# On antisymmetric infinitesimal conformal bialgebras

Yanyong Hong  Department of Mathematics, Hangzhou Normal University, Hangzhou
311121, PR China<EMAIL_ADDRESS>and  Chengming Bai  Chern Institute of
Mathematics & LPMC, Nankai University, Tianjin 300071, PR China
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we construct a bialgebra theory for associative conformal
algebras, namely antisymmetric infinitesimal conformal bialgebras. On the one
hand, it is an attempt to give conformal structures for antisymmetric
infinitesimal bialgebras. On the other hand, under certain conditions, such
structures are equivalent to double constructions of Frobenius conformal
algebras, which are associative conformal algebras that are decomposed into
the direct sum of another associative conformal algebra and its conformal dual
as $\mathbb{C}[\partial]$-modules such that both of them are subalgebras and
the natural conformal bilinear form is invariant. The coboundary case leads to
the introduction of associative conformal Yang-Baxter equation whose
antisymmetric solutions give antisymmetric infinitesimal conformal bialgebras.
Moreover, the construction of antisymmetric solutions of associative conformal
Yang-Baxter equation is given from $\mathcal{O}$-operators of associative
conformal algebras as well as dendriform conformal algebras.

###### Key words and phrases:

associative conformal algebra, dendriform conformal algebra, associative
conformal Yang-Baxter equation, $\mathcal{O}$-operator, Rota-Baxter operator

###### 2010 Mathematics Subject Classification:

16D20, 16D70, 17A30, 17B38

##  1\. Introduction

The theory of Lie conformal algebras appeared as a formal language describing
the algebraic properties of the operator product expansion in two-dimensional
conformal field theory ([16]). In particular, Lie conformal algebras turn out
to be valuable tools in studying vertex algebras and Hamiltonian formalism in
the theory of nonlinear evolution equations ([3]). Moreover, Lie conformal
algebras have close connections to infinite-dimensional Lie algebras
satisfying the locality property ([17]). The conformal analogues of
associative algebras, namely, associative conformal algebras naturally
appeared in the representation theory of Lie conformal algebras ([9]). They
were studied widely on the structure theory ([4, 5, 6, 8, 10, 11, 18, 20, 22,
26, 27, 28, 29, 30, 31, 32]) as well as representation theory ([7, 21, 23]).
We would like to point that there are the “conformal analogues” for certain
algebras besides Lie and associative algebras or the “conformal structures” of
these algebras such as left-symmetric conformal algebras ([13]) and Jordan
conformal algebras ([19]).

It is natural to extend such structures to the bialgebras, that is, consider
the conformal analogues of bialgebras. In the case of Lie bialgebras, Liberati
in [24] developed a theory of Lie conformal bialgebras including the
introduction of the notions of conformal classical Yang-Baxter equation,
conformal Manin triples and conformal Drinfeld’s doubles. Similarly, a theory
of left-symmetric conformal bialgebras was developed in [14], which are
equivalent to a class of special Lie conformal algebras named parakähler Lie
conformal algebras and the notion of conformal $S$-equation was introduced in
the coboundary case. Moreover, the operator forms of the conformal classical
Yang-Baxter equation and the conformal $S$-equation were studied in [12],
which shows that the antisymmetric solutions of the conformal classical Yang-
Baxter equation and the symmetric solutions of the conformal $S$-equation can
be interpreted in terms of a kind of operators called $\mathcal{O}$-operators
in the conformal sense.

But as far as we know, there is not a conformal analogue of “associative
bialgebras” yet. In fact, there are two kinds of “associative bialgebras”. One
is the usual bialgebras in the theory of Hopf algebras, which the coproducts
are homomorphisms of the products. Another is the infinitesimal bialgebras,
which the coproducts are “derivations” of the products in certain sense,
introduced by Joni and Rota in [15] in order to provide an algebraic framework
for the calculus of divided difference. In particular, for the latter, in the
case of antisymmetric infinitesimal (ASI) bialgebras which are called
“associative D-algebras” in [33] or “balanced infinitesimal bialgebras” in the
sense of the opposite algebras in [1], there is a systematic study in terms of
their equivalences with double constructions of Frobenius algebras as well as
their relationships with associative Yang-Baxter equation ([2]).

In this paper, we develop a “conformal” theory for antisymmetric infinitesimal
bialgebras, namely antisymmetric infinitesimal (ASI) conformal bialgebras. It
is also a bialgebra theory for associative conformal algebras. That is, the
following diagram is commutative:

associative algebrasassociative conformal algebrasASI bialgebrasASI conformal
bialgebrasconformal structuresconformal structuresbialgebra structures
bialgebra structures

We would like to point out that such an approach might not be available for
considering the conformal analogues of the usual (associative) bialgebras and
even Hopf algebras, but can help to shed light on further studies on the
latter.

The main idea is to extend the study of ASI bialgebras given in [2] to the
“conformal case”. Explicitly, we first introduce the notion of double
constructions of Frobenius conformal algebras as a conformal analogue of
double constructions of Frobenius algebras, which are associative conformal
algebras that are decomposed into the direct sum of another associative
conformal algebra and its conformal dual as a $\mathbb{C}[\partial]$-module
such that both of them are subalgebras and the natural conformal bilinear form
is invariant. Such structures are interpreted equivalently in terms of matched
pairs of associative conformal algebras which were introduced in [11]. Finally
the notion of antisymmetric infinitesimal (ASI) conformal bialgebras is
introduced as equivalent structures of the aforementioned matched pairs of
associative conformal algebras as well as the double constructions of
Frobenius conformal algebras. Note that the notion of ASI conformal bialgebras
is available for any associative conformal algebra, whereas the equivalence
with double constructions of Frobenius conformal algebras is available for the
associative conformal algebras which are finitely generated and free as
$\mathbb{C}[\partial]$-modules.

As in the case of ASI bialgebras, the definition of coboundary ASI conformal
bialgebra is introduced and its study is also meaningful. It leads to the
introduction of associative conformal Yang-Baxter equation as a conformal
analogue of the associative Yang-Baxter equation. In particular, its
antisymmetric solutions give ASI conformal bialgebras. The associative
conformal Yang-Baxter equation is interpreted in terms of its operator forms
by introducing the notion of $\mathcal{O}$-operators of associative conformal
algebras, especially an antisymmetric solution of the associative conformal
Yang-Baxter equation corresponds to the skew-symmetric part of a conformal
linear map $T$, where $T_{0}=T_{\lambda}\mid_{\lambda=0}$ is an
$\mathcal{O}$-operator in the conformal sense and moreover, an
$\mathcal{O}$-operator of an associative conformal algebra gives an
antisymmetric solution of associative conformal Yang-Baxter equation in a
semi-direct product associative conformal algebra. Furthermore, we introduce
the notion of dendriform conformal algebras and show that for a dendriform
conformal algebra which is finite and free as a $\mathbb{C}[\partial]$-module,
there is a natural $\mathcal{O}$-operator on the associated conformal
associative algebra. Therefore there are constructions of antisymmetric
solutions of associative conformal Yang-Baxter equation and hence ASI
conformal bialgebras from $\mathcal{O}$-operators of associative conformal
algebras as well as dendriform conformal algebras.

This paper is organized as follows. In Section 2, the notions of an
associative conformal algebra, its bimodule and a matched pair of associative
conformal algebras are recalled. In Section 3, we introduce the notion of
double constructions of Frobenius conformal algebras and study their
relationship with matched pairs of associative conformal algebras. In Section
4, the notion of ASI conformal bialgebras is introduced as (under certain
conditions) equivalent structures of the aforementioned matched pair of
associative conformal algebras as well as the double constructions of
Frobenius conformal algebras. Section 5 is devoted to studying the coboundary
case of ASI conformal bialgebras and the associative conformal Yang-Baxter
equation is introduced. In Section 6, we introduce the notions of
$\mathcal{O}$-operators of associative conformal algebras and dendriform
conformal algebras to construct (antisymmetric) solutions of associative
conformal Yang-Baxter equation and hence give ASI conformal bialgebras.

Throughout this paper, we denote by $\mathbb{C}$ the field of complex numbers.
All tensors over $\mathbb{C}$ are denoted by $\otimes$. We denote the identity
map by $I$. Moreover, if $A$ is a vector space, then the space of polynomials
of $\lambda$ with coefficients in $A$ is denoted by $A[\lambda]$.

##  2\. Preliminaries on associative conformal algebras

We recall the notions of an associative conformal algebra, its bimodule and a
matched pair of associative conformal algebras. The interested readers may
consult [16] and [11] for more details.

######  Definition 2.1.

A conformal algebra $R$ is a $\mathbb{C}[\partial]$-module endowed with a
$\mathbb{C}$-bilinear map $\cdot_{\lambda}\cdot:R\times R\rightarrow
R[\lambda]$, $(a,b)\mapsto a_{\lambda}b$ satisfying

$\displaystyle(\partial a)_{\lambda}b=-\lambda a_{\lambda}b,\quad
a_{\lambda}(\partial b)=(\partial+\lambda)a_{\lambda}b,\quad\forall\ a,b\in
R.~{}~{}~{}~{}~{}~{}\quad\text{(conformal sesquilinearity)}$ (1)

An associative conformal algebra $R$ is a conformal algebra with the
$\mathbb{C}$-bilinear map $\cdot_{\lambda}\cdot:R\times R\rightarrow
R[\lambda]$ satisfying

$\displaystyle(a_{\lambda}b)_{\lambda+\mu}c=a_{\lambda}(b_{\mu}c),\quad\forall\
a,b,c\in R.$ (2)

A conformal algebra is called finite if it is finitely generated as a
$\mathbb{C}[\partial]$-module. The rank of a conformal algebra $R$ is its rank
as a $\mathbb{C}[\partial]$-module. The notions of a homomorphism, an ideal
and a subalgebra of an associative conformal algebra are defined as usual.

######  Example 2.2.

Let $(A,\cdot)$ be an associative algebra. Then
$\text{Cur}(A)=\mathbb{C}[\partial]\otimes A$ is an associative conformal
algebra with the following $\lambda$-product:

$\displaystyle(p(\partial)a)_{\lambda}(q(\partial)b)=p(-\lambda)q(\lambda+\partial)(a\cdot
b),\;\;\forall\ \text{$p(\partial)$, $q(\partial)\in\mathbb{C}[\partial]$,
$a$, $b\in A$.}$

######  Definition 2.3.

A left module $M$ over an associative conformal algebra $A$ is a
$\mathbb{C}[\partial]$-module endowed with a $\mathbb{C}$-bilinear map
$A\times M\longrightarrow M[\lambda]$, $(a,v)\mapsto
a\rightharpoonup_{\lambda}v$, satisfying the following axioms $(\forall\
a,b\in A,v\in M)$:
(LM1)$\qquad\qquad(\partial a)\rightharpoonup_{\lambda}v=-\lambda
a\rightharpoonup_{\lambda}v,~{}~{}~{}a\rightharpoonup_{\lambda}(\partial
v)=(\partial+\lambda)(a\rightharpoonup_{\lambda}v),$
(LM2)$\qquad\qquad(a_{\lambda}b)\rightharpoonup_{\lambda+\mu}v=a\rightharpoonup_{\lambda}(b\rightharpoonup_{\mu}v).$
We denote it by $(M,\rightharpoonup_{\lambda})$.

A right module $M$ over an associative conformal algebra $A$ is a
$\mathbb{C}[\partial]$-module endowed with a $\mathbb{C}$-bilinear map
$M\times A\longrightarrow M[\lambda]$, $(v,a)\mapsto
v\leftharpoonup_{\lambda}a$, satisfying the following axioms $(\forall\ a,b\in
A,v\in M)$:
(RM1)$\qquad\qquad(\partial v)\leftharpoonup_{\lambda}a=-\lambda
v\leftharpoonup_{\lambda}a,~{}~{}~{}v\leftharpoonup_{\lambda}(\partial
a)=(\partial+\lambda)(v\leftharpoonup_{\lambda}a),$
(RM2)$\qquad\qquad(v\leftharpoonup_{\lambda}a)\leftharpoonup_{\lambda+\mu}b=v\leftharpoonup_{\lambda}(a_{\mu}b).$
We denote it by $(M,\leftharpoonup_{\lambda})$.

An $A$-bimodule is a triple
$(M,\rightharpoonup_{\lambda},\leftharpoonup_{\lambda})$ such that
$(M,\rightharpoonup_{\lambda})$ is a left $A$-module,
$(M,\leftharpoonup_{\lambda})$ is a right $A$-module, and they satisfy the
following condition

$\displaystyle(a\rightharpoonup_{\lambda}v)\leftharpoonup_{\lambda+\mu}b=a\rightharpoonup_{\lambda}(v\leftharpoonup_{\mu}b),\;\;\forall\
a,b\in A,v\in M.$ (3)

######  Definition 2.4.

Let $U$ and $V$ be two $\mathbb{C}[\partial]$-modules. A conformal linear map
from $U$ to $V$ is a $\mathbb{C}$-linear map $a:U\rightarrow V[\lambda]$,
denoted by $a_{\lambda}:U\rightarrow V$, such that
$[\partial,a_{\lambda}]=-\lambda a_{\lambda}$. Denote the vector space of all
such maps by $Chom(U,V)$. It has a canonical structure of a
$\mathbb{C}[\partial]$-module

$(\partial a)_{\lambda}=-\lambda a_{\lambda}.$

Define the conformal dual of a $\mathbb{C}[\partial]$-module $U$ as $U^{\ast
c}=Chom(U,\mathbb{C})$ where $\mathbb{C}$ is viewed as the trivial
$\mathbb{C}[\partial]$-module, that is,

$U^{\ast
c}=\\{a:U\rightarrow\mathbb{C}[\lambda]|\mathbb{C}\mbox{-linear~{}~{}and~{}~{}}a_{\lambda}(\partial
b)=\lambda a_{\lambda}b,\forall\ b\in U\\}.$

In the special case $U=V$, set $Cend(V)=Chom(V,V)$. If $V$ is a finite
$\mathbb{C}[\partial]$-module, then the $\mathbb{C}[\partial]$-module
$Cend(V)$ has a canonical structure of an associative conformal algebra
defined by

$\displaystyle(a_{\lambda}b)_{\mu}v=a_{\lambda}(b_{\mu-\lambda}v),\quad\forall\
a,b\in Cend(V),v\in V.$

Set $a\rightharpoonup_{\lambda}v=l_{A}(a)_{\lambda}v$ and
$v\leftharpoonup_{\lambda}a=r_{A}(a)_{-\lambda-\partial}v$. Then a structure
of a bimodule $M$ over an associative conformal algebra $A$ is the same as two
$\mathbb{C}[\partial]$-module homomorphisms $l_{A}$ and $r_{A}$ from $A$ to
$Cend(M)$ such that the following conditions hold:

$\displaystyle
l_{A}(a_{\lambda}b)_{\lambda+\mu}v=l_{A}(a)_{\lambda}(l_{A}(b)_{\mu}v),$ (4)
$\displaystyle
r_{A}(b)_{-\lambda-\mu-\partial}(r_{A}(a)_{-\lambda-\partial}v)=r_{A}(a_{\mu}b)_{-\lambda-\partial}v,$
(5) $\displaystyle
l_{A}(a)_{\lambda}(r_{A}(b)_{-\mu-\partial}v)=r_{A}(b)_{-\lambda-\mu-\partial}(l_{A}(a)_{\lambda}v),$
(6)

for all $a$, $b\in A$ and $v\in M$. We denote this bimodule by
$(M,l_{A},r_{A})$.

######  Proposition 2.5.

Let $(M,l_{A},r_{A})$ be a finite bimodule of an associative conformal algebra
$A$. Let $l_{A}^{\ast}$ and $r_{A}^{\ast}$ be two
$\mathbb{C}[\partial]$-module homomorphisms from $A$ to $Cend(M^{\ast c})$
defined by

$\displaystyle(l_{A}^{\ast}(a)_{\lambda}f)_{\mu}u=f_{\mu-\lambda}(l_{A}(a)_{\lambda}u),~{}(r_{A}^{\ast}(a)_{\lambda}f)_{\mu}u=f_{\mu-\lambda}(r_{A}(a)_{\lambda}u),\;\;\forall\
a\in A,f\in M^{\ast c},u\in M.$ (7)

Then $(M^{\ast c},r_{A}^{\ast},l_{A}^{\ast})$ is a bimodule of $A$.

###### Proof.

Let $a$, $b\in A$, $f\in M^{\ast c}$ and $u\in M$. Since

$\displaystyle(r_{A}^{\ast}(a)_{\lambda}(r_{A}^{\ast}(b)_{\mu}f))_{\nu}u$
$\displaystyle=$
$\displaystyle(r_{A}^{\ast}(b)_{\mu}f)_{\nu-\lambda}(r_{A}(a)_{\lambda}u)=f_{\nu-\lambda-\mu}(r_{A}(b)_{\mu}(r_{A}(a)_{\lambda}u))$
$\displaystyle=$ $\displaystyle
f_{\nu-\lambda-\mu}(r_{A}(a_{\lambda}b)_{\lambda+\mu}u)=(r_{A}^{\ast}(a_{\lambda}b)_{\lambda+\mu}f)_{\nu}u;$
$\displaystyle(l_{A}^{\ast}(b)_{-\lambda-\mu-\partial}(l_{A}^{\ast}(a)_{-\lambda-\partial}f))_{\nu}u$
$\displaystyle=$
$\displaystyle(l_{A}^{\ast}(a)_{\mu}f)_{\lambda+\mu}(l_{A}(b)_{\nu-\lambda-\mu}u)=f_{\lambda}(l_{A}(a)_{\mu}(l_{A}(b)_{\nu-\lambda-\mu}u))$
$\displaystyle=$ $\displaystyle
f_{\lambda}(l_{A}(a_{\mu}b)_{\nu-\lambda}u)=(l_{A}^{\ast}(a_{\mu}b)_{\nu-\lambda}f)_{\nu}u=(l_{A}^{\ast}(a_{\mu}b)_{-\lambda-\partial}f)_{\nu}u;$
$\displaystyle(r_{A}^{\ast}(a)_{\lambda}(l_{A}^{\ast}(b)_{-\mu-\partial}f))_{\nu}u$
$\displaystyle=$
$\displaystyle(l_{A}^{\ast}(b)_{-\lambda-\mu+\nu}f)_{\nu-\lambda}(r_{A}(a)_{\lambda}u)=f_{\mu}(l_{A}(b)_{\nu-\lambda-\mu}(r_{A}(a)_{\lambda}u))$
$\displaystyle=$ $\displaystyle
f_{\mu}(r_{A}(a)_{\lambda}(l_{A}(b)_{\nu-\lambda-\mu}u))=(r_{A}^{\ast}(a)_{\lambda}f)_{\lambda+\mu}(l_{A}(b)_{\nu-\lambda-\mu}u)$
$\displaystyle=$
$\displaystyle(l_{A}^{\ast}(b)_{-\lambda-\mu-\partial}(r_{A}^{\ast}(a)_{\lambda}f))_{\nu}u,$

we have

$\displaystyle
r_{A}^{\ast}(a)_{\lambda}(r_{A}^{\ast}(b)_{\mu}f)=r_{A}^{\ast}(a_{\lambda}b)_{\lambda+\mu}f,\;\;l_{A}^{\ast}(b)_{-\lambda-\mu-\partial}(l_{A}^{\ast}(a)_{-\lambda-\partial}f)=l_{A}^{\ast}(a_{\mu}b)_{-\lambda-\partial}f,$
$\displaystyle
r_{A}^{\ast}(a)_{\lambda}(l_{A}^{\ast}(b)_{-\mu-\partial}f)=l_{A}^{\ast}(b)_{-\lambda-\mu-\partial}(r_{A}^{\ast}(a)_{\lambda}f).$

Hence $(M^{\ast c},r_{A}^{\ast},l_{A}^{\ast})$ is a bimodule of $A$. ∎

######  Example 2.6.

Let $A$ be a finite associative conformal algebra. Define two
$\mathbb{C}[\partial]$-module homomorphisms $L_{A}$ and $R_{A}$ from $A$ to
$Cend(A)$ by $L_{A}(a)_{\lambda}b=a_{\lambda}b$ and
$R_{A}(a)_{\lambda}b=b_{-\lambda-\partial}a$ for all $a$, $b\in A$. Then
$(A,L_{A},R_{A})$ is a bimodule of $A$. Moreover, $(A^{\ast
c},R_{A}^{\ast},L_{A}^{\ast})$ is a bimodule of $A$.

######  Proposition 2.7.

([11, Proposition 4.4]) Let $A$ and $B$ be two associative conformal algebras.
Suppose that there are $\mathbb{C}[\partial]$-module homomorphisms $l_{A}$,
$r_{A}:A\rightarrow Cend(B)$ and $l_{B}$, $r_{B}:B\rightarrow Cend(A)$ such
that $(B,l_{A},r_{A})$ is a bimodule of $A$ and $(A,l_{B},r_{B})$ is a
bimodule of $B$ and they satisfy the following relations:

$\displaystyle
l_{A}(a)_{\lambda}(x_{\mu}y)=(l_{A}(a)_{\lambda}x)_{\lambda+\mu}y+l_{A}(r_{B}(x)_{-\lambda-\partial}a)_{\lambda+\mu}y,$
(8) $\displaystyle
r_{B}(x)_{-\lambda-\mu-\partial}(a_{\lambda}b)=a_{\lambda}(r_{B}(x)_{-\mu-\partial}b)+r_{B}(l_{A}(b)_{\mu}(x))_{-\lambda-\partial}a,$
(9) $\displaystyle
l_{B}(x)_{\lambda}(a_{\mu}b)=(l_{B}(x)_{\lambda}a)_{\lambda+\mu}b+l_{B}(r_{A}(a)_{-\lambda-\partial}x)_{\lambda+\mu}b,$
(10) $\displaystyle
r_{A}(l_{B}(y)_{\mu}a)_{-\lambda-\partial}x+x_{\lambda}(r_{A}(a)_{-\mu-\partial}y)=r_{A}(a)_{-\lambda-\mu-\partial}(x_{\lambda}y),$
(11) $\displaystyle
r_{A}(r_{B}(y)_{-\mu-\partial}(a))_{-\lambda-\partial}x+x_{\lambda}(l_{A}(a)_{\mu}y)=l_{A}(l_{B}(x)_{\lambda}a)_{\lambda+\mu}y+(r_{A}(a)_{-\lambda-\partial}x)_{\lambda+\mu}y,$
(12) $\displaystyle
a_{\lambda}(l_{B}(x)_{\mu}b)+r_{B}(r_{A}(b)_{-\mu-\partial}x)_{-\lambda-\partial}(a)=(r_{B}(x)_{-\lambda-\partial}a)_{\lambda+\mu}b+l_{B}(l_{A}(a)_{\lambda}x)_{\lambda+\mu}b,$
(13)

for all $a$, $b\in A$ and $x$, $y\in B$. Then there is an associative
conformal algebra structure on the $\mathbb{C}[\partial]$-module $A\oplus B$
given by

$\displaystyle(a+x)_{\lambda}(b+y)=(a_{\lambda}b+l_{B}(x)_{\lambda}b+r_{B}(y)_{-\lambda-\partial}a)+(x_{\lambda}y+l_{A}(a)_{\lambda}y+r_{A}(b)_{-\lambda-\partial}x),$
(14)

for all $a,b\in A$ and $x,y\in B$. We denote this associative conformal
algebra by $A\bowtie B$. $(A,B,l_{A},r_{A},l_{B},r_{B})$ satisfying the above
relations is called a matched pair  of associative conformal algebras.
Moreover, any associative conformal algebra $E=A\oplus B$ where the sum is the
direct sum of $\mathbb{C}[\partial]$-modules and $A$, $B$ are two associative
conformal subalgebras of $E$, is $A\bowtie B$ associated to some matched pair
of associative conformal algebras.

######  Remark 2.8.

If $l_{B}$, $r_{B}$ and the $\lambda$-product on $B$ are trivial in
Proposition 2.7, that is, $B$ is exactly a bimodule of $A$, then $A\bowtie B$
is the semi-direct product of $A$ and its bimodule $B$, which is denoted by
$A\ltimes_{l_{A},r_{A}}B$.

##  3\. Double constructions of Frobenius conformal algebras

We introduce the notion of double constructions of Frobenius conformal
algebras. The relationship between double constructions of Frobenius conformal
algebras and matched pairs of associative conformal algebras is investigated.

######  Definition 3.1.

Let $V$ be a $\mathbb{C}[\partial]$-module. A conformal bilinear form on $V$
is a $\mathbb{C}$-bilinear map $\langle\cdot,\cdot\rangle_{\lambda}:V\times
V\rightarrow\mathbb{C}[\lambda]$ satisfying

$\displaystyle\langle\partial a,b\rangle_{\lambda}=-\lambda\langle
a,b\rangle_{\lambda},~{}~{}~{}\langle a,\partial
b\rangle_{\lambda}=\lambda\langle a,b\rangle_{\lambda},\;\;\forall\ a,b\in V.$
(15)

A conformal bilinear form is called symmetric if $\langle
a,b\rangle_{\lambda}=\langle b,a\rangle_{-\lambda}$ for any $a$, $b\in V$.

Suppose that there is a conformal bilinear form on a
$\mathbb{C}[\partial]$-module $V$. Then we have a
$\mathbb{C}[\partial]$-module homomorphism $\varphi:V\longrightarrow V^{\ast
c},~{}~{}v\mapsto\varphi_{v}$ defined by

$(\varphi_{v})_{\lambda}w=\langle v,w\rangle_{\lambda},\quad\forall\ v,w\in
V.$

A conformal bilinear form is called non-degenerate if $\varphi$ gives an
isomorphism of $\mathbb{C}[\partial]$-modules between $V$ and $V^{\ast c}$.

######  Definition 3.2.

An associative conformal algebra $A$ is called a Frobenius conformal algebra
if there is a non-degenerate conformal bilinear form on $A$ such that

$\displaystyle\langle a_{\lambda}b,c\rangle_{\mu}=\langle
a,b_{\mu-\partial}c\rangle_{\lambda},\;\;\forall\ a,b,c\in A.$ (16)

A Frobenius conformal algebra $A$ is called symmetric if the conformal
bilinear form on $A$ is symmetric.

######  Example 3.3.

Let $A=\mathbb{C}[\partial]a\oplus\mathbb{C}[\partial]b$. Suppose that $A$ is
an associative conformal algebra with the following $\lambda$-product:

$\displaystyle
a_{\lambda}a=(\partial^{2}+\lambda\partial+\lambda^{2})b,~{}~{}a_{\lambda}b=b_{\lambda}a=b_{\lambda}b=0.$

Then $A$ is a Frobenius conformal algebra with the following conformal
bilinear form:

$\displaystyle\langle a,a\rangle_{\lambda}=\langle
b,b\rangle_{\lambda}=0,~{}~{}~{}\langle a,b\rangle_{\lambda}=\langle
b,a\rangle_{\lambda}=1.$

######  Example 3.4.

Let $(A,\langle\cdot,\cdot\rangle)$ be a Frobenius algebra, that is, $A$ is an
associative algebra with a non-degenerate bilinear form
$\langle\cdot,\cdot\rangle$ satisfying

$\langle ab,c\rangle=\langle a,bc\rangle,\;\;\forall\ a,b\in A.$

Then $(\text{Cur}(A),\langle\cdot,\cdot\rangle_{\lambda})$ is a Frobenius
conformal algebra with $\langle\cdot,\cdot\rangle_{\lambda}$ defined by

$\displaystyle\langle
p(\partial)a,q(\partial)b\rangle_{\lambda}=p(-\lambda)q(\lambda)\langle
a,b\rangle,~{}~{}~{}\forall\ \text{$p(\partial)$,
$q(\partial)\in\mathbb{C}[\partial]$, $a$, $b\in A$.}$

######  Definition 3.5.

If a Frobenius conformal algebra $A$ satisfies the following conditions:

  1. (1)

$A=B\oplus B^{\ast c}$ where the sum is the direct sum of
$\mathbb{C}[\partial]$-modules;

  2. (2)

$B$ and $B^{\ast c}$ are two associative conformal subalgebras of $A$;

  3. (3)

the conformal bilinear form on $A$ is naturally given by

$\displaystyle\langle
a+f,b+g\rangle_{\lambda}=f_{\lambda}(b)+g_{-\lambda}(a),~{}~{}~{}~{}\forall\
a,b\in B,f,g\in B^{\ast c},$ (17)

then $A$ is called a double construction of Frobenius conformal algebra
associated to $B$ and $B^{\ast c}$.

######  Theorem 3.6.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module. Suppose that there is an associative conformal
algebra structure on $A^{\ast c}$. Then there is a double construction of
Frobenius conformal algebra associated to $A$ and $A^{\ast c}$ if and only if
$(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast c}}^{\ast},L_{A^{\ast
c}}^{\ast})$ is a matched pair of associative conformal algebras.

###### Proof.

Suppose that $(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast
c}}^{\ast},L_{A^{\ast c}}^{\ast})$ is a matched pair of associative conformal
algebras. Then $A\bowtie A^{\ast c}$ is endowed with an associative conformal
algebra structure as follows.

$\displaystyle(a+f)_{\lambda}(b+g)=(a_{\lambda}b+R_{A^{\ast
c}}^{\ast}(f)_{\lambda}b+L_{A^{\ast
c}}^{\ast}(g)_{-\lambda-\partial}a)+(f_{\lambda}g+R_{A}^{\ast}(a)_{\lambda}g+L_{A}^{\ast}(b)_{-\lambda-\partial}f),$
(18)

for all $a,b\in A$ and $f,g\in A^{\ast c}$. By this $\lambda$-product, $A$ and
$A^{\ast c}$ are two subalgebras of $A\bowtie A^{\ast c}$.

Obviously, the conformal bilinear form on $A\bowtie A^{\ast c}$ given by (17)
is symmetric and non-degenerate. For all $a$, $b$, $c\in A$, and $f$, $g$,
$h\in A^{\ast c}$, we have

$\displaystyle\langle(a+f)_{\lambda}(b+g),c+h\rangle_{\mu}$ $\displaystyle=$
$\displaystyle\langle a_{\lambda}b+R_{A^{\ast
c}}^{\ast}(f)_{\lambda}b+L_{A^{\ast
c}}^{\ast}(g)_{-\lambda-\partial}a+f_{\lambda}g+R_{A}^{\ast}(a)_{\lambda}g+L_{A}^{\ast}(b)_{-\lambda-\partial}f,c+h\rangle_{\mu}$
$\displaystyle=$
$\displaystyle(f_{\lambda}g+R_{A}^{\ast}(a)_{\lambda}g+L_{A}^{\ast}(b)_{-\lambda-\partial}f)_{\mu}(c)+h_{-\mu}(a_{\lambda}b+R_{A^{\ast
c}}^{\ast}(f)_{\lambda}b+L_{A^{\ast c}}^{\ast}(g)_{-\lambda-\partial}a)$
$\displaystyle=$
$\displaystyle(f_{\lambda}g)_{\mu}c+g_{\mu-\lambda}(R_{A}(a)_{\lambda}c)+f_{\lambda}(L_{A}(b)_{\mu-\lambda}c)+h_{-\mu}(a_{\lambda}b)$
$\displaystyle+(R_{A^{\ast c}}(f)_{\lambda}h)_{\lambda-\mu}b+(L_{A^{\ast
c}}(g)_{\mu-\lambda}h)_{-\lambda}a$ $\displaystyle=$
$\displaystyle(f_{\lambda}g)_{\mu}c+g_{\mu-\lambda}(c_{-\lambda-\partial}a)+f_{\lambda}(b_{\mu-\lambda}c)+h_{-\mu}(a_{\lambda}b)+(h_{-\lambda-\partial}f)_{\lambda-\mu}b+(g_{\mu-\lambda}h)_{-\lambda}a$
$\displaystyle=$
$\displaystyle(f_{\lambda}g)_{\mu}c+g_{\mu-\lambda}(c_{-\mu}a)+f_{\lambda}(b_{\mu-\lambda}c)+h_{-\mu}(a_{\lambda}b)+(h_{-\mu}f)_{\lambda-\mu}b+(g_{\mu-\lambda}h)_{-\lambda}a,$

and

$\displaystyle\langle a+f,(b+g)_{\mu-\partial}(c+h)\rangle_{\lambda}$
$\displaystyle=$ $\displaystyle\langle a+f,b_{\mu-\partial}c+R_{A^{\ast
c}}^{\ast}(g)_{\mu-\partial}c+L_{A^{\ast
c}}^{\ast}(h)_{-\mu}b+g_{\mu-\partial}h+R_{A}^{\ast}(b)_{\mu-\partial}h+L_{A}^{\ast}(c)_{-\mu}g\rangle_{\lambda}$
$\displaystyle=$ $\displaystyle
f_{\lambda}(b_{\mu-\partial}c)+f_{\lambda}(R_{A^{\ast
c}}^{\ast}(g)_{\mu-\partial}c)+f_{\lambda}(L_{A^{\ast c}}^{\ast}(h)_{-\mu}b)$
$\displaystyle+(g_{\mu-\partial}h)_{-\lambda}a+(R_{A}^{\ast}(b)_{\mu-\partial}h)_{-\lambda}a+(L_{A}^{\ast}(c)_{-\mu}g)_{-\lambda}a$
$\displaystyle=$ $\displaystyle f_{\lambda}(b_{\mu-\lambda}c)+(R_{A^{\ast
c}}(g)_{\mu-\lambda}f)_{\mu}c+(L_{A^{\ast c}}(h)_{-\mu}f)_{\lambda-\mu}b$
$\displaystyle+(g_{\mu-\lambda}h)_{-\lambda}a+h_{-\mu}(R_{A}(b)_{\mu-\lambda}a)+g_{\mu-\lambda}(L_{A}(c)_{-\mu}a)$
$\displaystyle=f_{\lambda}(b_{\mu-\lambda}c)+(f_{\lambda}g)_{\mu}c+(h_{-\mu}f)_{\lambda-\mu}b+(g_{\mu-\lambda}h)_{-\lambda}a+h_{-\mu}(a_{\lambda}b)+g_{\mu-\lambda}(c_{-\mu}a).$

Hence this conformal bilinear form on $A\bowtie A^{\ast c}$ is invariant.
Therefore $A\bowtie A^{\ast c}$ is a double construction of Frobenius
conformal algebra associated to $A$ and $A^{\ast c}$.

Conversely, suppose that there is a double construction of Frobenius conformal
algebra associated to $A$ and $A^{\ast c}$. Therefore there is an associative
conformal algebra structure on $A\oplus A^{\ast c}$ associated to a matched
pair $(A,A^{\ast c},l_{A},r_{A},l_{A^{\ast c}},r_{A^{\ast c}})$. Note that in
$A\oplus A^{\ast c}$,

$\displaystyle a_{\lambda}f=r_{A^{\ast
c}}(f)_{-\lambda-\partial}a+l_{A}(a)_{\lambda}f,\;\;f_{\lambda}a=l_{A^{\ast
c}}(f)_{\lambda}a+r_{A}(a)_{-\lambda-\partial}f,\;\;\forall\ a\in A,f\in
A^{\ast c}.$

For all $a$, $b\in A$, and $f\in A^{\ast c}$, we have

$\displaystyle\langle{l_{A}(a)}_{\lambda}f,b\rangle_{\mu}$ $\displaystyle=$
$\displaystyle\langle a_{\lambda}f,b\rangle_{\mu}=\langle
a,f_{\mu-\partial}b\rangle_{\lambda}=\langle
f_{\mu-\lambda}b,a\rangle_{-\lambda}=\langle f,b_{-\mu}a\rangle_{\mu-\lambda}$
$\displaystyle=$ $\displaystyle
f_{\mu-\lambda}({R_{A}(a)}_{\mu-\partial}b)=(R_{A}^{\ast}(a)_{\lambda}f)_{\mu}b=\langle
R_{A}^{\ast}(a)_{\lambda}f,b\rangle_{\mu}.$

Since $\langle\cdot,\cdot\rangle_{\mu}$ is non-degenerate,
$l_{A}(a)_{\lambda}f=R_{A}^{\ast}(a)_{\lambda}f$ for all $a\in A$ and $f\in
A^{\ast c}$ and hence $l_{A}=R_{A}^{\ast}$. Similarly, we have

$r_{A}=L_{A}^{\ast},\;\;l_{A^{\ast c}}=R_{A^{\ast c}}^{\ast},\;\;r_{A^{\ast
c}}=L_{A^{\ast c}}^{\ast}.$

Thus $(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast c}}^{\ast},L_{A^{\ast
c}}^{\ast})$ is a matched pair of associative conformal algebras. ∎

######  Theorem 3.7.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module. Assume that there is an associative conformal
algebra structure on the $\mathbb{C}[\partial]$-module $A^{\ast c}$. Then
$(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast c}}^{\ast},L_{A^{\ast
c}}^{\ast})$ is a matched pair of associative conformal algebras if and only
if

$\displaystyle
R_{A}^{\ast}(a)_{\lambda}(f_{\mu}g)=(R_{A}^{\ast}(a)_{\lambda}f)_{\lambda+\mu}g+R_{A}^{\ast}(L_{A^{\ast
c}}^{\ast}(f)_{-\lambda-\partial}a)_{\lambda+\mu}g,$ (19) $\displaystyle
R_{A}^{\ast}(R_{A^{\ast
c}}^{\ast}(f)_{\lambda}a)_{\lambda+\mu}g+(L_{A}^{\ast}(a)_{-\lambda-\partial}f)_{\lambda+\mu}g=L_{A}^{\ast}(L_{A^{\ast
c}}^{\ast}(g)_{-\mu-\partial}(a))_{-\lambda-\partial}f+f_{\lambda}(R_{A}^{\ast}(a)_{\mu}g),$
(20)

for all $a\in A$ and $f$, $g\in A^{\ast c}$.

###### Proof.

Obviously, (19) is exactly (8) and (20) is exactly (12) when

$l_{A}=R_{A}^{\ast},\;\;r_{A}=L_{A}^{\ast},\;\;l_{B}=R_{A^{\ast
c}}^{\ast},\;\;r_{B}=L_{A^{\ast c}}^{\ast},$

in Proposition 2.7. Then the conclusion follows if we prove that (19), (9),
(10) and (11) are mutually equivalent, (20) and (13) are equivalent. As an
example, we give an explicit proof that (20) holds if and only if (13) holds.
The other cases can be proved similarly.

Let $\\{e_{1},\cdots,e_{n}\\}$ be a $\mathbb{C}[\partial]$-basis of $A$ and
$\\{e_{1}^{\ast},\cdots,e_{n}^{\ast}\\}$ be a dual
$\mathbb{C}[\partial]$-basis of $A^{\ast c}$ in the sense that
$(e_{j}^{\ast})_{\lambda}e_{i}=\delta_{ij}$. Set

${e_{i}}_{\lambda}e_{j}=\sum_{k=1}^{n}P_{k}^{ij}(\lambda,\partial)e_{k},\;\;{e_{i}^{\ast}}_{\lambda}e_{j}^{\ast}=\sum_{k=1}^{n}R_{k}^{ij}(\lambda,\partial)e_{k}^{\ast},$

where $P_{k}^{ij}(\lambda,\partial)$ and
$R_{k}^{ij}(\lambda,\partial)\in\mathbb{C}[\lambda,\partial]$. Since

$(L_{A}^{\ast}(e_{i})_{\lambda}e_{j}^{\ast})_{\mu}e_{k}={e_{j}^{\ast}}_{\mu-\lambda}({e_{i}}_{\lambda}e_{k})={e_{j}^{\ast}}_{\mu-\lambda}(\sum_{j=1}^{n}P_{j}^{ik}(\lambda,\partial)e_{j})=P_{j}^{ik}(\lambda,\mu-\lambda),$

we have

$\displaystyle
L_{A}^{\ast}(e_{i})_{\lambda}e_{j}^{\ast}=\sum_{k=1}^{n}P_{j}^{ik}(\lambda,-\lambda-\partial)e_{k}^{\ast}.$

Similarly, we have

$\displaystyle L_{A^{\ast
c}}^{\ast}(e_{i}^{\ast})_{\lambda}e_{j}=\sum_{k=1}^{n}R_{j}^{ik}(\lambda,-\lambda-\partial)e_{k},\;\;R_{A}^{\ast}(e_{i})_{\lambda}e_{j}^{\ast}=\sum_{k=1}^{n}P_{j}^{ki}(\partial,-\lambda-\partial)e_{k}^{\ast},$
$\displaystyle R_{A^{\ast
c}}^{\ast}(e_{i}^{\ast})_{\lambda}e_{j}=\sum_{k=1}^{n}R_{j}^{ki}(\partial,-\lambda-\partial)e_{k}.$

Therefore (20) holds for any $a\in A$, $f$, $g\in A^{\ast c}$ if and only if

$\displaystyle(R_{A}^{\ast}(R_{A^{\ast
c}}^{\ast}(e_{j}^{\ast})_{\lambda}e_{i})_{\lambda+\mu}e_{k}^{\ast}+(L_{A}^{\ast}(e_{i})_{-\lambda-\partial}e_{i}^{\ast})_{\lambda+\mu}e_{k}^{\ast}$
$\displaystyle-L_{A}^{\ast}(L_{A^{\ast
c}}^{\ast}(e_{k}^{\ast})_{-\mu-\partial}e_{i})_{-\lambda-\partial}e_{j}^{\ast}+{e_{j}^{\ast}}_{\lambda}(R_{A}^{\ast}(e_{i})_{\mu}e_{k}^{\ast}))_{\nu}e_{s}=0,~{}~{}~{}~{}\forall~{}~{}i,j,k,s,$

if and only if the following equation holds:

$\displaystyle\sum_{t=1}^{n}(R_{i}^{tj}(-\lambda-\mu,\mu)P_{k}^{st}(-\nu,\nu-\lambda-\mu)+R_{s}^{tk}(\lambda+\mu,-\nu)P_{j}^{it}(\mu,\lambda)$
$\displaystyle-
R_{i}^{kt}(\nu-\lambda-\mu,\mu)P_{j}^{ts}(\nu-\lambda,\lambda)-R_{s}^{jt}(\lambda,-\nu)P_{k}^{ti}(\lambda-\nu,\nu-\lambda-\mu))=0,~{}~{}~{}~{}\forall~{}~{}i,j,k,s.$
(21)

On the other hand, (13) holds for any $a$, $b\in A$, $x\in A^{\ast c}$ if and
only if

$\displaystyle{e_{j}^{\ast}}_{\nu}({e_{i}}_{\lambda}(R_{A^{\ast
c}}^{\ast}(e_{k}^{\ast})_{\mu}e_{s})+L_{A^{\ast
c}}^{\ast}(L_{A}^{\ast}(e_{s})_{-\mu-\partial}e_{k}^{\ast})_{-\lambda-\partial}e_{i}$
$\displaystyle-(L_{A^{\ast
c}}^{\ast}(e_{k}^{\ast})_{-\lambda-\partial}e_{i})_{\lambda+\mu}e_{s}-R_{A^{\ast
c}}^{\ast}(R_{A}^{\ast}(e_{i})_{\lambda}e_{k}^{\ast})_{\lambda+\mu}e_{s})=0,~{}~{}~{}~{}\forall~{}~{}i,j,k,s,$

if and only if the following equation holds:

$\displaystyle\sum_{t=1}^{n}(R_{i}^{tj}(-\lambda-\nu,\lambda)P_{k}^{st}(-\lambda-\mu-\nu,\mu)+R_{s}^{tk}(\lambda+\nu,-\lambda-\mu-\nu)P_{j}^{it}(\lambda,\nu)$
$\displaystyle-
R_{i}^{kt}(\mu,\lambda)P_{j}^{ts}(\lambda+\mu,\nu)-R_{s}^{jt}(\nu,-\lambda-\mu-\nu)P_{k}^{ti}(-\lambda-\mu,\mu))=0,~{}~{}~{}~{}\forall~{}~{}i,j,k,s.$
(22)

Note that (22) is exactly (21) when we replace $\lambda$ by $\mu$, $\nu$ by
$\lambda$, and $\mu$ by $\nu-\lambda-\mu$ in (22). Therefore (20) holds if and
only if (13) holds. ∎

##  4\. Antisymmetric infinitesimal conformal bialgebras

We introduce the notion of antisymmetric infinitesimal conformal bialgebras as
(under certain conditions) the equivalent structures of the matched pairs of
associative conformal algebras as well as double constructions of Frobenius
conformal algebras given in the previous section.

######  Definition 4.1.

An associative conformal coalgebra is a $\mathbb{C}[\partial]$-module $A$
endowed with a $\mathbb{C}[\partial]$-module homomorphism
$\Delta:A\longrightarrow A\otimes A$ such that

$(I\otimes\Delta)\Delta(a)=(\Delta\otimes I)\Delta(a),\;\;\forall\ a\in A,$
(23)

where the module action of $\mathbb{C}[\partial]$ on $A\otimes A$ is defined
as $\partial(a\otimes b)=(\partial a)\otimes b+a\otimes(\partial b)$ for any
$a$, $b\in A$.

######  Proposition 4.2.

Let $(A,\Delta)$ be a finite associative conformal coalgebra. Then $A^{\ast
c}=Chom(A,\mathbb{C})$ is endowed with an associative conformal algebra
structure with the following $\lambda$-product:

$\displaystyle(f_{\lambda}g)_{\mu}(a)=\sum
f_{\lambda}(a_{(1)})g_{\mu-\lambda}(a_{(2)})=(f\otimes
g)_{\lambda,\mu-\lambda}(\Delta(a)),\;\;\forall\ f,g\in A^{\ast c},$ (24)

where $\Delta(a)=\sum a_{(1)}\otimes a_{(2)}$ for all $a\in A$.

###### Proof.

By (24), the conformal sesquilinearity of the $\lambda$-product is naturally
satisfied. For all $a\in A$, $f$, $g$ and $h\in A^{\ast c}$, we have

$\displaystyle((f_{\lambda}g)_{\lambda+\mu}h-f_{\lambda}(g_{\mu}h))_{\nu}(a)$
$\displaystyle=\sum(f_{\lambda}g)_{\lambda+\mu}(a_{(1)})h_{\nu-\lambda-\mu}(a_{(2)})-\sum
f_{\lambda}(a_{(1)})(g_{\mu}h)_{\nu-\lambda}(a_{(2)})$ $\displaystyle=\sum
f_{\lambda}(a_{(11)})g_{\mu}(a_{(12)})h_{\nu-\lambda-\mu}(a_{(2)})-\sum
f_{\lambda}(a_{(1)})g_{\mu}(a_{(21)})h_{\nu-\lambda-\mu}(a_{(22)})$
$\displaystyle=(f\otimes g\otimes
h)_{\lambda,\mu,\nu-\lambda-\mu}(((\Delta\otimes
I)\Delta-(I\otimes\Delta)\Delta)(a))=0.$

Hence the conclusion holds. ∎

######  Proposition 4.3.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module, that is,
$A=\sum_{i=1}^{n}\mathbb{C}[\partial]e_{i}$, where $\\{e_{1},\cdots,e_{n}\\}$
is a $\mathbb{C}[\partial]$-basis of $A$. Then $A^{\ast
c}=Chom(A,\mathbb{C})=\sum_{i=1}^{n}\mathbb{C}[\partial]e_{i}^{\ast}$ is an
associative conformal coalgebra with the following coproduct:

$\displaystyle\Delta(f)=\sum_{i,j}f_{\mu}({e_{i}}_{\lambda}e_{j})(e_{i}^{\ast}\otimes
e_{j}^{\ast})|_{\lambda=\partial\otimes 1,\mu=-\partial\otimes
1-1\otimes\partial},$ (25)

where $\\{e_{1}^{\ast},\cdots,e_{n}^{\ast}\\}$ is a dual
$\mathbb{C}[\partial]$-basis of $A^{\ast c}$. More precisely, if
${e_{i}}_{\lambda}e_{j}=\sum_{k}P_{k}^{ij}(\lambda,\partial)e^{k}$, then

$\Delta(e_{k}^{\ast})=\sum_{i,j}Q_{k}^{ij}(\partial\otimes
1,1\otimes\partial)(e_{i}^{\ast}\otimes e_{j}^{\ast}),$

where $Q_{k}^{ij}(x,y)=P_{k}^{ij}(x,-x-y)$.

###### Proof.

By (25), $\Delta$ is a $\mathbb{C}[\partial]$-module homomorphism. By the
definition of $\Delta$, we have

$\displaystyle(I\otimes\Delta)\Delta(e_{k}^{\ast})-(\Delta\otimes
I)\Delta(e_{k}^{\ast})$
$\displaystyle=\sum_{i,j,l,r}Q_{k}^{ij}(\partial\otimes 1\otimes
1,1\otimes\partial\otimes 1+1\otimes 1\otimes\partial)$ $\displaystyle\times
Q_{j}^{lr}(1\otimes\partial\otimes 1,1\otimes
1\otimes\partial)(e_{i}^{\ast}\otimes e_{l}^{\ast}\otimes
e_{r}^{\ast})-\sum_{i,j,l,r}Q_{k}^{ij}(\partial\otimes 1\otimes
1+1\otimes\partial\otimes 1,1\otimes 1\otimes\partial)$ $\displaystyle\times
Q_{i}^{lr}(\partial\otimes 1\otimes 1,1\otimes\partial\otimes
1)(e_{l}^{\ast}\otimes e_{r}^{\ast}\otimes e_{j}^{\ast}).$

On the other hand, since
${e_{i}}_{\lambda}({e_{l}}_{\mu}e_{r})=({e_{i}}_{\lambda}e_{l})_{\lambda+\mu}e_{r}$,
we have

$\displaystyle\sum_{j}P_{j}^{lr}(\mu,\lambda+\partial)P_{k}^{ij}(\lambda,\partial)=\sum_{j}P_{j}^{il}(\lambda,-\lambda-\mu)P_{k}^{jr}(\lambda+\mu,\partial).$

Since $Q_{k}^{ij}(x,y)=P_{k}^{ij}(x,-x-y)$, we have

$\displaystyle\sum_{j}Q_{j}^{lr}(\mu,-\lambda-\mu-\partial)Q_{k}^{ij}(\lambda,-\lambda-\partial)=\sum_{j}Q_{j}^{il}(\lambda,\mu)Q_{k}^{jr}(\lambda+\mu,-\lambda-\mu-\partial).$
(26)

Set

$\lambda=\partial\otimes 1\otimes 1,\;\;\mu=1\otimes\partial\otimes
1,\;\;\partial=-\partial\otimes 1\otimes 1-1\otimes\partial\otimes 1-1\otimes
1\otimes\partial.$

Then by (26), we have

$\displaystyle\sum_{j}Q_{k}^{ij}(\partial\otimes 1\otimes
1,1\otimes\partial\otimes 1+1\otimes 1\otimes\partial)\cdot
Q_{j}^{lr}(1\otimes\partial\otimes 1,1\otimes 1\otimes\partial)$
$\displaystyle=\sum_{j}Q_{k}^{jr}(\partial\otimes 1\otimes
1+1\otimes\partial\otimes 1,1\otimes 1\otimes\partial)\cdot
Q_{j}^{il}(\partial\otimes 1\otimes 1,1\otimes\partial\otimes 1).$

Therefore for all $k\in\\{1,\cdots,n\\}$, we have

$\displaystyle(I\otimes\Delta)\Delta(e_{k}^{\ast})=(\Delta\otimes
I)\Delta(e_{k}^{\ast}).$

Hence the conclusion holds. ∎

######  Corollary 4.4.

Let $(A=\mathbb{C}[\partial]a,\Delta)$ be an associative conformal coalgebra
which is free of rank 1 as a $\mathbb{C}[\partial]$-module. Then
$\Delta(a)=ka\otimes a$ for some $k\in\mathbb{C}$.

###### Proof.

If $A=\mathbb{C}[\partial]a$ is an associative conformal algebra, then
$a_{\lambda}a=ka$ for some $k\in\mathbb{C}$. Therefore this conclusion follows
from Proposition 4.3. ∎

In the sequel, denote $\partial\otimes 1+1\otimes\partial$ by
$\partial^{\otimes^{2}}$ and $\partial\otimes 1\otimes
1+1\otimes\partial\otimes 1+1\otimes 1\otimes\partial$ by
$\partial^{\otimes^{3}}$. Moreover, for any vector space $V$, let
$\tau:V\otimes V\rightarrow V\otimes V$ be the flip map, that is,

$\tau(x\otimes y)=y\otimes x,\;\;\forall\ x,y\in V.$

######  Theorem 4.5.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module. Suppose there is another associative conformal
algebra structure on the $\mathbb{C}[\partial]$-module $A^{\ast c}$ obtained
from a $\mathbb{C}[\partial]$-module homomorphism $\Delta:A\rightarrow
A\otimes A$. Then $(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast
c}}^{\ast},L_{A^{\ast c}}^{\ast})$ is a matched pair of associative conformal
algebras if and only if $\Delta$ satisfies

$\displaystyle\Delta(a_{\lambda}b)=(I\otimes{L_{A}(a)}_{\lambda})\Delta(b)+({R_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes
I)\Delta(a),$ (27)
$\displaystyle({L_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes I-I\otimes
R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}})\Delta(a)+\tau({L_{A}(a)}_{\lambda}\otimes
I-I\otimes{R_{A}(a)}_{\lambda})\Delta(b)=0,$ (28)

for all $a$, $b\in A$.

###### Proof.

With the same assumption as that in the proof of Theorem 3.7 and by
Proposition 4.3, we have

$\displaystyle\Delta(e_{k})=\sum_{i,j}Q_{k}^{ij}(\partial\otimes
1,1\otimes\partial)e_{i}\otimes e_{j},~{}~{}{\rm
where}~{}~{}Q_{k}^{ij}(x,y)=R_{k}^{ij}(x,-x-y).$

Considering the coefficient of $e_{j}\otimes e_{k}$ in

$\displaystyle\Delta({e_{s}}_{\lambda}e_{i})=(I\otimes{L_{A}(e_{s})}_{\lambda})\Delta(e_{i})+({R_{A}(e_{i})}_{-\lambda-\partial\otimes
1-1\otimes\partial}\otimes I)\Delta(e_{s}),$

we have

$\displaystyle\sum_{i=1}^{n}P_{t}^{si}(\lambda,\partial\otimes
1+1\otimes\partial)R_{t}^{jk}(\partial\otimes
1,-1\otimes\partial-\partial\otimes 1)$
$\displaystyle=\sum_{i=1}^{n}R_{i}^{jt}(\partial\otimes
1,-\lambda-\partial\otimes
1-1\otimes\partial)P_{k}^{st}(\lambda,1\otimes\partial)$
$\displaystyle+\sum_{i=1}^{n}R_{s}^{tk}(-\lambda-1\otimes\partial,\lambda)P_{j}^{ti}(\lambda+1\otimes\partial,\partial\otimes
1).$ (29)

On the other hand, (19) holds for any $a\in A$, $f$, $g\in A^{\ast c}$ if and
only if

$\displaystyle(R_{A}^{\ast}(e_{i})_{\lambda}({e_{j}^{\ast}}_{\mu}e_{k}^{\ast}))_{\nu}e_{s}=((R_{A}(e_{i})_{\lambda}e_{j}^{\ast})_{\lambda+\mu}e_{k}^{\ast}+R_{A}(L_{A^{\ast
c}}^{\ast}(e_{j}^{\ast})_{-\lambda-\partial}e_{i})_{\lambda+\mu}e_{k}^{\ast})_{\nu}e_{s},~{}~{}~{}~{}\forall~{}~{}i,j,k,s,$

if and only if the following equation holds:

$\displaystyle\sum_{i=1}^{n}P_{t}^{si}(-\nu,\nu-\lambda)R_{t}^{jk}(\mu,\lambda-\nu)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}R_{i}^{jt}(\mu,\lambda)P_{k}^{st}(-\nu,\nu-\lambda-\mu)$
(30)
$\displaystyle+\sum_{i=1}^{n}R_{s}^{tk}(\lambda+\mu,-\nu)P_{j}^{ti}(-\lambda-\mu,\mu)),~{}~{}~{}~{}~{}\forall~{}~{}j,k,s,t.$

Obviously, (29) is exactly (30) by replacing $\lambda$ by $-\nu$,
$1\otimes\partial$ by $\nu-\lambda-\mu$, and $\partial\otimes 1$ by $\mu$ in
(29). Then (27) holds if and only if (19) holds. With a similar discussion,
(28) holds if and only if (20) holds. Then this conclusion follows from
Theorem 3.7. ∎

######  Definition 4.6.

Let $(A,\cdot_{\lambda}\cdot)$ be an associative conformal algebra and
$(A,\Delta)$ be an associative conformal coalgebra. If $\Delta$ satisfies
(27), then $(A,\cdot_{\lambda}\cdot,\Delta)$ is called an infinitesimal
conformal bialgebra. Moreover, if $\Delta$ also satisfies (28), then
$(A,\cdot_{\lambda}\cdot,\Delta)$ is called an antisymmetric infinitesimal
(ASI) conformal bialgebra.

######  Remark 4.7.

By the definition of infinitesimal bialgebra given in [1], the corresponding
definition of conformal version of infinitesimal bialgebra should be given as
follows. $(A,\circ_{\lambda})$ is an associative conformal algebra and
$(A,\Delta)$ is an associative conformal coalgebra satisfying

$\displaystyle\Delta(a\circ_{\lambda}b)=({L_{A}(a)}_{\lambda}\otimes
I)\Delta(b)+(I\otimes{R_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}})\Delta(a),\;\;\forall\
a,b\in A.$ (31)

We would like to point out that the definition of infinitesimal conformal
bialgebra given in Definition 4.6 is equivalent to this definition if we
replace $(A,\circ_{\lambda})$ by the opposite algebra of
$(A,\cdot_{\lambda}\cdot)$ in the sense that
$a\circ_{\lambda}b=b_{-\lambda-\partial}a$ for all $a$, $b\in A$.

Combining Theorems 3.6 and 4.5 together, we have the following conclusion.

######  Corollary 4.8.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module. Suppose there is another associative conformal
algebra structure on the $\mathbb{C}[\partial]$-module $A^{\ast c}$ obtained
from a $\mathbb{C}[\partial]$-module homomorphism $\Delta:A\rightarrow
A\otimes A$. Then the following conditions are equivalent:

  1. (1)

$(A,\cdot_{\lambda}\cdot,\Delta)$ is an ASI conformal bialgebra;

  2. (2)

There is a double construction of Frobenius conformal algebra associated to
$A$ and $A^{\ast c}$;

  3. (3)

$(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast c}}^{\ast},L_{A^{\ast
c}}^{\ast})$ is a matched pair of associative conformal algebras.

######  Proposition 4.9.

Let $(A,\cdot_{\lambda}\cdot,\Delta_{A})$ be a finite ASI conformal bialgebra,
where $A$ is free as a $\mathbb{C}[\partial]$-module. Then $(A^{\ast
c},\cdot_{\lambda}\cdot,\Delta_{A^{\ast c}})$ is also an ASI conformal
bialgebra, where $\cdot_{\lambda}\cdot$ and $\Delta_{A^{\ast c}}$ are on
$A^{\ast c}$ given by (24) and (25) respectively.

###### Proof.

It follows directly from Corollary 4.8 and the symmetry of the associative
conformal algebra $A$ and $A^{\ast c}$ in the double construction $A\bowtie
A^{\ast c}$ of Frobenius conformal algebra associated to $A$ and $A^{\ast c}$.
∎

At the end of this section, we present two examples of ASI conformal
bialgebras.

######  Example 4.10.

Let $(A,\cdot,\overline{\Delta})$ be an antisymmetric infinitesimal bialgebra
given in [2]. Then $\text{Cur}(A)$ has a natural ASI conformal bialgebra
structure defined by

$\displaystyle\Delta(p(\partial)a)=p(\partial\otimes
1+1\otimes\partial)\overline{\Delta}(a),\;\;\forall\
\text{$p(\partial)\in\mathbb{C}[\partial]$, $a\in A$.}$

######  Example 4.11.

Let $p(\lambda)\in\mathbb{C}[\lambda]$ and
$A=\mathbb{C}[\partial]a\oplus\mathbb{C}[\partial]b$ be a rank two associative
conformal algebra with the following $\lambda$-product:

$\displaystyle
a_{\lambda}a=p(\lambda+\partial)b,~{}~{}~{}a_{\lambda}b=b_{\lambda}a=b_{\lambda}b=0.$

Define a $\mathbb{C}[\partial]$-module homomorphism $\Delta:A\rightarrow
A\otimes A$ by

$\displaystyle\Delta(a)=a\otimes b,~{}~{}~{}\Delta(b)=b\otimes b.$

By a straightforward computation, $(A,\cdot_{\lambda}\cdot,\Delta)$ is an ASI
conformal bialgebra if and only if $p(\lambda)$ is an odd polynomial, i.e.,
$p(\lambda)=-p(-\lambda)$.

##  5\. Coboundary antisymmetric infinitesimal conformal bialgebras

We consider a special class of ASI conformal bialgebras which are called
coboundary ASI conformal bialgebras.

######  Definition 5.1.

For an ASI conformal bialgebra $(A,\cdot_{\lambda}\cdot,\Delta)$, if there
exists $r\in A\otimes A$ such that

$\displaystyle\Delta(a)=(I\otimes L_{A}(a)_{\lambda}-R_{A}(a)_{\lambda}\otimes
I)r|_{\lambda=-\partial^{\otimes^{2}}},\;\;\forall\ a\in A,$ (32)

then $(A,\cdot_{\lambda}\cdot,\Delta)$ is called coboundary.

For $r=\sum_{i}r_{i}\otimes l_{i}\in A\otimes A$, define

$\displaystyle r\bullet r=\sum_{i,j}r_{i}\otimes
r_{j}\otimes{l_{i}}_{\mu}l_{j}|_{\mu=\partial\otimes 1\otimes
1}-r_{i}\otimes{r_{j}}_{\mu}l_{i}\otimes
l_{j}|_{\mu=-\partial^{\otimes^{2}}\otimes 1}+{r_{i}}_{\mu}r_{j}\otimes
l_{i}\otimes l_{j}|_{\mu=1\otimes\partial\otimes 1}.$ (33)

######  Proposition 5.2.

Let $A$ be an associative conformal algebra and $r=\sum_{i}r_{i}\otimes
l_{i}\in A\otimes A$. Then the map defined by (32) gives an associative
conformal coalgebra $(A,\Delta)$ if and only if

$\displaystyle(I\otimes I\otimes
L_{A}(a)_{-\partial^{\otimes^{3}}}-R_{A}(a)_{-\partial^{\otimes^{3}}}\otimes
I\otimes I)(r\bullet r)=0,\;\;\forall\ a\in A.$ (34)

###### Proof.

By the definition of $\Delta$, we have

$\displaystyle(I\otimes\Delta)\Delta(a)$ $\displaystyle=$
$\displaystyle(I\otimes\Delta)(\sum_{i}r_{i}\otimes
a_{\lambda}l_{i}-\sum_{i}{r_{i}}_{-\lambda-\partial\otimes 1}a\otimes
l_{i})|_{\lambda={-\partial^{\otimes^{2}}}}$ $\displaystyle=$
$\displaystyle(I\otimes\Delta)(\sum_{i}r_{i}\otimes
a_{-\partial^{\otimes^{2}}}l_{i}-\sum_{i}{r_{i}}_{1\otimes\partial}a\otimes
l_{i})$ $\displaystyle=$
$\displaystyle\sum_{i}r_{i}\otimes\Delta(a_{-\partial^{\otimes^{2}}}l_{i})-\sum_{i}{r_{i}}_{1\otimes\partial\otimes
1+1\otimes 1\otimes\partial}a\otimes\Delta(l_{i})$ $\displaystyle=$
$\displaystyle\sum_{i,j}(r_{i}\otimes
r_{j}\otimes(a_{-\partial^{\otimes^{2}}\otimes
1}l_{i})_{\mu}l_{j}-r_{i}\otimes{r_{j}}_{-\mu-1\otimes\partial\otimes
1}(a_{-\partial^{\otimes^{2}}\otimes 1}l_{i})\otimes l_{j}$
$\displaystyle-({r_{i}}_{1\otimes\partial\otimes 1+1\otimes
1\otimes\partial}a\otimes
r_{j}\otimes{l_{i}}_{\mu}l_{j}-{r_{i}}_{1\otimes\partial\otimes 1+1\otimes
1\otimes\partial}a\otimes{r_{j}}_{-\mu-1\otimes\partial\otimes
1}{l_{i}}\otimes l_{j}))_{\mu={-1\otimes\partial^{\otimes^{2}}}}$
$\displaystyle=$ $\displaystyle\sum_{i,j}r_{i}\otimes r_{j}\otimes
a_{-\partial^{\otimes^{3}}}({l_{i}}_{\partial\otimes 1\otimes
1}l_{j})-\sum_{i,j}r_{i}\otimes{r_{j}}_{1\otimes
1\otimes\partial}(a_{-\partial^{\otimes^{2}}\otimes 1}l_{i})\otimes l_{j}$
$\displaystyle-\sum_{i,j}{r_{i}}_{1\otimes\partial^{\otimes^{2}}}a\otimes
r_{j}\otimes{l_{i}}_{-1\otimes\partial^{\otimes^{2}}}l_{j}+\sum_{i,j}{r_{i}}_{1\otimes\partial\otimes
1+1\otimes 1\otimes\partial}a\otimes{r_{j}}_{1\otimes
1\otimes\partial}l_{i}\otimes l_{j}$ $\displaystyle=$ $\displaystyle(1\otimes
1\otimes L_{A}(a)_{-\partial^{\otimes^{3}}})(\sum_{i,j}r_{i}\otimes
r_{j}\otimes{l_{i}}_{\partial\otimes 1\otimes
1}l_{j})-\sum_{i,j}r_{i}\otimes{r_{j}}_{1\otimes
1\otimes\partial}(a_{-\partial^{\otimes^{2}}\otimes 1}l_{i})\otimes l_{j}$
$\displaystyle-(R_{A}(a)_{-\partial^{\otimes^{3}}}\otimes 1\otimes
1)\sum_{i,j}(r_{i}\otimes r_{j}\otimes{l_{i}}_{\partial\otimes 1\otimes
1}l_{j}-r_{i}\otimes{r_{j}}_{1\otimes 1\otimes\partial}l_{i}\otimes l_{j}).$

Similarly, we have

$\displaystyle(\Delta\otimes I)\Delta(a)$ $\displaystyle=$
$\displaystyle(\Delta\otimes I)\sum_{i}(r_{i}\otimes
a_{-\partial^{\otimes^{2}}}l_{i}-{r_{i}}_{1\otimes\partial}a\otimes l_{i})$
$\displaystyle=$ $\displaystyle\sum_{i}(\Delta(r_{i})\otimes
a_{-\partial^{\otimes^{3}}}l_{i}-\Delta({r_{i}}_{1\otimes\partial}a)\otimes
l_{i})$ $\displaystyle=$
$\displaystyle\sum_{i,j}(r_{j}\otimes{r_{i}}_{-\partial^{\otimes^{2}}\otimes
1}l_{j}\otimes
a_{-\partial^{\otimes^{3}}}l_{i}-{r_{j}}_{1\otimes\partial\otimes
1}r_{i}\otimes l_{j}\otimes a_{-\partial^{\otimes^{3}}}l_{i})$
$\displaystyle-\sum_{i,j}(r_{j}\otimes({r_{i}}_{1\otimes
1\otimes\partial}a)_{-\partial^{\otimes^{2}}\otimes 1}l_{j}\otimes
l_{i}-{r_{j}}_{1\otimes\partial\otimes 1}({r_{i}}_{1\otimes
1\otimes\partial}a)\otimes l_{j}\otimes l_{i})$ $\displaystyle=$
$\displaystyle(I\otimes I\otimes
L_{A}(a)_{-\partial^{\otimes^{3}}})\sum_{i,j}(r_{i}\otimes{r_{j}}_{-\partial^{\otimes^{2}}\otimes
1}l_{i}\otimes l_{j}-{r_{i}}_{1\otimes\partial\otimes 1}r_{j}\otimes
l_{i}\otimes l_{j})$ $\displaystyle-\sum_{i,j}r_{i}\otimes{r_{j}}_{1\otimes
1\otimes\partial}(a_{-\partial^{\otimes^{2}}\otimes 1}l_{i})\otimes l_{j}$
$\displaystyle+(R_{A}(a)_{-\partial^{\otimes^{3}}}\otimes 1\otimes
1)\sum_{i,j}{r_{i}}_{1\otimes\partial\otimes 1}r_{j}\otimes l_{i}\otimes
l_{j}.$

Therefore $\Delta$ satisfies (23) if and only if (34) holds. ∎

######  Theorem 5.3.

Let $(A,\cdot_{\lambda}\cdot)$ be an associative conformal algebra and
$r=\sum_{i}r_{i}\otimes l_{i}\in A\otimes A$. Then the map defined by (32)
gives an associative conformal coalgebra $(A,\Delta)$ such that
$(A,\cdot_{\lambda}\cdot,\Delta)$ is an ASI conformal bialgebra if and only if
(34) and the following equation hold:

$\displaystyle(L_{A}(b)_{-\lambda-\partial^{\otimes^{2}}}\otimes I-I\otimes
R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}})(I\otimes
L_{A}(a)_{-\partial^{\otimes^{2}}}-R_{A}(a)_{-\partial^{\otimes^{2}}}\otimes
I)(r+\tau r)=0,$ (35)

for all $a$, $b\in A$.

###### Proof.

Let $a,b\in A$. We first check that (27) holds automatically. In fact, we have

$\displaystyle(I\otimes{L_{A}(a)}_{\lambda})\Delta(b)+({R_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes
I)\Delta(a)$ $\displaystyle=$
$\displaystyle(I\otimes{L_{A}(a)}_{\lambda})\sum_{i}(r_{i}\otimes
b_{-\partial^{\otimes^{2}}}l_{i}-{r_{i}}_{1\otimes\partial}b\otimes l_{i})$
$\displaystyle+(R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}}\otimes
I)\sum_{i}(r_{i}\otimes
a_{-\partial^{\otimes^{2}}}l_{i}-{r_{i}}_{1\otimes\partial}a\otimes l_{i})$
$\displaystyle=$ $\displaystyle\sum_{i}(r_{i}\otimes
a_{\lambda}(b_{-\partial^{\otimes^{2}}}l_{i})-{r_{i}}_{\lambda+1\otimes\partial}b\otimes
a_{\lambda}l_{i}$ $\displaystyle+{r_{i}}_{\lambda+1\otimes\partial}b\otimes
a_{\lambda}l_{i}-({r_{i}}_{1\otimes\partial}a)_{\lambda+1\otimes\partial}b\otimes
l_{i})$ $\displaystyle=$ $\displaystyle\sum_{i}(r_{i}\otimes
a_{\lambda}(b_{-\partial^{\otimes^{2}}}l_{i})-({r_{i}}_{1\otimes\partial}a)_{\lambda+1\otimes\partial}b\otimes
l_{i})$ $\displaystyle=$
$\displaystyle\sum_{i}(r_{i}\otimes(a_{\lambda}b)_{-\partial^{\otimes^{2}}}l_{i}-{r_{i}}_{1\otimes\partial}(a_{\lambda}b)\otimes
l_{i})$ $\displaystyle=$ $\displaystyle\Delta(a_{\lambda}b).$

Therefore (27) holds.

Obviously, we have

$\displaystyle({L_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes I-I\otimes
R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}})\Delta(a)$
$\displaystyle=({L_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes I-I\otimes
R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}})(I\otimes
L_{A}(a)_{-\partial^{\otimes^{2}}}-R_{A}(a)_{-\partial^{\otimes^{2}}}\otimes
I)r.$

Moreover, we have

$\displaystyle\tau({L_{A}(a)}_{\lambda}\otimes
I-I\otimes{R_{A}(a)}_{\lambda})\Delta(b)$ $\displaystyle=$
$\displaystyle\tau({L_{A}(a)}_{\lambda}\otimes
I-I\otimes{R_{A}(a)}_{\lambda})\sum_{i}(r_{i}\otimes
b_{-\partial^{\otimes^{2}}}l_{i}-{r_{i}}_{1\otimes\partial}b\otimes l_{i})$
$\displaystyle=$ $\displaystyle\tau\sum_{i}(a_{\lambda}r_{i}\otimes
b_{-\lambda-\partial^{\otimes^{2}}}l_{i}-a_{\lambda}({r_{i}}_{1\otimes\partial}b)\otimes
l_{i}-r_{i}\otimes(b_{-\partial^{\otimes^{2}}}l_{i})_{-\lambda-1\otimes\partial}a+{r_{i}}_{\lambda+1\otimes\partial}b\otimes{l_{i}}_{-\lambda-1\otimes\partial}a$
$\displaystyle=$
$\displaystyle\sum_{i}(b_{-\lambda-\partial^{\otimes^{2}}}l_{i}\otimes
a_{\lambda}r_{i}-l_{i}\otimes a_{\lambda}({r_{i}}_{\partial\otimes
1}b)-(b_{-\partial^{\otimes^{2}}}l_{i})_{-\lambda-\partial\otimes 1}a\otimes
r_{i}+{l_{i}}_{-\lambda-\partial\otimes
1}a\otimes{r_{i}}_{\lambda+\partial\otimes 1}b$ $\displaystyle=$
$\displaystyle\sum_{i}(b_{-\lambda-\partial^{\otimes^{2}}}l_{i}\otimes
a_{\lambda}r_{i}-l_{i}\otimes(a_{\lambda}r_{i})_{\lambda+\partial\otimes
1}b)-b_{-\lambda-\partial^{\otimes^{2}}}({l_{i}}_{1\otimes\partial}a)+{l_{i}}_{-\lambda-\partial\otimes
1}a\otimes{r_{i}}_{\lambda+\partial\otimes 1}b$ $\displaystyle=$
$\displaystyle({L_{A}(b)}_{-\lambda-\partial^{\otimes^{2}}}\otimes I-I\otimes
R_{A}(b)_{-\lambda-\partial^{\otimes^{2}}})(I\otimes
L_{A}(a)_{-\partial^{\otimes^{2}}}-R_{A}(a)_{-\partial^{\otimes^{2}}}\otimes
I)(\tau r).$

Therefore (28) holds if and only if (35) holds. Hence by Proposition 5.2, the
conclusion holds. ∎

######  Corollary 5.4.

Let $A$ be an associative conformal algebra and $r\in A\otimes A$. Suppose
that $r$ is antisymmetric. Then the map defined by (32) gives an associative
conformal coalgebra $(A,\Delta)$ such that $(A,\cdot_{\lambda}\cdot,\Delta)$
is an ASI conformal bialgebra if

$\displaystyle r\bullet r\equiv
0~{}~{}\text{mod}~{}~{}(\partial^{\otimes^{3}}).$ (36)

###### Proof.

If $r\bullet r\equiv 0~{}~{}\text{mod}~{}~{}(\partial^{\otimes^{3}})$, (34)
naturally holds by conformal sesquilinearity. Then this conclusion directly
follows from Theorem 5.3. ∎

######  Definition 5.5.

Let $A$ be an associative conformal algebra and $r\in A\otimes A$. (36) is
called associative conformal Yang-Baxter equation in $A$.

######  Remark 5.6.

In fact, the associative conformal Yang-Baxter equation (36) is regarded as a
conformal analogue of the associative Yang-Baxter equation in an associative
algebra ([2]).

######  Definition 5.7.

Let $(A,\cdot_{\lambda}\cdot,\Delta_{A})$ and
$(B,\cdot_{\lambda}\cdot,\Delta_{B})$ be two ASI conformal bialgebras. If
$\varphi:A\rightarrow B$ is a homomorphism of associative conformal algebras
satisfying

$\displaystyle(\varphi\otimes\varphi)\Delta_{A}(a)=\Delta_{B}(\varphi(a)),~{}~{}\forall\
a\in A,$ (37)

then $\varphi$ is called a homomorphism of ASI conformal bialgebras.

######  Theorem 5.8.

Let $(A,\cdot_{\lambda}\cdot,\Delta_{A})$ be a finite ASI conformal bialgebra
where $A$ is free as a $\mathbb{C}[\partial]$-module. Then there is a
canonical ASI conformal bialgebra structure on the
$\mathbb{C}[\partial]$-module $A\oplus A^{\ast c}$ such that the inclusions
$i_{1}:A\rightarrow A\oplus A^{\ast c}$ and $i_{2}:A^{\ast c}\rightarrow
A\oplus A^{\ast c}$ are homomorphisms of ASI conformal bialgebras. Here the
ASI conformal bialgebra structure on $A^{\ast c}$ is $(A^{\ast
c},\cdot_{\lambda}\cdot,-\Delta_{A^{\ast c}})$, where $\cdot_{\lambda}\cdot$
and $\Delta_{A^{\ast c}}$ on $A^{\ast c}$ are defined by (24) and (25)
respectively.

###### Proof.

Let $\\{e_{1},\cdots,e_{n}\\}$ be a $\mathbb{C}[\partial]$-basis of $A$ and
$\\{e_{1}^{\ast},\cdots,e_{n}^{\ast}\\}$ be the dual basis in $A^{\ast c}$.
Since $(A,\cdot_{\lambda}\cdot,\Delta_{A})$ is an ASI conformal bialgebra,
$(A,A^{\ast c},R_{A}^{\ast},L_{A}^{\ast},R_{A^{\ast c}}^{\ast},L_{A^{\ast
c}}^{\ast})$ is a matched pair of associative conformal algebras by Corollary
4.8, where the associative conformal algebra structure on $A^{\ast c}$ is
obtained from $\Delta_{A}$. Then by Proposition 2.7, there is an associative
conformal algebra structure on the $\mathbb{C}[\partial]$-module $A\oplus
A^{\ast c}$ associated with such a matched pair. Set

$\displaystyle{e_{i}}_{\lambda}e_{j}=\sum_{k}P_{k}^{ij}(\lambda,\partial)e_{k},~{}~{}~{}~{}\Delta_{A}(e_{j})=\sum_{i,k}R_{j}^{ik}(\partial\otimes
1,-\partial\otimes 1-1\otimes\partial)e_{i}\otimes e_{k},$ (38)

where $P_{k}^{ij}(\lambda,\partial)$,
$R_{j}^{ik}(\lambda,\partial)\in\mathbb{C}[\lambda,\partial]$. Then by
Proposition 2.7 again, we have

$\displaystyle{e_{i}^{\ast}}_{\lambda}e_{j}^{\ast}$ $\displaystyle=$
$\displaystyle\sum_{k}R_{k}^{ij}(\lambda,\partial)e_{k}^{\ast},$
$\displaystyle{e_{i}}_{\lambda}e_{j}^{\ast}$ $\displaystyle=$ $\displaystyle
R_{A}^{\ast}(e_{i})_{\lambda}e_{j}^{\ast}+L_{A^{\ast
c}}^{\ast}(e_{j}^{\ast})_{-\lambda-\partial}e_{i}=\sum_{k}P_{j}^{ki}(\partial,-\lambda-\partial)e_{k}^{\ast}+\sum_{k}R_{i}^{jk}(-\lambda-\partial,\lambda)e_{k},$
$\displaystyle{e_{i}^{\ast}}_{\lambda}e_{j}$ $\displaystyle=$ $\displaystyle
R_{A^{\ast c}}^{\ast}(e_{i}^{\ast})_{\lambda}e_{j}+L_{A^{\ast
c}}^{\ast}(e_{j})_{-\lambda-\partial}e_{i}^{\ast}=\sum_{k}R_{j}^{ki}(\partial,-\lambda-\partial)e_{k}+\sum_{k}P_{i}^{jk}(-\lambda-\partial,\lambda)e_{k}^{\ast}.$

Let $r=\sum_{i=1}^{n}e_{i}\otimes e_{i}^{\ast}\in(A\oplus A^{\ast
c})\otimes(A\oplus A^{\ast c})$ and define

$\Delta_{A\oplus A^{\ast c}}(a)=(I\otimes
L_{A}(a)_{\lambda}-R_{A}(a)_{\lambda}\otimes
I)r|_{\lambda=-\partial^{\otimes^{2}}},\;\;\forall\ a\in A\oplus A^{\ast c}.$

Note that

$\displaystyle r\bullet r$ $\displaystyle=$
$\displaystyle\sum_{i,j}(e_{i}\otimes
e_{j}\otimes{e_{i}^{\ast}}_{\mu}e_{j}^{\ast}|_{\mu=\partial\otimes 1\otimes
1}$ $\displaystyle-e_{i}\otimes{e_{j}}_{\mu}e_{i}^{\ast}\otimes
e_{j}^{\ast}|_{\mu=-\partial^{\otimes^{2}}\otimes 1}+{e_{i}}_{\mu}e_{j}\otimes
e_{i}^{\ast}\otimes e_{j}^{\ast}|_{\mu=1\otimes\partial\otimes 1})$
$\displaystyle=$ $\displaystyle\sum_{i,j,k}(R_{k}^{ij}(\partial\otimes
1\otimes 1,1\otimes 1\otimes\partial)e_{i}\otimes e_{j}\otimes
e_{k}^{\ast}-P_{i}^{kj}(1\otimes\partial\otimes 1,\partial\otimes 1\otimes
1)e_{i}\otimes e_{k}^{\ast}\otimes e_{j}^{\ast}$ $\displaystyle-
R_{j}^{ik}(\partial\otimes 1\otimes 1,-\partial^{\otimes^{2}}\otimes
1)e_{i}\otimes e_{k}\otimes e_{j}^{\ast}+P_{k}^{ij}(1\otimes\partial\otimes
1,\partial\otimes 1\otimes 1)e_{k}\otimes e_{i}^{\ast}\otimes e_{j}^{\ast})$
$\displaystyle=$ $\displaystyle\sum_{i,j,k}(R_{k}^{ij}(\partial\otimes
1\otimes 1,1\otimes 1\otimes\partial)-R_{k}^{ij}(\partial\otimes 1\otimes
1,-\partial^{\otimes^{2}}\otimes 1))e_{i}\otimes e_{i}\otimes e_{k}^{\ast}$
$\displaystyle\equiv$ $\displaystyle
0~{}~{}~{}\text{mod}~{}~{}~{}\partial^{\otimes^{3}}.$

Then (34) holds by conformal sesquilinearity. Moreover, for all $e_{i}\in A$,
we have

$\displaystyle(I\otimes
L_{A}(e_{i})_{-\partial^{\otimes^{2}}}-R_{A}(e_{i})_{-\partial^{\otimes^{2}}}\otimes
I)(r+\tau r)$ $\displaystyle=$ $\displaystyle\sum_{j}(I\otimes
L_{A}(e_{i})_{-\partial^{\otimes^{2}}}-R_{A}(e_{i})_{-\partial^{\otimes^{2}}}\otimes
I)(e_{j}\otimes e_{j}^{\ast}+e_{j}^{\ast}\otimes e_{j})$ $\displaystyle=$
$\displaystyle\sum_{j}(e_{j}\otimes{e_{i}}_{-\partial^{\otimes^{2}}}e_{j}^{\ast}+e_{j}^{\ast}\otimes{e_{i}}_{-\partial^{\otimes^{2}}}e_{j}-{e_{j}}_{1\otimes\partial}e_{i}\otimes
e_{j}^{\ast}-{e_{j}^{\ast}}_{1\otimes\partial}e_{i}\otimes e_{j})$
$\displaystyle=$
$\displaystyle\sum_{j,k}(P_{j}^{ki}(1\otimes\partial,\partial\otimes
1)e_{j}\otimes e_{k}^{\ast}+R_{i}^{jk}(\partial\otimes
1,-\partial^{\otimes^{2}})e_{j}\otimes
e_{k}+P_{k}^{ij}(-\partial^{\otimes^{2}},1\otimes\partial)e_{j}^{\ast}\otimes
e_{k}$ $\displaystyle-P_{k}^{ij}(1\otimes\partial,\partial\otimes
1)e_{k}\otimes e_{j}^{\ast}-R_{i}^{kj}(\partial\otimes
1,-\partial^{\otimes^{2}})e_{k}\otimes
e_{j}-P_{j}^{ik}(-\partial^{\otimes^{2}},1\otimes\partial)e_{k}^{\ast}\otimes
e_{j})=0.$

Similarly, for all $e_{i}^{*}\in A^{\ast c}$, we have

$(I\otimes
L_{A}(e_{i}^{\ast})_{-\partial^{\otimes^{2}}}-R_{A}(e_{i}^{\ast})_{-\partial^{\otimes^{2}}}\otimes
I)(r+\tau r)=0.$

Therefore (35) holds. Hence by Theorem 5.3, $\Delta_{A\oplus A^{\ast c}}$
gives an ASI conformal bialgebra structure on $A\oplus A^{\ast c}$.

Furthermore, for all $e_{i}\in A$, we have

$\displaystyle\Delta_{A\oplus A^{\ast c}}(e_{i})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}(I\otimes L_{A\oplus A^{\ast
c}}(e_{i})_{\lambda}-R_{A\oplus A^{\ast c}}(e_{i})_{\lambda}\otimes
I)(e_{j}\otimes e_{j}^{\ast})|_{\lambda=-\partial^{\otimes^{2}}}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{n}(e_{j}\otimes{e_{i}}_{-\partial^{\otimes^{2}}}e_{j}^{\ast}-{e_{j}}_{1\otimes\partial}e_{i}\otimes
e_{j}^{\ast})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}(e_{j}\otimes(\sum_{k}P_{j}^{ki}(1\otimes\partial,\partial\otimes
1)e_{k}^{\ast}+\sum_{k}R_{i}^{jk}(\partial\otimes
1,-\partial^{\otimes^{2}})e_{k})$
$\displaystyle-\sum_{k}P_{k}^{ji}(1\otimes\partial,\partial\otimes
1)e_{k}\otimes e_{j}^{\ast})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\sum_{k}R_{i}^{jk}(\partial\otimes
1,-\partial^{\otimes^{2}})e_{j}\otimes e_{k}=\Delta_{A}(e_{i}).$

Similarly, for all $e_{i}^{*}\in A^{\ast c}$, we have

$\Delta_{A\oplus A^{\ast c}}(e_{i}^{\ast})=-\Delta_{A^{\ast
c}}(e_{i}^{\ast}).$

Therefore $i_{1}:A\rightarrow A\oplus A^{\ast c}$ and $i_{2}:A^{\ast
c}\rightarrow A\oplus A^{\ast c}$ are homomorphisms of ASI conformal
bialgebras. Hence the proof is finished. ∎

##  6\. $\mathcal{O}$-operators of associative conformal algebras and
dendriform conformal algebras

We introduce the notion of $\mathcal{O}$-operators of associative conformal
algebras to interpret the associative conformal Yang-Baxter equation. In
particular, an $\mathcal{O}$-operator of an associative conformal algebra $A$
associated to a bimodule gives an antisymmetric solution of associative
conformal Yang-Baxter equation in a semi-direct product associative conformal
algebra. We also introduce the notion of dendriform conformal algebras to
construct $\mathcal{O}$-operators of their associated associative conformal
algebras and hence give (antisymmetric) solutions of associative conformal
Yang-Baxter equation.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module. Define a linear map $\varphi:A\otimes
A\rightarrow Chom(A^{\ast c},A)$ as

$\displaystyle\varphi(u\otimes
v)_{\lambda}(g)=g_{-\lambda-\partial^{A}}(u)v,\;\;\forall\ u,v\in A,g\in
A^{\ast c}.$ (39)

Here $\partial^{A}$ represents the action of $\partial$ on $A$. Obviously,
$\varphi$ is a $\mathbb{C}[\partial]$-module homomorphism. Similar to
Proposition 6.1 in [7], we show that $\varphi$ is a
$\mathbb{C}[\partial]$-module isomorphism.

Set $r=\sum_{i}r_{i}\otimes l_{i}\in A\otimes A$. By $\varphi$, we associate a
conformal linear map $T^{r}\in Chom(A^{\ast c},A)$ given by

$\displaystyle
T^{r}_{\lambda}(f)=\sum_{i}f_{-\lambda-\partial^{A}}(r_{i})l_{i},\;\;\forall\
f\in A^{\ast c}.$

For all $f\in A^{\ast c}$ and $a\in A$, we define $\langle
a,f\rangle_{\lambda}=\\{f,a\\}_{-\lambda}=f_{-\lambda}(a)$. Obviously,

$\displaystyle\langle\partial a,f\rangle_{\lambda}=-\lambda\langle
a,f\rangle_{\lambda}.$ (40)

We also define

$\langle a\otimes b\otimes c,f\otimes g\otimes
h\rangle_{(\lambda,\nu,\theta)}=\langle a,f\rangle_{\lambda}\langle
b,g\rangle_{\nu}\langle c,h\rangle_{\theta},\;\;\forall\ a,b,c\in A,f,g,h\in
A^{\ast c}.$

By Proposition 2.5, we have

$\displaystyle\langle a_{\lambda}b,f\rangle_{\mu}=\langle
b,L_{A}^{\ast}(a)_{\lambda}f\rangle_{\mu-\lambda},~{}~{}~{}\langle
b_{\mu-\lambda}a,f\rangle_{\mu}=\langle
b,R_{A}^{\ast}(a)_{\lambda}f\rangle_{\mu-\lambda},\;\;\forall a,b\in A,f\in
A^{\ast c}.$ (41)

######  Theorem 6.1.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module and $r\in A\otimes A$ be antisymmetric. Then $r$
is a solution of associative conformal Yang-Baxter equation if and only if
$T^{r}\in Chom(A^{\ast c},A)$ satisfies

$\displaystyle
T^{r}_{0}(f)_{\lambda}T^{r}_{0}(g)-T_{0}^{r}(R_{A}^{\ast}(T_{0}^{r}(f))_{\lambda}g)-T^{r}_{0}(L_{A}^{\ast}(T^{r}_{0}(g))_{-\lambda-\partial}f)=0,~{}~{}~{}~{}~{}\forall\
f,~{}g\in A^{\ast c}.$ (42)

###### Proof.

Since $r$ is antisymmetric, we have

$T^{r}_{\lambda}(f)=\sum_{i}\langle
r_{i},f\rangle_{\lambda+\partial^{A}}l_{i}=-\sum_{i}\langle
l_{i},f\rangle_{\lambda+\partial^{A}}r_{i},\;\;\forall\ f\in A^{\ast c}.$

Obviously, the fact that $r\bullet
r~{}~{}\text{mod}~{}~{}(\partial^{\otimes^{3}})=0$ holds if and only if the
following equation holds:

$\langle r\bullet r~{}~{}\text{mod}~{}~{}(\partial^{\otimes^{3}}),f\otimes
g\otimes h\rangle_{(\lambda,\eta,\nu)}=0,\;\;\forall\ f,g,h\in A^{\ast c},$

if and only if the following equation holds:

$\displaystyle\langle r\bullet r,f\otimes g\otimes
h\rangle_{(\lambda,\eta,\nu)}=0~{}~{}~{}\text{mod}~{}~{}(\lambda+\eta+\nu),\;\;\forall\
f,g,h\in A^{\ast c}.$ (43)

Let $f$, $g$, $h\in A^{\ast c}$. Then we have

$\displaystyle\langle\sum_{i,j}r_{i}\otimes
r_{j}\otimes{l_{i}}_{\mu}l_{j}|_{\mu=\partial\otimes 1\otimes 1},f\otimes
g\otimes h\rangle_{(\lambda,\eta,\nu)}$ $\displaystyle=$
$\displaystyle\sum_{i,j}\langle r_{i},f\rangle_{\lambda}\langle
r_{j},g\rangle_{\eta}\langle{l_{i}}_{-\lambda}l_{j},h\rangle_{\nu}=\langle(\sum_{i}\langle
r_{i},f\rangle_{\lambda}{l_{i}})_{-\lambda}(\sum_{j}\langle
r_{j},g\rangle_{\eta}l_{j}),h\rangle_{\nu}$ $\displaystyle=$
$\displaystyle\langle
T^{r}_{\lambda-\partial}(f)_{-\lambda}T^{r}_{\eta-\partial}(g),h\rangle_{\nu}=\langle
T^{r}_{0}(f)_{-\lambda}T^{r}_{\lambda+\eta+\nu}(g),h\rangle_{\nu},$

and

$\displaystyle\langle\sum_{i,j}r_{i}\otimes{r_{j}}_{\mu}l_{i}\otimes
l_{j}|_{\mu=-\partial^{\otimes^{2}}\otimes 1},f\otimes g\otimes
h\rangle_{(\lambda,\eta,\nu)}$ $\displaystyle=$
$\displaystyle\sum_{i,j}\langle
r_{i},f\rangle_{\lambda}\langle{r_{j}}_{\lambda+\eta}l_{i},g\rangle_{\eta}\langle
l_{j},h\rangle_{\nu}=\sum_{i,j}\langle{r_{j}}_{\lambda+\eta}(\langle
r_{i},f\rangle_{\lambda}l_{i}),g\rangle_{\eta}\langle l_{j},h\rangle_{\nu}$
$\displaystyle=$
$\displaystyle\sum_{j}\langle{r_{j}}_{\lambda+\eta}(T_{\lambda-\partial}^{r}(f)),g\rangle_{\eta}\langle
l_{j},h\rangle_{\nu}=\sum_{j}\langle{r_{j}},R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g\rangle_{\lambda+\eta}\langle
l_{j},h\rangle_{\nu}$ $\displaystyle=$
$\displaystyle\langle(\sum_{j}\langle{r_{j}},R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g\rangle_{\lambda+\eta}l_{j}),h\rangle_{\nu}=\langle
T_{\lambda+\eta-\partial}^{r}(R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g),h\rangle_{\nu}$
$\displaystyle=$ $\displaystyle\langle
T_{\lambda+\eta+\nu}^{r}(R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g),h\rangle_{\nu}.$

Similarly, we have

$\displaystyle\langle\sum_{i,j}{r_{i}}_{\mu}r_{j}\otimes l_{i}\otimes
l_{j}|_{\mu=1\otimes\partial\otimes 1},f\otimes g\otimes
h\rangle_{(\lambda,\eta,\nu)}=-\langle
T^{r}_{\lambda+\eta+\nu}(L_{A}^{\ast}(T^{r}_{0}(g))_{-\eta}f),h\rangle_{\nu}.$

Therefore (43) holds if and only if

$\displaystyle\langle
T^{r}_{0}(f)_{-\lambda}T^{r}_{\lambda+\eta+\nu}(g)-T_{\lambda+\eta+\nu}^{r}(R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g)-T^{r}_{\lambda+\eta+\nu}(L_{A}^{\ast}(T^{r}_{0}(g))_{-\eta}f),h\rangle_{\nu}$
$\displaystyle=0~{}~{}~{}\text{mod}~{}~{}~{}(\lambda+\eta+\nu),\;\;\forall\
f,g,h\in A^{\ast c}.$

It is straightforward that the above equality holds if and only if

$\displaystyle
T^{r}_{0}(f)_{-\lambda}T^{r}_{0}(g)-T_{0}^{r}(R_{A}^{\ast}(T_{0}^{r}(f))_{-\lambda}g)-T^{r}_{0}(L_{A}^{\ast}(T^{r}_{0}(g))_{\lambda-\partial}f)=0,\;\;\forall
f,g\in A^{\ast c}.$ (44)

Therefore the conclusion follows by replacing $\lambda$ by $-\lambda$ in (44).
∎

######  Corollary 6.2.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module, $r\in A\otimes A$ be antisymmetric and
$\Delta_{A}$ be the map defined by (32) through $r$. Suppose the associative
conformal algebra structure on $A^{\ast c}$ is obtained from $\Delta_{A}$. Let
$T^{r}\in\text{Chom}(A^{\ast c},A)$ be the element corresponding to $r$
through the isomorphism $A\otimes A\cong\text{Chom}(A^{\ast c},A)$. Then
$T_{0}^{r}:A^{\ast c}\rightarrow A$ is a homomorphism of associative conformal
algebras.

###### Proof.

Let $\\{e_{1},\cdots,e_{n}\\}$ be a $\mathbb{C}[\partial]$-basis of $A$ and
$\\{e_{1}^{\ast},\cdots,e_{n}^{\ast}\\}$ be the dual
$\mathbb{C}[\partial]$-basis. Set

${e_{i}}_{\lambda}e_{j}=\sum_{k}P_{ij}^{k}e_{k},\;\;r=\sum_{i,j}a_{ij}(\partial\otimes
1,1\otimes\partial)e_{i}\otimes e_{j},$

where $a_{ij}(\lambda,\mu)\in\mathbb{C}[\lambda,\mu]$. Then

$\displaystyle\Delta_{A}(e_{k})$ $\displaystyle=$ $\displaystyle(I\otimes
L_{A}(e_{k})_{\lambda}-R_{A}(a)_{\lambda}\otimes
1)r|_{\lambda=-\partial^{\otimes^{2}}}$ $\displaystyle=$
$\displaystyle\sum_{i,j}a_{ij}(\partial\otimes
1,\lambda+1\otimes\partial)e_{i}\otimes{e_{k}}_{\lambda}e_{j}-\sum_{i,j}a_{ij}(\lambda+\partial\otimes
1,1\otimes\partial){e_{i}}_{-\lambda-\partial}e_{k}\otimes e_{j}$
$\displaystyle=$ $\displaystyle\sum_{i,j,s}(a_{ij}(\partial\otimes
1,-\partial\otimes
1)P_{kj}^{s}(-\partial^{\otimes^{2}},1\otimes\partial)-a_{js}(-1\otimes\partial,1\otimes\partial)P_{jk}^{i}(1\otimes\partial,\partial\otimes
1))e_{i}\otimes e_{s}.$

Note that $T_{0}^{r}(e_{k}^{\ast})=\sum_{j}a_{kj}(-\partial,\partial)e_{j}$.
Then by Proposition 4.2,

$\displaystyle{e_{l}^{\ast}}_{\lambda}e_{t}^{\ast}$ $\displaystyle=$
$\displaystyle\sum_{j,k}(a_{lj}(\lambda,-\lambda)P_{kj}^{t}(\partial,-\lambda-\partial)-a_{jt}(\lambda+\partial,-\lambda-\partial)P_{jk}^{l}(-\lambda-\partial,\lambda))e_{k}^{\ast}$
$\displaystyle=$
$\displaystyle\sum_{j,k}(a_{lj}(\lambda,-\lambda){e_{t}^{\ast}}_{-\lambda-\partial^{A^{\ast
c}}}({e_{k}}_{\partial^{A^{\ast
c}}}e_{j})-a_{jt}(\lambda+\partial,-\lambda-\partial){e_{l}^{\ast}}_{\lambda}({e_{j}}_{-\lambda-\partial^{A^{\ast
c}}}e_{k}))e_{k}^{\ast}$ $\displaystyle=$
$\displaystyle\sum_{j,k}(a_{lj}(\lambda,-\lambda){e_{t}^{\ast}}_{-\lambda-\partial^{A^{\ast
c}}}({e_{k}}_{\partial^{A^{\ast
c}}}e_{j})+a_{tj}(-\lambda-\partial,\lambda+\partial){e_{l}^{\ast}}_{\lambda}({e_{j}}_{-\lambda-\partial^{A^{\ast
c}}}e_{k}))e_{k}^{\ast}$ $\displaystyle=$
$\displaystyle\sum_{k}({e_{t}^{\ast}}_{-\lambda-\partial^{A^{\ast
c}}}({e_{k}}_{\partial^{A^{\ast
c}}}T_{0}^{r}(e_{l}))+{e_{l}^{\ast}}_{\lambda}(T_{0}^{r}({e_{t}})_{-\lambda-\partial^{A^{\ast
c}}}e_{k}))e_{k}^{\ast}$ $\displaystyle=$ $\displaystyle
R_{A}^{\ast}(T_{0}^{r}(e_{l}))_{\lambda}e_{t}^{\ast}+L_{A}^{\ast}(T_{0}^{r}(e_{t}))_{-\lambda-\partial}e_{l}^{\ast}.$

Thus by Theorem 6.1, we have

${T_{0}^{r}(e_{l}^{\ast})}_{\lambda}{T_{0}^{r}(e_{t}^{\ast})}=T_{0}^{r}({e_{l}^{\ast}}_{\lambda}e_{t}^{\ast}),\;\;\forall\
l,t\in\\{1,\cdots,n\\}.$

Therefore $T_{0}^{r}:A^{\ast c}\rightarrow A$ is a homomorphism of associative
conformal algebras. ∎

Let $A$ be a Frobenius conformal algebra with a non-degenerate invariant
conformal bilinear form $\langle\cdot,\cdot\rangle_{\lambda}$, which is
finitely generated and free as a $\mathbb{C}[\partial]$-module. Define

$\displaystyle\langle a\otimes b,c\otimes d\rangle_{(\lambda,\mu)}=\langle
a,c\rangle_{\lambda}\langle b,d\rangle_{\mu},\;\;\forall\ a,b,c,d\in A.$ (45)

Let $r=\sum_{i}r_{i}\otimes l_{i}\in A\otimes A$. Define a linear map
$P^{r}:A\rightarrow A[\lambda]$ by

$\displaystyle\langle r,u\otimes v\rangle_{(\lambda,\mu)}=\langle
P_{\lambda-\partial}^{r}(u),v\rangle_{\mu},\;\;\forall u,v\in A.$

Obviously, $P^{r}\in\text{Cend}(A)$.

######  Corollary 6.3.

Let $A$ be a symmetric Frobenius conformal algebra which is finitely generated
and free as a $\mathbb{C}[\partial]$-module and $r\in A\otimes A$ be
antisymmetric. Then $r$ is a solution of associative conformal Yang-Baxter
equation in $A$ if and only if $P^{r}\in\text{Cend}(A)$ satisfies

$\displaystyle
P^{r}_{0}(a)_{\lambda}P^{r}_{0}(b)=P_{0}^{r}(P_{0}^{r}(a)_{\lambda}b)+P^{r}_{0}(a_{\lambda}P^{r}_{0}(b)),\;\;\forall\
a,b\in A.$ (46)

###### Proof.

Since $A$ has a non-degenerate symmetric invariant conformal bilinear form,
$\varphi:A\longrightarrow A^{\ast c},~{}~{}a\mapsto\varphi_{a}$ defined by

$(\varphi_{a})_{\lambda}b=\langle a,b\rangle_{\lambda},\quad\forall\ a,b\in
A,$

is an isomorphism of $\mathbb{C}[\partial]$-modules. By the definitions of
$T^{r}$ in Theorem 6.1 and $P^{r}$, we get $P^{r}=T^{r}\circ\varphi$.
Therefore $P^{r}_{0}=T^{r}_{0}\circ\varphi$. Since $\varphi$ is a
$\mathbb{C}[\partial]$-module isomorphism, for any $f$, $g\in A^{\ast c}$, we
assume that $\varphi(a)=f$ and $\varphi(b)=g$. Then (42) becomes

$\displaystyle
T^{r}_{0}(\varphi(a))_{\lambda}T^{r}_{0}(\varphi(b))-T_{0}^{r}(R_{A}^{\ast}(T_{0}^{r}(\varphi(a)))_{\lambda}\varphi(b))-T^{r}_{0}(L_{A}^{\ast}(T^{r}_{0}(\varphi(b)))_{-\lambda-\partial}\varphi(a))=0.$
(47)

Thus

$\displaystyle
P^{r}_{0}(a)_{\lambda}P^{r}_{0}(b)-T_{0}^{r}(R_{A}^{\ast}(P_{0}^{r}(a))_{\lambda}\varphi(b))-T^{r}_{0}(L_{A}^{\ast}(P^{r}_{0}(b))_{-\lambda-\partial}\varphi(a))=0,\;\;\forall\
a,b\in A.$ (48)

For all $a,b,c\in A$, we have

$\displaystyle(R_{A}^{\ast}(P_{0}^{r}(a))_{\lambda}\varphi(b))_{\mu}c$
$\displaystyle=$
$\displaystyle\varphi(b)_{\mu-\lambda}(R_{A}(P_{0}^{r}(a))_{\lambda}c)=\langle
b,R_{A}(P_{0}^{r}(a))_{\lambda}c\rangle_{\mu-\lambda}$ $\displaystyle=$
$\displaystyle\langle
b,c_{-\lambda-\partial}P_{0}^{r}(a)\rangle_{\mu-\lambda}=\langle
b_{\mu-\lambda}c,P_{0}^{r}(a)\rangle_{-\lambda}$ $\displaystyle=$
$\displaystyle\langle P_{0}^{r}(a),b_{\mu-\lambda}c\rangle_{\lambda}=\langle
P_{0}^{r}(a)_{\lambda}b,c\rangle_{\mu}.$

Therefore we have

$R_{A}^{\ast}(P_{0}^{r}(a))_{\lambda}\varphi(b)=\varphi(P_{0}^{r}(a)_{\lambda}b),\;\;\forall\
a,b\in A.$

Similarly, we have

$L_{A}^{\ast}(P^{r}_{0}(b))_{-\lambda-\partial}\varphi(a)=\varphi(a_{\lambda}P_{0}^{r}(b)),\;\;\forall\
a,b\in A.$

Hence (46) follows from (48).

Similarly, we also obtain (42) from (46) through $\varphi$. Then by Theorem
6.1, the conclusion holds. ∎

By the fact that the $T_{0}^{r}$ in Theorems 6.1 and the $P_{0}^{r}$ in
Corollary 6.3 are $\mathbb{C}[\partial]$-module homomorphisms, we present the
following notions.

######  Definition 6.4.

Let $A$ be an associative conformal algebra and $(M,l_{A},r_{A})$ be a
bimodule of $A$. A $\mathbb{C}[\partial]$-module homomorphism $T:M\rightarrow
A$ is called an $\mathcal{O}$-operator associated with $(M,l_{A},r_{A})$ if
$T$ satisfies

$\displaystyle
T(u)_{\lambda}T(v)=T(l_{A}(T(u))_{\lambda}v)+T(r_{A}(T(v))_{-\lambda-\partial}u),\;\;\forall\text{$u$,
$v\in M$.}$ (49)

In particular, an $\mathcal{O}$-operator $T:A\rightarrow A$ associated with
the bimodule $(A,L_{A},R_{A})$ is called a Rota-Baxter operator (of weight
zero) on $A$, that is, $T$ is a $\mathbb{C}[\partial]$-module homomorphism
satisfying

$\displaystyle
T(a)_{\lambda}T(b)=T(T(a)_{\lambda}b)+T(a_{\lambda}T(b)),\;\;\forall\
\text{$a$, $b\in A$.}$ (50)

######  Example 6.5.

Let $A$ be a finite associative conformal algebra which is free as a
$\mathbb{C}[\partial]$-module and $r\in A\otimes A$ be antisymmetric. Then $r$
is a solution of associative conformal Yang-Baxter equation if and only if
$T_{0}^{r}$ is an $\mathcal{O}$-operator associated with the bimodule
$(A^{\ast c},R_{A}^{\ast},L_{A}^{\ast})$. If in addition, $A$ is a symmetric
Frobenius conformal algebra, that is, $A$ has a non-degenerate symmetric
invariant conformal bilinear form, then $r$ is a solution of associative
conformal Yang-Baxter equation if and only if $P_{0}^{r}$ is a Rota-Baxter
operator (of weight zero) on $A$.

######  Example 6.6.

Let $A$ be an associative conformal algebra. The identity map $I$ is an
$\mathcal{O}$-operator associated with the bimodule $(A,L_{A},0)$ or
$(A,0,R_{A})$.

Let $(M,l_{A},r_{A})$ be a bimodule of an associative conformal algebra $A$.
Then $(M^{\ast c},r_{A}^{\ast},l_{A}^{\ast})$ is a bimodule of $A$ by
Proposition 2.5. Suppose that $M$ is a $\mathbb{C}[\partial]$-module of finite
rank. By Proposition 6.1 in [7], $M^{\ast c}\otimes A\cong Chom(M,A)$ as
$\mathbb{C}[\partial]$-modules through the isomorphism $\varphi$ defined as

$\varphi(f\otimes a)_{\lambda}v=f_{\lambda+\partial^{A}}(v)a,\;\;\forall\ a\in
A,v\in M,f\in M^{\ast c}.$

By the $\mathbb{C}[\partial]$-module actions on $M^{\ast c}\otimes A$, we also
get $M^{\ast c}\otimes A\cong A\otimes M^{\ast c}$ as
$\mathbb{C}[\partial]$-modules. Therefore as $\mathbb{C}[\partial]$-modules,
$Chom(M,A)\cong A\otimes M^{\ast c}$. Consequently, for any $T\in Chom(M,A)$,
we associate an $r_{T}\in A\otimes M^{\ast
c}\subset(A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast
c})\otimes(A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast c})$.

######  Theorem 6.7.

Let $A$ be a finite associative conformal algebra and $(M,l_{A},r_{A})$ be a
finite bimodule of $A$. Suppose that $A$ and $M$ are free as
$\mathbb{C}[\partial]$-modules. Let $T\in Chom(M,A)$ and $r_{T}\in A\otimes
M^{\ast c}\subset(A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast
c})\otimes(A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast c})$ be the element
corresponding to $T$ under the above correspondence. Then $r=r_{T}-\tau r_{T}$
is an antisymmetric solution of the associative conformal Yang-Baxter equation
in $A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast c}$ if and only if
$T_{0}=T_{\lambda}|_{\lambda=0}$ is an $\mathcal{O}$-operator associated with
the bimodule $(M,l_{A},r_{A})$.

###### Proof.

Let $\\{e_{1},\cdots,e_{n}\\}$ be a $\mathbb{C}[\partial]$-basis of $A$,
$\\{v_{1},\cdots,v_{m}\\}$ be a $\mathbb{C}[\partial]$-basis of $M$ and
$\\{v_{1}^{\ast},\cdots,v_{m}^{\ast}\\}$ be the dual
$\mathbb{C}[\partial]$-basis of $M^{\ast c}$. Assume that

$T_{\lambda}(v_{i})=\sum_{j=1}^{n}g_{ij}(\lambda,\partial)e_{j},\;\;\forall\
i=1,\cdots,m,$

where $g_{ij}(\lambda,\partial)\in\mathbb{C}[\lambda,\partial]$. Then we have

$\displaystyle
r_{T}=\sum_{j=1}^{n}\sum_{i=1}^{m}g_{ij}(-\partial^{\otimes^{2}},\partial\otimes
1)e_{j}\otimes v_{i}^{\ast}.$

Therefore we have

$\displaystyle r=\sum_{i,j}(g_{ij}(-\partial^{\otimes^{2}},\partial\otimes
1)e_{j}\otimes
v_{i}^{\ast}-g_{ij}(-\partial^{\otimes^{2}},1\otimes\partial)v_{i}^{\ast}\otimes
e_{j}).$

Moreover, by the definition of $(M^{\ast c},r_{A}^{\ast},l_{A}^{\ast})$, we
have

$\displaystyle
l_{A}^{\ast}(e_{i})_{\lambda}v_{j}^{\ast}=\sum_{k}{v_{j}^{\ast}}_{-\lambda-\partial}(l_{A}(e_{i})_{\lambda}v_{k})v_{k}^{\ast},~{}~{}~{}r_{A}^{\ast}(e_{i})_{\lambda}v_{j}^{\ast}=\sum_{k}{v_{j}^{\ast}}_{-\lambda-\partial}(r_{A}(e_{i})_{\lambda}v_{k})v_{k}^{\ast}.$

Then we get

$\displaystyle r\bullet r$ $\displaystyle\equiv$
$\displaystyle\sum_{i,j,k,l}(-g_{ij}(0,\partial\otimes 1\otimes
1)g_{kl}(0,-1\otimes\partial\otimes 1)e_{j}\otimes
v_{k}^{\ast}\otimes{v_{i}^{\ast}}_{\mu}e_{l}|_{\mu=\partial\otimes 1\otimes
1}$ $\displaystyle-g_{ij}(0,-\partial\otimes 1\otimes
1)g_{kl}(0,1\otimes\partial\otimes 1)v_{i}^{\ast}\otimes
e_{l}\otimes{e_{j}}_{\mu}v_{k}^{\ast}|_{\mu=\partial\otimes 1\otimes 1}$
$\displaystyle+g_{ij}(0,-\partial\otimes 1\otimes
1)g_{kl}(0,-1\otimes\partial\otimes 1)v_{i}^{\ast}\otimes
v_{k}^{\ast}\otimes{e_{j}}_{\mu}e_{l}|_{\mu=\partial\otimes 1\otimes 1}$
$\displaystyle-g_{ij}(0,\partial\otimes 1\otimes 1)g_{kl}(0,-1\otimes
1\otimes\partial)e_{j}\otimes{e_{l}}_{\mu}v_{i}^{\ast}\otimes
v_{k}^{\ast}|_{\mu=-\partial^{\otimes^{2}}\otimes 1}$
$\displaystyle+g_{ij}(0,-\partial\otimes 1\otimes 1)g_{kl}(0,-1\otimes
1\otimes\partial)v_{i}^{\ast}\otimes{e_{l}}_{\mu}e_{j}\otimes
v_{k}^{\ast}|_{\mu=-\partial^{\otimes^{2}}\otimes 1}$ $\displaystyle-
g_{ij}(0,-\partial\otimes 1\otimes 1)g_{kl}(0,1\otimes
1\otimes\partial)v_{i}^{\ast}\otimes{v_{k}^{\ast}}_{\mu}e_{j}\otimes
e_{l}|_{\mu=-\partial^{\otimes^{2}}\otimes 1}$
$\displaystyle+g_{ij}(0,-1\otimes\partial\otimes 1)g_{kl}(0,-1\otimes
1\otimes\partial){e_{j}}_{\mu}e_{l}\otimes v_{i}^{\ast}\otimes
v_{k}^{\ast}|_{\mu=1\otimes\partial\otimes 1}$ $\displaystyle-
g_{ij}(0,-1\otimes\partial\otimes 1)g_{kl}(0,1\otimes
1\otimes\partial){e_{j}}_{\mu}v_{k}^{\ast}\otimes v_{i}^{\ast}\otimes
e_{l}|_{\mu=1\otimes\partial\otimes 1}$ $\displaystyle-
g_{ij}(0,1\otimes\partial\otimes 1)g_{kl}(0,-1\otimes
1\otimes\partial){v_{i}^{\ast}}_{\mu}e_{l}\otimes e_{j}\otimes
v_{k}^{\ast}|_{\mu=1\otimes\partial\otimes
1})~{}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}$ $\displaystyle\equiv$
$\displaystyle\sum_{i,k}((-T_{0}(v_{i})\otimes
v_{k}^{\ast}\otimes{{v_{i}}^{\ast}}_{\mu}T_{0}(v_{k})-v_{i}^{\ast}\otimes
T_{0}(v_{k})\otimes T_{0}(v_{i})_{\mu}v_{k}^{\ast}$
$\displaystyle+v_{i}^{\ast}\otimes
v_{k}^{\ast}\otimes{T_{0}(v_{i})}_{\mu}T_{0}(v_{k}))|_{\mu=\partial\otimes
1\otimes 1}$ $\displaystyle+(-T_{0}(v_{i})\otimes
T_{0}(v_{k})_{\mu}v_{i}^{\ast}\otimes
v_{k}^{\ast}+v_{i}^{\ast}\otimes{T_{0}(v_{k})}_{\mu}T_{0}(v_{i})\otimes
v_{k}^{\ast}$ $\displaystyle-
v_{i}^{\ast}\otimes{v_{k}^{\ast}}_{\mu}T_{0}(v_{i})\otimes
T_{0}(v_{k}))|_{\mu=1\otimes 1\otimes\partial}$
$\displaystyle+(T_{0}(v_{i})_{\mu}T_{0}(v_{k})\otimes v_{i}^{\ast}\otimes
v_{k}^{\ast}-T_{0}(v_{i})_{\mu}v_{k}^{\ast}\otimes v_{i}^{\ast}\otimes
T_{0}(v_{k})$ $\displaystyle-{v_{i}^{\ast}}_{\mu}T_{0}(v_{k})\otimes
T_{0}(v_{i})\otimes v_{k}^{\ast})|_{\mu=1\otimes\partial\otimes
1}~{}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}.$

Since $T_{0}$ is a $\mathbb{C}[\partial]$-module homomorphism, we have

$\displaystyle\sum_{i,k}T_{0}(v_{i})\otimes
v_{k}^{\ast}\otimes{v_{i}^{\ast}}_{\mu}T_{0}(v_{k})|_{\mu=\partial\otimes
1\otimes 1}$ $\displaystyle=$ $\displaystyle\sum_{i,k}T_{0}(v_{i})\otimes
v_{k}^{\ast}\otimes
l_{A}^{\ast}(T_{0}(v_{k}))_{-\mu-\partial}v_{i}^{\ast}|_{\mu=\partial\otimes
1\otimes 1}$ $\displaystyle\equiv$ $\displaystyle\sum_{i,k}T_{0}(v_{i})\otimes
v_{k}^{\ast}\otimes l_{A}^{\ast}(T_{0}(v_{k}))_{1\otimes\partial\otimes
1}v_{i}^{\ast}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}$
$\displaystyle\equiv$ $\displaystyle\sum_{i,j,k}T_{0}(v_{i})\otimes
v_{k}^{\ast}\otimes{v_{i}^{\ast}}_{\partial\otimes 1\otimes
1}(l_{A}(T_{0}(v_{k}))_{1\otimes\partial\otimes
1}v_{j})v_{j}^{\ast}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}$
$\displaystyle\equiv$
$\displaystyle\sum_{i,j,k}T_{0}({v_{i}^{\ast}}_{\partial}(l_{A}(T_{0}(v_{k}))_{1\otimes\partial\otimes
1}v_{j})v_{i})\otimes v_{k}^{\ast}\otimes
v_{j}^{\ast}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}$
$\displaystyle\equiv$
$\displaystyle\sum_{j,k}T_{0}(l_{A}((T_{0}(v_{k}))_{\mu}v_{j})\otimes
v_{k}^{\ast}\otimes v_{j}^{\ast}|_{\mu=1\otimes\partial\otimes
1}~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}.$

Similarly, we have

$\displaystyle r\bullet r~{}~{}\text{$mod~{}~{}(\partial^{\otimes^{3}})$}$
$\displaystyle\equiv$
$\displaystyle\sum_{i,k}(({T_{0}(v_{k})}_{\mu}T_{0}(v_{i})-T_{0}(l_{A}(T(v_{k}))_{\mu}v_{i})-T_{0}(r_{A}(T_{0}(v_{i}))_{-\mu-\partial}v_{k})\otimes
v_{k}^{\ast}\otimes v_{i}^{\ast})|_{\mu=1\otimes\partial\otimes 1}$
$\displaystyle+(v_{i}^{\ast}\otimes({T_{0}(v_{k})}_{\mu}T_{0}(v_{i})-T_{0}(l_{A}(T(v_{k}))_{\mu}v_{i})-T_{0}(r_{A}(T_{0}(v_{i}))_{-\mu-\partial}v_{k})\otimes
v_{k}^{\ast})|_{\mu=1\otimes 1\otimes\partial}$
$\displaystyle+(v_{k}^{\ast}\otimes
v_{i}^{\ast}\otimes({T_{0}(v_{k})}_{\mu}T_{0}(v_{i})-T_{0}(l_{A}(T(v_{k}))_{\mu}v_{i})-T_{0}(r_{A}(T_{0}(v_{i}))_{-\mu-\partial}v_{k})\otimes
v_{k}^{\ast})|_{\mu=\partial\otimes 1\otimes 1}.$

Therefore $r$ is a solution of associative conformal Yang-Baxter equation in
the associative conformal algebra $A\ltimes_{r_{A}^{\ast},l_{A}^{\ast}}M^{\ast
c}$ if and only if

$\displaystyle{T_{0}(v_{k})}_{\mu}T_{0}(v_{i})=T_{0}(l_{A}(T(v_{k}))_{\mu}v_{i})+T_{0}(r_{A}(T_{0}(v_{i}))_{-\mu-\partial}v_{k}),\;\;\forall\
i,k\in\\{1,\cdots,m\\}.$

Thus this conclusion holds. ∎

At the end of this paper, we introduce a class of conformal algebras, namely,
dendriform conformal algebras, which are used to construct
$\mathcal{O}$-operators naturally and hence give solutions of associative
conformal Yang-Baxter equation.

######  Definition 6.8.

Let $A$ be a $\mathbb{C}[\partial]$-module with two bilinear products
$\prec_{\lambda}$ and $\succ_{\lambda}:A\times A\rightarrow A[\lambda]$. If
for all $a$, $b$, $c\in A$,

$\displaystyle(\partial a)\succ_{\lambda}b=-\lambda
a\succ_{\lambda}b,~{}~{}a\succ_{\lambda}(\partial
b)=(\partial+\lambda)(a\succ_{\lambda}b),$ (51) $\displaystyle(\partial
a)\prec_{\lambda}b=-\lambda a\prec_{\lambda}b,~{}~{}a\prec_{\lambda}(\partial
b)=(\partial+\lambda)(a\prec_{\lambda}b),$ (52)
$\displaystyle(a\prec_{\lambda}b)\prec_{\lambda+\mu}c=a\prec_{\lambda}(b\ast_{\mu}c),$
(53)
$\displaystyle~{}~{}(a\succ_{\lambda}b)\prec_{\lambda+\mu}c=a\succ_{\lambda}(b\prec_{\mu}c),$
(54)
$\displaystyle~{}~{}a\succ_{\lambda}(b\succ_{\mu}c)=(a\ast_{\lambda}b)\succ_{\lambda+\mu}c,$
(55)

where $a\ast_{\lambda}b=a\prec_{\lambda}b+a\succ_{\lambda}b$, then
$(A,\prec_{\lambda},\succ_{\lambda})$ is called a dendriform conformal
algebra.

######  Remark 6.9.

It is obvious that $(A,\prec_{\lambda},\succ_{\lambda})$ with
$\succ_{\lambda}$ being trivial (or $\prec_{\lambda}$ being trivial) is a
dendriform conformal algebra if and only if $(A,\prec_{\lambda})$ (or
$(A,\succ_{\lambda}))$ is an associative conformal algebra.

######  Example 6.10.

Let $(A,\prec,\succ)$ be a dendriform algebra ([25]). Then
$\text{Cur}(A)=\mathbb{C}[\partial]\otimes A$ is endowed a natural dendriform
conformal algebra as follows.

$\displaystyle a\prec_{\lambda}b=a\prec b,~{}~{}~{}a\succ_{\lambda}b=a\succ
b,~{}~{}~{}\forall a,b\in A.$ (56)

Moreover, $(\text{Cur}(A),\prec_{\lambda},\succ_{\lambda})$ is called a
current dendriform conformal algebra. It is straightforward that any
dendriform conformal algebra which is free and of rank one as a
$\mathbb{C}[\partial]$-module is current.

######  Proposition 6.11.

Let $(A,\prec_{\lambda},\succ_{\lambda})$ be a dendriform conformal algebra.
Define

$a\ast_{\lambda}b=a\prec_{\lambda}b+a\succ_{\lambda}b,\;\;\forall\ a,b\in A.$
(57)

Then $(A,\ast_{\lambda})$ is an associative conformal algebra. We call
$(A,\ast_{\lambda})$ the associated associative conformal algebra of
$(A,\prec_{\lambda},\succ_{\lambda})$ and
$(A,\prec_{\lambda},\succ_{\lambda})$ is called a compatible dendriform
conformal algebra structure on the associative conformal algebra
$(A,\ast_{\lambda})$.

###### Proof.

It is straightforward. ∎

Let $(A,\prec_{\lambda},\succ_{\lambda})$ be a dendriform conformal algebra.
Set

$\displaystyle{L_{\succ}(a)}_{\lambda}(b)=a\succ_{\lambda}b,\;\;{L_{\prec}(a)}_{\lambda}(b)=a\prec_{\lambda}b,$
$\displaystyle{R_{\succ}(a)}_{\lambda}b=b\succ_{-\lambda-\partial}a,\;\;{R_{\prec}(a)}_{\lambda}b=b\prec_{-\lambda-\partial}a,\;\;\forall\
a,b\in A.$

######  Proposition 6.12.

Let $(A,\prec_{\lambda},\succ_{\lambda})$ be a dendriform conformal algebra.
Then $(A,L_{\succ},R_{\prec})$ is a bimodule of the associated associative
conformal algebra $(A,\ast_{\lambda})$. Hence the identity $I$ is an
$\mathcal{O}$-operator of the associative conformal algebra
$(A,\ast_{\lambda})$ associated with $(A,L_{\succ},R_{\prec})$.

###### Proof.

It is straightforward. ∎

######  Proposition 6.13.

Let $A$ be an associative conformal algebra and $(M,l_{A},r_{A})$ be a
bimodule of $A$. Suppose $T:M\rightarrow A$ is an $\mathcal{O}$-operator
associated with $(M,l_{A},r_{A})$. Then the following $\lambda$-product

$\displaystyle
u\succ_{\lambda}v=l_{A}(T(u))_{\lambda}v,~{}~{}~{}u\prec_{\lambda}v=r_{A}(T(v))_{-\lambda-\partial}u,~{}~{}~{}~{}~{}u,~{}v\in
M,$ (58)

endows a dendriform conformal algebra structure on $M$. Therefore there is an
associated associative conformal algebra structure on $M$ and $T:M\rightarrow
A$ is a homomorphism of associative conformal algebras. Moreover, $T(M)\subset
A$ is an associative conformal subalgebra of $A$ and there is also a
dendriform conformal algebra structure on $T(M)$ defined by

$\displaystyle
T(u)\succ_{\lambda}T(v)=T(u\succ_{\lambda}v),~{}~{}T(u)\prec_{\lambda}T(v)=T(u\prec_{\lambda}v),~{}~{}~{}\text{$u$,
$v\in M$.}$ (59)

Furthermore, the associated associative conformal algebra on $T(M)$ is a
subalgebra of $A$ and $T:M\rightarrow A$ is a homomorphism of dendriform
conformal algebras.

###### Proof.

For all $u$, $v$, $w\in M$, we have

$\displaystyle(u\prec_{\lambda}v)\prec_{\lambda+\mu}w-u\prec_{\lambda}(v\prec_{\mu}w+v\succ_{\mu}w)$
$\displaystyle=$ $\displaystyle
r_{A}(T(w))_{-\lambda-\mu-\partial}(r_{A}(T(v))_{-\lambda-\partial}u)-r_{A}(r_{A}(T(w))_{-\mu-\partial}v)_{-\lambda-\partial}u-r_{A}(l_{A}(T(v))_{\mu}w)_{-\lambda-\partial}u$
$\displaystyle=$ $\displaystyle
r_{A}(T(v)_{\mu}T(w))_{-\lambda-\partial}u-r_{A}(r_{A}(T(w))_{-\mu-\partial}v)_{-\lambda-\partial}u-r_{A}(l_{A}(T(v))_{\mu}w)_{-\lambda-\partial}u$
$\displaystyle=$ $\displaystyle
r_{A}(T(v)_{\mu}T(w)-r_{A}(T(w))_{-\mu-\partial}v-l_{A}(T(v))_{\mu}w)_{-\lambda-\partial}u=0.$

Similarly, we have

$u\succ_{\lambda}(v\succ_{\mu}w)=(u\succ_{\lambda}v+u\prec_{\lambda}v)\succ_{\lambda+\mu}w,\;\;\forall\
u,v,w\in M.$

Moreover, since $(M,l_{A},r_{A})$ is a bimodule of $A$, we have

$(u\succ_{\lambda}v)\prec_{\lambda+\mu}w=u\succ_{\lambda}(v\prec_{\mu}w),\;\;\forall\
u,v,w\in M.$

Hence $(M,\prec_{\lambda},\succ_{\lambda})$ is a dendriform conformal algebra.
Moreover, the other conclusions follow straightforwardly. Therefore the
conclusion holds. ∎

######  Corollary 6.14.

Let $(A,\ast_{\lambda})$ be an associative conformal algebra. There is a
compatible dendriform conformal algebra structure on $A$ if and only if there
exists a bijective $\mathcal{O}$-operator $T:M\rightarrow A$ associated with
some bimodule $(M,l_{A},r_{A})$ of $A$.

###### Proof.

Suppose that there is a compatible dendriform conformal algebra structure
$(A,\succ_{\lambda},\prec_{\lambda})$ on $A$. Then by Proposition 6.12, the
identity map $I:A\rightarrow A$ is a bijective $\mathcal{O}$-operator of $A$
associated with $(A,L_{\succ},R_{\prec})$.

Conversely, suppose that there exists a bijective $\mathcal{O}$-operator
$T:M\rightarrow A$ of $(A,\ast_{\lambda})$ associated with a bimodule
$(M,l_{A},r_{A})$. Then by Proposition 6.13 with a straightforward checking,
we have

$\displaystyle
a\succ_{\lambda}b=T(l_{A}(a)_{\lambda}T^{-1}(b)),~{}~{}~{}a\prec_{\lambda}b=T({r_{A}(b)}_{-\lambda-\partial}T^{-1}(a)),\;\;\forall\
a,b\in A,$ (60)

defines a compatible dendriform conformal algebra structure on $A$. ∎

Finally, there is a construction of (antisymmetric) solutions of associative
conformal Yang-Baxter equation from dendriform conformal algebras.

######  Theorem 6.15.

Let $(A,\succ_{\lambda},\prec_{\lambda})$ be a finite dendriform conformal
algebra which is free as a $\mathbb{C}[\partial]$-module. Then

$\displaystyle r=\sum_{i=1}^{n}(e_{i}\otimes e_{i}^{\ast}-e_{i}^{\ast}\otimes
e_{i})$ (61)

is a solution of associative conformal Yang-Baxter equation in the associative
conformal algebra $A\ltimes_{R_{A}^{\ast},L_{A}^{\ast}}A^{\ast c}$, where
$\\{e_{1},\cdots,e_{n}\\}$ is a $\mathbb{C}[\partial]$-basis of $A$ and
$\\{e_{1}^{\ast},\cdots,e_{n}^{\ast}\\}$ is the dual
$\mathbb{C}[\partial]$-basis of $A^{\ast c}$.

###### Proof.

By Proposition 6.12, $T=I:A\rightarrow A$ is an $\mathcal{O}$-operator
associated with $(A,L_{\succ},R_{\prec})$. Then by Theorem 6.7, the conclusion
holds. ∎

Acknowledgments This work was supported by the National Natural Science
Foundation of China (11425104, 11931009), the Zhejiang Provincial Natural
Science Foundation of China (LY20A010022) and the Scientific Research
Foundation of Hangzhou Normal University (2019QDL012). C. Bai is also
supported by the Fundamental Research Funds for the Central Universities and
Nankai ZhiDe Foundation.

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