Conformal biderivations of loop $W(a,b)$ Lie conformal algebra

TL;DR

This paper investigates conformal biderivations within specific Lie conformal algebras, providing classifications and revealing that all biderivations on the Virasoro algebra are inner, advancing understanding of their algebraic structure.

Contribution

It classifies conformal biderivations for loop W(a,b), loop Virasoro, and Virasoro Lie conformal algebras, showing all on Virasoro are inner, which is a novel result.

Findings

All conformal biderivations on Virasoro Lie conformal algebra are inner.

Explicit descriptions of conformal biderivations for loop W(a,b) and loop Virasoro algebras.

Classification results contribute to the structural theory of Lie conformal algebras.

Abstract

In this paper, we study conformal biderivations of a Lie conformal algebra. First, we give the definition of conformal biderivation. Next, we determine the conformal biderivations of loop $W(a,b)$ Lie conformal algebra, loop Virasoro Lie conformal algebra and Virasoro Lie conformal algebra. Especially, all conformal biderivations on Virasoro Lie conformal algebra are inner conformal biderivations.

Equations

$$[{L_i}_{\lambda }{L}_j]=(\partial +2\lambda )L_{i+j},[{L_i}_{\lambda }{G}_j]=(\partial +(1-b)\lambda )G_{i+j},$$

$$[{G_i}_{\lambda }{L}_j]=-(b\partial +(b-1)\lambda )G_{i+j},[{G_i}_{\lambda }{G}_j]=0.$$

$$[L_{\lambda }L]=(\partial +2\lambda )L.$$

$$[{L_i}_{\lambda }{L}_j]=(\partial +2\lambda )L_{i+j}.$$

$$[{L_i}_{\lambda }{L}_j]=(\partial +2\lambda )L_{i+j},[{L_i}_{\lambda }{G}_j]=(\partial +(1-b)\lambda )G_{i+j},$$

$$[{G_i}_{\lambda }{L}_j]=-(b\partial +(b-1)\lambda )G_{i+j},[{G_i}_{\lambda }{G}_j]=0.$$

$${\varphi }_{\lambda }(x,y)=-{\varphi }_{-\partial -\lambda }(y,x);$$

$${\varphi }_{\lambda }(x,[y_{\mu }z])=[({\varphi }_{\lambda }(x,y){)}_{\lambda +\mu }z]+[y_{\mu }{\varphi }_{\lambda }(x,z)].$$

$${\varphi }_{\lambda +\mu }([x_{\mu }y],z)=[x_{\mu }{\varphi }_{\lambda }(y,z)]-[y_{\lambda }{\varphi }_{\mu }(x,z)].$$

$$-{\varphi }_{-\partial -\lambda -\mu }(z,[x_{\mu }y])=[x_{\mu }{\varphi }_{\lambda }(y,z)]-[y_{\lambda }{\varphi }_{\mu }(x,z)].$$

$$-{\varphi }_{-\partial -\lambda -\mu }(x,[y_{\mu }z])=[y_{\mu }{\varphi }_{\lambda }(z,x)]-[z_{\lambda }{\varphi }_{\mu }(y,x)].$$

$$-{\varphi }_{-\partial -\lambda -\mu }(x,[y_{\mu }z])=-[y_{\mu }{\varphi }_{-\partial -\lambda }(x,z)]+[z_{\lambda }{\varphi }_{-\partial -\mu }(x,y)].$$

$${\varphi }_{\lambda }(x,[y_{\mu }z])=[({\varphi }_{\lambda }(x,y){)}_{\lambda +\mu }z]+[y_{\mu }{\varphi }_{\lambda }(x,z)]$$

$$[({\varphi }_{\mu }(x,y){)}_{\mu +\gamma }[u_{\lambda }v]]=[[x_{\mu }y{]}_{\mu +\gamma }{\varphi }_{\lambda }(u,v)]$$

$${\varphi }_{\lambda +\mu }([x_{\mu }u],[y_{\gamma }v])$$

$$[({\varphi }_{\lambda }(u,y){)}_{\lambda +\gamma }[x_{\mu }v]]+[[x_{\mu }y{]}_{\mu +\gamma }{\varphi }_{\lambda }(u,v)]=[({\varphi }_{\mu }(x,y){)}_{\mu +\gamma }[u_{\lambda }v]]+[[u_{\lambda }y{]}_{\lambda +\gamma }{\varphi }_{\mu }(x,v)],$$

$$[({\varphi }_{\mu }(x,y){)}_{\mu +\gamma }[u_{\lambda }v]]-[[x_{\mu }y{]}_{\mu +\gamma }{\varphi }_{\lambda }(u,v)]=[({\varphi }_{\lambda }(u,y){)}_{\lambda +\gamma }[x_{\mu }v]]-[[u_{\lambda }y{]}_{\lambda +\gamma }{\varphi }_{\mu }(x,v)].$$

$${\Phi }_{\lambda ,\mu ,\gamma }(x,y;u,v)={\Phi }_{\mu ,\lambda ,\gamma }(u,y;x,v).$$

$${\Phi }_{\lambda ,\mu ,\gamma }(x,y;u,v)$$

$${\Phi }_{\lambda ,\mu ,\gamma }(x,y;u,v)=-{\Phi }_{\lambda ,\gamma ,\mu }(y,x;u,v)=-{\Phi }_{\gamma ,\lambda ,\mu }(u,x;y,v)={\Phi }_{\gamma ,\mu ,\lambda }(x,u;y,v).$$

$${\Phi }_{\lambda ,\mu ,\gamma }(x,y;u,v)$$

$${\Phi }_{\lambda ,\mu ,\gamma }(x,y;u,v)=0,$$

$${\varphi }_{\lambda }(L_i,L_j)=(\partial +2\lambda )\sum _{k\in \mathbb{Z}}{a}_{k-i-j}{L}_k,$$

$$[({\varphi }_{\mu }(L_i,L_j){)}_{\mu +\gamma }[{L_i}_{\lambda }{L}_j]]=[[{L_i}_{\mu }{L}_j{]}_{\mu +\gamma }{\varphi }_{\lambda }(L_i,L_j)],$$

$$(\partial +\mu +\gamma +2\lambda )\sum _{k\in \mathbb{Z}}{f}_{i,j}^k(-\mu -\gamma ,\mu )L_{i+j+k}=(\mu -\gamma )\sum _{k\in \mathbb{Z}}{f}_{i,j}^k(\partial +\mu +\gamma ,\lambda )L_{i+j+k},$$

$$(\partial +\mu +\gamma +2\lambda )f_{i,j}^k(-\mu -\gamma ,\mu )=(\mu -\gamma )f_{i,j}^k(\partial +\mu +\gamma ,\lambda ),$$

$$(\partial +\mu +\gamma +2\lambda )(a_0(\mu )+a_1(\mu )(-\mu -\gamma ))=(\mu -\gamma )(a_0(\lambda )+a_1(\lambda )(\partial +\mu +\gamma )).$$

$$a_0(\mu )+a_1(\mu )(-\mu -\gamma )=a_1(\lambda )(\mu -\gamma ),$$

$${\varphi }_{\lambda }(L_i,L_j)=(\partial +2\lambda )\sum _{k\in \mathbb{Z}}{a}_{k-i-j}{L}_k.$$

$${\varphi }_{\lambda }(L_i,L_j)=(\partial +2\lambda )(\sum _{k\in \mathbb{Z}}{a}_{k-i-j}{L}_k+{\delta }_{b+1,0}\sum _{k\in \mathbb{Z}}{b}_{k-i-j}{G}_k),$$

$${\varphi }_{\lambda }(G_i,G_j)=0,{\varphi }_{\lambda }(L_i,G_j)=(\partial +(1-b)\lambda )\sum _{k\in \mathbb{Z}}{a}_{k-i-j}{G}_k,$$

Full text

Key words: Lie conformal algebras, conformal biderivations, Virasoro Lie conformal algebra, loop Virasoro Lie conformal algebra, loop $W(a,b)$ Lie conformal algebra

Mathematics Subject Classification(2010): 16S70, 17A42, 17B10, 17B56, 17B70

000 Corresponding author(L. Chen): chenly640@nenu.edu.cn.000Supported by NNSF of China (Nos. 11771069 and 11301109), NNSF of Jilin province (No. 20170101048JC) and the project of jilin province department of education (No. JJKH20180005K).

1 Introduction

The notion of Lie conformal algebras was introduced by V. G. Kac as a formal language describing the singular part of the operator product expansion in conformal field theory. It is useful to research infinite dimensional Lie algebras satisfying the locality property. The structure theory and representation theory of some Lie conformal algebras have been extensively studied in [1, 6].

In recent years, biderivations have been aroused many scholar’s great interests. In [2, 3], Bre$\check{s}$ar et al. showed that all biderivations on commutative prime rings are inner biderivations and they determined the biderivations of semiprime rings. In [4], Zhao Kaiming proved all skew-symmetric biderivations on a perfect and centerless Lie algebra are inner biderivations. In [5, 8, 9, 10, 11, 12], authors give biderivations of specific examples of Lie algebras.

The main object investigated in this paper is the loop $W(a,b)$ Lie conformal algebra, denoted by $\rm CLW(a,b)$, which is a free $\mathbb{C}[\partial]$-module $\rm{CLW(a,b)}=(\bigoplus_{i\in\mathbb{Z}}\mathbb{C}[\partial]L_{i})\oplus(\bigoplus_{i\in\mathbb{Z}}\mathbb{C}[\partial]G_{i})$ with a $\mathbb{C}[\partial]$-basis $\{L_{i},G_{i}|i\in\mathbb{Z}\}$ satisfying the following $\lambda$-brackets:

[TABLE]

The relations between $W(a,b)$ and $\rm CLW(a,b)$ can be found in [7].

This paper is organized as follows. First, we give the definitions of conformal biderivation and inner conformal biderivation on a Lie conformal algebra. Next, we determine the conformal biderivations of loop $W(a,b)$ Lie conformal algebra, loop Virasoro Lie conformal algebra and Virasoro Lie conformal algebra. Especially, all conformal biderivations on Virasoro Lie conformal algebra are inner conformal biderivations.

Throughout this paper, all vector spaces, linear maps, and tensor products are over the complex field $\mathbb{C}$. In addition to the standard notations $\mathbb{Z}$ and $\mathbb{R}$, we use $\mathbb{Z}_{\geq 0}$ to denote the set of nonnegative integers.

2 Conformal biderivations of a Lie conformal algebra

The following notion was due to [1].

Definition 2.1**.**

A Lie conformal algebra is a $\mathbb{C}[\partial]$-module $\mathcal{R}$ endowed with a bilinear map $\mathcal{R}\times\mathcal{R}\rightarrow\mathcal{R}[\lambda],(a,b)\mapsto[a_{\lambda}b]$, called the $\lambda$-bracket, satisfying the following axioms ($x,y,z\in\mathcal{R}$):

1. $(1)$

Conformal sesquilinearity:$[(\partial x)_{\lambda}y]=-\lambda[x_{\lambda}y];$ 2. $(2)$

Skew-symmetry: $[x_{\lambda}y]=-[y_{-\partial-\lambda}x];$ 3. $(3)$

Jacobi identity: $[x_{\lambda}[y_{\mu}z]]=[[x_{\lambda}y]_{\lambda+\mu}z]+[y_{\mu}[x_{\lambda}z]].$

Example 2.2**.**

[6]Virasoro Lie conformal algebra $\rm Vir$ is a free $\mathbb{C}[\partial]$-module $\rm{Vir}=\mathbb{C}[\partial]L$, satisfying the following $\lambda$-bracket:

[TABLE]

Example 2.3**.**

[13] loop Virasoro Lie conformal algebra $\rm CW$ is a free $\mathbb{C}[\partial]$-module $\rm{CW}=\bigoplus_{i\in\mathbb{Z}}\mathbb{C}[\partial]L_{i}$ satisfying the following $\lambda$-brackets:

[TABLE]

Example 2.4**.**

[7] loop $W(a,b)$ Lie conformal algebra $\rm CLW(a,b)$ is a free $\mathbb{C}[\partial]$-module $\rm{CLW(a,b)}=(\bigoplus_{i\in\mathbb{Z}}\mathbb{C}[\partial]L_{i})\oplus(\bigoplus_{i\in\mathbb{Z}}\mathbb{C}[\partial]G_{i})$, satisfying the following $\lambda$-brackets:

[TABLE]

Definition 2.5**.**

[6] Let $U$ and $V$ be two ${\rm C}[\partial]$-modules. A conformal linear map from $U$ to $V$ is a $\mathbb{C}$-linear map $f:U\rightarrow V[\lambda],$ denoted by $f_{\lambda}:U\rightarrow V[\lambda]$, such that $f_{\lambda}(\partial x)=(\partial+\lambda)f_{\lambda}(x)$.

Moreover, Let $W$ also be a ${\rm C}[\partial]$-module. A conformal bilinear map from $U\times V$ to $W$ is a $\mathbb{C}$-bilinear map $f:U\times V\rightarrow W[\lambda]$, denoted by $f_{\lambda}:U\times V\rightarrow W[\lambda]$, such that $f_{\lambda}(\partial x,y)=-\lambda f_{\lambda}(x,y)$ and $f_{\lambda}(x,\partial y)=(\partial+\lambda)f_{\lambda}(x,y)$.

Definition 2.6**.**

Let $\mathcal{R}$ be a Lie conformal algebra. We call a conformal bilinear map $\phi_{\lambda}:\mathcal{R}\times\mathcal{R}\rightarrow\mathcal{R}[\lambda]$ skew-symmetric if it satisfies $\phi_{\lambda}(x,y)=-\phi_{-\partial-\lambda}(y,x)$.

Definition 2.7**.**

Let $\mathcal{R}$ be a Lie conformal algebra. We call a conformal bilinear map $\phi_{\lambda}:\mathcal{R}\times\mathcal{R}\rightarrow\mathcal{R}[\lambda]$ a conformal biderivation of $\mathcal{R}$ if it satisfies the following equations:

[TABLE]

Remark 2.8**.**

If $\phi_{\lambda}$ is a conformal biderivation of Lie conformal algebra $\mathcal{R}$, then $\phi_{\lambda}(x,\cdot)$ is a conformal derivation obviously.

Lemma 2.9**.**

Let $\phi_{\lambda}$ be a conformal biderivation of Lie conformal algebra $\mathcal{R}$. Then Eq.(2.2) is equivalent to

[TABLE]

Proof. .

If Eq.(2.3) satisfies, by conformal sesquilinearity, we have

[TABLE]

Replace $z,x,y$ by $x,y,z$, we obtain

[TABLE]

By conformal sesquilinearity, we get

[TABLE]

Replace $\lambda,\mu$ by $-\partial-\lambda-\mu,\mu$ respectively and by conformal sesquilinearity,

[TABLE]

satisfies. The reverse conclusion follows similarly. ∎

Definition 2.10**.**

Denote by ${\rm BDer}(\mathcal{R})$ the set of all conformal biderivations of $\mathcal{R}$.

Lemma 2.11**.**

If the map $\phi^{t}_{\lambda}:\mathcal{R}\times\mathcal{R}\rightarrow\mathcal{R}[\lambda]$, defined by $\phi^{t}_{\lambda}(x,y)=t[x_{\lambda}y]$ for all $x,y\in\mathcal{R}$, where $t\in\mathbb{C}$, then $\phi^{t}$ is a conformal biderivation of $\mathcal{R}$. We call this class conformal biderivations by inner conformal biderivations.

Proof. .

It is straightforward by the definition of conformal biderivation. ∎

Lemma 2.12**.**

Let $\phi_{\lambda}$ be a conformal biderivation of Lie conformal algebra $\mathcal{R}$. Then

[TABLE]

for any $x,y,u,v\in\mathcal{R}$.

Proof. .

On the one hand, using Eq.(2.3), we have

[TABLE]

On the other hand, using Eq.(2.2), we have

[TABLE]

Comparing two sides of the above equations, and using the Jacobi identity of Lie conformal algebra, we obtain

[TABLE]

it is equivalent to

[TABLE]

Now let $\Phi_{\lambda,\mu,\gamma}(x,y;u,v)=[(\phi_{\mu}(x,y))_{\mu+\gamma}[u_{\lambda}v]]-[[x_{\mu}y]_{\mu+\gamma}\phi_{\lambda}(u,v)]$. From the above equation, it follows at once that

[TABLE]

Obviously, by skew-symmetry and conformal sesquilinearity , we get

[TABLE]

For one thing, we have

[TABLE]

For another, we also have

[TABLE]

Then

[TABLE]

that is, $[(\phi_{\mu}(x,y))_{\mu+\gamma}[u_{\lambda}v])=[[x_{\mu}y]_{\mu+\gamma}\phi_{\lambda}(u,v)].$ ∎

Remark 2.13**.**

Let $\phi_{\lambda}$ be a conformal biderivation of Lie conformal algebra $\mathcal{R}$. If $[x_{\lambda}y]=0,$ then $\phi_{\lambda}(x,y)\in{\rm Z}[\mathcal{R}_{\lambda}\mathcal{R}]$.

3 Conformal biderivations of Lie conformal algebras $\rm CW$, $\rm Vir$ and $\rm CLW(a,b)$

Theorem 3.1**.**

Every conformal biderivation $\phi_{\lambda}$ on the loop Virasoro Lie conformal algebra $CW$ has the following forms:

[TABLE]

for some complex numbers set $\{a_{k},k\in\mathbb{Z}\}$.

Proof. .

Suppose that $\phi_{\lambda}(L_{i},L_{j})=\sum_{k\in\mathbb{Z}}f^{k}_{i,j}(\partial,\lambda)L_{k}$ is a conformal biderivation of the loop Virasoro Lie conformal algebra $\rm CW$.

Take $x=L_{i},y=L_{j},u=L_{i},v=L_{j}$ in Eq.(2.4), we get

[TABLE]

that is

[TABLE]

obviously, we have

[TABLE]

for any $i,j,k\in\mathbb{Z}$. Fix $i,j,k\in\mathbb{Z}$, considering the power of $\partial$, we have $f^{k}_{i,j}(\partial,\lambda)=a_{0}(\lambda)+a_{1}(\lambda)\partial$, substituting into the above equation, we get

[TABLE]

Considering the coefficients of $\partial$, we have

[TABLE]

it is easy to get $a_{1}(\lambda)=a_{1}\in\mathbb{C},a_{0}(\lambda)=2\lambda a_{1}$, so $f^{k}_{i,j}(\partial,\lambda)=a_{1}(\partial+2\lambda).$ So $\phi_{\lambda}(L_{i},L_{j})=(\partial+2\lambda)\sum_{k\in\mathbb{Z}}f^{k}_{i,j}L_{k}$, where $f^{k}_{i,j}\in\mathbb{C}$.

Take $x=L_{i},y=L_{j},u=L_{k},v=L_{l}$ in Eq.(2.4) again, we can get $f^{k}_{i,j}=f^{k+m+n-i-j}_{m,n},$ especially, $f^{k}_{0,0}=f^{k+m+n}_{m,n}$. Set $f^{k}_{0,0}:=a_{k}$, i.e. $\phi_{\mu}(L_{0},L_{0})=(\partial+2\lambda)\sum_{k\in\mathbb{Z}}a_{k}L_{k}$. Then

[TABLE]

Conversely, it is easy to prove that any skew-symmetric conformal bilinear map satisfying Eq.(3.1) is a conformal biderivation. ∎

Corollary 3.2**.**

Every conformal biderivation on the Virasoro Lie conformal algebra is an inner conformal biderivation.

Proof. .

By the proof of Theorem 3.1, we can get the conclusion immediately. ∎

Theorem 3.3**.**

Let $\mathcal{R}$ be the Lie conformal algebra $\rm CLW(a,b)$. Then a skew-symmetric conformal bilinear map $\phi_{\lambda}$ of $\mathcal{R}$ is a conformal biderivation if and only if $\phi_{\lambda}$ satisfies the following conditions:

[TABLE]

for some two complex numbers sets $\{a_{k},k\in\mathbb{Z}\}$ and $\{b_{k},k\in\mathbb{Z}\}$.

Proof. .

Let $\phi_{\lambda}$ be a conformal biderivation of Lie conformal algebra $\mathcal{R}$. By Remark (2.13), $\phi_{\lambda}(G_{i},G_{j})=0$ obviously. Suppose that

[TABLE]

Firstly, take $x=L_{i},y=L_{j},u=L_{i},v=G_{j}$ in Eq.(2.4), we get

[TABLE]

obviously, we have $s^{k}_{i,j}(\partial,\lambda)=0$ and

[TABLE]

for any $i,j,k\in\mathbb{Z}$. Considering the coefficients of $\partial$, we have $d^{k}_{i,j}(\partial,\lambda)=a_{0}(\lambda)+a_{1}(\lambda)\partial$, substituting into the above equation, we get

[TABLE]

Considering the coefficients of $\gamma$ again, we get $f^{k}_{i,j}(\partial,\lambda)=b_{0}(\lambda)+b_{1}(\lambda)\partial$, substituting into the above equation, we get

[TABLE]

observing the coefficients of $\gamma^{2}$, we have $b_{1}(\mu)=a_{1}(\lambda)=a_{1}\in\mathbb{C}$, substituting into the above equation, we get

[TABLE]

So we can set

[TABLE]

where $d^{k}_{i,j}\in\mathbb{C}$.

Secondly, take $x=L_{i},y=L_{j},u=L_{i},v=L_{j}$ in Eq.(2.4), we get

[TABLE]

it is easy to see that

[TABLE]

for any $i,j,k\in\mathbb{Z}$. Considering the power of $\partial$, we have $g^{k}_{i,j}(\partial,\lambda)=c_{0}(\lambda)+c_{1}(\lambda)\partial$, substituting into the above equation, we get

[TABLE]

Considering the coefficients of $\gamma^{3}$, we have

[TABLE]

$(i)$ If $b=1$, substituting in Eq.(3.2), we get

[TABLE]

observing the coefficients of $\partial$ in the above equation, we have $c_{0}(\lambda)=c_{1}(\lambda)=0$.

$(ii)$ If $b\neq 1$, by Eq.(3.3), we have $c_{1}(\lambda)=c_{1}\in\mathbb{C}$. Substituting in Eq.(3.2), we get

[TABLE]

let’s just think about the monomials only containing $\mu$, that is

[TABLE]

so we obtain $c_{0}(\mu)=2c_{1}\mu$. Substituting in Eq.(3.2) again, we get

[TABLE]

that is $c_{1}(b+1)=0$. Especially, if $b\neq-1$, then $c_{1}=0$.

From what has been discussed above, if

$(1)$ $b\neq-1$, we have

[TABLE]

$(2)$ $b=-1$, we have

[TABLE]

Finally, take $x=L_{i},y=L_{j},u=L_{k},v=L_{l}$ in Eq.(2.4), we can get $d^{k}_{i,j}=d^{k+m+n-i-j}_{m,n}$ and $g^{k}_{i,j}=g^{k+m+n-i-j}_{m,n}$. Especially, $d^{k}_{0,0}=d^{k+m+n}_{m,n}$ and $g^{k}_{0,0}=g^{k+m+n}_{m,n}$. Set $d^{k}_{0,0}:=a_{k},g^{k}_{0,0}:=b_{k}$, i.e. $\phi_{\lambda}(L_{0},L_{0})=(\partial+2\lambda)(\sum_{k\in\mathbb{Z}}a_{k}L_{k}+\delta_{b+1,0}\sum_{k\in\mathbb{Z}}b_{k}G_{k})$. Then

[TABLE]

Conversely, it is easy to prove that any skew-symmetric conformal bilinear map satisfying Eqs.(3.4), (3.5) and (3.6) is a conformal biderivation. ∎