# Degree Differences in the Eta Correspondences

Shu-Yen Pan  Department of Mathematics, National Tsing Hua University, Hsinchu
300, Taiwan<EMAIL_ADDRESS>
###### Abstract.

A sub-relation of the $\Theta$-correspondence called the _$\eta$
-correspondence_ is defined by Gurevich-Howe for a finite reductive dual pair
in stable range. In this paper we propose an extension of the correspondence
to general finite reductive dual pairs. Then we determine the domain the
correspondence and prove a formula on the difference of degrees in $q$ of two
irreducible characters paired by the correspondence.

###### Key words and phrases:

rank, unipotent character, reductive dual pair

###### 2010 Mathematics Subject Classification:

Primary: 20C33; Secondary: 22E50

###### Contents

  1. 1 Introduction
  2. 2 Degrees in $q$ of Unipotent Characters
  3. 3 Unipotent Characters of Symplectic or Orthogonal Groups
  4. 4 Unipotent Characters of Unitary Groups
  5. 5 Degree Differences in the Eta Correspondence

##  1\. Introduction

### 1.1.

Let $(\text{\bf G},\text{\bf G}^{\prime})$ be a reductive dual pair over a
finite field ${\mathbf{F}}_{q}$ of $q$ elements where $q$ is a power of an odd
prime. By restricting the character of the Weil representation to the dual
pair, we obtain a decomposition

$\omega_{\text{\bf G},\text{\bf G}^{\prime}}=\sum_{\rho\in{\mathcal{E}}(G),\
\rho^{\prime}\in{\mathcal{E}}(G^{\prime})}m_{\rho,\rho^{\prime}}\rho\otimes\rho^{\prime}$

where ${\mathcal{E}}(G)$ denotes the set of irreducible character of the
finite group $G$ of rational points of G. The subset

$\Theta_{\text{\bf G},\text{\bf G}^{\prime}}=\\{\,(\rho,\rho^{\prime})\mid
m_{\rho,\rho^{\prime}}\neq
0\,\\}\subset{\mathcal{E}}(G)\times{\mathcal{E}}(G^{\prime})$

gives a relation between ${\mathcal{E}}(G)$ and ${\mathcal{E}}(G^{\prime})$,
and is called the _$\Theta$ -correspondence_ for the dual pair $(\text{\bf
G},\text{\bf G}^{\prime})$. We say that
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})$ _occurs_ in the
$\Theta$-correspondence for $(\text{\bf G},\text{\bf G}^{\prime})$ if the set

$\Theta(\rho^{\prime})=\\{\,\rho\in{\mathcal{E}}(G)\mid
m_{\rho,\rho^{\prime}}\neq 0\,\\}$

is not empty.

For a non-negative integer $k$, let $\text{\bf G}_{k}$ denote one of the
following types of classical groups: a general linear group ${\rm GL}_{k}$, a
unitary group ${\rm U}_{k}$, a symplectic group ${\rm Sp}_{k}$ (only when $k$
is even), or an orthogonal group ${\rm O}^{\epsilon}_{k}$ where $\epsilon=+$
or $-$. For $\rho^{\prime}\in{\mathcal{E}}(G^{\prime})$, the _$\Theta$ -rank_
of $\rho^{\prime}$, denoted by $\Theta\text{\rm-rk}(\rho^{\prime})$, is
defined to be the smallest number $k$ such that $\rho^{\prime}\chi^{\prime}$
occurs in the $\Theta$-correspondence for some dual pair $(\text{\bf
G}_{k},\text{\bf G}^{\prime})$ and for some linear character $\chi^{\prime}$
of $G^{\prime}$ (cf. (3.11), (3.12) and (4.9)).

For a dual pair $(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$ is stable
range, Gurevich and Howe prove in [GH17] and [GH20] that for
$\rho\in{\mathcal{E}}(G_{k})$ there is a unique $\rho^{\prime}\in\Theta(\rho)$
such that $\Theta\text{\rm-rk}(\rho^{\prime})=k$. The character
$\rho^{\prime}$ is denoted by $\eta(\rho)$, and the mapping
$\rho\mapsto\eta(\rho)$ from
${\mathcal{E}}(G_{k})\rightarrow{\mathcal{E}}(G^{\prime}_{n})$ is injective
and is called the _$\eta$ -correspondence_ for the dual pair $(\text{\bf
G}_{k},\text{\bf G}^{\prime}_{n})$ in stable range.

In [Pan20a] and [Pan20b], a one-to-one sub-correspondence $\underline{\theta}$
of the $\Theta$-correspondence between ${\mathcal{E}}(G_{k})$ and
${\mathcal{E}}(G^{\prime}_{n})$ for a general dual pair $(\text{\bf
G}_{k},\text{\bf G}^{\prime}_{n})$ is defined (cf. (3.6)), and it is shown
that $\eta=\underline{\theta}$ if dual pair $(\text{\bf G}_{k},\text{\bf
G}^{\prime}_{n})$ is in stable range. Therefore, for general dual pair
$(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$ (i.e., not necessarily in
stable range), it might be reasonable to define
$\eta(\rho)=\underline{\theta}(\rho)$ on those $\rho\in{\mathcal{E}}(G_{k})$
such that $\underline{\theta}(\rho)$ is defined and $\Theta\text{\rm-
rk}(\rho)=k$.

For a non-negative integer $\ell$, a condition on irreducible characters
$\rho\in{\mathcal{E}}(G)$ called _$\ell$ -admissible for a dual pair
$(\text{\bf G},\text{\bf G}^{\prime})$_ is defined in Subsections 3.2, 4.2 and
5.1. Then we have our first main result which gives complete description of
the domain of the “$\eta$-correspondence” for a general dual pair:

######  Theorem 1.1.

Suppose that $0\leq k\leq n$.

  1. (i)

If $\rho\in{\mathcal{E}}(G_{k})$ is $(n-k)$-admissible for the dual pair
$(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$, then
$\underline{\theta}(\rho)$ is defined and $\Theta\text{\rm-
rk}(\underline{\theta}(\rho))=k$.

  2. (ii)

If $\rho^{\prime}\in{\mathcal{E}}(G^{\prime}_{n})$ is of $\Theta$-rank $k$,
then $\rho^{\prime}=\underline{\theta}(\rho)\chi^{\prime}$ for

     * •

some linear character $\chi^{\prime}$ of $G^{\prime}_{n}$; and

     * •

some group $\text{\bf G}_{k}$ and an irreducible character
$\rho\in{\mathcal{E}}(G_{k})$ which is $(n-k)$-admissible for the dual pair
$(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$.

######  Remark 1.2.

  1. (1)

A condition on an irreducible character $\rho^{\prime}\in{\mathcal{E}}({\rm
GL}_{n}(q))$ of $\Theta$-rank $k$ is given in [GH20] theorem 9.2.3.

  2. (2)

A similar condition on an irreducible character
$\rho^{\prime}\in{\mathcal{E}}({\rm U}_{n}(q))$ of $\Theta$-rank $k$ can be
found in [GLT20] theorem 3.9.

### 1.2.

For $\rho\in{\mathcal{E}}(G_{k})$, it is known that $\rho(1)$ is a polynomial
in $q$. Let $\deg_{q}(\rho)$ denote the degree of this polynomial. It turns
out that we have a very elegant formula on the difference between
$\deg_{q}(\rho)$ and $\deg_{q}(\underline{\theta}(\rho))$ when
$\Theta\text{\rm-rk}(\underline{\theta}(\rho))=k$:

######  Theorem 1.3.

Consider a dual pair $(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$ where
$k\leq n$. If $\rho\in{\mathcal{E}}(G_{k})$ is $(n-k)$-admissible for
$(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})$, then

$\deg_{q}(\underline{\theta}(\rho))=\deg_{q}(\rho)+\begin{cases}\frac{1}{2}k(n-k+1),&\text{if\/
$(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})=({\rm O}^{\epsilon}_{k},{\rm
Sp}_{n})$};\\\ \frac{1}{2}k(n-k-1),&\text{if\/ $(\text{\bf G}_{k},\text{\bf
G}^{\prime}_{n})=({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$};\\\
k(n-k),&\text{if\/ $(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})=({\rm
U}_{k},{\rm U}_{n})$}.\end{cases}$

######  Remark 1.4.

If $(\text{\bf G}_{k},\text{\bf G}^{\prime}_{n})=({\rm O}_{k}^{\epsilon},{\rm
Sp}_{n})$ and the dual pair is in the stable range, then the
$\eta$-correspondence and the $\underline{\theta}$-correspondence coincide,
and every $\rho\in{\mathcal{E}}(G_{k})$ is $(n-k)$-admissible (cf. Lemma 5.2
and Lemma 5.12). For this case, the above formula is already known in [GH17],
(1.11) and theorem 4.1.1.

An application of the above theorem is to determine the maximum and minimum
values of the set

$\\{\,\deg_{q}(\rho)\mid\rho\in{\mathcal{E}}(G_{n}),\ \Theta\text{\rm-
rk}(\rho)=k\,\\}$

where $\text{\bf G}_{n}={\rm Sp}_{n}$, ${\rm O}_{n}^{\epsilon}$ or ${\rm
U}_{n}$, and $k\leq n$. This problem will be investigated in another papers of
the author. A lower bound and an upper bound of the above set can be found in
[GLT19] and [GLT20].

### 1.3.

The contents of this article are as follows. In Section 2 we set up basic
notations and express the formula of $\deg_{q}(\rho)$ for any unipotent
characters in terms of Lusztig symbols. In Section 3, we investigate the
$\Theta$-correspondence and define $\ell$-admissibility on unipotent
characters for a dual pair of a symplectic group and an even orthogonal group.
Moreover, Theorem 1.1 and Theorem 1.3 are proved for this case. In Section 4,
we prove our main results for the cases of $\Theta$-correspondence of
unipotent characters for a dual pair of two unitary groups. We prove Theorem
1.1 and Theorem 1.3 completely via the Lusztig correspondence in the final
section.

##  2\. Degrees in $q$ of Unipotent Characters

###  2.1. Basic notations

Let G be a classical group defined over ${\mathbf{F}}_{q}$, and let
$G=\text{\bf G}(q)$ denote the finite group of rational points.

For a polynomial $f$ in $q$, let $\deg_{q}(f)$ denote the degree of the
polynomial $f$. Let $|G|_{p^{\prime}}$ denote the part of the order $|G|$
prime to $p$. Now $|G|_{p^{\prime}}$ is a polynomial in $q$ and it is well-
known that

$\deg_{q}(|G|_{p^{\prime}})=\begin{cases}\tfrac{1}{4}k(k+2),&\text{if
$\text{\bf G}={\rm Sp}_{k}$ or ${\rm O}_{k+1}$, $k$ even};\\\
\tfrac{1}{4}k^{2},&\text{if $\text{\bf G}={\rm O}^{\epsilon}_{k}$, $k$
even};\\\ \tfrac{1}{2}k(k+1),&\text{if $\text{\bf G}={\rm
U}_{k}$}.\end{cases}$

###  2.2. Symbols and unipotent characters

Let $\mu=[\mu_{1},\ldots,\mu_{k}]$ and $\mu=[\nu_{1},\ldots,\nu_{l}]$ be two
partitions. We may assume that $k=l$ by adding some $0$’s if necessary. Then
we denote

(2.1) $\mu\preccurlyeq\nu\quad\text{ if \
}\nu_{1}\geq\mu_{1}\geq\nu_{2}\geq\mu_{2}\geq\cdots\geq\nu_{k}\geq\mu_{k}.$

An ordered pair $\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$ of two partitions is
called a _bi-partition_ of $n$ if

$\left|\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}\right|:=|\mu|+|\nu|=n.$

The set of bi-partitions of $n$ is denoted by ${\mathcal{P}}_{2}(n)$. For a
bi-partition $\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$, we define its _transpose_ by
$\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}^{\rm t}=\genfrac{[}{]}{0.0pt}{}{\nu}{\mu}$.

A $\beta$-set is a finite subset $A=\\{a_{1},a_{2},\ldots,a_{m}\\}$ of non-
negative integers written strictly decreasingly, i.e.,
$a_{1}>a_{2}>\cdots>a_{m}$. A symbol $\Lambda=\binom{A}{B}$ is an ordered pair
of two $\beta$-sets. Let ${\mathcal{S}}$ denote the set of symbols
$\binom{A}{B}$ such that $0\not\in A\cap B$. For a symbol
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
we define its _defect_ and _rank_ by

(2.2) $\displaystyle\begin{split}{\rm def}(\Lambda)&=m_{1}-m_{2},\\\ {\rm
rk}(\Lambda)&=\sum_{i=1}^{m_{1}}a_{i}+\sum_{j=1}^{m_{2}}b_{j}-\left\lfloor\left(\frac{m_{1}+m_{2}-1}{2}\right)^{2}\right\rfloor.\end{split}$

We also define the _transpose_ of a symbol by $\binom{A}{B}^{\rm
t}=\binom{B}{A}$. There is a mapping $\Upsilon$ from the set of symbols to the
set of bi-partitions:

(2.3)
$\Upsilon\colon\binom{a_{1},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{m_{2}}}\mapsto\genfrac{[}{]}{0.0pt}{}{a_{1}-(m_{1}-1),a_{2}-(m_{1}-2),\ldots,a_{m_{1}-1}-1,a_{m_{1}}}{b_{1}-(m_{2}-1),b_{2}-(m_{2}-2),\ldots,b_{m_{2}-1}-1,b_{m_{2}}}.$

It is easy to check that

(2.4) $|\Upsilon(\Lambda)|=\begin{cases}{\rm rk}(\Lambda)-\tfrac{1}{4}({\rm
def}(\Lambda)-1)({\rm def}(\Lambda)+1),&\text{if ${\rm def}(\Lambda)$ is
odd};\\\ {\rm rk}(\Lambda)-\tfrac{1}{4}({\rm def}(\Lambda))^{2},&\text{if
${\rm def}(\Lambda)$ is even}.\end{cases}$

For a non-negative even integer $k$, we define a set of symbols associated to
a symplectic group or an even orthogonal group:

(2.5) $\displaystyle\begin{split}{\mathcal{S}}_{{\rm
O}^{+}_{k}}&=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=\tfrac{k}{2},\
{\rm def}(\Lambda)\equiv 0\pmod{4}\,\\};\\\ {\mathcal{S}}_{{\rm
Sp}_{k}}&=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=\tfrac{k}{2},\
{\rm def}(\Lambda)\equiv 1\pmod{4}\,\\};\\\ {\mathcal{S}}_{{\rm
O}^{-}_{k}}&=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=\tfrac{k}{2},\
{\rm def}(\Lambda)\equiv 2\pmod{4}\,\\}.\end{split}$

From [Lus77] it is known that there exists a bijection from
${\mathcal{S}}_{\text{\bf G}}$ to the set ${\mathcal{E}}(G)_{1}$ of unipotent
characters of $G$ denoted by $\Lambda\mapsto\rho_{\Lambda}$.

######  Remark 2.6.

The set ${\mathcal{S}}_{{\rm Sp}_{k}}$ of symbols considered here is slightly
different from the original one in [Lus77]. Please see [Pan21b] subsection 3.3
to see the equivalence of the two parametrizations.

Suppose all the entries $a_{1},\ldots,a_{m_{1}},b_{1},\ldots,b_{m_{2}}$ in a
symbol $\Lambda=\binom{a_{1},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{m_{2}}}$ are
$z_{1},z_{2},\ldots,z_{m}$ with $z_{1}\geq z_{2}\geq\cdots\geq z_{m}$ where
$m=m_{1}+m_{2}$. The following lemma is [Pan20a] lemma 2.12:

######  Lemma 2.7.

Let
$\Lambda=\binom{a_{1},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{\text{\bf
G}}$ where $\text{\bf G}={\rm Sp}_{k}$ or ${\rm O}^{\epsilon}_{k}$ and $k$ is
even. Then $\deg_{q}(\rho_{\Lambda})$ is equal to

$\sum_{i=1}^{m}(m-i)z_{i}-z_{i}(z_{i}+1)+\begin{cases}\frac{1}{4}k(k+2)-\frac{1}{24}(m-1)(m-3)(2m-1),&\text{if
$m$ is odd};\\\ \frac{1}{4}k^{2}-\frac{1}{24}m(m-2)(2m-5),&\text{if $m$ is
even}.\end{cases}$

###  2.3. Unipotent characters of a unitary group

For a symbol
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}$,
we define

(2.8) ${\rm rk}_{\rm
U}(\Lambda)=\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}+\frac{|m_{1}-m_{2}|}{2}-\frac{(m_{1}+m_{2})(m_{1}+m_{2}-2)}{2}.$

For a non-negative integer $k$, we let ${\mathcal{S}}_{{\rm U}_{k}}$ be the
set of symbols $\Lambda\in{\mathcal{S}}$ such that

  * •

${\rm def}(\Lambda)$ is either even and non-negative, or odd and negative;

  * •

${\rm rk}_{\rm U}(\Lambda)=k$.

Again, there is a parametrization from ${\mathcal{S}}_{{\rm U}_{k}}$ to
${\mathcal{E}}({\rm U}_{k}(q))_{1}$ denoted by $\Lambda\mapsto\rho_{\Lambda}$
(cf. [FS90] or [Pan20b]).

For a symbol
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
we define two $\beta$-sets:

(2.9)
$\displaystyle\begin{split}X^{0}=X^{0}_{\Lambda}&=\begin{cases}\\{2b_{1},2b_{2},\ldots,2b_{m_{2}}\\},&\text{if
${\rm def}(\Lambda)$ is even};\\\
\\{2a_{1},2a_{2},\ldots,2a_{m_{1}}\\},&\text{if ${\rm def}(\Lambda)$ is
odd},\end{cases}\\\
X^{1}=X^{1}_{\Lambda}&=\begin{cases}\\{2a_{1}+1,2a_{2}+1,\ldots,2a_{m_{1}}+1\\},&\text{if
${\rm def}(\Lambda)$ is even};\\\
\\{2b_{1}+1,2b_{2}+1,\ldots,2b_{m_{2}}+1\\},&\text{if ${\rm def}(\Lambda)$ is
odd}.\end{cases}\end{split}$

For a $\beta$-set $A$ or two $\beta$-sets $A,B$, we define the following
polynomials in $q$:

(2.10) $\displaystyle\begin{split}\Delta(A)&=\prod_{a,a^{\prime}\in A,\
a>a^{\prime}}(q^{a}-q^{a^{\prime}}),\\\ \Theta(A)&=\prod_{a\in
A}\prod_{h=1}^{a}(q^{h}-(-1)^{h}),\\\ \Xi(A,B)&=\prod_{a\in A,\ b\in
B}(q^{a}+q^{b}).\end{split}$

For
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{{\rm
U}_{k}}$, it is well known that (cf. for example, [Pan20b] proposition 2.15):

(2.11)
$\rho_{\Lambda}(1)=\frac{\Delta(X^{0})\Delta(X^{1})\Xi(X^{0},X^{1})|{\rm
U}_{k}(q)|_{p^{\prime}}}{\Theta(X^{0})\Theta(X^{1})q^{\binom{m_{1}+m_{2}-1}{2}+\binom{m_{1}+m_{2}-2}{2}+\cdots+\binom{2}{2}}}$

where $\binom{a}{b}$ denotes the binomial coefficients.

######  Lemma 2.12.

For $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$, write
$X_{\Lambda}=X^{0}_{\Lambda}\cup
X^{1}_{\Lambda}=\\{z_{1},z_{2},\ldots,z_{m}\\}$ such that
$z_{1}>z_{2}>\cdots>z_{m}$ where $m=m_{1}+m_{2}$. Then

$\deg_{q}(\rho_{\Lambda})=\sum_{i=1}^{m}(m-i)z_{i}-\sum_{i=1}^{m}\frac{z_{i}(z_{i}+1)}{2}+\frac{k(k+1)}{2}-\frac{m(m-1)(m-2)}{6}.$

###### Proof.

Because now $z_{1}>z_{2}>\cdots>z_{m}$, from (2.10) it is not difficult to see
that

$\displaystyle\deg_{q}(\Delta(X^{0})\Delta(X^{1})\Pi(X^{0},X^{1}))$
$\displaystyle=\sum_{(z_{i},z_{j})\in X\times X,\ i\neq
j}\max(z_{i},z_{j})=\sum_{i=1}^{m}(m-i)z_{i},$
$\displaystyle\deg_{q}(\Theta(X^{0})\Theta(X^{1}))$
$\displaystyle=\sum_{i=1}^{m}\frac{z_{i}(z_{i}+1)}{2}.$

Moreover, we have

$\displaystyle\deg_{q}(|{\rm U}_{k}(q)|_{p^{\prime}})$
$\displaystyle=\frac{k(k+1)}{2},$
$\displaystyle\binom{m-1}{2}+\binom{m-2}{2}+\cdots+\binom{2}{2}$
$\displaystyle=\frac{m(m-1)(m-2)}{6}.$

Then the lemma follows from (2.11) directly. ∎

##  3\. Unipotent Characters of Symplectic or Orthogonal Groups

###  3.1. Finite theta correspondence of unipotent characters

Let $(\text{\bf G},\text{\bf G}^{\prime})$ be a reductive dual pair of a
symplectic group and an even orthogonal group. For symbols
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ and
$\Lambda^{\prime}\in{\mathcal{S}}_{\text{\bf G}^{\prime}}$, we write
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$ and
$\Upsilon(\Lambda^{\prime})=\genfrac{[}{]}{0.0pt}{}{\mu^{\prime}}{\nu^{\prime}}$.
Now we define a relation on the set of symbols:

(3.1) $\displaystyle\begin{split}{\mathcal{B}}_{{\rm Sp}_{k},{\rm
O}^{+}_{n}}&=\\{\,(\Lambda,\Lambda^{\prime})\in{\mathcal{S}}_{{\rm
Sp}_{k}}\times{\mathcal{S}}_{{\rm O}^{+}_{n}}\mid\nu\preccurlyeq\mu^{\prime},\
\nu^{\prime}\preccurlyeq\mu,\ {\rm def}(\Lambda^{\prime})=-{\rm
def}(\Lambda)+1\,\\};\\\ {\mathcal{B}}_{{\rm Sp}_{k},{\rm
O}^{-}_{n}}&=\\{\,(\Lambda,\Lambda^{\prime})\in{\mathcal{S}}_{{\rm
Sp}_{k}}\times{\mathcal{S}}_{{\rm O}^{-}_{n}}\mid\mu\preccurlyeq\nu^{\prime},\
\mu^{\prime}\preccurlyeq\nu,\ {\rm def}(\Lambda^{\prime})=-{\rm
def}(\Lambda)-1\,\\}\end{split}$

where both $k,n$ are even. It is known that the unipotent characters are
preserved by the correspondence $\Theta_{\text{\bf G},\text{\bf G}^{\prime}}$
(cf. [AM93] theorem 3.5). The following result on the $\Theta$-correspondence
of unipotent characters is from [Pan19a] theorem 1.8:

######  Proposition 3.2.

Let $(\text{\bf G},\text{\bf G}^{\prime})$ be a reductive dual pair of a
symplectic group and an even orthogonal group. Then
$(\rho_{\Lambda},\rho_{\Lambda})\in\Theta_{\text{\bf G},\text{\bf
G}^{\prime}}$ if and only if
$(\Lambda,\Lambda^{\prime})\in{\mathcal{B}}_{\text{\bf G},\text{\bf
G}^{\prime}}$.

###  3.2. $\ell$-admissible symbols

Consider a dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$ where both
$n,k$ are even and $k\leq n$. Let $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$,
and write

(3.3)
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}},\qquad\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}.$

Then from (2.5) we can write

(3.4) ${\rm def}(\Lambda)=\begin{cases}4d,&\text{if $\text{\bf G}={\rm
O}^{+}_{k}$};\\\ 4d+1,&\text{if $\text{\bf G}={\rm Sp}_{k}$};\\\
4d+2,&\text{if $\text{\bf G}={\rm O}^{-}_{k}$}\end{cases}$

for some $d\in{\mathbb{Z}}$, and we define an integer

(3.5) $\tau=\begin{cases}\tfrac{n-k}{2}+2d,&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm O}^{+}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm
O}^{+}_{n})$};\\\ \tfrac{n-k}{2}-(2d+1),&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm O}^{-}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm
O}^{-}_{n})$}.\end{cases}$

Note that the number $\tau$ depends on $k,n$ and ${\rm def}(\Lambda)$. If
$\tau\geq 0$, as in [Pan20a], $\underline{\theta}(\Lambda)$ is defined to be
the symbol in ${\mathcal{S}}_{\text{\bf G}^{\prime}}$ such that

(3.6)
$\Upsilon(\underline{\theta}(\Lambda))=\begin{cases}\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}\cup\genfrac{[}{]}{0.0pt}{}{\tau}{-},&\text{if
$\epsilon=+$};\\\
\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}\cup\genfrac{[}{]}{0.0pt}{}{-}{\tau},&\text{if
$\epsilon=-$}.\end{cases}$

Moreover, if $\tau\geq\nu_{1}$ when $\epsilon=+$; or $\tau\geq\mu_{1}$ when
$\epsilon=-$, we have

$\underline{\theta}(\Lambda)=\begin{cases}\binom{\tau+m_{2},b_{1},b_{2},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}},&\text{if
$\epsilon=+$};\\\
\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{\tau+m_{1},a_{1},a_{2},\ldots,a_{m_{1}}},&\text{if
$\epsilon=-$}.\end{cases}$

Let $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ with $\Upsilon(\Lambda)$ given as
in (3.3) and ${\rm def}(\Lambda)$ given as in (3.4), and let $\ell$ be a non-
negative even integer. Then $\Lambda$ is called _$\ell$ -admissible_ for the
dual pair $(\text{\bf G},\text{\bf G}^{\prime})$ if the following condition
holds:

(3.7) $\begin{cases}\mu_{1}\leq\frac{\ell}{2}-2d\text{ and
}\nu_{1}\leq\frac{\ell}{2}+2d,&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm Sp},{\rm O}^{+}_{\rm even})$};\\\
\mu_{1}\leq\frac{\ell}{2}-2d-1\text{ and
}\nu_{1}\leq\frac{\ell}{2}+2d+1,&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm Sp},{\rm O}^{-}_{\rm even})$};\\\
\mu_{1}\leq\frac{\ell}{2}-2d+1\text{ and
}\nu_{1}\leq\frac{\ell}{2}+2d,&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm O}^{+}_{\rm even},{\rm Sp})$};\\\
\mu_{1}\leq\frac{\ell}{2}-2d-1\text{ and
}\nu_{1}\leq\frac{\ell}{2}+2d+2,&\text{if $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm O}^{-}_{\rm even},{\rm Sp})$}.\end{cases}$

A unipotent character $\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ is called
_$\ell$ -admissible_ if $\Lambda$ is $\ell$-admissible. It is obvious that if
$\Lambda$ is $\ell$-admissible, then it is also $\ell^{\prime}$-admissible for
any even $\ell^{\prime}\geq\ell$.

######  Example 3.8.

In this example we consider $0$-admissible symbols
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ for a dual pair $(\text{\bf
G},\text{\bf G}^{\prime})$. Write
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$
and
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.

  1. (1)

From (3.7), a symbol $\Lambda\in{\mathcal{S}}_{{\rm Sp}_{k}}$ of defect $4d+1$
is $0$-admissible for $({\rm Sp}_{k},{\rm O}^{+}_{n})$ if and only if
$\mu_{1}\leq-2d$ and $\nu_{1}\leq 2d$. Because $\mu_{1},\nu_{1}$ are non-
negative, we must have $d=0$ and hence $\mu_{1}=\nu_{1}=0$. Therefore,
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{0}{0}$, and then $k=0$.

  2. (2)

A symbol $\Lambda\in{\mathcal{S}}_{{\rm Sp}_{k}}$ of defect $4d+1$ is
$0$-admissible for $({\rm Sp}_{k},{\rm O}^{-}_{n})$ if and only if
$\mu_{1}\leq-2d-1$ and $\nu_{1}\leq 2d+1$. There is no integer $d$ such that
both $-2d-1$ and $2d+1$ are non-negative.

  3. (3)

A symbol $\Lambda\in{\mathcal{S}}_{{\rm O}^{+}_{k}}$ of defect $4d$ is
$0$-admissible for $({\rm O}^{+}_{k},{\rm Sp}_{n})$ if and only if
$\mu_{1}\leq-2d+1$ and $\nu_{1}\leq 2d$. The only possibility is $d=0$ and
hence $\mu_{1}\leq 1$ and $\nu_{1}\leq 0$, i.e.,

$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{1,1,\ldots,1}{0,0,\ldots,0}\in{\mathcal{P}}_{2}(k),\qquad\Lambda=\binom{k,k-1,\ldots,1}{k-1,k-2,\ldots,0}.$

  4. (4)

A symbol $\Lambda\in{\mathcal{S}}_{{\rm O}^{-}_{k}}$ of defect $4d+2$ is
$0$-admissible for $({\rm O}^{-}_{k},{\rm Sp}_{n})$ if and only if
$\mu_{1}\leq-2d-1$ and $\nu_{1}\leq 2d+2$. The only possibility is $d=-1$ and
hence $\mu_{1}\leq 1$ and $\nu_{1}\leq 0$, i.e.,

$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{1,1,\ldots,1}{0,0,\ldots,0}\in{\mathcal{P}}_{2}(k-1),\qquad\Lambda=\binom{k-1,k-2,\ldots,1}{k,k-1,\ldots,0}.$

######  Lemma 3.9.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
Sp}_{k},{\rm O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$
where both $n,k$ are even and $k\leq n$. Then every unipotent character
$\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ is $k$-admissible. Consequently, if
$(\text{\bf G},\text{\bf G}^{\prime})$ is in stable range, then every
unipotent character $\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ is
$(n-k)$-admissible.

###### Proof.

Let $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$, and write
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$
and
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{+}_{n})$, and then ${\rm def}(\Lambda)=4d+1$ for some $d\in{\mathbb{Z}}$.
Then we have

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-2d(2d+1)=\frac{k}{2}-2d-4d^{2}\leq\frac{k}{2}-2d,$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-2d(2d+1)=\frac{k}{2}+2d-(2d+1)^{2}\leq\frac{k}{2}+2d.$

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{-}_{n})$, and then ${\rm def}(\Lambda)=4d+1$ for some $d\in{\mathbb{Z}}$.
Then we have

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-2d(2d+1)\leq\frac{k}{2}-2d-1,$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-2d(2d+1)\leq\frac{k}{2}+2d+1.$

  3. (3)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{+}_{k},{\rm
Sp}_{n})$, and then ${\rm def}(\Lambda)=4d$ for some $d\in{\mathbb{Z}}$. Then
we have

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-4d^{2}\leq\frac{k}{2}-2d+1,$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-4d^{2}\leq\frac{k}{2}+2d.$

  4. (4)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{-}_{k},{\rm
Sp}_{n})$, and then ${\rm def}(\Lambda)=4d+2$ for some $d\in{\mathbb{Z}}$.
Then we have

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-(2d+1)^{2}\leq\frac{k}{2}-2d-1,$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{k}{2}-(2d+1)^{2}\leq\frac{k}{2}+2d+2.$

For all cases, we see that $\Lambda$ is $k$-admissible.

Now if $(\text{\bf G},\text{\bf G}^{\prime})$ is in stable range, then we have
$k\leq n-k$. Therefore, $\Lambda$ is also $(n-k)$-admissible. ∎

The following lemma means that if $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ is
$(n-k)$-admissible, then $\underline{\theta}(\Lambda)$ (cf. (3.5)) is defined
for the pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$.

######  Lemma 3.10.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
Sp}_{k},{\rm O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$
where both $n,k$ are even and $k\leq n$. If $\Lambda$ is $(n-k)$-admissible,
then $\tau\geq 0$.

###### Proof.

First, suppose that either

  * •

$(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm O}^{+}_{n})$ and
${\rm def}(\Lambda)=4d+1$, or

  * •

$(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{+}_{k},{\rm Sp}_{n})$ and
${\rm def}(\Lambda)=4d$

for some $d\in{\mathbb{Z}}$. If $\Lambda$ is $(n-k)$-admissible, then by (3.5)
and (3.7) we have $\tau=\frac{1}{2}(n-k)+2d\geq\nu_{1}\geq 0$. Next, suppose
that either

  * •

$(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm O}^{-}_{n})$ and
${\rm def}(\Lambda)=4d+1$, or

  * •

$(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{-}_{k},{\rm Sp}_{n})$ and
${\rm def}(\Lambda)=4d+2$

for some $d\in{\mathbb{Z}}$. If $\Lambda$ is $(n-k)$-admissible, then we have
$\tau=\frac{1}{2}(n-k)-2d-1\geq\mu_{1}\geq 0$. ∎

Now we recall the definition of _$\Theta$ -rank_ of
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})$, denoted by $\Theta\text{\rm-
rk}(\rho^{\prime})$, from [Pan21a] subsection 3.1:

  * •

If $\text{\bf G}^{\prime}$ is a symplectic group, we consider dual pairs
$({\rm O}^{\epsilon}_{k},\text{\bf G}^{\prime})$ and define

(3.11) $\Theta\text{\rm-
rk}(\rho^{\prime})=\min\\{\,k\mid\rho^{\prime}\in\Theta(\rho)\text{ for some
}\rho\in{\mathcal{E}}({\rm O}^{\epsilon}_{k}(q)),\ \epsilon=+\text{ or
}-\,\\}.$

  * •

If $\text{\bf G}^{\prime}$ is an orthogonal group, we consider dual pairs
$({\rm Sp}_{k},\text{\bf G}^{\prime})$ and define

(3.12) $\Theta\text{\rm-
rk}(\rho^{\prime})=\min\\{\,k\mid\rho^{\prime}\chi^{\prime}\in\Theta(\rho)\text{
for some }\rho\in{\mathcal{E}}({\rm Sp}_{k}(q))\\\ \text{ and some linear
character }\chi^{\prime}\in{\mathcal{E}}(G^{\prime})\,\\}.$

Note that an orthogonal group $G^{\prime}$ has at most four linear characters:
${\bf 1}$, ${\rm sgn}$, $\chi_{\text{\bf G}^{\prime}}$ and $\chi_{\text{\bf
G}^{\prime}}{\rm sgn}$ where $\chi_{\text{\bf G}^{\prime}}$ denotes the linear
character of order two given by the spinor norm.

######  Lemma 3.13.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
Sp}_{k},{\rm O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$
where both $n,k$ are even and $k\leq n$. If
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ is $(n-k)$-admissible, then
$\underline{\theta}(\rho_{\Lambda})$ is of $\Theta$-rank $k$.

###### Proof.

Write $\Lambda$ and $\Upsilon(\Lambda)$ as in (3.3). Suppose that
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ is $(n-k)$-admissible. Then $\tau\geq
0$ by Lemma 3.10, and so the unipotent character
$\underline{\theta}(\rho_{\Lambda})$ is defined.

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{+}_{n})$ and then ${\rm def}(\Lambda)=4d+1$ for some $d\in{\mathbb{Z}}$.
Now the condition $\nu_{1}\leq\frac{1}{2}(n-k)+2d$ means that
$\nu_{1}\leq\tau$ and then

$\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\tau,\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

By the result in [Pan19b] section 8, we see that the unipotent character
$\underline{\theta}(\rho_{\Lambda})\in{\mathcal{E}}({\rm O}^{+}_{n}(q))$ does
not occur in the $\Theta$-correspondence for the dual pair $({\rm
Sp}_{k^{\prime}},{\rm O}^{+}_{n})$ for any $k^{\prime}<k$. The condition
$\mu_{1}\leq\frac{n-k}{2}-2d$ means that $\mu_{1}+(m_{1}-1)\leq\tau+m_{2}$
from (3.5) and (3.6), i.e., $a_{1}\leq\tau+m_{2}$. This implies that the
unipotent character $\underline{\theta}(\rho_{\Lambda}){\rm
sgn}\in{\mathcal{E}}({\rm O}^{+}_{n}(q))$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm Sp}_{k^{\prime\prime}},{\rm
O}^{+}_{n})$ for any $k^{\prime\prime}<k$. Moreover, by [Pan21a] Lemma 2.16,
we see that both $\underline{\theta}(\rho_{\Lambda})\chi_{{\rm
O}^{+}_{n}},\underline{\theta}(\rho_{\Lambda})\chi_{{\rm O}^{+}_{n}}{\rm sgn}$
do not occur in the $\Theta$-correspondence for dual pair $({\rm
Sp}_{k^{\prime\prime\prime}},{\rm O}^{+}_{n})$ for any
$k^{\prime\prime\prime}<k$.

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{-}_{n})$ and then ${\rm def}(\Lambda)=4d+1$ for some $d\in{\mathbb{Z}}$.
Then the condition $\mu_{1}\leq\frac{n-k}{2}-2d-1$ means that
$\mu_{1}\leq\tau$ and then

$\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\tau,\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

Hence $\underline{\theta}(\rho_{\Lambda})\in{\mathcal{E}}({\rm O}^{-}_{n}(q))$
does not occur in the $\Theta$-correspondence for the dual pair $({\rm
Sp}_{k^{\prime}},{\rm O}^{-}_{n})$ for any $k^{\prime}<k$. The condition
$\nu_{1}\leq\frac{n-k}{2}+2d+1$ means that $b_{1}\leq\tau+m_{1}$. This implies
that the unipotent character $\underline{\theta}(\rho_{\Lambda}){\rm
sgn}\in{\mathcal{E}}({\rm O}^{-}_{n}(q))$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm Sp}_{k^{\prime\prime}},{\rm
O}^{-}_{n})$ for any $k^{\prime\prime}<k$. As in (1), we know that both
$\underline{\theta}(\rho_{\Lambda})\chi_{{\rm
O}^{-}_{n}},\underline{\theta}(\rho_{\Lambda})\chi_{{\rm O}^{-}_{n}}{\rm sgn}$
do not occur in the $\Theta$-correspondence for dual pair $({\rm
Sp}_{k^{\prime\prime\prime}},{\rm O}^{-}_{n})$ for any
$k^{\prime\prime\prime}<k$.

  3. (3)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{+}_{k},{\rm
Sp}_{n})$ and then ${\rm def}(\Lambda)=4d$ for some $d\in{\mathbb{Z}}$. Then
the conditions $\nu_{1}\leq\frac{n-k}{2}+2d$ means that $\nu_{1}\leq\tau$.
Then the unipotent character
$\underline{\theta}(\rho_{\Lambda})\in{\mathcal{E}}({\rm Sp}_{n}(q))$ does not
occur in the $\Theta$-correspondence for the dual pair $({\rm
O}^{+}_{k^{\prime}},{\rm Sp}_{n})$ for any even $k^{\prime}<k$. The condition
$\mu_{1}\leq\frac{n-k}{2}-2d+1$ means that $a_{1}\leq\tau+m_{2}$. This implies
that the unipotent character
$\underline{\theta}(\rho_{\Lambda})\in{\mathcal{E}}({\rm Sp}_{n}(q))$ does not
occur in the $\Theta$-correspondence for the dual pair $({\rm
O}^{-}_{k^{\prime\prime}},{\rm Sp}_{n})$ for any even $k^{\prime\prime}<k$.
From [Pan21a] lemma 2.13, we know that any unipotent character of ${\rm
Sp}_{n}(q)$ does not occur in the $\Theta$-correspondence for the dual pair
$({\rm O}_{k^{\prime\prime\prime}},{\rm Sp}_{n})$ for any odd
$k^{\prime\prime\prime}<n$.

  4. (4)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{-}_{k},{\rm
Sp}_{n})$ and then ${\rm def}(\Lambda)=4d+2$ for some $d\in{\mathbb{Z}}$. Then
condition $\mu_{1}\leq\frac{n-k}{2}-2d-1$ means that $\mu_{1}\leq\tau$, and
the condition $\nu_{1}\leq\frac{n-k}{2}+2d+2$ means that
$b_{1}\leq\tau+m_{2}$. The remaining proof is similar to that of (3).

For all four cases, we conclude that $\underline{\theta}(\rho_{\Lambda})$ is
of $\Theta$-rank $k$ from (3.11) and (3.12). ∎

######  Proposition 3.14.

Suppose that both $n,k$ are even and $k\leq n$. Then a unipotent character
$\rho^{\prime}\in{\mathcal{E}}({\rm O}^{\epsilon}_{n}(q))_{1}$ is of
$\Theta$-rank $k$ if and only if $\rho^{\prime}$ or $\rho^{\prime}{\rm sgn}$
is equal to $\underline{\theta}(\rho_{\Lambda})$ for some $(n-k)$-admissible
unipotent character $\rho_{\Lambda}\in{\mathcal{E}}({\rm Sp}_{k}(q))$.

###### Proof.

Let $\rho^{\prime}$ be a unipotent character of ${\rm O}^{\epsilon}_{n}(q)$.
If $\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ or
$\rho^{\prime}\cdot{\rm sgn}=\underline{\theta}(\rho_{\Lambda})$ for some
$(n-k)$-admissible $\Lambda\in{\mathcal{S}}_{{\rm Sp}_{k}}$, then
$\Theta\text{\rm-rk}(\rho^{\prime})=\Theta\text{\rm-rk}(\rho^{\prime}\cdot{\rm
sgn})=k$ by Lemma 3.13.

Now suppose that $\Theta\text{\rm-rk}(\rho^{\prime})=k$. Then from [Pan20a]
and the definition of $\Theta$-rank, we know that $\rho^{\prime}$ or
$\rho^{\prime}\cdot{\rm sgn}$ is equal to
$\rho_{\underline{\theta}(\Lambda)}=\underline{\theta}(\rho_{\Lambda})$ for
some $\Lambda\in{\mathcal{S}}_{{\rm Sp}_{k}}$. Without loss of generality, we
may assume that $\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$. So now we
need to show that $\Lambda$ is $(n-k)$-admissible. Write

$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}.$

Suppose that $\epsilon=+$ and ${\rm def}(\Lambda)=4d+1$ for some
$d\in{\mathbb{Z}}$. Then

$\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}\cup\genfrac{[}{]}{0.0pt}{}{\tau}{-}$

by (3.6). Note that ${\rm def}(\underline{\theta}(\Lambda))=-4d$. If $\Lambda$
is not $(n-k)$-admissible, by definition there are two possibilities:

  * •

$\nu_{1}>\frac{1}{2}(n-k)+2d$, or

  * •

$\mu_{1}>\frac{1}{2}(n-k)-2d$.

Suppose that $\nu_{1}>\frac{1}{2}(n-k)+2d=\tau$. Then the rank of the symbol
$\Lambda^{\prime}$ such that

$\Upsilon(\Lambda^{\prime})=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{2},\nu_{3},\ldots,\nu_{m_{2}}}\cup\genfrac{[}{]}{0.0pt}{}{-}{\tau}$

is less than the rank of $\Lambda$. Then we see that
$(\rho_{\Lambda^{\prime}},\rho_{\underline{\theta}(\Lambda)})$ occurs in the
correspondence $\Theta$ for a dual pair $({\rm Sp}_{k^{\prime}},{\rm
O}^{+}_{n})$ for some $k^{\prime}<k$.

Now $\rho^{\prime}\cdot{\rm sgn}=\rho_{\underline{\theta}(\Lambda)^{\rm t}}$
and ${\rm def}(\underline{\theta}(\Lambda)^{\rm t})=4d$. By the same argument
as above (with $d$ replaced by $-d$ and $\nu_{1}$ replaced by $\mu_{1}$), we
see that if $\mu_{1}>\frac{1}{2}(n-k)-2d$, then $\rho^{\prime}\cdot{\rm sgn}$
occurs in the $\Theta$-correspondence for a dual pair $({\rm
Sp}_{k^{\prime\prime}},{\rm O}^{+}_{n})$ for some $k^{\prime\prime}<k$.
Therefore we conclude that $\Theta\text{\rm-rk}(\rho^{\prime})<k$ if $\Lambda$
is not $(n-k)$-admissible.

The proof for the case that $\epsilon=-$ is similar. ∎

######  Proposition 3.15.

Suppose that both $n,k$ are even and $k\leq n$. Then a unipotent character
$\rho^{\prime}\in{\mathcal{E}}({\rm Sp}_{n}(q))_{1}$ is of $\Theta$-rank $k$
if and only if $\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ for some
$(n-k)$-admissible unipotent character $\rho_{\Lambda}\in{\mathcal{E}}({\rm
O}^{\epsilon}_{k}(q))$ and some $\epsilon=+$ or $-$.

###### Proof.

Let $\rho^{\prime}$ be a unipotent character of ${\rm Sp}_{n}(q)$. If
$\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ for some $(n-k)$-admissible
$\Lambda\in{\mathcal{S}}_{{\rm O}^{\epsilon}_{k}}$ for some $\epsilon=+$ or
$-$, then $\Theta\text{\rm-rk}(\rho^{\prime})=k$ by Lemma 3.13.

Now suppose that $\Theta\text{\rm-rk}(\rho^{\prime})=k$. Then we know that
$\rho^{\prime}=\rho_{\underline{\theta}(\Lambda)}=\underline{\theta}(\rho_{\Lambda})$
for some $\Lambda\in{\mathcal{S}}_{{\rm O}^{\epsilon}_{k}}$ and some
$\epsilon=+$ or $-$. Suppose that $\epsilon=+$ and ${\rm def}(\Lambda)=4d$ for
some $d\in{\mathbb{Z}}$. If $\Lambda$ is not $(n-k)$-admissible, then

  * •

$\mu_{1}>\frac{1}{2}(n-k)-2d+1$; or

  * •

$\nu_{1}>\frac{1}{2}(n-k)+2d$.

If $\nu_{1}>\frac{1}{2}(n-k)+2d=\tau$, then we know that
$\rho_{\underline{\theta}(\Lambda)}$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm O}^{+}_{k^{\prime}},{\rm Sp}_{n})$ for some even
$k^{\prime}<k$. If $\mu_{1}>\frac{1}{2}(n-k)-2d+1$, then we know that
$\rho_{\underline{\theta}(\Lambda)}$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm O}^{-}_{k^{\prime\prime}},{\rm Sp}_{n})$ for some even
$k^{\prime\prime}<k$. Therefore $\Lambda$ must be $(n-k)$-admissible. The
proof for case $\epsilon=-$ is similar. ∎

###  3.3. Degree difference for symplectic/orthogonal dual pairs

######  Proposition 3.16.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$
where both $k,n$ are even and $k\leq n$, $\epsilon=+$ or $-$. If
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ is $(n-k)$-admissible, then

$\deg_{q}(\underline{\theta}(\rho_{\Lambda}))=\deg_{q}(\rho_{\Lambda})+\begin{cases}\frac{1}{2}k(n-k+1),&\text{if\/
$(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{\epsilon}_{k},{\rm
Sp}_{n})$};\\\ \frac{1}{2}k(n-k-1),&\text{if\/ $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$}.\end{cases}$

###### Proof.

Let $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ be $(n-k)$-admissible, and write
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.
Let $z_{1},z_{2},\ldots,z_{m}$ denote the entries
$a_{1},\ldots,a_{m_{1}},b_{1},\ldots,b_{m_{2}}$ of $\Lambda$ where $z_{1}\geq
z_{2}\geq\cdots\geq z_{m}$ and $m=m_{1}+m_{2}$. By Lemma 3.10 and Lemma 3.13,
we know that $\underline{\theta}(\rho_{\Lambda})$ is defined and of
$\Theta$-rank $k$. Let
$\\{z^{\prime}_{1},z^{\prime}_{2},\ldots,z^{\prime}_{m+1}\\}$ denote the set
of entries of $\underline{\theta}(\Lambda)$ where $z^{\prime}_{1}\geq
z^{\prime}_{2}\geq\cdots\geq z^{\prime}_{m+1}$. From (2.2), we know that

(3.17) $\displaystyle\begin{split}\frac{k}{2}={\rm
rk}(\Lambda)&=\sum_{i=1}^{m}z_{i}-\left\lfloor\left(\frac{m-1}{2}\right)^{2}\right\rfloor,\\\
\frac{n}{2}={\rm
rk}(\underline{\theta}(\Lambda))&=\sum_{i=1}^{m+1}z^{\prime}_{i}-\left\lfloor\left(\frac{m}{2}\right)^{2}\right\rfloor.\end{split}$

Now we consider the following cases:

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{+}_{k},{\rm
Sp}_{n})$. Now by (3.1) and (3.6) we can write
$\underline{\theta}(\Lambda)=\binom{b^{\prime}_{1},b^{\prime}_{2},\ldots,b^{\prime}_{m_{2}+1}}{a_{1},a_{2},\ldots,a_{m_{1}}}$
for some $b^{\prime}_{i}$. The assumption that $\Lambda$ is $(n-k)$-admissible
implies that $b^{\prime}_{1}=\tau+m_{2}$, $b^{\prime}_{j+1}=b_{j}$ for
$j=1,2,\ldots,m_{2}$, and $b^{\prime}_{1}\geq a_{1}$. Hence
$z_{1}^{\prime}=b_{1}^{\prime}$ and $z^{\prime}_{i+1}=z_{i}$ for
$i=1,2,\ldots,m$. Now ${\rm def}(\Lambda)\equiv 0\pmod{4}$, so $m$ is even.
From (3.17), we obtain

$z^{\prime}_{1}=\frac{n}{2}+\left(\frac{m}{2}\right)^{2}-\frac{k}{2}-\left\lfloor\left(\frac{m-1}{2}\right)^{2}\right\rfloor=\frac{1}{2}(n-k+m).$

Now $\deg_{q}(|{\rm O}^{\epsilon}_{k}(q)|_{p^{\prime}})=\frac{1}{4}k^{2}$ and
$\deg_{q}(|{\rm Sp}_{n}(q)|_{p^{\prime}})=\frac{1}{4}n(n+2)$. Moreover,

$(m+1-(i+1))z^{\prime}_{i+1}-z^{\prime}_{i+1}(z^{\prime}_{i+1}+1)=(m-i)z_{i}-z_{i}(z_{i}+1)$

for $i=1,2,\ldots,m$. Then by Lemma 2.7, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda}))-\deg_{q}(\rho_{\Lambda})$
$\displaystyle=mz^{\prime}_{1}-z^{\prime}_{1}(z^{\prime}_{1}+1)-\tfrac{1}{4}m(m-2)+\tfrac{1}{4}n(n+2)-\tfrac{1}{4}k^{2}$
$\displaystyle=\tfrac{1}{2}m(n-k+m)-\tfrac{1}{4}(n-k+m)(n-k+m+2)-\tfrac{1}{4}m(m-2)+\tfrac{1}{4}n(n+2)-\tfrac{1}{4}k^{2}$
$\displaystyle=\tfrac{1}{2}k(n-k+1).$

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{-}_{k},{\rm
Sp}_{n})$. Now we can write
$\underline{\theta}(\Lambda)=\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{m_{1}+1}}$.
The assumption that $\Lambda$ is $(n-k)$-admissible implies that
$a^{\prime}_{1}=\tau+m_{1}$, $a^{\prime}_{j+1}=a_{j}$ for
$j=1,2,\ldots,m_{1}$, and $a^{\prime}_{1}\geq b_{1}$. Hence
$z_{1}^{\prime}=a_{1}^{\prime}$ and $z^{\prime}_{i+1}=z_{i}$ for
$i=1,2,\ldots,m$. Then the remaining argument is exactly the same as in the
above case.

  3. (3)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$. Now $\underline{\theta}(\Lambda)$ is written as in case
(1) if $\epsilon=+$; and as in case (2) if $\epsilon=-$. Now ${\rm
def}(\Lambda)\equiv 1\pmod{4}$, so $m$ is odd. The assumption that $\Lambda$
is $(n-k)$-admissible implies that

$z^{\prime}_{1}=\frac{n}{2}+\left\lfloor\left(\frac{m}{2}\right)^{2}\right\rfloor-\frac{k}{2}-\left(\frac{m-1}{2}\right)^{2}=\frac{1}{2}(n-k+m-1)$

and $z^{\prime}_{i+1}=z_{i}$ for $i=1,2,\ldots,m$. Then by Lemma 2.7, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda}))-\deg_{q}(\rho_{\Lambda})$
$\displaystyle=mz^{\prime}_{1}-z^{\prime}_{1}(z^{\prime}_{1}+1)-\tfrac{1}{4}(m-1)^{2}+\tfrac{1}{4}n^{2}-\tfrac{1}{4}k(k+2)$
$\displaystyle=\tfrac{1}{2}m(n-k+m-1)-\tfrac{1}{4}(n-k+m-1)(n-k+m+1)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad-\tfrac{1}{4}(m-1)^{2}+\tfrac{1}{4}n^{2}-\tfrac{1}{4}k(k+2)$
$\displaystyle=\tfrac{1}{2}k(n-k-1).$

∎

##  4\. Unipotent Characters of Unitary Groups

###  4.1. Finite theta correspondence of unipotent characters

Let $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm U}_{n})$ be a
reductive dual pair of two unitary groups. For symbols
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ and
$\Lambda^{\prime}\in{\mathcal{S}}_{\text{\bf G}^{\prime}}$, we write
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$ and
$\Upsilon(\Lambda^{\prime})=\genfrac{[}{]}{0.0pt}{}{\mu^{\prime}}{\nu^{\prime}}$.
Now we define ${\mathcal{B}}_{{\rm U}_{k},{\rm U}_{n}}$ as follows:

  * •

if $k+n$ is even, then

${\mathcal{B}}_{{\rm U}_{k},{\rm
U}_{n}}=\Biggl{\\{}\,(\Lambda,\Lambda^{\prime})\in{\mathcal{S}}_{{\rm
U}_{k}}\times{\mathcal{S}}_{{\rm U}_{n}}\mid\nu\preccurlyeq\mu^{\prime},\
\nu^{\prime}\preccurlyeq\mu,\\\ {\rm
def}(\Lambda^{\prime})=\begin{cases}0,&\text{if ${\rm def}(\Lambda)=0$};\\\
-{\rm def}(\Lambda)+1,&\text{if ${\rm def}(\Lambda)\neq
0$}\end{cases}\,\Biggr{\\}}.$

  * •

if $k+n$ is odd, then

${\mathcal{B}}_{{\rm U}_{k},{\rm
U}_{n}}=\\{\,(\Lambda,\Lambda^{\prime})\in{\mathcal{S}}_{{\rm
U}_{k}}\times{\mathcal{S}}_{{\rm U}_{n}}\mid\mu\preccurlyeq\nu^{\prime},\
\mu^{\prime}\preccurlyeq\nu,\ {\rm def}(\Lambda^{\prime})=-{\rm
def}(\Lambda)-1\,\\}.$

Here the relation $\mu\preccurlyeq\nu^{\prime}$ is given in (2.1). It is known
that the unipotent characters are preserved by the $\Theta_{\text{\bf
G},\text{\bf G}^{\prime}}$ (cf. [AM93] theorem 3.5). The following proposition
on the $\Theta$-correspondence of unipotent characters for a unitary dual pair
is rephrased from [AMR96] théorème 5.15 (see also [Pan19b] proposition 5.13):

######  Proposition 4.1.

Let $\rho_{\Lambda}\in{\mathcal{E}}({\rm U}_{k}(q))_{1}$ and
$\rho_{\Lambda^{\prime}}\in{\mathcal{E}}({\rm U}_{n}(q))_{1}$. Then
$(\rho_{\Lambda},\rho_{\Lambda^{\prime}})\in\Theta_{{\rm U}_{k},{\rm U}_{n}}$
if and only if $(\Lambda,\Lambda^{\prime})\in{\mathcal{B}}_{{\rm U}_{k},{\rm
U}_{n}}$.

###  4.2. $\ell$-admissible unipotent characters

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ such that $k\leq n$. Let $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$, and
write

$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}},\qquad\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}.$

Suppose that $(\Lambda,\Lambda^{\prime})\in{\mathcal{B}}_{{\rm U}_{k},{\rm
U}_{n}}$. Let $d=|{\rm def}(\Lambda)|$ and $d^{\prime}=|{\rm
def}(\Lambda^{\prime})|$. Define

(4.2)
$\tau=\frac{1}{2}\left[\left(n-\frac{d^{\prime}(d^{\prime}+1)}{2}\right)-\left(k-\frac{d(d+1)}{2}\right)\right].$

Then from the definition of ${\mathcal{B}}_{{\rm U}_{k},{\rm U}_{n}}$ we can
check that

(4.3) $\tau=\begin{cases}\frac{1}{2}(n-k+d),&\text{if $k+n+d$ is even};\\\
\frac{1}{2}(n-k-d-1),&\text{if $k+n+d$ is odd}.\end{cases}$

Note that $\tau$ is an integer depending on $k,n$ and ${\rm def}(\Lambda)$. If
$\tau\geq 0$, then $\underline{\theta}(\Lambda)$ is defined to be the symbol
in ${\mathcal{S}}_{{\rm U}_{n}}$ such that

$\Upsilon(\underline{\theta}(\Lambda))=\begin{cases}\Upsilon(\Lambda)^{\rm
t}\cup\genfrac{[}{]}{0.0pt}{}{\tau}{-},&\text{if $k+n$ is even};\\\
\Upsilon(\Lambda)^{\rm t}\cup\genfrac{[}{]}{0.0pt}{}{-}{\tau},&\text{if $k+n$
is odd}.\end{cases}$

For a non-negative integer $\ell$, a symbol $\Lambda\in{\mathcal{S}}_{{\rm
U}_{k}}$ is called _$\ell$ -admissible_ if the following conditions hold:

(4.4) $\begin{cases}\mu_{1}\leq\frac{1}{2}(\ell-d)\text{ and
}\nu_{1}\leq\frac{1}{2}(\ell+d),&\text{if $\ell$ is even, $d$ is even};\\\
\mu_{1}\leq\frac{1}{2}(\ell+d+1)\text{ and
}\nu_{1}\leq\frac{1}{2}(\ell-d-1),&\text{if $\ell$ is even, $d$ is odd};\\\
\mu_{1}\leq\frac{1}{2}(\ell-d-1)\text{ and
}\nu_{1}\leq\frac{1}{2}(\ell+d+1),&\text{if $\ell$ is odd, $d$ is even};\\\
\mu_{1}\leq\frac{1}{2}(\ell+d)\text{ and
}\nu_{1}\leq\frac{1}{2}(\ell-d),&\text{if $\ell$ is odd, $d$ is
odd}.\end{cases}$

It is clear that if $\Lambda$ is $\ell$-admissible, then it is also
$\ell^{\prime}$-admissible for any $\ell^{\prime}\geq\ell$. A unipotent
character $\rho_{\Lambda}\in{\mathcal{E}}({\rm U}_{k}(q))_{1}$ is called
_$\ell$ -admissible_ if $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is
$\ell$-admissible.

######  Lemma 4.5.

Every unipotent character $\rho_{\Lambda}\in{\mathcal{E}}({\rm U}_{k}(q))_{1}$
is $k$-admissible. Consequently, if the dual pair $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm U}_{k},{\rm U}_{n})$ is in stable range, then every
$\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ is $(n-k)$-admissible.

###### Proof.

Let $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$, $d=|{\rm def}(\Lambda)|$, and
write
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.

  1. (1)

Suppose that $k$ is even. Then

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{1}{2}(k-d(d+1))\leq\begin{cases}\frac{1}{2}(k-d),&\text{if
$d$ is even};\\\ \frac{1}{2}(k+d+1),&\text{if $d$ is odd},\end{cases}$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{1}{2}(k-d(d+1))\leq\begin{cases}\frac{1}{2}(k+d),&\text{if
$d$ is even};\\\ \frac{1}{2}(k-d-1),&\text{if $d$ is odd}.\end{cases}$

  2. (2)

Suppose that $k$ is odd. Then

$\displaystyle\mu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{1}{2}(k-d(d+1))\leq\begin{cases}\frac{1}{2}(k-d-1),&\text{if
$d$ is even};\\\ \frac{2}{2}(k+d),&\text{if $d$ is odd},\end{cases}$
$\displaystyle\nu_{1}$
$\displaystyle\leq|\Upsilon(\Lambda)|=\frac{1}{2}(k-d(d+1))\leq\begin{cases}\frac{1}{2}(k+d+1),&\text{if
$d$ is even};\\\ \frac{2}{2}(k-d),&\text{if $d$ is odd}.\end{cases}$

For two cases, $\Lambda$ is always $k$-admissible.

Now if $({\rm U}_{k},{\rm U}_{n})$ is in stable range, then we have $k\leq
n-k$. Therefore, $\Lambda$ is also $(n-k)$-admissible. ∎

######  Lemma 4.6.

For a non-negative integer $d$, let $\rho_{d}=\rho_{\Lambda_{d}}$ denote the
unipotent cuspidal character of ${\rm U}_{k}(q)$ where $k=\frac{1}{2}d(d+1)$.
Then $\Lambda_{d}$ is $\ell$-admissible for $\ell\geq d$ if $\ell-d$ is even;
and is $\ell$-admissible for $\ell\geq d+1$ if $\ell-d$ is odd.

###### Proof.

It is known that $\Lambda_{d}=\binom{d-1,d-2,\ldots,0}{-}$ if $d$ is even; and
$\Lambda_{d}=\binom{-}{d-1,d-2,\ldots,0}$ if $d$ is odd, and
$\Upsilon(\Lambda_{d})=\genfrac{[}{]}{0.0pt}{}{0}{0}$, i.e.,
$\mu_{1}=\nu_{1}=0$. Note that ${\rm def}(\Lambda_{d})=d$ if $d$ is even; and
${\rm def}(\Lambda_{d})=-d$ if $d$ is odd. Then the lemma follows from (4.4)
immediately. ∎

######  Example 4.7.

Let ${\rm St}_{{\rm U}_{k}}$ denote the Steinberg character of ${\rm
U}_{k}(q)$ where $k\geq 1$. It is known that $\deg_{q}({\rm St}_{{\rm
U}_{k}})=\frac{1}{2}k(k-1)$. Moreover, the symbol $\Lambda$ associated to
${\rm St}_{{\rm U}_{k}}$ satisfies

$\Lambda=\begin{cases}\binom{\frac{k-2}{2},\frac{k-4}{2},\ldots,0}{\frac{k}{2},\frac{k-2}{2},\ldots,1}\\\
\binom{\frac{k-1}{2},\frac{k-3}{2},\ldots,1}{\frac{k-1}{2},\frac{k-3}{2},\ldots,0}\end{cases}\
\Upsilon(\Lambda)=\begin{cases}\genfrac{[}{]}{0.0pt}{}{-}{1,1,\ldots,1},&\text{if
$k$ is even};\\\ \genfrac{[}{]}{0.0pt}{}{1,1,\ldots,1}{-},&\text{if $k$ is
odd}.\end{cases}$

Note that now ${\rm def}(\Lambda)=0$ if $k$ is even, ${\rm def}(\Lambda)=-1$
if $k$ is odd. Therefore we see that ${\rm St}_{{\rm U}_{k}}$ is
$1$-admissible, and hence it is $\ell$-admissible for any $\ell\geq 1$.

######  Lemma 4.8.

Consider the dual pair $({\rm U}_{k},{\rm U}_{n})$ where $k\leq n$. If
$\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is $(n-k)$-admissible, then $\tau\geq
0$.

###### Proof.

Write
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.
Suppose that $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is $(n-k)$-admissible.
First suppose that $k+n+d$ is even. The condition that $\Lambda$ is
$(n-k)$-admissible means that $(n-k)+d\geq 0$ from (4.4). This is exactly the
condition that $\tau\geq 0$.

Next, suppose that $k+n+d$ is odd. The condition that $\Lambda$ is
$(n-k)$-admissible means that $(n-k)-d-1\geq 0$ from (4.4). Then again we have
$\tau\geq 0$. ∎

Recall that from [Pan21a] subsection 3.1, the _$\Theta$ -rank_ of
$\rho^{\prime}\in{\mathcal{E}}({\rm U}_{n}(q))$ is defined by

(4.9) $\Theta\text{\rm-
rk}(\rho^{\prime})=\min\\{\,k\mid\rho^{\prime}\chi^{\prime}\in\Theta(\rho)\text{
for some }\rho\in{\mathcal{E}}({\rm U}_{k}(q))\\\ \text{and some linear
character }\chi^{\prime}\in{\mathcal{E}}({\rm U}_{n}(q))\,\\}.$

######  Lemma 4.10.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ where $k\leq n$. If $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is
$(n-k)$-admissible, then $\underline{\theta}(\rho_{\Lambda})$ is of
$\Theta$-rank $k$.

###### Proof.

Suppose that $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is $(n-k)$-admissible for
$({\rm U}_{k},{\rm U}_{n})$. From the previous lemma, we know that $\tau\geq
0$ and so $\underline{\theta}(\rho_{\Lambda})$ is defined. Let $d=|{\rm
def}(\Lambda)|$.

  1. (1)

Suppose that $k+n$ is even. Now we have $\tau\geq\nu_{1}$. Then

$\underline{\theta}(\Lambda)=\binom{\tau+m_{2},b_{1},b_{2},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}},\qquad\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\tau,\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

    1. (a)

Suppose that $d$ is even. Then the condition $\nu_{1}\leq\frac{1}{2}(n-k+d)$
implies that $\rho_{\underline{\theta}(\Lambda)}$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$
for any $k^{\prime}<k$ such that $k^{\prime}+n$ is even. The condition
$\mu_{1}\leq\frac{1}{2}(n-k-d)$ means that $2(\mu_{1}+m_{1}-1)\leq
2(\tau+m_{2})-1$, i.e.,

$\text{\rm rk}_{\rm
U}\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\leq\text{\rm
rk}_{\rm
U}\binom{a_{2},\ldots,a_{m_{1}}}{\tau+m_{2},b_{1},b_{2},\ldots,b_{m_{2}}}.$

Therefore the unipotent character $\underline{\theta}(\rho_{\Lambda})$ of
${\rm U}_{n}(q)$ does not occur in the $\Theta$-correspondence for the dual
pair $({\rm U}_{k^{\prime\prime}},{\rm U}_{n})$ for any $k^{\prime\prime}<k$
such that $k^{\prime\prime}+n$ is odd.

    2. (b)

Suppose that $d$ is odd. The condition $\nu_{1}\leq\frac{1}{2}(n-k-d-1)$
implies that $\underline{\theta}(\rho_{\Lambda})$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$
for any $k^{\prime}<k$ such that $k^{\prime}+n$ is even. The condition
$\mu_{1}\leq\frac{1}{2}(n-k+d+1)$ implies $\underline{\theta}(\rho_{\Lambda})$
does not occur in the $\Theta$-correspondence for the dual pair $({\rm
U}_{k^{\prime\prime}},{\rm U}_{n})$ for any $k^{\prime\prime}<k$ such that
$k^{\prime\prime}+n$ is odd.

  2. (2)

Suppose that $k+n$ is odd. Now we have $\tau\geq\mu_{1}$. Then

$\underline{\theta}(\Lambda)=\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{\tau+m_{1},a_{1},a_{2},\ldots,a_{m_{1}}},\qquad\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\tau,\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

    1. (a)

Suppose that $d$ is even. The condition $\mu_{1}\leq\frac{1}{2}(n-k-d-1)$
implies that $\underline{\theta}(\rho_{\Lambda})$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$
for any $k^{\prime}<k$ such that $k^{\prime}+n$ is even. The condition
$\nu_{1}\leq\frac{1}{2}(n-k+d+1)$ means $2(\nu_{1}+m_{2}-1)\leq
2(\tau+m_{1})+1$, i.e., .

$\text{\rm rk}_{\rm
U}\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\leq\text{\rm
rk}_{\rm
U}\binom{\tau+m_{1},a_{1},a_{2},\ldots,a_{m_{1}}}{b_{2},\ldots,b_{m_{2}}}.$

Therefore $\underline{\theta}(\rho_{\Lambda})$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime\prime}},{\rm
U}_{n})$ for any $k^{\prime\prime}<k$ such that $k^{\prime\prime}+n$ is odd.

    2. (b)

Suppose that $d$ is odd. The condition $\mu_{1}\leq\frac{1}{2}(n-k+d)$ implies
that $\underline{\theta}(\rho_{\Lambda})$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$
for any $k^{\prime}<k$ such that $k^{\prime}+n$ is even. The condition
$\nu_{1}\leq\frac{1}{2}(n-k-d)$ implies that
$\underline{\theta}(\rho_{\Lambda})$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm U}_{k^{\prime\prime}},{\rm
U}_{n})$ for any $k^{\prime\prime}<k$ such that $k^{\prime\prime}+n$ is odd.

For all cases, we conclude that $\underline{\theta}(\rho_{\Lambda})$ is of
$\Theta$-rank $k$. ∎

######  Proposition 4.11.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ where $k\leq n$. A unipotent character
$\rho^{\prime}\in{\mathcal{E}}({\rm U}_{n}(q))_{1}$ is of $\Theta$-rank $k$ if
and only if $\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ for some
$(n-k)$-admissible symbol $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$.

###### Proof.

Let $\rho^{\prime}$ be a unipotent character of ${\rm U}_{n}(q)$. Suppose that
$\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ for some $(n-k)$-admissible
$\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$. Then $\Theta\text{\rm-
rk}(\rho^{\prime})=k$ by Lemma 4.10.

Now we suppose that $\rho^{\prime}$ is of $\Theta$-rank $k$. From [Pan21a], we
know that $\rho^{\prime}=\underline{\theta}(\rho_{\Lambda})$ for some symbol
$\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$. We need to show that $\Lambda$ is
$(n-k)$-admissible. Write
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$
and let $d=|{\rm def}(\Lambda)|$. From [Pan19b] section 8, we can conclude the
following.

  1. (1)

Suppose that $k+n$ is even and $d$ is even.

    1. (a)

If $\mu_{1}>\frac{1}{2}(n-k-d)$, we know that
$\underline{\theta}(\rho_{\Lambda})$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$ for some $k^{\prime}<k$
such that $k^{\prime}+n$ is even.

    2. (b)

If $\nu_{1}>\frac{1}{2}(n-k+d)$, then $\underline{\theta}(\rho_{\Lambda})$
occurs in the $\Theta$-correspondence for the dual pair $({\rm
U}_{k^{\prime\prime}},{\rm U}_{n})$ for some $k^{\prime\prime}<k$ such that
$k^{\prime\prime}+n$ is odd.

  2. (2)

Suppose that $k+n$ is even and $d$ is odd.

    1. (a)

If $\mu_{1}>\frac{1}{2}(n-k+d+1)$, we know that
$\underline{\theta}(\rho_{\Lambda})$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$ for some $k^{\prime}<k$
such that $k^{\prime}+n$ is even.

    2. (b)

If $\nu_{1}>\frac{1}{2}(n-k-d-1)$, then $\underline{\theta}(\rho_{\Lambda})$
occurs in the $\Theta$-correspondence for the dual pair $({\rm
U}_{k^{\prime\prime}},{\rm U}_{n})$ for some $k^{\prime\prime}<k$ such that
$k^{\prime\prime}+n$ is odd.

  3. (3)

Suppose that $k+n$ is odd and $d$ is even.

    1. (a)

If $\mu_{1}>\frac{1}{2}(n-k-d-1)$, we know that
$\underline{\theta}(\rho_{\Lambda})$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$ for some $k^{\prime}<k$
such that $k^{\prime}+n$ is even.

    2. (b)

If $\nu_{1}>\frac{1}{2}(n-k+d+1)$, then $\underline{\theta}(\rho_{\Lambda})$
occurs in the $\Theta$-correspondence for the dual pair $({\rm
U}_{k^{\prime\prime}},{\rm U}_{n})$ for some $k^{\prime\prime}<k$ such that
$k^{\prime\prime}+n$ is odd.

  4. (4)

Suppose that $k+n$ is odd and $d$ is odd.

    1. (a)

If $\mu_{1}>\frac{1}{2}(n-k+d)$, we know that
$\underline{\theta}(\rho_{\Lambda})$ occurs in the $\Theta$-correspondence for
the dual pair $({\rm U}_{k^{\prime}},{\rm U}_{n})$ for some $k^{\prime}<k$
such that $k^{\prime}+n$ is even.

    2. (b)

If $\nu_{1}>\frac{1}{2}(n-k-d)$, then $\underline{\theta}(\rho_{\Lambda})$
occurs in the $\Theta$-correspondence for the dual pair $({\rm
U}_{k^{\prime\prime}},{\rm U}_{n})$ for some $k^{\prime\prime}<k$ such that
$k^{\prime\prime}+n$ is odd.

Therefore, for all cases, if $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is not
$(n-k)$-admissible, we will conclude that $\Theta\text{\rm-
rk}(\underline{\theta}(\rho_{\Lambda}))<k$. ∎

###  4.3. Degree differences for unitary dual pairs

######  Proposition 4.12.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ where $k\leq n$. If $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ is
$(n-k)$-admissible, then

$\deg_{q}(\underline{\theta}(\rho_{\Lambda}))=\deg_{q}(\rho_{\Lambda})+k(n-k).$

###### Proof.

Let $\Lambda\in{\mathcal{S}}_{{\rm U}_{k}}$ be $(n-k)$-admissible, and write
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}$.

  1. (1)

Suppose that $k+n$ is odd. The assumption that $\Lambda$ is $(n-k)$-admissible
means that

$\underline{\theta}(\Lambda)=\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{\tau+m_{1},a_{1},a_{2},\ldots,a_{m_{1}}},\qquad\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\tau,\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

    1. (a)

Suppose that $m_{1}+m_{2}$ is even. Then ${\rm def}(\Lambda)=m_{1}-m_{2}$ is
even and non-negative, and ${\rm
def}(\underline{\theta}(\Lambda))=m_{2}-m_{1}-1$ is odd and negative. By
(2.8), we have

$\displaystyle k$
$\displaystyle=\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}-\frac{1}{2}m(m-2)+\frac{1}{2}(m_{1}-m_{2})$
$\displaystyle n$
$\displaystyle=2(\tau+m_{1})+\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}-\frac{1}{2}(m+1)(m-1)+\frac{1}{2}(m_{1}-m_{2}+1)$

where $m=m_{1}+m_{2}$. Then we have

$2(\tau+m_{1})+1=n-k+m.$

Now by (2.9), we have

$\displaystyle X^{0}_{\Lambda}$
$\displaystyle=\\{2b_{1},2b_{2},\ldots,2b_{m_{2}}\\},$ $\displaystyle
X^{1}_{\Lambda}$ $\displaystyle=\\{2a_{1}+1,2a_{2}+1,\ldots,2a_{m_{1}}+1\\};$
$\displaystyle X^{0}_{\underline{\theta}(\Lambda)}$
$\displaystyle=\\{2b_{1},2b_{2},\ldots,2b_{m_{2}}\\},$ $\displaystyle
X^{1}_{\underline{\theta}(\Lambda)}$
$\displaystyle=\\{2(\tau+m_{1})+1,2a_{1}+1,2a_{2}+1,\ldots,2a_{m_{1}}+1\\}.$

Write

$\displaystyle X^{0}_{\Lambda}\cup
X^{1}_{\Lambda}=\\{z_{1},z_{2},\ldots,z_{m}\\},\qquad
X^{0}_{\underline{\theta}(\Lambda)}\cup
X^{1}_{\underline{\theta}(\Lambda)}=\\{z^{\prime}_{1},z^{\prime}_{2},\ldots,z^{\prime}_{m+1}\\}$

with $z_{1}>z_{2}>\cdots>z_{m}$ and
$z^{\prime}_{1}>z^{\prime}_{2}>\cdots>z^{\prime}_{m+1}$. Now
$z^{\prime}_{i+1}=z_{i}$ for $i=1,\ldots,m$, and $z^{\prime}_{1}=n-k+m$. Hence
by Lemma 2.12, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda}))-\deg_{q}(\rho_{\Lambda})$
$\displaystyle=mz^{\prime}_{1}-\tfrac{1}{2}z^{\prime}_{1}(z^{\prime}_{1}+1)+\tfrac{1}{2}n(n+1)-\tfrac{1}{2}k(k+1)-\tfrac{1}{2}m(m-1)$
$\displaystyle=\tfrac{1}{2}(n-k+m)(m-(n-k)-1)+\tfrac{1}{2}n(n+1)-\tfrac{1}{2}k(k+1)-\tfrac{1}{2}m(m-1)$
$\displaystyle=k(n-k).$

    2. (b)

Suppose that $m_{1}+m_{2}$ is odd. Then ${\rm def}(\Lambda)=m_{1}-m_{2}$ is
odd and negative, and ${\rm def}(\underline{\theta}(\Lambda))=m_{2}-m_{1}-1$
is even and non-negative. By (2.8), we have

$\displaystyle k$
$\displaystyle=\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}-\frac{1}{2}m(m-2)+\frac{1}{2}(m_{2}-m_{1})$
$\displaystyle n$
$\displaystyle=2(\tau+m_{1})+\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}-\frac{1}{2}(m+1)(m-1)+\frac{1}{2}(m_{2}-m_{1}-1)$

where $m=m_{1}+m_{2}$. Then we have

$2(\tau+m_{1})=n-k+m.$

Now by (2.9), we have

$\displaystyle X^{0}_{\Lambda}$
$\displaystyle=\\{2a_{1},2a_{2},\ldots,2a_{m_{1}}\\},$ $\displaystyle
X^{1}_{\Lambda}$ $\displaystyle=\\{2b_{1}+1,2b_{2}+1,\ldots,2b_{m_{2}}+1\\};$
$\displaystyle X^{0}_{\underline{\theta}(\Lambda)}$
$\displaystyle=\\{2(\tau+m_{1}),2a_{1},2a_{2},\ldots,2a_{m_{1}}\\},$
$\displaystyle X^{1}_{\underline{\theta}(\Lambda)}$
$\displaystyle=\\{2b_{1}+1,2b_{2}+1,\ldots,2b_{m_{2}}+1\\}.$

The remaining proof is the same as in case (a).

  2. (2)

Suppose that $k+n$ is even. The assumption that $\Lambda$ is
$(n-k)$-admissible means that

$\underline{\theta}(\Lambda)=\binom{\tau+m_{2},b_{1},b_{2},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}},\qquad\Upsilon(\underline{\theta}(\Lambda))=\genfrac{[}{]}{0.0pt}{}{\tau,\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}.$

Now the proof is similar to that in case (1). The only difference is that the
element $\tau+m_{1}$ is replaced by $\tau+m_{2}$.

∎

######  Example 4.13.

Consider the dual pair $({\rm U}_{k},{\rm U}_{n})$ with $k=\frac{1}{2}d(d+1)$
and $n=\frac{1}{2}(d+1)(d+2)$ where $d$ is a non-negative integer. Let
$\rho_{d}$ denote the unipotent cuspidal character of ${\rm U}_{k}$. It is
known that

$\deg_{q}(\rho_{d})=\frac{1}{24}(d-1)d(d+1)(3d+2).$

from [Lus77] (9.5.1). Note that $\rho_{d}=\rho_{\Lambda_{d}}$ where symbol
$\Lambda_{d}$ is given in the proof of Lemma 4.6. It is easy to check that
$\underline{\theta}(\rho_{d})=\rho_{d+1}$ for the dual pair $({\rm U}_{k},{\rm
U}_{n})$, and then

$\displaystyle\deg_{q}(\rho_{d})+k(n-k)$
$\displaystyle=\frac{1}{24}(d-1)d(d+1)(3d+2)+\frac{1}{2}d(d+1)^{2}$
$\displaystyle=\frac{1}{24}d(d+1)(d+2)(3d+5)$
$\displaystyle=\deg_{q}(\rho_{d+1}).$

##  5\. Degree Differences in the Eta Correspondence

###  5.1. Lusztig correspondence for symplectic or orthogonal groups

Let G be a symplectic group or an orthogonal group. For $s\in(G^{*})^{0}$, we
define

(5.1) $\displaystyle\begin{split}\text{\bf G}^{(0)}=\text{\bf
G}^{(0)}(s)&=\prod_{\langle\lambda\rangle\subset\\{\lambda_{1},\ldots,\lambda_{n}\\},\
\lambda\neq\pm 1}\text{\bf G}_{[\lambda]}(s);\\\ \text{\bf G}^{(-)}=\text{\bf
G}^{(-)}(s)&=\text{\bf G}_{[-1]}(s);\\\ \text{\bf G}^{(+)}=\text{\bf
G}^{(+)}(s)&=\begin{cases}(\text{\bf G}_{[1]}(s))^{*},&\text{if $\text{\bf G}$
is symplectic};\\\ \text{\bf
G}_{[1]}(s),&\text{otherwise},\end{cases}\end{split}$

where $\text{\bf G}_{[\lambda]}(s)$ is given in [AMR96] subsection 1.B (see
also [Pan19b] subsection 2.2). One can see that

  * •

if $\text{\bf G}={\rm O}^{\epsilon}_{k}$ where $k$ is even, then $\text{\bf
G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{k^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
O}^{\epsilon^{\prime}\epsilon}_{k^{(+)}}$ where $k^{(-)},k^{(+)}$ are even;

  * •

if $\text{\bf G}={\rm Sp}_{k}$, then $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{k^{(-)}}$, and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$
where $k^{(-)},k^{(+)}$ are even;

  * •

if $\text{\bf G}={\rm O}_{k}$ where $k$ is odd, then $\text{\bf G}^{(-)}={\rm
Sp}_{k^{(-)}}$, and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$ where
$k^{(-)},k^{(+)}$ are even,

for some $\epsilon^{\prime}=+$ or $-$ and some $k^{(-)},k^{(+)}$ all depending
on $s$. Note that $k^{(-)}+k^{(+)}\leq k$ for the first two cases above, and
$k^{(-)}+k^{(+)}\leq k-1$ if $\text{\bf G}={\rm O}_{k}$ is an odd orthogonal
group. Then we have a (modified) _Lusztig correspondence_

$\displaystyle\Xi_{s}\colon{\mathcal{E}}(G)_{s}$
$\displaystyle\rightarrow{\mathcal{E}}(G^{(0)}(s)\times G^{(-)}(s)\times
G^{(+)}(s))_{1}$ $\displaystyle\rho$
$\displaystyle\mapsto\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$

where $\Lambda^{(-)}\in{\mathcal{S}}_{\text{\bf G}^{(-)}}$ and
$\Lambda^{(+)}\in{\mathcal{S}}_{\text{\bf G}^{(+)}}$. The mapping $\Xi_{s}$ is
two-to-one, more precisely $\Xi_{s}(\rho)=\Xi_{s}(\rho\cdot{\rm sgn})$ when G
is an odd-orthogonal group; and $\Xi_{s}$ is a bijection otherwise.

Let $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{\epsilon}_{k},{\rm
Sp}_{n})$ or $({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$, and let $\ell$ be a non-
negative integer. Now we define the $\ell$-admissibility of
$\rho\in{\mathcal{E}}(G)$ according the following cases:

  1. (I)

Suppose that $\text{\bf G}={\rm Sp}_{k}$ where $k$ is even, $\ell$ is even.
Now $\text{\bf G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{k^{(-)}}$ for some
$\epsilon^{\prime}=+$ or $-$, and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$.
Then $\rho\in{\mathcal{E}}(G)$ is called _$\ell$ -admissible_ for the dual
pair $(\text{\bf G},\text{\bf G}^{\prime})$ if

     * •

$\rho_{\Lambda^{(-)}}$ is $\ell$-admissible for $(\text{\bf G}^{(-)},{\rm
Sp}_{\rm even})$ and

     * •

$\rho_{\Lambda^{(+)}}$ is $\ell$-admissible for $(\text{\bf G}^{(+)},{\rm
O}^{\epsilon^{\prime}}_{\rm even})$ given in (3.7).

  2. (II)

Suppose that $\text{\bf G}={\rm Sp}_{k}$ where $k$ is even, $\ell$ is odd. Now
$\text{\bf G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{k^{(-)}}$ for some
$\epsilon^{\prime}=+$ or $-$, and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$.
Then $\rho\in{\mathcal{E}}(G)$ is called _$\ell$ -admissible_ for the dual
pair $(\text{\bf G},\text{\bf G}^{\prime})$ if

     * •

$\rho_{\Lambda^{(-)}}$ is $(\ell-1)$-admissible for $(\text{\bf G}^{(-)},{\rm
Sp}_{\rm even})$ and

     * •

$\rho_{\Lambda^{(+)}}$ is $(\ell-1)$-admissible for $(\text{\bf G}^{(+)},{\rm
O}^{\epsilon^{\prime}}_{\rm even})$.

  3. (III)

Suppose that $\text{\bf G}={\rm O}^{\epsilon}_{k}$ where $k$ is even, $\ell$
is even. Then both $\text{\bf G}^{(-)},\text{\bf G}^{(+)}$ are even-orthogonal
groups. Then $\rho\in{\mathcal{E}}(G)$ is called _$\ell$ -admissible_ for the
dual pair $(\text{\bf G},\text{\bf G}^{\prime})$ if

     * •

$\rho_{\Lambda^{(-)}}$ is $\ell$-admissible for $(\text{\bf G}^{(-)},{\rm
Sp}_{\rm even})$ and

     * •

$\rho_{\Lambda^{(+)}}$ is $\ell$-admissible for $(\text{\bf G}^{(+)},{\rm
Sp}_{\rm even})$.

  4. (IV)

Suppose that $\text{\bf G}={\rm O}_{k}$ where $k$ is odd, $\ell$ is odd. Now
$\text{\bf G}^{(-)}={\rm Sp}_{k^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
Sp}_{k^{(+)}}$. Let $\rho\in{\mathcal{E}}(G)$. Because now $G\simeq{\rm
SO}_{k}(q)\times\\{\pm 1\\}$, we can write $\rho\simeq\rho_{1}\otimes\rho_{2}$
where $\rho_{1}\in{\mathcal{E}}({\rm SO}_{k}(q))$ and
$\rho_{2}\in{\mathcal{E}}(\\{\pm 1\\})$. Let $\epsilon^{\prime}=+$ of
$\rho_{2}=\bf 1$ and $\epsilon^{\prime}=-$ if $\rho_{2}={\rm sgn}$. Then
$\rho\in{\mathcal{E}}(G)$ is called _$\ell$ -admissible_ for the dual pair
$(\text{\bf G},\text{\bf G}^{\prime})$ if

     * •

$\rho_{\Lambda^{(-)}}$ is $(\ell+1)$-admissible for $(\text{\bf G}^{(-)},{\rm
O}^{\epsilon^{\prime}}_{\rm even})$ and

     * •

$\rho_{\Lambda^{(+)}}$ is $(\ell-1)$-admissible for $(\text{\bf G}^{(+)},{\rm
O}^{\epsilon^{\prime}}_{\rm even})$.

Note that the case when $\text{\bf G}={\rm O}^{\epsilon}_{k}$ and $k+\ell$ is
odd is undefined. It is clear that if $\rho$ is $\ell$-admissible then it is
also $\ell^{\prime}$-admissible (if it is defined) for any
$\ell^{\prime}\geq\ell$.

######  Lemma 5.2.

If the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$ or $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$ is in stable
range, then every irreducible character $\rho\in{\mathcal{E}}(G)$ is
$(n-k)$-admissible.

###### Proof.

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})$ is in stable range,
$\rho\in{\mathcal{E}}(G)_{s}$ for some $s$, and write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$ and $n$ is even. Now $n-k$ is even and $k\leq n-k$. Now
both $k^{(\varepsilon)}\leq k\leq n-k$, and so
$\rho_{\Lambda^{(\varepsilon)}}$ is $(n-k)$-admissible by Lemma 3.9 for both
$\varepsilon=+$ and $-$.

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm O}_{n})$
and $n$ is odd. Now $n-k$ is odd and $k\leq n-k-1$. Now both
$k^{(\varepsilon)}\leq k\leq n-k-1$, and so $\rho_{\Lambda^{(\varepsilon)}}$
is $(n-k-1)$-admissible by Lemma 3.9 for both $\varepsilon=+$ and $-$.

  3. (3)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ and $k$ is even. Now $n-k$ is even and $k\leq
n-k$. Now both $k^{(\varepsilon)}\leq k\leq n-k$, and so
$\rho_{\Lambda^{(\varepsilon)}}$ is $(n-k)$-admissible by Lemma 3.9 for both
$\varepsilon=+$ and $-$.

  4. (4)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}_{k},{\rm Sp}_{n})$
and $k$ is odd. Now $n-k$ is odd and $k\leq n-k$. Now both
$k^{(\varepsilon)}\leq k-1\leq n-k-1$, and so $\rho_{\Lambda^{(\varepsilon)}}$
is $(n-k-1)$-admissible by Lemma 3.9 for both $\varepsilon=+$ and $-$.

For all cases, we conclude that $\rho$ is $(n-k)$-admissible from the
definition above. ∎

Now Lemma 3.13 can be generalized to any irreducible characters:

######  Lemma 5.3.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$
where $k\leq n$ and $\epsilon=+$ or $-$. If $\rho\in{\mathcal{E}}(G)$ is
$(n-k)$-admissible, then $\underline{\theta}(\rho)$ is defined and of
$\Theta$-rank $k$.

###### Proof.

Suppose that $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$ is $(n-k)$-admissible.
Write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.
Now we consider the following cases:

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$ where both $k,n$ are even and $k\leq n$. By the definition
above, both $\rho_{\Lambda^{(-)}},\rho_{\Lambda^{(+)}}$ are
$(n-k)$-admissible. Now $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{k^{(-)}}$ and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$
for some $\epsilon^{\prime}=+$ or $-$ depending on $s$. Now $s$ is conjugate
to an element of the form $(t,1_{k^{(+)}})$ where $t$ does not have eigenvalue
$1$. Let $n^{(+)}$ be such that $n^{(+)}-k^{(+)}=n-k$ and so the unipotent
character $\rho_{\Lambda^{(+)}}$ is $(n^{(+)}-k^{(+)})$-admissible. Then
$\underline{\theta}(\rho_{\Lambda^{(+)}})=\rho_{\underline{\theta}(\Lambda^{(+)})}$
is defined for the dual pair $({\rm Sp}_{k^{(+)}},{\rm
O}^{\epsilon^{\prime}\epsilon}_{n^{(+)}})$. Hence $\underline{\theta}(\rho)$
is defined for the dual pair $(\text{\bf G},\text{\bf G}^{\prime})$ for some
$s^{\prime}\in G^{\prime*}$ via the following commutative diagram

(5.4) $\begin{CD}\rho @>{\underline{\theta}}>{}>\underline{\theta}(\rho)\\\
@V{\Xi_{s}}V{}V@V{}V{\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}@>{{\rm
id}\otimes{\rm
id}\otimes\underline{\theta}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\underline{\theta}(\Lambda^{(+)})}\\\
\end{CD}$

where $s^{\prime}$ is an element of the form $(t,1_{n^{(+)}})$ so that
$\text{\bf G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$, $\text{\bf
G}^{(-)}\simeq\text{\bf G}^{\prime(-)}$ and $(\text{\bf G}^{(+)},\text{\bf
G}^{\prime(+)})=({\rm Sp}_{k^{(+)}},{\rm
O}^{\epsilon^{\prime}\epsilon}_{n^{(+)}})$.

As in the proof of Lemma 3.13, the $(n^{(+)}-k^{(+)})$-admissibility of
$\rho_{\Lambda^{(+)}}$ implies that both unipotent characters
$\underline{\theta}(\rho_{\Lambda^{(+)}}),\underline{\theta}(\rho_{\Lambda^{(+)}}){\rm
sgn}$ of ${\rm O}^{\epsilon^{\prime}\epsilon}_{n^{(+)}}(q)$ do not occur in
the $\Theta$-correspondence for the dual pair $({\rm Sp}_{k^{\prime(+)}},{\rm
O}^{\epsilon^{\prime}\epsilon}_{n^{(+)}})$ for any $k^{\prime(+)}<k^{(+)}$.
Hence by [Pan19b] theorem 6.9 and remark 6.10, we see that both irreducible
characters $\underline{\theta}(\rho),\underline{\theta}(\rho){\rm sgn}$ of
${\rm O}^{\epsilon}_{n}(q)$ do not occur in the $\Theta$-correspondence for
the dual pair $({\rm Sp}_{k^{\prime}},{\rm O}^{\epsilon}_{n})$ for any
$k^{\prime}<k$.

We know that $\underline{\theta}(\rho)\chi_{{\rm
O}^{\epsilon}_{n}}\in{\mathcal{E}}(G^{\prime})_{-s^{\prime}}$, and if we write

$\Xi_{-s^{\prime}}(\underline{\theta}(\rho)\chi_{{\rm
O}^{\epsilon}_{n}})=\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}},$

then we have $\Lambda^{\prime(-)}=\underline{\theta}(\Lambda^{(+)})$ and
$\Lambda^{\prime(+)}=\Lambda^{(-)}$. By the same argument as in the previous
paragraph, the $(n^{(-)}-k^{(-)})$-admissibility of $\rho_{\Lambda^{(-)}}$
implies that both $\underline{\theta}(\rho)\chi_{{\rm
O}^{\epsilon}_{n}},\underline{\theta}(\rho)\chi_{{\rm O}^{\epsilon}_{n}}{\rm
sgn}$ of ${\rm O}^{\epsilon}_{n}(q)$ do not occur in the
$\Theta$-correspondence for the dual pair $({\rm Sp}_{k^{\prime\prime}},{\rm
O}^{\epsilon}_{n})$ for any $k^{\prime\prime}<k$.

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm O}_{n})$
where $k$ is even, $n$ is odd, and $k\leq n$. By definition, now both
$\rho_{\Lambda^{(-)}},\rho_{\Lambda^{(+)}}$ are $(n-k-1)$-admissible. Now
$\text{\bf G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{k^{(-)}}$ and $\text{\bf
G}^{(+)}={\rm Sp}_{k^{(+)}}$ for some $\epsilon^{\prime}=+$ or $-$ depending
on $s$. Note that now both $k^{(-)},k^{(+)}$ are even. Let $n^{(-)}$ be an
even integer such that $n^{(-)}-k^{(-)}=n-k-1$ and so the unipotent character
$\rho_{\Lambda^{(-)}}$ is $(n^{(-)}-k^{(-)})$-admissible. Then
$\underline{\theta}(\rho_{\Lambda^{(-)}})=\rho_{\underline{\theta}(\Lambda^{(-)})}$
is defined for the dual pair $({\rm O}^{\epsilon^{\prime}}_{k^{(-)}},{\rm
Sp}_{n^{(-)}})$. Hence $\underline{\theta}(\rho)$ is defined for the dual pair
$({\rm Sp}_{k},{\rm O}_{n})$ via the following commutative diagram

(5.5) $\begin{CD}\rho @>{\underline{\theta}}>{}>\underline{\theta}(\rho)\\\
@V{\Xi_{s}}V{}V@V{}V{\iota\circ\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}@>{{\rm
id}\otimes\underline{\theta}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\underline{\theta}(\Lambda^{(-)})}\otimes\rho_{\Lambda^{(-)}}\\\
\end{CD}$

where $s^{\prime}\in G^{\prime*}$ is an element so that $\text{\bf
G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$, $(\text{\bf G}^{(-)},\text{\bf
G}^{\prime(+)})=({\rm O}^{\epsilon^{\prime}}_{k^{(-)}},{\rm Sp}_{n^{(+)}})$,
and $\text{\bf G}^{(+)}\simeq\text{\bf G}^{\prime(-)}$, and
$\iota(\rho^{(0)}\otimes\rho^{(-)}\otimes\rho^{(+)})=\rho^{(0)}\otimes\rho^{(+)}\otimes\rho^{(-)}$.
The remaining argument is similar to that in case (1).

  3. (3)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ where both $k,n$ are even and $k\leq n$. Now
both $\rho_{\Lambda^{(-)}},\rho_{\Lambda^{(+)}}$ are $(n-k)$-admissible.
Moreover $\text{\bf G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{k^{(-)}}$ and
$\text{\bf G}^{(+)}={\rm O}^{\epsilon^{\prime}\epsilon}_{k^{(+)}}$ for some
$\epsilon^{\prime}=+$ or $-$ depending on $s$. Let $n^{(+)}$ be such that
$n^{(+)}-k^{(+)}=n-k$ and so the unipotent character $\rho_{\Lambda^{(+)}}$ is
$(n^{(+)}-k^{(+)})$-admissible. Then
$\underline{\theta}(\rho_{\Lambda^{(+)}})=\rho_{\underline{\theta}(\Lambda^{(+)})}$
is defined for the dual pair $({\rm Sp}_{k^{(+)}},{\rm
O}^{\epsilon^{\prime}\epsilon}_{n^{(+)}})$. Hence $\underline{\theta}(\rho)$
is defined for the dual pair $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$ via the
same diagram (5.4) where $s^{\prime}$ is an element so that $\text{\bf
G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$, $\text{\bf G}^{(-)}\simeq\text{\bf
G}^{\prime(-)}$ and $(\text{\bf G}^{(+)},\text{\bf G}^{\prime(+)})=({\rm
O}^{\epsilon^{\prime}\epsilon}_{k^{(+)}},{\rm Sp}_{n^{(+)}})$.

As in the proof of Lemma 3.13, the $(n^{(+)}-k^{(+)})$-admissibility of
$\rho_{\Lambda^{(+)}}$ implies that $\underline{\theta}(\rho_{\Lambda^{(+)}})$
of ${\rm Sp}_{n^{(+)}}(q)$ does not occur in the $\Theta$-correspondence for
the dual pair $({\rm O}^{\epsilon^{\prime\prime}}_{k^{\prime(+)}},{\rm
Sp}_{n^{(+)}})$ for any $\epsilon^{\prime\prime}=+$ or $-$ and any even
$k^{\prime(+)}<k^{(+)}$. Hence we see that the irreducible characters
$\underline{\theta}(\rho)$ of ${\rm Sp}_{n}(q)$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon^{\prime\prime}}_{k^{\prime}},{\rm Sp}_{n})$ for any
$\epsilon^{\prime\prime}=+$ or $-$ and any even integer $k^{\prime}$ such that
$k^{\prime}<k$.

Similarly, from diagram (5.5), the $(n^{(-)}-k^{(-)})$-admissibility of
$\rho_{\Lambda^{(-)}}$ implies that the $\underline{\theta}(\rho)$ of ${\rm
Sp}_{n}(q)$ does not occur in the $\Theta$-correspondence for the dual pair
$({\rm O}^{\epsilon^{\prime\prime}}_{k^{\prime\prime}},{\rm Sp}_{n})$ for any
$\epsilon^{\prime\prime}=+$ or $-$ and any odd integer $k^{\prime\prime}$ such
that $k^{\prime\prime}<k$.

  4. (4)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ where $k$ is odd, $n$ is even, and $k\leq n$.
By definition, now $\rho_{\Lambda^{(-)}}$ is $(n-k+1)$-admissible and
$\rho_{\Lambda^{(+)}}$ is $(n-k-1)$-admissible. Now $\text{\bf G}^{(-)}={\rm
Sp}_{k^{(-)}}$ and $\text{\bf G}^{(+)}={\rm Sp}_{k^{(+)}}$. Let $n^{(-)}$ be
such that $n^{(-)}-k^{(-)}=n-k+1$ and so the unipotent character
$\rho_{\Lambda^{(-)}}$ is $(n^{(-)}-k^{(-)})$-admissible. Then
$\underline{\theta}(\rho_{\Lambda^{(-)}})=\rho_{\underline{\theta}(\Lambda^{(-)})}$
is defined for the dual pair $({\rm Sp}_{k^{(-)}},{\rm
O}^{\epsilon^{\prime}}_{n^{(-)}})$. Hence $\underline{\theta}(\rho)$ is
defined for the dual pair $({\rm O}^{\epsilon}_{k},{\rm Sp}_{n})$ via the
commutative diagram

$\begin{CD}\rho @>{\underline{\theta}}>{}>\underline{\theta}(\rho)\\\
@V{\iota\circ\Xi_{s}}V{}V@V{}V{\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(+)}}\otimes\rho_{\Lambda^{(-)}}@>{{\rm
id}\otimes\underline{\theta}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\underline{\theta}(\Lambda^{(+)})}\otimes\rho_{\Lambda^{(+)}}\\\
\end{CD}$

where $s^{\prime}$ is an element so that $\text{\bf G}^{(0)}\simeq\text{\bf
G}^{\prime(0)}$, $(\text{\bf G}^{(+)},\text{\bf G}^{\prime(-)})=({\rm
Sp}_{k^{(+)}},{\rm O}^{\epsilon^{\prime}}_{n^{(-)}})$, and $\text{\bf
G}^{(-)}\simeq\text{\bf G}^{\prime(+)}$.

As in the proof as above, the $(n-k+1)$-admissibility of
$\rho_{\Lambda^{(-)}}$ implies that the irreducible characters
$\underline{\theta}(\rho)$ of ${\rm Sp}_{n}(q)$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon^{\prime\prime}}_{k^{\prime}},{\rm Sp}_{n})$ for any
$\epsilon^{\prime\prime}=+$ or $-$ and any odd integer $k^{\prime}$ such that
$k^{\prime}<k$. Moreover, the $(n-k-1)$-admissibility of
$\rho_{\Lambda^{(+)}}$ implies that the irreducible characters
$\underline{\theta}(\rho)$ of ${\rm Sp}_{n}(q)$ does not occur in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon^{\prime\prime}}_{k^{\prime\prime}},{\rm Sp}_{n})$ for any
$\epsilon^{\prime\prime}=+$ or $-$ and any even integer $k^{\prime\prime}$
such that $k^{\prime\prime}<k$.

For all cases, we conclude that $\Theta\text{\rm-
rk}(\underline{\theta}(\rho))=k$. ∎

######  Proposition 5.6.

Suppose that both $n,k$ are non-negative integers such that $k$ is even and
$k\leq n$. Then $\rho^{\prime}\in{\mathcal{E}}({\rm O}^{\epsilon}_{n}(q))$ is
of $\Theta$-rank $k$ if and only if there is a linear character $\chi$ of
${\rm O}^{\epsilon}_{n}(q)$ such that
$\rho^{\prime}\chi=\underline{\theta}(\rho)$ for some $(n-k)$-admissible
character $\rho\in{\mathcal{E}}({\rm Sp}_{k}(q))$.

###### Proof.

Let $\rho^{\prime}$ be an irreducible character of ${\rm
O}^{\epsilon}_{n}(q)$. Suppose that there is an $(n-k)$-admissible irreducible
character $\rho$ of ${\rm Sp}_{k}(q)$ and a linear character $\chi$ of ${\rm
O}^{\epsilon}_{n}(q)$ such that $\underline{\theta}(\rho)=\rho^{\prime}\chi$.
By Lemma 5.3, we know that $\Theta\text{\rm-rk}(\rho^{\prime}\chi)=k$ and
hence $\Theta\text{\rm-rk}(\rho^{\prime})=k$.

Conversely, suppose that $\rho^{\prime}\in{\mathcal{E}}({\rm
O}^{\epsilon}_{n}(q))$ such that $\Theta\text{\rm-rk}(\rho^{\prime})=k$. From
[Pan21a] we know that there exists a linear character $\chi$ of ${\rm
O}^{\epsilon}_{n}(q)$ such that $\rho^{\prime}\chi=\underline{\theta}(\rho)$
for some $\rho\in{\mathcal{E}}({\rm Sp}_{k}(q))$. Now we need to show that
$\rho$ is $(n-k)$-admissible. Suppose that $\rho\in{\mathcal{E}}({\rm
Sp}_{k}(q))_{s}$ for some $s$ and write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$
where $\Lambda^{(-)}\in{\mathcal{S}}_{{\rm O}^{\epsilon^{\prime}}_{k^{(-)}}}$
and $\Lambda^{(+)}\in{\mathcal{S}}_{{\rm Sp}_{k^{(+)}}}$ for some
$\epsilon^{\prime}=+$ or $-$ depending on $s$.

  1. (1)

Suppose that $n$ is even. If $\rho_{\Lambda^{(+)}}$ is not $(n-k)$-admissible,
then by the proof of Proposition 3.14 we see that $\rho^{\prime}\chi$ or
$\rho^{\prime}\chi{\rm sgn}$ have $\Theta$-rank less than $k$. Similarly, if
$\rho_{\Lambda^{(-)}}$ is not $(n-k)$-admissible, then
$\rho^{\prime}\chi\chi_{{\rm O}^{\epsilon}_{n}}$ or
$\rho^{\prime}\chi\chi_{{\rm O}^{\epsilon}_{n}}{\rm sgn}$ will have
$\Theta$-rank less than $k$.

  2. (2)

Suppose that $n$ is odd. If $\rho_{\Lambda^{(-)}}$ is not
$(n-k-1)$-admissible, then $\rho^{\prime}\chi$ or $\rho^{\prime}\chi{\rm sgn}$
will have $\Theta$-rank less than $k$. Similarly, if $\rho_{\Lambda^{(+)}}$ is
not $(n-k-1)$-admissible, then $\rho^{\prime}\chi\chi_{{\rm
O}^{\epsilon}_{n}}$ or $\rho^{\prime}\chi\chi_{{\rm O}^{\epsilon}_{n}}{\rm
sgn}$ will have $\Theta$-rank less than $k$.

Therefore, we conclude that $\rho$ must be $(n-k)$-admissible. ∎

######  Proposition 5.7.

Suppose that both $n,k$ are non-negative integers such that $n$ is even and
$k\leq n$. Then $\rho^{\prime}\in{\mathcal{E}}({\rm Sp}_{n}(q))$ is of
$\Theta$-rank $k$ is and only if $\rho^{\prime}=\underline{\theta}(\rho)$ for
some $(n-k)$-admissible character $\rho\in{\mathcal{E}}({\rm
O}^{\epsilon}_{k}(q))$ and some $\epsilon=+$ or $-$.

###### Proof.

Let $\rho^{\prime}$ be an irreducible character of ${\rm Sp}_{n}(q)$. Suppose
that there is an $(n-k)$-admissible irreducible character $\rho$ of ${\rm
O}^{\epsilon}_{k}(q)$ for $\epsilon=+$ or $-$ such that
$\underline{\theta}(\rho)=\rho^{\prime}$. By Lemma 5.3, we know that
$\Theta\text{\rm-rk}(\rho^{\prime})=k$.

Conversely, suppose that $\rho^{\prime}\in{\mathcal{E}}({\rm Sp}_{n}(q))$ has
$\Theta$-rank $k$. Then there is an orthogonal group ${\rm
O}^{\epsilon}_{k}(q)$ and $\rho\in{\mathcal{E}}({\rm O}^{\epsilon}_{k}(q))$
such that $\underline{\theta}(\rho)=\rho^{\prime}$. Suppose that
$\rho\in{\mathcal{E}}({\rm O}^{\epsilon}_{k}(q))_{s}$ for some $s$ and write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.

  1. (1)

Suppose that $k$ is even. Then $\Lambda^{(-)}\in{\mathcal{S}}_{{\rm
O}^{\epsilon^{\prime}}_{k^{(-)}}}$ and $\Lambda^{(+)}\in{\mathcal{S}}_{{\rm
O}^{\epsilon^{\prime}\epsilon}_{k^{(+)}}}$ for some $\epsilon^{\prime}=+$ or
$-$ depending on $s$. If $\rho_{\Lambda^{(+)}}$ is not $(n-k)$-admissible,
then $\rho^{\prime}$ will occur in the $\Theta$-correspondence for the dual
pair $({\rm O}^{\epsilon}_{k^{\prime}},{\rm Sp}_{n})$ for some even integer
$k^{\prime}<k$. Similarly, if $\rho_{\Lambda^{(-)}}$ is not
$(n-k)$-admissible, then $\rho^{\prime}$ will occur in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon}_{k^{\prime}},{\rm Sp}_{n})$ for some odd integer $k^{\prime}<k$.

  2. (2)

Suppose that $k$ is odd. Then $\Lambda^{(-)}\in{\mathcal{S}}_{{\rm
Sp}_{k^{(-)}}}$ and $\Lambda^{(+)}\in{\mathcal{S}}_{{\rm Sp}_{k^{(+)}}}$. If
$\rho_{\Lambda^{(+)}}$ is not $(n-k-1)$-admissible, then $\rho^{\prime}$ will
occur in the $\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon}_{k^{\prime}},{\rm Sp}_{n})$ for some odd integer $k^{\prime}<k$.
Similarly, if $\rho_{\Lambda^{(-)}}$ is not $(n-k+1)$-admissible, then
$\rho^{\prime}$ will occur in the $\Theta$-correspondence for the dual pair
$({\rm O}^{\epsilon}_{k^{\prime}},{\rm Sp}_{n})$ for some even integer
$k^{\prime}<k$.

Therefore, $\rho$ is $(n-k)$-admissible and so the proposition is proved. ∎

Now we generalize Proposition 3.16 to general irreducible characters.

######  Proposition 5.8.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$
where $k\leq n$. If $\rho\in{\mathcal{E}}(G)$ is $(n-k)$-admissible, then

$\deg_{q}(\underline{\theta}(\rho))=\deg_{q}(\rho)+\begin{cases}\frac{1}{2}k(n-k+1),&\text{if\/
$(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}^{\epsilon}_{k},{\rm
Sp}_{n})$};\\\ \frac{1}{2}k(n-k-1),&\text{if\/ $(\text{\bf G},\text{\bf
G}^{\prime})=({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$}.\end{cases}$

###### Proof.

For $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$, we write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.
From [Lus84] 4.23 (see also [DM91] remark 13.24), it is known that

(5.9)
$\rho(1)=\frac{|G|_{p^{\prime}}}{|G^{(0)}|_{p^{\prime}}|G^{(-)}|_{p^{\prime}}|G^{(+)}|_{p^{\prime}}}\rho^{(0)}(1)\rho_{\Lambda^{(-)}}(1)\rho_{\Lambda^{(+)}}(1).$

  1. (1)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$ or $({\rm Sp}_{k},{\rm O}^{\epsilon}_{n})$
where both $k,n$ are even. Now we have a commutative diagram

$\begin{CD}\rho @>{\underline{\theta}}>{}>\rho^{\prime}\\\
@V{\Xi_{s}}V{}V@V{}V{\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}@>{{\rm
id}\otimes{\rm
id}\otimes\underline{\theta}}>{}>\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}\\\
\end{CD}$

We know that $\text{\bf G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$,
$\rho^{(0)}=\rho^{\prime(0)}$ and $\text{\bf G}^{(-)}\simeq\text{\bf
G}^{\prime(-)}$, $\rho_{\Lambda^{(-)}}=\rho_{\Lambda^{\prime(-)}}$. Therefore
by (5.9), we have

$\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)=\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(+)}}))-\deg_{q}(\rho_{\Lambda^{(+)}})\\\
+\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})-\deg_{q}(|G^{\prime(+)}|_{p^{\prime}})+\deg_{q}(|G^{(+)}|_{p^{\prime}}).$

    1. (a)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
O}^{\epsilon}_{k},{\rm Sp}_{n})$. Then $\text{\bf G}^{(+)}={\rm
O}^{\epsilon}_{k-2l}$ and $\text{\bf G}^{\prime(+)}={\rm Sp}_{n-2l}$ for some
non-negative integer $l$. Hence by Proposition 3.16, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(+)}}))-\deg_{q}(\rho_{\Lambda^{(+)}})$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-2l-(k-2l)+1),$
$\displaystyle\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})$
$\displaystyle=\tfrac{1}{4}n(n+2)-\tfrac{1}{4}k^{2},$
$\displaystyle-\deg_{q}(|G^{\prime(+)}|_{p^{\prime}})+\deg_{q}(|G^{(+)}|_{p^{\prime}})$
$\displaystyle=-\tfrac{1}{4}(n-2l)(n-2l+21)+\tfrac{1}{4}(k-2l)^{2}.$

Therefore,

$\displaystyle\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-k+1)+\tfrac{1}{4}n(n+2)-\tfrac{1}{4}k^{2}-\tfrac{1}{4}(n-2l)(n-2l+2)+\tfrac{1}{4}(k-2l)^{2}$
$\displaystyle=\tfrac{1}{2}k(n-k+1).$

    2. (b)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm
O}^{\epsilon}_{n})$. Then $\text{\bf G}^{(+)}={\rm Sp}_{k-2l}$ and $\text{\bf
G}^{\prime(+)}={\rm O}^{\epsilon}_{n-2l}$ for some non-negative integer $l$.
Hence by Proposition 3.16, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(+)}}))-\deg_{q}(\rho_{\Lambda^{(+)}})$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-2l-(k-2l)-1),$
$\displaystyle\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})$
$\displaystyle=\tfrac{1}{4}n^{2}-\tfrac{1}{4}k(k+2),$
$\displaystyle-\deg_{q}(|G^{\prime(+)}|_{p^{\prime}})+\deg_{q}(|G^{(+)}|_{p^{\prime}})$
$\displaystyle=-\tfrac{1}{4}(n-2l)^{2}+\tfrac{1}{4}(k-2l)(k-2l+2).$

Therefore

$\displaystyle\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-k-1)+\tfrac{1}{4}n^{2}-\tfrac{1}{4}k(k+2)-\tfrac{1}{4}(n-2l)^{2}+\tfrac{1}{4}(k-2l)(k-2l+21)$
$\displaystyle=\tfrac{1}{2}k(n-k-1).$

  2. (2)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}_{k},{\rm Sp}_{n})$
where $k$ is odd and $n$ even; or $({\rm Sp}_{k},{\rm O}_{n})$ where $k$ is
even and $n$ is odd.

    1. (a)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm O}_{k},{\rm Sp}_{n})$
where $k$ is odd and $n$ is even. Now we have a commutative diagram

$\begin{CD}\rho @>{\underline{\theta}}>{}>\rho^{\prime}\\\
@V{\iota\circ\Xi_{s}}V{}V@V{}V{\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(+)}}\otimes\rho_{\Lambda^{(-)}}@>{{\rm
id}\otimes\underline{\theta}\otimes{\rm
id}}>{}>\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}\\\
\end{CD}$

We know that $\text{\bf G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$,
$\rho^{(0)}=\rho^{\prime(0)}$ and $\text{\bf G}^{(-)}\simeq\text{\bf
G}^{\prime(+)}$, $\rho_{\Lambda^{(-)}}=\rho_{\Lambda^{\prime(+)}}$. Therefore
by (5.9), we have

$\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)=\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(+)}}))-\deg_{q}(\rho_{\Lambda^{(+)}})\\\
+\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})-\deg_{q}(|G^{\prime(-)}|_{p^{\prime}})+\deg_{q}(|G^{(+)}|_{p^{\prime}}).$

Then $\text{\bf G}^{(+)}={\rm Sp}_{k-2l-1}$ and $\text{\bf G}^{\prime(-)}={\rm
O}^{\epsilon}_{n-2l}$ for some non-negative integer $l$. Hence by Proposition
3.16, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(-)}}))-\deg_{q}(\rho_{\Lambda^{(-)}})$
$\displaystyle=\tfrac{1}{2}(k-2l-1)(n-2l-(k-2l-1)-1),$
$\displaystyle\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})$
$\displaystyle=\tfrac{1}{4}n(n+2)-\tfrac{1}{4}(k-1)(k+1),$
$\displaystyle-\deg_{q}(|G^{\prime(-)}|_{p^{\prime}})+\deg_{q}(|G^{(+)}|_{p^{\prime}})$
$\displaystyle=-\tfrac{1}{4}(n-2l)^{2}+\tfrac{1}{4}(k-1-2l)(k+1-2l).$

Therefore

$\displaystyle\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)$
$\displaystyle=\tfrac{1}{2}(k-2l-1)(n-k)+\tfrac{1}{4}n(n+2)-\tfrac{1}{4}(k-1)(k+1)$
$\displaystyle\qquad\qquad\qquad\qquad-\tfrac{1}{4}(n-2l)^{2}+\tfrac{1}{4}(k-1-2l)(k+1-2l)$
$\displaystyle=\tfrac{1}{2}k(n-k+1).$

    2. (b)

Suppose that $(\text{\bf G},\text{\bf G}^{\prime})=({\rm Sp}_{k},{\rm O}_{n})$
where $k$ is even and $n$ is odd. Now we have a commutative diagram

$\begin{CD}\rho @>{\underline{\theta}}>{}>\rho^{\prime}\\\
@V{\Xi_{s}}V{}V@V{}V{\iota\circ\Xi_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}@>{{\rm
id}\otimes\underline{\theta}\otimes{\rm
id}}>{}>\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(+)}}\otimes\rho_{\Lambda^{\prime(-)}}\\\
\end{CD}$

We know that $\text{\bf G}^{(0)}\simeq\text{\bf G}^{\prime(0)}$,
$\rho^{(0)}=\rho^{\prime(0)}$ and $\text{\bf G}^{(+)}\simeq\text{\bf
G}^{\prime(-)}$, $\rho_{\Lambda^{(+)}}=\rho_{\Lambda^{\prime(-)}}$. Therefore
by (5.9), we have

$\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)=\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(-)}}))-\deg_{q}(\rho_{\Lambda^{(-)}})\\\
+\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})-\deg_{q}(|G^{\prime(+)}|_{p^{\prime}})+\deg_{q}(|G^{(-)}|_{p^{\prime}}).$

Then $\text{\bf G}^{(-)}={\rm O}^{\epsilon}_{k-2l}$ and $\text{\bf
G}^{\prime(+)}={\rm Sp}_{n-2l-1}$ for some non-negative integer $l$. Hence by
Proposition 3.16, we have

$\displaystyle\deg_{q}(\underline{\theta}(\rho_{\Lambda^{(-)}}))-\deg_{q}(\rho_{\Lambda^{(-)}})$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-2l-1-(k-2l)+1),$
$\displaystyle\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})$
$\displaystyle=\tfrac{1}{4}(n-1)(n+1)-\tfrac{1}{4}k(k+2),$
$\displaystyle-\deg_{q}(|G^{\prime(+)}|_{p^{\prime}})+\deg_{q}(|G^{(-)}|_{p^{\prime}})$
$\displaystyle=-\tfrac{1}{4}(n-1-2l)(n+1-2l)+\tfrac{1}{4}(k-2l)^{2}.$

Therefore

$\displaystyle\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)$
$\displaystyle=\tfrac{1}{2}(k-2l)(n-k)+\tfrac{1}{4}(n-1)(n+1)-\tfrac{1}{4}k(k+2)$
$\displaystyle\qquad\qquad\qquad\quad-\tfrac{1}{4}(n-1-2l)(n+1-2l)+\tfrac{1}{4}(k-2l)^{2}$
$\displaystyle=\tfrac{1}{2}k(n-k-1).$

∎

###  5.2. Lusztig correspondence for unitary groups

Let $\text{\bf G}={\rm U}_{k}$. For $s\in G^{*}={\rm U}_{k}(q)$, we can write

(5.10) $C_{G^{*}}(s)=\prod_{i=1}^{r}\prod_{j=1}^{t_{i}}{\rm
GL}_{k_{ij}}^{(-1)^{i}}(q^{i})$

for some non-negative integers $r,t_{i},k_{ij}$ such that
$\sum_{i=1}^{r}\sum_{j=1}^{t_{i}}ik_{ij}=k$ where ${\rm GL}^{+1}_{k_{ij}}={\rm
GL}^{+}_{k_{ij}}:={\rm GL}_{k_{ij}}$ and ${\rm GL}^{-1}_{k_{ij}}={\rm
GL}^{-}_{k_{ij}}:={\rm U}_{k_{ij}}$. For $i=1$, we can let $t_{1}=q+1$ and
each $j=1,\ldots,t_{1}$ corresponds an eigenvalue
$\lambda_{j}\in\overline{\text{\bf f}}_{q}$ of $s$ such that
$\lambda_{j}^{q+1}=1$. Let $\rho\in{\mathcal{E}}(G)_{s}$ and for the Lusztig
correspondence
${\mathfrak{L}}_{s}\colon{\mathcal{E}}(G)_{1}\rightarrow{\mathcal{E}}(C_{G^{*}}(s))_{1}$
we can write

(5.11)
${\mathfrak{L}}_{s}(\rho)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{(ij)}$

for some $\rho^{(ij)}\in{\mathcal{E}}({\rm
GL}_{k_{ij}}^{(-1)^{i}}(q^{i}))_{1}$. Note that $\rho^{(1j)}$ is a unipotent
character of ${\rm U}_{k_{1j}}(q)$.

Now for a non-negative integer $\ell$, an irreducible character
$\rho\in{\mathcal{E}}({\rm U}_{k}(q))$ is called _$\ell$ -admissible_ if each
$\rho^{(1j)}$ is $\ell$-admissible for $j=1,\ldots,t_{1}$. It is clear that if
$\rho$ is $\ell$-admissible, then $\rho$ is also $\ell^{\prime}$-admissible
for any $\ell^{\prime}\geq\ell$, and $\rho\chi$ is also $\ell$-admissible for
any linear character $\chi$ of ${\rm U}_{k}(q)$.

######  Lemma 5.12.

If the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ is in stable range, then every irreducible character
$\rho\in{\mathcal{E}}(G)$ is $(n-k)$-admissible.

###### Proof.

Suppose that the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm
U}_{k},{\rm U}_{n})$ is in stable range, and let $\rho\in{\mathcal{E}}(G)_{s}$
for some $s$. Write
${\mathfrak{L}}_{s}(\rho)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{(ij)}$
as above. Let $k_{ij}$ be given as in (5.10). Then $k_{1j}\leq k\leq n-k$ and
hence $\rho^{(1j)}$ is $(n-k)$-admissible by Lemma 4.5 for each
$j=1,\ldots,t_{1}$. Hence $\rho$ is $(n-k)$-admissible by the definition
above. ∎

######  Lemma 5.13.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ where $k\leq n$. If $\rho\in{\mathcal{E}}(G)$ is $(n-k)$-admissible,
then $\underline{\theta}(\rho)$ is defined and of $\Theta$-rank $k$.

###### Proof.

Let $\rho$ be an $(n-k)$-admissible character of ${\rm U}_{k}(q)$. Suppose
that $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$. Let $\text{\bf
G}^{(1)}=U_{k_{1,1}}$ be the component in (5.10) corresponding the eigenvalue
$1$, and let $\text{\bf G}^{(0)}$ be the product of the other components. So
we have $C_{G^{*}}(s)=G^{(0)}\times G^{(1)}$ and
${\mathfrak{L}}_{s}(\rho)=\rho^{(0)}\otimes\rho^{(1)}$ where
$\rho^{(1)}=\rho^{(1,1)}$ and $\rho^{(0)}$ is the tensor product of the
components $\rho^{(ij)}$ with $(i,j)\neq(1,1)$.

Let $n_{1,1}$ be such that $n_{1,1}-k_{1,1}=n-k$, so the unipotent character
$\rho^{(1,1)}$ is $(n_{1,1}-k_{1,1})$-admissible. Then
$\underline{\theta}(\rho^{(1)})$ is defined for the dual pair $({\rm
U}_{k_{1,1}},{\rm U}_{n_{1,1}})$ by Lemma 4.8. Hence
$\underline{\theta}(\rho)$ is defined for the dual pair $({\rm U}_{k},{\rm
U}_{n})$ via the following commutative diagram

(5.14) $\begin{CD}\rho @>{\underline{\theta}}>{}>\rho^{\prime}\\\
@V{{\mathfrak{L}}_{s}}V{}V@V{}V{{\mathfrak{L}}_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho^{(1)}@>{{\rm
id}\otimes\underline{\theta}}>{}>\rho^{(0)}\otimes\underline{\theta}(\rho^{(1)})\\\
\end{CD}$

where $s^{\prime}=(s,1_{n-k})\in{\rm U}_{k}(q)\times{\rm
U}_{n-k}(q)\subset{\rm U}_{n}(q)$.

Let $\chi$ be a linear character of ${\rm U}_{n}(q)$ and suppose that
$\underline{\theta}(\rho)\chi$ is in ${\mathcal{E}}({\rm
U}_{n}(q))_{s^{\prime\prime}}$ for some $s^{\prime\prime}$. Note that if we
write

${\mathfrak{L}}_{s^{\prime}}(\underline{\theta}(\rho))=\bigotimes_{i=1}^{r^{\prime}}\bigotimes_{j=1}^{t^{\prime}_{i}}\rho^{\prime(ij)},\qquad{\mathfrak{L}}_{s^{\prime\prime}}(\underline{\theta}(\rho)\chi)=\bigotimes_{i=1}^{r^{\prime\prime}}\bigotimes_{j=1}^{t^{\prime\prime}_{i}}\rho^{\prime\prime(ij)},$

then we know that two ordered set of the components

$\\{\rho^{\prime(1,1)},\rho^{\prime(1,2)},\ldots,\rho^{\prime(1t^{\prime}_{1})}\\}\quad\text{and}\quad\\{\rho^{\prime\prime(1,1)},\rho^{\prime\prime(1,2)},\ldots,\rho^{\prime\prime(1t^{\prime\prime}_{1})}\\}$

are different by a permutation. Now the assumption that each $\rho^{(1j)}$ is
$(n-k)$-admissible means that $\underline{\theta}(\rho)\chi$ of ${\rm
U}_{n}(q)$ does not occurs in the $\Theta$-correspondence for the dual pair
$({\rm U}_{k^{\prime}},{\rm U}_{n})$ for any $k^{\prime}<k$. This implies that
$\Theta\text{\rm-rk}(\underline{\theta}(\rho))=k$. ∎

######  Proposition 5.15.

Let $n,k$ be non-negative integers such that $k\leq n$. Then
$\rho^{\prime}\in{\mathcal{E}}({\rm U}_{n}(q))$ is of $\Theta$-rank $k$ if and
only if there is a linear character $\chi$ of ${\rm U}_{n}(q)$ such that
$\rho^{\prime}\chi=\underline{\theta}(\rho)$ for some $(n-k)$-admissible
character $\rho\in{\mathcal{E}}({\rm U}_{k}(q))$.

###### Proof.

The proof is similar to those of Proposition 5.6 or Proposition 5.7. ∎

Now we generalize Proposition 4.12 to general irreducible characters.

######  Proposition 5.16.

Consider the dual pair $(\text{\bf G},\text{\bf G}^{\prime})=({\rm U}_{k},{\rm
U}_{n})$ where $k\leq n$. If an irreducible character
$\rho\in{\mathcal{E}}({\rm U}_{k}(q))$ is $(n-k)$-admissible, then

$\deg_{q}(\underline{\theta}(\rho))=\deg_{q}(\rho)+k(n-k).$

###### Proof.

Let $\rho$ be an $(n-k)$-admissible irreducible character of ${\rm U}_{k}(q)$.
Suppose that $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$ and write
${\mathfrak{L}}_{s}(\rho)=\rho^{(0)}\otimes\rho^{(1)}$. Let
$\rho^{\prime}=\underline{\theta}(\rho)\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$
for some $s^{\prime}$ and write
${\mathfrak{L}}_{s^{\prime}}(\rho^{\prime})=\rho^{\prime(0)}\otimes\rho^{\prime(1)}$
as above. Now we have a commutative diagram

$\begin{CD}\rho @>{\underline{\theta}}>{}>\rho^{\prime}\\\
@V{{\mathfrak{L}}_{s}}V{}V@V{}V{{\mathfrak{L}}_{s^{\prime}}}V\\\
\rho^{(0)}\otimes\rho^{(1)}@>{{\rm
id}\otimes\underline{\theta}}>{}>\rho^{\prime(0)}\otimes\rho^{\prime(1)}\\\
\end{CD}$

We know that

$\displaystyle\deg_{q}(\underline{\theta}(\rho))$
$\displaystyle=|G^{\prime}|_{p^{\prime}}+\deg_{q}(\rho^{\prime(0)})+\deg_{q}(\rho^{\prime(1)})-|G^{\prime(0)}|_{p^{\prime}}-|G^{\prime(1)}|_{p^{\prime}},$
$\displaystyle\deg_{q}(\rho)$
$\displaystyle=|G|_{p^{\prime}}+\deg_{q}(\rho^{(0)})+\deg_{q}(\rho^{(1)})-|G^{(0)}|_{p^{\prime}}-|G^{(1)}|_{p^{\prime}}.$

Now we have $\text{\bf G}^{(1)}={\rm U}_{k-l}$ and $\text{\bf
G}^{\prime(1)}={\rm U}_{n-l}$ for some non-negative integer $l\leq k$,
$G^{(0)}\simeq G^{\prime(0)}$, $\rho^{(0)}=\rho^{\prime(0)}$. Then by
Proposition 4.12, we have

$\displaystyle\deg_{q}(\rho^{\prime(1)})-\deg_{q}(\rho^{(1)})$
$\displaystyle=(k-l)(n-l-(k-l))=(k-l)(n-k),$
$\displaystyle\deg_{q}(|G^{\prime}|_{p^{\prime}})-\deg_{q}(|G|_{p^{\prime}})$
$\displaystyle=\tfrac{1}{2}n(n+1)-\tfrac{1}{2}k(k+1),$
$\displaystyle-\deg_{q}(|G^{\prime(1)}|_{p^{\prime}})+\deg_{q}(|G^{(1)}|_{p^{\prime}})$
$\displaystyle=-\tfrac{1}{2}(n-l)(n-l+1)+\tfrac{1}{2}(k-l)(k-l+1).$

Therefore

$\displaystyle\deg_{q}(\underline{\theta}(\rho))-\deg_{q}(\rho)$
$\displaystyle=(k-l)(n-k)+\tfrac{1}{2}n(n+1)-\tfrac{1}{2}k(k+1)-\tfrac{1}{2}(n-l)(n-l+1)+\tfrac{1}{2}(k-l)(k-l+1)$
$\displaystyle=k(n-k).$

∎

###  5.3. Proofs of the main results

###### Proof of Theorem 1.1.

Part (i) is Lemma 5.3 and Lemma 5.13. Part (ii) is Proposition 5.6,
Proposition 5.7 and Proposition 5.15. ∎

###### Proof of Theorem 1.3.

The theorem is Proposition 5.8 and Proposition 5.16. ∎

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