# On the associated variety of a highest weight Harish-Chandra module

Zhanqiang Bai  School of Mathematical Sciences, Soochow University, Suzhou
215006, Jiangsu, China<EMAIL_ADDRESS>,  Markus Hunziker  Department of
Mathematics, Baylor University, Waco, TX 76798, USA  Markus
<EMAIL_ADDRESS>,  Xun Xie  School of Mathematics and Statistics, Beijing
Institute of Technology, Beijing 100081, China<EMAIL_ADDRESS>and  Roger
Zierau  Mathematics Department, Oklahoma State University, Stillwater, OK
74078, USA<EMAIL_ADDRESS>
###### Abstract.

We prove a simple formula that calculates the associated variety of a highest
weight Harish-Chandra module directly from its highest weight. We also give a
formula for the Gelfand–Kirillov dimension of highest weight Harish-Chandra
module which is uniform across Cartan types and is valid for arbitrary
infinitesimal character.

###### Key words and phrases:

Highest weight Harish-Chandra module, associated variety, Springer
representations, Gelfand–Kirillov dimension, category $\mathcal{O}$

###### 2010 Mathematics Subject Classification:

Primary 22E47; Secondary 17B10

###### Contents

  1. 1 Introduction
  2. 2 Posets of positive noncompact roots
  3. 3 Associated varieties and highest weight Harish-Chandra modules
  4. 4 Proof of Theorems 1.2 and 1.4
  5. 5 Another approach
  6. A Hasse diagrams and distinguished antichains

##  1\. Introduction

###  1.1. Highest weight Harish-Chandra modules

Let $G_{\mathbb{R}}$ be a simple Lie group. It follows from the work of
Harish-Chandra ([14, 15]) that there exist infinite-dimensional highest weight
Harish-Chandra modules for $G_{\mathbb{R}}$ if and only if $G_{\mathbb{R}}$ is
of Hermitian type. Our main result, Theorem 1.2, is a very simple formula for
the associated varieties of highest weight Harish-Chandra modules. Therefore,
we assume that $G_{\mathbb{R}}$ is of Hermitian type. We also assume that
$G_{\mathbb{R}}$ has finite center.

Standard Lie theory tells us that ${\mathfrak{g}}_{\mathbb{R}}={\EuScript
Lie}(G_{\mathbb{R}})$ has a Cartan decomposition
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{k}}_{\mathbb{R}}\oplus{\mathfrak{p}}_{\mathbb{R}}$
and the subgroup $K_{\mathbb{R}}$ of $G_{\mathbb{R}}$ with Lie algebra
${\mathfrak{k}}_{\mathbb{R}}$ is compact. Therefore, $K_{\mathbb{R}}$ has a
complexification $K$. Writing the complexified Cartan decomposition as
${\mathfrak{g}}={\mathfrak{k}}\oplus{\mathfrak{p}}$ (since $G_{\mathbb{R}}$ is
of Hermitian type) there is a decomposition
${\mathfrak{p}}={\mathfrak{p}}^{+}\oplus{\mathfrak{p}}^{-}$ into nonzero
irreducible $K$-subrepresentations. This gives a triangular decomposition
${\mathfrak{g}}={\mathfrak{p}}^{-}\oplus{\mathfrak{k}}\oplus{\mathfrak{p}}^{+}$
and a maximal parabolic subalgebra
${\mathfrak{q}}={\mathfrak{k}}\oplus{\mathfrak{p}}^{+}$. It follows from
Harish-Chandra that a highest weight Harish-Chandra module is the irreducible
quotient of a Verma module and the highest weight $\lambda$ is integral for
the roots in ${\mathfrak{k}}$ (with respect to a Cartan subalgebra
${\mathfrak{h}}$ of ${\mathfrak{g}}$ contained in ${\mathfrak{k}}$). See §3.2
for precisely which Verma modules have irreducible quotients occurring as
highest weight Harish-Chandra modules.

###  1.2. Associated varieties

The associated variety of a $({\mathfrak{g}},K)$-module, as defined in [26],
is a union of (closures of) nilpotent $K$-orbits in
$({\mathfrak{g}}/{\mathfrak{k}})^{*}$. When $L(\lambda)$ is a highest weight
Harish-Chandra module for $G_{\mathbb{R}}$, the associated variety is
contained in $({\mathfrak{g}}/{\mathfrak{q}})^{*}\simeq{\mathfrak{p}}^{+}$. It
is well-known that the closures of the $K$-orbits in ${\mathfrak{p}}^{+}$ form
a chain

(1.1)
$\\{0\\}=\overline{\mathcal{O}_{0}}\subsetneq\overline{\mathcal{O}_{1}}\subsetneq\cdots\subsetneq\overline{\mathcal{O}_{r}}={\mathfrak{p}}^{+}\\!,$

where $r$ is the real rank of ${\mathfrak{g}}_{\mathbb{R}}$
($=\operatorname{rank}(G_{\mathbb{R}}/K_{\mathbb{R}})$). It follows that

(1.2) $\operatorname{AV}(L(\lambda))=\overline{{\mathcal{O}}_{k}},\text{ for
some }k=0,1,\dots,r.$

The purpose of this paper is to express this integer $k=k(\lambda)$ in terms
of the highest weight $\lambda$ in a simple way that is uniform across Cartan
types and is valid for arbitrary infinitesimal character. In the special case
when $L(\lambda)$ is the Harish-Chandra module of a _unitary_ representation
of $G_{\mathbb{R}}$ this was accomplished in [2] by using previous work of
Joseph [20].

###  1.3. Statement of the main result

To state our main result precisely, we need some notation for sets of roots.
Let $\Delta$ and $\Delta({\mathfrak{k}})$ denote the root systems of
$({\mathfrak{g}},{\mathfrak{h}})$ and $({\mathfrak{k}},{\mathfrak{h}})$,
respectively. Let $\Delta^{+}$ be a positive system of $\Delta$ containing
$\Delta({\mathfrak{p}}^{+})$ and define
$\Delta^{+}({\mathfrak{k}})=\Delta({\mathfrak{k}})\cap\Delta^{+}$. Let
$\Lambda^{+}({\mathfrak{k}})$ denote the lattice of integral and dominant
weights for $\Delta({\mathfrak{k}})$. We view all sets of roots as partially
ordered sets via the usual partial ordering, where $\alpha\leq\beta$ means
that $\beta-\alpha$ is nonnegative integer linear combination of positive
roots.

######  Definition 1.1.

For $\lambda\in\Lambda^{+}({\mathfrak{k}})$, define the _diagram_ of $\lambda$
as the set

(1.3)
$Y_{\lambda}:=\\{\alpha\in\Delta({\mathfrak{p}}^{+})\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}_{\leq
0}\\},$

viewed as a subposet of
$\Delta_{\lambda}({\mathfrak{p}}^{+}):=\Delta_{\lambda}\cap\Delta({\mathfrak{p}}^{+})$,
where
$\Delta_{\lambda}:=\\{\alpha\in\Delta\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}\\}$
is the integral root system associated to $\lambda$ as usual.

An _antichain_ in a poset is a subset consisting of pairwise noncomparable
elements. The _width_ of a poset is the cardinality of maximal antichain in
the poset. Our main result states that the associated variety of a highest
weight Harish-Chandra module $L(\lambda)$ is completely determined by the
width of $Y_{\lambda}$.

######  Theorem 1.2.

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$ and
$\operatorname{AV}(L(\lambda))=\overline{\mathcal{O}_{k(\lambda)}}$. Let
$m=\operatorname{width}(Y_{\lambda})$. Then $k(\lambda)$ is given as follows.

  1. (a)

If $\Delta$ is simply laced and $\lambda$ is integral, then $k(\lambda)=m$.

  2. (b)

If $\Delta$ is non-simply laced and $\lambda$ is integral, then

$k(\lambda)=\begin{cases}2m,&\text{if }m<\frac{r+1}{2}\\\ r,&\text{if
}m=\frac{r+1}{2}.\end{cases}$

  3. (c)

If $\Delta$ is non-simply laced and $\lambda$ is half-integral, then

$k(\lambda)=\begin{cases}2m+1,&\text{if }m<\frac{r}{2}\\\ r,&\text{if
}m=\frac{r}{2}.\end{cases}$

  4. (d)

In all other cases $k(\lambda)=r$.

At first glance, it may seem difficult to compute
$\operatorname{width}(Y_{\lambda})$. However, this is not the case.

######  Example 1.3.

Suppose $G_{\mathbb{R}}$ is the metaplectic double cover of the symplectic
group $Sp(2n,\mathbb{R})$ and $L(\lambda)$ is a highest weight Harish-Chandra
module with half-integral highest weight $\lambda$. Using the conventions for
roots and weights as in [12], we then have
$\lambda+\rho=(t_{1},\ldots,t_{n})$, where the entries $t_{i}$ are half-
integers such that $t_{1}>\cdots>t_{n}$. Since
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}+\varepsilon_{j}\mid 1\leq
i<j\leq n\\}\cup\\{2\varepsilon_{i}\mid 1\leq i\leq n\\}$ and two short roots
$\varepsilon_{i_{1}}+\varepsilon_{j_{1}},\varepsilon_{i_{2}}+\varepsilon_{j_{2}}\in\Delta({\mathfrak{p}}^{+})$
are incomparable if and only if either $i_{1}<i_{2}<j_{2}<j_{1}$ or
$i_{2}<i_{1}<j_{1}<j_{2}$, it follows that $\operatorname{width}(Y_{\lambda})$
is the maximal nonnegative integer $m$ for which there exists a sequence of
indices $1\leq i_{1}<i_{2}<\cdots<i_{m}<j_{m}<\cdots<j_{2}<j_{1}\leq n$ such
that

$t_{i_{1}}+t_{j_{1}}\leq 0\ ,\ t_{i_{2}}+t_{j_{2}}\leq 0\ ,\ \ldots\ ,\
t_{i_{m}}+t_{j_{m}}\leq 0.$

Since the sequence $(t_{1},\ldots,t_{n})$ is strictly decreasing, we may
furthermore assume that $j_{k}=n-k+1$ for $k=1,\ldots,m$. This leads to the
following simple algorithm to find $\operatorname{width}(Y_{\lambda})$ with a
running time that is linear in $n$, which is optimal. Set $j_{1}=n$ and let
$i_{1}$ be the smallest index such that $i_{1}<j_{1}$ and
$t_{i_{1}}+t_{j_{1}}\leq 0$. If no such $i_{1}$ exists, then $m=0$. Next set
$j_{2}=n-1$ and let $i_{2}$ be the smallest index such that
$i_{1}<i_{2}<j_{2}$ and $t_{i_{2}}+t_{j_{2}}\leq 0$. If no such $i_{2}$
exists, then $m=1$. Continue doing this until it is no longer possible which
happens after at most $[n/2]$ steps. For a concrete example, suppose

$\lambda+\rho=(\underbracket{\frac{25}{2},\frac{23}{2},\frac{19}{2},\underbracket{\frac{15}{2},\frac{13}{2},\frac{11}{2},\underbracket{\frac{9}{2},\underbracket{-\frac{3}{2},-\frac{7}{2}},-\frac{9}{2}},-\frac{17}{2}}}).$

Here brackets were drawn from the $j_{1}$-th entry to the $i_{1}$-th entry,
from the $j_{2}$-th entry to the $i_{2}$-th entry, and from the $j_{3}$-th
entry to the $i_{3}$-th entry. We see that
$\operatorname{width}(Y_{\lambda})=3$ and the theorem says that
$\operatorname{AV}(L(\lambda))=\overline{\mathcal{O}_{7}}$.

###  1.4. Gelfand–Kirillov dimension

Our main result has another interpretation in terms of the Gelfand–Kirillov
(GK) dimension of Harish-Chandra modules. The GK dimension of a Harish-Chandra
module is equal to the dimension of its associated variety. Since the
associated variety of a highest weight Harish-Chandra module $L(\lambda)$ is
one of the varieties in the chain (1.1), it follows that the associated
variety of $L(\lambda)$ is determined by the GK dimension of $L(\lambda)$. The
following theorem gives $\operatorname{GKdim}(L(\lambda))$ when $\lambda$ is
nonintegral. One sees that in this case $\Delta_{\lambda}({\mathfrak{p}}^{+})$
is either empty or is isomorphic to a poset $\Delta({\mathfrak{p}}^{\prime+})$
corresponding to a noncompact Hermitian symmetric space
$G_{\mathbb{R}}^{\prime}/K_{\mathbb{R}}^{\prime}$, which in fact has simply
laced root system. This follows from Cartan’s classification of Hermitian
symmetric spaces and the work of Jakobsen [17] on the posets of noncompact
roots (see the diagrams in [17, Appendix]).

Let ${\mathcal{O}}_{k}^{\prime},k=0,1,\dots,r^{\prime}$, be the
$K^{\prime}$-orbits in ${\mathfrak{p}}^{\prime+}$ (analogous to the chain
described in 1.1). Set $\delta=\\#(\Delta^{+})-\\#(\Delta_{\lambda}^{+})$,
which equals
$\\#(\Delta({\mathfrak{p}}^{+}))-\\#(\Delta_{\lambda}({\mathfrak{p}}^{+}))$
since $\lambda$ is $\Delta({\mathfrak{k}})$-integral.

######  Theorem 1.4.

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module and assume that
the highest weight $\lambda$ is nonintegral. If $m=0pt(Y_{\lambda})$, then

$\operatorname{GKdim}(L(\lambda))=\delta+\dim({\mathcal{O}}_{m}^{\prime}).$

It turns out that $\Delta_{\lambda}({\mathfrak{p}}^{+})$ is nonempty (for
nonintegral $\lambda$) only when $\Delta$ is non-simply laced and $\lambda$ is
half-integral, see [12, Lem. 3.17]. When
$\Delta_{\lambda}({\mathfrak{p}}^{+})=\emptyset$, then
${\mathcal{O}}_{m}^{\prime}=\\{0\\}$ and
$\delta=\\#(\Delta({\mathfrak{p}}^{+}))=\dim({\mathfrak{p}}^{+})$, so
$\operatorname{AV}(L(\lambda))=\overline{{\mathcal{O}}_{r}}={\mathfrak{p}}^{+}$.

###  1.5. About the proofs

Theorems 1.2 and 1.4 are proved in Section 4. Theorem 1.2 is first proved when
$\lambda+\rho$ is regular integral; it suffices to assume the infinitesimal
character is $\rho$. The argument is to first observe that if
$0pt(Y_{\lambda})=m$, then
$\operatorname{AV}(L(\lambda))\supseteq\overline{{\mathcal{O}}_{m}}$. A
counting argument is then used. The general theory of Harish-Chandra cells
gives a lower bound on the size of a cell in terms of the Springer
correspondence. This, along with the number of $L(\lambda)$ (of infinitesimal
character $\rho$) for which $0pt(Y_{\lambda})=m$, completes the proof. The
number with a given width is calculated in Section 2 using some combinatorics
of the posets. The necessary background on Harish-Chandra cells and some
specific data on the Springer correspondence is given in Section 3. The case
of arbitrary integral infinitesimal character follows from the translation
principle, illustrating that our definition of the diagram $Y_{\lambda}$ is
‘correct’. This leaves the nonintegral case. When $\Delta$ is non-simply laced
and $\lambda$ is half-integral, there is a group $G_{\mathbb{R}}^{\prime}$, as
described above. Lusztig’s formula for the GK dimension proves Theorem 1.4 and
this reduces the proof of Theorem 1.2 to the integral case for
$G_{\mathbb{R}}^{\prime}$. In the other nonintegral cases $L(\lambda)$ is
precisely a (full) generalized Verma module for the parabolic
${\mathfrak{q}}$; the associated variety is then easily seen to be
$\overline{{\mathcal{O}}_{m}}={\mathfrak{p}}^{+}$.

Another proof of Theorem 1.2 is given in Section 5. In [4], Bai–Xiao–Xie used
Robinson–Schensted insertion algorithm to compute the Gelfand–Kirillov
dimensions of highest weight modules for classical type Lie algebras and cell
decompositions of maximal parabolic subgroups of Hermitian type for
exceptional type Lie algebras. Then, we translate these results into the world
of antichains of posets, which gives us another proof for Theorem 1.2.

###  1.6. Connections with other work

In 1992, Joseph [20] described the associated varieties of unitary highest
weight modules in a case-by-case fashion. Nishiyama–Ochiai–Taniguchi [24] and
Enright–Willenbring [13] independently gave some characterizations of the
associated varieties of unitary highest weight modules appearing in the dual
pair settings. More recently, Bai–Hunziker [2] found a uniform formula for the
GK dimensions and associated varieties of all unitary highest weight modules.
For $G_{\mathbb{R}}=Sp(2n,\mathbb{R})$, Barchini–Zierau [8] computed the
characteristic cycles of highest weight Harish-Chandra modules for regular
integral infinitesimal character and associated varieties are given implicitly
by an inductive procedure.

##  2\. Posets of positive noncompact roots

###  2.1.

Let
${\mathfrak{g}}={\mathfrak{p}}^{-}\oplus{\mathfrak{k}}\oplus{\mathfrak{p}}^{+}$
be the triangular decomposition corresponding to an irreducible Hermitian
symmetric space (of noncompact type) as in the introduction. Therefore,
${\mathfrak{q}}={\mathfrak{k}}\oplus{\mathfrak{p}}^{+}$ is a maximal parabolic
subalgebra of ${\mathfrak{g}}$ with Levi subgroup ${\mathfrak{k}}$ and abelian
nilradical ${\mathfrak{p}}^{+}\\!$. We fix a Cartan subalgebra
${\mathfrak{h}}$ of ${\mathfrak{g}}$ that is contained in ${\mathfrak{k}}$ and
a Borel subalgebra ${\mathfrak{b}}={\mathfrak{h}}\oplus{\mathfrak{n}}$ of
${\mathfrak{g}}$ such that ${\mathfrak{b}}\subset{\mathfrak{q}}$. Let $\Delta$
and $\Delta({\mathfrak{k}})$ denote the root systems of
$({\mathfrak{g}},{\mathfrak{h}})$ and $({\mathfrak{k}},{\mathfrak{h}})$,
respectively, and set
$\Delta^{+}=\\{\alpha\in\Delta\mid{\mathfrak{g}}_{\alpha}\subseteq{\mathfrak{n}}\\}$,
$\Delta^{+}({\mathfrak{k}})=\Delta({\mathfrak{k}})\cap\Delta^{+}$, and
$\Delta({\mathfrak{p}}^{+})=\\{\alpha\in\Delta\mid{\mathfrak{g}}_{\alpha}\subset{\mathfrak{p}}^{+}\\}=\Delta^{+}\setminus\Delta^{+}({\mathfrak{k}})$.
The roots in $\Delta^{+}({\mathfrak{k}})$ are called the positive compact
roots and the roots in $\Delta({\mathfrak{p}}^{+})$ are called the positive
noncompact roots. Let $\Pi$ denote the set of simple roots for $\Delta^{+}$.
Exactly one simple root is in $\Delta({\mathfrak{p}}^{+})$. As usual, $\rho$
is half the sum of the positive roots. Denote by $W$ the Weyl group of
$\Delta$.

Define
$\Lambda^{+}({\mathfrak{k}}):=\\{\lambda\in{\mathfrak{h}}^{*}\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}_{\geq
0}\text{ for all }\alpha\in\Delta^{+}({\mathfrak{k}})\\}$. Note that
$\Lambda^{+}({\mathfrak{k}})$ is the set of highest weights of irreducible
finite-dimensional representations of ${\mathfrak{k}}$. We consider arbitrary
$\lambda\in\Lambda^{+}({\mathfrak{k}})$. When $\lambda$ is not integral we
need to consider the integral root system
$\Delta_{\lambda}=\\{\alpha\in\Delta\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}\\}$
and its Weyl group $W_{\lambda}$.

We view all sets of roots as partially ordered sets with the usual partial
ordering, where $\alpha\leq\beta$ means that $\beta-\alpha$ is a sum of roots
in $\Delta^{+}$. In what follows, the set of positive noncompact roots
$\Delta({\mathfrak{p}}^{+})$, viewed as a poset, will play a fundamental role.

######  Lemma 2.1 (Jakobsen [17, Lem. 4.1]).

Let $\alpha\in\Delta({\mathfrak{p}}^{+})$, let $\pi_{1},\ldots,\pi_{k}$ be any
distinct elements of $\Pi\cap\Delta({\mathfrak{k}})$, and assume that
$\alpha+\pi_{i}\in\Delta({\mathfrak{p}}^{+})$ for all $i=1,\ldots,k$. Then
$k\leq 2$. Furthermore, if $k=2$, then $\pi_{1}$ and $\pi_{2}$ are orthogonal
and $\alpha+\pi_{1}+\pi_{2}\in\Delta({\mathfrak{p}}^{+})$. $\Box$

In light of this lemma, the Hasse diagram of $\Delta({\mathfrak{p}}^{+})$ is
an upward planar graph of order dimension two and hence can be drawn on a two-
dimensional orthogonal lattice that has been rotated by a 45-degree angle.
That is, we draw the Hasse diagram as a graph with vertices labelled by the
positive noncompact roots and a directed edge from $\beta$ to $\beta^{\prime}$
when $\beta^{\prime}-\beta$ is a simple (necessarily compact) root; the edge
is sometimes labelled by the simple compact root. The unique minimal element
of the poset is the simple noncompact root (drawn at the bottom) and the
unique maximal element, which we denote by $\theta$, is the highest root
(drawn at the top). These diagrams appear in the appendix of Jakobsen’s paper
[17]. We have included the Hasse diagrams for the simply-laced types in the
appendix.

###  2.2. Lower-order ideals

A subset $Y\subseteq\Delta({\mathfrak{p}}^{+})$ is called a _lower-order
ideal_ if, for $\alpha\in\Delta({\mathfrak{p}}^{+})$ and $\beta\in Y$,
$\alpha\leq\beta$ implies that $\alpha\in Y$.

For $\lambda\in\Lambda^{+}({\mathfrak{k}})$, define the _diagram_ of $\lambda$
is the set

(2.1)
$Y_{\lambda}:=\\{\alpha\in\Delta({\mathfrak{p}}^{+})\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}_{\leq
0}\\},$

viewed as a subposet of $\Delta({\mathfrak{p}}^{+})$.

######  Lemma 2.2.

When $\lambda$ is integral, the poset $Y_{\lambda}$ is a lower order ideal of
$\Delta({\mathfrak{p}}^{+})$.

###### Proof.

Suppose $\alpha\in\Delta({\mathfrak{p}}^{+})$, $\beta\in Y_{\lambda}$ and
$\alpha\leq\beta$. Since $\alpha,\beta\in\Delta({\mathfrak{p}}^{+})$ and
$\alpha\leq\beta$ it follows that $\beta-\alpha$ is a sum of roots in
$\Delta^{+}({\mathfrak{k}})$. Since $\lambda\in\Lambda^{+}({\mathfrak{k}})$,
it follows that $\alpha\in Y_{\lambda}$. ∎

It is easy to describe the lower order ideals in $\Delta({\mathfrak{p}}^{+})$.
Define

(2.2) $\mathcal{W}:=\\{w\in W\mid-w\rho\text{ is
$\Delta^{+}({\mathfrak{k}})$-dominant}\\}.$

It follows that $\mathcal{W}=\\{w\in
W\mid{\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}\ \subseteq\
{\mathfrak{p}}^{+}\\}$. Note that the cardinality of $\mathcal{W}$ is
$\\#\mathcal{W}=\\#(W/W({\mathfrak{k}}))$, where $W({\mathfrak{k}})$ is the
Weyl group of ${\mathfrak{k}}$.

######  Proposition 2.3.

There is a bijection

(2.3) $\mathcal{W}\longleftrightarrow\\{\mbox{lower order ideals of
$\Delta({\mathfrak{p}}^{+})$}\\}$

given by $w\mapsto Y_{-w\rho-\rho}$. Furthermore, for any $w\in\mathcal{W}$,
we have

(2.4)
$Y_{-w\rho-\rho}=\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}).$

###### Proof.

This is an easy consequence of [11, §3]. ∎

######  Example 2.4.

Suppose ${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{u}}(4,3)$ and
$\lambda+\rho=-w\rho=(2,1,-1,-2\mid 3,0,-3)$. Then

${\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}=\left\\{\displaystyle\left(\begin{array}[]{cccc|ccc}\cdot&\cdot&\cdot&\cdot&*&\cdot&\cdot\\\
\cdot&\cdot&\cdot&\cdot&*&\cdot&\cdot\\\ \cdot&\cdot&\cdot&\cdot&*&*&\cdot\\\
\cdot&\cdot&\cdot&\cdot&*&*&\cdot\\\
\hline\cr\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\\
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\\
\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\end{array}\right)\right\\}$and$Y_{\lambda}=$$\alpha_{4}=\varepsilon_{4}-\varepsilon_{5}.$32156

It is useful to know how to associate an element of $\mathcal{W}$ to a diagram
$Y_{\lambda}$. Given $\lambda\in\Lambda^{+}({\mathfrak{k}})$ that is integral
(for $\Delta$), we show how to determine $w\in\mathcal{W}$ so that
$Y_{\lambda}=Y_{-w\rho-\rho}$. Let
$C^{-}:=\\{y\in{\mathfrak{h}}^{*}\mid\langle y\,,\alpha\rangle<0,\text{ for
all }\alpha\in\Delta^{+}\\}$, the negative Weyl chamber. When $\lambda+\rho$
is regular there is a unique $w\in W$ such that $w^{-1}(\lambda+\rho)\in
C^{-}$. This is the correct $w$ (since $w^{-1}(\lambda+\rho)$ and $-\rho$ lie
in the same Weyl chamber, which implies $\lambda+\rho$ and $-w\rho$ are also
in the same chamber). However, when $\lambda+\rho$ is singular the situation
is more subtle. The following proposition shows how to determine the correct
$w$. The importance lies in Proposition 4.1.

######  Proposition 2.5.

Suppose that $\lambda\in\Lambda^{+}({\mathfrak{k}})$ and $\lambda$ is
integral. Let $w\in W$ be a minimal length element of $W$ such that
$w^{-1}(\lambda+\rho)$ is in the closure of $C^{-}$. Then the following hold.

  1. (1)

$w\in\mathcal{W}$.

  2. (2)

If $\alpha\in\Delta^{+}$ and $\langle w^{-1}(\lambda+\rho)\,,\alpha\rangle=0$,
then $w\alpha>0$.

  3. (3)

$Y_{\lambda}=Y_{-w\rho-\rho}$.

Such $w$ is unique.

###### Proof.

Since $w^{-1}(\lambda+\rho)$ and $-\rho$ both lie in the closure of $C^{-}$,
$\lambda+\rho$ and $-w\rho$ lie in the same (closure of) Weyl chamber.
Therefore, $-w\rho\in\Lambda^{+}({\mathfrak{k}})$, so $w\in\mathcal{W}$.
Suppose that $\langle w^{-1}(\lambda+\rho)\,,\alpha\rangle=0$ for some
positive root $\alpha$. Then
$(ws_{\alpha})^{-1}(\lambda+\rho)=s_{\alpha}(w^{-1}(\lambda+\rho))=w^{-1}(\lambda+\rho)$
(where $s_{\alpha}$ is the reflection in $\alpha$). By the minimal length
hypothesis for $w$,
$\operatorname{length}(w)<\operatorname{length}(ws_{\alpha})$, which implies
$w\alpha>0$.

The inclusion $Y_{\lambda}\supseteq Y_{-w\rho-\rho}$ is clear since
$\lambda+\rho$ and $-w\rho$ lie in the same closure of Weyl chamber.
Therefore, suppose that $\beta\in Y_{\lambda}$. If
$\langle\lambda+\rho\,,\beta\rangle\neq 0$, then
$\langle-w\rho\,,\beta\rangle<0$. If $\langle\lambda+\rho\,,\beta\rangle=0$
there are two possibilities. Case (i): $w^{-1}\beta>0$. Then
$\langle-w\rho\,,\beta\rangle=-\langle\rho\,,w^{-1}\beta\rangle<0$, therefore
$\beta\in Y_{-w\rho-\rho}$. Case (ii): $w^{-1}\beta<0$. Then
$\alpha:=-w^{-1}\beta\in\Delta^{+}$ is orthogonal to $w^{-1}(\lambda+\rho)$,
However, part (2) says that $-\beta=w\alpha>0$, which is a contradiction, so
Case (ii) does not occur. Part (3) is now proved.

Uniqueness follows from (3). ∎

###  2.3. Antichains and width of lower-order ideals

An _antichain_ in any poset $Y$ is a subset of $Y$ consisting of pairwise
noncomparable elements. The _width_ of the poset, denoted by $0pt(Y)$, is the
greatest cardinality of antichains in $Y$.

######  Remark 2.6.

Example 2.4 illustrates the fact that for ${\mathfrak{su}}(p,q)$ the width of
$Y_{\lambda}=Y_{-w\rho-\rho}$ is the maximal rank of a matrix in
${\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}$. See the proof of Lemma
4.7 for the analogous statement when
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{sp}}(2n,\mathbb{R})$.

Recall that two roots $\gamma_{1},\gamma_{2}\in\Delta$ are called _strongly
orthogonal_ if neither $\gamma_{1}\pm\gamma_{2}$ is a root. When
$\gamma_{1},\gamma_{2}\in\Delta({\mathfrak{p}}^{+})$, $\gamma_{1}+\gamma_{2}$
is never a root (as ${\mathfrak{p}}^{+}$ is abelian), therefore, they are
strongly orthogonal if and only if $\gamma_{1}-\gamma_{2}$ is a root. Thus,
$\gamma_{1},\gamma_{2}$ are not comparable implies they are strongly
orthogonal. (The converse does not hold; for example, in
${\mathfrak{so}}(2,2n-1)$,
$(\varepsilon_{1}+\varepsilon_{n})-(\varepsilon_{1}-\varepsilon_{n})$ is a sum
of roots, but is not a root.) The following lemma therefore holds.

######  Lemma 2.7.

Every antichain of roots in $\Delta({\mathfrak{p}}^{+})$ is a set of pairwise
strongly orthogonal roots. $\square$

######  Corollary 2.8.

For every $w\in\mathcal{W}$, $\operatorname{width}(Y_{-w\rho-\rho})\leq r$,
where $r$ is the real rank of ${\mathfrak{g}}_{\mathbb{R}}$.

###### Proof.

It is well-known (see [19]) that $r$ is the cardinality of a maximal set of
mutually strongly orthogonal roots in $\Delta({\mathfrak{p}}^{+})$. ∎

###  2.4. The number of lower-order ideals of a given width

The number of lower order ideals of width $m$ is determined in this section.
This is important for our counting arguments in Section 4.

######  Proposition 2.9.

For $0\leq m\leq r$, the number of lower-order ideals in
$\Delta({\mathfrak{p}}^{+})$ of given width $m$ is given by the last column of
Table 1.

Table 1. The number of lower-order ideals of $\Delta({\mathfrak{p}}^{+})$ of a given width ${\mathfrak{g}}_{\mathbb{R}}$ | $\operatorname{card}(\mathcal{W})$ | $\operatorname{card}\\{w\in\mathcal{W}\mid\operatorname{width}(Y_{-w\rho-\rho})=m\\}$
---|---|---
$\mathfrak{su}(p,q)$ | ${p+q\choose p}$ | ${p+q\choose m}-{p+q\choose m-1}$
$\mathfrak{sp}(2n,\mathbb{R})$ | $2^{n}$ |  ${n+1\choose m}$ if $m<\frac{n+1}{2}$,
|  |  $\frac{1}{2}{n+1\choose m}$ if $m=\frac{n+1}{2}$, and
|  |  $0$ if $m>\frac{n+1}{2}$
$\mathfrak{so}^{*}(2n)$ | $2^{n-1}$ |  ${n\choose m}$ if $m<\frac{n}{2}$ and
|  |  $\frac{1}{2}{n\choose m}$ if $m=\frac{n}{2}$
$\mathfrak{so}(2,2n-1)$ | $2n$ |  $1$, $2n-1$, $0$
$\mathfrak{so}(2,2n-2)$ | $2n$ |  $1$, $n$, $n-1$
$\mathfrak{e}_{6(-14)}$ | 27 |  $1$, $6$, $20$
$\mathfrak{e}_{7(-25)}$ | 56 |  $1$, $7$, $27$, $21$

The proof of this proposition is a case-by-case argument.

####  2.4.1. Proof for
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{u}}(p,q)$

Write $n=p+q$. Take $\Delta^{+}$ to be the usual positive system of roots for
Type $A_{n-1}$, so $\Delta^{+}=\\{\varepsilon_{i}-\varepsilon_{j}\mid 1\leq
i<j\leq n\\}$ and
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}-\varepsilon_{j}\mid 1\leq i\leq
p<j\leq n\\}$.

Suppose that $\beta_{1}=\varepsilon_{i_{1}}-\varepsilon_{j_{1}}$ and
$\beta_{2}=\varepsilon_{i_{2}}-\varepsilon_{j_{2}}$ are noncomparable roots in
$\Delta({\mathfrak{p}}^{+})$. Then $i_{1}\leq p<j_{1}$ and $i_{2}\leq p<j_{2}$
and we may assume that $i_{1}<i_{2}$. Then
$\beta_{1}-\beta_{2}=(\varepsilon_{i_{1}}-\varepsilon_{i_{2}})+(\varepsilon_{j_{2}}-\varepsilon_{j_{1}})$.
The first is a positive root, so the second is necessarily a negative root
(otherwise $\beta_{1}\geq\beta_{2}$), so $j_{1}<j_{2}$. We may conclude that
antichains take the form

$\\{\varepsilon_{i_{s}}-\varepsilon_{j_{s}}\mid s=1,2,\dots,m\\},\text{ where
}i_{1}<i_{2}<\dots i_{m}\leq p<j_{1}<j_{2}<\dots<j_{m}.$

If $Y$ has width $m$, then $Y$ contains a maximal antichain of the above form.
Since $Y$ is a lower order ideal we may subtract positive compact roots and
stay in $Y$, therefore $Y$ contains the antichain

(2.5) $\\{\varepsilon_{p-s+1}-\varepsilon_{p+s}\mid s=1,2,\dots,m\\},$

that is $i_{1}<i_{2}<\dots i_{m}=p<j_{1}<j_{2}<\dots<j_{m}$ are consecutive.
(It is useful to view this antichain in the Hasse diagram for
$\Delta({\mathfrak{p}}^{+})$: it is the horizontal row of nodes $m-1$ rows up
from the minimal root. See the Appendix for the Hasse diagram for
${\mathfrak{s}}{\mathfrak{u}}(4,3)$.) Note that $m$ is necessarily less than
$r=\min\\{p,q\\}$. We conclude that _all_ lower order ideals $Y$ of width $m$
have a maximal antichain in common, namely the one in (2.5).

Let $\wp_{p,q}(m)=\\#\\{w\in\mathcal{W}\mid 0pt(Y_{-w\rho-\rho})=m\\}$. (We
interpret this to be $0$ if $p$ or $q$ is $0$.) Notice that $\wp_{p,q}$ is
symmetric in $p,q$. We assume $p\leq q$.

Claim: $\wp_{p,q}(m)=\begin{cases}\wp_{p-1,q}(m)+\wp_{p,q-1}(m-1),&\text{if
}m<p\\\ \wp_{p,q-1}(p)+\wp_{p,q-1}(p-1),&\text{if }m=p.\end{cases}$

For our argument it is useful to keep in mind the Hasse diagram for
$\Delta({\mathfrak{p}}^{+})$. Note that
$\rho=(\frac{n-1}{2},\frac{n-3}{2},\dots,-\frac{n-1}{2})$ and for
$w\in\mathcal{W}$ the entries of $-w\rho$ are the same as those in $\rho$ and
the first $p$ entries decrease as do the last $q$ entries. Therefore, $-w\rho$
takes the form

(a) $\,(\frac{n-1}{2},\dots\,|\,\dots)\text{ or (b)
}\,(\dots\dots\,|\,\frac{n-1}{2},\dots).$

To prove the claim we consider the contributions to $\wp_{p,q}(m)$ from the
two forms of $-w\rho$. The contribution from (a) is $\wp_{p-1,q}(m)$, since no
roots $\varepsilon_{1}-\varepsilon_{j}$ appear in $Y_{-w\rho-\rho}$. The
contribution from (b) is

$\begin{cases}\wp_{p,q-1}(p-1),&\text{ if }m<p\\\
\wp_{p,q-1}(p)+\wp_{p,q-1}(p-1),&\text{ if }m=p.\end{cases}$

This is easily understood from the Hasse diagram of
$\Delta({\mathfrak{p}}^{+})$. The claim follows since $\wp_{p-1,q}(m)=0$ when
$m=p$.

Now we show that for $p\leq q$

$\wp_{p,q}(m)=\binom{p+q}{m}-\binom{p+q}{m-1},\text{ for }m=1,2,\dots,p,$

using induction on $n=p+q$. When $n=2$ this is immediate. There are several
cases.

  1. (i)

Case $m<p$:

$\displaystyle\wp_{p,q}(m)$
$\displaystyle=\wp_{p-1,q}(m)+\wp_{p,q-1}(m-1),\text{ by the claim,}$
$\displaystyle=\left(\binom{n-1}{m}-\binom{n-1}{m-1}\right)+\left(\binom{n-1}{m-1}-\binom{n-1}{m-2}\right),\text{
by induction,}$ $\displaystyle=\binom{n-1}{m}-\binom{n-1}{m-2}$
$\displaystyle=\binom{n}{m}-\binom{n}{m-1},\text{ an easily checked
identity}.$

  2. (ii)

Case $m=p<q$:

$\displaystyle\wp_{p,q}(p)$
$\displaystyle=\wp_{p,q-1}(p)+\wp_{p,q-1}(p-1),\text{ by the claim,}$
$\displaystyle=\left(\binom{n-1}{p}-\binom{n-1}{p-1}\right)+\left(\binom{n-1}{p-1}-\binom{n-1}{p-2}\right),\text{
by induction,}$ $\displaystyle=\binom{n-1}{p}-\binom{n-1}{p-2}$
$\displaystyle=\binom{n}{p}-\binom{n}{p-1},\text{ same identity}.$

  3. (iii)

Case $m=p=q$:

$\displaystyle\wp_{p,p}(p)$
$\displaystyle=\wp_{p,p-1}(p)+\wp_{p,p-1}(p-1),\text{ by the claim,}$
$\displaystyle=\wp_{p,p-1}(p)+\wp_{p-1,p}(p-1),\text{ by symmetry,}$
$\displaystyle=\binom{2p-1}{p-1}-\binom{2p-1}{p-2},\text{ first term is $0$,}$
$\displaystyle=\binom{2p-1}{p}-\binom{2p-1}{p-2}$
$\displaystyle=\binom{2p}{p}-\binom{2p}{p-1},\text{ same identity}.$

####  2.4.2. Proof for
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}^{*}(2n)$

This case is is quite similar to the ${\mathfrak{su}}(p,q)$ case above. The
set of positive compact roots is
$\Delta^{+}({\mathfrak{k}})=\\{\varepsilon_{i}-\varepsilon_{j}\mid 1\leq
i<j\leq n\\}$ and the poset of positive noncompact roots is
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}+\varepsilon_{j}\mid 1\leq
i<j\leq n\\}$. Suppose $\gamma_{1}=\varepsilon_{i_{1}}+\varepsilon_{j_{1}}$
and $\gamma_{2}=\varepsilon_{i_{2}}+\varepsilon_{j_{2}}$ are two noncomparable
roots in $\Delta({\mathfrak{p}}^{+})$. We may assume that
$i_{1}<i_{2},i_{1}<j_{1}$ and $i_{2}<j_{2}$. Then in the expression

$\gamma_{1}-\gamma_{2}=(\varepsilon_{i_{1}}-\varepsilon_{i_{2}})+(\varepsilon_{j_{1}}-\varepsilon_{j_{2}})$

the first term is a positive root, therefore $\gamma_{1},\gamma_{2}$ are
noncomparable if and only if the second term is a negative root. This holds if
and only if $j_{2}<j_{1}$ (so $i_{1}<i_{2}<j_{2}<j_{1}$). We may conclude from
this that the antichains of length $m$ in any lower order ideal are

$\\{\varepsilon_{i_{s}}+\varepsilon_{j_{s}}\mid s=1,\dots,m\\},\text{ where
}i_{1}<\dots<i_{m}<j_{m}<\dots,<j_{1}.$

For $w\in\mathcal{W}$, $-w\rho$ has decreasing entries, so if
$0pt(Y_{-w\rho-\rho})=m$ then we may assume that
$i_{1}<\dots<i_{m}<j_{m}<\dots,j_{1}$ are _consecutive_ ending in $j_{1}=n$:
$\\{\varepsilon_{n-2m+s}+\varepsilon_{n+1-s}\mid s=1,\dots,m\\}.$ Therefore,
whenever a lower order ideal has width $m$ it contains this maximal antichain,
which appears in the Hasse diagram as the horizontal row of nodes appearing
$2m-2$ rows up from the bottom. For $n=6$, the Hasse diagram is shown in the
appendix.

Let $\wp_{n}(m)$ be the number of $w\in\mathcal{W}$ such that
$0pt(Y_{-w\rho-\rho})=m$. Since
$\operatorname{rank}_{\mathbb{R}}({\mathfrak{so}}^{*}(2n))=\lfloor\frac{n}{2}\rfloor$,
$\wp_{n}(m)=0$ when $m>\lfloor\frac{n}{2}\rfloor$.

######  Claim 2.10.

We have

$\wp_{n}(m)=\begin{cases}\binom{n}{m},&\text{if }m<\frac{n}{2}\\\
\frac{1}{2}\binom{n}{n/2},&\text{if }m=\frac{n}{2}.\end{cases}$

###### Proof of claim.

Since $\rho=(n-1,\dots 2,1,0)$ and $-w\rho$ has decreasing entries, $-w\rho$
has the form

$\text{(a) }(n-1,\dots\dots)\quad\text{ or\quad(b) }(\dots\dots,-(n-1)).$

As in the ${\mathfrak{su}}(p,q)$ case we count the contributions to
$\wp_{n}(m)$ from the two cases. It is again useful to consider the Hasse
diagram. The contribution from case (a) is $\wp_{n-1}(m)$, since
$\varepsilon_{1}+\varepsilon_{j}$ is never in $Y_{-w\rho-\rho}$. The
contribution from case (b) is $\wp_{n-1}(m-1)$, when $m\neq\frac{n-1}{2}$, and
is $\wp_{n-1}(m-1)+\wp_{n-1}(m)$, when $m=\frac{n-1}{2}$. Thus, we have

$\wp_{n}(m)=\begin{cases}\wp_{n-1}(m)+\wp_{n-1}(m-1),&\text{if
}m\neq\frac{n-1}{2}\\\ \wp_{n-1}(m)+\wp_{n-1}(m-1)+\wp_{n-1}(m),&\text{if
}m=\frac{n-1}{2}.\end{cases}$

We now prove the claim by induction on $n$. There are three cases.

  * (i)

$\displaystyle m<\frac{n-1}{2}$:

$\wp_{n}(m)=\wp_{n-1}(m)+\wp_{n-1}(m-1)=\binom{n-1}{m}+\binom{n-1}{m-1}=\binom{n}{m}.$

  * (ii)

$\displaystyle m=\frac{n-1}{2}$:

$\displaystyle\wp_{n}(m)$
$\displaystyle=2\wp_{n-1}\left(\frac{n-1}{2}\right)+\wp_{n-1}\left(\frac{n-1}{2}-1\right)$
$\displaystyle=\binom{n-1}{(n-1)/2}+\binom{n-1}{(n-3)/2}=\binom{n}{(n-1)/2}.$

  * (iii)

$\displaystyle m=\frac{n}{2}$:

$\wp_{n}(m)=\wp_{n-1}\left(\frac{n}{2}\right)+\wp_{n-1}\left(\frac{n}{2}-1\right)=0+\wp_{n-1}\left(\frac{n}{2}-1\right)=\binom{n-1}{n/2-1}=\frac{1}{2}\binom{n}{n/2}.$

∎

####  2.4.3. Proof for
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{p}}(2n,\mathbb{R})$

The poset $\Delta({\mathfrak{p}}^{+})$ is isomorphic to
$\Delta({\mathfrak{p}}^{\prime+})$, where
${\mathfrak{g}}^{\prime}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}^{*}(2n+2)$.
Therefore, this case is already done. Note that the real rank of
${\mathfrak{s}}{\mathfrak{p}}(2n,\mathbb{R})$ is $n$ whereas the real rank of
${\mathfrak{s}}{\mathfrak{o}}^{*}(2n+2)$ is $\lfloor\frac{n+1}{2}\rfloor$.
This explains the entry $0$ for $m>\frac{n+1}{2}$ in the third column of the
Table 1.

Using the isomorphism with the poset $\Delta({\mathfrak{p}}^{\prime+})$ and
the description of antichains given in §2.4.2, we see that if $0pt(Y)=m$, then
there is an maximal antichain having one of the forms

$\displaystyle\left\\{\varepsilon_{n-m+1-s}+\varepsilon_{n-m+1+s}\mid
s=1,\dots,m-1\right\\}\cup\left\\{2\varepsilon_{n-m+1}\right\\}\text{ or }$
$\displaystyle\left\\{\varepsilon_{n-m+1-s}+\varepsilon_{n-m+s}\mid
s=1,\dots,m\right\\}.$

####  2.4.4. Proof for
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}(2,2n-1)$

The poset of $\Delta({\mathfrak{p}}^{+})$ is a linear graph. The result is
immediate because $\\#\Delta({\mathfrak{p}}^{+})=2n-1$.

####  2.4.5. Proof for
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}(2,2n-2)$

The poset $\Delta({\mathfrak{p}}^{+})$ in this case is quite simple and is
illustrated below. Noting that the only antichain of width $2$ is
$\\{\varepsilon_{1}-\varepsilon_{2},\varepsilon_{1}+\varepsilon_{2}\\}$, the
lower order ideals of widths $0,1$ and $2$ are easily counted. We get
$\wp_{n}(0)=1,\wp_{n}(1)=n,$ and $\wp_{n}(2)=n-1$. For $n=6$, the Hasse
diagram of $\Delta({\mathfrak{p}}^{+})$ is shown in the appendix.

####  2.4.6. Proof for ${\mathfrak{g}}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$

The poset $\Delta({\mathfrak{p}}^{+})$ for
${\mathfrak{g}}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$ is shown in the Appendix.
A lower-order ideal $Y\subseteq\Delta({\mathfrak{p}}^{+})$ has width $\geq 1$
if and only if $\alpha_{1}\in\Delta({\mathfrak{p}}^{+})$ and width $\geq 2$ if
and only if $\beta_{1},\beta_{2}\in\Delta({\mathfrak{p}}^{+})$. A quick
inspection shows that there are exactly $6$ lower-order ideals of
$\Delta({\mathfrak{p}}^{+})$ that contain $\alpha_{1}$, but not both
$\beta_{1}$ and $\beta_{2}$. Thus, there are exactly $6$ posets of
$\Delta({\mathfrak{p}}^{+})$ that have width $1$. Since the total number of
lower-order ideals of $\Delta({\mathfrak{p}}^{+})$ is
$\operatorname{card}(\mathcal{W})=27$, the number of lower-order ideals of
$\Delta({\mathfrak{p}}^{+})$ width $2$ is $27-1-6=20$.

####  2.4.7. Proof for ${\mathfrak{g}}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$

The poset $\Delta({\mathfrak{p}}^{+})$ for
${\mathfrak{g}}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$ is shown in the Appendix.
A lower-order ideal $Y\subseteq\Delta({\mathfrak{p}}^{+})$ has width $\geq 1$
if and only if $\alpha_{7}\in\Delta({\mathfrak{p}}^{+})$, width $\geq 2$ if
and only if $\beta_{1},\beta_{2}\in\Delta({\mathfrak{p}}^{+})$, and width
$\geq 3$ if and only if
$\gamma_{1},\gamma_{2},\gamma_{3}\in\Delta({\mathfrak{p}}^{+})$. Again, a
quick inspection shows that there are exactly $7$ lower-order ideals of
$\Delta({\mathfrak{p}}^{+})$ that contain $\alpha_{1}$, but not both
$\beta_{1}$ and $\beta_{2}$. A more tedious inspection shows that there are
exactly $27$ lower-order ideals of $\Delta({\mathfrak{p}}^{+})$ that contain
$\beta_{1}$ and $\beta_{2}$ but not all three roots $\gamma_{1}$,
$\gamma_{2}$, and $\gamma_{3}$. Thus, there are exactly $7$ lower-order ideals
of $\Delta({\mathfrak{p}}^{+})$ of width $1$, exactly $27$ lower-order ideals
of width $2$, and $56-1-7-27=21$ lower-order ideals of width $3$.

This concludes the proof of Proposition 2.9. $\Box$

###  2.5. Half-integral case

We need to consider highest weight Harish-Chandra modules $L(\lambda)$ when
$\lambda$ is half-integral. As we shall see, all the information needed is
obtained by reducing to the integral case.

Let $\lambda\in\Lambda^{+}({\mathfrak{k}})$ be half-integral. This means that
$\langle\lambda+\rho\,,\beta^{\vee}\rangle\in\frac{1}{2}+\mathbb{Z}$, for some
$\beta\in\Delta({\mathfrak{p}}^{+})$. We consider the integral root system

$\Delta_{\lambda}=\\{\alpha\in\Delta\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle\in\mathbb{Z}\\}.$

Let $W_{\lambda}$ denote the Weyl group of $\Delta_{\lambda}$ and fix the
positive system $\Delta_{\lambda}^{+}:=\Delta^{+}\cap\Delta_{\lambda}$. Let
${\mathfrak{g}}_{\lambda}$ be the complex semisimple Lie algebra with root
system $\Delta_{\lambda}$. Then there is a triangular decomposition

(2.6)
${\mathfrak{g}}_{\lambda}={\mathfrak{p}}_{\lambda}^{-}+{\mathfrak{k}}+{\mathfrak{p}}_{\lambda}^{+},$

where
$\Delta({\mathfrak{p}}_{\lambda}^{\pm}):=\Delta({\mathfrak{p}}^{\pm})\cap\Delta_{\lambda}$.
Note that $\Delta({\mathfrak{k}})\subset\Delta_{\lambda}$, since $\lambda$ is
$\Delta({\mathfrak{k}})$-integral, so
${\mathfrak{k}}_{\lambda}\simeq{\mathfrak{k}}$. Therefore, we may apply §2.2
and §2.3 to $\Delta({\mathfrak{p}}_{\lambda}^{+})\subset\Delta_{\lambda}^{+}$
in place of $\Delta({\mathfrak{p}}^{+})\subset\Delta^{+}$. In particular,
$Y_{\lambda}$ is a lower order ideal in $\Delta_{\lambda}$ and defining
$\mathcal{W}_{\lambda}$ in $W_{\lambda}$ there is a bijection
$\mathcal{W}_{\lambda}\leftrightarrow\\{\text{lower order ideals in
}\Delta({\mathfrak{p}}_{\lambda}^{+})\\}$. Note that for
$\alpha,\beta\in\Delta({\mathfrak{p}}_{\lambda}^{+})$, $\alpha\leq\beta$ is
the _same_ in $\Delta$ (with respect to $\Delta^{+}$) as in $\Delta_{\lambda}$
(with respect to $\Delta_{\lambda}^{+}$), since $\beta-\alpha$ is a sum of
compact roots and $\Delta({\mathfrak{k}})\subset\Delta_{\lambda}$. Therefore,
$Y_{\lambda}$ is the same whether defined in $\Delta$ or in
$\Delta_{\lambda}$.

There are two examples we return to in §4.4.

  * (a)

${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{sp}}(2n,\mathbb{R}),n\geq 2$. If
$\lambda=(\lambda_{1},\dots,\lambda_{n})$ is half-integral, then
$\lambda_{j}\in\frac{1}{2}+\mathbb{Z}$, all $j$. Therefore, the integral root
system is

$\Delta_{\lambda}=\\{\varepsilon_{j}\pm\varepsilon_{k}\mid 1\leq j<k\leq
n\\},$

a root system of type $D_{n}$. Observe that the complex Lie algebra
${\mathfrak{g}}_{\lambda}$ corresponding to $\Delta_{\lambda}$ is of type
$D_{n}$ and has a real form
${\mathfrak{g}}_{\lambda,\mathbb{R}}\simeq{\mathfrak{so}}^{*}(2n)$ of
Hermitian type. The set of positive noncompact roots is
$\Delta({\mathfrak{p}}_{\lambda}^{+})=\Delta({\mathfrak{p}}^{+})\cap\Delta_{\lambda}=\\{\varepsilon_{j}+\varepsilon_{k}:1\leq
j<k\leq n\\}$.

  * (b)

${\mathfrak{g}}={\mathfrak{so}}(2,2n-1),n\geq 2$. Then

$\Delta_{\lambda}=\\{\pm\varepsilon_{1}\\}\cup\Delta({\mathfrak{k}}),$

a root system of type $A_{1}\times B_{n-1}$. The corresponding Lie Algebra
${\mathfrak{g}}_{\lambda}$ has a real form
${\mathfrak{g}}_{\lambda,\mathbb{R}}\simeq{\mathfrak{sl}}(2,\mathbb{R})\times{\mathfrak{so}}(2n-1)$
of Hermitian type.

##  3\. Associated varieties and highest weight Harish-Chandra modules

In this section we review some background information on associated varieties,
cells and the Springer correspondence.

###  3.1. Associated varieties and cells for $({\mathfrak{g}},K)$-modules

Assume that $G_{\mathbb{R}}$ is a connected semisimple Lie group with finite
center. Let
${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{k}}_{\mathbb{R}}\oplus{\mathfrak{p}}_{\mathbb{R}}$
be a Cartan decomposition; since $G_{\mathbb{R}}$ has finite center, there is
a maximal compact subgroup $K_{\mathbb{R}}$ with Lie algebra
${\mathfrak{k}}_{\mathbb{R}}$. Let
${\mathfrak{g}}={\mathfrak{k}}\oplus{\mathfrak{p}}$ be the complexified Cartan
decomposition and $K$ the connected complex algebraic group containing
$K_{\mathbb{R}}$ with Lie algebra ${\mathfrak{k}}$. Any irreducible
$K_{\mathbb{R}}$ representation extends to an algebraic representation of $K$.
Therefore, $({\mathfrak{g}},K)$ is a pair as in [28, §1] and the Harish-
Chandra module of an admissible representation of $G_{\mathbb{R}}$ is a
$({\mathfrak{g}},K)$-module. The action of the algebraic group $K$ is needed
for definition of associated variety. We follow the definition of associated
variety of a finitely generated $({\mathfrak{g}},K)$-module as given in [28].
Note that we are including finite covers of linear groups, such as the
metaplectic cover of $Sp(2n,\mathbb{R})$ which occurs for half integral
highest weights. However, groups with infinite center, such as the universal
cover of $G_{\mathbb{R}}=Sp(2n,\mathbb{R})$, are not included in our
discussion.

Suppose that $({\mathfrak{g}},K)$ is a pair as above and $X$ is a
$({\mathfrak{g}},K)$-module of finite length. The associated variety of $X$ is
defined by starting with a $K$-invariant ‘good’ filtration $\\{X_{n}\\}$ of
$X$. Then $\underline{\operatorname{Gr}}\,X$ is a module for
$\underline{\operatorname{Gr}}\,\mathcal{U}({\mathfrak{g}})\simeq
S({\mathfrak{g}})\simeq P({\mathfrak{g}}^{*})$, and by $K$-invariance of the
filtration, is a module for $S({\mathfrak{g}}/{\mathfrak{k}})$. The associated
variety of $X$ is defined as the support of
$\underline{\operatorname{Gr}}\,X$:

$\operatorname{AV}(X):=\operatorname{supp}(\underline{\operatorname{Gr}}\,X)\subseteq({\mathfrak{g}}/{\mathfrak{k}})^{*}.$

Identifying $({\mathfrak{g}}/{\mathfrak{k}})^{*}$ with ${\mathfrak{p}}$ by the
Killing form, $\operatorname{AV}(X)$ is a $K$-stable subvariety of
${\mathfrak{p}}$. In fact,
$\operatorname{AV}(X)\subseteq{\mathfrak{p}}\cap{\mathcal{N}}$, where
${\mathcal{N}}$ is the nilpotent cone in ${\mathfrak{g}}$. Therefore,

(3.1) $\operatorname{AV}(X)=\cup\,\overline{{\mathcal{O}}_{\alpha}},$

a union of closures of nilpotent $K$-orbits in ${\mathfrak{p}}$ and we write
this union in a way that no $\overline{{\mathcal{O}}}_{\alpha}$ is contained
in any other.

When $X$ is irreducible, the ${\mathcal{O}}_{\alpha}$ occurring in (3.1) are
all equidimensional and all have the same $G$-saturation, that is, there is a
complex nilpotent orbit ${\mathcal{O}}^{\mathbb{C}}$ such that

(3.2) ${\mathcal{O}}^{\mathbb{C}}=G\cdot{\mathcal{O}}_{\alpha},\text{ for all
}\alpha.$

It is a fact that $\overline{{\mathcal{O}}^{\mathbb{C}}}$ is the associated
variety of the annihilator of $X$. See [28] for details.

Suppose $X$ and $Y$ are irreducible Harish-Chandra modules having
infinitesimal character of Harish-Chandra parameter
$\Lambda\in{\mathfrak{h}}^{*}$. We say that $X\lesssim Y$ when $X$ occurs as a
subquotient of $Y\otimes F$, with $F$ some finite-dimensional
$G$-subrepresentation of the tensor algebra of ${\mathfrak{g}}$. We say $X\sim
Y$ when $X\lesssim Y$ and $Y\lesssim X$; this generates an equivalence
relation on the set of irreducible representations of infinitesimal character
$\Lambda$. The _Harish-Chandra cells_ are defined to be the equivalence
classes. Since $X\lesssim Y$ implies
$\operatorname{AV}(X)\subseteq\operatorname{AV}(Y)$, we see that the
associated variety is constant on any given cell.

Now assume that $\Lambda$ is regular and integral and assume also that
$G_{\mathbb{R}}$ is linear. Then the coherent continuation representation of
the Weyl group $W$ of ${\mathfrak{g}}$ is defined on the Grothendieck group of
Harish-Chandra modules of infinitesimal character $\Lambda$. This gives a
representation of $W$ for each cell as follows. Fix an irreducible $X_{0}$
(infinitesimal character $\Lambda$) and let ${\EuScript C}$ be the cell of
$X_{0}$, that is, the cell containing $X_{0}$. Set

$W_{{\EuScript C}}:=\operatorname{span}_{\mathbb{C}}\\{X\mid X\lesssim
X_{0}\\}\text{ and }U_{{\EuScript
C}}:=\operatorname{span}_{\mathbb{C}}\\{X\mid X\lesssim X_{0}\text{ and
}X\nsim X_{0}\\}$

Then both $W_{\EuScript C}$ and $U_{\EuScript C}$ are $W$-stable under
coherent continuation. The _cell representation_ for ${\EuScript C}$ is the
quotient $V_{\EuScript C}:=W_{\EuScript C}/U_{\EuScript C}$. We may view

$V_{\EuScript C}=\sum_{X\in{\EuScript C}}\mathbb{C}\cdot X$

(but keeping in mind that it is really a quotient). See [6] for cells and [26,
Ch. 7] for coherent continuation.

Much is known about cells and the cell representations. We have already
mentioned that the associated variety is constant on cells, as is the
associated variety of annihilator. An important property of the cell
representations is that if $\overline{{\mathcal{O}}^{\mathbb{C}}}$ is the
associated variety of annihilator, as in (3.2), then the $W$ representation
$\pi({\mathcal{O}}^{\mathbb{C}},1)$ corresponding to
${\mathcal{O}}^{\mathbb{C}}$ under the Springer correspondence occurs in
$V_{\EuScript C}$ ([27, Cor. 14.11]) and is a special representation (in the
sense of Lusztig).

What makes this useful for us is that there are easy combinatorial procedures
known (for the classical groups) and tables (for the exceptional groups) that
(1) give the Springer correspondence and (2) determine which
$W$-representations are special. See [10, §13.1-13.3]. The Atlas of Lie Groups
software ([1]) also gives this information for Lie algebras of rank up to at
least 8.

Most of the relevant data for us is contained in Table 2.

###  3.2. Highest weight Harish-Chandra modules

A _highest weight Harish-Chandra module_ for our group $G_{\mathbb{R}}$
(having finite center) is a $({\mathfrak{g}},K)$-module (for the pair
described in 3.1) having a vector $x_{\lambda}^{+}\in X$ satisfying

  1. (i)

$x_{\lambda}^{+}$ is a weight vector for ${\mathfrak{h}}$, a compact Cartan
subalgebra of ${\mathfrak{g}}$, of weight $\lambda\in{\mathfrak{h}}^{*}$,

  2. (ii)

${\mathfrak{n}}\cdot x_{\lambda}^{+}=0$, where ${\mathfrak{n}}$ is the
nilradical of some Borel subalgebra
${\mathfrak{b}}={\mathfrak{h}}\oplus{\mathfrak{n}}\subset{\mathfrak{g}}$, and

  3. (iii)

$X=\mathcal{U}({\mathfrak{g}})\cdot x_{\lambda}^{+}$.

Suppose $X$ is an infinite-dimensional irreducible highest weight Harish-
Chandra module for $G_{\mathbb{R}}$. Then, by [14, § 1-2], $G_{\mathbb{R}}$ is
of Hermitian type. In this case, there is a triangular decomposition
${\mathfrak{g}}={\mathfrak{p}}^{-}+{\mathfrak{k}}+{\mathfrak{p}}^{+}$ as in
§2.1. It is also shown in [14] that ${\mathfrak{h}}$ and ${\mathfrak{b}}$ in
(i) and (ii) are given by our choices in §2.1. Therefore,
$\lambda\in\Lambda^{+}({\mathfrak{k}})$ and $X$ contains the $K$-type
$F(\lambda)$ having highest weight $\lambda$. Furthermore,

$X=\mathcal{U}({\mathfrak{g}})x_{\lambda}^{+}=\mathcal{U}({\mathfrak{p}}^{-})F(\lambda).$

We conclude that $X$ is the unique irreducible quotient of the generalized
Verma module
$\mathcal{U}({\mathfrak{g}})\\!\\!\underset{\;{\mathfrak{k}}+{\mathfrak{p}}^{+}}{\otimes}\\!F(\lambda)$.
Therefore, $X$ is also the irreducible quotient $L(\lambda)$ of the full Verma
module

(3.3)
$M(\lambda)=\mathcal{U}({\mathfrak{g}})\underset{{\mathfrak{b}}}{\otimes}\mathbb{C}_{\lambda}.$

The infinitesimal character is $\lambda+\rho$.

We recall that if $X=L(\lambda)$ is the irreducible quotient of a Verma module
for our choice of ${\mathfrak{b}}$ (as in (3.3)), then
$\lambda\in\Lambda^{+}({\mathfrak{k}})$ and
$\langle\lambda+\rho\,,\beta\rangle\in\mathbb{R}$, for all $\beta\in\Delta$
(see [12, p. 100]).

Our condition that $G_{\mathbb{R}}$ has finite center implies, by [21], that
$X=L(\lambda)$ is the Harish-Chandra module of a continuous representation of
$G_{\mathbb{R}}$.

###  3.3. Associated varieties of highest weight Harish-Chandra modules

Suppose $X$ is any irreducible highest weight Harish-Chandra module for
$G_{\mathbb{R}}$ of highest weight $\lambda$. Then (as in §3.2)
$X=\mathcal{U}({\mathfrak{p}}^{-})F(\lambda)$. Using the usual filtration of
the enveloping algebra we obtain a filtration

(3.4) $X_{n}:=\mathcal{U}_{n}({\mathfrak{p}}^{-})F(\lambda).$

One easily checks that it is a good filtration and is $K$-invariant. It is
also ${\mathfrak{k}}+{\mathfrak{p}}_{+}$-invariant (so also
${\mathfrak{b}}$-invariant). We conclude that
$\operatorname{AV}(X)\subseteq\left({\mathfrak{g}}/{\mathfrak{k}}+{\mathfrak{p}}^{+}\right)^{*}\simeq{\mathfrak{p}}^{+}$.

It is well-known that the $K$-orbits in ${\mathfrak{p}}^{+}$ have a
particularly nice form. Letting
$r=\operatorname{rank}_{\mathbb{R}}({\mathfrak{g}}_{\mathbb{R}})$, there are
$r+1$ such orbits. They may be written as follows. Let
$\\{\gamma_{1},\dots,\gamma_{r}\\}$ be a maximal set of mutually strongly
orthogonal long roots in ${\mathfrak{p}}^{+}$ (necessarily of cardinality
$r$). Any two sets of mutually strongly orthogonal long roots having the same
cardinality are conjugate under the Weyl group $W({\mathfrak{k}})$. The
$K$-orbits in ${\mathfrak{p}}^{+}$ are

(3.5) $\begin{split}&{\mathcal{O}}_{0}=\\{0\\}\text{ and}\\\
&{\mathcal{O}}_{k}=K\cdot(X_{\gamma_{1}}+\cdots+X_{\gamma_{k}}),\,k=1,2,\dots,r,\end{split}$

where $X_{\gamma}$ is a root vector for $\gamma$. Note that the choice (and
numbering) of strongly orthogonal roots doesn’t matter, by the
$W({\mathfrak{k}})$-conjugacy mentioned above. These orbits satisfy the
inclusions

$\overline{{\mathcal{O}}_{k}}\subseteq\overline{{\mathcal{O}}_{k+1}}.$

We conclude that the associated variety of any highest weight Harish-Chandra
module is the closure of exactly one ${\mathcal{O}}_{k}$.

The following table collects relevant data on the $K$-orbits in
${\mathfrak{p}}^{+}$.

Table 2. Data on orbits and the Springer correspondence ${\mathfrak{g}}_{\mathbb{R}}$ | $\dim(\mathcal{O}_{k})$ | label of $\mathcal{O}_{k}^{\mathbb{C}}$ | special? | $\dim(\pi(\mathcal{O}_{k}^{\mathbb{C}},1))$
---|---|---|---|---
$\mathfrak{su}(p,q)$ | $k(p+q-k)$ | $[2^{k},1^{p+q-2k}]$ | yes | ${p+q\choose p}-{p+q\choose k-1}$
$\mathfrak{sp}(2n,\mathbb{R})$ | $\frac{1}{2}k(2n-k+1)$ | $[2^{k},1^{2n-2k}]$ | yes |  ${n\choose\frac{k}{2}}$ if $k$ is even
|  |  | no |  ${n\choose\frac{k-1}{2}}$ if $k$ is odd, $k<n$
|  |  | yes |  ${n\choose\frac{n-1}{2}}$ if $k=n$ is odd, $k=n$
$\mathfrak{so}^{*}(2n)$ | $k(2n-2k-1)$ | $[2^{2k},1^{2n-4k}]$ | yes |  ${n\choose k}$ if $k\not=\frac{n}{2}$
|  |  | yes |  $\frac{1}{2}{n\choose\frac{n}{2}}$ if $k=\frac{n}{2}$
$\mathfrak{so}(2,2n-1)$ | $2n-2$ | $[2^{2},1^{2n-3}]$ | no | $n-1$
| $2n-1$ | $[3,1^{2n-2}]$ | yes | $n$
$\mathfrak{so}(2,2n-2)$ | $2n-3$ | $[2^{2},1^{2n-4}]$ | yes | $n$
| $2n-2$ | $[3,1^{2n-3}]$ | yes | $n-1$
$\mathfrak{e}_{6(-14)}$ | $0$ | $0$ | yes | $1$
| $11$ | $A_{1}$ | yes | $6$
| $16$ | $2A_{1}$ | yes | $20$
$\mathfrak{e}_{7(-25)}$ | $0$ | $0$ | yes | $1$
| $17$ | $A_{1}$ | yes | $7$
| $26$ | $2A_{1}$ | yes | $27$
| $27$ | $(3A_{1})^{\prime\prime}$ | yes | $21$

######  Remark 3.1.

By Jantzen’s irreducibility criterion or by [12, Lem. 3.17], the generalized
Verma module
$\mathcal{U}({\mathfrak{g}})\\!\underset{\;{\mathfrak{k}}+{\mathfrak{p}}^{+}}{\otimes}F(\lambda)$
can be reducible only when $\lambda$ is integral or half-integral. In the
integral case we may assume that $G_{\mathbb{R}}$ is linear and in the half-
integral case we may assume that $G_{\mathbb{R}}$ is the double cover of a
linear group.

######  Remark 3.2.

If $X$ is an _irreducible_ generalized Verma module
$\mathcal{U}({\mathfrak{g}})\\!\\!\underset{\;{\mathfrak{k}}+{\mathfrak{p}}^{+}}{\otimes}\\!F(\lambda)$
(as in the previous remark as well as in the case of holomorphic discrete
series representations), then it follows easily from the definitions that
$\operatorname{AV}(X)=\overline{{\mathcal{O}}_{r}}={\mathfrak{p}}^{+}$. One
may also see from the definitions that if one member of a cell is a highest
weight Harish-Chandra module, then the cell consists entirely of highest
weight Harish-Chandra modules. Also, if the associated variety of a Harish-
Chandra module $X$ is contained in ${\mathfrak{p}}^{+}$, then $X$ is a highest
weight module ([7, Prop. B1]). If
$\operatorname{AV}(X)={\mathcal{O}}_{0}=\\{0\\}$, then $X$ is finite-
dimensional, so ${\EuScript C}_{0}=\\{\mathbb{C}\\}$ is the only cell (for
infinitesimal character $\rho$) with associated variety ${\mathcal{O}}_{0}$,

See [24] for more on associated varieties of highest weight Harish-Chandra
modules.

Now consider any highest weight module $L(\lambda)$. We may apply the the
theory of associated varieties of highest weight modules (as in [9] and [20]).
This tells us that the associated varieties are unions of orbital varieties,
which we write as

$\nu(w):=\overline{B\cdot({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}})},\quad
w\in W.$

As in [9] we write $L_{w}=L(-w\rho-\rho),w\in W$. Then the support of $L_{w}$
is the Schubert variety $Z_{w}=\overline{B\cdot w{\mathfrak{b}}}$ in the flag
variety ${\mathfrak{B}}$ of $G$. By [9, Prop. 6.11 ] the associated variety of
$L_{w}$ contains the moment map image
$\mu(\overline{T_{\mathfrak{B}}^{*}Z_{w}})=\nu(w)$, so

(3.6) $\operatorname{AV}(L_{w})\supset\nu(w).$

(However, the associated variety is sometimes larger and of greater
dimension.)

Our first goal is a formula for associated varieties when $L_{w}$ is a Harish-
Chandra module, that is, when $w\in\mathcal{W}$. Our description of
$\operatorname{AV}(L_{w})$ may be given in terms of some ${\mathcal{O}}_{k}$
_or_ in terms of orbital varieties. Theorem 1.2 expresses the associated
variety in terms of $K$-orbits in ${\mathfrak{p}}^{+}$, however the orbital
variety picture will allow us to use the combinatorics of posets discussed in
Section 2. The following lemma puts this into perspective.

######  Lemma 3.3 (see [7]).

  1. (1)

For each $j=0,1,\dots,r,$ we have $\overline{{\mathcal{O}}_{j}}=\nu(w)$ for
some $w\in\mathcal{W}$.

  2. (2)

If $\nu(w)\subseteq{\mathfrak{p}}^{+}$, then
$\nu(w)=\overline{{\mathcal{O}}_{j}}$, for some $j=0,1,\dots,r$, and $w$ is
necessarily in $\mathcal{W}$.

The proof of Theorem 1.2 will be largely a counting argument using Tables 1
and 2 along with the following important fact:

(3.7)
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=m\implies\operatorname{AV}(L_{w})\supset\overline{{\mathcal{O}}_{m}}.$

This fact follows from (3.6), (2.7) and (3.5).

##  4\. Proof of Theorems 1.2 and 1.4

Theorem 1.2 is proved first for integral infinitesimal character. We begin by
showing that it suffices to prove the theorem for infinitesimal character
$\rho$. The proof for $X=L_{w}~{}(w\in\mathcal{W})$ is then carried out
separately for the simply laced and non-simply laced cases. Then we consider
the half-integral case. We first prove Theorem 1.4, then reduce the half-
integral case to the integral case.

###  4.1. The translation principle

The reduction of the proof of Theorem 1.2 from integral infinitesimal
character to infinitesimal character $\rho$ is accomplished using the
translation principle. This principle is well-known; we use [16, Ch. 7] as a
convenient reference; see also [18]. A translation functor $T$ applied to $X$
is tensoring by a finite-dimensional ${\mathfrak{g}}$-representation followed
by projection to the constituents of a generalized infinitesimal character.
Therefore, one has $\operatorname{AV}(T(X))\subseteq\operatorname{AV}(X)$. The
following proposition is essentially contained in [3, Cor. 3.3]; we give the
proof as it plays a key role for us.

######  Proposition 4.1.

Let $L(\lambda)$ be a highest weight Harish-Chandra module having integral
infinitesimal character $\lambda+\rho$. Choose $w\in\mathcal{W}$ such that
$Y_{\lambda}=Y_{-w\rho-\rho}$ (as in Prop. 2.3). The following hold.

  1. (1)

$\lambda+\rho$ and $-w\rho$ lie in the closure of the same Weyl chamber,
namely, $C=\\{y\in{\mathfrak{h}}^{*}:\langle y\,,w\alpha\rangle<0,\text{ for
}\alpha\in\Delta^{+}\\}$.

  2. (2)

There is a translation functor $T$ such that $T(L_{w})=L(\lambda)$.

  3. (3)

$\operatorname{AV}(L(\lambda))=\operatorname{AV}(L_{w})$.

###### Proof.

(1) As $L(\lambda)$ and $L_{w}$ are highest weight Harish-Chandra modules
$\lambda+\rho$ and $-w\rho$ are $\Delta^{+}({\mathfrak{k}})$-dominant. Since
$Y_{\lambda}=Y_{-w\rho-\rho}$, $\lambda+\rho$ and $-w\rho$ are nonnegative on
the same roots in $\Delta({\mathfrak{p}})$.

(2) The translation functor $T$ is tensoring by the finite-dimensional
${\mathfrak{g}}$-representation of extreme weight $\lambda+\rho+w\rho$, then
projecting to infinitesimal character $\lambda+\rho$. When $\lambda+\rho$ is
nonsingular the statement is easy. In the singular case $T(L_{w})$ is
potentially $0$. However, our choice of $w$ guarantees $T(L_{w})=L(\lambda)$.
This follows from [16, Thm. 7.9] as follows. One needs
$\lambda+\rho\in\widehat{C}$, the upper closure of the chamber $C$. Part (1)
says that $\lambda+\rho\in\overline{C}$, so suppose that
$\langle\lambda+\rho\,,\alpha\rangle=0$. For $\lambda+\rho\in\widehat{C}$ we
need $w\alpha>0$. This is (2) of Proposition 2.5.

(3) The translation functor $T$ goes from infinitesimal character $\rho$ to
infinitesimal character $\lambda+\rho$. There is also a translation functor
$T^{\prime}$ going the other way. Since $T(L_{w})=L(\lambda)$, we have
$\operatorname{AV}(L(\lambda))\subseteq\operatorname{AV}(L_{w})$. For the
opposite inclusion, we use the adjointness property of translation functors
[16, §7.2]:

$\operatorname{Hom}_{\mathfrak{g}}(X,T^{\prime}(Y))\simeq\operatorname{Hom}_{\mathfrak{g}}(T(X),Y).$

With $X=L_{w}$ and $Y=L(\lambda)$, we get $L_{w}\subseteq
T^{\prime}(L(\lambda))$, so
$\operatorname{AV}(L_{w})\subset\operatorname{AV}(L(\lambda))$. ∎

To complete the reduction to infinitesimal character $\rho$, observe that we
have $\operatorname{AV}(L_{w})=\operatorname{AV}(L(\lambda))$ and
$0pt(Y_{\lambda})=0pt(Y_{-w\rho-\rho})$.

###  4.2. Simply laced case

Assume $\Delta$ is simply laced. We assume the infinitesimal character is
integral, otherwise the highest weight module is irreducible and the
associated variety is $\overline{{\mathcal{O}}_{r}}={\mathfrak{p}}^{+}$.
Theorem 1.2 follows from the following Proposition.

######  Proposition 4.2.

Suppose $\Delta$ is simply laced and let $w\in\mathcal{W}$.

  1. (1)

For $k=0,1,\dots,r$, the set $\\{L_{w}\mid w\in\mathcal{W}\text{ and
}\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{k}}\\}$ is a cell and the
corresponding cell representation is irreducible.

  2. (2)

If $0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k,$ then
$\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{k}}$.

  3. (3)

The cardinality of a maximal set of strongly orthogonal roots in
$\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}})$ is the width.

  4. (4)

$\operatorname{AV}(L_{w})=\nu(w).$

######  Remark 4.3.

Each of these statements is often false in the non-simply laced cases.

We shall use the following lemma without proof.

######  Lemma 4.4.

Suppose $\Delta$ is simply laced. For each $k=0,1,\dots,r$ there exists a
highest weight Harish-Chandra module $L_{w}$ having
$\overline{{\mathcal{O}}_{k}}$ as associated variety.

###### Proof.

For classical ${\mathfrak{g}}_{\mathbb{R}}$, a case-by-case argument shows
that all $\overline{{\mathcal{O}}_{k}},k=0,1,\dots,r$, occur as associated
variety of some $A_{\mathfrak{q}}(\lambda)$ representation that is a highest
weight module. In this case, writing the Levi decomposition of
${\mathfrak{q}}$ as ${\mathfrak{l}}+{\mathfrak{u}}$, the associated variety is
$K\cdot{\mathfrak{u}}\cap{\mathfrak{p}}$, which is easily computed. For the
exceptional cases we may argue as follows. There exist
$A_{\mathfrak{q}}(\lambda)$ modules with associated varieties
$\overline{{\mathcal{O}}_{k}}$ for $k\neq 1$ (e.g., the holomorphic discrete
series for $k=r$). When $k=1$ this is not the case; there exists no unitary
representation of regular integral infinitesimal character having associated
variety $\overline{{\mathcal{O}}_{1}}$. It is known (see [29]) that there
exist unitary ladder representations. These have associated varieties
$\overline{{\mathcal{O}}_{1}}$ (by computation of GK dimension directly from
the $K$-types) and have integral infinitesimal character. The infinitesimal
character is however singular. The translation principle produces a
$({\mathfrak{g}},K)$-module of infinitesimal character $\rho$ having the same
associated variety (as in the proof of Proposition 4.1). Alternatively,
Corollary 4.7 of [25] may be applied to find $w\in\mathcal{W}$ so that
$\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{1}}$. Yet another
alternative is to use the atlas software to view all cell decompositions (as
$W$-representations) for cells having associated variety of annihilator
$\overline{{\mathcal{O}}_{k}^{\mathbb{C}}}$. ∎

The lemma implies there exists a cell ${\EuScript C}_{k}$ of associated
variety $\overline{{\mathcal{O}}_{k}}$, for each $k=0,1,\dots,r$. We know by
§3.1 that $\\#{\EuScript
C}_{k}\geq\dim(\pi({\mathcal{O}}_{k}^{\mathbb{C}},1))$. Let
$M_{k}:=\\{w\in\mathcal{W}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}$ and
let $m_{k}$ be the cardinality of $M_{k}$. Our Tables 1 and 2 show that

(4.1) $m_{k}=\dim(\pi({\mathcal{O}}_{k}^{\mathbb{C}},1)),\quad k=0,1,\dots,r.$

The following example illustrates the argument of our proof of the
proposition.

######  Example 4.5.

Consider ${\mathfrak{e}}_{6(-14)}$. The cardinality of $\mathcal{W}$ is 27 and
$r=2$. So there are three $K$-orbits ${\mathcal{O}}_{0},{\mathcal{O}}_{1}$ and
${\mathcal{O}}_{2}$ in ${\mathfrak{p}}^{+}$. Here is the data for
${\mathfrak{e}}_{6}$ (using the notation of [10] where $\phi_{a,b}$ has
dimension $a$ and degree $b$).

$i$ | $\dim({\mathcal{O}}_{i})$ | $\pi({\mathcal{O}}_{i}^{\mathbb{C}},1)$ | ${\mathcal{O}}_{i}^{\mathbb{C}}$
---|---|---|---
$0$ | $0$ | $\phi_{1,36}$ | $0$
$1$ | $11$ | $\phi_{6,25}$ | $A_{1}$
$2$ | $16$ | $\phi_{20,20}$ | $2A_{1}$

There are three cells ${\EuScript C}_{0}=\\{\mathbb{C}\\},{\EuScript C}_{1}$
and ${\EuScript C}_{2}$. We have $\\#{\EuScript C}_{0}=1$ and by the table
$\\#{\EuScript C}_{1}\geq 6$ and $\\#{\EuScript C}_{2}\geq 20$, but
$\\#\mathcal{W}=27$, so equality holds and we conclude (a) there are no other
cells and (b) each cell representation is irreducible. In Table 1 we have
$m_{0}=1,m_{1}=6$ and $m_{2}=20$. By (3.7), the only candidate for ${\EuScript
C}_{0}$ is $L_{w}$ with $0pt(w)=0$ (so, of course, $L_{w}=\mathbb{C}$ and $w$
is the long element of $W$). The only candidates for ${\EuScript C}_{1}$
correspond to width $0$ or $1$ (by (3.7)), but width $0$ is already accounted
for. Therefore ${\EuScript C}_{1}=\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=1\\}$. The
remaining $20$ elements of $\mathcal{W}$ have
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=2$ and must
give us all of ${\EuScript C}_{2}$.

###### Proof of the proposition.

Let ${\EuScript C}_{k}$ be a cell with associated variety
$\overline{{\mathcal{O}}_{k}}$ for each $k$. We know that
$m_{k}=\dim(\pi({\mathcal{O}}_{k}^{\mathbb{C}},1))$, by our tables, and
$m_{k}\leq\\#{\EuScript C}_{k}$. Since $\sum
m_{k}=\\#\mathcal{W}=\sum\\#{\EuScript C}_{k}$ (obviously), we conclude

$m_{k}=\\#{\EuScript C}_{k}.$

By (4.1), part (1) is proved. Now (3.7) implies

(4.2) $\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))\geq
k\\}\subseteq\bigcup_{j\geq k}{\EuScript C}_{j},\text{ all }k.$

Then

(4.3) $\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}={\EuScript
C}_{k},\text{ for all }k=0,1,\dots,r,$

by the following induction argument. Use downward induction on $k$. Take $k=r$
in (4.2) to get $\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=r\\}\subseteq{\EuScript
C}_{r}$, but the two sets have the same cardinality, so are equal. For the
inductive hypothesis assume $\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=j\\}={\EuScript
C}_{j}$ holds for $j\geq k+1$ and suppose
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k$. Then
$L_{w}\in{\EuScript C}_{j}$ for some $j\geq k$. By induction,
$L_{w}\notin{\EuScript C}_{j}$ for any $j\geq k+1$, so $L_{w}\in{\EuScript
C}_{k}$. Again we get an inclusion $\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}\subseteq{\EuScript
C}_{k}$ for which both sides have the same cardinality, so equality holds and
(2) is proved.

For part (3), let

$\displaystyle S_{k}:=\\{w\in$
$\displaystyle\mathcal{W}\mid\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}})\text{
contains a maximal set of}$ $\displaystyle\text{ strongly orthogonal roots
having cardinality }k\\}$

and let

$M_{k}:=\\{w\in\mathcal{W}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}.$

Then $M_{k}\subseteq\cup_{j\geq k}S_{j}$. An argument very similar to the
induction above shows that $M_{k}=S_{k}$, for all $k$.

If $0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k$,
then, by part (2), Lemma 2.7 and (3.6) we have

$\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{k}}\subseteq\nu(w)\subset\operatorname{AV}(L_{w}),$

and part (4) follows. ∎

###  4.3. Non-simply laced cases: integral infinitesimal character

The combinatorics in the simply vs. non-simply laced cases has a difference
that is worth pointing out. As an example, consider ${\mathfrak{g}}$ of type
$B_{3}$. The Hasse diagram for
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{1}\\}\cup\\{\varepsilon_{1}\pm\varepsilon_{2},\varepsilon_{1}\pm\varepsilon_{3}\\}$
a linear chain. There are five lower order ideals of width $1$. Two contain a
pair of strongly orthogonal long roots and three contain just one; the width
is not the number of strongly orthogonal roots. So (3) of Proposition 4.2
fails. It turns out that for all five
$\operatorname{AV}(L_{w})=K\cdot(X_{\varepsilon_{1}-\varepsilon_{2}}+X_{\varepsilon_{1}+\varepsilon_{2}})=\overline{{\mathcal{O}}_{2}}$.
So part (2) fails in type $B_{3}$ for some $w\in\mathcal{W}$. We shall see
that (1) also fails. Similarly, these statements fail for types $B_{n}$ and
$C_{n}$, $n\geq 2$.

We prove the following for ${\mathfrak{g}}$ of type $B_{n}$ and $C_{n}$,
$n\geq 2$. Theorem 1.2 will follow (for all integral infinitesimal
characters).

######  Claim 4.6.

Let $w\in\mathcal{W}$ and
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k$. Then
the following hold.

  1. (1)

If ${\mathfrak{g}}$ is of type $B_{n}$, then
$\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{2k}}$.

  2. (2)

If ${\mathfrak{g}}$ is of type $C_{n}$, then

$\operatorname{AV}(L_{w})=\begin{cases}\overline{{\mathcal{O}}_{2k}},&2k\neq
n+1\\\ \overline{{\mathcal{O}}_{n}},&2k=n+1.\end{cases}$

###### Proof of (1).

In this case $\\#\mathcal{W}=2n$ and $r=2$. Table 2 says that
$\pi({\mathcal{O}}_{0}^{\mathbb{C}},1)$ and
$\pi({\mathcal{O}}_{2}^{\mathbb{C}},1)$ are special and
$\pi({\mathcal{O}}_{1}^{\mathbb{C}},1)$ is not special. Therefore, only
$\overline{{\mathcal{O}}_{0}}$ and $\overline{{\mathcal{O}}_{2}}$ can occur as
associated varieties. We know that ${\EuScript C}_{0}=\\{\mathbb{C}\\}$ , so
the remaining $2n-1$ $L_{w}$’s have associated variety
$\overline{{\mathcal{O}}_{2}}$. For each of these $w$’s the width is $1$ (by
Table 1). This proves (1) of the claim. We also conclude there is just one
cell of associated variety $\overline{{\mathcal{O}}_{2}}$, since such a cell
has size at least $\dim(\pi({\mathcal{O}}_{k}^{\mathbb{C}},1))=n$. ∎

###### Proof of (2).

This case is more involved; it requires more than the counting arguments. Here
$\\#\mathcal{W}=2^{n}$ and $r=n$. The $\pi({\mathcal{O}}_{k}^{\mathbb{C}},1)$
are special when either $k$ is even or $k=n$. We begin with a lemma about the
width of lower order ideals in $\Delta({\mathfrak{p}}^{+})$ for type $C_{n}$.

Consider the usual realization of
${\mathfrak{g}}={\mathfrak{sp}}(2n,\mathbb{C})$. Then ${\mathfrak{p}}^{+}$ is
identified with the symmetric $n\times n$ matrices
$\operatorname{Sym}(n,\mathbb{C})$. From the description of the $K$-orbits in
${\mathfrak{p}}^{+}$ (given in (3.5)) it follows that

$\displaystyle{\mathcal{O}}_{j}=\\{x\in\operatorname{Sym}(n,\mathbb{C})\mid\operatorname{rank}(x)=j\\},$
therefore (4.4)
$\displaystyle\overline{{\mathcal{O}}_{j}}=\\{x\in\operatorname{Sym}(n,\mathbb{C})\mid\operatorname{rank}(x)\leq
j\\}.$

∎

######  Lemma 4.7.

For type $C_{n}$, if
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k,$ then
$\nu(w)=\overline{{\mathcal{O}}_{2k}}$ or $\overline{{\mathcal{O}}_{2k-1}}$,
for $k=0,1,\dots,\lfloor\frac{n}{2}\rfloor$.

###### Proof.

Suppose $0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k$,
then $\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}})$ contains
$2\varepsilon_{n},2\varepsilon_{n-1},\dots,$ $2\varepsilon_{n-k+1}$ and does
not contain any roots $\varepsilon_{i}+\varepsilon_{j}$ with $i\leq k\leq j$.
Therefore, ${\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}$ is contained
in the space of matrices

$\left[\begin{array}[]{c|c}0&B\\\ \hline\cr
B^{t}&D\end{array}\right]\subseteq\operatorname{Sym}(n,\mathbb{C}),\text{ with
}D\in\operatorname{Sym}(k,\mathbb{C}).$

These matrices all have rank at most $2k$, so
$\nu(w)\subseteq\overline{{\mathcal{O}}_{2k}}$.

In §2.4.3 we gave the form of a maximal chain when the width is $k$.
Therefore, ${\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}$ contains
either

$\sum_{s=1}^{k-1}X_{\varepsilon_{n-k+1-s}+\varepsilon_{n-k+1+s}}+X_{2\varepsilon_{n-k+1}}\quad\text{
or }\quad\sum_{s=1}^{k}X_{\varepsilon_{n-k+1-s}+\varepsilon_{n-k+s}},$

matrices of rank either $2k-1$ or $2k$. So
$\nu(w)\supseteq\overline{{\mathcal{O}}_{2k-1}}$.

Therefore
$\overline{{\mathcal{O}}_{2k-1}}\subseteq\nu(w)\subseteq\overline{{\mathcal{O}}_{2k}}$.
∎

We need several known facts contained in [8] about the cell structure for type
$C_{n}$, which we collect here.

######  Lemma 4.8.

Let ${\mathfrak{g}}$ be of type $C_{n}$.

  * (a)

The highest weight Harish-Chandra modules of infinitesimal character $\rho$
are partitioned into cells:

$\displaystyle{\EuScript C}_{2k},\text{ having associated variety
}\overline{{\mathcal{O}}_{2k}},\ k=0,1,\dots,\lfloor n/2\rfloor\text{ and }$
$\displaystyle{\EuScript C}_{n},\text{ having associated variety
}\overline{{\mathcal{O}}_{n}},\text{ when $n$ is odd.}$

  * (b)

These cells have cardinalities

$\displaystyle\\#{\EuScript
C}_{2k}=\dim{\pi({\mathcal{O}}_{2k}^{\mathbb{C}},1)}+\dim{\pi({\mathcal{O}}_{2k-1}^{\mathbb{C}},1)}\text{
and }\\#{\EuScript
C}_{n}=\dim(\pi({\mathcal{O}}_{\frac{n+1}{2}}^{\mathbb{C}},1)).$

###### Proof.

For part (b), see [8, Prop. 3.7] and the preceding discussion in [8].

Now we prove the claim by showing

$\\{L_{w}:0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}=\begin{cases}{\EuScript
C}_{2k},&k=0,1,\dots,\lfloor\frac{n}{2}\rfloor\\\ {\EuScript
C}_{n},&k=\frac{n+1}{2}.\end{cases}$

Table 1 along with part (a) of the lemma imply

(4.5)
$\begin{split}&m_{k}=\dim{\pi({\mathcal{O}}_{2k}^{\mathbb{C}},1)}+\dim{\pi({\mathcal{O}}_{2k-1}^{\mathbb{C}},1)}=\\#{\EuScript
C}_{2k}\text{ and}\\\
&m_{\frac{n+1}{2}}=\dim(\pi({\mathcal{O}}_{\frac{n+1}{2}}^{\mathbb{C}},1))=\\#{\EuScript
C}_{n},\ n\text{ odd}.\end{split}$

If $\operatorname{AV}(L_{w})\subset\overline{{\mathcal{O}}_{2k}}$, then
$\nu(w)\subset\overline{{\mathcal{O}}_{2k}}$. Therefore, by Lemma 4.7,
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))\leq k$. It
follows that

$\bigcup_{j\leq k}{\EuScript C}_{2j}\subset\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))\leq
k\\}=\bigcup_{j\leq k}\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=j\\},$

for $k=0,1,\dots,\lfloor\frac{n}{2}\rfloor$. Equality holds by
(LABEL:eqn:numbers-cn). Part (2) of the claim now follows by induction on $k$
(for $k=0,1,\dots,\lfloor\frac{n}{2}\rfloor).$ The $k=0$ case is clear. If

$\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=j\\}={\EuScript
C}_{2j},\text{ for }j<k$

then

$\\{L_{w}\mid
0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=k\\}=\bigcup_{j\leq
k}{\EuScript C}_{2j}\smallsetminus\bigcup_{j<k}{\EuScript C}_{2j}={\EuScript
C}_{2k}.$

Finally, when $n$ is odd, for the remaining $L_{w}$’s we have
$0pt(\Delta({\mathfrak{n}}\cap\operatorname{Ad}(w){\mathfrak{n}}))=\frac{n+1}{2}$
and $\operatorname{AV}(L_{w})=\overline{{\mathcal{O}}_{n}}$. ∎

###  4.4. Half-integral case

Theorem 1.2 will now be proved for highest weight Harish-Chandra modules
having half-integral highest weight. As mentioned in the introduction, when
$\lambda$ is half-integral then either $L(\lambda)$ is irreducible (in which
case the associated variety is
$\overline{{\mathcal{O}}_{r}}={\mathfrak{p}}^{+}$) _or_
$\Delta=\Delta({\mathfrak{g}},{\mathfrak{h}})$ is non-simply laced. We assume
for the remainder of this section that $\Delta$ is non-simply laced and
$\lambda$ is half-integral.

The two cases to consider are ${\mathfrak{sp}}(2n,\mathbb{R})$ and
${\mathfrak{so}}(2,2n-1)$. The integral root systems $\Delta_{\lambda}$ and
corresponding Lie algebras ${\mathfrak{g}}_{\lambda}$ are described in §2.5.
Recall that $W_{\lambda}$ denotes the Weyl group of $\Delta_{\lambda}$ and we
have fixed the positive system
$\Delta_{\lambda}^{+}=\Delta^{+}\cap\Delta_{\lambda}$. Let $\Pi_{\lambda}$ be
the simple roots for this positive system. The Lie algebra
${\mathfrak{g}}_{\lambda}$ has a real form of Hermitian type and we my choose
corresponding group $G_{\lambda,\mathbb{R}}$ to be either $SO^{*}(2n)$ or
$SL(2,\mathbb{R})$. Then, since $\lambda$ is integral for $\Delta_{\lambda}$,
$L(\lambda)$ is an Harish-Chandra module for $G_{\lambda,\mathbb{R}}$.
Furthermore, letting $K_{\lambda}$ be the complexification of a maximal
compact subgroup, the associated varieties of highest weight Harish-Chandra
modules are the closures of the $K_{\lambda}$-orbits in
${\mathfrak{p}}_{\lambda}^{+}$, which we denote by
${\mathcal{O}}_{\lambda,0},{\mathcal{O}}_{\lambda,1},\dots$.

In our proof of Theorem 1.2 we first determine the Gelfand–Kirillov dimension
of $L(\lambda)$. This is done by applying Lusztig’s formula for the
Gelfand–Kirillov dimension given in terms of his $a$-function. Since the
possible associated varieties are the $\overline{{\mathcal{O}}_{k}}$,
$k=0,1,\dots,r$, which form an increasing sequence, the Gelfand–Kirillov
dimension determines the associated variety. The key fact is that the
$a$-function depends only on the Coxeter system $(W,\Pi)$, therefore we are
able to relate the Gelfand–Kirillov dimension of $L(\lambda)$ to the
Gelfand–Kirillov dimension of a highest weight Harish-Chandra module for
${\mathfrak{g}}_{\lambda,\mathbb{R}}$, then we may apply the integral case of
Theorem 1.2 proved in the preceding sections.

For Lusztig’s $a$-function see [22], [23], [5] and [3]. However, details about
the $a$-function are not needed here. Lusztig’s formula for the
Gelfand–Kirillov dimension may be stated as follows. If $\lambda+\rho$ is
regular and $w\in W_{\lambda}$ is chosen so that $w^{-1}(\lambda+\rho)$ is
antidominant, then

$\operatorname{GKdim}(L(\lambda))=\\#\Delta^{+}-a_{\lambda}(w),$ where
$a_{\lambda}$ is Lusztig’s $a$-function for $(W_{\lambda},\Pi_{\lambda})$. In
particular (4.6a) $\operatorname{GKdim}(L_{w})=\\#\Delta^{+}-a(w),$ where $a$
is the $a$-function for $(W,\Pi)$.

When $\lambda+\rho$ is singular, then choose $w$ to be the minimal length
element of $W_{\lambda}$ for which $w^{-1}(\lambda+\rho)$ is antidominant,
then

(4.6b) $\operatorname{GKdim}(L(\lambda))=\\#\Delta^{+}-a_{\lambda}(w).$

See [3, Prop. 1.1] for this formula.

######  Proposition 4.9.

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module and $\lambda$
is half-integral. If $0pt(Y_{\lambda})=m$, then

$\operatorname{GKdim}(L(\lambda))=(\\#(\Delta({\mathfrak{p}}^{+}))-\\#\Delta({\mathfrak{p}}_{\lambda}^{+}))+\dim({\mathcal{O}}_{\lambda,m}).$

###### Proof.

Choose $w\in W_{\lambda}$ to be the element of minimal length such that
$w^{-1}(\lambda+\rho)$ is antidominant. Then, by (4.6b),

$\operatorname{GKdim}(L(\lambda))=\\#(\Delta^{+})-a_{\lambda}(w),w\in
W_{\lambda}.$

Write $L_{w}^{\prime}$ for the irreducible highest weight module for
${\mathfrak{g}}_{\lambda}$ of highest weight
$-w\rho_{\lambda}-\rho_{\lambda}$, $w\in W_{\lambda}$. Then formula (4.6a)
(applied to ${\mathfrak{g}}_{\lambda}$) gives

$\operatorname{GKdim}(L_{w}^{\prime})=\\#(\Delta_{\lambda}^{+})-a_{\lambda}(w).$

Therefore,

(4.7)
$\begin{split}\operatorname{GKdim}(L(\lambda))&=(\\#(\Delta^{+})-\\#(\Delta_{\lambda}^{+}))+\operatorname{GKdim}(L_{w}^{\prime})\\\
&=(\\#(\Delta({\mathfrak{p}}^{+}))-\\#(\Delta({\mathfrak{p}}_{\lambda}^{+})))+\operatorname{GKdim}(L_{w}^{\prime}),\text{as
}\Delta({\mathfrak{k}})\subset\Delta_{\lambda}.\end{split}$

Our choice of $w\in W_{\lambda}$ guarantees that
$Y_{\lambda}=Y_{-w\rho_{\lambda}-\rho_{\lambda}}$ (by Proposition 2.5 applied
to $\Delta_{\lambda}$). Therefore, we have that
$0pt(Y_{-w\rho_{\lambda}-\rho_{\lambda}})=0pt(Y_{\lambda})=m$. By the integral
case we know that
$\operatorname{AV}(L_{w}^{\prime})=\overline{{\mathcal{O}}_{\lambda,m}}$ and
the proposition follows from (4.7). ∎

Theorem 1.2 now follows easily for the half-integral case by comparing
dimensions of orbits using Table 2.

Assume $0pt(Y_{\lambda})=m$.

(a) ${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{sp}}(2n,\mathbb{R}),n\geq 2$. The
real rank is $r=n$.

$\displaystyle\operatorname{GKdim}(L(\lambda))$
$\displaystyle=(n(n+1)-n(n-1))+\dim({\mathcal{O}}_{\lambda,m})$
$\displaystyle=2n+m(2n-2m-1),\text{ by Table \ref{tab:springer}},$
$\displaystyle=(n-m)(2m+1)$
$\displaystyle=\begin{cases}\dim({\mathcal{O}}_{2m+1}),&m<\frac{n}{2}\\\
\dim({\mathcal{O}}_{n}),&m=\frac{n}{2}.\end{cases}$

(b) ${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{so}}(2,2n-1),n\geq 2$. The real
rank is $r=2$.

$\displaystyle\operatorname{GKdim}(L(\lambda))$
$\displaystyle=((2n-1)-1)+\dim({\mathcal{O}}_{\lambda,m^{\prime}})$
$\displaystyle=2n-2+m$
$\displaystyle=\begin{cases}\dim({\mathcal{O}}_{1}),&m=0\\\
\dim({\mathcal{O}}_{2}),&m=1(=\frac{r}{2}).\end{cases}$

Therefore,

$\operatorname{AV}(L(\lambda))=\begin{cases}{\mathcal{O}}_{2m+1},&m<\frac{r}{2}\\\
{\mathcal{O}}_{r},&m=\frac{r}{2}\end{cases}$

in both cases.

##  5\. Another approach

In this section, we will give another proof for Theorem 1.2. In [3] and [4],
Bai–Xie and Bai–Xiao–Xie have found an algorithm to compute the
Gelfand–Kirillov dimensions of highest weight modules of classical Lie
algebras. We recall their results and will give another proof for our Theorem
1.2.

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
$x_{i}\in\Gamma$. We say $q=(q_{1},\cdots,q_{N})$ is the dual partition of a
partition $p=(p_{1},\cdots,p_{N})$ and write $q=p^{t}$ if $q_{i}$ is the
length of $i$-th column of the Young diagram $p$. Let $p(x)$ be the shape of
the Young tableau $Y(x)$ obtained by applying Robinson–Schensted algorithm to
$x\in\mathrm{Seq}_{n}(\Gamma)$. For convenience, we set $q(x)=p(x)^{t}$.

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. $p_{i}^{\mathrm{odd}}$) be the
number of even (resp. odd) boxes in the $i$-th row of the Young diagram $p$.
One can easily check that

(5.1)
$p_{i}^{\mathrm{ev}}=\begin{cases}\left\lceil\frac{p_{i}}{2}\right\rceil&\text{
if }i\text{ is odd},\\\ \left\lfloor\frac{p_{i}}{2}\right\rfloor&\text{ if
}i\text{ is even},\end{cases}\quad
p_{i}^{\mathrm{odd}}=\begin{cases}\left\lfloor\frac{p_{i}}{2}\right\rfloor&\text{
if }i\text{ is odd},\\\ \left\lceil\frac{p_{i}}{2}\right\rceil&\text{ if
}i\text{ is even}.\end{cases}$

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

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

For $x=(x_{1},x_{2},\cdots,x_{n})\in\mathrm{Seq}_{n}(\Gamma)$, 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 5.1 ([3, Thm. 6.4]).

Let $L(\lambda)$ be a highest weight Harish-Chandra module of $G_{\mathbb{R}}$
with $\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(p,q)$ and
$\lambda+\rho=(t_{1},\cdots,t_{n})\in\mathfrak{h}^{*}$. Set
$q=q(\lambda)=(q_{1},\cdots,q_{n})$. Then
$AV(L(\lambda))=\overline{\mathcal{O}_{k^{\prime}(\lambda)}}$ with

$k^{\prime}(\lambda)=\begin{cases}q_{2}&\text{ if }\lambda\ \text{is
integral},\\\ \min\\{p,q\\}&\text{ otherwise}.\end{cases}$

######  Proposition 5.2 ([4, Thm. 6.2]).

Let $L(\lambda)$ be a highest weight Harish-Chandra module of $G_{\mathbb{R}}$
(classical Hermitian type) with
$\lambda+\rho=(t_{1},\cdots,t_{n})\in\mathfrak{h}^{*}$. Set
$q=q(\lambda^{-})=(q_{1},\cdots,q_{2n})$. Then
$AV(L(\lambda))=\overline{\mathcal{O}_{k^{\prime}(\lambda)}}$ with
$k^{\prime}(\lambda)$ given as follows.

  * (1)

${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{p}}(2n,\mathbb{R})$ with
$n\geq 2$. Then

$k^{\prime}(\lambda)=\begin{cases}2q_{2}^{\mathrm{odd}}&\text{ if
}\lambda_{1}\in\mathbb{Z},\\\ 2q_{2}^{\mathrm{ev}}+1&\text{ if
}\lambda_{1}\in\frac{1}{2}+\mathbb{Z},\\\ n&\text{ otherwise}.\end{cases}$

Note that $q_{2}^{\mathrm{odd}}=[\frac{q_{2}+1}{2}]$ and
$q_{2}^{\mathrm{ev}}=[\frac{q_{2}}{2}]$.

  * (2)

${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}^{*}(2n)$ with $n\geq
4$. Then

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

  * (3)

${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}(2,2n-1)$ with $n\geq
3$. Then

$k^{\prime}(\lambda)=\begin{cases}0&\text{ if
}\lambda_{1}-\lambda_{2}\in\mathbb{Z},\lambda_{1}>\lambda_{2},\\\ 1&\text{ if
}\lambda_{1}-\lambda_{2}\in\frac{1}{2}+\mathbb{Z},\lambda_{1}>0,\\\ 2&\text{
otherwise}.\end{cases}$

  * (4)

${\mathfrak{g}}_{\mathbb{R}}={\mathfrak{s}}{\mathfrak{o}}(2,2n-2)$ with $n\geq
4$. Then

$k^{\prime}(\lambda)=\begin{cases}0&\text{ if
}\lambda_{1}-\lambda_{2}\in\mathbb{Z},\lambda_{1}>\lambda_{2},\\\ 1&\text{ if
}\lambda_{1}-\lambda_{2}\in\mathbb{Z},-|\lambda_{n}|<\lambda_{1}\leq\lambda_{2}\\\
2&\text{ otherwise}.\end{cases}$

Let $\Delta$ be the root system of $E_{6}$ or $E_{7}$. As usual $\Delta$ can
be realized as a subset of $\mathbb{R}^{8}$. The simple roots are

$\displaystyle\alpha_{1}$
$\displaystyle=\frac{1}{2}(\varepsilon_{1}-\varepsilon_{2}-\varepsilon_{3}-\varepsilon_{4}-\varepsilon_{5}-\varepsilon_{6}-\varepsilon_{7}+\varepsilon_{8}),$
$\displaystyle\alpha_{2}$ $\displaystyle=\varepsilon_{1}+\varepsilon_{2},$
$\displaystyle\alpha_{k}$
$\displaystyle=\varepsilon_{k-1}-\varepsilon_{k-2}\text{ with }3\leq k\leq n,$

where $n=6,7$. Here we use notations in [12]. Let $\Pi$ denote the set of
simple roots in $\Delta^{+}$. Exactly one root in $\Pi$ is in
$\Delta({\mathfrak{p}}^{+})$. In particular, for $\mathfrak{e}_{6(-14)}$, this
simple root is $\alpha_{1}$; for $\mathfrak{e}_{7(-25)}$, this simple root is
$\alpha_{7}$.

For an integral weight $\lambda\in\mathfrak{h}^{*}$, recall that
$\Delta_{\lambda}^{+}=\\{\alpha\in\Delta^{+}\mid\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0\\}$.

######  Proposition 5.3 ([3, Thm. 7.1]).

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

  * (1)

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$, we have

$k^{\prime}(\lambda)=\begin{cases}0&\text{ if }\Delta_{\lambda}^{+}\cap
A_{1}\neq\emptyset,\\\ 1&\text{ if }\Delta_{\lambda}^{+}\cap
A_{2}\neq\emptyset\text{ and }\Delta_{\lambda}^{+}\cap A_{1}=\emptyset,\\\
2&\text{ if }\Delta_{\lambda}^{+}\cap A_{2}=\emptyset,\end{cases}$

where
$A_{2}=\\{\frac{1}{2}(\pm(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}-\varepsilon_{4})-\varepsilon_{5}-\varepsilon_{6}-\varepsilon_{7}+\varepsilon_{8})\\}$
and $A_{1}=\\{\alpha_{1}\\}$.

  * (2)

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$, we have

$k^{\prime}(\lambda)=\begin{cases}0&\text{ if }\Delta_{\lambda}^{+}\cap
A_{1}\neq\emptyset,\\\ 1&\text{ if }\Delta_{\lambda}^{+}\cap
A_{2}\neq\emptyset\text{ and }\Delta_{\lambda}^{+}\cap A_{1}=\emptyset,\\\
2&\text{ if }\Delta_{\lambda}^{+}\cap A_{3}\neq\emptyset\text{ and
}\Delta_{\lambda}^{+}\cap A_{2}=\emptyset,\\\ 3&\text{ if
}\Delta_{\lambda}^{+}\cap A_{3}=\emptyset,\end{cases}$

where
$A_{3}=\\{\frac{1}{2}(\pm(\varepsilon_{1}-\varepsilon_{2}-\varepsilon_{3}+\varepsilon_{4})-\varepsilon_{5}+\varepsilon_{6}-\varepsilon_{7}+\varepsilon_{8}),\varepsilon_{5}+\varepsilon_{6}\\}$,
$A_{2}=\\{\pm\varepsilon_{1}+\varepsilon_{6}\\}$ and
$A_{1}=\\{-\varepsilon_{5}+\varepsilon_{6}\\}$.

######  Lemma 5.4.

Suppose $\mu=(t_{1},...,t_{n},s_{1},...,s_{m})$ is a $(n,m)$-dominant sequence
with $t_{n}\leq s_{1}$. Then applying the Schensted insertion algorithm to
$\mu$. We can get a Young tableau $Y(\mu)$ with two columns. And
$c_{2}(Y(\mu))$ is precisely the largest integer $k$ for which we have

$t_{n-k+1}\leq s_{1},...,t_{n-1}\leq s_{k-1},t_{n}\leq s_{k}.$

The integer $k$ is also equal to the maximum integer $k_{1}$ for which there
exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{k_{1}}\leq n,1\leq l_{1}<l_{2}<...<l_{k_{1}}\leq m,$

such that

(5.2) $t_{i_{1}}\leq s_{l_{1}},...,t_{i_{k_{1}}}\leq s_{l_{k_{1}}}.$

###### Proof.

In [3], a white-black pair model was constructed to compute $c_{2}(Y(\mu))$.
From this model and [3, Prop. 5.3 and Thm. 5.6], we know
$k(\mu):=c_{2}(Y(\mu))$ is equal to the maximum integer $k$ for which we can
divide the sequence $(s_{1},...,s_{m})$ into several parts such that
$t_{n}\leq s_{j_{k}}$ and $t_{n}>s_{j_{k}+1}$,…,$t_{n-k+1}\leq s_{j_{1}}$ and
$t_{n-k+1}>s_{j_{1}+1}$, with $1\leq j_{1}<j_{2}<...<j_{k}\leq m$. Note that
$j_{k}$ is the first index sucht $t_{n}\leq s_{j}$.

We denote a maximum integer $k_{1}$ for which there exists a sequence of
indices

$1\leq i_{1}<i_{2}<...<i_{k_{1}}\leq n,1\leq l_{1}<l_{2}<...<l_{k_{1}}\leq m,$

such that

(5.3) $t_{i_{1}}\leq s_{l_{1}},...,t_{i_{k_{1}}}\leq s_{l_{k_{1}}}.$

So we have $k\leq k_{1}$. We suppose $k<k_{1}$.

Since $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $1\leq i<j\leq n$, we can make a new
choice for these indices such that

$i_{1}=n-k_{1}+1,i_{2}=n-k_{1},...,i_{k_{1}}=n.$

Then we still have the condition (5.3):

$t_{n-k_{1}+1}\leq s_{l_{1}},...,t_{n}\leq s_{l_{k_{1}}}.$

From our white-black pair model, we can choose $l_{k_{1}}=j_{k}$ since
$t_{n}\leq s_{j_{k}}$ and $t_{n}>s_{j_{k}+1}$ (i.e., $j_{k}$ is the first
index such that $t_{n}\leq s_{j}$). Similarly we can choose
$l_{k_{1}-1}=j_{k-1}$,…,$l_{k_{1}-k+1}=j_{1}$. Then we choose
$l_{k_{1}-k}=j^{\prime}_{1}$ such that $t_{n-k}\leq
s_{j^{\prime}_{1}}=s_{l_{k_{1}-k}}$ and $t_{n-k}>s_{j^{\prime}_{1}+1}$. We
continue this process until we choose $l_{1}=j^{\prime}_{k_{1}-k}$ such that
$t_{n-k_{1}+1}\leq s_{j^{\prime}_{k_{1}-k}}=s_{l_{1}}$ and
$t_{n-k_{1}+1}>s_{j^{\prime}_{k_{1}-k+1}}$. Then from the definition of $k$,
we must have $k_{1}\leq k$, which is a contradiction. So we have $k=k_{1}$.

Since $s_{i}-s_{j}\in\mathbb{Z}_{>0}$ for $1\leq i<j\leq m$, we make a new
choice such that

$j_{1}=1,...,j_{k}=k.$

Then we get

$t_{n-k+1}\leq s_{1},...,t_{n-1}\leq s_{k-1},t_{n}\leq s_{k}.$

Since this process is invertible, we can see that $k$ is precisely the largest
number satisfying the above condition.

∎

######  Corollary 5.5.

Suppose $\lambda=(t_{1},...,t_{n},-t_{n},...,-t_{1})$ is a $(n,n)$-dominant
sequence with $t_{n}\leq-t_{n}$. Then applying the Schensted insertion
algorithm to $\lambda$. We can get a Young tableau $Y_{\lambda}$ with two
columns. And $c_{2}(Y_{\lambda})$ is precisely the largest integer $t$ for
which we have

$t_{n-t+1}\leq-t_{n},t_{n-t+2}\leq-t_{n-1}...,t_{n-1}\leq-t_{n-t+2},t_{n}\leq-
t_{n-t+1}.$

Now we will give our proof for Theorem 1.2 by the algorithms in [3, 4]. In the
following we denote $m(\lambda):=\operatorname{width}(Y_{\lambda})$. We will
show that the values of $k^{\prime}(\lambda)$ in Proposition 5.1, Proposition
5.2 and Proposition 5.3 will equal to the values of $k(\lambda)$ given in
Theorem 1.2.

###  5.1. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(p,q)$

We use the notations from [12]. Suppose $L(\lambda)$ is a highest weight
Harish-Chandra ($\mathfrak{g}$,$K$)-module with highest weight $\lambda$. We
denote $\lambda+\rho=(t_{1},...,t_{n})$. Suppose $\lambda$ is integral. Then
from [12], we have $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $1\leq i\leq p$,
$p+1\leq j\leq n$. From [3], we say that $(t_{1},...,t_{n})$ is
$(p,q)$-dominant.

By using Schensted insertion algorithm for $(t_{1},...,t_{n})$, we can get a
Young tableau $Y_{\lambda}$ which consists of at most two columns. Now we
suppose the number of entries in the second column of $Y_{\lambda}$ is
$c_{2}(Y_{\lambda})=q_{2}$, then $c_{1}(Y_{\lambda})=n-q_{2}$. From
Proposition 5.1, we know $AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}}}.$

Then from Lemma 5.4, we know $q_{2}$ is equal to the maximum integer $t$ for
which there exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{t}\leq p,~{}p+1\leq j_{1}<j_{2}<...<j_{t}\leq n,$

such that

$t_{i_{1}}\leq t_{j_{1}},...,t_{i_{t}}\leq t_{j_{t}}.$

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(p,q)$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}-\varepsilon_{j}|1\leq i\leq
p,~{}p+1\leq j\leq n\\}$. So $m=q_{2}$ is just our $m(\lambda)$ in Theorem
1.2.

Thus
$AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}}}=\overline{\mathcal{O}_{m}}=\overline{\mathcal{O}_{m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

When $\lambda$ is non-integral, we will have $N(\lambda)=L(\lambda)$ by [12,
Lem. 3.17]. So $AV(L(\lambda))=\overline{\mathcal{O}_{r}}$.

###  5.2. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(2n,\mathbb{R})$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{n})$. Suppose
$\lambda$ is integral. Then from [12], we have $t_{i}-t_{j}\in\mathbb{Z}_{>0}$
for $1\leq i<j\leq n$. From [3], we say that
$(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$ is $(n,n)$-dominant.

By using Schensted insertion algorithm for
$(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$, we can get a Young
tableau $Y(\lambda^{-})$ which consists of at most two columns. Now we suppose
the number of entries in the second column of $Y(\lambda^{-})$ is
$c_{2}(Y(\lambda^{-}))=q_{2}$, then $c_{1}(Y(\lambda^{-}))=2n-q_{2}$. From
Proposition 5.2, we know

$AV(L(\lambda))=\overline{\mathcal{O}_{2q_{2}^{\mathrm{odd}}}}=\left\\{\begin{array}[]{ll}\overline{\mathcal{O}_{q_{2}+1}},&\hbox{if
$q_{2}$ is odd,}\\\ \overline{\mathcal{O}_{q_{2}}},&\hbox{if $q_{2}$ is
even.}\end{array}\right.$

Then from Lemma 5.4, we know $q_{2}$ is equal to the maximum integer $t$ for
which there exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{t}\leq n,n\geq j_{1}>j_{2}>...>j_{t}\geq 1,$

such that

$t_{i_{1}}\leq-t_{j_{1}},...,t_{i_{t}}\leq-t_{j_{t}}.$

Since $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $1\leq i<j\leq n$, we can make a new
choice for these indices such that

$i_{1}=j_{t},i_{2}=j_{t-1},...,i_{t}=j_{1}.$

Thus we have

$t_{i_{1}}\leq-t_{i_{t}},t_{i_{2}}\leq-t_{i_{t-1}},...,t_{i_{t}}\leq-
t_{i_{1}}.$

When $q_{2}=t=2t^{\prime}+1$ is an odd number, we will have

$t_{i_{1}}+t_{i_{t}}\leq 0,t_{i_{2}}+t_{i_{t-1}}\leq
0,...,t_{i_{t^{\prime}}}+t_{i_{t^{\prime}+2}}\leq 0,t_{i_{t^{\prime}+1}}\leq
0.$

From from Lemma 5.4, we know $m=t^{\prime}$ is the maximum integer for which
there exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{m}<l<j_{m}<...<j_{1}\leq n,$

such that

$t_{i_{1}}+t_{j_{1}}\leq 0,...,t_{i_{m}}+t_{j_{m}}\leq 0,t_{l}\leq 0.$

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(2n,\mathbb{R})$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}+\varepsilon_{j}|1\leq i<j\leq
n\\}\cup\\{2\varepsilon_{l}|1\leq l\leq n\\}$. So $m+1$ is just our
$m(\lambda)$ in Theorem 1.2.

Thus $q_{2}+1=t+1=2(m+1)$ and
$AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}+1}}=\overline{\mathcal{O}_{2(m+1)}}=\overline{\mathcal{O}_{2m(\lambda)}}==\overline{\mathcal{O}_{k(\lambda)}}$.

When $q_{2}=t=2t^{\prime}$ is an even number, we will have

$t_{i_{1}}+t_{i_{t}}\leq 0,t_{i_{2}}+t_{i_{t-1}}\leq
0,...,t_{i_{t^{\prime}}}+t_{i_{t^{\prime}+1}}\leq 0.$

From the definition of $t$, we know $m=t^{\prime}$ is the maximum integer for
which there exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{m}<l<j_{m}<...<j_{1}\leq n,$

such that

$t_{i_{1}}+t_{j_{1}}\leq 0,...,t_{i_{m}}+t_{j_{m}}\leq 0,t_{l}\leq 0.$

So $m$ is just our $m(\lambda)$ in Theorem 1.2.

Thus $q_{2}=t=2m$ and
$AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}}}=\overline{\mathcal{O}_{2m}}=\overline{\mathcal{O}_{2m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

Now we suppose $\lambda$ is half-integral. Then from [12], we have
$t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $1\leq i<j\leq n$ and
$t_{k}-\frac{1}{2}\in\mathbb{Z}$ for $1\leq k\leq n$. By using Schensted
insertion algorithm for $(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$,
we can get a Young tableau $Y(\lambda^{-})$ which consists of at most two
columns. Now we suppose the number of entries in the second column of
$Y(\lambda^{-})$ is $c_{2}(Y(\lambda^{-}))=q_{2}$, then
$c_{1}(Y(\lambda^{-}))=2n-q_{2}$. From Proposition 5.2, we know

$AV(L(\lambda))=\overline{\mathcal{O}_{2q_{2}^{\mathrm{ev}}+1}}=\left\\{\begin{array}[]{ll}\overline{\mathcal{O}_{q_{2}}},&\hbox{if
$q_{2}$ is odd,}\\\ \overline{\mathcal{O}_{q_{2}+1}},&\hbox{if $q_{2}$ is
even.}\end{array}\right.$

When $q_{2}=t=2t^{\prime}+1$ is an odd number, similarly to the integral case,
we will have

$t_{i_{1}}+t_{i_{t}}\leq 0,t_{i_{2}}+t_{i_{t-1}}\leq
0,...,t_{i_{t^{\prime}}}+t_{i_{t^{\prime}+2}}\leq 0,t_{i_{t^{\prime}+1}}\leq
0,$

and $m=t^{\prime}$ is the maximum integer for which there exists a sequence of
indices

$1\leq i_{1}<i_{2}<...<i_{m}<l<j_{m}<...<j_{1}\leq n,$

such that

$t_{i_{1}}+t_{j_{1}}\leq 0,...,t_{i_{m}}+t_{j_{m}}\leq 0,t_{l}\leq 0.$

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(2n,\mathbb{R})$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}+\varepsilon_{j}|1\leq i<j\leq
n\\}\cup\\{2\varepsilon_{l}|1\leq l\leq n\\}$. Since $t_{l}\notin\mathbb{Z}$,
$m$ is just our $m(\lambda)$ in Theorem 1.2.

Thus $q_{2}=t=2m+1$ and
$AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}}}=\overline{\mathcal{O}_{2m+1}}=\overline{\mathcal{O}_{2m(\lambda)+1}}=\overline{\mathcal{O}_{k(\lambda)}}$.

When $q_{2}=t=2t^{\prime}$ is an even number, similarly to above case, we will
have $m=m(\lambda)=t^{\prime}$ and $q_{2}+1=2m+1$. Thus
$AV(L(\lambda))=\overline{\mathcal{O}_{q_{2}+1}}=\overline{\mathcal{O}_{2m+1}}=\overline{\mathcal{O}_{2m(\lambda)+1}}=\overline{\mathcal{O}_{k(\lambda)}}$.

###  5.3. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}^{*}(2n)$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{n})$. Then from [12,
Lem. 3.17], we may suppose $\lambda$ is integral (otherwise $N(\lambda)$ will
be irreducible). Also we have $t_{1}+t_{2}\in\mathbb{Z}$ and
$t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $1\leq i<j\leq n$. Then we have

$t_{1}>t_{2}>...>t_{n-1}>t_{n},-t_{n}>-t_{n-1}>...>-t_{1}.$

From [3], we say that $(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$ is
$(n,n)$-dominant. When $t_{n}\geq 0$, $L(\lambda)$ will be finite-dimensional.
We only consider infinite-dimensional modules. Thus we suppose $t_{n}<0$.

By using Schensted insertion algorithm for
$(t_{1},\dots,t_{n},-t_{n},-t_{n-1},\ldots,-t_{2},-t_{1})$, we can get a Young
tableau $Y(\lambda^{-})$ which consists of at most two columns. Now we suppose
the number of entries in the second column of $Y(\lambda^{-})$ is
$c_{2}(Y(\lambda^{-}))=q_{2}$, then $c_{1}(Y(\lambda^{-}))=2n-q_{2}$. From
Proposition 5.2, we know

$AV(L(\lambda))=\overline{\mathcal{O}_{[\frac{q_{2}}{2}]}}=\left\\{\begin{array}[]{ll}\overline{\mathcal{O}_{\frac{q_{2}-1}{2}}},&\hbox{if
$q_{2}$ is odd,}\\\ \overline{\mathcal{O}_{\frac{q_{2}}{2}}},&\hbox{if $q_{2}$
is even.}\end{array}\right.$

Then similar to the case of ${\mathfrak{s}}{\mathfrak{p}}(2n,\mathbb{R})$, we
know $q_{2}$ is equal to the maximum integer $t$ for which there exists a
sequence of indices

$1\leq i_{1}<i_{2}<...<i_{t}\leq n,$

such that

$t_{i_{1}}\leq-t_{i_{t}},t_{i_{2}}\leq-t_{i_{t-1}},...,t_{i_{t}}\leq-
t_{i_{1}}.$

When $q_{2}=t=2t^{\prime}+1$ is an odd number, we will have

$t_{i_{1}}+t_{i_{t}}\leq 0,t_{i_{2}}+t_{i_{t-1}}\leq
0,...,t_{i_{t^{\prime}}}+t_{i_{t^{\prime}+2}}\leq 0,t_{i_{t^{\prime}+1}}\leq
0.$

From the definition of $t$, we know $m=t^{\prime}$ is the maximum integer for
which there exists a sequence of indices

$1\leq i_{1}<i_{2}<...<i_{m}<j_{m}<...<j_{1}\leq n,$

such that

$t_{i_{1}}+t_{j_{1}}\leq 0,...,t_{i_{m}}+t_{j_{m}}\leq 0.$

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}^{*}(2n)$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{i}+\varepsilon_{j}|1\leq i<j\leq
n\\}$. So $m=t^{\prime}=\frac{q_{2}-1}{2}$ is just our $m(\lambda)$ in Theorem
1.2. Thus
$AV(L(\lambda))=\overline{\mathcal{O}_{\frac{q_{2}-1}{2}}}=\overline{\mathcal{O}_{m}}=\overline{\mathcal{O}_{m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

When $q_{2}=t=2t^{\prime}$ is an even number, similarly we will have
$AV(L(\lambda))=\overline{\mathcal{O}_{\frac{q_{2}}{2}}}=\overline{\mathcal{O}_{m}}=\overline{\mathcal{O}_{m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

###  5.4. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-1)$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{n})$. Suppose
$\lambda$ is integral. Then from [12], we have $t_{n}>0$,
$t_{1}-t_{2}\in\mathbb{Z}$, $2t_{k}\in\mathbb{Z}$ 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
$t_{2}>t_{3}>...>t_{n}>-t_{n}>...>-t_{2}$. When $t_{1}>t_{2}$, $L(\lambda)$
will be finite-dimensional and $V(L(\lambda))=\overline{\mathcal{O}}_{0}$. We
only consider infinite-dimensional modules. Thus we suppose $t_{1}\leq t_{2}$.

By using Schensted insertion algorithm for
$(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$, we can get a Young
tableau $Y(\lambda^{-})$ which consists of at most three columns. From the
construction process, we can see that $Y(\lambda^{-})$ will be a Young tableau
consisting of two columns with $c_{1}(Y(\lambda^{-}))=2n-2$ and
$c_{2}(Y(\lambda^{-}))=2$ when $t_{2}\geq t_{1}>-t_{n}$, and it will be a
Young tableau consisting of three columns with $c_{1}(Y(\lambda^{-}))=2n-2$,
$c_{2}(Y(\lambda^{-}))=1$ and $c_{3}(Y(\lambda^{-}))=1$ when $t_{1}\leq-
t_{n}$.

From Proposition 5.2, we have $AV(L(\lambda))=\overline{\mathcal{O}_{2}}$ for
both cases.

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-1)$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{1}\pm\varepsilon_{j}|2\leq i<j\leq
n\\}\cup\\{\varepsilon_{l}|1\leq l\leq n\\}$. When $t_{2}\geq t_{1}>-t_{n}$,
we have $t_{1}-t_{2}\leq 0$. So $m=1$ is just our $m(\lambda)$ in Theorem 1.2
and
$AV(L(\lambda))=\overline{\mathcal{O}_{2}}=\overline{\mathcal{O}_{2m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

When $t_{1}\leq-t_{n}$, we have $t_{1}+t_{n}\leq 0$. So $m=1$ is just our
$m(\lambda)$ in Theorem 1.2 and
$AV(L(\lambda))=\overline{\mathcal{O}_{2}}=\overline{\mathcal{O}_{2m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

Now we suppose $\lambda$ is half-integral. Then from Enright–Howe–Wallach
[12], 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 $t_{2}>t_{3}>...>t_{n}>-t_{n}>...>-t_{2}$.

From Proposition 5.2, we have
$AV(L(\lambda))=\left\\{\begin{array}[]{ll}\overline{\mathcal{O}_{1}},&\hbox{if
$t_{1}>0$,}\\\ \overline{\mathcal{O}_{2}},&\hbox{if $t_{1}\leq
0$.}\end{array}\right.$

When $t_{1}>0$, we find that $m(\lambda)=0$. So
$AV(L(\lambda))=\overline{\mathcal{O}_{1}}=\overline{\mathcal{O}_{2m(\lambda)+1}}=\overline{\mathcal{O}_{k(\lambda)}}$.

When $t_{1}\leq 0$, we have $m(\lambda)=1$ and $2m(\lambda)+1=3>2$. We still
have
$AV(L(\lambda))=\overline{\mathcal{O}_{2}}=\overline{\mathcal{O}_{k(\lambda)}}$.

###  5.5. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-2)$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{n})$. Then from [12,
Lem. 3.17], we may suppose $\lambda$ is integral (otherwise $N(\lambda)$ will
be irreducible). Also we have $t_{1}-t_{2}\in\mathbb{Z}$, $t_{n-1}-|t_{n}|>0$
and $t_{i}-t_{j}\in\mathbb{Z}_{>0}$ for $2\leq i<j\leq n$. Then we have

$t_{2}>t_{3}>...>t_{n-1}>|t_{n}|\geq 0\geq-|t_{n}|>-t_{n-1}>...>-t_{2}.$

When $t_{1}>t_{2}$, $L(\lambda)$ will be finite-dimensional and
$AV(L(\lambda))=\overline{\mathcal{O}_{0}}$. We only consider infinite-
dimensional modules. Thus we suppose $t_{1}\leq t_{2}$.

By using Schensted insertion algorithm for
$(t_{1},...,t_{n},-t_{n},-t_{n-1},...,-t_{2},-t_{1})$, we can get a Young
tableau $Y(\lambda^{-})$ which consists of at most four columns.

From Proposition 5.2, when $-|t_{n}|<t_{1}\leq t_{2}$, then
$q(\lambda^{-})=(2n-2,2)$ or $(2n-3,3)$, we get
$AV(L(\lambda))=\overline{\mathcal{O}_{1}}$.

For $\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-2)$, we know
$\Delta({\mathfrak{p}}^{+})=\\{\varepsilon_{1}\pm\varepsilon_{j}|2\leq i<j\leq
n\\}$. For $t_{n}>0$ and $t_{2}\geq t_{1}>-t_{n}$, we have $t_{1}-t_{2}\leq 0$
and $t_{1}+t_{n}>0$. For $t_{n}\leq 0$ and $t_{2}\geq t_{1}>t_{n}$, we have
$t_{1}-t_{2}\leq 0$ and $t_{1}-t_{n}>0$. So for both cases, $m=1$ is just our
$m(\lambda)$ in Theorem 1.2 and
$AV(L(\lambda))=\overline{\mathcal{O}_{1}}=\overline{\mathcal{O}_{m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

From Proposition 5.2, when $t_{1}\leq-|t_{n}|$, then
$q(\lambda^{-})=(2n-2,1^{2})$ or $(2n-3,1^{3})$, one has
$AV(L(\lambda))=\overline{\mathcal{O}_{2}}$.

When $t_{1}\leq-t_{n}<0$, we have $t_{1}+t_{n}\leq 0$ and $t_{1}-t_{n}\leq 0$.
When $t_{1}\leq t_{n}\leq 0$, we have $t_{1}-t_{n}\leq 0$ and $t_{1}+t_{n}\leq
0$. So for both cases, $m=2$ is just our $m(\lambda)$ in Theorem 1.2 and
$AV(L(\lambda))=\overline{\mathcal{O}_{2}}=\overline{\mathcal{O}_{m(\lambda)}}=\overline{\mathcal{O}_{k(\lambda)}}$.

###  5.6. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{8})$. Then from [12,
Lem. 3.17], we may suppose $\lambda$ is integral (otherwise $N(\lambda)$ will
be irreducible). Also we have $|t_{1}|<t_{2}<...<t_{5}$,
$t_{i}-t_{j}\in\mathbb{Z}$ and $2t_{i}\in\mathbb{Z}$ for all $1\leq i,j\leq
5$.

The Hasse diagram of $\Delta({\mathfrak{p}}^{+})$ for
$\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$ is shown in the Appendix.

The highest root is
$\theta=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+3\alpha_{4}+2\alpha_{5}+\alpha_{6}=\frac{1}{2}(1,1,1,1,1,-1,-1,1)$.

We denote

$\beta_{1}=\alpha_{1}+\alpha_{3}+\alpha_{4}+\alpha_{5}=\frac{1}{2}(-1,-1,-1,1,-1,-1,-1,1)$

and

$\beta_{2}=\alpha_{1}+\alpha_{3}+\alpha_{4}+\alpha_{2}=\frac{1}{2}(1,1,1,-1,-1,-1,-1,1).$

From Proposition 5.3, we have$A_{1}=\\{\alpha_{1}\\}$ and
$A_{2}=\\{\beta_{1},\beta_{2}\\}$.

If $\Delta_{\lambda}^{+}\cap A_{1}\neq\emptyset$, we will have
$\langle\lambda+\rho\,,\alpha_{1}^{\vee}\rangle>0$. Since $\alpha_{1}$ is the
minimal noncompact root in $\Delta({\mathfrak{p}}^{+})$, we will have
$\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0$ for all
$\alpha\in\Delta({\mathfrak{p}}^{+})$ and $m(\lambda)=0$. Thus $L(\lambda)$
will be finite-dimensional and $AV(L(\lambda))=\overline{\mathcal{O}_{0}}$. So
$k(\lambda)=m(\lambda)=0$.

If $\Delta_{\lambda}^{+}\cap A_{1}=\emptyset$ and $\Delta_{\lambda}^{+}\cap
A_{2}\neq\emptyset$, we will have
$\langle\lambda+\rho\,,\alpha_{1}^{\vee}\rangle\leq 0$ and there is at least
one root of $A_{2}$ satisfying $\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0$.
So we have $m(\lambda)\geq 1$. From the diagram of
$\Delta({\mathfrak{p}}^{+})$ for
$\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}$, we find that if an
antichain $A\subseteq\Delta({\mathfrak{p}}^{+})$ has $m(\lambda)=2$, it must
contain $\beta_{1}$ and $\beta_{2}$ since they are the smallest incompatible
pair in $\Delta({\mathfrak{p}}^{+})$. So we must have $m(\lambda)=1$ since
$\Delta_{\lambda}^{+}\cap A_{2}\neq\emptyset$. Thus $k(\lambda)=m(\lambda)=1$.

If $\Delta_{\lambda}^{+}\cap A_{2}=\emptyset$, $\beta_{1}$ and $\beta_{2}$
will satisfy that $\langle\lambda+\rho\,,\alpha^{\vee}\rangle\leq 0$. Thus, we
have $m(\lambda)\geq 2$. The equality must hold since $m(\lambda)\leq 2$ by
the diagram of $\Delta({\mathfrak{p}}^{+})$. So $k(\lambda)=m(\lambda)=2$.

###  5.7. $\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$

Suppose $L(\lambda)$ is a highest weight Harish-Chandra module with highest
weight $\lambda$. We denote $\lambda+\rho=(t_{1},...,t_{8})$. Then from [12,
Lem. 3.17], we may suppose $\lambda$ is integral (otherwise $N(\lambda)$ will
be irreducible). Also we have $|t_{1}|<t_{2}<...<t_{5}$,
$t_{i}-t_{j}\in\mathbb{Z}$ and $2t_{i}\in\mathbb{Z}$ for all $1\leq i,j\leq
5$.

The Hasse diagram of $\Delta({\mathfrak{p}}^{+})$ for
$\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$ is shown in the Appendix.

The highest root is
$\theta=2\alpha_{1}+2\alpha_{2}+3\alpha_{3}+4\alpha_{4}+3\alpha_{5}+2\alpha_{6}+\alpha_{7}$.

We denote

$\beta_{1}=\alpha_{7}+\alpha_{3}+\alpha_{4}+\alpha_{5}+\alpha_{6}=(-1,0,0,0,0,1,0,0),$
$\beta_{2}=\alpha_{7}+\alpha_{2}+\alpha_{4}+\alpha_{5}+\alpha_{6}=(1,0,0,0,0,1,0,0),$
$\gamma_{1}=\beta_{1}+\alpha_{2}+\alpha_{4}+\alpha_{1}+\alpha_{3}=\frac{1}{2}(1,-1,-1,1,-1,1,-1,1),$
$\gamma_{2}=\beta_{1}+\alpha_{1}+\alpha_{2}+\alpha_{4}+\alpha_{5}=\frac{1}{2}(1,-1,-1,1,-1,1,-1,1),$

and

$\gamma_{3}=\beta_{1}+\alpha_{2}+\alpha_{4}+\alpha_{5}+\alpha_{6}=(0,0,0,0,1,1,0,0).$

From Proposition 5.3, we have $A_{1}=\\{\alpha_{7}\\}$,
$A_{2}=\\{\beta_{1},\beta_{2}\\}$, and
$A_{3}=\\{\gamma_{1},\gamma_{2},\gamma_{3}\\}$.

If $\Delta_{\lambda}^{+}\cap A_{1}\neq\emptyset$, we will have
$\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0$. Since $\alpha_{7}$ is the
minimal noncompact root in $\Delta({\mathfrak{p}}^{+})$, we will have
$\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0$ for all
$\alpha\in\Delta({\mathfrak{p}}^{+})$ and $m(\lambda)=0$. Thus $L(\lambda)$
will be finite-dimensional and $AV(L(\lambda))=\overline{\mathcal{O}_{0}}$. So
$k(\lambda)=m(\lambda)=0$.

If $\Delta_{\lambda}^{+}\cap A_{1}=\emptyset$ and $\Delta_{\lambda}^{+}\cap
A_{2}\neq\emptyset$, we will have
$\langle\lambda+\rho\,,\alpha_{7}^{\vee}\rangle\leq 0$ and there is at least
one root of $A_{2}$ satisfying $\langle\lambda+\rho\,,\alpha^{\vee}\rangle>0$.
So we have $m(\lambda)\geq 1$. From the diagram of
$\Delta({\mathfrak{p}}^{+})$ for
$\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$, we find that if an
antichain $A\subseteq\Delta({\mathfrak{p}}^{+})$ has $m(\lambda)=2$, it must
contain $\beta_{1}$ and $\beta_{2}$ since they are the smallest incompatible
pair in $\Delta({\mathfrak{p}}^{+})$. So we must have $m(\lambda)=1$ since
$\Delta_{\lambda}^{+}\cap A_{2}\neq\emptyset$. Thus $k(\lambda)=m(\lambda)=1$.

If $\Delta_{\lambda}^{+}\cap A_{2}=\emptyset$ and $\Delta_{\lambda}^{+}\cap
A_{3}\neq\emptyset$, $\beta_{1}$ and $\beta_{2}$ will satisfy that
$\langle\lambda+\rho\,,\alpha^{\vee}\rangle\leq 0$. Thus, we have
$m(\lambda)\geq 2$. From the diagram of $\Delta({\mathfrak{p}}^{+})$ for
$\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}$, we find that if an
antichain $A\subseteq\Delta({\mathfrak{p}}^{+})$ has $m(\lambda)=3$, it must
contain $\gamma_{1}$, $\gamma_{2}$ and $\gamma_{3}$ since they are the
smallest incompatible triples in $\Delta({\mathfrak{p}}^{+})$. So we must have
$m(\lambda)=2$ since $\Delta_{\lambda}^{+}\cap A_{3}\neq\emptyset$. Thus
$k(\lambda)=m(\lambda)=2$.

If $\Delta_{\lambda}^{+}\cap A_{3}=\emptyset$, then $\gamma_{1}$, $\gamma_{2}$
and $\gamma_{3}$ will satisfy that
$\langle\lambda+\rho\,,\alpha^{\vee}\rangle\leq 0$. Thus, we have
$m(\lambda)\geq 3$. The equality must hold since $m(\lambda)\leq 3$ by the
diagram of $\Delta({\mathfrak{p}}^{+})$. So $k(\lambda)=m(\lambda)=3$.

This finishes our second proof of Theorem 1.2. $\Box$

##  Appendix A Hasse diagrams and distinguished antichains

The figures below show the Hasse diagrams of $\Delta({\mathfrak{p}}^{+})$ for
${\mathfrak{g}}_{\mathbb{R}}$ in the simply-laced types. Let $c$ be the
constant given in [12, Table on p. 115] and shown below. Then the roots in the
distinguished antichains
$A_{k}=\\{\alpha\in\Delta({\mathfrak{p}}^{+}):\operatorname{ht}(\alpha)=(k-1)c+1\\}$,
$k=1,2,\ldots,r$, are labelled so that $A_{1}=\\{\alpha_{*}\\}$,
$A_{2}=\\{\beta_{1},\beta_{2}\\}$, and
$A_{3}=\\{\gamma_{1},\gamma_{2},\gamma_{3}\\}$.

${\mathfrak{s}}{\mathfrak{u}}(p,q)$, here shown for $p=4$, $q=3$:

$c=1$$\beta_{1}$$\beta_{2}$$\gamma_{1}$$\gamma_{2}$$\gamma_{3}$$\alpha_{4}=\varepsilon_{4}-\varepsilon_{5}$$\theta=\varepsilon_{1}-\varepsilon_{7}=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}+\alpha_{5}+\alpha_{6}$32156

${\mathfrak{s}}{\mathfrak{o}}(2,2n-2)$, here shown for $n=6$:

$c=n-2=4$$\beta_{1}$$\beta_{2}$23456432$\alpha_{1}=\varepsilon_{1}-\varepsilon_{2}$$\theta=\varepsilon_{1}+\varepsilon_{2}=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+2\alpha_{4}+\alpha_{5}+\alpha_{6}$

${\mathfrak{s}}{\mathfrak{o}}^{*}(2n)$, here shown for $n=6$:

$c=2$$\beta_{1}$$\beta_{2}$$\gamma_{1}$$\gamma_{2}$$\gamma_{3}$43215432$\alpha_{6}=\varepsilon_{5}+\varepsilon_{6}$$\theta=\varepsilon_{1}+\varepsilon_{2}=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+2\alpha_{4}+\alpha_{5}+\alpha_{6}$

$\mathfrak{e}_{6(-14)}$:

$c=3$$\beta_{1}$$\beta_{2}$$\alpha_{1}$$\theta=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+3\alpha_{4}+2\alpha_{5}+\alpha_{6}$34562435542

$\mathfrak{e}_{7(-25)}$:

$c=4$6543124534265431$\alpha_{7}$$\beta_{1}$$\beta_{2}$$\gamma_{1}$$\gamma_{2}$$\gamma_{3}$$\theta=2\alpha_{1}+2\alpha_{2}+3\alpha_{3}+4\alpha_{4}+2\alpha_{6}+\alpha_{7}$

### Acknowledgments

Z. Bai was supported 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). X. Xie was supported by the
National Natural Science Foundation of China (No. 12171030 and No. 11801031).

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