Non-standard Verma type modules for $\mathfrak{q}(n)^{(2)}$

TL;DR

This paper investigates non-standard Verma modules over the Kac-Moody queer Lie superalgebra rak{q}(n)^{(2)}, providing conditions for irreducibility and classifying certain irreducible modules within related Heisenberg superalgebras.

Contribution

It introduces new criteria for irreducibility of non-standard Verma modules and classifies irreducible diagonal b-graded modules over specific Heisenberg superalgebras.

Findings

Identified sufficient conditions for module irreducibility.

Classified all irreducible diagonal b-graded modules in certain superalgebras.

Extended understanding of module structures over rak{q}(n)^{(2)}.

Abstract

We study non-standard Verma type modules over the Kac-Moody queer Lie superalgebra $\mathfrak{q}(n)^{(2)}$. We give a sufficient condition under which such modules are irreducible. We also give a classification of all irreducible diagonal $\mathbb{Z}$-graded modules over certain Heisenberg Lie superalgebras contained in $\mathfrak{q}(n)^{(2)}$.

Equations

$$M_{\hat{\mathfrak{b}}}(\lambda ):={\mathrm{Ind}}_{\hat{\mathfrak{b}}}^{\hat{\mathfrak{g}}}{\mathbb{C}}_{\lambda }.$$

$$\mathfrak{q}:=\{\left(\begin{array}[]{c|c}A&B\\ \hline\cr B&A\end{array}\right)\mid A\in\mathfrak{gl}(n+1),B\in\mathfrak{sl}(n+1)\}.$$

$$\hat{\mathfrak{g}}=L(\mathfrak{sl}(n))\oplus \mathbb{C}K\oplus \mathbb{C}D,{\hat{\mathfrak{g}}}_0=\mathfrak{sl}(n)\otimes \mathbb{C}[t^{\pm 2}]\oplus \mathbb{C}K\oplus \mathbb{C}D,\text{and}{\hat{\mathfrak{g}}}_1=\mathfrak{sl}(n)\otimes t\mathbb{C}[t^{\pm 2}],$$

$$[x(k),y(m)]=(xy-yx)(k+m),$$

$$[x(k),y(m)]=\iota (xy+yx)(k+m)+2{\delta }_{-k,m}\mathrm{tr}(xy)K,$$

$$(xy-yx):=[x,y{]}_0,\text{ and }\iota (xy+yx):=[x,y{]}_1.$$

$$[x(k),y(m)]=[x,y{]}_0(k+m),\text{ and }[x(k),y(m)]=[x,y{]}_1(k+m)+2{\delta }_{-k,m}\mathrm{tr}(xy)K,$$

$$\hat{\mathfrak{h}}:=\mathfrak{h}\otimes 1\oplus \mathbb{C}K\oplus \mathbb{C}D$$

$$[h,g_{{\epsilon }_i-{\epsilon }_j}{]}_1=({\epsilon }_i+{\epsilon }_j)(h)g_{{\epsilon }_i-{\epsilon }_j},\text{ for all }h.$$

$$[h,g_{\alpha }{]}_1=\bar{\alpha }(h)g_{\alpha },\text{ for all }h\in \mathfrak{h}.$$

$$[e_{{\alpha }_i},f_{{\alpha }_j}{]}_1=g_{{\alpha }_i+{\alpha }_j},$$

$$[e_{\alpha },f_{\alpha }{]}_1=\iota (h_{\alpha }^{\prime }),$$

$$\Delta (\mathfrak{a}):=\{\alpha \in \Delta \mid {\hat{\mathfrak{g}}}_{\alpha }\subseteq \mathfrak{a}\}.$$

$${\hat{\mathcal{H}}}_0=\sum _{r\in \mathbb{Z}}\mathfrak{h}\otimes t^{2r}\oplus \mathbb{C}K,{\hat{\mathcal{H}}}_1=\sum _{r\in \mathbb{Z}}\mathfrak{h}\otimes t^{2r+1}.$$

$$[x(2r+1),y(-2r-1)]=2xy,[x(2r+1),y(2s+1)]=\iota (2xy)(2(r+s+1))$$

$$\Delta (X{)}^{+}$$

$$\cup \{\alpha +k\delta \mid \alpha \in \dot{\Delta }(X)\cup \{0\},k\in {\mathbb{Z}}_{>0}\}\cup \dot{\Delta }(X{)}^{+}.$$

$$\mathfrak{m}(X):=\mathfrak{m}(X{)}^{-}\oplus \mathfrak{h}\oplus \mathfrak{m}(X{)}^{+},\mathfrak{m}(X{)}^{\pm }:=\underset{\alpha \in \dot{\Delta }(X{)}^{\pm }}{⨁}{\mathfrak{g}}_{\alpha }.$$

$$\mathfrak{u}(X{)}^{\pm }:=\underset{\alpha \in {\dot{\Delta }}^{\pm }\setminus \dot{\Delta }(X{)}^{\pm }}{⨁}{\mathfrak{g}}_{\alpha }.$$

$$\mathfrak{g}=\mathfrak{u}(X{)}^{-}\oplus \mathfrak{m}(X)\oplus \mathfrak{u}(X{)}^{+}\text{and}\hat{\mathfrak{g}}=L(\mathfrak{u}(X{)}^{-})\oplus \hat{\mathfrak{m}}(X)\oplus L(\mathfrak{u}(X{)}^{+}),$$

$$\mathfrak{k}(X):=\mathfrak{m}(X{)}^{-}\oplus {\mathfrak{h}}_X\oplus \mathfrak{m}(X{)}^{+},\text{ where }{\mathfrak{h}}_X:=\underset{\alpha \in \dot{\Delta }(X{)}^{+}}{⨁}[{\mathfrak{g}}_{-\alpha },{\mathfrak{g}}_{\alpha }].$$

$$\hat{\mathfrak{k}}(X):=L(\mathfrak{k}(X))\oplus \mathbb{C}K\oplus \mathbb{C}D\oplus {\mathfrak{h}}^X$$

$$\hat{\mathfrak{k}}(X)=\hat{\mathfrak{k}}(X{)}^{-}\oplus \hat{\mathfrak{h}}\oplus \hat{\mathfrak{k}}(X{)}^{+},\hat{\mathfrak{k}}(X{)}^{\pm }=(\mathfrak{k}(X)\otimes t^{\pm 1}\mathbb{C}[t^{\pm 1}])\oplus \mathfrak{m}(X{)}^{\pm }.$$

$$\hat{\mathfrak{m}}(X)={\mathfrak{h}}^X\otimes t^{-1}\mathbb{C}[t^{-1}]\oplus (\hat{\mathfrak{k}}(X{)}^{-}\oplus \hat{\mathfrak{h}}\oplus \hat{\mathfrak{k}}(X{)}^{+})\oplus {\mathfrak{h}}^X\otimes t\mathbb{C}[t]$$

$$\hat{\mathfrak{g}}=(L(\mathfrak{u}(X{)}^{-})\oplus {\mathfrak{h}}^X\otimes t^{-1}\mathbb{C}[t^{-1}])\oplus \hat{\mathfrak{k}}(X{)}^{-}\oplus \hat{\mathfrak{h}}\oplus \hat{\mathfrak{k}}(X{)}^{+}\oplus ({\mathfrak{h}}^X\otimes t\mathbb{C}[t]\oplus L(\mathfrak{u}(X{)}^{+})).$$

$$\mathcal{H}(X):=({\mathfrak{h}}^X\otimes t^{-1}\mathbb{C}[t^{-1}])\oplus ({\mathfrak{h}}^X\oplus \mathbb{C}K)\oplus ({\mathfrak{h}}^X\otimes t\mathbb{C}[t])$$

$$[{\mathcal{H}}_0(X),\hat{\mathfrak{k}}(X)]=[\mathcal{H}(X),\hat{\mathfrak{k}}(X{)}_0]=0,\text{ but }[{\mathcal{H}}_1(X),\hat{\mathfrak{k}}(X{)}_1]\ne 0.$$

$$\mathcal{H}(X)=\mathcal{H}(X{)}^{-}\oplus ({\mathfrak{h}}^X\oplus \mathbb{C}K)\oplus \mathcal{H}(X{)}^{+},$$

$$\mathcal{H}(X{)}^{\pm }:={\mathfrak{h}}^X\otimes t^{\pm 1}\mathbb{C}[t^{\pm 1}],$$

$$\mathcal{S}(X):=\mathbf{U}(\mathcal{H}(X{)}_0^{-}),$$

$$\hat{\mathfrak{m}}(X)=\hat{\mathfrak{m}}(X{)}^{-}\oplus \hat{\mathfrak{h}}\oplus \hat{\mathfrak{m}}(X{)}^{+},\text{where}\hat{\mathfrak{m}}(X{)}^{\pm }=\mathcal{H}(X{)}^{\pm }\oplus \hat{\mathfrak{k}}(X{)}^{\pm }$$

Full text

Introduction

Kac-Moody algebras and their representations play a very important role in many areas of mathematics and physics. The "super" version of these algebras was introduced in [Kac77]. Affine Kac-Moody superalgebras are those of finite growth. Affine symmetrizable superalgebras were described in [Ser11] and [vdL89]. Theory of Verma type modules for affine Lie superalgebras was developed in [ERF09] and [CF18]. In particular, given a Borel subsuperalgebra $\hat{\mathfrak{b}}$ of the affine Lie superalgebra $\hat{\mathfrak{g}}$ and a $1$-dimensional representation $\mathbb{C}_{\lambda}$ of $\hat{\mathfrak{b}}$ for some weight $\lambda$ of the Cartan subalgebra of $\hat{\mathfrak{g}}$, one can construct the Verma type module

[TABLE]

This module admits a unique maximal proper submodule, and thus, a unique simple quotient. The Verma type module is non-standard if $\hat{\mathfrak{b}}$ does not contain all positive root subspaces for some basis of the root system of $\hat{\mathfrak{g}}$. In the case the finite-dimensional Lie superalgebra associated to $\hat{\mathfrak{g}}$ is a contragredient Lie superalgebra, all Borel subsuperalgebras of $\hat{\mathfrak{g}}$ were described in [CF18], see also [DFG09]. The paper [CF18] also gives a criterion for the irreducibility for non-standard Verma type module.

Non-symmetrizable affine Lie superalgebras were classified in [HS07]. In particular, this classification includes a degenerate family of affine Lie superalgebras, series $\mathfrak{q}(n)^{(2)}$. These superalgebras are twisted affinizations of queer Lie superalgebras $\mathfrak{q}(n)$. Structure of Verma modules (= standard Verma type modules) over the twisted affine superalgebra $\mathfrak{q}(n)^{(2)}$ with $n\geq 3$ was studied in [GS08]. The current paper advances the theory of Verma type modules for the affine queer Lie superalgebra. We establish sufficient conditions for the irreducibility of all non-standard Verma type modules (Theorem 4.1 and Theorem 4.7). We also consider modules induced from analogs of Heisenberg subsuperalgebra and give a criterion of their irreducibility (Theorem 3.9, Corollary 4.8).

Notation The ground field is $\mathbb{C}$. All vector spaces, algebras, and tensor products are considered to be over $\mathbb{C}$, unless otherwise stated. For a vector space $V$ we denote by $\Lambda(V)$ its Grassmann algebra (i.e., its exterior algebra). For any Lie superalgebra $\mathfrak{a}$ we let $\mathbf{U}(\mathfrak{a})$ denote its universal enveloping algebra.

1. Preliminaries

Let $\mathfrak{q}=\mathfrak{q}(n)$ for $n\geq 3$, be the queer Lie superalgebra, that is,

[TABLE]

Let $\mathfrak{q}_{0}$ and $\mathfrak{q}_{1}$ be the even and odd parts of $\mathfrak{q}$, respectively. Choose a Cartan subalgebra $\mathfrak{h}_{\mathfrak{q}}=\mathfrak{h}_{0}\oplus\mathfrak{h}_{1}$ of $\mathfrak{q}$ (i.e. $\mathfrak{h}_{0}$ a Cartan subalgebra of $\mathfrak{q}_{0}$) and let $\mathfrak{q}=\mathfrak{h}_{\mathfrak{q}}\oplus\left(\bigoplus_{\alpha\in\dot{\Delta}}\mathfrak{q}^{\alpha}\right)$ be the root space decomposition of $\mathfrak{q}$, where $\mathfrak{q}^{\alpha}$ denotes the root space associated to the root $\alpha\in\dot{\Delta}\subseteq\mathfrak{h}_{0}^{*}$. Recall that every root of $\dot{\Delta}$ is both even and odd, meaning that, for any $\alpha\in\dot{\Delta}$, $\mathfrak{q}^{\alpha}\cap\mathfrak{q}_{i}\neq 0$, for $i=0,1$. Recall also that $\dot{\Delta}=\dot{\Delta}_{0}=\dot{\Delta}_{1}=\{\varepsilon_{i}-\varepsilon_{j}\mid i\neq j\}$.

Although neither $\mathfrak{q}$ nor its affinization $\mathfrak{q}^{(1)}$ are Kac-Moody Lie superalgebras, i.e. admit a set of simple generators, after a twist of $\mathfrak{q}^{(1)}$ by an involution we obtain a regular quasisimple Kac-Moody superalgebra $\hat{\mathfrak{g}}:=\mathfrak{q}^{(2)}$ (see [Ser11]). As a super vector space we have that

[TABLE]

where for any Lie superalgebra $\mathfrak{k}$, $L(\mathfrak{k}):=\mathfrak{k}\otimes\mathbb{C}[t^{1},t^{-1}]$ is its associated loop superalgebra, $K$ is a central element, and, for all $x(k):=x\otimes t^{k}\in L(\mathfrak{k})$ with $x\in\mathfrak{k}$ and $k\in\mathbb{Z}$, we have $[D,x(k)]=kx(k)$. Let $\mathfrak{g}=\mathfrak{sl}(n)$. Then for any $x,y\in\mathfrak{g}$, the bracket of $\hat{\mathfrak{g}}$ is given as follows:

[TABLE]

if $km$ is even; and if we define $\iota:\mathfrak{gl}(n)\to\mathfrak{sl}(n)$ by $x\mapsto x-\frac{\operatorname{tr}(x)}{n}I_{n}$ where $I_{n}$ is the $n\times n$ identity matrix, then

[TABLE]

if $km$ is odd. Notice that $K$ does not lie in $[\hat{\mathfrak{g}}_{0},\hat{\mathfrak{g}}]$, but it lies in $[\hat{\mathfrak{g}}_{1},\hat{\mathfrak{g}}_{1}]$. For convenience we set

[TABLE]

Hence, in this notation we have that

[TABLE]

if $km$ is even/odd, respectively.

Remark 1.1**.**

Notice that if we assume $m\in 2\mathbb{Z}$, then the bracket between any two elements $x(m),y(k)\in L(\mathfrak{g})$ reduces to the bracket in the loop Lie algebra $L(\mathfrak{g})$.

Fix a Cartan subalgebra of $\hat{\mathfrak{g}}$

[TABLE]

where $\mathfrak{h}$ is the Cartan subalgebra of diagonal matrices in $\mathfrak{g}$, and for each $\alpha\in\dot{\Delta}$, choose $f_{\alpha}\in\mathfrak{g}^{-\alpha}$, $e_{\alpha}\in\mathfrak{g}^{\alpha}$ and $h_{\alpha}\in\mathfrak{h}$ such that $[f_{\alpha},e_{\alpha}]_{0}=h_{\alpha}$.

Notice that, for $g_{\varepsilon_{i}-\varepsilon_{j}}\in\mathfrak{g}^{\varepsilon_{i}-\varepsilon_{j}}$, we have

[TABLE]

For simplicity, if $\alpha=\varepsilon_{i}-\varepsilon_{j}$, then we set ${\bar{\alpha}}:=\varepsilon_{i}+\varepsilon_{j}$. Thus, in this notation, we have that

[TABLE]

Moreover, if $\alpha_{i}\neq-\alpha_{j}$, then

[TABLE]

where $g_{\alpha_{i}+\alpha_{j}}=0$ if $\alpha_{i}+\alpha_{j}\notin\dot{\Delta}$, $g_{\alpha_{i}+\alpha_{j}}=f_{\alpha_{i}+\alpha_{j}}$ if $\alpha_{i}+\alpha_{j}\in\dot{\Delta}^{-}$ and $g_{\alpha_{i}+\alpha_{j}}=e_{\alpha_{i}+\alpha_{j}}$ if $\alpha_{i}+\alpha_{j}\in\dot{\Delta}^{+}$. Finally, for $\alpha=\varepsilon_{i}-\varepsilon_{j}$ we have

[TABLE]

where $h_{\alpha}^{\prime}=E_{i,i}+E_{j,j}$.

If we identify $K$ with $(1/n)I_{n}$, then $\mathfrak{h}\otimes 1\oplus\mathbb{C}K$ can be identified with the Cartan subalgebra of diagonal matrices of $\mathfrak{gl}(n)$. Let $H_{1},\ldots,H_{n}$ denote the standard basis of it (i.e. $H_{i}=E_{ii}$). The root system of $\hat{\mathfrak{g}}$ with respect to $\hat{\mathfrak{h}}$ is given by $\Delta=\{\alpha+k\delta,\ m\delta\mid\alpha\in\dot{\Delta},\ k\in\mathbb{Z},\ m\in\mathbb{Z}\setminus\{0\}\}$. Moreover, $p(\alpha+k\delta)=p(k)$ and $p(m\delta)=p(m)$, where $p(k)$ denotes the parity of $k$, and by abuse of notation we are denoting the parity of a root $\beta$ also by $p(\beta)$. Finally, for a subalgebra $\mathfrak{a}\subseteq\hat{\mathfrak{g}}$ we set

[TABLE]

Consider the subalgebra $\widehat{\mathcal{H}}=\widehat{\mathcal{H}}_{0}\oplus\widehat{\mathcal{H}}_{1}$ generated by the imaginary root spaces of $\hat{\mathfrak{g}}$. Then

[TABLE]

Notice that the center of $\widehat{\mathcal{H}}$ equals to $\widehat{\mathcal{H}}_{0}$, the odd part $\widehat{\mathcal{H}}_{1}$ is spanned by $\{(H_{i}-H_{i+1})(2r+1)\mid r\in\mathbb{Z}\}$ and the relations in $\widehat{\mathcal{H}}_{1}$ are given by

[TABLE]

for $r+s+1\neq 0$. In particular, differently from the case of basic classical Lie superalgebras, the subalgebra $\widehat{\mathcal{H}}$ is not isomorphic to a Heisenberg algebra.

2. Generalized Verma type modules

Since the root system $\Delta$ of $\hat{\mathfrak{g}}$ is the same as that of $\widehat{\mathfrak{sl}}(n)$, the sets of positive roots of $\Delta$ are obtained in the same way: fix $\Pi\subseteq\dot{\Delta}$ a set of simple roots, pick a subset $X\subseteq\Pi$, and let $\mathcal{W}$ denote the Weyl group of $\widehat{\mathfrak{sl}}(n)$. Let $\dot{\Delta}^{+}=\langle\Pi\rangle_{\mathbb{Z}_{>0}}\cap\dot{\Delta}$, $\dot{\Delta}(X)^{+}=\langle X\rangle_{\mathbb{Z}_{>0}}\cap\dot{\Delta}$, and $\dot{\Delta}(X)=\langle X\rangle_{\mathbb{Z}}\cap\dot{\Delta}$. Associated to $X$ we define

[TABLE]

Then $\Delta(X)^{+}$ is a set of positive roots of $\Delta$, and up to $\mathcal{W}\times\{\pm 1\}$-conjugation, every set of positive roots is of this form for some set of simple roots $\Pi$ and some subset $X\subseteq\Pi$.

Consider the following subalgebras associated to $X$:

1. [TABLE]
2. [TABLE]

Thus

[TABLE]

where $\hat{\mathfrak{m}}(X)=L(\mathfrak{m}(X))\oplus\mathbb{C}K\oplus\mathbb{C}D$.

Consider now the subalgebra

[TABLE]

Then $\mathfrak{m}(X)=\mathfrak{k}(X)\oplus\mathfrak{h}^{X}$, where $\mathfrak{h}^{X}:=\{h\in\mathfrak{h}\mid\alpha(h)=0,\ \forall\alpha\in\dot{\Delta}(X)\}$ is the center of $\mathfrak{m}(X)$. Set

[TABLE]

with standard triangular decomposition

[TABLE]

In particular, we have that

[TABLE]

and

[TABLE]

Remark 2.1**.**

1. (a)

Differently from the case of basic classical Lie superalgebras (this includes all simple Lie algebras), the imaginary subalgebra

[TABLE]

is not a Heisenberg algebra. Another difference (from the Lie algebra case) is that we do not have that $[\mathcal{H}(X),\hat{\mathfrak{k}}(X)]=0$. In fact,

[TABLE]

Compare also with the isotropic case of [CF18]. 2. (b)

Heisenberg algebras admit a family of triangular decompositions parametrized by maps $\varphi:\mathbb{N}\to\{\pm\}^{d}$, where $d$ is a certain dimension. It is worth noting that the algebra $\mathcal{H}(X)$ does not admit such decompositions, except the trivial ones (i.e. when $\varphi(i)=(+,\ldots,+)$ for all $i\in\mathbb{N}$, or $\varphi(i)=(-,\ldots,-)$ for all $i\in\mathbb{N}$).

Consider the triangular decomposition of $\mathcal{H}(X)$

[TABLE]

where

[TABLE]

and define $\mathcal{H}(X)_{i}^{\pm}:=\mathcal{H}(X)_{i}\cap\mathcal{H}(X)^{\pm}$, for $i\in\mathbb{Z}_{2}$. Then we have a commutative algebra

[TABLE]

and we let $\mathcal{S}(X)^{+}$ denote the augmentation ideal of $\mathcal{S}(X)$.

Consider the triangular decompositions

[TABLE]

and

[TABLE]

Fix the subalgebra $\hat{\mathfrak{b}}(X):=\hat{\mathfrak{h}}\oplus\hat{\mathfrak{g}}(X)^{+}$ of $\hat{\mathfrak{g}}$. Notice that $\hat{\mathfrak{g}}(X)^{+}\cap\hat{\mathfrak{m}}(X)=\hat{\mathfrak{m}}(X)^{+}$, $\hat{\mathfrak{g}}(X)^{+}\cap\hat{\mathfrak{k}}(X)=\hat{\mathfrak{k}}(X)^{+}$, and $\hat{\mathfrak{g}}(X)^{+}\cap\mathcal{H}(X)=\mathcal{H}(X)^{+}$. In what follows, we fix a set $X\subseteq\Pi$ and we drop the $X$ from the notation above (for instance, we write $\hat{\mathfrak{m}}^{+}$ instead of writing $\hat{\mathfrak{m}}(X)^{+}$).

Let $\lambda\in\hat{\mathfrak{h}}^{*}$, $\hat{\mathfrak{s}}\in\{\hat{\mathfrak{g}},\hat{\mathfrak{m}},\hat{\mathfrak{k}},\mathcal{H}\}$, and $\hat{\mathfrak{r}}=\hat{\mathfrak{s}}\cap\hat{\mathfrak{b}}$. Then we define the Verma $\hat{\mathfrak{s}}$-module

[TABLE]

where $\mathbb{C}v_{\lambda}$ is the $\hat{\mathfrak{r}}$-module whose action of $\hat{\mathfrak{h}}$ is determined by $\lambda$ and the action of the nilpotent radical of $\hat{\mathfrak{r}}$ is trivial. The unique irreducible quotient of $M(\hat{\mathfrak{s}},\lambda)$ will be denoted $L(\hat{\mathfrak{s}},\lambda)$. Also, for $\hat{\mathfrak{s}},\hat{\mathfrak{r}}$ such that either $\hat{\mathfrak{r}}=\hat{\mathfrak{k}}$ and $\hat{\mathfrak{s}}=\hat{\mathfrak{m}}$, or $\hat{\mathfrak{r}}=\hat{\mathfrak{m}}$ and $\hat{\mathfrak{s}}=\hat{\mathfrak{g}}$, and an $\hat{\mathfrak{r}}$-module $N$ we define the module

[TABLE]

where $\mathcal{H}^{+}$ is assumed to act trivially on $N$ if $\hat{\mathfrak{r}}=\hat{\mathfrak{k}}$ and $\hat{\mathfrak{s}}=\hat{\mathfrak{m}}$, and $L(\mathfrak{u}^{+})$ is assumed to act trivially on $N$ if $\hat{\mathfrak{r}}=\hat{\mathfrak{m}}$ and $\hat{\mathfrak{s}}=\hat{\mathfrak{g}}$. Notice that

[TABLE]

Using the terminology of [Fut97], the module $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};N)$ is called a generalized Verma type module, or a generalized Imaginary Verma module. When $N$ is an irreducible weight $\hat{\mathfrak{r}}$-module, $M(\hat{\mathfrak{s}},\hat{\mathfrak{r}};N)$ admits a unique irreducible quotient which will be denoted by $L(\hat{\mathfrak{s}},\hat{\mathfrak{r}};N)$.

3. Irreducible $\mathcal{H}$-modules

Consider the triangular decomposition

[TABLE]

Then we have the following character formula

[TABLE]

Notice that the subalgebra $\mathcal{S}$ lies in the center of $\mathbf{U}(\mathcal{H})$ and acts freely on $M(\mathcal{H},\lambda)$. Then any ideal $J$ of $\mathcal{S}$ defines the $\mathcal{H}$-submodule $JM(\mathcal{H},\lambda)$ of $M(\mathcal{H},\lambda)$. On the other direction, for any $\mathcal{H}$-submodule $N\subseteq M(\mathcal{H},\lambda)$ we define an ideal $J_{N}$ of $\mathcal{S}$ by requiring the equality:

[TABLE]

In other words, $J_{N}=\{a\in\mathcal{S}\mid av_{\lambda}\in N\}$.

Let $D_{\delta}^{X}$ be the matrix determined by the pairing

[TABLE]

and consider $\det D_{\delta}^{X}$ as an element of the symmetric algebra $S(\hat{\mathfrak{h}})$.

Example 3.1**.**

If $n=3$ and $X=\{\varepsilon_{1}-\varepsilon_{2}\}$, then $\mathfrak{h}^{X}=\mathbb{C}h^{1}$, where $h^{1}=H_{1}+H_{2}-2H_{3}$. In particular, $\det D_{\delta}^{X}=2(H_{1}+H_{2}-2H_{3})^{2}$. If $X=\emptyset$ and $n\geq 3$, then $\mathfrak{h}^{X}=\mathfrak{h}$, and $\det D_{\delta}^{X}=2^{n-1}H_{1}\cdots H_{n}(\frac{1}{H_{1}}+\cdots+\frac{1}{H_{n}})$ (see [GS08]).

Proposition 3.2**.**

The $\mathcal{H}$-module $M(\mathcal{H},\lambda)$ is reducible. If $\det D_{\delta}^{X}(\lambda)\neq 0$, then there is a bijection between submodules of $M(\mathcal{H},\lambda)$ and ideals of $\mathcal{S}$. In particular, $L(\mathcal{H},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})$ as vector spaces.

Proof.

The fact that $\mathcal{H}_{0}^{-}$ is in the center of $\mathcal{H}$ implies that any ideal $J$ of $\mathcal{S}$ defines a submodule of $M(\mathcal{H},\lambda)$, namely, $JM(\mathcal{H},\lambda)$. Thus the first statement follows.

Now, if we assume $\det D_{\delta}^{X}(\lambda)\neq 0$, then we can use similar arguments to those of [GS08, Proposition 3] to prove that there is a bijection between ideals of $\mathcal{S}$ and submodules of $M(\mathcal{H},\lambda)$. Namely, let $M=M(\mathcal{H},\lambda)$, and let $N$ be a submodule of $M$. We claim that $N=J_{N}M$. Indeed, let $jm\in J_{N}M$. Then, writing $m=uv_{\lambda}$ with $u\in\mathbf{U}(\mathcal{H}^{-})$, we get that

[TABLE]

Thus $J_{N}M\subseteq N$. In order to prove the other inclusion, we consider the canonical projection $\pi:M\to V:=M/J_{N}M$, $W=\pi(N)$, and $R=\mathcal{S}/J_{N}$. Notice that $V$ is free as an $R$-module, and that $W\cap Rv_{\lambda}=\pi(N\cap\mathcal{S}v_{\lambda})=\pi(J_{N}v_{\lambda})=0$. Now we suppose that $W\neq 0$ to get a contradiction.

Let $h^{1},\ldots,h^{t}$ be any fixed basis of $\mathfrak{h}^{X}$, and set $X_{i,m}:=h^{i}(m)$, and $Y_{i,m}:=h^{i}(-m)$. Recall from the commutation relations of $\mathcal{H}$ that $[X_{i,m},Y_{k,m}]=[X_{i,0},Y_{k,0}]$, and since we are assuming that $\det D_{\delta}^{X}(\lambda)\neq 0$, we may consider that the basis elements $h^{1},\ldots,h^{t}$ were chosen so that $\lambda([X_{i,j},Y_{k,j}])=\delta_{i,k}$. Notice that the elements $X_{i,j}$ for $i=1,\ldots,r$ and $m\geq 0$ form a basis for $\mathcal{H}_{1}^{-}$. In particular, if we let $X_{i,m}\geq X_{k,n}$ if $m\geq n$ or $m=n$ and $i\geq k$, then the monomials $X_{i_{1},m_{1}}\cdots X_{i_{s},m_{s}}$ with $X_{i_{1},m_{1}}>\cdots>X_{i_{s},m_{s}}$ form a basis $B$ of $V$ over $R$.

Since we are assuming $W\neq 0$, and since $W\cap Rv_{\lambda}=0$, we can choose a nonzero $v\in W$ such that the maximal $X_{i,m}$ that occurs in the expression of $v$ as a linear combination of elements of $B$ is minimal among all nonzero vectors of $W$. Now we write $v=X_{i,m}w+u$ for nonzero $w$ and $u$ such that all factors occurring in $w$ and $u$ are less than $X_{i,m}$. Thus

[TABLE]

as $[X_{i,m},Y_{i,m}]$ is in the center of $\mathcal{H}$ and it acts as $\lambda([X_{i,m},Y_{k,m}])=1$ on $v_{\lambda}$. But this implies $0\neq w\in W$, and all factors occurring in $w$ are less than $X_{i,m}$, which is a contradiction. ∎

Corollary 3.3**.**

Suppose that $\det D_{\delta}^{X}(\lambda)\neq 0$. Then we have the character formula

[TABLE]

Proof.

This follows from the isomorphism of vector spaces $L(\mathcal{H},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})$. ∎

1. Modules for Heisenberg Lie superalgebra

In this section we consider the special case where $X=\emptyset$, and, in particular, $\mathfrak{h}^{X}=\mathfrak{h}$ and $\mathcal{H}=\mathcal{H}(X)=L(\mathfrak{h})\oplus\mathbb{C}K$.

Define

[TABLE]

It is clear that $\mathcal{H}_{0}^{\prime}$ is an ideal of $\mathcal{H}$, and $K\notin\mathcal{H}_{0}^{\prime}$. Define

[TABLE]

Lemma 3.4**.**

Let $\pi:\mathcal{H}\to\widetilde{\mathcal{H}}$ be the canonical projection. Then there exists a basis $\{h^{1},\ldots,h^{n-1}\}$ of $\mathfrak{h}$ such that $\pi(h^{i}h^{j})=\delta_{ij}K$.

Proof.

The set $\{H_{1}+\cdots+H_{i}-iH_{i+1}\mid 1\leq i\leq n-1\}$ is a basis of $\mathfrak{h}$ such that $H^{i}H^{j}\in\mathfrak{h}$ if and only if $i\neq j$. Then a suitable normalization of this basis gives the required one. ∎

Now we have the following result:

Proposition 3.5**.**

$\widetilde{\mathcal{H}}$* is an infinite dimensional Heisenberg Lie superalgebra such that*

[TABLE]

as vector spaces, where $[h\otimes t^{2r+1},h^{\prime}\otimes t^{-2r-1}]$ is a multiple of $K$ and $[h\otimes t^{2r+1},h^{\prime}\otimes t^{2s+1}]=0$ for all $h,h^{\prime}\in\mathfrak{h}$ and all integer $r,s$ with $r+s+1\neq 0$.

Fix a basis $h^{1},\ldots,h^{n-1}$ of $\mathfrak{h}$ as in Lemma 3.4, and let $\varphi:\mathbb{N}\to\{\pm\}^{n-1}$ be a map of sets. Then $\varphi$ induces a triangular decomposition on $\widetilde{\mathcal{H}}$:

[TABLE]

where

[TABLE]

and $\varphi(n)=(\varphi(n)_{1},\ldots,\varphi(n)_{n-1})$. The Verma module associated to such a decomposition is called the $\varphi$-Verma module and it is denoted by $M_{\varphi}(\widetilde{\mathcal{H}},a)$, where $a\in\mathbb{C}$ is the value of $K$ on $M_{\varphi}(\widetilde{\mathcal{H}},a)$. The module $M_{\varphi}(\widetilde{\mathcal{H}},a)$ is isomorphic (as a vector space) to $\mathbf{U}(\widetilde{\mathcal{H}}_{\varphi}^{-})$ which is nothing but the Grassmann algebra $\Lambda(\widetilde{\mathcal{H}}_{\varphi}^{-})$. Finally let $L_{\varphi}(\widetilde{\mathcal{H}},a)$ denote the unique irreducible quotient of $M_{\varphi}(\widetilde{\mathcal{H}},a)$.

Remark 3.6**.**

Notice that every $\widetilde{\mathcal{H}}$-module can (and will) be regarded as an $\mathcal{H}$-module via the canonical projection $\mathcal{H}\twoheadrightarrow\widetilde{\mathcal{H}}$.

Corollary 3.7**.**

If $\lambda(\mathfrak{h})=0$, then the action of $\mathcal{H}$ on $L(\mathcal{H},\lambda)$ factors through the epimorphism

[TABLE]

In particular, if $\lambda(K):=a\neq 0$, then $L(\mathcal{H},\lambda)\cong M_{\varphi}(\widetilde{\mathcal{H}},a)$ as $\mathcal{H}$-modules, where $\varphi(i)=(+,\ldots,+)$ for all $i\in\mathbb{N}$ (i.e. $M_{\varphi}(\widetilde{\mathcal{H}},a)$ is nothing but the standard Verma module of $\widetilde{\mathcal{H}}$).

Proof.

We have $(\mathfrak{h}\otimes t^{2}\mathbb{C}[t])L(\mathcal{H},\lambda)=0$, since $\mathfrak{h}\otimes t^{2}\mathbb{C}[t]$ is in the center of $\mathcal{H}$ and it acts trivially on $v_{\lambda}$. Next, $\mathfrak{h}\otimes t^{-2}\mathbb{C}[t^{-1}]$ is contained in the maximal ideal $\mathcal{S}^{+}$ of $\mathcal{S}$, and then, by Proposition 3.2, we must have $(\mathfrak{h}\otimes t^{-2}\mathbb{C}[t^{-1}])L(\mathcal{H},\lambda)=0$. Finally, since $\lambda(\mathfrak{h})=0$, we conclude that $\mathcal{H}_{0}^{\prime}L(\mathcal{H},\lambda)=0$ and the first statement follows.

Using similar arguments as those of [BBFK13, Proposition 3.3] one easily shows that $M_{\varphi}(\widetilde{\mathcal{H}},a)$ is an irreducible $\widetilde{\mathcal{H}}$-module if and only if $a\neq 0$. Thus the result follows. ∎

Let $N$ be an irreducible $\mathcal{H}$-module such that $\mathfrak{h}N=0$. We are interested in the case when $N$ is $\mathbb{Z}$-graded. Then we can define the action of $D$ on $N$ by $D|_{N_{i}}=i\operatorname{Id}$. Notice that under such conditions $\mathcal{H}_{0}^{\prime}$ must act trivially on $N$ (indeed, $N$ is irreducible and $\mathbb{Z}$-graded, $\mathcal{H}_{0}^{\prime}$ is central in $\mathcal{H}$, $\mathfrak{h}N=0$ and any element of $\mathfrak{h}\otimes t^{2r}$ with $r\in\mathbb{Z}^{\times}$ have degree different from [math]).

Set $x_{k}^{j}=h^{j}\otimes t^{k}$, $k\in\mathbb{Z}$, $j=1,\ldots,n-1$, so that

[TABLE]

and $[x_{2r+1}^{j},x_{2s-1}^{i}]=\delta_{ij}\delta_{r,-s}K$, after suitable rescaling (see Lemma 3.4 and Proposition 3.5). Also set

[TABLE]

Since $K$ is central and $N$ is irreducible, we have that $K$ acts on $N=\sum_{i\in\mathbb{Z}}N_{i}$ via multiplication by some $a\in\mathbb{C}$. Assume that $a\neq 0$, and fix a nonzero $v\in N_{i}$ for some $i$. Then

[TABLE]

that is, $d_{2r+1}^{j}$ is diagonalizable on $N_{i}$ and has eigenvalues $a$ or [math]. Now we have:

Lemma 3.8**.**

If $d_{2r+1}^{j}v=av$, then $x_{-2r-1}^{j}v=0$. On the other hand, if $d_{2r+1}^{j}v=0$, then $x_{2r+1}^{j}v=0$.

Proof.

The fact that $x^{j}_{-2r-1}d^{j}_{2r+1}=0$ implies the first statement. For the second statement observe that $d_{2r+1}^{j}v=0$ implies $x_{2r+1}^{j}x_{-2r-1}^{j}v=av$. Hence the result follows. ∎

A non-zero $\mathbb{Z}$-graded $\mathcal{H}$-module $N$ is diagonal if all $d_{2r+1}^{j}$ are simultaneously diagonalizable for $r\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$. Let $N_{i}$ be a graded component of a diagonal $\mathbb{Z}$-graded $\mathcal{H}$-module $N$. We associate to $N_{i}$ a $t$-tuple $(\mu^{1},\ldots,\mu^{n-1})$ of infinite sequences $\mu^{j}=(\mu_{2r+1}^{j})$ consisting of the eigenvalues $\mu_{2r+1}^{j}$ of $d_{2r+1}^{j}$, $r\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$. In what follows we classify all diagonal irreducible modules with trivial action of $\mathfrak{h}$, and we describe their structure.

Theorem 3.9**.**

Let $N$ be an irreducible diagonal $\mathbb{Z}$-graded $\mathcal{H}$-module, such that $\mathfrak{h}N=0$ and $Kv=av$ for some $a\in\mathbb{C}$ and all $v\in N$. Then the following hold:

1. (a)

$\mathcal{H}_{0}^{\prime}$* acts trivially on $N$, which is irreducible $\widetilde{\mathcal{H}}$-module;* 2. (b)

If $v\in N$ is a nonzero homogeneous element, then $v$ is $\varphi_{\mu}$-highest vector, where $\varphi_{\mu}$ is determined by the eigenvalues of $d_{2r+1}^{j}$ on $v$, and $N\simeq L_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)$ up to a shift of gradation. In particular, if $a\neq 0$, then $N\simeq M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)$ up to a shift of gradation; 3. (c)

If $a=0$, then $N$ is the trivial $1$-dimensional module. 4. (d)

If $a\neq 0$, then $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)$ has finite dimensional graded components if and only if $\varphi_{\mu}$ differs from $\varphi_{\nu}$ only in finitely many places, where $\nu_{2k+1}^{j}=0$ for all $k\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$, or $\nu_{2k+1}^{j}\neq 0$ for all $k\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$.

Proof.

Part(a): this follows from the fact that $N$ is irreducible and $\mathbb{Z}$-graded, $\mathcal{H}_{0}^{\prime}$ is central and its elements have degree different from [math]. Part(b): let $N_{i}\neq 0$ such that all $d_{2r+1}^{j}$ are simultaneously diagonalizable with eigenvalues $\mu_{2r+1}^{j}$. Set $\mu^{j}=(\mu_{2r+1}^{j})$, $r\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$. By Lemma 3.8, each $(\mu^{1},\ldots,\mu^{n-1})$ defines a function $\varphi_{\mu}:\mathbb{N}\to\{\pm\}^{n-1}$, where $\varphi_{\mu}(k)_{j}=+$ if $\mu_{2k+1}^{j}=0$ and $\varphi_{\mu}(k)_{j}=-$ if $\mu_{2k+1}^{j}=a$. Then $v$ is a $\varphi_{\mu}$-highest vector and $N\simeq L_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)$ up to a shift of gradation. Part(c) is clear. Part(d): without loss of generality we assume that $\nu_{2k+1}^{j}=0$ for all $k\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$. Clearly, $M_{\varphi_{\nu}}(\widetilde{\mathcal{H}},\lambda)$ has finite dimensional graded components. Suppose that $\varphi_{\mu}$ differs from $\varphi_{\nu}$ only in $s$ places. Consider a nonzero $\varphi_{\mu}$-highest vector $v$. If $w=x_{2k+1}^{j}v\neq 0$ for some $k\geq 0$ and $j=1,\ldots,n-1$, then $x_{2k+1}^{j}w=0$ and thus $w$ is a $\varphi_{\mu^{\prime}}$-highest vector where $\varphi_{\mu^{\prime}}$ differs from $\varphi_{\nu}$ in $s-1$ places. Continuing we find a $\varphi_{\nu}$-highest vector in $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)$. Since $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)$ is irreducible when $a\neq 0$ we conclude that $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)\simeq M_{\varphi_{\nu}}(\widetilde{\mathcal{H}},\lambda)$ and hence it has finite dimensional graded components. Conversely, assume that $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)$ has finite dimensional graded components and let $v$ be a nonzero $\varphi_{\mu}$-highest vector. Denote by $\Omega_{\mu}$ the subset of odd integers defined as follows: $k\in\Omega_{\mu}$ if $x_{k}^{j}v\neq 0$ for at least one $j=1,\ldots,n-1$. A sequence $(k_{1},\ldots,k_{r})$ of $\Omega_{\mu}$ is called cycle if $\sum_{i=1}^{r}k_{i}=0$. Suppose $\Omega$ contains infinitely many positive as well as negative odd integers. Then one can form infinitely many cycles. Each such cycle $(k_{1},\ldots,k_{r})$ yields a basis element $\Pi_{i=1}^{r}x_{k_{i}}^{j_{i}}v$ of $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},\lambda)$ which is a contradiction. Hence, $\Omega$ contains only finitely many positive or only finitely negative odd integers. This means that $\varphi_{\mu}$ differs from $\varphi_{\nu}$ only in finitely many places, where $\nu_{2k+1}^{j}=0$ for all $k\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$, or $\nu_{2k+1}^{j}\neq 0$ for all $k\in\mathbb{Z}_{\geq 0}$, $j=1,\ldots,n-1$. ∎

Remark 3.10**.**

We conjecture that any irreducible $\mathbb{Z}$-graded $\widetilde{\mathcal{H}}$-module is diagonal.

We also have the following isomorphism criterion.

Proposition 3.11**.**

We have that $M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)\simeq M_{\varphi_{\mu^{\prime}}}(\widetilde{\mathcal{H}},a^{\prime})$ (up to a shift of gradation) if and only if $a=a^{\prime}$ and $\varphi_{\mu}$ and $\varphi_{\mu^{\prime}}$ differ only in finitely many places.

Proof.

The condition $a=a^{\prime}$ is clear. Assume that for some $r$ and $j$, $d_{2r+1}^{j}$ has an eigenvector $v\in M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)$ with eigenvalue $\mu_{2r+1}^{j}=a$. Set $w=x_{2r+1}^{j}v\neq 0$. Then $x_{2r+1}^{j}w=0$ and hence $w$ is a $\varphi_{\nu}$-highest vector where $\nu_{2k+1}^{i}=\mu_{2k+1}^{i}$ if $k\neq r$ or $i\neq j$, while $\nu_{2r+1}^{j}=0$. We have $M_{\varphi_{\nu}}(\widetilde{\mathcal{H}},a)\simeq M_{\varphi_{\mu}}(\widetilde{\mathcal{H}},a)$. Similarly, we can change finitely many nonzeros $\mu$’s to zeros.

Conversely, if we have the isomorphism, then one can obtain a $\varphi_{\mu^{\prime}}$-highest weight vector by finitely many actions of elements $x_{\pm(2r+1)}^{j}$ on a $\varphi_{\mu}$-highest weight vector. This implies the statement. ∎

4. Irreducibility of generalized Verma type modules

In this section we prove our main result which is the following theorem.

Theorem 4.1**.**

1. (a)

$M(\hat{\mathfrak{m}},\hat{\mathfrak{k}};L(\hat{\mathfrak{k}},\lambda))$* and $M(\hat{\mathfrak{s}},\lambda)$ are reducible for any $\hat{\mathfrak{s}}\in\{\hat{\mathfrak{g}},\hat{\mathfrak{m}},\hat{\mathfrak{k}},\mathcal{H}\}$.* 2. (b)

If $\det D_{\delta}^{X}(\lambda)\neq 0$, then there is a bijection between submodules of $M(\hat{\mathfrak{m}},\hat{\mathfrak{k}};L(\hat{\mathfrak{k}},\lambda))$ and ideals of $\mathcal{S}$. 3. (c)

If $\det D_{\delta}^{X}(\lambda)\neq 0$, then $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$ is irreducible.

The next two results imply Theorem 4.1 items (a) and (b).

Corollary 4.2**.**

$M(\hat{\mathfrak{s}},\lambda)$* is reducible for any $\hat{\mathfrak{s}}\in\{\hat{\mathfrak{g}},\hat{\mathfrak{m}},\hat{\mathfrak{k}},\mathcal{H}\}$.*

Proof.

It follows from Proposition 3.2. ∎

Proposition 4.3**.**

Let $M=M(\hat{\mathfrak{m}},\hat{\mathfrak{k}};L(\hat{\mathfrak{k}},\lambda))$, $L=L(\hat{\mathfrak{k}},\lambda)$, and assume that $\det D_{\delta}^{X}(\lambda)\neq 0$. Then there is a bijection between submodules of $M$ and ideals of $\mathcal{S}$; $\mathcal{S}^{+}M$ is a maximal proper submodule of $M$; and $L(\hat{\mathfrak{m}},\hat{\mathfrak{k}};L)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L$ as vector spaces.

Proof.

Let $J$ be an ideal of $\mathcal{S}$. Since $[\mathcal{H}_{0},\hat{\mathfrak{m}}]=0$, it is clear that $JM$ defines a submodule of $M$. In the other direction, for a submodule $N\subseteq M$, we consider the ideal $J_{N}\subseteq\mathcal{S}$ such that $N\cap\mathcal{S}L=J_{N}L$. We claim that $J_{N}=\{a\in\mathcal{S}\mid av_{\lambda}\in N\}$. Indeed, let $a\in J_{N}$, and write an arbitrary $v\in L$ as $uv_{\lambda}$ for some $u\in\mathbf{U}(\hat{\mathfrak{k}}^{-})$. Then we have $av=auv_{\lambda}=uav_{\lambda}\in N$, and hence $J_{N}L\subseteq N\cap\mathcal{S}L$. For the other inclusion, write a general element $v=\sum_{i=1}^{m}a_{i}v_{i}\in N\cap\mathcal{S}L$ with $a_{i}\in\mathcal{S}$ and assume that $v_{1},\ldots,v_{m}\in L$ are linearly independent. The fact that $[\mathcal{H}_{0},\hat{\mathfrak{k}}]=0$ along with the fact that $L$ is a simple $\hat{\mathfrak{k}}$-module with countable dimension allow us to apply Jacobson Density Theorem to find, for each $i=1,\ldots,m$, an element $u_{i}\in\mathbf{U}(\hat{\mathfrak{k}}^{+})$ for which $u_{i}v=a_{i}v_{\lambda}\in N\cap\mathcal{S}L$. In particular, $a_{i}\in J_{N}$ for every $i$, and the claim is proved.

Now we claim that $N=J_{N}M$. Indeed, let $jm\in J_{N}M$. Then, writing $m=ul$ with $u\in\mathbf{U}(\mathcal{H}^{-})$ and $l\in L$, we get that

[TABLE]

Thus $J_{N}M\subseteq N$. For the other inclusion, consider the canonical projection $\pi:M\to V:=M/J_{N}M$, $W=\pi(N)$, and $R=\mathcal{S}/J_{N}$. Notice that $V$ is free as an $R$-module, and that $W\cap RL=\pi(N\cap\mathcal{S}L)=\pi(J_{N}L)=0$. Now if we suppose that $W\neq 0$, then we can use the fact that $\det D_{\delta}^{X}(\lambda)\neq 0$, and that $[\mathcal{H}_{0},\hat{\mathfrak{m}}]=0$, to get a contradiction just as in the proof of Proposition 3.2. Thus $W=0$ and the proof is complete. ∎

Corollary 4.4**.**

If $\det D_{\delta}^{X}(\lambda)\neq 0$, then $L(\hat{\mathfrak{m}},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L(\hat{\mathfrak{k}},\lambda)$ as vector spaces.

Proof.

This follows from $L(\hat{\mathfrak{m}},\lambda)\cong L(\hat{\mathfrak{m}},\hat{\mathfrak{k}};L(\hat{\mathfrak{k}},\lambda))$ and Proposition 4.3. ∎

Corollary 4.5**.**

Suppose that $\det D_{\delta}^{X}(\lambda)\neq 0$. Then we have the character formula

[TABLE]

where $\Delta(X)_{{\rm re},0}^{+}$ denotes the set of real positive even roots of $\hat{\mathfrak{k}}$.

Proof.

This follows from [GS08] along with the fact that $L(\hat{\mathfrak{m}},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L(\hat{\mathfrak{k}},\lambda)$. ∎

From now on we assume that

[TABLE]

Before proving the irreducibility of $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$, we introduce an ordered basis of $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$. Recall that for a subalgebra $\mathfrak{a}\subseteq\hat{\mathfrak{g}}$ we defined $\Delta(\mathfrak{a})=\{\alpha\in\Delta\mid\hat{\mathfrak{g}}_{\alpha}\subseteq\mathfrak{a}\}$. Let $B(\mathfrak{u}^{-})=\{f_{i}\in\mathfrak{g}_{\alpha_{i}}\mid\alpha_{i}\in\dot{\Delta}(\mathfrak{u}^{-})\}$ be a basis of $\mathfrak{u}^{-}$ such that

[TABLE]

Now we order the basis $B(L(\mathfrak{u}^{-}))=\{f_{i}(m)\mid m\in\mathbb{Z}\}$ of $L(\mathfrak{u}^{-})$ so that

1. (a)

if $m$ is odd and $n$ is even, then $f_{i}(m)<f_{j}(n)$, 2. (b)

if $m,n$ are both even or both odd, then $f_{i}(m)<f_{j}(n)$ if $m<n$, or $m=n$ and $f_{i}<f_{j}$.

For $r\geq 1$ and $(\mathbf{i},2\mathbf{m},\mathbf{p})=(i_{1},\ldots,i_{r},m_{1},\ldots,m_{r},p_{1},\ldots,p_{r})\in\mathbb{Z}_{t}^{r}\times 2\mathbb{Z}^{r}\times\mathbb{Z}_{\geq 0}^{r}$, we set $f_{\mathbf{i},2\mathbf{m},\mathbf{p}}:=f_{i_{1}}(m_{1})^{p_{1}}\cdots f_{i_{r}}(m_{r})^{p_{r}}\in\mathbf{U}(L(\mathfrak{u}^{-})_{0})$ and we define $\deg f_{\mathbf{i},2\mathbf{m},\mathbf{p}}:=\sum p_{i}$. For monomials of the different degree we let $f_{\mathbf{i},2\mathbf{m},\mathbf{p}}<f_{\mathbf{i}^{\prime},2\mathbf{m}^{\prime},\mathbf{p}^{\prime}}$ if $\deg f_{\mathbf{i},2\mathbf{m},\mathbf{p}}<\deg f_{\mathbf{i}^{\prime},2\mathbf{m}^{\prime},\mathbf{p}^{\prime}}$; for monomials of same degree we define $f_{\mathbf{i},2\mathbf{m},\mathbf{p}}<f_{\mathbf{i}^{\prime},2\mathbf{m}^{\prime},\mathbf{p}^{\prime}}$ if $({\mathbf{i},2\mathbf{m},\mathbf{p}})<({\mathbf{i}^{\prime},2\mathbf{m}^{\prime},\mathbf{p}^{\prime}})$, where the latter order is the reverse lexicographical order. This provides us a totally ordered basis $B(\mathbf{U}(L(\mathfrak{u}^{-})_{0}))=\{f_{\mathbf{i},2\mathbf{m},\mathbf{p}}:=f_{i_{1}}(m_{1})^{p_{1}}\cdots f_{i_{r}}(m_{r})^{p_{r}}\}$ of $\mathbf{U}(L(\mathfrak{u}^{-})_{0})$. For $r\geq 1$ and $(\mathbf{i},\mathbf{m})=(i_{1},\ldots,i_{r},m_{1},\ldots,m_{r})\in\mathbb{Z}_{t}^{r}\times(2\mathbb{Z}^{r}+1)$, we set $f_{\mathbf{i},\mathbf{m}}:=f_{i_{1}}(m_{1})\cdots f_{i_{r}}(m_{r})$ and we define $\deg f_{\mathbf{i},\mathbf{m}}:=r$. For monomials of the different degree we let $f_{\mathbf{i},\mathbf{m}}<f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}$ if $\deg f_{\mathbf{i},\mathbf{m}}<\deg f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}$; for monomials of same degree we define $f_{\mathbf{i},\mathbf{m}}<f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}$ if $({\mathbf{i},\mathbf{m}})<({\mathbf{i}^{\prime},\mathbf{m}^{\prime}})$, where the latter order is the reverse lexicographical order. Finally, we let $f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}<f_{\mathbf{i},2\mathbf{m},\mathbf{p}}$ for all such monomials. By PBW Theorem, we have that $B(\mathbf{U}(L(\mathfrak{u}^{-})))=\{f_{\mathbf{i},2\mathbf{m},\mathbf{p}}f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}\}$ is a totally ordered basis of $\mathbf{U}(L(\mathfrak{u}^{-}))$.

Let $h_{1},\ldots,h_{t}$ be a basis of $\mathfrak{h}^{X}$. Then $H_{i,m}:=h_{i}(-m)$ for $i=1,\ldots,t$ and $m\in\{2\mathbb{Z}_{\geq 0}+1\}$ form a basis for $\mathcal{H}_{1}^{-}$. In particular, if we let $H_{i,m}\geq H_{k,n}$ if $m\geq n$ or $m=n$ and $i\geq k$, then the monomials $H_{i_{1},m_{1}}\cdots H_{i_{s},m_{s}}$ with $H_{i_{1},m_{1}}>\cdots>H_{i_{s},m_{s}}$ form a basis $B(\mathcal{H}_{1}^{-})$ of $\Lambda(\mathcal{H}_{1}^{-})$.

Since we are assuming $\det D_{\delta}^{X}(\lambda)\neq 0$, Corollary 4.4 implies that $L(\hat{\mathfrak{m}},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L(\hat{\mathfrak{k}},\lambda)$ as vector spaces. Let $\{v_{i}\mid i\in I\}$ be an ordered basis of $L(\hat{\mathfrak{m}},\lambda)$, where the order is induced by the order of $\Lambda(\mathcal{H}_{1}^{-})$. We say $f_{\mathbf{i},\mathbf{m},\mathbf{p}}f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}v_{i}<f_{\mathbf{i}_{1},\mathbf{m}_{1},\mathbf{p}_{1}}f_{\mathbf{i}_{1}^{\prime},\mathbf{m}_{1}^{\prime}}v_{j}$ if $f_{\mathbf{i},\mathbf{m},\mathbf{p}}f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}<f_{\mathbf{i}_{1},\mathbf{m}_{1},\mathbf{p}_{1}}f_{\mathbf{i}_{1}^{\prime},\mathbf{m}_{1}^{\prime}}$ or if $f_{\mathbf{i},\mathbf{m},\mathbf{p}}f_{\mathbf{i}^{\prime},\mathbf{m}^{\prime}}=f_{\mathbf{i}_{1},\mathbf{m}_{1},\mathbf{p}_{1}}f_{\mathbf{i}_{1}^{\prime},\mathbf{m}_{1}^{\prime}}$ and $i<j$. Finally, for an element

[TABLE]

we define

[TABLE]

For the next result recall that $L(\hat{\mathfrak{m}},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L(\hat{\mathfrak{k}},\lambda)$ as vector spaces when $\det(D_{\delta}^{X}(\lambda))\neq 0$. Also recall that for $\alpha_{i}\in\dot{\Delta}$ we have a triple $f_{i}\in\mathfrak{g}^{-\alpha_{i}}$, $e_{i}\in\mathfrak{g}^{\alpha_{i}}$, $h_{i}\in\mathfrak{h}$ such that $[f_{i},e_{i}]_{0}=h_{i}$.

Lemma 4.6**.**

Let ${\bar{f}}={\bar{f}}_{0}{\bar{f}}_{1}=f_{i_{1}}(m_{1})^{p_{1}}\cdots f_{i_{r}}(m_{r})^{p_{r}}f_{i_{1}^{\prime}}(m_{1}^{\prime})\cdots f_{i_{r^{\prime}}^{\prime}}(m_{r^{\prime}}^{\prime})\in B(\mathbf{U}(L(\mathfrak{u}^{-})))$, $v\in L(\hat{\mathfrak{m}},\lambda)$ be a nonzero vector, and assume that all factors occurring in ${\bar{f}}$ are simple. For any such factor $f_{i_{l}}$, we consider $e_{i_{l}}\in\mathfrak{n}^{+}=\mathfrak{m}^{+}\oplus\mathfrak{u}^{+}$. If $\det D_{\delta}^{X}(\lambda)\neq 0$, then the following hold:

1. (a)

If $\deg{\bar{f}}_{1}=0$, then there is $0\gg m_{l}\in\{2\mathbb{Z}+1\}$ or $0\ll m\in\{2\mathbb{Z}+1\}$ such that

[TABLE] 2. (b)

If $\deg{\bar{f}}_{1}\geq 1$, then there is $0\gg m\in 2\mathbb{Z}$ or $0\ll m\in 2\mathbb{Z}$ such that

[TABLE]

Proof.

We prove part (b) first, as part (a) follows from it. Choose $0\gg m\in 2\mathbb{Z}$ or $0\ll m\in 2\mathbb{Z}$ such that $h_{i_{l}}(m+m_{j}^{\prime})$ is in $B(\mathcal{H}_{1}^{-})$. Since $\operatorname{ad}(e_{i_{l}}(m))$ is a derivation of even degree, we have that

[TABLE]

where the first equivalence follows from the fact that $\operatorname{ad}(h_{i_{l}}(m+m_{j}))$ is an even derivation, $\operatorname{ad}(h_{i_{j}}(m+m_{j}^{\prime}))$ is an odd derivation, and $f_{i_{\xi}^{\prime}}(m_{\xi}^{\prime})$ is an odd element for any $m_{\xi}^{\prime}$; and the second equivalence follows from the fact that $h_{i_{l}}(m+m_{j})v=0$ for all $1\leq j\leq r$, since either $h_{i_{l}}(m+m_{j})\in\mathcal{H}_{0}^{+}$ that implies $h_{i_{l}}(m+m_{j})v=0$, or $h_{i_{l}}(m+m_{j})\in\mathcal{S}^{+}$, and hence $h_{i_{l}}(m+m_{j})v$ lies in the maximal proper submodule of $M(\hat{\mathfrak{k}},\lambda)$.

For part (a), we notice that the second parentheses above does not appear in the expression of $e_{a_{l}}(m){\bar{f}}v$. Moreover, despite the fact that $\operatorname{ad}(e_{i_{l}}(m))$ and $\operatorname{ad}(h_{i_{l}}(m+m_{k}))$ are odd derivations (as $m\in\{2\mathbb{Z}+1\}$ and $m_{j}\in 2\mathbb{Z}$ for all $1\leq j\leq r$), they behave as regular derivations when applied on factors of ${\bar{f}}_{0}$, since $m_{j}\in 2\mathbb{Z}$ for all $1\leq j\leq r$. Thus the proof follows from the above equation. ∎

We now state our key result.

Theorem 4.7**.**

If $\det D_{\delta}^{X}(\lambda)\neq 0$, then $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$ is irreducible.

Proof.

We claim that any non-trivial submodule $N$ of $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$ intersects $L(\hat{\mathfrak{m}},\lambda)$ non-trivially. Assuming that the claim holds, the result follows from the simplicity of $L(\hat{\mathfrak{m}},\lambda)$.

To prove the claim, let $0\neq v\in N_{\mu}$, and let ${\bar{f}}_{\max}x_{\max}=f_{\mathbf{a},2\mathbf{b},\mathbf{c}}f_{\mathbf{a}^{\prime},\mathbf{b}^{\prime}}x_{d}$ be the maximal monomial occurring in $v$. We now reduce the proof to the case where all factors $f_{i_{j}}$ of maximal degree monomials occurring in $v$ are simple root vectors. Indeed, consider all factors $f_{i_{j}}$ that occur in the monomials of maximal degree of $v$, and let $f_{i_{k}}$ be the minimal among them (i.e. its associated root $\alpha_{i_{k}}$ is such that $|\alpha_{i_{k}}|$ is maximal among them). Let ${\bar{f}}_{\min}x_{\min}=f_{\mathbf{d},2\mathbf{g},\mathbf{k}}f_{\mathbf{d}^{\prime},\mathbf{g}^{\prime}}h_{\mathbf{d}^{\prime\prime},\mathbf{g}^{\prime\prime}}x_{\min}={\bar{f}}_{0,\min}{\bar{f}}_{1,\min}x_{\min}$ be an element (occurring in $v$) of maximal degree having $f_{i_{k}}$ as a factor, and let $z\in\mathfrak{n}^{+}=\mathfrak{m}^{+}\oplus\mathfrak{u}^{+}$ be such that $0\neq[z,f_{i_{k}}]\in\mathfrak{u}^{-}$ (such $z$ exists by [Cox94, Lemma 4.2]). Let $J_{\min}$ the set of indexes $j$ for which $f_{i_{j}}$ is a factor of ${\bar{f}}_{\min}$ and $[z,f_{i_{j}}]\in\mathfrak{u}^{-}$. Let $0\gg m\in 2\mathbb{Z}$ (if $z\in\mathfrak{u}^{+}$) or $0\ll m\in 2\mathbb{Z}$ (if $z\in\mathfrak{m}^{+}$) (here $m\ll 0$ (resp. $m\gg 0$) means $m$ so that for every fixed $j$, $m+m_{j}\notin\{g_{k},\ g_{l}^{\prime}\mid 1\leq k\leq r,\ 1\leq l\leq r^{\prime}\}$). Then, using that $\operatorname{ad}(z(m))$ is an even derivation, we obtain

[TABLE]

where $k+d^{\prime}=\deg{\bar{f}}_{\min}$. Now if $S_{1}$ denote this summation, then it is nonzero since $[z,f_{i_{k}}]_{0}\neq 0$ and $m+m_{j}\notin\{g_{k},\ g_{l}^{\prime}\mid 1\leq k\leq r,\ 1\leq l\leq r^{\prime}\}$. Moreover, if ${\bar{f}}x={\bar{f}}_{0}{\bar{f}}_{1}x$ is a different monomial occurring in $v$, then, similarly we have that

[TABLE]

where $p+r=\deg{\bar{f}}$. Since ${\bar{f}}_{\min}$ has maximal degree among monomials in $v$, we have that $p+r\leq k+d^{\prime}$. Hence, if $T_{1}$ is the summation above, then $S_{1}\notin\operatorname{LinSpan}(T_{1})+\mathbf{U}(L(\mathfrak{u}^{-}))_{(p-1)}\otimes L(\hat{\mathfrak{m}},\lambda)$, since this could happen only if $p+r=k+d^{\prime}$; $\mathbb{C}[z,f_{i_{j}}]_{\ell}=\mathbb{C}[z,f_{i_{l}}]_{\ell}$ for $\ell=0,1$; $m_{j}=g_{l}$; ${\bar{f}}_{\min}^{\hat{j}}={\bar{f}}^{\hat{l}}$; and $x_{\min}=x$. But this would imply ${\bar{f}}_{\min}={\bar{f}}$, and $x_{\min}=x$, which contradicts the fact that ${\bar{f}}x\neq{\bar{f}}_{\min}x_{\min}$.

We may now assume that factors of all maximal degree monomials occurring in $v$ are simple. In particular, this is the case for

[TABLE]

Moreover, we may also assume that $\deg f_{1,\max}\geq 1$ (as otherwise the proof is the same as that of [Cox94, Proposition 4.5], using Lemma 4.6 and a suitable $e\in\{2\mathbb{Z}+1\}$ in his notation). By Lemma 4.6, for each simple root factor $f_{a_{l}}$ of ${\bar{f}}_{\max}$, there is $0\gg m\in 2\mathbb{Z}$ or $0\ll m\in 2\mathbb{Z}$ for which

[TABLE]

Finally, it is not hard to prove that for any fixed index $l$, the summand

[TABLE]

is not in the $\operatorname{LinSpan}$ of the remaining monomials occurring in $e_{l}(m)v$. Therefore, $e_{l}(m)v\neq 0$, the maximal monomial occurring in $e_{l}(m)v$ has degree less than that of the maximal monomial occurring in $v$, and thus the result follows by induction. ∎

Applying Theorem 4.7 in the case $X=\emptyset$ gives:

Corollary 4.8**.**

If $\det D_{\delta}^{\emptyset}(\lambda)\neq 0$, then $M(\hat{\mathfrak{g}},\mathcal{H};L(\mathcal{H},\lambda))$ is irreducible.

Remark 4.9**.**

Notice that differently from the other cases studied in the literature, we do not need the central charge to be nonzero in order to have $M(\hat{\mathfrak{g}},\hat{\mathfrak{m}};L(\hat{\mathfrak{m}},\lambda))$ to be irreducible (compare with [Cox94, Fut94, CF18]). This is due to the fact that the central element $K$ does not play a role in the action of the imaginary subalgebra $\mathcal{H}$ on $L(\hat{\mathfrak{m}},\lambda)$. On the other hand, the condition $\det D_{\delta}^{X}(\lambda)\neq 0$ is essential in our context. Without this condition we do not necessarily have that $L(\hat{\mathfrak{m}},\lambda)\cong\Lambda(\mathcal{H}_{1}^{-})\otimes_{\mathbb{C}}L(\hat{\mathfrak{k}},\lambda)$ as vector spaces (see Corollary 4.4).

Acknowledgment. Second author is grateful to the University of California Berkeley for hospitality. The authors are very grateful to Vera Serganova for very stimulating discussions.