On the annihilator variety of a highest weight Harish-Chandra module

Zhanqiang Bai Department of Mathematical Sciences, Soochow University, Suzhou 215006, China [email protected] and Jing Jiang School of mathematical Sciences, Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice, East China Normal University, Shanghai 200241, China [email protected]

Abstract.

Let $G$ be a Hermitian type Lie group with maximal compact subgroup $K$ . Let $L(\lambda)$ be a highest weight Harish-Chandra module of $G$ with the infinitesimal character $\lambda$ . By using some combinatorial algorithm, we obtain a description of the annihilator variety of $L(\lambda)$ . As an application, when $L(\lambda)$ is unitarizable, we prove that the Gelfand-Kirillov dimension of $L(\lambda)$ only depends on the value of $z=(\lambda,\beta^{\vee})$ , where $\beta$ is the highest root.

Key words and phrases:

associated variety; Gelfand-Kirillov dimension; Young tableau; nilpotent orbit

2020 Mathematics Subject Classification:

22E47, 17B10

^†^†*Corresponding author: [email protected]

1. Introduction

Let $G$ be a simple Lie group with complexified Lie algebra $\mathfrak{g}$ and $K$ be a maximal compact subgroup with complexified Lie algebra $\mathfrak{k}$ such that $G/K$ is a Hermitian symmetric space. We have a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ and a decomposition $\mathfrak{p}=\mathfrak{p}^{+}\oplus\mathfrak{p}^{-}$ into nonzero irreducible $K$ -subrepresentations. Then we have a triangular decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}^{+}\oplus\mathfrak{p}^{-}$ and $\mathfrak{q}=\mathfrak{k}\oplus\mathfrak{p}^{+}$ is a maximal parabolic subalgebra of $\mathfrak{g}$ . Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ . Denote by $\Phi$ the root system associated with $(\mathfrak{g},\mathfrak{h})$ , fix a positive system $\Phi^{+}$ , let $\mathbb{C}_{\lambda-\rho}$ be a finite-dimensional irreducible $\mathfrak{k}$ -module with highest weight $\lambda-\rho\in\mathfrak{h}^{*}$ , where $\rho$ is half the sum of positive roots. It can also be viewed as a $\mathfrak{q}$ -module with trivial $\mathfrak{p}^{+}$ -action. The corresponding generalized Verma module $N(\lambda)$ is defined by

N(\lambda):=U(\mathfrak{g})\otimes_{U(\mathfrak{q})}\mathbb{C}_{\lambda-\rho}.

The simple quotient of $N(\lambda)$ is denoted by $L(\lambda)$ , which is a highest weight module with highest weight $\lambda-\rho$ . From [EHW83], we call $L(\lambda)$ a highest weight Harish-Chandra module. The classification of unitarizable highest weight Harish-Chandra modules is given in [EHW83] and [Jak83] independently.

For a $U(\mathfrak{g})$ -module $M$ , let $I(M)=\mathrm{Ann}(M)$ be its annihilator ideal in $U(\mathfrak{g})$ and $J(M)$ be the corresponding graded ideal in $S(\mathfrak{g})=\text{gr}(U(\mathfrak{g}))$ . The zero set of $J(M)$ in the dual vector space $\mathfrak{g}^{*}$ of $\mathfrak{g}$ is called the annihilator variety of $M$ , which is also called the associated variety of $I(M)$ . We usually denote it by $V(\mathrm{Ann}(M))$ . The study of associated varieties of primitive ideals (or annihilator varieties of highest weight modules) is a very important problem since it is closely related with many research fields, such as representations of Weyl groups, Kazhdan-Lusztig cells and representations of Lie groups, see for example \citesBB,BV82,BMSZ-ABCD,GSK. Borho-Brylinski [BB82] proved that the associated variety of a primitive ideal $I$ with a fixed regular integral infinitesimal character is the Zariski closure of the nilpotent orbit in $\mathfrak{g}^{*}$ attached to $I$ , via the Springer correspondence. Joseph [Jos85] extended this result to a primitive ideal with a general infinitesimal character.

Recently Bai-Ma-Wang [BMW23] found some simple algorithms to characterize the nilpotent orbit appearing in the annihilator variety of a highest weight module for all classical type Lie algebras by using the Robinson-Schensted algorithm. In this paper, our first goal is to give the explicit description of $V(\mathrm{Ann}(L(\lambda)))$ for all highest weight Harish-Chandra modules of classical types by using the partition algorithm in [BMW23]. See Theorem 3.3, Theorem 4.1, Theorem 4.3, Theorem 5.1 and Theorem 5.3.

The Gelfand-Kirillov (GK) dimension is an important invariant to measure the size of an infinite-dimensional $U(\mathfrak{g})$ -module $M$ , see [Jos78] and [Vog78]. From Joseph [Jos78], we know that $\dim V(\mathrm{Ann}(L(\lambda)))=2\mathrm{GKdim}L(\lambda)$ . When there is only one nilpotent orbit with the dimension $2\mathrm{GKdim}L(\lambda)$ , we can determine this nilpotent orbit appearing in $V(\mathrm{Ann}(L(\lambda)))$ by the Gelfand-Kirillov dimension of $L(\lambda)$ . In [BXX23], Bai-Xiao-Xie computed the Gelfand-Kirillov dimensions of highest weight Harish-Chandra modules of exceptional types. Then we can determine $V(\mathrm{Ann}(L(\lambda)))$ by the Gelfand-Kirillov dimension of $L(\lambda)$ since there is only one nilpotent orbit with the dimension $2\mathrm{GKdim}L(\lambda)$ for these cases. See Theorem 6.1 and Theorem 6.2. Except for the cases of $SU(p,q)$ , $Sp(n,\mathbb{R})$ and $SO^{*}(2n)$ , our characterization of $V(\mathrm{Ann}(L(\lambda)))$ is given in a function of $z=(\lambda,\beta^{\vee})$ .

Our second goal of this paper is to prove that when $L(\lambda)$ is a unitary highest weight module with highest weight $\lambda-\rho=\lambda_{0}+z\xi$ , where $z=(\lambda,\beta^{\vee})$ , $(\lambda_{0}+\rho,\beta)=0$ , $\xi$ is orthogonal to the root system $\Phi(\mathfrak{k})$ and $(\xi,\beta^{\vee})=1$ , then ${\rm GKdim}\>L(\lambda)$ is independent of the selection of $\lambda_{0}$ . In other words, ${\rm GKdim}\>L(\lambda)$ only depends on the value of $z=(\lambda,\beta^{\vee})$ . This result was firstly proved in [BH15] by using previous results in [Jos92]. In our paper, we use our characterization of $V(\mathrm{Ann}(L(\lambda)))$ to give a new proof for this result. See Theorem 7.2.

The paper is organized as follows. In §2, we give some necessary preliminaries about Gelfand-Kirillov dimension, associated variety, annihilator variety and partition algorithm for highest weigh modules. In §3-6, we give the characterization of the annihilator variety $V(\mathrm{Ann}(L(\lambda)))$ for a highest weight Harish-Chandra module of type $A_{n-1}$ , type $B_{n}$ and $C_{n}$ , type $D_{n}$ , type $E_{6}$ and $E_{7}$ respectively. In §7, we prove that when $L(\lambda)$ is a unitary highest weight module, then ${\rm GKdim}\>L(\lambda)$ only depends on the value of $z=(\lambda,\beta^{\vee})$ .

2. Preliminaries

In this section, we will give some brief preliminaries on GK dimension, associated variety, annihilator variety and partition algorithm for highest weight modules. See \citesVo78,VJ,BMW for more details.

2.1. Gelfand-Kirillov dimension and associated variety

Let $\mathfrak{g}$ be a simple complex Lie algebra and $\mathfrak{h}$ be a Cartan subalgebra. Let $M$ be a finitely generated $U(\mathfrak{g})$ -module. Fix a finite dimensional generating subspace $M_{0}$ of $M$ . Let $U_{n}(\mathfrak{g})$ be the standard filtration of $U(\mathfrak{g})$ . Set $M_{n}=U_{n}(\mathfrak{g})\cdot M_{0}$ and ${\rm gr}(M)=\bigoplus\limits_{n=0}^{\infty}{\rm gr}_{n}M,$ where ${\rm gr}_{n}M=M_{n}/{M_{n-1}}$ . Thus ${\rm gr}(M)$ is a graded module of ${\rm gr}(U(\mathfrak{g}))\simeq S(\mathfrak{g})$ .

Definition 2.1.

The Gelfand-Kirillov dimension of $M$ is defined by

\operatorname{GKdim}M=\varlimsup\limits_{n\rightarrow\infty}\frac{\log\dim(U_{n}(\mathfrak{g})M_{0})}{\log n}.

Definition 2.2 ([Vog91]).

The associated variety of $M$ is defined by

V(M):=\{X\in\mathfrak{g}^{*}\mid f(X)=0{\rm~{}for~{}all~{}}f\in\operatorname{Ann}_{S(\mathfrak{g})}(\operatorname{gr}M)\}.

The above two definitions are independent of the choice of $M_{0}$ , and $\dim V(M)={\rm GKdim}\>M$ (e.g., [NOT01]).

Definition 2.3.

Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra. Let $I$ be a two-sided ideal in $U(\mathfrak{g})$ . Then ${\rm gr}(U(\mathfrak{g})/I)\simeq S(\mathfrak{g})/\text{gr}I$ is a graded $S(\mathfrak{g})$ -module. Its annihilator is ${\rm gr}I$ . We define its associated variety by

V(I):=V(U(\mathfrak{g})/I)=\{X\in\mathfrak{g}^{*}\mid p(X)=0\ \mbox{for all $p\in{\rm gr}I$}\}.

2.2. Associated varieties of highest weight Harish-Chandra modules

In [BH15], Bai-Hunziker have found the formula of the Gelfand-Kirillov dimensions of unitary highest weight modules. We adopt their notations here.

Let $(-,-):\mathfrak{h}\times\mathfrak{h}^{*}\to\mathbb{C}$ be the canonical pairing. Let $\beta$ denote the highest positive root. Let $\rho$ be the half sum of positive roots. The dual Coxeter number $h^{\lor}$ of the root system $\Phi$ is defined by

h^{\lor}:=(\rho,\beta^{\lor})+1.

where $\beta^{\lor}$ is the coroot of $\beta$ . Equivalently, $h^{\lor}$ can be defined as the sum of coefficients of highest short roots in $\Phi^{+}$ when it is written as a sum of simple roots. Let $r$ be the $\mathbb{R}$ -rank of $G$ , i.e., the dimension of a Cartan subgroup of the (real) group $G$ . And we have the following table.

Table 1. Some constants associated to

G/K

$G$	$r$	$c$	$h^{\vee}-1$
$SU(p,q)$ , $n=p+q-1$	$\min\{p,q\}$	$1$	$n$
$Sp(n,\mathbb{R})$	$n$	$1/2$	$n$
$SO^{*}(2n)$ , $n=2m$	$m$	$2$	$2n-3$
$SO^{*}(2n)$ , $n=2m+1$	$m$	$2$	$2n-3$
$SO(2,2n-1)$	$2$	$n-3/2$	$2n-2$
$SO(2,2n-2)$	$2$	$n-2$	$2n-3$
$E_{6(-14)}$	$2$	$3$	$11$
$E_{7(-25)}$	$3$	$4$	$17$

They also have proved that if $L(\lambda)$ is a unitary highest weight module with highest weight $\lambda-\rho$ , then there is an integer $0\leqslant k(\lambda)\leqslant r$ such that the associated variety of $L(\lambda)$ is $\overline{\mathcal{O}}_{k(\lambda)}$ . We list the two crucial theorems.

Proposition 2.4 ([BH15, Thm. 1.1 and Cor. 5.2]).

Suppose $L(\lambda)$ is a unitary highest weight module with highest weight $\lambda-\rho$ . Let $0\leqslant k(\lambda)\leqslant r$ be the integer such that $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ . Then $k(\lambda)=r$ if $-(\lambda-\rho,\beta^{\vee})>c(r-1)$ and $k(\lambda)=\frac{-(\lambda-\rho,\beta^{\vee})}{c}$ if $-(\lambda-\rho,\beta^{\vee})\leq c(r-1)$ . Also we have

\operatorname{dim}\overline{\mathcal{O}}_{k(\lambda)}=k(\lambda)\left(h^{\vee}-1\right)-k(\lambda)(k(\lambda)-1)c.

In particular, $\operatorname{dim}\mathfrak{p}^{+}=r\left(h^{\vee}-1\right)-r(r-1)c$ .

Proposition 2.5 ([BH15, Thm. 6.2]).

Suppose $L(\lambda)$ is a unitary highest weight module with highest weight $\lambda-\rho$ . Define $z(\lambda):=(\lambda,\beta^{\vee})$ and $z_{k}=(\rho,\beta^{\vee})-kc$ . Then

{\rm GKdim}\>L(\lambda)=\left\{\begin{array}[]{ll}rz_{r-1},&\textnormal{if}\>z(\lambda)<z_{r-1},\\ kz_{k-1},&\textnormal{if}\>z(\lambda)=z_{k}{\rm~{}with~{}}1\leqslant k\leqslant r-1,\\ 0,&\textnormal{if}\>z(\lambda)=z_{0}.\end{array}\right.

In [BXX23], Bai-Xiao-Xie have given the description of the associated variety of $L(\lambda)$ , we recall some notations and results.

For a totally ordered set $\Gamma$ , we denote by $\mathrm{Seq}_{n}(\Gamma)$ the set of sequences $x=(x_{1},x_{2},\cdots,x_{n})$ of length $n$ with $x_{i}\in\Gamma$ . In our paper, we usually take $\Gamma$ to be $\mathbb{Z}$ or a coset of $\mathbb{Z}$ in $\mathbb{C}$ . By using the Robinson-Schensted algorithm for $x\in\mathrm{Seq}_{n}(\Gamma)$ , we can get a Young tableau $P(x)$ . Denote ${\bf p}(x)=\mathrm{sh}(P(x))=[p_{1},p_{2},...,p_{N}]$ , where $p_{i}$ is the number of boxes in the $i$ -th row of $P(x)$ .

Definition 2.6.

Fix a positive integer $n$ , a partition of $n$ is a decreasing sequence ${\bf p}=[p_{1},p_{2},...,p_{n}]$ of nonnegative integers such that $\sum_{1\leq i\leq n}p_{i}=n$ . We say ${\bf q}=[q_{1},q_{2},\cdots,q_{N}]$ is the dual partition of a partition ${\bf p}=[p_{1},p_{2},\cdots,p_{N}]$ and write ${\bf q}={\bf p}^{t}$ if $q_{i}$ is the length of $i$ -th column of the Young tableau $P$ with shape ${\bf p}$ .

For a Young diagram $P$ , use $(k,l)$ to denote the box in the $k$ -th row and the $l$ -th column. We say the box $(k,l)$ is even (resp. odd) if $k+l$ is even (resp. odd). Let $p_{i}^{\mathrm{ev}}$ (resp. $p_{i}^{\mathrm{od}}$ ) be the number of even (resp. odd) boxes in the $i$ -th row of the Young diagram $P$ . One can easily check that

(2.7)

p_{i}^{\mathrm{ev}}=\begin{cases}\left\lceil\frac{p_{i}}{2}\right\rceil,&\text{ if }i\text{ is odd},\\ \left\lfloor\frac{p_{i}}{2}\right\rfloor,&\text{ if }i\text{ is even},\end{cases}\quad p_{i}^{\mathrm{od}}=\begin{cases}\left\lfloor\frac{p_{i}}{2}\right\rfloor,&\text{ if }i\text{ is odd},\\ \left\lceil\frac{p_{i}}{2}\right\rceil,&\text{ if }i\text{ is even}.\end{cases}

Here for $a\in\mathbb{R}$ , $\lfloor a\rfloor$ is the largest integer $n$ such that $n\leq a$ , and $\lceil a\rceil$ is the smallest integer $n$ such that $n\geq a$ . For convenience, we set

{\bf p}^{\mathrm{ev}}=(p_{1}^{\mathrm{ev}},p_{2}^{\mathrm{ev}},\cdots)\quad\mbox{and}\quad{\bf p}^{\mathrm{od}}=(p_{1}^{\mathrm{od}},p_{2}^{\mathrm{od}},\cdots).

Example 2.8.

Let ${\bf p}=[6,5,4,3,2,1]$ be the shape of the Young diagram $P$ . Then odd and even boxes in $P$ are marked as follows.

Then ${\bf p}^{\mathrm{ev}}=(3,2,2,1,1,0)$ and ${\bf p}^{\mathrm{od}}=(3,3,2,2,1,1)$ .

For convenience, $x=(x_{1},x_{2},\cdots,x_{n})\in\mathrm{Seq}_{n}(\Gamma)$ , set

\displaystyle{x}^{-}=

\displaystyle(x_{1},x_{2},\cdots,x_{n-1},x_{n},-x_{n},-x_{n-1},\cdots,-x_{2},-x_{1}).

Proposition 2.9 ([BXX23, Thm. 6.1, Thm. 6.2]).

Let $L(\lambda)$ be a highest weight Harish-Chandra module of $G$ with highest weight $\lambda-\rho$ and $\lambda=(t_{1},\cdots,t_{n})\in\mathfrak{h}^{*}$ . Set ${\bf q}={\bf q}(\lambda)=(q_{1},q_{2},\cdots,q_{2n})={\bf p}(\lambda)^{t}$ when $G$ is of type $A$ and ${\bf q}={\bf q}(\lambda^{-})=(q_{1},q_{2},\cdots,q_{2n})={\bf p}(\lambda^{-})^{t}$ when $G$ is of type $B,C$ or $D$ . Then $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ with $k(\lambda)$ given as follows.

(1)

$G=SU(p,n-p)$ . Then

k(\lambda)=\left\{\begin{array}[]{ll}q_{2},&\textnormal{~{}if~{}$\lambda$~{}is~{}integral},\\ \min\{p,q\},&\textnormal{~{}otherwise}.\end{array}\right.

(2)

$G=Sp(n,\mathbb{R})$ with $n\geq 2$ . Then

k(\lambda)=\begin{cases}2q_{2}^{\mathrm{od}},&\textnormal{ if }t_{1}\in\mathbb{Z},\\ 2q_{2}^{\mathrm{ev}}+1,&\textnormal{ if }t_{1}\in\frac{1}{2}+\mathbb{Z},\\ n,&\textnormal{ otherwise. }\end{cases}

(3)

$G=SO^{*}(2n)$ with $n\geq 4$ . Then

k(\lambda)=\begin{cases}q_{2}^{\mathrm{ev}},&\textnormal{ if }t_{1}\in\frac{1}{2}\mathbb{Z},\\ \left\lfloor\frac{n}{2}\right\rfloor,&\textnormal{ otherwise. }\end{cases}

(4)

$G=SO(2,2n-1)$ with $n\geq 3$ . Then

k(\lambda)=\begin{cases}0,&\textnormal{ if }t_{1}-t_{2}\in\mathbb{Z},t_{1}>t_{2},\\ 1,&\textnormal{ if }t_{1}-t_{2}\in\frac{1}{2}+\mathbb{Z},t_{1}>0,\\ 2,&\textnormal{ otherwise. }\end{cases}

(5)

$G=SO(2,2n-2)$ with $n\geq 4$ . Then

k(\lambda)=\begin{cases}0,&\textnormal{ if }t_{1}-t_{2}\in\mathbb{Z},t_{1}>t_{2},\\ 1,&\textnormal{ if }t_{1}-t_{2}\in\mathbb{Z},-\left|t_{n}\right|<t_{1}\leq t_{2},\\ 2,&\textnormal{ otherwise. }\end{cases}

Proposition 2.10 ([BXX23, Thm. 7.1]).

Let $L(\lambda)$ be a highest weight Harish-Chandra module of $E_{6(-14)}$ or $E_{7(-25)}$ . If $\lambda\in\mathfrak{h}^{\ast}$ is not integral, then $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ with $k(\lambda)=2$ (for $E_{6(-14)}$ ) or 3 (for $E_{7(-25)}$ ). In the case when $\lambda$ is integral, $k(\lambda)$ is given as follows.

(1)

If $G=E_{6(-14)}$ , then

k(\lambda)=\left\{\begin{array}[]{ll}\vspace{1ex}0,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{2}\neq\emptyset$},\\ \vspace{1ex}1,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{1}\neq\emptyset$~{}and~{}$\Psi_{\lambda}^{+}\cap S_{2}=\emptyset$},\\ 2,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{1}=\emptyset$}.\end{array}\right.

Here $S_{1}=\big{\{}\frac{1}{2}(\pm(e_{1}+e_{2}+e_{3}-e_{4})-e_{5}-e_{6}-e_{7}+e_{8})\}=\big{\{}\beta_{1},\beta_{2}\}$ , $S_{2}=\big{\{}\frac{1}{2}(e_{1}-e_{2}-e_{3}-e_{4}-e_{5}-e_{6}-e_{7}+e_{8})\}=\big{\{}\alpha_{1}\}$ and $\Psi_{\lambda}^{+}=\big{\{}\alpha\in\Phi^{+}|(\lambda,\alpha^{\vee})>0\}$ .

(2)

If $G=E_{7(-25)}$ , then

k(\lambda)=\left\{\begin{array}[]{ll}\vspace{1ex}0,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{3}\neq\emptyset$},\\ \vspace{1ex}1,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{2}\neq\emptyset$~{}and~{}$\Psi_{\lambda}^{+}\cap S_{3}=\emptyset$},\\ \vspace{1ex}2,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{1}\neq\emptyset$~{}and~{}$\Psi_{\lambda}^{+}\cap S_{2}=\emptyset$},\\ \vspace{1ex}3,&\textnormal{~{}if~{}$\Psi_{\lambda}^{+}\cap S_{1}=\emptyset$}.\end{array}\right.

Here $S_{1}=\big{\{}\frac{1}{2}(\pm(e_{1}-e_{2}-e_{3}+e_{4})-e_{5}+e_{6}-e_{7}+e_{8}),e_{5}+e_{6}\}=\big{\{}\beta_{3},\beta_{4},\beta_{5}\}$ , $S_{2}=\big{\{}\pm e_{1}+e_{6}\}=\big{\{}\beta_{6},\beta_{7}\}$ , $S_{3}=\big{\{}-e_{5}+e_{6}\}=\big{\{}\beta_{8}\}$ and $\Psi_{\lambda}^{+}=\big{\{}\alpha\in\Phi^{+}|(\lambda,\alpha^{\vee})>0\}$ .

2.3. Annihilator variety

For a $U(\mathfrak{g})$ -module $M$ , let $I(M)=\mathrm{Ann}(M)$ be its annihilator ideal in $U(\mathfrak{g})$ and $J(M)$ be the corresponding graded ideal in $S(\mathfrak{g})=\text{gr}(U(\mathfrak{g}))$ . The zero set of $J(M)$ in the dual vector space $\mathfrak{g}^{\ast}$ is called the annihilator variety of $M$ , and we denote it by $V(\mathrm{Ann}(M))$ . Let $G$ be a connected semisimple finite dimensional complex algebraic group with Lie algebra $\mathfrak{g}$ and $W$ be the Weyl group of $\mathfrak{g}$ . We use $L_{w}$ to denote the simple highest weight $\mathfrak{g}$ -module of highest weight $-w\rho-\rho$ with $w\in W$ . We denote $I_{w}=\mathrm{Ann}(L_{w})$ , then by [BB85] $V(I_{w})=\overline{\mathcal{O}}_{w}$ is irreducible, where ${\mathcal{O}}_{w}$ is a nilpotent coadjoint orbit in $\mathfrak{g}^{\ast}$ .

Proposition 2.11 ([Jos85]).

Let $\mathfrak{g}$ be a reductive Lie algebra and $I$ be a primitive ideal in $U(\mathfrak{g})$ .Then $V(I)$ is the closure of a single nilpotent coadjoint orbit $\mathcal{O}_{I}$ in $\mathfrak{g}^{*}$ . In particular, for a highest weight module $L(\lambda)$ , we have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{\mathrm{Ann}(L(\lambda))}$ .

From [Jos78, Prop. 2.7] and [Vog78], we have $\dim V(I_{w})=2\mathrm{GKdim}(L_{w})$ . In general, we have the following corollary.

Corollary 2.12.

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ for some $0\leq k(\lambda)\leq r$ . Then $\dim{\mathcal{O}}_{\mathrm{Ann}(L(\lambda))}=2\dim{\mathcal{O}}_{k(\lambda)}=2\dim V(L(\lambda))$ .

In [BMW23] the authors give the annihilator varieties of classical Lie algebras, and we recall the notions and algorithms here.

For $\lambda\in\mathfrak{h}^{*}$ , write $\lambda=\left(t_{1},\ldots,t_{n}\right)=\sum\limits_{i=1}^{n}t_{i}e_{i}$ , where $t_{i}\in\mathbb{C}$ and $\left\{e_{i}\mid 1\leq i\leq n\right\}$ is the canonical basis of the Euclidean space $\mathbb{R}^{n}$ . We associate to $\lambda$ a set $S(\lambda)$ of some Young tableaux as follows. Let $\lambda_{Y}:t_{i_{1}},t_{i_{2}},\ldots,t_{i_{r}}$ be a maximal subsequence of $t_{1},t_{2},\ldots,t_{n}$ such that $t_{i_{k}},1\leq k\leq r$ are congruent to each other by $\mathbb{Z}$ . Then the Young tableau $P\left(\lambda_{Y}\right)$ associated to the subsequence $\lambda_{Y}$ by using RS algorithm is a Young tableau in $S(\lambda)$ .

In case of $\mathfrak{g}=\mathfrak{so}(2n+1,\mathbb{C})$ , $\mathfrak{sp}(n,\mathbb{C})$ and $\mathfrak{so}(2n,\mathbb{C})$ , we need some more notations before we give the descriptions of $V(\mathrm{Ann}(L(\lambda)))$ .

Define $[\lambda]$ to be the set of maximal subsequence $\lambda_{Y}$ of $\lambda$ such that any two entries of $\lambda_{Y}$ have an integral difference or sum. In this case, we set $[\lambda]_{1}$ (resp. $[\lambda]_{2}$ ) to be the subset of $[\lambda]$ consisting of sequences with all entries belonging to $\mathbb{Z}$ (resp. $\frac{1}{2}+\mathbb{Z}$ ), Since there is at most one element in $[\lambda]_{1}$ and $[\lambda]_{2}$ , we denote them by $\lambda_{(0)}$ and $\lambda_{(\frac{1}{2})}$ . Set $[\lambda]_{1,2}=[\lambda]_{1}\cup[\lambda]_{2}$ and $[\lambda]_{3}=[\lambda]\setminus[\lambda]_{1,2}$ . For $\lambda_{Y}\in[\lambda]_{1,2}$ we can get a Young tableau $P(\lambda^{-}_{Y})$ and $P(\tilde{\lambda}_{Y})$ for $\lambda_{Y}\in[\lambda]_{3}$ . We use $P{\stackrel{{\scriptstyle c}}{{\sqcup}}}Q$ to denote a new Young tableau whose columns are the union of columns of the Young tableaux $P$ and $Q$ . We use $X$ to denote the corresponding type of Lie algebras, i.e., $X=B,C$ or $D$ . Then we have the following result.

Proposition 2.13 ([BMW23], Thm. 5.4, 6.5, 6.6 and 7.14).

Suppose $\lambda\in\mathfrak{h}^{\ast}$ , the annihilator variety of $L(\lambda)$ is given as follows.

(1)

If $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ , then

$V(\operatorname{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}(\lambda)},$

where ${\bf p}(\lambda)$ is the partition of the Young tableau

$P(\lambda)=\underset{P\left(\lambda_{Y}\right)\in S(\lambda)}{\stackrel{{\scriptstyle c}}{{\sqcup}}}P\left(\lambda_{Y}\right).$

(2)

If $\mathfrak{g}=\mathfrak{so}(2n+1,\mathbb{C}),\mathfrak{sp}(n,\mathbb{C})$ or $\mathfrak{so}(2n,\mathbb{C})$ , $\lambda\in\mathfrak{h}^{*}$ and $[\lambda]=[\lambda]_{1}\cup[\lambda]_{2}\cup[\lambda]_{3}$ with $[\lambda]_{3}=\{{\lambda}_{{Y}_{1}},\dots,{\lambda}_{{Y}_{m}}\}$ . Let

(a)

$\mathbf{p}_{0}$ be the $X$ -type special partition associated to $[\lambda]_{1}$ ;
(b)

${\mathbf{p}}_{\frac{1}{2}}$ be the $C$ -type special partition (for type $B$ ) or $C$ -type metaplectic special partition (for types $C$ and $D$ ) associated to $[\lambda]_{2}$ ;
(c)

$\mathbf{p}_{i}$ be the $A$ -type partition associated to $\tilde{\lambda}_{Y_{i}}$ .

Let ${\bf p}_{{}_{X}}(\lambda)$ be the $X$ -collapse of

\mathbf{d}_{\lambda}:=\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{\mathbf{p}}_{\frac{1}{2}}{\stackrel{{\scriptstyle c}}{{\sqcup}}}({{\stackrel{{\scriptstyle c}}{{\sqcup}}}}_{i}2\mathbf{p}_{i}).

Then we have

V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}_{{}_{X}}(\lambda)}.

Remark 2.14.

From [BMW23, §6], we know that $\mathbf{p}_{0}$ will be the partition $[1]$ in the case of type $B$ if $[\lambda]_{1}=\emptyset$ .

Note that a $C$ -type metaplectic special partition ${\bf p}$ means that ${\bf p}=(({\bf d}^{t})_{D})^{t}$ , where ${\bf d}$ is a special partition of type $D$ . From [BMW23], we can get a $C$ -type metaplectic special partition from the Young tableau $P(\lambda_{(\frac{1}{2})}^{-})$ by using the H-algorithm defined in [BMW23, Chap. 8]. We can also get a special partition from the Young tableau $P(\lambda^{-})$ or $P(\lambda_{(\frac{1}{2})}^{-})$ by using the H-algorithm.

Roughly speaking, the H-algorithm is equivalent to the following: during the process of collapse and expansion, we can not move the odd boxes for types B and C (resp. even boxes for type D) and the moving even box can not meet another even box in the same row (resp. the moving odd box can not meet another odd box in the same row for type D).

Special partitions are given by the following characterizations.

Proposition 2.15 ([CM93, Thm. 5.1.1, 5.1.2, 5.1.3 $\&$ 5.1.4 and Prop. 6.3.7]).

The nilpotent orbits of classical types can be identified with some partitions as follows:

(1)

Nilpotent orbits in $\mathfrak{sl}(n,\mathbb{C})$ are in one-to-one correspondence with the set of partitions of $n$ . Every partition is special.
(2)

Nilpotent orbits in $\mathfrak{so}{(2n+1,\mathbb{C})}$ are in one-to-one correspondence with the set of partitions of $2n+1$ in which even parts occur with even multiplicity. A partition $\bf q$ of type $B$ is special if and only if its dual partition ${\bf q}^{t}$ is a partition of type $B$ .
(3)

Nilpotent orbits in $\mathfrak{sp}{(n,\mathbb{C})}$ are in one-to-one correspondence with the set of partitions of $2n$ in which odd parts occur with even multiplicity. A partition $\bf q$ of type $C$ is special if and only if its dual partition ${\bf q}^{t}$ is a partition of type $C$ .
(4)

Nilpotent orbits in $\mathfrak{so}{(2n,\mathbb{C})}$ are in one-to-one correspondence with the set of partitions of $2n$ in which even parts occur with even multiplicity, except that each “very even” partition ${\bf d}$ (consisting of only even parts) correspond to two orbits, denoted by $\mathcal{O}^{I}_{\bf d}$ and $\mathcal{O}^{II}_{\bf d}$ . A partition $\bf q$ of type $D$ is special if and only if its dual partition ${\bf q}^{t}$ is a partition of type $C$ .

3. Annihilator varieties for type $A_{n-1}$

In this section, we consider the case of type $A_{n-1}$ .

Definition 3.1.

For $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ , we say $\lambda\in\mathfrak{h}^{\ast}$ is $(p,q)$ -dominant if $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for all $i,j$ such that $1\leq i<j\leq p$ and $p+1\leq i<j\leq p+q$ , where $\lambda=(t_{1},t_{2},\dots,t_{n})$ . In particular, $t_{1}>t_{2}>\dots>t_{p}$ and $t_{p+1}>t_{p+2}>\dots>t_{p+q}$ .

Proposition 3.2 ([BX19, Thm. 5.2]).

For $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ , assume that $\lambda\in\mathfrak{h}^{\ast}$ is $(p,q)$ -dominant.

(1)

If $t_{p}-t_{p+1}\in\mathbb{Z}$ , that is, $\lambda$ is an integral weight, then $P(\lambda)$ is a Young tableau with at most two columns. And in this case ${\rm GKdim}\>L(\lambda)=m(n-m)$ where $m$ is the number of entries in the second column of $P(\lambda$ ).
(2)

If $t_{p}-t_{p+1}\notin\mathbb{Z}$ , then $S(\lambda)$ consists of two Young tableaux with single column, and in this case ${\rm GKdim}\>L(\lambda)=pq$ .

Theorem 3.3.

Let $L(\lambda)$ be a Harish-Chandra module of $SU(p,n-p)$ with highest weight $\lambda-\rho\in\mathfrak{h}^{*}$ . Suppose $q_{2}$ is the length of second column of the Young tableau $P(\lambda)$ . Then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=\left\{\begin{array}[]{ll}[2^{q_{2}},1^{n-2q_{2}}],&\textnormal{~{}if~{}$\lambda$~{}is~{}integral},\\ [2^{r},1^{n-2r}],&\textnormal{~{}otherwise}.\end{array}\right.

Proof.

From [EHW83], we know that $\lambda\in\mathfrak{h}^{\ast}$ is $(p,n-p)$ -dominant.

(1)

When $\lambda$ is not integral, $S(\lambda)$ has two Young tableaux and these two Young tableaux have only one column, so

${\bf p}(\lambda_{Y_{1}})=[1^{p}]\text{~{}and~{}}{\bf p}(\lambda_{Y_{2}})=[1^{n-p}].$

By Proposition 2.13, ${\bf p}={\bf p}(\lambda)={\bf p}(\lambda_{Y_{1}}){\stackrel{{\scriptstyle c}}{{\sqcup}}}{\bf p}(\lambda_{Y_{2}})=[2^{r},1^{n-2r}]$ is the corresponding partition for $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ , where $r=\min\{p,n-p\}$ .
(2)

When $\lambda$ is integral, $P(\lambda)$ is a Young tableau with at most two columns by Proposition 3.2. Thus by Proposition 2.13, we have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ , where ${\bf p}={\bf p}(\lambda)=[2^{q_{2}},1^{n-2q_{2}}]$ with $q_{2}$ being the length of second column of the Young tableau $P(\lambda)$ .

So far, we have completed the proof. ∎

Corollary 3.4.

Keep notations as above. When $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ , we will have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=[2^{k(\lambda)},1^{n-2k(\lambda)}].

4. Annihilator varieties for types $B_{n}$ and $C_{n}$

In this section, we consider the case of types $B_{n}$ and $C_{n}$ .

Theorem 4.1.

Let $L(\lambda)$ be a Harish-Chandra module of $SO(2,2n-1)$ with highest weight $\lambda-\rho=\lambda_{0}+z\xi\in\mathfrak{h}^{*}$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=\left\{\begin{array}[]{ll}[1^{2n+1}],&\textnormal{~{}if~{}$z\in\lambda_{2}-\lambda_{1}+\mathbb{Z}_{\geq 0}$},\\ [2^{2},1^{2n-3}],&\textnormal{~{}if~{}$z>-\lambda_{1}-n+\frac{1}{2}\textnormal{~{}and~{}}z\in\frac{1}{2}+\mathbb{Z}$},\\ [3,1^{2n-2}],&\textnormal{~{}otherwise}.\end{array}\right.

Proof.

From [EHW83] we know that $\lambda-\rho=\lambda_{0}+z\xi$ with $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , $\lambda_{2}\geq\lambda_{3}\geq\cdots\geq\lambda_{n}\geq 0$ , $\lambda_{i}-\lambda_{j}\in\mathbb{Z}\textnormal{~{}and~{}}2\lambda_{i}\in\mathbb{N}$ for $2\leq i,\>j\leq n$ . By the normalization $(\lambda_{0}+\rho,\beta)=0$ , we have $\lambda_{1}+\lambda_{2}=-2n+2$ .

(1)

When $\lambda$ is integral, we have $\xi=(1,0,\dots,0)$ and

$\displaystyle\lambda=(z+\lambda_{1}+n-\frac{1}{2},\lambda_{2}+n-\frac{3}{2},\cdots,\lambda_{n}+\frac{1}{2}):=(t_{1},...,t_{n}).$

Thus we have $t_{2}>t_{3}>\cdots>t_{n-1}>t_{n}>0$ .

When $t_{1}>t_{2}$ (equivalently $z\in\lambda_{2}-\lambda_{1}+\mathbb{Z}_{\geq 0}$ ), $L(\lambda)$ will be finite-dimensional and $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with the partition ${\bf p}=[1^{2n+1}]$ corresponding to the trivial orbit of type $B_{n}$ .

When $t_{1}\leq t_{2}$ , by using the Robinson-Schensted algorithm for $\lambda^{-}$ , we can get a Young tableau $P(\lambda^{-})$ which consists of at most three columns. When $t_{2}\geq t_{1}>-t_{n}$ , from the construction process, we can see that $P(\lambda^{-})$ will be a Young tableau consisting of two columns with $c_{1}(P(\lambda^{-}))=2n-2$ and $c_{2}(P(\lambda^{-}))=2$ . Thus ${\bf p}(\lambda^{-})=[2,2,1^{2n-4}]$ . In this case, the special partition of type $B$ corresponding to ${\bf p}(\lambda^{-})$ is ${\bf p}=[3,1^{2n-2}]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ by Proposition 2.13. When $t_{1}\leq-t_{n}$ , $P(\lambda^{-})$ will be a Young tableau consisting of three columns with $c_{1}(P(\lambda^{-}))=2n-2$ , $c_{2}(P(\lambda^{-}))=1$ and $c_{3}(P(\lambda^{-}))=1$ . Thus ${\bf p}(\lambda^{-})=[3,1^{2n-3}]$ . In this case, the special partition of type $B$ corresponding to ${\bf p}(\lambda^{-})$ is ${\bf p}=[3,1^{2n-2}]$ .

(2)

When $\lambda$ is half integral, we have $t_{n}>0$ , $t_{1}-t_{2}\in\frac{1}{2}+\mathbb{Z}$ , $2t_{k}\in\mathbb{Z}$ for $1\leq k\leq n$ and $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $2\leq i<j\leq n$ . Then we have $t_{2}>t_{3}>...>t_{n}>-t_{n}>...>-t_{2}$ . We denote $\lambda_{Y_{1}}=(t_{2},...,t_{n})$ and $\lambda_{Y_{2}}=(t_{1})$ . Thus $\lambda_{Y_{1}},\lambda_{Y_{2}}\in[\lambda]_{1,2}$ . By using Robinson-Schensted algorithm, we can get a Young tableau $P(\lambda_{Y_{1}}^{-})$ with shape ${\bf p}(\lambda_{Y_{1}}^{-})=[1^{2n-2}]$ and a Young tableau $P(\lambda_{Y_{2}}^{-})$ with shape

{\bf p}(\lambda_{Y_{2}}^{-})=\left\{\begin{array}[]{ll}[2],&\textnormal{~{}if~{}}t_{1}\leq 0,\\ [1,1],&\textnormal{~{}if~{}}t_{1}>0.\end{array}\right.

If $t_{1}\in\frac{1}{2}+\mathbb{Z}$ (equivalently $z\in\frac{1}{2}+\mathbb{Z}$ ), we will have $\lambda_{Y_{1}}\in[\lambda]_{1}$ and $\lambda_{Y_{2}}\in[\lambda]_{2}$ . From Proposition 2.13, we have $\mathbf{p}_{0}=[1^{2n-1}]$ and

\mathbf{p}_{\frac{1}{2}}=\left\{\begin{array}[]{ll}[1,1],&\textnormal{~{}if~{}}t_{1}>0,\\ [2],&\textnormal{~{}if~{}}t_{1}\leq 0.\end{array}\right.

Therefore by Proposition 2.13, we have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{\mathbf{p}}_{\frac{1}{2}})_{B}=\left\{\begin{array}[]{ll}[2,2,1^{2n-3}],&\textnormal{~{}if~{}}t_{1}>0,\\ [3,1^{2n-2}],&\textnormal{~{}if~{}}t_{1}\leq 0.\end{array}\right.

If $t_{1}\in\mathbb{Z}$ (equivalently $z\in\frac{1}{2}+\mathbb{Z}$ ), we will have $\lambda_{Y_{1}}\in[\lambda]_{2}$ and $\lambda_{Y_{2}}\in[\lambda]_{1}$ . From Proposition 2.13, we have $\mathbf{p}_{\frac{1}{2}}=[1^{2n-2}]$ and

\mathbf{p}_{0}=\left\{\begin{array}[]{ll}[1,1,1],&\textnormal{~{}if~{}}t_{1}>0,\\ [3],&\textnormal{~{}if~{}}t_{1}\leq 0.\end{array}\right.

Therefore by Proposition 2.13, we have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{\mathbf{p}}_{\frac{1}{2}})_{B}=\left\{\begin{array}[]{ll}[2,2,1^{2n-3}],&\textnormal{~{}if~{}}t_{1}>0,\\ [3,1^{2n-2}],&\textnormal{~{}if~{}}t_{1}\leq 0.\end{array}\right.

Note that $t_{1}>0$ is equivalent to $z>-\lambda_{1}-n+\frac{1}{2}$ .

(3)

When $\lambda$ is not integral or half integral, we have $\lambda_{Y_{1}}=(t_{2},\cdots,t_{n})\in[\lambda]_{1,2}$ and $\lambda_{Y_{2}}=(t_{1})\in[\lambda]_{3}$ . By using Robinson-Schensted algorithm, we can get a Young tableau $P(\lambda_{Y_{1}}^{-})$ with shape ${\bf p}(\lambda_{Y_{1}}^{-})=[1^{2n-2}]$ and a Young tableau $P(\lambda_{Y_{2}})$ with shape $[1]$ . Thus we have $\mathbf{p}_{0}=[1^{2n-1}]$ or ${\mathbf{p}}_{\frac{1}{2}}=[1^{2n-2}]$ , and $\mathbf{p}_{1}=[1]$ . Therefore by Proposition 2.13, we have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{2\mathbf{p}}_{1})_{B}=[3,1^{2n-2}]$ or ${\bf p}=([1]{\stackrel{{\scriptstyle c}}{{\sqcup}}}\mathbf{p}_{\frac{1}{2}}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{2\mathbf{p}}_{1})_{B}=[3,1^{2n-2}]$ .

This finishes the proof.

∎

Remark 4.2.

From [EHW83, Lem. 3.17], we know that the generalized Verma module $N(\lambda)$ will be irreducible when $\lambda$ is not integral or half integral. Thus we will have $V(\mathrm{Ann}(L(\lambda)))=V(\mathrm{Ann}(N(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=[3,1^{2n-2}]$ , which is just the Richardson nilpotent orbit $G\cdot\mathfrak{p}^{+}$ , see [BZ17].

Theorem 4.3.

Let $L(\lambda)$ be a Harish-Chandra module of $Sp(n,\mathbb{R})$ with highest weight $\lambda-\rho\in\mathfrak{h}^{*}$ . Suppose $q_{2}$ is the length of second column of the Young tableau $P(\lambda^{-})$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=\left\{\begin{array}[]{ll}[2^{2q_{2}^{\mathrm{od}}},1^{2n-4q_{2}^{\mathrm{od}}}],&\textnormal{~{}if~{}}z\in\mathbb{Z},\\ [2^{2q_{2}^{\mathrm{ev}}+1},1^{2n-4q_{2}^{\mathrm{ev}}-2}],&\textnormal{~{}if~{}}z\in\frac{1}{2}+\mathbb{Z},\\ [2^{n}],&\textnormal{~{}otherwise.}\\ \end{array}\right.

Proof.

From [EHW83] we know that $\lambda-\rho=\lambda_{0}+z\xi$ with $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , $\lambda_{i}-\lambda_{j}\in\mathbb{N}$ for $i<j$ . By the normalization $(\lambda_{0}+\rho,\beta)=0$ , we have $\lambda_{1}=-n$ .

Firstly we suppose that $n$ is even. Then we have the follows.

(1)

When $\lambda$ is integral, we have $\xi=(\underbrace{1,1,\dots,1}_{n})$ and

\lambda=(\lambda_{1}+z+n,\lambda_{2}+z+n-1,\cdots,\lambda_{n}+z+1):=(t_{1},t_{1},\cdots,t_{n}).

Thus we have $t_{1}>t_{2}>\cdots>t_{n-1}>t_{n}$ .

When $t_{1}\in\mathbb{Z}$ (equivalently $z\in\mathbb{Z}$ ), by using the Robinson-Schensted algorithm for $\lambda^{-}$ , we can get a Young tableau $P(\lambda^{-})$ which consists of at most two columns with $c_{2}(P(\lambda^{-}))=q_{2}$ and $c_{1}(P(\lambda^{-}))=2n-q_{2}$ . Thus ${\bf p}(\lambda^{-})=[2^{q_{2}},1^{2n-2q_{2}}]$ . In this case the special partition of type $C$ corresponding to ${\bf p}(\lambda^{-})$ is

{\bf p}=\left\{\begin{array}[]{ll}[2^{q_{2}},1^{2n-2q_{2}}],&\textnormal{~{}if~{}}q_{2}=2k,\\ [2^{q_{2}+1},1^{2n-2q_{2}-2}],&\textnormal{~{}if~{}}q_{2}=2k+1.\end{array}\right.

Equivalently, ${\bf p}=[2^{2q_{2}^{\mathrm{od}}},1^{2n-4q_{2}^{\mathrm{od}}}]$ by (2.7). Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ by Proposition 2.13.

(2)

When $\lambda$ is half integral (equivalently $z\in\frac{1}{2}+\mathbb{Z}$ ), we have $\lambda\in[\lambda]_{2}$ . Similarly we can get a Young tableau $P(\lambda^{-})$ which consists of at most two columns with $c_{2}(P(\lambda^{-}))=q_{2}$ and $c_{1}(P(\lambda^{-}))=2n-q_{2}$ . Thus ${\bf p}(\lambda^{-})=[2^{q_{2}},1^{2n-2q_{2}}]$ . In this case the $C$ -type metaplectic special partition ${\mathbf{p}}_{\frac{1}{2}}$ corresponding to ${\bf p}(\lambda^{-})$ is

{\mathbf{p}}_{\frac{1}{2}}=[2^{2q_{2}^{\mathrm{ev}}+1},1^{2n-4q_{2}^{\mathrm{ev}}-2}]=\left\{\begin{array}[]{ll}[2^{q_{2}+1},1^{2n-2q_{2}-2}],&\textnormal{~{}if~{}}q_{2}=2k,\\ [2^{q_{2}},1^{2n-2q_{2}}],&\textnormal{~{}if~{}}q_{2}=2k+1.\end{array}\right.

by (2.7).

Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=({\mathbf{p}}_{\frac{1}{2}})_{C}={\mathbf{p}}_{\frac{1}{2}}$ by Proposition 2.13.

(3)

When $\lambda$ is not integral or half integral, we have $\lambda\in[\lambda]_{3}$ . By Proposition 2.13, we have ${\bf p}_{1}=p(\tilde{\lambda})=p(\lambda)=[1^{n}]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(2\mathbf{p}_{1})_{C}=[2^{n}]$ by Proposition 2.13.

When $n$ is odd, the argument is similar to the case when $n$ is even. We omit the details here.

This finishes the proof. ∎

Corollary 4.4.

Keep notations as above. When $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ , we will have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=[2^{k(\lambda)},1^{2n-2k(\lambda)}].

Remark 4.5.

From [EHW83, Lem. 3.17], we know that the generalized Verma module $N(\lambda)$ will be irreducible when $\lambda$ is not integral or half integral. Thus we will have $V(\mathrm{Ann}(L(\lambda)))=V(\mathrm{Ann}(N(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=[2^{n}]$ , which is just the Richardson nilpotent orbit $G\cdot\mathfrak{p}^{+}$ , see [BZ17].

5. Annihilator varieties for type $D_{n}$

In this section, we consider the case of type $D_{n}$ .

Theorem 5.1.

Let $L(\lambda)$ be a Harish-Chandra module of $SO(2,2n-2)$ with highest weight $\lambda-\rho=\lambda_{0}+z\xi\in\mathfrak{h}^{*}$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=\left\{\begin{array}[]{ll}[1^{2n}],&\textnormal{~{}if~{}}z\in\lambda_{2}-\lambda_{1}+\mathbb{Z}_{\geq 0},\\ [2^{2},1^{2n-4}],&\textnormal{~{}if~{}}z+\lambda_{1}-\lambda_{2}\in\mathbb{Z}\textnormal{~{}and~{}}-|\lambda_{n}|-\lambda_{1}-n+1<z\leq\lambda_{2}-\lambda_{1}-1,\\ [3,1^{2n-3}],&\textnormal{~{}otherwise.}\\ \end{array}\right.

Proof.

From [EHW83] we know that $\lambda-\rho=\lambda_{0}+z\xi$ with $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , $\lambda_{2}\geq\lambda_{3}\geq\cdots\lambda_{n-1}\geq|\lambda_{n}|$ , $\lambda_{i}-\lambda_{j}\in\mathbb{N}\textnormal{~{}and~{}}2\lambda_{i}\in\mathbb{N}$ for $2\leq i<j\leq n$ . By the normalization $(\lambda_{0}+\rho,\beta)=0$ , we have $\lambda_{1}+\lambda_{2}=-2n+3$ .

(1)

When $\lambda$ is integral, we have $\xi=(1,0,\dots,0)$ and

$\displaystyle\lambda=(z+\lambda_{1}+n-1,\lambda_{2}+n-2,\cdots,\lambda_{n-1}+1,\lambda_{n}):=(t_{1},...,t_{n}).$

Thus we have $t_{2}>t_{3}>\cdots>t_{n-1}>|t_{n}|$ .

When $t_{1}>t_{2}$ (equivalently $z\in\lambda_{2}-\lambda_{1}+\mathbb{Z}_{\geq 0}$ ), $L(\lambda)$ will be finite-dimensional and $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with the partition ${\bf p}=[1^{2n}]$ corresponding to the trivial orbit of type $D_{n}$ .

When $t_{1}\leq t_{2}$ , by using the Robinson-Schensted algorithm for $\lambda^{-}$ , we can get a Young tableau $P(\lambda^{-})$ which consists of at most four columns. When $-|t_{n}|<t_{1}\leq t_{2}$ (equivalently $-|\lambda_{n}|-\lambda_{1}-n+1<z\leq\lambda_{2}-\lambda_{1}-1$ ), from the construction process, we can see that $P(\lambda^{-})$ will be a Young tableau consisting of two columns with shape ${\bf p}(\lambda^{-})=[2^{2},1^{2n-4}]$ or $[2^{3},1^{2n-6}]$ . In this case, the special partition of type $D$ corresponding to ${\bf p}(\lambda^{-})$ is ${\bf p}=[2^{2},1^{2n-4}]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ by Proposition 2.13.

When $t_{1}\leq-|t_{n}|$ , $P(\lambda^{-})$ will be a Young tableau with shape ${\bf p}(\lambda^{-})=[3,1^{2n-3}]$ or ${\bf p}(\lambda^{-})=[4,1^{2n-4}]$ . In this case, the special partition of type $D$ corresponding to ${\bf p}(\lambda^{-})$ is ${\bf p}=[3,1^{2n-3}]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ by Proposition 2.13.

(2)

When $\lambda$ is half integral, we have $\lambda_{Y_{1}}=(t_{2},\cdots,t_{n})\in[\lambda]_{1}$ and $\lambda_{Y_{2}}=(t_{1})\in[\lambda]_{2}$ (when $z\in\frac{1}{2}+\mathbb{Z}$ ), or $\lambda_{Y_{1}}\in[\lambda]_{2}$ and $\lambda_{Y_{2}}=(t_{1})\in[\lambda]_{1}$ (when $z\in\frac{1}{2}+\mathbb{Z}$ ). Then we can get a Young tableau $P(\lambda_{Y_{1}}^{-})$ with shape ${\bf p}(\lambda_{Y_{1}}^{-})=[1^{2n-2}]$ or $[2,1^{2n-4}]$ , and a Young tableau $P(\lambda_{Y_{2}}^{-})$ with shape

{\bf p}(\lambda_{Y_{2}}^{-})=\left\{\begin{array}[]{ll}[2],&\textnormal{~{}if~{}}t_{1}\leq 0,\\ [1,1],&\textnormal{~{}if~{}}t_{1}>0.\end{array}\right.

(a)

If $t_{1}\in\frac{1}{2}+\mathbb{Z}$ (equivalently $z\in\frac{1}{2}+\mathbb{Z}$ ), we will have $\lambda_{Y_{1}}\in[\lambda]_{1}$ and $\lambda_{Y_{2}}\in[\lambda]_{2}$ . From Proposition 2.13, we have $\mathbf{p}_{0}=[1^{2n-2}]$ and the $C$ -type metaplectic special partition ${\mathbf{p}}_{\frac{1}{2}}$ corresponding to $p(\lambda_{Y_{2}}^{-})$ is ${\mathbf{p}}_{\frac{1}{2}}=[2]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{\mathbf{p}}_{\frac{1}{2}})_{B}=[3,1^{2n-3}]$ by Proposition 2.13.
(b)

If $t_{1}\in\mathbb{Z}$ (equivalently $z\in\frac{1}{2}+\mathbb{Z}$ ), we will have $\lambda_{Y_{1}}\in[\lambda]_{2}$ and $\lambda_{Y_{2}}\in[\lambda]_{1}$ . From Proposition 2.13, the $C$ -type metaplectic special partition corresponding to $p(\lambda_{Y_{1}}^{-})$ is $\mathbf{p}_{\frac{1}{2}}=[2,1^{2n-4}]$ and $\mathbf{p}_{0}=[1,1]$ if $t_{1}>0$ or $\mathbf{p}_{0}=[2]$ if $t_{1}\leq 0$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{\mathbf{p}}_{\frac{1}{2}})_{B}=[3,1^{2n-3}]$ by Proposition 2.13.

(3)

When $\lambda$ is not integral or half integral, we have $\lambda_{Y_{1}}=(t_{2},\cdots,t_{n})\in[\lambda]_{1,2}$ and $\lambda_{Y_{2}}=(t_{1})\in[\lambda]_{3}$ (equivalently $z\notin\frac{1}{2}\mathbb{Z}$ ). Then we can get a Young tableau $P(\lambda_{Y_{1}}^{-})$ with shape ${\bf p}(\lambda_{Y_{1}}^{-})=[1^{2n-2}]$ or $[2,1^{2n-4}]$ , and a Young tableau $P(\lambda_{Y_{2}})$ with shape $[1]$ .
1. (a)
  
  If $\lambda_{Y_{1}}\in[\lambda]_{1}$ , we will have $\mathbf{p}_{0}=[1^{2n-2}]$ . From $P(\lambda_{Y_{2}})$ , we can get a partition ${\bf p}_{1}=[1]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(\mathbf{p}_{0}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{2\mathbf{p}}_{1})_{B}=[3,1^{2n-3}]$ by Proposition 2.13.
2. (b)
  
  If $\lambda_{Y_{1}}\in[\lambda]_{2}$ , the $C$ -type metaplectic special partition corresponding to ${\bf p}(\lambda_{Y_{1}}^{-})$ is $\mathbf{p}_{\frac{1}{2}}=[2,1^{2n-4}]$ . From $P(\lambda_{Y_{2}})$ , we can get a partition ${\bf p}_{1}=[1]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=(\mathbf{p}_{\frac{1}{2}}{\stackrel{{\scriptstyle c}}{{\sqcup}}}{2\mathbf{p}}_{1})_{B}=[3,1^{2n-3}]$ by Proposition 2.13.

This finishes the proof.

∎

Remark 5.2.

From [EHW83, Lem. 3.17], we know that the generalized Verma module $N(\lambda)$ will be irreducible when $\lambda$ is not integral. Thus we will have $V(\mathrm{Ann}(L(\lambda)))=V(\mathrm{Ann}(N(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with ${\bf p}=[3,1^{2n-3}]$ , which is just the Richardson nilpotent orbit $G\cdot\mathfrak{p}^{+}$ , see [BZ17].

Theorem 5.3.

Let $L(\lambda)$ be a Harish-Chandra module of $SO^{\ast}(2n)$ with highest weight $\lambda-\rho=\lambda_{0}+z\xi\in\mathfrak{h}^{*}$ . Suppose $q_{2}$ is the length of second column of the Young tableau $P(\lambda^{-})$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=\left\{\begin{array}[]{ll}[2^{2q_{2}^{\mathrm{ev}}},1^{2n-4q_{2}^{\mathrm{ev}}}],&\textnormal{~{}if~{}}z\in\mathbb{Z},\\ [2^{n}],&\textnormal{~{}if~{}}n\in 2\mathbb{Z},\\ [2^{n-1},1,1],&\textnormal{~{}if~{}}n\in 2\mathbb{Z}+1.\end{array}\right.

Proof.

From [EHW83] we know that $\lambda-\rho=\lambda_{0}+z\xi$ with $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ , and $\lambda_{i}-\lambda_{j}\in\mathbb{N}$ for $i\leq j$ . By the normalization $(\lambda_{0}+\rho,\beta)=0$ , we have $\lambda_{1}+\lambda_{2}=-2n+3$ and $2\lambda_{i}\in\mathbb{Z}$ for $1\leq i\leq n$ .

(1)

When $\lambda$ is integral (equivalently $z\in\mathbb{Z}$ ), we have $\xi=(\underbrace{\frac{1}{2},\frac{1}{2},\dots,\frac{1}{2}}_{n})$ and

\lambda=(\frac{1}{2}z+\lambda_{1}+n-1,\frac{1}{2}z+\lambda_{2}+n-2,\cdots,\frac{1}{2}z+\lambda_{n-1}+1,\frac{1}{2}z+\lambda_{n}):=(t_{1},t_{2},\cdots,t_{n}).

Thus we have $t_{1}>t_{2}>\cdots>t_{n}$ .

By using the Robinson-Schensted algorithm for $\lambda^{-}$ , we can get a Young tableau $P(\lambda^{-})$ which consists of at most two columns with $c_{1}(P(\lambda^{-}))=2n-q_{2}$ and $c_{2}(P(\lambda^{-}))=q_{2}$ . Thus ${\bf p}(\lambda^{-})=[2^{q_{2}},1^{2n-2q_{2}}]$ . In this case the special partition of type $D$ corresponding to ${\bf p}(\lambda^{-})$ is ${\bf p}=[2^{2q_{2}^{\mathrm{ev}}},1^{2n-4q_{2}^{\mathrm{ev}}}]$ by (2.7). Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ by Proposition 2.13.

(2)

When $\lambda$ is not integral (equivalently $z\notin\mathbb{Z}$ ), we will have $\lambda\in[\lambda]_{3}$ . From Proposition 2.13 we have ${\bf p}_{1}=p(\lambda_{Y_{1}})=[1^{n}]$ . Therefore $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=(2\mathbf{p}_{1})_{D}=\left\{\begin{array}[]{ll}[2^{n}],&\textnormal{~{}if~{}}n\in 2\mathbb{Z},\\ [2^{n-1},1,1],&\textnormal{~{}if~{}}n\in 2\mathbb{Z}+1.\end{array}\right.

by Proposition 2.13.

∎

Corollary 5.4.

Keep notations as above. When $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ , we will have $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}$ with

{\bf p}=[2^{2k(\lambda)},1^{2n-4k(\lambda)}].

Remark 5.5.

From [EHW83, Lem. 3.17], we know that the generalized Verma module $N(\lambda)$ will be irreducible when $\lambda$ is not integral. Thus when $n$ is even, we will have

V(\mathrm{Ann}(L(\lambda)))=V(\mathrm{Ann}(N(\lambda)))=\overline{\mathcal{O}}_{{\bf p}}

with ${\bf p}=[2^{n}]$ , which is just the Richardson nilpotent orbit $G\cdot\mathfrak{p}^{+}$ , see [BZ17].

When $G=SO^{\ast}(2n)$ , from Proposition 2.15, we know that each very even partition ${\bf p}$ corresponds to two special nilpotent orbits $\overline{\mathcal{O}}_{{\bf p}}^{I}$ and $\overline{\mathcal{O}}_{{\bf p}}^{II}$ . Note that in $SO^{\ast}(2n)$ , ${\bf p}=[2^{n}]$ is a very even partition when $n$ is even, which corresponds to two nilpotent orbits. In fact, the Levi subalgebra $\mathfrak{k}$ of our parabolic subalgebra $\mathfrak{q}=\mathfrak{k}\oplus\mathfrak{p}^{+}$ is of type $I$ by [CM93, Lem. 7.3.2]. From [CM93, Lem. 7.3.3], we know that the numeral of $\mathcal{O}$ appeared in $V(\mathrm{Ann}(L(\lambda)))$ is $I$ if $n$ is even. In other words, we have

V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{[2^{n}]}^{I}\text{~{}when~{} $n$ ~{}is~{} even}.

6. Annihilator varieties for types $E_{6}$ and $E_{7}$

In this section, we consider annihilator varieties of highest weight Harish-Chandra modules for two exceptional groups of type $E_{6(-14)}$ and $E_{7(-25)}$ . We recall some notations from [Car85] and [CM93].

Let $\mathfrak{g}$ be a semisimple Lie algebra. Let $\mathfrak{l}$ be a Levi subalgebra of $\mathfrak{g}$ and $\mathfrak{p}_{\mathfrak{l}}$ be a distinguished parabolic subalgebra of the semisimple algebra $[\mathfrak{l},\mathfrak{l}]$ . From [CM93, Thm. 8.2.12], we know that a nilpotent orbit is corresponding to an ordered pair $(\mathfrak{l},\mathfrak{p}_{\mathfrak{l}})$ . From [BC76], we denote this orbit by the label $X_{N}(a_{i})$ , where $X_{N}$ is the Cartan type of the semisimple part of $\mathfrak{l}$ and $i$ is the number of simple roots in any Levi subalgebra of $\mathfrak{p}_{\mathfrak{l}}$ . If $i=0$ , one writes $X_{N}$ rather than $X_{N}(a_{0})$ . When $\mathfrak{g}$ is of type $E_{7}$ , it has two non-conjugate isomorphic Levi subalgebras. One subalgebra is chosen arbitrarily and labeled with a prime, and the other one has a double prime.

In the following tables we give four pieces of information about the nilpotent orbits which will be used: Bala-Carter label, dimension, fundamental group and whether the orbit is special or not. The Bala-Carter label $\mathcal{K}=sX_{N}$ denotes $s$ copies of $X_{N}$ .

Table 2. Some nilpotent orbits in type

E_{6}

${\rm label}~{}\mathcal{K}$	$\dim\mathcal{O}$	$\pi_{1}(\mathcal{O})$	${\rm special~{}orbit}$
$0$	$0$	$1$	yes
$A_{1}$	$22$	$1$	${\rm yes}$
$2A_{1}$	$32$	$1$	${\rm yes}$
$3A_{1}$	$40$	$1$	${\rm no}$
$A_{2}$	$42$	$S_{2}$	${\rm yes}$

Table 3. Some nilpotent orbits in type

E_{7}

${\rm label}~{}\mathcal{K}$	$\dim\mathcal{O}$	$\pi_{1}(\mathcal{O})$	${\rm special~{}orbit}$
$0$	$0$	$1$	yes
$A_{1}$	$34$	$1$	${\rm yes}$
$2A_{1}$	$52$	$1$	${\rm yes}$
$(3A_{1})^{\prime\prime}$	$54$	$\mathbb{Z}/2\mathbb{Z}$	${\rm yes}$
$(3A_{1})^{\prime}$	$64$	$1$	${\rm no}$
$A_{2}$	$66$	$S_{2}$	${\rm yes}$

Theorem 6.1.

Let $L(\lambda)$ be a highest weight Harish-Chandra module of the exceptional Lie group $E_{6(-14)}$ with highest weight $\lambda-\rho=\lambda_{0}+z\xi\in\mathfrak{h}^{*}$ and $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{8})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{\mathcal{K}}$ with

\mathcal{K}=\left\{\begin{array}[]{ll}\vspace{1ex}0,&\textnormal{~{}if~{}}z>-(\lambda_{0},\alpha_{1})-1\textnormal{~{}and~{}}\lambda\textnormal{~{}is~{}integral},\\ \vspace{1ex}A_{1},&\textnormal{~{}if~{}}-4+\min\{-(\lambda_{0},\beta_{1,2})\}<z\leq-(\lambda_{0},\alpha_{1})-1\textnormal{~{}and~{}}\lambda\textnormal{~{}is~{}integral},\\ \vspace{1ex}2A_{1},&\textnormal{~{}otherwise.}\\ \end{array}\right.

Here $\beta_{1,2}$ means the root $\beta_{1}$ and $\beta_{2}$ in Proposition 2.10 (1).

Proof.

From [EHW83] we know that $\lambda-\rho=\lambda_{0}+z\xi$ with $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{8})$ , $|\lambda_{1}|\leq\lambda_{2}\leq\cdots\lambda_{5}$ , $\lambda_{i}-\lambda_{j}\in\mathbb{Z}$ and $2\lambda_{i}\in\mathbb{Z}$ for $i,j\leq 5$ . By the normalization $(\lambda_{0}+\rho,\beta)=0$ , we have $(\lambda_{0},\beta)=-11$ .

(1)
When $\lambda$ is integral, we have $\xi=(0,0,0,0,0,-\frac{2}{3},-\frac{2}{3},\frac{2}{3})$
1. (a)
  
  When $\Psi_{\lambda}^{+}\cap S_{2}\neq\emptyset$ , equivalently $(\lambda,\alpha_{1}^{\vee})>0$ , we will have $k(\lambda)=0$ by Proposition 2.10. By Corollary 2.12 and Table 2, we can get that the label of the corresponding nilpotent orbit is $\mathcal{K}=0$ .
  
  Hence $(\lambda,\alpha_{1}^{\vee})>0$ if and only if $z>-(\lambda_{0},\alpha_{1})-1$ . In this case the label $\mathcal{K}=0$ .
2. (b)
  
  When $\Psi_{\lambda}^{+}\cap S_{1}=\emptyset$ , equivalently, $(\lambda,\beta_{i}^{\vee})\leq 0$ for $i=1,2$ , we will have $k(\lambda)=2$ by Proposition 2.10. By Corollary 2.12 and Table 2, we can get that the label of the corresponding nilpotent orbit is $\mathcal{K}=2A_{1}$ .
  
  Hence $(\lambda,\beta_{i}^{\vee})\leq 0$ if and only if $z\leq-4+\min\{-(\lambda_{0},\beta_{i})\}$ for $i=1,2$ . In this case the label $\mathcal{K}=2A_{1}$ .
3. (c)
  
  When $\Psi_{\lambda}^{+}\cap S_{1}\neq\emptyset$ and $\Psi_{\lambda}^{+}\cap S_{2}=\emptyset$ , equivalently, $z>-4+\min\{-(\lambda_{0},\beta_{i})\}$ and $z\leq-(\lambda_{0},\alpha_{1})-1$ , we will have $k(\lambda)=2$ by Proposition 2.10. By Corollary 2.12 and Table 2, we can get that the label of the corresponding nilpotent orbit is $\mathcal{K}=A_{1}$ .
  
  Hence when $-4+\min\{-(\lambda_{0},\beta_{i})\}<z\leq-(\lambda_{0},\alpha_{1})-1$ , the label $\mathcal{K}=A_{1}$ .
(2)

When $\lambda$ is non-integral, we will have $k(\lambda)=2$ by Proposition 2.10. By Corollary 2.12 and Table 2, we can get that the label of the corresponding nilpotent orbit is $\mathcal{K}=2A_{1}$ .

∎

Theorem 6.2.

Let $L(\lambda)$ be a highest weight Harish-Chandra module of the exceptional Lie group $E_{7(-25)}$ with highest weight $\lambda-\rho=\lambda_{0}+z\xi\in\mathfrak{h}^{*}$ and $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ . Denote $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{8})$ , then $V(\mathrm{Ann}(L(\lambda)))=\overline{\mathcal{O}}_{\mathcal{K}}$ with

\mathcal{K}=\left\{\begin{array}[]{ll}\vspace{1ex}0,&\textnormal{if~{}}z>-(\lambda_{0},\beta_{8})-1\textnormal{~{}and~{}}\lambda\textnormal{~{}is~{}integral},\\ \vspace{1ex}A_{1},&\textnormal{if~{}}-(\lambda_{0},\beta_{6,7})-5<z\leq-(\lambda_{0},\beta_{8})-1\textnormal{~{}and~{}}\lambda\textnormal{~{}is~{}integral},\\ \vspace{1ex}2A_{1},&\textnormal{if~{}}-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}<z\leq-(\lambda_{0},\beta_{6,7})-5\textnormal{~{}and~{}}\lambda\textnormal{~{}is~{}integral},\\ (3A_{1})^{\prime\prime},&\textnormal{otherwise.}\\ \end{array}\right.

Here $\beta_{3,4,5}$ means the root $\beta_{3}$ , $\beta_{4}$ and $\beta_{5}$ in Proposition 2.10 (2). Similarly $\beta_{6,7}$ means the root $\beta_{6}$ and $\beta_{7}$ in Proposition 2.10 (2).

Proof.

The argument is similar to the case of $E_{6(-14)}$ . So we omit the details here. ∎

7. The Gelfand-Kirillov dimension of a unitary highest weight module

Proposition 2.5 indicates that if $L(\lambda)$ is a unitary highest weight module, then ${\rm GKdim}\>L(\lambda)$ only depends on the value $z=(\lambda,\beta^{\vee})$ . The proof of Proposition 2.5 used some results from Joseph [Jos92]. In this section, we will prove this result by another method.

Lemma 7.1 ([Bai+24, Lem. 5.4]).

Suppose $\mu=(t_{1},\cdots,t_{n},s_{1},\cdots,s_{m})$ is a $(n,m)$ -dominant sequence with $t_{n}\leq s_{1}$ . Then applying R-S algorithm to $\mu$ , we can get a Young tableau $P(\mu)$ with two columns, and the number of boxes in the second column is precisely the largest integer $k$ for which we have

t_{n-k+1}\leq s_{1},\>t_{n-k+2}\leq s_{2},\>\cdots,\>t_{n-1}\leq s_{k-1},\>t_{n}\leq s_{k}.

Now we recall two root systems $Q(\lambda_{0})$ and $R(\lambda_{0})$ given in [EHW83]. Their constructions are as follows. Let $\Phi_{c}(\lambda_{0})=\{\alpha\in\Phi(\mathfrak{k})|(\lambda_{0},\alpha)=0\}$ . Consider the subroot system $\Psi_{1}$ of $\Phi$ , which is generated by $\pm\beta$ and $\Phi_{c}(\lambda_{0})$ . Let $Q(\lambda_{0})$ be the simple component of $\Psi_{1}$ which contains $-\beta.$ If $\Phi$ has two root lengths and there exist short roots $\alpha^{\prime}\in\Phi(\mathfrak{k})$ which are not orthogonal to $Q(\lambda_{0})$ and satisfy $(\lambda_{0},\alpha^{\prime\vee})=1$ , then let $\Psi_{2}$ be the root system generated by $\pm\beta$ , $\Phi_{c}(\lambda_{0})$ and all such $\alpha$ . Let $R(\lambda_{0})$ be the simple component of $\Psi_{2}$ which contains $-\beta$ . If $\Phi$ has only one root length or no such $\alpha$ exists, we let $R(\lambda_{0})=Q(\lambda_{0})$ .

Theorem 7.2.

When $L(\lambda)$ is a unitary highest weight module and $z=z_{k}=(\rho,\beta^{\vee})-kc$ , we will have $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ , where $k(\lambda)=r$ if $k\geq r$ and $k(\lambda)=k$ if $0\leq k\leq r-1$ .

When $L(\lambda)$ is a unitary highest weight module, we will have $V(L(\lambda))=\overline{\mathcal{O}}_{k(\lambda)}$ with $k(\lambda)$ given in Proposition 2.9. From [EHW83], we can write $\lambda-\rho=\lambda_{0}+z\xi$ , where $\xi$ is the fundamental weight perpendicular to the compact roots such that $(\xi,\beta^{\vee})=1$ and $\lambda_{0}=(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$ with $(\lambda_{0}+\rho,\beta^{\vee})=0$ .

Now $z=z(\lambda)=(\lambda,\beta^{\vee})=z_{k}=(\rho,\beta^{\vee})-kc$ , we want to prove that $k(\lambda)=r$ if $k\geq r$ and $k(\lambda)=k$ if $0\leq k\leq r-1$ .

In the following, we will give the proof of our Theorem 7.2 in a case-by-case way. The idea of our proof is uniform. For $G=SU(p,q),Sp(n,\mathbb{R})$ and $SO^{*}(2n)$ , we use Lemma 7.1 to prove our result. For other cases, we use our characterizations of annihilator varieties of $L(\lambda)$ since they are given in a distribution of the value of $z=z(\lambda)=(\lambda,\beta^{\vee})$ .

7.1. Proof for $G=SU(p,q)$

For $SU(p,q)$ with $p+q=n$ , from [EHW83] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(p^{\prime},q^{\prime})$ with $p^{\prime}\leq p$ and $q^{\prime}\leq q$ . Then we have $\lambda_{1}=\lambda_{2}=\cdots=\lambda_{p^{\prime}}>\lambda_{p^{\prime}+1}$ and $\lambda_{n}=\lambda_{n-1}=\cdots=\lambda_{n-q^{\prime}+1}<\lambda_{n-q^{\prime}}$ .

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq\max\{p^{\prime},q^{\prime}\}$ or $z\in\mathbb{Z}$ with $z\leq p^{\prime}+q^{\prime}-1$ . In this case, we have $c=1$ , $2\rho=(n-1,n-3,...,-n+3,-n+1)$ , $(\rho,\beta^{\vee})=n-1$ , $\lambda_{1}-\lambda_{n}+n-1=0$ and $n\xi=(q,...,q,-p,...,-p)$ with $p$ copies of $q$ and $q$ copies of $-p$ .

We define an equivalent relation, such that $\lambda\sim\lambda^{\prime}$ if and only if $q_{2}(\lambda)=q_{2}(\lambda^{\prime})$ where $q_{2}(\lambda)$ denotes the number of boxes in the second column of the Young tableau $P(\lambda)$ . We denote $e=(1,1,...,1)$ with $n$ copies of $1$ and $e_{0}=(1,...,1,0,...,0)$ with $p$ copies of $1$ and $q$ copies of $0$ . By using R-S algorithm, we have

\lambda=\lambda_{0}+z\xi+\rho\sim\lambda_{0}+z\xi+\rho+\frac{zp}{n}e=\lambda_{0}+\rho+ze_{0}\sim\lambda_{0}+\rho+ze_{0}+\frac{n-1}{2}e.

Thus

	$\displaystyle\lambda\sim\lambda_{0}+\rho+ze_{0}+\frac{n-1}{2}e$
	$\displaystyle=(\lambda_{n}+z,\lambda_{n}+z-1,...,\lambda_{n}+z-p^{\prime}+1,\lambda_{p^{\prime}+1}+z+n-p^{\prime}-1,...,\lambda_{p}+z+n-p,$
	$\displaystyle\quad\quad\lambda_{p+1}+n-p-1,...,\lambda_{n-q^{\prime}}+q^{\prime},\lambda_{n}+q^{\prime}-1,...,\lambda_{n}+1,\lambda_{n})$
	$\displaystyle:=(s_{1},...,s_{p},s_{p+1},...,s_{n}).$

We may assume $p\leq q$ . Then $(s_{1},...,s_{p})$ is a strictly decreasing sequence smaller that $\lambda_{n}+z$ and $(s_{p+1},...,s_{n})$ is a strictly decreasing sequence larger than $\lambda_{n}$ .

Thus we have

\left\{\begin{array}[]{ll}s_{p+m}\geq s_{n}+q-m,&{\rm~{}if~{}}\>1\leq m\leq q\\ s_{i}\leq s_{n}+z-i+1,&{\rm~{}if~{}}\>1\leq i\leq p.\end{array}\right.

Note that $z_{k}=(\rho,\beta^{\vee})-kc=n-1-k$ , so we have

\{z_{k}\mid 0\leq k\leq r=p\}=\{n-1,n-2,\dots,q,q-1\}.

Now suppose that $L(\lambda)$ is a unitary highest weight module, then we have the follows.

(1)

When $p^{\prime}+q^{\prime}-1\leq q-1$ , we will have $z=z_{k}\leq p^{\prime}+q^{\prime}-1\leq q-1=z_{r}$ . So we can get

\left\{\begin{array}[]{ll}s_{1}=s_{n}+z\leq s_{n}+q-1\leq s_{p+1},\\ s_{2}\leq s_{p+2},\\ \quad\quad\vdots\\ s_{p^{\prime}}\leq s_{n}+z+1-p^{\prime}\leq s_{n}+q-p^{\prime}\leq s_{p+p^{\prime}},\\ s_{p^{\prime}+1}\leq s_{n}+z-(p^{\prime}+1)+1=s_{n}+z-p^{\prime}\leq s_{n}+q-1-p^{\prime}\leq s_{p+p^{\prime}+1},\\ \quad\quad\vdots\\ s_{p}\leq s_{2p}.\end{array}\right.

Thus we have $k(\lambda)=q_{2}(\lambda)\geq p$ by Lemma 7.1 and Proposition 2.9. Since $k(\lambda)\leq r=\min\{p,q\}=p$ , we must have $k(\lambda)=p=k$ .

(2)

When $p^{\prime}+q^{\prime}-1\geq q$ and $q\leq z=z_{k}=n-1-k\leq p^{\prime}+q^{\prime}-1$ for some $1\leq k\leq r-1=p-1$ , similarly we can get

$\left\{\begin{array}[]{ll}s_{p+m}\geq s_{n}+q-m,&1\leq m\leq q\\ s_{i}\leq s_{n}+z-i+1,&1\leq i\leq p.\end{array}\right.$

When $z-q+2=p-k+1\leq i\leq p$ , we have $q-z+i-1=k+i-p\geq 1$ and

$s_{i}\leq s_{n}+z-i+1=s_{n}+q-(q-z+i-1)\leq s_{p+q-z+i-1}=s_{k+i}.$

So

$\left\{\begin{array}[]{ll}s_{p-k+1}\leq s_{p+1},\\ \quad\quad\vdots\\ s_{p}\leq s_{p+q-z+p-1}=s_{p+n-z-1}=s_{p+k}.\end{array}\right.$

Thus we have $k(\lambda)=q_{2}(\lambda)\geq k$ by Lemma 7.1.

If $q_{2}(\lambda)>k$ , we will have $s_{p-k}\leq s_{p+1},\cdots,s_{p}\leq s_{p+k+1}$ . From $z=z_{k}=n-1-k\leq p^{\prime}+q^{\prime}-1$ , we can get

$n-k\leq p^{\prime}+q^{\prime}\Rightarrow q-q^{\prime}+p-k\leq p^{\prime}.$

Now we choose a positive integer $1\leq i_{0}\leq k$ such that

$i_{0}\geq q-q^{\prime}~{}(\Leftrightarrow p+1+i_{0}>n-q^{\prime})\text{~{}and~{}}p-k+i_{0}\leq p^{\prime}.$

Then we have

$s_{p-k+i_{0}}\leq s_{p+1+i_{0}}=\lambda_{n}+n-(p+1+i_{0})=\lambda_{n}+n-p-1-i_{0}.$

On the other hand, we have

$s_{p-k+i_{0}}=\lambda_{n}+z-(p-k+i_{0})+1=\lambda_{n}+n-p-i_{0}.$

Therefore we get contradiction! So we must have $k(\lambda)=q_{2}(\lambda)=k$ .
(3)

When $p^{\prime}+q^{\prime}-1\geq q$ and $z=z_{k}=n-1-k\leq q-1$ , by similar arguments as in (1) we can show that $k(\lambda)=p=k$ .

7.2. Proof for $G=Sp(n,\mathbb{R})$

For $Sp(n,\mathbb{R})$ , from [EHW83] the root system $Q(\lambda_{0})$ is of type $\mathfrak{sp}(q,\mathbb{R})$ and $R(\lambda_{0})$ is of type $\mathfrak{sp}(r,\mathbb{R})$ with $r\geq q$ . Then from [EHW83] we have $\lambda_{1}=\cdots=\lambda_{q}=-n>\lambda_{q+1}\geq\cdots\geq\lambda_{n}$ and $\lambda_{q+1}=\cdots=\lambda_{r}=\lambda_{1}-1=-n-1>\lambda_{r+1}$ . From Table 1, we have $c=\frac{1}{2}$ and $(\rho,\beta^{\vee})=n$ .

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq\frac{1}{2}(r+1)$ or $2z\in\mathbb{Z}$ with $z\leq\frac{1}{2}(q+r)$ . Thus

	$\displaystyle\lambda=\lambda_{0}+z\xi+\rho=$	$\displaystyle(z,z-1,\cdots,z-q+1,z-q-1,\cdots,z-r,$
		$\displaystyle z+\lambda_{r+1}+n-r,\cdots,z+\lambda_{n}+1)$
	$\displaystyle:=$	$\displaystyle(s_{1},s_{2},\cdots,s_{q},s_{q+1},\cdots,s_{r},s_{r+1},\cdots,s_{n}),$

then $(s_{1},s_{2},\cdots,s_{n})$ and $(-s_{n},-s_{n-1},\cdots,-s_{1})$ are strictly decreasing sequences.

Note that $z_{k}=(\rho,\beta^{\vee})-kc=n-\frac{k}{2}$ , so we have

\{z_{k}\mid 0\leq k\leq r=n\}=\left\{n,n-\frac{1}{2},\cdots,\frac{n}{2}+\frac{1}{2},\frac{n}{2}\right\}.

Now suppose that $L(\lambda)$ is a unitary highest weight module, then we have the follows.

(1)

When $n$ is even and $\frac{1}{2}(q+r)\leq\frac{1}{2}(n-1),\text{~{}in ~{}other ~{}words},z\leq\frac{1}{2}(n-1)=z_{n+1}$ , we will have

\left\{\begin{array}[]{ll}s_{m}\leq-s_{1}+n-m,&{\rm~{}if~{}}\>1\leq m\leq q,\\ s_{l}\leq-s_{1}+2z-l+1,&{\rm~{}if~{}}\>q+1\leq l\leq n,\\ -s_{n-i}\geq-s_{1}+n-i-1,&{\rm~{}if~{}}\>0\leq i\leq n-q-1,\\ -s_{j}=-s_{1}+j-1,&{\rm~{}if~{}}\>1\leq j\leq q.\end{array}\right.

Then we have the follows.

\left\{\begin{array}[]{ll}&-s_{n}\geq-s_{1}+n-1\geq s_{1},\\ &-s_{n-1}\geq-s_{1}+n-2\geq s_{2},\\ &\quad\quad\vdots\\ &-s_{q+1}\geq-s_{1}+q-1\geq s_{n-q},\\ &-s_{q}=-s_{1}+q-1=-s_{1}+n-1-(n-q+1)+1\\ &\quad\quad\geq-s_{1}+2z-(n-q+1)+1\geq s_{n-q+1},\\ &\quad\quad\vdots\\ &-s_{2}=-s_{1}+1=-s_{1}+n-1-n+2\geq-s_{1}+2z-(n-1)+1\geq s_{n-1},\\ &-s_{1}=-s_{1}+n-1-n+1\geq-s_{1}+2z-n+1\geq s_{n}.\end{array}\right.

Recall that

2q_{2}^{\mathrm{od}}=\begin{cases}q_{2}+1,&\text{ if }q_{2}\text{ is odd},\\ q_{2},&\text{ if }q_{2}\text{ is even},\end{cases}\quad 2q_{2}^{\mathrm{ev}}+1=\begin{cases}q_{2},&\text{ if }q_{2}\text{ is odd},\\ q_{2}+1,&\text{ if }q_{2}\text{ is even}.\end{cases}

Thus $k(\lambda)\geq q_{2}(\lambda)\geq n$ by Lemma 7.1 and Proposition 2.9. We must have $k(\lambda)=n$ since $k(\lambda)\leq r=n$ .

(2)

When $z\in\mathbb{Z}$ and $\frac{n}{2}\leq z=z_{k}=n-\frac{k}{2}\leq\frac{1}{2}(q+r)$ for $1\leq k\leq n$ , we will have

\left\{\begin{array}[]{ll}s_{m}\geq-s_{1}+n-m+1,&{\rm~{}if~{}}\>1\leq m\leq q,\\ s_{l}\leq-s_{1}+2z-l+1,&{\rm~{}if~{}}\>q+1\leq l\leq n,\\ -s_{n-i}\geq-s_{1}+n-i-1,&{\rm~{}if~{}}\>0\leq i\leq n-q-1,\\ -s_{j}=-s_{1}+j-1,&{\rm~{}if~{}}\>1\leq j\leq q.\end{array}\right.

(a)

When $q+1\leq n-k+2\leq n$ (equivalently $2\leq k\leq n-q+1$ ) and $n$ is even, we have the follows.

(7.3)

\displaystyle\begin{cases}&s_{n-(k-1)+1}\leq-s_{1}+2z-(n-k+2)+1=-s_{1}+n-1\leq-s_{n},\\ &s_{n-(k-1)+2}\leq-s_{1}+2z-(n-k+3)+1=-s_{1}+n-2\leq-s_{n-1},\\ &\quad\quad\vdots\\ &s_{n-1}\leq-s_{1}+2z-(n-1)+1=-s_{1}+n-(k-3)-1\leq-s_{n-k+3},\\ &s_{n}\leq-s_{1}+2z-n+1\leq-s_{n-(k-2)}=-s_{n-k+2}.\end{cases}

Thus we have $q_{2}(\lambda)\geq k-1$ by Lemma 7.1.

If $q_{2}(\lambda)\geq k+1$ , by Lemma 7.1 we will have $s_{n-k+i}\leq-s_{n-i}$ for $0\leq i\leq k$ . From $z=z_{k}=n-\frac{k}{2}\leq\frac{q+r}{2}$ , we can get $2n-k\leq q+r$ . Now we choose a positive integer $0\leq i_{0}\leq k$ such that

n-k+i_{0}\leq q\text{~{}and~{}}n-i_{0}\leq r.

(i)

When $n-k+i_{0}\leq q\text{~{}and~{}}n-i_{0}\leq q$ , we have

$s_{n-k+i_{0}}=z-(n-k+i_{0})+1\leq-s_{n-i_{0}}=-(z-(n-i_{0})+1),$

which implies that $1\leq-1$ . Therefore we get a contradiction!
(ii)

When $n-k+i_{0}\leq q\text{~{}and~{}}q<n-i_{0}\leq r$ , we have

$s_{n-k+i_{0}}=z-(n-k+i_{0})+1\leq-s_{n-i_{0}}=-(z-(n-i_{0})),$

which implies that $1\leq 0$ . Therefore we get a contradiction!

To sum up, $k+1$ does not satisfy Lemma 7.1 and we must have $k-1\leq q_{2}(\lambda)\leq k$ .

If $q_{2}(\lambda)=k-1$ , $z=z_{k}=n-\frac{k}{2}\in\mathbb{Z}$ implies that $k$ is even. Thus $q_{2}(\lambda)=k-1$ is odd and $k(\lambda)=2q_{2}^{\mathrm{od}}=q_{2}+1=k$ by Proposition 2.9.

If $q_{2}(\lambda)=k$ , by Proposition 2.9 we have $k(\lambda)=2q_{2}^{\mathrm{od}}=q_{2}=k$ since $q_{2}(\lambda)=k$ is even.

Similarly the result is the same when $n$ is odd, and we omit the process here.

(b)

When $2\leq n-k+2\leq q$ (equivalently $n-q+2\leq k\leq n$ ) and $n$ is even, we have the follows.

(7.4)

\displaystyle\begin{cases}&-s_{n-(k-1)+1}=-s_{1}+n-k+1=-s_{1}+2z-n+1\geq s_{n},\\ &-s_{n-(k-1)+2}=-s_{1}+n-k+2=-s_{1}+2z-(n-1)+1\geq s_{n-1},\\ &\quad\quad\vdots\\ &-s_{n-1}\geq-s_{1}+n-2=-s_{1}+2z-(n-k+3)+1\geq s_{n-k+3},\\ &-s_{n}\geq-s_{1}+n-1=-s_{1}+2z-(n-k+2)+1\geq s_{n-k+2}.\end{cases}

Then we have $q_{2}(\lambda)\geq k-1$ by Lemma 7.1. Similarly in this case, we can obtain $k(\lambda)=k$ by Proposition 2.9. The result is the same when $n$ is odd, and we omit the process here.

(c)

When $z=z_{1}=n-\frac{1}{2}\leq\frac{1}{2}(q+r)\leq n$ , we will have $q=n-1<r=n$ or $q=r=n$ . Thus $\lambda$ is half-integral and $\lambda_{1}=\cdots=\lambda_{n-1}=-n>\lambda_{n}=-n-1$ or $\lambda_{1}=\cdots=\lambda_{n}=-n$ . So $s_{n-1}=\frac{3}{2}>s_{n}=n-\frac{1}{2}+\lambda_{n}+1=-\frac{1}{2}$ or $\frac{1}{2}$ . By using R-S algorithm, we have $q_{2}(\lambda)=1$ if $s_{n}=-\frac{1}{2}$ and $q_{2}(\lambda)=0$ if $s_{n}=\frac{1}{2}$ . Therefore $k(\lambda)=1=k$ by Proposition 2.9.
(d)

When $z=z_{0}=n\leq\frac{1}{2}(q+r)\leq n$ , we will have $q=r=n$ . Thus $\lambda$ will be dominant integral since $\lambda_{1}=\cdots=\lambda_{n}=-n$ and $s_{n}=n+\lambda_{n}+1=1$ . Therefore $k(\lambda)=0=k$ by Proposition 2.9.

(3)

When $z\in\frac{1}{2}+\mathbb{Z}$ and $\frac{n}{2}\leq z=z_{k}=n-\frac{k}{2}\leq\frac{1}{2}(q+r)$ for $1\leq k\leq n$ , we still have $q_{2}(\lambda)=k-1$ by (7.3) and (7.4). In this case $q_{2}(\lambda)=k-1$ and $k(\lambda)=2q_{2}^{\mathrm{ev}}+1=k$ by Proposition 2.9.

To sum up, we have $k(\lambda)=k$ , where $k(\lambda)$ is given in Proposition 2.9. Therefore we have completed the proof of the case when $G=Sp(n,\mathbb{R})$ .

7.3. Proof for $G=SO^{\ast}(2n)$

For $SO^{\ast}(2n)$ , from [EHW83] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,q)$ with $1\leq q\leq n-1$ or $\mathfrak{so}^{\ast}(2p)$ with $3\leq p\leq n$ . From Table 1, we have $c=2$ and $(\rho,\beta^{\vee})=2n-3$ . $(\lambda_{0}+\rho,\beta^{\vee})=0$ implies that $\lambda_{1}+\lambda_{2}=-2n+3$ .

(1)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,q)$ with $1\leq q\leq n-1$ , we have

\lambda_{1}>\lambda_{2}=\lambda_{3}=\cdots=\lambda_{q+1}>\lambda_{q+2}\geq\cdots\geq\lambda_{n}.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq q$ . Thus we can write

	$\displaystyle\lambda$	$\displaystyle=\lambda_{0}+z\xi+\rho$
		$\displaystyle=(\frac{1}{2}z+\lambda_{1}+n-1,\frac{1}{2}z+\lambda_{2}+n-2,\cdots,\frac{1}{2}z+\lambda_{q+1}+n-q-1,$
		$\displaystyle\quad\>\frac{1}{2}z+\lambda_{q+2}+n-q-2,\cdots,\frac{1}{2}z+\lambda_{n})$
		$\displaystyle=(\frac{1}{2}z-\lambda_{2}-n+2,\frac{1}{2}z+\lambda_{2}+n-2,\cdots,\frac{1}{2}z+\lambda_{2}+n-q-1,$
		$\displaystyle\quad\>\frac{1}{2}z+\lambda_{q+2}+n-q-2,\cdots,\frac{1}{2}z+\lambda_{n})$
		$\displaystyle:=(t_{1},\cdots,t_{q+1},t_{q+2},\cdots,t_{n}).$

So $(t_{1},t_{2},\cdots,t_{n})$ and $(-t_{n},-t_{n-1},\cdots,-t_{1})$ are strictly decreasing sequences, and it is not difficult to get

\left\{\begin{array}[]{ll}t_{1}>-t_{1}+z+1,\\ t_{i}\leq-t_{1}+z+2-i,&{\rm~{}if~{}}\>2\leq i\leq n,\\ -t_{j}\geq-t_{1}+j-1,&{\rm~{}if~{}}\>1\leq j\leq n.\end{array}\right.

When $L(\lambda)$ is a unitary highest weight module and $n$ is even, we have the follows.

(a)

When $z<q\leq n-1=z_{r-1}$ , we have $L(\lambda)=N(\lambda)$ by [EHW83]. Thus $k(\lambda)=\frac{n}{2}=k$ .

(b)

When $z=q<n-1=z_{r-1}$ , we have

\displaystyle\begin{cases}&t_{1}=\frac{1}{2}z-\lambda_{2}-n+2\leq\frac{1}{2}z-\lambda_{n}-n+2\leq-\frac{1}{2}z-\lambda_{n}=-t_{n},\\ &t_{2}=\frac{1}{2}z+\lambda_{2}+n-2=\frac{1}{2}z-\lambda_{1}-2n+3+n-2\leq-\frac{1}{2}z-\lambda_{n-1}-1=-t_{n-1},\\ &\quad\quad\vdots\\ &t_{n}\leq-t_{1}.\end{cases}

Thus $k(\lambda)=q^{\mathrm{ev}}_{2}=[\frac{q_{2}}{2}]\geq\frac{n}{2}$ by Lemma 7.1 and Proposition 2.9. We must have $k(\lambda)=\frac{n}{2}=r$ since $k(\lambda)\leq r=[\frac{n}{2}]=\frac{n}{2}$ and $n$ is even.

(c)

When $z=z_{k}=2n-3-2k=q\leq n-1$ for $1\leq k\leq r-1=\frac{n}{2}-1$ , we will have $z=z_{r-1}=n-1=q$ and $\lambda_{1}>\lambda_{2}=\lambda_{3}=\cdots=\lambda_{n}$ . Thus we can get

\displaystyle\begin{cases}&t_{2}=\frac{1}{2}z+\lambda_{2}+n-2=\frac{1}{2}z-\lambda_{1}-n+1\leq-\frac{1}{2}z-\lambda_{n}=-t_{n},\\ &\quad\quad\vdots\\ &t_{n}\leq-t_{2}.\end{cases}

Thus $q_{2}(\lambda)\geq n-1$ by Lemma 7.1. Note that

t_{1}=\frac{1}{2}z-\lambda_{2}-n+2=-\frac{1}{2}z-\lambda_{n}+1=-t_{n}+1>-t_{n}.

So we can not have $q_{2}(\lambda)=n$ . Thus $q_{2}(\lambda)=n-1$ and $k(\lambda)=q^{\mathrm{ev}}_{2}=[\frac{q_{2}}{2}]\geq\frac{n}{2}$ by Lemma 7.1 and Proposition 2.9.

Thus $k(\lambda)=q^{\mathrm{ev}}_{2}=[\frac{q_{2}}{2}]=k$ by Lemma 7.1 and Proposition 2.9.

The argument is similar when $n$ is odd, so we omit the details here. Therefore, when $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,q)$ with $1\leq q\leq n-1$ , we have $k(\lambda)=k$ .

(2)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{so}^{\ast}(2p)$ with $3\leq p\leq n$ , we have

\lambda_{1}=\lambda_{2}=\cdots=\lambda_{p}=-n+\frac{3}{2}>\lambda_{p+1}\geq\cdots\geq\lambda_{n}.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if

z\leq\begin{cases}p,&\text{ if }p\text{ is odd},\\ p-1,&\text{ if }p\text{ is even},\end{cases}

or $z=2p-3-2j\leq 2p-3$ for some integer $0\leq j\leq[\frac{p}{2}]-2$ .

Thus we can write

	$\displaystyle\lambda$	$\displaystyle=\lambda_{0}+z\xi+\rho$
		$\displaystyle=(\frac{1}{2}z+\lambda_{1}+n-1,\cdots,\frac{1}{2}z+\lambda_{p}+n-p,\frac{1}{2}z+\lambda_{p+1}+n-p-1,\cdots,\frac{1}{2}z+\lambda_{n})$
		$\displaystyle=(\frac{1}{2}z+\frac{1}{2},\cdots,\frac{1}{2}z+\frac{3}{2}-p,\frac{1}{2}z+\lambda_{p+1}+n-p-1,\cdots,\frac{1}{2}z+\lambda_{n})$
		$\displaystyle:=(t_{1},\cdots,t_{p},t_{p+1},\cdots,t_{n}).$

So $(t_{1},t_{2},\cdots,t_{n})$ and $(-t_{n},-t_{n-1},\cdots,-t_{1})$ are strictly decreasing sequences, and it is not difficult to get

\left\{\begin{array}[]{ll}t_{i}\leq-t_{1}+z+2-i,&{\rm~{}if~{}}\>1\leq i\leq n,\\ -t_{j}\geq-t_{1}+j-1,&{\rm~{}if~{}}\>1\leq j\leq n.\end{array}\right.

Suppose $p$ and $n$ are even. Then we have the follows.

(a)

When $z<p-1\leq n-1=z_{r-1}$ , by similar arguments as in the case of $\mathfrak{su}(1,q)$ we have $q_{2}(\lambda)=n$ . Thus $k(\lambda)=q^{\mathrm{ev}}_{2}=[\frac{q_{2}}{2}]=[\frac{n}{2}]=r$ by Lemma 7.1 and Proposition 2.9.

(b)

When $z\geq p-1$ and $z=z_{k}=2n-3-2k\leq 2p-3$ for some $1\leq k\leq r-1=\left\lfloor\frac{n}{2}\right\rfloor-1$ , we have

\displaystyle\begin{cases}&t_{n-(2k+1)+1}\leq-t_{1}+z+2-(n-2k)=-t_{1}+z-n+2k+2\leq-t_{n},\\ &t_{n-(2k+1)+2}\leq-t_{1}+z+2-(n-2k+1)=-t_{1}+n-2\leq-t_{n-1},\\ &\quad\quad\vdots\\ &t_{n-1}\leq-t_{1}+z+2-(n-1)=-t_{1}+n-2k\leq-t_{n-(2k+1)+2},\\ &t_{n}\leq-t_{1}+z+2-n=-t_{1}+n-2k-1\leq-t_{n-(2k+1)+1}.\end{cases}

Thus we have $q_{2}(\lambda)\geq 2k+1$ by Lemma 7.1.

If $q_{2}(\lambda)\geq 2k+2$ , by Lemma 7.1 we will have $t_{n-2k-2+i}\leq-t_{n-i+1}$ for $1\leq i\leq 2k+2$ . From $z=z_{k}=2n-3-2k\leq 2p-3$ , we can get

2n-2k-1<2p\Rightarrow n-2k-2+n+1<2p.

Now we choose a positive integer $1\leq i_{0}\leq 2k+2$ such that

n+1-i_{0}\leq p~{}\text{~{}and~{}}n-2k-2+i_{0}\leq p.

Then we have

	$\displaystyle t_{n-2k-2+i_{0}}=\frac{1}{2}z+\frac{1}{2}-(n-2k-2+i_{0}-1)$
	$\displaystyle\leq-t_{n+1-i_{0}}=-(\frac{1}{2}z+\frac{1}{2}-(n+1-i_{0}-1)),$

which implies that

z-2n+2k+4\leq 0\Rightarrow 1\leq 0.

Therefore we get contradiction! So we must have $q_{2}(\lambda)=2k+1$ .

Thus $k(\lambda)=q^{\mathrm{ev}}_{2}=[\frac{q_{2}}{2}]=k$ by Lemma 7.1 and Proposition 2.9.

The argument is similar when $n$ is odd, so we omit the details here.

7.4. Proof for $G=SO(2,2n-2)$

For $SO(2,2n-2)$ , from [EHW83] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq n-1$ or $\mathfrak{so}(2,2n-2)$ . From Table 1, we have $c=n-2$ and $(\rho,\beta^{\vee})=2n-3$ . $(\lambda_{0}+\rho,\beta^{\vee})=0$ implies that $\lambda_{1}+\lambda_{2}=-2n+3$ .

(1)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq n-1$ , we have

$|\lambda_{n}|\leq\lambda_{n-1}\leq\cdots\leq\lambda_{p+2}<\lambda_{p+1}=\lambda_{p}=\cdots=\lambda_{2}\in\frac{1}{2}\mathbb{Z}_{>0}.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq p$ . From Theorem 5.1, we have $k(\lambda)=2$ if and only if $\lambda$ is not integral or

$z\leq-|\lambda_{n}|-\lambda_{1}-n+1=-|\lambda_{n}|+\lambda_{2}+2n-3-n+1=-|\lambda_{n}|+\lambda_{2}+n-2.$

Now when $z<n-1=z_{1}$ , $z$ will be not integral, or $z\in\mathbb{Z}$ and $z\leq n-2\leq-|\lambda_{n}|+\lambda_{2}+n-2$ . Thus we have $k(\lambda)=2$ . Note that when $p\leq n-2$ , we always have $z\leq p\leq n-1$ .

Now when $z=n-1=z_{1}$ , we must have $p=n-1$ since $z\leq p\leq n-1$ . Then we have $\lambda_{2}=\cdots=\lambda_{n-1}=|\lambda_{n}|\in\frac{1}{2}\mathbb{Z}_{>0}$ . From Theorem 5.1, we have $k(\lambda)=1$ if and only if $\lambda$ is integral and

$-|\lambda_{n}|-\lambda_{1}-n+1=-|\lambda_{n}|+\lambda_{2}+n-2<z\leq\lambda_{2}-\lambda_{1}-1,$

equivalently we have

$n-2<z\leq 2\lambda_{2}+2n-4.$

Thus when $z=n-1=z_{1}$ , we will have $k(\lambda)=1$ .
(2)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{so}(2,2n-2)$ , we have

$\lambda_{2}=\lambda_{3}=\cdots=\lambda_{n}=0.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq n-1=z_{1}$ or $z=2n-3=z_{0}$ . In this case, we have $\lambda_{1}=-2n+3$ .

From Theorem 5.1, we have $k(\lambda)=2$ if and only if $\lambda$ is not integral or

$z\leq-|\lambda_{n}|-\lambda_{1}-n+1=n-2.$

Now when $z<n-1=z_{1}$ , $z$ will be not integral, or $z\in\mathbb{Z}$ and $z\leq n-2$ . Thus we have $k(\lambda)=2$ .

From Theorem 5.1, we have $k(\lambda)=1$ if and only if $\lambda$ is integral and

$-|\lambda_{n}|-\lambda_{1}-n+1<z\leq\lambda_{2}-\lambda_{1}-1,$

equivalently we have

$n-2<z\leq 2n-4.$

Thus when $z=n-1=z_{1}$ , we will have $k(\lambda)=1$ .

From Theorem 5.1, we have $k(\lambda)=0$ if and only if $\lambda$ is integral and

$z\geq\lambda_{2}-\lambda_{1}=2n-3.$

Thus when $z=2n-3=z_{0}$ , we will have $k(\lambda)=0$ .

Now we complete the proof for all cases.

7.5. Proof for $G=SO(2,2n-1)$

For $SO(2,2n-1)$ , from [EHW83] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq n-1$ or $\mathfrak{so}(2,2n-1)$ , or $Q(\lambda_{0})=\mathfrak{su}(1,n-1)$ and $R(\lambda_{0})=\mathfrak{so}(2,2n-1)$ . From Table 1, we have $c=n-\frac{3}{2}$ and $(\rho,\beta^{\vee})=2n-2$ . $(\lambda_{0}+\rho,\beta^{\vee})=0$ implies that $\lambda_{1}+\lambda_{2}=-2n+2$ .

(1)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq n-1$ , we have

$0\leq\lambda_{n}\leq\lambda_{n-1}\leq\cdots\leq\lambda_{p+2}<\lambda_{p+1}=\lambda_{p}=\cdots=\lambda_{2}.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq p$ .

From Theorem 4.1, we have $k(\lambda)=2$ if and only if $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<\lambda_{2}-\lambda_{1}$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq-\lambda_{1}-n+\frac{1}{2}$ .

Now $z\leq p<n-\frac{1}{2}=z_{1}$ , we will have $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<n-\frac{1}{2}<2n-2\leq\lambda_{2}-\lambda_{1}$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq n-\frac{3}{2}<\lambda_{2}+n-\frac{3}{2}=-\lambda_{1}-n+\frac{1}{2}$ . Thus we have $k(\lambda)=2$ by Theorem 4.1.
(2)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{so}(2,2n-1)$ , we have

$\lambda_{2}=\lambda_{3}=\cdots=\lambda_{n}=0.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq n-\frac{1}{2}=z_{1}$ or $z=2n-2=z_{0}$ . In this case, we have $\lambda_{1}=-2n+2$ .

From Theorem 4.1, we have $k(\lambda)=2$ if and only if $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<\lambda_{2}-\lambda_{1}=2n-2$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq-\lambda_{1}-n+\frac{1}{2}=n-\frac{3}{2}$ .

Now when $z<n-\frac{1}{2}=z_{1}$ , we will have $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<n-\frac{1}{2}<2n-2$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq n-\frac{3}{2}$ . Thus we have $k(\lambda)=2$ .

From Theorem 4.1, we have $k(\lambda)=1$ if and only if $z\in\frac{1}{2}+\mathbb{Z}$ and $z>-\lambda_{1}-n+\frac{1}{2}=n-\frac{3}{2}$ . Thus when $z=z_{1}=n-\frac{1}{2}>n-\frac{3}{2}$ , we will have $k(\lambda)=1$ .

From Theorem 4.1, we have $k(\lambda)=0$ if and only if $z\in\mathbb{Z}$ and $z\geq\lambda_{2}-\lambda_{1}=2n-2$ . Thus when $z=z_{0}=2n-2$ , we will have $k(\lambda)=0$ .
(3)

When $Q(\lambda_{0})=\mathfrak{su}(1,n-1)$ and $R(\lambda_{0})=\mathfrak{so}(2,2n-1)$ , we have

$\lambda_{2}=\lambda_{3}=\cdots=\lambda_{n}=\frac{1}{2}.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq n-\frac{1}{2}$ .

From Theorem 4.1, we have $k(\lambda)=2$ if and only if $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<\lambda_{2}-\lambda_{1}=2n-1$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq-\lambda_{1}-n+\frac{1}{2}=n-1$ .

Now when $z<n-\frac{1}{2}=z_{1}$ , we will have $z\notin\frac{1}{2}\mathbb{Z}$ , or $z\in\mathbb{Z}$ and $z<n-\frac{1}{2}<2n-1$ , or $z\in\frac{1}{2}+\mathbb{Z}$ and $z\leq n-\frac{3}{2}<n-1$ . Thus we have $k(\lambda)=2$ .

From Theorem 4.1, we have $k(\lambda)=1$ if and only if $z\in\frac{1}{2}+\mathbb{Z}$ and $z>-\lambda_{1}-n+\frac{1}{2}=n-1$ . Thus when $z=z_{1}=n-\frac{1}{2}>n-1$ , we will have $k(\lambda)=1$ .

Now we complete the proof of all cases.

7.6. Proof for $G=E_{6(-14)}$

For $E_{6(-14)}$ , from [EHW83, §12] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq 5$ , $\mathfrak{so}(2,8)$ or $\mathfrak{e}_{6(-14)}$ . For all cases, we have $\lambda_{6}=\lambda_{7}=-\lambda_{8}:=b$ by [Bou02]. From Table 1, we have $c=3$ and $(\rho,\beta^{\vee})=11$ . $(\lambda_{0}+\rho,\beta^{\vee})=0$ implies that $\lambda_{1}+\cdots+\lambda_{5}-\lambda_{6}-\lambda_{7}+\lambda_{8}=-22$ .

(1)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,5)$ , we have

$-\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}:=a\in\frac{1}{2}\mathbb{Z}_{\geq 0}.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 5$ . In this case, we have $3a-3b=-22$ since $\lambda_{1}+\cdots+\lambda_{5}-\lambda_{6}-\lambda_{7}+\lambda_{8}=-22$ .

From Theorem 6.1, we have

$-(\lambda_{0},\beta_{1,2})=\frac{1}{2}(a+3b)=\frac{1}{2}(a+3a+22)=2a+11$

and $k(\lambda)=2$ if and only if $\lambda$ is not integral or $z\in\mathbb{Z}$ and

$z\leq-4+\min\{-(\lambda_{0},\beta_{1,2})\}=7+2a.$

Now we have $z\leq 5<7+2a$ , so $k(\lambda)=2$ .

The proof for other cases of type $\mathfrak{su}(1,p)$ with $1\leq p\leq 4$ is similar, and we omit the details here.

(2)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{so}(2,8)$ , we have

\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=0<\lambda_{5}\in\mathbb{Z}_{>0}.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 4$ or $z=7$ . In this case, we have $\lambda_{5}-3b=-22$ .

From Theorem 6.1, we have

-(\lambda_{0},\beta_{1,2})=\frac{1}{2}(a+3b)=\frac{1}{2}(\lambda_{5}+3b)=\frac{1}{2}(\lambda_{5}+\lambda_{5}+22)=\lambda_{5}+11

and $k(\lambda)=2$ if and only if $\lambda$ is not integral or $z\in\mathbb{Z}$ and

z\leq-4+\min\{-(\lambda_{0},\beta_{1,2})\}=7+\lambda_{5}.

Now we have $z\leq 7<7+\lambda_{5}$ , so $k(\lambda)=2$ .

(3)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{e}_{6(-14)}$ , we have

$\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0.$

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 8=z_{1}$ or $z=11=z_{0}$ . In this case, we have $3b=22$ .

From Theorem 6.1, we have

$-(\lambda_{0},\beta_{1,2})=\frac{1}{2}(0+3b)=11$

and $k(\lambda)=2$ if and only if $\lambda$ is not integral or $z\in\mathbb{Z}$ and

$z\leq-4+\min\{-(\lambda_{0},\beta_{1,2})\}=7.$

Now when $z<8=z_{1}$ , $z$ will be not integral, or $z\in\mathbb{Z}$ and $z\leq 7$ . Thus we have $k(\lambda)=2$ .

From Theorem 6.1, we have

$-(\lambda_{0},\alpha_{1})=\frac{1}{2}(0+3b)=11$

and $k(\lambda)=1$ if and only if $z\in\mathbb{Z}$ and

$7=-4+\min\{-(\lambda_{0},\beta_{1,2})\}<z\leq-(\lambda_{0},\alpha_{1})-1=10.$

Thus when $z=z_{1}=8$ , we will have $k(\lambda)=1$ .

From Theorem 6.1, we have $k(\lambda)=0$ if and only if $z\in\mathbb{Z}$ and

$z>-(\lambda_{0},\alpha_{1})-1=10.$

Thus when $z=z_{0}=11$ , we will have $k(\lambda)=0$ .

Now we complete the proof for all cases.

7.7. Proof for $G=E_{7(-25)}$

For $E_{7(-25)}$ , from [EHW83, §13] the root system $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,p)$ with $1\leq p\leq 6$ , $\mathfrak{so}(2,10)$ or $\mathfrak{e}_{7(-25)}$ . For all cases, we have $\lambda_{7}=-\lambda_{8}$ by [Bou02]. From Table 1, we have $c=4$ and $(\rho,\beta^{\vee})=17$ . $(\lambda_{0}+\rho,\beta^{\vee})=0$ implies that $\lambda_{8}=-\frac{17}{2}$ .

(1)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{su}(1,6)$ , we have

0<\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}:=a\in\frac{1}{2}\mathbb{Z}\text{~{}and~{}}\lambda_{8}-(\sum\limits_{i=2}^{7}\lambda_{i})+\lambda_{1}=0.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 6$ . In this case we have $\lambda_{6}=-17-3a$ .

From Theorem 6.2 we have $k(\lambda)=3$ if and only if $\lambda$ is not integral or $z\in\mathbb{Z}$ and

z\leq-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}=8+2a.

Now we have $z\leq 6<8+2a$ , so $k(\lambda)=3$ .

The proof for other cases of type $\mathfrak{su}(1,p)$ with $1\leq p\leq 5$ is similar, and we omit the details here.

(2)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{so}(2,10)$ , we have

\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=0<\lambda_{5}\in\mathbb{N}^{\ast}\text{~{}and~{}}\lambda_{8}-(\sum\limits_{i=2}^{7}\lambda_{i})+\lambda_{1}=0.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 5=z_{3}$ or $z=9=z_{2}$ . In this case $\lambda_{5}+\lambda_{6}=\lambda_{8}-\lambda_{7}=-17$ .

From Theorem 6.2 we have

-(\lambda_{0},\beta_{3,4})=-\frac{1}{2}(-\lambda_{5}+\lambda_{6}-\lambda_{7}+\lambda_{8})=\lambda_{5}+17,-(\lambda_{0},\beta_{5})=-(\lambda_{5}+\lambda_{6})=17

and $k(\lambda)=3$ if and only if $\lambda$ is not integral or $z\in\mathbb{Z}$ and

z\leq-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}=8.

Now when $z<9=z_{2}$ , $z$ will be not integral, or $z\in\mathbb{Z}$ and $z\leq 8$ . Thus we have $k(\lambda)=3$ .

From Theorem 6.2 we have $k(\lambda)=2$ if and only if $z\in\mathbb{Z}$ and

8=-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}<z\leq-(\lambda_{0},\beta_{6,7})-5=\lambda_{5}+12.

Thus when $z=9=z_{2}$ , we will have $k(\lambda)=2$ .

(3)

When $Q(\lambda_{0})=R(\lambda_{0})$ is of type $\mathfrak{e}_{7(-25)}$ , we have

\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0\text{~{}and~{}}\lambda_{8}-(\sum\limits_{i=2}^{7}\lambda_{i})+\lambda_{1}=0.

From [EHW83] we know that $L(\lambda)$ is unitarizable if and only if $z\leq 9=z_{2}$ or $z=13=z_{1}$ or $z=17=z_{0}$ . In this case we have $\lambda_{6}=-17$ .

From Theorem 6.2 we have $k(\lambda)=3$ if and only if $z$ is not integral or $z\in\mathbb{Z}$ and

z\leq-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}=8.

Now when $z<9=z_{2}$ , $z$ will be not integral, or $z\in\mathbb{Z}$ and $z\leq 8$ , hence we have $k(\lambda)=3$ .

From Theorem 6.2 we have $k(\lambda)=2$ if and only if $z\in\mathbb{Z}$ and

8=-9+\min\{-(\lambda_{0},\beta_{3,4,5})\}<z\leq-(\lambda_{0},\beta_{6,7})-5=12.

Thus when $z=9=z_{2}$ , we will have $k(\lambda)=2$ .

From Theorem 6.2 we have $k(\lambda)=1$ if and only if $z\in\mathbb{Z}$ and

12=-(\lambda_{0},\beta_{6,7})-5<z\leq-(\lambda_{0},\beta_{8})-1=16.

Thus when $z=13=z_{1}$ , we will have $k(\lambda)=1$ .

From Theorem 6.2 we have $k(\lambda)=0$ if and only if $z\in\mathbb{Z}$ and

z\geq-(\lambda_{0},\beta_{8})-1=16.

Thus when $z=17=z_{0}$ , we will have $k(\lambda)=0$ .

So far, we have completed the proof of all cases of Theorem 7.2.

Acknowledgments

Z. Bai was supported in part by the National Natural Science Foundation of China (No. 12171344) and the National Key $\textrm{R}\,\&\,\textrm{D}$ Program of China (No. 2018YFA0701700 and No. 2018YFA0701701).

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On the annihilator variety of a highest weight Harish-Chandra module

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

2. Preliminaries

2.1. Gelfand-Kirillov dimension and associated variety

Definition 2.1.

Definition 2.2 ([Vog91]).

Definition 2.3.

2.2. Associated varieties of highest weight Harish-Chandra modules

Proposition 2.4 ([BH15, Thm. 1.1 and Cor. 5.2]).

Proposition 2.5 ([BH15, Thm. 6.2]).

Definition 2.6.

Example 2.8.

Proposition 2.9 ([BXX23, Thm. 6.1, Thm. 6.2]).

Proposition 2.10 ([BXX23, Thm. 7.1]).

2.3. Annihilator variety

Proposition 2.11 ([Jos85]).

Corollary 2.12.

Proposition 2.13 ([BMW23], Thm. 5.4, 6.5, 6.6 and 7.14).

Remark 2.14.

Proposition 2.15 ([CM93, Thm. 5.1.1, 5.1.2, 5.1.3 &\& 5.1.4 and Prop. 6.3.7]).

3. Annihilator varieties for type An−1A_{n-1}

Definition 3.1.

Proposition 3.2 ([BX19, Thm. 5.2]).

Theorem 3.3.

Proof.

Corollary 3.4.

4. Annihilator varieties for types BnB_{n} and CnC_{n}

Theorem 4.1.

Proof.

Remark 4.2.

Theorem 4.3.

Proof.

Corollary 4.4.

Remark 4.5.

5. Annihilator varieties for type DnD_{n}

Theorem 5.1.

Proof.

Remark 5.2.

Theorem 5.3.

Proof.

Corollary 5.4.

Remark 5.5.

6. Annihilator varieties for types E6E_{6} and E7E_{7}

Theorem 6.1.

Proof.

Theorem 6.2.

Proof.

7. The Gelfand-Kirillov dimension of a unitary highest weight module

Lemma 7.1 ([Bai+24, Lem. 5.4]).

Theorem 7.2.

7.1. Proof for G=S​U​(p,q)G=SU(p,q)

7.2. Proof for G=S​p​(n,ℝ)G=Sp(n,\mathbb{R})

7.3. Proof for G=S​O∗​(2​n)G=SO^{\ast}(2n)

7.4. Proof for G=S​O​(2,2​n−2)G=SO(2,2n-2)

7.5. Proof for G=S​O​(2,2​n−1)G=SO(2,2n-1)

7.6. Proof for G=E6​(−14)G=E_{6(-14)}

7.7. Proof for G=E7​(−25)G=E_{7(-25)}

Acknowledgments

References

Proposition 2.15 ([CM93, Thm. 5.1.1, 5.1.2, 5.1.3 $\&$ 5.1.4 and Prop. 6.3.7]).

3. Annihilator varieties for type $A_{n-1}$

4. Annihilator varieties for types $B_{n}$ and $C_{n}$

5. Annihilator varieties for type $D_{n}$

6. Annihilator varieties for types $E_{6}$ and $E_{7}$

7.1. Proof for $G=SU(p,q)$

7.2. Proof for $G=Sp(n,\mathbb{R})$

7.3. Proof for $G=SO^{\ast}(2n)$

7.4. Proof for $G=SO(2,2n-2)$

7.5. Proof for $G=SO(2,2n-1)$

7.6. Proof for $G=E_{6(-14)}$

7.7. Proof for $G=E_{7(-25)}$