A Lie conformal algebra of Block type

Lamei Yuan

Academy of Fundamental and Interdisciplinary Science,

Harbin Institute of Technology, Harbin 150080, China

<EMAIL_ADDRESS>
Abstract: The aim of this paper is to study a Lie conformal algebra of Block
type. In this paper, conformal derivation, conformal module of rank 1 and low-
dimensional comohology of the Lie conformal algebra of Block type are studied.
Also, the vertex Poisson algebra structure associated with the Lie conformal
algebra of Block type is constructed.

Keywords:  Lie conformal algebra, vertex Lie algebra, cohomology, vertex
Poisson algebra

MR(2000) Subject Classification: 17B65, 17B69

##  1 Introduction

The notion of Lie conformal algebra, introduced by Kac [9], encode an
axiomatic description of the operator product expansions of chiral fields in
conformal field theory. It is a powerful tool for the study of infinite-
dimensional Lie (super)algebras, associative algebras and their
representations. Lie conformal algebras have been extensively studied,
including the classification problem [5, 6], cohomology theory [2, 12] and
representation theory [3].

The Lie conformal algebras are closely related to vertex algebras. Primc [11]
introduced and studied a notion of vertex Lie algebra, which is a special case
of a more general notion of local vertex Lie algebra [4]. As it was explained
in [10], the notion of Lie conformal algebra and the notion of vertex Lie
algebra are equivalent. In this paper, we shall use Lie conformal algebra and
vertex Lie algebra synonymously.

With the notion of vertex Lie algebra, one arrives at the notion of vertex
Poisson algebra, which is a combination of a differential algebra structure
and a vertex Lie algebra structure, satisfying a natural compatibility
condition. The symmetric algebra of a vertex Lie algebra is naturally a vertex
Poisson algebra [7]. A general construction theorem of vertex Poisson algebras
was given in [10]. Applications of vertex Poisson algebras to the theory of
integrable systems were studied in [1].

In the present paper, we study a nonsimple Lie conformal algebra of infinite
rank, which is endowed with a ${\mathbb{C}}[\partial]$-basis
$\\{J_{i}|i\in\mathbb{Z}^{+}\\}$, such that

$\displaystyle[J_{i\,\lambda}J_{j}]=((i+1)\partial+(i+j+2)\lambda)J_{i+j},\
\mbox{for}\ i,j\in\mathbb{Z}^{+}.$ (1.1)

The corresponding formal distribution Lie algebra is a Block type Lie algebra,
which is the associated graded Lie algebra of the filtered Lie algebra
$W_{1+\infty}$ [13, 14, 15, 16, 18]. Thus we call this Lie conformal algebra a
Lie conformal algebra of Block type and denote it by ${\mathcal{B}}$ in this
paper. It is a conformal subalgebra of ${\rm gr}\,gc_{1}$ studied in [17]. In
addition, it contains the Virasoro conformal algebra $\rm
Vir={\mathbb{C}}[\partial]J_{0}$ with
$[J_{0\,\lambda}J_{0}]=(\partial+2\lambda)J_{0}$ as a subalgebra.

The paper is organized as follows. In Section 2, we recall the notions of Lie
conformal algebra and vertex Lie algebra. In Section 3, we study conformal
derivations of the Lie conformal algebra of Block type ${\mathcal{B}}$. In
Section 4, we recall the notions of conformal module and comohology of Lie
conformal algebras. Then we study conformal module of rank 1 and low-
dimensional comohology of ${\mathcal{B}}$ with coefficients in
${\mathcal{B}}$-modules. In Section 5, we equip a vertex Lie algebra structure
$(Y_{-},\partial)$ with ${\mathcal{B}}$ and establish an association of a
vertex Poisson algebra structure to the vertex Lie algebra
$({\mathcal{B}},Y_{-},\partial)$.

##  2 Preliminaries

Throughout this paper, all vector spaces and tensor products are over the
complex field $\mathbb{C}$. We use notations $\mathbb{Z}$ for the set of
integers and $\mathbb{Z}^{+}$ for the set of nonnegative integers.

######  Definition 2.1.

A Lie conformal algebra $\mathcal{R}$ is a ${\mathbb{C}}[\partial]$-module
with a ${\mathbb{C}}$-bilinear map,

$\displaystyle\mathcal{R}\otimes\mathcal{R}\rightarrow{\mathbb{C}}[\lambda]\otimes\mathcal{R},\
\ \ a\otimes b\mapsto[a_{\lambda}b],$

called the $\lambda$-bracket, and satisfying the following axioms
($a,b,c\in\mathcal{R}$),

$\displaystyle\mbox{(conformal\ sesquilinearity)}\ \ [\partial a_{\lambda}b]$
$\displaystyle=$ $\displaystyle-\lambda[a_{\lambda}b],\ [a_{\lambda}\partial
b]=(\partial+\lambda)[a_{\lambda}b],$ (2.1) $\displaystyle\mbox{(skew-
symmetry)}\ \ \ \ {[a_{\lambda}b]}$ $\displaystyle=$
$\displaystyle-[b_{-\lambda-\partial}a],$ (2.2) $\displaystyle\mbox{(Jacobi \
identity)}\ \ {[a_{\lambda}[b_{\mu}c]]}$ $\displaystyle=$
$\displaystyle[[a_{\lambda}b]_{\lambda+\mu}c]+[b_{\mu}[a_{\lambda}c]].$ (2.3)

If we consider the expansion

$\displaystyle[a_{\lambda}b]=\mbox{$\sum\limits_{j\in\mathbb{Z}^{+}}$}\frac{\lambda^{j}}{j!}(a_{(j)}b),$
(2.4)

the coefficients of $\frac{\lambda^{j}}{j!}$ are called the $j$-product
satisfying ${a_{(n)}b}=0$ for $n$ sufficiently large, and the axioms
(2.1)–(2.3) can be written in terms of them as follows:

$\displaystyle{\partial a_{(n)}b}$ $\displaystyle=$ $\displaystyle-
na_{(n-1)}b,\ \ {a_{(n)}\partial b}=\partial(a_{(n)}b)+na_{(n-1)}b,$ (2.5)
$\displaystyle{{a_{(n)}b}}$ $\displaystyle=$
$\displaystyle-\mbox{$\sum\limits_{i\in\mathbb{Z}^{+}}$}(-1)^{n+i}\frac{1}{i!}\partial^{i}b_{(n+i)}a,$
(2.6) $\displaystyle{a_{(m)}b_{(n)}c}$ $\displaystyle=$
$\displaystyle{b_{(n)}a_{(m)}c}+\mbox{$\sum\limits_{i=0}^{m}$}\mbox{${m\choose
i}$}(a_{(i)}b)_{(m+n-i)}c.$ (2.7)

In terms of generating functions, Mirko Primc in [11] presented an equivalent
definition of a Lie conformal algebra under the name of vertex Lie algebra
(see also [10]). Let $V$ be any vector space. Following [11], for a formal
series

$\displaystyle
f(x_{1},\cdots,x_{n})=\mbox{$\sum\limits_{m_{1},\cdots,m_{n}\in\mathbb{Z}}$}u(m_{1},\cdots,m_{n})x_{1}^{-m_{1}-1}\cdots
x_{n}^{-m_{n}-1}\in V[[x_{1}^{\pm 1},\cdots,x_{n}^{\pm 1}]],$

we set

$\displaystyle{\rm
Sing}f(x_{1},\cdots,x_{n})=\mbox{$\sum\limits_{m_{1},\cdots,m_{n}\in\mathbb{Z}^{+}}$}u(m_{1},\cdots,m_{n})x_{1}^{-m_{1}-1}\cdots
x_{n}^{-m_{n}-1}.$ (2.8)

Clearly, for $1\leq i\leq n$,

$\displaystyle\frac{\partial}{\partial x_{i}}{\rm
Sing}f(x_{1},\cdots,x_{n})={\rm Sing}\frac{\partial}{\partial
x_{i}}f(x_{1},\cdots,x_{n}).$ (2.9)

######  Definition 2.2.

A vertex Lie algebra is a vector space $A$ equipped with a linear operator
$\partial$ called the derivation and a linear map

$\displaystyle Y_{-}(\cdot,z):A\rightarrow z^{-1}({\rm End\,}A)[[z^{-1}]],\ \
\ a\mapsto Y_{-}(a,z)=\mbox{$\sum_{n\geq 0}$}a_{(n)}z^{-n-1},$

satisfying the following conditions for $a,b\in A$, $n\in\mathbb{Z}^{+}$:

$\displaystyle{a_{(n)}b}=0\ \ \mbox{for}\ n\ \mbox{sufficiently large},$
(2.10) $\displaystyle{[\partial,Y_{-}(a,z)]}={Y_{-}(\partial
a,z)}=\frac{d}{dz}Y_{-}(a,z),$ (2.11) $\displaystyle{Y_{-}(a,z)b}={\rm
Sing}\big{(}e^{z\partial}Y_{-}(b,-z)a\big{)},$ (2.12)

and the half Jacobi identity holds:

$\displaystyle{\rm
Sing}\Big{(}z_{0}^{-1}\delta(\frac{z_{1}-z_{2}}{z_{0}})Y_{-}(a,z_{1})Y_{-}(b,z_{2})-z_{0}^{-1}\delta(\frac{z_{2}-z_{1}}{-z_{0}})Y_{-}(b,z_{1})Y_{-}(a,z_{2})\Big{)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \hskip 56.9055pt={\rm
Sing}\Big{(}z_{2}^{-1}\delta(\frac{z_{1}-z_{0}}{z_{2}})Y_{-}(Y_{-}(a,z_{0})b,z_{2})\Big{)}.$
(2.13)

Relation (2.12) is called the half skew-symmetry. It was shown in [11] that
the half Jacobi identity (2.2) amounts to the following half commutator
formula:

$\displaystyle Y_{-}(a,z_{1})Y_{-}(b,z_{2})-Y_{-}(b,z_{1})Y_{-}(a,z_{2})={\rm
Sing}\Big{(}\mbox{$\sum_{i\in\mathbb{Z}^{+}}$}(z_{1}-z_{2})^{-i-1}Y_{-}(a_{(i)}b,z_{2})\Big{)}.$
(2.14)

As it was explained in [10, Remark 2.6], the notion of Lie conformal algebra
is equivalent to the notion of vertex Lie algebra. We often denote a vertex
Lie algebra by $(A,Y_{-},\partial)$ and refer to $(Y_{-},\partial)$ as the
vertex Lie algebra structure. A vertex Lie algebra $(A,Y_{-},\partial)$ is
said to be free, if $A$ is a free ${\mathbb{C}}[\partial]$-module over a
vector space $V$, namely,
$A={\mathbb{C}}[\partial]V\cong{\mathbb{C}}[\partial]\otimes_{{\mathbb{C}}}V.$

##  3 Conformal derivation

Let $\mathcal{C}$ denote the ring $\mathbb{C}[\partial]$ of polynomials in the
indeterminate $\partial$.

######  Definition 3.1.

Let $V$ and $W$ be two $\mathcal{C}$-modules. A linear map
$\phi:V\rightarrow\mathcal{C}[\lambda]\otimes_{\mathcal{C}}W$, denoted by
$\phi_{\lambda}:V\rightarrow W$, is called a conformal linear map, if

$\displaystyle\phi_{\lambda}(\partial v)=(\partial+\lambda)(\phi_{\lambda}v),\
\mbox{for}\ v\in V.$ (3.1)

The space of conformal linear maps between $\mathcal{C}$-modules $V$ and $W$
is denoted by ${\rm Chom}(V,W)$ and it can be made into an
$\mathcal{C}$-module via

$\displaystyle(\partial\phi)_{\lambda}v=-\lambda\phi_{\lambda}v,\ \mbox{for}\
v\in V.$

######  Definition 3.2.

Let $\mathcal{R}$ be a Lie conformal algebra. A conformal linear map
$d_{\lambda}:\mathcal{R}\rightarrow\mathcal{R}$ is called a conformal
derivation of $\mathcal{R}$ if

$\displaystyle
d_{\lambda}[a_{\mu}b]=[(d_{\lambda}a)_{\lambda+\mu}b]+[a_{\mu}(d_{\lambda}b)],\
\mbox{for}\ a,b\in\mathcal{R}.$ (3.2)

The space of all conformal derivations of $\mathcal{R}$ is denoted by ${\rm
CDer}(\mathcal{R})$. For any $a\in\mathcal{R}$, one can define a conformal
derivation $({\rm ad}\,a)_{\lambda}:\mathcal{R}\rightarrow\mathcal{R}$ by
$({\rm ad}\,a)_{\lambda}b=[a_{\lambda}b]$ for $b\in\mathcal{R}$. Such
conformal derivation is called inner. Denote by ${\rm CInn}(\mathcal{R})$ the
space of all conformal inner derivations of $\mathcal{R}$.

######  Proposition 3.3.

Every conformal derivation of the Lie conformal algebra ${\mathcal{B}}$ is
inner.

Proof. Let $d$ be any conformal derivation of ${\mathcal{B}}$. Denote
$L=J_{0}$. Assume that there exists a finite subset
$I=\\{i_{1},\cdots,i_{n}\\}\subseteq\mathbb{Z}^{+}$ such that
$d_{\lambda}L=\mbox{$\sum_{j=1}^{n}$}f_{i_{j}}(\partial,\lambda)J_{i_{j}}$,
where $f_{i_{j}}(\partial,\lambda)\in\mathbb{C}[\partial,\lambda]$. Condition
(3.2) requires
$d_{\lambda}([L_{\mu}L])=[L_{\mu}(d_{\lambda}L)]+[(d_{\lambda}L)_{\lambda+\mu}L].$
This is equivalent to

$\displaystyle(\partial+\lambda+2\mu)\mbox{$\sum\limits_{j=1}^{n}$}f_{i_{j}}(\partial,\lambda)J_{i_{j}}-\mbox{$\sum\limits_{j=1}^{n}$}(\partial+(i_{j}+2)\mu)f_{i_{j}}(\partial+\mu,\lambda)J_{i_{j}}$
$\displaystyle\ \ \ \
=\mbox{$\sum\limits_{j=1}^{n}$}((i_{j}+1)\partial+(i_{j}+2)(\lambda+\mu))f_{i_{j}}(-\lambda-\mu,\lambda)J_{i_{j}}.$
(3.3)

For each $j$,

$\displaystyle(\partial+\lambda+2\mu)f_{i_{j}}(\partial,\lambda)-(\partial+(i+2)\mu)f_{i_{j}}(\partial+\mu,\lambda)$
$\displaystyle\ \ \ \
=((i_{j}+1)\partial+(i_{j}+2)(\lambda+\mu))f_{i_{j}}(-\lambda-\mu,\lambda).$
(3.4)

Write
$f_{i_{j}}(\lambda,\partial)=\sum_{k=0}^{m}a_{i_{j},k}(\lambda)\partial^{k}$
with $a_{i_{j},m}(\lambda)\neq 0$. Then, assuming $m>1$, if we equate terms of
degree $m$ in $\partial$, we have $(\lambda-i-m\mu)a_{i_{j},m}(\lambda)=0$ and
thus $a_{i_{j},m}(\lambda)=0$. This contradicts $a_{i_{j},m}(\lambda)\neq 0$.
Thus ${\rm deg}_{\partial}f_{i_{j}}(\lambda,\partial)\leq 1$, and
$f_{i_{j}}(\lambda,\partial)=a_{i_{j},0}(\lambda)+a_{i_{j},1}(\lambda)\partial$.
Substituting it into (3) gives
$a_{i_{j},0}(\lambda)=\frac{i_{j}+2}{i_{j}+1}\lambda a_{i_{j},1}(\lambda)$.
Therefore,

$\displaystyle
d_{\lambda}L=\mbox{$\sum\limits_{j=1}^{n}$}\frac{a_{i_{j},1}(\lambda)}{i_{j}+1}((i_{j}+1)\partial+(i_{j}+2)\lambda)J_{i_{j}}.$

Replacing $d_{\lambda}$ by $d_{\lambda}-({\rm ad}\,h)_{\lambda}$ with
$h=\mbox{$\sum_{j=1}^{n}$}\frac{a_{i_{j},1}(-\partial)}{i_{j}+1}J_{i_{j}}$, we
get $d_{\lambda}(L)=0$.

For $k>0$, assume that
$d_{\lambda}J_{k}=\sum_{i=1}^{l}f_{k_{i}}(\partial,\lambda)J_{k_{i}}.$
Applying $d_{\lambda}$ to $[L_{\mu}J_{k}]=(\partial+(k+2)\mu)J_{k}$ and using
$d_{\lambda}(L)=0$, we obtain

$\displaystyle(\partial+\lambda+(k+2)\mu)\mbox{$\sum\limits_{i=1}^{l}$}f_{k_{i}}(\partial,\lambda)J_{k_{i}}=\mbox{$\sum\limits_{i=1}^{l}$}(\partial+(k_{i}+2)\mu)f_{k_{i}}(\partial+\mu,\lambda)J_{k_{i}},$
(3.5)

and thus

$\displaystyle(\partial+\lambda+(k+2)\mu)f_{k_{i}}(\partial,\lambda)=(\partial+(k_{i}+2)\mu)f_{k_{i}}(\partial+\mu,\lambda),\
\ \mbox{for}\ 1\leq i\leq l.$ (3.6)

Comparing the highest degree of $\lambda$ gives
$f_{k_{i}}(\partial,\lambda)=0$ for $1\leq i\leq l$. Hence
$d_{\lambda}(J_{k})=0$ for $k>0$. This concludes the proof. $\Box$

##  4 Cohomology

######  Definition 4.1.

A module $M$ over a Lie conformal algebra $\mathcal{R}$ is a
$\mathbb{C}[\partial]$-module endowed with a bilinear map

$\displaystyle\mathcal{R}\otimes M\rightarrow M[[\lambda]],\ \ a\otimes
v\mapsto a_{\lambda}v$

such that ($a,b\in\mathcal{R}$, $v\in M$)

$\displaystyle
a_{\lambda}(b_{\mu}v)-b_{\mu}(a_{\lambda}v)=[a_{\lambda}b]_{\lambda+\mu}v,$
(4.1) $\displaystyle(\partial a)_{\lambda}v=-\lambda a_{\lambda}v,\ \ \
a_{\lambda}(\partial v)=(\partial+\lambda)a_{\lambda}v,$ (4.2)

If $a_{\lambda}v\in M[\lambda]$ for all $a\in\mathcal{R}$, $v\in M$, then the
$\mathcal{R}$-module $M$ is said to be conformal. If $M$ is finitely generated
as $\mathbb{C}[\partial]$-module, then $M$ is simply called finite.

Since we only consider conformal modules, we will simply shorten the term
“conformal module” to “module”. The one-dimensional vector space
${\mathbb{C}}$ can be viewed as a module (called the trivial module) over any
conformal algebra $\mathcal{R}$ with both the action of $\partial$ and the
action of $\mathcal{R}$ being zero. In addition, for a fixed nonzero complex
constant $a$, there is a natural ${\mathbb{C}}[\partial]$-module
${\mathbb{C}}_{a}$, which is the one-dimensional vector space ${\mathbb{C}}$
such that $\partial v=av$ for $v\in{\mathbb{C}}_{a}$. Then ${\mathbb{C}}_{a}$
becomes an $\mathcal{R}$-module on which all elements of $\mathcal{R}$ act by
zero.

For the Virasoro conformal algebra Vir, it was proved in [3] that all free
nontrivial Vir-modules of rank $1$ are the following ones
$(\Delta,\alpha\in\mathbb{C})$:

$\displaystyle M_{\Delta,\alpha}=\mathbb{C}[\partial]v,\ \
L_{\lambda}v=(\partial+\alpha+\Delta\lambda)v.$ (4.3)

The module $M_{\Delta,\alpha}$ is irreducible if and only if $\Delta\neq 0$,
the module $M_{0,\alpha}$ contains a unique nontrivial submodule
$(\partial+\alpha)M_{0,\alpha}$ isomorphic to $M_{1,\alpha}$, and the modules
$M_{\Delta,\alpha}$ with $\Delta\neq 0$ exhaust all finite irreducible
nontrivial conformal Vir-modules.

######  Proposition 4.2.

All free nontrivial ${\mathcal{B}}$-modules of rank $1$ are as follows
$(\Delta,\alpha\in\mathbb{C})$:

$\displaystyle M_{\Delta,\alpha}=\mathbb{C}[\partial]v,\
J_{0_{\lambda}}v=(\partial+\alpha+\Delta\lambda)v,\ J_{i}\,{{}_{\lambda}}v=0,\
\mbox{for}\ i>0.$

Proof. By (4.3), $J_{0\,\lambda}v=(\partial+\alpha+\Delta\lambda)v$ for some
$\Delta,\alpha\in\mathbb{C}$. By [17, Lemma 5.1], we can suppose that $k$ is
the smallest nonnegative integer such that $J_{k}\,{}_{\lambda}v\neq 0$,
$J_{k+1}\,{}_{\lambda}v=0$. Assume $k>0$ and write
$J_{k}\,{}_{\lambda}v=g(\lambda,\partial)v$, for some
$g(\lambda,\partial)\in{\mathbb{C}}[\lambda,\partial]$. Since
$[J_{k}\,{}_{\lambda}J_{k}]_{\lambda+\mu}v=0$,

$\displaystyle
g(\lambda,\partial)g(\mu,\lambda+\partial)=g(\mu,\partial)g(\lambda,\mu+\partial).$
(4.4)

This implies ${\rm deg}_{\lambda}g(\lambda,\partial)+{\rm
deg}_{\partial}g(\lambda,\partial)={\rm deg}_{\lambda}g(\lambda,\partial)$.
Thus ${\rm deg}_{\partial}g(\lambda,\partial)=0$. Then we have
$g(\lambda,\partial)=g(\lambda)$ for some
$g(\lambda)\in{\mathbb{C}}[\lambda]$. The fact that
$[J_{0\,\lambda}J_{k}]_{\lambda+\mu}v=((k+1)\lambda-\mu)J_{k}{{}_{\lambda+\mu}}v$
yields

$\displaystyle((k+1)\lambda-\mu)g(\lambda+\mu)v=(\partial+\alpha+\Delta\lambda)g(\mu)v-(\partial+\mu+\alpha+\Delta\lambda)g(\mu)v=-\mu
g(\mu)v,$

which gives $g(\mu)=0$. Hence, $J_{k}\,{}_{\lambda}v=0$, a contradiction. Thus
$k=0$ and $J_{1}\,{}_{\lambda}v=0$. It follows immediately that
$J_{i}\,{}_{\lambda}v=0$ for all $i\geq 1$. $\Box$

In the following we study cohomology of the Lie conformal algebra
${\mathcal{B}}$ with coefficients in ${\mathcal{B}}$-modules ${\mathbb{C}}$,
${\mathbb{C}}_{a}$ and $M_{\Delta,\alpha}$, respectively. For completeness, we
shall present the definition of cohomology of Lie conformal algebras given in
[2].

######  Definition 4.3.

An $n$-cochain ($n\in\mathbb{Z}^{+}$) of a Lie conformal algebra $\mathcal{R}$
with coefficients in an $\mathcal{R}$-module $M$ is a ${\mathbb{C}}$-linear
map

$\displaystyle\gamma:\mathcal{R}^{\otimes n}\rightarrow
M[\lambda_{1},\cdots,\lambda_{n}],\ \ \ a_{1}\otimes\cdots\otimes
a_{n}\mapsto\gamma_{\lambda_{1},\cdots,\lambda_{n}}(a_{1},\cdots,a_{n})$

satisfying

(1) $\gamma_{\lambda_{1},\cdots,\lambda_{n}}(a_{1},\cdots,\partial
a_{i},\cdots,a_{n})=-\lambda_{i}\gamma_{\lambda_{1},\cdots,\lambda_{n}}(a_{1},\cdots,a_{n})$
(conformal antilinearity),

(2) $\gamma$ is skew-symmetric with respect to simultaneous permutations of
$a_{i}$’s and $\lambda_{i}$’s, namely,

$\displaystyle\ \ \ \ \ \ \ \ \
\gamma_{\lambda_{1},\cdots,\lambda_{i-1},\lambda_{i+1},\lambda_{i},\lambda_{i+2},\cdots,\lambda_{n}}(a_{1},\cdots,a_{i-1},a_{i+1},a_{i},\alpha_{i+2},\cdots,a_{n})$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
=-\gamma_{\lambda_{1},\cdots,\lambda_{i-1},\lambda_{i},\lambda_{i+1},\lambda_{i+2},\cdots,\lambda_{n}}(a_{1},\cdots,a_{i-1},a_{i},a_{i+1},a_{i+2},\cdots,a_{n}).$
(4.5)

As usual, let $\mathcal{R}^{\otimes 0}={\mathbb{C}}$, so that a $0$-cochain is
an element of $M$. Denote by ${\tilde{C}}^{n}(\mathcal{R},M)$ the set of all
$n$-cochains. The differential $d$ of an $n$-cochain $\gamma$ is defined by

$\displaystyle(d\gamma)_{\lambda_{1},\cdots,\lambda_{n+1}}(a_{1},\cdots,a_{n+1})$
$\displaystyle=\mbox{$\sum\limits_{i=1}^{n+1}$}(-1)^{i+1}a_{i\,{\lambda_{i}}}\gamma_{\lambda_{1},\cdots,\hat{\lambda_{i}},\cdots,\lambda_{n+1}}(a_{1},\cdots,\hat{a_{i}},\cdots,a_{n+1})$
$\displaystyle+\mbox{$\sum\limits_{i,j=1;i<j}^{n+1}$}(-1)^{i+j}\gamma_{\lambda_{i}+\lambda_{j},\lambda_{1},\cdots,\hat{\lambda_{i}},\cdots,\hat{\lambda_{j}},\cdots,\lambda_{n+1}}\big{(}[a_{i\,{\lambda_{i}}}a_{j}],a_{1},\cdots,\hat{a_{i}},\cdots,\hat{a_{j}},\cdots,a_{n+1}\big{)},$
(4.6)

where $\gamma$ is extended linearly over the polynomials in $\lambda_{i}$. In
particular, if $\gamma\in M$ is a $0$-cochain, then
$(d\gamma)_{\lambda}(a)=a_{\lambda}\gamma$.

It is proved in [2] that the operator $d$ preserves the space of cochains and
$d^{2}=0$. Thus the cochains of a Lie conformal algebra $\mathcal{R}$ with
coefficients in $\mathcal{R}$-module $M$ form a complex, which is called the
basic complex and will be denoted by

$\displaystyle\tilde{C}^{\bullet}(\mathcal{R},M)=\mbox{$\bigoplus\limits_{n\in\mathbb{Z}^{+}}$}\tilde{C}^{n}(\mathcal{R},M).$

Moreover, define a ${\mathbb{C}}[\partial]$-module structure on
$\tilde{C}^{\bullet}(\mathcal{R},M)$ by

$\displaystyle(\partial\gamma)_{\lambda_{1},\cdots,\lambda_{n}}(a_{1},\cdots,a_{n})=(\partial_{M}+\mbox{$\sum\limits_{i=1}^{n}$}\lambda_{i})\gamma_{\lambda_{1},\cdots,\lambda_{n}}(a_{1},\cdots,a_{n}),$
(4.7)

where $\partial_{M}$ denotes the action of $\partial$ on $M$. Then
$d\partial=\partial d$ and thus
$\partial\tilde{C}^{\bullet}(\mathcal{R},M)\subset\tilde{C}^{\bullet}(\mathcal{R},M)$
forms a subcomplex. The quotient complex

$\displaystyle
C^{\bullet}(\mathcal{R},M)=\tilde{C}^{\bullet}(\mathcal{R},M)/\partial\tilde{C}^{\bullet}(\mathcal{R},M)=\mbox{$\bigoplus\limits_{n\in\mathbb{Z}^{+}}$}C^{n}(\mathcal{R},M)$

is called the reduced complex.

######  Definition 4.4.

The basis cohomology $\tilde{H}^{\bullet}(\mathcal{R},M)$ of a Lie conformal
algebra $\mathcal{R}$ with coefficients in $\mathcal{R}$-module $M$ is the
cohomology of the basis complex $\tilde{C}^{\bullet}(\mathcal{R},M)$ and the
(reduced) cohomology $H^{\bullet}(\mathcal{R},M)$ is the cohomology of the
reduced complex $C^{\bullet}(\mathcal{R},M)$.

######  Remark 4.5.

The basic cohomology $\tilde{H}^{\bullet}(\mathcal{R},M)$ is naturally a
${\mathbb{C}}[\partial]$-module, whereas the reduced cohomology
$H^{\bullet}(\mathcal{R},M)$ is a complex vector space.

For a $q$-cochain $\gamma\in{\tilde{C}}^{q}(\mathcal{R},M)$, we call $\gamma$
a $q$-cocycle if $d(\gamma)=0$; a $q$-coboundary or a trivial $q$-cocycle if
there is a $(q-1)$-cochain $\phi\in\tilde{C}^{q-1}(\mathcal{R},M)$ such that
$\gamma=d(\phi)$. Two cochains $\gamma_{1}$ and $\gamma_{2}$ are called
equivalent if $\gamma_{1}-\gamma_{2}$ is a coboundary. Denote by
$\tilde{D}^{q}(\mathcal{R},M)$ and $\tilde{B}^{q}(\mathcal{R},M)$ the spaces
of $q$-cocycles and $q$-boundaries, respectively. By Definition 4.4,

$\displaystyle{\rm\tilde{H}}^{q}(\mathcal{R},M)=\tilde{D}^{q}(\mathcal{R},M)/\tilde{B}^{q}(\mathcal{R},M)=\\{\mbox{equivalent
classes of $q$-cocycles}\\}.$

The main results of this section are the following theorem.

######  Theorem 4.6.

For the Lie conformal algebra $\mathcal{B}$, the following statements hold.

  * (1)

For the trivial module ${\mathbb{C}}$, we have

$\displaystyle{\rm
dim\,\tilde{H}}^{q}(\mathcal{B},{\mathbb{C}})=\left\\{\begin{array}[]{ll}1&{\mbox{i}f}\
q=0,\\\ 0&{\mbox{i}f}\ q=1,{\mbox{o}r}\ 2,\end{array}\right.$

and

$\displaystyle{\rm
dim\,H}^{q}(\mathcal{B},{\mathbb{C}})=\left\\{\begin{array}[]{ll}1&{\mbox{i}f}\
q=0,{\mbox{o}r}\ 2,\\\ 0&{\mbox{i}f}\ q=1.\end{array}\right.$

  * (2)

If $a\neq 0$, then ${\rm dim\,H}^{\bullet}({\mathcal{B}},{\mathbb{C}}_{a})=0$.

  * (3)

If $\alpha\neq 0$, then ${\rm
dim\,H}^{\bullet}(\mathcal{B},M_{\Delta,\alpha})=0$.

Proof.  (1) Since a $0$-cochain $\gamma$ is an element of ${\mathbb{C}}$ and
$(d\gamma)_{\lambda}(a)=a_{\lambda}\gamma=0$ for $a\in\mathcal{B}$, we have
$\tilde{D}^{0}(\mathcal{B},{\mathbb{C}})=\tilde{C}^{0}(\mathcal{B},{\mathbb{C}})={\mathbb{C}}$
and $\tilde{B}^{0}(\mathcal{B},{\mathbb{C}})=0$. Thus $\tilde{\rm
H}^{0}(\mathcal{B},{\mathbb{C}})=\tilde{D}^{0}(\mathcal{B},{\mathbb{C}})/\tilde{B}^{0}(\mathcal{B},{\mathbb{C}})={\mathbb{C}}$,
and ${\rm H}^{0}(\mathcal{B},{\mathbb{C}})={\mathbb{C}}$ because
$\partial\tilde{C}^{0}(\mathcal{B},{\mathbb{C}})=\partial{\mathbb{C}}=0$.

Let $\gamma\in\tilde{C}^{1}(\mathcal{B},{\mathbb{C}})$ and
$d\gamma\in\partial\tilde{C}^{2}(\mathcal{B},{\mathbb{C}})$, namely, there is
$\phi\in\tilde{C}^{2}(\mathcal{B},{\mathbb{C}})$ such that
$d(\gamma)=\partial\phi$. By (4), (4.7) and $\partial{\mathbb{C}}=0$,

$\displaystyle\gamma_{\lambda_{1}+\lambda_{2}}([a_{\lambda_{1}}b])=-(d\gamma)_{\lambda_{1},\lambda_{2}}(a,b)=-(\partial\phi)_{\lambda_{1},\lambda_{2}}(a,b)=-(\lambda_{1}+\lambda_{2})\phi_{\lambda_{1},\lambda_{2}}(a,b),\
\ a,b\in{\mathcal{B}},$ (4.10)

By (1.1), (4.10) and Definition 4.3 (1),

$\displaystyle((i+1)\lambda_{1}-\lambda_{2})\gamma_{\lambda_{1}+\lambda_{2}}(J_{i})=-(\lambda_{1}+\lambda_{2})\phi_{\lambda_{1},\lambda_{2}}(J_{0},J_{i}),\
i\geq 0.$ (4.11)

Setting $\lambda=\lambda_{1}+\lambda_{2}$ in (4.11) gives

$\displaystyle((i+1)\lambda-(i+2)\lambda_{2})\gamma_{\lambda}(J_{i})=-\lambda\phi_{\lambda_{1},\lambda_{2}}(L,J_{i}),\
i\geq 0,$

which implies that $\gamma_{\lambda}(J_{i})$ is divisible by $\lambda$. We can
define a $1$-cochain
$\gamma^{\prime}\in\tilde{C}^{1}({\mathcal{B}},{\mathbb{C}})$ by

$\displaystyle\gamma^{\prime}_{\lambda}(J_{i})=\lambda^{-1}\gamma_{\lambda}(J_{i}),\
\mbox{for}\ i\geq 0.$ (4.12)

Since $\partial{\mathbb{C}}=0$,
$\gamma=\partial\gamma^{\prime}\in\partial\tilde{C}^{1}({\mathcal{B}},{\mathbb{C}})$.
Hence ${\rm H}^{1}({\mathcal{B}},{\mathbb{C}})=0$. If $\gamma$ is a 1-cocycle,
namely, $\phi=0$ in (4.10), then (4.11) gives $\gamma=0$. Thus
${\rm\tilde{H}}^{1}({\mathcal{B}},{\mathbb{C}})=0$.

Let $\psi\in\tilde{D}^{2}({\mathcal{B}},{\mathbb{C}})$ be a 2-cocycle. We have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(d\psi)_{\lambda_{1},\lambda_{2},\lambda_{3}}(J_{i},J_{0},J_{0})|_{\lambda_{3}=0}$
(4.13) $\displaystyle=$
$\displaystyle-(\lambda_{1}-(i+1)\lambda_{2})\psi_{\lambda_{1}+\lambda_{2},\lambda_{3}}(J_{i},J_{0})|_{\lambda_{3}=0}+(\lambda_{1}-(i+1)\lambda_{3})\psi_{\lambda_{1}+\lambda_{3},\lambda_{2}}(J_{i},J_{0})|_{\lambda_{3}=0}$
$\displaystyle-(\lambda_{2}-\lambda_{3})\psi_{\lambda_{2}+\lambda_{3},\lambda_{1}}(J_{0},J_{i})|_{\lambda_{3}=0}$
$\displaystyle=$
$\displaystyle-((\lambda_{1}-(i+1)\lambda_{2})\psi_{\lambda_{1}+\lambda_{2},0}(J_{i},J_{0})+(\lambda_{1}+\lambda_{2})\psi_{\lambda_{1},\lambda_{2}}(J_{i},J_{0}).$

Setting $\lambda=\lambda_{1}+\lambda_{2}$ in (4.13) gives
$((\lambda-(i+2)\lambda_{2})\psi_{\lambda,0}(J_{i},J_{0})=\lambda\psi_{\lambda_{1},\lambda_{2}}(J_{i},J_{0})$.
Thus $\psi_{\lambda,0}(J_{i},J_{0})$ is divisible by $\lambda$. Define a
$1$-cochain $f$ by

$\displaystyle
f_{\lambda_{1}}(J_{i})=\lambda_{1}^{-1}\psi_{\lambda_{1},\lambda}(J_{i},J_{0})|_{\lambda=0},\
\mbox{for}\ i\geq 0.$

Set $\gamma=\psi+df$, which is also a 2-cocycle. For all $i\geq 0$,

$\displaystyle\gamma_{\lambda_{1},\lambda}(J_{i},J_{0})|_{\lambda=0}=\psi_{\lambda_{1},\lambda}(J_{i},J_{0})|_{\lambda=0}-\lambda_{1}f_{\lambda_{1}}(J_{i})=0.$
(4.14)

By (4.14) and (4.13) with $\gamma$ in place of $\psi$, we have
$(\lambda_{1}+\lambda_{2})\gamma_{\lambda_{1},\lambda_{2}}(J_{i},J_{0})=0.$
Therefore
$\gamma_{\lambda_{1},\lambda_{2}}(J_{i},J_{0})=0=\gamma_{\lambda_{1},\lambda_{2}}(J_{i},J_{0})$.
With this,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(d\gamma)_{\lambda_{1},\lambda_{2},\lambda}(J_{0},J_{i},J_{k})|_{\lambda=0}$
(4.15) $\displaystyle=$
$\displaystyle-\gamma_{\lambda_{1}+\lambda_{2},\lambda}([J_{0\,{\lambda_{1}}}J_{i}],J_{k})|_{\lambda=0}+\gamma_{\lambda_{1}+\lambda,\lambda_{2}}([J_{0\,{\lambda_{1}}}J_{k}],J_{i})|_{\lambda=0}-\gamma_{\lambda_{2}+\lambda,\lambda_{1}}([J_{i\,{\lambda_{2}}}J_{k}],J_{0})|_{\lambda=0}$
$\displaystyle=$
$\displaystyle-((i+1)\lambda_{1}-\lambda_{2})\gamma_{\lambda_{1}+\lambda_{2},0}(J_{i},J_{k})-(k+1)\lambda_{1}\gamma_{\lambda_{2},\lambda_{1}}(J_{i},J_{k}).$

Setting $\lambda_{1}=0$ in (4.15) gives
$\gamma_{\lambda_{2},0}(J_{i},J_{k})=0$ and thus
$\gamma_{\lambda_{2},\lambda_{1}}(J_{i},J_{k})=0$. This proves $\gamma=0$.
Hence ${\rm\tilde{H}}^{2}({\mathcal{B}},{\mathbb{C}})=0$.

It remains to compute ${\rm H}^{2}({\mathcal{B}},{\mathbb{C}})$. Following
[12], we define a linear map $\sigma\gamma:\mathcal{B}^{\otimes
q}\rightarrow{\mathbb{C}}[\lambda_{1},\cdots,\lambda_{q-1}]$ for $q\geq 2$ by

$\displaystyle(\sigma\gamma)(\ a_{1}\otimes\cdots\otimes
a_{n})=\gamma_{\lambda_{1},\cdots,\lambda_{q}}(a_{1},\cdots,a_{q})|_{\lambda_{q}=-\lambda_{1}-\cdots-\lambda_{q-1}},\
\ a_{1},\cdots,a_{q}\in{\mathcal{B}}.$ (4.16)

We define $\sigma\gamma=\gamma$ if $q=0$ and
$\sigma\gamma(a_{1})=\gamma_{\lambda_{1}}(a_{1})|_{\lambda_{1}=0}$ if $q=1$.
Set $C^{\prime
q}(\mathcal{B},{\mathbb{C}})=\\{\sigma\gamma|\gamma\in{\tilde{C}}^{q}(\mathcal{B},{\mathbb{C}})\\}$.
Obviously, $\sigma:{\tilde{C}}^{q}(\mathcal{B},{\mathbb{C}})\rightarrow
C^{\prime q}(\mathcal{B},{\mathbb{C}})$ is a surjective map. If
$\gamma\in\partial{\tilde{C}}^{q}(\mathcal{B},{\mathbb{C}})=(\mbox{$\sum_{i=1}^{q}$}\lambda_{i})\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}})$,
then $\sigma\gamma=0$. That is, $\sigma$ factors to a map
$\sigma:C^{q}(\mathcal{B},{\mathbb{C}})\rightarrow C^{\prime
q}(\mathcal{B},{\mathbb{C}})$.

We claim that $\sigma:C^{q}(\mathcal{B},{\mathbb{C}})\rightarrow C^{\prime
q}(\mathcal{B},{\mathbb{C}})$ is an isomorphism as vector spaces. In fact, if
$\sigma\gamma=0$ for a $q$-cochain $\gamma$, then, by (4.16),
$\gamma_{\lambda_{1},\cdots,\lambda_{q}}(a_{1},\cdots,a_{q})$ as a polynomial
in $\lambda_{q}$ has a root
$\lambda_{q}=-\mbox{$\sum_{i=1}^{q-1}$}\lambda_{i}$, namely, it is divided by
$\mbox{$\sum_{i=1}^{q}$}\lambda_{i}$. Thus we get a $q$-cochain

$\displaystyle\gamma^{\prime}_{\lambda_{1},\cdots,\lambda_{q}}(a_{1},\cdots,a_{q})=(\mbox{$\sum_{i=1}^{q}$}\lambda_{i})^{-1}\gamma_{\lambda_{1},\cdots,\lambda_{q}}(a_{1},\cdots,a_{q}),$
(4.17)

and
$\gamma=(\mbox{$\sum_{i=1}^{q}$}\lambda_{i})\gamma^{\prime}\in\partial{\tilde{C}}^{q}(\mathcal{B},{\mathbb{C}})$,
which proves that $\sigma$ is injective. Hence the claim is true.

In the following we can identify $C^{q}(\mathcal{B},{\mathbb{C}})$ with
$C^{\prime q}(\mathcal{B},{\mathbb{C}})$. We still call an element in
$C^{\prime q}(\mathcal{B},{\mathbb{C}})$ a reduced $q$-cochain. By defining
the operator $d^{\prime}:C^{\prime q}(\mathcal{B},{\mathbb{C}})\rightarrow
C^{\prime q+1}(\mathcal{B},{\mathbb{C}})$ by $d^{\prime}(\sigma\gamma)=\sigma
d\gamma$, we have similar notions of reduced $q$-cocycle and $q$-coboundary.
For convenience, we will abbreviate
$\gamma_{\lambda_{1},\cdots,\lambda_{q}}(a_{1},\cdots,a_{q})|_{\lambda_{q}=-\lambda_{1}-\cdots-\lambda_{q-1}}$
to $\gamma_{\lambda_{1},\cdots,\lambda_{q-1}}(a_{1},\cdots,a_{q})$.

Let $\psi^{\prime}=\sigma\psi\in C^{\prime 2}({\mathcal{B}},{\mathbb{C}})$ be
a reduced 2-cochain. By (4) and (4.16),

$\displaystyle(d\psi^{\prime})_{\lambda_{1},\lambda_{2}}(a_{1},a_{2},a_{3})$
$\displaystyle\ \ \ \ \ \ \
\,=-\psi^{\prime}_{\lambda_{1}+\lambda_{2}}([a_{1\,\lambda_{1}}a_{2}],a_{3})+\psi^{\prime}_{-\lambda_{2}}([a_{1\,\lambda_{1}}a_{3}],a_{2})-\psi^{\prime}_{-\lambda_{1}}([a_{2\,\lambda_{2}}a_{3}],a_{1}),$
(4.18)

for $a_{1},a_{2},a_{3}\in{\mathcal{B}}$. Define a reduce 1-cochain
$f^{\prime}=\sigma f\in C^{\prime 1}({\mathcal{B}},{\mathbb{C}})$ by

$\displaystyle
f^{\prime}(J_{i})=(i+2)^{-1}\frac{d}{d\lambda}\psi^{\prime}_{\lambda}(J_{i},J_{0})|_{\lambda=0},\
\mbox{for}\ i\geq 0.$ (4.19)

Note that $f^{\prime}(a)=f_{\lambda}(a)|_{\lambda=0}$. Thus $f^{\prime}$ is
simply a linear function from ${\mathcal{B}}$ to ${\mathbb{C}}$, satisfying
$f^{\prime}(\partial a)=f_{\lambda}(\partial a)|_{\lambda=0}=-\lambda
f_{\lambda}(a)|_{\lambda=0}=0$, and

$\displaystyle(df^{\prime})_{\lambda}(a_{1},a_{2})=-f^{\prime}([a_{1\,\lambda}a_{2}]),\
\mbox{for}\ a_{1},\ a_{2}\in{\mathcal{B}}.$ (4.20)

If $\psi^{\prime}$ is a reduced 2-cocycle, then
$\gamma^{\prime}=\psi^{\prime}+df^{\prime}$ is a reduced 2-cocycle, equivalent
to $\psi^{\prime}.$ By (4.19) and (4.20),

$\displaystyle\frac{d}{d\lambda}\gamma^{\prime}_{\lambda}(J_{i},J_{0})|_{\lambda=0}=0,\
\mbox{for}\ i\geq 0.$ (4.21)

This, along with (4) and (4.5), gives

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\partial}{\partial\lambda}(d\gamma^{\prime})_{\lambda_{1},\lambda}(J_{i},J_{k},J_{0})|_{\lambda=-\lambda_{1}}$
(4.22) $\displaystyle=$
$\displaystyle\frac{\partial}{\partial\lambda}\big{(}-\gamma^{\prime}_{\lambda_{1}+\lambda}([J_{i\,{\lambda_{1}}}J_{k}],J_{0})+\gamma^{\prime}_{-\lambda}([J_{i\,{\lambda_{1}}}J_{0}],J_{k})-\gamma^{\prime}_{-\lambda_{1}}([J_{k\,{\lambda}}J_{0}],J_{i})\big{)}|_{\lambda=-\lambda_{1}}$
$\displaystyle=$
$\displaystyle\frac{\partial}{\partial\lambda}\big{(}((i+1)(\lambda+\lambda_{1})+\lambda_{1})\gamma^{\prime}_{-\lambda}(J_{i},J_{k})-((k+1)(\lambda+\lambda_{1})+\lambda)\gamma^{\prime}_{-\lambda_{1}}(J_{k},J_{i})\big{)}|_{\lambda=-\lambda_{1}}.$
$\displaystyle=$
$\displaystyle(i+k+3)\gamma^{\prime}_{\lambda_{1}}(J_{i},J_{k})-\lambda_{1}\frac{\partial}{\partial\lambda_{1}}\gamma^{\prime}_{\lambda_{1}}(J_{i},J_{k}).$

Thus,

$\displaystyle\gamma^{\prime}_{\lambda}(J_{i},J_{k})=c_{i,k}\lambda^{i+k+3},\
\mbox{for some}\ c_{i,k}\in{\mathbb{C}}.$ (4.23)

By (4) with $\gamma^{\prime}$ in place of $\psi^{\prime}$ and (4.23),

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\gamma^{\prime}_{\lambda_{1}+\lambda_{2}}([J_{i\,\lambda_{1}}J_{j}],J_{k})+\gamma^{\prime}_{-\lambda_{2}}([J_{i\,\lambda_{1}}J_{k}],J_{j})-\gamma^{\prime}_{-\lambda_{1}}([J_{j\,\lambda_{2}}J_{k}],J_{i})$
(4.24) $\displaystyle=$
$\displaystyle-((j+1)\lambda_{1}-(i+1)\lambda_{2})c_{i+j,k}(\lambda_{1}+\lambda_{2})^{i+j+k+3}$
$\displaystyle+((i+1)\lambda_{2}+(i+k+2)\lambda_{1})c_{i+k,j}(-\lambda_{2})^{i+j+k+3}$
$\displaystyle-((j+1)\lambda_{1}+(j+k+2)\lambda_{2})c_{j+k,i}(-\lambda_{1})^{i+j+k+3}.$

Taking $i=j=0$ in (4.24), we get

$\displaystyle
c_{0,k}(\lambda_{1}-\lambda_{2})(\lambda_{1}+\lambda_{2})^{k+3}=c_{k,0}\big{(}(\lambda_{2}+(k+2)\lambda_{1})(-\lambda_{2})^{k+3}-(\lambda_{1}+(k+2)\lambda_{2})(-\lambda_{1})^{k+3}\big{)}.$
(4.25)

Setting $\lambda_{2}=0$ in (4.25) gives $c_{0,k}=(-1)^{k}c_{k,0}$. Comparing
coefficients of $\lambda_{1}^{2}\lambda_{2}^{k+2}$ in (4.25) gives
$c_{0,k}=c_{k,0}=0$ for $k\geq 1$. Setting $i=0$ in (4.24) and comparing
coefficients of $\lambda_{1}^{j+k+4}$, we obtain $c_{j,k}=0$ for $j,k\geq 1$.
Thus, by (4.23), there exists a nonzero complex number $c$, such that

$\displaystyle\gamma^{\prime}_{\lambda}(J_{0},J_{0})=c\lambda^{3},\ \
\gamma^{\prime}_{\lambda}(J_{i},J_{j})=0,\ \mbox{for}\ i,\ j\geq 1.$ (4.26)

Therefore, ${\rm dim\,H}^{2}({\mathcal{B}},{\mathbb{C}})=1$, which proves (1).

(2) Define an operator
$\tau:\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}}_{a})\rightarrow\tilde{C}^{q-1}({\mathcal{B}},{\mathbb{C}}_{a})$
by

$\displaystyle(\tau\gamma)_{\lambda_{1},\cdots,\lambda_{q-1}}(a_{1},\cdots,a_{q-1})=(-1)^{q-1}\gamma_{\lambda_{1},\cdots,\lambda_{q-1},\lambda}(a_{1},\cdots,a_{q-1},J_{0})|_{\lambda=0},$
(4.27)

for $a_{1},\cdots,a_{q-1}\in{\mathcal{B}}$. By the fact that
$\partial\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}}_{a})=(a+\sum_{i=1}^{q}\lambda_{i})\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}}_{a})$
and (4.27),

$\displaystyle((d\tau+\tau
d)\gamma)_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})$
$\displaystyle=$
$\displaystyle\big{(}\mbox{$\sum_{i=1}^{q}$}\lambda_{i}\big{)}\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})$
(4.28) $\displaystyle\equiv$
$\displaystyle-a\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})\
(\mbox{mod}\ \partial\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}}_{a}).\ \ $

Suppose that $\gamma$ is a $q$-cochain such that
$d\gamma\in\partial\tilde{C}^{q+1}({\mathcal{B}},{\mathbb{C}}_{a})$, namely,
there is a $(q+1)$-cochain $\phi$ such that
$d\gamma=(a+\sum_{i=1}^{q+1}\lambda_{i})\phi$. By (4.27), $\tau
d\gamma=(a+\sum_{i=1}^{q}\lambda_{i})\tau\phi\in\partial\tilde{C}^{q}({\mathcal{B}},{\mathbb{C}}_{a})$.
By(4.28), $\gamma\equiv-d(a^{-1}\tau\gamma)$ is a reduced coboundary because
$a\neq 0$. Thus ${\rm H}^{q}(\mathcal{B},{\mathbb{C}}_{a})=0$ for $q\geq 0$.
This proves (2).

(3) Note that
$\partial\tilde{C}^{q}({\mathcal{B}},M_{\Delta,\alpha})=(\partial+\sum_{i=1}^{q}\lambda_{i})\tilde{C}^{q}({\mathcal{B}},M_{\Delta,\alpha})$.
Similarly to the proof of (2), we define an operator
$\kappa:C^{q}({\mathcal{B}},M_{\Delta,\alpha})\rightarrow
C^{q-1}({\mathcal{B}},M_{\Delta,\alpha})$ by

$\displaystyle(\kappa\gamma)_{\lambda_{1},\cdots,\lambda_{q-1}}(a_{1},\cdots,a_{q-1})=(-1)^{q-1}\gamma_{\lambda_{1},\cdots,\lambda_{q-1},\lambda}(a_{1},\cdots,a_{q-1},J_{0})|_{\lambda=0},$

for $a_{1},\cdots,a_{q-1}\in{\mathcal{B}}$. By Theorem 4.2,

$\displaystyle((d\kappa+\kappa
d)\gamma)_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})$
$\displaystyle\ \ \
=J_{0\,\lambda}\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})|_{\lambda=0}+\big{(}\mbox{$\sum\limits_{i=1}^{q}$}\lambda_{i}\big{)}\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})$
$\displaystyle\ \ \
=\big{(}\partial+\alpha+\mbox{$\sum\limits_{i=1}^{q}$}\lambda_{i}\big{)}\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})$
$\displaystyle\ \ \
\equiv\alpha\gamma_{\lambda_{1},\cdots,\lambda_{q}}(J_{n_{1}},\cdots,J_{n_{q}})\
(\mbox{mod}\ \partial\tilde{C}^{q}({\mathcal{B}},M_{\Delta,\alpha})).$ (4.29)

If $\gamma$ is a reduced $q$-cocycle, then there is a $(q+1)$-cochain
$\varphi$ such that
$d\gamma=\partial\varphi=(\partial+\sum_{i=1}^{q+1}\lambda_{i})\varphi$. In
this case, $\kappa
d\gamma=(\partial+\sum_{i=1}^{q}\lambda_{i})\kappa\phi\in\partial\tilde{C}^{q}({\mathcal{B}},M_{\Delta,\alpha})$.
It follows from (4) that $\gamma\equiv d(\alpha^{-1}\kappa\gamma)$ is a
reduced $q$-coboundary, since we assume $\alpha\neq 0$. Hence ${\rm
H}^{q}(\mathcal{B},M_{\Delta,\alpha})=0$ for $q\geq 0$.

This completes the proof of Theorem 4.6. $\Box$

######  Corollary 4.7.

There is a unique nontrivial universal central extension
$\tilde{\mathcal{B}}={\mathcal{B}}\oplus{\mathbb{C}}{\mathfrak{c}}$ of the Lie
conformal algebra ${\mathcal{B}}$, satisfying

$\displaystyle[J_{0\,\lambda}J_{0}]$ $\displaystyle=$
$\displaystyle(\partial+2\lambda)J_{0}+\lambda^{3}{\mathfrak{c}},$
$\displaystyle{[J_{i\,\lambda}J_{j}]}$ $\displaystyle=$
$\displaystyle((i+1)\partial+(i+j+2)\lambda)J_{i+j},\ \mbox{for}\ i,j>0.$

######  Remark 4.8.

The formal distribution Lie algebra corresponding to $\tilde{\mathcal{B}}$ is
a well-known Lie algebra of Block type studied in [16, 18].

##  5 Vertex Poisson algebra structure associated to ${\mathcal{B}}$

Denote $V=\mbox{$\bigoplus_{i\in\mathbb{Z}^{+}}{\mathbb{C}}J_{i}$}$. Thus the
Lie conformal algebra of Block type ${\mathcal{B}}$ is a free
${\mathbb{C}}[\partial]$-module over $V$. By (2.4), the $\lambda$-bracket
(1.1) is equivalent to the following $j$-products

$\displaystyle J_{i\,{(0)}}J_{k}=(i+1)\partial J_{i+k},\ \
J_{i\,{(1)}}J_{k}=(i+k+2)J_{i+k},\ \ J_{i\,{(n)}}J_{k}=0,$ (5.1)

for $i,k\in\mathbb{Z}^{+}$, $n\geq 2.$ Define a linear map $Y_{-}(\cdot,z)$
from $V$ to $z^{-1}({\rm End\,}V)[[z^{-1}]]$ by

$\displaystyle
Y_{-}(J_{i},z)=\mbox{$\sum\limits_{n\in\mathbb{Z}^{+}}$}J_{i\,{(n)}}z^{-n-1},\
\mbox{for}\ i\geq 0.$ (5.2)

From (5.1) and (5.2), we have

$\displaystyle Y_{-}(J_{i},z)J_{k}=(i+1)\partial
J_{i+k}z^{-1}+(i+k+2)J_{i+k}z^{-2},\ \mbox{for}\ i,k\in\mathbb{Z}^{+}.$ (5.3)

Extending the map $Y_{-}(\cdot,z)$ to the whole
${\mathcal{B}}={\mathbb{C}}[\partial]\otimes_{\mathbb{C}}V$ by

$\displaystyle
Y_{-}(f(\partial)J_{i},z)\big{(}\partial^{m}J_{k}\big{)}=f(d/dz)\mbox{$\sum\limits_{l=0}^{m}$}(-1)^{l}\partial^{m-l}(d/dz)^{l}Y_{-}(J_{i},z)J_{k},$
(5.4)

then $(\mathcal{B},Y_{-},\partial)$ forms a free vertex Lie algebra. Indeed,
(5.3) and (5.4) guarantee axioms (2.10) and (2.11). It suffices to check
(2.12) and (2.14) on the generators. For $i,j\in\mathbb{Z}^{+}$,

$\displaystyle{\rm Sing}(e^{z\partial}Y_{-}(J_{j},-z)J_{i})$ $\displaystyle=$
$\displaystyle(j+1)\partial
J_{i+j}(-z)^{-1}+(i+j+2)J_{i+j}z^{-2}+(i+j+2)\partial J_{i+j}z^{-1}$ (5.5)
$\displaystyle=$ $\displaystyle(i+1)\partial
J_{i+j}z^{-1}+(i+j+2)J_{i+j}z^{-2}$ $\displaystyle=$ $\displaystyle
Y_{-}(J_{i},z)J_{j},$

which proves (2.12). Furthermore, for any $i,j,k\in\mathbb{Z}^{+}$,

$\displaystyle Y_{-}(J_{i},z_{1})Y_{-}(J_{j},z_{2})J_{k}$ $\displaystyle\ \
=(j+1)(i+1)\partial^{2}J_{i+j+k}z_{1}^{-1}z_{2}^{-1}+(j+1)(2i+j+k+3)\partial
J_{i+j+k}z_{1}^{-2}z_{2}^{-1}$ $\displaystyle\ \ \ \ \
+2(j+1)(i+j+k+2)J_{i+j+k}z_{1}^{-3}z_{2}^{-1}+(i+1)(j+k+2)\partial
J_{i+j+k}z_{1}^{-1}z_{2}^{-2}$ $\displaystyle\ \ \ \ \
+(j+k+2)(i+j+k+2)J_{i+j+k}z_{1}^{-2}z_{2}^{-2},$ (5.6)

and

$\displaystyle{\rm Sing}\big{(}\mbox{$\sum_{n\geq
0}$}(z_{1}-z_{2})^{-n-1}Y_{-}(J_{i\,{(n)}}J_{j},z_{2})\big{)}J_{k}$
$\displaystyle\ ={\rm
Sing}\big{(}(z_{1}-z_{2})^{-1}Y_{-}(J_{i\,{(0)}}J_{j},z_{2})+(z_{1}-z_{2})^{-2}Y_{-}(J_{\,i{(1)}}J_{j},z_{2})\big{)}J_{k}$
$\displaystyle\ ={\rm Sing}\big{(}(i+1)(z_{1}-z_{2})^{-1}Y_{-}(\partial
J_{i+j},z_{2})+(i+j+2)(z_{1}-z_{2})^{-2}Y_{-}(J_{i+j},z_{2})\big{)}J_{k}$
$\displaystyle\ =-(i+1)(i+j+1)\partial
J_{i+j+k}(z_{1}^{-1}z_{2}^{-2}+z_{1}^{-2}z_{2}^{-1})+(i+j+2)(i+j+1)\partial
J_{i+j+k}z_{1}^{-1}z_{2}^{-2}$ $\displaystyle\ \ \ \
-2(i+1)(i+j+k+2)J_{i+j+k}(z_{1}^{-1}z_{2}^{-3}+z_{1}^{-2}z_{2}^{-2}+z_{1}^{-3}z_{2}^{-1})$
$\displaystyle\ \ \ \
+(i+j+2)(i+j+k+2)J_{i+j+k}(z_{1}^{-2}z_{2}^{-2}+2z_{1}^{-3}z_{2}^{-1})$
$\displaystyle\
=Y_{-}(J_{i},z_{1})Y_{-}(J_{j},z_{2})J_{k}-Y_{-}(J_{j},z_{1})Y_{-}(J_{i},z_{2})J_{k},$
(5.7)

where the last equality follows from (5). This proves (2.14) and thus
$(\mathcal{B},Y_{-},\partial)$ is a vertex Lie algebra, called the vertex Lie
algebra of Block type.

######  Remark 5.1.

By [8], the vertex Lie algebra $(\mathcal{B},Y_{-},\partial)$ is equivalent to
a Novikov algebra $(V,\circ)$ with the operation $\circ$ defined by

$\displaystyle J_{i}\circ J_{j}=(j+1)J_{i+j},\ \mbox{for}\ \ i,j\geq 0.$ (5.8)

Moreover, the commutator on $(V,\circ)$ of the form

$\displaystyle[J_{i},J_{j}]=(j-i)J_{i+j},\ \ \mbox{for}\ \ i,j\geq 0,$ (5.9)

make $V$ into a Lie algebra. The Lie algebra $(V,[\cdot,\cdot])$ is a
subalgebra of the one-sided Witt algebra $W_{1}^{+}$ with a basis
$\\{J_{i}|i\geq-1\\}$, which occurs in the study of conformal field theory.

By a differential algebra, we mean a commutative associative algebra $A$ (with
$1$) equipped with a derivation $\partial$, denoted by $(A,\partial)$. A
subset $U$ of $A$ is said to generate $A$ as a differential algebra if
$\partial^{n}U$ for $n\geq 0$ generate $A$ as an algebra.

The following notion of vertex Poisson algebra is due to [7].

######  Definition 5.2.

A vertex Poisson algebra is a differential algebra $(A,\partial)$ equipped
with a vertex Lie algebra structure $(Y_{-},\partial)$ such that for $a,b,c\in
A$,

$\displaystyle Y_{-}(a,z)(bc)=(Y_{-}(a,z)b)c+b(Y_{-}(a,z)c),$ (5.10)

where $Y_{-}(a,z)=\mbox{$\sum_{n\geq 0}$}a_{n}z^{-n-1}$.

In terms of components, relation (5.10) is equivalent to

$\displaystyle a_{n}(bc)=(a_{n}b)c+b(a_{n}c)\ \mbox{for}\ a,b,c\in A,\ n\geq
0,$ (5.11)

namely, these $a_{n}$ are derivations of $A$. Therefore,

$\displaystyle Y_{-}(a,z)\in z^{-1}({\rm\,Der}A)[[z^{-1}]],\ \ \mbox{for}\
a\in A,$ (5.12)

which implies $Y_{-}(a,z)1=0$. Then $Y_{-}(1,z)a=0$ by (2.12), namely,
$Y_{-}(1,z)=0.$

Let $U$ be a vector space. Denote by $A=S({\mathbb{C}}[\partial]\otimes U)$
the symmetric algebra over the free ${\mathbb{C}}[\partial]$-module
${\mathbb{C}}[\partial]\otimes U$. The operator $\partial$ can be uniquely
extended to a derivation of $A$. Then $(A,\partial)$ forms a differential
algebra, which is referred to as the free differential algebra over $U$.
Following [10], a week pre-vertex Poisson structure on $A$ is a linear map
$Y_{-}^{0}$ from $U\times U$ to $z^{-1}A[z^{-1}]$ such that

$\displaystyle Y_{-}^{0}(a,z)b={\rm Sing}(e^{z\partial}Y_{-}^{0}(b,-z)a),\
\mbox{for}\ a,b\in A.$ (5.13)

By [10, Proposition 3.10], the operator $Y_{-}^{0}$ can be uniquely extended
to a linear map

$\displaystyle Y_{-}:A\rightarrow{\rm Hom}(A,z^{-1}A[z^{-1}]),\ \ \ a\mapsto
Y_{-}(a,z)=\mbox{$\sum\limits_{n\in\mathbb{Z}^{+}}$}a_{n}z^{-n-1}$

such that for $a,b\in A$,

$\displaystyle Y_{-}(a,z)$ $\displaystyle=$ $\displaystyle\mbox{$\sum_{n\geq
0}$}a_{n}z^{-n-1}\in z^{-1}({\rm Der}A)[[z^{-1}]],$ (5.14)
$\displaystyle{[\partial,Y_{-}(a,z)]}$ $\displaystyle=$
$\displaystyle{Y_{-}(\partial a,z)}=\frac{d}{dz}Y_{-}(a,z),$ (5.15)
$\displaystyle{Y_{-}(a,z)b}$ $\displaystyle=$ $\displaystyle{\rm
Sing}\big{(}e^{z\partial}Y_{-}(b,-z)a\big{)}.$ (5.16)

The following result, due to [10, Theorem 3.11], gives a general construction
of vertex Poisson algebras from free differential algebras.

######  Theorem 5.3.

Let $A$ be the free differential algebra over $U$ and let $Y_{-}^{0}$ be a
week pre-vertex Poisson structure on $A$ such that for $u,v,w\in U$ from an
ordered basis of $U$ with $u\leq v\leq w$,

$\displaystyle\tilde{Y}_{-}^{0}(u,z_{1})Y_{-}^{0}(v,z_{2})w-\tilde{Y}_{-}^{0}(v,z_{2})Y^{0}(u,z_{1})w={\rm
Sing}\big{(}e^{z_{2}\partial}\tilde{Y}_{-}^{0}(w,-z_{2})Y_{-}^{0}(u,z_{1}-z_{2})v\big{)},$
(5.17)

where $\tilde{Y}_{-}^{0}(u,z)\in z^{-1}({\rm Der})[[z^{-1}]]$ is uniquely
determined by

$\displaystyle\tilde{Y}_{-}^{0}(u,z)e^{z_{1}\partial}v=e^{z_{1}\partial}e^{-z_{1}\frac{d}{dz}}Y_{-}^{0}(u,z)v,\
\mbox{for}\ {v\in U}.$ (5.18)

Then $Y_{-}^{0}$ uniquely extends to a vertex Poisson structure $Y_{-}$ on
$A$.

The following result is an application of the above theorem.

######  Proposition 5.4.

Let $({\mathcal{B}},Y_{-},\partial)$ be the vertex Lie algebra of Block type
and $A=S({\mathcal{B}})$ the free differential algebra over
$V=\mbox{$\bigoplus\limits_{i\in\mathbb{Z}^{+}}$}{\mathbb{C}}J_{i}$. Define

$\displaystyle Y_{-}^{0}(J_{i},z)J_{j}=(i+1)\partial
J_{i+j}z^{-1}+(i+j+2)J_{i+j}z^{-2},\ \mbox{for}\ i,j\geq 0.$ (5.19)

Then $Y_{-}^{0}$ uniquely extends to a vertex Poisson algebra structure on
$A$.

Proof. By (5.3) and (5.19), $Y_{-}^{0}(J^{i},z)J^{j}=Y_{-}(J^{i},z)J^{j}$ for
$i,j\in\mathbb{Z}^{+}$. Thus $Y_{-}^{0}$ satisfies (5.13) and so it is a week
pre-vertex Poisson structure on $A$. By (5.18) and (5.19), for
$i,j,k\in\mathbb{Z}^{+}$,

$\displaystyle\tilde{Y}_{-}^{0}(J_{i},z_{1})Y_{-}^{0}(J_{j},z_{2})J_{k}$
$\displaystyle\ \ \ \ \ \
=(j+1)(i+1)\partial^{2}J_{i+j+k}z_{1}^{-1}z_{2}^{-1}+(j+1)(2i+j+k+3)\partial
J_{i+j+k}z_{1}^{-2}z_{2}^{-1}$ $\displaystyle\ \ \ \ \ \ \ \ \
\,+2(j+1)(i+j+k+2)J_{i+j+k}z_{1}^{-3}z_{2}^{-1}+(i+1)(j+k+2)\partial
J_{i+j+k}z_{1}^{-1}z_{2}^{-2}$ $\displaystyle\ \ \ \ \ \ \ \ \
\,+(j+k+2)(i+j+k+2)J_{i+j+k}z_{1}^{-2}z_{2}^{-2}.$ (5.20)

By (5.18)–(5),

$\displaystyle{\rm
Sing}\big{(}e^{z_{2}\partial}\tilde{Y}_{-}^{0}(J_{k},-z_{2})Y_{-}^{0}(J_{i},z_{1}-z_{2})J_{j}\big{)}$
$\displaystyle=$ $\displaystyle{\rm
Sing}\big{(}e^{z_{2}\partial}Y_{-}^{0}(J_{k},-z_{2})\big{(}(i+1)\partial
J_{i+j}(z_{1}-z_{2})^{-1}+(i+j+2)J_{i+j}(z_{1}-z_{2})^{-2}\big{)}\big{)}$
$\displaystyle=$ $\displaystyle{\rm Sing}\big{(}(i+1)Y_{-}(\partial
J_{i+j},z_{2})J_{k}(z_{1}-z_{2})^{-1}+(i+j+2)Y_{-}(J_{i+j},z_{2})J_{k}(z_{1}-z_{2})^{-2}\big{)}$
$\displaystyle=$ $\displaystyle{\rm Sing}\big{(}-(i+1)\big{(}(i+j+1)\partial
J_{i+j+k}z_{2}^{-2}+2(i+j+k+2)z_{2}^{-3}J_{i+j+k}\big{)}(z_{1}-z_{2})^{-1}$
$\displaystyle+(i+j+2)\big{(}(i+j+1)\partial
J_{i+j+k}z_{2}^{-1}+(i+j+k+2)z_{2}^{-2}J_{i+j+k}\big{)}(z_{1}-z_{2})^{-2}\big{)}$
$\displaystyle=$ $\displaystyle-(i+1)(i+j+1)\partial
J_{i+j+k}(z_{2}^{-2}z_{1}^{-1}+z_{1}^{-2}z_{2}^{-1})+(i+j+2)(i+j+1)\partial
J_{i+j+k}z_{2}^{-1}z_{1}^{-2}$
$\displaystyle-2(i+1)(i+j+k+2)J_{i+j+k}(z_{2}^{-3}z_{1}^{-1}+z_{2}^{-2}z_{1}^{-2}+z_{2}^{-1}z_{1}^{-3})$
$\displaystyle+(i+j+2)(i+j+k+2)J_{i+j+k}(z_{2}^{-2}z_{1}^{-2}+2z_{1}^{-3}z_{2}^{-1})$
$\displaystyle=$
$\displaystyle\tilde{Y}_{-}^{0}(L^{i},z_{1})Y_{-}^{0}(L^{j},z_{2})L^{k}-\tilde{Y}_{-}^{0}(L^{j},z_{2})Y_{-}^{0}(L^{i},z_{1})L^{k}.$

This proves the result by Theorem 5.3. $\Box$

Acknowledgements  This work was supported by National Natural Science
Foundation grants of China (11301109) and the Research Fund for the Doctoral
Program of Higher Education (20132302120042).

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