtimes\pi_\rho^c$ form
a continuous family of irreducible hermitian representations. Consider the point $\tfrac12<x_1=x_2<\alpha$ of the above region.
After switching $x_1$ to $-x_1$, the unitary parabolic reduction implies that this representation is not unitarizable (since $[x_1]\times[-x_1]$ is not unitarizable).
Therefore, the whole family is non-unitarizable.
This completes the proof of the
lemma\end{proof}

We infer

\begin{corollary}
Let $\pi$ be a weakly real irreducible subquotient of $\theta_1\times\dots\times\theta_k\rtimes\s$,
where $\theta_i\in\Cusp $, $k\leq 3$. Then, $\pi$ is unitarizable if and only if all $X_{\rho_i}(\pi)$ in the
Jantzen decomposition of $\pi$ are unitarizable.
\qed
\end{corollary}


Having in mind the results of \cite{MR2046512}, \cite{MR2767523}, \cite{MR3969882} and the present paper,
we conjecture that the above corollary holds for any $k$, namely

\begin{conjecture}
Jantzen decomposition preserves unitarizability in both directions.
\end{conjecture}


\endinput
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