#  $U(\mathfrak{h})$-free modules over the Lie algebras of differential
operators

Munayim Dilxat  College of Mathematics and System Sciences, Xinjiang
University, Urumqi 830046, Xinjiang, China<EMAIL_ADDRESS>,  Shoulan Gao
Department of Mathematics, Huzhou University, Zhejiang Huzhou, 313000, China
<EMAIL_ADDRESS>,  Dong Liu  College of Mathematics and System
Sciences, Xinjiang University, Urumqi 830046, Xinjiang, China; Department of
Mathematics, Huzhou University, Zhejiang Huzhou, 313000, China
<EMAIL_ADDRESS>and  Limeng Xia  Institute of Applied System Analysis,
Jiangsu University, Jiangsu Zhenjiang, 212013, China<EMAIL_ADDRESS>
⋆ D. Liu, Corresponding author

Abstract. In this paper, we consider some non-weight modules over the Lie
algebra of Weyl type. First, we determine the modules whose restriction to
$U(\mathfrak{h})$ are free of rank $1$ over the Lie algebra of differential
operators on the circle. Then we determine the necessary and sufficient
conditions for the tensor products of quasi-finite highest weight modules and
$U(\mathfrak{h})$-free modules to be irreducible, and obtain that any two such
tensor products are isomorphic if and only if the corresponding highest weight
modules and $U(\mathfrak{h})$-free modules are isomorphic. Finally, we extend
such results to the Lie algebras of differential operators in the general
case.

Key words: Virasoro algebra; differential operator; polynomial module; Weyl
algebra

MSC: 17B10,17B65, 17B68.

##  1\. Introduction

Extended symmetries have played an important role in conformal field theories.
This leads one to study $W$ algebras, which are higher-spin extensions of the
Virasoro algebra, and their superanalogues, if one takes supersymmetry into
account, cf. [5, 10] and the vast literatures therein. A fundamental example
of $W$ algebras, now known as the $W_{1+\infty}$ algebra, appears as the limit
of the $W_{N}$ algebras, as $N$ goes to $\infty$ [18]. It is well known that
the $W_{1+\infty}$ algebra can also be interpreted as the central extension of
the Lie algebra of differential operators on the circle [11, 19]. Weight
modules over $W_{1+\infty}$ were studied sufficiently and have appeared in
various models of two-dimensional quantum field theory and integrable systems
(see [7, 11, 8, 20], 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, J. Nilsson [16] constructed a class of
${\mathfrak{s}l}_{n}$-modules that are free of rank one when restricted to the
enveloping algebra of its Cartan subalgebra. This kind of non-weight modules,
which are called $U(\mathfrak{h})$-free modules, have been extensively
studied. In [16] and a subsequent paper [17], J. Nilsson showed that a finite
dimensional simple Lie algebra has nontrivial $U(\mathfrak{h})$-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 $n\geq 1$. And the idea was
exploited and generalized to consider modules over infinite dimensional Lie
algebras, such as the Witt algebra (the Virasoro algebra) [15, 21], the
twisted Heisenberg-Virasoro algebra and the $W(2,2)$ algebra [6], and so on.
The aim of this paper is to determine such modules for the Lie algebra
$W_{1+\infty}$ motivated by [6] and [21]. We get a kind of new simple modules
over $W_{1+\infty}$ but they are not modules of the corresponding associative
algebras (see (3.1)). Moreover, such results can be extended to the Lie
algebras of differential operators on the Laurent polynomial rings of multi-
variables.

The paper is organized as follows. In Section 2, we recall some necessary
definitions and preliminary results. In Section 3, we determine all module
structures on $U(\mathfrak{h})$, see Theorem 3.2. In Section 4, we give the
necessary and sufficient conditions for the tensor product of a quasi-finite
highest weight module $L(h,c)$ and a $U(\mathfrak{h})$-free
$\mathcal{D}$-module $\Omega(\lambda,\epsilon)$ to be irreducible, see Theorem
4.1. Furthermore, we show that two such tensor product modules are isomorphic
if and only if the corresponding quasi-finite highest weight modules and
$U(\mathfrak{h})$-free modules are isomorphic, see Theorem 4.2. Consequently,
we obtain a lot of new irreducible non-weight modules over the Lie algebra of
differential operators. In Section 5, we construct such modules for the Lie
algebras of differential operators in the general case.

Throughout this paper, denote by $\mathbb{Z},$ $\mathbb{Z}_{+},$ $\mathbb{N}$,
$\mathbb{C}$ and $\mathbb{C}^{*}$ the sets of integers, positive integers,
non-negative integers, complex numbers and non-zero complex numbers,
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
${\mathbb{C}}$, ${\mathcal{D}}_{ass}=\hbox{Diff}\,{\mathbb{C}}[\,t,t^{-1}\,]$
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:

$(t^{m_{1}}D^{n_{1}})(t^{m_{2}}D^{n_{2}})=\sum_{i=0}^{n_{1}}{n_{1}\choose
i}\,m_{2}^{i}\,t^{m_{1}+m_{2}}D^{n_{1}+n_{2}-i}$

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

$\bigl{[}t^{m_{1}}D^{n_{1}},t^{m_{2}}D^{n_{2}}\,\bigr{]}=\sum_{i=0}^{n_{1}}{n_{1}\choose
i}\,m_{2}^{i}\,t^{m_{1}+m_{2}}D^{n_{1}+n_{2}-i}-\sum_{j=0}^{n_{2}}{n_{2}\choose
j}\,m_{1}^{j}\,t^{m_{1}+m_{2}}D^{n_{1}+n_{2}-j}$

for all $m_{1},\;m_{2}\in\mathbb{Z},\;n_{1},\;n_{2}\in\mathbb{N}$.

######  Lemma 2.1.

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

Li [12] proved $H^{2}({\mathcal{D}},{\mathbb{C}})=1$ (also see [11]). More
precisely, we have the following result.

######  Lemma 2.2.

[11, 12] Any non-trivial $2$-cocycle on ${\mathcal{D}}$ is equivalent to
$\phi$:

$\phi(t^{m_{1}}D^{n_{1}},t^{m_{2}}D^{n_{2}})=\begin{cases}&0,\qquad\hbox{\it
if}\quad m_{1}=0,\\\ &(-1)^{n_{1}+1}\,\delta_{m_{1}+m_{2},0}\,{1\over
2}\,\sum_{i=1}^{m_{1}}\,(m_{1}-i)^{n_{1}}\,i^{n_{2}},\qquad\hbox{\it if}\quad
m_{1}>0,\\\ &(-1)^{n_{1}}\,\delta_{m_{1}+m_{2},0}\,{1\over
2}\,\sum_{i=m_{1}}^{-1}\,(m_{1}-i)^{n_{1}}\,i^{n_{2}},\qquad\hbox{\it if}\quad
m_{1}<0.\end{cases}$

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 $\phi$ with a central element $C$. $\widehat{\mathcal{D}}$ is
$\mathbb{Z}$-graded by $\widehat{\mathcal{D}}_{i}={\rm
Span}_{\mathbb{C}}\\{t^{i}D^{n}\mid n\geq 0\\}\oplus\delta_{i,0}\mathbb{C}C$
for $i\in\mathbb{Z}$.

A ${\widehat{\mathcal{D}}}$-module $V$ is called a highest (resp. lowest)
weight module, if there exists a nonzero $v\in V_{\lambda}$ 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 [11].

Next, we define the Verma module, which is a highest weight module with the
free action of $U(\widehat{\mathcal{D}}_{-})$ on the highest weight vector
(see [9]). More precisely, for any $c\in\mathbb{C}$ and
$h=(h_{1},h_{2},\ldots),h_{i}\in\mathbb{C},i\in\mathbb{Z}_{+}$, let
$\mathbb{C}{\bf 1}$ be the one-dimensional module over the subalgebra
${\widehat{\mathcal{D}}}_{+}\oplus{\widehat{\mathcal{D}}}_{0}$ defined by

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

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

$M(h,c)=U(\widehat{\mathcal{D}})\otimes_{U(\widehat{\mathcal{D}}_{+}+\widehat{\mathcal{D}}_{0})}\mathbb{C}{\bf
1}.$

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

Introduce the generating series

$\Delta_{h}(x)=-\sum^{\infty}_{n=0}\frac{x^{n}}{n!}h_{n},$

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
$p(x)e^{\alpha x}$, where $p(x)$ is a polynomial and $\alpha\in\mathbb{C}$.

The following results were given in [11]:

######  Theorem 2.3.

[11] For any $c\in\mathbb{C}$ and $h\in{\mathcal{D}}_{0}^{*}$, $L(h,c)$ is
quasi-finite if and only if

$\Delta_{h}(x)=\frac{\phi(x)}{e^{x}-1},$

where $\phi(x)$ is a quasipolynomial such that $\phi(0)=0$.

###  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, $\mathcal{W}=\mbox{Der}\,{\mathbb{C}}[t,t^{-1}]$ (also denote 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:

$\displaystyle[L_{m},L_{n}]=(n-m)L_{m+n},$ $\displaystyle[I_{m},I_{n}]=0,$
$\displaystyle[L_{m},I_{n}]=nI_{m+n},\forall m,n\in\mathbb{Z}.$

Its universal central extension is called the twisted Heisenberg-Virasoro
algebra (see [3, 4, 13, 14]).

###  2.3. $U(\mathfrak{h})$-free modules

In this subsection, we recall some modules which are free of rank $1$ when
regarded as a $\mathbb{C}[L_{0}]$-module over the Virasoro algebra.

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

For $\lambda\in\mathbb{C}^{*},\alpha\in\mathbb{C},$ denote by
$\Omega(\lambda,\alpha)=\mathbb{C}[x]$ the polynomial algebra over
$\mathbb{C}$. In [15], the Virasoro algebra $U({\rm Vir})$-module structure on
$\Omega(\lambda,\alpha)$ is given by

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

By [15], we know that $\Omega(\lambda,\alpha)$ is simple if and only if
$\lambda,\alpha\in\mathbb{C}^{*}.$ If $\alpha=0,$ then $\Omega(\lambda,0)$ has
a simple submodule $x\Omega(\lambda,0)$ with codimension $1.$

######  Theorem 2.4.

[21] Let $M$ be a $U({\rm Vir})$-module such that the restriction of $U({\rm
Vir})$ to $U(\mathbb{C}L_{0})$ is free of rank $1$, that is,
$M=U(\mathbb{C}L_{0})v$ for some torsion-free $v\in M$. Then
$M\cong\Omega(\lambda,\alpha)$ for some
$\alpha\in\mathbb{C},\lambda\in\mathbb{C}^{*}$.

The module $\Omega(\lambda,\alpha)$ can be naturally made into a
$\mathcal{D}_{1}$-module (or a module over the twisted Heisenberg-Virasoro
algebra) as follows. For $\lambda\in\mathbb{C}^{*},\alpha,\beta\in\mathbb{C},$
denote by $\Omega(\lambda,\alpha,\beta)=\mathbb{C}[x]$ the polynomial algebra
over $\mathbb{C}$. We can define the $\mathcal{D}_{1}$-module structure on
$\Omega(\lambda,\alpha,\beta)$ as follows

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

where $f(x)\in\mathbb{C}[x]$, $m\in\mathbb{Z}$. By [6], it is easy to know
that $\Omega(\lambda,\alpha,\beta)$ is simple if and only if $\alpha\neq 0$ or
$\beta\neq 0$. If $\alpha=\beta=0$, then $\Omega(\lambda,0,0)$ has a simple
submodule $x\Omega(\lambda,0,0)$ with codimension $1.$

######  Theorem 2.5.

[6] Let $M$ be a $\mathcal{D}_{1}$-module such that the restriction of
$U(\mathcal{D}_{1})$ to $U(\mathbb{C}L_{0})$ is free of rank $1$, that is,
$M=U(\mathbb{C}L_{0})v$ 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 $\alpha\neq 0$ or $\beta\neq 0$. If $\alpha=\beta=0$, then
$\Omega(\lambda,0,0)$ has a simple submodule $x\Omega(\lambda,0,0)$ with
codimension $1.$

##  3\. $U(\mathfrak{h})$-free modules over the Lie algebra of differential
operators

In this section, we mainly study $U(\mathfrak{h})$-free modules over the Lie
algebra of differential operators on the circle based on such researches in
[6]. We classify all $\mathcal{D}$-modules such that the restriction of
$U(\mathcal{D})$ to $U(\mathbb{C}D)$ is free of rank $1$ and get some new
irreducible $\mathcal{D}$-modules.

For $\lambda\in\mathbb{C}^{*},\epsilon=0,1$, denote by
$\Omega(\lambda,\epsilon)=\mathbb{C}[x]$ the polynomial algebra over
$\mathbb{C}$. We can define the actions of $\mathcal{D}$ on
$\Omega(\lambda,\epsilon)$ as follows

$t^{m}D^{n}f(x)=\beta^{1-n}\lambda^{m}(x-\epsilon m)^{n}f(x-m),$ (3.1)

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

######  Remark 1.

In the case of $\epsilon=1$, the module $\Omega(\lambda,1)$ is also a
$\mathcal{D}_{ass}$-module. From [15] we see that if $V$ is an irreducible
$\mathcal{D}_{ass}$-module on which $\mathbb{C}[t,t^{-1}]$ is torsion, then
$V\cong\Omega(\lambda,1)$.

######  Lemma 3.1.

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

###### Proof.

For $f(x)\in\mathbb{C}[x],m,m_{1}\in\mathbb{Z},n,n_{1}\in{\mathbb{N}}$,

$\displaystyle[t^{m}D^{n},t^{m_{1}}D^{n_{1}}]f(x)=(t^{m}D^{n})(t^{m_{1}}D^{n_{1}})f(x)-(t^{m_{1}}D^{n_{1}})(t^{m}D^{n})f(x)$
$\displaystyle=$ $\displaystyle\lambda^{m+m_{1}}\beta^{2-n_{1}-n}(x-\epsilon
m)^{n}(x-m-\epsilon m_{1})^{n_{1}}f(x-m-m_{1})$
$\displaystyle-\lambda^{m+m_{1}}\beta^{2-n_{1}-n}(x-\epsilon
m_{1})^{n_{1}}(x-m_{1}-\epsilon m)^{n}f(x-m-m_{1}).$

On the other side,

$\displaystyle[t^{m}D^{n},t^{m_{1}}D^{n_{1}}]f(x)$ $\displaystyle=$
$\displaystyle(\sum_{i=0}^{n}{n\choose
i}m_{1}^{i}t^{m+m_{1}}D^{n+n_{1}-i}-\sum_{j=0}^{n_{1}}{n_{1}\choose
j}m^{j}t^{m+m_{1}}D^{n+n_{1}-j})f(x)$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n}{n\choose
i}\beta^{1-n-n_{1}+i}\lambda^{m+m_{1}}m_{1}^{i}(x-\epsilon m-\epsilon
m_{1})^{n+n_{1}-i}f(x-m_{1}-m)$ $\displaystyle-\sum_{j=0}^{n_{1}}{n_{1}\choose
j}\beta^{1-n-n_{1}+j}\lambda^{m+m_{1}}m^{j}(x-\epsilon m-\epsilon
m_{1})^{n+n_{1}-j}f(x-m_{1}-m)$ $\displaystyle=$
$\displaystyle\beta^{1-n-n_{1}}\lambda^{m+m_{1}}(x-\epsilon
m+(\beta-\epsilon)m_{1})^{n}(x-\epsilon m-\epsilon m_{1})^{n_{1}}f(x-m_{1}-m)$
$\displaystyle-\beta^{1-n-n_{1}}\lambda^{m+m_{1}}(x-\epsilon
m_{1}+(\beta-\epsilon)m)^{n_{1}}(x-\epsilon m-\epsilon
m_{1})^{n}f(x-m_{1}-m).$

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

$\displaystyle(x-\epsilon m+(\beta-\epsilon)m_{1})^{n}(x-\epsilon m-\epsilon
m_{1})^{n_{1}}$ $\displaystyle-(x-\epsilon
m_{1}+(\beta-\epsilon)m)^{n_{1}}(x-\epsilon m-\epsilon m_{1})^{n}$
$\displaystyle=$ $\displaystyle\beta(x-\epsilon m)^{n}(x-m-\epsilon
m_{1})^{n_{1}}-\beta(x-\epsilon m_{1})^{n_{1}}(x-m_{1}-\epsilon m)^{n},$

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

It is clear that $\Omega(\lambda,\epsilon)$ is irreducible since $\lambda\neq
0$. ∎

Now we can give the main result of this section.

######  Theorem 3.2.

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

###### Proof.

Let $M=U(\mathbb{C}D)v=\mathbb{C}[D]v$ be a $U(\mathcal{D})$-module such that
the restriction of $U(\mathcal{D})$ to $U(\mathbb{C}D)$ is free of rank $1$.
By Theorem $2$ in [6], we have

$t^{m}Df(x)=\lambda^{m}(x-m\alpha)f(x-m),\quad\,t^{m}f(x)=\beta\lambda^{m}f(x-m)$
(3.2)

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

Claim 1. $\beta\neq 0$.

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

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

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

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

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

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

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

$\displaystyle\beta\lambda^{m}(\frac{1}{\beta}x^{2}+bx+c)f(x-m)-\beta[\frac{1}{\beta}(x-m)^{2}+b(x-m)+c]f(x-m)$
$\displaystyle=$ $\displaystyle 2m\lambda^{m}(x-\alpha
m)f(x-m)+m^{2}\beta\lambda^{m}f(x-m),$

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

$\beta=2\alpha-1,\quad\,b=0.$ (3.3)

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

$\displaystyle 2mt^{m}D^{2}\cdot f(x)$ $\displaystyle=$ $\displaystyle
D^{2}\cdot t^{m}D\cdot f(x)-t^{m}D\cdot D^{2}\cdot f(x)-m^{2}t^{m}D\cdot f(x)$
$\displaystyle=$ $\displaystyle
D^{2}\cdot(\lambda^{m}(x-m\alpha)f(x-m))-(t^{m}D)\cdot(g(x)f(x))-m^{2}\lambda^{m}(x-m\alpha)f(x-m)$
$\displaystyle=$
$\displaystyle\lambda^{m}(x-m\alpha)f(x-m)\big{(}g(x)-g(x-m)-m^{2}\big{)}$
$\displaystyle=$ $\displaystyle 2m\beta^{-1}\lambda^{m}(x-m\alpha)^{2}f(x-m).$

Therefore,

$t^{m}D^{2}f(x)=\beta^{-1}\lambda^{m}(x-\alpha m)^{2}f(x-m),\quad\,m\neq 0.$
(3.4)

Similarly, by $[D^{2},t^{m}D^{2}]=2mt^{m}D^{3}+m^{2}t^{m}D^{2}$ and (3.4), we
can obtain

$t^{m}D^{3}f(x)=\beta^{-2}\lambda^{m}(x-\alpha m)^{3}f(x-m),\quad\,m\neq 0.$
(3.5)

Since $[t^{-1}D^{2},tD^{2}]=4D^{3}$, using (3.4), we get

$\displaystyle 4D^{3}\cdot f(x)$ $\displaystyle=$ $\displaystyle
t^{-1}D^{2}\cdot tD^{2}\cdot f(x)-tD^{2}\cdot t^{-1}D^{2}\cdot f(x)$
$\displaystyle=$ $\displaystyle
t^{-1}D^{2}\cdot(\beta^{-1}\lambda(x-\alpha)^{2}f(x-1))-tD^{2}\cdot(\beta^{-1}\lambda^{-1}(x+\alpha)^{2}f(x+1))$
$\displaystyle=$
$\displaystyle\beta^{-2}(x+\alpha)^{2}(x-\alpha+1)^{2}f(x)-\beta^{-2}(x-\alpha)^{2}(x+\alpha-1)^{2}f(x)$
$\displaystyle=$ $\displaystyle 4\beta^{-2}x(x^{2}-\alpha^{2}+\alpha)f(x).$

Then

$D^{3}\cdot f(x)=\beta^{-2}x(x^{2}-\alpha^{2}+\alpha)f(x).$ (3.6)

Using (3.2) and (3.6), we can get

$\displaystyle D^{3}\cdot tD\cdot f(x)-tD\cdot D^{3}\cdot f(x)$ (3.7)
$\displaystyle=$
$\displaystyle\beta^{-2}\lambda(x-\alpha)(3x^{2}-3x+1-\alpha^{2}+\alpha)f(x-1).$

Using (3.2)–(3.5), we have

$\displaystyle(3tD^{3}+3tD^{2}+tD)\cdot f(x)$ (3.8) $\displaystyle=$
$\displaystyle\lambda(x-\alpha)[3\beta^{-2}(x-\alpha)^{2}+3\beta^{-1}(x-\alpha)+1]f(x-1)$
$\displaystyle=$
$\displaystyle\beta^{-2}\lambda(x-\alpha)[3x^{2}-3x+1+\alpha^{2}-\alpha]f(x-1).$

Since $[D^{3},tD]=3tD^{3}+3tD^{2}+tD$, by (3.7) and (3.8), we can deduce that

$\alpha(\alpha-1)=0.$ (3.9)

Combining (3.3) and (3.9), we have $\alpha=0,\beta=-1$ or $\alpha=\beta=1$.
That is,

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

Using (3.2), (3.4) and (3.9), we have

$\displaystyle(tD^{2})\cdot(t^{-1}D)\cdot f(x)-(t^{-1}D)\cdot(tD^{2})\cdot
f(x)$ (3.10) $\displaystyle=$
$\displaystyle(tD^{2})(\lambda^{-1}(x+\alpha)f(x+1))-(t^{-1}D)(\beta^{-1}\lambda(x-\alpha)^{2}f(x-1))$
$\displaystyle=$
$\displaystyle\beta^{-1}(x-\alpha)^{2}(x+\alpha-1)f(x)-\beta^{-1}(x+\alpha)(x+1-\alpha)^{2}f(x)$
$\displaystyle=$
$\displaystyle\beta^{-1}(-3x^{2}+(2\alpha-1)x-\alpha+\alpha^{2})f(x)$
$\displaystyle=$ $\displaystyle(-3\beta^{-1}x^{2}+x)f(x),$

and

$(-3D^{2}+D)\cdot f(x)=(-3\beta^{-1}x^{2}-3c+x)f(x).$ (3.11)

Since $[tD^{2},t^{-1}D]=-3D^{2}+D$, by (3.10) and (3.11), it is easy to see
that

$c=0.$

Furthermore, (3.4) holds for all $m\in\mathbb{Z}$, that is,

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

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

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

$\displaystyle D^{2}\cdot(t^{m}D^{k})\cdot f(x)-(t^{m}D^{k})\cdot D^{2}\cdot
f(x)$ (3.12) $\displaystyle=$ $\displaystyle
D^{2}\cdot(\beta^{1-k}\lambda^{m}(x-m\alpha)^{k}f(x-m))-(t^{m}D^{k})\cdot(\beta^{-1}x^{2}f(x))$
$\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}x^{2}(x-m\alpha)^{k}f(x-m)-\beta^{-k}\lambda^{m}(x-m\alpha)^{k}(x-m)^{2}f(x-m)$
$\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}[x^{2}-(x-m)^{2}](x-m\alpha)^{k}f(x-m)$
$\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}m(2x-m)(x-m\alpha)^{k}f(x-m).$

Since $[D^{2},t^{m}D^{k}]=m^{2}t^{m}D^{k}+2mt^{m}D^{k+1}$, by (3.12), we get

$\displaystyle 2mt^{m}D^{k+1}\cdot f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}m(2x-m)(x-m\alpha)^{k}f(x-m)-m^{2}(t^{m}D^{k})\cdot
f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}m(2x-m)(x-m\alpha)^{k}f(x-m)-m^{2}\beta^{1-k}\lambda^{m}(x-m\alpha)^{k}f(x-m)$
$\displaystyle=$
$\displaystyle\beta^{-k}\lambda^{m}m(2x-2m\alpha)(x-m\alpha)^{k}f(x-m)$
$\displaystyle=$ $\displaystyle
2m\beta^{1-(k+1)}\lambda^{m}(x-m\alpha)^{k+1}f(x-m).$

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

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

We shall prove the result by induction on $n$ in $D^{n}$. Suppose that
$D^{n}f(x)=\beta^{1-n}x^{n}f(x)$ for any $n\leq k$. By
$[t^{-1}D^{k},tD^{2}]=(k+2)D^{k+1}+\frac{1}{2}(k+1)(k-2)D^{k}+\sum_{i=3}^{k}{k\choose
i}D^{k+2-i},$ we get

$\displaystyle(k+2)D^{k+1}\cdot f(x)$ (3.13) $\displaystyle=$
$\displaystyle(t^{-1}D^{k})(tD^{2})\cdot f(x)-(tD^{2})(t^{-1}D^{k})\cdot f(x)$
$\displaystyle-\frac{1}{2}(k+1)(k-2)\beta^{1-k}x^{k}f(x)-\sum_{i=3}^{k}{k\choose
i}\beta^{i-k-1}x^{k+2-i}f(x)$ $\displaystyle=$
$\displaystyle(t^{-1}D^{k})(\beta^{-1}\lambda(x-\alpha)^{2}f(x-1))-(tD^{2})(\beta^{1-k}\lambda^{-1}(x+\alpha)^{k}f(x+1))$
$\displaystyle-\frac{1}{2}(k+1)(k-2)\beta^{1-k}x^{k}f(x)-\sum_{i=3}^{k}{k\choose
i}\beta^{i-k-1}x^{k+2-i}f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}[(x+\alpha)^{k}(x+1-\alpha)^{2}-(x-\alpha)^{2}(x-1+\alpha)^{k}]f(x)$
$\displaystyle-\frac{1}{2}(k+1)(k-2)\beta^{1-k}x^{k}f(x)-\sum_{i=3}^{k}{k\choose
i}\beta^{i-k-1}x^{k+2-i}f(x).$

If $\alpha=0$ in (3.13), we have

$\displaystyle(k+2)D^{k+1}\cdot f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}\left(x^{k}(x+1)^{2}-x^{2}(x-1)^{k}-\frac{1}{2}(k+1)(k-2)\beta
x^{k}-\sum_{i=3}^{k}{k\choose i}\beta^{i-1}x^{k+2-i}\right)f(x)$
$\displaystyle=$
$\displaystyle\beta^{-k}[x^{k+2}+2x^{k+1}+x^{k}-\sum_{i=0}^{k}{k\choose
i}\beta^{i}x^{k+2-i}+\frac{1}{2}(k+1)(k-2)x^{k}$
$\displaystyle-\sum_{i=3}^{k}{k\choose i}\beta^{i-1}x^{k+2-i}]f(x)$
$\displaystyle=$
$\displaystyle\beta^{-k}\left((k+2)x^{k+1}-\sum_{i=3}^{k}{k\choose
i}\beta^{i}x^{k+2-i}-\sum_{i=3}^{k}{k\choose
i}\beta^{i-1}x^{k+2-i}\right)f(x)$ $\displaystyle=$
$\displaystyle\beta^{1-(k+1)}(k+2)x^{k+1}f(x).$

If $\alpha=1$ in (3.13), we have

$\displaystyle(k+2)D^{k+1}\cdot f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}\left(x^{2}(x+1)^{k}-x^{k}(x-1)^{2}-\frac{1}{2}(k+1)(k-2)x^{k}-\sum_{i=3}^{k}{k\choose
i}x^{k+2-i}\right)f(x)$ $\displaystyle=$
$\displaystyle\beta^{-k}\left(\sum_{i=3}^{k}{k\choose
i}x^{k+2-i}+(k+2)x^{k+1}-\sum_{i=3}^{k}{k\choose i}x^{k+2-i}\right)f(x)$
$\displaystyle=$ $\displaystyle\beta^{1-(k+1)}(k+2)x^{k+1}f(x).$

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

By Claim $4$ and Claim $5$, we obtain

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

The theorem holds. ∎

######  Corollary 3.3.

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

###### Proof.

It follows from Theorem 3.2 and Theorem 2.4. ∎

##  4\. Tensor Products

In this section, we will obtain a class of irreducible non-weight modules over
$\mathcal{D}$ by taking the tensor products of $\Omega(\lambda,\epsilon)$ with
quasifinite highest weight modules and determine the necessary and sufficient
conditions for two irreducible tensor products to be isomorphic.

######  Theorem 4.1.

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

###### Proof.

Let $W=\Omega(\lambda,\epsilon)\otimes V$. It is clear that, for any $v\in V$,
there is a positive integer $K(v)$ such that $t^{m}D^{n}(v)=0$ for all $m\geq
K(v)$.

Suppose $M$ is a nonzero submodule of $W$. It suffices to show that $W=M$.
Take a nonzero $w=\sum_{j=0}^{s}x^{j}\otimes v_{j}\in W$ such that $v_{j}\in
V$, $v_{s}\neq 0$ and $s$ is minimal.

Claim 1. $s=0$.

Let $K={\rm max}\,\\{K(v_{j}):j=0,1,...,s\\}$. Using $t^{m}D(v_{j})=0$ for all
$m\geq K$ and $j=0,1,...,s$, we can deduce that

$\lambda^{-m}t^{m}D(w)=\sum_{j=0}^{s}(x-\epsilon m)(x-m)^{j}\otimes v_{j}\in
M,\,\forall m\geq K.$

Write the right hand side as

$\sum_{j=0}^{s+1}m^{j}w_{j}\in M,\quad\,\forall m\geq K,$

where $w_{j}\in W$ are independent of $m$. In particular,
$w_{s+1}=\epsilon(-1)^{s-1}\otimes v_{s}$. Taking $m=K,K+1,...,K+s$, we see
that the coefficient matrix of $w_{i}$ is a Vandermonde matrix. So each
$w_{i}\in M$. In particular, $w_{s+1}=\epsilon(-1)^{s-1}\otimes v_{s}\in M$.
Consequently, $s=0$.

Claim 2. $W=M$.

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

$\displaystyle t^{m}D(x^{i}\otimes v)$ $\displaystyle=$
$\displaystyle(\lambda^{m}(x-\epsilon m)(x-m)^{i})\otimes v$ $\displaystyle=$
$\displaystyle(\lambda^{m}(x-m)^{i+1})\otimes
v+(\lambda^{m}m(1-\epsilon)(x-m)^{i})\otimes v,$

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

Next, we will determine the necessary and sufficient conditions for two
irreducible tensor products defined in Theorem 4.1 to be isomorphic.

######  Theorem 4.2.

Let $\lambda,\lambda^{\prime}\in\mathbb{C}^{\ast}$ and
$\epsilon,\epsilon^{\prime}\in\\{0,1\\}$, and let $V$ and $V^{\prime}$ be two
irreducible quasifinite highest weight modules over $\mathcal{D}$. Then
$W=\Omega(\lambda,\epsilon)\otimes V$ and
$W^{\prime}=\Omega(\lambda^{\prime},\epsilon^{\prime})\otimes V^{\prime}$ are
isomorphic as $\mathcal{D}$-modules if and only if
$(\lambda,\epsilon)=(\lambda^{\prime},\epsilon^{\prime})$ and $V\cong
V^{\prime}$ as $\mathcal{D}$-modules.

###### Proof.

The sufficiency of the theorem is obvious. We only need to prove the
necessity. Let $\varphi$ be an isomorphism from $W$ to $W^{\prime}$ as
$\mathcal{D}$-modules. Take a nonzero element $1\otimes v\in W$. Suppose

$\varphi(1\otimes v)=\sum_{j=0}^{k}x^{j}\otimes w_{j},$

where $w_{j}\in V^{\prime}$ with $w_{k}\neq 0$. There is a positive integer
$K=K(v)$ such that $t^{m}D\cdot v=t^{m}D\cdot w_{j}=0$ for all $m\geq K$ and
$0\leq j\leq k$. For any $m,m^{\prime}\geq K$, we have

$(\lambda^{-m}t^{m}D-\lambda^{-m^{\prime}}t^{m^{\prime}}D)(1\otimes
v)=\epsilon(m^{\prime}-m)(1\otimes v),$

which suggests

$\displaystyle\epsilon(m^{\prime}-m)(\sum_{j=0}^{k}x^{j}\otimes
w_{j})=(\lambda^{-m}t^{m}D-\lambda^{-m^{\prime}}t^{m^{\prime}}D)(\sum_{j=0}^{k}x^{j}\otimes
w_{j})$ (4.1) $\displaystyle=$
$\displaystyle\sum_{j=0}^{k}\left[(\frac{\lambda^{\prime}}{\lambda})^{m}(x-\epsilon^{\prime}m)(x-m)^{j}-(\frac{\lambda^{\prime}}{\lambda})^{m^{\prime}}(x-\epsilon^{\prime}m^{\prime})(x-m^{\prime})^{j}\right]\otimes
w_{j}.$

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 $m,m^{\prime}\geq K.$ So

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

and (4.1) becomes

$\epsilon(m^{\prime}-m)(\sum_{j=0}^{k}x^{j}\otimes
w_{j})=\sum_{j=0}^{k}((x-\epsilon^{\prime}m)(x-m)^{j}-(x-\epsilon^{\prime}m^{\prime})(x-m^{\prime})^{j})\otimes
w_{j},$ (4.2)

where $m,m^{\prime}\geq K$. If $k>0$, for $\epsilon^{\prime}=1$, by the
coefficient of $m^{k+1}$ in (4.2), we get
$(-1)^{k+1}\epsilon^{\prime}(1\otimes w_{k})=0$, a contradiction. If $k>0$,
for $\epsilon^{\prime}=0$, by the leading coefficient of $m$ in (4.2), it is
easy to see that $k=1$ and $\epsilon(m^{\prime}-m)(1\otimes w_{0}+x\otimes
w_{1})=(m^{\prime}-m)x\otimes w_{1}$ for all $m,m^{\prime}\geq K$, which is
impossible. Therefore, $k=0$ and $\epsilon(m^{\prime}-m)(1\otimes
w_{k})=\epsilon^{\prime}(m^{\prime}-m)(1\otimes w_{k})$ for all
$m,m^{\prime}\geq K$, which forces

$\epsilon=\epsilon^{\prime}.$

Thus there is a bijection $\tau:V\longrightarrow V^{\prime}$ such that

$\varphi(1\otimes v)=1\otimes\tau(v),\quad\,\forall v\in V.$

Since $\varphi(t^{m}D^{n}\cdot(1\otimes v))=t^{m}D^{n}\cdot(\varphi(1\otimes
v))=t^{m}D^{n}\cdot(1\otimes\tau(v)),$ we have

$\displaystyle\beta^{1-n}\lambda^{m}\varphi((x-\varphi m)^{n}\otimes
v)+\varphi(1\otimes(t^{m}D^{n}\cdot v))$ $\displaystyle=$
$\displaystyle\beta^{1-n}\lambda^{m}(x-\varphi
m)^{n}\otimes\tau(v)+1\otimes(t^{m}D^{n}\cdot\tau(v)),\quad\,m\in\mathbb{Z}.$

Hence $\beta^{1-n}\lambda^{m}\varphi((x-\epsilon m)^{n}\otimes
v)=\beta^{1-n}\lambda^{m}(x-\epsilon m)^{n}\otimes\tau(v)$ for $m\geq K.$ So

$\varphi((x-\epsilon m)^{n}\otimes v)=(x-\epsilon
m)^{n}\otimes\tau(v),\quad\,m\geq K.$

Then it is easy to deduce that

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

By $\varphi(t^{m}D^{n}\cdot(1\otimes v))=t^{m}D^{n}\cdot(\varphi(1\otimes
v))$, we can obtain

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

So

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

Clearly, for $c\in\mathbb{C}$, $\varphi(c(1\otimes v))=c(\varphi(1\otimes v))$
implies that $\tau(cv)=c\tau(v)$. Thus $\tau:V\longrightarrow V^{\prime}$ is a
$\mathcal{D}$-module isomorphism and $V\cong V^{\prime}$. The proof is
finished. ∎

##  5\. Differential operator algebra on ${\mathbb{C}}[t_{1}^{\pm
1},\ldots,t_{\nu}^{\pm 1}]$

For any positive integer $\nu$, let $\mathcal{A}:={\mathbb{C}}[t_{1}^{\pm
1},\ldots,t_{\nu}^{\pm 1}]$ be the Laurent polynomial algebras. Then we have
the Lie algebras $\mathcal{W}_{\nu}={\rm Der}(\mathcal{A})$ and the Lie
algebra ${\mathcal{D}}_{\nu}={\rm Diff}(\mathcal{A})$. These Lie algebras 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
$\underline{m}=(m_{1},\ldots,m_{\nu})\in{\mathbb{Z}}^{\nu},\underline{n}=(n_{1},\ldots,n_{\nu})\in{\mathbb{N}}^{\nu}$,
set $t^{\underline{m}}=t_{1}^{m_{1}}\ldots
t_{\nu}^{m_{\nu}},D^{\underline{n}}=D_{1}^{n_{1}}\ldots D_{\nu}^{n_{\nu}}$,
where $D_{i}=t_{i}\frac{d}{dt_{i}}$ for all $i=1,2,\ldots,\nu$. We can write
the Lie brackets in $\mathcal{W}_{\nu}$ and $\mathcal{D}_{\nu}$ as follows:

$\displaystyle[t^{\underline{r}}D_{i},t^{\underline{s}}D_{j}]$
$\displaystyle=$ $\displaystyle
s_{i}t^{\underline{r}+\underline{s}}D_{j}-r_{j}t^{\underline{r}+\underline{s}}D_{i},$
$\displaystyle\bigl{[}\,t^{\underline{r}}D^{\underline{m}},t^{\underline{s}}D^{\underline{n}}\,\bigr{]}$
$\displaystyle=$
$\displaystyle\sum_{i_{1}=0}^{m_{1}}\ldots\sum_{i_{\nu}=0}^{m_{\nu}}{m_{1}\choose
i_{1}}\ldots{m_{\nu}\choose
i_{\nu}}\,\underline{s}^{\underline{i}}\,t^{\underline{r}+\underline{s}}D^{m+n-i}$
(5.1) $\displaystyle-$
$\displaystyle\sum_{j_{1}=0}^{n_{1}}\ldots\sum_{j_{\nu}=0}^{n_{\nu}}{n_{1}\choose
i_{1}}{n_{2}\choose i_{2}}\ldots{n_{\nu}\choose
i_{\nu}}\,\underline{r}^{\underline{j}}\,t^{\underline{r}+\underline{s}}D^{m+n-j}$

for all
$\underline{r},\underline{s}\in\mathbb{Z}^{\nu},\underline{m},\underline{n}\in\mathbb{N}^{\nu},\underline{i}=(i_{1},\ldots,i_{\nu}),\underline{j}=(j_{1},\ldots,j_{\nu})$.

It is known that $\mathfrak{h}_{\nu}={\rm
span}_{\mathbb{C}}\\{D_{1},D_{2},\ldots,D_{\nu}\\}$ is the Cartan subalgebra
of $\mathcal{W}_{\nu}$ and $\mathcal{D}_{\nu}$.

######  Lemma 5.1.

The Lie algebra $\mathcal{D}_{\nu}$ is generated by
$\\{t_{i},t_{i}^{-1},D_{i}D_{j}\mid i,j=1,2,\ldots,\nu\\}$.

###### Proof.

It follows by Lemma 2.1 and (5.1). ∎

In [21], a class of $\mathcal{W}_{\nu}$-modules were defined as follows, where
$1\leq\nu<\infty$. For any $\alpha\in\mathbb{C}$ and
$\Lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{\nu})\in(\mathbb{C}^{*})^{n}$,
denote by $\Omega(\Lambda,\alpha)=\mathbb{C}[x_{1},x_{2},\ldots,$ $x_{\nu}]$
the polynomial algebra over $\mathbb{C}$ in commuting indeterminates
$x_{1},x_{2},\ldots,x_{\nu}.$ The action of $\mathcal{W}_{\nu}$ on
$\Omega(\Lambda,\alpha)$ is defined by

$t^{\underline{m}}D_{i}\cdot f(x_{1},\ldots,x_{\nu})=\Lambda^{m}(x_{i}-\alpha
m_{i})f(x_{1}-m_{1},\ldots,x_{\nu}-m_{\nu}),$ (5.2)

where
$\underline{m}=(m_{1},m_{2},\ldots,m_{\nu})\in\mathbb{Z}^{\nu},f(x_{1},\ldots,x_{\nu})\in\mathcal{A},\Lambda^{\underline{m}}=\lambda_{1}^{m_{1}}\lambda_{2}^{m_{2}}\ldots\lambda_{\nu}^{m_{\nu}},$
$i=1,2,\ldots,\nu.$ By [21], we know that $\Omega(\Lambda,\alpha)$ is
irreducible if and only if $\alpha\neq 0.$ The following result was given in
[21].

######  Theorem 5.2.

[21] Let $M$ be a $\mathcal{W}_{\nu}$-module which is a free
$U(\mathfrak{h}_{\nu})$-module of rank $1$. Then
$M\cong\Omega(\Lambda,\alpha)$ for some $\Lambda\in(\mathbb{C}^{*})^{\nu}$ and
some $\alpha\in\mathbb{C}.$

For $\epsilon=0,1$,
$\Lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{\nu})\in(\mathbb{C}^{*})^{\nu}$,
we define a $\mathcal{D}_{\nu}$-module $\Omega(\Lambda,\epsilon)$ as follows:
Denote by $\Omega(\Lambda,\epsilon)=\mathbb{C}[x_{1},x_{2},\ldots,$ $x_{\nu}]$
the polynomial algebra over $\mathbb{C}$ in commuting indeterminates
$x_{1},x_{2},\ldots,x_{\nu}$. The action of $\mathcal{D}_{\nu}$ on
$\Omega(\Lambda,\epsilon)$ is defined by

$t^{\underline{m}}D^{\underline{n}}\cdot
f(x_{1},\ldots,x_{\nu})=\beta^{1-\sum_{i=1}^{\nu}n_{i}}\Lambda^{m}\Pi_{j=1}^{\nu}(x_{j}-\epsilon
m_{j})^{n_{j}}f(x_{1}-m_{1},\ldots,x_{\nu}-m_{\nu})$

for any $f(x_{1},\ldots,x_{\nu})\in\mathcal{A}$, where
$\beta=(-1)^{1-\epsilon}$,
$\underline{m}=(m_{1},m_{2},\ldots,m_{\nu})\in\mathbb{Z}^{\nu},\underline{n}=(n_{1},n_{2},\ldots,n_{\nu})\in{\mathbb{N}}^{\nu},$
$i=1,2,\ldots,\nu$.

######  Lemma 5.3.

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

###### Proof.

It follows by Lemma 3.1 and some direct calculations. ∎

######  Theorem 5.4.

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

###### Proof.

By Theorem 3.2, we get

$t_{i}^{m_{i}}D_{i}^{n_{i}}\cdot
f(x_{1},\ldots,x_{\nu})=\beta^{1-n_{i}}\lambda_{i}^{m_{i}}(x_{i}-\epsilon
m_{i})^{n_{i}}f(x_{1},\ldots,x_{i}-m_{i},\ldots,x_{\nu})$ (5.3)

for any $f(x_{1},\ldots,x_{\nu})\in\mathcal{A}$, where $\epsilon\in\\{0,1\\}$,
$\beta=(-1)^{1-\epsilon}$, $m_{i}\in{\mathbb{Z}}$, and
$\lambda_{i}\in\mathbb{C}^{*}$, $i=1,\ldots,\nu$.

Moreover, in this case, (5.2) becomes

$t^{\underline{m}}D_{i}\cdot
f(x_{1},\ldots,x_{\nu})=\Lambda^{\underline{m}}(x_{i}-\epsilon
m_{i})f(x_{1}-m_{1},\ldots,x_{i}-m_{i},\ldots,x_{\nu}-m_{\nu})$ (5.4)

for any $f(x_{1},\ldots,x_{\nu})\in\mathcal{A}$, where
$\underline{m}=(m_{1},m_{2},\ldots,m_{\nu})\in\mathbb{Z}^{\nu}$,
$\Lambda^{\underline{m}}=\lambda_{1}^{m_{1}}\ldots\lambda_{\nu}^{m_{\nu}}$,
$i=1,2,\ldots,\nu$.

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 (5.3) and (5.4), we get

$t^{\underline{m}}D_{i}^{2}\cdot
f(x_{1},\ldots,x_{\nu})=\beta^{-1}\Lambda^{\underline{m}}(x_{i}-\epsilon
m_{i})^{2}f(x_{1}-m_{1},\ldots,x_{i}-m_{i},\ldots,x_{\nu}-m_{\nu}).$ (5.5)

For any $1\leq i\neq j\leq\nu$, by
$[t_{i}^{-1}D_{i}^{2},t_{i}D_{j}]=2D_{i}D_{j}+D_{j}$, using (5.4) and (5.5),
we can obtain

$D_{i}D_{j}\cdot
f(x_{1},\ldots,x_{\nu})=\beta^{-1}x_{i}x_{j}f(x_{1},\ldots,x_{\nu}),\
\text{for all}\ f(x_{1},\ldots,x_{\nu})\in\mathcal{A}.$

By Lemma 5.1, 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 (Theorem 3.2 and Theorem 5.4). Certainly, the
study of non-weight module over the Lie algebra of differential operators is
still in its infancy.

Acknowledgments

This work is partially supported by the NNSF (Nos. 12071405, 11971315,
11871249,12171155), and is partially supported by Xinjiang Uygur Autonomous
Region graduate scientific research innovation project (No. XJ2021G021).

Data availability Data sharing not applicable to this paper.

Conflict of interest The authors declared that they have no conflicts of
interest to this work.

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