Non-Weight Modules over the N = 1 Heisenberg–Virasoro Superalgebra

Abstract

In this paper, we mainly study free $U\left(\mathfrak{h}\right)$-modules over the $N=1$ Heisenberg–Virasoro superalgebra. We construct a family of non-weight modules that are free of rank 2 when regarded as modules over the Cartan subalgebra. Moreover, we classify the free $U\left(\mathfrak{h}\right)$-modules of rank 2 over the $N = 1$ Heisenberg–Virasoro superalgebra and provide the necessary and sufficient conditions for such $\mathfrak{g}$-modules to be isomorphic.

1. Introduction

The $N = 1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$ is a type of superconformal current algebra that corresponds to 2-dimensional quantum field theories with both chiral and superconformal symmetries [1,2]. It can also be viewed as an extension of the Beltrami algebra with supersymmetry presented in [3]. Recently, there has been plenty of research on the representation of the $N = 1$ Heisenberg–Virasoro superalgebra, including studies on Verma modules [4,5], cuspidal modules [6], Whittaker modules, and smooth modules [7].

$U\left(\mathfrak{h}\right)$-modules, which are free of finite rank when restricted to the Cartan subalgebra, over the simple Lie algebra ${\mathfrak{sl}}_n$, were first constructed by Nilsson in [8]. Since then, many authors have considered these modules over various infinite dimensional Lie (super)algebras, such as the Witt algebra $W_n$ in [9], the Lie algebra of differential operators in [10], the super–Virasoro algebra and the $N = 2$ superconformal algebra in [11,12], the basic Lie superalgebra in [13], the Kac–Moody Lie algebra in [14], and the super-BMS ${ }_3$ algebra in [15]. Based on these research works, the aim of this paper was to determine free $U\left(\mathfrak{h}\right)$-modules for the $N = 1$ Heisenberg–Virasoro superalgebra. Specifically, we constructed and studied a family of non-weight modules over the $N = 1$ Heisenberg–Virasoro superalgebra, which are free of rank 2 when regarded as modules over the Cartan subalgebra. We also provided the necessary and sufficient conditions in order for these modules to be simple. We also provided the necessary and sufficient conditions for the $\mathfrak{g}$-modules to be isomorphic.

This paper is arranged as follows. The definition of the $N = 1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$ and some preliminary results for the super Virasoro algebra and the twisted Heisenberg–Virasoro algebra are introduced in Section 2. A family of non-weight modules, which are free of rank 2 when restricted to the Cartan subalgebra, are constructed over the $N = 1$ Heisenberg–Virasoro superalgebra in Section 3. The last section is devoted to classifying free $U\left(\mathfrak{h}\right)$-modules of rank 2 over the $N = 1$ Heisenberg–Virasoro superalgebra (Theorem 3 below).

Throughout this paper, $\mathbb{C}$, ${\mathbb{C}}^{*}$, $\mathbb{Z}$, and ${\mathbb{Z}}_{+}$ are denoted as sets of complex numbers, nonzero complex numbers, integers, and positive integers, respectively. All algebras are defined in the complex number field $\mathbb{C}$. We used $U\left(L\right)$ to denote the universal enveloping algebra of a Lie algebra L.

2. Basics

In this section, we recall some notations about the $N = 1$ Heisenberg–Virasoro superalgebra and its subalgebras.

Definition 1.

The $N = 1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}={\mathfrak{g}}_{\overline{0}}\oplus {\mathfrak{g}}_{\overline{1}}={\mathrm{span}}_{\mathbb{C}}\{L_n,I_n,G_{n-\frac{1}{2}}$, $Q_{n-\frac{1}{2}},C_1,C_2,C_3\mid n\in \mathbb{Z}\}$ is a Lie superalgebra with the following commutation relations:

$$\begin{array}{ccc}& & [L_m,L_n]=(m-n)L_{m+n}+{\delta }_{m+n,0}\frac{m^3-m}{12}{C}_1,\hfill \\ & & [L_m,I_n]=-n{I}_{m+n}+{\delta }_{m+n,0}(m^2+m)C_2,\hfill \\ & & [I_m,I_n]=m{\delta }_{m+n,0}{C}_3,\hfill \\ & & [L_m,G_p]=(\frac{m}{2}-p)G_{m+p},[L_m,Q_p]=-(\frac{m}{2}+p)Q_{m+p},\hfill \\ & & [I_m,G_p]=m{Q}_{m+p},[I_m,Q_p]=0,\hfill \\ & & {[G_p,G_q]}_{+}=2L_{p+q}+\frac{1}{3}(p^2-\frac{1}{4}){\delta }_{p+q,0}{C}_1,\hfill \\ & & {[G_p,Q_q]}_{+}=I_{p+q}+(2p+1){\delta }_{p+q,0}{C}_2,{[Q_p,Q_q]}_{+}={\delta }_{p+q,0}{C}_3,\hfill \\ & & [\mathfrak{g},C_1]=[\mathfrak{g},C_2]=[\mathfrak{g},C_3]=0,\forall m,n\in \mathbb{Z},p,q\in \mathbb{Z}+\frac{1}{2},\hfill \end{array}$$

where

$${\mathfrak{g}}_{\overline{0}}={\mathrm{span}}_{\mathbb{C}}\{L_m,I_m,C_1,C_2,C_3\mid m\in \mathbb{Z}\}\mathrm{and}{\mathfrak{g}}_{\overline{1}}={\mathrm{span}}_{\mathbb{C}}\{G_r,Q_r\mid r\in \mathbb{Z}+\frac{1}{2}\}.$$

The Lie superalgerba $\mathfrak{g}$ has a subalgebra spanned by $\{L_n,I_n,$$C_1,C_2,C_3\mid n\in \mathbb{Z}\}$, which is called the twisted Heisenberg–Virasoro algebra $\mathcal{D}$. It also has a subalgebra spanned by $\{L_n,G_r,C_1\mid n\in \mathbb{Z},r\in \mathbb{Z}+\frac{1}{2}\}$, which is called the $N=1$ Neveu–Schwarz superconformal algebra $\mathcal{N}S$. $\mathfrak{C}={\mathrm{span}}_{\mathbb{C}}\{I_0,C_1,C_2,C_3\}$ is the center of $\mathfrak{g}$.

First, we recall some modules over $\mathcal{D}$ constructed in [16].

For $\lambda \in {\mathbb{C}}^{*},\alpha ,\beta \in \mathbb{C}$, the $\mathcal{D}$-module $\Omega (\lambda ,\alpha ,\beta )$ was first constructed in [16]. As a vector space, $\Omega (\lambda ,\alpha ,\beta )=\mathbb{C}\left[t\right]$ is the polynomial algebra over $\mathbb{C}$, and the module actions are given by

$$L_{m}f\left(t\right)={\lambda }^m(t+m\alpha )f(t+m),I_{m}f\left(t\right)=\beta {\lambda }^{m}f(t+m),C_{i}f\left(t\right)=0,$$

where $f\left(t\right)\in \mathbb{C}\left[t\right]$, $m\in \mathbb{Z}$ and $i=1,2,3$.

$\Omega (\lambda ,\alpha ,\beta )$ is simple if and only if $\alpha \ne 0$ or $\beta \ne 0$. If $\alpha =\beta =0$, then $\Omega (\lambda ,0,0)$ has a simple submodule $t\Omega (\lambda ,0,0)$ with codimension 1 [16].

Theorem 1

([16]). Let M be a $U\left(\mathcal{D}\right)$-module, such that the restriction of $U\left(\mathcal{D}\right)$ to $U\left(\mathbb{C}{L}_0\right)$ is free of rank 1, that is, $M=U\left(\mathbb{C}{L}_0\right)v$ for some torsion-free $v\in M$. Then, $M\cong \Omega (\lambda ,\alpha ,\beta )$ for some $\alpha ,\beta \in \mathbb{C},\lambda \in {\mathbb{C}}^{*}$. Moreover, M is simple if and only if $\alpha \ne 0$ or $\beta \ne 0$. If $\alpha =\beta =0$, then $\Omega (\lambda ,0,0)$ has a simple submodule $t\Omega (\lambda ,0,0)$ with codimension 1.

Next, we recall some modules over $\mathcal{N}S$ constructed in [12].

For $\lambda \in {\mathbb{C}}^{*},c\in \mathbb{C}$, the $\mathcal{N}S$-module ${\Omega }_{\mathcal{N}S}(\lambda ,c)$, which is free of rank 2 when regarded as a $\mathbb{C}{L}_0$-module, was constructed in [12]. As a vector space

$${\Omega }_{\mathcal{N}S}(\lambda ,c)={\Omega }_{\mathcal{N}S}{(\lambda ,c)}_{\overline{0}}\oplus {\Omega }_{\mathcal{N}S}{(\lambda ,c)}_{\overline{1}}=\mathbb{C}\left[t\right]\oplus \mathbb{C}\left[z\right]$$

is the polynomial algebra over $\mathbb{C}$, which is an $\mathcal{N}S$-module, and the module actions are provided by

$$\begin{array}{ccc}& & L_{m}f\left(t\right)={\lambda }^m(t+mc)f(t+m),L_{m}g\left(z\right)={\lambda }^m(z+m(c+\frac{1}{2}))g(z+m),\hfill \\ & & G_{r}f\left(t\right)={\lambda }^{r-\frac{1}{2}}f(z+r),G_{r}g\left(z\right)={\lambda }^{r+\frac{1}{2}}(t+2rc)g(t+r),C_{1}f\left(t\right)=C_{1}g\left(z\right)=0,\hfill \end{array}$$

where $f\left(t\right)\in \mathbb{C}\left[t\right]$, $g\left(z\right)\in \mathbb{C}\left[z\right]$, $m\in \mathbb{Z}$, $r\in \mathbb{Z}+\frac{1}{2}$.

From [12] (Corollary 2.12, Proposition 2.13 ), we know that ${\Omega }_{\mathcal{N}S}(\lambda ,c)$ is simple if and only if $c\ne 0$. If $c=0$, then ${\Omega }_{\mathcal{N}S}(\lambda ,0)$ has a unique proper submodule $\Gamma =t\left({\Omega }_{\mathcal{N}S}{(\lambda ,c)}_{\overline{0}}\right)\oplus {\Omega }_{\mathcal{N}S}{(\lambda ,c)}_{\overline{1}}$, and ${\Omega }_{\mathcal{N}S}(\lambda ,0)/\Gamma$ is a 1-dimensional trivial $\mathcal{N}S$-module.

Theorem 2

([12]). Let $\lambda ,{\lambda }^{\prime }\in {\mathbb{C}}^{*},c,c^{\prime }\in \mathbb{C}$. Then, ${\Omega }_{\mathcal{N}S}(\lambda ,c)\cong {\Omega }_{\mathcal{N}S}({\lambda }^{\prime },c^{\prime })$ as $\mathcal{N}S$-modules if and only if $\lambda ={\lambda }^{\prime }$ and $c=c^{\prime }$.

Inspired by the above resuts, we will determine all irreducible $\mathfrak{g}$-modules, which are free of rank 2 when regarded as a $U\left(\mathfrak{h}\right)$-module in this paper, where $\mathfrak{h}=\mathbb{C}{L}_0$ is the canonical Cartan subalgebra (modulo the center) of $\mathfrak{g}$.

3. A Family of Non-Weight Modules over the $\mathit{N}=\mathbf{1}$ Heisenberg–Virasoro Superalgebra

In this section, we construct the free $U\left(\mathfrak{h}\right)$-modules of rank 2 over the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$.

Let $\mathbb{C}\left[t\right]$ and $\mathbb{C}\left[z\right]$ be the polynomial algebras in indeterminates t and z, respectively. For $\lambda \in {\mathbb{C}}^{*}$, $\alpha ,\beta \in \mathbb{C}$, $\Omega (\lambda ,\alpha ,\beta )=\mathbb{C}\left[t\right]\oplus \mathbb{C}\left[z\right]$ is a ${\mathbb{Z}}_2$-graded vector space with $\Omega {(\lambda ,\alpha ,\beta )}_{\overline{0}}=\mathbb{C}\left[t\right]$ and $\Omega {(\lambda ,\alpha ,\beta )}_{\overline{1}}=\mathbb{C}\left[z\right]$. For any $m\in \mathbb{Z},p\in \mathbb{Z}+\frac{1}{2}$, $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$, we define the action of $\mathfrak{g}$ on $\Omega (\lambda ,\alpha ,\beta )$ as follows:

$$\begin{array}{ccc}\hfill & & L_{m}f\left(t\right)={\lambda }^m(t+m\alpha )f(t+m),L_{m}f\left(z\right)={\lambda }^m(z+m(\alpha +\frac{1}{2}))f(z+m),\hfill \end{array}$$

(1)

$$\begin{array}{ccc}\hfill & & I_{m}f\left(t\right)=\beta {\lambda }^{m}f(t+m),I_{m}f\left(z\right)=\beta {\lambda }^{m}f(z+m),\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill & & G_{p}f\left(t\right)={\lambda }^{p-\frac{1}{2}}f(z+p),G_{p}f\left(z\right)={\lambda }^{p+\frac{1}{2}}(t+2p\alpha )f(t+p),\hfill \end{array}$$

(3)

$$\begin{array}{ccc}\hfill & & Q_{p}f\left(t\right)=0,Q_{p}f\left(z\right)={\lambda }^{p+\frac{1}{2}}\beta f(t+p),\hfill \end{array}$$

(4)

$$\begin{array}{ccc}\hfill & & C_{i}f\left(t\right)=C_{i}f\left(z\right)=0,i=1,2,3.\hfill \end{array}$$

(5)

Then, $\Omega (\lambda ,\alpha ,\beta )$ is a $\mathfrak{g}$-module under the action of (1) ∼ (5).

Proposition 1.

For $\lambda \in {\mathbb{C}}^{*}$, $\alpha ,\beta \in \mathbb{C}$, $\Omega (\lambda ,\alpha ,\beta )$ is a $\mathfrak{g}$-module under the action defined by (1)∼(5). Moreover, $\Omega (\lambda ,\alpha ,\beta )$ is free of rank 2 as a module over $\mathbb{C}\left[L_0\right]$.

Proof. 

For any $m,n\in \mathbb{Z}$, $p,q\in \mathbb{Z}+\frac{1}{2}$, $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$, by [12,16] we only need to check the following relations.

It follows from (1) and (4) that

$$\begin{array}{c}\hfill L_m{Q}_{p}f\left(t\right)-Q_p{L}_{m}f\left(t\right)=-{\lambda }^m{Q}_p(t+m\alpha )f(t+m)=0=-(\frac{m}{2}+p)Q_{m+p}f\left(t\right)\end{array}$$

and

$$\begin{array}{ccc}& & L_m{Q}_{p}f\left(z\right)-Q_p{L}_{m}f\left(z\right)={\lambda }^{p+\frac{1}{2}}\beta L_{m}f(t+p)-{\lambda }^m{Q}_p(z+m(\alpha +\frac{1}{2}))f(z+m)\hfill \\ & & ={\lambda }^{m+p+\frac{1}{2}}\beta (t+m\alpha )f(t+m+p)-{\lambda }^{m+p+\frac{1}{2}}\beta (t+p+m(\alpha +\frac{1}{2}))f(t+m+p)\hfill \\ & & =-(\frac{m}{2}+p)Q_{m+p}f\left(z\right).\hfill \end{array}$$

It follows from (2) and (3) that

$$\begin{array}{ccc}& & I_m{G}_{p}f\left(t\right)-G_p{I}_{m}f\left(t\right)={\lambda }^{p-\frac{1}{2}}{I}_{m}f(z+p)-{\lambda }^m\beta G_{p}f(t+m)\hfill \\ & & ={\lambda }^{m+p-\frac{1}{2}}\beta f(z+p+m)-{\lambda }^{m+p-\frac{1}{2}}\beta f(z+p+m)\hfill \\ & & =m{Q}_{m+p}f\left(t\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & I_m{G}_{p}f\left(z\right)-G_p{I}_{m}f\left(z\right)={\lambda }^{p+\frac{1}{2}}{I}_m(t+2p\alpha )f(t+p)-{\lambda }^m\beta G_{p}f(z+m)\hfill \\ & & ={\lambda }^{m+p+\frac{1}{2}}\beta (t+m+2p\alpha )f(t+m+p)-{\lambda }^{m+p+\frac{1}{2}}\beta (t+2p\alpha )f(t+m+p)\hfill \\ & & ={\lambda }^{m+p+\frac{1}{2}}\beta mf(t+m+p)\hfill \\ & & =m{Q}_{m+p}f\left(z\right).\hfill \end{array}$$

It follows from (2) and (4) that

$$\begin{array}{c}\hfill I_m{Q}_{p}f\left(t\right)-Q_p{I}_{m}f\left(t\right)=0\end{array}$$

and

$$\begin{array}{ccc}& & I_m{Q}_{p}f\left(z\right)-Q_p{I}_{m}f\left(z\right)={\lambda }^{p+\frac{1}{2}}\beta I_{m}f(t+p)-{\lambda }^m\beta Q_{p}f(z+m)\hfill \\ & & ={\lambda }^{m+p+\frac{1}{2}}{\beta }^{2}f(t+p+m)-{\lambda }^{m+p+\frac{1}{2}}{\beta }^{2}f(t+p+m)\hfill \\ & & =0.\hfill \end{array}$$

It follows from (3) and (4) that

$$\begin{array}{ccc}& & G_p{Q}_{q}f\left(t\right)+Q_q{G}_{p}f\left(t\right)={\lambda }^{p-\frac{1}{2}}{Q}_{q}f(z+p)={\lambda }^{p+q}\beta f(t+q+p)\hfill \\ & & =(I_{p+q}+(2p+1){\delta }_{p+q,0}{C}_2)f\left(t\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & G_p{Q}_{q}f\left(z\right)+Q_q{G}_{p}f\left(z\right)={\lambda }^{q+\frac{1}{2}}\beta G_{p}f(t+q)={\lambda }^{q+p}\beta f(z+q+p)\hfill \\ & & =(I_{p+q}+(2p+1){\delta }_{p+q,0}{C}_2)f\left(z\right).\hfill \end{array}$$

It follows from (4) that

$$\begin{array}{ccc}& & Q_p{Q}_{q}f\left(t\right)+Q_q{Q}_{p}f\left(t\right)={\delta }_{p+q,0}{C}_{3}f\left(t\right),\hfill \\ & & Q_p{Q}_{q}f\left(z\right)+Q_q{Q}_{p}f\left(z\right)={\delta }_{p+q,0}{C}_{3}f\left(z\right).\hfill \end{array}$$

Hence, $\Omega (\lambda ,\alpha ,\beta )$ is a $\mathfrak{g}$-module. We find that $\Omega (\lambda ,\alpha ,\beta )$ is free of rank 2 as a module over $\mathbb{C}\left[L_0\right]$, when $m=0$ in (1). □

4. Rank 2 Free $\mathit{U}\left(\mathfrak{h}\right)$-Modules

In order to classify the free $U\left(\mathfrak{h}\right)$-modules over the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$, the following lemmas are provided.

Lemma 1

([15]). Let $\mathfrak{G}={\mathfrak{G}}_{\overline{0}}\oplus {\mathfrak{G}}_{\overline{1}}$ be a Lie superalgebra. $\mathfrak{H}$ is a Cartan subalgebra of $\mathfrak{G}$ with $\mathfrak{H}\subseteq {\mathfrak{G}}_{\overline{0}}$, and ${[{\mathfrak{G}}_{\overline{1}},{\mathfrak{G}}_{\overline{1}}]}_{+}={\mathfrak{G}}_{\overline{0}}$. Then there do not exist $\mathfrak{G}$-modules which are free of rank 1 as $U\left(\mathfrak{H}\right)$-modules.

Recall that the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$ has a Cartan subalgebra $\mathfrak{h}={\mathrm{span}}_{\mathbb{C}}\left\{L_0\right\}$, which lies in the even part. Thus, $U\left(\mathfrak{h}\right)=\mathbb{C}\left[L_0\right]$. As the Lie superalgebra $\mathfrak{g}$ is generated by odd elements $G_r,Q_r$, where $r\in \mathbb{Z}+\frac{1}{2}$, $\mathfrak{g}$ does not have non-trivial modules that are pure even or pure odd. Consequently, free $U\left(\mathfrak{h}\right)$-modules of rank 1 do not exist. Next, we classify the free $U\left(\mathfrak{h}\right)$-modules of rank 2 over the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$.

Let $\mathfrak{M}={\mathfrak{M}}_{\overline{0}}\oplus {\mathfrak{M}}_{\overline{1}}$ be the free $U\left(\mathfrak{h}\right)$-modules of rank 2 over the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$. Let $\mathfrak{v}$ and $\mathfrak{w}$ be two homogeneous basis elements of $\mathfrak{M}$. If $\mathfrak{v}$ and $\mathfrak{w}$ have the same parity, we obtain

$$G_{\pm \frac{1}{2}}\mathfrak{v}=G_{\pm \frac{1}{2}}\mathfrak{w}=0.$$

Then,

$$L_{\pm 1}\mathfrak{v}=G_{\pm \frac{1}{2}}^2\mathfrak{v}=G_{\pm \frac{1}{2}}^2\mathfrak{w}=0,L_0\mathfrak{v}=\frac{1}{2}[L_1,L_{-1}]\mathfrak{v}=0,L_0\mathfrak{w}=\frac{1}{2}[L_1,L_{-1}]\mathfrak{w}=0,$$

a contradiction. So, $\mathfrak{v}$ and $\mathfrak{w}$ are not in the same parities. Thus, we set $\mathfrak{v}\in {\mathfrak{M}}_{\overline{0}}$ and $\mathfrak{w}\in {\mathfrak{M}}_{\overline{1}}$. As a vector space,

$${\mathfrak{M}}_{\overline{0}}=\mathbb{C}\left[t\right]\mathfrak{v}\mathit{and}{\mathfrak{M}}_{\overline{1}}=\mathbb{C}\left[z\right]\mathfrak{w}.$$

The following results are necessary for future use.

Lemma 2.

For any $v\in \mathfrak{M}$, $i\in {\mathbb{Z}}_{+}$, and $m\in \mathbb{Z},p\in \mathbb{Z}+\frac{1}{2}$, we have

$$L_m{L}_0^{i}v={(L_0+m)}^i{L}_{m}v,I_m{L}_0^{i}v={(L_0+m)}^i{I}_{m}v,$$

$$G_p{L}_0^{i}v={(L_0+p)}^i{G}_{p}v,Q_p{L}_0^{i}v={(L_0+p)}^i{Q}_{p}v.$$

Proof. 

This follows from direct calculations. □

As the even part of the $N=1$ Heisenberg–Virasoro superalgebra $\mathfrak{g}$ is isomorphic to the twisted Heisenberg–Virasoro algebra $\mathcal{D}$, we can regard both ${\mathfrak{M}}_{\overline{0}}$ and ${\mathfrak{M}}_{\overline{1}}$ as $\mathcal{D}$-modules. According to Theorem 1, there exist $\lambda ,{\lambda }_1\in {\mathbb{C}}^{*}$, $\alpha ,\beta ,{\alpha }_1,{\beta }_1\in \mathbb{C}$, $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$ such that

$$\begin{array}{c}\hfill L_{m}f\left(t\right)\mathfrak{v}=L_{m}f\left(L_0\right)\mathfrak{v}=f(L_0+m)L_m\mathfrak{v}={\lambda }^m(t+m\alpha )f(t+m)\mathfrak{v},\end{array}$$

(6)

$$\begin{array}{c}\hfill L_{m}f\left(z\right)\mathfrak{w}=L_{m}f\left(L_0\right)\mathfrak{w}=f(L_0+m)L_m\mathfrak{w}={\lambda }_1^m(z+m{\alpha }_1)f(z+m)\mathfrak{w},\end{array}$$

(7)

$$\begin{array}{c}\hfill I_{m}f\left(t\right)\mathfrak{v}=I_{m}f\left(L_0\right)\mathfrak{v}=f(L_0+m)I_m\mathfrak{v}=\beta {\lambda }^{m}f(t+m)\mathfrak{v},\end{array}$$

(8)

$$\begin{array}{c}\hfill I_{m}f\left(z\right)\mathfrak{w}=I_{m}f\left(L_0\right)\mathfrak{w}=f(L_0+m)I_m\mathfrak{w}={\beta }_1{\lambda }_1^{m}f(z+m)\mathfrak{w},\end{array}$$

(9)

$$\begin{array}{c}\hfill C_i\mathfrak{v}=C_i\mathfrak{w}=0,i=1,2,3.\end{array}$$

(10)

Lemma 3.

Keep the same notations as above. Then, $\lambda ={\lambda }_1$ and there exist $c\in {\mathbb{C}}^{*}$ such that one of the following two cases occurs.

(1)

${\alpha }_1=\alpha +\frac{1}{2}$, $G_{\frac{1}{2}}\mathfrak{v}=c\mathfrak{w}$ and $G_{\frac{1}{2}}\mathfrak{w}=\frac{1}{c}\lambda (z+\alpha )\mathfrak{v}$.

(2)

${\alpha }_1=\alpha -\frac{1}{2}$, $G_{\frac{1}{2}}\mathfrak{v}=\frac{1}{c}\lambda (z+\alpha -\frac{1}{2})\mathfrak{w}$ and $G_{\frac{1}{2}}\mathfrak{w}=c\mathfrak{v}$.

Proof. 

Assume $G_{\frac{1}{2}}\mathfrak{v}=f\left(z\right)\mathfrak{w}$ and $G_{\frac{1}{2}}\mathfrak{w}=g\left(t\right)\mathfrak{v}$. From ${[G_{\frac{1}{2}},G_{\frac{1}{2}}]}_{+}\mathfrak{v}=2L_1\mathfrak{v}$, we obtain

$$\begin{array}{c}\hfill G_{\frac{1}{2}}^2\mathfrak{v}=G_{\frac{1}{2}}f\left(z\right)\mathfrak{w}=G_{\frac{1}{2}}f\left(L_0\right)\mathfrak{w}=f(L_0+\frac{1}{2})G_{\frac{1}{2}}\mathfrak{w}=f(t+\frac{1}{2})g\left(t\right)\mathfrak{v}\end{array}$$

and

$$\begin{array}{c}\hfill L_1\mathfrak{v}=\lambda (t+\alpha )\mathfrak{v}.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill f(t+\frac{1}{2})g\left(t\right)=\lambda (t+\alpha ),\end{array}$$

(11)

which implies $f\left(t\right)=c$ and $g\left(t\right)=\frac{1}{c}\lambda (t+\alpha )$ or $g\left(t\right)=c$ and $f\left(t\right)=\frac{1}{c}\lambda (t+\alpha -\frac{1}{2})$.

Similarly, from ${[G_{\frac{1}{2}},G_{\frac{1}{2}}]}_{+}\mathfrak{w}=2L_1\mathfrak{w}$, we have

$$\begin{array}{c}\hfill G_{\frac{1}{2}}^2\mathfrak{w}=G_{\frac{1}{2}}g\left(t\right)\mathfrak{v}=G_{\frac{1}{2}}g\left(L_0\right)\mathfrak{v}=g(L_0+\frac{1}{2})G_{\frac{1}{2}}\mathfrak{v}=g(z+\frac{1}{2})f\left(z\right)\mathfrak{w}\end{array}$$

and

$$\begin{array}{c}\hfill L_1\mathfrak{w}={\lambda }_1(z+{\alpha }_1)\mathfrak{w}.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill g(z+\frac{1}{2})f\left(z\right)={\lambda }_1(z+{\alpha }_1).\end{array}$$

(12)

Case 1. If $f\left(t\right)=c$ and $g\left(t\right)=\frac{1}{c}\lambda (t+\alpha )$, we obtain

$$g(z+\frac{1}{2})f\left(z\right)=\lambda (z+\frac{1}{2}+\alpha )={\lambda }_1(z+{\alpha }_1).$$

This implies that

$$\lambda ={\lambda }_1,{\alpha }_1=\alpha +\frac{1}{2}.$$

Case 2. If $g\left(t\right)=c$ and $f\left(t\right)=\frac{1}{c}\lambda (t+\alpha -\frac{1}{2})$, we obtain

$$g(z+\frac{1}{2})f\left(z\right)=\lambda (z+\alpha -\frac{1}{2})={\lambda }_1(z+{\alpha }_1).$$

This implies that

$$\lambda ={\lambda }_1,{\alpha }_1=\alpha -\frac{1}{2}.$$

We complete the proof. □

From Lemma 3, up to a parity, we can assume ${\alpha }_1=\alpha +\frac{1}{2}$, $G_{\frac{1}{2}}\mathfrak{v}=\mathfrak{w}$ and $G_{\frac{1}{2}}\mathfrak{w}=\lambda (z+\alpha )\mathfrak{v}$ without a loss of generality.

Lemma 4.

For any $p\in \mathbb{Z}+\frac{1}{2}$, $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$, we have

(1)

$G_{p}f\left(t\right)\mathfrak{v}={\lambda }^{p-\frac{1}{2}}f(z+p)\mathfrak{w}$.

(2)

$G_{p}f\left(z\right)\mathfrak{w}={\lambda }^{p+\frac{1}{2}}(t+2p\alpha )f(t+p)\mathfrak{v}$.

Proof. 

We first show that part $\left(1\right)$ and part $\left(2\right)$ hold for $p=\frac{1}{2}$. From Lemmas 2 and 3, we obtain

$$\begin{array}{c}\hfill G_{\frac{1}{2}}f\left(t\right)\mathfrak{v}=G_{\frac{1}{2}}f\left(L_0\right)\mathfrak{v}=f(L_0+\frac{1}{2})G_{\frac{1}{2}}\mathfrak{v}=f(z+\frac{1}{2})\mathfrak{w},\end{array}$$

(13)

$$\begin{array}{c}\hfill G_{\frac{1}{2}}f\left(z\right)\mathfrak{w}=G_{\frac{1}{2}}f\left(L_0\right)\mathfrak{w}=f(L_0+\frac{1}{2})G_{\frac{1}{2}}\mathfrak{w}=\lambda (t+\alpha )f(t+\frac{1}{2})\mathfrak{w}.\end{array}$$

(14)

For any $p\in \mathbb{Z}+\frac{1}{2}$ and $p\ne \frac{3}{2}$, from (6), (7), (13), (14), we have

$$\begin{array}{ccc}& & (\frac{1}{2}p-\frac{3}{4})G_{p}f\left(t\right)\mathfrak{v}=[L_{p-\frac{1}{2}},G_{\frac{1}{2}}]f\left(t\right)\mathfrak{v}\hfill \\ & & =L_{p-\frac{1}{2}}{G}_{\frac{1}{2}}f\left(t\right)\mathfrak{v}-G_{\frac{1}{2}}{L}_{p-\frac{1}{2}}f\left(t\right)\mathfrak{v}\hfill \\ & & =L_{p-\frac{1}{2}}f(z+\frac{1}{2})\mathfrak{w}-G_{\frac{1}{2}}{\lambda }^{p-\frac{1}{2}}(t+(p-\frac{1}{2})\alpha )f(t+p-\frac{1}{2})\mathfrak{v}\hfill \\ & & ={\lambda }^{p-\frac{1}{2}}(z+(p-\frac{1}{2})(\alpha +\frac{1}{2}))f(z+p)\mathfrak{w}-{\lambda }^{p-\frac{1}{2}}(z+\frac{1}{2}+(p-\frac{1}{2})\alpha )f(z+p)\mathfrak{w}\hfill \\ & & =(\frac{1}{2}p-\frac{3}{4}){\lambda }^{p-\frac{1}{2}}f(z+p)\mathfrak{v}\hfill \end{array}$$

and

$$\begin{array}{ccc}& & (\frac{1}{2}p-\frac{3}{4})G_{p}f\left(z\right)\mathfrak{w}=[L_{p-\frac{1}{2}},G_{\frac{1}{2}}]f\left(z\right)\mathfrak{w}\hfill \\ & & =L_{p-\frac{1}{2}}{G}_{\frac{1}{2}}f\left(z\right)\mathfrak{w}-G_{\frac{1}{2}}{L}_{p-\frac{1}{2}}f\left(z\right)\mathfrak{w}\hfill \\ & & =L_{p-\frac{1}{2}}\lambda (t+\alpha )f(t+\frac{1}{2})\mathfrak{v}-G_{\frac{1}{2}}{\lambda }^{p-\frac{1}{2}}(z+(p-\frac{1}{2})(\alpha +\frac{1}{2}))f(z+p-\frac{1}{2})\mathfrak{w}\hfill \\ & & ={\lambda }^{p-\frac{1}{2}}(t+(p-\frac{1}{2})\alpha )\lambda (t+p+\alpha -\frac{1}{2})f(t+p)\mathfrak{v}\hfill \\ & & -\lambda (t+\alpha ){\lambda }^{p-\frac{1}{2}}(t+\frac{1}{2}+(p-\frac{1}{2})(\alpha +\frac{1}{2}))f(t+p)\mathfrak{v}\hfill \\ & & =(\frac{1}{2}p-\frac{3}{4}){\lambda }^{p+\frac{1}{2}}(t+2p\alpha )f(t+p)\mathfrak{v}.\hfill \end{array}$$

Part $\left(1\right)$ and part $\left(2\right)$ hold for any $p\in \mathbb{Z}+\frac{1}{2}$ and $p\ne \frac{3}{2}$.

Furthermore, we have

$$\begin{array}{ccc}& & \frac{3}{2}{G}_{\frac{3}{2}}f\left(t\right)\mathfrak{v}=[L_2,G_{-\frac{1}{2}}]f\left(t\right)\mathfrak{v}\hfill \\ & & =L_2{G}_{-\frac{1}{2}}f\left(t\right)\mathfrak{v}-G_{-\frac{1}{2}}{L}_{2}f\left(t\right)\mathfrak{v}\hfill \\ & & =L_2{\lambda }^{-1}f(z-\frac{1}{2})\mathfrak{w}-G_{-\frac{1}{2}}{\lambda }^2(t+2\alpha )f(t+2)\mathfrak{v}\hfill \\ & & =\lambda (z+2(\alpha +\frac{1}{2}))f(z+\frac{3}{2})\mathfrak{w}-\lambda (z-\frac{1}{2}+2\alpha )f(z+\frac{3}{2})\mathfrak{w}\hfill \\ & & =\frac{3}{2}\lambda f(z+\frac{3}{2})\mathfrak{w}\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \frac{3}{2}{G}_{\frac{3}{2}}f\left(z\right)\mathfrak{w}=[L_2,G_{-\frac{1}{2}}]f\left(z\right)\mathfrak{w}=L_2{G}_{-\frac{1}{2}}f\left(z\right)\mathfrak{w}-G_{-\frac{1}{2}}{L}_{2}f\left(z\right)\mathfrak{w}\hfill \\ & & =L_2(t-\alpha )f(t-\frac{1}{2})\mathfrak{v}-G_{-\frac{1}{2}}{\lambda }^2(z+2(\alpha +\frac{1}{2}))f(z+2)\mathfrak{w}\hfill \\ & & ={\lambda }^2(t+2\alpha )(t+2-\alpha )f(t+\frac{3}{2})\mathfrak{v}-{\lambda }^2(t-\alpha )(t-\frac{1}{2}+2(\alpha +\frac{1}{2}))f(t+\frac{3}{2})\mathfrak{v}\hfill \\ & & =\frac{3}{2}{\lambda }^2(t+3\alpha )f(t+\frac{3}{2})\mathfrak{v}.\hfill \end{array}$$

Part $\left(1\right)$ and part $\left(2\right)$ also hold for $p=\frac{3}{2}$. We complete the proof. □

Lemma 5.

For any $p\in \mathbb{Z}+\frac{1}{2}$, $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$, we have

(1)

$Q_{p}f\left(t\right)\mathfrak{v}=0$.

(2)

$Q_{p}f\left(z\right)\mathfrak{w}={\lambda }^{p+\frac{1}{2}}\beta f(t+p)\mathfrak{v}$.

Proof. 

From (8), Lemma 4 and $[I_m,G_{p-m}]\mathfrak{v}=m{Q}_p\mathfrak{v}$ with $m\ne 0$, we have

$$\begin{array}{ccc}& & Q_{p}f\left(t\right)\mathfrak{v}=\frac{1}{m}[I_m,G_{p-m}]f\left(t\right)\mathfrak{v}\hfill \\ & & =\frac{1}{m}{I}_m{G}_{p-m}f\left(t\right)\mathfrak{v}-\frac{1}{m}{G}_{p-m}{I}_{m}f\left(t\right)\mathfrak{v}\hfill \\ & & =\frac{1}{m}{I}_m{\lambda }^{p-m-\frac{1}{2}}f(z+p-m)\mathfrak{w}-\frac{1}{m}{G}_{p-m}{\lambda }^m\beta f(t+m)\mathfrak{v}\hfill \\ & & =\frac{1}{m}{\lambda }^{p-\frac{1}{2}}{\beta }_{1}f(z+p)\mathfrak{w}-\frac{1}{m}{\lambda }^{p-\frac{1}{2}}\beta f(z+p)\mathfrak{w}\hfill \\ & & =\frac{1}{m}{\lambda }^{p-\frac{1}{2}}({\beta }_1-\beta )f(z+p)\mathfrak{w}.\hfill \end{array}$$

Arbitrariness of m, we obtain $\beta ={\beta }_1$. Thus,

$$\begin{array}{c}\hfill Q_{p}f\left(t\right)\mathfrak{v}=0.\end{array}$$

(15)

From (15), (9), Lemma 4, and $[I_m,G_{p-m}]\mathfrak{w}=m{Q}_p\mathfrak{w}$ with $m\ne 0$, we have

$$\begin{array}{ccc}& & Q_{p}f\left(z\right)\mathfrak{w}=\frac{1}{m}[I_m,G_{p-m}]f\left(z\right)\mathfrak{w}\hfill \\ & & =\frac{1}{m}{I}_m{G}_{p-m}f\left(z\right)\mathfrak{w}-\frac{1}{m}{G}_{p-m}{I}_{m}f\left(z\right)\mathfrak{w}\hfill \\ & & =\frac{1}{m}{I}_m{\lambda }^{p-m+\frac{1}{2}}(t+2(p-m)\alpha )f(t+p-m)\mathfrak{v}-\frac{1}{m}{G}_{p-m}{\lambda }^m{\beta }_{1}f(z+m)\mathfrak{w}\hfill \\ & & =\frac{1}{m}{\lambda }^{p+\frac{1}{2}}\beta (t+m+2(p-m)\alpha )f(t+p)\mathfrak{v}-\frac{1}{m}{\lambda }^{p+\frac{1}{2}}{\beta }_1(t+2(p-m)\alpha )f(t+p)\mathfrak{v}\hfill \\ & & ={\lambda }^{p+\frac{1}{2}}\beta f(t+p)\mathfrak{v}.\hfill \end{array}$$

Thus, we obtain

$$\begin{array}{c}\hfill Q_{p}f\left(z\right)\mathfrak{w}={\lambda }^{p+\frac{1}{2}}\beta f(t+p)\mathfrak{v}.\end{array}$$

(16)

We complete the proof. □

Now, we present the main result of this section, which provides a complete classification of free $U\left(\mathfrak{h}\right)$-module of rank 2 over $\mathfrak{g}$, as well as the proof that follows from (6), (7), (8), (9), and Lemmas 4 and 5.

Theorem 3.

Let $\mathfrak{M}$ be an $\mathfrak{g}$-module, such that the restriction of $\mathfrak{M}$ as a $U\left(\mathfrak{h}\right)$-module is free of rank 2. Then, up to a parity, $\mathfrak{M}\cong \Omega (\lambda ,\alpha ,\beta )$ for some $\lambda \in {\mathbb{C}}^{*}$, $\alpha ,\beta \in \mathbb{C}$ with the $\mathfrak{g}$-module structure defined as in (1)∼(5).

Proof. 

The assertion follows from (6), (7), (8), (9), and Lemmas 4 and 5. □

Theorem 4.

Let $\mathfrak{g}$ be the $N=1$ Heisenberg–Virasoro superalgebra, $\lambda \in {\mathbb{C}}^{*},\alpha ,\beta \in \mathbb{C}$. Let

$$\Gamma =t(\Omega {(\lambda ,0,0)}_{\overline{0}})\oplus \Omega {(\lambda ,0,0)}_{\overline{1}}.$$

Then, the following statements hold.

(1)

$\Omega (\lambda ,\alpha ,\beta )$ is simple if and only if $\alpha \ne 0$ or $\beta \ne 0$.

(2)

$\Omega (\lambda ,0,0)$ has a unique proper submodule Γ, and $\Omega (\lambda ,0,0)/\Gamma$ is a 1-dimensional trivial $\mathfrak{g}$-module.

(3)

$\Gamma \cong \Pi (\Omega (\lambda ,\frac{1}{2},0))$, so that Γ is an irreducible $\mathfrak{g}$-module.

Proof. 

Let $V=V_{\overline{0}}\oplus V_{\overline{1}}$ be a nonzero submodule of $\Omega (\lambda ,\alpha ,\beta )$. By

$$G_{p}f\left(z\right)={\lambda }^{p+\frac{1}{2}}(t+2p\alpha )f(t+p)\mathit{or}{Q}_{p}f\left(z\right)={\lambda }^{p+\frac{1}{2}}\beta f(t+p),$$

we obtain $V_{\overline{0}}\ne 0$. If $V_{\overline{0}}=\Omega {(\lambda ,\alpha ,\beta )}_{\overline{0}}$, we obtain $V_{\overline{1}}=\Omega {(\lambda ,\alpha ,\beta )}_{\overline{1}}$ by $G_{p}f\left(t\right)={\lambda }^{p-\frac{1}{2}}f(z+p)$. So, $V=\Omega (\lambda ,\alpha ,\beta )$. Which implies that $\Omega (\lambda ,\alpha ,\beta )$ is a simple $\mathfrak{g}$-module, and it can be deduced that $\Omega {(\lambda ,\alpha ,\beta )}_{\overline{0}}$ is a simple ${\mathfrak{g}}_{\overline{0}}$-module.

Case 1. $\alpha =\beta =0$.

In this case, $\Gamma$ is a proper submodule of $\Omega (\lambda ,0,0)$. Next, we suppose $V=V_{\overline{0}}\oplus V_{\overline{1}}$ is an arbitrary nonzero proper submodule of $\Omega (\lambda ,0,0)$, then $V_{\overline{0}}=t\Omega {(\lambda ,0,0)}_{\overline{0}}$ by Lemma 2.3 (2) in [12]. Furthermore, from $G_{p}f\left(t\right)={\lambda }^{p-\frac{1}{2}}f(z+p)$, we obtain $V_{\overline{1}}=\Omega {(\lambda ,0,0)}_{\overline{1}}$. Thus, $V=\Gamma$ and $\Omega (\lambda ,0,0)/\Gamma$ is a 1-dimensional trivial $\mathfrak{g}$-module

Case 2. $\alpha \ne 0$.

As $\Omega {(\lambda ,\alpha ,\beta )}_{\overline{0}}$ is a simple ${\mathfrak{g}}_{\overline{0}}$-module by Lemma 2.3 (2) in [12], it follows that $\Omega (\lambda ,\alpha ,\beta )$ is a simple $\mathfrak{g}$-module.

Case 3. $\beta \ne 0$.

Let $V=V_{\overline{0}}\oplus V_{\overline{1}}$ be a nonzero submodule of $\Omega (\lambda ,\alpha ,\beta )$. From $Q_{p}f\left(z\right)={\lambda }^{p+\frac{1}{2}}\beta f(t+p)$, we obtain $V_{\overline{0}}\ne 0$. If $V_{\overline{0}}=\Omega {(\lambda ,\alpha ,\beta )}_{\overline{0}}$, we obtain $V_{\overline{1}}=\Omega {(\lambda ,\alpha ,\beta )}_{\overline{1}}$ from $G_{p}f\left(t\right)={\lambda }^{p-\frac{1}{2}}f(z+p)$. Consequently, $V=\Omega (\lambda ,\alpha ,\beta )$. This implies that $\Omega (\lambda ,\alpha ,\beta )$ is a simple $\mathfrak{g}$-module.

For part (3), we define a linear map

$$\phi :\Gamma \to \Pi (\Omega (\lambda ,\frac{1}{2},0))$$

such that

$$\phi \left(tf\left(t\right)\right)=\frac{1}{\sqrt{\lambda }}f\left(z\right),\phi \left(f\left(z\right)\right)=\sqrt{\lambda }f\left(t\right)$$

(17)

for $f\left(t\right)\in \mathbb{C}\left[t\right]$ and $f\left(z\right)\in \mathbb{C}\left[z\right]$. From this easy calculation, we obrtain

$$\begin{array}{ccc}\hfill \phi \left(L_{m}tf\left(t\right)\right)& =& {\lambda }^m\phi \left(t(t+m)f(t+m)\right)=\frac{{\lambda }^m}{\sqrt{\lambda }}(z+m)f(z+m)=\frac{1}{\sqrt{\lambda }}{L}_{m}f\left(z\right)\hfill \\ & =& L_m\phi \left(tf\left(t\right)\right);\hfill \\ \hfill \phi \left(I_{m}tf\left(t\right)\right)& =& I_m\phi \left(tf\left(t\right)\right);\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \phi \left(G_{p}tf\left(t\right)\right)& =& {\lambda }^{p-\frac{1}{2}}\phi \left((z+p)f(z+p)\right)={\lambda }^{p-\frac{1}{2}}\sqrt{\lambda }(t+p)f(t+p)=\frac{1}{\sqrt{\lambda }}{G}_{p}f\left(z\right)\hfill \\ & =& G_p\phi \left(tf\left(t\right)\right);\hfill \\ \hfill \phi \left(Q_{p}tf\left(t\right)\right)& =& Q_p\phi \left(tf\left(t\right)\right).\hfill \end{array}$$

Similar arguments yield the following

$$\begin{array}{ccc}& & \phi \left(L_{m}f\left(z\right)\right)=L_m\phi \left(f\left(z\right)\right);\phi \left(I_{m}f\left(z\right)\right)=I_m\phi \left(f\left(z\right)\right);\hfill \\ & & \phi \left(G_{p}f\left(z\right)\right)=G_p\phi \left(f\left(z\right)\right);\phi \left(Q_{p}f\left(z\right)\right)=Q_p\phi \left(f\left(z\right)\right).\hfill \end{array}$$

Thus, $\phi$ is a $\mathfrak{g}$-module isomorphism, and $\Gamma$ is irreducible. □

Similar arguments as the proof of ([12] [Theorem 2.7]) yield the following result for the $N=1$ Heisenberg–Virasoro superalgebra.

Theorem 5.

For $\lambda ,{\lambda }^{\prime }\in {\mathbb{C}}^{*},\alpha ,{\alpha }^{\prime },\beta ,{\beta }^{\prime }\in \mathbb{C}$, we have $\Omega (\lambda ,\alpha ,\beta )\cong \Omega ({\lambda }^{\prime },{\alpha }^{\prime },{\beta }^{\prime })$, if and only if $\lambda ={\lambda }^{\prime }$, $\alpha ={\alpha }^{\prime }$ and $\beta ={\beta }^{\prime }$.

In conclusion, a new simple type of modules over the $N=1$ Heisenberg–Virasoro superalgebra, which are non-weight modules, are constructed by classical methods in this paper (Theorems 3 and 4). Certainly, the study of non-weight modules over the $N=1$ Heisenberg–Virasoro superalgebra is still in its infancy.