Representations of Super $W(2,2)$ algebra $\mathfrak{L}$

TL;DR

This paper investigates the representation theory of the super W(2,2) algebra, classifying its irreducible modules, analyzing conjugate-linear anti-involutions, and identifying unitary modules of intermediate series.

Contribution

It provides the first classification of irreducible modules of intermediate series and characterizes unitary modules for the super W(2,2) algebra.

Findings

No mixed irreducible modules exist for ${rak{L}}$

Complete classification of irreducible modules of intermediate series

Determination of conjugate-linear anti-involution and unitary modules

Abstract

In paper, we study the representation theory of super $W(2,2)$ algebra ${\mathfrak{L}}$. We prove that ${\mathfrak{L}}$ has no mixed irreducible modules and give the classification of irreducible modules of intermediate series. We determinate the conjugate-linear anti-involution of ${\mathfrak{L}}$ and give the unitary modules of intermediate series.

Equations

$$\displaystyle\begin{array}[]{lllll}&[L_{m},L_{n}]=(m-n)L_{m+n},&[L_{m},I_{n}]=(m-n)I_{m+n},\vspace*{8pt}\\ &[L_{m},H_{n}]=(\frac{m}{2}-n)H_{m+n},&[G_{m},G_{n}]=I_{m+n},\vspace*{8pt}\\ &[L_{m},G_{n}]=(\frac{m}{2}-n)G_{m+n},&[I_{m},G_{n}]=(m-2n)H_{m+n}.\end{array}$$

$$M=\underset{\lambda \in \mathbb{C}}{⨁}{M}_{\lambda },M_{\lambda }=\{x\in M|L_{0}x=-\lambda x\},$$

$$\{L_1,L_2,I_1,G_1,H_1,L_{-1},L_{-2},I_{-1},G_{-1},H_{-1}\}$$

$$V=({\cap }_{A\in \{L,I,G,H\}}\ker \{A_1:M_{\lambda }⟶M_{\lambda +1}\})\cap ({\cap }_{A\in \{L,I,G,H\}}\ker \{A_j:M_{\lambda }⟶M_{\lambda +j}\}),$$

$$L_{k}V=0k=1,j,j+1,j+2,\cdots$$

$$A_{k}V=0k=1,2,3,\cdots ,A\in \{I,G,H\}.$$

$$W={\cap }_{A\in \{L,I,G,H\}}\ker \{A_{-1}:L_{2}V⟶M_{\mu }\}.$$

$$[L_{-1},I_1]=-2I_0,[L_{-1},G_1]=-\frac{3}{2}{G}_0,[L_{-1},H_1]=-\frac{3}{2}{H}_0,$$

$$M=\underset{\lambda \in \mathbb{C}}{⨁}{M}_{\lambda },M_{\lambda }=\{x\in M|L_{0}x=-\lambda x\},$$

$$M_{\overline{i}}=\underset{\lambda \in \mathbb{C}}{⨁}{M}_{\overline{i}}^{\lambda },M_{\overline{i}}^{\lambda }=\{x\in M|L_{0}x=-\lambda x\},$$

$$A_{a,b}:$$

$$A(\alpha ):$$

$$B(\beta ):$$

$$L_i{u}_j=(a-j+ib)u_{i+j},$$

$$[L_m,I_n]=(m-n)I_{m+n}.$$

$$(a-n-t+bm)f(n,t)v_{m+n+t}-(a-t+bm)f(n,m+t)v_{m+n+t}$$

$$=(m-n)f(m+n,t)v_{m+n+t},$$

$$(a-n-t+bm)f(n,t)-(a-t+bm)f(n,m+t)=(m-n)f(m+n,t).$$

$$(a+bm)f(n,m)=(a-n+bm)f(n,0)-(m-n)f(m+n,0).$$

$$(a+(b-1)m-t)f(m+t)-(a-t+bm)f(m,m+t)=0.$$

$$(a+bm)f(m,0)=(a+(b-1)m)f(m,0).$$

$$(a+bm)f(m,-m)=(a+(b+1)m)f(m,0).$$

$$(a-bn)f(n,-n)=(a-(b+1)n)f(n,0)+2nf(0,0).$$

$$f(m,0)=\frac{a+bm}{a}f(0,0),m\ne 0.$$

$$(a-m+bm)(f(0,m)-f(0,2m))=mf(m,m).$$

$$(a+bm)f(0,m)=(a+bm)f(0,0)-mf(m,0).$$

$$f(0,m)=\frac{a-m}{a}f(0,0).$$

$$f(m,m)=\frac{a-m+bm}{a}f(0,0).$$

$$f(n,m)=\frac{f(0,0)(a+bn-m)}{a}.$$

$$f(0,-p)=0$$

$$(m+n+t)f(n,t)-(m+t)f(n,m+t)=(n-m)f(m+n,t).$$

$$mf(n,m)=(m+n)f(n,0)+(m-n)f(m+n,0).$$

Full text

Representations of Super $W(2,2)$ algebra $\mathfrak{L}$

Hao Wang1), Huanxia Fa2), Junbo Li2)

*1)*Wu Wen-Tsun Key Laboratory of Mathematics and School of Mathematical Sciences,

University of Science and Technology of China, Hefei 230026, China

*2)*School of Mathematics and Statistics, Changshu Institute of Technology, Changshu 215500, China

Abstract. In paper, we study the representation theory of super $W(2,2)$ algebra ${\mathfrak{L}}$. We prove that ${\mathfrak{L}}$ has no mixed irreducible modules and give the classification of irreducible modules of intermediate series. We determinate the conjugate-linear anti-involution of ${\mathfrak{L}}$ and give the unitary modules of intermediate series.

Key words: conjugate-linear anti-involution, Harish-Chandra module, mixed weight module, modules of intermediate series, unitary representation.

*MR(2000) Subject Classification: * 17B10, 17B65, 17B68.

1 Introduction

It is well known that the Virasoro algebra (named after the physicist Miguel Angel Virasoro) is a very important infinite dimensional Lie algebra and is widely used in conformal field theory and string theory. After that much attention has been paid to the Virasoro type Lie algebras and superalgebras (which contains the Virasoro algebra as their subalgebras), including their constructions, structures and representations. The $W$-algebra $W(2,2)$ is certainly a Virasoro type Lie algebra, which plays important rolls in many areas of mathematics and physics (It was introduced by Zhang and Dong in [13] for the study of classification of vertex operator algebras generated by vectors of weight $2$). It possesses a basis $\{L_{m},\,I_{m}|m\in\mathbb{Z}\}$ as a vector space over the complex field $\mathbb{C}$, with the Lie brackets $[L_{m},L_{n}]=(m-n)L_{m+n}$, $[L_{m},I_{n}]=(m-n)I_{m+n}$, $[I_{m},I_{n}]=0$. Structures and representations of $W(2,2)$ are extensively investigated in many references, such as [2], [4], [5], [7], [8] and [15].

Some Lie superalgebras with $W$-algebra $W(2,2)$ as their even parts were constructed in [9] as an application of the classification of Balinsky-Novikov super-algebras with dimension $2|2$.In this paper we consider the infinite dimensional Lie super $W(2,2)$-algebra over the algebraic closed field $\mathbb{C}$ (for convenience, we denote it $\mathfrak{L}$), with the following non-vanishing brackets:

[TABLE]

Finally we would like to make some remarks. We observe that many papers are forced on the Virasoro type Lie superalgebras which contain the super Virasoro algebra as their subalgebra (e.g., Refs. [3, 12]), especially the $N=2$ super Virasoro algebras. It is easy to find that the algebra $\mathfrak{L}$ doesn’t contain the super Virasoro Lie algebra as it’s subalgebra, so the methods developed there are not applicable to $\mathfrak{L}$. All these make the study of $\mathfrak{L}$ more challengeable and attractive, we need to find some new methods to handel these problems.

2 Weight module of ${\mathfrak{L}}$ with a finite dimensional weight space

2.1 Preliminaries and main results

First we introduce some standard concepts. Denote $\mathfrak{h}={\rm Span}_{\mathbb{C}}\{L_{0}\}$ the Cartan subalgebra of $\mathfrak{L}$. The module of $\mathfrak{L}$ $M$ is called $\mathfrak{h}$-diagonalisable, if $M$ have the following decomposition:

[TABLE]

$M_{\lambda}$ is called the weight space of $\lambda$. Denote ${\rm Supp}(M)=\{\lambda\in\mathbb{C}\,|\,M_{\lambda}\neq 0\}$ the support set of $M$. It is obvious if $M$ is a irreducible weight module of $\mathfrak{L}$, then there exists $\lambda\in\mathbb{C}$, such that ${\rm Supp}(M)\subset\lambda+\mathbb{Z}$.

The weight module $M$ of $\mathfrak{L}$ is called Harish-Chandra, if all weight spaces of $M$ is finite dimensional. The module $M$ of $\mathfrak{L}$ is called mixed, if there exists $\lambda,\,\mu\in{\rm Supp}(M)$, such that ${\rm dim}M_{\lambda}=\infty$, ${\rm dim}M_{\mu}<\infty$.

Our main result of this section is the following theorem.

Theorem 2.1**.**

If $M$ is a irreducible weight module of $\mathfrak{L}$ and there exists $\lambda\in\mathbb{C}$, such that ${\rm dim}M_{\lambda}=\infty$, then ${\rm Supp}(M)=\lambda+\mathbb{Z}$, and for each $k\in\mathbb{Z}$ we have ${\rm dim}M_{\lambda+k}=\infty$.

From the above theorem, we immediately get the following two facts.

Corollary 2.2**.**

If $M$ is a irreducible weight module of $\mathfrak{L}$, with a finite dimensional weight space, then $M$ ia a Harish-Chandra module.

Corollary 2.3**.**

There exists no mixed module of $\mathfrak{L}$.

2.2 Proof of theorem 2.1

In this subsection, $M$ is always a irreducible weight module of $\mathfrak{L}$. Notice that

[TABLE]

is a set of generators of $\mathfrak{L}$, we have the following proposition.

Proposition 2.4**.**

If there exists $\mu\in\mathbb{C}$, $0\neq v\in M_{\mu}$ such that $L_{1}v=L_{2}v=I_{1}v=G_{1}v=H_{1}v=0$ or $L_{-1}v=L_{-2}v=I_{-1}v=G_{-1}v=H_{-1}v=0$, then $M$ is a Harish-Chandra module.

Lemma 2.5**.**

If $\rm{dim}M_{\lambda}=\infty$, then there exists at most one $i\in\mathbb{Z}$, such that $\rm{dim}M_{\lambda+i}<\infty$.

Proof. We prove it with reduction to absurdity. Without lose of generality, assume $\rm{dim}M_{\lambda+1}<\infty$, $\rm{dim}M_{\lambda+j}<\infty$, in whick $j\in\mathbb{Z}$, $j\geq 3$. (If $j=2$, we immediately get a contradiction from 2.4).

Denote

[TABLE]

It is a subspace of $M_{\lambda}$. $\rm{dim}M_{\lambda}=\infty$, $\rm{dim}M_{\lambda+1}<\infty$, together with $\rm{dim}M_{\lambda+j}<\infty$ imply $\rm{dim}V=\infty$. Act both sides of (4) of $V$, we get

[TABLE]

and

[TABLE]

If there exists $0\neq v\in V$ such that $L_{2}v=0$, Proposition 2.4 implies $M$ is a Harish-Chandra module, contradicts with $\rm{dim}M_{\lambda}=\infty$ . Now we can assume for all $0\neq v\in V$, $L_{2}v\neq 0$. this implies $\rm{dim}L_{2}V=\infty$.

Denote $W=\cap_{A\in\{L,\,I,\,G,\,H\}}\ker\{A_{-1}:L_{2}V\longrightarrow M_{\lambda+1}\}$, it is a subspace of $M_{\lambda+2}$. Since $\dim L_{2}V=\infty$, $\dim M_{\lambda+1}<\infty$, we know $\rm{dim}W=\infty$, this means $w=L_{2}v\ (0\neq v\in V)$ such that $A_{-1}w=0$, in which $A\in\{\,L,\,I,\,G,\,H\,\}$. For $k\geq j$ $(\geq 3)$, we have $L_{k}L_{2}v=(k-2)L_{k+2}v+L_{2}L_{k}v=0$. So $L_{-1}w=L_{k}w=0$ holds for all $k=1,\,j,\,j+1,\,j+2,\,j+3,\,\cdots$. Since $[L_{-1},L_{k}]=-(1+k)L_{k-1}\neq 0$ for all $k\geq 1$, it is obvious to see $L_{k}w=0$, $k=1,\,2,\,3,\,\cdots$, This implies $A_{k}w=0$, $k=1,\,2,\,3,\,\cdots$, $A\in\{I,\,G,\,H\}$. From Proposition 2.4, we get that $M$ is a Harish-Chandra module, contradicts with $\dim M_{\lambda}=\infty$. ∎

Now we can assume for $i\in\mathbb{Z}^{*}$, $\dim M_{\mu}<\infty$, $\dim M_{\mu+i}=\infty$.

Lemma 2.6**.**

If $0\neq v\in M_{\mu-1}$ satisfies $L_{1}v=I_{1}v=G_{1}v=H_{1}v=0$, then:

(1)  $A_{k}v=0$.

(2)  $A_{k}L_{2}v=0$, $\forall$ $k=1,\,2,\,3,\,\cdots,\,A\in\{I,\,G,\,H\}$.

Proof. (1) Since $[L_{1},A_{k}]=(\frac{1}{2}-k)A_{k+1}\neq 0$ holds for $k=1,\,2,\,3,\,\cdots$, $A\in\{G,\,H\}$, By induction on $k$, we have for $k=1,\,2,\,3,\,\cdots$, $G_{k}v=H_{k}v=0$. From $[G_{1},G_{1}]=I_{2}$ and $k=2,\,3,\,\cdots$, $[L_{1},I_{k}]=(1-k)I_{k+1}\neq 0$, Act both sides of the above equation on $v$, we have $I_{k}v=0$ holds for $k=1,\,2,\,3,\,\cdots$.

(2)Holds immediately from (1) and $A_{k}L_{2}v=[A_{k},L_{2}]v+L_{2}A_{k}v$. ∎

Proof of theorem 2.1  Denote $V=\cap_{A\in\{L,I,G,H\}}\ker\{A_{1}:M_{\mu-1}\longrightarrow M_{\mu}\}$. Since $\rm{dim}M_{\mu-1}=\infty$, $\rm{dim}M_{\mu}<\infty$, $\rm{dim}V=\infty$. For arbitrary $0\neq v\in V$, if $L_{2}v=0$, Proposition 2.4 implies $M$ is a Harish-Chandra module, contradicts with the existence of a infinite weight space with $M$.

So for each $0\neq v\in V$,  $L_{2}v\neq 0$, this implies $\rm{dim}L_{2}V=\infty$. Denote

[TABLE]

Since $\rm{dim}L_{2}V=\infty$, $\rm{dim}M_{\mu}<\infty$, we have $\rm{dim}W=\infty$. This implies there exists $w=L_{2}v$ ($0\neq v\in V$) such that for $A\in\{L,\,I,\,G,\,H\}$, $A_{-1}w=0$. For $A\in\{G,\,H\}$ we have $[L_{-1},\,A_{k}]=(-\frac{1}{2}-k)A_{k-1}\neq 0$, Act both sides of the above equation on $w$, we know for $k=-1,\,-2,\,-3,\,\cdots$, $A_{k}w=0$. Since for $k=-2,\,-3,\,-4,\,\cdots$, $[L_{-1},I_{k}]=(-1-k)I_{k-1}\neq 0$, and $[G_{-1},G_{-1}]=I_{-2}$, Similar arguments as above, we have for $k=-1,\,-2,\,-3,\,\cdots$, $I_{k}w=0$.

Act both sides of

[TABLE]

on $w$, Lemma 2.6 implies for $k\in\mathbb{Z},A\in\{I,G,H\}$, $A_{k}w=0$. Since $M$ is a irreducible weight module of $\mathfrak{L}$, we have $M=\mathfrak{U}(\mathfrak{L})w$ (In which $\mathfrak{U}(\mathfrak{L})$ is the universal enveloping algebra of $\mathfrak{L}$). Denote L the vector space generated by $\{L_{m}\,|\,m\in\mathbb{Z}\}$ over $\mathbb{C}$ (Obviously L is a centerless Virasoro Lie algebra ). The above discussion implies $M=\mathfrak{U}(\textbf{L})w$, which means $M$ is a irreducible weight module of centerless Virasoro algebra. From the already known result: Virasoro owns no mixed weight weight module (to be concrete, one can see [11]), We have $M$ is a Harish-Chandra modules, contradicts with the assumption that $M$ has infinite dimensional weight space. This complete the proof of our main theorem. ∎

3 Intermediate series module of ${\mathfrak{L}}$

3.1 Preliminaries and main results

Denote $\mathfrak{h}={\rm Span}_{\mathbb{C}}\{L_{0}\}$ the Cartan subalgebra of $\mathfrak{L}$. Weight module $M$ of $\mathfrak{L}$ is called $\mathfrak{h}$-diagonalisable , if $M$ has the decomposition:

[TABLE]

$M_{\lambda}$ is a weight space of weight $\lambda$. Denote ${\rm Supp}(M)=\{\lambda\in\mathbb{C}\,|\,M_{\lambda}\neq 0\}$ the support set of $M$, a $\mathfrak{L}$-module $M=M_{\overline{0}}\oplus M_{\overline{1}}$ is called the intermediate series module of $\mathfrak{L}$, in which

[TABLE]

if $M$ is irreducible and dim$M_{\overline{i}}^{\lambda}\preceq 1$, for $i=0$ or $1$.

First we recall some already known results about intermediate series modules of Virasoro algebra. (To be concrete, one can see [10]):

Theorem 3.1**.**

The intermediate series module $M$ of Virasoro algebra must be one the following modules $A_{a,b}$, $A(\alpha)$, $B(\beta)$, or their quotient modules:

[TABLE]

In which $i,j\in\mathbb{Z}$, $a,b,\alpha,\beta\in\mathbb{C}$. $A_{a,b}$, $A(\alpha)$, $B(\beta)$ as vector spaces over $\mathbb{C}$, own {$u_{k}$ — $k\in\mathbb{Z}$}* as a basis.*

The main result of this section is the following theorem.

Theorem 3.2**.**

If $M$ is a intermediate series module of $\mathfrak{L}$, it is also the intermediate series module of Virasoro algebra. (In other words, $\forall k\in\mathbb{Z}$, $I_{k},G_{k},H_{k}$ act trivally on $M$).

3.2 Proof of Theorem 3.2

We prove the theorem 3.2 by several lemmas. First we introduce some concepts. Assume $M_{\overline{0}}$, $M_{\overline{1}}$ (modules of the Virasoro algebra) are in forms of $A_{a,b}$, $M_{\overline{0}}=\rm{Span}_{\mathbb{C}}\{u_{k}\,|\,k\in\mathbb{Z}\}$, $M_{\overline{1}}=\rm{Span}_{\mathbb{C}}\{v_{k}\,|\,k\in\mathbb{Z}\}$. satisfy:

[TABLE]

in which $i,j\in\mathbb{Z}$, $a,a^{\prime},b,b^{\prime},f(i,j),\widetilde{f}(i,j),g(i,j),\widetilde{g}(i,j),h(i,j),\widetilde{h}(i,j)\in\mathbb{C}$.

Lemma 3.3**.**

$f(i,j)=\widetilde{f}(i,j)=0$, $\forall i,j\in\mathbb{Z}$.

Proof.  We prove this lemma case by case.

Case $1$  $a+mb\neq 0$ $\forall m\in\mathbb{Z}$.

Act both sides of

[TABLE]

on $u_{t}$, we have:

[TABLE]

This implies:

[TABLE]

In (10), take $t=0$:

[TABLE]

In (10), take $m=n$:

[TABLE]

In (12), take $t=0$:

[TABLE]

In (12), take $t=-m$:

[TABLE]

In (11), take $m=-n$:

[TABLE]

Associate (14) with (15):

[TABLE]

In (10), take $t=m,n=0$:

[TABLE]

In (11), take $n=0$:

[TABLE]

Associate (16) with (18):

[TABLE]

From (17) and (19):

[TABLE]

Take (16) together with (19), (11) becomes:

[TABLE]

holds for all $m,n\in\mathbb{Z}$. Act both sides of $[I_{m},I_{n}]=0$ on $u_{k}$, we have $\frac{f(0,0)}{a}=0$. From the above discussion we know for all $m,n\in\mathbb{Z}$, $f(m,n)=0$.

Case $2$ $a\neq 0$, $0\in\{a+bn\,|\,n\in\mathbb{Z}\}$.

Under this assumption we know $b\neq 0,-1$ and there exists $p\in\mathbb{Z}^{*}$ such that $a=bp$. Similar arguments as in case $1$: $f(m,n)=0$ $\forall n\neq-p$. together with $\eqref{ISM-eq-4}$:

[TABLE]

In (10) take $t=-p,m=0$, together with (22): $f(n,-p)=0$ $\forall n\neq 0$. From the above discussion, we know: $f(m,n)=0$ $\forall m,n\in\mathbb{Z}$.

Case $3$ $a\in\mathbb{Z}$.

Since $A_{a,b}\cong A_{0,b}$, without lose of generality, we assume $a=0$.

Case $3.1$ $b\neq 0,-1$. Similar arguments as in $1$, we have: $f(m,n)=0$ holds $\forall m,n\in\mathbb{Z}$.

Case $3.2$ $b=-1$.

In this situation, $\eqref{ISM-eq-2}$ and $\eqref{ISM-eq-3}$ turn into the following formula:

[TABLE]

[TABLE]

In (23) take $n=t=0$:

[TABLE]

this implies:

[TABLE]

In (23) take $t=m,m=-n$:

[TABLE]

In (24) replace $m$ with $m-n$:

[TABLE]

Take together (24)–(27), we have:

[TABLE]

In (28) take $m=n\neq 0$, we get $f(0,0)=0$. $\eqref{ISM-eq-20}$ becomes:

[TABLE]

By induction:

[TABLE]

Associative (28) with (24):

[TABLE]

From (30) and (31):

[TABLE]

Replaced in (23) we have $c=0$, in other words: $f(m,n)=d(m+n)$. Together with $[I_{m},I_{n}]=0$ we know: $f(m,n)=0$ $\forall m,n\in\mathbb{Z}$.

Case $3.3$ $b=0$.

In this situation (10) becomes:

[TABLE]

In (33), take $t=0,m=-n$:

[TABLE]

In (33), take $t=0$:

[TABLE]

In (33), take $t=-m$, together with (35):

[TABLE]

In (33), take $t=1$:

[TABLE]

In (37), take $m=-1$, associate with (34) we get:

[TABLE]

By induction on (38), we have:

[TABLE]

(36) and (39) imply $c_{1}=c_{2}$. From (38) we have $f(n,m)=mc$ holds for all $m,n\in\mathbb{Z}$. Act both sides of $[I_{m,I_{n}}]=0$ on $u_{k}$, we get $c=0$, This implies for each $m,n\in\mathbb{Z}$, $f(m,n)=0$.

Similarly, we have: $\widetilde{f}(m,n)=0$ $\forall m,n\in\mathbb{Z}$. ∎

Lemma 3.3 and the Lie bracket $[I_{m},G_{n}]=H_{m+n}$ imply the following lemma.

Lemma 3.4**.**

$h(m,n)=\widetilde{h}(m,n)=0$, $\forall m,n\in\mathbb{Z}$.

Lemma 3.3 and the Lie bracket $[G_{m},G_{n}]=I_{m+n}$ imply the following lemma.

Lemma 3.5**.**

[TABLE]

$\forall m,n,k\in\mathbb{Z}$**

Lemma 3.6**.**

At least one of $g(m,n)$ $\widetilde{g}(m,n)$ is [math].

Proof. Suppose not, assume neither $g(m,n)$ nor $\widetilde{g}(m,n)$ is [math].

Act both sides of $[L_{0},G_{n}]=-nG_{n}$ on $u_{k}$, we know:

[TABLE]

$\forall n,k\in\mathbb{Z}$. Under the assumption $g(m,n)\not\equiv 0$, we get $a=a^{\prime}$. Act both sides of $[L_{i},G_{j}]=(\frac{i}{2}-j)G_{i+j}$ on $u_{k}$:

[TABLE]

In (40), take $i=2j$:

[TABLE]

In (40) take $i=-2j$:

[TABLE]

(42) times $a-(k-j)+2jb^{\prime}$, and replace the last item of (42) by (41):

[TABLE]

In which $t=b^{\prime 2}-{(b+\frac{1}{2})}^{2}$. Similarly: for (40), let $j=-j,i=2j$, $j=-j,i=-2j$:

[TABLE]

Together with (3.2) and (44):

[TABLE]

From (45) we have:

[TABLE]

(40) implies: there sxists $k_{0}\in\mathbb{Z}$ such that $g(1,k_{0})\neq 0$ (If not, (40) implies $g(m,n)\equiv 0$, contradicts with our assumption). Together with (46), we have:

[TABLE]

In the following, we determine $g(m,n),\widetilde{g}(m,n)$ case by case.

Case $1$  $b^{\prime}=b+\frac{1}{2}$.

Assume $a-k+2b\neq 0$, $\forall k\in\mathbb{Z}$. In (41), take $j=1$:

[TABLE]

In (40), take $j=1$, $i,k$ are even integers:

[TABLE]

Associative (41) with $b^{\prime}=b+\frac{1}{2}$:

[TABLE]

Obviously we have $x_{0}=x_{1}$. Similar arguments following (46), we have:

[TABLE]

In which $d_{1}\in\mathbb{C}^{*}$ is a constant.

Now we assume $a-k^{\prime}+2b=0$, $\forall k^{\prime}\in\mathbb{Z}$. Together with (41) we know: $(a-k+2b)g(1,k)=(a-k+2b)g(1,k+2)$. This implies:

[TABLE]

[TABLE]

Together with (40):

[TABLE]

[TABLE]

Select some $k,j\in\mathbb{Z}$, satisfy $a-k+2jb\neq 0,k\preceq k^{\prime},k+j-1\preceq k^{\prime},k+2j>k^{\prime},k+3j-1>k^{\prime}$, one of $k,j$ is even and the other is odd. From (41) we know:

[TABLE]

this implies: $x_{0}=y_{0},x_{1}=y_{1}$. Similar arguments, we have: $g(m,n)=d_{1}$, $\forall m,n\in\mathbb{Z}$, . In which $d_{1}\in\mathbb{C}^{*}$.

Case $2$  $b^{\prime}=-(b+\frac{1}{2})$.

In (40), take $i=2j$:

[TABLE]

In (40), take $i=-2j$:

[TABLE]

The second equation on the right side of (57) follows from (56) with the replacement $kj$ is replaced by $k-2$. This implies: $g(j,k)=g(-j,k)$ holds $\forall\,j,k\in\mathbb{Z}$. Together with (40):

[TABLE]

This implies:

[TABLE]

In (56), take $j=1$ and replace $k$ by $k+2$:

[TABLE]

Similarly, we get an expression of $g(1,k-4)$ in form of (60). IN (59), take $i=4$, together with (60):

[TABLE]

From (56) and the similar arguments following (46), we know the coefficient of $g(1,k)$ is [math]. This implies: either $b=-1$ or $-\frac{1}{2}$. Since $b=-\frac{1}{2},b^{\prime}=0$ is just in case $1$, we only need to consider $b=-1,b^{\prime}=\frac{1}{2}$. In (41), take $j=1$, we know $(a-k)g(1,k)$ $k$ is a constant $\forall k$. Assume:

[TABLE]

In (59), take $i=j=1$, we have $x_{0}=x_{1}$. In other words, $(a-k)g(1,k)$ is a constant. If for $k_{1}\in\mathbb{Z}$, $a-k_{1}=0$ holds, then $\forall k\neq k_{1}$:

[TABLE]

This ensures $g(m,n)\equiv 0$, contradicts with our assumption. From the discussion above, we have $(a-k)g(1,k)=d_{2}$, in which $d_{2}\in\mathbb{C}^{*}$. In (40), take $j=1$:

[TABLE]

this implies $g(i,k)=g(1,k)$ holds for all $k\in\mathbb{Z},k\neq 3$. In (40), take $j=2,i=1$:

[TABLE]

this implies: $g(3,k)=g(1,k)$. Now, we have $g(i,j)=\frac{d_{2}}{a-j}$, in which $d_{2}\in\mathbb{C}$.

Case $3$  $b^{\prime}=-b-\frac{3}{2}$.

Similar arguments as in case $2$, we have $2jg(j,k)=2jg(j,k-2j)$, together with:

[TABLE]

[TABLE]

This implies $b=-\frac{3}{2}$ or $-1$. The case $b=-1,b^{\prime}=-\frac{1}{2}$ is just in case $1$. So we only need to consider $b=-\frac{3}{2},b^{\prime}=0$. In (41), take $j=1$:

[TABLE]

Similar arguments as in case $2$, we know: $(a-k-1)g(1,k)$ is a constant. If there exists $k_{1}\in\mathbb{Z}$ such that $a-K_{1}-1=0$, then for all $k\neq k_{1}$, we have:

[TABLE]

this ensures $g(m,n)\equiv 0$, contradicts with our assumption. Therefor, $g(1,k)=\frac{d_{3}}{a-k-1}$, in which $d_{3}\in\mathbb{C}^{*}$. In (40), take $j=1$:

[TABLE]

This implies:

[TABLE]

In (40), take $i=1,j=2$:

[TABLE]

This implies:

[TABLE]

in which $d_{3}\in\mathbb{C}^{*}$.

Case $4$  $b^{\prime}=b-\frac{1}{2}$.

Similar as in case $1$, we have: $\widetilde{g}(m,n)=d^{\prime}_{1}$.

Take together all the cases $1--4$ and make the similar argument for $\widetilde{g}(m,n)$, we have:

[TABLE]

[TABLE]

In which $i,j\in\mathbb{Z},d_{1},d^{\prime}_{1},d_{2},d^{\prime}_{2},d_{3},d^{\prime}_{3}\in\mathbb{C}^{*}$. Obviously only the two cases are possible:

[TABLE]

associate with (74) and lemma 3.5, we have: either $g(m,n)\equiv 0$ or $\widetilde{g}(m,n)\equiv 0$, this contradicts with our assumption. This complete the proof of our lemma. ∎

For $M_{\overline{0}},M_{\overline{1}}$ in forms of $A_{\alpha},B_{\beta}$, the conclusion of the relative coefficients are same as above via similar arguments. Without lose of generality, we can assume $g(m,n)\equiv 0$. From the above discussion we know, the coefficients defined is (8) satisfy:

[TABLE]

Proof of theorem 3.2: If $M=M_{\overline{0}}\oplus M_{\overline{1}}$ is the irreducible intermediate series module of $\mathfrak{L}$. From the irreducibility and (3.2), we know $M_{\overline{1}}=0$. This means $I_{k},G_{k},H_{k}$ act trivially $M$, $\forall k\in\mathbb{Z}$. This complete the proof of theorem 3.2. ∎

4 Unitary representation of ${\mathfrak{L}}$

Denote $\mathfrak{h}=\rm{Span}_{\mathbb{C}}\{L_{0}\}$ the Cartan subalgebra of $\mathfrak{L}$.

4.1 Conjugate linear anti-involution of $\mathfrak{L}$

Denote $\mathbf{A}=\rm{Span}_{\mathbb{C}}\{A_{k}\,|\,k\in\mathbb{Z}\}$, in which $\mathbf{A}\in\{\mathbf{L},\mathbf{I},\mathbf{G},\mathbf{H}\}$, $A\in\{L,I,G,H\}$. We have the following lemma.

Lemma 4.1**.**

$\mathbf{I}\oplus\mathbf{G}\oplus\mathbf{H}$* is the unique maximal ideal of $\mathfrak{L}$.*

Proof. From the Lie super bracket of $\mathfrak{L}$, by direct calculation we know:

• (1)

$\mathbf{I}\oplus\mathbf{G}\oplus\mathbf{H}$ is an ideal of $\mathfrak{L}$.

• (2)

None of $\mathfrak{L}$ś ideal contain an nonzero element of $\mathbf{L}$.

This complete the proof of this lemma.

Definition 4.2**.**

$(1)$* A map $\theta$: $\mathfrak{L}\rightarrow\mathfrak{L}$ is called the conjugate linear anti-involution of $\mathfrak{L}$ if $\theta$ satisfies the following conditions:*

1. C1:

$\theta(x+y)=\theta(x)+\theta(y)$, 2. C2:

$\theta(\alpha x)=\overline{\alpha}x$, 3. C3:

$\theta([x,y])=[\theta(y),\theta(x)]$, 4. C4:

${\theta}^{2}=\rm{id}$.

in which $x,y\in\mathfrak{L},\alpha\in\mathbb{C}$, $\rm{id}$ is the identity map of $\mathfrak{L}$.

$(2)$* An $\mathfrak{L}$–module $M$ is called the unitary $\mathfrak{L}$–module, If there exists a positive defined Hermitian form $\langle,\rangle$ on $M$ such that $\forall u,v\in M,x\in\mathfrak{L}$:*

[TABLE]

The already known results about the conjugate linear anti-involution of the Virasoro algebra is listed in the following: (one can see [1]):

Theorem 4.3**.**

The conjugate linear anti-involution of the Virasoro algebra is one of the following.

1. (1)

${\theta}^{+}_{\alpha}(L_{m})={\alpha}^{m}L_{-m}$, in which $\alpha\in{\mathbb{R}}^{*}$. 2. (2)

${\theta}^{-}_{\alpha}(L_{m})=-{\alpha}^{m}L_{m}$, in which $\alpha\in S^{1}$, the set of all length $1$ complex nunber .

Lemma 4.4**.**

If $\theta$ is an arbitrary conjugate linear anti-involution of $\mathfrak{L}$, we have:

1. (1)

$\theta(\mathbf{I}+\mathbf{G}+\mathbf{H})=\mathbf{I}+\mathbf{G}+\mathbf{H}$. 2. (2)

$\theta(\mathfrak{h})=\mathfrak{h}$.

Proof. $(1)$ $\forall x\in\mathfrak{L},y\in\mathbf{I}+\mathbf{G}+\mathbf{H}$, From equation

[TABLE]

and the fact $\mathbf{I}+\mathbf{G}+\mathbf{H}$ is an ideal of $\mathfrak{L}$, we have: $\theta(\mathbf{I}+\mathbf{G}+\mathbf{H})$ is an ideal of $\mathfrak{L}$. once more, we get:

[TABLE]

(If not, From the ideal $\theta(\mathbf{I}+\mathbf{G}+\mathbf{H})$ we get another maximal ideal of $\mathbf{I}+\mathbf{G}+\mathbf{H}$, contradicts with the fact that $\mathbf{I}+\mathbf{G}+\mathbf{H}$ is the unique maximal ideal of $\mathfrak{L}$).

[TABLE]

Since ${\theta}^{2}=\rm{id}$, together with (76) and (77), we have:

[TABLE]

$(2)$ From $C3$ of the definition 4.2, we have:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

This implies $\theta(L_{0})$ acts diagonally on $\mathfrak{L}$. We get $\theta(L_{0})\in\mathfrak{h}$. Since $\theta$ is non-degenerate, together with $\rm{dim}(\mathfrak{h})=1$, we know: $\theta(\mathfrak{h})=\mathfrak{h}$. ∎

Suppose $\theta$ is an arbitrary conjugate linear anti-involution of $\mathfrak{L}$ , assume:

[TABLE]

Only finitely many summands on the right side of the above four equations are non zero, all the coefficients are lie in $\mathbb{C}$.

Since $\mathbf{I}+\mathbf{G}+\mathbf{H}$ is the maximal ideal of $\mathfrak{L}$, and

[TABLE]

Here the notation Vir means the centerless Virasoro algebra. From theorem 4.3 we know $\theta(L_{n})$ must be one of the following two forms:

Lemma 4.5**.**

1. (a)

${\theta}^{+}_{\alpha}(L_{m})={\alpha}^{m}L_{-m}+\sum_{k}b_{m,k}^{L}I_{k}+\sum_{k}c_{m,k}^{L}G_{k}+\sum_{k}d_{m,k}^{L}H_{k}$, in which $\alpha\in{\mathbb{R}}^{*}$. 2. (b)

${\theta}^{-}_{\alpha}(L_{m})=-{\alpha}^{m}L_{m}+\sum_{k}b_{m,k}^{L}I_{k}+\sum_{k}c_{m,k}^{L}G_{k}+\sum_{k}d_{m,k}^{L}H_{k}$, in which $\alpha\in S^{1}$, the set of all complex numbers of length $1$.

We determinate $\theta$ case by case.

Case $1$   $\theta(L_{n})$ has the form $(a)$ in 4.5.

From $(2)$ of Lemma 4.4 we know $\theta(L_{0})=L_{0}$. Since

[TABLE]

together with (4.1), we have:

[TABLE]

Compare the coefficients of the summands respectively in equation (4.1), we have:

[TABLE]

This implies:

[TABLE]

Similar arguments as (79), for $\theta(A_{m})$ we have:

[TABLE]

In which $A\in\{I,G,M\}$. Since

[TABLE]

together with (4.1), replace (82) in the above equation, compare the coefficients of each summands after the concrete calculation, we get:

[TABLE]

[TABLE]

[TABLE]

In (85), take $m=n\neq 0$, we know $c_{k,-k}^{L}=0$, $k\neq 0$. Again in (85), take $m=-n=1$, we get $c_{0,0}^{L}=0$. Take together, we have:

[TABLE]

Now, (84) becomes:

[TABLE]

By induction on (88):

[TABLE]

(86) becomes:

[TABLE]

By induction on (90):

[TABLE]

From (87) (89) and (91), we get:

[TABLE]

From the super Lie bracket in $\mathfrak{L}$, together with (1) in lemma 4.4 and (82), we can assume:

[TABLE]

Plug (4.1) into

[TABLE]

compare the coefficients of each summands respectively, we have:

[TABLE]

[TABLE]

By induction on (94), we get:

[TABLE]

In (95), take $m=n\neq 0$:

[TABLE]

Again in (95), take $m=-n=1$, we have:

[TABLE]

this implies:

[TABLE]

Associative with (96) and (97), we get:

[TABLE]

Plug (4.1) into

[TABLE]

compare the coefficients of each summands respectively, we have:

[TABLE]

[TABLE]

[TABLE]

(99) implies:

[TABLE]

By induction on (100), we get:

[TABLE]

By induction on (101):

[TABLE]

Associate (102) (103) (104), we get:

[TABLE]

Plug (4.1) into

[TABLE]

compare the coefficients of each summands respectively, we have:

[TABLE]

from this we know:

[TABLE]

this implies:

[TABLE]

Plug (4.1) into

[TABLE]

compare the coefficients of each summands respectively, we have:

[TABLE]

Plug (4.1) into

[TABLE]

compare the coefficients of each summands respectively, together together with (109), we have:

[TABLE]

Associate (92), (98), (105), (108), (109), (100), we have:

[TABLE]

From (111) and C4 of the definition 4.2, we get:

[TABLE]

Case $2$  $\theta(L_{n})$ has the form $(b)$ in 4.5.

Similar arguments as in case $1$, we have the following results:

[TABLE]

In which

[TABLE]

∎

From the discussion above, we have the following theorem:

Theorem 4.6**.**

The conjugate linear anti-involution of $\mathfrak{L}$ must lies in one of the following forms:

1. (1)

Given by (111) and (112). 2. (2)

Given by (113) and (114).

4.2 Irreducible intermediate series unitary module of $\mathfrak{L}$

The unitary representation of the Virasoro algebra, owns the following already known results: (To be concrete, one can see [1]).

Theorem 4.7**.**

The irreducible unitary module of the Virasoro algebra is either the highest or lowest weight module, or isomorphism to the intermediate series module $A_{a,b}$. In which $a\in\mathbb{R},b\in\frac{1}{2}+\rm{i}\mathbb{R}$.

From the discussion of the irreducible intermediate series module of $\mathfrak{L}$ discussed in the previous subsection: the irreducible intermediate series module $M$ of $\mathfrak{L}$ must also be the irreducible intermediate series module of the Virasoro algebra, in other word, $\forall k\in\mathbb{Z}$, $I_{k},G_{k},H_{k}$ act trivially on $M$. From the discussion above, we have the following theorem:

Theorem 4.8**.**

The irreducible intermediate series unitary module of $\mathfrak{L}$ is isomorphism to the irreducible intermediate series module of form $A_{a,b,0,0,0}$, in which $a\in\mathbb{R},b\in\frac{1}{2}+\rm{i}\mathbb{R}$.