Structure of a class of Lie conformal algebras of Block type

TL;DR

This paper investigates the structure of a class of infinite rank Lie conformal algebras of Block type, focusing on their derivations, biderivations, and cohomologies, thus advancing understanding of their algebraic properties.

Contribution

It provides a complete analysis of conformal derivations, biderivations, and specific cohomologies for the Lie conformal algebras (p), a class recently introduced.

Findings

Determined all conformal derivations of (p).

Characterized conformal biderivations of (p).

Computed certain second cohomology groups of (p).

Abstract

Let $p$ be a nonzero complex number. Recently, a class of infinite rank Lie conformal algebras $\mathfrak{B}(p)$ was introduced in [13]. In this paper, we study the structure theory of this class of Lie conformal algebras. Specifically, we completely determine the conformal derivations, the conformal biderivations and certain second cohomologies of $\mathfrak{B}(p)$.

Equations

$$[\partial a_{\lambda }b]=-\lambda [a_{\lambda }b],[a_{\lambda }\partial b]=(\partial +\lambda )[a_{\lambda }b](\mathrm{conformal}\mathrm{sesquilinearity}),$$

$$[a_{\lambda }b]=-[b_{-\lambda -\partial }a](\mathrm{skew}-\mathrm{commutativity}),$$

$$[a_{\lambda }[b_{\mu }c]]=[[a_{\lambda }b{]}_{\lambda +\mu }c]+[b_{\mu }[a_{\lambda }c]](\mathrm{Jacobi}\mathrm{identity}),$$

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

$$Vir={\mathrm{span}}_{\mathbb{C}[\partial ]}\{p^{-1}{L}_0\},$$

$$\mathcal{H}\mathcal{V}={\mathrm{span}}_{\mathbb{C}[\partial ]}\{p^{-1}{L}_0,L_{-p}\},$$

$$[a_{(m)},b_{(n)}]=\sum_{k\in\mathbb{\mathbb{Z}^{+}}}\left(\!\!\begin{array}[]{ll}m\\ k\end{array}\!\!\right)(a_{(k)}b)_{(m+n-k)},\ \ (\partial a)_{(n)}=-na_{(n-1)},$$

$$[L_{i,m},L_{j,n}]=((j+p)(m+1)-(i+p)(n+1))L_{i+j,m+n}.$$

$$D_{\lambda }(\partial a)=(\partial +\lambda )(D_{\lambda }a),$$

$$D_{\lambda }[a_{\mu }b]=[(D_{\lambda }a{)}_{\lambda +\mu }b]+[a_{\mu }(D_{\lambda }b)],$$

$$(\mathbf{ad}a{)}_{\lambda }b=[a_{\lambda }b],b\in R.$$

$$D_{\lambda }(L_0)=\sum _{j\in {\mathbb{Z}}^{+}}{a}_j(\lambda ,\partial )L_j$$

$$\displaystyle{}\begin{array}[]{lll}&p(\partial+\lambda+2\mu)a_{j}(\lambda,\partial)\vspace{8pt}\\ =\!\!\!\!&((j+p)\partial+(j+2p)(\lambda+\mu))a_{j}(\lambda,-\lambda-\mu)+(p\partial+(j+2p)\mu)a_{j}(\lambda,\partial+\mu).\end{array}$$

$$(j+p)c_{j,0}(\lambda )=(j+2p)\lambda c_{j,1}(\lambda ).$$

$${\bar{D}}_{\lambda }(L_0)=0.$$

$$[L_0{}_{\mu }{\bar{D}}_{\lambda }(L_j)]=(p(\partial +\lambda )+(j+2p)\mu ){\bar{D}}_{\lambda }(L_j).$$

$$(p\partial +(k+2p)\mu )e_{j,k}(\lambda ,\partial +\mu )=(p(\partial +\lambda )+(j+2p)\mu )e_{j,k}(\lambda ,\partial ).$$

$$D_{\lambda }(L_j)=(\mathbf{ad}\gamma {)}_{\lambda }(L_j),j\in {\mathbb{Z}}^{+}.$$

$$({\tilde{D}}_{\lambda }(L_0)=c{L}_{-p},$$

$$(p(\partial +\lambda )+(j+2p)\mu )d_{j,l}(\lambda ,\partial )$$

$$D_{\lambda }^p(L_j)=(j+p)L_{-p+j},j\in {\mathbb{Z}}^{+},p\in {\mathbb{Z}}^{-}.$$

$$D_{\lambda }(L_j)=(\mathbf{ad}(\bar{\gamma }+\tilde{\gamma }){)}_{\lambda }(L_j)+p^{-1}c{D}_{\lambda }^p(L_j),j\in {\mathbb{Z}}^{+},p\in {\mathbb{Z}}^{-}.$$

$$\displaystyle{\rm Der}(\mathfrak{B}(p))=\left\{\begin{array}[]{lllllllll}{\rm Inn}(\mathfrak{B}(p))&{\rm if}\ \ p\notin{\mathbb{Z}^{-}},\vspace{5pt}\\ {\rm Inn}(\mathfrak{B}(p))\oplus\mathbb{C}D_{\lambda}^{p}&{\rm if}\ \ p\in\mathbb{Z}^{-}.\end{array}\right.$$

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

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

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

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

$$d_{\lambda }^c(x,y)=c[x_{\lambda }y].$$

$$d_{\lambda }(L_i,L_j)=\sum _{k\in {\mathbb{Z}}^{+}}{f}_{i,j}^k(\partial ,\lambda )L_k$$

$$\displaystyle\begin{array}[]{ll}&(\overline{m}(\partial+\mu+\gamma)+(\overline{m}+\overline{n})\lambda)\displaystyle\sum_{k\in\mathbb{Z}^{+}}(\overline{k}\partial+(\overline{m}+\overline{n}+k)(\mu+\gamma))f_{i,j}^{k}(-\mu-\gamma,\mu)L_{k+m+n}\vspace{8pt}\\ =\!\!\!\!&(\overline{j}\mu-\overline{i}\gamma)\displaystyle\sum_{k\in\mathbb{Z}^{+}}(\overline{i+j}\partial+(\overline{i}+\overline{j}+k)(\mu+\gamma))f_{m,n}^{k}(\partial+\mu+\gamma,\lambda)L_{k+i+j},\end{array}$$

$$\displaystyle\begin{array}[]{ll}&(\overline{i}(\partial+\mu+\gamma)+(\overline{i}+\overline{j})\lambda)(\overline{k}\partial+(\overline{i}+\overline{j}+k)(\mu+\gamma))f_{i,j}^{k}(-\mu-\gamma,\mu)\vspace{8pt}\\ =\!\!\!\!&(\overline{j}\mu-\overline{i}\gamma)(\overline{i+j}\partial+(\overline{i}+\overline{j}+k)(\mu+\gamma))f_{i,j}^{k}(\partial+\mu+\gamma,\lambda).\end{array}$$

$$d_{\lambda }(L_i,L_j)=d_{\lambda }^c(L_i,L_j)$$

Full text

[TABLE]

for any $a,b,c\in R$. If there exists a finite generating subset $S\subset R$ such that $S$ generates $R$ as a $\mathbb{C}[\partial]$-module, then we call that $R$ is a finite rank Lie conformal algebra. Otherwise, it is called infinite.

What we mainly consider in this paper is a class of infinite Lie conformal algebras $\mathfrak{B}(p)=\oplus_{i\in\mathbb{Z}^{+}}\mathbb{C}[\partial]L_{i}$ with $p$ being a nonzero complex number, satisfying the following $\lambda$-brackets:

[TABLE]

The Lie conformal algebra $\mathfrak{B}(p)$ contains a Lie conformal subalgebra

[TABLE]

which is isomorphic to the Virasoro Lie conformal algebra [5]. Moreover, if $p$ is a negative integer, then $\mathfrak{B}(p)$ has another Lie conformal subalgebra

[TABLE]

which is isomorphic to Heisenberg-Virasoro conformal algebra [17]. In particular, the case $\mathfrak{B}(1)$ is a maximal subalgebra of the associated graded conformal algebra ${\rm gr}gc_{1}$ of the filtered algebra $gc_{1}$ [14]. Besides, for any integer $n\geq 1$, $\mathfrak{B}(-n)$ contains a series of finite Lie conformal quotient algebras (see [13], Section 2.2).

[TABLE]

where $a_{(k)}b$ is called the $k$-product, given by $[a_{\lambda}b]=\sum_{k\in\mathbb{Z}^{+}}\frac{\lambda^{k}}{k!}a_{(k)}b$.

Recall from [13] that the annihilation algebra $\mathcal{A}(\mathfrak{B}(p))$ of $\mathfrak{B}(p)$ is spanned by $\{L_{i,m}\,|\,i\in\mathbb{Z}^{+},m\in\mathbb{Z}_{\geq-1}\}$ with relations

[TABLE]

The Lie algebra $\mathcal{A}(\mathfrak{B}(p))$ is related to the Lie algebra $\mathcal{B}(q)$ of Block type [11, 12], Hence, this Lie conformal algebra $\mathfrak{B}(p)$ is called a Lie conformal algebra of Block type[13].

**3.  Conformal derivations of $\mathfrak{B}(p)$ **

Recall [5] that a $\mathbb{C}$-linear map $D_{\lambda}:\mathfrak{B}(p)\rightarrow\mathfrak{B}(p)[\lambda]$ is called a conformal derivation of $\mathfrak{B}(p)$ if

[TABLE]

for any $a,b\in\mathfrak{B}(p)$. Denote by ${\rm Der}(\mathfrak{B}(p))$ the space of all conformal derivations. For any $a\in\mathfrak{B}(p)$, one can define a conformal derivation $({\rm\bf ad}a)_{\lambda}$ of $\mathfrak{B}(p)$ by

[TABLE]

All derivations of this kind are called inner. Denoted by ${\rm Inn}(\mathfrak{B}(p))$ the space of all inner derivations.

Let $D_{\lambda}\in{\rm Der}(\mathfrak{B}(p))$. Suppose that

[TABLE]

for some $a_{j}(\lambda,\partial)\in\mathbb{C}[\lambda,\partial]$.

Next, we shall discuss the coefficients $a_{j}(\lambda,\partial)$.

By applying $D_{\lambda}$ to $[{L_{0}}_{\mu}L_{0}]=p(\partial+2\mu)L_{0}$ and considering the coefficients of $L_{j}$, we can obtain

[TABLE]

Set $a_{j}(\lambda,\partial)=\sum_{k=0}^{m}c_{j,k}(\lambda)\partial^{k}$ with $c_{j,m}(\lambda)\neq 0$, where $c_{j,k}(\lambda)\in\mathbb{C}[\lambda]$. Assume that $m>1$. Then comparing the coefficients of $\partial^{m}$ in (3.3), one can get $(p\lambda-(j+mp)\mu)c_{j,m}(\lambda)=0$, which gives $c_{j,m}(\lambda)=0$ since $p\neq 0$. This contradicts with our assumption. Thus, one can set $a_{j}(\lambda,\partial)=c_{j,0}(\lambda)+c_{j,1}(\lambda)\partial$. By substituting this into the equation (3.3) and then comparing the coefficients of $\partial$, we can obtain

[TABLE]

(1) The case $p\notin\mathbb{Z}^{-}$.

In this case, from (3.4), we have $c_{j,0}(\lambda)=\frac{j+2p}{j+p}\lambda c_{j,1}(\lambda)$. Thus $a_{j}(\lambda,\partial)=(\partial+\frac{j+2p}{j+p}\lambda)c_{j,1}(\lambda)$. Setting $\beta_{j}=\frac{c_{j,1}(-\partial)}{j+p}L_{j}$, then we can check that $({\rm\bf ad}\beta_{j})_{\lambda}(L_{0})=a_{j}(\lambda,\partial)$. Hence, by taking $\gamma=\sum_{j\in\mathbb{Z}^{+}}\beta_{j}$ and letting $\bar{D}_{\lambda}=D_{\lambda}-({\rm\bf ad}\gamma)_{\lambda}$, we can obtain

[TABLE]

Now, by using $\bar{D}_{\lambda}(L_{0})=0$ and applying $\bar{D}_{\lambda}$ to $[L_{0}\,{}_{\mu}L_{j}]=(p\partial+(j+2p)\mu)L_{j}$ with $j>0$, one has

[TABLE]

Set $\bar{D}_{\lambda}(L_{j})=\sum_{k\in\mathbb{Z}^{+}}e_{j,k}(\lambda,\partial)L_{k}$ with $e_{j,k}\in\mathbb{C}[\lambda,\partial]$. Then considering the coefficients of $L_{k}$ in (3.6), one has

[TABLE]

Taking $\mu=0$ in the above equation, we have $p\lambda e_{j,k}(\lambda,\partial)=0$. Since $p\neq 0$, we have $e_{j,k}(\lambda,\partial)=0$ for any $k\in\mathbb{Z}^{+}$ and $0\neq j\in\mathbb{Z}^{+}$. Thus $\bar{D}_{\lambda}(L_{j})=0$ for $0\neq j\in\mathbb{Z}^{+}$. This togethers with (3.5) gives $\bar{D}_{\lambda}(L_{j})=0$ for any $j\in\mathbb{Z}^{+}$, namely,

[TABLE]

(2) The case $p\in\mathbb{Z}^{-}$.

In this case, it follows from (3.4) and $p\neq 0$ that $c_{-p,1}(\lambda)=0$. Thus, there exists some $c(\lambda)\in\mathbb{C}[\lambda]$ such that $a_{-p}(\lambda,\partial)=c(\lambda)$. Denote by $\bar{\gamma}=\sum_{-p\neq j\in\mathbb{Z}^{+}}\beta_{j}$, where $\beta_{j}=\frac{c_{j,1}(-\partial)}{j+p}L_{j}$. Then, from the same arguments as in Case (1), we can get $\bar{D}_{\lambda}(L_{0})=c(\lambda)L_{-p}$, where $\bar{D}_{\lambda}=D_{\lambda}-{(\rm\bf ad}\bar{\gamma})_{\lambda}$. Let $g(\lambda)=c(\lambda)-c(0)$, then $g(\lambda)$ divided by $\lambda$. Thus, from setting $\tilde{\gamma}=p^{-1}g(-\partial)(-\partial)^{-1}L_{-p}$, we can get

[TABLE]

where $\tilde{D}_{\lambda}=\bar{D}_{\lambda}-{(\rm\bf ad}\tilde{\gamma})_{\lambda}$, $c=c(0)$. Now, set $\tilde{D}_{\lambda}(L_{j})=\sum_{l\in\mathbb{Z}^{+}}d_{j,l}(\lambda,\partial)L_{l}$ with $d_{j,l}\in\mathbb{C}[\lambda,\partial]$ and $j>0$. Then, by applying $\tilde{D}_{\lambda}$ to $[L_{0}\,{}_{\mu}L_{j}]=(p\partial+(j+2p)\mu)L_{j}$ and comparing the coefficients of $L_{l}$, one has

[TABLE]

From discussions in Case (1), we get $d_{j,l}=0$ for $l\neq-p+j$. Now assume that $l=-p+j$. Then, taking $l=-p+j$ and $\mu=0$, we can get $d_{j,-p+j}(\lambda,\partial)=cp^{-1}(j+p)$. Thus, we obtain $\tilde{D}_{\lambda}(L_{j})=cp^{-1}(j+p)L_{-p+j}$ with $j>0$. Combining this with (3.8), we get $\tilde{D}_{\lambda}(L_{j})=cp^{-1}(j+p)L_{-p+j}$ for any $j\in\mathbb{Z^{+}}$. Now, Define a $\mathbb{C}$-linear map $D_{\lambda}^{p}$ as follows:

[TABLE]

One can check that $D_{\lambda}^{p}$ is a non-inner conformal derivation of $\mathfrak{B}(p)$. Thus, we obtain

[TABLE]

Hence, from (3.7) and (3.10), the following theorem is immediate.

[TABLE]

**4.  Conformal biderivations of $\mathfrak{B}(p)$ **

In the section, we shall study the conformal biderivations of $\mathfrak{B}(p)$. First we list some definitions introduced in [6, 18].

[TABLE]

Furthermore, we call $\phi_{\lambda}$ skew-symmetric if

[TABLE]

[TABLE]

One can see that if $d_{\lambda}$ is a conformal biderivation, then $d_{\lambda}(x,\cdot)$ is a conformal derivation of $\mathfrak{B}(p)$. From [18], we get that if $d_{\lambda}$ is a biderivation, then the identity

[TABLE]

holds for any $x,y,u,v\in\mathfrak{B}(p)$.

Fix a complex number $c$ and consider the map ${d^{c}_{\lambda}}:R\times R\rightarrow R[\lambda]$ defined by

[TABLE]

Then, one can immediately see that $d^{c}_{\lambda}$ is a conformal biderivation. We call the conformal biderivation $d_{\lambda}^{c}$ the inner conformal biderivation.

Let $d_{\lambda}$ be a conformal biderivation of $\mathfrak{B}(p)$, then one can set

[TABLE]

for some $f^{k}_{i,j}(\partial,\lambda)\in\mathbb{C}[\partial,\lambda]$.

In the following parts of this section, we shall describe the coefficients $f_{i,j}^{k}(\partial,\lambda)$.

Letting $x=L_{i},y=L_{j},u=L_{m},v=L_{n}$ in (4.3), we have

[TABLE]

where $\overline{x}=x+p$, $x\in\{i,j,m,n,i+j,k\}$

Letting $m=i,n=j$ and comparing the coefficients of $L_{k+i+j}$, we have

[TABLE]

[TABLE]

for any $i,j\in\mathbb{Z}^{+}$.

[TABLE]

By comparing the coefficients of $\gamma$ in (4.12), we have

[TABLE]

Then $c_{i,j}^{k,1}(\mu)=c_{i,j}^{k,1}(\lambda)$ for $k=i+j$, which forces $c_{i,j}^{k,1}(\mu)\in\mathbb{C}$. Thus we can write $c_{i,j}^{k,1}$ instead of $c_{i,j}^{k,1}(\mu)$. Using this in (4.12), we have $c_{i,j}^{k,0}(\mu)=(\overline{i})^{-1}(\overline{i}+\overline{j})c_{i,j}^{k,1}\mu$. If $k\neq i+j$, then from (4.13), we have $c_{i,j}^{k,1}(\mu)=0$. Together this with (4.12), we get $c_{i,j}^{k,0}(\mu)=0$. Hence, we can set

[TABLE]

for any $k\in\mathbb{Z}^{+}$, where $c_{i,j}=(\bar{i})^{-1}c_{i,j}^{k,1}\in\mathbb{C}$.

Letting $m=n=0$ in (4.8) and using (4.14), one can get

[TABLE]

Taking $\mu=-\gamma$ in (4.15), we have

[TABLE]

which gives $c_{i,j}=c_{0,0}$ since $p>0$. Thus, one can set $f_{i,j}^{k}(\partial,\mu)=\delta_{k,i+j}c(\overline{i}\partial+(\overline{i}+\overline{j})\mu)$ for any $i,j\in\mathbb{Z^{+}}$, where $c=c_{0,0}$. Hence, the lemma follows from (4.5). ∎

[TABLE]

for any $i,j\in\mathbb{Z}^{+}$.

(1) $\overline{i}\neq 0$, $\overline{i+j}\neq 0$. If $\overline{k}\neq 0$, then considering the coefficients of $\partial$ in (4.11) and using the same argument as in Lemma 4.4, we can obtain the equations (4.14)–(4.16). For $\overline{i}+\overline{j}\neq 0$, one can get $c_{i,j}=c_{0,0}$ from (4.16). For $\overline{i}+\overline{j}=0$, by taking $\mu=0$ in (4.15), we have $\overline{i}(\partial+\gamma+2\lambda)\overline{i+j}(c_{i,j}-c_{0,0})\gamma\partial=0$, which forces $c_{i,j}=c_{0,0}$. Thus $d_{\lambda}(L_{i},L_{j})=d_{\lambda}^{\bar{c}}(L_{i},L_{j})$, where $\bar{c}=c_{0,0}$. If $\overline{k}=0$, then considering the power of $\partial$ in (4.11), one can set $f_{i,j}^{-p}(\partial,\lambda)=c^{-p}_{i,j}(\lambda)$ for some $c^{-p}_{i,j}(\lambda)\in\mathbb{C}[\lambda]$. Using this in (4.11) and comparing the coefficients of $\gamma^{2}$, we have $\overline{i}(c^{-p}_{i,j}(\mu)+c^{-p}_{i,j}(\lambda))\overline{i+j}=0$, which forces $c^{-p}_{i,j}(\lambda)=0$ since $\overline{i},\overline{i+j}\neq 0$. This gives $f_{i,j}^{-p}(\partial,\lambda)=0$. Thus, in this case we have

[TABLE]

for some $\bar{c}\in\mathbb{C}$ and any $i,j\in\mathbb{Z}^{+}$.

(2) $\overline{i}\neq 0$, $\overline{i+j}=0$. In this case, if $\overline{k}\neq 0$, then from (4.11), one can set $f_{i,j}^{k}(\partial,\lambda)=e^{k,0}_{i,j}(\lambda)+e^{k,1}_{i,j}(\lambda)\partial+e^{k,2}_{i,j}(\lambda)\partial^{2}$. Using this in (4.11), we can get

[TABLE]

where $\bar{\partial}=\partial+\mu+\gamma$. Then comparing the coefficients of $\gamma^{4}$ in the above equation, we can get $\bar{i}\bar{k}(e^{k,2}_{i,j}(\mu)+e^{k,2}_{i,j}(\lambda)=0$, which forces $e^{k,2}_{i,j}(\lambda)=0$ since $\bar{i}\bar{k}\neq 0$. Using this and comparing the coefficients of $\gamma^{3}$ in (4.20), we can set $e^{k,1}_{i,j}(\mu)=e^{k,1}_{i,j}\in\mathbb{C}$, substituting into (4.20) and comparing the coefficients of $\partial^{2}$, one can deduce $e^{k,1}_{i,j}=e^{k,0}_{i,j}(\mu)=0$. Thus we have $f_{i,j}^{k}(\partial,\lambda)=\delta_{k,-p}f_{i,j}^{-p}(\partial,\lambda)$. By using this in (4.8) and take $m=n=0$, we can get $(\partial+\mu+\gamma+2\lambda)f_{i,j}^{-p}(-\mu-\gamma,\mu)(\mu+\gamma)=0$, which forces $f_{i,j}^{-p}(\partial,\mu)=0$. Thus, $f_{i,j}^{k}(\partial,\lambda)=0$ for any $i,j,k\in\mathbb{Z}^{+}$, namely,

[TABLE]

(3) $\overline{i}=0$, $\overline{j}\neq 0$. In this case, letting $\mu=-\gamma$ in (4.11) and using $\overline{j}\neq 0$, one can obtain

[TABLE]

Assume that $j\neq 0$. If $\overline{k}=0$, then $f_{-p,j}^{-p}(\partial,\lambda)=0$ from (4.22). If $\overline{k}\neq 0$, then from (4.22), we can safely set $f_{-p,j}^{k}(\partial,\lambda)=\delta_{k,-p+j}d_{-p,j}\lambda$ for some $d_{-p,j}\in\mathbb{C}$. This together with taking $i=m=-p$ in (4.8), then considering the coefficients of $\partial$, we have $j((n+p)d_{-p,j}-(j+p)d_{-p,n})\lambda\mu=0$ for any $0\neq n\in\mathbb{Z}^{+}$, which gives $d_{-p,j}=\tilde{c}(j+p)$ for some $\tilde{c}\in\mathbb{C}$ since $j\neq 0$. Hence,

[TABLE]

Now, by taking $i=-p$, $m=-p+1$ in (4.8) and using (4.14) and (4.23), then comparing the coefficients of $\partial^{2}$, we have $j(j+p)(\bar{c}-\tilde{c})\mu=0$ for any $0\neq j\in\mathbb{Z}^{+}$, which forces $\tilde{c}=\bar{c}$. Now suppose that $j=0$. Since $d_{\lambda}(L_{0},L_{-p})=0$ by (4.21), we have $d_{\lambda}(L_{-p},L_{0})=0$. Thus, in this case we have

[TABLE]

for any $j\in\mathbb{Z}^{+}$, where $\bar{c}$ is defined in (4.17).

(4) $\overline{i}=0$, $\overline{j}=0$. By using $[{L_{-p}}_{\lambda}{L_{-p}}]=0$ in (4.3), we get $d_{\lambda}(L_{-p},L_{-p})$ is in the center of $[\mathfrak{B}(p)_{\lambda}\mathfrak{B}(p)]$. Thus we have

[TABLE]

Hence, the lemma holds from (4.17), (4.21), (4.24) and (4.25). ∎

The following theorem follows immediately from Lemmas 4.4 and 4.5.

(1)* If $p\notin{\mathbb{Z}^{-}}$, then $d_{\lambda}$ is an inner conformal biderivation.*

(2)* If $p\in\mathbb{Z}^{-}$, then, for any $i,j\in\mathbb{Z}^{+}$, there exists $c\in\mathbb{C}$ such that*

[TABLE]

where $d_{\lambda}^{c}$ is defined in (4.4). Note that $p^{-1}L_{0}$ generates the Virasoro conformal algebra. Thus, every conformal biderivation on the Virasoro Lie conformal algebra is inner. Besides, if ${d}_{\lambda}$ is a conformal biderivation of $\mathcal{HV}$, then

[TABLE]

**5.  Second Cohomologies of $\mathfrak{B}(p)$ **

First, let us recall some definitions from [6].

[TABLE]

satisfying the following conditions:

[TABLE]

Denote by $C^{2}(A,M)$ the space of all $2$-$\lambda$-brackets on $A$ with coefficients in $M$. Recall from [6] that there is a differential $d$ on the space $C^{2}(A,M)$, and the equation $dc=0$ can be written as follows:

[TABLE]

for every $a,b,z\in A$.

Let $\phi:A\rightarrow M$ be a $\mathbb{C}[\partial]$-module homomorphism. Then it follows from [6] that exact elements $c=d\phi$ are of the form

[TABLE]

Hence, the second cohomology can be defined as follows.

[TABLE]

We say that the elements of $H^{2}(A,M)$ are $2$-cocycles, and call $c\in H^{2}(A,M)$ is trivial if there exists some $d\phi$ defined in (5.6) such that $c=d\phi$.

[TABLE]

Let $M$ be a nontrivial free conformal module of rank one over $\mathfrak{B}(p)$. One can see that the one-dimensional vector space $\mathbb{C}$ can be regarded as a module (called a trivial module) over $\mathfrak{B}(p)$ with the action of $\partial$,$L_{i}$ being zero. Recall from [13] that there are some $\Delta,\alpha,\beta\in\mathbb{C}$ such that

(1) If $p\neq-1$, then $M\cong M_{\Delta,\alpha}$, where

[TABLE]

(2) If $p=-1$, then $M\cong M_{\Delta,\alpha,\beta}$, where

[TABLE]

Furthermore, the module $M_{\Delta,\alpha}$ (resp., $M_{\Delta,\alpha,\beta}$) is irreducible if and only if $\Delta\neq 0$ (resp., $\Delta\neq 0$ or $\beta\neq 0$).

In the following sections, we shall study the second cohomologies $H^{2}(\mathfrak{B}(p),\mathbb{C})$, $H^{2}(\mathfrak{B}(p),M_{\Delta,{\alpha}})$ with $p\neq-1$ and $H^{2}(\mathfrak{B}(-1),M_{\Delta,{\alpha},\beta})$ for $\alpha\neq 0$, respectively.

5.1.  Second Cohomology of $\mathfrak{B}(p)$ with trivial coefficients

In this section, we shall compute the second cohomology $H^{2}(\mathfrak{B}(p),\mathbb{C})$, where $\partial$, $L_{i}$ $(i\in\mathbb{Z}^{+})$ act by zero on $\mathbb{C}$.

Assume that $c\in C^{2}(\mathfrak{B}(p),\mathbb{C})$ be a $2$-$\lambda$-bracket. Since $\partial$ acts by zero on $\mathbb{C}$, we can write (5.5) and (5.6) as follows:

[TABLE]

Then, replacing $a,b,c$ by $L_{i},L_{j},L_{k}$ in (5.8), respectively, we have

[TABLE]

which gives

[TABLE]

[TABLE]

where $a\in\mathbb{C}$ and $d\phi$ is as in (5.9).

[TABLE]

Since $\{{L_{0}}\,_{\lambda}L_{i}\}_{d\phi}=-\lambda(\frac{d}{d\,_{\lambda}}\{{L_{0}}\,_{\lambda}L_{i}\}_{c}|_{\lambda=0})$ by (5.9), replacing $c$ with $c+d\phi$, we have

[TABLE]

Letting $i=0$ and taking the derivative of both sides of (5.12) with respect to $\lambda$, we get

[TABLE]

Note that $\{{{L_{0}}}\,_{0}L_{k}\}_{{d\phi}}=0$. Thus, by letting $i=j=\lambda=0$ in (5.12) and using $p>0$, one can deduce $\{{L_{0}}\,_{0}L_{k}\}_{c}=0$ for all $k\in\mathbb{Z}^{+}$. Thus, taking $\lambda=0$ in (5.16) and using (5.13), we have

[TABLE]

Set $\{{L_{j}}\,_{\mu}{L_{k}}\}_{{c}}=\sum_{l=0}^{m}h_{l}(j,k)\mu^{l}\in\mathbb{C[\mu]}$. Assume that $m\geq 3$. Then from the above equation, we can get

[TABLE]

which gives

[TABLE]

Fix $j$ and $k$. If $j+k=0$, then (5.19) and (5.20) imply that $\{{L_{0}}\,_{\lambda}L_{0}\}_{c}=h_{3}(0,0)\mu^{3}$. Now, let $j+k\neq 0$. Assume that $(l_{0}-3)p=j+k$ for some $4\leq l_{0}\leq m$. Then, it follows from (5.19) and (5.20) that $\{{L_{j}}\,_{\lambda}L_{k}\}_{c}=\delta_{j+k,(l_{0}-3)p}h_{l_{0}}(j,k)\mu^{l_{0}}$. Then, letting $i=0$ and $j+k=(l_{0}-3)p$ in (5.12), we have

[TABLE]

By considering the coefficients of $\lambda^{2}\mu^{l_{0}-1}$ and $\lambda^{3}\mu^{l_{0}-2}$ in the above equation, we get $(2j+(3-l_{0})p)h_{l_{0}}(j,k)=0$ and $(3j+(5-l_{0})p)h_{l_{0}}(j,k)=0$, respectively, which forces $h_{l_{0}}(j,k)=0$ for any $j+k>0$. Hence, the proof follows by setting $a=h_{3}(0,0)$.

∎

Now we assume that $p<0$.

Letting $k=0$ in (5.12), one has

[TABLE]

On the other hand, by taking $i=j=0$ in (5.12), we have

[TABLE]

Set $\{{L_{0}}\,_{\lambda}{L_{k}}\}_{c}=\sum_{i=0}^{n}g_{i}(k)\lambda^{i}\in\mathbb{C}[\lambda]$ with $g_{n}(k)\neq 0$. If $n>3$, then considering the coefficients of $\lambda^{n}$ in (5.25), we have \big{(}(n-3)p-k\big{)}g_{n}(k)=0, which forces $g_{n}(k)=0$ since $n>3$. Thus, we can set $\{{L_{0}}\,_{\lambda}{L_{k}}\}_{c}=g_{0}(k)+g_{1}(k)\lambda+g_{2}(k)\lambda^{2}+g_{3}(k)\lambda^{3}.$ By substituting this into (5.25) and then considering the power of $\lambda$, one can deduce

[TABLE]

[TABLE]

(1) The case $p\notin\mathbb{Z}^{-}$, $2p\notin\mathbb{Z}^{-}$.

From (5.26) that $\{{L_{0}}\,_{\lambda}{L_{k}}\}_{c}=g_{1}(k)\lambda+\delta_{k,0}g_{3}(k)\lambda^{3}$. Now, define a $\mathbb{C}[\partial]$-linear map $\phi:\mathfrak{B}(p)\rightarrow\mathbb{C}$,

[TABLE]

for any $k\in\mathbb{Z}^{+}$. Then replacing $c$ by $c+d\phi$, one can set

[TABLE]

Assume that $i+j\neq 0$. Then using (5.27) in (5.24), we have

[TABLE]

Since $p,2p\notin\mathbb{Z}^{-}$, by considering the power of $\lambda$ in the above equation, we can set $\{{L_{i}}\,_{\lambda}{L_{j}}\}_{c}=f(i,j)\in\mathbb{C}$. Thus, it follows from (5.28) that $(i+j+3p)f(i,j)=0$, which gives

[TABLE]

Suppose that $3p\in\mathbb{Z}^{-}$. Set $\tilde{p}=-3p-1$. Then by taking $i=1$ and $k=\tilde{p}-j\geq 0$ in (5.12), we can get

[TABLE]

Note that $p\notin\mathbb{Z}^{-}$. Thus from the above identities that $f(j+1,\tilde{p}-j)=(j+1)f(1,\tilde{p})$, which implies $f(1,\tilde{p})=-(3p)^{-1}f(-3p,0)$. Thus, $f(1,\tilde{p})=0$ since $f(-3p,0)=-\{{L_{0}}\,_{-\lambda}L_{-3p}\}_{c}=0$ by (5.27). Hence, $f(i,j)=f((i-1)+1,\tilde{p}-(i-1))=if(1,\tilde{p})=0$ for any $i+j=-3p$ with $i\neq 0$. Combing this with (5.27) and (5.29), we have $\{{L_{i}}\,_{\lambda}{L_{j}}\}_{c}=\delta_{i+j,0}a_{1}\lambda^{3}$, where $a_{1}=g_{3}(0)$.

(2) The case $p\notin\mathbb{Z}^{-}$, $2p\in\mathbb{Z}^{-}$.

It follows from (5.26) that

[TABLE]

Define a $\mathbb{C}[\partial]$-linear map $\phi:\mathfrak{B}(p)\rightarrow\mathbb{C}$ as follows:

[TABLE]

Then, replacing $c$ by $c+d\phi$, one can set

[TABLE]

If $i+j\neq-2p,0$, then from (5.24), we can get (5.28). Thus, if $i\neq-2p$, then from (5.28), we also get $\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}\in\mathbb{C}$ since $p\notin\mathbb{Z}^{-}$. Using this in (5.28), we have $(i+j+3p)\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}=0$. Since $p\notin\mathbb{Z}^{-}$ but $2p\in\mathbb{Z}^{-}$, we have $3p\notin\mathbb{Z}^{-}$. Thus, in case $i\neq-2p$, we get $\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}=0$. For $i=-2p$, by taking $i=-2p$ in (5.28), we immediately get $\{L_{-2p}{\,{}_{\lambda}}L_{j}\}_{c}=0$. Thus, we obtain $\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}=0$ for $i+j\neq-2p$.

If $i+j=-2p$, then from (5.24) and (5.33), we can get

[TABLE]

Note that $p\notin\mathbb{Z}^{-}$. Thus, by considering the power of $\lambda$ in above equation, one can easily deduce $\{{L_{-2p-j}}\,_{\lambda}{L_{j}}\}_{c}=a_{2}(j+p)+a_{3}\lambda$, where $a_{2}=-p^{-1}g_{0}(-2p)$, $a_{3}=g_{1}(-2p)$.

(3) The case $p\in\mathbb{Z}^{-}$.

Recall the result in (5.26),

[TABLE]

Define $\phi$ as those given in (5.30). Then, replacing $c$ by $c+d\phi$, one can set

[TABLE]

(i) Case $i+j=-2p$. If $j\neq-p$, then from (5.36), we can get $\{{L_{-2p-j}}\,_{\lambda}{L_{j}}\}_{c}=a_{2}(j+p)+a_{3}\lambda$. If $j=-p$, then $i=-p$. Thus, taking $j=-p$ in (5.36), one can easily deduce $\{L_{-p}{\,{}_{\lambda}}L_{-p}\}_{c}=b\lambda$ for some $b\in\mathbb{C}$. Now, setting $i=-p$ and $k=-p-1$ in (5.12), then comparing the coefficients of $\lambda^{2}$, we can get $(p+1)(b-a_{3})=0$, which gives $b=a_{3}$ for $p\neq-1$. Thus, for any $j\in\mathbb{Z}^{+}$, we get

[TABLE]

(ii) Case $i+j=-p$. In this case, from (5.24) that

[TABLE]

By considering the power of $\lambda$ in the above equation, we can set $\{{L_{-p-j}}\,_{\lambda}{L_{j}}\}_{c}=h_{0}(j)+h_{1}(j)\lambda+h_{2}(j)\lambda^{2}$. Using this in (5.42), we can deduce $h_{0}(j)=h_{2}(j)=h_{2}(j)=g_{2}(-p)=0$. Thus, $\{{L_{i}}\,_{\lambda}{L_{j}}\}_{c}=0$ for any $i+j=-p$.

(iii) Case $i+j=0$. It follows from (5.37) that $\{{L_{0}}\,_{\lambda}{L_{0}}\}_{c}=a_{1}\lambda^{3}$, where $a_{1}=g_{3}(0)$.

(iv) Case $i+j\neq-2p,-p,0$. In this case, we also have (5.28). Thus, If $j\neq-p$ and $i\neq-2p$, then we can get (5.29), which gives $\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}=0$. One the other hand, using (5.28), one can easily deduce $\{L_{i}{\,{}_{\lambda}}L_{j}\}_{c}=0$ for $j=-p$ and $i\neq-2p$, or $j\neq-p$ and $i=-2p$, respectively. Besides, by taking $j=-p$ and $i=-2p$ in (5.24), we can get $\{L_{-2p}{\,{}_{\lambda}}L_{-p}\}_{c}\in\mathbb{C}$. Combing this with setting $j=-p$, $k=-2p-i$, $i=1$ and $2$ in (5.12), respectively, one can check that $(1+p)\{L_{-2p}{\,{}_{\lambda}}L_{-p}\}_{c}=(2+p)\{L_{-2p}{\,{}_{\lambda}}L_{-p}\}_{c}=0$, which forces $\{L_{-2p}{\,{}_{\lambda}}L_{-p}\}_{c}=0$. This completes the proof.

∎

The following statements follow from straightforward verifications.

(1) For any $0\neq p\in\mathbb{C}$, the $2$-$\lambda$-bracket $\alpha$ defined by

[TABLE]

is a nontrivial 2-cocycle.

(2) If $2p\in\mathbb{Z}^{-}$, then the following $2$-$\lambda$-brackets $\beta$, $\tilde{\beta}$ and $\bar{\beta}$ defined by (all other terms are vanishing)

[TABLE]

are three nontrivial 2-cocycles.

[TABLE]

(1) If $2p\notin\mathbb{Z}^{-}$, then $\{{L_{j}}_{\,\lambda}L_{k}\}_{c}=\delta_{j+k,0}a\lambda^{3}$ for any $i,j\in\mathbb{Z}^{+}$, namely, $c=a\alpha$.

(2) If $2p\in\mathbb{Z^{-}}$ with $p\neq-1$, then $\{L_{i}{{}_{\,\lambda}}L_{j}\}_{c}=\delta_{i+j,0}a_{1}\lambda^{3}+\delta_{i+j,-2p}(a_{2}(j+p)+a_{3}\lambda)$ for any $i,j\in\mathbb{Z}^{+}$, which imply $c=a_{1}\alpha+a_{2}\beta+a_{3}\bar{\beta}$.

(3) If $p=-1$, then (all other terms are vanishing)

[TABLE]

which give $c=a_{1}\alpha+a_{2}\beta+a_{3}\bar{\beta}+b\tilde{\beta}$. This completes the proof. ∎

5.2.  Second cohomology of $\mathfrak{B}(p)$ $(resp.,\mathfrak{B}(-1))$ with coefficients in $M_{\Delta,{\alpha}}(resp.,M_{\Delta,{\alpha},\beta})$

In this section, we only consider the case $\alpha\neq 0$.

Assume that $p\neq-1$. Take $c\in C^{2}(\mathfrak{B}(p),M_{\Delta,{\alpha}})$ and suppose that $\{{L_{i}}\,_{\lambda}{L_{j}}\}_{c}=f_{i,j}(\partial,\lambda)v\in\mathbb{C}[\lambda]\otimes M_{\Delta,{\alpha}}$. Then by taking $a=L_{0}$, $\lambda=0$, $b=L_{j}$ and $z=L_{k}$ with $jk\neq 0$ in (5.5), respectively, we have

[TABLE]

Letting $a=b=z=L_{0}$ and $\lambda=0$ in (5.5), one has

[TABLE]

Letting $a=b=L_{0}$, $z=L_{k}$ with $k\neq 0$ and $\lambda=0$ in (5.5), one can obtain

[TABLE]

Note that $f_{k,0}(\partial,\mu)=-f_{0,k}(\partial,-\mu-\partial)$. Now, define a $\mathbb{C}[\partial]$-linear map $\phi$ from $\mathfrak{B}(p)$ to $M_{\Delta,{\alpha}}=\mathbb{C}[\partial]v$ as $\phi(L_{i})={\alpha p}^{-1}f_{0,i}(\partial,0)v$. Then replacing $c$ by $c-d\phi$ and using (5.47)–(5.49), one can obtain that $\{{{L_{i}}\,_{\lambda}}L_{j}\}_{c}=0$ for any $i,j\in\mathbb{Z}^{+}$. Thus, we get $H^{2}(\mathfrak{B}({p}),M_{\Delta,{\alpha}})=0$.

Next, we study the second cohomology of $\mathfrak{B}(-1)$ with coefficients in $M_{\Delta,{\alpha},\beta}$.

Let $\bar{c}\in C^{2}(\mathfrak{B}(-1),M_{\Delta,{\alpha},\beta})$ and set $\{{L_{i\,_{\lambda}}}{L_{j}}\}_{\bar{c}}=g_{i,j}(\partial,\lambda)v\in\mathbb{C}[\lambda]\otimes M_{\Delta,{\alpha},\beta}$. Then taking $a=b=L_{0}$, $z=L_{1}$ and $\lambda=0$ in (5.5), we have

[TABLE]

Letting $a=L_{1}$, $b=L_{0}$, $z=L_{k}$ with $k\neq 0,1$ and $\mu=0$ in (5.5), we have

[TABLE]

Using $g_{1,0}(-\lambda,\lambda)=-g_{0,1}(-\lambda,0)$, taking $a=z=L_{1}$, $b=L_{0}$ and $\mu=0$ in (5.5), we have

[TABLE]

Define a $\mathbb{C}[\partial]$-linear map $\varphi$ from $\mathfrak{B}(-1)$ to $M_{\Delta{\alpha},\beta}=\mathbb{C}[\partial]v$ as $\varphi(L_{i})={\alpha}^{-1}g_{0,i}(\partial,0)v$. Note that $g_{i,j}(\partial,\lambda)$ also satisfy the identities (5.47)–(5.49) for $i,j>1$. These together with (5.50)–(5.52) imply that $\{{{L_{i}}\,_{\lambda}}L_{j}\}_{\bar{c}}=\{{{L_{i}}\,_{\lambda}}L_{j}\}_{d\varphi}$ for any $i,j\in\mathbb{Z}^{+}$. Hence, we have $H^{2}(\mathfrak{B}(-1),M_{\Delta,{\alpha},\beta})=0$.

The following theorem is immediate.