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A characterization of unitarity of some highest weight Harish-Chandra modules

Zhanqiang Bai Department of Mathematical Sciences, Soochow University, Suzhou, Jiangsu, China [email protected]  and  Markus Hunziker Department of Mathematics, Baylor University, Waco, Texas, USA [email protected]
Abstract.

Let L(λ)L(\lambda) be a highest weight Harish-Chandra module with highest weight λ\lambda. When the associated variety of L(λ)L(\lambda) is not maximal, that is, not equal to the nilradical of the corresponding parabolic subalgebra, we prove that the unitarity of L(λ)L(\lambda) can be determined by a simple condition on the value of z=(λ+ρ,β)z=(\lambda+\rho,\beta^{\vee}), where ρ\rho is half the sum of positive roots and β\beta is the highest root. In the proof, certain distinguished antichains of positive noncompact roots play a key role.

By using these antichains, we are also able to provide a uniform formula for the Gelfand–Kirillov dimension of all highest weight Harish-Chandra modules, generalizing our previous result for the case of unitary highest weight Harish-Chandra modules.

Key words and phrases:
Unitary highest weight module, associated variety, Gelfand–Kirillov dimension
2020 Mathematics Subject Classification:
22E47, 17B10

1. Introduction

Let GG_{\mathbb{R}} be a connected non-compact simple Lie group with finite center, and let KK_{\mathbb{R}} be a maximal compact subgroup. From the work of Harish-Chandra (see comments in [BHXZ, §3.2]), it follows that infinite-dimensional highest weight Harish-Chandra modules for GG_{\mathbb{R}} exist if and only if (GG_{\mathbb{R}}, KK_{\mathbb{R}}) is a Hermitian symmetric pair. The problem of determining when a highest weight Harish-Chandra module is unitarizable has been extensively studied by various authors (see, for example, the references in [EHW]). The full classification was independently completed in [EHW] and [Ja1], though the classification itself is rather intricate. In this paper, we provide a simple and uniform characterization of unitarity for Harish-Chandra modules with a given associated variety, expressed in terms of the highest weight (see Theorem 1.1).

From now, we assume that (G,K)(G_{\mathbb{R}},K_{\mathbb{R}}) is a Hermitian symmetric pair. We denote by KK the complexification of the compact group KK_{\mathbb{R}} and by (𝔤,𝔨)(\mathfrak{g},\mathfrak{k}) the complexified Lie algebras of (G,K)(G_{\mathbb{R}},K_{\mathbb{R}}). Then we have the usual decompositition 𝔤=𝔭𝔨𝔭+\mathfrak{g}=\mathfrak{p}^{-}\oplus\mathfrak{k}\oplus\mathfrak{p}^{+} of 𝔤\mathfrak{g} as a KK-representation. Let 𝔥𝔨\mathfrak{h}\subseteq\mathfrak{k} be a Cartan subalgebra. Then 𝔥\mathfrak{h} is a Cartan subalgebra of 𝔤\mathfrak{g}. Let Δ\Delta and Δ(𝔨)\Delta(\mathfrak{k}) denote the root systems of (𝔤,𝔥)(\mathfrak{g},\mathfrak{h}) and (𝔨,𝔥)(\mathfrak{k},\mathfrak{h}), respectively. Let Δ+\Delta^{+} be the positive system of Δ\Delta, and define Δ+(𝔨)=Δ(𝔨)Δ+\Delta^{+}(\mathfrak{k})=\Delta(\mathfrak{k})\cap\Delta^{+} and Δ(𝔭+)=Δ+Δ+(𝔨)\Delta(\mathfrak{p}^{+})=\Delta^{+}\setminus\Delta^{+}(\mathfrak{k}). Let β\beta denote the unique maximal noncompact root of Δ+\Delta^{+}. Now choose ζ𝔥\zeta\in\mathfrak{h}^{*} so that ζ\zeta is orthogonal to Δ(𝔨)\Delta(\mathfrak{k}) and (ζ,β\zeta,\beta^{\vee})=1. Let λ𝔥\lambda\in\mathfrak{h}^{*} be Δ+(𝔨)\Delta^{+}(\mathfrak{k})-dominant integral and F(λ)F(\lambda) be the irreducible 𝔨\mathfrak{k}-module with highest weight λ\lambda. By letting the nilradical act by zero, we may consider F(λ)F(\lambda) as a module of the parabolic subalgebra 𝔮=𝔨+𝔭+\mathfrak{q}=\mathfrak{k}+\mathfrak{p}^{+}. Then we define:

N(λ)=U(𝔤)U(𝔮)F(λ).N(\lambda)=U(\mathfrak{g})\otimes_{U(\mathfrak{q})}F(\lambda).

Let L(λ)L(\lambda) denote the irreducible quotient of N(λ)N(\lambda), which is a highest weight module of 𝔤\mathfrak{g}.

From [EHW], L(λ)L(\lambda) is a highest weight Harish-Chandra module if and only if λΛ+(𝔨)\lambda\in\Lambda^{+}(\mathfrak{k}), where

Λ+(𝔨)={λ𝔥λ is Δ+(𝔨)-dominant  integral}.\Lambda^{+}(\mathfrak{k})=\{\lambda\in\mathfrak{h}^{*}\mid\lambda\text{~{}is $\Delta^{+}(\mathfrak{k})$-dominant~{} integral}\}.

Write ρ\rho for half the sum of positive roots in Δ+\Delta^{+}. Then we can write λ=λ0+zζ\lambda=\lambda_{0}+z\zeta, with λ0𝔥\lambda_{0}\in\mathfrak{h}^{*} such that (λ0+ρ,β\lambda_{0}+\rho,\beta)=0, and z=(λ+ρ,β)z=(\lambda+\rho,\beta^{\vee})\in\mathbb{R}.

The associated variety of a highest weight Harish-Chandra module is known to be the closure of a single KK-orbit in 𝔭+\mathfrak{p}^{+} (see §3 for more details). Furthermore, the closures of the KK-orbits in 𝔭+\mathfrak{p}^{+} form a linear chain of varieties

{0}=𝒪0¯𝒪1¯𝒪r1¯𝒪r¯=𝔭+,\{0\}={\overline{\mathcal{O}_{0}}}\subset\overline{\mathcal{O}_{1}}\subset\cdots\subset\overline{\mathcal{O}_{r-1}}\subset\overline{\mathcal{O}_{r}}=\mathfrak{p}^{+},

where rr is the \mathbb{R}-rank of GG_{\mathbb{R}}, i.e., the dimension of a Cartan subgroup of the group GG_{\mathbb{R}}, which is also equal to the rank of the symmetric space G/KG_{\mathbb{R}}/K_{\mathbb{R}}. Therefore, if L(λ)L(\lambda) is a highest weight Harish-Chandra module with highest weight λ\lambda, then there is an integer 0k(λ)r0\leq k(\lambda)\leq r such that the associated variety of L(λ)L(\lambda) is 𝒪k(λ)¯\overline{\mathcal{O}_{k(\lambda)}}.

Denote

(1.1) zk=(ρ,β)+uk=(ρ,β)kc,z_{k}=(\rho,\beta^{\vee})+u_{k}=(\rho,\beta^{\vee})-kc,

for 0kr0\leq k\leq r. Here cc is a real number associated with the Hermitian type Lie group GG_{\mathbb{R}}, see Table 1.

In this paper, we will prove the following result.

Theorem 1.1.

Let λΛ+(𝔨)\lambda\in\Lambda^{+}(\mathfrak{k}) and suppose AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0kr10\leq k\leq r-1. Then

L(λ)L(\lambda) is unitarizable if and only if (λ+ρ,β)=zk(\lambda+\rho,\beta^{\vee})=z_{k}.
Remark 1.2.

Note that when AV(L(λ))=𝒪r¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{r}}, it may happen that there is more than one point such that L(λ)L(\lambda) is unitarizable.

In [BH], we have found a uniform formula for the Gelfand–Kirillov dimensions of unitary highest weight Harish-Chandra modules. Now we want to generalize our formula to all highest weight Harish-Chandra modules.

Definition 1.3.

For 1kr11\leq k\leq r-1, define:

Ak={αΔ(𝔭+)ht(α)=kc+1}.A_{k}=\{\alpha\in\Delta(\mathfrak{p}^{+})\mid\operatorname{ht}(\alpha)=k\left\lceil c\right\rceil+1\}.

Here c\left\lceil c\right\rceil denotes the smallest integer nn such that ncn\geq c.

Definition 1.4.

Let λ0Λ+(𝔨)\lambda_{0}\in\Lambda^{+}(\mathfrak{k}).

  • (a)

    If 𝔤\mathfrak{g} is of type ADEADE, for 0kr10\leq k\leq r-1, define

    zk(λ0):=min{zαAk such that (λ0+zζ+ρ,α)>0}.z_{k}(\lambda_{0}):=\min\{z\in\mathbb{Z}\mid\mbox{$\exists~{}\alpha\in A_{k}$ such that $(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0$}\}.
  • (b)

    If 𝔤\mathfrak{g} is of type BCBC, for 0kr10\leq k\leq r-1, define

    zk(λ0):={min{zαAk such that (λ0+zζ+ρ,α)>0},if k is evenmin{z12+αAk such that (λ0+zζ+ρ,α)>0},if k is odd.z_{k}(\lambda_{0}):=\begin{cases}\min\{z\in\mathbb{Z}\mid\mbox{$\exists~{}\alpha\in A_{k}$ such that $(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0$}\},&\mbox{if $k$ is even}\\ \min\{z\in{\frac{1}{2}}+\mathbb{Z}\mid\mbox{$\exists~{}\alpha\in A_{k}$ such that $(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0$}\},&\mbox{if $k$ is odd.}\\ \end{cases}

In the special case when λ0=(ρ,β)ζ\lambda_{0}=-(\rho,\beta^{\vee})\zeta, we also write zkz_{k} instead of zk(λ0)z_{k}(\lambda_{0}) (This coincides with our definition of zkz_{k} in (1.1)).

Our new formula for the Gelfand–Kirillov dimensions of all highest weight Harish-Chandra modules is as follows.

Theorem (Theorem 6.3).

Suppose λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is a reduction point. Then

  • (a)

    If 𝔤\mathfrak{g} is of type ADEADE, then zz\in\mathbb{Z} and

    GKdimL(λ)={rzr1,if z<zr1(λ0)kzk1,if zk(λ0)z<zk1(λ0), where 1kr10,if z0(λ0)z.\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}(\lambda_{0})$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-1}(\lambda_{0})$, where $1\leq k\leq r-1$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})\leq z\in\mathbb{Z}$}.\\ \end{cases}
  • (b)

    If 𝔤\mathfrak{g} is of type BCBC, then zz\in\mathbb{Z} or z12+z\in\frac{1}{2}+\mathbb{Z} and

    GKdimL(λ)={rzr1,if z<zr1(λ0)kzk1,if zk(λ0)z<zk2(λ0), where 2kr1 and either z and k is even or z12+ and k is oddz0,if z1(λ0)z12+0,if z0(λ0)z.\displaystyle\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}(\lambda_{0})$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-2}(\lambda_{0})$, where $2\leq k\leq r-1$ and}\\ &\mbox{\quad either $z\in\mathbb{Z}$ and $k$ is even or $z\in\frac{1}{2}+\mathbb{Z}$ and $k$ is odd}\\ z_{0},&\mbox{if $z_{1}(\lambda_{0})\leq z\in\frac{1}{2}+\mathbb{Z}$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})\leq z\in\mathbb{Z}$}.\\ \end{cases}

2. Distinguished antichains of positive noncompact roots

Note that for any integer 1hht(β)1\leq h\leq\operatorname{ht}(\beta), the set

Δ(𝔭+)h:={αΔ(𝔭+)ht(α)=h}\Delta(\mathfrak{p}^{+})_{h}:=\{\alpha\in\Delta(\mathfrak{p}^{+})\mid\operatorname{ht}(\alpha)=h\}

is an antichain in Δ(𝔭+)\Delta(\mathfrak{p}^{+}).

Let Π\Pi denote the set of simple roots in Δ+\Delta^{+}.

Lemma 2.1 ([Ja1, Lemma 4.1]).

Let αΔ(𝔭+)\alpha\in\Delta(\mathfrak{p^{+}}), let π1,,πk\pi_{1},\ldots,\pi_{k} be distinct elements of ΠΔ(𝔨)\Pi\cap\Delta(\mathfrak{k}), and assume that α+πiΔ(𝔭+)\alpha+\pi_{i}\in\Delta(\mathfrak{p^{+}}) for i=1,,ki=1,\ldots,k. Then k2k\leq 2. Furthermore, if k=2k=2, then π1π2\pi_{1}\perp\pi_{2} and α+π1+π2Δ(𝔭+)\alpha+\pi_{1}+\pi_{2}\in\Delta(\mathfrak{p^{+}}). \Box

In light of this lemma, the Hasse diagram of Δ(𝔭+)\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 4545-degree angle.

Example 2.2.

Let 𝔤=𝔰𝔲(3,2)\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(3,2). Then we have

𝔭+={()}\mathfrak{p}^{+}=\{\left(\begin{array}[]{ccc|cc}\cdot&\cdot&\cdot&\,*&*\\[-4.0pt] \cdot&\cdot&\cdot&\,*&*\\[-4.0pt] \cdot&\cdot&\cdot&\,*&*\\[-4.0pt] \hline\cr\cdot&\cdot&\,\cdot&\cdot&\cdot\\[-4.0pt] \cdot&\cdot&\,\cdot&\cdot&\cdot\end{array}\right)\}Δ(𝔭+)=\Delta(\mathfrak{p}^{+})=α3=e3e4\alpha_{3}=e_{3}-e_{4}β=e1e5\beta=e_{1}-e_{5}214

The antichains in Δ(𝔭+)\Delta(\mathfrak{p}^{+}) are given in Appendix.

A subset YΔ(𝔭+)Y\subseteq\Delta({\mathfrak{p}}^{+}) is called a lower-order ideal if, for αΔ(𝔭+)\alpha\in\Delta({\mathfrak{p}}^{+}) and βY\beta\in Y, αβ\alpha\leq\beta implies that αY\alpha\in Y.

Definition 2.3.

For λΛ+(𝔨)\lambda\in\Lambda^{+}({\mathfrak{k}}), define the diagram of λ\lambda as the set

(2.1) Yλ:={αΔ(𝔭+)(λ+ρ,α)0},Y_{\lambda}:=\{\alpha\in\Delta({\mathfrak{p}}^{+})\mid({\lambda+\rho},{\alpha^{\vee}})\in\mathbb{Z}_{\leq 0}\},

viewed as a subposet of Δλ(𝔭+):=ΔλΔ(𝔭+)\Delta_{\lambda}({\mathfrak{p}}^{+}):=\Delta_{\lambda}\cap\Delta({\mathfrak{p}}^{+}), where Δλ:={αΔ(λ+ρ,α)}\Delta_{\lambda}:=\{\alpha\in\Delta\mid({\lambda+\rho},{\alpha^{\vee}})\in\mathbb{Z}\} is the integral root system associated to λ\lambda.

By [BHXZ, Lem 2.2], the poset YλY_{\lambda} is a lower order ideal of Δ(𝔭+)\Delta({\mathfrak{p}}^{+}) when λ\lambda is integral.

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. Suppose λΛ+(𝔨)\lambda\in\Lambda^{+}(\mathfrak{k}), we use m=m(λ)m=m(\lambda) to denote the width of YλY_{\lambda}.

By inspection of the Hasse diagram of Δ(𝔭+)\Delta(\mathfrak{p}^{+}), we have the following lemma.

Lemma 2.4.

Suppose YΔ(𝔭+)Y\subseteq\Delta(\mathfrak{p}^{+}) be a lower order ideal. If mm is the width of YY, then there exists an antichain AYA\subseteq Y of length mm such that all the roots in AA have the same height.

Lemma 2.5.

Let λΛ+(𝔨)\lambda\in\Lambda^{+}(\mathfrak{k}). If AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0kr10\leq k\leq r-1, then (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 for some αAk\alpha\in A_{k}.

Recall that Ak=Δ(𝔭+)kc+1A_{k}=\Delta(\mathfrak{p}^{+})_{k\left\lceil c\right\rceil+1} is an antichain in Δ(𝔭+)\Delta(\mathfrak{p}^{+}).

Proof.

Suppose AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0kr10\leq k\leq r-1.

When Δ\Delta is simply-laced, by our main theorem in [BHXZ], λ\lambda is integral and m(λ)=km(\lambda)=k. Assume that (λ+ρ,α)0(\lambda+\rho,\alpha^{\vee})\leq 0 for all αAk\alpha\in A_{k}. Since AkA_{k} is an antichain of length k+1k+1, it would follow that m(λ)k+1m(\lambda)\geq k+1. Contradiction.

When Δ\Delta is non-simply-laced and k=2lk=2l is even, by our main theorem in [BHXZ], λ\lambda is integral and 2m(λ)=k2m(\lambda)=k. Assume that (λ+ρ,α)0(\lambda+\rho,\alpha^{\vee})\leq 0 for all αAk\alpha\in A_{k}. Since AkA_{k} is an antichain of length l+1l+1, it would follow that m(λ)l+1m(\lambda)\geq l+1. Then 2m(λ)2l+2=k+2>k2m(\lambda)\geq 2l+2=k+2>k. Contradiction!

When Δ\Delta is non-simply-laced and k=2l+1k=2l+1 is odd, by our main theorem in [BHXZ], λ\lambda is half-integral and 2m(λ)+1=k2m(\lambda)+1=k. Assume that (λ+ρ,α)0(\lambda+\rho,\alpha^{\vee})\leq 0 for all αAk\alpha\in A_{k}. Since AkA_{k} is an antichain of length l+1l+1, it would follow that m(λ)l+1m(\lambda)\geq l+1. Then 2m(λ)+12l+3=k+2>k2m(\lambda)+1\geq 2l+3=k+2>k. Contradiction!

3. Gelfand–Kirillov dimension and associated variety

In this section, we will recall some preliminaries on Gelfand–Kirillov dimensions and associated varieties of highest weight modules. See [Vo78, Vo91] for more details.

Let MM be a finite generated U(𝔤)U(\mathfrak{g})-module. Fix a finite dimensional generating space M0M_{0} of MM. Let Un(𝔤)U_{n}(\mathfrak{g}) be the standard filtration of U(𝔤)U(\mathfrak{g}). Set Mn=Un(𝔤)M0M_{n}=U_{n}(\mathfrak{g})\cdot M_{0} and gr(M)=n=0grnM,\text{gr}(M)=\bigoplus\limits_{n=0}^{\infty}\text{gr}_{n}M, where grnM=Mn/Mn1\text{gr}_{n}M=M_{n}/{M_{n-1}}. Thus gr(M)\text{gr}(M) is a graded module of gr(U(𝔤))S(𝔤)\text{gr}(U(\mathfrak{g}))\simeq S(\mathfrak{g}).

The Gelfand–Kirillov dimension of MM is defined by

GKdimM=limn¯logdim(Un(𝔤)M0)logn.\operatorname{GKdim}M=\overline{\lim\limits_{n\rightarrow\infty}}\frac{\log\dim(U_{n}(\mathfrak{g})M_{0})}{\log n}.

The associated variety of MM is defined by

AV(M):={X𝔤f(X)=0 for all fAnnS(𝔤)(grM)}.\mathrm{AV}(M):=\{X\in\mathfrak{g}^{*}\mid f(X)=0\text{ for all~{}}f\in\operatorname{Ann}_{S(\mathfrak{g})}(\operatorname{gr}M)\}.

These two definitions are independent of the choice of M0M_{0}, and dimV(M)=GKdimM\dim V(M)=\operatorname{GKdim}M (e.g., [NOT]). If M0M_{0} is 𝔞\mathfrak{a}-invariant for a subalgebra 𝔞𝔤\mathfrak{a}\subset\mathfrak{g}, then

(3.1) AV(M)(𝔤/𝔞).\mathrm{AV}(M)\subset(\mathfrak{g}/\mathfrak{a})^{*}.

When M=L(λ)M=L(\lambda) is a highest weight Harish-Chandra module, we can choose M0M_{0} to be the finite dimensional U(𝔨)U(\mathfrak{k})-module generated by λ\mathbb{C}_{\lambda}. Then M0M_{0} is 𝔨𝔭+\mathfrak{k}\oplus\mathfrak{p}^{+}-invariant. In view of (3.1),

AV(L(λ))(𝔤/(𝔨𝔭+))(𝔭)𝔭+,\mathrm{AV}(L(\lambda))\subset(\mathfrak{g}/(\mathfrak{k}\oplus\mathfrak{p}^{+}))^{*}\simeq(\mathfrak{p}^{-})^{*}\simeq\mathfrak{p}^{+},

where the last isomorphism is induced from the Killing form. As shown in [Vo91], the associated variety AV(M)\mathrm{AV}(M) is also KK-invariant. In fact, Yamashita [Hir01] proved that AV(M)\mathrm{AV}(M) must be one of 𝒪k¯\overline{\mathcal{O}_{k}}.

Lemma 3.1.

Let L(λ)L(\lambda) be a highest weight Harish-Chandra module. Then

AV(L(λ))=𝒪k(λ)¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k(\lambda)}}

for some 0k(λ)r0\leq k(\lambda)\leq r.

We have the following table from [EHW]:

𝔤\mathfrak{g}_{\mathbb{R}} rr cc (ρ,β)(\rho,\beta^{\vee})
𝔰𝔲(p,np)\mathfrak{su}(p,n-p) min{p,np}\min\{p,n-p\} 11 n1n-1
𝔰𝔭(n,)\mathfrak{sp}(n,\mathbb{R}) nn 1/21/2 nn
𝔰𝔬(2n)\mathfrak{so}^{*}(2n) [n/2][n/2] 22 2n32n-3
𝔰𝔬(2,2n1)\mathfrak{so}(2,2n-1) 22 n3/2n-3/2 2n22n-2
𝔰𝔬(2,2n2)\mathfrak{so}(2,2n-2) 22 n2n-2 2n32n-3
𝔢6(14)\mathfrak{e}_{6(-14)} 22 33 1111
𝔢7(25)\mathfrak{e}_{7(-25)} 33 44 1717
Table 1. Some constants of Lie groups of Hermitian type

In [BH], we have found a uniform expression for the GK dimensions and associated varieties of unitary highest weight Harish-Chandra modules.

Proposition 3.2 ([BH]).

Suppose L(λ)L(\lambda) is a unitary highest weight Harish-Chandra module with highest weight λ\lambda. We denote z=z(λ)=(λ+ρ,β)z=z(\lambda)=(\lambda+\rho,\beta^{\vee}), then

GKdimL(λ)={rzr1,if z<zr1kzk1,if z=zk=(ρ,β)kc,1kr10,if z=z0=(ρ,β).\displaystyle\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}$}\\ kz_{k-1},&\mbox{if $z=z_{k}=(\rho,\beta^{\vee})-kc,1\leq k\leq r-1$}\\ 0,&\mbox{if $z=z_{0}=(\rho,\beta^{\vee})$}.\end{cases}

Denote k=k(λ):=(λ,β)ck=k(\lambda):=-\frac{(\lambda,\beta^{\vee})}{c}. Then

  1. (1)

    If k>r1k>r-1, we have GKdimL(λ)=rzr1=12dim(G/K).\operatorname{GKdim}L(\lambda)=rz_{r-1}=\frac{1}{2}\dim(G/K).

  2. (2)

    If 0kr10\leq k\leq r-1, then kk is a non-negative integer and

    GKdimL(λ)=k((ρ,β)(k1)c)=kzk1=dim𝒪k(λ)¯.\operatorname{GKdim}L(\lambda)=k((\rho,\beta^{\vee})-(k-1)c)=kz_{k-1}=\dim\overline{\mathcal{O}_{k(\lambda)}}.

The associated variety of L(λ)L(\lambda) is 𝒪k(λ)¯\overline{\mathcal{O}_{k(\lambda)}}.

4. Proof of the main theorem: simply-laced cases

In this section, we assume that Δ\Delta is simply-laced.

For any integer 0kr10\leq k\leq r-1, AkA_{k} is the antichain in Δ(𝔭+)\Delta(\mathfrak{p}^{+}) such that |Ak|=k+1|A_{k}|=k+1 and Ak=Δ(𝔭+)hA_{k}=\Delta(\mathfrak{p}^{+})_{h} with hh minimal.

To prove our Theorem 1.1, we need the following useful lemma.

Lemma 4.1.

Let τΛ+(𝔨)\tau\in\Lambda^{+}(\mathfrak{k}). If λ0=τ(τ+ρ,β)ζ\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta and λ=λ0+zζ\lambda=\lambda_{0}+z\zeta, then, for any αΔ(𝔭+)h\alpha\in\Delta(\mathfrak{p}^{+})_{h},

(λ+ρ,α)=z(ρ,β)+h(τ,βα).(\lambda+\rho,\alpha^{\vee})=z-(\rho,\beta^{\vee})+h-(\tau,\beta^{\vee}-\alpha^{\vee}).
Proof.

Let αΔ(𝔭+)h\alpha\in\Delta(\mathfrak{p}^{+})_{h}. Since Δ\Delta is simply-laced, (ζ,α)=1(\zeta,\alpha^{\vee})=1 and (ρ,α)=h(\rho,\alpha^{\vee})=h. Thus,

(λ+ρ,α)\displaystyle(\lambda+\rho,\alpha^{\vee}) =(λ0+zζ+ρ,α)\displaystyle=(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})
=(τ,α)(τ+ρ,β)(ζ,α)+z(ζ,α)+(ρ,α)\displaystyle=(\tau,\alpha^{\vee})-(\tau+\rho,\beta^{\vee})(\zeta,\alpha^{\vee})+z(\zeta,\alpha^{\vee})+(\rho,\alpha^{\vee})
=(τ,α)(τ+ρ,β)+z+h\displaystyle=(\tau,\alpha^{\vee})-(\tau+\rho,\beta^{\vee})+z+h
=(τ,αβ)(ρ,β)+z+h\displaystyle=(\tau,\alpha^{\vee}-\beta^{\vee})-(\rho,\beta^{\vee})+z+h
=z(ρ,β)+h(τ,βα).\displaystyle=z-(\rho,\beta^{\vee})+h-(\tau,\beta^{\vee}-\alpha^{\vee}).

Now we can prove our Theorem 1.1. The idea is very simple. From Lemma 2.5, since AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}}, we have (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 for some αAk=Δ(𝔭+)kc+1\alpha\in A_{k}=\Delta(\mathfrak{p^{+}})_{k\left\lceil c\right\rceil+1}. Then ht(α)=kc+1\operatorname{ht}(\alpha)=kc+1. So if (λ+ρ,β)=z=zk=(ρ,β)kc(\lambda+\rho,\beta^{\vee})=z=z_{k}=(\rho,\beta^{\vee})-kc, we will have

(λ+ρ,α)\displaystyle(\lambda+\rho,\alpha^{\vee}) =z(ρ,β)+h(τ,βα)\displaystyle=z-(\rho,\beta^{\vee})+h-(\tau,\beta^{\vee}-\alpha^{\vee})
=(ρ,β)kc(ρ,β)+kc+1(τ,βα)\displaystyle=(\rho,\beta^{\vee})-kc-(\rho,\beta^{\vee})+kc+1-(\tau,\beta^{\vee}-\alpha^{\vee})
=1(τ,βα)>0.\displaystyle=1-(\tau,\beta^{\vee}-\alpha^{\vee})>0.

This condition is very restrictive. In the following, we will give a case-by-case discussion for this condition, which will imply the unitarity of L(λ)L(\lambda).

4.1. Case 𝔤=𝔰𝔲(p,q)\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(p,q)

For 0kr1=min{p,q}10\leq k\leq r-1=\min\{p,q\}-1,

Ak=Δ(𝔭+)k+1={εpk+iεp+i+1=[0,,0pk+i1,1,,1k+1,0,,0qi1]0ik}.A_{k}=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{p-k+i}-\varepsilon_{p+i+1}=[\,\underbrace{0,\ldots,0}_{p-k+i-1},\underbrace{1,\ldots,1}_{k+1},\underbrace{0,\ldots,0}_{q-i-1}\,]\mid 0\leq i\leq k\}.

(Here and in the following we write [n1,n2,]:=n1α1+n2α2+[n_{1},n_{2},\ldots]:=n_{1}\alpha_{1}+n_{2}\alpha_{2}+\cdots, where α1,α2,\alpha_{1},\alpha_{2},\ldots are the simple roots.)

Fix τ=a1ω1++ap1ωp1+bq1ωp+1++b1ωp+q1Λ+(𝔨)\tau=a_{1}\omega_{1}+\cdots+a_{p-1}\omega_{p-1}+b_{q-1}\omega_{p+1}+\cdots+b_{1}\omega_{p+q-1}\in\Lambda^{+}(\mathfrak{k}). Then

λ0=τ(p+q1+a1+ap1+bq1++b1)ζ.\lambda_{0}=\tau-(p+q-1+a_{1}+\cdots a_{p-1}+b_{q-1}+\cdots+b_{1})\zeta.

For α=εpk+iεp+i+1\alpha=\varepsilon_{p-k+i}-\varepsilon_{p+i+1},

(λ0+zζ+ρ,α)=(z(p+qk2))(a1++apk+i1+bqi1++b1).(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})=(z-(p+q-k-2))-(a_{1}+\cdots+a_{p-k+i-1}+b_{q-i-1}+\cdots+b_{1}).

Now suppose λ=λ0+zζ\lambda=\lambda_{0}+z\zeta such that AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0kr10\leq k\leq r-1 and z=zk=p+qk1z=z_{k}=p+q-k-1. By Lemma 2.5, (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 for some αAk\alpha\in A_{k}. For α=εpk+iεp+i+1\alpha=\varepsilon_{p-k+i}-\varepsilon_{p+i+1}, by Lemma 4.1 we have

(λ+ρ,α)=1(a1++apk+i1+bqi1++b1)(\lambda+\rho,\alpha^{\vee})=1-(a_{1}+\cdots+a_{p-k+i-1}+b_{q-i-1}+\cdots+b_{1})

and hence (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 implies a1==apk+i1=0a_{1}=\cdots=a_{p-k+i-1}=0 and b1==bqi1=0b_{1}=\cdots=b_{q-i-1}=0. It follows that λ\lambda is of the form

λ=(k,,k1,,,;,,k,0,,01).\lambda=(\underbrace{-k,\dots,-k}_{\geq 1},\underbrace{*,\ldots,*\,;*,\ldots,*}_{k},\underbrace{0,\ldots,0}_{\geq 1}).

By [KV], L(λ)L(\lambda) is unitary.

4.2. Case 𝔤=𝔰𝔬(2n)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}^{*}(2n)

For 1kr1=[n2]11\leq k\leq r-1=[\frac{n}{2}]-1,

Ak\displaystyle A_{k} =Δ(𝔭+)2k+1={[0,,0n2k2,1,,12k,0,1],\displaystyle=\Delta(\mathfrak{p}^{+})_{2k+1}=\{[\,\underbrace{0,\ldots,0}_{n-2k-2},\underbrace{1,\ldots,1}_{2k},0,1\,],
[0,,0n2k1+i,1,,12k2i1,2,,2i,1,1]0ik1},\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}\quad\quad\quad\quad\quad\quad\quad[\,\underbrace{0,\ldots,0}_{n-2k-1+i},\underbrace{1,\ldots,1}_{2k-2i-1},\underbrace{2,\ldots,2}_{i},1,1\,]\mid 0\leq i\leq k-1\},
A0\displaystyle A_{0} =Δ(𝔭+)1={[0,0,,0,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[0,0,\dots,0,1]\}.

Fix τ=a1ω1+a2ω2++an1ωn1Λ+(𝔨)\tau=a_{1}\omega_{1}+a_{2}\omega_{2}+\cdots+a_{n-1}\omega_{n-1}\in\Lambda^{+}(\mathfrak{k}). Then, since β=[1,2,,2,1,1]\beta=[1,2,\cdots,2,1,1],

λ0=τ(τ+ρ,β)ζ=τ(2n3+a1+2a2++2an2+an1)ζ.\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta=\tau-(2n-3+a_{1}+2a_{2}+\cdots+2a_{n-2}+a_{n-1})\zeta.

Suppose AV(L(λ))=𝒪0¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{0}} and z(λ)=z0=2n3z(\lambda)=z_{0}=2n-3. By Lemma 2.5 and Lemma 4.1,

1(a1+2a2++2an2+an1)>01-(a_{1}+2a_{2}+\cdots+2a_{n-2}+a_{n-1})>0

and hence a1=a2==an1=0a_{1}=a_{2}=\cdots=a_{n-1}=0. It follows that λ=0\lambda=0 (trivial representation).

Now suppose λ=λ0+zζ\lambda=\lambda_{0}+z\zeta such that AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 1kr11\leq k\leq r-1 and z=zk=2n32kz=z_{k}=2n-3-2k. By Lemma 2.5, (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 for some αAk\alpha\in A_{k}. So if n2k>2n-2k>2, by Lemma 4.1 we have

1(a1+2a2++2an2k2+an2k1++an2+an1)>01-(a_{1}+2a_{2}+\cdots+2a_{n-2k-2}+a_{n-2k-1}+\cdots+a_{n-2}+a_{n-1})>0

or

1(a1+2a2++2an2k1+i+an2k+i++ani2)>0.1-(a_{1}+2a_{2}+\cdots+2a_{n-2k-1+i}+a_{n-2k+i}+\cdots+a_{n-i-2})>0.

In the first case, a1=a2==an1=0a_{1}=a_{2}=\cdots=a_{n-1}=0, and hence λ=2kζ\lambda=-2k\zeta (kk-th Wallach representation).

In the second case, a1=a2==ani2=0a_{1}=a_{2}=\cdots=a_{n-i-2}=0. It follows that

λ=\displaystyle\lambda= ani1ωni1++an1ωn1\displaystyle a_{n-i-1}\omega_{n-i-1}+\cdots+a_{n-1}\omega_{n-1}
(2k+2ani1++2an2+an1)ζ\displaystyle-(2k+2a_{n-i-1}+\cdots+2a_{n-2}+a_{n-1})\zeta
=\displaystyle= ani1ωni1++an1ωn1\displaystyle a_{n-i-1}\omega_{n-i-1}+\cdots+a_{n-1}\omega_{n-1}
(2k+2ani1++2an2+an1)ωn.\displaystyle-(2k+2a_{n-i-1}+\cdots+2a_{n-2}+a_{n-1})\omega_{n}.

So λ\lambda is of the form

λ=(k,,kni1,,,i+1).\lambda=(\underbrace{-k,\dots,-k}_{n-i-1},\underbrace{*,\ldots,*}_{i+1}).

By [EW], [EHW] or [DES], L(λ)L(\lambda) is unitary.

So if n2k=2n-2k=2, by Lemma 4.1 we have

1(a2++an1)>01-(a_{2}+\cdots+a_{n-1})>0

or

1(a1+2a2++2an2k1+i+an2k+i++ani2)>0.1-(a_{1}+2a_{2}+\cdots+2a_{n-2k-1+i}+a_{n-2k+i}+\cdots+a_{n-i-2})>0.

In the first case, a2==an1=0a_{2}=\cdots=a_{n-1}=0, and hence λ=a1ω1+(2na1)ζ\lambda=a_{1}\omega_{1}+(2-n-a_{1})\zeta (unitary reduction point).

In the second case, a1=a2==ani2=0a_{1}=a_{2}=\cdots=a_{n-i-2}=0. It follows that

λ=\displaystyle\lambda= ani1ωni1++an1ωn1\displaystyle a_{n-i-1}\omega_{n-i-1}+\cdots+a_{n-1}\omega_{n-1}
(2k+2ani1++2an2+an1)ζ\displaystyle-(2k+2a_{n-i-1}+\cdots+2a_{n-2}+a_{n-1})\zeta
=\displaystyle= ani1ωni1++an1ωn1\displaystyle a_{n-i-1}\omega_{n-i-1}+\cdots+a_{n-1}\omega_{n-1}
(2k+2ani1++2an2+an1)ωn.\displaystyle-(2k+2a_{n-i-1}+\cdots+2a_{n-2}+a_{n-1})\omega_{n}.

So λ\lambda is of the form

λ=(k,,kni1,,,i+1).\lambda=(\underbrace{-k,\dots,-k}_{n-i-1},\underbrace{*,\ldots,*}_{i+1}).

By [EW], [EHW] or [DES], L(λ)L(\lambda) is unitary.

4.3. Case 𝔤=𝔰𝔬(2,2n2)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-2)

For 0kr1=21=10\leq k\leq r-1=2-1=1,

A0\displaystyle A_{0} =Δ(𝔭+)1={ε1ε2=[1,0,,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{\varepsilon_{1}-\varepsilon_{2}=[1,0,...,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)n1={ε1εn=[1,,1,0],ε1+εn=[1,,1,0,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{n-1}=\{\varepsilon_{1}-\varepsilon_{n}=[1,\dots,1,0],\varepsilon_{1}+\varepsilon_{n}=[1,\dots,1,0,1]\}.

Fix τ=a2ω2++an1ωn1+anωnΛ+(𝔨)\tau=a_{2}\omega_{2}+\cdots+a_{n-1}\omega_{n-1}+a_{n}\omega_{n}\in\Lambda^{+}(\mathfrak{k}). Then, since β=ε1+ε2=[1,2,,2,1,1]\beta=\varepsilon_{1}+\varepsilon_{2}=[1,2,\cdots,2,1,1],

λ0=τ(τ+ρ,β)ζ=τ(2n3+2a2++2an2+an1+an)ζ.\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta=\tau-(2n-3+2a_{2}+\cdots+2a_{n-2}+a_{n-1}+a_{n})\zeta.

Suppose AV(L(λ))=𝒪0¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{0}} and z(λ)=z0=2n3z(\lambda)=z_{0}=2n-3. By Lemma 2.5 and Lemma 4.1,

1(2a2++2an2+an1+an)>01-(2a_{2}+\cdots+2a_{n-2}+a_{n-1}+a_{n})>0

and hence a2==an=0a_{2}=\cdots=a_{n}=0. It follows that λ=0\lambda=0 (trivial representation).

Suppose AV(L(λ))=𝒪1¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{1}} and z(λ)=z1=n1z(\lambda)=z_{1}=n-1. By Lemma 2.5 and Lemma 4.1,

1(a2++an1)>01-(a_{2}+\cdots+a_{n-1})>0

or

1(a2++an2+an)>0.1-(a_{2}+\cdots+a_{n-2}+a_{n})>0.

In the first case, we have a2==an1=0a_{2}=\cdots=a_{n-1}=0. Hence λ=anωn(n2+an)ζ\lambda=a_{n}\omega_{n}-(n-2+a_{n})\zeta (unitary reduction point).

In the second case, we have a2==an2=an=0a_{2}=\cdots=a_{n-2}=a_{n}=0. Hence λ=an1ωn1(n2+an1)ζ\lambda=a_{n-1}\omega_{n-1}-(n-2+a_{n-1})\zeta (unitary reduction point).

4.4. Case 𝔤=𝔢6(14)\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}

By inspection of the Hasse diagram of Δ(𝔭+)\Delta(\mathfrak{p}^{+}),

A0\displaystyle A_{0} =Δ(𝔭+)1={[1,0,0,0,0,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[1,0,0,0,0,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)4={[1,0,1,1,1,0],[1,1,1,1,0,0]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{4}=\{[1,0,1,1,1,0],[1,1,1,1,0,0]\}.

Fix τ=a2ω2+a3ω3+a4ω4+a5ω5+a6ω6\tau=a_{2}\omega_{2}+a_{3}\omega_{3}+a_{4}\omega_{4}+a_{5}\omega_{5}+a_{6}\omega_{6}. Then, since β=[1,2,2,3,2,1]\beta=[1,2,2,3,2,1],

λ0=τ(τ+ρ,β)ζ=τ(11+2a2+2a3+3a4+2a5+a6)ζ.\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta=\tau-(11+2a_{2}+2a_{3}+3a_{4}+2a_{5}+a_{6})\zeta.

Suppose AV(L(λ))=𝒪0¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{0}} and z(λ)=z0=11z(\lambda)=z_{0}=11. By Lemma 2.5 and Lemma 4.1,

1(2a2+2a3+3a4+2a5+a6)>01-(2a_{2}+2a_{3}+3a_{4}+2a_{5}+a_{6})>0

and hence a2=a3=a4=a5=a6=0a_{2}=a_{3}=a_{4}=a_{5}=a_{6}=0. It follows that λ=0\lambda=0 (trivial representation).

Suppose AV(L(λ))=𝒪1¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{1}} and z(λ)=z1=8z(\lambda)=z_{1}=8. By Lemma 2.5 and Lemma 4.1,

1(2a2+a3+2a4+a5+a6)>0or 1(a2+a3+2a4+2a5+a6)>0.1-(2a_{2}+a_{3}+2a_{4}+a_{5}+a_{6})>0\ \mbox{or}\ 1-(a_{2}+a_{3}+2a_{4}+2a_{5}+a_{6})>0.

In either case, a2=a3=a4=a5=a6=0a_{2}=a_{3}=a_{4}=a_{5}=a_{6}=0 and hence λ=3ζ\lambda=-3\zeta (1st Wallach representation).

4.5. Case 𝔤=𝔢7(25)\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}

By inspection of the Hasse diagram of Δ(𝔭+)\Delta(\mathfrak{p}^{+}),

A0\displaystyle A_{0} =Δ(𝔭+)1={[0,0,0,0,0,0,1]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[0,0,0,0,0,0,1]\},
A1\displaystyle A_{1} =Δ(𝔭+)5={[0,0,1,1,1,1,1],[0,1,0,1,1,1,1]},\displaystyle=\Delta(\mathfrak{p}^{+})_{5}=\{[0,0,1,1,1,1,1],[0,1,0,1,1,1,1]\},
A2\displaystyle A_{2} =Δ(𝔭+)9={[1,1,2,2,1,1,1],[1,1,1,2,2,1,1],[0,1,1,2,2,2,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{9}=\{[1,1,2,2,1,1,1],[1,1,1,2,2,1,1],[0,1,1,2,2,2,1]\}.

Fix τ=a1ω2+a2ω2+a3ω3+a4ω4+a5ω5+a6ω6\tau=a_{1}\omega_{2}+a_{2}\omega_{2}+a_{3}\omega_{3}+a_{4}\omega_{4}+a_{5}\omega_{5}+a_{6}\omega_{6}. Then, since β=[2,2,3,4,3,2,1]\beta=[2,2,3,4,3,2,1],

λ0=τ(17+2a1+2a2+3a3+4a4+3a5+2a6)ζ.\lambda_{0}=\tau-(17+2a_{1}+2a_{2}+3a_{3}+4a_{4}+3a_{5}+2a_{6})\zeta.

Suppose AV(L(λ))=𝒪0¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{0}} and z(λ)=z0=17z(\lambda)=z_{0}=17. By Lemma 2.5 and Lemma 4.1,

1(2a1+2a2+3a3+4a4+3a5+2a6)>01-(2a_{1}+2a_{2}+3a_{3}+4a_{4}+3a_{5}+2a_{6})>0

and hence a1=a2=a3=a4=a5=a6=0a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=a_{6}=0. It follows that λ=0\lambda=0 (trivial representation).

Suppose AV(L(λ))=𝒪1¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{1}} and z(λ)=z1=13z(\lambda)=z_{1}=13. By Lemma 2.5 and Lemma 4.1,

1(2a1+2a2+2a3+3a4+2a5+a6)>0or 1(2a1+a2+3a3+3a4+2a5+a6)>0.1-(2a_{1}+2a_{2}+2a_{3}+3a_{4}+2a_{5}+a_{6})>0\ \mbox{or}\ 1-(2a_{1}+a_{2}+3a_{3}+3a_{4}+2a_{5}+a_{6})>0.

In either case, a1=a2=a3=a4=a5=a6=0a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=a_{6}=0 and hence λ=4ω7\lambda=-4\omega_{7} (1st Wallach representation).

Suppose AV(L(λ))=𝒪2¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{2}} and z(λ)=z2=9z(\lambda)=z_{2}=9. By Lemma 2.5 and Lemma 4.1,

1(a1+a2+a3+2a4+2a5+a6>0,\displaystyle 1-(a_{1}+a_{2}+a_{3}+2a_{4}+2a_{5}+a_{6}>0,
1(a1+a2+2a3+2a4+a5+a6)>0,\displaystyle 1-(a_{1}+a_{2}+2a_{3}+2a_{4}+a_{5}+a_{6})>0,
or 1(2a1+a2+2a3+2a4+a5)>0.\displaystyle 1-(2a_{1}+a_{2}+2a_{3}+2a_{4}+a_{5})>0.

In the first two cases, a1=a2=a3=a4=a5=a6=0a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=a_{6}=0 and hence λ=8ζ\lambda=-8\zeta (2nd Wallach representation). In the third case, a1=a2=a3=a4=a5=0a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=0 and λ=a6ω6+(2a68)ζ\lambda=a_{6}\omega_{6}+(-2a_{6}-8)\zeta (unitary reduction point).

5. Proof of the main theorem: non-simply-laced cases

In this section, we assume that Δ\Delta is not simply-laced.

For any integer 0kr10\leq k\leq r-1, AkA_{k} is the antichain in Δ(𝔭+)\Delta(\mathfrak{p}^{+}) such that |Ak|=[k2]+1|A_{k}|=[\frac{k}{2}]+1 and Ak=Δ(𝔭+)hA_{k}=\Delta(\mathfrak{p}^{+})_{h} with hh minimal.

The proof of Theorem 1.1 in non-simply-laced cases is similar to the simply-laced cases. We need Lemma 2.5 and the following useful lemma in the computation.

Lemma 5.1.

Let τΛ+(𝔨)\tau\in\Lambda^{+}(\mathfrak{k}). If λ0=τ(τ+ρ,β)ζ\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta and λ=λ0+zζ\lambda=\lambda_{0}+z\zeta, then, for any αΔ(𝔭+)h\alpha\in\Delta(\mathfrak{p}^{+})_{h},

(λ+ρ,α)=z(ρ,β)+(ρ,α)(τ,βα) when α is a long root,(\lambda+\rho,\alpha^{\vee})=z-(\rho,\beta^{\vee})+(\rho,\alpha^{\vee})-(\tau,\beta^{\vee}-\alpha^{\vee})\text{~{}when $\alpha$ is a long root},

and

(λ+ρ,α)=2z2(ρ,β)+(ρ,α)(τ,2βα) when α is a short root.(\lambda+\rho,\alpha^{\vee})=2z-2(\rho,\beta^{\vee})+(\rho,\alpha^{\vee})-(\tau,2\beta^{\vee}-\alpha^{\vee})\text{~{}when $\alpha$ is a short root}.
Proof.

Let αΔ(𝔭+)h\alpha\in\Delta(\mathfrak{p}^{+})_{h}. Since Δ\Delta is not simply-laced, (ζ,α)=1(\zeta,\alpha^{\vee})=1 if α\alpha is a long root and (ζ,α)=2(\zeta,\alpha^{\vee})=2 if α\alpha is a short root. (ρ,α)=h(\rho,\alpha^{\vee})=h. Thus, if α\alpha is a long root, we have

(λ+ρ,α)\displaystyle(\lambda+\rho,\alpha^{\vee}) =(λ0+zζ+ρ,α)\displaystyle=(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})
=(τ,α)(τ+ρ,β)(ζ,α)+z(ζ,α)+(ρ,α)\displaystyle=(\tau,\alpha^{\vee})-(\tau+\rho,\beta^{\vee})(\zeta,\alpha^{\vee})+z(\zeta,\alpha^{\vee})+(\rho,\alpha^{\vee})
=(τ,α)(τ+ρ,β)+z+(ρ,α)\displaystyle=(\tau,\alpha^{\vee})-(\tau+\rho,\beta^{\vee})+z+(\rho,\alpha^{\vee})
=(τ,αβ)(ρ,β)+z+(ρ,α)\displaystyle=(\tau,\alpha^{\vee}-\beta^{\vee})-(\rho,\beta^{\vee})+z+(\rho,\alpha^{\vee})
=z(ρ,β)+(ρ,α)(τ,βα).\displaystyle=z-(\rho,\beta^{\vee})+(\rho,\alpha^{\vee})-(\tau,\beta^{\vee}-\alpha^{\vee}).

If α\alpha is a short root, we have

(λ+ρ,α)\displaystyle(\lambda+\rho,\alpha^{\vee}) =(λ0+zζ+ρ,α)\displaystyle=(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})
=(τ,α)(τ+ρ,β)(ζ,α)+z(ζ,α)+(ρ,α)\displaystyle=(\tau,\alpha^{\vee})-(\tau+\rho,\beta^{\vee})(\zeta,\alpha^{\vee})+z(\zeta,\alpha^{\vee})+(\rho,\alpha^{\vee})
=(τ,α)2(τ+ρ,β)+2z+(ρ,α)\displaystyle=(\tau,\alpha^{\vee})-2(\tau+\rho,\beta^{\vee})+2z+(\rho,\alpha^{\vee})
=(τ,α2β)2(ρ,β)+2z+(ρ,α)\displaystyle=(\tau,\alpha^{\vee}-2\beta^{\vee})-2(\rho,\beta^{\vee})+2z+(\rho,\alpha^{\vee})
=2z2(ρ,β)+(ρ,α)(τ,2βα).\displaystyle=2z-2(\rho,\beta^{\vee})+(\rho,\alpha^{\vee})-(\tau,2\beta^{\vee}-\alpha^{\vee}).

5.1. 𝔤=𝔰𝔭(n,)\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(n,\mathbb{R})

For 0kr1=n10\leq k\leq r-1=n-1,

if k=2m is even,Ak\displaystyle\text{if $k=2m$ is even},A_{k} =Δ(𝔭+)k+1={2εnm,εnmi+εnm+i1im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{2\varepsilon_{n-m},\varepsilon_{n-m-i}+\varepsilon_{n-m+i}\mid 1\leq i\leq m\}
={[0,,0nm1i,1,,12i,2,,2mi,1]0im},\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-1-i},\underbrace{1,\ldots,1}_{2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\},
if k=2m+1 is odd,Ak\displaystyle\text{if $k=2m+1$ is odd},A_{k} =Δ(𝔭+)k+1={εnm1i+εnm+i0im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{n-m-1-i}+\varepsilon_{n-m+i}\mid 0\leq i\leq m\}
={[0,,0nm2i,1,,11+2i,2,,2mi,1]0im}.\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-2-i},\underbrace{1,\ldots,1}_{1+2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\}.

Fix τ=a1ω1+a2ω2++an1ωn1Λ+(𝔨)\tau=a_{1}\omega_{1}+a_{2}\omega_{2}+\cdots+a_{n-1}\omega_{n-1}\in\Lambda^{+}(\mathfrak{k}). Then, since β=[2,,2,1]\beta=[2,\cdots,2,1],

λ0=τ(τ+ρ,β)ζ=τ(n+a1+a2++an2+an1)ζ.\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta=\tau-(n+a_{1}+a_{2}+\cdots+a_{n-2}+a_{n-1})\zeta.

Now suppose λ=λ0+zζ\lambda=\lambda_{0}+z\zeta such that AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 1kr11\leq k\leq r-1 and z=zk=nk2z=z_{k}=n-\frac{k}{2}. By Lemma 2.5, (λ+ρ,α)>0(\lambda+\rho,\alpha^{\vee})>0 for some αAk\alpha\in A_{k}. So if k=2mk=2m is even, by Lemma 5.1 we have

1(a1+a2++anm1)>01-(a_{1}+a_{2}+\cdots+a_{n-m-1})>0

or

2(2a1++2anm1i+anmi++an1m+i)>0.2-(2a_{1}+\cdots+2a_{n-m-1-i}+a_{n-m-i}+\cdots+a_{n-1-m+i})>0.

In the first case, a1=a2==anm1=0a_{1}=a_{2}=\cdots=a_{n-m-1}=0 and hence

λ\displaystyle\lambda =anmωnm++an1ωn1(anm++an1)ζ12ζ\displaystyle=a_{n-m}\omega_{n-m}+\cdots+a_{n-1}\omega_{n-1}-(a_{n-m}+\cdots+a_{n-1})\zeta-\frac{1}{2}\zeta
=(0,,0nm,,,m)12ζ.\displaystyle=(\underbrace{0,\ldots,0}_{n-m},\underbrace{*,\ldots,*}_{m})-\frac{1}{2}\zeta.

By [EHW] or [EW], L(λ)L(\lambda) is unitary.

In the second case, a1=a2==anm1i=0a_{1}=a_{2}=\cdots=a_{n-m-1-i}=0 and anmi==an1m+i=0a_{n-m-i}=\cdots=a_{n-1-m+i}=0 or at most one of {anmi,,an1m+i}\{a_{n-m-i},\cdots,a_{n-1-m+i}\} equals to 11 with the rest queal to 0. Hence

λ\displaystyle\lambda =anmiωnmi++an1ωn1(anmi++an1)ζ12ζ\displaystyle=a_{n-m-i}\omega_{n-m-i}+\cdots+a_{n-1}\omega_{n-1}-(a_{n-m-i}+\cdots+a_{n-1})\zeta-\frac{1}{2}\zeta
=(0,,0nmi,1,,1,,,mim+i)12ζ.\displaystyle=(\underbrace{0,\ldots,0}_{\geq n-m-i},\underbrace{-1,\cdots,-1,\overbrace{*,\ldots,*}^{m-i}}_{\leq m+i})-\frac{1}{2}\zeta.

By [EHW] or [EW], L(λ)L(\lambda) is unitary.

If k=2m+1k=2m+1 is odd, we have

2(2a1++2anm2i+anm1i++an1m+i)>0,2-(2a_{1}+\cdots+2a_{n-m-2-i}+a_{n-m-1-i}+\cdots+a_{n-1-m+i})>0,

then the arguments are similar to the above case.

5.2. 𝔤=𝔰𝔬(2,2n1)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-1)

For 0kr1=21=10\leq k\leq r-1=2-1=1,

A0\displaystyle A_{0} =Δ(𝔭+)1={ε1ε2=[1,0,,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{\varepsilon_{1}-\varepsilon_{2}=[1,0,...,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)n={ε1=[1,1,,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{n}=\{\varepsilon_{1}=[1,1,\dots,1]\}.

Fix τ=a2ω2++an1ωn1+anωnΛ+(𝔨)\tau=a_{2}\omega_{2}+\cdots+a_{n-1}\omega_{n-1}+a_{n}\omega_{n}\in\Lambda^{+}(\mathfrak{k}). Then, since β=ε1+ε2=[1,2,,2]\beta=\varepsilon_{1}+\varepsilon_{2}=[1,2,\cdots,2],

λ0=τ(τ+ρ,β)ζ=τ(2n2+2a2++2an1+an)ζ.\lambda_{0}=\tau-(\tau+\rho,\beta^{\vee})\zeta=\tau-(2n-2+2a_{2}+\cdots+2a_{n-1}+a_{n})\zeta.

Suppose AV(L(λ))=𝒪0¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{0}} and z(λ)=z0=2n2z(\lambda)=z_{0}=2n-2. By Lemma 2.5 and Lemma 5.1,

1(2a2++2an1+an)>01-(2a_{2}+\cdots+2a_{n-1}+a_{n})>0

and hence a2==an=0a_{2}=\cdots=a_{n}=0. It follows that λ=0\lambda=0 (trivial representation).

Suppose AV(L(λ))=𝒪1¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{1}} and z(λ)=z1=n12z(\lambda)=z_{1}=n-\frac{1}{2}. By Lemma 2.5 and Lemma 5.1,

2(2a2++2an1+an)>02-(2a_{2}+\cdots+2a_{n-1}+a_{n})>0

and hence a2==an=0a_{2}=\cdots=a_{n}=0 or a2==an1=an1=0a_{2}=\cdots=a_{n-1}=a_{n}-1=0.

In the first case, it follows that λ=(n32)ζ\lambda=-(n-\frac{3}{2})\zeta (1st Wallach representation).

In the second case, we have λ=ωn(n12)ζ\lambda=\omega_{n}-(n-\frac{1}{2})\zeta (unitary reduction point).

6. A uniform formula for the Gelfand–Kirillov dimension

In our previous paper [BH], we found a uniform formula for the Gelfand–Kirillov dimensions of all unitary highest weight modules. Now we will give a new formula for the Gelfand–Kirillov dimensions of all highest weight Harish-Chandra modules.

We recall the definition of zk(λ0)z_{k}(\lambda_{0}) in the introduction. Then we have the following lemma.

Lemma 6.1.

Suppose 𝔤\mathfrak{g} is of type ADEADE. For z=(λ+ρ,β)z=(\lambda+\rho,\beta^{\vee})\in\mathbb{Z}, we have

zk(λ0)zm(λ)k.z_{k}(\lambda_{0})\leq z\iff m(\lambda)\leq k.
Proof.

It is easy to verify (case-by-case) that

m(λ)k+1Ak{αΔ(𝔭+)(λ+ρ,α)0}.m(\lambda)\geq k+1\iff A_{k}\subset\{\alpha\in\Delta(\mathfrak{p}^{+})\mid(\lambda+\rho,\alpha^{\vee})\in\mathbb{Z}_{\leq 0}\}.

For zz\in\mathbb{Z}, we have

zk(λ0)z\displaystyle z_{k}(\lambda_{0})\leq z αAk such that (λ0+zζ+ρ,α)>0\displaystyle\iff\exists~{}\alpha\in A_{k}\text{~{}such~{}that~{}}(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0
Ak{αΔ(𝔭+)(λ+ρ,α)0}\displaystyle\iff A_{k}\nsubseteq\{\alpha\in\Delta(\mathfrak{p}^{+})\mid(\lambda+\rho,\alpha^{\vee})\leq 0\}
m(λ)<k+1=|Ak|\displaystyle\iff m(\lambda)<k+1=|A_{k}|
m(λ)k.\displaystyle\iff m(\lambda)\leq k.

We recall the main theorem in [BHXZ].

Proposition 6.2 ([BHXZ]).

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

  1. (a)

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

  2. (b)

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

    k(λ)={2m,if m<r+12r,if m=r+12.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 Δ\Delta is non-simply laced and λ\lambda is half-integral, then

    k(λ)={2m+1,if m<r2r,if m=r2.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(λ)=rk(\lambda)=r.

In the following, we give the new formula for the Gelfand–Kirillov dimensions of all highest weight Harish-Chandra modules.

Theorem 6.3.

Suppose λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is a reduction point. Then

  • (a)

    If 𝔤\mathfrak{g} is of type ADEADE, then zz\in\mathbb{Z} and

    GKdimL(λ)={rzr1,if z<zr1(λ0)kzk1,if zk(λ0)z<zk1(λ0), where 1kr10,if z0(λ0)z.\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}(\lambda_{0})$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-1}(\lambda_{0})$, where $1\leq k\leq r-1$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})\leq z\in\mathbb{Z}$}.\end{cases}
  • (b)

    If 𝔤\mathfrak{g} is of type BCBC, then zz\in\mathbb{Z} or z12+z\in\frac{1}{2}+\mathbb{Z} and

    GKdimL(λ)={rzr1,if z<zr1(λ0)kzk1,if zk(λ0)z<zk2(λ0), where 2kr1 and either z and k is even or z12+ and k is oddz0,if z1(λ0)z12+0,if z0(λ0)z.\displaystyle\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}(\lambda_{0})$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-2}(\lambda_{0})$, where $2\leq k\leq r-1$ and}\\ &\mbox{\quad either $z\in\mathbb{Z}$ and $k$ is even or $z\in\frac{1}{2}+\mathbb{Z}$ and $k$ is odd}\\ z_{0},&\mbox{if $z_{1}(\lambda_{0})\leq z\in\frac{1}{2}+\mathbb{Z}$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})\leq z\in\mathbb{Z}$}.\\ \end{cases}
Proof.

In the following, suppose 0kr10\leq k\leq r-1.

(a) In type ADEADE, if λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is a reduction point, then zz\in\mathbb{Z} by [EHW].

Thus, we have

zk(λ0)z\displaystyle z_{k}(\lambda_{0})\leq z αAk:(λ0+zζ+ρ,α)>0\displaystyle\iff\exists~{}\alpha\in A_{k}:(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0
Ak{αΔ(𝔭+)(λ+ρ,α)0}\displaystyle\iff A_{k}\nsubseteq\{\alpha\in\Delta(\mathfrak{p}^{+})\mid(\lambda+\rho,\alpha^{\vee})\leq 0\}
m(λ)<k+1=|Ak|\displaystyle\iff m(\lambda)<k+1=|A_{k}|
m(λ)k.\displaystyle\iff m(\lambda)\leq k.

The formula then follows from our main theorem (in type ADEADE) in [BHXZ].

(b) In type BCBC, if λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is a reduction point, then z12z\in\frac{1}{2}\mathbb{Z} by [EHW].

When λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is integral, then zz\in\mathbb{Z}. Thus, when k=2mk=2m is even, we have

zk(λ0)z\displaystyle z_{k}(\lambda_{0})\leq z αAk:(λ0+zζ+ρ,α)>0\displaystyle\iff\exists~{}\alpha\in A_{k}:(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0
Ak{αΔ(𝔭+)(λ+ρ,α)0}\displaystyle\iff A_{k}\nsubseteq\{\alpha\in\Delta(\mathfrak{p}^{+})\mid(\lambda+\rho,\alpha^{\vee})\leq 0\}
m(λ)<m+1=|Ak|\displaystyle\iff m(\lambda)<m+1=|A_{k}|
m(λ)m\displaystyle\iff m(\lambda)\leq m
2m(λ)2m=k.\displaystyle\iff 2m(\lambda)\leq 2m=k.

When λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is half-integral, then z12+z\in\frac{1}{2}+\mathbb{Z}. Thus, when k=2m+1k=2m+1 is odd, we have

zk(λ0)z\displaystyle z_{k}(\lambda_{0})\leq z αAk:(λ0+zζ+ρ,α)>0\displaystyle\iff\exists~{}\alpha\in A_{k}:(\lambda_{0}+z\zeta+\rho,\alpha^{\vee})>0
Ak{αΔ(𝔭+)(λ+ρ,α)0}\displaystyle\iff A_{k}\nsubseteq\{\alpha\in\Delta(\mathfrak{p}^{+})\mid(\lambda+\rho,\alpha^{\vee})\leq 0\}
m(λ)<m+1=|Ak|\displaystyle\iff m(\lambda)<m+1=|A_{k}|
m(λ)m\displaystyle\iff m(\lambda)\leq m
2m(λ)+12m+1=k.\displaystyle\iff 2m(\lambda)+1\leq 2m+1=k.

In particular, z1(λ0)z2m(λ)+12m+1=1z_{1}(\lambda_{0})\leq z\iff 2m(\lambda)+1\leq 2m+1=1. When λ=λ0+zζ\lambda=\lambda_{0}+z\zeta is half-integral, we also know that L(λ)L(\lambda) is not finite-dimensional. Thus AV(L(λ))=𝒪k(λ)¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k(\lambda)}} with k(λ)=2m(λ)+11.k(\lambda)=2m(\lambda)+1\geq 1.

The formula then follows from our main theorem (in type BCBC) in [BHXZ].

The following result was firstly proved in [BH]. Now we give a new proof.

Corollary 6.4.

Suppose L(λ)L(\lambda) is a unitary highest weight Harish-Chandra module with highest weight λ\lambda. We denote z=z(λ)=(λ+ρ,β)z=z(\lambda)=(\lambda+\rho,\beta^{\vee}), then

GKdimL(λ)={rzr1,if z<zr1kzk1,if z=zk=(ρ,β)kc,1kr10,if z=z0=(ρ,β).\displaystyle\operatorname{GKdim}L(\lambda)=\begin{cases}rz_{r-1},&\mbox{if $z<z_{r-1}$}\\ kz_{k-1},&\mbox{if $z=z_{k}=(\rho,\beta^{\vee})-kc,1\leq k\leq r-1$}\\ 0,&\mbox{if $z=z_{0}=(\rho,\beta^{\vee})$}.\end{cases}
Proof.

First we suppose that L(λ)L(\lambda) is a unitary highest weight Harish-Chandra module and z=(λ+ρ,β)=zkz=(\lambda+\rho,\beta^{\vee})=z_{k} for some 0kr10\leq k\leq r-1.

From Yamashita [Ya-94] we have AV(L(λ))=𝒪k(λ)¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k(\lambda)}} for some 0k(λ)r0\leq k(\lambda)\leq r. From Theorem 1.1, we will have z=(λ+ρ,β)=zk(λ)z=(\lambda+\rho,\beta^{\vee})=z_{k(\lambda)} since L(λ)L(\lambda) is unitarizable. Thus we must have z=zk=zk(λ)z=z_{k}=z_{k(\lambda)}, which implies that k(λ)=kk(\lambda)=k. So we must have GKdimL(λ)=dim𝒪k=kzk1\operatorname{GKdim}L(\lambda)=\dim\mathcal{O}_{k}=kz_{k-1}.

Now we suppose that L(λ)L(\lambda) is a unitary highest weight Harish-Chandra module and z=(λ+ρ,β)<zr1z=(\lambda+\rho,\beta^{\vee})<z_{r-1}. From Yamashita [Ya-94] we still have AV(L(λ))=𝒪k(λ)¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k(\lambda)}} for some 0k(λ)r0\leq k(\lambda)\leq r. If k(λ)r1k(\lambda)\leq r-1, by Theorem 1.1 we will have z=(λ+ρ,β)=zk(λ)z=(\lambda+\rho,\beta^{\vee})=z_{k(\lambda)} since L(λ)L(\lambda) is unitarizable. From our assumption, we will have z=zk(λ)<zr1z=z_{k(\lambda)}<z_{r-1}, which implies that

(ρ,β)k(λ)c<(ρ,β)(r1)cr1<k(λ)r1.(\rho,\beta^{\vee})-k(\lambda)c<(\rho,\beta^{\vee})-(r-1)c\Rightarrow r-1<k(\lambda)\leq r-1.

This is a contradiction! So we must have k(λ)=rk(\lambda)=r and GKdimL(λ)=dim𝒪r=rzr1\operatorname{GKdim}L(\lambda)=\dim\mathcal{O}_{r}=rz_{r-1}. ∎

Example 6.5.

Let 𝔤=𝔰𝔲(4,3)\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(4,3) and let L(λ)L(\lambda) be a highest weight Harish-Chandra module with highest weight λ=λ0+zζ\lambda=\lambda_{0}+z\zeta. Here λ0=(0,0,0,20,8,6,6)\lambda_{0}=(0,0,0,-20,8,6,6), ζ=(37,37,37,37,47,47,47)\zeta=(\frac{3}{7},\frac{3}{7},\frac{3}{7},\frac{3}{7},-\frac{4}{7},-\frac{4}{7},-\frac{4}{7}) and ρ=(3,2,1,0,1,2,3)\rho=(3,2,1,0,-1,-2,-3). From [EHW], we know the unitary reduction points correspond to z=3z=3 and 44. For 0kr1=min{p,q}1=20\leq k\leq r-1=\min\{p,q\}-1=2, we know

Ak=Δ(𝔭+)k+1={εpk+iεp+i+1=[0,,0pk+i1,1,,1k+1,0,,0qi1]0ik}.A_{k}=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{p-k+i}-\varepsilon_{p+i+1}=[\,\underbrace{0,\ldots,0}_{p-k+i-1},\underbrace{1,\ldots,1}_{k+1},\underbrace{0,\ldots,0}_{q-i-1}\,]\mid 0\leq i\leq k\}.

So A0={ε4ε5}A_{0}=\{\varepsilon_{4}-\varepsilon_{5}\}, A1={ε3ε5,ε4ε6}A_{1}=\{\varepsilon_{3}-\varepsilon_{5},\varepsilon_{4}-\varepsilon_{6}\}, A2={ε2ε5,ε3ε6,ε4ε7}A_{2}=\{\varepsilon_{2}-\varepsilon_{5},\varepsilon_{3}-\varepsilon_{6},\varepsilon_{4}-\varepsilon_{7}\}.

We write λ0+zζ+ρ=(3+37z,2+37z,1+37z,20+37z,747z,447z,347z)\lambda_{0}+z\zeta+\rho=(3+\frac{3}{7}z,2+\frac{3}{7}z,1+\frac{3}{7}z,-20+\frac{3}{7}z,7-\frac{4}{7}z,4-\frac{4}{7}z,3-\frac{4}{7}z). Thus we have z0(λ0)=28z_{0}(\lambda_{0})=28, z1(λ0)=7z_{1}(\lambda_{0})=7, and z2(λ0)=4z_{2}(\lambda_{0})=4. So

(6.1) GKdimL(λ)\displaystyle\operatorname{GKdim}L(\lambda) ={3z2,if z<z2(λ0)=4kzk1,if zk(λ0)z<zk1(λ0), where 1k20,if z0(λ0)=28z.\displaystyle=\begin{cases}3z_{2},&\mbox{if $z<z_{2}(\lambda_{0})=4$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-1}(\lambda_{0})$, where $1\leq k\leq 2$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})=28\leq z\in\mathbb{Z}$}.\\ \end{cases}
(6.2) ={12,if z<z2(λ0)=410,if z2(λ0)=4z<z1(λ0)=76,if z1(λ0)=7z<z0(λ0)=280,if z0(λ0)=28z.\displaystyle=\begin{cases}12,&\mbox{if $z<z_{2}(\lambda_{0})=4$}\\ 10,&\mbox{if $z_{2}(\lambda_{0})=4\leq z<z_{1}(\lambda_{0})=7$}\\ 6,&\mbox{if $z_{1}(\lambda_{0})=7\leq z<z_{0}(\lambda_{0})=28$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})=28\leq z\in\mathbb{Z}$}.\\ \end{cases}

For this given λ0\lambda_{0}, when AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0k20\leq k\leq 2, from Theorem 1.1, we have

L(λ) is unitarizable if and only if (λ+ρ,β)=zk=n1k=6k.L(\lambda)\text{ ~{}is unitarizable if and only if ~{}}(\lambda+\rho,\beta^{\vee})=z_{k}=n-1-k=6-k.

So from the above equation (6.2), we have

L(λ) is unitarizable if and only if (λ+ρ,β)=z=z2=4.L(\lambda)\text{ ~{}is unitarizable if and only if ~{}}(\lambda+\rho,\beta^{\vee})=z=z_{2}=4.

Note that z=3z=3 is a unitary reduction point with AV(L(λ))=𝒪3¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{3}}, which is not included in our Theorem 1.1.

Example 6.6.

Let 𝔤=𝔰𝔭(6,)\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(6,\mathbb{R}) and let L(λ)L(\lambda) be a highest weight Harish-Chandra module with highest weight λ=λ0+zζ\lambda=\lambda_{0}+z\zeta. Here λ0=(6,6,6,6,10,15)\lambda_{0}=(-6,-6,-6,-6,-10,-15), ζ=(1,1,1,1,1,1)\zeta=(1,1,1,1,1,1) and ρ=(6,5,4,3,2,1)\rho=(6,5,4,3,2,1). From [EHW], we know the unitary reduction points correspond to z=2.5,3,3.5z=2.5,3,3.5 and 44. For 0kr1=n1=50\leq k\leq r-1=n-1=5, we know

if k=2m is even,Ak\displaystyle\text{if $k=2m$ is even},A_{k} =Δ(𝔭+)k+1={2εnm,εnmi+εnm+i1im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{2\varepsilon_{n-m},\varepsilon_{n-m-i}+\varepsilon_{n-m+i}\mid 1\leq i\leq m\}
={[0,,0nm1i,1,,12i,2,,2mi,1]0im},\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-1-i},\underbrace{1,\ldots,1}_{2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\},
if k=2m+1 is odd,Ak\displaystyle\text{if $k=2m+1$ is odd},A_{k} =Δ(𝔭+)k+1={εnm1i+εnm+i0im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{n-m-1-i}+\varepsilon_{n-m+i}\mid 0\leq i\leq m\}
={[0,,0nm2i,1,,11+2i,2,,2mi,1]0im}.\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-2-i},\underbrace{1,\ldots,1}_{1+2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\}.

So A0={2ε6}A_{0}=\{2\varepsilon_{6}\}, A2={2ε5,ε4+ε6}A_{2}=\{2\varepsilon_{5},\varepsilon_{4}+\varepsilon_{6}\}, A4={2ε4,ε3+ε5,ε2+ε6}A_{4}=\{2\varepsilon_{4},\varepsilon_{3}+\varepsilon_{5},\varepsilon_{2}+\varepsilon_{6}\}, A1={ε5+ε6}A_{1}=\{\varepsilon_{5}+\varepsilon_{6}\}, A3={ε4+ε5,ε3+ε6}A_{3}=\{\varepsilon_{4}+\varepsilon_{5},\varepsilon_{3}+\varepsilon_{6}\}, A5={ε3+ε4,ε2+ε5,ε1+ε6}A_{5}=\{\varepsilon_{3}+\varepsilon_{4},\varepsilon_{2}+\varepsilon_{5},\varepsilon_{1}+\varepsilon_{6}\}.

We write λ0+zζ+ρ=(z,z1,z2,z3,z8,z14)\lambda_{0}+z\zeta+\rho=(z,z-1,z-2,z-3,z-8,z-14). Thus we have z0(λ0)=15z_{0}(\lambda_{0})=15, z2(λ0)=9z_{2}(\lambda_{0})=9, z4(λ0)=4z_{4}(\lambda_{0})=4, z1(λ0)=11.5z_{1}(\lambda_{0})=11.5, z3(λ0)=6.5z_{3}(\lambda_{0})=6.5, z5(λ0)=3.5z_{5}(\lambda_{0})=3.5. So

(6.3) GKdimL(λ)\displaystyle\operatorname{GKdim}L(\lambda) ={6z5,if z<z5(λ0)=3.5kzk1,if zk(λ0)z<zk2(λ0), where 2k5 and either z and k is even or z12+ and k is odd6,if z1(λ0)=11.5z12+0,if z0(λ0)=15z.\displaystyle=\begin{cases}6z_{5},&\mbox{if $z<z_{5}(\lambda_{0})=3.5$}\\ kz_{k-1},&\mbox{if $z_{k}(\lambda_{0})\leq z<z_{k-2}(\lambda_{0})$, where $2\leq k\leq 5$ and}\\ &\mbox{\quad either $z\in\mathbb{Z}$ and $k$ is even or $z\in\frac{1}{2}+\mathbb{Z}$ and $k$ is odd}\\ 6,&\mbox{if $z_{1}(\lambda_{0})=11.5\leq z\in\frac{1}{2}+\mathbb{Z}$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})=15\leq z\in\mathbb{Z}$}.\\ \end{cases}
(6.4) ={21,if z<z5(λ0)=3.520,if z5(λ0)=3.5z<z3(λ0)=6.5, where z12+18,if z4(λ0)=4z<z2(λ0)=9, where z15,if z3(λ0)=6.5z<z1(λ0)=11.5, where z12+11,if z2(λ0)=9z<z0(λ0)=15, where z6,if z1(λ0)=11.5z12+0,if z0(λ0)=15z.\displaystyle=\begin{cases}21,&\mbox{if $z<z_{5}(\lambda_{0})=3.5$}\\ 20,&\mbox{if $z_{5}(\lambda_{0})=3.5\leq z<z_{3}(\lambda_{0})=6.5$, where $z\in\frac{1}{2}+\mathbb{Z}$}\\ 18,&\mbox{if $z_{4}(\lambda_{0})=4\leq z<z_{2}(\lambda_{0})=9$, where $z\in\mathbb{Z}$}\\ 15,&\mbox{if $z_{3}(\lambda_{0})=6.5\leq z<z_{1}(\lambda_{0})=11.5$, where $z\in\frac{1}{2}+\mathbb{Z}$}\\ 11,&\mbox{if $z_{2}(\lambda_{0})=9\leq z<z_{0}(\lambda_{0})=15$, where $z\in\mathbb{Z}$}\\ 6,&\mbox{if $z_{1}(\lambda_{0})=11.5\leq z\in\frac{1}{2}+\mathbb{Z}$}\\ 0,&\mbox{if $z_{0}(\lambda_{0})=15\leq z\in\mathbb{Z}$}.\\ \end{cases}

For this given λ0\lambda_{0}, when AV(L(λ))=𝒪k¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{k}} with 0k50\leq k\leq 5, from Theorem 1.1, we have

L(λ) is unitarizable if and only if (λ+ρ,β)=z=zk=6k2.L(\lambda)\text{ ~{}is unitarizable if and only if ~{}}(\lambda+\rho,\beta^{\vee})=z=z_{k}=6-\frac{k}{2}.

So from the above equation (6.4), we have

L(λ) is unitarizable if and only if (λ+ρ,β)=z=z4=4,or z=z5=3.5.L(\lambda)\text{ ~{}is unitarizable if and only if ~{}}(\lambda+\rho,\beta^{\vee})=z=z_{4}=4,\text{or~{}}z=z_{5}=3.5.

Note that z=3z=3 and z=2.5z=2.5 are two unitary reduction points with AV(L(λ))=𝒪6¯\mathrm{AV}(L(\lambda))=\overline{\mathcal{O}_{6}}, which are not included in our Theorem 1.1.

Acknowledgments

Z. Bai was supported by the National Natural Science Foundation of China (No. 12171344).

7. Appendix

The diagrams of Δ(𝔭+)\Delta{(\mathfrak{p}^{+})}.

𝔤=𝔰𝔲(p,q)\mathfrak{g}_{\mathbb{R}}=\mathfrak{su}(p,q):

αp=εpεp+1\alpha_{p}=\varepsilon_{p}-\varepsilon_{p+1}p+1p+1p+2p+2εpεn\varepsilon_{p}-\varepsilon_{n}β=ε1εn=α1+α2++αn1\beta=\varepsilon_{1}-\varepsilon_{n}=\alpha_{1}+\alpha_{2}+\dots+\alpha_{n-1}

For 0kr1=min{p,q}10\leq k\leq r-1=\min\{p,q\}-1,

Ak=Δ(𝔭+)k+1={εpk+iεp+i+1=[0,,0pk+i1,1,,1k+1,0,,0qi1]0ik}.A_{k}=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{p-k+i}-\varepsilon_{p+i+1}=[\,\underbrace{0,\ldots,0}_{p-k+i-1},\underbrace{1,\ldots,1}_{k+1},\underbrace{0,\ldots,0}_{q-i-1}\,]\mid 0\leq i\leq k\}.

𝔤=𝔰𝔬(2n)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}^{*}(2n):

αn=εn1+εn\alpha_{n}=\varepsilon_{n-1}+\varepsilon_{n}β=ε1+ε2=α1+2α2++2αn2+αn1+αn\beta=\varepsilon_{1}+\varepsilon_{2}=\alpha_{1}+2\alpha_{2}+\dots+2\alpha_{n-2}+\alpha_{n-1}+\alpha_{n}n2n-2n3n-3\ddots1n1n-1n2n-2\cdot\cdot\cdot2

For 1kr1=[n2]11\leq k\leq r-1=[\frac{n}{2}]-1,

Ak\displaystyle A_{k} =Δ(𝔭+)2k+1={[0,,0n2k2,1,,12k,0,1],\displaystyle=\Delta(\mathfrak{p}^{+})_{2k+1}=\{[\,\underbrace{0,\ldots,0}_{n-2k-2},\underbrace{1,\ldots,1}_{2k},0,1\,],
[0,,0n2k1+i,1,,12k2i1,2,,2i,1,1]0ik1},\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}\quad\quad\quad\quad\quad\quad\quad[\,\underbrace{0,\ldots,0}_{n-2k-1+i},\underbrace{1,\ldots,1}_{2k-2i-1},\underbrace{2,\ldots,2}_{i},1,1\,]\mid 0\leq i\leq k-1\},
A0\displaystyle A_{0} =Δ(𝔭+)1={[0,0,,0,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[0,0,\dots,0,1]\}.

𝔤=𝔰𝔬(2,2n2)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-2):

β1\beta_{1}β2\beta_{2}23\ddotsn1n-1nn\cdot\cdot\cdot32α1=ε1ε2\alpha_{1}=\varepsilon_{1}-\varepsilon_{2}β=ε1+ε2=α1+2α2+2α3+2α4+α5+α6\beta=\varepsilon_{1}+\varepsilon_{2}=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+2\alpha_{4}+\alpha_{5}+\alpha_{6}
A0\displaystyle A_{0} =Δ(𝔭+)1={ε1ε2=[1,0,,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{\varepsilon_{1}-\varepsilon_{2}=[1,0,...,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)n1={ε1εn=[1,,1,0],ε1+εn=[1,,1,0,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{n-1}=\{\varepsilon_{1}-\varepsilon_{n}=[1,\dots,1,0],\varepsilon_{1}+\varepsilon_{n}=[1,\dots,1,0,1]\}.

𝔤=𝔢6(14)\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{6(-14)}:

α1\alpha_{1}β=α1+2α2+2α3+3α4+2α5+α6\beta=\alpha_{1}+2\alpha_{2}+2\alpha_{3}+3\alpha_{4}+2\alpha_{5}+\alpha_{6}β1\beta_{1}β2\beta_{2}34562435542
A0\displaystyle A_{0} =Δ(𝔭+)1={[1,0,0,0,0,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[1,0,0,0,0,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)4={[1,0,1,1,1,0],[1,1,1,1,0,0]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{4}=\{[1,0,1,1,1,0],[1,1,1,1,0,0]\}.

𝔤=𝔢7(25)\mathfrak{g}_{\mathbb{R}}=\mathfrak{e}_{7(-25)}:

6543124534265431α7\alpha_{7}β1\beta_{1}β2\beta_{2}γ1\gamma_{1}γ2\gamma_{2}γ3\gamma_{3}β=2α1+2α2+3α3+4α4+3α5+2α6+α7\beta=2\alpha_{1}+2\alpha_{2}+3\alpha_{3}+4\alpha_{4}+3\alpha_{5}+2\alpha_{6}+\alpha_{7}
A0\displaystyle A_{0} =Δ(𝔭+)1={[0,0,0,0,0,0,1]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{[0,0,0,0,0,0,1]\},
A1\displaystyle A_{1} =Δ(𝔭+)5={[0,0,1,1,1,1,1],[0,1,0,1,1,1,1]},\displaystyle=\Delta(\mathfrak{p}^{+})_{5}=\{[0,0,1,1,1,1,1],[0,1,0,1,1,1,1]\},
A2\displaystyle A_{2} =Δ(𝔭+)9={[1,1,2,2,1,1,1],[1,1,1,2,2,1,1],[0,1,1,2,2,2,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{9}=\{[1,1,2,2,1,1,1],[1,1,1,2,2,1,1],[0,1,1,2,2,2,1]\}.

𝔤=𝔰𝔬(2,2n1)\mathfrak{g}_{\mathbb{R}}=\mathfrak{so}(2,2n-1):

α1\alpha_{1}22n1n-1nnnnn1n-13322β=ε1+ε2\beta=\varepsilon_{1}+\varepsilon_{2}
A0\displaystyle A_{0} =Δ(𝔭+)1={ε1ε2=[1,0,,0]},\displaystyle=\Delta(\mathfrak{p}^{+})_{1}=\{\varepsilon_{1}-\varepsilon_{2}=[1,0,...,0]\},
A1\displaystyle A_{1} =Δ(𝔭+)n={ε1=[1,1,,1]}.\displaystyle=\Delta(\mathfrak{p}^{+})_{n}=\{\varepsilon_{1}=[1,1,\dots,1]\}.

𝔤=𝔰𝔭(n,)\mathfrak{g}_{\mathbb{R}}=\mathfrak{sp}(n,\mathbb{R}):

αn=2εn\alpha_{n}=2\varepsilon_{n}β=2ε1\beta=2\varepsilon_{1}n-1n-2\ddots1n-1n-2\cdot\cdot\cdot1

For 0kr1=n10\leq k\leq r-1=n-1,

if k=2m is even,Ak\displaystyle\text{if $k=2m$ is even},A_{k} =Δ(𝔭+)k+1={2εnm,εnmi+εnm+i1im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{2\varepsilon_{n-m},\varepsilon_{n-m-i}+\varepsilon_{n-m+i}\mid 1\leq i\leq m\}
={[0,,0nm1i,1,,12i,2,,2mi,1]0im},\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-1-i},\underbrace{1,\ldots,1}_{2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\},
if k=2m+1 is odd,Ak\displaystyle\text{if $k=2m+1$ is odd},A_{k} =Δ(𝔭+)k+1={εnm1i+εnm+i0im}\displaystyle=\Delta(\mathfrak{p}^{+})_{k+1}=\{\varepsilon_{n-m-1-i}+\varepsilon_{n-m+i}\mid 0\leq i\leq m\}
={[0,,0nm2i,1,,11+2i,2,,2mi,1]0im}.\displaystyle=\{[\,\underbrace{0,\ldots,0}_{n-m-2-i},\underbrace{1,\ldots,1}_{1+2i},\underbrace{2,\ldots,2}_{m-i},1\,]\mid 0\leq i\leq m\}.

References