# Structures of BiHom-Poisson algebras

Sylvain Attan Département de Mathématiques, Université d’Abomey-Calavi 01 BP
4521, Cotonou 01, Bénin. E.mail<EMAIL_ADDRESS>Ismail Laraiedh
Departement of Mathematics, Faculty of Sciences, Sfax University, BP 1171,
3000 Sfax, Tunisia. E.mail<EMAIL_ADDRESS>and Departement of
Mathematics, College of Sciences and Humanities - Kowaiyia, Shaqra University,
Kingdom of Saudi Arabia. E.mail<EMAIL_ADDRESS>
###### Abstract

This paper gives some constructions results and examples of BiHom-Poisson
algebras. Next, BiHom-flexible algebras are defined and it is shown that
admissible BiHom-Poisson algebras are BiHom-flexible. Furthermore, generalized
derivations of Bihom-Poisson algebras are introduced and some their basic
properties are given. Finally, BiHom-Poisson modules and several constructions
of these notions are obtained.

2010 Mathematics Subject Classification: 17A20, 17A30, 17B05, 17B10, 17B65 .
Keywords: BiHom-Poisson algebras, BiHom-Poisson admissible, generalized
derivation,
BiHom-Poisson modules.

##  1 Introduction

A Poisson algebra $(P,\\{\cdot,\cdot\\},\mu)$ consists of a commutative
associative algebra $(A,\mu)$
together with a Lie structure $\\{\cdot,\cdot\\},$ satisfying the Leibniz
identity:

$\\{\mu(x,y),z\\}=\mu(\\{x,z\\},y)+\mu(x,\\{y,z\\})$

These algebras firstly appeared in the work of Siméon-Denis Poisson two
centuries ago when he was studying the three-body problem in celestial
mechanics. Since then, Poisson algebras have shown to be connected to many
areas of mathematics and physics. Indeed, in mathematics, Poisson algebras
play a fundamental role in Poisson geometry [34], quantum groups [8],[11] and
deformation of commutative associative algebras [13] whereas in physics,
Poisson algebras represent a major part of deformation quantization [17],
Hamiltonian mechanics [3] and topological field theories [33]. Poisson-like
structures are also used in the study of vertex operator algebras [12].

Algebras where the identities defining the structure are twisted by a
homomorphism are called Hom-algebras. Hom-type algebras appeared in the
Physics literature of the 1990’s, when looking for quantum deformations of
some algebras of vector fields, like Witt and Virasoro algebras ([2], [24]).
It was observed that algebras obtained by deforming certain Lie algebras no
longer satisfied the Jacobi identity, but a modified version of it involving a
homomorphism. An axiomatization of this type of algebras (called Hom-Lie
algebras) was given in [21], [25]. The associative counterpart of Hom-Lie
algebras (called Hom-associative algebras) has been introduced in [31], where
it was proved also that the commutator bracket defined by the multiplication
in a Hom-associative algebra gives rise to a Hom-Lie algebra.

A BiHom-algebra is an algebra in such a way that the identities defining the
structure are twisted by two homomorphisms $\alpha$ and $\beta$. This class of
algebras was introduced from a categorical approach in [18] as an extension of
the class of Hom-algebras. These algebraic structures include BiHom-
associative algebras, BiHom-Jordan algebras, BiHom-alternative algebras and
BiHom-Lie algebras. BiHom-Poisson algebras was first introduced in [29] and
studied in [1] where in partiular, the concept of module of BiHom-Poisson
algebra is introduced and admissible BiHom-Poisson algebras, and only these
BiHom-algebras, are shown to give rise to BiHom-Poisson algebras via
polarization. More applications of BiHom-type can be found in [19, 20, 9, 10,
26, 28, 30].

The purpose of this paper is further to study BiHom-Poisson algebras. The
paper is organized as follows. Section 2 contains some important basic notions
and notations related to BiHom-algebras, BiHom-Poisson algebras and modules
over BiHom-associative algebras. Section 3 presents some constructions results
of BiHom-Poisson algebras, BiHom-flexible structures and admissible BiHom-
Poisson algebras. In section 4 we give some basic properties concerning
derivation algebras, quasiderivation algebras and generalized derivation
algebras of Bihom-Poisson algebras. In section 5, we introduce and give some
properties of BiHom-Poisson modules. Next, we prove that from a given BiHom-
Poisson modules, a sequence of this kind of modules can be constructed and we
then define the semi-direct product of BiHom-Poisson algebras. Finally, we
define the first and second cohomology spaces of BiHom-Poisson algebras.

Throughout this paper $\mathbb{K}$ is an algebraically closed field of
characteristic $0$ and $A$ is a $\mathbb{K}$-vector space.

##  2 Preliminaries

This section contains necessary important basic notions and notations which
will be used in next sections. For the map $\mu:A^{\otimes 2}\longrightarrow
A,$ we will sometimes $\mu(a\otimes b)$ as $\mu(a,b)$ or $ab$ for $a,b\in A$
and if $V$ is another vector space, $\tau_{1}:A\otimes V\longrightarrow
V\otimes A$ (resp. $\tau_{2}:V\otimes A\longrightarrow A\otimes V$) denote the
twist isomorphism $\tau_{1}(a\otimes v)=v\otimes a$ (resp. $\tau_{2}(v\otimes
a)=a\otimes v$).

######  Definition 2.1.

A BiHom-module is a pair $(M,\alpha_{M},\beta_{M})$ consisting of a
$\mathbb{K}$-module $M$ and a linear self-maps
$\alpha_{M},\beta_{M}:M\longrightarrow M$ such that
$\alpha_{M}\beta_{M}=\beta_{M}\alpha_{M}.$ A morphism
$f:(M,\alpha_{M},\beta_{M})\rightarrow(N,\alpha_{N},\beta_{N})$ of BiHom-
modules is a linear map $f:M\longrightarrow N$ such that
$f\alpha_{M}=\alpha_{N}f$ and $f\beta_{M}=\beta_{N}f.$

######  Definition 2.2.

[15] A BiHom-algebra is a quadruple $(A,\mu,\alpha,\beta)$ in which
$(A,\alpha,\beta)$ is a BiHom-module, $\mu:A^{\otimes 2}\rightarrow A$ is a
linear map. The BiHom-algebra $(A,\mu,\alpha;\beta)$ is said to be
multiplicative if $\alpha\circ\mu=\mu\circ\alpha^{\otimes 2}$ and
$\beta\circ\mu=\mu\circ\beta^{\otimes 2}$ (multiplicativity).

######  Definition 2.3.

  1. 1.

A BiHom-algebra $(A,\mu,\alpha,\beta)$ is said to be BiHom-commutative if

$\mu(\beta(x),\alpha(y))=\mu(\beta(y),\alpha(x)),~{}\forall x,y\in A$

.

  2. 2.

A BiHom-associative algebra [15] is a multiplicative Bihom-algebra
$(A,\mu,\alpha,\beta)$ satisfying the following BiHom-associativity condition:

$as_{A}(x,y,z):=\mu(\mu(x,y),\beta(z))-\mu(\alpha(x),\mu(y,z))=0,\ \text{for
all}\ x,y,z\in A.\ $ (2.1)

  3. 3.

A BiHom-Lie algebra [15] is a multiplicative Bihom-algebra
$(A,\\{\cdot,\cdot\\},\alpha,\beta)$ satisfying the BiHom-skew-symmetry and
the BiHom-Jacobi identities i.e.

$\displaystyle\\{\beta(x),\alpha(y)\\}=-\\{\beta(y),\alpha(x)\\}$
$\displaystyle\circlearrowleft_{x,w,z}\\{\beta^{2}(x),\\{\beta(y),\alpha(z)\\}\\}=0$
(2.2)

where $\circlearrowleft_{x,y,z}$ denotes the summation over the cyclic
permutation on $x,y,z.$

Clearly, a Hom-associative algebra $(A,\mu,\alpha)$ can be regarded as a
BiHom-associative algebra $(A,\mu,\alpha,\alpha)$.

######  Definition 2.4.

[29] A BiHom-Poisson algebra consists of a vector space $A,$ two bilinear maps
$\mu,\\{\cdot,\cdot\\}:A^{\otimes 2}\longrightarrow A,$ linear maps
$\alpha,\beta:A\longrightarrow A$ such that

  1. 1.

$(A,\mu,\alpha,\beta)$ is a BiHom-commutative BiHom-associative algebra,

  2. 2.

$(A,\\{\cdot,\cdot\\},\alpha,\beta)$ is a BiHom-Lie algebra,

  3. 3.

the BiHom-Leibniz identity

$\displaystyle\\{\alpha\beta(x),\mu(y,z)\\}=\mu(\\{\beta(x),y\\},\beta(z))+\mu(\beta(y),\\{\alpha(x),z\\})$
(2.3)

is satisfied for all $x,y,z\in A.$

In a BiHom-Poisson algebra $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$, the
operations $\mu$ and $\\{\cdot,\cdot\\}$ are called the BiHom-associative
product and the BiHom-Poisson bracket, respectively.

######  Remark 2.5.

A non-BiHom-commutative BiHom-Poisson algebra is a BiHom-Poisson algebra
without the BiHom-commutativity assumption [1]. These Bihom-algebras are
called BiHom-Poisson algebras [29].

######  Definition 2.6.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ a BiHom-Poisson algebra. A
subspace $H$ of $A$ is called

  1. 1.

A BiHom-subalgebra of $A$ if

$\alpha(H)\subseteq H,~{}\beta(H)\subseteq H,~{}\mu(H,H)\subseteq
H~{}\textrm{and}~{}\\{H,H\\}\subseteq H.$

  2. 2.

A left-BiHom ideal of $A$ if

$\alpha(H)\subseteq H,~{}\beta(H)\subseteq H,~{}\mu(A,H)\subseteq
H~{}\textrm{and}~{}\\{A,H\\}\subseteq H.$

  3. 3.

A right-BiHom ideal of $A$ if

$\alpha(H)\subseteq H,~{}\beta(H)\subseteq H,~{}\mu(H,A)\subseteq
H~{}\textrm{and}~{}\\{H,A\\}\subseteq H.$

  4. 4.

A two sided BiHom-ideal if $H$ is both a left and a right BiHom-ideal of $A$.

Note that, if $\alpha$ and $\beta$ are bijective, then the notion of left-
BiHom ideals is equivalent to the one of right-BiHom ideals.

######  Definition 2.7.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a Bihom-Poisson algebra. If
$Z(A)=\\{x\in A|~{}\\{x,y\\}=\mu(x,y)=0,~{}~{}\forall y\in A\\}$, then $Z(A)$
is called the centralizer of $A$.

######  Definition 2.8.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ a BiHom-Poisson algebra.

  1. 1.

$A$ is multiplicative if

$\alpha\\{\cdot,\cdot\\}=\\{\cdot,\cdot\\}\alpha^{\otimes
2},~{}~{}\beta\\{\cdot,\cdot\\}=\\{\cdot,\cdot\\}\beta^{\otimes
2}\quad\text{and}\quad\alpha\mu=\mu\alpha^{\otimes
2},~{}~{}\beta\mu=\mu\beta^{\otimes 2}.$

  2. 2.

$(A,\mu,\alpha,\beta)$ is said to be regular if $\alpha$ and $\beta$ are
algebra automorphisms.

  3. 3.

$(A,\mu,\alpha,\beta)$ is said to be involutive if $\alpha$ and $\beta$ are
two involutions, that is $\alpha^{2}=\beta^{2}=id$.

  4. 4.

Let
$(A^{\prime},\\{\cdot,\cdot\\}^{\prime},\mu^{\prime},\alpha^{\prime},\beta^{\prime})$
be another BiHom-Poisson algebra. A weak morphism $f\colon A\to A^{\prime}$ is
a linear map such that

$f\\{\cdot,\cdot\\}=\\{\cdot,\cdot\\}^{\prime}f^{\otimes
2}\quad\text{and}\quad f\mu=\mu^{\prime}f^{\otimes 2}.$

A morphism $f\colon A\to A^{\prime}$ is a weak morphism such that
$f\alpha=\alpha^{\prime}f$ and $f\beta=\beta^{\prime}f$.

Note that a $5$-tuple $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ is
multiplicative if and only if the twisting map $\alpha,\beta\colon A\to A$ are
morphisms.

Denote by $\Gamma_{f}=\\{x+f(x);~{}~{}x\in A\\}\subset A\oplus A^{\prime}$ the
graph of a linear map $f:A\longrightarrow A^{{}^{\prime}}$.

######  Definition 2.9.

Let $(A,\mu,\alpha,\beta)$ be any BiHom-algebra.

  1. 1.

A BiHom-module $(V,\phi,\psi)$ is called an $A$-bimodule if it comes equipped
with a left and a right structures maps on $V$ that is morphisms
$\rho_{l}:(A\otimes
V,\alpha\otimes\phi,\beta\otimes\psi)\rightarrow(V,\phi,\psi),~{}a\otimes
v\mapsto a.v$ and $\rho_{r}:(V\otimes
A,\phi\otimes\alpha,\psi\otimes\psi)\rightarrow(V,\phi,\psi),~{}v\otimes
a\mapsto v.a$ of Bihom-modules.

  2. 2.

A morphism
$f:(V,\phi,\psi,\rho_{l},\rho_{r})\rightarrow(W,\phi^{\prime},\psi^{\prime},\rho_{l}^{\prime},\rho_{r}^{\prime})$
of $A$-bimodules is a morphism of the underlying BiHom-modules such that

$f\circ\rho_{l}=\rho_{l}^{\prime}\circ(Id_{A}\otimes
f)~{}~{}and~{}~{}f\circ\rho_{r}=\rho_{r}^{\prime}\circ(f\otimes Id_{A}).$

That yields the commutative diagrams

$\textstyle{A\otimes
V\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Id_{A}\otimes
f}$$\scriptstyle{\rho_{l}}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{A\otimes
W\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{l}^{\prime}}$$\textstyle{W}$
$\textstyle{V\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f\otimes
Id_{A}}$$\scriptstyle{\rho_{r}}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{W\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{r}^{\prime}}$$\textstyle{W}$

Now, let consider the following notions for BiHom-associative algebras.

######  Definition 2.10.

Let $(A,\mu,\alpha,\beta)$ be a BiHom-associative algebra,
$(L,\\{\cdot,\cdot\\},\alpha,\beta)$ be a Hom-Lie algebra and $(V,\phi,\psi)$
be a BiHom-module. Then

  1. 1.

A left BiHom-associative $A$-module structure on $V$ [15] consists of a
morphism $\rho_{l}:A\otimes V\longrightarrow V$ of BiHom-modules, such that

$\displaystyle\rho_{l}\circ(\alpha\otimes\rho_{l})=\rho_{l}\circ(\mu\otimes\psi).$
(2.4)

In terms of diagram, we have

$\textstyle{A\otimes A\otimes
V\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu\otimes\psi}$$\scriptstyle{\alpha\otimes\rho_{l}}$$\textstyle{A\otimes
V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{l}}$$\textstyle{A\otimes
V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{l}}$$\textstyle{V}$

  2. 2.

A right BiHom-associative $A$-module structure on $V$ [15] consists of a
morphism $\rho_{r}:V\otimes A\longrightarrow V$ of BiHom-modules, such that

$\displaystyle\rho_{r}\circ(\phi\otimes\mu)=\rho_{r}\circ(\rho_{r}\otimes\beta).$
(2.5)

In terms of diagram, we have

$\textstyle{V\otimes A\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi\otimes\mu}$$\scriptstyle{\rho_{r}\otimes\beta}$$\textstyle{V\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{r}}$$\textstyle{V\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{r}}$$\textstyle{V}$

  3. 3.

A left BiHom-Lie $L$-module structure on $V$ [1][36] consists of a structure
map $\rho:L\otimes V\longrightarrow V$ such that

$\displaystyle\rho(\\{\beta(x),y\\},\psi(v))$ $\displaystyle=$
$\displaystyle\rho(\alpha\beta(x),\rho(y,v))-\rho(\beta(y),\rho(\alpha(x),v))$
(2.6)

##  3 BiHom-Poisson algebras

###  3.1 Constructions of BiHom-Poisson algebras

In this subsection, we provide some constructions results of BiHom-Poisson
algebras.

######  Proposition 3.1.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra and
$I$ be a two-sided BiHom-ideal of $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$.
Then $(A/I,[\cdot,\cdot],\overline{\mu},\overline{\alpha},\overline{\beta})$
is a BiHom-Poisson algebra where
$[\overline{x},\overline{y}]=\overline{\\{x,y\\}},~{}\overline{\mu}(\overline{x},\overline{y})=\overline{\mu(x,y)}$,
$\overline{\alpha}(\overline{x})=\overline{\alpha(x)}$ and
$\overline{\beta}(\overline{x})=\overline{\beta(x)}$, for all
$\overline{x},\overline{y}\in A/I$

###### Proof.

We only prove item 1. of definition 2.4, item 2. and item 3. are being proved
similarly

For all $\overline{x},\overline{y},\overline{z}\in A/I$ we have

$\begin{array}[]{llllll}as_{A/I}(\overline{x},\overline{y},\overline{z})&=&\overline{\mu}(\overline{\mu}(\overline{x},\overline{y}),\overline{\beta}(\overline{z}))-\overline{\mu}(\overline{\alpha}(\overline{x}),\overline{\mu}(\overline{y},\overline{z}))\\\\[5.69046pt]
&=&\overline{\mu(\mu(x,y),\beta(z))-\mu(\alpha(x),\mu(y,z))}\\\\[5.69046pt]
&=&\overline{0}~{}(by~{}BiHom-associativity~{}of~{}A).\end{array}$

Then $(A/I,\overline{\mu},\overline{\alpha},\overline{\beta})$ is a BiHom-
associative algebra.

$\begin{array}[]{llll}\overline{\mu}(\overline{\beta}(\overline{x}),\overline{\alpha}(\overline{y}))&=&\overline{\mu(\beta(x),\alpha(y))}\\\\[5.69046pt]
&=&\overline{\mu(\beta(y),\alpha(x))}~{}(by~{}BiHom-
commutativity~{}ofA).\\\\[5.69046pt]
&=&\overline{\mu}(\overline{\beta}(\overline{y}),\overline{\alpha}(\overline{x}))\end{array}$

Then $(A/I,\overline{\mu},\overline{\alpha},\overline{\beta})$ is a BiHom-
commutative algebra. ∎

######  Proposition 3.2.

[1] Let $(A,\mu,\alpha,\beta)$ be a regular BiHom-associative algebra. Then
$A^{-}=(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ is a regular non-BiHom-
commutative BiHom-Poisson algebra, where
$\\{\cdot,\cdot\\}=\mu-\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau$,

where for all $x,y\in A,~{}~{}\tau(x\otimes y)=y\otimes x$.

######  Example 3.3.

Consider the 2-dimensional regular BiHom-associative algebras
$(A,\mu,\alpha,\beta)$ with a basis $(e_{1},e_{2}),$ (see [15]) defined by

$\begin{array}[]{llllll}\alpha(e_{1})=e_{1},&&\alpha(e_{2})=\frac{b(1-a)}{a}e_{1}+ae_{2},\\\
\beta(e_{1})=e_{1},&&\beta(e_{2})=be_{1}+(1-a)e_{2},\\\
\mu(e_{1},e_{1})=e_{1},&&\mu(e_{1},e_{2})=be_{1}+(1-a)e_{2},\\\
\mu(e_{2},e_{1})=\frac{b(1-a)}{a}e_{1}+ae_{2},&&\mu(e_{2},e_{2})=\frac{b}{a}e_{2},\end{array}$

where $a$, $b$ are parameters in $\mathbb{K}$, with $a\neq 0,1$. Using the
Proposition 3.2, the 5-tuple $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ is a
regular non-BiHom-commutative BiHom-Poisson algebra where,
$\\{\cdot,\cdot\\}=\mu-\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau$
and

$\begin{array}[]{llllll}\alpha^{-1}(e_{1})=e_{1},&&\alpha^{-1}(e_{2})=\frac{b(a-1)}{a^{2}}e_{1}+\frac{1}{a}e_{2},\\\
\beta^{-1}(e_{1})=e_{1},&&\beta^{-1}(e_{2})=\frac{b}{a-1}e_{1}+\frac{1}{1-a}e_{2}.\end{array}$

######  Proposition 3.4.

[29] Let $(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{A},\beta_{A})$ and
$(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{B},\beta_{B})$ be two BiHom-Poisson
algebras. Then there exists a BiHom-Poissson algebra $(A\oplus
B,\\{\cdot,\cdot\\},\mu,\alpha=\alpha_{A}+\alpha_{B},\beta=\beta_{A}+\beta_{B}),$
where the bilinear maps $\\{\cdot,\cdot\\},\mu:(A\oplus B)^{\times
2}\longrightarrow(A\oplus B)$ are given by

$\begin{array}[]{llll}\mu(a_{1}+b_{1},a_{2}+b_{2})=\mu_{A}(a_{1},a_{2})+\mu_{B}(b_{1},b_{2}),\\\
\\{a_{1}+b_{1},a_{2}+b_{2}\\}=\\{a_{1},a_{2}\\}_{A}+\\{b_{1},b_{2}\\}_{B},\forall\
a_{1},a_{2}\in A,\ \forall\ b_{1},b_{2}\in B.\par\end{array}$

and the linear maps
$\beta=\beta_{A}+\beta_{B},~{}\alpha=\alpha_{A}+\alpha_{B}:(A\oplus
B)\longrightarrow(A\oplus B)$ are given by

$\begin{array}[]{lll}(\alpha_{A}+\alpha_{B})(a+b)&=&\alpha_{A}(a)+\alpha_{B}(b),\\\
(\beta_{A}+\beta_{B})(a+b)&=&\beta_{A}(a)+\beta_{B}(b),~{}~{}\forall\ (a,b)\in
A\times B.\end{array}$

###### Proof.

It is easy to see that $(A\oplus
B,\mu,\alpha_{A}+\alpha_{B},\beta_{A}+\beta_{B})$ is a BiHom-associative and
BiHom-commutative algebra and $(A\oplus
B,\\{\cdot,\cdot\\},\alpha_{A}+\alpha_{B},\beta_{A}+\beta_{B})$ is a BiHom-Lie
algebra. Then $(A\oplus
B,\\{\cdot,\cdot\\},\mu,\alpha_{A}+\alpha_{B},\beta_{A}+\beta_{B})$ is a
BiHom-Poisson algebra. ∎

######  Proposition 3.5.

Let $(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{1},\beta_{1})$ and
$(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{2},\beta_{2})$ be two BiHom-Poisson
algebras and $\varphi:A\rightarrow B$ be a linear map. Then $\varphi$ is a
morphism from $(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{1},\beta_{1})$ to
$(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{2},\beta_{2})$ if and only if its
graph $\Gamma_{\varphi}$ is a BiHom-subalgebra of $(A\oplus
B,\\{\cdot,\cdot\\},\mu,\alpha,\beta).$

###### Proof.

Let
$\varphi:(A,\mu_{A},\alpha_{1},\beta_{1})\longrightarrow(B,\mu_{B},\alpha_{2},\beta_{2})$
be a morphism of BiHom-Poisson algebras. Then for all $u,v\in A,$

$\begin{array}[]{llllllllll}\mu(u+\varphi(u),v+\varphi(v))=(\mu_{A}(u,v)+\mu_{B}(\varphi(u),\varphi(v)))=(\mu_{A}(u,v)+\varphi(\mu_{A}(u,v))),\\\
\\{u+\varphi(u),v+\varphi(v)\\}=\\{u,v\\}_{A}+\\{\varphi(u),\varphi(v))\\}_{B}=\\{u,v\\}_{A}+\varphi(\\{u,v\\}_{A})\end{array}$

Thus the graph $\Gamma_{\varphi}$ is closed under the operations $\mu$ and
$\\{\cdot,\cdot\\}$. Furthermore since
$\varphi\circ\alpha_{1}=\alpha_{2}\circ\varphi$, we have
$(\alpha_{1}\oplus\alpha_{2})(u,\varphi(u))=(\alpha_{1}(u),\alpha_{2}\circ\varphi(u))=(\alpha_{1}(u),\varphi\circ\alpha_{1}(u)).$
In the same way, we have
$(\beta_{1}\oplus\beta_{2})(u,\varphi(u))=(\beta_{1}(u),\beta_{2}\circ\varphi(u))=(\beta_{1}(u),\varphi\circ\beta_{1}(u)),$
which implies that $\Gamma_{\varphi}$ is closed under
$\alpha_{1}\oplus\alpha_{2}$ and $\beta_{1}\oplus\beta_{2}$ Thus
$\Gamma_{\varphi}$ is a BiHom-subalgebra of $(A\oplus
B,\\{\cdot,\cdot\\},\mu,\alpha,\beta).$
Conversely, if the graph $\Gamma_{\varphi}\subset A\oplus B$ is a BiHom-
subalgebra of $(A\oplus B,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ then we have

$\begin{array}[]{llllll}\mu(u+\varphi(u)),v+\varphi(v))=(\mu_{A}(u,v)+\mu_{B}(\varphi(u),\varphi(v)))\in\Gamma_{\varphi},\\\
\\{u+\varphi(u)),v+\varphi(v)\\}=\\{u,v\\}_{A}+\\{\varphi(u),\varphi(v)\\}_{B}\in\Gamma_{\varphi}\par\end{array}$

which implies that

$\begin{array}[]{llll}\mu_{B}(\varphi(u),\varphi(v))=\varphi(\mu_{A}(u,v)),\\\
\\{\varphi(u),\varphi(v)\\}_{B}=\varphi(\\{u,v\\}_{A}).\end{array}$

Furthermore,
$(\alpha_{1}\oplus\alpha_{2})(\Gamma_{\varphi})\subset\Gamma_{\varphi},~{}(\beta_{1}\oplus\beta_{2})(\Gamma_{\varphi})\subset\Gamma_{\varphi}$
implies

$(\alpha_{1}\oplus\alpha_{2})(u,\varphi(u))=(\alpha_{1}(u),\alpha_{2}\circ\varphi(u))\in\Gamma_{\varphi},$
$(\beta_{1}\oplus\beta_{2})(u,\varphi(u))=(\beta_{1}(u),\beta_{2}\circ\varphi(u))\in\Gamma_{\varphi}.$

which is equivalent to the condition
$\alpha_{2}\circ\varphi(u)=\varphi\circ\alpha_{1}(u),$ i.e.
$\alpha_{2}\circ\varphi=\varphi\circ\alpha_{1}$. Similarly,
$\beta_{2}\circ\varphi=\varphi\circ\beta_{1}$. Therefore, $\varphi$ is a
morphism of BiHom-Poisson algebras. ∎

######  Theorem 3.6.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a (non-BiHom-commutative)
BiHom-Poisson algebra and $\alpha^{\prime},\beta^{\prime}\colon A\to A$ be
endomorphisms of $A$ such that any two of the maps
$\alpha,\beta,\alpha^{\prime},\beta^{\prime}$ commute. Then

$A_{\alpha^{\prime},\beta^{\prime}}=(A,\\{\cdot,\cdot\\}_{\alpha^{\prime},\beta^{\prime}}=\\{\cdot,\cdot\\}\circ(\alpha^{\prime}\otimes\beta^{\prime}),\mu_{\alpha^{\prime},\beta^{\prime}}=\mu\circ(\alpha^{\prime}\otimes\beta^{\prime}),\alpha\alpha^{\prime},\beta\beta^{\prime})$

is also a (non-BiHom-commutative) BiHom-Poisson algebra. Moreover suppose that
$(B,\\{\cdot,\cdot\\}^{\prime},\mu^{\prime},\gamma,\delta)$ is another BiHom-
Poisson algebra and $\gamma^{\prime},\delta^{\prime}$ be endomorphisms of $B$
such that any two of the maps $\gamma,\delta,\gamma^{\prime},\delta^{\prime}$
commute. If
$f:(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)\rightarrow(B,\\{\cdot,\cdot\\}^{\prime},\mu^{\prime},\gamma,\delta)$
is a morphism such that $f\circ\alpha^{\prime}=\gamma^{\prime}\circ f$ and
$f\circ\beta^{\prime}=\delta^{\prime}\circ f$, then
$f:A_{\alpha^{\prime},\beta^{\prime}}\rightarrow
B_{\gamma^{\prime},\delta^{\prime}}$ is also a morphism.

###### Proof.

Let give the proof in BiHom-commutativity case. We only prove item 1. of
Definition 2.4, item 2. and item 3. can be proved similarly.

For all $x,y,z\in A$ we have

$\begin{array}[]{llll}&&as_{A_{\alpha^{\prime},\beta^{\prime}}}(x,y,z)\\\
&=&\mu_{\alpha^{\prime},\beta^{\prime}}(\mu_{\alpha^{\prime},\beta^{\prime}}(x,y),\beta\beta^{\prime}(z))-\mu_{\alpha^{\prime},\beta^{\prime}}(\alpha\alpha^{\prime}(x),\mu_{\alpha^{\prime},\beta^{\prime}}(y,z))\\\
&=&\mu_{\alpha^{\prime},\beta^{\prime}}(\mu(\alpha^{\prime}(x),\beta^{\prime}(y)),\beta\beta^{\prime}(z))-\mu_{\alpha^{\prime},\beta^{\prime}}(\alpha\alpha^{\prime}(x),\mu(\alpha^{\prime}(y),\beta^{\prime}(z)))\\\
&=&\mu(\mu(\alpha^{\prime
2}(x),\alpha^{\prime}\beta^{\prime}(y)),\beta\beta^{\prime
2}(z))-\mu(\alpha\alpha^{\prime
2}(x),\mu(\alpha^{\prime}\beta^{\prime}(y),\beta^{\prime 2}(z)))\\\
&=&as_{A}(\alpha^{\prime 2}(x),\alpha^{\prime}\beta^{\prime}(y),\beta^{\prime
2}(z))=0~{}(BiHom-associativity~{}condition~{}of~{}A).\end{array}$

Then
$(A,\mu_{\alpha^{\prime},\beta^{\prime}},\alpha\alpha^{\prime},\beta\beta^{\prime})$
is a BiHom-associative algebra.

Now, for all $x,y\in A$ we have:

$\begin{array}[]{llllllll}\mu_{\alpha^{\prime},\beta^{\prime}}(\beta\beta^{\prime}(x),\alpha\alpha^{\prime}(y))&=&\mu(\alpha^{\prime}\beta^{\prime}\beta(x),\alpha\alpha^{\prime}\beta^{\prime}(y))\\\
&=&\mu(\beta(\alpha^{\prime}\beta^{\prime}(x)),\alpha(\alpha^{\prime}\beta^{\prime}(y)))\\\
&=&\mu(\beta(\alpha^{\prime}\beta^{\prime}(y)),\alpha(\alpha^{\prime}\beta^{\prime}(x)))~{}(BiHom-
commutativity~{}in~{}A).\\\
&=&\mu_{\alpha^{\prime},\beta^{\prime}}(\beta\beta^{\prime}(y),\alpha\alpha^{\prime}(x)).\end{array}$

Then
$(A,\mu_{\alpha^{\prime},\beta^{\prime}},\alpha\alpha^{\prime},\beta\beta^{\prime})$
is a BiHom-commutative algebra.

The second part is proved as follows: $\forall x,y\in A$

$\begin{array}[]{llllllll}f\\{x,y\\}_{\alpha^{\prime},\beta^{\prime}}&=&f\\{\alpha^{\prime}(x),\beta^{\prime}(y)\\}\\\
&=&\\{f\alpha^{\prime}(x),f\beta^{\prime}(y)\\}^{\prime}\\\
&=&\\{\gamma^{\prime}f(x),\delta^{\prime}f(y)\\}^{\prime}\\\
&=&\\{f(x),f(y)\\}^{\prime}_{\gamma^{\prime},\delta^{\prime}}.\end{array}$

In the same way we have
$f\mu_{\alpha^{\prime},\beta^{\prime}}(x,y)=\mu^{\prime}_{\gamma^{\prime},\delta^{\prime}}(f(x),f(y)).$

This finishes the proof. ∎

Taking $\alpha^{\prime}=\alpha^{k},~{}\beta^{\prime}=\beta^{l}$, yields the
following statement.

######  Corollary 3.7.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a (non-BiHom-commutative)
BiHom-Poisson algebra. Then

$A_{\alpha^{k},\beta^{l}}=(A,\\{\cdot,\cdot\\}_{\alpha^{k},\beta^{l}}=\\{\cdot,\cdot\\}\circ(\alpha^{k}\otimes\beta^{l}),\mu_{\alpha^{k},\beta^{l}}=\mu\circ(\alpha^{k}\otimes\beta^{l}),\alpha^{k+1},\beta^{l+1})$

is also a (non-BiHom-commutative) BiHom-Poisson algebra.

Taking $\alpha=\beta=id$, yields the following statement.

######  Corollary 3.8.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a (non-commutative) Poisson algebra and
$\alpha,\beta\colon A\to A$ be two endomorphisms such that
$\alpha\beta=\beta\alpha$. Then

$A_{\alpha,\beta}=(A,\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\cdot,\cdot\\}\circ(\alpha\otimes\beta),\mu_{\alpha,\beta}=\mu\circ(\alpha\otimes\beta),\alpha,\beta)$

is also a (non-BiHom-commutative) BiHom-Poisson algebra.

######  Definition 3.9.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a non-commutative Poisson algebra.

  1. 1.

Given two commuting morphisms $\alpha,\beta\colon A\to A$, the triple
$A^{\prime}_{\alpha,\beta}=(A,\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\cdot,\cdot\\}\circ(\alpha\otimes\beta),\mu_{\alpha,\beta}=\mu\circ(\alpha\otimes\beta))$
is called the $(\alpha,\beta)$-twisting of $A$. A twisting of $A$ is a
$(\alpha,\beta)$-twisting of $A$ for some morphisms $\alpha,\beta\colon A\to
A$.

  2. 2.

The $(\alpha,\beta)$-twisting $A^{\prime}_{\alpha,\beta}$ of $A$ is called
trivial if

$\\{\cdot,\cdot\\}_{\alpha,\beta}=0=\mu_{\alpha,\beta}.$

$A^{\prime}_{\alpha,\beta}$ is called non-trivial if either
$\\{\cdot,\cdot\\}\neq 0$ or $\mu_{\alpha,\beta}\neq 0$.

  3. 3.

$A$ is called rigid if every twisting of $A$ is either trivial or isomorphic
to $A$.

######  Proposition 3.10.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a non-commutative Poisson algebra. Suppose
there exists two commuting morphisms $\alpha,\beta\colon A\to A$ such that
either:

  1. 1.

$\mu_{\alpha,\beta}=\mu\circ(\alpha\otimes\beta)$ is not BiHom-associative or

  2. 2.

$\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\cdot,\cdot\\}\circ(\alpha\otimes\beta)$
does not satisfy the BiHom-Jacobi identity.

Then $A$ is not rigid.

###### Proof.

The $(\alpha,\beta)$-twisting $A^{\prime}_{\alpha,\beta}$ is non-trivial,
since otherwise $\mu_{\alpha,\beta}$ would be BiHom-associative and
$\\{\cdot,\cdot\\}_{\alpha,\beta}$ would satisfy the BiHom-Jacobi identity.
For the same reason, the $(\alpha,\beta)$-twisting $A^{\prime}_{\alpha,\beta}$
cannot be isomorphic to $A$. ∎

######  Remark 3.11.

Let $(\mathfrak{g},[\cdot,\cdot])$ be a finite-dimensional Lie algebra, and
let $(S(\mathfrak{g}),\mu)$ be its symmetric algebra. If
$\\{e_{i}\\}_{i=1}^{n}$ is a basis of $\mathfrak{g}$, then $S(\mathfrak{g})$
is the polynomial algebra $\mathbb{K}[e_{1},\ldots,e_{n}]$. Suppose the
structure constants for $\mathfrak{g}$ are given by

$[e_{i},e_{j}]=\sum_{k=1}^{n}c_{ij}^{k}e_{k}.$

Then the symmetric algebra $S(\mathfrak{g})$ becomes a Poisson algebra with
the Poisson bracket

$\\{F,G\\}=\frac{1}{2}\sum_{i,j,k=1}^{n}c_{ij}^{k}e_{k}\left(\frac{\partial
F}{\partial e_{i}}\frac{\partial G}{\partial e_{j}}-\frac{\partial F}{\partial
e_{j}}\frac{\partial G}{\partial e_{i}}\right)$ (3.7)

for $F,G\in S(\mathfrak{g})$. This Poisson algebra structure on
$S(\mathfrak{g})$ is called the linear Poisson structure. Note that
$\\{e_{i},e_{j}\\}=[e_{i},e_{j}]$.

######  Example 3.12 ($S(\mathfrak{sl}(2))$ is not rigid).

In this example, we show that the symmetric algebra
$(S(\mathfrak{sl}(2)),\mu)$ on the Lie algebra $\mathfrak{sl}(2)$, equipped
with the linear Poisson structure (3.7), is not rigid in the sense of
Definition 3.9.

The Lie algebra $\mathfrak{sl}(2)$ has a basis $\\{e,f,h\\}$, with respect to
which the Lie bracket is given by

$[h,e]=2e,\quad[h,f]=-2f,\quad[e,f]=h.$

To show that $S(\mathfrak{sl}(2))=\mathbb{K}\langle e,f,h\rangle$ is not
rigid, consider the Lie algebra morphisms
$\alpha,\beta\colon\mathfrak{sl}(2)\to\mathfrak{sl}(2)$ given by

$\begin{array}[]{llllllllll}\alpha(e)&=&\lambda
e,\quad\alpha(f)&=&\lambda^{-1}f,\quad\alpha(h)&=&h,\\\ \beta(e)&=&\gamma
e,\quad\beta(f)&=&\gamma^{-1}f,\quad\beta(h)&=&h.\end{array}$

where $\lambda,\gamma\in\mathbb{K}$ is a fixed scalar with
$\lambda,\gamma\not=0,1$. Denote by $\alpha,\beta\colon S(\mathfrak{sl}(2))\to
S(\mathfrak{sl}(2))$ the extended maps, which is a Poisson algebra morphisms.
By Proposition 3.10, the Poisson algebra $S(\mathfrak{sl}(2))$ is not rigid if
$\mu_{\alpha,\beta}=\mu\circ(\alpha\otimes\beta)$ is not BiHom-associative. We
have

$\begin{split}\mu_{\alpha,\beta}(\mu_{\alpha,\beta}(e,h),h)-\mu_{\alpha,\beta}(e,\mu_{\alpha,\beta}(h,h))&=\alpha^{2}(e)\alpha\beta(h)\beta(h)-\alpha(e)\alpha\beta(h)\beta^{2}(h)\\\
&=(\lambda^{2}-\lambda)eh^{2},\end{split}$

which is not $0$ in the symmetric algebra $S(\mathfrak{sl}(2))$ because
$\lambda\not=0,1$. Therefore, $\mu_{\alpha,\beta}$ is not BiHom-associative,
and the linear Poisson structure on $S(\mathfrak{sl}(2))$ is not rigid. ∎

###  3.2 BiHom-flexibles structures and admissible BiHom-Poisson algebras

In this subsection, we introduce BiHom-flexible algebras and prove that
admissible BiHom-Poisson algebras are BiHom-flexible.

######  Definition 3.1.

Let $(A,\mu,\alpha,\beta)$ be a BiHom-algebra. Then $A$ is called BiHom-
flexible algebra if for any $x,y\in A$

$\mu(\mu(\beta^{2}(x),\alpha\beta(y)),\beta\alpha^{2}(x))-\mu(\alpha\beta^{2}(x),\mu(\alpha\beta(y),\alpha^{2}(x))=0.$
(3.8)

######  Remark 3.2.

  1. 1.

If $\alpha=\beta=Id,$ then $(A,\mu,\alpha,\beta)$ is reduced to a flexible
algebra $(A,\mu).$

  2. 2.

If $(A,\mu,\alpha)$ is a Hom-flexible algebra [31], then
$(A,\mu,\alpha,\alpha)$ is a BiHom-flexible algebra. Conversely, if
$(A,\mu,\alpha,\beta)$ is a BiHom-flexible algebra such $\alpha$ is injective
and $\alpha=\beta$ then, $(A,\mu,\alpha)$ is a Hom-flexible algebra.

######  Lemma 3.3.

Let $A=(A,\mu,\alpha,\beta)$ be a BiHom-algebra. The following assertions are
equivalent

  1. 1.

$A$ is BiHom-flexible.

  2. 2.

For any $x,y\in A,\quad as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(x))=0$.

  3. 3.

For any $x,y,z$

$\displaystyle
as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))+as_{A}(\beta^{2}(z),\alpha\beta(y),\alpha^{2}(x))=0$
(3.9)

###### Proof.

The equivalence of the first two assertions follows from the definition. The
assertion $as_{A}(\beta^{2}(x-z),\alpha\beta(y),\alpha^{2}(x-z))=0$ holds by
definition and it is equivalent to
$as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))+as_{A}(\beta^{2}(z),\alpha\beta(y),\alpha^{2}(x))=0$
by linearity. ∎

It is easy to prove the following

######  Proposition 3.13.

Let $(A,\mu)$ be a flexible algebra, $\alpha,\beta:A\rightarrow A$ be two
commuting morphisms.. Then the Bihom-algebra
$(A,\mu_{\alpha,\beta}=\mu(\alpha\otimes\beta),\alpha,\beta)$ is BiHom-
flexible.

###### Proof.

Let $(A,\mu)$ be a flexible algebra, $\alpha$ and $\beta$ be morphisms of
$(A,\mu)$. Then for all $x,y\in A,$

$\displaystyle\mu_{\alpha,\beta}(\mu_{\alpha,\beta}(\beta^{2}(x),\alpha\beta(y)),\beta\alpha^{2}(x))-\mu_{\alpha,\beta}(\alpha\beta^{2}(x),\mu_{\alpha,\beta}(\alpha\beta(y),\alpha^{2}(x)))$
$\displaystyle=\mu_{\alpha,\beta}(\mu(\alpha\beta^{2}(x),\alpha\beta^{2}(y)),\beta\alpha^{2}(x))-\mu_{\alpha,\beta}(\alpha\beta^{2}(x),\mu(\alpha^{2}\beta(y),\beta\alpha^{2}(x)))$
$\displaystyle=\mu(\mu(\alpha^{2}\beta^{2}(x),\alpha^{2}\beta^{2}(y)),\beta^{2}\alpha^{2}(x))-\mu(\alpha^{2}\beta^{2}(x),\mu(\alpha^{2}\beta^{2}(y),\beta^{2}\alpha^{2}(x)))=0.$

Hence the Bihom-algebra
$(A,\mu_{\alpha,\beta}=\mu(\alpha\otimes\beta),\alpha,\beta)$ is BiHom-
flexible. ∎

######  Corollary 3.4.

Any BiHom-associative algebra is BiHom-flexible.

######  Proposition 3.5.

A BiHom-algebra $A=(A,\mu,\alpha,\beta)$ is BiHom-flexible if and only if

$\\{\alpha\beta(x),\alpha\beta(y)\diamond\alpha^{2}(z)\\}-\\{\beta^{2}(x),\alpha\beta(y)\\}\diamond\alpha^{2}\beta(z)-\alpha\beta^{2}(y)\diamond\\{\alpha\beta(x),\alpha^{2}(z)\\}=0,$
(3.10)

where
$\\{\cdot,\cdot\\}=\frac{1}{2}(\mu-\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau))$
and
$\diamond=\frac{1}{2}(\mu+\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau)$.

###### Proof.

Since
$\\{\cdot,\cdot\\}=\frac{1}{2}(\mu-\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau))$
and
$\diamond=\frac{1}{2}(\mu+\mu\circ(\alpha^{-1}\beta\otimes\alpha\beta^{-1})\circ\tau)$,
by expansion in terms of $\mu$

$\begin{array}[]{llllll}&&4\left(\\{\alpha\beta(x),\alpha\beta(y)\diamond\alpha^{2}(z)\\}-\\{\beta^{2}(x),\alpha\beta(y)\\}\diamond\alpha^{2}\beta(z)-\alpha\beta^{2}(y)\diamond\\{\alpha\beta(x),\alpha^{2}(z)\\}\right)\\\
&=&\mu(\alpha\beta^{2}(x)),\mu(\alpha\beta(y),\alpha^{2}(z)))-\mu(\mu(\beta^{2}(y),\alpha\beta(z)),\beta\alpha^{2}(x)))+\mu(\alpha\beta^{2}(x)),\mu(\alpha\beta(z),\alpha^{2}(y)))\\\
&-&\mu(\mu(\beta^{2}(z),\alpha\beta(y)),\beta\alpha^{2}(x)))-\mu(\mu(\beta^{2}(x),\alpha\beta(y)),\beta\alpha^{2}(z)))+\mu(\alpha\beta^{2}(z)),\mu(\alpha\beta(x),\alpha^{2}(y)))\\\
&+&\mu(\mu(\beta^{2}(y),\alpha\beta(x)),\beta\alpha^{2}(z)))-\mu(\alpha\beta^{2}(z)),\mu(\alpha\beta(y),\alpha^{2}(x)))-\mu(\alpha\beta^{2}(y)),\mu(\alpha\beta(x),\alpha^{2}(z)))\\\
&+&\mu(\mu(\beta^{2}(x),\alpha\beta(z)),\beta\alpha^{2}(y)))+\mu(\alpha\beta^{2}(y)),\mu(\alpha\beta(z),\alpha^{2}(x)))-\mu(\mu(\beta^{2}(z),\alpha\beta(x)),\beta\alpha^{2}(y)))\\\
&=&-as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))-as_{A}(\beta^{2}(z),\alpha\beta(y),\alpha^{2}(x))-as_{A}(\beta^{2}(x),\alpha\beta(z),\alpha^{2}(y))\\\
&-&as_{A}(\beta^{2}(y),\alpha\beta(z),\alpha^{2}(x))+as_{A}(\beta^{2}(y),\alpha\beta(x),\alpha^{2}(z))+as_{A}(\beta^{2}(z),\alpha\beta(x),\alpha^{2}(y))\\\
&=&0~{}(By~{}Lemma\eqref{flex_lemma}).\end{array}$

Conversely, assume we have the condition (3.10). By setting $x=z$, one gets
$as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(x))=0$. Therefore $A$ is BiHom-
flexible. ∎

Let’s give the notion of an admissible BiHom-Poisson algebras.

######  Definition 3.14.

[1] Let $(A,\mu,\alpha,\beta)$ be a BiHom-algebra. Then $A$ is called an
admissible BiHom-Poisson algebra if it satisfies

$\displaystyle
as_{A}(\beta(x),\alpha(y),\alpha^{2}(z))=\frac{1}{3}\Big{\\{}\mu(\mu(\beta(x),\alpha\beta(z)),\alpha^{2}(y)))-\mu(\mu(\beta^{2}(z),\alpha(x)),\alpha^{2}(y))$
$\displaystyle+\mu(\mu(\beta(y),\alpha\beta(z)),\alpha^{2}(x))-\mu(\mu(\beta(y),\alpha(x)),\beta\alpha^{2}(z))\Big{\\}}$
(3.11)

It is observed that if $\alpha$ and $\beta$ are inversible, then (3.14) is
equivalent to

$\displaystyle
as_{A}(x,y,z)=\frac{1}{3}\Big{\\{}\mu(\mu(x,\alpha^{-1}\beta(z)),\alpha(y)))-\mu(\mu(\alpha^{-2}\beta^{2}(z),\alpha\beta^{-1}(x)),\alpha(y))$
$\displaystyle+\mu(\mu(\alpha^{-1}\beta(y),\alpha^{-1}\beta(z)),\alpha^{2}\beta^{-1}(x))-\mu(\mu(\alpha^{-1}\beta(y),\alpha^{-1}\beta(x)),\beta(z))\Big{\\}}$
(3.12)

######  Proposition 3.15.

Every admissible BiHom-Poisson algebra $(A,\mu,\alpha,\beta)$ is BiHom-
flexible, i.e.,

$as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))+as_{A}(\beta^{2}(z),\alpha\beta(y),\alpha^{2}(x))=0$
(3.13)

for all $x,y,z\in A$.

###### Proof.

The required identity (3.13) follows immediately from the defining identity
(3.14), in which the right-hand side is anti-symmetric in $x$ and $z$.

$\begin{array}[]{llllll}&&as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))+as_{A}(\beta^{2}(z),\alpha\beta(y),\alpha^{2}(x))\\\
&=&\frac{1}{3}\Big{\\{}\mu(\mu(\beta^{2}(x),\alpha\beta(z)),\beta\alpha^{2}(y)))-\mu(\mu(\beta^{2}(z),\alpha\beta(x)),\beta\alpha^{2}(y)))\\\
&&+\mu(\mu(\beta^{2}(y),\alpha\beta(z)),\beta\alpha^{2}(x)))-\mu(\mu(\beta^{2}(y),\alpha\beta(x)),\beta\alpha^{2}(z)))\\\
&&+\mu(\mu(\beta^{2}(z),\alpha\beta(x)),\beta\alpha^{2}(y)))-\mu(\mu(\beta^{2}(x),\alpha\beta(z)),\beta\alpha^{2}(y)))\\\
&&+\mu(\mu(\beta^{2}(y),\alpha\beta(x)),\beta\alpha^{2}(z)))-\mu(\mu(\beta^{2}(y),\alpha\beta(z)),\beta\alpha^{2}(x)))\Big{\\}}=0.\end{array}$
∎

Next we observe that in an admissible BiHom-Poisson algebra the cyclic sum of
the BiHom-associator is trivial.

######  Proposition 3.16.

Let $(A,\mu,\alpha,\beta)$ be an admissible BiHom-Poisson algebra. Then

$S_{A}(x,y,z):=as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))+as_{A}(\beta^{2}(y),\alpha\beta(z),\alpha^{2}(x))+as_{A}(\beta^{2}(z),\alpha\beta(x),\alpha^{2}(y))=0$
(3.14)

for all $x,y,z\in A$.

###### Proof.

Using the defining identity (3.14), we have:

$\begin{array}[]{llllll}as_{A}(\beta^{2}(x),\alpha\beta(y),\alpha^{2}(z))&=&\frac{1}{3}\Big{(}\mu(\mu(\beta^{2}(x),\alpha\beta(z)),\beta\alpha^{2}(y)))-\mu(\mu(\beta^{2}(z),\alpha\beta(x)),\beta\alpha^{2}(y)))\\\
&&+\mu(\mu(\beta^{2}(y),\alpha\beta(z)),\beta\alpha^{2}(x)))-\mu(\mu(\beta^{2}(y),\alpha\beta(x)),\beta\alpha^{2}(z)))\Big{)}\\\
&=&-\frac{1}{3}\Big{(}\mu(\mu(\beta^{2}(z),\alpha\beta(y)),\beta\alpha^{2}(x)))-\mu(\mu(\beta^{2}(y),\alpha\beta(z)),\beta\alpha^{2}(x)))\\\
&&+\mu(\mu(\beta^{2}(x),\alpha\beta(y)),\beta\alpha^{2}(z)))-\mu(\mu(\beta^{2}(x),\alpha\beta(z)),\beta\alpha^{2}(y)))\Big{)}\\\
&&+\frac{1}{3}\Big{(}\mu(\mu(\beta^{2}(x),\alpha\beta(y)),\beta\alpha^{2}(z)))-\mu(\mu(\beta^{2}(y),\alpha\beta(x)),\beta\alpha^{2}(z)))\\\
&&+\mu(\mu(\beta^{2}(z),\alpha\beta(y)),\beta\alpha^{2}(x)))-\mu(\mu(\beta^{2}(z),\alpha\beta(x)),\beta\alpha^{2}(y)))\Big{)}\\\
&=&-as_{A}(\beta^{2}(z),\alpha\beta(x),\alpha^{2}(y))+as_{A}(\beta^{2}(x),\alpha\beta(z),\alpha^{2}(y))\\\
&=&-as_{A}(\beta^{2}(z),\alpha\beta(x),\alpha^{2}(y))-as_{A}(\beta^{2}(y),\alpha\beta(z),\alpha^{2}(x))~{}(by~{}(\ref{homflexible})).\end{array}$

Therefore, we conclude that $S_{A}=0$. ∎

##  4 Derivations of BiHom-Poisson algebras

In this section, we introduce and study derivations, generalized derivations
and quasiderivations of BiHom-Poisson algebras.

######  Definition 4.1.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. A
linear map $D:A\rightarrow A$ is called an $(\alpha^{k},\beta^{l})$-derivation
of $A$ if it satisfies

  1. 1.

$D\circ\alpha=\alpha\circ D,\;D\circ\beta=\beta\circ D;$

  2. 2.

$D(\\{x,y\\})=\\{\alpha^{k}\beta^{l}(x),D(y)\\}+\\{D(x),\alpha^{k}\beta^{l}(y)\\}$;

  3. 3.

$D(\mu(x,y))=\mu(\alpha^{k}\beta^{l}(x),D(y))+\mu(D(x),\alpha^{k}\beta^{l}(y))$,

for all $x,y\in A$.

We denote by $Der(A):=\displaystyle{\bigoplus_{k\geq
0}}\displaystyle{\bigoplus_{l\geq 0}}Der_{(\alpha^{k},\beta^{l})}(A)$, where
$Der_{(\alpha^{k},\beta^{l})}(A)$ is the set of all
$(\alpha^{k},\beta^{l})$-derivations of $A$. Obviously, $Der(A)$ is a
subalgebra of $End(A)$.

######  Lemma 4.2.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. We
define a subspace $\mathcal{W}$ of $End(A)$ by $\mathcal{W}=\\{w\in
End(A)|~{}w\circ\alpha=\alpha\circ w~{}and~{}w\circ\beta=\beta\circ w\\}$ and
$\sigma_{1},\sigma_{2}:\mathcal{W}\rightarrow\mathcal{W}$ linear maps
satisfying $\sigma_{1}(w)=\alpha\circ w$ and $\sigma_{2}(w)=\beta\circ w$.
Then a quadruple $(\mathcal{W},[\cdot,\cdot],\sigma_{1},\sigma_{2})$, where
the multiplication
$[\cdot,\cdot]:\mathcal{W}\times\mathcal{W}\rightarrow\mathcal{W}$ is defined
for $w_{1},w_{2}\in\mathcal{W}$ by

$[w_{1},w_{2}]=w_{1}\circ w_{2}-w_{2}\circ w_{1},$

is a BiHom-Lie algebra.

###### Proof.

For any $w_{1},w_{2},w_{3}\in\mathcal{W},~{}~{}k_{1},k_{2}\in\mathbb{K},$ we
have

$[w_{1},w_{1}]=w_{1}\circ w_{1}-w_{1}\circ w_{1}=0,$

$\begin{array}[]{lll}&&[k_{1}w_{1}+k_{2}w_{2},w_{3}]=(k_{1}w_{1}+k_{2}w_{2})w_{3}-w_{3}(k_{1}w_{1}+k_{2}w_{2})\\\
&=&k_{1}(w_{1}w_{3}-w_{3}w_{1})+k_{2}(w_{2}w_{3}-w_{3}w_{2})=k_{1}[w_{1},w_{3}]+k_{2}[w_{2},w_{3}],\end{array}$

$\begin{array}[]{lll}&&[\sigma_{2}(w_{1}),\sigma_{1}(w_{2})]=[\beta(w_{1}),\alpha(w_{2})]=\alpha\beta(w_{1}w_{2}-w_{2}w_{1})=-\alpha\beta(w_{2}w_{1}-w_{1}w_{2})\\\
&=&-[\sigma_{2}(w_{2}),\sigma_{1}(w_{1})]\end{array}$

$\begin{array}[]{llllllll}&&[\sigma_{2}^{2}(w_{1}),[\sigma_{2}(w_{2}),\sigma_{1}(w_{3})]]+[\sigma_{2}^{2}(w_{3}),[\sigma_{2}(w_{1}),\sigma_{1}(w_{2})]]\\\
&&+[\sigma_{2}^{2}(w_{2}),[\sigma_{2}(w_{3}),\sigma_{1}(w_{1})]]\\\
&=&\beta^{3}\alpha w_{1}w_{2}w_{3}-\beta^{3}\alpha
w_{1}w_{3}w_{2}-\beta^{3}\alpha w_{2}w_{3}w_{1}+\beta^{3}\alpha
w_{3}w_{2}w_{1}\\\ &&+\beta^{3}\alpha w_{2}w_{3}w_{1}-\beta^{3}\alpha
w_{2}w_{1}w_{3}-\beta^{3}\alpha w_{3}w_{1}w_{2}+\beta^{3}\alpha
w_{1}w_{3}w_{2}\\\ &&+\beta^{3}\alpha w_{3}w_{1}w_{2}-\beta^{3}\alpha
w_{3}w_{2}w_{1}-\beta^{3}\alpha w_{1}w_{2}w_{3}+\beta^{3}\alpha
w_{2}w_{1}w_{3}\\\ &=&0.\end{array}$

Then $(\mathcal{W},[\cdot,\cdot],\sigma_{1},\sigma_{2})$ is a Bihom-Lie
algebra. ∎

######  Theorem 4.3.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. For
any $D\in Der_{(\alpha^{k},\beta^{l})}(A)$ and $D^{{}^{\prime}}\in
Der_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$, define their commutator
$[D,D^{{}^{\prime}}]$ as usual:

$[D,D^{{}^{\prime}}]=D\circ D^{{}^{\prime}}-D^{{}^{\prime}}\circ D.$

Then $[D,D^{{}^{\prime}}]\in
Der_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$.

###### Proof.

It is sufficient to prove
$[Der_{(\alpha^{k},\beta^{l})}(A),Der_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)]\subseteq
Der_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime})}}(A)$. It is easy to check
that $[D,D^{\prime}]\circ\alpha=\alpha\circ[D,D^{\prime}]$ and
$[D,D^{\prime}]\circ\beta=\beta\circ[D,D^{\prime}]$.

For any $x,y\in A$, we have

$\begin{array}[]{lllll}&&[D,D^{\prime}](\\{x,y\\})\\\ &=&D\circ
D^{\prime}(\\{x,y\\})-D^{\prime}\circ D(\\{x,y\\})\\\
&=&D(\\{D^{\prime}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\}+\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D^{\prime}(y)\\})\\\
&&-D^{\prime}(\\{D^{\prime}(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D(y)\\})\\\
&=&D(\\{D^{\prime}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\})+D((\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D^{\prime}(y)\\})\\\
&&-D^{\prime}(\\{D(x)(x),\alpha^{k}\beta^{l}(y)\\})-D^{\prime}(\\{\alpha^{k}\beta^{l}(x),D(y)\\})\\\
&=&\\{D\circ
D^{\prime}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}+\\{\alpha^{k}\beta^{l}\circ
D^{\prime}(x),D\circ\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\}\\\
&&+\\{D\circ\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),\alpha^{k}\beta^{l}\circ
D^{\prime}(y)\\}+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D\circ
D^{\prime}(y)\\}\\\ &&-\\{D^{\prime}\circ
D(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}\circ
D(x),D^{\prime}\circ\alpha^{k}\beta^{l}(y)\\}\\\
&&-\\{D^{\prime}\circ\alpha^{k}\beta^{l}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}\circ
D(y)\\}-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D^{\prime}\circ
D(y)\\}.\end{array}$

Similarly, we can check that

$\displaystyle[D,D^{\prime}](\mu(x,y))\mu([D,D^{\prime}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))+\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D,D^{\prime}](y)).$

It follows that $[D,D^{\prime}]\in
Der_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$.∎

######  Definition 4.4.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. $D\in
End(A)$ is said to be a generalized $(\alpha^{k},\beta^{l})$-derivation of
$A$, if there exists two endomorphisms $D^{\prime},D^{\prime\prime}\in End(A)$
such that

  1. 1.

$D\circ\alpha=\alpha\circ D,\;D\circ\beta=\beta\circ D\;$;

  2. 2.

$D^{{}^{\prime}}\circ\alpha=\alpha\circ
D^{{}^{\prime}},\;D^{\prime}\circ\beta=\beta\circ D^{{}^{\prime}}\;$;

  3. 3.

$D^{{}^{\prime\prime}}\circ\alpha=\alpha\circ
D^{{}^{\prime\prime}},\;D^{{}^{\prime\prime}}\circ\beta=\beta\circ
D^{{}^{\prime\prime}}$;

  4. 4.

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D^{\prime}(y)\\}=D^{\prime\prime}(\\{x,y\\})$;

  5. 5.

$\mu(D(x),\alpha^{k}\beta^{l}(y))+\mu(\alpha^{k}\beta^{l}(x),D^{{}^{\prime}}(y))=D^{\prime\prime}(\mu(x,y))$,

for all $x,y\in A$.

The set of generalized $(\alpha^{k},\beta^{l})$-derivations of $A$ is ${\rm
GDer}_{(\alpha^{k},\beta^{l})}(A)$ and we denote

${\rm GDer}(A):=\bigoplus_{k\geq 0}\bigoplus_{l\geq 0}{\rm
GDer}_{(\alpha^{k},\beta^{l})}(A).$

######  Definition 4.5.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. $D\in
End(A)$ is said to be an $(\alpha^{k},\beta^{l})$-quasiderivation of $A$, if
there exists endomorphisms $D^{\prime},D^{\prime\prime}\in End(A)$ such that

  1. 1.

$D\circ\alpha=\alpha\circ D,\;D\circ\beta=\beta\circ D\;$;

  2. 2.

$D^{{}^{\prime}}\circ\alpha=\alpha\circ
D^{{}^{\prime}},\;D^{{}^{\prime}}\circ\beta=\beta\circ D^{{}^{\prime}}\;$;

  3. 3.

$D^{{}^{\prime\prime}}\circ\alpha=\alpha\circ
D^{{}^{\prime\prime}},\;D^{{}^{\prime\prime}}\circ\beta=\beta\circ
D^{{}^{\prime\prime}}$;

  4. 4.

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D(y)\\}=D{{}^{\prime}}(\\{x,y\\}),$

  5. 5.

$\mu(D(x),\alpha^{k}\beta^{l}(y))+\mu(\alpha^{k}\beta^{l}(x),D(y))=D^{{}^{\prime\prime}}(\mu(x,y)),$

for all $x,y\in A$.

We then define

${\rm QDer}(A):=\bigoplus_{k\geq 0}\bigoplus_{l\geq 0}{\rm
QDer}_{(\alpha^{k},\beta^{l})}(A).$

######  Definition 4.6.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. A
linear map $D:A\rightarrow A$ is called an $(\alpha^{k},\beta^{l})$-centroid
of $A$ if it satisfies

  1. 1.

$D\circ\alpha=\alpha\circ D,~{}D\circ\beta=\beta\circ D$;

  2. 2.

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}=\\{\alpha^{k}\beta^{l}(x),D(y)\\}=D(\\{x,y\\})$;

  3. 3.

$\mu(D(x),\alpha^{k}\beta^{l}(y))=\mu(\alpha^{k}\beta^{l}(x),D(y))=D(\mu(x,y)),\quad\forall
x,y\in A$.

We set

$C(A):=\bigoplus_{k\geq 0}\bigoplus_{l\geq 0}C_{(\alpha^{k},\beta^{l})}(A).$

######  Definition 4.7.

The $(\alpha^{k},\beta^{l})$-quasicentroid of a Bihom-Poisson algebra
$(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ denoted by ${\rm
QC}_{(\alpha^{k},\beta^{l})}(A)$ is the set of linear maps $D$ such that

  1. 1.

$D\circ\alpha=\alpha\circ D,~{}D\circ\beta=\beta\circ D$;

  2. 2.

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}=\\{\alpha^{k}\beta^{l}(x),D(y)\\}$;

  3. 3.

$\mu(D(x),\alpha^{k}\beta^{l}(y))=\mu(\alpha^{k}\beta^{l}(x),D(y)),\quad\forall
x,y\in A$.

We set

${\rm QC}(A):=\bigoplus_{k\geq 0}\bigoplus_{l\geq 0}{\rm
QC}_{(\alpha^{k},\beta^{l})}(A).$

######  Remark 4.8.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. Then
$C(A)\subseteq QC(A).$

######  Definition 4.9.

A linear map $D$ is called an $(\alpha^{k},\beta^{l})$-central derivation of
$A$ if it satisfies

  1. 1.

$D\circ\alpha=\alpha\circ D,~{}D\circ\beta=\beta\circ D$;

  2. 2.

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}=D(\\{x,y\\})=0;$

  3. 3.

$\mu(D(x),\alpha^{k}\beta^{l}(y))=D(\mu(x,y))=0,\quad\forall x,y\in A$.

The set of $(\alpha^{k},\beta^{l})$-central derivations is denoted by ${\rm
ZDer}_{(\alpha^{k},\beta^{l})}(A)$ and we set

${\rm ZDer}(A):=\bigoplus_{k\geq 0}\bigoplus_{l\geq 0}{\rm
ZDer}_{(\alpha^{k},\beta^{l})}(A).$

######  Remark 4.10.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. Then

$\displaystyle ZDer(A)\subseteq Der(A)\subseteq QDer(A)\subseteq
GDer(A)\subseteq End(A).$

######  Proposition 4.11.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra. Then
the following statements hold:

  1. 1.

$GDer(A)$, $QDer(A)$ and $C(A)$ are BiHom-subalgebras of
$(\mathcal{W},[\cdot,\cdot],\sigma_{1},\sigma_{2})$;

  2. 2.

$ZDer(A)$ is a BiHom-ideal of $Der(A)$.

###### Proof.

We only prove that $GDer(A)$ is a subalgebra of $\mathcal{W}.$ The case of
$QDer(L)$ and $C(L)$ is similar.

  1. 1.

Suppose that $D_{1}\in GDer_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
GDer_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$,

$\displaystyle\\{\sigma_{1}(D_{1})(x),\alpha^{k+1}\beta^{l}(y)\\}=\\{\alpha_{1}\circ
D_{1}(x),\alpha^{k+1}\beta^{l}(y)\\}=\alpha(\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\})$
$\displaystyle=\alpha(D_{1}^{{}^{\prime\prime}}(\\{x,y\\})-\\{\alpha^{k}\beta^{l}(x),D_{1}^{{}^{\prime}}(y)\\})=\sigma_{1}(D_{1}^{{}^{\prime\prime}})(\\{x,y\\})-\\{\alpha^{k+1}\beta^{l}(x),\sigma_{1}(D_{1}^{{}^{\prime}})(y)\\}.$

Since
$\sigma_{1}(D_{1}^{{}^{\prime\prime}}),\sigma_{1}(D_{1}^{{}^{\prime}})\in
End(A)$, we have $\sigma_{1}(D_{1})\in GDer_{(\alpha^{k+1},\beta^{l})}(A)$,

$\displaystyle\\{\sigma_{2}(D_{1})(x),\alpha^{k}\beta^{l+1}(y)\\}=\\{\beta\circ
D_{1}(x),\alpha^{k}\beta^{l+1}(y)\\}=\beta(\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\})$
$\displaystyle=\beta(D_{1}^{{}^{\prime\prime}}(\\{x,y\\})-\\{\alpha^{k}\beta^{l}(x),D_{1}^{{}^{\prime}}(y)\\})=\sigma_{2}(D_{1}^{{}^{\prime\prime}})(\\{x,y\\})-\\{\alpha^{k}\beta^{l+1}(x),\sigma_{2}(D_{1}^{{}^{\prime}})(y)\\}.$

Since
$\sigma_{2}(D_{1}^{{}^{\prime\prime}}),\sigma_{2}(D_{1}^{{}^{\prime}})\in
End(A)$, we have $\sigma_{2}(D_{1})\in GDer_{(\alpha^{k},\beta^{l+1})}(A)$.

$\displaystyle\\{\\{D_{1},D_{2}\\}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=\\{D_{1}\circ
D_{2}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}-\\{D_{2}\circ
D_{1}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=D_{1}^{{}^{\prime\prime}}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\})-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}^{{}^{\prime}}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}-D_{2}^{{}^{\prime\prime}}(\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\})$
$\displaystyle+\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}^{{}^{\prime}}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=D_{1}^{{}^{\prime\prime}}(D_{2}^{{}^{\prime\prime}}(\\{x,y\\})-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}^{{}^{\prime}}(y)\\})-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}^{{}^{\prime}}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=D_{1}^{{}^{\prime\prime}}\circ
D_{2}^{{}^{\prime\prime}}(\\{x,y\\})-D_{1}^{{}^{\prime\prime}}(\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}^{{}^{\prime}}(y)\\})-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}^{{}^{\prime}}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}$
$\displaystyle-D_{2}^{{}^{\prime\prime}}\circ
D_{1}^{{}^{\prime\prime}}(\\{x,y\\})+D_{2}^{{}^{\prime\prime}}(\\{\alpha^{k}\beta^{l}(x),D_{1}^{{}^{\prime}}(y)\\})+\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}^{{}^{\prime}}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=D_{1}^{{}^{\prime\prime}}\circ
D_{2}^{{}^{\prime\prime}}(\\{x,y\\})-\\{D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x)),\alpha^{k}\beta^{l}(D_{2}^{{}^{\prime}}(y))\\}-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}^{{}^{\prime}}(D_{2}^{{}^{\prime}}(y))\\}$
$\displaystyle-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}^{{}^{\prime}}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}-D_{2}^{{}^{\prime\prime}}\circ
D_{1}^{{}^{\prime\prime}}(\\{x,y\\})+\\{D_{2}(\alpha^{k}\beta^{l}(x)),\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}^{{}^{\prime}}(y))\\}$
$\displaystyle+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{2}^{{}^{\prime}}(D_{1}^{{}^{\prime}}(y))\\}+\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}^{{}^{\prime}}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=D_{1}^{{}^{\prime\prime}}\circ
D_{2}^{{}^{\prime\prime}}(\\{x,y\\})-D_{2}^{{}^{\prime\prime}}\circ
D_{1}^{{}^{\prime\prime}}(\\{x,y\\})-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}^{{}^{\prime}}(D_{2}^{{}^{\prime}}(y))\\}$
$\displaystyle+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{2}^{{}^{\prime}}(D_{1}^{{}^{\prime}}(y))\\}$
$\displaystyle=\\{D_{1}^{{}^{\prime\prime}},D_{2}^{{}^{\prime\prime}}\\}(\\{x,y\\})-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),\\{D_{1}^{{}^{\prime}},D_{2}^{{}^{\prime}}\\}(y)\\}.$

Since
$\\{D_{1}^{{}^{\prime\prime}},D_{2}^{{}^{\prime\prime}}\\},\\{D_{1}^{{}^{\prime}},D_{2}^{{}^{\prime}}\\}\in
End(A)$, we have $\\{D_{1},D_{2}\\}\in
GDer_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$.

Similarly, we have $\mu(D_{1},D_{2})\in
GDer_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$.

Therefore, $GDer(A)$ is a BiHom-subalgebra of
$(\mathcal{W},[\cdot,\cdot],\sigma_{1},\sigma_{2})$.

  2. 2.

Suppose that $D_{1}\in ZDer_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
Der_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$,

$\sigma_{1}(D_{1})(\mu(x,y))=\alpha\circ D_{1}(\mu(x,y))=0.$
$\sigma_{1}(D_{1})(\mu(x,y))=\alpha\circ
D_{1}(\mu(x,y))=\alpha(\mu(D_{1}(x),\alpha^{k}\beta^{l}(y)))=\mu(\sigma_{1}(D_{1})(x),\alpha^{k+1}\beta^{l}(y)).$

Similarly,
$\sigma_{1}(D_{1})(\\{x,y\\})=\\{\sigma_{1}(D_{1})(x),\alpha^{k+1}\beta^{l}(y)\\}.$

Hence, $\sigma_{1}(D_{1})\in ZDer_{(\alpha^{k+1},\beta^{l})}(A)$.

In the same way, we have, $\sigma_{2}(D_{1})\in
ZDer_{(\alpha^{k},\beta^{l+1})}(A)$. Next, we have

$\displaystyle[D_{1},D_{2}](\mu(x,y))$ $\displaystyle=D_{1}\circ
D_{2}(\mu(x,y))-D_{2}\circ D_{1}(\mu(x,y))$
$\displaystyle=D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))+\mu(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}(y)))=0.$
$\displaystyle[D_{1},D_{2}](\mu(x,y))$ $\displaystyle=D_{1}\circ
D_{2}(\mu(x,y))-D_{2}\circ D_{1}(\mu(x,y))$
$\displaystyle=D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))+\mu(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}(y)))-D_{2}(\mu(D_{1}(x),\alpha^{k}\beta^{l}(y)))$
$\displaystyle=\mu(D_{1}(D_{2}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))+\mu(D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x)),\alpha^{k}\beta^{l}(D_{2}(y)))-\mu(D_{2}(D_{1}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))$
$\displaystyle-\mu(\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}(\alpha^{k}\beta^{l}(y)))$
$\displaystyle=\mu([D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)).$

Similarly,
$[D_{1},D_{2}](\\{x,y\\})=\\{[D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}=0$.

Hence, we have $[D_{1},D_{2}]\in
ZDer_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$. Therefore, $ZDer(A)$
is a BiHom-ideal of $Der(A)$.

∎

######  Lemma 4.12.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra, then
the following statements hold:

  1. 1.

$[Der(A),C(A)]\subseteq C(A)$.

  2. 2.

$[QDer(A),QC(A)]\subseteq QC(A)$.

  3. 3.

$[QC(A),QC(A)]\subseteq QDer(A)$.

  4. 4.

$C(A)\subseteq QDer(A)$.

  5. 5.

$QDer(A)+QC(A)\subseteq GDer(A)$.

  6. 6.

$C(A)\circ Der(A)\subseteq Der(A)$.

###### Proof.

  1. 1.

Suppose that $D_{1}\in Der_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
C_{(\alpha^{k^{\prime}}\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$, we
have

$\displaystyle[D_{1},D_{2}](\\{x,y\\})$ $\displaystyle=D_{1}\circ
D_{2}(\\{x,y\\})-D_{2}\circ D_{1}(\\{x,y\\})$
$\displaystyle=D_{1}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\})-D_{2}(\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D_{1}(y)\\})$
$\displaystyle=\\{D_{1}(D_{2}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}+\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}-\\{D_{2}(D_{1}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle-\\{D_{2}(\alpha^{k}\beta^{l}(x)),\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(y))\\}$
$\displaystyle=\\{[D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}.$

Similarly, we have
$[D_{1},D_{2}](\\{x,y\\})=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}](y)\\}$.

In the same way, we have
$[D_{1},D_{2}](\mu(x,y))=\mu([D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))\\\
=\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}](y))$.

Hence, $[D_{1},D_{2}]\in C_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$.
Therefore $[D_{1},D_{2}]\in C(A)$.

  2. 2.

Suppose that $D_{1}\in QDer_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
QC_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$, we
have

$\displaystyle\\{[D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=\\{D_{1}\circ
D_{2}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}-\\{D_{2}\circ
D_{1}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=D_{1}^{{}^{\prime}}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\})-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}$
$\displaystyle-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=D_{1}^{{}^{\prime}}(\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}(y)\\})-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}$
$\displaystyle-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=\\{D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x)),\alpha^{k}\beta^{l}(D_{2}(y))\\}+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\}$
$\displaystyle-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\}-\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}$
$\displaystyle=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\}-\\{D_{2}(\alpha^{k}\beta^{l}(x)),\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(y))\\}$
$\displaystyle=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\}-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{2}(D_{1}(y))\\}$
$\displaystyle=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}](y)\\}.$

Similarly,
$\mu([D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))=\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}](y)).$

Hence, we have $[D_{1},D_{2}]\in
QC_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$. So
$[QDer(A),QC(A)]\subseteq QC(A)$.

  3. 3.

Suppose that $D_{1}\in QC_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
QC_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$, we
have

$\displaystyle\\{[D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=\\{D_{1}\circ
D_{2}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}-\\{D_{2}\circ
D_{1}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}$
$\displaystyle=\\{\alpha^{k}\beta^{l}(D_{2}(x)),D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))\\}-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(x)),D_{2}(\alpha^{k}\beta^{l}(y))\\}$
$\displaystyle=\\{D_{2}(\alpha^{k}\beta^{l}(x)),\alpha^{k^{\prime}}\beta^{l^{\prime}}(D_{1}(y))\\}-\\{D_{1}(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x)),\alpha^{k}\beta^{l}(D_{2}(y))\\}$
$\displaystyle=\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{2}(D_{1}(y))\\}-\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\}$
$\displaystyle=-\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}](y)),$

i.e.,
$\\{[D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}]\\}=0$.

Similarly,
$\mu([D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))+\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),[D_{1},D_{2}])=0$.
Hence, we have $[D_{1},D_{2}]\in
QDer_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$, which implies that
$[QC(A),QC(A)]\subseteq QDer(A)$.

  4. 4.

Suppose that $D\in C_{(\alpha^{k},\beta^{l})}(A)$. Then for any $x,y\in A$, we
have

$D(\\{x,y\\})=\\{D(x),\alpha^{k}\beta^{l}(y)\\}=\\{\alpha^{k}\beta^{l}(x),D(y)\\}.$

Hence, we have

$\\{D(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D(y)\\}=2D(\\{x,y\\}),$

Similarly,
$\mu(D(x),\alpha^{k}\beta^{l}(y))+\mu(\alpha^{k}\beta^{l}(x),D(y))=2D(\mu(x,y)),$

which implies that $D\in QDer_{(\alpha^{k},\beta^{l})}(A)$. So $C(A)\subseteq
QDer(A)$.

  5. 5.

In fact. Let $D_{1}\in QDer_{(\alpha^{k},\beta^{l})}(A),D_{2}\in
QC_{(\alpha^{k},\beta^{l})}(A)$. Then there exist
$D^{\prime}_{1},D^{\prime\prime}_{1}\in End(A)$, for any $x,y\in A$, we have

$\displaystyle\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D_{1}(y)\\}=D_{1}^{\prime}(\\{x,y\\}),$
$\displaystyle\mu(D_{1}(x),\alpha^{k}\beta^{l}(y))+\mu(\alpha^{k}\beta^{l}(x),D_{1}(y))=D_{1}^{\prime\prime}(\mu(x,y)).$

Thus, for any $x,y\in A$, we have

$\displaystyle\\{(D_{1}+D_{2})(x),\alpha^{k}\beta^{l}(y)\\}$ $\displaystyle=$
$\displaystyle\\{D_{1}(x),\alpha^{k}\beta^{l}(y)\\}+\\{D_{2}(x),\alpha^{k}\beta^{l}(y)\\}$
$\displaystyle=$ $\displaystyle
D_{1}^{\prime}(\\{x,y\\})-\\{\alpha^{k}\beta^{l}(x),D_{1}(y)\\}+\\{\alpha^{k}\beta^{l}(x),D_{2}(y)\\}$
$\displaystyle=$ $\displaystyle
D_{1}^{\prime}(\\{x,y\\})-\\{\alpha^{k}\beta^{l}(x),(D_{1}-D_{2})(y)\\},$

and

$\displaystyle\mu((D_{1}+D_{2})(x),\alpha^{k}\beta^{l}(y))$ $\displaystyle=$
$\displaystyle\mu(D_{1}(x),\alpha^{k}\beta^{l}(y))+\mu(D_{2}(x),\alpha^{k}\beta^{l}(y)\\}$
$\displaystyle=$ $\displaystyle
D_{1}^{\prime\prime}(\mu(x,y))-\mu(\alpha^{k}\beta^{l}(x),D_{1}(y))+\mu(\alpha^{k}\beta^{l}(x),D_{2}(y))$
$\displaystyle=$ $\displaystyle
D_{1}^{\prime\prime}(\mu(x,y))-\mu(\alpha^{k}\beta^{l}(x),(D_{1}-D_{2})(y)),$

Therefore, $D_{1}+D_{2}\in GDer_{(\alpha^{k},\beta^{l})}(A)$.

  6. 6.

Suppose that $D_{1}\in C_{(\alpha^{k},\beta^{l})}(A)$, $D_{2}\in
Der_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$. Then for any $x,y\in A$,
we have

$\displaystyle D_{1}\circ D_{2}(\\{x,y\\})$
$\displaystyle=D_{1}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y)\\}+\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}(y)\\})$
$\displaystyle=\\{D_{1}(D_{2}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y)\\}+\\{\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))\\},$
$\displaystyle D_{1}\circ D_{2}(\mu(x,y))$
$\displaystyle=D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y))+\mu(\alpha^{k^{\prime}}\beta^{l^{\prime}}(x),D_{2}(y)))$
$\displaystyle=\mu(D_{1}(D_{2}(x)),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y))+\mu(\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(x),D_{1}(D_{2}(y))),$

which implies that $D_{1}\circ D_{2}\in
Der_{(\alpha^{k+k^{\prime}},\beta^{l+l^{\prime}})}(A)$. So $C(A)\circ
Der(A)\subseteq Der(A)$.

∎

######  Theorem 4.13.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra,
$\alpha$ and $\beta$ surjections, then $[C(A),QC(A)]\subseteq End(A,Z(A))$.
Moreover, if $Z(A)=\\{0\\}$, then $[C(A),QC(A)]=\\{0\\}$.

###### Proof.

For any $D_{1}\in C_{(\alpha^{k},\beta^{l})}(A),D_{2}\in
QC_{(\alpha^{k^{\prime}},\beta^{l^{\prime}})}(A)$ and $x,y\in A$, since
$\alpha$ and $\beta$ are surjections, there exist $y^{\prime}\in A$ such that
$y=\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime})$, we have

$\displaystyle\\{[D_{1},D_{2}](x),y\\}$ $\displaystyle=$
$\displaystyle\\{\\{D_{1},D_{2}\\}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime})\\}$
$\displaystyle=$
$\displaystyle\\{D_{1}D_{2}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime})\\}-\\{D_{2}D_{1}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime})\\}$
$\displaystyle=$ $\displaystyle
D_{1}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})\\})-\\{\alpha^{k^{\prime}}\beta^{l^{\prime}}D_{1}(x),D_{2}\alpha^{k}\beta^{l}(y^{\prime})\\}$
$\displaystyle=$ $\displaystyle
D_{1}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})\\})-D_{1}(\\{D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})\\})$
$\displaystyle=$ $\displaystyle 0,$

and

$\displaystyle\mu([D_{1},D_{2}](x),y)$ $\displaystyle=$
$\displaystyle\mu([D_{1},D_{2}](x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime}))$
$\displaystyle=$
$\displaystyle\mu(D_{1}D_{2}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime}))-\mu(D_{2}D_{1}(x),\alpha^{k+k^{\prime}}\beta^{l+l^{\prime}}(y^{\prime}))$
$\displaystyle=$ $\displaystyle
D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})))-\mu(\alpha^{k^{\prime}}\beta^{l^{\prime}}D_{1}(x),D_{2}\alpha^{k}\beta^{l}(y^{\prime}))$
$\displaystyle=$ $\displaystyle
D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})))-D_{1}(\mu(D_{2}(x),\alpha^{k^{\prime}}\beta^{l^{\prime}}(y^{\prime})))$
$\displaystyle=$ $\displaystyle 0.$

So $[D_{1},D_{2}](x)\in Z(A)$ and therefore $[C(A),QC(A)]\subseteq
End(A,Z(A))$. Moreover, if $Z(A)=\\{0\\}$, then it is easy to see that
$[C(A),QC(A)]=\\{0\\}$. ∎

##  5 BiHom-Poisson modules

First, let recall the following.

######  Definition 5.1.

Let $(A,\\{\cdot,\cdot\\}_{A},\mu_{A})$ be a Poisson algebra. Then a left
Poisson module structure on a left $A$-module over $A$ is linear maps
$\mu_{M},\\{\cdot,\cdot\\}_{M}:A\otimes M\longrightarrow M$ such that

$\displaystyle\mu_{M}(a,\mu_{M}(b,m))=\mu_{M}(\mu_{A}(a,b),m)$ (5. 1)
$\displaystyle\\{\\{a,b\\}_{A},m\\}_{M}=\\{a,\\{b,m\\}_{M}\\}_{M}-\\{b,\\{a,m\\}_{M}\\}_{M}$
(5. 2)
$\displaystyle\\{a,\mu_{M}(b,m)\\}_{M}=\mu_{M}(\\{a,b\\}_{A},m)+\mu_{M}(b,\\{a,m\\}_{M})$
(5. 3)
$\displaystyle\\{\mu_{A}(a,b),m\\}_{M}=\mu_{M}(a,\\{b,m\\}_{M})+\mu_{M}(b,\\{a,m\\}_{M})$
(5. 4)

for any $a,b\in A$ and $m\in M.$

######  Remark 5.2.

In [27], Poisson algebras are defined without the associativity assumption and
then, left Poisson modules are defined without the identity (5\. 1). In a
similar way, basing on the definition above, one can defined a right Poisson
module.

######  Definition 5.3.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra.

  1. 1.

A left BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$ with
structure maps $\lambda:A\otimes V\longrightarrow V$ and $\rho:A\otimes
V\longrightarrow V$ such that the following equalities hold:

$\displaystyle\lambda(\alpha(x),\lambda(y,v))$ $\displaystyle=$
$\displaystyle\lambda(\mu(x,y),\psi(v))$ (5. 5)
$\displaystyle\rho(\\{\beta(x),y\\},\psi(v))$ $\displaystyle=$
$\displaystyle\rho(\alpha\beta(x),\rho(y,v))-\rho(\beta(y),\rho(\alpha(x),v))$
(5. 6) $\displaystyle\rho(\alpha\beta(x),\lambda(y,v))$ $\displaystyle=$
$\displaystyle\lambda(\\{\beta(x),y\\},\psi(v))+\lambda(\beta(y),\rho(\alpha(x),v))$
(5. 7) $\displaystyle\rho(\mu(\beta(x),y),\psi(v))$ $\displaystyle=$
$\displaystyle\lambda(\alpha\beta(x),\rho(y,v))+\lambda(\beta(y),\rho(\alpha(x),v))$
(5. 8)

  2. 2.

A right BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$ with
structure maps $\wedge:V\otimes A\longrightarrow V$ and $\delta:V\otimes
A\longrightarrow V$ such that the following equalities hold:

$\displaystyle\wedge(\wedge(v,x),\beta(y))$ $\displaystyle=$
$\displaystyle\wedge(\phi(v),\mu(x,y))$ (5. 9)
$\displaystyle\delta(\phi(v),\\{x,\alpha(y)\\})$ $\displaystyle=$
$\displaystyle\delta(\delta(v,x),\alpha\beta(y))-\delta(\delta(v,\beta(x)),\alpha(y))$
(5. 10) $\displaystyle\delta(\wedge(v,x),\alpha\beta(y))$ $\displaystyle=$
$\displaystyle\delta(\phi(v),\\{x,\alpha(y)\\})+\wedge(\delta(v,\beta(x)),\alpha(y))$
(5. 11) $\displaystyle\delta(\phi(v),\mu(x,\alpha(y)))$ $\displaystyle=$
$\displaystyle\wedge(\delta(v,x),\alpha\beta(y))+\wedge(\delta(v,\beta(x)),\alpha(y))$
(5. 12)

######  Remark 5.4.

  1. 1.

A left BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$ with
structure maps $\lambda:A\otimes V\longrightarrow V$ and $\rho:A\otimes
V\longrightarrow V$ such that $(V,\phi,\psi,\lambda)$ is a left Bihom-
associative $A$-module, $(V,\phi,\psi,\rho)$ is a left BiHom-Lie $A$-module
[1][yong] and (5\. 7) and (5\. 8) hold. Similarly a right BiHom-Poisson
$A$-module is a BiHom-module $(V,\phi,\psi)$ with structure maps
$\wedge:V\otimes A\longrightarrow V$ and $\delta:V\otimes A\longrightarrow V$
such that $(V,\phi,\psi,\wedge)$ is a right Bihom-associative $A$-module,
$(V,\phi,\psi,\delta)$ is a right BiHom-Lie $A$-module and (5\. 11) and (5\.
12) hold.

  2. 2.

If $\alpha=\beta=Id$ and $\phi=\psi=Id$, we recover a left (respectively a
right) Poisson module. Thus if $(A,\\{\cdot,\cdot\\},\mu)$ is a Poisson
algebra and $V$ is a left Poisson $A$-module in the usual sense, then
$(V,Id_{V},Id_{V})$ is a left BiHom-Poisson $\mathbb{A}$-module where
$\mathbb{A}=(A,\\{\cdot,\cdot\\},\mu,Id_{A},Id_{A})$ is a BiHom-Poisson
algebra.

The following results allow to give some examples of left BiHom-Poisson
$A$-modules.

######  Proposition 5.5.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a regular BiHom-Poisson
algebra. Then $(A,\alpha,\beta)$ is a left BiHom-Poisson $A$-module where the
structure maps are $\lambda(a,b)=\mu(a,b)$ and $\rho(a,b)=\\{a,b\\}$. More
generally, if $B$ is a left BiHom-ideal of
$(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$, then $(B,\alpha,\beta)$ is a left
BiHom-Poisson $A$-module where the structure maps are $\lambda(a,x)=\mu(a,x),\
\rho(x,a)=\\{x,a\\}$ for all $x\in B$ and $(a,b)\in A^{\times 2}$.

###### Proof.

The fact that $\lambda$ and $\rho$ are structure maps follows from the
multiplicativity of $\alpha$ and $\beta$ with respect to $\mu$ and
$\\{\cdot,\cdot\\}.$ Next, observe that from the BiHom-commutativity of $\mu$
and the Bihom-skew-symmetry of $\\{\cdot,\cdot\\}$ that
$\mu(x,y)=\mu(\alpha^{-1}\beta(y),\alpha\beta^{-1}(x))$ and
$\\{x,y\\}=-\\{\alpha^{-1}\beta(y),\alpha\beta^{-1}(x)\\}$ for all $x,y\in A.$
Now, pick $(x,y,v)\in A^{\times 3}$ then, we have by the BiHom-associativity

$\displaystyle\lambda(\alpha(x),\lambda(y,v))=\mu(\alpha(x),\mu(y,v))=\mu(\mu(x,y),\beta(v))=\lambda(\mu(x,y),\beta(v))$

Next, compute (5\. 6) using the BiHom-Jacobi identity in the third line, as
follows

$\displaystyle\rho(\\{\beta(x),y\\},\beta(v))=\\{\\{\beta(x),y\\},\beta(v)\\}=\\{\alpha^{-1}\beta(\beta(v)),\alpha\beta^{-1}\\{\alpha^{-1}\beta(y),\alpha\beta^{-1}(\beta(x))\\}\\}$
$\displaystyle=\\{\beta^{2}(\alpha^{-1}(v)),\\{y,\alpha^{2}\beta^{-1}(x)\\}\\}=\\{\beta^{2}(\alpha^{-1}(v)),\\{\beta(\beta^{-1}(y)),\alpha(\alpha\beta^{-1}(x))\\}\\}$
$\displaystyle=-\\{\beta^{2}(\beta^{-1}(y)),\\{\beta(\alpha\beta^{-1}(x)),\alpha(\alpha^{-1}(v))\\}\\}-\\{\beta^{2}(\alpha\beta^{-1}(x)),\\{\beta(\alpha^{-1}(v)),\alpha(\beta^{-1}(y))\\}\\}$
$\displaystyle=-\\{\beta(y),\\{\alpha(x),v)\\}\\}-\\{\alpha\beta(x),\\{\beta\alpha^{-1}(v),\alpha\beta^{-1}(y)\\}\\}=-\\{\beta(y),\\{\alpha(x),v)\\}\\}$
$\displaystyle+\\{\alpha\beta(x),\\{y,v\\}\\}=-\rho(\beta(y),\rho(\alpha(x),v))+\rho(\alpha\beta(x),\rho(y,v))$

Similarly, using (2.3), we compute

$\displaystyle\rho(\alpha\beta(x),\lambda(y,v))=\\{\alpha\beta(x),\mu(y,v)\\}=\mu(\\{\beta(x),y\\},\beta(v))+\mu(\beta(y),\\{\alpha(x),v\\})$
$\displaystyle=\lambda(\\{\beta(x),y\\},\beta(v))+\lambda(\beta(y),\rho(\alpha(x),v))\mbox{\
which is (\ref{lbhm3})}$

Finally, we obtain (5\. 8) as follows

$\displaystyle\rho(\mu(\beta(x),y),\beta(v))=\\{\mu(\beta(x),y),\beta(v)\\}=-\\{\alpha^{-1}\beta(\beta(v)),\alpha\beta^{-1}\mu(\beta(x),y)\\}$
$\displaystyle=-\\{\alpha\beta(\alpha^{-2}\beta(v)),\mu(\alpha(x),\alpha\beta^{-1}(y))\\}=-\mu(\\{\beta(\alpha^{-2}\beta(v)),\alpha(x)\\},\beta(\beta^{-1}\alpha(y)))$
$\displaystyle-\mu(\beta(\alpha(x)),\\{\alpha(\alpha^{-2}\beta(v)),\alpha\beta^{-1}(y)\\})\mbox{\
( by (\ref{BHL}) )}$
$\displaystyle=-\mu(\\{\beta^{2}\alpha^{-2}(v),\alpha(x)\\},\alpha(y))-\mu(\beta\alpha(x),\\{\alpha^{-1}\beta(v),\alpha\beta^{-1}(y)\\})$
$\displaystyle=-\mu(\alpha^{-1}\beta(\alpha(y)),\alpha\beta^{-1}\\{\beta^{2}\alpha^{-2}(v),\alpha(x)\\})+\mu(\alpha\beta(x),\\{y,v\\})$
$\displaystyle=-\mu(\beta(y),\\{\beta\alpha^{-1}(v),\alpha\beta^{-1}(\alpha(x))\\})+\mu(\alpha\beta(x),\\{y,v\\})$
$\displaystyle=\mu(\beta(y),\\{\alpha(x),v\\})+\mu(\alpha\beta(x),\\{y,v\\})=\lambda(\beta(y),\rho(\alpha(x),v))+\lambda(\alpha\beta(x),\rho(y,v)).$

Hence $(A,\alpha,\beta)$ is a left BiHom-Poisson $A$-module. Similarly, we
prove that more generally, any two-sided BiHom-ideal $(B,\alpha,\beta)$ of
$(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ is a left BiHom-Poisson $A$-module. ∎

######  Remark 5.6.

The analogous of Proposition 5.5 can be proved for right BiHom-Poisson
algebras.

More generally, we prove:

######  Proposition 5.7.

If
$f:(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{A},\beta_{A})\longrightarrow(B,\\{\cdot,\cdot\\}_{A},\mu_{B},\alpha_{B},\beta_{B})$
is a morphism of BiHom-Poisson algebras and $\alpha_{B}$ and $\beta_{B}$ are
invertible then, $(B,\alpha_{B},\beta_{B})$ becomes a left BiHom-Poisson
$A$-module via $f$, i.e, the structure maps are defined as
$\lambda(a,b)=\mu_{B}(f(a),b)$ and $\rho(a,b)=\\{f(a),b\\}_{B}$ for all
$(a,b)\in A\times B$.

###### Proof.

The fact that $\lambda$ and $\rho$ are structure maps follows from the
multiplicativity of $\alpha_{B}$ and $\beta_{B}$ with respect to $\mu_{B}$ and
$\\{\cdot,\cdot\\}_{B}.$ Next, observe that from the BiHom-commutativity of
$\mu_{B}$ and the Bihom-skew-symmetry of $\\{\cdot,\cdot\\}_{B}$ that
$\mu_{B}(b_{1},b_{2})=\mu_{B}(\alpha^{-1}\beta(b_{2}),\alpha\beta^{-1}(b_{1}))$
and
$\\{b_{1},b_{2}\\}_{B}=-\\{\alpha_{B}^{-1}\beta_{B}(b_{2}),\alpha_{B}\beta_{B}^{-1}(b_{1})\\}_{B}$
for all $b_{1},b_{2}\in B.$ Now, pick $(x,y)\in A^{\times 2}$ and $v\in B$
then sine $f$ is a morphism of BiHom-algebras, we have by the BiHom-
associativity in $B$

$\displaystyle\lambda(\alpha_{A}(x),\lambda(y,v))=\mu_{B}(f\alpha_{A}(x),\mu_{B}(f(y),v))=\mu_{B}(\alpha_{B}(f(x)),\mu_{B}(f(y),v))$
$\displaystyle=\mu_{B}(f\mu_{B}(x,y),\beta_{B}(v))=\lambda(\mu_{B}(x,y),\beta_{B}(v))$

Next, compute (5\. 6) using the BiHom-Jacobi identity in the third line, as
follows

$\displaystyle\rho(\\{\beta_{A}(x),y\\}_{A},\beta_{B}(v))=\\{f\\{\beta(x),y\\}_{A},\beta(v)\\}_{B}=\\{\\{\beta_{B}(f(x)),f(y)\\}_{B},\beta_{B}(v)\\}_{B}$
$\displaystyle=\\{\alpha_{B}^{-1}\beta_{B}(\beta_{B}(v)),\alpha_{B}\beta_{B}^{-1}\\{\alpha_{B}^{-1}\beta_{B}f(y),\alpha_{B}\beta_{B}^{-1}(\beta_{B}f(x))\\}_{B}\\}_{B}$
$\displaystyle=\\{\beta_{B}^{2}(\alpha_{B}^{-1}(v)),\\{f(y),\alpha_{B}^{2}\beta_{B}^{-1}f(x)\\}_{B}\\}_{B}$
$\displaystyle=\\{\beta_{B}^{2}(\alpha_{B}^{-1}f(v)),\\{\beta_{B}(\beta_{B}^{-1}f(y)),\alpha_{B}(\alpha_{B}\beta_{B}^{-1}f(x))\\}_{B}\\}_{B}$
$\displaystyle=-\\{\beta_{B}^{2}(\beta_{B}^{-1}f(y)),\\{\beta_{B}(\alpha_{B}\beta_{B}^{-1}f(x)),\alpha_{B}(\alpha_{B}^{-1}(v))\\}_{B}\\}_{B}$
$\displaystyle-\\{\beta_{B}^{2}(\alpha_{B}\beta_{B}^{-1}f(x)),\\{\beta_{B}(\alpha_{B}^{-1}(v)),\alpha_{B}(\beta_{B}^{-1}f(y))\\}_{B}\\}_{B}$
$\displaystyle=-\\{\beta_{B}(f(y)),\\{\alpha_{B}(f(x)),v)\\}_{B}\\}_{B}-\\{\alpha_{B}\beta_{B}(f(x)),\\{\beta_{B}\alpha_{B}^{-1}(v),\alpha_{B}\beta_{B}^{-1}(f(y))\\}_{B}\\}_{B}$
$\displaystyle=-\\{\beta_{B}(f(y)),\\{\alpha_{B}(f(x)),v)\\}_{B}\\}_{B}+\\{\alpha_{B}\beta_{B}(f(x)),\\{f(y),v\\}_{B}\\}_{B}$
$\displaystyle=-\\{f(\beta_{A}(y)),\\{f(\alpha_{A}(x)),v)\\}_{B}\\}_{B}+\\{f(\alpha_{A}\beta_{A}(x)),\\{f(y),v\\}_{B}\\}_{B}$
$\displaystyle=-\rho(\beta(y),\rho(\alpha(x),v))+\rho(\alpha\beta(x),\rho(y,v))$

Similarly, using (2.3) for $B$ and $f$ is a morphism, we compute

$\displaystyle\rho(\alpha_{A}\beta_{A}(x),\lambda(y,v))=\\{f(\alpha_{A}\beta_{A}(x)),\mu_{B}(f(y),v)\\}_{B}=\\{\alpha_{B}\beta_{B}(f(x)),\mu_{B}(f(y),v)\\}_{B}$
$\displaystyle=\mu_{B}(\\{\beta_{B}(f(x)),f(y)\\},\beta_{B}(v))+\mu_{B}(\beta_{B}(f(y)),\\{\alpha_{B}(f(x)),v\\}_{B})$
$\displaystyle=\mu_{B}(f\\{\beta_{A}(x),y\\}_{A},\beta_{B}(v))+\mu_{B}(f(\beta_{A}(y)),\\{f(\alpha_{B}(x)),v\\}_{B})$
$\displaystyle=\lambda(\\{\beta_{A}(x),y\\}_{A},\beta_{B}(v))+\lambda(\beta_{A}(y),\rho(\alpha_{A}(x),v))\mbox{\
which is (\ref{lbhm3})}$

Finally, using $f$ is a morphism we obtain (5\. 8) as follows

$\displaystyle\rho(\mu_{A}(\beta_{A}(x),y),\beta_{B}(v))=\\{f\mu_{A}(\beta_{A}(x),y),\beta_{B}(v)\\}_{B}$
$\displaystyle=-\\{\alpha_{B}^{-1}\beta_{B}(\beta_{B}(v)),\alpha_{B}\beta_{B}^{-1}f\mu_{A}(\beta_{A}(x),y)\\}_{B}$
$\displaystyle=-\\{\alpha_{B}\beta_{B}(\alpha_{B}^{-2}\beta_{B}(v)),\mu_{B}(\alpha_{B}(f(x)),\alpha_{B}\beta_{B}^{-1}(f(y)))\\}_{B}$
$\displaystyle=-\mu_{B}(\\{\beta_{B}(\alpha_{B}^{-2}\beta_{B}(v)),\alpha_{B}(f(x))\\}_{B},\beta_{B}(\beta_{B}^{-1}\alpha_{B}f(y)))$
$\displaystyle-\mu_{B}(\beta_{B}(\alpha_{B}f(x)),\\{\alpha_{B}(\alpha_{B}^{-2}\beta_{B}(v)),\alpha_{B}\beta_{B}^{-1}(f(y))\\})\mbox{\
( by (\ref{BHL}) in $B$ )}$
$\displaystyle=-\mu_{B}(\\{\beta_{B}^{2}\alpha_{B}^{-2}(v),\alpha_{B}(f(x))\\}_{B},\alpha_{B}(f(y)))$
$\displaystyle-\mu_{B}(\beta_{B}\alpha_{B}(f(x)),\\{\alpha_{B}^{-1}\beta_{B}(v),\alpha_{B}\beta_{B}^{-1}(f(y))\\}_{B})$
$\displaystyle=-\mu_{B}(\alpha_{B}^{-1}\beta_{B}(\alpha_{B}f(y)),\alpha_{B}\beta_{B}^{-1}\\{\beta_{B}^{2}\alpha_{B}^{-2}(v),\alpha_{B}(f(x))\\}_{B})+\mu_{B}(\alpha_{B}\beta_{B}(f(x)),\\{f(y),v\\}_{B})$
$\displaystyle=-\mu_{B}(\beta_{B}(f(y)),\\{\beta_{B}\alpha_{B}^{-1}(v),\alpha_{B}\beta_{B}^{-1}(\alpha_{B}f(x))\\})+\mu_{B}(\alpha_{B}\beta_{B}(f(x)),\\{f(y),v\\}_{B})$
$\displaystyle=\mu_{B}(\beta_{B}(f(y)),\\{\alpha_{B}(f(x)),v\\}_{B})+\mu_{B}(\alpha_{B}\beta_{B}(f(x)),\\{f(y),v\\}_{B})$
$\displaystyle=\mu_{B}(f(\beta_{A}(y)),\\{f(\alpha_{A}(x)),v\\}_{B})+\mu_{B}(f(\alpha_{A}\beta_{A}(x)),\\{f(y),v\\}_{B})$
$\displaystyle=\lambda(\beta_{A}(y),\rho(\alpha_{A}(x),v))+\lambda(\alpha_{A}\beta_{A}(x),\rho(y,v))$

Hence $(B,\alpha_{B},\beta_{B})$ is a left BiHom-Poisson $A$-module. ∎

Similarly, we can prove:

######  Proposition 5.8.

If
$f:(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{A},\beta_{A})\longrightarrow(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{B},\beta_{B})$
is a morphism of BiHom-Poisson algebras and $\alpha_{B}$ and $\beta_{B}$ are
invertible then, $(B,\alpha_{B},\beta_{B})$ becomes a right BiHom-Poisson
$A$-module via $f$, i.e, the structure maps are defined as
$\lambda(b,a)=\mu_{B}(b,f(a))$ and $\rho(b,a)=\\{b,f(a)\\}_{B}$ for all
$(a,b)\in A\times B$.

As the case of BiHom-alternative and BiHom-Jordan algebras [7], in order to
give another example of left BiHom-Poisson modules, let us consider the
following:

######  Definition 5.9.

An abelian extension of BiHom-Poisson algebras is a short exact sequence of
BiHom-Poisson algebras

$0\longrightarrow(V,\alpha_{V},\beta_{V})\stackrel{{\scriptstyle\mbox{i}}}{{\longrightarrow}}(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{A},\beta_{A})\stackrel{{\scriptstyle\mbox{$\pi$}}}{{\longrightarrow}}(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{B},\beta_{B})\longrightarrow
0$

where $(V,\alpha_{V},\beta_{B})$ is a trivial BiHom-Poisson algebra, $i$ and
$\pi$ are morphisms of BiHom-algebras. Furthermore, if there exists a morphism
$s:(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{B},\beta_{B})\longrightarrow\\\
(A,\\{\cdot,\cdot\\}_{A},\mu_{A},\alpha_{A},\beta_{A})$ such that $\pi\circ
s=id_{B}$ then the abelian extension is said to be split and $s$ is called a
section of $\pi.$

######  Example 5.10.

Given an abelian extension as in the previous definition, the BiHom-module
$(V,\alpha_{V},\beta_{V})$ inherits a structure of a left BiHom-Poisson
$B$-module and the actions of the BiHom-algebra
$(B,\\{\cdot,\cdot\\}_{B},\mu_{B},\alpha_{B},\beta_{B})$ on $V$ are as
follows. For any $x\in B,$ there exist $\tilde{x}\in A$ such that
$x=\pi(\tilde{x}).$ Let $x$ acts on $v\in V$ by
$\lambda(x,v):=\mu_{A}(\tilde{x},i(v))$ and
$\rho(x,v):=\\{\tilde{x},i(v)\\}_{A}.$ These are well-defined, as another lift
$\tilde{x^{\prime}}$ of $x$ is written
$\tilde{x^{\prime}}=\tilde{x}+v^{\prime}$ for some $v^{\prime}\in V$ and thus
$\lambda(x,v)=\mu_{A}(\tilde{x},i(v))=\mu_{A}(\tilde{x^{\prime}},i(v))$ and
$\rho(x,v)=\\{\tilde{x},i(v)\\}_{A}=\\{\tilde{x^{\prime}},i(v)\\}_{A}$ because
$V$ is trivial. The actions property follow from the BiHom-Poisson identities.
In case these actions of $B$ on $V$ are trivial, one speaks of a central
extension.

The next result allow to construct a sequence of left BiHom-Poisson modules
from a given one.

######  Proposition 5.11.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra and
$V_{\phi,\psi}=(V,\phi,\psi)$ be a left BiHom-Poisson $A$-module with the
structure maps $\lambda$ and $\rho$. Then for each $n,m\in\mathbb{N},$ the
maps

$\displaystyle\lambda^{(n,m)}=\lambda\circ(\alpha^{n}\beta^{m}\otimes Id_{V})$
(5. 13) $\displaystyle\rho^{(n,m)}=\rho\circ(\alpha^{n}\beta^{m}\otimes
Id_{V})$ (5. 14)

give the BiHom-module $(V,\phi,\psi)$ the structure of a left BiHom-Poisson
$A$-module that we denote by $V_{\phi,\psi}^{(n,m)}.$

###### Proof.

Since the structure map $\lambda$ is a morphism of BiHom-modules, we get:

$\displaystyle\phi\lambda^{(n,m)}=\phi\rho_{l}\circ(\alpha^{n}\beta^{m}\otimes
Id_{V})\mbox{ {( by (\ref{nmj1}) )}}$ $\displaystyle=$
$\displaystyle\lambda\circ(\alpha^{n+1}\beta^{m}\otimes\phi)=\lambda\circ(\alpha^{n}\beta^{m}\otimes
Id_{V})\circ(\alpha\otimes\phi)=\lambda^{(n,m)}\circ(\alpha\otimes\phi)$

Similarly, we get that
$\psi\lambda^{(n,m)}=\lambda^{(n,m)}\circ(\beta\otimes\psi),\
\phi\rho^{(n,m)}=\rho^{(n,m)}\circ(\phi\otimes\alpha)$ and
$\psi\rho^{(n,m)}=\rho^{(n,m)}\circ(\psi\otimes\beta).$ Thus $\rho_{l}^{(n)}$
and $\rho_{r}^{(n)}$ are morphisms of BiHom-modules. First, pick $(x,y)\in
A^{\times 2}$ and $v\in V,$ then using (5\. 5) in the second line for
$V_{\phi,\psi},$ we get

$\displaystyle\lambda^{(n,m)}(\alpha(x),\lambda^{(n,m)}(y,v))=\lambda(\alpha^{n+1}\beta^{m}(x),\lambda(\alpha^{n}\beta^{m}(y),v))$
$\displaystyle=\lambda(\mu(\alpha^{n}\beta^{m}(x),\alpha^{n}\beta^{m}(y)),\psi(v))=\lambda^{(n,m)}(\mu(x,y),\psi(v))$

Secondly, we compute

$\displaystyle\rho^{(n,m)}(\\{\beta(x),y\\},\psi(v))=\rho(\\{\alpha^{n}\beta^{m+1}(x),\beta(y)\\},\psi(v))$
$\displaystyle=\rho(\alpha\beta\alpha^{n}\beta^{m}(x),\rho(\alpha^{n}\beta^{m}(y),v))-\rho(\beta\alpha^{n}\beta^{m}(y),\rho(\alpha\alpha^{n}\beta^{m}(x),v))\mbox{
( by (\ref{lbhm2}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\lambda^{(n,m)}(\alpha\beta(x),\lambda^{(n,m)}(y,v))-\lambda^{(n,m)}(\beta(y),\lambda^{(n,m)}(\alpha(y),v))$

Next, we obtain

$\displaystyle\rho^{(n,m)}(\beta\alpha(x),\lambda^{(n,m)}(y,v))=\rho(\alpha\beta\alpha^{n}\beta^{m}(x),\lambda(\alpha^{n}\beta^{m}(y),v))$
$\displaystyle=\lambda\\{\beta\alpha^{n}\beta^{m}(x),\alpha^{n}\beta^{m}(y)\\},\psi(v))+\lambda(\beta\alpha^{n}\beta^{(}y),\rho(\alpha\alpha^{n}\beta^{m}(x),v))\mbox{
( by (\ref{lbhm3}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\lambda^{(n,m)}(\\{\beta(x),y\\},\psi(v))+\lambda^{(n,m)}(\beta(y),\rho^{(n,m)}(\alpha(x),v))$

Finally, we compute

$\displaystyle\rho^{(n,m)}(\mu(\beta(x),y),\psi(v))=\rho(\mu(\beta\alpha^{n}\beta^{m}(x),\alpha^{n}\beta^{m}(y)),\psi(v))$
$\displaystyle=\lambda(\alpha\beta\alpha^{n}\beta^{m}(x),\rho(\alpha^{n}\beta^{m}(y),v))+\lambda(\beta\alpha^{n}\beta^{m}(y),\rho(\alpha\alpha^{n}\beta^{m}(x),v))\mbox{
( by (\ref{lbhm3}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\lambda^{(n,m)}(\alpha\beta(x),\rho^{(n,m)}(y,v))+\lambda^{(n,m)}(\beta(y),\rho^{(n,m)}(\alpha(x),v)).$

Hence, $V_{\phi,\psi}^{(n,m)}$ is a left BiHom-Poisson $A$-module. ∎

Proposition 5.11 reads for the case of right BiHom-Poisson module as:

######  Proposition 5.12.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra and
$V_{\phi,\psi}=(V,\phi,\psi)$ be a right BiHom-Poisson $A$-module with the
structure maps $\wedge$ and $\delta$. Then for each $n,m\in\mathbb{N},$ the
maps

$\displaystyle\wedge^{(n,m)}=\wedge\circ(Id_{V}\otimes\alpha^{n}\beta^{m})$
(5. 15)
$\displaystyle\delta^{(n,m)}=\delta\circ(Id_{V}\otimes\alpha^{n}\beta^{m})$
(5. 16)

give the BiHom-module $(V,\phi,\psi)$ the structure of a right BiHom-Poisson
$A$-module that we denote by $V_{\phi,\psi}^{(n,m)}.$

######  Theorem 5.13.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a Poisson algebra, $V$ be a left Poisson
$A$-module with the structure maps $\lambda$, $\rho$ and $\alpha,\beta$ be
endomorphisms of the Jordan algebra $A$ and $\phi,\psi$ be linear self-maps of
$V$ such that $\phi\circ\lambda=\lambda\circ(\alpha\otimes\phi),$
$\phi\circ\rho=\rho\circ(\alpha\otimes\phi),$
$\psi\circ\lambda=\lambda\circ(\beta\otimes\psi)$ and
$\psi\circ\rho=\rho\circ(\beta\otimes\psi)$. Write $A_{\alpha,\beta}$ for the
BiHom-Poisson algebra
$(A,\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\cdot,\cdot\\}(\alpha\otimes\beta),\mu_{\alpha,\beta}=\mu(\alpha\otimes\beta),\alpha,\beta)$
and $V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:

$\displaystyle\tilde{\lambda}=\lambda\circ(\alpha\beta\otimes\psi)\mbox{ and
}\tilde{\rho}=\rho\circ(\alpha\beta\otimes\psi)$ (5. 17)

give the BiHom-module $V_{\phi,\psi}$ the structure of a left BiHom-Poisson
$A_{\alpha,\beta}$-module.

###### Proof.

It is clear that $\tilde{\lambda}$ and $\tilde{\rho}$ are morphisms of BiHom-
modules. Netx, pick $(x,y)\in A^{\times 2}$ and $v\in V,$ then using (5\. 5)
for $V_{\phi,\psi}$ in the second line we get:

$\displaystyle\tilde{\lambda}(\alpha(x),\tilde{\lambda}(y,v))=\lambda(\alpha^{2}\beta(x),\lambda(\alpha\beta^{2}(y),\psi^{2}(v)))$
$\displaystyle=\lambda(\mu(\alpha^{2}\beta(x),\alpha\beta^{2}(y)),\psi^{2}(v))=\lambda(\alpha\beta\mu(\alpha(x),\beta(y)),\psi^{2}(v))=\tilde{\lambda}(\mu_{\alpha,\beta}(x,y),\psi(v))$

Secondly, we compute

$\displaystyle\tilde{\rho}(\\{\beta(x),y\\}_{\alpha,\beta},\psi(v))=\rho(\\{\alpha^{2}\beta^{2}(x),\alpha\beta^{2}(y)\\},\psi^{2}(v))$
$\displaystyle=\rho(\alpha^{2}\beta^{2}(x),\rho(\alpha\beta^{2}(y),\psi^{2}(v)))-\rho(\alpha\beta^{2}(y),\rho(\alpha^{2}\beta^{2}(x),\psi^{2}(v))\mbox{
( by (\ref{lbhm2}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\tilde{\lambda}(\alpha\beta(x),\tilde{\lambda}(y,v))-\tilde{\lambda}(\beta(y),\tilde{\lambda}(\alpha(y),v))$

Next, we obtain

$\displaystyle\tilde{\rho}(\beta\alpha(x),\tilde{\lambda}(y,v))=\rho(\alpha^{2}\beta(x),\lambda(\alpha\beta^{2}(y),\psi^{2}(v)))$
$\displaystyle=\lambda\\{\alpha^{2}\beta(x),\alpha\beta^{2}(y)\\},\psi^{2}(v))+\lambda(\alpha\beta^{2}(y),\rho(\alpha^{2}\beta(x),\psi^{2}(v)))\mbox{
( by (\ref{lbhm3}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\tilde{\lambda}(\\{\beta(x),y\\}_{\alpha,\beta},\psi(v))+\tilde{\lambda}(\beta(y),\tilde{\rho}(\alpha(x),v))$

Finally, we compute

$\displaystyle\tilde{\rho}(\mu_{\alpha,\beta}(\beta(x),y),\psi(v))=\rho(\mu(\alpha^{2}\beta^{2}(x),\alpha\beta^{2}(y)),\psi^{2}(v))$
$\displaystyle=\lambda(\alpha^{2}\beta^{2}(x),\rho(\alpha\beta^{2}(y),\psi^{2}(v)))+\lambda(\alpha\beta^{2}(y),\rho(\alpha^{2}\beta^{2}(x),\psi^{2}(v)))\mbox{
( by (\ref{lbhm3}) in $V_{\phi,\psi}$ )}$
$\displaystyle=\tilde{\lambda}(\alpha\beta(x),\tilde{\rho}(y,v))+\tilde{\lambda}(\beta(y),\tilde{\rho}(\alpha(x),v)))$

Hence, $V_{\phi,\psi}^{(n,m)}$ is a left BiHom-Poisson $A$-module. ∎

######  Corollary 5.14.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a Poisson algebra, $V$ be a left Poisson
$A$-module with the structure maps $\lambda$ and $\rho$, $\alpha,\ \beta$ be
endomorphisms of the Poisson algebra $A$ and $\phi,\psi$ be linear self-maps
of $V$ such that $\phi\circ\lambda=\lambda\circ(\alpha\otimes\phi),$
$\phi\circ\rho=\rho\circ(\alpha\otimes\phi),$
$\psi\circ\lambda=\lambda\circ(\beta\otimes\psi)$ and
$\psi\circ\rho=\rho\circ(\alpha\otimes\psi)$.
Write $A_{\alpha,\beta}$ for the BiHom-Poisson algebra
$(A,\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\alpha\otimes\beta\\},\mu_{\alpha,\beta}=\mu(\alpha\otimes\beta),\alpha,\beta)$
and $V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:

$\displaystyle\tilde{\lambda}^{(n,m)}=\lambda\circ(\alpha^{n+1}\beta^{m+1}\otimes\psi)\mbox{
and }\tilde{\rho}^{(n,m)}=\rho\circ(\alpha^{n+1}\beta^{m+1}\otimes\psi)$ (5.
18)

give the BiHom-module $V_{\phi,\psi}$ the structure of a left BiHom-Poisson
$A_{\alpha,\beta}$-module for all $n,m\in\mathbb{N}$.

###### Proof.

The proof follows from Proposition 5.11 and Theorem 5.15. ∎

Similarly, we can prove the following result which is the analogous of Theorem
5.15 for right BiHom-Poisson modules.

######  Theorem 5.15.

Let $(A,\\{\cdot,\cdot\\},\mu)$ be a Poisson algebra, $V$ be a right Poisson
$A$-module with the structure maps $\lambda$, $\rho$ and $\alpha,\beta$ be
endomorphisms of the Jordan algebra $A$ and $\phi,\psi$ be linear self-maps of
$V$ such that $\phi\circ\lambda=\lambda\circ(\alpha\otimes\phi),$
$\phi\circ\rho=\rho\circ(\alpha\otimes\phi),$
$\psi\circ\lambda=\lambda\circ(\beta\otimes\psi)$ and
$\psi\circ\rho=\rho\circ(\beta\otimes\psi)$. Write $A_{\alpha,\beta}$ for the
BiHom-Poisson algebra
$(A,\\{\cdot,\cdot\\}_{\alpha,\beta}=\\{\cdot,\cdot\\}(\alpha\otimes\beta),\mu_{\alpha,\beta}=\mu(\alpha\otimes\beta),\alpha,\beta)$
and $V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:

$\displaystyle\tilde{\wedge}=\wedge\circ(\phi\otimes\alpha\beta)\mbox{ and
}\tilde{\delta}=\rho\circ(\phi\otimes\alpha\beta)$ (5. 19)

give the BiHom-module $V_{\phi,\psi}$ the structure of a right BiHom-Poisson
$A_{\alpha,\beta}$-module.

In the case of Poisson algebras, we can form semidirect products when given a
left (or a right) module. Similarly, we have

######  Theorem 5.16.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom-Poisson algebra and
$(V,\phi,\psi)$ be a left $A$-module with the structure maps $\lambda$ and
$\rho$. Then $(A\oplus V,[\dot{,}\dot{]},\ast,\tilde{\alpha},\tilde{\beta})$
is a BiHom-Poisson algebra where $\ast,[\cdot,\cdot]:(A\oplus V)^{\otimes
2}\longrightarrow A\oplus V,~{}$
$(a+u)\ast(b+v):=\mu(a,b)+\lambda(a,v)+\lambda(\alpha^{-1}\beta(b),\psi^{-1}\phi(u)),$
$[a+u,b+v]:=\\{a,b\\}+\rho(a,v)-\rho(\alpha^{-1}\beta(b),\psi^{-1}\phi(u))$
and $\tilde{\alpha},\tilde{\beta}:A\oplus V\longrightarrow A\oplus V,$
$\tilde{\alpha}(a+u):=\alpha(a)+\phi(u)$ and
$\tilde{\beta}(a+u):=\beta(a)+\psi(u)$ called the semidirect product of the
BiHom-Poisson $(A,\mu,\\{\cdot,\cdot\\},\alpha,\beta)$ and $(V,\phi,\psi).$

###### Proof.

Clearly, $\tilde{\alpha}$ and $\tilde{\beta}$ are multiplicative with respect
to $\ast$ and $[\cdot,\cdot].$ Next

$\displaystyle\tilde{\beta}(a+u)\ast\tilde{\alpha}(b+v)=(\beta(a)+\psi(u))\ast(\alpha(b)+\phi(v))$
$\displaystyle=\mu(\beta(a),\alpha(b))+\lambda(\beta(a),\phi(v))+\lambda(\beta(b),\phi(u)$
$\displaystyle=\mu(\beta(b),\alpha(a))+\lambda(\beta(b),\phi(u)+\lambda(\beta(a),\phi(v))\mbox{
( by the BiHom-commutativity of $\mu$ )}$
$\displaystyle=\tilde{\beta}(b+v)\ast\tilde{\alpha}(u+v).$ (5. 20)

Next, pick $(a,b,c)\in A^{\times 2}$ and $(u,v,w)\in V^{\times 2},$ then

$\displaystyle\Big{(}(a+u)\ast(b+v)\Big{)}\ast\bar{\beta}(c+w)=\Big{(}\mu(a,b)+\lambda(a,v)+\lambda(\alpha^{-1}\beta(b),\psi^{-1}\phi(u))\Big{)}\ast(\beta(c)+\psi(w))$
$\displaystyle=\mu(\mu(a,b),\beta(c))+\lambda(\mu(a,b),\psi(w))+\lambda(\alpha^{-1}\beta^{2}(c),\lambda(\beta^{-1}\alpha(a),\phi\psi^{-1}(v))$
$\displaystyle+\lambda(\alpha^{-1}\beta^{2}(c),\lambda(b,(\psi^{-1}\phi)^{2}(u)))\mbox{
( using $\lambda$ is a morphism )}$
$\displaystyle=\mu(\alpha(a),\mu(b,c))+\lambda(\mu(a,b),\psi(w))+\lambda(\mu((\alpha^{-1}\beta)^{2}(c),\beta^{-1}\alpha(a)),\phi(v))$
$\displaystyle+\lambda(\mu((\alpha^{-1}\beta)^{2}(c),b),\psi^{-1}\phi^{2}(u))\mbox{
( using BiHom-associativity and (\ref{lbhm1}) )}$ (5. 21)
$\displaystyle=\mu(\alpha(a),\mu(b,c))+\lambda(\mu(a,b),\psi(w))+\lambda(\mu(a,\beta\alpha^{-1}(c)),\phi(v))$
$\displaystyle+\lambda(\mu(\alpha^{-1}\beta(b),\alpha^{-1}\beta(c),\psi^{-1}\phi^{2}(u))\mbox{
( using BiHom-commutativity )}$

Similarly, we prove that

$\displaystyle\bar{\alpha}(a+u)\ast\Big{(}(b+v)\ast(c+w)\Big{)}=\mu(\alpha(a),\mu(b,c))+\lambda(\mu(a,b),\psi(w))+\lambda(\mu(a,\beta\alpha^{-1}(c)),\phi(v))$
$\displaystyle+\lambda(\mu(\alpha^{-1}\beta(b),\alpha^{-1}\beta(c),\psi^{-1}\phi^{2}(u))$

Hence $(A\oplus V,\ast,\bar{\alpha},\bar{\beta})$ is a BiHom-commutative
BiHom-associative algebra. Now, observe that $(A\oplus
V,[\cdot,\cdot],\bar{\alpha},\bar{\beta})$ is a BiHom-Lie algebra (Proposition
4.9, [15] ). Finally, let $(a,b,c)\in A^{\times 2}$ and $(u,v,w)\in V^{\times
2}.$ Then

$\displaystyle[\bar{\alpha}\bar{\beta}(a+u),(b+v)\ast(c+w)]=[\alpha\beta(a)+\phi\psi(u),\mu(b,c)+\lambda(b,w)+\lambda(\alpha^{-1}\beta(),\psi^{-1}\phi(v)]$
$\displaystyle=\\{\alpha\beta(a),\mu(b,c)\\}+\rho(\alpha\beta(a),\lambda(b,w))+\rho(\alpha\beta(a),\lambda(\alpha^{-1}\beta(c),\psi^{-1}\phi(v)))$
$\displaystyle-\rho(\mu(\beta\alpha^{-1}(b),\beta\alpha^{-1}(c)),\psi\psi^{-1}\phi^{2}(u))=\mu(\\{\beta(a),b\\},\beta(c))$
$\displaystyle+\mu(\beta(c),\\{\alpha(a),c\\})+\lambda(\\{\beta(a),b\\},\psi(w))+\lambda(\beta(b),\rho(\alpha(a),w))+\lambda(\\{\beta(a),\alpha^{-1}\beta(c)\\},\phi(v))$
$\displaystyle+\lambda(\beta^{2}\alpha^{-1}(c),\rho(\alpha(a),\psi^{-1}\phi(v)))-\lambda(\beta(b),\rho(\beta\alpha^{-1}(c),\psi^{-1}\phi^{2}(u)))$
$\displaystyle-\lambda(\beta^{2}\alpha^{-1}(c),\rho(b,\psi^{-1}\phi^{2}(u)))\mbox{
( by (\ref{BHL}}),(\ref{lbhm3}),(\ref{lbhm4}))$
$\displaystyle=\Big{(}\mu(\\{\beta(a),b\\},\beta(c))+\lambda(\\{\beta(a),b\\},\psi(w))+\lambda(\beta^{2}\alpha^{-1}(c),\rho(\alpha(a),\psi^{-1}\phi(v)))$
$\displaystyle-\lambda(\beta^{2}\alpha^{-1}(c),\rho(b,\psi^{-1}\phi^{2}(u)))\Big{)}+\Big{(}\mu(\beta(c),\\{\alpha(a),c\\})+\lambda(\beta(b),\rho(\alpha(a),w))$
$\displaystyle-\lambda(\beta(b),\rho(\beta\alpha^{-1}(c),\psi^{-1}\phi^{2}(u)))+\lambda(\\{\beta(a),\alpha^{-1}\beta(c)\\},\phi(v))\Big{)}\mbox{
( rearranging terms )}$
$\displaystyle=[\bar{\beta}(a+u),(b+v)]\ast\bar{\beta}(c+w)+\bar{\beta}(b+v)\ast[\bar{\alpha}(a+u),(c+w)]$

Hence, the conclusion follows. ∎

######  Remark 5.17.

Consider the split null extension $A\oplus V$ determined by the left BiHom-
Poisson module $(V,\phi,\psi)$ for the BiHom-Poisson algebra
$(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ in the previous theorem. Write
elements $a+v$ of $A\oplus V$ as $(a,v).$ Then there is an injective
homomorphism of BiHom-modules $i:V\rightarrow A\oplus V$ given by $i(v)=(0,v)$
and a surjective homomorphism of BiHom-modules $\pi:A\oplus V\rightarrow A$
given by $\pi(a,v)=a.$ Moreover, $i(V)$ is a two-sided BiHom-ideal of $A\oplus
V$ such that $A\oplus V/i(V)\cong A$. On the other hand, there is a morphism
of BiHom-algebras $\sigma:A\rightarrow A\oplus V$ given by $\sigma(a)=(a,0)$
which is clearly a section of $\pi.$ Hence, we obtain the abelian split exact
sequence of BiHom-Poisson algebras and $(V,\phi,\psi)$ is a left BiHom-Poisson
module for $A$ via $\pi.$

######  Definition 5.18.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a BiHom Poisson algebra. A
skew-symmetric $n$-linear map $f:\underbrace{A\times\cdots\times
A}_{n~{}times}\rightarrow A$ that is a derivation in each argument is called
an $n$-BiHom-cochain, if it satisfies

$\begin{array}[]{llllllllllll}f(\alpha(x_{1}),\cdots,\alpha(x_{n}))&=&\alpha\circ
f(x_{1},\cdots,x_{n}),\\\ f(\beta(x_{1}),\cdots,\beta(x_{n}))&=&\beta\circ
f(x_{1},\cdots,x_{n}).\end{array}$

The set of $n$-Hom-cochains is denoted by $C_{\alpha,\beta}^{n}(A,A)$, for
$n\geq 1$.

######  Definition 5.19.

Let $(A,\\{\cdot,\cdot\\},\mu,\alpha,\beta)$ be a regular BiHom Poisson
algebra. For $n=1,2,$ the coboundary operator
$\delta^{n}:C_{\alpha,\beta}^{n}(A,A)\rightarrow C_{\alpha,\beta}^{n+1}(A,A)$
is defined as follows:

$\displaystyle\delta^{1}f(x,y)=\\{\alpha(x),f(y)\\}-\\{f(x),\alpha(y)\\}-f(\\{\alpha^{-1}\beta(x),y\\})$
(5. 22)
$\displaystyle\delta^{2}f(x,y,z)=\\{\alpha\beta(x),f(y,z)\\}-\\{\alpha\beta(y),f(x,z)\\}+\\{\alpha\beta(y),f(x,z)\\}$
$\displaystyle-f(\\{\alpha^{-1}\beta(x),y\\},\beta(z))+f(\\{\alpha^{-1}\beta(x),z\\},\beta(y))-f(\\{\alpha^{-1}\beta(y),z\\},\beta(x))$
(5. 23)

######  Lemma 5.20.

The coboundary operators $\delta^{i}$ are well defined, for $i=1,2$.

###### Proof.

For any $x,y,z\in A$ we have:

$\displaystyle\delta^{1}f(\alpha(x),\alpha(y))=\\{\alpha^{2}(x),f\alpha(y)\\}-\\{f\alpha(x),\alpha^{2}(y)\\}-f(\\{\alpha^{-1}\beta\alpha(x),\alpha
y)\\})$ $\displaystyle=\\{\alpha^{2}(x),\alpha f(y)\\}-\\{\alpha
f(x),\alpha^{2}(y)\\}-f(\\{\alpha\alpha^{-1}\beta(x),\alpha(y)\\})=\alpha\circ\delta^{1}f(x,y)$

and

$\displaystyle\delta^{2}f(\alpha(x),\alpha(y),\alpha(z))=\\{\alpha\beta\alpha(x),f(\alpha(y),\alpha(z))\\}-\\{\alpha\beta\alpha(y),f(\alpha(x),\alpha(z))\\}$
$\displaystyle+\\{\alpha\beta\alpha(y),f(\alpha(x),\alpha(z))\\}-f(\\{\alpha^{-1}\beta\alpha(x),\alpha(y)\\},\beta\alpha(z))+f(\\{\alpha^{-1}\beta\alpha(x),\alpha(z)\\},\beta\alpha(y))$
$\displaystyle-f(\\{\alpha^{-1}\beta\alpha(y),\alpha(z)\\},\beta\alpha(x))=\\{\alpha\beta\alpha(x),\alpha
f(y,z)\\}-\\{\alpha\beta\alpha(y),\alpha f(x,z)\\}$
$\displaystyle+\\{\alpha\beta\alpha(y),\alpha
f(x,z)\\}-f(\alpha(\\{\alpha^{-1}\beta(x),y\\}),\alpha\beta(z))+f(\alpha(\\{\alpha^{-1}\beta()x,z\\}),\alpha\beta(y))$
$\displaystyle-f(\alpha(\\{\alpha^{-1}\beta(y),z\\}),\alpha\beta(x))=\alpha\circ\delta^{2}f(x,y,z)$

In the same way, we obtain $\delta^{1}f\circ\beta=\beta\circ\delta^{1}f$ and
$\delta^{2}f\circ\beta=\beta\circ\delta^{2}f.$

Then $\delta^{i}$ are well defined, for $i=1,2$. ∎

######  Theorem 5.21.

With notations as above, we have

$\delta^{2}\circ\delta^{1}=0.$

###### Proof.

Let $f\in C_{\alpha,\beta}^{1}(A,A)$ and $(x,y,z)\in A^{\times 3}$ then, we
have:

$\displaystyle\delta^{2}\circ\delta^{1}f(x,y,z)=\\{\alpha\beta(x),\delta^{1}f(y,z)\\}-\\{\alpha\beta(y),\delta^{1}f(x,z)\\}+\\{\alpha\beta(y),\delta^{1}f(x,z)\\}$
$\displaystyle-\delta^{1}f(\\{\alpha^{-1}\beta(x),y\\},\beta(z))+\delta^{1}f(\\{\alpha^{-1}\beta(x),z\\},\beta(y))-\delta^{1}f(\\{\alpha^{-1}\beta(y),z\\},\beta(x))$
$\displaystyle=$
$\displaystyle\\{\alpha\beta(x),\\{\alpha(y),f(z)\\}\\}-\\{\alpha\beta(x),\\{\alpha(z),f(y)\\}\\}-\\{\alpha\beta(x),f(\\{\alpha^{-1}\beta(y),z\\})\\}$
$\displaystyle-\\{\alpha\beta(y),\\{\alpha(x),f(z)\\}\\}+\\{\alpha\beta(y),\\{\alpha(z),f(x)\\}\\}+\\{\alpha\beta(y),f(\\{\alpha^{-1}\beta(x),z\\})\\}$
$\displaystyle+\\{\alpha\beta(z),\\{\alpha(x),f(y)\\}\\}-\\{\alpha\beta(z),\\{\alpha(y),f(x)\\}\\}-\\{\alpha\beta(z),f(\\{\alpha^{-1}\beta(x),y\\})\\}$
$\displaystyle-\\{\\{\beta(x),\alpha(y)\\},f\beta(z)\\}+\\{\alpha\beta(z),f(\\{\alpha^{-1}\beta(x),y\\}),\\}+f(\\{\\{\alpha^{-2}\beta^{2}(x),\alpha^{-1}\beta(y)\\},\beta(z)\\})$
$\displaystyle+\\{\\{\beta(x),\alpha(z)\\},f\beta(y)\\}-\\{\alpha\beta(y),f(\\{\alpha^{-1}\beta(x),z\\}),\\}-f(\\{\\{\alpha^{-2}\beta^{2}(x),\alpha^{-1}\beta(z)\\},\beta(y)\\})$
$\displaystyle-\\{\\{\beta(y),\alpha(z)\\},f\beta(x)\\}+\\{\alpha\beta(x),f(\\{\alpha^{-1}\beta(y),z\\}),\\}+f(\\{\\{\alpha^{-2}\beta^{2}(y),\alpha^{-1}\beta(z)\\},\beta(x)\\})$
$\displaystyle=$
$\displaystyle\\{\beta^{2}(\alpha\beta^{-1}(x)),\\{\beta(\alpha\beta^{-1}(y)),\alpha(\alpha^{-1}f(z))\\}\\}+\\{\beta^{2}(\alpha\beta^{-1}(x)),\\{\beta(\alpha^{-1}f(y)),\alpha(\beta\alpha^{-1}(z))\\}\\}$
$\displaystyle-\\{\alpha\beta(x),f(\\{\alpha^{-1}\beta(y),z\\})\\}-\\{\beta^{2}(\alpha\beta^{-1}(y)),\\{\beta(\alpha^{-1}f(z)),\alpha(\beta\alpha^{-1}(x))\\}\\}$
$\displaystyle+\\{\beta^{2}(\alpha\beta^{-1}(y)),\\{\beta(\beta^{-1}\alpha(z)),\alpha(\alpha^{-1}f(x))\\}\\}+\\{\alpha\beta(y),f(\\{\alpha^{-1}\beta(x),z\\})\\}$
$\displaystyle+\\{\beta^{2}(\alpha\beta^{-1}(z)),\\{\beta(\beta^{-1}\alpha(x)),\alpha(\alpha^{-1}f(y))\\}\\}+\\{\beta^{2}(\alpha\beta^{-1}(z)),\\{\beta(\alpha^{-1}f(x)),\alpha(\alpha\beta^{-1}(y))\\}\\}$
$\displaystyle-\\{\alpha\beta(z),f(\\{\alpha^{-1}\beta(x),y\\})\\}+\\{\beta^{2}(\alpha^{-1}f(z)),\\{\beta(\alpha\beta^{-1}(x)),\alpha(\beta^{-1}\alpha(y))\\}\\}$
$\displaystyle+\\{\alpha\beta(z),f(\\{\alpha^{-1}\beta(x),y\\}),\\}-f(\\{\beta^{2}(\alpha^{-1}(z)),\\{\beta(\alpha^{-1}(x)),\alpha(\alpha^{-1}(y))\\}\\})$
$\displaystyle+\\{\beta^{2}(\alpha^{-1}f(y)),\\{\beta(\beta^{-1}\alpha(z)),\alpha(\beta^{-1}\alpha(x))\\}\\}-\\{\alpha\beta(y),f(\\{\alpha^{-1}\beta(x),z\\}),\\}$
$\displaystyle-f(\\{\beta^{2}(\alpha^{-1}(y)),\\{\beta(\alpha^{-1}(z)),\alpha(\alpha^{-1}(x))\\}\\})+\\{\beta^{2}(\alpha^{-1}f(x)),\\{\beta(\alpha\beta^{-1}(y)),\alpha(\alpha\beta^{-1}(z))\\}\\}$
$\displaystyle+\\{\alpha\beta(x),f(\\{\alpha^{-1}\beta(y),z\\}),\\}-f(\\{\beta^{2}(\alpha^{-1}(x)),\\{\beta(\alpha^{-1}(y)),\alpha(\alpha^{-1}(z))\\}\\})$
( since $\\{u,v\\}=-\\{\beta\alpha^{-1}(v),\alpha\beta^{-1}(u)\\}$ $\forall
u,v\in A$) $\displaystyle=0\mbox{ ( by the BiHom-Jaobi identity) }.$

∎

For $n=1,2,$ the map $f\in C_{\alpha,\beta}^{n}(A,A)$ is called an $n$-BiHom-
cocycle $\delta^{n}f=0.$ We denote the subspace spanned by $n$-Bihom-cocycles
by $Z_{\alpha,\beta}^{n}(A,A)$ and
$B_{\alpha,\beta}^{n}(A,A)=\delta^{n-1}C_{\alpha,\beta}^{n-1}(A,A).$ Since
$\delta^{2}\circ\delta^{1}=0,$ $B_{\alpha,\beta}^{2}(A,A)$ is a subspace of
$Z_{\alpha,\beta}^{2}(A,A).$ Hence we can define a cohomology space
$H_{\alpha,\beta}^{2}(A,A)$ of as the factor space
$Z_{\alpha,\beta}^{2}(A,A)/B_{\alpha,\beta}^{2}(A,A).$

## References

  * [1] H. Adimi, H. Amri, S. Mabrouk, A. Makhlouf, (Non-BiHom-Commutative) BiHom-Poisson algebras, arXiv:2008.04597vi [math.RA].
  * [2] N. Aizawa, H. Sato, _$q$ -deformation of the Virasoro algebra with central extension_, Phys. Lett. B 256 (1991), 185–190.
  * [3] V. I. Arnold,  Mathematical methods of classical mechanics, Grad. Texts in Math. 60, Springer, Berlin, 1978.
  * [4] S. Attan, Some characterizations of color Hom-Poisson algebras, Hacettepe J. of Mathematics and Statistics, $\bf 47$ no 6 (2018), 1552-1563.
  * [5] S. Attan, H. Hounnon and B. Kpamègan, Hom-Jordan and Hom-alternative bimodules. Extracta Mathematicae 35(2020), 69-97.
  * [6] S. Attan, Structures and bimodules of simple Hom-alternative algebras. arXiv:1908.08711v1 [math.RA].
  * [7] S. Attan and I. Laraiedh, Constructions and bimodules of BiHom-Jordan and BiHom-alternative algebras, arXiv:2008.0702v1 [math.RA].
  * [8] V. Chari and A. N. Pressley,  A guide to quantum groups, Cambridge Univ. Press, Cambridge, 1994.
  * [9] Chtioui, T., Mabrouk, S., Makhlouf, A.: BiHom-alternative, BiHom-Malcev and BiHom-Jordan algebas, arXiv:1811.10394v1 [math.RA]
  * [10] Chtioui, T., Mabrouk, S., A. Makhlouf.: BiHom-pre-alternative algebras and BiHom-altenative quadri-algebas, arXiv:1903.03994v1 [math.RA]
  * [11] V. G. Drinfel’d,  Quantum groups, in: Proc. ICM (Berkeley, 1986), p.798-820, AMS, Providence, RI, 1987.
  * [12] E. Frenkel and D. Ben-Zvi,  Vertex algebras and algebraic curves, Math Surveys and Monographs 88, 2nd ed., AMS, Providence, RI, 2004.
  * [13] M. Gerstenhaber, On the deformation of rings and algebras, Ann. Math., 79, 59-103, 1964.
  * [14] M. Goze and E. Remm, Poisson algebras in terms of non-associative algebras, J. Alg. 320 (2008) 294-317.
  * [15] G. Graziani, A. Makhlouf, C. Menini and F. Panaite, BiHom-Associative Algebras, BiHom-Lie Algebras and BiHom-Bialgebras. SIGMA Symmetry Integrability Geom. Methods Appl. 11 (2015), Paper 086, 34 pp.
  * [16] M. Goze, and E. Remm,  Poisson algebras in terms of non-associative algebras, J. Algebra., 320 (1), 294-317, 2008.
  * [17] M. Kontsevich,  Deformation quantization of Poisson manifolds, Lett. Math. Phys., 66, 157-216, 2003.
  * [18] G. Graziani, A. Makhlouf, C. Menini, F. Panaite,  BiHom-associative algebras, BiHom-Lie algebras and BiHom-bialgebras,  Symmetry, Integrability and geometry, SIGMA 11 (2015), 086, 34 pages.
  * [19] S. Guo, S. Wang, On split regular BiHom-Leibniz superalgebras. arXiv:1903.12474.
  * [20] S. Guo, X. Zhang and S. Wang, The construction and deformation of BiHom-Novikov algebras. J. Geom. Phys. 132 (2018), 460-472.
  * [21] J. T. Hartwig, D. Larsson, S. D. Silvestrov, _Deformations of Lie algebras using $\sigma$-derivations_, J. Algebra 295 (2006), 314–361.
  * [22] J. T. Hartwig, D. Larsson and S. D. Silvestrov, Deformations of Lie algebras using $\sigma$-derivations. J. Algebra. 292 (2006), 314-361.
  * [23] A. B. Hassine, S. Mabrouk, O. Ncib, $3$-BiHom-Lie superalgebras induced by BiHom-Lie superalgebras.arXiv:1905.04518.
  * [24] N. Hu, _$q$ -Witt algebras, $q$-Lie algebras, $q$-holomorph structure and representations_, Algebra Colloq. 6 (1999), 51–70.
  * [25] D. Larsson, S. D. Silvestrov, _Quasi-hom-Lie algebras, central extensions and $2$-cocycle-like identities_, J. Algebra 288 (2005), 321–344.
  * [26] Laraiedh, I.: Construction and $T^{\ast}$-extensions of $3$-BiHom-Lie superalgebras, arXiv:2004.08823.
  * [27] J. Lü , X. Wang , G. Zhuang, Universal enveloping algebras of Poisson Hopf algebras, Journal of Algebra 426 (2015) 92-136.
  * [28] J. Li, L. Chen, The construction of $3$-Bihom-Lie algebras, arXiv:2001.10348.
  * [29] Xiaojie Li, BiHom-Poisson Algebra and Its Application, Int. J. Alg. 13 (2019), 73-81.
  * [30] L. Liu, A. Makhlouf, A. C. Menini , F. Panaite,Rota-Baxter operators on BiHom-associative algebras and related structures, Colloq. Math, (2020).
  * [31] A. Makhlouf, S.D. Silvestrov, _Hom-algebras structures_ , J. Gen. Lie Theory Appl. 2 (2008), 51–64.
  * [32] M. Markl and E. Remm, Algebras with one operation including Poisson and other Lie-admissible algebras, J. Alg. 299 (2006) 171-189.
  * [33] P. Schaller and T. Strobl, Poisson structure induced (topological) field theories, Mod. Phys. Lett. A 9, 3129-3136, 1994.
  * [34] I. Vaisman,  Lectures on the geometry of Poisson manifolds, Birkhäuser, Basel, 1994.
  * [35] D. Yau, Non-commutative Hom-Poisson algebras, e-Print arXiv:1010.3408 (2010).
  * [36] C. Yongsheng, and H. Qi, Representations of Bihom-Lie algebras. arXiv:1610.04302 (2016).