$U\left(\mathfrak{h}\right)$-Free Modules over the Lie Algebras of Differential Operators

Abstract

This paper mainly considers a class of non-weight modules over the Lie algebra of the Weyl type. First, we construct the $U\left(\mathfrak{h}\right)$-free modules of rank one over the differential operator algebra. Then, we characterize the tensor products of these kind of modules and the quasi-finite highest weight modules. Finally, we undertake such research for the differential operator algebra of multi-variables.

1. Introduction

W-algebras, which are higher-spin extensions of the Virasoro algebra and their superanalogues, have played an important role in conformal field theories [1,2]. The Lie algebra $W_{1+\infty }$, which is a fundamental example of W-algebras, appears as the limit of the $W_N$ algebra, as N goes to ∞ [3]. It is well known that $W_{1+\infty }$ can also be realized as the universal central extension of the Lie algebra of differential operators on the circle [4,5]. Weight modules over $W_{1+\infty }$ were studied sufficiently and have important applications in quantum field theory and integrable systems (see [4,6,7,8], etc.). However, there are few papers on non-weight modules over $W_{1+\infty }$ up to now.

Recently, many authors constructed various non-weight modules over some Lie algebras. In particular, a class of ${\mathfrak{s}l}_n$-modules that are free of rank one when restricted to the enveloping algebra of its Cartan subalgebra were constructed in [9]. These kind of non-weight modules, which are called $U\left(\mathfrak{h}\right)$-free modules, have been extensively studied. It was proven in [9,10] that a finite dimensional simple Lie algebra has nontrivial $U\left(\mathfrak{h}\right)$-free modules if and only if it is of the special linear algebra $\mathfrak{sl}(n+1,\mathbb{C})$ or the symplectic algebra $\mathfrak{sp}(2n,\mathbb{C})$ for some $n\ge 1$. Such research was exploited and generalized to study modules over many infinite dimensional Lie algebras, such as the Witt algebra (the Virasoro algebra) [11,12], the twisted Heisenberg–Virasoro algebra and the $W(2,2)$ algebra [13], and so on. The aim of this paper is to determine $U\left(\mathfrak{h}\right)$-free modules for the Lie algebra $W_{1+\infty }$ motivated by [12,13]. We get a type of new simple modules over $W_{1+\infty }$, but they are not modules of the corresponding associative algebras (see (1) in Section 3). Moreover, such results can be extended to the Lie algebras of differential operators on the Laurent polynomial rings of multi-variables.

The paper is arranged as follows. In Section 2, we recall some necessary definitions and preliminary results. In Section 3, we determine all module structures on $U\left(\mathfrak{h}\right)$ (see Theorem 4 below). In Section 4, we give a necessary and sufficient condition for the tensor product of a quasi-finite highest weight module $L(h,c)$ and a $U\left(\mathfrak{h}\right)$-free $\mathcal{D}$-module $\mathsf{\Omega }(\lambda ,ϵ)$ to be irreducible (see Theorem 5 below). Furthermore, we show that two such tensor product modules are isomorphic if and only if the corresponding quasi-finite highest weight modules and $U\left(\mathfrak{h}\right)$-free modules are isomorphic (see Theorem 6 below). Consequently, we obtain a lot of new irreducible non-weight modules over $W_{1+\infty }$. In Section 5, we construct such modules for the Lie algebras of differential operators in the general case.

Throughout this paper, the sets of integers, positive integers, non-negative integers, complex numbers and non-zero complex numbers are denoted by $\mathbb{Z},$${\mathbb{Z}}_{+},$$\mathbb{N}$, $\mathbb{C}$ and ${\mathbb{C}}^{*}$, respectively. All algebras and modules are over the complex number field $\mathbb{C}$.

2. Basics

In this section, we recall some necessary definitions and preliminary results.

2.1. The Lie Algebra of Differential Operators

Let $\mathbb{C}[t,t^{-1}]$ be the Laurent polynomial ring over $\mathbb{C}$, ${\mathcal{D}}_{ass}=\mathrm{Diff}\mathbb{C}[t,t^{-1}]$ the associative algebra of all differential operators over $\mathbb{C}[t,t^{-1}]$, which has a basis $\{t^m{D}^n\mid m\in \mathbb{Z},n\in \mathbb{N}\}$ with multiplications:

$$\left(t^{m_1}{D}^{n_1}\right)\left(t^{m_2}{D}^{n_2}\right)=\sum _{i=0}^{n_1}\left(\genfrac{}{}{0pt}{}{n_1}{i}\right)m_2^i{t}^{m_1+m_2}{D}^{n_1+n_2-i}$$

for all $m_1,m_2\in \mathbb{Z},n_1,n_2\in \mathbb{N}$, where $D=t\frac{d}{dt}$.

Let $\mathcal{D}$ be the Lie algebra of ${\mathcal{D}}_{ass}$ under Lie bracket given by

$$\left[t^{m_1}{D}^{n_1},t^{m_2}{D}^{n_2}\right]=\sum _{i=0}^{n_1}\left(\genfrac{}{}{0pt}{}{n_1}{i}\right)m_2^i{t}^{m_1+m_2}{D}^{n_1+n_2-i}-\sum _{j=0}^{n_2}\left(\genfrac{}{}{0pt}{}{n_2}{j}\right)m_1^j{t}^{m_1+m_2}{D}^{n_1+n_2-j}$$

for all $m_1,m_2\in \mathbb{Z},n_1,n_2\in \mathbb{N}$.

Lemma 1

([14]). The Lie algebra $\mathcal{D}$ is generated by $\{t,t^{-1},D^2\}$.

Li [15] proved $H^2(\mathcal{D},\mathbb{C})=1$ (see also [4]). More precisely, we have the following result.

Lemma 2

([4,15]). Any non-trivial 2-cocycle on $\mathcal{D}$ is equivalent to ϕ:

$$\varphi (t^{m_1}{D}^{n_1},t^{m_2}{D}^{n_2})=\left\{\begin{array}{cc}& 0,\mathit{if}{m}_1=0,\hfill \\ \hfill & {(-1)}^{n_1+1}{\delta }_{m_1+m_2,0}\frac{1}{2}\sum _{i=1}^{m_1}{(m_1-i)}^{n_1}{i}^{n_2},\mathit{if}{m}_1>0,\hfill \\ \hfill & {(-1)}^{n_1}{\delta }_{m_1+m_2,0}\frac{1}{2}\sum _{i=m_1}^{-1}{(m_1-i)}^{n_1}{i}^{n_2},\mathit{if}{m}_1<0.\hfill \end{array}\right.$$

Let $\widehat{\mathcal{D}}$ (or $W_{1+\infty }$) denote the universal (one-dimensional) central extension of the Lie algebra $\mathcal{D}$ by the above 2-cocycle $\varphi$ with a central element C. $\widehat{\mathcal{D}}$ is $\mathbb{Z}$-graded by ${\widehat{\mathcal{D}}}_i={\mathrm{Span}}_{\mathbb{C}}\{t^i{D}^n\mid n\ge 0\}\oplus {\delta }_{i,0}\mathbb{C}C$ for $i\in \mathbb{Z}$.

A $\widehat{\mathcal{D}}$-module V is called a highest (resp. lowest) weight module, if there exists a nonzero element $v\in V_{\lambda }$ such that

(1) $\mathbb{C}v$ is a one dimensional ${\widehat{\mathcal{D}}}_0$-module;

(2) ${\widehat{\mathcal{D}}}_{+}v=0$ (resp. ${\widehat{\mathcal{D}}}_{-}v=0$), where ${\widehat{\mathcal{D}}}_{+}={\sum }_{i>0}{\widehat{\mathcal{D}}}_i$, ${\widehat{\mathcal{D}}}_{-}={\sum }_{i<0}{\widehat{\mathcal{D}}}_i$.

Here we shall note that the highest weight module defined as above is not always quasi-finite. Quasi-finite highest or lowest weight modules were studied in [4].

Next, we define the Verma module, which is a highest weight module with the free action of $U\left({\widehat{\mathcal{D}}}_{-}\right)$ on the highest weight vector (see [16]). More precisely, for any $c\in \mathbb{C}$ and $h=(h_1,h_2,\dots ),h_i\in \mathbb{C},i\in {\mathbb{Z}}_{+}$, let $\mathbb{C}\mathbf{1}$ be the one-dimensional module over the subalgebra ${\widehat{\mathcal{D}}}_{+}\oplus {\widehat{\mathcal{D}}}_0$ defined by

$${\widehat{\mathcal{D}}}_{+}\mathbf{1}=0,C\mathbf{1}=c\mathbf{1},D^k\mathbf{1}=h_k\mathbf{1}.$$

Then we get the induced $\widehat{\mathcal{D}}$-module, called the Verma module:

$$M(h,c)=U\left(\widehat{\mathcal{D}}\right){\otimes }_{U({\widehat{\mathcal{D}}}_{+}+{\widehat{\mathcal{D}}}_0)}\mathbb{C}\mathbf{1}.$$

Clearly, $M(h,c)$ has a unique maximal submodule $J(h,c)$, and denoted by $L(h,c)$ the corresponding simple quotient module. For a nonzero weight vector $u^{\prime }\in M(h,c)$, if ${\widehat{\mathcal{D}}}_{+}{u}^{\prime }=0$, then $u^{\prime }$ is called a singular vector. Certainly, $J(h,c)$ is generated by all homogenous singular vectors in $M(h,c)$ not in $\mathbb{C}\mathbf{1}$, and $M(h,c)$ is irreducible if and only if $M(h,c)$ does not contain any other singular vectors besides those in $\mathbb{C}\mathbf{1}$.

Introduce the generating series

$${\mathsf{\Delta }}_h\left(x\right)=-\sum _{n=0}^{\infty }\frac{x^n}{n!}{h}_n,$$

which is a formal power series. A formal power series is called a quasipolynomial if it is a finite linear combination of functions of the form $p\left(x\right)e^{\alpha x}$, where $p\left(x\right)$ is a polynomial and $\alpha \in \mathbb{C}$.

The following results were given in [4]:

Theorem 1

([4]). For any $c\in \mathbb{C}$ and $h=(h_1,h_2,\dots ),h_i\in \mathbb{C},i\in {\mathbb{Z}}_{+}$, $L(h,c)$ is quasi-finite if and only if

$${\mathsf{\Delta }}_h\left(x\right)=\frac{\varphi \left(x\right)}{e^x-1},$$

where $\varphi \left(x\right)$ is a quasipolynomial such that $\varphi \left(0\right)=0$.

2.2. The Subalgebras of the Lie Algebra of Differential Operators

The Lie algebra $\mathcal{D}$ has a subalgebra, which is called the Witt algebra, $\mathcal{W}=Der\mathbb{C}[t,t^{-1}]$ (also denoted by Vect($S^1$), the Lie algebra of all vector fields on the circle). The Virasoro algebra Vir is the universal central extension of $\mathcal{W}$, with a basis $\{L_n=t^{n+1}\frac{d}{dt},C|n\in \mathbb{Z}\}$ and relations

$$[L_m,L_n]=(n-m)L_{m+n}+{\delta }_{n,-m}\frac{m^3-m}{12}C,$$

$$[C,L_m]=0,\forall m,n\in \mathbb{Z}.$$

The Lie algebra $\mathcal{D}$ has also a subalgebra ${\mathcal{D}}_1$ generated by $\{I_m:=t^m,L_m:=t^{m}D\mid m\in \mathbb{Z}\}$ with the following relations:

$$\begin{array}{ccc}\hfill & & [L_m,L_n]=(n-m)L_{m+n},\hfill \\ & & [I_m,I_n]=0,\hfill \\ & & [L_m,I_n]=n{I}_{m+n},\forall m,n\in \mathbb{Z}.\hfill \end{array}$$

Its universal central extension is called the twisted Heisenberg–Virasoro algebra (see [17,18,19,20,21]).

2.3. $U\left(\mathfrak{h}\right)$-Free Modules

In this subsection, we recall some $U\left(\mathfrak{h}\right)$-free modules of rank one over some sublagebras of $\mathcal{D}$.

Let L be an associative or Lie algebra and R be a subspace of L. For a L-module V, if there exists a nonzero $f\in R$ such that $fv=0$ for some nonzero $v\in V$, then V is called R-torsion. Otherwise, V is called R-torsion-free.

Let $\lambda \in {\mathbb{C}}^{*},\alpha \in \mathbb{C}$, $\mathbb{C}\left[x\right]$ be the polynomial algebra over $\mathbb{C}$. Define the actions of the Virasoro algebra on $\mathbb{C}\left[x\right]$ as follows

$$L_{m}f\left(x\right)={\lambda }^m(x-m\alpha )f(x-m),Cf\left(x\right)=0,\forall m\in \mathbb{Z},f\left(x\right)\in \mathbb{C}\left[x\right].$$

Then, $\mathbb{C}\left[x\right]$ becomes a $\mathrm{Vir}$-module, which is denoted by $\mathsf{\Omega }(\lambda ,\alpha )$. By [11], $\mathsf{\Omega }(\lambda ,\alpha )$ is simple if and only if $\lambda ,\alpha \in {\mathbb{C}}^{*}$. $\mathsf{\Omega }(\lambda ,0)$ has a simple submodule $x\mathsf{\Omega }(\lambda ,0)$ with codimension one.

Theorem 2

([12]). Let M be a $\mathrm{Vir}$-module such that the restriction of $U\left(\mathrm{Vir}\right)$ to $U\left(\mathbb{C}{L}_0\right)$ is free of rank one. Then, $M\cong \mathsf{\Omega }(\lambda ,\alpha )$ for some $\alpha \in \mathbb{C},\lambda \in {\mathbb{C}}^{*}$.

The module $\mathsf{\Omega }(\lambda ,\alpha )$ can be naturally made into a ${\mathcal{D}}_1$-module (or a module over the twisted Heisenberg–Virasoro algebra) as follows. Let $\lambda \in {\mathbb{C}}^{*},\alpha ,\beta \in \mathbb{C}$, $\mathbb{C}\left[x\right]$ be the polynomial algebra over $\mathbb{C}$. Define the actions of ${\mathcal{D}}_1$ on $\mathbb{C}\left[x\right]$ by

$$L_{m}f\left(x\right)={\lambda }^m(x-m\alpha )f(x-m),I_{m}f\left(x\right)=\beta {\lambda }^{m}f(x-m),$$

where $f\left(x\right)\in \mathbb{C}\left[x\right]$, $m\in \mathbb{Z}$.

In this case, $\mathbb{C}\left[x\right]$ becomes a ${\mathcal{D}}_1$-module. Moreover, $\mathsf{\Omega }(\lambda ,\alpha ,\beta )$ is simple if and only if $\alpha \ne 0$ or $\beta \ne 0$ (see [13]). $\mathsf{\Omega }(\lambda ,0,0)$ has a simple submodule $x\mathsf{\Omega }(\lambda ,0,0)$ with codimension one.

Theorem 3

([13]). Let M be a ${\mathcal{D}}_1$-module such that the restriction of $U\left({\mathcal{D}}_1\right)$ to $U\left(\mathbb{C}{L}_0\right)$ is free of rank one. Then, $M\cong \mathsf{\Omega }(\lambda ,\alpha ,\beta )$ for some $\alpha ,\beta \in \mathbb{C},\lambda \in {\mathbb{C}}^{*}$.

3. $\mathbf{U}\left(\mathfrak{h}\right)$-Free Modules over the Lie Algebra of Differential Operators

In this section, we mainly study $U\left(\mathfrak{h}\right)$-free modules over the Lie algebra of differential operators on the circle based on the research found in [13]. We classify all $\mathcal{D}$-modules such that the restriction of $U\left(\mathcal{D}\right)$ to $U\left(\mathbb{C}D\right)$ is free of rank one and get some new irreducible $\mathcal{D}$-modules.

For $\lambda \in {\mathbb{C}}^{*},ϵ=0,1$, define the actions of $\mathcal{D}$ on $\mathsf{\Omega }(\lambda ,ϵ):=\mathbb{C}\left[x\right]$ as follows

$$t^m{D}^{n}f\left(x\right)={\beta }^{1-n}{\lambda }^m{(x-ϵm)}^{n}f(x-m),$$

(1)

where $\beta ={(-1)}^{1-ϵ}$, $f\left(x\right)\in \mathbb{C}\left[x\right]$ and $m\in \mathbb{Z},n\in \mathbb{N}$.

Lemma 3.

$\mathsf{\Omega }(\lambda ,ϵ)$ is an irreducible $\mathcal{D}$-module.

Proof. 

For $f\left(x\right)\in \mathbb{C}\left[x\right],m,m_1\in \mathbb{Z},n,n_1\in \mathbb{N}$,

$$\begin{array}{ccc}& & [t^m{D}^n,t^{m_1}{D}^{n_1}]f\left(x\right)=\left(t^m{D}^n\right)\left(t^{m_1}{D}^{n_1}\right)f\left(x\right)-\left(t^{m_1}{D}^{n_1}\right)\left(t^m{D}^n\right)f\left(x\right)\hfill \\ & =& {\lambda }^{m+m_1}{\beta }^{2-n_1-n}{(x-ϵm)}^n{(x-m-ϵm_1)}^{n_1}f(x-m-m_1)\hfill \\ & & -{\lambda }^{m+m_1}{\beta }^{2-n_1-n}{(x-ϵm_1)}^{n_1}{(x-m_1-ϵm)}^{n}f(x-m-m_1).\hfill \end{array}$$

On the other side,

$$\begin{array}{ccc}& & [t^m{D}^n,t^{m_1}{D}^{n_1}]f\left(x\right)\hfill \\ & =& (\sum _{i=0}^n\left(\genfrac{}{}{0pt}{}{n}{i}\right)m_1^i{t}^{m+m_1}{D}^{n+n_1-i}-\sum _{j=0}^{n_1}\left(\genfrac{}{}{0pt}{}{n_1}{j}\right)m^j{t}^{m+m_1}{D}^{n+n_1-j})f\left(x\right)\hfill \\ & =& \sum _{i=0}^n\left(\genfrac{}{}{0pt}{}{n}{i}\right){\beta }^{1-n-n_1+i}{\lambda }^{m+m_1}{m}_1^i{(x-ϵm-ϵm_1)}^{n+n_1-i}f(x-m_1-m)\hfill \\ & & -\sum _{j=0}^{n_1}\left(\genfrac{}{}{0pt}{}{n_1}{j}\right){\beta }^{1-n-n_1+j}{\lambda }^{m+m_1}{m}^j{(x-ϵm-ϵm_1)}^{n+n_1-j}f(x-m_1-m)\hfill \\ & =& {\beta }^{1-n-n_1}{\lambda }^{m+m_1}{(x-ϵm+(\beta -ϵ)m_1)}^n{(x-ϵm-ϵm_1)}^{n_1}f(x-m_1-m)\hfill \\ & & -{\beta }^{1-n-n_1}{\lambda }^{m+m_1}{(x-ϵm_1+(\beta -ϵ)m)}^{n_1}{(x-ϵm-ϵm_1)}^{n}f(x-m_1-m).\hfill \end{array}$$

So $\mathsf{\Omega }(\lambda ,ϵ)$ becomes a $\mathcal{D}$-module since

$$\begin{array}{ccc}& & {(x-ϵm+(\beta -ϵ)m_1)}^n{(x-ϵm-ϵm_1)}^{n_1}\hfill \\ & & -{(x-ϵm_1+(\beta -ϵ)m)}^{n_1}{(x-ϵm-ϵm_1)}^n\hfill \\ & =& \beta {(x-ϵm)}^n{(x-m-ϵm_1)}^{n_1}-\beta {(x-ϵm_1)}^{n_1}{(x-m_1-ϵm)}^n,\hfill \end{array}$$

where $\beta =1$ if $ϵ=1$, and $\beta =-1$ if $ϵ=0$.

It is clear that $\mathsf{\Omega }(\lambda ,ϵ)$ is irreducible since $\lambda \ne 0$. □

Remark 1.

In the case of $ϵ=1$, the module $\mathsf{\Omega }(\lambda ,1)$ is also a ${\mathcal{D}}_{ass}$-module. From [11] we see that if V is an irreducible ${\mathcal{D}}_{ass}$-module on which $\mathbb{C}[t,t^{-1}]$ is torsion, then $V\cong \mathsf{\Omega }(\lambda ,1)$.

Now we can give the main theorem of this paper.

Theorem 4.

Let M be a $\mathcal{D}$-module such that the restriction of $U\left(\mathcal{D}\right)$ to $U\left(\mathbb{C}D\right)$ is free of rank one. Then, $M\cong \mathsf{\Omega }(\lambda ,ϵ)$ for some $\lambda \in {\mathbb{C}}^{*}$ and $ϵ\in \{0,1\}$.

Proof. 

Let $M=U\left(\mathbb{C}D\right)v=\mathbb{C}\left[D\right]v$ be a $U\left(\mathcal{D}\right)$-module, where $v\in M$ is torsion-free. By Theorem 3, we have

$$t^{m}Df\left(x\right)={\lambda }^m(x-m\alpha )f(x-m),t^{m}f\left(x\right)=\beta {\lambda }^{m}f(x-m)$$

(2)

for some $\alpha ,\beta \in \mathbb{C}$, where $f\left(x\right)\in \mathbb{C}\left[x\right]$, $m\in \mathbb{Z}$.

Claim 1.$\beta \ne 0$.

In fact, if $\beta =0$, by $[D^2,t^m]=2t^{m}D+t^m$, we have $t^{m}Df\left(x\right)=0$ for any $m\in \mathbb{Z}$. In this case, M becomes a trivial $\mathcal{D}$-module.

Claim 2.$D^{2}f\left(x\right)=(\frac{1}{\beta }{x}^2+bx+c)f\left(x\right)$ for some $b,c\in \mathbb{C}$.

For any $f\left(x\right)\in \mathbb{C}\left[x\right]$, $D^{2}f\left(D\right)v=f\left(D\right)D^{2}v$. Now we suppose that $D^{2}v=g\left(D\right)v$ for some $g\left(x\right)\in \mathbb{C}\left[x\right]$, then $D^{2}f\left(x\right)=g\left(x\right)f\left(x\right)$ for any $f\left(x\right)\in \mathbb{C}\left[x\right]$. By $[D^2,t]=2tD+t$, we get $D^2·t·f\left(x\right)-t·D^2·f\left(x\right)=(2tD+t)·f\left(x\right)$. Then we can obtain $\beta \left[g\right(x)-g(x-1\left)\right]=2x+\beta -2\alpha .$ It is easy to deduce that

$$g\left(x\right)=\frac{1}{\beta }{x}^2+bx+c$$

for some $b,c\in \mathbb{C}$.

Claim 3.$b=c=0$, $\alpha \in \{0,1\}$, and $\beta ={(-1)}^{1-\alpha }$.

By $[D^2,t^m]=2m{t}^{m}D+m^2{t}^m$, we have

$$\begin{array}{ccc}& & \beta {\lambda }^m(\frac{1}{\beta }{x}^2+bx+c)f(x-m)-\beta [\frac{1}{\beta }{(x-m)}^2+b(x-m)+c]f(x-m)\hfill \\ & =& 2m{\lambda }^m(x-\alpha m)f(x-m)+m^2\beta {\lambda }^{m}f(x-m),\hfill \end{array}$$

which forces that $m^2+bm=(2\alpha -\beta )m^2$ for all $m\in \mathbb{Z}$. So

$$\beta =2\alpha -1,b=0.$$

(3)

By $[D^2,t^{m}D]=m^2{t}^{m}D+2m{t}^m{D}^2$, we can get

$$\begin{array}{ccc}& & 2m{t}^m{D}^2·f\left(x\right)\hfill \\ & =& D^2·t^{m}D·f\left(x\right)-t^{m}D·D^2·f\left(x\right)-m^2{t}^{m}D·f\left(x\right)\hfill \\ & =& D^2·\left({\lambda }^m(x-m\alpha )f(x-m)\right)-\left(t^{m}D\right)·\left(g\left(x\right)f\left(x\right)\right)-m^2{\lambda }^m(x-m\alpha )f(x-m)\hfill \\ & =& {\lambda }^m(x-m\alpha )f(x-m)\left(g\left(x\right)-g(x-m)-m^2\right)\hfill \\ & =& 2m{\beta }^{-1}{\lambda }^m{(x-m\alpha )}^{2}f(x-m).\hfill \end{array}$$

Therefore,

$$t^m{D}^{2}f\left(x\right)={\beta }^{-1}{\lambda }^m{(x-\alpha m)}^{2}f(x-m),m\ne 0.$$

(4)

Similarly, by $[D^2,t^m{D}^2]=2m{t}^m{D}^3+m^2{t}^m{D}^2$ and (4), we can obtain

$$t^m{D}^{3}f\left(x\right)={\beta }^{-2}{\lambda }^m{(x-\alpha m)}^{3}f(x-m),m\ne 0.$$

(5)

Since $[t^{-1}{D}^2,t{D}^2]=4D^3$, using (4), we get

$$\begin{array}{ccc}\hfill 4D^3·f\left(x\right)& =& t^{-1}{D}^2·t{D}^2·f\left(x\right)-t{D}^2·t^{-1}{D}^2·f\left(x\right)\hfill \\ & =& t^{-1}{D}^2·({\beta }^{-1}\lambda {(x-\alpha )}^{2}f(x-1))-t{D}^2·({\beta }^{-1}{\lambda }^{-1}{(x+\alpha )}^{2}f(x+1))\hfill \\ & =& {\beta }^{-2}{(x+\alpha )}^2{(x-\alpha +1)}^{2}f\left(x\right)-{\beta }^{-2}{(x-\alpha )}^2{(x+\alpha -1)}^{2}f\left(x\right)\hfill \\ & =& 4{\beta }^{-2}x(x^2-{\alpha }^2+\alpha )f\left(x\right).\hfill \end{array}$$

Then

$$D^3·f\left(x\right)={\beta }^{-2}x(x^2-{\alpha }^2+\alpha )f\left(x\right).$$

(6)

Using (2) and (6), we can get

$$\begin{array}{ccc}& & D^3·tD·f\left(x\right)-tD·D^3·f\left(x\right)\hfill \\ & =& {\beta }^{-2}\lambda (x-\alpha )(3x^2-3x+1-{\alpha }^2+\alpha )f(x-1).\hfill \end{array}$$

(7)

Using (2)–(5), we have

$$\begin{array}{ccc}\hfill & & (3t{D}^3+3t{D}^2+tD)·f\left(x\right)\hfill \\ & =& \lambda (x-\alpha )[3{\beta }^{-2}{(x-\alpha )}^2+3{\beta }^{-1}(x-\alpha )+1]f(x-1)\hfill \\ & =& {\beta }^{-2}\lambda (x-\alpha )[3x^2-3x+1+{\alpha }^2-\alpha ]f(x-1).\hfill \end{array}$$

(8)

Since $[D^3,tD]=3t{D}^3+3t{D}^2+tD$, by (7) and (8), we can deduce that

$$\alpha (\alpha -1)=0.$$

(9)

Combining (3) and (9), we have $\alpha =0,\beta =-1$ or $\alpha =\beta =1$.

That is,

$$\alpha \in \{0,1\},\beta ={(-1)}^{1-\alpha }.$$

Using (2), (4) and (9), we have

$$\begin{array}{ccc}\hfill & & \left(t{D}^2\right)·\left(t^{-1}D\right)·f\left(x\right)-\left(t^{-1}D\right)·\left(t{D}^2\right)·f\left(x\right)\hfill \\ & =& \left(t{D}^2\right)\left({\lambda }^{-1}(x+\alpha )f(x+1)\right)-\left(t^{-1}D\right)({\beta }^{-1}\lambda {(x-\alpha )}^{2}f(x-1))\hfill \\ & =& {\beta }^{-1}{(x-\alpha )}^2(x+\alpha -1)f\left(x\right)-{\beta }^{-1}(x+\alpha ){(x+1-\alpha )}^{2}f\left(x\right)\hfill \\ & =& {\beta }^{-1}(-3x^2+(2\alpha -1)x-\alpha +{\alpha }^2)f\left(x\right)\hfill \\ & =& (-3{\beta }^{-1}{x}^2+x)f\left(x\right),\hfill \end{array}$$

(10)

and

$$(-3D^2+D)·f\left(x\right)=(-3{\beta }^{-1}{x}^2-3c+x)f\left(x\right).$$

(11)

Since $[t{D}^2,t^{-1}D]=-3D^2+D$, by (10) and (11), it is easy to see that

$$c=0.$$

Furthermore, (4) holds for all $m\in \mathbb{Z}$, that is,

$$t^m{D}^{2}f\left(x\right)={\beta }^{-1}{\lambda }^m{(x-\alpha m)}^{2}f(x-m),\forall m\in \mathbb{Z}.$$

Claim 4. For $m\ne 0$, $t^m{D}^n·f\left(x\right)={\beta }^{1-n}{\lambda }^m{(x-m\alpha )}^{n}f(x-m)$ for all $n\in \mathbb{N}$.

Now we shall prove the result by induction on n in $t^m{D}^n$ with $m\ne 0$. Suppose that $t^m{D}^{n}f\left(x\right)={\beta }^{1-n}{\lambda }^m{(x-m\alpha )}^{n}f(x-m)$ for any $n\le k$. Then,

$$\begin{array}{ccc}\hfill & & D^2·\left(t^m{D}^k\right)·f\left(x\right)-\left(t^m{D}^k\right)·D^2·f\left(x\right)\hfill \\ & =& D^2·({\beta }^{1-k}{\lambda }^m{(x-m\alpha )}^{k}f(x-m))-\left(t^m{D}^k\right)·\left({\beta }^{-1}{x}^{2}f\left(x\right)\right)\hfill \\ & =& {\beta }^{-k}{\lambda }^m{x}^2{(x-m\alpha )}^{k}f(x-m)-{\beta }^{-k}{\lambda }^m{(x-m\alpha )}^k{(x-m)}^{2}f(x-m)\hfill \\ & =& {\beta }^{-k}{\lambda }^m[x^2-{(x-m)}^2]{(x-m\alpha )}^{k}f(x-m)\hfill \\ & =& {\beta }^{-k}{\lambda }^{m}m(2x-m){(x-m\alpha )}^{k}f(x-m).\hfill \end{array}$$

(12)

Since $[D^2,t^m{D}^k]=m^2{t}^m{D}^k+2m{t}^m{D}^{k+1}$, by (12), we get

$$\begin{array}{ccc}& & 2m{t}^m{D}^{k+1}·f\left(x\right)\hfill \\ & =& {\beta }^{-k}{\lambda }^{m}m(2x-m){(x-m\alpha )}^{k}f(x-m)-m^2\left(t^m{D}^k\right)·f\left(x\right)\hfill \\ & =& {\beta }^{-k}{\lambda }^{m}m(2x-m){(x-m\alpha )}^{k}f(x-m)-m^2{\beta }^{1-k}{\lambda }^m{(x-m\alpha )}^{k}f(x-m)\hfill \\ & =& {\beta }^{-k}{\lambda }^{m}m(2x-2m\alpha ){(x-m\alpha )}^{k}f(x-m)\hfill \\ & =& 2m{\beta }^{1-(k+1)}{\lambda }^m{(x-m\alpha )}^{k+1}f(x-m).\hfill \end{array}$$

Therefore, $t^m{D}^{k+1}·f\left(x\right)={\beta }^{1-(k+1)}{\lambda }^m{(x-m\alpha )}^{k+1}f(x-m)$ for $m\ne 0$.

Claim 5.$D^n·f\left(x\right)={\beta }^{1-n}{x}^{n}f\left(x\right)$ for all $n\in \mathbb{N}$.

We shall prove the result by induction on n in $D^n$. Suppose that $D^{n}f\left(x\right)={\beta }^{1-n}{x}^{n}f\left(x\right)$ for any $n\le k$. By $[t^{-1}{D}^k,t{D}^2]=(k+2)D^{k+1}+\frac{1}{2}(k+1)(k-2)D^k+{\sum }_{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right)D^{k+2-i},$ we get

$$\begin{array}{ccc}\hfill & & (k+2)D^{k+1}·f\left(x\right)\hfill \\ & =& \left(t^{-1}{D}^k\right)\left(t{D}^2\right)·f\left(x\right)-\left(t{D}^2\right)\left(t^{-1}{D}^k\right)·f\left(x\right)\hfill \\ & & -\frac{1}{2}(k+1)(k-2){\beta }^{1-k}{x}^{k}f\left(x\right)-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-k-1}{x}^{k+2-i}f\left(x\right)\hfill \\ & =& \left(t^{-1}{D}^k\right)\left({\beta }^{-1}\lambda {(x-\alpha )}^{2}f(x-1)\right)-\left(t{D}^2\right)\left({\beta }^{1-k}{\lambda }^{-1}{(x+\alpha )}^{k}f(x+1)\right)\hfill \\ & & -\frac{1}{2}(k+1)(k-2){\beta }^{1-k}{x}^{k}f\left(x\right)-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-k-1}{x}^{k+2-i}f\left(x\right)\hfill \\ & =& {\beta }^{-k}[{(x+\alpha )}^k{(x+1-\alpha )}^2-{(x-\alpha )}^2{(x-1+\alpha )}^k]f\left(x\right)\hfill \\ & & -\frac{1}{2}(k+1)(k-2){\beta }^{1-k}{x}^{k}f\left(x\right)-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-k-1}{x}^{k+2-i}f\left(x\right).\hfill \end{array}$$

(13)

If $\alpha =0$ in (13), we have

$$\begin{array}{ccc}& & (k+2)D^{k+1}·f\left(x\right)\hfill \\ & =& {\beta }^{-k}\left(x^k{(x+1)}^2-x^2{(x-1)}^k-\frac{1}{2}(k+1)(k-2)\beta x^k-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-1}{x}^{k+2-i}\right)f\left(x\right)\hfill \\ & =& {\beta }^{-k}[x^{k+2}+2x^{k+1}+x^k-\sum _{i=0}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^i{x}^{k+2-i}+\frac{1}{2}(k+1)(k-2)x^k\hfill \\ & & -\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-1}{x}^{k+2-i}]f\left(x\right)\hfill \\ & =& {\beta }^{-k}\left((k+2)x^{k+1}-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^i{x}^{k+2-i}-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right){\beta }^{i-1}{x}^{k+2-i}\right)f\left(x\right)\hfill \\ & =& {\beta }^{1-(k+1)}(k+2)x^{k+1}f\left(x\right).\hfill \end{array}$$

If $\alpha =1$ in (13), we have

$$\begin{array}{ccc}& & (k+2)D^{k+1}·f\left(x\right)\hfill \\ & =& {\beta }^{-k}\left(x^2{(x+1)}^k-x^k{(x-1)}^2-\frac{1}{2}(k+1)(k-2)x^k-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right)x^{k+2-i}\right)f\left(x\right)\hfill \\ & =& {\beta }^{-k}\left(\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right)x^{k+2-i}+(k+2)x^{k+1}-\sum _{i=3}^k\left(\genfrac{}{}{0pt}{}{k}{i}\right)x^{k+2-i}\right)f\left(x\right)\hfill \\ & =& {\beta }^{1-(k+1)}(k+2)x^{k+1}f\left(x\right).\hfill \end{array}$$

Therefore, $t^m{D}^{k+1}·f\left(x\right)={\beta }^{1-(k+1)}{\lambda }^m{(x-m\alpha )}^{k+1}f(x-m).$ Hence the claim holds.

By Claim 4 and Claim 5, we obtain

$$t^m{D}^n·f\left(x\right)={\beta }^{1-n}{\lambda }^m{(x-m\alpha )}^{n}f(x-m),\forall m\in \mathbb{Z},n\in \mathbb{N}.$$

The theorem holds. □

Corollary 1.

Let M be a $\widehat{\mathcal{D}}$-module such that the restriction of $U\left(\widehat{\mathcal{D}}\right)$ to $U\left(\mathbb{C}D\right)$ is free of rank one. Then, $M\cong \mathsf{\Omega }(\lambda ,ϵ)$ for some $\lambda \in {\mathbb{C}}^{*}$ and $ϵ=0,1$, where C acts trivially on M.

Proof. 

It follows from Theorems 2 and 4. □

4. Tensor Products

In this section, we study the tensor products of $\mathsf{\Omega }(\lambda ,ϵ)$ with quasi-finite highest weight modules, and then get a class of simple non-weight $\mathcal{D}$-modules.

Theorem 5.

Let $\lambda \in {\mathbb{C}}^{*}$ and $ϵ\in \{0,1\}$. Let V be an irreducible quasi-finite highest weight module over $\mathcal{D}$. Then $\mathsf{\Omega }(\lambda ,ϵ)\otimes V$ is an irreducible $\mathcal{D}$-module.

Proof. 

Let $W=\mathsf{\Omega }(\lambda ,ϵ)\otimes V$. For any $v\in V$, there is a positive integer $K\left(v\right)$ such that $t^m{D}^n\left(v\right)=0$ for all $m\ge K\left(v\right)$ since V is quasi-finite.

For any nonzero submodule M of W, we shall prove that $W=M$. Choose a nonzero element $w={\sum }_{j=0}^s{x}^j\otimes v_j\in W$ such that $v_j\in V$, $v_s\ne 0$ and s is minimal.

Claim 6.$s=0$.

Let $K=\max \{K\left(v_j\right):j=0,1,\dots ,s\}$. Using $t^{m}D\left(v_j\right)=0$ for any $m\ge K$ and $j=0,1,\dots ,s$, we can get

$${\lambda }^{-m}{t}^{m}D\left(w\right)=\sum _{j=0}^s(x-ϵm){(x-m)}^j\otimes v_j\in M,\forall m\ge K.$$

Write the right hand side as

$$\sum _{j=0}^{s+1}{m}^j{w}_j\in M,\forall m\ge K,$$

where $w_j\in W$ are independent of m and $w_{s+1}=ϵ{(-1)}^{s-1}\otimes v_s$. Setting $m=K,K+1,\dots ,K+s$, we can see that the coefficient matrix of $w_i$ is a Vandermonde matrix. So we get each $w_i\in M$. Moreover, $w_{s+1}=ϵ{(-1)}^{s-1}\otimes v_s\in M$. In this case $s=0$.

Claim 7.$W=M$.

By Claim 6, we know that $1\otimes v\in M$ for some nonzero $v\in V$. By induction on i and using

$$\begin{array}{ccc}\hfill t^{m}D(x^i\otimes v)& =& \left({\lambda }^m(x-ϵm){(x-m)}^i\right)\otimes v\hfill \\ & =& \left({\lambda }^m{(x-m)}^{i+1}\right)\otimes v+\left({\lambda }^{m}m(1-ϵ){(x-m)}^i\right)\otimes v,\hfill \end{array}$$

where $m\ge K\left(v\right)$, $i\in \mathbb{N}$, we deduce that $x^i\otimes v\in M$ for all $i\in \mathbb{N}$, i.e., $\mathsf{\Omega }(\lambda ,ϵ)\otimes v\subset M$. Let X be a maximal subspace of V such that $\mathsf{\Omega }(\lambda ,ϵ)\otimes X\subset M$. We know that $X\ne 0$. Clearly, X is a nonzero submodule of V. Since V is irreducible, we obtain that $X=V$. Therefore, $M=W$ and M is irreducible. □

Theorem 6.

For $\lambda ,{\lambda }^{\prime }\in {\mathbb{C}}^{*}$ and $ϵ,{ϵ}^{\prime }\in \{0,1\}$, let V and $V^{\prime }$ be two irreducible quasi-finite highest weight modules over $\mathcal{D}$. Then $\mathsf{\Omega }(\lambda ,ϵ)\otimes V\cong \mathsf{\Omega }({\lambda }^{\prime },{ϵ}^{\prime })\otimes V^{\prime }$ as $\mathcal{D}$-modules if and only if $(\lambda ,ϵ)=({\lambda }^{\prime },{ϵ}^{\prime })$ and $V\cong V^{\prime }$.

Proof. 

The sufficiency of the theorem is clear. We just prove the necessity.

Let $\phi :\mathsf{\Omega }(\lambda ,ϵ)\otimes V⟶\mathsf{\Omega }({\lambda }^{\prime },{ϵ}^{\prime })\otimes V^{\prime }$ be an isomorphism as $\mathcal{D}$-modules. Choose a nonzero element $1\otimes v\in \mathsf{\Omega }(\lambda ,ϵ)\otimes V$. Assume that

$$\phi (1\otimes v)=\sum _{j=0}^k{x}^j\otimes w_j,$$

where $w_j\in V^{\prime }$ with $w_k\ne 0$. There exists $K=K\left(v\right)\in {\mathbb{Z}}_{+}$ such that $t^{m}D·v=t^{m}D·w_j=0$ for any $m\ge K$ and $0\le j\le k$. So

$$({\lambda }^{-m}{t}^{m}D-{\lambda }^{-m^{\prime }}{t}^{m^{\prime }}D)(1\otimes v)=ϵ(m^{\prime }-m)(1\otimes v),\forall m,m^{\prime }\ge K,$$

which suggests

$$\begin{array}{ccc}\hfill & & ϵ(m^{\prime }-m)(\sum _{j=0}^k{x}^j\otimes w_j)=({\lambda }^{-m}{t}^{m}D-{\lambda }^{-m^{\prime }}{t}^{m^{\prime }}D)(\sum _{j=0}^k{x}^j\otimes w_j)\hfill \\ & =& \sum _{j=0}^k\left[{(\frac{{\lambda }^{\prime }}{\lambda })}^m(x-{ϵ}^{\prime }m){(x-m)}^j-{(\frac{{\lambda }^{\prime }}{\lambda })}^{m^{\prime }}(x-{ϵ}^{\prime }{m}^{\prime }){(x-m^{\prime })}^j\right]\otimes w_j.\hfill \end{array}$$

(14)

Then we can get that $\left[{(\frac{{\lambda }^{\prime }}{\lambda })}^m-{(\frac{{\lambda }^{\prime }}{\lambda })}^{m^{\prime }}\right](x^{k+1}\otimes w_k)=0$ for all $m,m^{\prime }\ge K.$ So

$$\lambda ={\lambda }^{\prime }$$

and (14) becomes

$$ϵ(m^{\prime }-m)(\sum _{j=0}^k{x}^j\otimes w_j)=\sum _{j=0}^k((x-{ϵ}^{\prime }m){(x-m)}^j-(x-{ϵ}^{\prime }{m}^{\prime }){(x-m^{\prime })}^j)\otimes w_j,$$

(15)

where $m,m^{\prime }\ge K$. If $k>0$, for ${ϵ}^{\prime }=1$, by the coefficient of $m^{k+1}$ in (15), we get ${(-1)}^{k+1}{ϵ}^{\prime }(1\otimes w_k)=0$, a contradiction. If $k>0$, for ${ϵ}^{\prime }=0$, by the leading coefficient of m in (15), it is easy to see that $k=1$ and $ϵ(m^{\prime }-m)(1\otimes w_0+x\otimes w_1)=(m^{\prime }-m)x\otimes w_1$ for all $m,m^{\prime }\ge K$, which is impossible. Therefore, $k=0$ and $ϵ(m^{\prime }-m)(1\otimes w_k)={ϵ}^{\prime }(m^{\prime }-m)(1\otimes w_k)$ for all $m,m^{\prime }\ge K$, which forces

$$ϵ={ϵ}^{\prime }.$$

Then, there exists a bijection $\tau :V⟶V^{\prime }$ such that

$$\phi (1\otimes v)=1\otimes \tau \left(v\right),\forall v\in V.$$

Since $\phi (t^m{D}^n·(1\otimes v))=t^m{D}^n·\left(\phi (1\otimes v)\right)=t^m{D}^n·(1\otimes \tau \left(v\right)),$ we have

$$\begin{array}{ccc}\hfill & & {\beta }^{1-n}{\lambda }^m\phi ({(x-\phi m)}^n\otimes v)+\phi (1\otimes (t^m{D}^n·v))\hfill \\ & =& {\beta }^{1-n}{\lambda }^m{(x-\phi m)}^n\otimes \tau \left(v\right)+1\otimes (t^m{D}^n·\tau \left(v\right)),m\in \mathbb{Z}.\hfill \end{array}$$

Hence ${\beta }^{1-n}{\lambda }^m\phi ({(x-ϵm)}^n\otimes v)={\beta }^{1-n}{\lambda }^m{(x-ϵm)}^n\otimes \tau \left(v\right)$ for $m\ge K.$ So

$$\phi ({(x-ϵm)}^n\otimes v)={(x-ϵm)}^n\otimes \tau \left(v\right),m\ge K.$$

Then it is easy to deduce that

$$\phi (t^m{D}^n\left(1\right)\otimes v)=t^m{D}^n\left(1\right)\otimes \tau \left(v\right),m\in \mathbb{Z}.$$

By $\phi (t^m{D}^n·(1\otimes v))=t^m{D}^n·\left(\phi (1\otimes v)\right)$, we can obtain

$$\phi (1\otimes t^m{D}^n\left(v\right))=1\otimes t^m{D}^n\left(\tau \left(v\right)\right).$$

So

$$\tau \left(t^m{D}^n\left(v\right)\right)=t^m{D}^n\left(\tau \left(v\right)\right),\forall m\in \mathbb{Z},v\in V.$$

Clearly, for $c\in \mathbb{C}$, $\phi \left(c\right(1\otimes v\left)\right)=c\left(\phi \right(1\otimes v\left)\right)$ implies that $\tau \left(cv\right)=c\tau \left(v\right)$. Then $\tau :V⟶V^{\prime }$ becomes an isomorphism as $\mathcal{D}$-modules and $V\cong V^{\prime }$. The proof is finished. □

5. Differential Operator Algebra on $\mathbb{C}[{\mathbf{t}}_{\mathbf{1}}^{\pm \mathbf{1}},\dots ,{\mathbf{t}}_{\mathbf{\nu }}^{\pm \mathbf{1}}]$

For any positive integer $\nu$, let $\mathcal{A}:=\mathbb{C}[t_1^{\pm 1},\dots ,t_{\nu }^{\pm 1}]$ be the Laurent polynomial algebra. The Lie algebra ${\mathcal{W}}_{\nu }=\mathrm{Der}\left(\mathcal{A}\right)$ and the Lie algebra ${\mathcal{D}}_{\nu }=\mathrm{Diff}\left(\mathcal{A}\right)$ are known as the Witt algebra and the differential operator algebra of rank $\nu$. We know that ${\mathcal{W}}_1$ is the centerless Virasoro algebra and ${\mathcal{D}}_1=\mathcal{D}$ is the Lie algebra of the differential operators.

For $\underline{m}=(m_1,\dots ,m_{\nu })\in {\mathbb{Z}}^{\nu },\underline{n}=(n_1,\dots ,n_{\nu })\in {\mathbb{N}}^{\nu }$, set $t^{\underline{m}}=t_1^{m_1}\dots t_{\nu }^{m_{\nu }},D^{\underline{n}}=D_1^{n_1}\dots D_{\nu }^{n_{\nu }}$, where $D_i=t_i\frac{d}{d{t}_i}$ for all $i=1,2,\dots ,\nu$. We can write the Lie brackets in ${\mathcal{W}}_{\nu }$ and ${\mathcal{D}}_{\nu }$ as follows:

$$\begin{array}{ccc}\hfill [t^{\underline{r}}{D}_i,t^{\underline{s}}{D}_j]& =& s_i{t}^{\underline{r}+\underline{s}}{D}_j-r_j{t}^{\underline{r}+\underline{s}}{D}_i,\hfill \\ \hfill \left[t^{\underline{r}}{D}^{\underline{m}},t^{\underline{s}}{D}^{\underline{n}}\right]& =& \sum _{i_1=0}^{m_1}\dots \sum _{i_{\nu }=0}^{m_{\nu }}\left(\genfrac{}{}{0pt}{}{m_1}{i_1}\right)\dots \left(\genfrac{}{}{0pt}{}{m_{\nu }}{i_{\nu }}\right){\underline{s}}^{\underline{i}}{t}^{\underline{r}+\underline{s}}{D}^{m+n-i}\hfill \\ & -& \sum _{j_1=0}^{n_1}\dots \sum _{j_{\nu }=0}^{n_{\nu }}\left(\genfrac{}{}{0pt}{}{n_1}{j_1}\right)\left(\genfrac{}{}{0pt}{}{n_2}{j_2}\right)\dots \left(\genfrac{}{}{0pt}{}{n_{\nu }}{j_{\nu }}\right){\underline{r}}^{\underline{j}}{t}^{\underline{r}+\underline{s}}{D}^{m+n-j}\hfill \end{array}$$

(16)

for all $\underline{r},\underline{s}\in {\mathbb{Z}}^{\nu },\underline{m},\underline{n}\in {\mathbb{N}}^{\nu },\underline{i}=(i_1,\dots ,i_{\nu }),\underline{j}=(j_1,\dots ,j_{\nu })$.

It is known that ${\mathfrak{h}}_{\nu }={\mathrm{span}}_{\mathbb{C}}\{D_1,D_2,\dots ,D_{\nu }\}$ is the Cartan subalgebra of ${\mathcal{W}}_{\nu }$ and ${\mathcal{D}}_{\nu }$.

Lemma 4.

The Lie algebra ${\mathcal{D}}_{\nu }$ is generated by $\{t_i,t_i^{-1},D_i{D}_j\mid i,j=1,2,\dots ,\nu \}$.

Proof. 

It follows by Lemma 1 and (16). □

For $\alpha \in \mathbb{C}$, $1\le \nu <\infty$, $\mathsf{\Lambda }=({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{\nu })\in {\left({\mathbb{C}}^{*}\right)}^n$, Tan-Zhao in [12] defined a class of ${\mathcal{W}}_{\nu }$-modules, denoted by $\mathsf{\Omega }(\mathsf{\Lambda },\alpha )=\mathbb{C}[x_1,x_2,\dots ,$ $x_{\nu }]$ the polynomial algebra over $\mathbb{C}$ in commuting indeterminates $x_1,x_2,\dots ,x_{\nu }$ as follows:

$$t^{\underline{m}}{D}_i·f(x_1,\dots ,x_{\nu })={\mathsf{\Lambda }}^m(x_i-\alpha m_i)f(x_1-m_1,\dots ,x_{\nu }-m_{\nu }),$$

(17)

where $\underline{m}=(m_1,m_2,\dots ,m_{\nu })\in {\mathbb{Z}}^{\nu },f(x_1,\dots ,x_{\nu })\in \mathbb{C}[x_1,x_2,\dots ,x_{\nu }],{\mathsf{\Lambda }}^{\underline{m}}={\lambda }_1^{m_1}{\lambda }_2^{m_2}\dots {\lambda }_{\nu }^{m_{\nu }},$ $i=1,2,\dots ,\nu$.

By [12], $\mathsf{\Omega }(\mathsf{\Lambda },\alpha )$ is simple if and only if $\alpha \ne 0.$ The following result was given in [12].

Theorem 7

([12]). Let M be a ${\mathcal{W}}_{\nu }$-module which is a free $U\left({\mathfrak{h}}_{\nu }\right)$-module of rank one. Then, $M\cong \mathsf{\Omega }(\mathsf{\Lambda },\alpha )$ for some $\mathsf{\Lambda }\in {\left({\mathbb{C}}^{*}\right)}^{\nu }$ and some $\alpha \in \mathbb{C}.$

For $ϵ=0,1$, $\mathsf{\Lambda }=({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{\nu })\in {\left({\mathbb{C}}^{*}\right)}^{\nu }$, we define a ${\mathcal{D}}_{\nu }$-module $\mathsf{\Omega }(\mathsf{\Lambda },ϵ):=\mathbb{C}[x_1,x_2,\dots ,x_{\nu }]$ as follows:

$$t^{\underline{m}}{D}^{\underline{n}}·f(x_1,\dots ,x_{\nu })={\beta }^{1-{\sum }_{i=1}^{\nu }{n}_i}{\mathsf{\Lambda }}^m{\Pi }_{j=1}^{\nu }{(x_j-ϵm_j)}^{n_j}f(x_1-m_1,\dots ,x_{\nu }-m_{\nu })$$

for any $f(x_1,\dots ,x_{\nu })\in \mathbb{C}[x_1,x_2,\dots ,x_{\nu }]$, where $\beta ={(-1)}^{1-ϵ}$, $\underline{m}=(m_1,m_2,\dots ,m_{\nu })\in {\mathbb{Z}}^{\nu },\underline{n}=(n_1,n_2,\dots ,n_{\nu })\in {\mathbb{N}}^{\nu },$ $i=1,2,\dots ,\nu$.

Lemma 5.

$\mathsf{\Omega }(\mathsf{\Lambda },ϵ)$ is an irreducible ${\mathcal{D}}_{\nu }$-module.

Proof. 

It follows by Lemma 3 and some direct calculations. □

Theorem 8.

Let M be a ${\mathcal{D}}_{\nu }$-module which is a free $U\left({\mathfrak{h}}_{\nu }\right)$-module of rank one. Then, $M\cong \mathsf{\Omega }(\mathsf{\Lambda },ϵ)$ for some $\mathsf{\Lambda }\in {\left({\mathbb{C}}^{*}\right)}^{\nu },ϵ=0,1$.

Proof. 

By Theorem 4, we get

$$t_i^{m_i}{D}_i^{n_i}·f(x_1,\dots ,x_{\nu })={\beta }^{1-n_i}{\lambda }_i^{m_i}{(x_i-ϵm_i)}^{n_i}f(x_1,\dots ,x_i-m_i,\dots ,x_{\nu })$$

(18)

for any $f(x_1,\dots ,x_{\nu })\in \mathbb{C}[x_1,x_2,\dots ,x_{\nu }]$, where $ϵ\in \{0,1\}$, $\beta ={(-1)}^{1-ϵ}$, $m_i\in \mathbb{Z}$, and ${\lambda }_i\in {\mathbb{C}}^{*}$, $i=1,\dots ,\nu$.

Moreover, in this case, (17) becomes

$$t^{\underline{m}}{D}_i·f(x_1,\dots ,x_{\nu })={\mathsf{\Lambda }}^{\underline{m}}(x_i-ϵm_i)f(x_1-m_1,\dots ,x_i-m_i,\dots ,x_{\nu }-m_{\nu })$$

(19)

for any $f(x_1,\dots ,x_{\nu })\in \mathbb{C}[x_1,x_2,\dots ,x_{\nu }]$, where $\underline{m}=(m_1,m_2,\dots ,m_{\nu })\in {\mathbb{Z}}^{\nu }$, ${\mathsf{\Lambda }}^{\underline{m}}={\lambda }_1^{m_1}\dots {\lambda }_{\nu }^{m_{\nu }}$, $i=1,2,\dots ,\nu$.

By $[t^{\underline{m}}{D}_i,D_i^2]=-2m_i{t}^{\underline{m}}{D}_i^2-m_i^2{t}^{\underline{m}}{D}_i$, using (18) and (19), we get

$$t^{\underline{m}}{D}_i^2·f(x_1,\dots ,x_{\nu })={\beta }^{-1}{\mathsf{\Lambda }}^{\underline{m}}{(x_i-ϵm_i)}^{2}f(x_1-m_1,\dots ,x_i-m_i,\dots ,x_{\nu }-m_{\nu }).$$

(20)

For any $1\le i\ne j\le \nu$, by $[t_i^{-1}{D}_i^2,t_i{D}_j]=2D_i{D}_j+D_j$, using (19) and (20), we can obtain

$$D_i{D}_j·f(x_1,\dots ,x_{\nu })={\beta }^{-1}{x}_i{x}_{j}f(x_1,\dots ,x_{\nu }),\mathrm{for}\mathrm{all}f(x_1,\dots ,x_{\nu })\in \mathbb{C}[x_1,x_2,\dots ,x_{\nu }].$$

By Lemma 4, the theorem holds. □

In conclusion, a kind of new simple modules over the Lie algebra of differential operators, which are non-weight modules, are constructed by classical methods in this paper (Theorems 4 and 8). Certainly, the study of non-weight modules over the Lie algebra of differential operators is still in its infancy.