# On Theta Ranks of Irreducible Characters of a Finite Classical Group

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

In this paper, we interpret the theta rank of an irreducible character of a
finite classical group in terms of the data from the Lusztig classification.
Then we prove the following two results: (1) the agreement of the $U$-rank and
the theta rank for any irreducible character of low Theta rank; (2) the
agreement of the parabolic asymptotic rank and the theta rank for any
irreducible character.

###### Key words and phrases:

theta rank, theta correspondence, eta correspondence, reductive dual pair

###### 2010 Mathematics Subject Classification:

Primary: 20C33

###### Contents

  1. 1 Introduction
  2. 2 Theta Correspondence and Lusztig Correspondence
  3. 3 $\Theta$-ranks of Irreducible Characters
  4. 4 $\Theta$-ranks and Other Ranks

##  1\. Introduction

### 1.1.

Let $\text{\bf G}=\text{\bf G}_{n}$ be a finite symplectic group ${\rm
Sp}_{2n}$, a finite orthogonal group ${\rm O}^{+}_{2n}$, ${\rm O}^{-}_{2n+2}$,
${\rm O}_{2n+1}$, or a finite unitary group ${\rm U}_{2n}$, ${\rm U}_{2n+1}$
defined over a finite field $\text{\bf f}_{q}$ where $q$ is a power of an odd
prime. Let $G$ denote the finite group of rational points of G. For an
irreducible character $\rho$ of $G_{n}$, several notions of “rank” attached to
$\rho$ are defined in [GH20], namely the _$U$ -rank_ (denoted by $U\text{\rm-
rk}(\rho)$), the _asymptotic rank_ (denoted by $A\text{\rm-rk}(\rho)$), and
the _tensor rank_ (denoted by $\otimes\text{\rm-rk}(\rho)$). It is known that
$0\leq U\text{\rm-rk}(\rho)\leq n$. An irreducible character $\rho$ of $G_{n}$
is said to be _of low $U$-rank_ if $U\text{\rm-rk}(\rho)<n$ (resp.
$U\text{\rm-rk}(\rho)<n-1$) if $G_{n}$ is a symplectic group or a unitary
group (resp. an orthogonal group).

Because the tensor rank of $\rho$ can be interpreted as the first occurrence
in the theta correspondence for some dual pair involving $G$, in this article
the tensor rank will be called the _$\Theta$ -rank_ and denoted by
$\Theta\text{\rm-rk}(\rho)$ (cf. Subsection 3.1). Note that $\Theta\text{\rm-
rk}(\rho)$ is always even if $G$ is an orthogonal group. A similar notion of
the $\Theta$-rank which is called _level_ and denoted by
${\mathfrak{l}}(\rho)$ is also defined in [GLT20] and [GLT19]. Due to
different normalizations, we can see that

${\mathfrak{l}}(\rho)=\begin{cases}\Theta\text{\rm-rk}(\rho),&\text{if $G$ is
a symplectic group or a unitary group;}\\\ \frac{1}{2}\Theta\text{\rm-
rk}(\rho),&\text{if $G$ is an orthogonal group.}\end{cases}$

Moreover, we will modify the definition of the asymptotic ranks and define the
_parabolic asymptotic rank_ (denoted by $\overline{A}\text{\rm-rk}(\rho)$, cf.
Subsection 4.1).

Of course, it is expected these ranks are closely related. In fact, one can
see that

$U\text{\rm-rk}(\rho)\leq A\text{\rm-rk}(\rho)\leq\overline{A}\text{\rm-
rk}(\rho)\leq\otimes\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho)$

(cf. (4.10)) for any $\rho\in{\mathcal{E}}(G)$. Moreover, Gurevich-Howe make
the following conjecture in [GH20]:

######  Conjecture 1.1 (Gurevich-Howe).

Let $\rho$ be an irreducible character of a finite classical group $G$. Then

  1. (i)

$A\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho)$;

  2. (ii)

if $\rho$ is of low $U$-rank, then $U\text{\rm-rk}(\rho)=\Theta\text{\rm-
rk}(\rho)$.

As indicated in [GH20], if the conjecture is true, the theta correspondence
provides a method to construct an irreducible character of a given $U$-rank or
$A$-rank. When $G$ is a general linear group, (i) of the conjecture is already
proved in [GH20] theorem 11.4. One of the purposes of the article is to
provide similar results for other classical groups.

### 1.2.

Let ${\mathcal{E}}(G)$ denote the set of irreducible characters of $G$. It is
known that the set ${\mathcal{E}}(G)$ partitions into subsets
${\mathcal{E}}(G)_{s}$, called _Lusztig series_ , indexed by the set of
conjugacy classes of semisimple elements $s$ in the connected component of the
dual group $G^{*}$ of $G$ (cf. [Lus84]). Elements in ${\mathcal{E}}(G)_{1}$
are called _unipotent_. It is known by Lusztig that unipotent characters
$\rho=\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ are parametrized by the
equivalence classes of _symbols_

$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{\text{\bf
G}}$

satisfying certain conditions where $\\{a_{i}\\}$, $\\{b_{j}\\}$ are finite
sets of non-negative integers written in decreasing order, i.e,
$a_{1}>a_{2}>\cdots>a_{m_{1}}$ and $b_{1}>b_{2}>\cdots>b_{m_{2}}$ (cf.
[Lus77]). Note that by replacing $\Lambda$ by another symbol in its
equivalence class if necessary, we can always assume that both rows
$\\{a_{i}\\},\\{b_{j}\\}$ of $\Lambda$ are non-empty. Our first result (cf.
Proposition 3.4) is to interpret the $\Theta$-rank of a unipotent character
$\rho_{\Lambda}$ in terms of the invariants of symbols:

(1.2) $\Theta\text{\rm-rk}(\rho_{\Lambda})=\begin{cases}\min\\{2\,{\rm
rk}(\Lambda^{\rm t}\smallsetminus\textstyle\binom{-}{a_{1}}),2\,{\rm
rk}(\Lambda^{\rm t}\smallsetminus\textstyle\binom{b_{1}}{-})\\},&\text{if
$\text{\bf G}$ is ${\rm Sp}$ or ${\rm O}$};\\\ \min\\{{\rm rk}_{\rm
U}(\Lambda^{\rm t}\smallsetminus\textstyle\binom{-}{a_{1}}),{\rm rk}_{\rm
U}(\Lambda^{\rm t}\smallsetminus\textstyle\binom{b_{1}}{-})\\}&\text{if
$\text{\bf G}$ is ${\rm U}$}\end{cases}$

where

$\displaystyle\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}^{\rm
t}$
$\displaystyle=\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}$
$\displaystyle{\rm rk}(\Lambda)$
$\displaystyle=\sum_{i=1}^{m_{1}}a_{i}+\sum_{j=1}^{m_{2}}b_{j}-\left\lfloor\left(\frac{m_{1}+m_{2}-1}{2}\right)^{2}\right\rfloor,$
$\displaystyle{\rm rk}_{\rm U}(\Lambda)$
$\displaystyle=\sum_{i=1}^{m_{1}}2a_{i}+\sum_{j=1}^{m_{2}}2b_{j}+\frac{|m_{1}-m_{2}|}{2}-\frac{(m_{1}+m_{2})(m_{1}+m_{2}-2)}{2}.$

From (1.2), we see that $\Theta\text{\rm-rk}(\rho_{\Lambda})$ is always even
if G is a symplectic group or an orthogonal group.

### 1.3.

Next we want to investigate the $\Theta$-rank of a general irreducible
character. First suppose G is a symplectic group or an orthogonal group. For
$s\in(G^{*})^{0}$, we can define $\text{\bf G}^{(0)},\text{\bf
G}^{(-)},\text{\bf G}^{(+)}$ (cf. Subsection 2.4) where $\text{\bf G}^{(0)}$
is a product of unitary groups or general linear groups, and $\text{\bf
G}^{(-)},\text{\bf G}^{(+)}$ are given as follows:

  * •

if $\text{\bf G}={\rm O}^{\epsilon}_{2n}$, then $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
O}^{\epsilon^{\prime}\epsilon}_{2n^{(+)}}$;

  * •

if $\text{\bf G}={\rm Sp}_{2n}$, then $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
Sp}_{2n^{(+)}}$;

  * •

if $\text{\bf G}={\rm O}_{2n+1}$, then $\text{\bf G}^{(-)}={\rm
Sp}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm Sp}_{2n^{(+)}}$,

for some $\epsilon^{\prime}=\pm$, and $n^{(-)},n^{(+)}$ depending on $s$. Then
we have a (modified) _Lusztig correspondence_

$\Xi_{s}\colon{\mathcal{E}}(G)_{s}\rightarrow{\mathcal{E}}(G^{(0)}\times
G^{(-)}\times G^{(+)})_{1}.$

The mapping $\Xi_{s}$ is two-to-one: $\Xi_{s}(\rho)=\Xi_{s}(\rho\cdot{\rm
sgn})$ when G is an odd-orthogonal group; and $\Xi_{s}$ is bijective
otherwise. Then we can write

$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$

for $\rho\in{\mathcal{E}}(G)_{s}$ where
$\rho^{(0)}\in{\mathcal{E}}(G^{(0)})_{1}$,
$\Lambda^{(-)}\in{\mathcal{S}}_{\text{\bf G}^{(-)}}$, and
$\Lambda^{(+)}\in{\mathcal{S}}_{\text{\bf G}^{(+)}}$.

Next suppose G is a unitary group. As described in Subsection 3.4, for the
Lusztig correspondence
${\mathfrak{L}}_{s}\colon{\mathcal{E}}(G)_{1}\rightarrow{\mathcal{E}}(C_{G^{*}}(s))_{1}$
we can write

(1.3)
${\mathfrak{L}}_{s}(\rho)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{(ij)}.$

Note that $t_{1}=q+1$ and $\rho^{(1,j)}$ is a unipotent character of ${\rm
U}_{n_{1,j}}(q)$ for $j=1,\ldots,q+1$. Now we have the following formula (cf.
Proposition 3.12 and Proposition 3.17) for the $\Theta$-rank of a general
irreducible character in terms of data from the Lusztig classification:

######  Theorem 1.4.

Let $\rho\in{\mathcal{E}}(G)$.

  1. (i)

If $\text{\bf G}={\rm Sp}_{2n}$, then

$\Theta\text{\rm-rk}(\rho)=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(+)}}),2n-2n^{(-)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(-)}})+1\\}.$

  2. (ii)

If $\text{\bf G}={\rm O}^{\epsilon}_{2n}$ or ${\rm O}_{2n+1}$, then

$\Theta\text{\rm-rk}(\rho)=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(+)}}),2n-2n^{(-)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(-)}})\\}.$

  3. (iii)

If $\text{\bf G}={\rm U}_{n}$, then

$\Theta\text{\rm-rk}(\rho)=\min\\{n-n_{1,j}+\Theta\text{\rm-
rk}(\rho^{(1,j)})\mid j=1,\ldots,q+1\\}.$

### 1.4.

After we analyze several properties of the $\Theta$-ranks, the following
result can be derived easily from the result on $\eta$-correspondence for a
reductive dual pair in stable range by Gurevich-Howe in [GH20]:

######  Theorem 1.5.

Let $\rho\in{\mathcal{E}}(G_{n})$ where $\text{\bf G}_{n}$ is ${\rm Sp}_{2n}$,
${\rm O}^{+}_{2n}$, ${\rm O}^{-}_{2n+2}$, ${\rm O}_{2n+1}$, ${\rm U}_{2n}$, or
${\rm U}_{2n+1}$. If $\Theta\text{\rm-rk}(\rho)\leq n$, then

$U\text{\rm-rk}(\rho)=A\text{\rm-rk}(\rho)=\overline{A}\text{\rm-
rk}(\rho)=\Theta\text{\rm-rk}(\rho).$

Finally, by using the same arguments in [GH20] section 11, we can show the
agreement of the parabolic asymptotic rank and the theta rank:

######  Theorem 1.6.

Let $\rho\in{\mathcal{E}}(G)$ where $G$ is a symplectic group, an orthogonal
group or a unitary group. Then

$\overline{A}\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho).$

Theorems 1.5 and 1.6 provide a partial answer for Conjecture 1.1.

### 1.5.

The contents of this article are as follows. In Section 2 we first review some
results on the Lusztig’s classification on unipotent characters and the
Lusztig correspondence. Then we prove several results on the first occurrence
for theta correspondence. In Section 3, we define the $\Theta$-rank of an
irreducible character via the theta correspondence. Then we investigate
several properties of the $\Theta$-ranks and prove Theorem 1.4. We discuss the
relation of $\Theta$-ranks and other ranks in the final section. In
particular, Theorem 1.5 and Theorem 1.6 are proved.

##  2\. Theta Correspondence and Lusztig Correspondence

###  2.1. Basic notations

Let $\text{\bf f}_{q}$ be a finite field of $q$ elements where $q$ is a power
of an odd prime $p$, and let $\overline{\text{\bf f}}_{q}$ denote a fixed
algebraic closure of $\text{\bf f}_{q}$.

Let G be a symplectic group, an orthogonal group or a unitary group defined
over $\text{\bf f}_{q}$, and let $G$ denote the finite group of rational
points. The character of an irreducible representation of $G$ is called an
_irreducible character_ of $G$, and the set of all irreducible characters of
$G$ is denoted by ${\mathcal{E}}(G)$. The character of a one-dimensional
representation is also called a _linear character_. It is proved by Lusztig
that the set ${\mathcal{E}}(G)$ admits a partition

${\mathcal{E}}(G)=\bigcup_{(s)\subset(G^{*})^{0}}{\mathcal{E}}(G)_{s}$

indexed by conjugacy classes $(s)$ of semisimple simple elements $s$ in the
connected component $(G^{*})^{0}$ of the dual group $G^{*}$ of $G$. Each
${\mathcal{E}}(G)_{s}$ is called a _Lusztig series_ , and the elements in
${\mathcal{E}}(G)_{1}$ are called _unipotent_. The trivial character of $G$ is
denoted by $\bf 1$.

###  2.2. Unipotent characters of a symplectic group or an orthogonal group

The set of _partitions_ of $n$ is denoted by ${\mathcal{P}}(n)$. If
$\mu\in{\mathcal{P}}(n)$, we also write $|\mu|=n$. An ordered pair
$\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$ of two partitions $\mu,\nu$, 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 ${\mathcal{P}}_{2}(n)$.

Now we recall the notations of symbols from [Lus77]. Note that for our
convenience, the definition here is slightly different from the original one.
A _$\beta$ -set_ $A=\\{a_{1},a_{2},\ldots,a_{m}\\}$ is a finite subset
(possibly empty) of non-negative integers written in strictly decreasing
order, i.e., $a_{1}>a_{2}>\cdots>a_{m}$. A _symbol_

$\Lambda=\binom{A}{B}=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$

is an ordered pair of two $\beta$-sets $A,B$. On the set of symbols we define
an equivalence relation “$\sim$” generated by

$\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\sim\binom{a_{1}+1,a_{2}+1\ldots,a_{m_{1}}+1,0}{b_{1}+1,b_{2}+1,\ldots,b_{m_{2}}+1,0}.$

The set of equivalence classes of symbols is denoted by ${\mathcal{S}}$.

For a symbol $\Lambda=\binom{a_{1},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{m_{2}}}$,
we define its _defect_ and _rank_ by

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

It is not difficult to see that two equivalent symbols have the same defect
and the same rank. For a symbol $\Lambda=\binom{A}{B}$ we define its
_transpose_ $\Lambda^{\rm t}=\binom{B}{A}$. It is obvious that ${\rm
rk}(\Lambda^{\rm t})={\rm rk}(\Lambda)$ and ${\rm def}(\Lambda^{\rm t})=-{\rm
def}(\Lambda)$. We define a mapping $\Upsilon$ from symbols to bi-partitions
by

(2.1)
$\Upsilon\colon\binom{a_{1},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{m_{2}}}\mapsto\genfrac{[}{]}{0.0pt}{}{a_{1}-(m_{1}-1),a_{2}-(m_{1}-2),\ldots,a_{m_{1}-1}-1,a_{m_{1}}}{b_{1}-(m_{2}-1),b_{2}-(m_{2}-2),\ldots,b_{m_{2}-1}-1,b_{m_{2}}}.$

The image $\Upsilon(\Lambda)$ in fact depends only on the equivalence class of
$\Lambda$. Moreover, it is easy to check that

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

For a symplectic group or an orthogonal group G, we define the following sets
of (equivalence classes of) symbols:

$\displaystyle{\mathcal{S}}_{{\rm O}^{+}_{2n}}$
$\displaystyle=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=n,\ {\rm
def}(\Lambda)\equiv 0\pmod{4}\,\\};$ $\displaystyle{\mathcal{S}}_{{\rm
Sp}_{2n}}$ $\displaystyle=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm
rk}(\Lambda)=n,\ {\rm def}(\Lambda)\equiv 1\pmod{4}\,\\};$
$\displaystyle{\mathcal{S}}_{{\rm O}^{-}_{2n}}$
$\displaystyle=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=n,\ {\rm
def}(\Lambda)\equiv 2\pmod{4}\,\\};$ $\displaystyle{\mathcal{S}}_{{\rm
SO}_{2n+1}}={\mathcal{S}}_{{\rm O}_{2n+1}}$
$\displaystyle=\\{\,\Lambda\in{\mathcal{S}}\mid{\rm rk}(\Lambda)=n,\ {\rm
def}(\Lambda)\equiv 3\pmod{4}\,\\}.$

The following are modified from the results proved by Lusztig (cf. [Lus77]):

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n},{\rm O}^{\epsilon}_{2n},{\rm
SO}_{2n+1}$ where $\epsilon=+$ or $-$. There is a parametrization
${\mathcal{E}}(G)_{1}$ by ${\mathcal{S}}_{\text{\bf G}}$, so a unipotent
character $\rho$ of $G$ is written as $\rho=\rho_{\Lambda}$ for
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$. If $\text{\bf G}={\rm
O}^{\epsilon}_{2n}$, it is known that $\rho_{\Lambda}\cdot{\rm
sgn}=\rho_{\Lambda^{\rm t}}$.

  2. (2)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$. Then a unipotent character
$\rho\in{\mathcal{E}}(G)_{1}$ can be written as $\rho=\rho_{\Lambda}$ or
$\rho=\rho_{\Lambda}\cdot{\rm sgn}$ for $\Lambda\in{\mathcal{S}}_{{\rm
SO}_{2n+1}}$

###  2.3. Unipotent characters of a unitary group

In this subsection, suppose that G is a unitary group. For a symbol $\Lambda$,
we define

(2.2) ${\rm rk}_{\rm U}(\Lambda)=2|\Upsilon(\Lambda)|+\tfrac{1}{2}|{\rm
def}(\Lambda)|(|{\rm def}(\Lambda)|+1).$

It is easy to check that

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

Then we let ${\mathcal{S}}_{{\rm U}_{n}}$ be the set of (equivalence classes
of) symbols $\Lambda\in{\mathcal{S}}$ such that

  * •

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

  * •

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

Again, there is a parametrization ${\mathcal{E}}({\rm U}_{n}(q))_{1}$ by
${\mathcal{S}}_{{\rm U}_{n}}$ (cf. [FS90] or [Pan20b]), i.e., for
$\rho\in{\mathcal{E}}({\rm U}_{n}(q))_{1}$, we can write $\rho=\rho_{\Lambda}$
for a unique $\Lambda\in{\mathcal{S}}_{{\rm U}_{n}}$.

###  2.4. Lusztig correspondence

The following result is proved by Lusztig (cf. [DM91] theorem 13.23, remark
13.24):

######  Proposition 2.4 (Lusztig).

There is a bijection

${\mathfrak{L}}_{s}\colon{\mathcal{E}}(G)_{s}\rightarrow{\mathcal{E}}(C_{G^{*}}(s))_{1}$

such that $R^{\text{\bf G}}_{\text{\bf T}^{*},s}\mapsto\varepsilon_{\text{\bf
G}}\varepsilon_{C_{\text{\bf G}^{*}}(s)}R^{C_{\text{\bf G}^{*}}(s)}_{\text{\bf
T}^{*},1}$ where $C_{G^{*}}(s)$ denotes the centralizer in $G^{*}$ of $s$,
$\varepsilon_{\text{\bf G}}=(-1)^{\kappa}$ and $\kappa$ is the relative rank
of G.

The bijection ${\mathfrak{L}}_{s}$ called a _Lusztig correspondence_.

######  Remark 2.5.

Clearly, ${\rm def}(\Lambda)\equiv 1\pmod{4}$ if and only if ${\rm
def}(\Lambda^{\rm t})\equiv 3\pmod{4}$. The the Lusztig correspondence
${\mathfrak{L}}_{1}\colon{\mathcal{E}}({\rm
Sp}_{2n}(q))_{1}\rightarrow{\mathcal{E}}({\rm SO}_{2n+1}(q))_{1}$ is a
bijection given by $\rho_{\Lambda}\mapsto\rho_{\Lambda^{\rm t}}$.

For $s\in(G^{*})^{0}$, we define

(2.6)
$\displaystyle\begin{split}G^{(0)}=G^{(0)}(s)&=\prod_{\langle\lambda\rangle\subset\\{\lambda_{1},\ldots,\lambda_{n}\\},\
\lambda\neq\pm 1}G_{[\lambda]}(s);\\\ G^{(-)}=G^{(-)}(s)&=G_{[-1]}(s);\\\
G^{(+)}=G^{(+)}(s)&=\begin{cases}(G_{[1]}(s))^{*},&\text{if $G$ is
symplectic};\\\ G_{[1]}(s),&\text{otherwise},\end{cases}\end{split}$

where $G_{[\lambda]}(s)$ is given in [AMR96] subsection 1.B (see also [Pan19b]
subsection 2.2). Hence we have a bijection

${\mathcal{E}}(C_{G^{*}}(s))_{1}\simeq\begin{cases}{\mathcal{E}}(G^{(0)}\times
G^{(-)}\times G^{(+)})_{1}\times\\{\pm 1\\},&\text{if $G$ is odd-
orthogonal};\\\ {\mathcal{E}}(G^{(0)}\times G^{(-)}\times
G^{(+)})_{1},&\text{otherwise}.\end{cases}$

It is known that $\text{\bf G}^{(0)}$ is a product of unitary groups or
general linear groups, and $\text{\bf G}^{(-)},\text{\bf G}^{(+)}$ are given
as follows:

  * •

if $\text{\bf G}={\rm O}^{\epsilon}_{2n}$, then $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
O}^{\epsilon^{\prime}\epsilon}_{2n^{(+)}}$ for some $\epsilon^{\prime}=+$ or
$-$;

  * •

if $\text{\bf G}={\rm Sp}_{2n}$, then $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm
Sp}_{2n^{(+)}}$ for some $\epsilon^{\prime}=+$ or $-$;

  * •

if $\text{\bf G}={\rm O}_{2n+1}$, then $\text{\bf G}^{(-)}={\rm
Sp}_{2n^{(-)}}$, and $\text{\bf G}^{(+)}={\rm Sp}_{2n^{(+)}}$;

  * •

if $\text{\bf G}={\rm U}_{n}$, then $\text{\bf G}^{(-)}={\rm U}_{n^{(-)}}$,
and $\text{\bf G}^{(+)}={\rm U}_{n^{(+)}}$,

where $n^{(+)}=n^{(+)}(s)$ (resp. $n^{(-)}=n^{(-)}(s)$) denotes the
multiplicity of $+1$ (resp. $-1$) in the set
$\\{\lambda_{1},\ldots,\lambda_{n}\\}$ of “eigenvalues” of $s$.

So now a Lusztig correspondence ${\mathfrak{L}}_{s}$ induces a mapping

$\Xi_{s}\colon{\mathcal{E}}(G)_{s}\rightarrow{\mathcal{E}}(G^{(0)}(s)\times
G^{(-)}(s)\times G^{(+)}(s))_{1}$

called the _modified Lusztig correspondence_. The mapping $\Xi_{s}$ is two-to-
one and $\Xi_{s}(\rho)=\Xi_{s}(\rho\cdot{\rm sgn})$ when G is an odd-
orthogonal group; and $\Xi_{s}$ is a bijection otherwise. We can write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho^{(-)}\otimes\rho^{(+)}$. Because now
$\rho^{(\varepsilon)}$ where $\varepsilon=+$ or $-$ is a unipotent character
of a symplectic group, an even-orthogonal group, or a unitary group, we can
write $\rho^{(\varepsilon)}=\rho_{\Lambda^{(\varepsilon)}}$ for some symbol
$\Lambda^{(\varepsilon)}\in{\mathcal{S}}_{\text{\bf G}^{(\varepsilon)}}$. Now
$\rho\mapsto\Lambda^{(\varepsilon)}$ is a mapping from ${\mathcal{E}}(G)_{s}$
to ${\mathcal{S}}_{\text{\bf G}^{(\varepsilon)}}$ and
$\Lambda^{(\varepsilon)}$ is also denoted by $\Lambda^{(\varepsilon)}(\rho)$,
and hence we can write

$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}.$

Note that when G is an orthogonal group, $\rho\in{\mathcal{E}}(G)_{s}$ if and
only if $\rho\cdot{\rm sgn}\in{\mathcal{E}}(G)_{s}$.

######  Lemma 2.7.

Let $G$ be an orthogonal group and $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$.
Then

$\Lambda^{(\varepsilon)}(\rho\cdot{\rm
sgn})=\begin{cases}\Lambda^{(\varepsilon)}(\rho)^{\rm t},&\text{if $G$ is an
even orthogonal group};\\\ \Lambda^{(\varepsilon)}(\rho),&\text{if $G$ is an
odd orthogonal group}.\end{cases}$

###### Proof.

This is obvious from the definition in (2.6). ∎

Recall that a linear character $\chi_{\text{\bf G}}$ of order two is defined
when G is an orthogonal group or a unitary group (cf. [Pan16] subsection 3.3).
Then there exists a bijection
${\mathcal{E}}(G)_{s}\rightarrow{\mathcal{E}}(G)_{-s}$ given by
$\rho\mapsto\rho\chi_{\text{\bf G}}$.

######  Lemma 2.8.

Let $G$ be an orthogonal group or a unitary group, and let $s$ be a semisimple
element in $(G^{*})^{0}$. Then $G^{(0)}(-s)\simeq G^{(0)}(s)$,
$G^{(-)}(-s)\simeq G^{(+)}(s)$ and $G^{(+)}(-s)\simeq G^{(-)}(s)$.

###### Proof.

This is obvious from the definition in (2.6). ∎

######  Lemma 2.9.

Let $G$ be an orthogonal group, and let $\rho\in{\mathcal{E}}(G)$. Then
$\Lambda^{(\varepsilon)}(\rho\chi_{\text{\bf
G}})=\Lambda^{(-\varepsilon)}(\rho)$ for $\varepsilon=+$ or $-$.

###### Proof.

Let $\rho\in{\mathcal{E}}(G)_{s}$ for some semisimple $s\in(G^{*})^{0}$. It is
known that $\chi_{\text{\bf G}}R^{\text{\bf G}}_{\text{\bf
T}^{*},s}=R^{\text{\bf G}}_{\text{\bf T}^{*},-s}$ for any rational maximal
torus $\text{\bf T}^{*}$ in $\text{\bf G}^{*}$ containing $s$. Then from Lemma
2.8 we have $\text{\bf G}^{(-)}(-s)\simeq\text{\bf G}^{(+)}(s)$ and $\text{\bf
G}^{(+)}(-s)\simeq\text{\bf G}^{(-)}(s)$. If we write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho^{(-)}\otimes\rho^{(+)}$ and
$\Xi_{-s}(\rho\chi_{\text{\bf
G}})=\rho^{\prime(0)}\otimes\rho^{\prime(-)}\otimes\rho^{\prime(+)}$, then we
have $\rho^{\prime(-)}=\rho^{(+)}$ and $\rho^{\prime(+)}=\rho^{(-)}$. This
means that $\Lambda^{(-)}(\rho\chi_{\text{\bf G}})=\Lambda^{(+)}(\rho)$ and
$\Lambda^{(+)}(\rho\chi_{\text{\bf G}})=\Lambda^{(-)}(\rho)$. ∎

###  2.5. Lusztig correspondence and parabolic induction

Let $\text{\bf G}_{n}$ denote the group of split rank $n$ in its Witt series,
i.e., $\text{\bf G}_{n}$ is one of ${\rm Sp}_{2n}$, ${\rm O}^{+}_{2n}$, ${\rm
O}^{-}_{2n+2}$, ${\rm O}_{2n+1}$, ${\rm U}_{2n}$, or ${\rm U}_{2n+1}$. For
$m\geq n$, let ${\mathcal{E}}^{G_{m}}_{G_{n}}(\rho)$ denote the set of
irreducible components of ${\rm Ind}_{P_{n}}^{G_{m}}(\rho\otimes\bf 1)$ where
$P_{n}$ is a parabolic subgroup of $G_{m}$ whose Levi factor is isomorphic to
$G_{n}\times T_{m-n}$ and $T_{m-n}$ is the direct product of $m-n$ copies of
${\rm GL}_{1}(q)$ if $\text{\bf G}_{n}$ is a symplectic group or an orthogonal
group; $T_{m-n}$ is the direct product of $m-n$ copies of ${\rm
GL}_{1}(q^{2})$ if $\text{\bf G}_{n}$ is a unitary group. For $m<n$ and
$\rho\in{\mathcal{E}}(G_{n})$, let ${\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)$
denote the set of $\rho^{\prime}\in{\mathcal{E}}(G_{m})$ such that
$\rho\in{\mathcal{E}}_{G_{m}}^{G_{n}}(\rho^{\prime})$.

Let $\rho\in{\mathcal{E}}(G_{n})$ and
$\rho_{m}\in{\mathcal{E}}^{G_{m}}_{G_{n}}(\rho)$ for some $m\geq n$. Because

${\rm Ind}^{G_{m}}_{P_{n}}(R_{\text{\bf T}^{*},s}^{\text{\bf
G}_{n}}\otimes{\bf 1})=R^{\text{\bf G}_{m}}_{\text{\bf T}^{*}\times\text{\bf
T}_{m-n},s_{m}}$

where $s_{m}=(s,1)\in(G^{*}_{n})^{0}\times T_{m-n}\subset(G^{*}_{m})^{0}$, we
see that $\langle\rho_{m},R^{\text{\bf G}_{m}}_{\text{\bf
T}^{*}\times\text{\bf T}_{m-n},s_{m}}\rangle\neq 0$ if and only if
$\langle\rho,R^{\text{\bf G}_{n}}_{\text{\bf T}^{*},s}\rangle\neq 0$. This
means that if $\rho\in{\mathcal{E}}(G_{n})_{s}$ for some $s$, then
$\rho_{m}\in{\mathcal{E}}(G_{m})_{s_{m}}$. In particular, if $\rho$ is
unipotent, then every element in ${\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)$ is
unipotent.

######  Lemma 2.10.

Suppose that $m\geq n$, $\rho\in{\mathcal{E}}(G_{n})_{s}$ for some $s$, and
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho^{(-)}\otimes\rho^{(+)}$. Then

${\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)=\\{\,\rho_{m}\in{\mathcal{E}}(G_{m})_{s_{m}}\mid\Xi_{s_{m}}(\rho_{m})=\rho^{(0)}\otimes\rho^{(-)}\otimes\rho_{m}^{(+)},\
\rho^{(+)}_{m}\in{\mathcal{E}}_{G_{n^{(+)}}}^{G_{m^{(+)}}}(\rho^{(+)})\,\\}$

where $m^{(+)}=n^{(+)}+(m-n)$.

###### Proof.

Let $\rho\in{\mathcal{E}}(G_{n})_{s}$. If
$\rho_{m}\in{\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)$, then from above we see that
$\rho_{m}\in{\mathcal{E}}(G_{m})_{s_{m}}$ where
$s_{m}=(s,1)\in(G^{*}_{n})^{0}\times T_{m-n}\subset(G_{m}^{*})^{0}$. Then
$\text{\bf G}^{(0)}(s_{m})=\text{\bf G}^{(0)}(s)$, $\text{\bf
G}^{(-)}(s_{m})=\text{\bf G}^{(-)}(s)$, and $\text{\bf
G}^{(+)}(s_{m})=\text{\bf G}_{m^{(+)}}$ where $m^{(+)}=n^{(+)}+(m-n)$. It is
known that $R^{\text{\bf G}_{m}}_{\text{\bf T}^{*}\times\text{\bf
T}_{m-n},s_{m}}={\rm Ind}_{P_{n}}^{G_{m}}(R^{G_{n}}_{\text{\bf
T}^{*},s}\otimes{\bf 1})$ where $P_{n}$ is a parabolic subgroup of $G_{m}$
whose Levi factor is isomorphic to $G_{n}\times T_{m-n}$. Then the lemma
follows easily. ∎

Suppose that $\rho\in{\mathcal{E}}(G_{n})_{1}$. We write $\rho=\rho_{\Lambda}$
($\rho=\rho_{\Lambda}$ or $\rho=\rho_{\Lambda}\cdot{\rm sgn}$ if $\text{\bf
G}_{n}={\rm O}_{2n+1}$) for $\Lambda\in{\mathcal{S}}_{\text{\bf G}_{n}}$. From
the Littlewood-Richardson rule, we know that an element
$\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$ is of the form
$\rho_{\Lambda^{\prime}}$ ($\rho_{\Lambda^{\prime}}$ or
$\rho_{\Lambda^{\prime}}\cdot{\rm sgn}$ if $\text{\bf G}_{n}={\rm O}_{2n+1}$)
where $\Lambda^{\prime}\in{\mathcal{S}}_{\text{\bf G}_{n+1}}$ such that ${\rm
def}(\Lambda^{\prime})={\rm def}(\Lambda)$ and $\Upsilon(\Lambda^{\prime})$ is
obtained from $\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{\mu}{\nu}$ by adding
a box to some row (including an empty row) of $\mu$ or $\nu$.

######  Example 2.11.

Suppose that $\rho\in{\mathcal{E}}({\rm Sp}_{2n})_{s}$ and
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$
such that $\rho_{\Lambda^{(+)}}$ is cuspidal. Hence $n^{(+)}=t(t+1)$ for some
non-negative integer $t$ and

$\Lambda^{(+)}=\begin{cases}\binom{t,t-1,\ldots,1,0}{-},&\text{if $t$ is
even};\\\ \binom{-}{t,t-1,\ldots,1,0},&\text{if $t$ is odd}.\end{cases}$

Hence the parabolic induced character ${\rm Ind}_{P_{n}}^{{\rm
Sp}_{2(n+1)}}(\rho\otimes{\bf 1})$ is the sum of two irreducible characters

$\Xi_{(s,1)}^{-1}(\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda_{1}})+\Xi_{(s,1)}^{-1}(\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda_{2}})$

where $\Xi_{(s,1)}$ is the bijection

$\Xi_{(s,1)}\colon{\mathcal{E}}({\rm
Sp}_{2(n+1)}(q))_{(s,1)}\rightarrow{\mathcal{E}}(G^{(0)}\times
G^{(-)}\times{\rm Sp}_{2(t(t+1)+1)}(q))_{1}$

and

$\\{\Lambda_{1},\Lambda_{2}\\}=\begin{cases}\bigl{\\{}\binom{t+1,t-1,t-2,\ldots,1,0}{-},\binom{t+1,t,t-1,\ldots,1,0}{1}\bigr{\\}},&\text{if
$t$ is even};\\\
\bigl{\\{}\binom{-}{t+1,t-1,t-2,\ldots,1,0},\binom{1}{t+1,t,t-1,\ldots,1,0}\bigr{\\}},&\text{if
$t$ is odd}.\end{cases}$

###  2.6. Finite Howe correspondences on unipotent characters

First we recall the concept of finite reductive dual pairs from [How]. Let
$(\text{\bf G}^{\prime},\text{\bf G})$ be a finite reductive dual pair of one
of the following types:

  1. (I)

$({\rm U}_{n^{\prime}},{\rm U}_{n})$;

  2. (II)

$({\rm O}_{2n^{\prime}}^{\epsilon},{\rm Sp}_{2n})$ or $({\rm
Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$ where $\epsilon=+$ or $-$;

  3. (III)

$({\rm O}_{2n^{\prime}+1},{\rm Sp}_{2n})$ or $({\rm Sp}_{2n^{\prime}},{\rm
O}_{2n+1})$

where $n^{\prime},n$ are non-negative integers. By restricting the Weil
character with respect to a nontrivial additive character $\psi$ of $\text{\bf
f}_{q}$ to the dual pair $(\text{\bf G}^{\prime},\text{\bf G})$, we obtain a
decomposition

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

where $m_{\rho^{\prime},\rho}=0,1$. Then we have a relation between
${\mathcal{E}}(G^{\prime})$ and ${\mathcal{E}}(G)$:

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

called the _$\Theta$ -correspondence_ for the dual pair $(\text{\bf
G}^{\prime},\text{\bf G})$. For $\rho^{\prime}\in{\mathcal{E}}(G^{\prime})$,
we write

$\Theta(\rho^{\prime})=\Theta_{\text{\bf
G}}(\rho^{\prime})=\\{\,\rho\in{\mathcal{E}}(G)\mid(\rho^{\prime},\rho)\in\Theta_{\text{\bf
G}^{\prime},\text{\bf G}}\,\\}.$

It is well known that for cases (I) and (II) above, the unipotent characters
are preserved by $\Theta$-correspondence, i.e., when
$(\rho^{\prime},\rho)\in\Theta_{\text{\bf G}^{\prime},\text{\bf G}}$, then
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{1}$ if and only if
$\rho\in{\mathcal{E}}(G)_{1}$ (cf. [AM93]). For these two cases, the unipotent
part $\omega_{\text{\bf G}^{\prime},\text{\bf G},1}$ of the Weil character
$\omega_{\text{\bf G}^{\prime},\text{\bf G}}^{\psi}$ has the decomposition

$\omega_{\text{\bf G}^{\prime},\text{\bf
G},1}=\sum_{\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{1},\
\rho\in{\mathcal{E}}(G)_{1}}m_{\rho^{\prime},\rho}\rho^{\prime}\otimes\rho=\sum_{(\Lambda^{\prime},\Lambda)\in{\mathcal{B}}_{\text{\bf
G}^{\prime},\text{\bf G}}}\rho_{\Lambda^{\prime}}\otimes\rho_{\Lambda}$

where the relation ${\mathcal{B}}_{\text{\bf G}^{\prime},\text{\bf G}}$ are
given in [Pan19a] and [Pan19b].

###  2.7. Howe correspondence and Lusztig correspondence

The following result on the commutativity between the (probably chosen)
modified Lusztig correspondence and the Howe correspondence is from [Pan19b]
theorem 6.9 and theorem 7.9:

######  Proposition 2.12.

Let $(\text{\bf G}^{\prime},\text{\bf G})$ be a finite reductive dual pair,
and let $\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$,
$\rho\in{\mathcal{E}}(G)_{s}$ for some $s^{\prime},s$. Write
$\Xi_{s^{\prime}}(\rho^{\prime})=\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}$
and
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.

  1. (i)

Suppose that $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}^{\epsilon}_{2n^{\prime}},{\rm Sp}_{2n})$ or $({\rm U}_{n^{\prime}},{\rm
U}_{n})$. Then $(\rho^{\prime},\rho)$ occurs in $\Theta_{\text{\bf
G}^{\prime},\text{\bf G}}$ if and only if

     * •

$\text{\bf G}^{\prime(0)}\simeq\text{\bf G}^{(0)}$ and
$\rho^{\prime(0)}=\rho^{(0)}$;

     * •

$\text{\bf G}^{\prime(-)}\simeq\text{\bf G}^{(-)}$ and
$\rho_{\Lambda^{\prime(-)}}=\rho_{\Lambda^{(-)}}$;

     * •

$(\rho_{\Lambda^{\prime(+)}},\rho_{\Lambda^{(+)}})$ occurs in
$\Theta_{\text{\bf G}^{\prime(+)},\text{\bf G}^{(+)}}$,

i.e., the diagram

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes{\rm id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf
G}^{(+)}}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

commutes.

  2. (ii)

Suppose that $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}_{2n^{\prime}+1},{\rm Sp}_{2n})$. Then $(\rho^{\prime},\rho)$ occurs in
$\Theta_{\text{\bf G}^{\prime},\text{\bf G}}$ if and only if

     * •

$\text{\bf G}^{\prime(0)}\simeq\text{\bf G}^{(0)}$ and
$\rho^{\prime(0)}=\rho^{(0)}$;

     * •

$(\rho_{\Lambda^{\prime(+)}},\rho_{\Lambda^{(-)}})$ occurs in
$\Theta_{\text{\bf G}^{\prime(+)},\text{\bf G}^{(-)}}$,

     * •

$\text{\bf G}^{\prime(-)}\simeq\text{\bf G}^{(+)}$ and
$\rho_{\Lambda^{\prime(-)}}=\rho_{\Lambda^{(+)}}$;

i.e., the diagram

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\iota\circ\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(+)}}\otimes\rho_{\Lambda^{\prime(-)}}@>{{\rm
id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf G}^{(-)}}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

commutes where
$\iota(\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}})=\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(+)}}\otimes\rho_{\Lambda^{\prime(-)}}$.

Recall that an irreducible character $\rho$ of $G={\rm Sp}_{2n}(q)$ is called
_pseudo-unipotent_ if $\rho\in{\mathcal{E}}(G)_{s}$ such that $C_{{\rm
SO}_{2n+1}}(s)={\rm O}^{\epsilon}_{2n}$ for some $\epsilon=+$ or $-$. Now from
the proposition we have several lemmas concerning the first occurrences of
unipotent or pseudo-unipotent characters.

######  Lemma 2.13.

Let $\rho$ be an irreducible unipotent character of ${\rm Sp}_{2n}(q)$.
Suppose that $\rho$ occurs in the $\Theta$-correspondence for some dual pair
$({\rm O}_{2n^{\prime}+1},{\rm Sp}_{2n})$. Then $n^{\prime}\geq n$.

###### Proof.

Consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}_{2n^{\prime}+1},{\rm Sp}_{2n})$ and the following commutative diagram:

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\iota\circ\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(+)}}\otimes\rho_{\Lambda^{\prime(-)}}@>{{\rm
id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf G}^{(-)}}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Because $\rho$ is a unipotent character of ${\rm Sp}_{2n}(q)$, we see that
$\text{\bf G}^{(0)}={\rm U}_{0}$, $\text{\bf G}^{(-)}={\rm O}_{0}^{+}$, and
$\text{\bf G}^{(+)}={\rm Sp}_{2n}$. This means by Proposition 2.12 that
$\text{\bf G}^{\prime(-)}={\rm Sp}_{2n}$ and hence $n^{\prime}\geq n$. ∎

######  Lemma 2.14.

Let $\rho$ be an irreducible pseudo-unipotent character of ${\rm Sp}_{2n}(q)$.
Suppose that $\rho$ occurs in the $\Theta$-correspondence for some dual pair
$({\rm O}^{\epsilon}_{2n^{\prime}},{\rm Sp}_{2n})$. Then $n^{\prime}\geq n$.

###### Proof.

Consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}^{\epsilon}_{2n^{\prime}},{\rm Sp}_{2n})$ and the following commutative
diagram:

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes{\rm id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf
G}^{(+)}}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Because $\rho$ is a pseudo-unipotent character of ${\rm Sp}_{2n}(q)$, we see
that $\text{\bf G}^{(0)}={\rm U}_{0}$, $\text{\bf G}^{(-)}={\rm
O}_{2n}^{\epsilon^{\prime}}$ for some $\epsilon^{\prime}=+$ or $-$, and
$\text{\bf G}^{(+)}={\rm Sp}_{0}$. This means by Proposition 2.12 that
$\text{\bf G}^{\prime(-)}={\rm O}^{\epsilon^{\prime}}_{2n}$ and hence
$n^{\prime}\geq n$. ∎

######  Lemma 2.15.

Let $\rho$ be an irreducible unipotent character of ${\rm
O}^{\epsilon}_{2n}(q)$. Suppose that $\rho\chi_{{\rm O}^{\epsilon}_{2n}}$
occurs in the correspondence for the dual pair $({\rm Sp}_{2n^{\prime}},{\rm
O}^{\epsilon}_{2n})$. Then $n^{\prime}\geq n$.

###### Proof.

Consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$ and the following commutative
diagram:

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\chi_{\text{\bf G}}\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes{\rm id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf
G}^{(+)}}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Because $\rho$ is a unipotent character of ${\rm O}^{\epsilon}_{2n}(q)$, now
$\rho\chi_{\text{\bf G}}$ is in ${\mathcal{E}}({\rm
O}_{2n}^{\epsilon}(q))_{-1}$ and $\text{\bf G}^{(0)}={\rm U}_{0}$, $\text{\bf
G}^{(-)}={\rm O}^{\epsilon}_{2n}$, and $\text{\bf G}^{(+)}={\rm O}^{+}_{0}$.
This means by Proposition 2.12 that $\text{\bf G}^{\prime(-)}={\rm
O}^{\epsilon}_{2n}$ and hence $n^{\prime}\geq n$. ∎

######  Lemma 2.16.

Let $\rho$ be an irreducible unipotent character of ${\rm O}_{2n+1}(q)$.
Suppose that $\rho\chi_{{\rm O}_{2n+1}}$ occurs in the correspondence for some
dual pair $({\rm Sp}_{2n^{\prime}},{\rm O}_{2n+1})$. Then $n^{\prime}\geq n$.

###### Proof.

Consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
Sp}_{2n^{\prime}},{\rm O}_{2n+1})$ and the following commutative diagram:

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\chi_{\text{\bf G}}\\\
@V{\Xi_{s^{\prime}}}V{}V@V{}V{\iota\circ\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes\Theta_{\text{\bf G}^{\prime(-)},\text{\bf G}^{(+)}}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(+)}}\otimes\rho_{\Lambda^{(-)}}\end{CD}$

Because $\rho$ is a unipotent character of ${\rm O}_{2n+1}(q)$, now
$\rho\chi_{\text{\bf G}}$ is in ${\mathcal{E}}({\rm O}_{2n+1}(q))_{-1}$ and
$\text{\bf G}^{(0)}={\rm U}_{0}$, $\text{\bf G}^{(-)}={\rm Sp}_{2n}$, and
$\text{\bf G}^{(+)}={\rm Sp}_{0}$. This means by Proposition 2.12 that
$\text{\bf G}^{\prime(+)}={\rm Sp}_{2n}$ and hence $n^{\prime}\geq n$. ∎

######  Lemma 2.17.

Let $\rho$ be an irreducible unipotent character of ${\rm U}_{n}(q)$, and let
$\chi$ be a non-trivial linear character of ${\rm U}_{n}(q)$. Suppose that
$\rho\chi$ occurs in the correspondence for some dual pair $({\rm
U}_{n^{\prime}},{\rm U}_{n})$. Then $n^{\prime}\geq n$.

###### Proof.

Consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
U}_{n^{\prime}},{\rm U}_{n})$ and the following commutative diagram:

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\chi\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes{\rm id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf
G}^{(+)}}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Now $\rho$ is unipotent and $\chi$ is a nontrivial linear character, we see
that $\rho\chi$ is in ${\mathcal{E}}(G)_{s}$ where $s$ does not have
eigenvalue $1$, i.e., $\text{\bf G}^{(+)}={\rm U}_{0}$. Because now $\text{\bf
G}^{\prime(0)}=\text{\bf G}^{(0)}$ and $\text{\bf G}^{\prime(-)}=\text{\bf
G}^{(-)}$ by Proposition 2.12, we conclude that $n^{\prime}\geq n$. ∎

##  3\. $\Theta$-ranks of Irreducible Characters

###  3.1. Definition of $\Theta$-ranks

Let $G$ be a symplectic group, an orthogonal group or a unitary group. For
$\rho\in{\mathcal{E}}(G)$, the _$\Theta$ -rank_ of $\rho$, denoted by
$\Theta\text{\rm-rk}(\rho)$, is defined as follows:

  1. (1)

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

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

  2. (2)

If $G$ is an orthogonal group, we consider dual pairs $({\rm
Sp}_{2k},\text{\bf G})$ and define

$\Theta\text{\rm-
rk}(\rho)=\min\\{\,2k\mid\rho\chi\in\Theta(\rho^{\prime})\text{ for some
}\rho^{\prime}\in{\mathcal{E}}({\rm Sp}_{2k}(q))\\\ \text{ and some linear
character }\chi\in{\mathcal{E}}(G)\,\\}.$

So the $\Theta$-rank of an irreducible character of an orthogonal group is
always even. Note that for a non-trivial orthogonal group, there are four
linear characters: $\bf 1$, ${\rm sgn}$, $\chi_{\text{\bf G}}$, and ${\rm
sgn}\chi_{\text{\bf G}}$.

  3. (3)

If $G$ is a unitary group, we consider dual pairs $({\rm U}_{k},\text{\bf G})$
and define

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

From the above definition, we see that $\Theta\text{\rm-
rk}(\rho)=\Theta\text{\rm-rk}(\rho\chi)$ for any linear character $\chi$ of
$G$.

######  Example 3.1.

By convention, we have $\omega_{\text{\bf G}^{\prime},\text{\bf
G}}^{\psi}={\bf 1}_{\text{\bf G}^{\prime}}\otimes{\bf 1}_{\text{\bf G}}$ when
$(\text{\bf G}^{\prime},\text{\bf G})=({\rm Sp}_{0},{\rm O}^{\epsilon}_{n})$,
$({\rm O}_{0}^{+},{\rm Sp}_{2n})$ or $({\rm U}_{0},{\rm U}_{n})$. Therefore
linear characters are the only irreducible characters of $G$ of $\Theta$-rank
0.

######  Remark 3.2.

Let $\rho$ be an irreducible character of $G$.

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. Let $n_{0}^{\epsilon}(\rho)$ denote
the minimum of $2n^{\prime}$ such that $\rho$ occurs in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{\epsilon}_{2n^{\prime}},{\rm Sp}_{2n})$. It is known from [Pan19b] that

$n^{+}_{0}(\rho)+n^{-}_{0}(\rho)\leq 4n+2.$

Because now both $n^{+}_{0}(\rho),n^{-}_{0}(\rho)$ are even, we see that at
least one of $n^{+}_{0}(\rho),n^{-}_{0}(\rho)$ is less than or equal to $2n$.
Moreover, $\Theta\text{\rm-
rk}(\rho)\leq\min\\{n^{+}_{0}(\rho),n^{-}_{0}(\rho)\\}$ by the definition of
$\Theta$-ranks, so we have $0\leq\Theta\text{\rm-rk}(\rho)\leq 2n$.

  2. (2)

Suppose that $\text{\bf G}={\rm O}^{\epsilon}_{2n}$ or ${\rm O}_{2n+1}$. Let
$n_{0}(\rho)$ denote the minimum of $2n^{\prime}$ such that $\rho$ occurs in
the $\Theta$-correspondence for the dual pair $({\rm
Sp}_{2n^{\prime}},\text{\bf G})$. It is known that

$n_{0}(\rho)+n_{0}(\rho\cdot{\rm sgn})\leq\begin{cases}4n,&\text{if $\text{\bf
G}={\rm O}^{\epsilon}_{2n}$};\\\ 4n+2,&\text{if $\text{\bf G}={\rm
O}_{2n+1}$}.\end{cases}$

Now both $n_{0}(\rho),n_{0}(\rho\cdot{\rm sgn})$ are even. It again implies
that $0\leq\Theta\text{\rm-rk}(\rho)\leq 2n$.

  3. (3)

Suppose that $\text{\bf G}={\rm U}_{n}$. Let $n_{0}^{+}(\rho)$ (resp.
$n_{0}^{-}(\rho)$) denote the minimum of $2n^{\prime}$ (resp. $2n^{\prime}+1$)
such that $\rho$ occurs in the $\Theta$-correspondence for the dual pair
$({\rm U}_{2n^{\prime}},{\rm U}_{n})$ (resp. $({\rm U}_{2n^{\prime}+1},{\rm
U}_{n})$). It is known that

$n^{+}_{0}(\rho)+n^{-}_{0}(\rho)\leq 2n+1.$

This implies that $0\leq\Theta\text{\rm-rk}(\rho)\leq n$.

###  3.2. $\Theta$-ranks of unipotent or pseudo-unipotent characters

Let
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{\text{\bf
G}}$ where G is a symplectic group, an orthogonal group, or a unitary group.
Because
$\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\sim\binom{a_{1}+1,a_{2}+1,\ldots,a_{m_{1}}+1,0}{b_{1}+1,b_{2}+1,\ldots,b_{m_{2}}+1,0}$
and ${\rm rk}(\Lambda)={\rm rk}(\Lambda^{\prime})$, ${\rm rk}_{\rm
U}(\Lambda)={\rm rk}_{\rm U}(\Lambda^{\prime})$ if
$\Lambda\sim\Lambda^{\prime}$, we may assume that both rows
$\\{a_{i}\\},\\{b_{i}\\}$ are non-empty. Then we define

(3.3) $\Theta\text{\rm-rk}(\Lambda)=\begin{cases}\min\\{2{\rm rk}(\Lambda^{\rm
t}\smallsetminus\textstyle\binom{-}{a_{1}}),2{\rm rk}(\Lambda^{\rm
t}\smallsetminus\textstyle\binom{b_{1}}{-})\\},&\text{if $\text{\bf G}$ is
${\rm Sp}$ or ${\rm O}$};\\\ \min\\{{\rm rk}_{\rm U}(\Lambda^{\rm
t}\smallsetminus\textstyle\binom{-}{a_{1}}),{\rm rk}_{\rm U}(\Lambda^{\rm
t}\smallsetminus\textstyle\binom{b_{1}}{-})\\}&\text{if $\text{\bf G}$ is
${\rm U}$}\end{cases}$

where $\Lambda^{\rm
t}\smallsetminus\textstyle\binom{-}{a_{1}}=\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},\ldots,a_{m_{1}}}$,
etc. Note that $\Theta\text{\rm-rk}(\Lambda)$ depends only on the equivalence
class of $\Lambda$.

######  Proposition 3.4.

Let G be a symplectic group, an orthogonal group, or a unitary group, and let
$\rho\in{\mathcal{E}}(G)_{1}$. Write $\rho=\rho_{\Lambda}$
($\rho=\rho_{\Lambda}$ or $\rho=\rho_{\Lambda}\cdot{\rm sgn}$ if $\text{\bf
G}={\rm O}_{2n+1}$) for some symbol $\Lambda\in{\mathcal{S}}_{\text{\bf G}}$.
Then

$\Theta\text{\rm-rk}(\rho_{\Lambda})=\Theta\text{\rm-rk}(\Lambda).$

###### Proof.

Let
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{\text{\bf
G}}$ and $\rho=\rho_{\Lambda}$.

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. By Lemma 2.13, to compute
$\Theta\text{\rm-rk}(\rho_{\Lambda})$ we only need to consider the dual pairs
$({\rm O}^{\epsilon}_{2n^{\prime}},{\rm Sp}_{2n})$ for $\epsilon=+$ or $-$,
i.e., we do not need to consider the first occurrence of $\rho_{\Lambda}$ in
the correspondence for the dual pair $({\rm O}_{2n^{\prime}+1},{\rm
Sp}_{2n})$. From [Pan19b] section 8, we know that the minimum of $2n^{\prime}$
such that $\rho_{\Lambda}$ occurs in the $\Theta$-correspondence for the dual
pair $({\rm O}^{+}_{2n^{\prime}},{\rm Sp}_{2n})$ is equal to $2{\rm
rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}$.
Similarly, the minimum of $2n^{\prime}$ such that $\rho_{\Lambda}$ occurs in
the $\Theta$-correspondence for the dual pair $({\rm O}^{-}_{2n^{\prime}},{\rm
Sp}_{2n})$ is equal to $2{\rm
rk}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}$.

  2. (2)

Suppose that $\text{\bf G}={\rm O}^{\epsilon}_{2n}$. By Lemma 2.15, if
$\rho_{\Lambda}\chi_{\text{\bf G}}$ or $\rho_{\Lambda}\cdot{\rm
sgn}\chi_{\text{\bf G}}$ occurs in the $\Theta$-correspondence for the dual
pair $({\rm Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$, then $n^{\prime}\geq
n$. So we need only to consider the first occurrences of $\rho_{\Lambda}$ and
$\rho_{\Lambda^{\rm t}}=\rho_{\Lambda}\cdot{\rm sgn}$ in the correspondence
for the dual pair $({\rm Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$. We know
that the minimum of $2n^{\prime}$ such that $\rho_{\Lambda}$ occurs in the
$\Theta$-correspondence for the dual pair $({\rm Sp}_{2n^{\prime}},{\rm
O}^{\epsilon}_{2n})$ is equal to $2{\rm
rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}$ when
$\epsilon=+$; and equal to $2{\rm
rk}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}$ when
$\epsilon=-$. Similarly, the minimum of $2n^{\prime}$ such that
$\rho_{\Lambda}\cdot{\rm sgn}$ occurs in the $\Theta$-correspondence for the
dual pair $({\rm Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$ is equal to
$2{\rm rk}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}$
when $\epsilon=+$; and equal to $2{\rm
rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}$ when
$\epsilon=-$.

  3. (3)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$. By Lemma 2.16, we know that if
$\rho_{\Lambda}\chi_{\text{\bf G}}$ or $\rho_{\Lambda}\cdot{\rm
sgn}\chi_{\text{\bf G}}$ occurs in the $\Theta$-correspondence for the dual
pair $({\rm Sp}_{2n^{\prime}},{\rm O}_{2n+1})$, then $n^{\prime}\geq n$. So we
need only to consider the first occurrences of $\rho_{\Lambda}$ and
$\rho_{\Lambda}\cdot{\rm sgn}$ in the correspondence for the dual pair $({\rm
Sp}_{2n^{\prime}},{\rm O}_{2n+1})$.

Now we have a commutative diagram

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho_{\Lambda}\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\iota\circ\Xi_{1}}V\\\
{\bf 1}\otimes\rho_{\Lambda^{\prime(-)}}\otimes{\bf 1}@>{{\rm
id}\otimes\Theta_{\text{\bf G}^{\prime(-)},\text{\bf G}^{(+)}}\otimes{\rm
id}}>{}>{\bf 1}\otimes\rho_{\Lambda^{\rm t}}\otimes{\bf 1}\end{CD}$

Now $\text{\bf G}^{(+)}={\rm Sp}_{2n}$ and $\text{\bf G}^{\prime(-)}={\rm
O}^{+}_{2n^{\prime}}$, so the minimum of $2n^{\prime}$ such that
$\rho_{\Lambda}$ occurs in the $\Theta$-correspondence for the dual pair
$({\rm Sp}_{2n^{\prime}},{\rm O}_{2n+1})$ is equal to the minimum of
$2n^{\prime\prime}$ such that $\rho_{\Lambda^{\rm t}}$ occurs in the
$\Theta$-correspondence for the dual pair $({\rm
O}^{+}_{2n^{\prime\prime}},{\rm Sp}_{2n})$, and hence equal to $2{\rm
rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}$.

Moreover, $\Xi_{1}(\rho_{\Lambda}\cdot{\rm sgn})=\Xi_{1}(\rho)$ and now
$\text{\bf G}^{(+)}={\rm Sp}_{2n}$ and $\text{\bf G}^{\prime(-)}={\rm
O}^{-}_{2n^{\prime}}$, so the minimum of $2n^{\prime}$ such that
$\rho_{\Lambda}\cdot{\rm sgn}$ occurs in the $\Theta$-correspondence for the
dual pair $({\rm Sp}_{2n^{\prime}},{\rm O}_{2n+1})$ is equal to the minimum of
$2n^{\prime\prime}$ such that $\rho_{\Lambda^{\rm t}}\in{\mathcal{E}}({\rm
Sp}_{2n}(q))_{1}$ occurs in the $\Theta$-correspondence for the dual pair
$({\rm O}^{-}_{2n^{\prime\prime}},{\rm Sp}_{2n})$ and hence equal to $2{\rm
rk}\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{2},b_{3},\ldots,b_{m_{2}}}$.

  4. (4)

Suppose that $\text{\bf G}={\rm U}_{n}$. By Lemma 2.17, to compute
$\Theta\text{\rm-rk}(\rho_{\Lambda})$, we do not need to consider the first
occurrences of $\rho_{\Lambda}\chi$ for any nontrivial linear character $\chi$
of $G$. Now the minimum of $n^{\prime}$ such that $\rho_{\Lambda}$ occurs in
the $\Theta$-correspondence for the dual pair $({\rm U}_{n^{\prime}},{\rm
U}_{n})$ is equal to ${\rm rk}_{\rm
U}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}$ if
$n^{\prime}+n$ is even; and is equal to ${\rm rk}_{\rm
U}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}$ if
$n^{\prime}+n$ is odd.

∎

The following two corollaries follow from the proof of the above proposition
immediately.

######  Corollary 3.5.

The $\Theta$-rank of an irreducible unipotent character of ${\rm Sp}_{2n}(q)$
is always even.

Note that if $\rho_{\Lambda}\in{\mathcal{E}}({\rm O}_{2n+1}(q))$, then
$\rho_{\Lambda^{\rm t}}\in{\mathcal{E}}({\rm Sp}_{2n}(q))$.

######  Corollary 3.6.

If $\rho_{\Lambda}\in{\mathcal{E}}({\rm O}_{2n+1}(q))_{1}$, then
$\Theta\text{\rm-rk}(\rho_{\Lambda})=\Theta\text{\rm-
rk}(\rho_{\Lambda}\cdot{\rm sgn})=\Theta\text{\rm-rk}(\rho_{\Lambda^{\rm
t}})$.

######  Example 3.7.

It is clear from the result in [AM93] and Lemma 2.15 that the $\Theta$-ranks
of cuspidal unipotent characters are described as follows:

  1. (1)

The $\Theta$-rank of the unique cuspidal unipotent character of ${\rm
Sp}_{2d(d+1)}(q)$ is $2d^{2}$.

  2. (2)

The $\Theta$-ranks of the two cuspidal unipotent characters of ${\rm
O}^{\epsilon}_{2d^{2}}(q)$ (for $d\geq 1$) are $2d(d-1)$ where $\epsilon=+$ if
$d$ is even, and $\epsilon=-$ if $d$ is odd.

  3. (3)

The $\Theta$-ranks of the two cuspidal unipotent characters of ${\rm
O}_{2d(d+1)+1}(q)$ are $2d^{2}$.

  4. (4)

The $\Theta$-rank of the unique cuspidal unipotent character of ${\rm
U}_{\frac{1}{2}d(d+1)}(q)$ is $\frac{1}{2}d(d-1)$.

######  Example 3.8.

Now we want to compute the $\Theta$-rank of a Steinberg character of $G$.

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. The Steinberg character ${\rm
St}_{{\rm Sp}_{2n}}$ is the unipotent character associated to the symbol
$\binom{n,n-1,\ldots,0}{n,n-1,\ldots,1}$. Therefore,

$\Theta\text{\rm-rk}({\rm St}_{{\rm Sp}_{2n}})=\min\\{2{\rm
rk}\textstyle\binom{n,n-1,\ldots,1}{n-1,n-2,\ldots,0},2{\rm
rk}\binom{n-1,n-2,\ldots,1}{n,n-1,\ldots,0}\\}=2n,$

i.e., ${\rm St}_{{\rm Sp}_{2n}}$ is a character of maximal $\Theta$-rank.

  2. (2)

Suppose that $\text{\bf G}={\rm O}^{+}_{2n}$ and $n\geq 1$. The two Steinberg
characters $\rho,\rho\cdot{\rm sgn}$ are the unipotent characters associated
to the symbols $\binom{n,n-1,\ldots,1}{n-1,n-2,\ldots,0}$ or
$\binom{n-1,n-2,\ldots,0}{n,n-1,\ldots,1}$. Therefore

$\Theta\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho\cdot{\rm
sgn})=\min\\{2{\rm
rk}\textstyle\binom{n-1,n-2,\ldots,0}{n-1,n-2,\ldots,1},2{\rm
rk}\binom{n-2,n-3,\ldots,0}{n,n-1,\ldots,1}\\}=2n-2.$

Suppose that $\text{\bf G}={\rm O}^{-}_{2n}$ and $n\geq 1$. The Steinberg
characters are the unipotent characters associated to the symbols
$\binom{n,n-1,\ldots,0}{n-1,n-2,\ldots,1}$ or
$\binom{n-1,n-2,\ldots,1}{n,n-1,\ldots,0}$. Therefore,

$\Theta\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho\cdot{\rm
sgn})=\min\\{2{\rm
rk}\textstyle\binom{n-1,n-2,\ldots,1}{n-1,n-2,\ldots,0},2{\rm
rk}\binom{n-2,n-3,\ldots,1}{n,n-1,\ldots,0}\\}=2n-2.$

  3. (3)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$. The two Steinberg characters
$\rho,\rho\cdot{\rm sgn}$ of $G$ are the two unipotent characters associated
to the symbol $\binom{n,n-1,\ldots,1}{n,n-1,\ldots,0}$. Therefore, by the same
argument in the symplectic case, we obtain $\Theta\text{\rm-
rk}(\rho)=\Theta\text{\rm-rk}(\rho\cdot{\rm sgn})=2n$.

  4. (4)

Suppose that $\text{\bf G}={\rm U}_{n}$ and $n\geq 1$.

    1. (a)

Suppose that $n$ is even. The Steinberg character of $G$ is the unipotent
character associated to the symbol
$\Lambda=\binom{\frac{n}{2}-1,\frac{n}{2}-2,\ldots,0}{\frac{n}{2},\frac{n}{2}-1,\ldots,1}$
and $\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{-}{1,1,\ldots,1}$
($\frac{n}{2}$ copies of “$1$”). Then

$\Theta\text{\rm-rk}({\rm St}_{{\rm U}_{n}})={\rm rk}_{\rm
U}\textstyle\binom{\frac{n}{2}-1,\frac{n}{2}-2,\ldots,1}{\frac{n}{2}-1,\frac{n}{2}-2,\ldots,0}=2\left(\frac{n}{2}-1\right)+\frac{1\cdot
2}{2}=n-1.$

    2. (b)

Suppose that $n$ is odd. The Steinberg character of $G$ is the unipotent
character associated to the symbol
$\Lambda=\binom{\frac{n-1}{2},\frac{n-3}{2},\ldots,1}{\frac{n-1}{2},\frac{n-3}{2},\ldots,0}$
and $\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{1,1,\ldots,1}{-}$
($\frac{n-1}{2}$ copies of “$1$”). Then

$\Theta\text{\rm-rk}({\rm St}_{{\rm U}_{n}})={\rm rk}_{\rm
U}\textstyle\binom{\frac{n-1}{2},\frac{n-3}{2},\ldots,0}{\frac{n-3}{2},\frac{n-5}{2},\ldots,1}=2\left(\frac{n-1}{2}\right)+\frac{0\cdot
1}{2}=n-1.$

######  Lemma 3.9.

Let G be ${\rm Sp}_{2n},{\rm O}^{+}_{2n}$, or ${\rm O}_{2n+1}$, and let
$\rho\in{\mathcal{E}}(G)_{1}$. If $\Theta\text{\rm-rk}(\rho)=2n$, then

$\begin{cases}\text{$\rho$ is a Steinberg character},&\text{if\/ $\text{\bf
G}={\rm Sp}_{2n},{\rm O}_{2n+1}$};\\\ n=0,&\text{if\/ $\text{\bf G}={\rm
O}^{+}_{2n}$}.\end{cases}$

Moreover, ${\rm O}^{-}_{2n}$ does not have any unipotent character of
$\Theta$-rank $2n$.

###### Proof.

Let $\rho=\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$ ($\rho=\rho_{\Lambda}$ or
$\rho=\rho_{\Lambda}\cdot{\rm sgn}$ if $\text{\bf G}={\rm O}_{2n+1}$) such
that $\Theta\text{\rm-rk}(\rho)=2n$, and write

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

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. By Proposition 3.4, we need both
${\rm rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}=n$
and ${\rm
rk}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}=n$, and
then
$\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}\in{\mathcal{S}}_{{\rm
O}^{+}_{2n}}$ and
$\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}\in{\mathcal{S}}_{{\rm
O}^{-}_{2n}}$. So we need $\mu_{1}=0$ and $\nu_{1}=1$, thus
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{-}{1,1,\ldots,1}$ ($n$ copies of
“$1$”). That is, $\rho_{\Lambda}$ is the Steinberg character of ${\rm
Sp}_{2n}(q)$.

  2. (2)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$. By Corollary 3.6, we know that
$\Theta\text{\rm-rk}(\rho_{\Lambda})=2n$ if and only if $\Theta\text{\rm-
rk}(\rho_{\Lambda^{\rm t}})=2n$. Now by (1), $\rho_{\Lambda^{\rm t}}$ is the
Steinberg character of ${\rm Sp}_{2n}(q)$, hence $\rho_{\Lambda}$ is a
Steinberg character of ${\rm O}_{2n+1}(q)$.

  3. (3)

Suppose that $\text{\bf G}={\rm O}^{+}_{2n}$. By Proposition 3.4, we need both
${\rm rk}\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}=n$
and ${\rm
rk}\binom{b_{2},b_{3},\ldots,b_{m_{2}}}{a_{1},a_{2},\ldots,a_{m_{1}}}=n$, and
then
$\binom{b_{1},b_{2},\ldots,b_{m_{2}}}{a_{2},a_{3},\ldots,a_{m_{1}}}\in{\mathcal{S}}_{{\rm
Sp}_{2n}}$ and
$\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{2},b_{3},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{{\rm
Sp}_{2n}}$. So we need $\mu_{1}=0$ and $\nu_{1}=0$, thus
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{-}{-}$, i.e., $\rho_{\Lambda}$ is a
cuspidal unipotent character. But from Example 3.7, we must have $n=0$.

  4. (4)

Suppose that $\text{\bf G}={\rm O}^{-}_{2n}$ (where $n\geq 1$). Now
$|\Upsilon(\Lambda)|=n-1$ and the minimum value of ${\rm rk}(\Lambda^{\rm
t}\smallsetminus\binom{-}{a_{1}})$ and ${\rm rk}(\Lambda^{\rm
t}\smallsetminus\binom{b_{1}}{-})$ is less than or equal to $n-1$. This means
that the maximum value of $\Theta\text{\rm-rk}(\rho)$ for
$\rho\in{\mathcal{E}}({\rm O}^{-}_{2n}(q))_{1}$ is at most $2(n-1)$.

∎

######  Lemma 3.10.

Let $\rho$ be an irreducible pseudo-unipotent character of ${\rm Sp}_{2n}(q)$
and write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.
Then

$\Theta\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho_{\Lambda^{(-)}})+1,$

in particular, $\Theta\text{\rm-rk}(\rho)$ is odd.

###### Proof.

Let $\rho$ be an irreducible pseudo-unipotent character of $G={\rm
Sp}_{2n}(q)$. Then by Lemma 2.14, to determine $\Theta\text{\rm-rk}(\rho)$, we
only need to consider the first occurrence of $\rho$ for the dual pair
$(\text{\bf G}^{\prime},\text{\bf G})=({\rm O}_{2n^{\prime}+1},{\rm
Sp}_{2n})$. Now we have $\rho\in{\mathcal{E}}(G)_{s}$ such that $\text{\bf
G}^{(0)}={\rm U}_{0}$, $\text{\bf G}^{(-)}={\rm O}^{\epsilon^{\prime}}_{2n}$
for some $\epsilon^{\prime}$, and $\text{\bf G}^{(+)}={\rm Sp}_{0}$. Consider
the diagram

$\begin{CD}\rho^{\prime}@>{}>{}>\rho\\\ @V{}V{}V@V{}V{}V\\\ {\bf
1}\otimes\rho_{\Lambda^{\prime(+)}}\otimes{\bf 1}@>{}>{}>{\bf
1}\otimes\rho_{\Lambda^{(-)}}\otimes{\bf 1}\end{CD}$

Then $\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$ such that
$\text{\bf G}^{\prime(0)}={\rm U}_{0}$, $\text{\bf G}^{\prime(-)}={\rm
Sp}_{2n^{\prime}}$ and $\text{\bf G}^{(+)}={\rm Sp}_{0}$. Now $\rho$ of ${\rm
Sp}_{2n}(q)$ occurs in the correspondence for the dual pair $({\rm
O}_{2n^{\prime}+1},{\rm Sp}_{2n})$ if and only if $\rho_{\Lambda^{(-)}}$ of
${\rm O}^{\epsilon^{\prime}}_{2n}(q)$ occurs in the correspondence for the
dual pair $({\rm Sp}_{2n^{\prime}},{\rm O}^{\epsilon^{\prime}}_{2n})$. Thus
the lemma is proved. ∎

######  Lemma 3.11.

Let $\rho\in{\mathcal{E}}({\rm U}_{n}(q))_{1}$. If $\Theta\text{\rm-
rk}(\rho)=n$, then $n=0$.

###### Proof.

Suppose that $\rho=\rho_{\Lambda}\in{\mathcal{E}}(G)_{1}$. By the same
argument in the proof of Lemma 3.9, we must have
$\Upsilon(\Lambda)=\genfrac{[}{]}{0.0pt}{}{0}{0}$, i.e., $\rho_{\Lambda}$ is
cuspidal. This means that ${\rm rk}_{\rm U}(\Lambda)=\frac{1}{2}d(d+1)$ and
$\Theta\text{\rm-rk}(\Lambda)=\frac{1}{2}d(d-1)$ for some non-negative integer
$d$. The equality $n=\frac{1}{2}d(d+1)=\frac{1}{2}d(d-1)$ implies that
$n=d=0$. ∎

###  3.3. $\Theta$-ranks for symplectic groups or orthogonal groups

Now we can describe the $\Theta$-rank of a general irreducible character of a
symplectic group or an orthogonal group in terms of its data under the
modified Lusztig correspondence.

######  Proposition 3.12.

Let G be ${\rm Sp}_{2n}$, ${\rm O}^{\epsilon}_{2n}$, or ${\rm O}_{2n+1}$, and
let $\rho\in{\mathcal{E}}(G)$. Suppose that $\rho\in{\mathcal{E}}(G)_{s}$ for
some $s$ and write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$.
Then $\Theta\text{\rm-rk}(\rho)$ is equal to

$\begin{cases}\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\Lambda^{(+)}),2n-2n^{(-)}+\Theta\text{\rm-
rk}(\Lambda^{(-)})+1\\},&\text{if\/ $\text{\bf G}={\rm Sp}_{2n}$};\\\
\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\Lambda^{(+)}),2n-2n^{(-)}+\Theta\text{\rm-
rk}(\Lambda^{(-)})\\},&\text{if\/ $\text{\bf G}={\rm O}^{\epsilon}_{2n},{\rm
O}_{2n+1}$}\end{cases}$

where $n^{(-)},n^{(+)}$ are given in Subsection 2.4, and $\Theta\text{\rm-
rk}(\Lambda^{(+)}),\Theta\text{\rm-rk}(\Lambda^{(-)})$ are given in (3.3).

###### Proof.

First suppose that $\rho^{\prime}\otimes\rho$ occurs in the correspondence for
the dual pair $(\text{\bf G}^{\prime},\text{\bf G})$ for some
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$ and write
$\Xi_{s^{\prime}}(\rho^{\prime})=\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}$.

  1. (1)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. First we consider the dual pair
$(\text{\bf G}^{\prime},\text{\bf G})=({\rm O}^{\epsilon}_{2n^{\prime}},{\rm
Sp}_{2n})$ for some $n^{\prime}$. Then we have a commutative diagram

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(-)}}\otimes\rho_{\Lambda^{\prime(+)}}@>{{\rm
id}\otimes{\rm id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf
G}^{(+)}}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Now we have $\text{\bf G}^{\prime(-)}={\rm
O}^{\epsilon^{\prime}\epsilon}_{2n^{\prime(-)}}$ and $\text{\bf
G}^{\prime(+)}={\rm O}^{\epsilon^{\prime}}_{2n^{\prime(+)}}$; $\text{\bf
G}^{(-)}={\rm O}^{\epsilon^{\prime}\epsilon}_{2n^{(-)}}$ and $\text{\bf
G}^{(+)}={\rm Sp}_{2n^{(+)}}$ for some $\epsilon^{\prime}$ depending on $s$.
Because now $G^{\prime(0)}=G^{(0)}$ and $G^{\prime(-)}=G^{(-)}$, we have
$2n^{\prime}-2n^{\prime(+)}=2n-2n^{(+)}$, i.e.,
$2n^{\prime}=2n-2n^{(+)}+2n^{\prime(+)}$. Now $n$ and $n^{(+)}$ are fixed and
the minimal value of $2n^{\prime(+)}$ (for both possible $\epsilon=+$ or $-$)
is equal to $\Theta\text{\rm-rk}(\Lambda^{(+)})$.

Next we consider the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}_{2n^{\prime}+1},{\rm Sp}_{2n})$. Then we have a commutative diagram

$\begin{CD}\rho^{\prime}@>{\Theta_{\text{\bf G}^{\prime},\text{\bf
G}}}>{}>\rho\\\ @V{\iota\circ\Xi_{s^{\prime}}}V{}V@V{}V{\Xi_{s}}V\\\
\rho^{\prime(0)}\otimes\rho_{\Lambda^{\prime(+)}}\otimes\rho_{\Lambda^{\prime(-)}}@>{{\rm
id}\otimes\Theta_{\text{\bf G}^{\prime(+)},\text{\bf G}^{(-)}}\otimes{\rm
id}}>{}>\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}\end{CD}$

Now we have $\text{\bf G}^{\prime(-)}={\rm Sp}_{2n^{\prime(-)}}$ and
$\text{\bf G}^{\prime(+)}={\rm Sp}_{2n^{\prime(+)}}$; $\text{\bf G}^{(-)}={\rm
O}^{\epsilon^{\prime}}_{2n^{(-)}}$ and $\text{\bf G}^{(+)}={\rm
Sp}_{2n^{(+)}}$ for some $\epsilon^{\prime}$. Because now
$G^{\prime(0)}=G^{(0)}$ and $G^{\prime(-)}=G^{(+)}$, we have
$2n^{\prime}-2n^{\prime(+)}=2n-2n^{(-)}$, i.e.,
$2n^{\prime}+1=2n-2n^{(-)}+2n^{\prime(+)}+1$. Now $n$ and $n^{(-)}$ are fixed
and the minimal value of $2n^{\prime(+)}$ is equal to $\Theta\text{\rm-
rk}(\Lambda^{(-)})$.

From the definition of $\Theta$-rank and Proposition 3.4, we see that

$\displaystyle\Theta\text{\rm-rk}(\rho)$
$\displaystyle=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(+)}}),2n-2n^{(-)}+\Theta\text{\rm-
rk}(\rho_{\Lambda^{(-)}})+1\\}$
$\displaystyle=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\Lambda^{(+)}),2n-2n^{(-)}+\Theta\text{\rm-rk}(\Lambda^{(-)})+1\\}.$

  2. (2)

Suppose that $\text{\bf G}={\rm O}^{\epsilon}_{2n}$. Now we consider the dual
pair $({\rm Sp}_{2n^{\prime}},{\rm O}^{\epsilon}_{2n})$ and we have a
commutative diagram

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

Now we have $\text{\bf G}^{\prime(-)}(s^{\prime})={\rm
O}^{\epsilon^{\prime}\epsilon}_{2n^{\prime(-)}}$ and $\text{\bf
G}^{\prime(+)}(s^{\prime})={\rm Sp}_{2n^{\prime(+)}}$; $\text{\bf
G}^{(-)}(s)={\rm O}^{\epsilon^{\prime}\epsilon}_{2n^{(-)}}$ and $\text{\bf
G}^{(+)}(s)={\rm O}^{\epsilon^{\prime}}_{2n^{(+)}}$ for some
$\epsilon^{\prime}$.

Now we need consider the occurrences for four irreducible characters $\rho$,
$\rho\cdot{\rm sgn}$, $\rho\chi_{\text{\bf G}}$ and $\rho\chi_{\text{\bf
G}}\cdot{\rm sgn}$. Note that by Lemma 2.9, we have
$\Lambda^{(\varepsilon)}(\rho\cdot{\rm
sgn})=(\Lambda^{(\varepsilon)}(\rho))^{\rm t}$ and
$\Lambda^{(\varepsilon)}(\rho\chi_{\text{\bf
G}})=\Lambda^{(-\varepsilon)}(\rho)$ for $\varepsilon=+$ or $-$. By the
similar argument in (1), we conclude that

$\Theta\text{\rm-rk}(\rho)=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\Lambda^{(+)}),2n-2n^{(-)}+\Theta\text{\rm-rk}(\Lambda^{(-)})\\}.$

  3. (3)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$. Now consider the dual pair $({\rm
Sp}_{2n^{\prime}},{\rm O}_{2n+1})$ and we have a commutative diagram

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

Now we have $\text{\bf G}^{\prime(-)}(s^{\prime})={\rm
O}^{\epsilon^{\prime}\epsilon}_{2n^{\prime(-)}}$ and $\text{\bf
G}^{\prime(+)}(s^{\prime})={\rm Sp}_{2n^{\prime(+)}}$; $\text{\bf
G}^{(-)}(s)={\rm Sp}_{2n^{(-)}}$ and $\text{\bf G}^{(+)}(s)={\rm
Sp}_{2n^{(+)}}$ for some $\epsilon^{\prime}$. Then by the similar argument in
(2), we conclude that

$\Theta\text{\rm-rk}(\rho)=\min\\{2n-2n^{(+)}+\Theta\text{\rm-
rk}(\Lambda^{(+)}),2n-2n^{(-)}+\Theta\text{\rm-rk}(\Lambda^{(-)})\\}.$

∎

######  Example 3.13.

There are two irreducible characters $\rho_{1},\rho_{2}$ of degree
$\frac{1}{2}(q^{n}+1)$ of ${\rm Sp}_{2n}(q)$ for $n\geq 1$. They are in the
Lusztig series ${\mathcal{E}}(G)_{s}$ such that $\text{\bf G}^{(0)}={\rm
U}_{0}$, $\text{\bf G}^{(-)}={\rm O}^{+}_{2n}$, $\text{\bf G}^{(+)}={\rm
Sp}_{0}$, i.e., $n^{(-)}=n$ and $n^{(+)}=0$. Moreover, we know that
$\Lambda^{(-)}=\binom{n}{0}$ or $\binom{0}{n}$, and
$\Lambda^{(+)}=\binom{0}{-}$. Therefore

$\displaystyle\Theta\text{\rm-rk}(\rho_{1})=\Theta\text{\rm-rk}(\rho_{2})$
$\displaystyle=\min\\{2n-0+\Theta\text{\rm-
rk}(\textstyle\binom{0}{-}),2n-2n+\Theta\text{\rm-
rk}(\textstyle\binom{n}{0})+1\\}$ $\displaystyle=\min\\{2n,1\\}=1.$

In fact, both $\rho_{1},\rho_{2}$ are pseudo-unipotent, and each of
$(\rho_{1})_{\Lambda^{(-)}},(\rho_{2})_{\Lambda^{(-)}}$ is the trivial
character or the ${\rm sgn}$ character of ${\rm O}^{+}_{2n}(q)$. So the above
result is also a direct consequence of Lemma 3.10.

######  Corollary 3.14.

Let $\text{\bf G}={\rm Sp}_{2n},{\rm O}_{2n+1}$, or ${\rm O}^{\epsilon}_{2n}$,
and let $\rho\in{\mathcal{E}}(G)_{s}$ for some $s$. Then $\Theta\text{\rm-
rk}(\rho)=2n$ if and only if

$\begin{cases}\text{$n^{(-)}=0$ and $\rho_{\Lambda^{(+)}}$ is a Steinberg
character},&\text{if\/ $\text{\bf G}={\rm Sp}_{2n}$};\\\ \text{both
$\rho_{\Lambda^{(-)}},\rho_{\Lambda^{(+)}}$ are Steinberg
characters},&\text{if\/ $\text{\bf G}={\rm O}_{2n+1}$};\\\ \text{both
$n^{(-)}=n^{(+)}=0$},&\text{if\/ $\text{\bf G}={\rm
O}^{\epsilon}_{2n}$}.\end{cases}$

###### Proof.

Suppose that $\text{\bf G}={\rm Sp}_{2n}$. By Proposition 3.12, we know that
$\Theta\text{\rm-rk}(\rho)=2n$ if and only if $\Theta\text{\rm-
rk}(\rho_{\Lambda^{(-)}})=2n^{(-)}$ and $\Theta\text{\rm-
rk}(\rho_{\Lambda^{(+)}})=2n^{(+)}$. Now $\text{\bf G}^{(-)}={\rm
O}_{2n^{(-)}}$ and $G^{(+)}={\rm Sp}_{2n^{(+)}}$, so we need $n^{(-)}=0$ and
$\rho^{(+)}$ to be the Steinberg character by Lemma 3.9. The proofs for two
other cases are similar. ∎

We provide some necessary condition in Remark 3.2 on the possible values of
$\Theta\text{\rm-rk}(\rho)$ for $\rho\in{\mathcal{E}}(G)$. Now we also have
the sufficient condition:

######  Proposition 3.15.

Let $\text{\bf G}={\rm O}^{\epsilon}_{2n}$, ${\rm Sp}_{2n}$ or ${\rm
O}_{2n+1}$ over $\text{\bf f}_{q}$.

  1. (i)

Suppose that $\text{\bf G}={\rm O}^{\epsilon}_{2n}$.

    1. (a)

For each even integer $k$ such that $0\leq k<2n$, there exists an irreducible
character $\rho\in{\mathcal{E}}({\rm O}^{\epsilon}_{2n}(q))$ such that
$\Theta\text{\rm-rk}(\rho)=k$.

    2. (b)

If $2n\geq 4$, or $q>3$, or $\epsilon=-$, then there exists an irreducible
character $\rho\in{\mathcal{E}}({\rm O}^{\epsilon}_{2n}(q))$ such that
$\Theta\text{\rm-rk}(\rho)=2n$.

    3. (c)

There is no irreducible character of ${\rm O}^{+}_{2}(3)$ of $\Theta$-rank
$2$.

  2. (ii)

For each integer $k$ such that $0\leq k\leq 2n$, there exists an irreducible
character $\rho\in{\mathcal{E}}({\rm Sp}_{2n}(q))$ such that $\Theta\text{\rm-
rk}(\rho)=k$.

  3. (iii)

For each even integer $k$ such that $0\leq k\leq 2n$, there exists an
irreducible character $\rho\in{\mathcal{E}}({\rm O}_{2n+1}(q))$ such that
$\Theta\text{\rm-rk}(\rho)=k$.

###### Proof.

  1. (1)

Suppose that $\text{\bf G}={\rm O}^{-}_{2n}$. The assertion is trivial when
$k=0$, so we will assume that $k$ is an even integer such that $2\leq k\leq
2n$.

    1. (a)

Suppose that $2\leq k<2n$. Let
$\Lambda_{k}=\binom{n-1,\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-2,\frac{k}{2}-3,\ldots,0}\in{\mathcal{S}}_{{\rm
O}^{-}_{2n}}$. Then we have
$\Upsilon(\Lambda_{k})=\genfrac{[}{]}{0.0pt}{}{-}{n-\frac{k}{2}-1,1,\ldots,1}\in{\mathcal{P}}_{2}(n-1)$.
Then
$\binom{\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-2,\frac{k}{2}-3,\ldots,0}\in{\mathcal{S}}_{{\rm
Sp}_{k}}$. Moreover, $\binom{-}{n,2,0}\in{\mathcal{S}}_{{\rm Sp}_{2n+2}}$ if
$k=2$;
$\binom{\frac{k}{2}-3,\frac{k}{2}-4,\ldots,0}{n-1,\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
Sp}_{2n+2}}$ if $k\geq 4$. Then by Proposition 3.4, $\Theta\text{\rm-
rk}(\rho_{\Lambda_{k}})=\min\\{k,2n+2\\}=k$.

    2. (b)

It is known that there exists a semisimple element $s\in G^{*}={\rm
O}^{-}_{2n}(q)$ such that $C_{G^{*}}(s)\simeq{\rm U}_{1}(q^{n})$. That is,
$n^{(-)}=n^{(+)}=0$, and then any $\rho\in{\mathcal{E}}(G)_{s}$ has
$\Theta\text{\rm-rk}(\rho)=2n$ by Proposition 3.12.

  2. (2)

Suppose that $\text{\bf G}={\rm O}_{2n}^{+}$ and $k$ is an even integer such
that $0\leq k\leq 2n$.

    1. (a)

Suppose that $k<2n$. Let
$\Lambda_{k}=\binom{n,\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2},\frac{k}{2}-1,\ldots,0}\in{\mathcal{S}}_{{\rm
O}^{+}_{2n}}$. Then we have
$\Upsilon(\Lambda_{k})=\genfrac{[}{]}{0.0pt}{}{n-\frac{k}{2},1,\ldots,1}{-}\in{\mathcal{P}}_{2}(n)$,
$\binom{\frac{k}{2},\frac{k}{2}-1,\ldots,0}{\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
Sp}_{k}}$, and
$\binom{n,\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-1,\frac{k}{2}-2,\ldots,0}\in{\mathcal{S}}_{{\rm
Sp}_{2n}}$. Hence by Proposition 3.4, $\Theta\text{\rm-
rk}(\rho_{\Lambda_{k}})=\min\\{k,2n\\}=k$.

    2. (b)

Suppose that $q>3$ or $n\geq 2$. It is known that there exists a semisimple
element $s\in G^{*}={\rm O}^{+}_{2n}(q)$ such that $C_{G^{*}}(s)\simeq{\rm
GL}_{1}(q^{n})$. That is, $n^{(-)}=n^{(+)}=0$, and then any
$\rho\in{\mathcal{E}}(G)_{s}$ has $\Theta\text{\rm-rk}(\rho)=2n$ by
Proposition 3.12.

    3. (c)

Now ${\rm O}^{+}_{2}(3)$ is a finite group of $4$ element. Each irreducible
character of ${\rm O}^{+}_{2}(3)$ is a linear character and hence has
$\Theta$-rank $0$.

  3. (3)

Suppose that $\text{\bf G}={\rm Sp}_{2n}$ and $k$ is an integer such that
$0\leq k\leq 2n$.

    1. (a)

Suppose that $k$ is even and $0\leq k<2n$. Let
$\Lambda_{k}=\binom{n,\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-1,\frac{k}{2}-2,\ldots,0}\in{\mathcal{S}}_{{\rm
Sp}_{2n}}$ and then
$\Upsilon(\Lambda_{k})=\genfrac{[}{]}{0.0pt}{}{n-\frac{k}{2},1,\ldots,1}{-}\in{\mathcal{P}}_{2}(n)$.
Now
$\binom{\frac{k}{2}-1,\frac{k}{2}-2,\ldots,0}{\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
O}^{+}_{2k}}$,
$\binom{\frac{k}{2}-2,\frac{k}{2}-3,\ldots,0}{n,\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
O}^{-}_{2n+2}}$. Thus by Proposition 3.4, $\Theta\text{\rm-
rk}(\rho_{\Lambda_{k}})=\min\\{k,2n+2\\}=k$.

    2. (b)

Suppose that $k=2n$. Let
$\Lambda_{k}=\binom{n,n-1,\ldots,0}{n,n-1,\ldots,1}\in{\mathcal{S}}_{{\rm
Sp}_{2n}}$. Then
$\Upsilon(\Lambda_{k})=\genfrac{[}{]}{0.0pt}{}{-}{1,1,\ldots,1}\in{\mathcal{P}}_{2}(n)$.
Now $\binom{n,n-1,\ldots,1}{n-1,n-2,\ldots,0}\in{\mathcal{S}}_{{\rm
O}^{+}_{2n}}$ and
$\binom{n-1,n-2,\ldots,1}{n,n-1,\ldots,0}\in{\mathcal{S}}_{{\rm O}^{-}_{2n}}$.
Thus $\Theta\text{\rm-rk}(\rho_{\Lambda_{k}})=2n$.

    3. (c)

Suppose that $k$ is odd. Then $k-1$ is even and $0\leq k-1<2n$. Let $s$ be an
semisimple element in $G^{*}$ such that $\text{\bf G}^{(0)}={\rm U}_{0}$,
$\text{\bf G}^{(-)}={\rm O}^{+}_{2n}$ and $\text{\bf G}^{(+)}={\rm Sp}_{0}$.
Let $\rho\in{\mathcal{E}}(G)_{s}$ such that $\Xi_{s}(\rho)={\bf
1}\otimes\rho_{\Lambda^{(-)}}\otimes{\bf 1}$ where
$\Lambda^{(-)}\in{\mathcal{S}}_{{\rm O}^{+}_{2n}}$ is the symbol
$\Lambda_{k-1}$ given in (2.a). Now $\rho$ is pseudo-unipotent (cf. Subsection
2.7), and by Lemma 3.10, $\Theta\text{\rm-rk}(\rho)=\Theta\text{\rm-
rk}(\rho_{\Lambda_{k-1}})+1=k-1+1=k$.

  4. (4)

Suppose that $\text{\bf G}={\rm O}_{2n+1}$ and $k$ is an even integer such
that $0\leq k\leq 2n$. Let $\rho=\rho_{\Lambda}$ where $\Lambda^{\rm t}$ is
equal to the symbol $\Lambda_{k}\in{\mathcal{S}}_{{\rm Sp}_{2n}}$ given in
(3.a) and (3.b). Then by Corollary 3.6, we have $\Theta\text{\rm-
rk}(\rho)=\Theta\text{\rm-rk}(\rho_{\Lambda^{\rm t}})=k$.

∎

###  3.4. $\Theta$-ranks for unitary groups

In this subsection, we consider $\text{\bf G}={\rm U}_{n}$. For $s\in G^{*}$,
we can write

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

for some non-negative integers $r,t_{i},n_{ij}$ such that
$\sum_{i=1}^{r}\sum_{j=1}^{t_{i}}in_{ij}=n$ where ${\rm GL}^{+1}_{n_{ij}}={\rm
GL}^{+}_{n_{ij}}={\rm GL}_{n_{ij}}$ and ${\rm GL}^{-1}_{n_{ij}}={\rm
GL}^{-}_{n_{ij}}={\rm U}_{n_{ij}}$. 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
$\lambda_{j}^{q+1}=1$. Then for the Lusztig correspondence
${\mathfrak{L}}_{s}\colon{\mathcal{E}}(G)_{1}\rightarrow{\mathcal{E}}(C_{G^{*}}(s))_{1}$
we can write

(3.16)
${\mathfrak{L}}_{s}(\rho)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{(ij)}$

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

######  Proposition 3.17.

Let $\rho\in{\mathcal{E}}({\rm U}_{n}(q))_{s}$ for some $s$, and keep the
above notations. Then

$\Theta\text{\rm-rk}(\rho)=\min\\{n-n_{1,j}+\Theta\text{\rm-
rk}(\rho^{(1,j)})\mid j=1,\ldots,q+1\\}.$

###### Proof.

Write ${\mathfrak{L}}_{s}(\rho)$ as in (3.16). Suppose that $\text{\bf
G}^{(+)}(s)={\rm U}_{n_{1,1}}$. Then the minimal value of $n^{\prime}$ such
that $\rho$ occurs in the $\Theta$-correspondence for the pair $({\rm
U}_{n^{\prime}},{\rm U}_{n})$ is equal to $n-n_{1,1}+\Theta\text{\rm-
rk}(\rho^{(1,1)})$. Suppose that $\chi$ is a linear character of ${\rm
U}_{n}(q)$. Then $\rho\chi\in{\mathcal{E}}(G)_{s^{\prime}}$ where
$s^{\prime}=sa$ for some $a\in\text{\bf f}_{q^{2}}$ such that $a^{q+1}=1$.
Then we can write

$C_{G^{*}}(s^{\prime})=\prod_{i=1}^{r}\prod_{j=1}^{t_{i}}{\rm
GL}_{n^{\prime}_{ij}}^{(-1)^{i}}(q^{i})\quad\text{and}\quad{\mathfrak{L}}_{s^{\prime}}(\rho\chi)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{\prime(ij)}$

where $\\{{\rm U}_{n^{\prime}_{1,1}},\ldots{\rm U}_{n^{\prime}_{1,t_{1}}}\\}$
is a permutation of $\\{{\rm U}_{n_{1,1}},\ldots{\rm U}_{n_{1,t_{1}}}\\}$ and
$\\{\rho^{\prime(1,1)},\ldots,\rho^{\prime(1,t_{1})}\\}$ are a permutation of
$\\{\rho^{(1,1)},\ldots,\rho^{(1,t_{1})}\\}$. Moreover, for any $1\leq j\leq
t_{1}$ there exists a linear character $\chi$ of ${\rm U}_{n}(q)$ such that
$\rho^{\prime(1,1)}=\rho^{(1,j)}$. Hence the proposition follows from the
definition of the $\Theta$-rank. ∎

######  Remark 3.18.

The same result of this proposition can be found in [GLT20] theorem 3.9.

######  Corollary 3.19.

Let $\rho\in{\mathcal{E}}({\rm U}_{n}(q))$. Then $\Theta\text{\rm-rk}(\rho)=n$
if and only if $n_{1,j}=0$ for each $j=1,\ldots,q+1$.

###### Proof.

By Proposition 3.17, $\Theta\text{\rm-rk}(\rho)=n$ implies that
$\Theta\text{\rm-rk}(\rho^{(1,j)})=n_{1,j}$ for each $j=1,\ldots,q+1$. Then by
Lemma 3.11, we conclude that $n_{1,j}=0$ for each $j=1,\ldots,q+1$. ∎

######  Proposition 3.20.

For each integer $k$ such that $k=0$ if $n=0,1$; or $0\leq k\leq n$ if $n\geq
2$, there exists an irreducible character $\rho\in{\mathcal{E}}({\rm
U}_{n}(q))$ such that $\Theta\text{\rm-rk}(\rho)=k$. In particular, there is
no $\rho\in{\mathcal{E}}({\rm U}_{1}(q))$ of $\Theta$-rank $1$.

###### Proof.

The case for $n=0$ is trivial. For $n=1$, every irreducible character of ${\rm
U}_{1}(q)$ is a linear character and hence of $\Theta$-rank $0$. Now we assume
$n\geq 2$.

  1. (1)

Suppose that $0\leq k<n$.

    1. (a)

Suppose that $n+k$ is even. Let

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

      1. (i)

Suppose that $n$ is even. Now
$\binom{\frac{k}{2},\frac{k}{2}-1,\ldots,0}{\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
U}_{k}}$ and
$\binom{\frac{k}{2}-1,\frac{k}{2}-2,\ldots,0}{\frac{n}{2},\frac{k}{2},\frac{k}{2}-1,\ldots,1}\in{\mathcal{S}}_{{\rm
U}_{n+1}}$. Hence $\Theta\text{\rm-rk}(\rho_{\Lambda_{k}})=\min\\{k,n+1\\}=k$.

      2. (ii)

Suppose that $n$ is odd. Now
$\binom{\frac{k+1}{2},\frac{k-1}{2},\ldots,0}{\frac{k-1}{2},\frac{k-3}{2},\ldots,1}\in{\mathcal{S}}_{{\rm
U}_{k}}$ and
$\binom{\frac{k-1}{2},\frac{k-3}{2},\ldots,0}{\frac{n-1}{2},\frac{k-1}{2},\frac{k-3}{2},\ldots,1}\in{\mathcal{S}}_{{\rm
U}_{n-1}}$. Hence $\Theta\text{\rm-rk}(\rho_{\Lambda_{k}})=\min\\{k,n-1\\}=k$.

    2. (b)

Suppose that $n+k$ is odd. Let

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

      1. (i)

Suppose that $n$ is even. Now
$\binom{\frac{k-1}{2},\frac{k-3}{2},\ldots,1}{\frac{k-1}{2},\frac{k-3}{2},\ldots,0}\in{\mathcal{S}}_{{\rm
U}_{k}}$ and
$\binom{\frac{n}{2},\frac{k-1}{2},\frac{k-3}{2},\ldots,1}{\frac{k-1}{2},\frac{k-3}{2},\ldots,0}\in{\mathcal{S}}_{{\rm
U}_{n}}$. Hence $\Theta\text{\rm-rk}(\rho_{\Lambda_{k}})=\min\\{k,n\\}=k$.

      2. (ii)

Suppose that $n$ is odd. Now
$\binom{\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-1,\frac{k}{2}-2,\ldots,0}\in{\mathcal{S}}_{{\rm
U}_{k}}$ and
$\binom{\frac{n-1}{2},\frac{k}{2},\frac{k}{2}-1,\ldots,1}{\frac{k}{2}-2,\frac{k}{2}-3,\ldots,0}\in{\mathcal{S}}_{{\rm
U}_{n+2}}$. Hence $\Theta\text{\rm-rk}(\rho_{\Lambda_{k}})=\min\\{k,n+2\\}=k$.

  2. (2)

Suppose that $k=n$.

    1. (a)

If $n$ is even, let $\rho\in{\mathcal{E}}(G)_{s}$ such that $C_{G^{*}}(s)={\rm
GL}_{1}(q^{n})$.

    2. (b)

If $n$ is odd and $n\geq 3$, let $\rho\in{\mathcal{E}}(G)_{s}$ such that
$C_{G^{*}}(s)={\rm U}_{1}(q^{n})$.

For each case, such a semisimple element $s$ exists, and by Corollary 3.19, we
have $\Theta\text{\rm-rk}(\rho)=n$.

∎

###  3.5. $\Theta$-ranks and parabolic induction

Let $\text{\bf G}_{n}$ be one of ${\rm Sp}_{2n}$, ${\rm O}^{+}_{2n}$, ${\rm
O}^{-}_{2n+2}$, ${\rm O}_{2n+1}$, ${\rm U}_{2n}$, or ${\rm U}_{2n+1}$. For
$\rho\in{\mathcal{E}}(G_{n})$ and an integer $m\geq 0$, recall that a subset
${\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)\subset{\mathcal{E}}(G_{m})$ is defined in
Subsection 2.5.

######  Lemma 3.21.

Let $\rho\in{\mathcal{E}}(G_{n})_{1}$ and $m\geq n$. Then

$\Theta\text{\rm-rk}(\rho)=\min\\{\,\Theta\text{\rm-
rk}(\rho^{\prime})\mid\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)\,\\}$

###### Proof.

Let $\rho$ be a unipotent character of $G_{n}$. It is clear that we need only
to show that $\Theta\text{\rm-rk}(\rho^{\prime})\geq\Theta\text{\rm-rk}(\rho)$
for any $\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$, and there
exists $\rho_{0}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$ such that
$\Theta\text{\rm-rk}(\rho_{0})=\Theta\text{\rm-rk}(\rho)$.

Now we can write $\rho=\rho_{\Lambda}$ ($\rho=\rho_{\Lambda}$ or
$\rho=\rho_{\Lambda}\cdot{\rm sgn}$ if $\text{\bf G}={\rm O}_{2n+1}$) for some
symbol
$\Lambda=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\in{\mathcal{S}}_{\text{\bf
G}_{n}}$. Consider the following reductive dual pairs $(\text{\bf
G}^{\prime}_{n^{\prime}},\text{\bf G}_{n})$:

  1. (a)

$({\rm O}^{+}_{2n^{\prime}},{\rm Sp}_{2n})$, $({\rm Sp}_{2n^{\prime}},{\rm
O}^{+}_{2n})$, $({\rm U}_{2n^{\prime}},{\rm U}_{2n})$, $({\rm
U}_{2n^{\prime}+1},{\rm U}_{2n+1})$;

  2. (b)

$({\rm O}^{-}_{2n^{\prime}},{\rm Sp}_{2n})$, $({\rm Sp}_{2n^{\prime}},{\rm
O}^{-}_{2n+2})$, $({\rm U}_{2n^{\prime}+1},{\rm U}_{2n})$, $({\rm
U}_{2n^{\prime}},{\rm U}_{2n+1})$.

Let $\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$. Then
$\rho^{\prime}$ is also unipotent and we write
$\rho^{\prime}=\rho_{\Lambda^{\prime}}$
($\rho^{\prime}=\rho_{\Lambda^{\prime}}$ or
$\rho^{\prime}=\rho_{\Lambda^{\prime}}\cdot{\rm sgn}$ if $\text{\bf
G}_{n}={\rm O}_{2n+1}$) where $\Lambda^{\prime}$ is equal to one of the
following:

$\textstyle\binom{a_{1},\ldots,a_{i-1},a_{i}+1,a_{i+1},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}},\binom{a_{1}+1,a_{2}+1,\ldots,a_{m_{1}}+1,1}{b_{1}+1,b_{2}+1,\ldots,b_{m_{2}}+1,0},\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},\ldots,b_{i-1},b_{i}+1,b_{i+1},\ldots,b_{m_{2}}},\binom{a_{1}+1,a_{2}+1,\ldots,a_{m_{1}}+1,0}{b_{1}+1,b_{2}+1,\ldots,b_{m_{2}}+1,1}$

for some index $i$. By Proposition 3.4, it is clear that $\Theta\text{\rm-
rk}(\rho_{\Lambda^{\prime}})=\Theta\text{\rm-
rk}(\Lambda^{\prime})\geq\Theta\text{\rm-rk}(\Lambda)=\Theta\text{\rm-
rk}(\rho_{\Lambda})$. Now define $\Lambda_{0}\in{\mathcal{S}}_{\text{\bf
G}_{n+1}}$ by

$\Lambda_{0}=\begin{cases}\binom{a_{1}+1,a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}},&\text{if
$(\text{\bf G}^{\prime}_{n^{\prime}},\text{\bf G}_{n})$ is in case (a)};\\\
\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1}+1,b_{2},\ldots,b_{m_{2}}},&\text{if
$(\text{\bf G}^{\prime}_{n^{\prime}},\text{\bf G}_{n})$ is in case
(b)}.\end{cases}$

Then $\rho_{0}:=\rho_{\Lambda_{0}}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$
and by Proposition 3.4, we know that $\Theta\text{\rm-
rk}(\rho_{\Lambda_{0}})=\Theta\text{\rm-rk}(\rho_{\Lambda})$. ∎

From the proof of the lemma, we have the following result:

######  Corollary 3.22.

Let $\rho\in{\mathcal{E}}(G_{n})$ and
$\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$. Then either
$\Theta\text{\rm-rk}(\rho^{\prime})=\Theta\text{\rm-rk}(\rho)$ or
$\Theta\text{\rm-rk}(\rho^{\prime})=\Theta\text{\rm-rk}(\rho)+2$.

Then we generalize Lemma 3.21 to any irreducible characters.

######  Lemma 3.23.

Let $\rho\in{\mathcal{E}}(G_{n})$ and $m\geq n$. Then

$\Theta\text{\rm-rk}(\rho)=\min\\{\,\Theta\text{\rm-
rk}(\rho^{\prime})\mid\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{m}}(\rho\chi),\
\chi\text{ a linear character of }G_{n}\,\\}$

###### Proof.

As in the proof of Lemma 3.21, we need only to show that $\Theta\text{\rm-
rk}(\rho^{\prime})\geq\Theta\text{\rm-rk}(\rho)$ for any
$\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho\chi)$ for any linear
character $\chi$, and there exists
$\rho_{0}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho\chi)$ for some linear
character $\chi$ such that $\Theta\text{\rm-rk}(\rho_{0})=\Theta\text{\rm-
rk}(\rho)$.

Now we suppose that $\rho\in{\mathcal{E}}(G_{n})_{s}$ for some $s$. When
$\text{\bf G}_{n}$ is a symplectic group or an orthogonal group, we write
$\Xi_{s}(\rho)=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}}$
for some $\Lambda^{(\varepsilon)}\in{\mathcal{S}}_{\text{\bf
G}^{(\varepsilon)}}$ where $\varepsilon=\pm$.

  1. (1)

Suppose that $\text{\bf G}_{n}={\rm Sp}_{2n}$ and $\Theta\text{\rm-rk}(\rho)$
is even. By Proposition 3.12, we see that $\Theta\text{\rm-
rk}(\rho)=2n-2n^{(+)}+\Theta(\rho_{\Lambda^{(+)}})$. Let
$\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$. Then we know that
$\rho^{\prime}\in{\mathcal{E}}(G_{n+1})_{s_{n+1}}$ where
$s_{n+1}=(s,1)\in(G_{n}^{*})^{0}\times T_{1}\subset(G^{*}_{n+1})^{0}$. Now
$\Xi_{s_{n+1}}(\rho^{\prime})=\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{\prime(+)}}$
where
$\rho_{\Lambda^{\prime(+)}}\in{\mathcal{E}}^{G^{(+)}_{n^{(+)}+1}}_{G^{(+)}_{n^{(+)}}}(\rho_{\Lambda^{(+)}})$
by Lemma 2.10. Then by Lemma 3.21,

$\displaystyle\Theta\text{\rm-
rk}(\rho^{\prime})=2(n+1)-2(n^{(+)}+1)+\Theta\text{\rm-
rk}(\rho_{\Lambda^{\prime(+)}})$ $\displaystyle\geq
2n-2n^{(+)}+\Theta\text{\rm-rk}(\rho_{\Lambda^{(+)}})$
$\displaystyle=\Theta\text{\rm-rk}(\rho).$

Next let $\Lambda^{(+)}_{0}$ be defined as $\Lambda_{0}$ in the proof of Lemma
3.21, and let

$\rho_{0}=\Xi^{-1}_{s_{n+1}}(\rho^{(0)}\otimes\rho_{\Lambda^{(-)}}\otimes\rho_{\Lambda^{(+)}_{0}}).$

Then we have $\rho_{0}\in{\mathcal{E}}_{G_{n}}^{G_{n+1}}(\rho)$ and
$\Theta\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho_{0})$

  2. (2)

Suppose that $\text{\bf G}_{n}={\rm Sp}_{2n}$ and $\Theta\text{\rm-rk}(\rho)$
is odd. Then by Proposition 3.12, we have $\Theta\text{\rm-
rk}(\rho)=2n-2n^{(-)}+\Theta(\rho_{\Lambda^{(-)}})+1$. The remaining proof is
similar to case (1) with only “$+$” replaced by “$-$”.

  3. (3)

Suppose that $\text{\bf G}_{n}$ is an orthogonal group. By twisting a linear
character $\chi$ of $\text{\bf G}_{n}$, by Lemma 2.9 and Proposition 3.12, we
may assume $\Theta\text{\rm-
rk}(\rho)=2n-2n^{(+)}+\Theta(\rho_{\Lambda^{(+)}})$. Then the remaining proof
is similar to case (1).

Now suppose that $\text{\bf G}_{n}$ is a unitary group. Write
${\mathfrak{L}}_{s}(\rho)=\bigotimes_{i=1}^{r}\bigotimes_{j=1}^{t_{i}}\rho^{(ij)}$
as in (3.16). By Proposition 3.17, after twisting a linear character $\chi$ of
$\text{\bf G}_{n}$ if necessary, we may assume that $\Theta\text{\rm-
rk}(\rho)=n-n_{1,1}+\Theta(\rho^{(1,1)})$ where the notations are as in
Subsection 3.4. Then the remaining proof is similar to case (1), again. ∎

##  4\. $\Theta$-ranks and Other Ranks

###  4.1. $U$-rank, $A$-rank, $\overline{A}$-rank, and $\otimes$-rank

First we recall the definition of the $U$-rank and the asymptotic rank of an
irreducible character $\rho$ of a classical group $G=G(V)$ given in [GH17] and
[GH20] where $V$ is the standard space with an associated form over $\text{\bf
f}_{q}$ (if $G$ is a symplectic group or an orthogonal group) or $\text{\bf
f}_{q^{2}}$ (if $G$ is a unitary group) on which $G$ acts. Suppose that we
have a Witt decomposition

$V=X\oplus V_{0}\oplus Y$

where $X,Y$ are totally isotropic and dual to each other, and $V_{0}$ is
anisotropic. The group

$U=\\{\,g\in G(V)\mid(g-1)(X\oplus V_{0})=0,\ (g-1)(Y)\subset X\,\\}$

is an abelian subgroup of $G$. Then we have a decomposition

$\rho|_{U}=\sum_{\phi\in{\mathcal{E}}(U)}m_{\phi}\phi$

for $m_{\phi}\in{\mathbb{N}}\cup\\{0\\}$. We say that $\rho|_{U}$ _contains_
$\phi$ if $m_{\phi}\neq 0$. Now each character $\phi$ of $U$ is naturally
assigned a rank.

The irreducible character $\rho$ is said to be of _$U$ -rank_ $k$, denoted by
$U\text{\rm-rk}(\rho)=k$, if the restriction $\rho|_{U}$ contains a character
$\phi$ of rank $k$, but contains no character of higher rank. From the
definition, we see that $0\leq U\text{\rm-rk}(\rho)\leq n$ where
$n=\dim(X)=\dim(Y)$. Moreover, it is clear that $U\text{\rm-
rk}(\rho)=U\text{\rm-rk}(\rho\chi)$ for any linear character $\chi$ of $G$
because $\chi|_{U}$ is always trivial. An irreducible character
$\rho\in{\mathcal{E}}(G_{n})$ is said to be of _low $U$-rank_ if

$U\text{\rm-rk}(\rho)<\begin{cases}n,&\text{if $\text{\bf G}_{n}={\rm
Sp}_{2n}$, ${\rm U}_{2n}$ or ${\rm U}_{2n+1}$};\\\ n-1,&\text{if $\text{\bf
G}_{n}={\rm O}^{+}_{2n}$, ${\rm O}^{-}_{2n+2}$, or ${\rm
O}_{2n+1}$}.\end{cases}$

An irreducible character $\rho$ of $G_{n}$ is said to have _asymptotic rank_
less than or equal to $k$, if for all sufficiently large $N$, there is a
character $\rho_{N}\in{\mathcal{E}}(G_{N})$ such that $U\text{\rm-
rk}(\rho_{N})\leq k<N$ and $\rho$ occurs in $\rho_{N}|_{G_{n}}$. The
asymptotic rank of $\rho$ is denoted by $A\text{\rm-rk}(\rho)$. As indicated
in [GH20], it is known that

(4.1) $U\text{\rm-rk}(\rho)\leq A\text{\rm-rk}(\rho)$

for any $\rho\in{\mathcal{E}}(G)$.

To study the representation theory of reductive groups, parabolic induction is
usually more convenient than usual induction, so we modified the definition of
asymptotic rank to define the parabolic asymptotic rank as follows. An
irreducible character $\rho$ of $G_{n}$ is said to have _parabolic asymptotic
rank_ less than or equal to $k$, if for all sufficiently large $N$, there is a
character $\rho_{N}\in{\mathcal{E}}_{G_{n}}^{G_{N}}(\rho)$ such that
$U\text{\rm-rk}(\rho_{N})\leq k<N$. The parabolic asymptotic rank of $\rho$ is
denoted by $\overline{A}\text{\rm-rk}(\rho)$. It is clear from the definition
that

(4.2) $A\text{\rm-rk}(\rho)\leq\overline{A}\text{\rm-rk}(\rho)$

for any $\rho\in{\mathcal{E}}(G)$.

######  Lemma 4.3.

Let $\rho$ be an irreducible character of $G_{n}$ and $m\geq n$. Then

$\overline{A}\text{\rm-rk}(\rho)=\min\\{\,\overline{A}\text{\rm-
rk}(\rho^{\prime})\mid\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)\,\\}.$

###### Proof.

Let $\rho^{\prime}\in{\mathcal{E}}_{G_{n}}^{G_{m}}(\rho)$ and suppose that
$\overline{A}\text{\rm-rk}(\rho^{\prime})=k_{1}$. This means for all
sufficiently large $N$, there exists
$\rho_{N}\in{\mathcal{E}}^{G_{N}}_{G_{m}}(\rho^{\prime})$ such that
$U\text{\rm-rk}(\rho_{N})=k_{1}<N$. Clearly, $\rho_{N}$ is in
${\mathcal{E}}_{G_{n}}^{G_{N}}(\rho)$, so we have $\overline{A}\text{\rm-
rk}(\rho)\leq k_{1}$.

On the other hand, suppose that $\overline{A}\text{\rm-rk}(\rho)=k_{2}$. Then
for all sufficiently large $N$, there exists
$\rho_{N}\in{\mathcal{E}}^{G_{N}}_{G_{n}}(\rho)$ such that $U\text{\rm-
rk}(\rho_{N})=k_{2}<N$. For $m\geq n$, we know that

${\mathcal{E}}^{G_{N}}_{G_{n}}(\rho)=\bigcup_{\rho^{\prime}\in{\mathcal{E}}^{G_{m}}_{G_{n}}(\rho)}{\mathcal{E}}^{G_{N}}_{G_{m}}(\rho^{\prime}),$

i.e., $\rho_{N}\in{\mathcal{E}}^{G_{N}}_{G_{m}}(\rho^{\prime})$ for some
$\rho^{\prime}\in{\mathcal{E}}^{G_{m}}_{G_{n}}(\rho)$. Then we have
$\overline{A}\text{\rm-rk}(\rho^{\prime})\leq k_{2}$ for some
$\rho^{\prime}\in{\mathcal{E}}^{G_{m}}_{G_{n}}(\rho)$. Thus the lemma is
proved. ∎

According to [GH20] definition 6.4.1, the _tensor rank_ of an irreducible
character $\rho\in{\mathcal{E}}(G)$, denoted by $\otimes\text{\rm-rk}(\rho)$,
is defined as follows:

  1. (1)

Suppose that G is a symplectic group. An irreducible character
$\rho\in{\mathcal{E}}(G)$ has tensor rank less than or equal to $k$ if $\rho$
occurs in

$(\omega_{{\rm O}^{\epsilon_{1}}_{1},\text{\bf G}}\otimes\omega_{{\rm
O}^{\epsilon_{2}}_{1},\text{\bf G}}\otimes\cdots\otimes\omega_{{\rm
O}^{\epsilon_{k}}_{1},\text{\bf G}})|_{G}$

for some $\epsilon_{i}=+$ or $-$ and $i=1,\ldots,k$.

  2. (2)

Suppose that G is an orthogonal group. An irreducible character
$\rho\in{\mathcal{E}}(G)$ has tensor rank less than or equal to $2k$ if
$\rho\chi$ occurs in $(\omega_{{\rm Sp}_{2},\text{\bf G}})^{\otimes k}|_{G}$
for some linear character $\chi$ of $G$.

  3. (3)

Suppose that G is a unitary group. An irreducible character
$\rho\in{\mathcal{E}}(G)$ has tensor rank less than or equal to $k$ if
$\rho\chi$ occurs in $(\omega_{{\rm U}_{1},\text{\bf G}})^{\otimes k}|_{G}$
for some linear character $\chi$ of $G$.

Here $\omega^{\otimes k}$ means the tensor product
$\omega\otimes\cdots\otimes\omega$ of $k$ copies of $\omega$. Because

$\displaystyle(\omega_{{\rm O}^{\epsilon_{1}}_{1},\text{\bf
G}}\otimes\omega_{{\rm O}^{\epsilon_{2}}_{1},\text{\bf
G}}\otimes\cdots\otimes\omega_{{\rm O}^{\epsilon_{k}}_{1},\text{\bf G}})|_{G}$
$\displaystyle=(\omega_{{\rm O}^{\epsilon}_{k},\text{\bf G}})|_{G}$
$\displaystyle(\omega_{{\rm Sp}_{2},\text{\bf G}})^{\otimes k}|_{G}$
$\displaystyle=(\omega_{{\rm Sp}_{2k},\text{\bf G}})|_{G}$
$\displaystyle(\omega_{{\rm U}_{1},\text{\bf G}})^{\otimes k}|_{G}$
$\displaystyle=(\omega_{{\rm U}_{k},\text{\bf G}})|_{G}$

for some $\epsilon=+$ or $-$, we see that

(4.4) $\otimes\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho)$

for any $\rho\in{\mathcal{E}}(G)$.

###  4.2. $\eta$\- and $\underline{\theta}$-correspondence

The following fundamental result is from [GH20]:

######  Proposition 4.5 (Gurevich-Howe).

Let $(\text{\bf G}^{\prime},\text{\bf G})$ be a dual pair in stable range,
i.e., it is of one of the following types:

  1. (i)

$({\rm Sp}_{k},{\rm O}^{+}_{2n})$, $({\rm Sp}_{k},{\rm O}^{-}_{2n+2})$ or
$({\rm Sp}_{k},{\rm O}_{2n+1})$ with $k$ even and $k\leq n$;

  2. (ii)

$({\rm O}_{k}^{\epsilon},{\rm Sp}_{2n})$ with $k\leq n$;

  3. (iii)

$({\rm U}_{k},{\rm U}_{2n})$ or $({\rm U}_{k},{\rm U}_{2n+1})$ with $k\leq n$.

Then for $\rho^{\prime}\in{\mathcal{E}}(G^{\prime})$ there is a unique
irreducible character $\eta(\rho^{\prime})\in\Theta_{\text{\bf
G}}(\rho^{\prime})$ of $U$-rank $k$, and all other elements in
$\Theta_{\text{\bf G}}(\rho^{\prime})$ have $U$-rank less than $k$.
Furthermore, the mapping
$\eta\colon{\mathcal{E}}(G^{\prime})\rightarrow{\mathcal{E}}(G)$ is injective.

The mapping $\eta\colon{\mathcal{E}}(G^{\prime})\rightarrow{\mathcal{E}}(G)$
is called the _$\eta$ -correspondence_ for the dual pair $(\text{\bf
G}^{\prime},\text{\bf G})$ in stable range. In [Pan20a] and [Pan20b], a sub-
relation of $\Theta$, called the _$\underline{\theta}$ -correspondence_, is
defined as follows. Let $(\text{\bf G}^{\prime},\text{\bf G})$ be one of the
following types of dual pairs:

  1. (I)

$({\rm O}^{+}_{2n^{\prime}},{\rm Sp}_{2n})$ or $({\rm Sp}_{2n^{\prime}},{\rm
O}^{+}_{2n})$

  2. (II)

$({\rm O}^{-}_{2n^{\prime}},{\rm Sp}_{2n})$ or $({\rm Sp}_{2n^{\prime}},{\rm
O}^{-}_{2n})$

  3. (III)

$({\rm U}_{2n^{\prime}},{\rm U}_{2n})$ or $({\rm U}_{2n^{\prime}+1},{\rm
U}_{2n+1})$

  4. (IV)

$({\rm U}_{2n^{\prime}},{\rm U}_{2n+1})$ or $({\rm U}_{2n^{\prime}+1},{\rm
U}_{2n})$

Let $\Lambda^{\prime}\in{\mathcal{S}}_{\text{\bf G}^{\prime}}$ and
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ such that
$(\Lambda^{\prime},\Lambda)\in{\mathcal{B}}_{\text{\bf G}^{\prime},\text{\bf
G}}$. Then we know that
$|\Upsilon(\Lambda)|=|\Upsilon(\Lambda^{\prime})|+\tau$ where $\tau$ is given
by

$\tau=\begin{cases}n-n^{\prime}+\left\lfloor\frac{{\rm
def}(\Lambda)}{2}\right\rfloor,&\text{for case (I);}\\\
n-n^{\prime}-\left\lfloor\frac{{\rm def}(\Lambda)}{2}\right\rfloor,&\text{for
case (II);}\\\ n-n^{\prime}+|{\rm def}(\Lambda)|,&\text{for case (III);}\\\
n-n^{\prime}-|{\rm def}(\Lambda)|-1,&\text{for case (IV).}\\\ \end{cases}$

For $\Lambda^{\prime}\in{\mathcal{S}}_{\text{\bf G}^{\prime}}$ such that
$\tau\geq 0$, we define $\underline{\theta}(\Upsilon(\Lambda^{\prime}))$ by

$\underline{\theta}\left(\genfrac{[}{]}{0.0pt}{}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}\right)=\begin{cases}\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}\cup\genfrac{[}{]}{0.0pt}{}{\tau}{-},&\text{for
cases (I),(III) ;}\\\
\genfrac{[}{]}{0.0pt}{}{\nu_{1},\nu_{2},\ldots,\nu_{m_{2}}}{\mu_{1},\mu_{2},\ldots,\mu_{m_{1}}}\cup\genfrac{[}{]}{0.0pt}{}{-}{\tau},&\text{for
cases (II),(IV).}\end{cases}$

Then we define $\underline{\theta}(\Lambda^{\prime})$ to be the symbol
$\Lambda\in{\mathcal{S}}_{\text{\bf G}}$ such that
$\Upsilon(\Lambda)=\underline{\theta}(\Upsilon(\Lambda^{\prime}))$. Hence we
obtain a relation $\underline{\theta}$ between ${\mathcal{E}}(G^{\prime})_{1}$
and ${\mathcal{E}}(G)_{1}$ by letting
$\underline{\theta}(\rho_{\Lambda^{\prime}})=\rho_{\underline{\theta}(\Lambda^{\prime})}$.

Next we extend the domain of $\underline{\theta}$ by letting it compatible
with the Lusztig correspondence, i.e.,

  1. (1)

for $(\text{\bf G}^{\prime},\text{\bf G})$ being of one cases
(I),(II),(III),(IV) above,
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$ and
$\rho\in{\mathcal{E}}(G)_{s}$, we have the following commutative diagram:

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

  2. (2)

for the dual pair $(\text{\bf G}^{\prime},\text{\bf G})=({\rm
O}_{2n^{\prime}+1},{\rm Sp}_{2n})$,
$\rho^{\prime}\in{\mathcal{E}}(G^{\prime})_{s^{\prime}}$ and
$\rho\in{\mathcal{E}}(G)_{s}$, we have the following commutative diagram

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

We know that $\underline{\theta}$ is a sub-correspondence of $\Theta$, and we
can also define the _$\underline{\theta}$ -rank_ of $\rho\in{\mathcal{E}}(G)$
by the same way as $\Theta$-rank. It is know that

(4.6) $\underline{\theta}\text{\rm-rk}(\rho^{\prime})=\Theta\text{\rm-
rk}(\rho^{\prime})$

for any $\rho\in{\mathcal{E}}(G^{\prime})$ by [Pan20a] lemma 6.11.

######  Proposition 4.7.

Let $(\text{\bf G}^{\prime},\text{\bf G})$ be a reductive dual pair in stable
range. Then $\eta=\underline{\theta}$.

###### Proof.

If $(\text{\bf G}^{\prime},\text{\bf G})$ is a symplectic/orthogonal dual pair
in stable range, this proposition is [Pan20a] theorem 6.16. Moreover, it is
not difficult to see that the proof in [Pan20a] works for a unitary dual pair
in table range. ∎

######  Lemma 4.8.

Let $\rho\in{\mathcal{E}}(G_{n})$. If $\Theta\text{\rm-rk}(\rho)\leq n$, then
$U\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho)$.

###### Proof.

Suppose that $\Theta\text{\rm-rk}(\rho)=k\leq n$. So by (4.6) we have
$\underline{\theta}\text{\rm-rk}(\rho)=k\leq n$, i.e.,
$\rho=\underline{\theta}(\rho^{\prime})$ for some
$\rho^{\prime}\in{\mathcal{E}}({\rm O}^{\epsilon}_{k}(q))$ for some
$\epsilon=+$ or $-$. Now the dual pair $({\rm O}^{\epsilon}_{k},{\rm
Sp}_{2n})$ is in stable range, so the two mappings $\eta$ and
$\underline{\theta}$ coincides by Proposition 4.7, i.e.,
$\rho=\eta(\rho^{\prime})$ and hence $U\text{\rm-rk}(\rho)=k$ by Proposition
4.5. ∎

######  Corollary 4.9.

We have $\overline{A}\text{\rm-rk}(\rho)\leq\Theta\text{\rm-rk}(\rho)$ for any
$\rho\in{\mathcal{E}}(G)$.

###### Proof.

Let $\text{\bf G}=\text{\bf G}_{n}$, $\rho\in{\mathcal{E}}(G)$ and suppose
that $\Theta\text{\rm-rk}(\rho)=k$. So now $\rho^{\prime}\otimes\rho$ occurs
in the $\Theta$-correspondence for some dual pair $(G^{\prime},G_{n})$. Then
we know that, for any $N\geq n$, $\rho^{\prime}\otimes\rho_{N}$ occurs in the
$\Theta$-correspondence for the dual pair $(G^{\prime},G_{N})$ for some
$\rho_{N}\in{\mathcal{E}}_{G_{n}}^{G_{N}}(\rho)$. By definition, we have
$\Theta\text{\rm-rk}(\rho_{N})\leq k$, and by Lemma 3.23, we have
$\Theta\text{\rm-rk}(\rho_{N})\geq\Theta\text{\rm-rk}(\rho)=k$. Therefore, we
conclude that $\Theta\text{\rm-rk}(\rho_{N})=k$. When $N\geq k$, by Lemma 4.8,
we see that $U\text{\rm-rk}(\rho_{N})=k$. Then by the definition of
$\overline{A}$-rank, we conclude that $\overline{A}\text{\rm-rk}(\rho)\leq
U\text{\rm-rk}(\rho_{N})=k$. Therefore, $\overline{A}\text{\rm-
rk}(\rho)\leq\Theta\text{\rm-rk}(\rho)$. ∎

###### Proof of Theorem 1.5.

Now from (4.1), (4.2), (4.4) and Corollary 4.9, we can summarize the relation
between these various ranks as

(4.10) $U\text{\rm-rk}(\rho)\leq A\text{\rm-
rk}(\rho)\leq\overline{A}\text{\rm-rk}(\rho)\leq\otimes\text{\rm-
rk}(\rho)=\Theta\text{\rm-rk}(\rho)$

for any $\rho\in{\mathcal{E}}(G)$. Then Theorem 1.5 follows from Lemma 4.8
immediately. ∎

###  4.3. Agreement of $\overline{A}$-ranks and $\Theta$-ranks

It is proved by Howe-Gurevich in [GH20] theorem 11.4 that $A\text{\rm-
rk}(\rho)=\otimes\text{\rm-rk}(\rho)$ for any $\rho\in{\mathcal{E}}({\rm
GL}_{n}(q))$. Now we want to apply their method to other classical groups.

Let $G=G_{n}=G(V)$ and $V=X_{n}\oplus V_{0}\oplus Y_{n}$ be a complete
polarization of $V$ where $V_{0}$ is anisotropic. Let
$\\{x_{1},x_{2},\ldots,x_{n}\\}$ be a basis of $X_{n}$, and let
$\\{y_{1},y_{2},\ldots,y_{n}\\}$ be the corresponding dual basis of $Y_{n}$,
i.e., $\langle x_{i},y_{j}\rangle=\delta_{ij}$ where $\delta_{ij}$ is the
Kronecker delta symbol. For $1\leq k\leq n$, let $X_{k}$ (resp. $Y_{k}$)
denote the space spanned by $\\{x_{1},x_{2},\ldots,x_{k}\\}$ (resp.
$\\{y_{1},y_{2},\ldots,y_{k}\\}$). So we have the decomposition $V=X_{k}\oplus
V_{n-k}\oplus Y_{k}$ where $V_{n-k}$ is a space of Witt index $n-k$. Define

$\displaystyle H_{k}$ $\displaystyle=\\{\,g\in G(V)\mid(g-1)(X_{k})=0,\
(g-1)(Y_{k})=0\,\\}.$

Then $H_{k}$ is a subgroup of $G(V)$, $H_{k}\subset H_{k-1}$ for each $k$, and
$H_{0}=G_{n}$.

######  Lemma 4.11.

Let $\rho\in{\mathcal{E}}(G_{n})$ be of low $U$-rank $k$, i.e.,

$U\text{\rm-rk}(\rho)=k<\begin{cases}n,&\text{if\/ $\text{\bf G}_{n}$ is ${\rm
Sp}_{2n}$, ${\rm U}_{2n}$ or ${\rm U}_{2n+1}$};\\\ n-1,&\text{if\/ $\text{\bf
G}_{n}$ is ${\rm O}^{+}_{2n}$, ${\rm O}^{-}_{2n+2}$ or ${\rm
O}_{2n+1}$}.\end{cases}$

Then $\rho|_{H_{k}}$ contains a linear character.

###### Proof.

Suppose that $\rho\in{\mathcal{E}}(G_{n})$ is of low $U$-rank $k$. We define
the following unipotent subgroups of $G(V)$:

$\displaystyle U_{1}$ $\displaystyle=\\{\,g\in G(V)\mid(g-1)(X_{k})=0,\
(g-1)(V_{n-k}\oplus Y_{k})\subset X_{k}\,\\},$ $\displaystyle U_{2}$
$\displaystyle=\\{\,g\in G(V)\mid(g-1)(X_{k}\oplus V_{n-k})=0,\
(g-1)(Y_{k})\subset X_{k}\,\\},$ $\displaystyle U_{3}$
$\displaystyle=U_{1}\cap H_{k}.$

It is not difficult to see that $U_{1}\simeq U_{2}\times U_{3}$, and $U_{1}$
is normalized by $H_{k}$. Define the subgroup $U_{4}$ of $G(V)$ as follows:

  * •

If $G(V)$ is a symplectic group, then $k<n$, and let $U_{4}$ be the subgroup
of elements $g\in G(V)$ such that

    * –

$(g-1)y_{k+1}=ax_{k+1}$, for $a\in\text{\bf f}_{q}$, and

    * –

$(g-1)(X_{k+1}\oplus V_{n-k-1}\oplus Y_{k})=0$.

  * •

If $G(V)$ is an orthogonal group, then $k<n-1$, and let $U_{4}$ be the
subgroup of elements $g\in G(V)$ such that

    * –

$(g-1)y_{k+2}=ax_{k+1}$ and $(g-1)y_{k+1}=-ax_{k+2}$, for $a\in\text{\bf
f}_{q}$, and

    * –

$(g-1)(X_{k+2}\oplus V_{n-k-2}\oplus Y_{k})=0$.

  * •

If $G(V)$ is a unitary group, then $k<n$, and let $U_{4}$ be the subgroup of
elements $g\in G(V)$ such that

    * –

$(g-1)y_{k+1}=ax_{k+1}$, for $a\in\text{\bf f}_{q^{2}}$ satisfying
$\overline{a}=-a$, and

    * –

$(g-1)(X_{k+1}\oplus V_{n-k-1}\oplus Y_{k})=0$.

Here $a\mapsto\overline{a}$ is the nontrivial automorphism of $\text{\bf
f}_{q^{2}}$ over $\text{\bf f}_{q}$.

Finally we define the subgroup $U_{5}$ of $G(V)$ as follows:

  * •

If $G(V)$ is a symplectic group or a unitary group, let $U_{5}$ be the
subgroup of elements $g\in G(V)$ such that

    * –

$(g-1)(X_{k+1}\oplus V_{n-k-1})=0$ and

    * –

$(g-1)(Y_{k+1})\subset X_{k+1}$.

  * •

If $G(V)$ is an orthogonal group, let $U_{5}$ be the subgroup of elements
$g\in G(V)$ such that

    * –

$(g-1)(X_{k+2}\oplus V_{n-k-2})=0$ and

    * –

$(g-1)(Y_{k+2})\subset X_{k+2}$.

There exists a linear character $\psi$ of $U_{1}$ of rank $k$ and trivial on
$U_{3}$, so $\psi|_{U_{2}}$ still has rank $k$. It is easy to see that
$h(u_{2},u_{3})h^{-1}=(u_{2},u_{3}^{\prime})$ for $h\in H_{k}$, $u_{2}\in
U_{2}$ and $u_{3}\in U_{3}$. Because now $\psi$ is trivial on $U_{3}$, $\psi$
is invariant under the conjugation by $H_{k}$. Because now $U\text{\rm-
rk}(\rho)=k$, $\rho|_{U_{1}}$ contains a nontrivial $\psi$-eigenspace, denoted
by $\rho_{\psi}$. Because $\psi$ is invariant under the conjugation of
$H_{k}$, the space $\rho_{\psi}$ is invariant under the action of
$\rho|_{H_{k}}$. Let $\sigma$ denote the representation of $H_{k}$ on the
space $\rho_{\psi}$.

Now we claim that the representation $\sigma$ of $H_{k}$ is trivial on the
commutator subgroup of $H_{k}$. Suppose that the claim is not true, i.e.,
$\sigma$ is not trivial on the commutator subgroup of $H_{k}$ Then $\sigma$ is
not trivial on $U_{4}$, because the normal subgroup of $H_{k}$ generated by
$U_{4}$ is the full commutator subgroup of $H_{k}$. The space $\rho_{\psi}$ is
decomposed into eigenspaces for $U_{4}$. Let $\rho_{\psi}^{1}$ be the
eigenspace corresponding to a non-trivial linear character $\lambda$ of
$U_{4}$. The group $U_{5}$ is commutative and contains both $U_{2}$ and
$U_{4}$. So the space $\rho_{\psi}^{1}$ is decomposed into eigenspaces of
$U_{5}$. The character of $U_{5}$ which corresponding a nontrivial eigenspace
in $\rho_{\psi}^{1}$ of $U_{5}$ has rank greater than $k$ and we get a
contradiction.

Now the representation $\sigma$ of $H_{k}$ is trivial on the commutator
subgroup of $H_{k}$, we see that $\sigma$ is a sum of linear characters of
$H_{k}$. ∎

Recall that under the _Shrödinger model_ of the Weil representation $\omega$
(cf. [GH17]), the group $G(V)$ acts on the space $L^{2}({\rm Hom}(X_{k},V))$
of functions on ${\rm Hom}(X_{k},V)$ by

$(\omega(g)f)(T)=\chi_{k}(g)f(g^{-1}\circ T)$

for $g\in G(V)$ and $T\in{\rm Hom}(X_{k},V)$ where $\chi_{k}$ is a linear
character of $G(V)$ of order at most $2$. Recall that
$L^{2}(G/H_{k},\chi_{k}^{-1})$ denotes the space of functions $f$ on $G$ such
that $f(gh)=\chi_{k}^{-1}(h)f(g)$ for $g\in G$ and $h\in H_{k}$. Then $G(V)$
acts on the space $L^{2}(G/H_{k},\chi_{k})$ by

$(\pi(g)f)(x)=f(g^{-1}x)$

for $g,x\in G(V)$.

Fix an element $T_{0}\in{\rm Hom}(X_{k},V)$ such that the image of $T_{0}$ is
equal to $X_{k}$. Note that for $h\in H_{k}$, we have $h\circ T_{0}=T_{0}$
from the definition of $H_{k}$. Then it is not difficult to see that the
mapping

$\phi\colon L^{2}({\rm Hom}(X_{k},V))\rightarrow L^{2}(G/H_{k},\chi_{k})$

given by $\phi(f)(x)=\chi_{k}(x^{-1})f(x\circ T_{0})$ is a surjective linear
transformation. Moreover, we have

$\displaystyle\pi(g^{\prime})(\phi(f))(g)$
$\displaystyle=\phi(f)(g^{\prime-1}g)=\chi_{k}(g^{-1}g^{\prime})f(g^{\prime-1}g\circ
T)$ $\displaystyle\phi(\omega(g^{\prime})f)(g)$
$\displaystyle=\chi_{k}(g^{-1})(\omega(g^{\prime})f)(g\circ
T)=\chi_{k}(g^{-1}g^{\prime})f(g^{\prime-1}g\circ T),$

i.e., $\phi$ commutes the two actions $\omega$ and $\pi$ of $G$. Therefore
$\pi$ can be regarded as a subrepresentation of $\omega$.

######  Lemma 4.12.

Let $\rho\in{\mathcal{E}}(G_{n})$ and $0\leq k\leq n$. If $\rho|_{H_{k}}$
contains a linear character, then $\Theta\text{\rm-rk}(\rho)\leq 2k$.

###### Proof.

Suppose that $\rho\in{\mathcal{E}}(G_{n})$ and $\rho|_{H_{k}}$ contains a
linear character. Since a linear character of $H_{k}$ is the restriction of a
linear character of $G_{n}$, after twisting a linear character of $G_{n}$, we
may assume that $\rho|_{H_{k}}$ contains the character $\chi_{k}^{-1}$. By
Frobenius reciprocity, $\rho$ is contained in
$L^{2}(G_{n}/H_{k},\chi_{k}^{-1})$ under the action $\pi$ of left translation
of $G_{n}$. Then by the above discussion, we see that $\rho$ is contained in
the Schöndinger model $L^{2}({\rm Hom}(X_{k},V))$ of the Weil representation
$\omega$. This means that $\rho$ occurs in the $\Theta$-correspondence for the
dual pair $(\text{\bf G}^{\prime},\text{\bf G})$ where $\text{\bf G}^{\prime}$
is the group acting on a space $V^{\prime}$ which has a Witt decomposition
with a trivial anisotropic kernel and a maximal totally isotropic space
isomorphic to $X_{k}$, i.e., $\dim(V^{\prime})=2k$. This concludes that
$\Theta\text{\rm-rk}(\rho)\leq 2k$. ∎

Now we have an analogous result of [GH20] proposition 11.2.

######  Corollary 4.13.

If $\rho\in{\mathcal{E}}(G_{n})$ is of $U$-rank $k\leq\frac{n}{2}$, then
$U\text{\rm-rk}(\rho)=\Theta\text{\rm-rk}(\rho)$.

###### Proof.

Suppose that $U\text{\rm-rk}(\rho)=k\leq\frac{n}{2}$. Then by Lemma 4.11 and
Lemma 4.12 we have $\Theta\text{\rm-rk}(\rho)\leq 2k\leq n$. Then by Lemma
4.8, we conclude that $\Theta\text{\rm-rk}(\rho)=U\text{\rm-rk}(\rho)=k$. ∎

###### Proof of Theorem 1.6.

Let $\rho\in{\mathcal{E}}(G_{n})$ where $\text{\bf G}_{n}$ is ${\rm Sp}_{2n}$,
${\rm O}^{+}_{2n}$, ${\rm O}^{-}_{2n+2}$, ${\rm O}_{2n+1}$, ${\rm U}_{2n}$, or
${\rm U}_{2n+1}$. Suppose that $\overline{A}\text{\rm-rk}(\rho)=k_{1}$ and
$\Theta\text{\rm-rk}(\rho)=k_{2}$. We know that $k_{1}\leq k_{2}$ by Corollary
4.9 and $k_{2}\leq 2n+2$ from Remark 3.2. Then by definition of parabolic
asymptotic rank we have

(4.14) $k_{1}=\min\\{\,U\text{\rm-
rk}(\rho_{N})\mid\rho_{N}\in{\mathcal{E}}_{G_{n}}^{G_{N}}(\rho\chi),\
\chi\text{ linear},\ N>\\!>n\,\\},$

and by Lemma 3.23 we know that

(4.15) $k_{2}=\min\\{\,\Theta\text{\rm-
rk}(\rho_{N})\mid\rho_{N}\in{\mathcal{E}}_{G_{n}}^{G_{N}}(\rho\chi),\
\chi\text{ linear},\ N>\\!>n\,\\}.$

Let $N$ be sufficiently larger than $n$, in particular, we assume that $N\geq
4n+4$. Then by (4.14) there is
$\rho_{1}\in{\mathcal{E}}_{G_{n}}^{G_{N}}(\rho\chi_{1})$ for some linear
character $\chi_{1}$ of $G_{n}$ such that $U\text{\rm-rk}(\rho_{1})=k_{1}\leq
k_{2}\leq 2n+2\leq\frac{N}{2}$. Then by Lemma 4.12, we have $\Theta\text{\rm-
rk}(\rho_{1})\leq 2k_{1}\leq N$. Then by Lemma 4.8, we have $\Theta\text{\rm-
rk}(\rho_{1})=U\text{\rm-rk}(\rho_{1})=k_{1}$. Then (4.15) implies that

$k_{2}=\Theta\text{\rm-rk}(\rho)\leq\Theta\text{\rm-
rk}(\rho_{1})=k_{1}=\overline{A}\text{\rm-rk}(\rho)\leq k_{2}.$

Thus the theorem is proved. ∎

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