\betaegin{document}

\title{Structures of BiHom-Poisson algebras}

\alphauthor{ Sylvain Attan\thanks{D\'{e}partement de Math\'{e}matiques, Universit\'{e} d'Abomey-Calavi
01 BP 4521, Cotonou 01, B\'{e}nin. E.mail: syltane2010@yahoo.fr}\alphand Ismail Laraiedh\thanks{Departement of Mathematics, Faculty of Sciences, Sfax University, BP 1171, 3000 Sfax, Tunisia. E.mail:
Ismail.laraiedh@gmail.com and Departement of Mathematics, College of Sciences and Humanities - Kowaiyia, Shaqra University,
Kingdom of Saudi Arabia. E.mail:
ismail.laraiedh@su.edu.sa}}

\muaketitle

\betaegin{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.
\end{abstract}
{\betaf 2010 Mathematics Subject Classification:} 17A20, 17A30,  17B05, 17B10, 17B65 .
\\
{\betaf Keywords:} BiHom-Poisson algebras, BiHom-Poisson admissible, generalized
derivation,\\ BiHom-Poisson modules.
\muathrm{e}ction{Introduction}
A Poisson algebra $(P,\{\cdot,\cdot\}, \muu)$ consists of a commutative associative algebra
$(A,\muu)$\\ together with a Lie structure $\{\cdot,\cdot\},$ satisfying the Leibniz identity:
$$\{\muu(x,y),z\}=\muu(\{x,z\},y)+\muu(x,\{y,z\})$$
 These algebras firstly appeared in the work of Sim\'eon-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 \cite{iv}, quantum groups \cite{vca},\cite{vgd} and deformation of commutative associative algebras \cite{mg} whereas in physics, Poisson algebras represent a major part of deformation quantization \cite{mk}, Hamiltonian mechanics \cite{via} and topological field theories \cite{pct}. Poisson-like structures are also used in the study of vertex operator algebras \cite{efd}.

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
(\cite{AizawaSaito}, \cite{Hu}). 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 \cite{JDS}, \cite{DS}. The associative counterpart of Hom-Lie algebras
(called Hom-associative algebras) has been introduced in \cite{ms}, 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
$\alphalpha$ and $\betaeta$. This class of algebras was introduced from a categorical approach in \cite{Gra-Mak-Men-Pan} 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 \cite{XLi} and  studied in 
\cite{hadimi} 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 \cite{S S, Guo1, chtioui1,chtioui2, ismail, Li C, Liu2}.

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 $\muathbb{K}$ is an algebraically closed field of characteristic $0$ and $A$ is a $\muathbb{K}$-vector space.
\muathrm{e}ction{Preliminaries}

This section contains necessary important basic notions and notations which will be used in next sections.
For the map $\muu: A^{\otimes 2}\lambdaongrightarrow A,$ we will sometimes $\muu(a\otimes b)$ as $\muu(a,b)$ or $ab$  for $a,b\in A$ and if $V$ is another vector space, $\tau_1: A\otimes V\lambdaongrightarrow V\otimes A$ (resp. $\tau_2: V\otimes A\lambdaongrightarrow A\otimes V$) denote the twist isomorphism $\tau_1(a\otimes v)=v\otimes a$ (resp.
$\tau_2(v\otimes a)=a\otimes v$).
\betaegin{defn}
A BiHom-module is a pair $(M,\alphalpha_M,\betaeta_M)$ consisting of a $\muathbb{K}$-module $M$ and
a linear self-maps $\alphalpha_M, \betaeta_M: M\lambdaongrightarrow M$  such that $\alphalpha_M\betaeta_M=\betaeta_M\alphalpha_M.$ A morphism\\
$f: (M,\alphalpha_M, \betaeta_M)\rhoightarrow (N,\alphalpha_N,\betaeta_N)$ of BiHom-modules is a linear map
 $f: M\lambdaongrightarrow N$ such that $f\alphalpha_M=\alphalpha_N f$ and
 $f\betaeta_M=\betaeta_N f.$
\end{defn}
\betaegin{defn}\cite{GRAZIANI}
A BiHom-algebra is a quadruple $(A,\muu,\alphalpha,\betaeta)$ in which $(A,\alphalpha,\betaeta)$ is a BiHom-module, $\muu : A^{\otimes 2} \rhoightarrow A$ is a linear map.
The BiHom-algebra $(A,\muu,\alphalpha;\betaeta)$ is said to be  multiplicative if $\alphalpha\circ\muu=\muu\circ\alphalpha^{\otimes 2}$ and $\betaeta\circ\muu=\muu\circ\betaeta^{\otimes 2}$ (multiplicativity).
\end{defn}
\betaegin{defn}
\betaegin{enumerate}
\item A BiHom-algebra $(A,\muu,\alphalpha,\betaeta)$ is said to be BiHom-commutative if
$$\muu(\betaeta(x),\alphalpha(y))=\muu(\betaeta(y),\alphalpha(x)),~\forall x,y\in A$$.
\item A BiHom-associative algebra \cite{GRAZIANI} is a multiplicative Bihom-algebra $(A,\muu,\alphalpha,\betaeta)$
satisfying the following BiHom-associativity condition:
\betaegin{equation}\lambdaabel{BiHom ass identity}
    as_{A}(x,y,z):=\muu(\muu(x,y),\betaeta(z))-\muu(\alphalpha(x),\muu(y,z))=0,\ \text{for all}\ x,y,z \in A.\
\end{equation}
\item A BiHom-Lie algebra \cite{GRAZIANI} is a multiplicative Bihom-algebra $(A,\{\cdot,\cdot\},\alphalpha,\betaeta)$ satisfying the  BiHom-skew-symmetry and the BiHom-Jacobi identities i.e.
\betaegin{eqnarray}
&&\{\betaeta(x),\alphalpha(y)\}=-\{\betaeta(y),\alphalpha(x)\}  \nonumber\\
&&\circlearrowleft_{x,w,z}\{\betaeta^2(x),\{\betaeta(y),\alphalpha(z)\}\}=0 \lambdaabel{BiHJacob}
\end{eqnarray}
where $\circlearrowleft_{x,y,z}$ denotes the summation over the cyclic permutation on $x,y,z.$
\end{enumerate}
\end{defn}
Clearly, a Hom-associative algebra $(A,\muu,\alphalpha)$ can be regarded as a BiHom-associative algebra $(A,\muu,\alphalpha,\alphalpha)$.
\betaegin{defn}\lambdaabel{si}\cite{ XLi}
A BiHom-Poisson algebra consists of a  vector space $A,$ two  bilinear maps
$\muu, \{\cdot,\cdot\}: A^{\otimes2}\lambdaongrightarrow A,$  linear maps $\alphalpha,\betaeta: A\lambdaongrightarrow A$
 such that
\betaegin{enumerate}
\item
$(A,\muu,\alphalpha, \betaeta)$ is a BiHom-commutative BiHom-associative algebra,
\item
$(A,\{\cdot,\cdot\},\alphalpha, \betaeta)$ is a BiHom-Lie algebra,
\item
the BiHom-Leibniz identity
\betaegin{eqnarray}
\{\alphalpha\betaeta(x),\muu(y,z)\}=\muu(\{\betaeta(x),y\},\betaeta(z))+\muu(\betaeta(y),\{\alphalpha(x),z\})\lambdaabel{BHL}
\end{eqnarray}
is satisfied for all $x,y,z\in A.$
\end{enumerate}
\end{defn}
  In a BiHom-Poisson algebra $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$, the operations $\muu$ and $\{\cdot,\cdot\}$ are called the
  BiHom-associative product and the BiHom-Poisson bracket, respectively.
\betaegin{rmk}
  A non-BiHom-commutative BiHom-Poisson algebra is a BiHom-Poisson algebra without the BiHom-commutativity assumption \cite{hadimi}. These Bihom-algebras are called BiHom-Poisson algebras \cite{XLi}.
  \end{rmk}
  \betaegin{defn}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ a BiHom-Poisson algebra. A subspace $H$ of $A$ is called
\betaegin{enumerate}
\item
A BiHom-subalgebra of $A$ if \[\alphalpha(H)\sigmaubseteq H,~\betaeta(H)\sigmaubseteq H,~\muu(H,H)\sigmaubseteq H ~\textrm{and} ~\{H,H\}\sigmaubseteq H.\]
\item
A left-BiHom ideal of $A$ if \[\alphalpha(H)\sigmaubseteq H,~\betaeta(H)\sigmaubseteq H,~\muu(A,H)\sigmaubseteq H~\textrm{and} ~\{A,H\}\sigmaubseteq H.\]
\item
A right-BiHom ideal of $A$ if \[\alphalpha(H)\sigmaubseteq H,~\betaeta(H)\sigmaubseteq H,~\muu(H,A)\sigmaubseteq H~\textrm{and} ~\{H,A\}\sigmaubseteq H.\]
\item
A two sided BiHom-ideal if $H$ is both a left and a right BiHom-ideal of $A$.
  \end{enumerate}
  Note that, if $\alphalpha$ and $\betaeta$ are bijective, then the notion of left-BiHom ideals is equivalent to the one of right-BiHom ideals.
\betaegin{defn}
Let $(A,\{\cdot,\cdot\}, \muu, \alphalpha,\betaeta)$ be a Bihom-Poisson algebra. If $Z(A) = \{x \in A |~ \{x,y\}=\muu(x, y) = 0,~~\forall y \in A\}$, then $Z(A)$ is called the centralizer of $A$.
\end{defn}
  \end{defn}
\betaegin{defn}
\lambdaabel{def:morphism}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ a BiHom-Poisson algebra.
\betaegin{enumerate}
\item
$A$ is multiplicative if
\[
\alphalpha\{\cdot,\cdot\} = \{\cdot,\cdot\} \alphalpha^{\otimes 2},~~\betaeta\{\cdot,\cdot\} = \{\cdot,\cdot\} \betaeta^{\otimes 2} \quad\text{and}\quad \alphalpha\muu = \muu\alphalpha^{\otimes 2},~~\betaeta\muu = \muu\betaeta^{\otimes 2}.
\]
  \item $(A,\muu,\alphalpha,\betaeta)$ is said to be regular if $\alphalpha$ and $\betaeta$ are algebra automorphisms.
  \item $(A,\muu,\alphalpha,\betaeta)$ is said to be involutive if $\alphalpha$ and $\betaeta$ are two involutions,  that is $\alphalpha^2=\betaeta^2=id$.
\item
Let $(A',\{\cdot,\cdot\}',\muu',\alphalpha',\betaeta')$ be another BiHom-Poisson algebra. A weak morphism $f \colon A \to A'$ is a linear map such that
\[
f\{\cdot,\cdot\} = \{\cdot,\cdot\}' f^{\otimes 2} \quad\text{and}\quad
f\muu = \muu'f^{\otimes 2}.
\]
A morphism $f \colon A \to A'$ is a weak morphism such that $f\alphalpha = \alphalpha'f$ and $f\betaeta=\betaeta' f$.
\end{enumerate}
\end{defn}

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

Denote by $\Gamma_{f}=\{x+f(x);~~x\in A\}\sigmaubset A\oplus A'$  the graph of a linear map $f:A\lambdaongrightarrow A^{'}$.
\betaegin{defn} Let $(A,\muu,\alphalpha,\betaeta)$ be any BiHom-algebra.
\betaegin{enumerate}
\item 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
$\muathrm{h}o_{l}: (A\otimes V, \alphalpha\otimes\phi,\betaeta\otimes\psi)\rhoightarrow (V,\phi,\psi),~a\otimes v\muapsto a.v$ and $\muathrm{h}o_{r}:(V\otimes A,\phi\otimes\alphalpha,\psi\otimes\psi)\rhoightarrow (V,\phi,\psi),~v\otimes a\muapsto v.a$ of Bihom-modules.
\item
A morphism $f:(V,\phi,\psi,\muathrm{h}o_{l},\muathrm{h}o_{r})\rhoightarrow (W,\phi',\psi',\muathrm{h}o_{l}',\muathrm{h}o_{r}')$ of $A$-bimodules is a morphism of the underlying BiHom-modules such that
$$f\circ\muathrm{h}o_{l}=\muathrm{h}o_{l}'\circ(Id_{A}\otimes f)~~and~~f\circ\muathrm{h}o_{r}=\muathrm{h}o_{r}'\circ( f \otimes Id_{A}).$$
That yields the commutative diagrams
  $$
\xymatrix{
 A\otimes V \alphar[d]_{  Id_{A}\otimes f}\alphar[rr]^{\muathrm{h}o_l}
                && V  \alphar[d]^{f}  \\
 A\otimes W \alphar[rr]^{\muathrm{h}o_l'}
                && W}\quad\xymatrix{
 V\otimes A \alphar[d]_{ f\otimes Id_{A} }\alphar[rr]^{\muathrm{h}o_r}
                && V  \alphar[d]^{f}  \\
 W\otimes A \alphar[rr]^{\muathrm{h}o_r'}
                && W  }
$$

\end{enumerate}
\end{defn}

Now, let consider the following notions for BiHom-associative algebras.
\betaegin{defn}
Let $(A,\muu,\alphalpha,\betaeta)$ be a BiHom-associative algebra, 
$(L,\{\cdot,\cdot\},\alphalpha,\betaeta)$ be a Hom-Lie algebra
 and $(V, \phi,\psi)$ be a BiHom-module. Then
\betaegin{enumerate}
\item A left BiHom-associative $A$-module structure on $V$ \cite{GRAZIANI} consists of a morphism $\muathrm{h}o_{l}: A\otimes V\lambdaongrightarrow V$ of BiHom-modules, such that
\betaegin{eqnarray}\lambdaabel{00}
\muathrm{h}o_{l}\circ(\alphalpha\otimes\muathrm{h}o_{l})=\muathrm{h}o_{l}\circ(\muu\otimes\psi).\lambdaabel{LeftAssMod}
\end{eqnarray}
In terms of diagram, we have
  $$
\xymatrix{
 A\otimes A\otimes V \alphar[d]_{ \muu\otimes\psi}\alphar[rr]^{\alphalpha\otimes\muathrm{h}o_{l}}
                && A\otimes V  \alphar[d]^{\muathrm{h}o_{l}}  \\
 A\otimes V \alphar[rr]^{\muathrm{h}o_{l}}
                && V  }$$
\item A right BiHom-associative $A$-module structure on $V$ \cite{GRAZIANI} consists of a morphism $\muathrm{h}o_{r}: V\otimes A\lambdaongrightarrow V$ of BiHom-modules, such that
\betaegin{eqnarray}\lambdaabel{11}
\muathrm{h}o_{r}\circ(\phi\otimes\muu)=\muathrm{h}o_{r}\circ(\muathrm{h}o_{r}\otimes\betaeta).\lambdaabel{RightAssMod}
\end{eqnarray}
In terms of diagram, we have
 $$
\xymatrix{
V\otimes A\otimes A  \alphar[d]_{ \phi\otimes\muu}\alphar[rr]^{\muathrm{h}o_{r}\otimes\betaeta}
                && V\otimes A  \alphar[d]^{\muathrm{h}o_{r}}  \\
 V\otimes A \alphar[rr]^{\muathrm{h}o_{r}}
                && V  }$$
\item A left BiHom-Lie $L$-module structure on $V$ \cite{hadimi}\cite{cyong} consists of a structure map $\muathrm{h}o: L\otimes V\lambdaongrightarrow V$ 
such that 
\betaegin{eqnarray}
\muathrm{h}o(\{\betaeta(x),y \},\psi(v))&=&\muathrm{h}o(\alphalpha\betaeta(x),\muathrm{h}o(y,v))-\muathrm{h}o(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))\lambdaabel{Li1}
\end{eqnarray}
\end{enumerate}
\end{defn}
\muathrm{e}ction{BiHom-Poisson algebras}

\sigmaubsection{Constructions of BiHom-Poisson algebras}
In this subsection, we provide
some constructions results of BiHom-Poisson algebras.
\betaegin{prop}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a BiHom-Poisson algebra and $I$ be a two-sided BiHom-ideal of $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$.
Then $(A/I,[\cdot,\cdot],\overline{\muu},\overline{\alphalpha},\overline{\betaeta})$ is a BiHom-Poisson algebra where $[\overline{x},\overline{y}] =\overline{\{x,y\}} ,~\overline{\muu}(\overline{x},\overline{y})=\overline{\muu(x,y)}$,
$\overline{\alphalpha}(\overline{x})=\overline{\alphalpha(x)}$  and $\overline{\betaeta}(\overline{x})=\overline{\betaeta(x)}$, for all $\overline{x},\overline{y}\in A/I$
\betaegin{proof}
We only prove item 1. of definition \rhoef{si}, item 2. and item 3. are being proved similarly

For all $\overline{x},\overline{y},\overline{z}\in A/I$ we have
$$\betaegin{array}{llllll}
as_{A/I}(\overline{x},\overline{y},\overline{z})&=&\overline{\muu}(\overline{\muu}(\overline{x},\overline{y}),\overline{\betaeta}(\overline{z}))-\overline{\muu}(\overline{\alphalpha}(\overline{x}),\overline{\muu}(\overline{y},\overline{z}))\\[0.2cm]
&=&\overline{\muu(\muu(x,y),\betaeta(z))-\muu(\alphalpha(x),\muu(y,z))}\\[0.2cm]
&=&\overline{0}~(by~ BiHom-associativity ~of ~A).
\end{array}$$
Then $(A/I,\overline{\muu},\overline{\alphalpha},\overline{\betaeta})$ is a BiHom-associative algebra.
$$\betaegin{array}{llll}
\overline{\muu}(\overline{\betaeta}(\overline{x}),\overline{\alphalpha}(\overline{y}))
&=&\overline{\muu(\betaeta(x),\alphalpha(y))}\\[0.2cm]&=&\overline{\muu(\betaeta(y),\alphalpha(x))}~(by~BiHom-commutativity ~of A).\\[0.2cm]
&=&\overline{\muu}(\overline{\betaeta}(\overline{y}),\overline{\alphalpha}(\overline{x}))
\end{array}$$Then $(A/I,\overline{\muu},\overline{\alphalpha},\overline{\betaeta})$ is a BiHom-commutative algebra.
\end{proof}

\end{prop}
\betaegin{prop}\lambdaabel{adimi}\cite{hadimi}
Let $(A, \muu, \alphalpha,\betaeta)$ be a regular BiHom-associative algebra. Then\\
$A^{-} = (A, \{\cdot,\cdot \}, \muu, \alphalpha,\betaeta)$
is a regular non-BiHom-commutative BiHom-Poisson algebra, where $\{\cdot,\cdot\}=\muu-\muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau$,

where for all $x,y\in A,~~\tau(x\otimes y)=y\otimes x$.
\end{prop}
\betaegin{ex} \lambdaabel{e1}
Consider the 2-dimensional regular BiHom-associative algebras $(A,\muu,\alphalpha,\betaeta)$ with a basis $(e_1,e_2),$ (see \cite{GRAZIANI}) defined by

$$\betaegin{array}{llllll}
  \alphalpha(e_1)=e_1,&&\alphalpha(e_2)=\frac{b(1-a)}{a}e_1+a e_2, \\
 \betaeta(e_1)=e_1,&& \betaeta(e_2)=be_1+(1-a) e_2, \\
  \muu(e_1,e_1)=e_1, && \muu(e_1,e_2)= be_1+(1-a) e_2, \\
 \muu(e_2,e_1)=\frac{b(1-a)}{a}e_1+ae_2, && \muu(e_2,e_2)=\frac{b}{a}e_2,
\end{array}$$
where $a$, $b$ are parameters in $\muathbb{K}$, with $a\neq 0,1$.
Using the Proposition \rhoef{adimi}, the 5-tuple $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ is a regular  non-BiHom-commutative BiHom-Poisson algebra where, $\{\cdot,\cdot\}=\muu-\muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau$ and
$$\betaegin{array}{llllll}
 \alphalpha^{-1}(e_1)=e_1,&&\alphalpha^{-1}(e_2)=\frac{b(a-1)}{a^{2}}e_1+\frac{1}{a} e_2, \\
 \betaeta^{-1}(e_1)=e_1,&& \betaeta^{-1}(e_2)=\frac{b}{a-1}e_1+\frac{1}{1-a} e_2.
\end{array}$$

\end{ex}

\betaegin{prop}\lambdaabel{bylv}\cite{ XLi}
 Let $(A,\{\cdot,\cdot\}_{A},\muu_A,\alphalpha_{A},\betaeta_{A})$ and
$(B,\{\cdot,\cdot\}_{B},\muu_B,\alphalpha_{B},\betaeta_{B})$ be two BiHom-Poisson algebras. Then there exists a BiHom-Poissson algebra $(A\oplus B,\{\cdot,\cdot\},\muu, \alphalpha=\alphalpha_{A}+\alphalpha_{B},\betaeta=\betaeta_{A}+\betaeta_{B}),$ where the bilinear maps $\{\cdot,\cdot\},\muu:(A\oplus B)^{\times 2}\lambdaongrightarrow (A\oplus B)$ are given by
 $$\betaegin{array}{llll}
 \muu(a_1+b_1,a_2+b_2)=\muu_A(a_1,a_2)+\muu_B(b_1,b_2),\nonumber\\
 \{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.

 \end{array}$$
 and the linear maps $\betaeta=\betaeta_{A}+\betaeta_{B},~\alphalpha=\alphalpha_{A}+\alphalpha_{B}: (A\oplus B)\lambdaongrightarrow (A\oplus B)$ are given by
$$ \betaegin{array}{lll}
(\alphalpha_{A}+\alphalpha_{B})(a+b)&=& \alphalpha_{A}(a)+\alphalpha_{B}(b),\\
(\betaeta_{A}+\betaeta_{B})(a+b)&=& \betaeta_{A}(a)+\betaeta_{B}(b),~~ \forall \ (a,b)\in A\times B.
 \end{array}$$
 \end{prop}
\betaegin{proof}
It is easy to see that $(A\oplus B,\muu, \alphalpha_{A}+\alphalpha_{B},\betaeta_{A}+\betaeta_{B})$ is a BiHom-associative and BiHom-commutative algebra and
$(A\oplus B,\{\cdot,\cdot\}, \alphalpha_{A}+\alphalpha_{B},\betaeta_{A}+\betaeta_{B})$ is a BiHom-Lie algebra. Then $(A\oplus B,\{\cdot,\cdot\},\muu, \alphalpha_{A}+\alphalpha_{B},\betaeta_{A}+\betaeta_{B})$ is a BiHom-Poisson algebra.
\end{proof}
\betaegin{prop}
Let $(A,\{\cdot,\cdot\}_{A},\muu_A,\alphalpha_{1},\betaeta_{1})$ and $(B,\{\cdot,\cdot\}_{B},\muu_B,\alphalpha_{2},\betaeta_{2})$ be two BiHom-Poisson algebras and $\varphi: A\rhoightarrow B$ be a linear map. Then
$\varphi$ is a morphism from $(A,\{\cdot,\cdot\}_{A},\muu_A,\alphalpha_{1},\betaeta_{1})$ to $(B,\{\cdot,\cdot\}_{B},\muu_B,\alphalpha_{2},\betaeta_{2})$  if and only if its graph
$\Gamma_{\varphi}$ is a BiHom-subalgebra of $(A\oplus B, \{\cdot,\cdot\},\muu, \alphalpha,\betaeta).$
\end{prop}
\betaegin{proof}
  Let $\varphi: (A, \muu_A, \alphalpha_{1},\betaeta_{1})\lambdaongrightarrow (B, \muu_B ,\alphalpha_{2},\betaeta_{2})$ be a morphism of BiHom-Poisson algebras.
Then for all $u, v\in A,$
$$ \betaegin{array}{llllllllll}\muu(u+\varphi(u),v+\varphi(v))=(\muu_A(u,v)+\muu_B(\varphi(u),\varphi(v)))=(\muu_A(u,v)+\varphi(\muu_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}
$$
Thus the graph $\Gamma_{\varphi}$ is closed under the operations $\muu$ and $\{\cdot,\cdot\}$.
Furthermore since $\varphi\circ\alphalpha_{1}=\alphalpha_{2}\circ\varphi$, we have $(\alphalpha_{1}\oplus\alphalpha_{2})(u, \varphi(u)) = (\alphalpha_{1}(u),
\alphalpha_{2}\circ\varphi(u)) = (\alphalpha_{1}(u), \varphi\circ\alphalpha_{1}(u)).$ In the same way, we have $(\betaeta_{1}\oplus\betaeta_{2})(u, \varphi(u)) = (\betaeta_{1}(u),
\betaeta_{2}\circ\varphi(u)) = (\betaeta_{1}(u), \varphi\circ\betaeta_{1}(u)),$
which implies that $\Gamma_{\varphi}$ is closed under $\alphalpha_{1}\oplus\alphalpha_{2}$ and $\betaeta_{1}\oplus\betaeta_{2}$ Thus $\Gamma_{\varphi}$ is a BiHom-subalgebra of
$(A\oplus B,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta).$\\
Conversely, if the graph $\Gamma_{\varphi}\sigmaubset A\oplus B$ is a
BiHom-subalgebra of
$(A\oplus B,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ then we have
$$\betaegin{array}{llllll}\muu(u+ \varphi(u)), v+ \varphi(v))=(\muu_A(u, v) + \muu_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}

\end{array}
$$
which implies that
$$\betaegin{array}{llll}\muu_B(\varphi(u), \varphi(v))=\varphi(\muu_A(u, v)),\\
\{\varphi(u), \varphi(v)\}_{B}=\varphi(\{u, v\}_{A}).
\end{array}
$$
Furthermore, $(\alphalpha_{1}\oplus\alphalpha_{2})(\Gamma_{\varphi})\sigmaubset\Gamma_{\varphi},~(\betaeta_{1}\oplus\betaeta_{2})(\Gamma_{\varphi})\sigmaubset\Gamma_{\varphi}$ implies
$$(\alphalpha_{1}\oplus\alphalpha_{2})(u, \varphi(u))=(\alphalpha_{1}(u), \alphalpha_{2}\circ\varphi(u)) \in\Gamma_{\varphi},$$
$$(\betaeta_{1}\oplus\betaeta_{2})(u, \varphi(u))=(\betaeta_{1}(u), \betaeta_{2}\circ\varphi(u)) \in\Gamma_{\varphi}.$$
which is equivalent to the condition $\alphalpha_{2}\circ\varphi(u)=\varphi\circ\alphalpha_{1}(u),$ i.e. $ \alphalpha_{2}\circ\varphi=\varphi\circ\alphalpha_{1}$. Similarly, $ \betaeta_{2}\circ\varphi=\varphi\circ\betaeta_{1}$. Therefore, $\varphi$ is a
morphism of BiHom-Poisson algebras.
\end{proof}

\betaegin{thm}
\lambdaabel{thm:twist}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a (non-BiHom-commutative) BiHom-Poisson algebra and $\alphalpha',\betaeta' \colon A \to A$ be
endomorphisms of $A$ such that any two of the maps $\alphalpha,\betaeta,\alphalpha',\betaeta'$ commute. Then
\[
A_{\alphalpha',\betaeta'} = (A,\{\cdot,\cdot\}_{\alphalpha',\betaeta'} =\{\cdot,\cdot\}\circ(\alphalpha'\otimes\betaeta'),\muu _{\alphalpha',\betaeta'} = \muu\circ(\alphalpha'\otimes\betaeta'),\alphalpha\alphalpha',\betaeta\betaeta')
\]
is also a (non-BiHom-commutative) BiHom-Poisson algebra. Moreover suppose that\\ $(B,\{\cdot,\cdot\}', \muu', \gamma,\deltaelta)$ is another BiHom-Poisson algebra and $\gamma',\deltaelta'$ be
endomorphisms of $B$ such that any two of the maps $\gamma,\deltaelta,\gamma',\deltaelta'$ commute.
If $f : (A,\{\cdot,\cdot\},\muu, \alphalpha,\betaeta)\rhoightarrow (B,\{\cdot,\cdot\}', \muu', \gamma,\deltaelta)$
is a morphism such that $f\circ\alphalpha'=\gamma'\circ f$ and $f\circ\betaeta'=\deltaelta'\circ f$, then $f : A_{\alphalpha',\betaeta'}\rhoightarrow B_{\gamma',\deltaelta'}$ is also a morphism.
\end{thm}

\betaegin{proof} Let give the proof in BiHom-commutativity case.
We only prove item 1. of Definition \rhoef{si}, item 2. and item 3. can be proved similarly.

For all $x,y,z\in A$ we have
$$\betaegin{array}{llll}
&&as_{A_{\alphalpha',\betaeta'}}(x,y,z)\\
&=&\muu_{\alphalpha',\betaeta'}(\muu_{\alphalpha',\betaeta'}(x,y),\betaeta\betaeta'(z))-\muu_{\alphalpha',\betaeta'}(\alphalpha\alphalpha'(x),\muu_{\alphalpha',\betaeta'}(y,z))\\
&=&\muu_{\alphalpha',\betaeta'}(\muu(\alphalpha'(x),\betaeta'(y)),\betaeta\betaeta'(z))-\muu_{\alphalpha',\betaeta'}(\alphalpha\alphalpha'(x),\muu(\alphalpha'(y),\betaeta'(z)))\\
&=&\muu(\muu(\alphalpha'^{2}(x),\alphalpha'\betaeta'(y)),\betaeta\betaeta'^{2}(z))-\muu(\alphalpha\alphalpha'^{2}(x),\muu(\alphalpha'\betaeta'(y),\betaeta'^{2}(z)))\\
&=&as_{A}(\alphalpha'^{2}(x),\alphalpha'\betaeta'(y),\betaeta'^{2}(z))=0~(BiHom-associativity~condition~of~A).
\end{array}$$
Then $(A,\muu _{\alphalpha',\betaeta'},\alphalpha\alphalpha',\betaeta\betaeta')$ is a BiHom-associative algebra.

Now, for all $x,y\in A$ we have:
$$\betaegin{array}{llllllll}
\muu_{\alphalpha',\betaeta'}(\betaeta\betaeta'(x),\alphalpha\alphalpha'(y))&=&\muu(\alphalpha'\betaeta'\betaeta(x),\alphalpha\alphalpha'\betaeta'(y))\\
&=&\muu(\betaeta(\alphalpha'\betaeta'(x)),\alphalpha(\alphalpha'\betaeta'(y)))\\
&=&\muu(\betaeta(\alphalpha'\betaeta'(y)),\alphalpha(\alphalpha'\betaeta'(x)))~(BiHom-commutativity~in~A).\\
&=&\muu_{\alphalpha',\betaeta'}(\betaeta\betaeta'(y),\alphalpha\alphalpha'(x)).
\end{array}$$
Then $(A,\muu _{\alphalpha',\betaeta'},\alphalpha\alphalpha',\betaeta\betaeta')$ is a BiHom-commutative algebra.

The second part is proved as follows: $\forall x,y\in A$
$$\betaegin{array}{llllllll}
f\{x,y\}_{\alphalpha',\betaeta'}&=&f\{\alphalpha'(x),\betaeta'(y)\}\\
&=&\{f\alphalpha'(x),f\betaeta'(y)\}'\\
&=&\{\gamma'f(x),\deltaelta' f(y)\}'\\
&=&\{f(x),f(y)\}'_{\gamma',\deltaelta'}.
\end{array}$$
In the same way we have
$f\muu_{\alphalpha',\betaeta'}(x,y)=\muu'_{\gamma',\deltaelta'}(f(x),f(y)).$

This finishes the proof.
\end{proof}
Taking $\alphalpha'=\alphalpha^{k},~\betaeta'=\betaeta^{l}$, yields the following statement.
\betaegin{cor}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a (non-BiHom-commutative) BiHom-Poisson algebra. Then
\[
A_{\alphalpha^{k},\betaeta^{l}} = (A,\{\cdot,\cdot\}_{\alphalpha^{k},\betaeta^{l}} =\{\cdot,\cdot\}\circ(\alphalpha^{k}\otimes\betaeta^{l}),\muu _{\alphalpha^{k},\betaeta^{l}} = \muu\circ(\alphalpha^{k}\otimes\betaeta^{l}),\alphalpha^{k+1},\betaeta^{l+1})
\]
is also a (non-BiHom-commutative) BiHom-Poisson algebra.
\end{cor}
Taking $\alphalpha=\betaeta=id$, yields the following statement.
\betaegin{cor}
Let $(A,\{\cdot,\cdot\},\muu)$ be a (non-commutative) Poisson algebra and $\alphalpha,\betaeta \colon A \to A$ be two
endomorphisms such that $\alphalpha\betaeta=\betaeta\alphalpha$. Then
\[
A_{\alphalpha,\betaeta} = (A,\{\cdot,\cdot\}_{\alphalpha,\betaeta} =\{\cdot,\cdot\}\circ(\alphalpha\otimes\betaeta),\muu _{\alphalpha,\betaeta} = \muu\circ(\alphalpha\otimes\betaeta),\alphalpha,\betaeta)
\]
is also a (non-BiHom-commutative) BiHom-Poisson algebra.
\end{cor}
\betaegin{defn}
\lambdaabel{def:deform}
Let $(A,\{\cdot,\cdot\},\muu)$ be a non-commutative Poisson algebra.
\betaegin{enumerate}
\item
Given two commuting morphisms $\alphalpha,\betaeta \colon A \to A$, the triple $A'_{\alphalpha,\betaeta}=(A,\{\cdot,\cdot\}_{\alphalpha,\betaeta} =\{\cdot,\cdot\}\circ(\alphalpha\otimes\betaeta),\muu _{\alphalpha,\betaeta} = \muu\circ(\alphalpha\otimes\betaeta))$ is called the $(\alphalpha,\betaeta)$-twisting of $A$.  A twisting of $A$ is a $(\alphalpha,\betaeta)$-twisting of $A$ for some morphisms $\alphalpha,\betaeta \colon A \to A$.
\item
The $(\alphalpha,\betaeta)$-twisting $A'_{\alphalpha,\betaeta}$ of $A$ is called trivial if
\[
\{\cdot,\cdot\}_{\alphalpha,\betaeta}= 0 = \muu_{\alphalpha,\betaeta}.
\]
$A'_{\alphalpha,\betaeta}$ is called non-trivial if either $\{\cdot,\cdot\} \neq 0$ or $\muu_{\alphalpha,\betaeta} \neq 0$.
\item
$A$ is called rigid if every twisting of $A$ is either trivial or isomorphic to $A$.
\end{enumerate}
\end{defn}

\betaegin{prop}
\lambdaabel{prop:nonrigidity}
Let $(A,\{\cdot,\cdot\},\muu)$ be a non-commutative Poisson algebra.  Suppose there exists two commuting morphisms $\alphalpha,\betaeta \colon A \to A$ such that either:
\betaegin{enumerate}
\item
$\muu_{\alphalpha,\betaeta} = \muu\circ(\alphalpha\otimes\betaeta)$ is not BiHom-associative or
\item
$\{\cdot,\cdot\}_{\alphalpha,\betaeta}=\{\cdot,\cdot\}\circ(\alphalpha\otimes\betaeta)$ does not satisfy the BiHom-Jacobi identity.
\end{enumerate}
Then $A$ is not rigid.
\end{prop}

\betaegin{proof}
The $(\alphalpha,\betaeta)$-twisting $A'_{\alphalpha,\betaeta}$ is non-trivial, since otherwise $\muu_{\alphalpha,\betaeta}$ would be BiHom-associative and $\{\cdot,\cdot\}_{\alphalpha,\betaeta}$ would satisfy the BiHom-Jacobi identity. For the same reason, the $(\alphalpha,\betaeta)$-twisting $A'_{\alphalpha,\betaeta}$ cannot be isomorphic to $A$.
\end{proof}
\betaegin{rmk}
\lambdaabel{ex:linear}
Let $(\muathfrak{g},[\cdot,\cdot])$ be a finite-dimensional Lie algebra, and let $(S(\muathfrak{g}),\muu)$ be its symmetric algebra.  If $\{e_i\}_{i=1}^n$ is a basis of $\muathfrak{g}$, then $S(\muathfrak{g})$ is the polynomial algebra $\muathbb{K}[e_1,\lambdadots,e_n]$.  Suppose the structure constants for $\muathfrak{g}$ are given by
\[
[e_i,e_j] = \sigmaum_{k=1}^n c_{ij}^k e_k.
\]
Then the symmetric algebra $S(\muathfrak{g})$ becomes a Poisson algebra with the Poisson bracket
\betaegin{equation}
\lambdaabel{FG}
\{F,G\} = \frac{1}{2} \sigmaum_{i,j,k=1}^n c_{ij}^k e_k \lambdaeft(\frac{\partial F}{\partial e_i}\frac{\partial G}{\partial e_j} - \frac{\partial F}{\partial e_j}\frac{\partial G}{\partial e_i}\rhoight)
\end{equation}
for $F,G \in S(\muathfrak{g})$.  This Poisson algebra structure on $S(\muathfrak{g})$ is called the linear Poisson structure.  Note that $\{e_i,e_j\} = [e_i,e_j]$.

\end{rmk}

\betaegin{ex}[\textbf{$S(\muathfrak{sl}(2))$ is not rigid}]
In this example, we show that the symmetric algebra $(S(\muathfrak{sl}(2)),\muu)$ on the Lie algebra $\muathfrak{sl}(2)$, equipped with the linear Poisson structure \eqref{FG}, is not rigid in the sense of Definition \rhoef{def:deform}.

The Lie algebra $\muathfrak{sl}(2)$ 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 $S(\muathfrak{sl}(2)) = \muathbb{K}\lambdaangle e,f,h\rhoangle$ is not rigid, consider the Lie algebra morphisms $\alphalpha,\betaeta \colon \muathfrak{sl}(2) \to \muathfrak{sl}(2)$ given by
\[
\betaegin{array}{llllllllll}
\alphalpha(e) &=& \lambdaambda e,\quad \alphalpha(f) &=& \lambdaambda^{-1}f, \quad \alphalpha(h) &=& h,\\
\betaeta(e) &=& \gamma e,\quad \betaeta(f) &=& \gamma^{-1}f, \quad \betaeta(h) &=& h.
\end{array}
\]
where $\lambdaambda,\gamma \in \muathbb{K}$ is a fixed scalar with $\lambdaambda,\gamma \not= 0,1$.  Denote by $\alphalpha,\betaeta \colon S(\muathfrak{sl}(2)) \to S(\muathfrak{sl}(2))$ the extended maps, which is a Poisson algebra morphisms.  By Proposition \rhoef{prop:nonrigidity}, the Poisson algebra $S(\muathfrak{sl}(2))$ is not rigid if $\muu_{\alphalpha,\betaeta} = \muu\circ(\alphalpha\otimes\betaeta)$ is not BiHom-associative.  We have
\[
\betaegin{split}
\muu_{\alphalpha,\betaeta}(\muu_{\alphalpha,\betaeta}(e,h),h) - \muu_{\alphalpha,\betaeta}(e,\muu_{\alphalpha,\betaeta}(h,h))
&= \alphalpha^2(e)\alphalpha\betaeta(h)\betaeta(h) - \alphalpha(e)\alphalpha\betaeta(h)\betaeta^2(h)\\
&= (\lambdaambda^2 - \lambdaambda)eh^2,
\end{split}
\]
which is not $0$ in the symmetric algebra $S(\muathfrak{sl}(2))$ because $\lambdaambda \not= 0,1$.  Therefore, $\muu_{\alphalpha,\betaeta}$ is not BiHom-associative, and the linear Poisson structure on $S(\muathfrak{sl}(2))$ is not rigid.
\qed
\end{ex}
\sigmaubsection{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.
\betaegin{definition}
Let $ (A, \muu, \alphalpha,\betaeta)$ be a BiHom-algebra. Then $A$   is called BiHom-flexible algebra if for any $x, y \in A$
\betaegin{equation} \lambdaabel{flex_eq}
\muu(\muu(\betaeta^{2}(x), \alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(x))-\muu(\alphalpha\betaeta^2(x),\muu(\alphalpha\betaeta(y),\alphalpha^2(x))= 0.
\end{equation}
\end{definition}
\betaegin{remark}
\betaegin{enumerate}
\item If $\alphalpha=\betaeta=Id,$ then $(A, \muu, \alphalpha,\betaeta)$ is reduced to a
flexible algebra $(A, \muu).$
\item If $(A,\muu,\alphalpha)$ is a Hom-flexible algebra \cite{ms}, then
$(A,\muu,\alphalpha, \alphalpha)$ is a BiHom-flexible algebra. Conversely, if
$(A,\muu,\alphalpha, \betaeta)$ is a BiHom-flexible algebra such $\alphalpha$ is injective
and $\alphalpha=\betaeta$ then, $(A,\muu,\alphalpha)$ is a Hom-flexible algebra.
\end{enumerate}
\end{remark}
\betaegin{lemma} \lambdaabel{flex_lemma} Let $A = (A, \muu, \alphalpha,\betaeta)$ be a BiHom-algebra. The following assertions are equivalent
\betaegin{enumerate}
\item $A$ is BiHom-flexible.
\item For any $x, y \in A,\quad as_{A}(\betaeta^{2}(x), \alphalpha\betaeta(y), \alphalpha^{2}(x)) = 0$.
\item For any $x, y, z$
\betaegin{eqnarray}
\lambdaabel{homflexible}
as_{A}(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z)) + as_{A}(\betaeta^{2}(z),\alphalpha\betaeta(y),\alphalpha^{2}(x)) = 0
\end{eqnarray}
\end{enumerate}
\end{lemma}

\betaegin{proof}
The equivalence of the first two assertions follows from the definition. The  assertion  $as_{A}(\betaeta^{2}(x - z), \alphalpha\betaeta(y), \alphalpha^{2}(x - z)) = 0$ holds by definition and it is equivalent to $as_{A}(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z)) + as_{A}(\betaeta^{2}(z),\alphalpha\betaeta(y),\alphalpha^{2}(x)) = 0$ by linearity.
\end{proof}
It is easy to prove the following
\betaegin{prop}
Let $(A,\muu)$ be a flexible algebra, $\alphalpha,\betaeta:A\rhoightarrow A$ be two commuting morphisms.. Then the Bihom-algebra $(A,\muu_{\alphalpha,\betaeta}=\muu(\alphalpha\otimes \betaeta),\alphalpha, \betaeta)$
is BiHom-flexible.
\end{prop}
\betaegin{proof}
Let $(A,\muu)$ be a flexible algebra, $\alphalpha$ and $\betaeta$ be morphisms of
$(A,\muu)$. Then for all $x,y\in A,$
\betaegin{eqnarray}
&&\muu_{\alphalpha,\betaeta}(\muu_{\alphalpha,\betaeta}(\betaeta^{2}(x), \alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(x))-\muu_{\alphalpha,\betaeta}(\alphalpha\betaeta^2(x),\muu_{\alphalpha,\betaeta}(\alphalpha\betaeta(y),\alphalpha^2(x)))\nonumber\\
&&=\muu_{\alphalpha,\betaeta}(\muu(\alphalpha\betaeta^{2}(x), \alphalpha\betaeta^2(y)),\betaeta\alphalpha^{2}(x))-\muu_{\alphalpha,\betaeta}(\alphalpha\betaeta^2(x),\muu(\alphalpha^2\betaeta(y),\betaeta\alphalpha^2(x)))\nonumber\\
&&=\muu(\muu(\alphalpha^2\betaeta^{2}(x), \alphalpha^2\betaeta^2(y)),\betaeta^2\alphalpha^{2}(x))-\muu(\alphalpha^2\betaeta^2(x),\muu(\alphalpha^2\betaeta^2(y),\betaeta^2\alphalpha^2(x)))=0.\nonumber
\end{eqnarray}
Hence the Bihom-algebra $(A,\muu_{\alphalpha,\betaeta}=\muu(\alphalpha\otimes \betaeta),\alphalpha, \betaeta)$ is BiHom-flexible.
\end{proof}
\betaegin{corollary}
Any BiHom-associative algebra is BiHom-flexible.
\end{corollary}
\betaegin{proposition} \lambdaabel{flex_Leibniz_Proposition} A BiHom-algebra $A = (A, \muu, \alphalpha,\betaeta)$ is BiHom-flexible if and only if
\betaegin{equation} \lambdaabel{flex_Leibniz}
\{\alphalpha\betaeta(x),\alphalpha\betaeta(y) \deltaiamond \alphalpha^{2}(z)\} - \{\betaeta^{2}(x),\alphalpha\betaeta(y)\} \deltaiamond \alphalpha^{2}\betaeta(z) - \alphalpha\betaeta^{2}(y) \deltaiamond \{\alphalpha\betaeta(x),\alphalpha^{2}(z)\}=0,
\end{equation}
where $\{\cdot,\cdot\} = \frac{1}{2}(\muu - \muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau))$ and $\deltaiamond = \frac{1}{2}(\muu + \muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau)$.
\end{proposition}

\betaegin{proof}
Since $\{\cdot,\cdot\} = \frac{1}{2}(\muu - \muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau))$ and $\deltaiamond = \frac{1}{2}(\muu + \muu\circ(\alphalpha^{-1}\betaeta\otimes\alphalpha\betaeta^{-1})\circ\tau)$, by expansion in terms of $\muu$
$$\betaegin{array}{llllll}
&&4 \lambdaeft(\{\alphalpha\betaeta(x),\alphalpha\betaeta(y) \deltaiamond \alphalpha^{2}(z)\} - \{\betaeta^{2}(x),\alphalpha\betaeta(y)\} \deltaiamond \alphalpha^{2}\betaeta(z) - \alphalpha\betaeta^{2}(y) \deltaiamond \{\alphalpha\betaeta(x),\alphalpha^{2}(z)\}\rhoight)\\
&=&\muu(\alphalpha\betaeta^{2}(x)),\muu(\alphalpha\betaeta(y),\alphalpha^{2}(z)))
-\muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(x)))
+\muu(\alphalpha\betaeta^{2}(x)),\muu(\alphalpha\betaeta(z),\alphalpha^{2}(y))) \\
&-&\muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(x)))
-\muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(z)))
+\muu(\alphalpha\betaeta^{2}(z)),\muu(\alphalpha\betaeta(x),\alphalpha^{2}(y)))\\
&+&\muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(z)))
-\muu(\alphalpha\betaeta^{2}(z)),\muu(\alphalpha\betaeta(y),\alphalpha^{2}(x)))
-\muu(\alphalpha\betaeta^{2}(y)),\muu(\alphalpha\betaeta(x),\alphalpha^{2}(z)))\\
&+&\muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(y)))
+\muu(\alphalpha\betaeta^{2}(y)),\muu(\alphalpha\betaeta(z),\alphalpha^{2}(x)))
-\muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(y)))\\
&=& -as_A(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z)) - as_A(\betaeta^{2}(z),\alphalpha\betaeta(y),\alphalpha^{2}(x)) - as_A(\betaeta^{2}(x),\alphalpha\betaeta(z),\alphalpha^{2}(y))\\
&-& as_A(\betaeta^{2}(y),\alphalpha\betaeta(z),\alphalpha^{2}(x))  + as_A(\betaeta^{2}(y),\alphalpha\betaeta(x),\alphalpha^{2}(z)) + as_A(\betaeta^{2}(z),\alphalpha\betaeta(x),\alphalpha^{2}(y))\\
&=&0~(By~ Lemma \eqref{flex_lemma}).
\end{array}$$
Conversely, assume we have the condition \eqref{flex_Leibniz}. By setting $x = z$, one
gets\\ $as_{A}(\betaeta^{2}(x), \alphalpha\betaeta(y), \alphalpha^{2}(x)) = 0$. Therefore $A$ is BiHom-flexible.
\end{proof}
Let's give the notion of an admissible BiHom-Poisson algebras.

\betaegin{defn}\cite{hadimi}
\lambdaabel{def:admissible}
Let $(A,\muu,\alphalpha,\betaeta)$ be a BiHom-algebra.  Then $A$ is called an admissible BiHom-Poisson algebra if it satisfies
\betaegin{eqnarray}\lambdaabel{admissibility}
&&as_A(\betaeta(x),\alphalpha(y),\alphalpha^{2}(z)) = \frac{1}{3}\Big\lambdabrace\muu(\muu(\betaeta(x),\alphalpha\betaeta(z)),\alphalpha^{2}(y))) - \muu(\muu(\betaeta^{2}(z),\alphalpha(x)),\alphalpha^{2}(y))\nonumber\\
&&+ \muu(\muu(\betaeta(y),\alphalpha\betaeta(z)),\alphalpha^{2}(x)) - \muu(\muu(\betaeta(y),\alphalpha(x)),\betaeta\alphalpha^{2}(z))\Big\rhobrace
\end{eqnarray}
\end{defn}
It is observed that if $\alphalpha$ and $\betaeta$ are inversible, then
(\rhoef{admissibility}) is equivalent to
 \betaegin{eqnarray}
 &&as_A(x,y,z) = \frac{1}{3}\Big\lambdabrace\muu(\muu(x,\alphalpha^{-1}\betaeta(z)),\alphalpha(y))) - \muu(\muu(\alphalpha^{-2}\betaeta^{2}(z),\alphalpha\betaeta^{-1}(x)),\alphalpha(y))\nonumber\\
&&+ \muu(\muu(\alphalpha^{-1}\betaeta(y),\alphalpha^{-1}\betaeta(z)),\alphalpha^2\betaeta^{-1}(x)) - \muu(\muu(\alphalpha^{-1}\betaeta(y),\alphalpha^{-1}\betaeta(x)),\betaeta(z))\Big\rhobrace
 \end{eqnarray}
\betaegin{prop}
\lambdaabel{lem:flexible}
Every admissible BiHom-Poisson algebra $(A,\muu,\alphalpha,\betaeta)$ is BiHom-flexible, i.e.,
\betaegin{equation}
\lambdaabel{homflexible}
as_A(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z)) + as_A(\betaeta^{2}(z),\alphalpha\betaeta(y),\alphalpha^{2}(x)) = 0
\end{equation}
for all $x,y,z \in A$.
\end{prop}

\betaegin{proof}
The required identity \eqref{homflexible} follows immediately from the defining identity \eqref{admissibility}, in which the right-hand side is anti-symmetric in $x$ and $z$.

$\betaegin{array}{llllll}&&as_A(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z)) + as_A(\betaeta^{2}(z),\alphalpha\betaeta(y),\alphalpha^{2}(x))\\&=& \frac{1}{3}\Big\lambdabrace\muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(y))) - \muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(y)))\\&&+ \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(x))) -\muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(z)))\\&&+\muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(y))) - \muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(y)))\\&&+ \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(z))) - \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(x)))\Big\rhobrace=0.\end{array}$
\end{proof}

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

\betaegin{prop}
\lambdaabel{lem:cyclichomass}
Let $(A,\muu,\alphalpha,\betaeta)$ be an admissible BiHom-Poisson algebra.  Then
\betaegin{equation}
\lambdaabel{S}
S_A(x,y,z): = as_{A}(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z))  + as_{A}(\betaeta^{2}(y),\alphalpha\betaeta(z),\alphalpha^{2}(x))  + as_{A}(\betaeta^{2}(z),\alphalpha\betaeta(x),\alphalpha^{2}(y))  = 0
\end{equation}
for all $x,y,z \in A$.
\end{prop}

\betaegin{proof}
Using the defining identity \eqref{admissibility}, we have:
\[
\betaegin{array}{llllll}
as_A(\betaeta^{2}(x),\alphalpha\betaeta(y),\alphalpha^{2}(z))
&=& \frac{1}{3}\Big(\muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(y)))- \muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(y)))\\&&+ \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(x))) - \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(z)))\Big)\\
&=& -\frac{1}{3}\Big(\muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(x))) - \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(x)))\\&&+ \muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(z))) - \muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(z)),\betaeta\alphalpha^{2}(y)))\Big)\\
&&+\frac{1}{3}\Big(\muu(\muu(\betaeta^{2}(x),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(z))) - \muu(\muu(\betaeta^{2}(y),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(z)))\\&&+ \muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(y)),\betaeta\alphalpha^{2}(x))) - \muu(\muu(\betaeta^{2}(z),\alphalpha\betaeta(x)),\betaeta\alphalpha^{2}(y)))\Big)\\
&=& - as_A(\betaeta^{2}(z),\alphalpha\betaeta(x),\alphalpha^{2}(y)) + as_A(\betaeta^{2}(x),\alphalpha\betaeta(z),\alphalpha^{2}(y))\\
&= &- as_A(\betaeta^{2}(z),\alphalpha\betaeta(x),\alphalpha^{2}(y)) - as_A(\betaeta^{2}(y),\alphalpha\betaeta(z),\alphalpha^{2}(x))~(by~(\rhoef{homflexible})).
\end{array}
\]
Therefore, we conclude that $S_A = 0$.
\end{proof}

 \muathrm{e}ction{Derivations  of BiHom-Poisson algebras}
\deltaef\alpharabic{section}. \alpharabic{equation}{\alpharabic{section}. \alpharabic{equation}}
\muathrm{e}tcounter{equation} {0}

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

\betaegin{defn}Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  be a BiHom-Poisson algebra.
 A  linear map $D: A\rhoightarrow A$  is called an  $(\alpha^k,\beta^{l})$-derivation of $A$  if it satisfies
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,\; D \circ \betaeta = \betaeta \circ D; $
\item $D(\{x, y\})=\{\alpha^{k}\beta^{l}(x), D(y)\}+\{D(x), \alpha^{k}\beta^{l}(y)\}$;
\item $D(\muu(x,y))=\muu(\alpha^{k}\beta^{l}(x), D(y))+\muu(D(x),\alpha^{k}\beta^{l}(y))$,
\end{enumerate}
for all $x,y\in A$.
\end{defn}

We denote by $Der(A):=\deltaisplaystyle{\betaigoplus_{k\geq 0}}\deltaisplaystyle{\betaigoplus_{l\geq 0}}Der_{(\alpha^k,\beta^{l})}(A)$, 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 $End(A)$.
\betaegin{lem}\lambdaabel{le:2.7}
Let $(A,\{\cdot,\cdot\}, \muu, \alphalpha,\betaeta)$ be a BiHom-Poisson algebra. We define a subspace $\muathcal{W}$ of $End(A)$ by $\muathcal{W} = \{w \in End(A) |~ w \circ \alphalpha = \alphalpha \circ w~and~w \circ \betaeta = \betaeta \circ w\}$ and $\sigmaigma_{1},\sigmaigma_{2} : \muathcal{W} \rhoightarrow \muathcal{W}$  linear maps satisfying $\sigmaigma_{1}(w) = \alphalpha \circ w$ and $\sigmaigma_{2}(w) = \betaeta \circ w$. Then
a quadruple $(\muathcal{W}, [\cdot,\cdot], \sigmaigma_{1},\sigmaigma_{2})$, where the multiplication $[\cdot,\cdot] : \muathcal{W} \times \muathcal{W} \rhoightarrow \muathcal{W}$ is defined for
$w_{1}, w_{2} \in \muathcal{W}$ by
 \[   [w_{1}, w_{2}]= w_{1} \circ w_{2} - w_{2} \circ w_{1},\]
    is a BiHom-Lie algebra.
\end{lem}
\betaegin{proof}
For any $w_{1},w_{2},w_{3}\in\muathcal{W},~~k_{1},k_{2}\in\muathbb{K},$ we have

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

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

$\betaegin{array}{lll}
&&[\sigmaigma_{2}(w_{1}),\sigmaigma_{1}(w_{2})]
=[\betaeta(w_{1}),\alphalpha(w_{2})]
=\alphalpha\betaeta (w_{1} w_{2}-w_{2}w_{1})
=-\alphalpha\betaeta (w_{2} w_{1}-w_{1}w_{2})\\
&=&-[\sigmaigma_{2}(w_{2}),\sigmaigma_{1}(w_{1})]
\end{array}$\\\\

$\betaegin{array}{llllllll}
&&[\sigmaigma_{2}^{2}(w_{1}),[\sigmaigma_{2}(w_{2}),\sigmaigma_{1}(w_{3})]]+[\sigmaigma_{2}^{2}(w_{3}),[\sigmaigma_{2}(w_{1}),\sigmaigma_{1}(w_{2})]]\\
&&+[\sigmaigma_{2}^{2}(w_{2}),[\sigmaigma_{2}(w_{3}),\sigmaigma_{1}(w_{1})]]\\
&=&\betaeta^{3}\alphalpha w_{1}w_{2}w_{3}-\betaeta^{3}\alphalpha w_{1}w_{3}w_{2}-\betaeta^{3}\alphalpha w_{2}w_{3}w_{1}+\betaeta^{3}\alphalpha w_{3}w_{2}w_{1}\\&&+
\betaeta^{3}\alphalpha w_{2}w_{3}w_{1}-\betaeta^{3}\alphalpha w_{2}w_{1}w_{3}-\betaeta^{3}\alphalpha w_{3}w_{1}w_{2}+\betaeta^{3}\alphalpha w_{1}w_{3}w_{2}\\&&+
\betaeta^{3}\alphalpha w_{3}w_{1}w_{2}-\betaeta^{3}\alphalpha w_{3}w_{2}w_{1}-\betaeta^{3}\alphalpha w_{1}w_{2}w_{3}+\betaeta^{3}\alphalpha w_{2}w_{1}w_{3}\\
&=&0.
\end{array}$

Then $(\muathcal{W}, [\cdot,\cdot], \sigmaigma_{1},\sigmaigma_{2})$  is a Bihom-Lie algebra.
\end{proof}

\betaegin{thm}\lambdaabel{le:2.9}Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  be a BiHom-Poisson algebra.
For any $D \in Der_{(\alphalpha^{k},\betaeta^{l})}(A)$ and $D^{'} \in Der_{(\alphalpha^{k'},\betaeta^{l'})}(A)$, define their commutator $[D, D^{'}]$ as usual:
\[[D, D^{'}] = D \circ D^{'} - D^{'} \circ D.\]
Then $[D, D^{'}] \in Der_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$.
\end{thm}
\betaegin{proof}
It is sufficient to prove $[Der_{(\alpha^k,\beta^l)}(A), Der_{(\alpha^{k'},\beta^{l'})}(A)]\sigmaubseteq Der_{(\alpha^{k+k'},\beta^{l+l')}}(A)$. It is easy to check that $[D, D']\circ \alpha=\alpha\circ [D, D']$ and $[D, D']\circ \beta=\beta\circ [D, D']$.

For any $x,y\in A$, we have
$$\betaegin{array}{lllll}
&&[D,D'](\{x,y\})\\
&=&D\circ D'(\{x,y\})-D'\circ D(\{x,y\})\\
&=&D(\{D'(x),\alphalpha^{k'}\betaeta^{l'}(y)\}+\{\alphalpha^{k'}\betaeta^{l'}(x),D'(y)\})\\
&&-D'(\{D'(x), \alphalpha^{k}\betaeta^{l}(y)\}+\{\alphalpha^{k}\betaeta^{l}(x),D(y)\})\\
&=&D(\{D'(x),\alphalpha^{k'}\betaeta^{l'}(y)\})+D((\{\alphalpha^{k'}\betaeta^{l'}(x),D'(y)\})\\
&&-D'(\{D(x)(x),\alphalpha^{k}\betaeta^{l}(y)\})-D'(\{\alphalpha^{k}\betaeta^{l}(x),D(y)\})\\
&=&\{D\circ D'(x),\alphalpha^{k+k'}\betaeta^{l+l'}(y)\}+\{\alphalpha^{k}\betaeta^{l}\circ D'(x),D\circ\alphalpha^{k'}\betaeta^{l'}(y)\}\\
&&+\{D\circ\alphalpha^{k'}\betaeta^{l'}(x),\alphalpha^{k}\betaeta^{l}\circ D'(y)\}+\{\alphalpha^{k+k'}\betaeta^{l+l'}(x),D\circ D'(y)\}\\
&&-\{D'\circ D(x),\alphalpha^{k+k'}\betaeta^{l+l'}(y)\}-\{\alphalpha^{k'}\betaeta^{l'}\circ D(x),D'\circ \alphalpha^{k}\betaeta^{l}(y)\}\\
&&-\{D'\circ\alphalpha^{k}\betaeta^{l}(x),\alphalpha^{k'}\betaeta^{l'}\circ D(y)\}-\{\alphalpha^{k+k'}\betaeta^{l+l'}(x),D'\circ D(y)\}.
\end{array}$$

Similarly, we can check that
\betaegin{eqnarray*}
[D, D'](\muu(x, y))
\muu([D, D'](x), \alpha^{k+k'}\beta^{l+l'}(y))+\muu(\alpha^{k+k'}\beta^{l+l'}(x), [D,D'](y)).
\end{eqnarray*}
It follows that $[D, D']\in Der_{(\alpha^{k+k'},\beta^{l+l'})}(A)$.\end{proof}
\betaegin{defn}
Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  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
$D', D''\in End(A)$ such that
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,\; D \circ \betaeta = \betaeta \circ D\; $;
\item $D^{'} \circ \alphalpha = \alphalpha \circ D^{'},\; D' \circ \betaeta = \betaeta \circ D^{'}\;$;
\item $D^{''} \circ \alphalpha = \alphalpha \circ D^{''},\;D^{''} \circ \betaeta = \betaeta \circ D^{''}$;
\item $\{D(x), \alpha^{k}\beta^l (y)\}+\{\alpha^{k}\beta^ l (x), D'(y)\}=D''(\{x, y\})$;
\item $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) + \muu(\alphalpha^{k}\betaeta^{l}(x), D^{'}(y)) = D''(\muu(x, y))$,
\end{enumerate}
for all $x,y\in A$.

The set of generalized $(  \alphalpha^{k},\betaeta^{l}) $-derivations of $ A $ is $ {\rhom GDer}_{(  \alphalpha^{k},\betaeta^{l})}(A) $ and we denote  \[{\rhom GDer}(A):=\betaigoplus_{k \geq 0}\betaigoplus_{l \geq 0} {\rhom GDer}_{(\alphalpha^{k},\betaeta^{l})}(A).\]
\end{defn}
\betaegin{defn}
Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  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
$D', D''\in End(A)$ such that
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,\; D \circ \betaeta = \betaeta \circ D\; $;
\item $D^{'} \circ \alphalpha = \alphalpha \circ D^{'},\; D^{'} \circ \betaeta = \betaeta \circ D^{'}\;$;
\item $D^{''} \circ \alphalpha = \alphalpha \circ D^{''},\;D^{''} \circ \betaeta = \betaeta \circ D^{''}$;
\item $\{D(x), \alpha^{k}\beta^l (y)\}+\{\alpha^{k}\beta^ l (x), D(y)\}=D{'}(\{x, y\}),$
\item $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) + \muu(\alphalpha^{k}\betaeta^{l}(x), D(y)) = D^{''}(\muu(x, y)),$
\end{enumerate}
for all $x,y\in A$.
\end{defn}

We then define
\[{\rhom QDer}(A):=\betaigoplus_{k \geq 0}\betaigoplus_{l \geq 0} {\rhom QDer}_{(  \alphalpha^{k},\betaeta^{l})}(A).\]

\betaegin{defn}Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  be a BiHom-Poisson algebra.
 A  linear map $D: A\rhoightarrow A$  is called an  $(\alphalpha^{k},\betaeta^{l})$-centroid of $A$  if it satisfies
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,~D \circ \betaeta = \betaeta \circ D$;
\item $\{D(x), \alphalpha^{k}\betaeta^{l}(y)\} = \{\alphalpha^{k}\betaeta^{l}(x), D(y)\} = D(\{x, y\})$;
\item $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) = \muu(\alphalpha^{k}\betaeta^{l}(x), D(y)) = D(\muu(x, y)),\quad\forall x, y \in A$.
\end{enumerate}
\end{defn}
We set
$$C(A):=\betaigoplus_{k\geq 0}\betaigoplus_{l\geq 0} C_{(\alphalpha^{k},\betaeta^{l})}(A).$$
\betaegin{defn}
The $(\alphalpha^{k},\betaeta^{l}) $-quasicentroid of a Bihom-Poisson algebra $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ denoted by $ {\rhom QC}_{(  \alphalpha^{k},\betaeta^{l})} (A)$ is the set of linear maps $ D $ such that
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,~D \circ \betaeta = \betaeta \circ D$;
\item $\{D(x), \alphalpha^{k}\betaeta^{l}(y)\} = \{\alphalpha^{k}\betaeta^{l}(x), D(y)\}$;
\item $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) = \muu(\alphalpha^{k}\betaeta^{l}(x), D(y)),\quad\forall x, y \in A$.
\end{enumerate}
\end{defn}
We set
\[{\rhom QC}(A):=\betaigoplus_{k \geq 0}\betaigoplus_{l \geq 0} {\rhom QC}_{(\alphalpha^{k},\betaeta^{l})}(A).\]

\betaegin{rmk}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$  be a BiHom-Poisson algebra.
Then $C(A)\sigmaubseteq QC(A).$
\end{rmk}

\betaegin{defn}
A linear map $ D $ is called an $(  \alphalpha^{k},\betaeta^{l}) $-central derivation of $ A $ if it satisfies
\betaegin{enumerate}
\item $D \circ \alphalpha = \alphalpha \circ D,~D \circ \betaeta = \betaeta\circ D$;
\item $\{D(x), \alphalpha^{k}\betaeta^{l}(y)\} = D(\{x, y\}) = 0;$
\item $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) = D(\muu(x, y)) = 0,\quad\forall x, y \in A$.
\end{enumerate}
\end{defn}
The set of $(\alphalpha^{k},\betaeta^{l}) $-central derivations is denoted by $ {\rhom ZDer}_{(  \alphalpha^{k},\betaeta^{l})}(A) $ and we set
\[{\rhom ZDer}(A):=\betaigoplus_{k \geq 0}\betaigoplus_{l \geq 0} {\rhom ZDer}_{(  \alphalpha^{k},\betaeta^{l})}(A).\]

\betaegin{rmk}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$  be a BiHom-Poisson algebra.
Then
\betaegin{eqnarray*}
ZDer(A)\sigmaubseteq Der(A)\sigmaubseteq QDer(A)\sigmaubseteq GDer(A) \sigmaubseteq End(A).
\end{eqnarray*}
\end{rmk}

\betaegin{prop}
Let $(A,\{\cdot,\cdot\},\muu,\alpha,\beta)$  be a BiHom-Poisson algebra.
 Then the following statements hold:
\betaegin{enumerate}
\item $GDer(A)$, $QDer(A)$ and $C(A)$ are BiHom-subalgebras of $(\muathcal{W}, [\cdot,\cdot], \sigmaigma_{1},\sigmaigma_{2})$;
\item $ZDer(A)$ is a BiHom-ideal of $Der(A)$.
\end{enumerate}
\end{prop}
\betaegin{proof}
 We only prove that $GDer(A)$ is a subalgebra of $\muathcal{W}.$ The case of $ QDer(L)$ and $C(L)$ is similar.
\betaegin{enumerate}
\item
Suppose that $D_{1} \in GDer_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in GDer_{(\alphalpha^{k'},\betaeta^{l'})}(A)$. Then for any $x, y \in A$,
\betaegin{align*}
&\{\sigmaigma_{1}(D_{1})(x), \alphalpha^{k + 1}\betaeta^{l}(y)\} = \{\alphalpha_{1} \circ D_{1}(x), \alphalpha^{k + 1}\betaeta^{l}(y)\} = \alphalpha(\{D_{1}(x), \alphalpha^{k}\betaeta^{l}(y)\})\\
&= \alphalpha(D_{1}^{''}(\{x, y\}) - \{\alphalpha^{k}\betaeta^{l}(x), D_{1}^{'}(y)\}) = \sigmaigma_{1}(D_{1}^{''})(\{x, y\}) - \{\alphalpha^{k + 1}\betaeta^{l}(x), \sigmaigma_{1}(D_{1}^{'})(y)\}.
\end{align*}
Since $\sigmaigma_{1}(D_{1}^{''}), \sigmaigma_{1}(D_{1}^{'}) \in End(A)$, we have $\sigmaigma_{1}(D_{1}) \in GDer_{(\alphalpha^{k + 1},\betaeta^{l})}(A)$,
\betaegin{align*}
&\{\sigmaigma_{2}(D_{1})(x), \alphalpha^{k}\betaeta^{l+1}(y)\} = \{\betaeta \circ D_{1}(x), \alphalpha^{k}\betaeta^{l+1}(y)\} = \betaeta(\{D_{1}(x), \alphalpha^{k}\betaeta^{l}(y)\})\\
&= \betaeta(D_{1}^{''}(\{x, y\}) - \{\alphalpha^{k}\betaeta^{l}(x), D_{1}^{'}(y)\}) = \sigmaigma_{2}(D_{1}^{''})(\{x, y\}) - \{\alphalpha^{k}\betaeta^{l+1}(x), \sigmaigma_{2}(D_{1}^{'})(y)\}.
\end{align*}
Since $\sigmaigma_{2}(D_{1}^{''}), \sigmaigma_{2}(D_{1}^{'}) \in End(A)$, we have $\sigmaigma_{2}(D_{1}) \in GDer_{(\alphalpha^{k},\betaeta^{l+1})}(A)$.
\betaegin{align*}
&\{\{D_{1}, D_{2}\}(x), \alphalpha^{k+k'}\betaeta^{l+l'}(y)\}\\
&= \{D_{1} \circ D_{2}(x), \alphalpha^{k+k'}\betaeta^{l+l'}(y)\} - \{D_{2} \circ D_{1}(x), \alphalpha^{k+k'}\betaeta^{l+l'}(y)\}\\
&= D_{1}^{''}(\{D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)\}) - \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}^{'}(\alphalpha^{k'}\betaeta^{l'}(y))\} - D_{2}^{''}(\{D_{1}(x), \alphalpha^{k}\betaeta^{l}(y)\})\\ &+ \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}^{'}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= D_{1}^{''}(D_{2}^{''}(\{x, y\}) - \{\alphalpha^{k'}\betaeta^{l'}(x), D_{2}^{'}(y)\}) -\{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}^{'}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= D_{1}^{''} \circ D_{2}^{''}(\{x, y\}) - D_{1}^{''}(\{\alphalpha^{k'}\betaeta^{l'}(x), D_{2}^{'}(y)\}) - \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}^{'}(\alphalpha^{k'}\betaeta^{l'}(y))\}\\
&- D_{2}^{''} \circ D_{1}^{''}(\{x, y\}) + D_{2}^{''}(\{\alphalpha^{k}\betaeta^{l}(x), D_{1}^{'}(y)\}) + \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}^{'}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= D_{1}^{''} \circ D_{2}^{''}(\{x, y\}) - \{D_{1}(\alphalpha^{k'}\betaeta^{l'}(x)), \alphalpha^{k}\betaeta^{l}(D_{2}^{'}(y))\} - \{\alphalpha^{k +k'}\betaeta^{l+l'}(x), D_{1}^{'}(D_{2}^{'}(y))\}\\
&- \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}^{'}(\alphalpha^{k'}\betaeta^{l'}(y))\} - D_{2}^{''} \circ D_{1}^{''}(\{x, y\}) + \{D_{2}(\alphalpha^{k}\betaeta^{l}(x)), \alphalpha^{k'}\betaeta^{l'}(D_{1}^{'}(y))\}\\
&+ \{\alphalpha^{k+k'}\betaeta^{l+l'}(x), D_{2}^{'}(D_{1}^{'}(y))\} + \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}^{'}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= D_{1}^{''} \circ D_{2}^{''}(\{x, y\}) - D_{2}^{''} \circ D_{1}^{''}(\{x, y\}) - \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}^{'}(D_{2}^{'}(y))\}\\
&+ \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{2}^{'}(D_{1}^{'}(y))\}\\
&= \{D_{1}^{''}, D_{2}^{''}\}(\{x, y\}) - \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), \{D_{1}^{'}, D_{2}^{'}\}(y)\}.
\end{align*}
Since $\{D_{1}^{''}, D_{2}^{''}\}, \{D_{1}^{'}, D_{2}^{'}\} \in End(A)$, we have $\{D_{1}, D_{2}\} \in GDer_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$.

Similarly, we have $\muu(D_{1}, D_{2}) \in GDer_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$.

Therefore, $GDer(A)$ is a BiHom-subalgebra of $(\muathcal{W}, [\cdot,\cdot], \sigmaigma_{1},\sigmaigma_{2})$.
\item
Suppose that $D_{1} \in ZDer_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in Der_{(\alphalpha^{k'},\betaeta^{l'})}(A)$. Then for any $x, y \in A$,
\[\sigmaigma_{1}(D_{1})(\muu(x, y)) = \alphalpha \circ D_{1}(\muu(x, y)) = 0.\]
\[\sigmaigma_{1}(D_{1})(\muu(x, y)) = \alphalpha \circ D_{1}(\muu(x, y)) = \alphalpha(\muu(D_{1}(x), \alphalpha^{k}\betaeta^{l}(y))) = \muu(\sigmaigma_{1}(D_{1})(x), \alphalpha^{k + 1}\betaeta^{l}(y)).\]
Similarly, $\sigmaigma_{1}(D_{1})(\{x, y\})= \{\sigmaigma_{1}(D_{1})(x), \alphalpha^{k + 1}\betaeta^{l}(y)\}.$

Hence, $\sigmaigma_{1}(D_{1}) \in ZDer_{(\alphalpha^{k+1 },\betaeta^{l})}(A)$.

In the same way, we have, $\sigmaigma_{2}(D_{1}) \in ZDer_{(\alphalpha^{k},\betaeta^{l+1})}(A)$. Next, we have
\betaegin{align*}
&[D_{1}, D_{2}](\muu(x, y))\\
&= D_{1} \circ D_{2}(\muu(x, y)) - D_{2} \circ D_{1}(\muu(x, y))\\
&= D_{1}(\muu(D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)) + \muu(\alphalpha^{k'}\betaeta^{l'}(x), D_{2}(y))) = 0.
\end{align*}
\betaegin{align*}
&[D_{1}, D_{2}](\muu(x, y))\\
&= D_{1} \circ D_{2}(\muu(x, y)) - D_{2} \circ D_{1}(\muu(x, y))\\
&= D_{1}(\muu(D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)) + \muu(\alphalpha^{k'}\betaeta^{l'}(x), D_{2}(y))) - D_{2}(\muu(D_{1}(x), \alphalpha^{k}\betaeta^{l}(y)))\\
&= \muu(D_{1}(D_{2}(x)), \alphalpha^{k+k'}\betaeta^{l+l'}(y)) + \muu(D_{1}(\alphalpha^{k'}\betaeta^{l'}(x)), \alphalpha^{k}\betaeta^{l}(D_{2}(y))) - \muu(D_{2}(D_{1}(x)), \alphalpha^{k +k'}\betaeta^{l+l'}(y))\\
&- \muu(\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}(\alphalpha^{k}\betaeta^{l}(y)))\\
&= \muu([D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)).
\end{align*}
Similarly, $[D_{1}, D_{2}](\{x, y\})=\{[D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}=0$.

Hence, we have $[D_{1}, D_{2}] \in ZDer_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$.
Therefore, $ZDer(A)$ is a BiHom-ideal of $Der(A)$.
\end{enumerate}
\end{proof}

\betaegin{lem}
Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  be a BiHom-Poisson algebra,
 then the following statements hold:
\betaegin{enumerate}
\item $[Der(A), C(A)]\sigmaubseteq C(A)$.

\item $[QDer(A), QC(A)]\sigmaubseteq QC(A)$.

\item
$[QC(A), QC(A)]\sigmaubseteq QDer(A)$.

\item
$C(A)\sigmaubseteq QDer(A)$.

\item
$QDer(A)+QC(A)\sigmaubseteq GDer(A)$.

\item
$C(A) \circ Der(A) \sigmaubseteq Der(A)$.
\end{enumerate}
\end{lem}
\betaegin{proof}
\betaegin{enumerate}
\item
Suppose that $D_{1} \in Der_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in C_{(\alphalpha^{k'}\betaeta^{l'})}(A)$. Then for any $x, y \in A$, we have
\betaegin{align*}
&[D_{1}, D_{2}](\{x, y\})\\
&= D_{1} \circ D_{2}(\{x, y\}) - D_{2} \circ D_{1}(\{x, y\})\\
&= D_{1}(\{D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)\}) - D_{2}(\{D_{1}(x), \alphalpha^{k}\betaeta^{l}(y)\} + \{\alphalpha^{k}\betaeta^{l}(x), D_{1}(y)\})\\
&= \{D_{1}(D_{2}(x)), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\} + \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\} - \{D_{2}(D_{1}(x)), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}\\
&- \{D_{2}(\alphalpha^{k}\betaeta^{l}(x)), \alphalpha^{k'}\betaeta^{l'}(D_{1}(y))\}\\
&= \{[D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}.
\end{align*}
Similarly, we have $[D_{1}, D_{2}](\{x, y\}) = \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}](y)\}$.

In the same way, we have
$[D_{1}, D_{2}](\muu(x, y))= \muu([D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y))\\ = \muu(\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}](y))$.

Hence, $[D_{1}, D_{2}] \in C_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$. Therefore $[D_{1}, D_{2}] \in C(A)$.

\item
 Suppose that $D_{1} \in QDer_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in QC_{(\alphalpha^{k'},\betaeta^{l'})}(A)$. Then for any $x, y \in A$, we have
\betaegin{align*}
&\{[D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}\\
&= \{D_{1} \circ D_{2}(x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\} - \{D_{2} \circ D_{1}(x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}\\
&= D_{1}^{'}(\{D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)\}) - \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\} \\&- \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= D_{1}^{'}(\{\alphalpha^{k'}\betaeta^{l'}(x), D_{2}(y)\}) - \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\}\\& - \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= \{D_{1}(\alphalpha^{k'}\betaeta^{l'}(x)), \alphalpha^{k}\betaeta^{l}(D_{2}(y))\} + \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\} \\&- \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\}
- \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\} - \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\}\\
&= \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\} - \{D_{2}(\alphalpha^{k}\betaeta^{l}(x)), \alphalpha^{k'}\betaeta^{l'}(D_{1}(y))\}\\
&= \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\} - \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{2}(D_{1}(y))\}\\
&= \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}](y)\}.
\end{align*}
Similarly, $\muu([D_{1}, D_{2}](x), \alphalpha^{k + k'}\betaeta^{l+l'}(y))= \muu(\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}](y)).$

Hence, we have $[D_{1}, D_{2}] \in QC_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$. So $[QDer(A), QC(A)] \sigmaubseteq QC(A)$.

\item
Suppose that $D_{1} \in QC_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in QC_{(\alphalpha^{k'},\betaeta^{l'})}(A)$. Then for any $x, y \in A$, we have
\betaegin{align*}
&\{[D_{1}, D_{2}] (x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}\\
&= \{D_{1} \circ D_{2}(x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\} - \{D_{2} \circ D_{1}(x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\}\\
&= \{\alphalpha^{k}\betaeta^{l}(D_{2}(x)), D_{1}(\alphalpha^{k'}\betaeta^{l'}(y))\} - \{\alphalpha^{k'}\betaeta^{l'}(D_{1}(x)), D_{2}(\alphalpha^{k}\betaeta^{l}(y))\}\\
&= \{D_{2}(\alphalpha^{k}\betaeta^{l}(x)), \alphalpha^{k'}\betaeta^{l'}(D_{1}(y))\} - \{D_{1}(\alphalpha^{k'}\betaeta^{l'}(x)), \alphalpha^{k}\betaeta^{l}(D_{2}(y))\}\\
&= \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{2}(D_{1}(y))\} - \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\}\\
&= -\muu(\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}] (y)),
\end{align*}
i.e., $\{[D_{1}, D_{2}] (x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\} + \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}]\} = 0$.

Similarly, $\muu([D_{1}, D_{2}] (x), \alphalpha^{k + k'}\betaeta^{l+l'}(y)) + \muu(\alphalpha^{k + k'}\betaeta^{l+l'}(x), [D_{1}, D_{2}] ) = 0$.
Hence, we have $[D_{1}, D_{2}]\in QDer_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$, which implies that $[QC(A), QC(A)] \sigmaubseteq QDer(A)$.
\item
Suppose that $D \in C_{(\alphalpha^{k},\betaeta^{l})}(A)$. Then for any $x, y \in A$, we have
\[D(\{x, y\}) = \{D(x), \alphalpha^{k}\betaeta^{l}(y)\} = \{\alphalpha^{k}\betaeta^{l}(x), D(y)\}.\]

Hence, we have
\[\{D(x), \alphalpha^{k}\betaeta^{l}(y)\} + \{\alphalpha^{k}\betaeta^{l}(x), D(y)\} = 2D(\{x, y\}),\]
Similarly, $\muu(D(x), \alphalpha^{k}\betaeta^{l}(y)) + \muu(\alphalpha^{k}\betaeta^{l}(x), D(y)) = 2D(\muu(x, y)),$

which implies that $D \in QDer_{(\alphalpha^{k},\betaeta^{l})}(A)$. So $C(A) \sigmaubseteq QDer(A)$.
\item
In fact. Let $D_1\in QDer_{(\alpha^{k},\beta^{l})}(A), D_2\in QC_{(\alpha^{k},\beta^l)}(A)$. Then there exist $D'_1, D''_1\in End(A)$, for any $x,y\in A$, we have
\betaegin{eqnarray*}
&&\{D_1(x), \alpha^{k}\beta^{l}(y)\}+\{\alpha^{k}\beta^{l}(x), D_1(y)\}=D_1'(\{x, y\}),\\
&&\muu(D_1(x),\alpha^{k}\beta^{l}(y))+\muu(\alpha^{k}\beta^{l}(x), D_1(y))
=D_1''(\muu(x,y)).
\end{eqnarray*}
Thus, for any $x, y\in A$, we have
\betaegin{eqnarray*}
\{(D_1+D_2)(x), \alpha^{k}\beta^{l}(y)\}&=&\{D_1(x), \alpha^{k}\beta^{l}(y)\}+\{ D_2(x), \alpha^{k}\beta^{l}(y)\}\\
&=& D_1'(\{x, y\})-\{\alpha^{k}\beta^{l}(x), D_1(y)\}+\{ \alpha^{k}\beta^{l}(x),D_2(y)\}\\
&=& D_1'(\{x, y\})-\{\alpha^{k}\beta^{l}(x), (D_1-D_2)(y)\},
\end{eqnarray*}
and
\betaegin{eqnarray*}
\muu((D_1+D_2)(x), \alpha^{k}\beta^{l}(y))&=&\muu(D_1(x), \alpha^{k}\beta^{l}(y))+\muu( D_2(x), \alpha^{k}\beta^{l}(y)\}\\
&=& D_1''(\muu(x, y))-\muu(\alpha^{k}\beta^{l}(x), D_1(y))+\muu( \alpha^{k}\beta^{l}(x),D_2(y))\\
&=& D_1''(\muu(x, y))-\muu(\alpha^{k}\beta^{l}(x), (D_1-D_2)(y)),
\end{eqnarray*}
Therefore, $D_1+D_2\in GDer_{(\alpha^{k},\beta^{l})}(A)$.

\item
Suppose that $D_{1} \in C_{(\alphalpha^{k},\betaeta^{l})}(A)$, $D_{2} \in Der_{(\alphalpha^{k'},\betaeta^{l'})}(A)$. Then for any $x, y \in A$, we have
\betaegin{align*}
&D_{1} \circ D_{2}(\{x, y\})\\
&= D_{1}(\{D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)\} + \{\alphalpha^{k'}\betaeta^{l'}(x), D_{2}(y)\})\\
&= \{D_{1}(D_{2}(x)), \alphalpha^{k + k'}\betaeta^{l+l'}(y)\} + \{\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))\},
\end{align*}
\betaegin{align*}
&D_{1} \circ D_{2}(\muu(x, y))\\
&= D_{1}(\muu(D_{2}(x), \alphalpha^{k'}\betaeta^{l'}(y)) + \muu(\alphalpha^{k'}\betaeta^{l'}(x), D_{2}(y)))\\
&= \muu(D_{1}(D_{2}(x)), \alphalpha^{k + k'}\betaeta^{l+l'}(y)) + \muu(\alphalpha^{k + k'}\betaeta^{l+l'}(x), D_{1}(D_{2}(y))),
\end{align*}
which implies that $D_{1} \circ D_{2} \in Der_{(\alphalpha^{k + k'},\betaeta^{l+l'})}(A)$. So $C(A) \circ Der(A) \sigmaubseteq Der(A)$.
\end{enumerate}
\end{proof}
\muedskip
\betaegin{thm}
Let $(A,\{\cdot,\cdot\},\muu, \alpha,\beta)$  be a BiHom-Poisson algebra, $\alpha$ and $\beta$  surjections,
 then $[C(A), QC(A)]\sigmaubseteq End(A,Z(A))$.
  Moreover, if $Z(A)=\{0\}$, then  $[C(A), QC(A)]=\{0\}$.
\end{thm}

\betaegin{proof}
For any $D_1\in C_{(\alpha^{k},\beta^{l})}(A), D_2\in QC_{(\alpha^{k'},\beta^{l'})}(A)$ and $x,y\in A$, since $\alpha$ and $\beta$ are surjections, there exist $y'\in A$ such that $y=\alpha^{k+k'}\beta^{l+l'}(y')$, we have
\betaegin{eqnarray*}
\{[D_1, D_2](x), y\}&=&\{\{D_1, D_2\}(x), \alpha^{k+k'}\beta^{l+l'}(y')\}\\
&=& \{D_1D_2(x), \alpha^{k+k'}\beta^{l+l'}(y')\}-\{D_2D_1(x), \alpha^{k+k'}\beta^{l+l'}(y')\}\\
&=& D_1(\{D_2(x), \alpha^{k'}\beta^{l'}(y')\})-\{\alpha^{k'}\beta^{l'}D_1(x), D_2\alpha^{k}\beta^{l}(y')\}\\
&=& D_1(\{D_2(x), \alpha^{k'}\beta^{l'}(y')\})-D_1(\{D_2(x), \alpha^{k'}\beta^{l'}(y')\})\\
&=& 0,
\end{eqnarray*}
and
\betaegin{eqnarray*}
\muu([D_1, D_2](x), y)&=&\muu([D_1, D_2](x), \alpha^{k+k'}\beta^{l+l'}(y'))\\
&=& \muu(D_1D_2(x), \alpha^{k+k'}\beta^{l+l'}(y'))-\muu(D_2D_1(x), \alpha^{k+k'}\beta^{l+l'}(y'))\\
&=& D_1(\muu(D_2(x), \alpha^{k'}\beta^{l'}(y')))-\muu(\alpha^{k'}\beta^{l'}D_1(x), D_2\alpha^{k}\beta^{l}(y'))\\
&=& D_1(\muu(D_2(x), \alpha^{k'}\beta^{l'}(y')))-D_1(\muu(D_2(x), \alpha^{k'}\beta^{l'}(y')))\\
&=& 0.
\end{eqnarray*}
So $[D_1,D_2](x)\in Z(A)$ and therefore $[C(A), QC(A)]\sigmaubseteq End(A,Z(A))$.
 Moreover, if $Z(A)=\{0\}$, then  it is easy to see that $[C(A), QC(A)]=\{0\}$.
\end{proof}
\muathrm{e}ction{BiHom-Poisson modules}
First, let recall the following.
\betaegin{defn}
Let $(A,\{\cdot,\cdot\}_A, \muu_A )$ be a Poisson algebra. Then a left Poisson module structure on a
left $A$-module over $A$ is  linear maps
$\muu_M,\{\cdot,\cdot\}_M: A\otimes M\lambdaongrightarrow M$
such that
\betaegin{eqnarray}
\muu_M(a,\muu_M(b,m))=\muu_M(\muu_A(a,b),m)\lambdaabel{wid}\\
\{\{a,b\}_A,m\}_M=\{a,\{b,m\}_M \}_M-\{b,\{a,m\}_M\}_M \\
\{a,\muu_M(b,m)\}_M=\muu_M(\{a, b\}_A,m)+ \muu_M(b,\{a, m\}_M) \\
\{\muu_A(a,b), m\}_M =\muu_M(a,\{b, m\}_M)+\muu_M(b,\{a, m\}_M)
\end{eqnarray}
for any $a, b\in A$ and $m\in M.$
\end{defn}
\betaegin{rmk}
 In \cite{LWZ}, Poisson algebras are defined without the associativity assumption and then, left Poisson modules are defined without the identity (\rhoef{wid}).
In a similar way, basing on the definition above, one can defined a right Poisson module.
\end{rmk}
\betaegin{defn}
Let $(A, \{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a BiHom-Poisson algebra.
\betaegin{enumerate}
\item A left BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$  with  structure maps  $\lambdaambda:A\otimes V\lambdaongrightarrow V$  and  $\muathrm{h}o:A\otimes V\lambdaongrightarrow V$ such that the following equalities hold:
\betaegin{eqnarray}
\lambdaambda(\alphalpha(x),\lambdaambda(y,v))&=&\lambdaambda(\muu(x,y),\psi(v))\lambdaabel{lbhm1}\\
\muathrm{h}o(\{\betaeta(x),y \},\psi(v))&=&\muathrm{h}o(\alphalpha\betaeta(x),\muathrm{h}o(y,v))-\muathrm{h}o(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))\lambdaabel{lbhm2}\\
\muathrm{h}o(\alphalpha\betaeta(x),\lambdaambda(y,v))&=& \lambdaambda(\{\betaeta(x),y\},\psi(v))+\lambdaambda(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))\lambdaabel{lbhm3}\\
\muathrm{h}o(\muu(\betaeta(x),y),\psi(v))&=&\lambdaambda(\alphalpha\betaeta(x),\muathrm{h}o(y,v))+\lambdaambda(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))
\lambdaabel{lbhm4}
\end{eqnarray}
\item A right BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$  with  structure maps  $\wedge:V\otimes A\lambdaongrightarrow V$  and  $\deltaelta:V\otimes A\lambdaongrightarrow V$ such that the following equalities hold:
\betaegin{eqnarray}
\wedge(\wedge(v,x),\betaeta(y))&=&\wedge(\phi(v),\muu(x,y))\lambdaabel{ebhm1}\\
\deltaelta(\phi(v),\{x,\alphalpha(y)\})&=&\deltaelta(\deltaelta(v,x),\alphalpha\betaeta(y))-\deltaelta(\deltaelta(v,\betaeta(x)),\alphalpha(y))\lambdaabel{rbhm2}\\
\deltaelta(\wedge(v,x),\alphalpha\betaeta(y))&=&\deltaelta(\phi(v),\{x,\alphalpha(y)\})+
\wedge(\deltaelta(v,\betaeta(x)),\alphalpha(y))\lambdaabel{rbhm3}\\
\deltaelta(\phi(v),\muu(x,\alphalpha(y)))&=&\wedge(\deltaelta(v,x),\alphalpha\betaeta(y))+\wedge(
\deltaelta(v,\betaeta(x)),\alphalpha(y))\lambdaabel{rbhm4}
\end{eqnarray}
\end{enumerate}
\end{defn}
\betaegin{rmk}
\betaegin{enumerate}
\item A left BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$  with  structure maps  $\lambdaambda:A\otimes V\lambdaongrightarrow V$  and  $\muathrm{h}o:A\otimes V\lambdaongrightarrow V$ such that $(V,\phi,\psi,\lambdaambda)$ is a left Bihom-associative $A$-module, $(V,\phi,\psi,\muathrm{h}o)$ is a left BiHom-Lie $A$-module 
\cite{hadimi}\cite{yong} and (\rhoef{lbhm3}) and (\rhoef{lbhm4}) hold.  Similarly a right BiHom-Poisson $A$-module is a BiHom-module $(V,\phi,\psi)$  with  structure maps  $\wedge:V\otimes A\lambdaongrightarrow V$  and  $\deltaelta:V\otimes A\lambdaongrightarrow V$ such that $(V,\phi,\psi,\wedge)$ is a right Bihom-associative $A$-module, $(V,\phi,\psi,\deltaelta)$ is a right BiHom-Lie $A$-module and (\rhoef{rbhm3}) and (\rhoef{rbhm4}) hold.
\item If $\alphalpha=\betaeta=Id$ and $\phi=\psi=Id$, we recover a left (respectively a right) Poisson module. Thus if $(A,\{\cdot,\cdot\}, \muu)$ 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 $\muathbb{A}$-module where $\muathbb{A}=(A, \{\cdot,\cdot\}, \muu,  Id_{A},
Id_{A})$ is a BiHom-Poisson algebra.
\end{enumerate}
\end{rmk}
The following results allow to give some examples of left BiHom-Poisson $A$-modules.
\betaegin{prop}\lambdaabel{prop1}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a regular BiHom-Poisson algebra. Then $(A,\alphalpha,\betaeta)$ is a left BiHom-Poisson $A$-module where the structure maps are $\lambdaambda(a,b)=\muu(a,b)$ and $\muathrm{h}o(a,b)=\{a,b\}$. More generally, if $B$ is a left  BiHom-ideal of $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$, then $(B,\alphalpha,\betaeta)$ is a left BiHom-Poisson
$A$-module where the structure maps are $\lambdaambda(a,x)=\muu(a,x),\ \muathrm{h}o(x,a)=\{x,a\}$ for all $x\in B$ and $(a,b)\in A^{\times 2}$.
\end{prop}
\betaegin{proof}
The fact that $\lambdaambda$ and $\muathrm{h}o$ are structure maps follows from the multiplicativity of $\alphalpha$ and  $\betaeta$ with respect to $\muu$ and $\{\cdot,\cdot\}.$
Next, observe that from the BiHom-commutativity of $\muu$ and the Bihom-skew-symmetry of
$\{\cdot,\cdot\}$ that  $\muu(x,y)=\muu(\alphalpha^{-1}\betaeta(y),\alphalpha\betaeta^{-1}(x))$ and
$\{x,y\}=-\{\alphalpha^{-1}\betaeta(y),\alphalpha\betaeta^{-1}(x)\}$ for all $x,y\in A.$ Now, pick $(x,y,v)\in A^{\times 3}$ then, we have by the BiHom-associativity
\betaegin{eqnarray}
&&\lambdaambda(\alphalpha(x),\lambdaambda(y,v))=\muu(\alphalpha(x),\muu(y,v))=\muu(\muu(x,y),\betaeta(v))=
\lambdaambda(\muu(x,y),\betaeta(v))\nonumber
\end{eqnarray}
Next, compute (\rhoef{lbhm2}) using the BiHom-Jacobi identity in the third line, as follows
\betaegin{eqnarray}
&&\muathrm{h}o(\{\betaeta(x),y\},\betaeta(v))=\{\{\betaeta(x),y\},\betaeta(v)\}=
\{\alphalpha^{-1}\betaeta(\betaeta(v)),\alphalpha\betaeta^{-1}\{\alphalpha^{-1}\betaeta(y),\alphalpha\betaeta^{-1}(\betaeta(x))\}\}\nonumber\\
&&=\{\betaeta^2(\alphalpha^{-1}(v)),\{y,\alphalpha^2\betaeta^{-1}(x)\}\}=
\{\betaeta^2(\alphalpha^{-1}(v)),\{\betaeta(\betaeta^{-1}(y)),\alphalpha(\alphalpha\betaeta^{-1}(x))\}\}
\nonumber\\
&&=-\{\betaeta^2(\betaeta^{-1}(y)),\{\betaeta(\alphalpha\betaeta^{-1}(x)),\alphalpha(\alphalpha^{-1}(v))\}\}
-\{\betaeta^2(\alphalpha\betaeta^{-1}(x)),\{\betaeta(\alphalpha^{-1}(v)),\alphalpha(\betaeta^{-1}(y))\}\}
\nonumber\\
&&=-\{\betaeta(y),\{\alphalpha(x),v)\}\}
-\{\alphalpha\betaeta(x),\{\betaeta\alphalpha^{-1}(v),\alphalpha\betaeta^{-1}(y)\}\}=-\{\betaeta(y),\{\alphalpha(x),v)\}\}\nonumber\\
&&+\{\alphalpha\betaeta(x),\{y,v\}\}=-\muathrm{h}o(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))+\muathrm{h}o(\alphalpha\betaeta(x),\muathrm{h}o(y,v))\nonumber
\end{eqnarray}
Similarly, using (\rhoef{BHL}), we compute
\betaegin{eqnarray}
&&\muathrm{h}o(\alphalpha\betaeta(x),\lambdaambda(y,v))=\{\alphalpha\betaeta(x),\muu(y,v)\}
=\muu(\{\betaeta(x),y\},\betaeta(v))+\muu(\betaeta(y),\{\alphalpha(x),v\})\nonumber\\
&&=\lambdaambda(\{\betaeta(x),y\},\betaeta(v))+\lambdaambda(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))
 \mubox{\ which is (\rhoef{lbhm3})} \nonumber
\end{eqnarray}
Finally, we obtain (\rhoef{lbhm4}) as follows
\betaegin{eqnarray}
&& \muathrm{h}o(\muu(\betaeta(x),y),\betaeta(v))=\{\muu(\betaeta(x),y),\betaeta(v)\}=
-\{\alphalpha^{-1}\betaeta(\betaeta(v)),\alphalpha\betaeta^{-1}\muu(\betaeta(x),y)\}\nonumber\\
&&=-\{\alphalpha\betaeta(\alphalpha^{-2}\betaeta(v)),\muu(\alphalpha(x),\alphalpha\betaeta^{-1}(y))\}
=-\muu(\{\betaeta(\alphalpha^{-2}\betaeta(v)),\alphalpha(x)\},\betaeta(\betaeta^{-1}\alphalpha(y)))
\nonumber\\
&&-\muu(\betaeta(\alphalpha(x)),\{\alphalpha(\alphalpha^{-2}\betaeta(v)),\alphalpha\betaeta^{-1}(y)\})
\mubox{\  ( by (\rhoef{BHL})  )} \nonumber\\
&&=-\muu(\{\betaeta^2\alphalpha^{-2}(v),\alphalpha(x)\},\alphalpha(y))
-\muu(\betaeta\alphalpha(x),\{\alphalpha^{-1}\betaeta(v),\alphalpha\betaeta^{-1}(y)\})\nonumber\\
&&=-\muu(\alphalpha^{-1}\betaeta(\alphalpha(y)),\alphalpha\betaeta^{-1}\{\betaeta^2\alphalpha^{-2}(v),\alphalpha(x)\})+\muu(\alphalpha\betaeta(x),\{y,v\})\nonumber\\
&&=-\muu(\betaeta(y),\{\betaeta\alphalpha^{-1}(v),\alphalpha\betaeta^{-1}(\alphalpha(x))\})+\muu(\alphalpha\betaeta(x),\{y,v\})\nonumber\\
&&=\muu(\betaeta(y),\{\alphalpha(x),v\})+\muu(\alphalpha\betaeta(x),\{y,v\})
=\lambdaambda(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))+\lambdaambda(\alphalpha\betaeta(x),\muathrm{h}o(y,v)).
\nonumber
\end{eqnarray}
Hence $(A,\alphalpha,\betaeta)$ is a left BiHom-Poisson $A$-module.
Similarly, we prove that more generally, any two-sided BiHom-ideal $(B,\alphalpha,\betaeta)$  of $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ is a left BiHom-Poisson
$A$-module.
\end{proof}
\betaegin{rmk}
The analogous of Proposition \rhoef{prop1} can be proved for right BiHom-Poisson algebras.
\end{rmk}
More generally, we prove:
\betaegin{prop}
If $f:(A,\{\cdot,\cdot\}_A,\muu_A, \alphalpha_{A},\betaeta_{A})\lambdaongrightarrow(B,\{\cdot,\cdot\}_A, \muu_B,\alphalpha_{B},\betaeta_{B})$ is a morphism of BiHom-Poisson algebras and $\alphalpha_B$ and $\betaeta_B$ are invertible then, $(B,\alphalpha_{B},\betaeta_{B})$ becomes a
left BiHom-Poisson $A$-module via $f$, i.e, the structure maps are defined as $\lambdaambda(a,b)=\muu_{B}(f(a),b)$ and $\muathrm{h}o(a,b)=\{f(a),b\}_B$ for all
$(a,b)\in A\times B$.
\end{prop}
\betaegin{proof}
The fact that $\lambdaambda$ and $\muathrm{h}o$ are structure maps follows from the multiplicativity of $\alphalpha_B$ and  $\betaeta_B$ with respect to $\muu_B$ and $\{\cdot,\cdot\}_B.$
Next, observe that from the BiHom-commutativity of $\muu_B$ and the Bihom-skew-symmetry of
$\{\cdot,\cdot\}_B$ that $\muu_B(b_1,b_2)=\muu_B(\alphalpha^{-1}\betaeta(b_2),\alphalpha\betaeta^{-1}(b_1))$ and
$\{b_1,b_2\}_B=-\{\alphalpha_B^{-1}\betaeta_B(b_2),\alphalpha_B\betaeta_B^{-1}(b_1)\}_B$ 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 $B$
\betaegin{eqnarray}
&&\lambdaambda(\alphalpha_A(x),\lambdaambda(y,v))=\muu_B(f\alphalpha_A(x),\muu_B(f(y),v))=
\muu_B(\alphalpha_B(f(x)),\muu_B(f(y),v))\nonumber\\
&&=\muu_B(f\muu_B(x,y),\betaeta_B(v))=
\lambdaambda(\muu_B(x,y),\betaeta_B(v))\nonumber
\end{eqnarray}
Next, compute (\rhoef{lbhm2}) using the BiHom-Jacobi identity in the third line, as follows
\betaegin{eqnarray}
&&\muathrm{h}o(\{\betaeta_A(x),y\}_A,\betaeta_B(v))=\{f\{\betaeta(x),y\}_A,\betaeta(v)\}_B=
\{\{\betaeta_B(f(x)),f(y)\}_B,\betaeta_B(v)\}_B\nonumber\\
&&=\{\alphalpha_B^{-1}\betaeta_B(\betaeta_B(v)),\alphalpha_B\betaeta_B^{-1}\{\alphalpha_B^{-1}\betaeta_B f(y),\alphalpha_B\betaeta_B^{-1}(\betaeta_B f(x))\}_B\}_B\nonumber\\
&&=\{\betaeta_B^2(\alphalpha_B^{-1}(v)),\{f(y),\alphalpha_B^2\betaeta_B^{-1}f(x)\}_B\}_B\nonumber\\
&&=
\{\betaeta_B^2(\alphalpha_B^{-1}f(v)),\{\betaeta_B(\betaeta_B^{-1}f(y)),\alphalpha_B(\alphalpha_B\betaeta_B^{-1}f(x))\}_B\}_B
\nonumber\\
&&=-\{\betaeta_B^2(\betaeta_B^{-1}f(y)),\{\betaeta_B(\alphalpha_B\betaeta_B^{-1}f(x)),\alphalpha_B(\alphalpha_B^{-1}(v))\}_B\}_B\nonumber\\
&&-\{\betaeta_B^2(\alphalpha_B\betaeta_B^{-1}f(x)),\{\betaeta_B(\alphalpha_B^{-1}(v)),\alphalpha_B(\betaeta_B^{-1}f(y))\}_B\}_B
\nonumber\\
&&=-\{\betaeta_B(f(y)),\{\alphalpha_B(f(x)),v)\}_B\}_B
-\{\alphalpha_B\betaeta_B(f(x)),\{\betaeta_B\alphalpha_B^{-1}(v),\alphalpha_B\betaeta_B^{-1}(f(y))\}_B\}_B\nonumber\\
&&=-\{\betaeta_B(f(y)),\{\alphalpha_B(f(x)),v)\}_B\}_B+\{\alphalpha_B\betaeta_B(f(x)),\{f(y),v\}_B\}_B\nonumber\\
&&=-\{f(\betaeta_A(y)),\{f(\alphalpha_A(x)),v)\}_B\}_B+\{f(\alphalpha_A\betaeta_A(x)),\{f(y),v\}_B\}_B\nonumber\\
&&=-\muathrm{h}o(\betaeta(y),\muathrm{h}o(\alphalpha(x),v))+\muathrm{h}o(\alphalpha\betaeta(x),\muathrm{h}o(y,v))\nonumber
\end{eqnarray}
Similarly, using (\rhoef{BHL}) for $B$ and $f$ is a morphism, we compute
\betaegin{eqnarray}
&&\muathrm{h}o(\alphalpha_A\betaeta_A(x),\lambdaambda(y,v))=\{f(\alphalpha_A\betaeta_A(x)),\muu_B(f(y),v)\}_B
=\{\alphalpha_B\betaeta_B(f(x)),\muu_B(f(y),v)\}_B\nonumber\\
&&=\muu_B(\{\betaeta_B(f(x)),f(y)\},\betaeta_B(v))+\muu_B(\betaeta_B(f(y)),\{\alphalpha_B(f(x)),v\}_B)\nonumber\\
&&=\muu_B(f\{\betaeta_A(x),y\}_A,\betaeta_B(v))+\muu_B(f(\betaeta_A(y)),\{f(\alphalpha_B(x)),v\}_B)\nonumber\\
&&=\lambdaambda(\{\betaeta_A(x),y\}_A,\betaeta_B(v))+\lambdaambda(\betaeta_A(y),\muathrm{h}o(\alphalpha_A(x),v))
 \mubox{\ which is (\rhoef{lbhm3})} \nonumber
\end{eqnarray}
Finally, using $f$ is a morphism we obtain (\rhoef{lbhm4}) as follows
\betaegin{eqnarray}
&& \muathrm{h}o(\muu_A(\betaeta_A(x),y),\betaeta_B(v))=\{f\muu_A(\betaeta_A(x),y),\betaeta_B(v)\}_B\nonumber\\
&&=-\{\alphalpha_B^{-1}\betaeta_B(\betaeta_B(v)),\alphalpha_B\betaeta_B^{-1}f\muu_A(\betaeta_A(x),y)\}_B\nonumber\\
&&=-\{\alphalpha_B\betaeta_B(\alphalpha_B^{-2}\betaeta_B(v)),\muu_B(\alphalpha_B(f(x)),\alphalpha_B\betaeta_B^{-1}(f(y)))\}_B\nonumber\\
&&=-\muu_B(\{\betaeta_B(\alphalpha_B^{-2}\betaeta_B(v)),\alphalpha_B(f(x))\}_B,\betaeta_B(\betaeta_B^{-1}\alphalpha_B f(y)))
\nonumber\\
&&-\muu_B(\betaeta_B(\alphalpha_B f(x)),\{\alphalpha_B(\alphalpha_B^{-2}\betaeta_B(v)),\alphalpha_B\betaeta_B^{-1}(f(y))\})
\mubox{\  ( by (\rhoef{BHL})  in $B$ )} \nonumber\\
&&=-\muu_B(\{\betaeta_B^2\alphalpha_B^{-2}(v),\alphalpha_B(f(x))\}_B,\alphalpha_B(f(y)))\nonumber\\
&&-\muu_B(\betaeta_B\alphalpha_B(f(x)),\{\alphalpha_B^{-1}\betaeta_B(v),\alphalpha_B\betaeta_B^{-1}(f(y))\}_B)\nonumber\\
&&=-\muu_B(\alphalpha_B^{-1}\betaeta_B(\alphalpha_B f(y)),\alphalpha_B\betaeta_B^{-1}\{\betaeta_B^2\alphalpha_B^{-2}(v),\alphalpha_B(f(x))\}_B)+\muu_B(\alphalpha_B\betaeta_B(f(x)),\{f(y),v\}_B)\nonumber\\
&&=-\muu_B(\betaeta_B(f(y)),\{\betaeta_B\alphalpha_B^{-1}(v),\alphalpha_B\betaeta_B^{-1}(\alphalpha_B f(x))\})+\muu_B(\alphalpha_B\betaeta_B(f(x)),\{f(y),v\}_B)\nonumber\\
&&=\muu_B(\betaeta_B(f(y)),\{\alphalpha_B(f(x)),v\}_B)+\muu_B(\alphalpha_B\betaeta_B(f(x)),\{f(y),v\}_B)\nonumber\\
&&=\muu_B(f(\betaeta_A(y)),\{f(\alphalpha_A(x)),v\}_B)+\muu_B(f(\alphalpha_A\betaeta_A(x)),\{f(y),v\}_B)\nonumber\\
&&=\lambdaambda(\betaeta_A(y),\muathrm{h}o(\alphalpha_A(x),v))+\lambdaambda(\alphalpha_A\betaeta_A(x),\muathrm{h}o(y,v))
\nonumber
\end{eqnarray}
Hence $(B,\alphalpha_{B},\betaeta_{B})$ is a
left BiHom-Poisson $A$-module.
\end{proof}
Similarly, we can prove:
\betaegin{prop}
If $f:(A,\{\cdot,\cdot\}_A, \muu_A, \alphalpha_{A},\betaeta_{A})\lambdaongrightarrow(B,\{\cdot,\cdot\}_B, \muu_B, \alphalpha_{B},\betaeta_{B})$ is a morphism of BiHom-Poisson algebras and $\alphalpha_B$ and $\betaeta_B$ are invertible then, $(B,\alphalpha_{B},\betaeta_{B})$ becomes a
right BiHom-Poisson $A$-module via $f$, i.e, the structure maps are defined as $\lambdaambda(b,a)=\muu_{B}(b,f(a))$ and $\muathrm{h}o(b,a)=\{b,f(a)\}_B$ for all
$(a,b)\in A\times B$.
\end{prop}
As the case of BiHom-alternative and BiHom-Jordan algebras \cite{LA}, in order to give another example of left BiHom-Poisson modules, let us consider the following:
\betaegin{defn} An abelian extension of BiHom-Poisson algebras is a short exact sequence of BiHom-Poisson algebras
$$0\lambdaongrightarrow(V,\alphalpha_V,\betaeta_V)\sigmatackrel{\mubox{i}}{\lambdaongrightarrow} (A,\{\cdot,\cdot\}_A, \muu_A, \alphalpha_A,\betaeta_A)\sigmatackrel{\mubox{$\pi$}}{\lambdaongrightarrow} (B,\{\cdot,\cdot\}_B, \muu_B, \alphalpha_B,\betaeta_B)\lambdaongrightarrow 0$$
 where $(V,\alphalpha_V,\betaeta_B)$ is a trivial BiHom-Poisson algebra, $i$ and $\pi$ are morphisms of BiHom-algebras. Furthermore,  if there exists a morphism
$s :(B,\{\cdot,\cdot\}_B, \muu_B, \alphalpha_B,\betaeta_B)\lambdaongrightarrow\\ (A,\{\cdot,\cdot\}_A, \muu_A, \alphalpha_A, \betaeta_A)$ such that $\pi\circ s = id_B$ then the abelian extension is said to be split and $s$ is called a section of $\pi.$
\end{defn}
\betaegin{ex}
Given an abelian extension as in the previous definition,  the BiHom-module
$(V,\alphalpha_V, \betaeta_V)$ inherits a structure of a left BiHom-Poisson $B$-module and the actions of the BiHom-algebra $(B,\{\cdot,\cdot\}_B,\muu_B, \alphalpha_B,\betaeta_B)$  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 $\lambdaambda(x, v):=\muu_A(\tilde{x},i(v))$ and $\muathrm{h}o(x, v):=\{\tilde{x},i(v)\}_A.$
These are well-defined, as another lift $\tilde{x'}$
of $x$ is written $\tilde{x'}=\tilde{x}+v'$ for some $v'\in V$ and thus
$\lambdaambda(x,v)=\muu_A(\tilde{x},i(v))=\muu_A(\tilde{x'},i(v))$ and $\muathrm{h}o(x,v)=\{\tilde{x},i(v)\}_A=\{\tilde{x'},i(v)\}_A$ 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.
\end{ex}
The next result allow to construct a sequence of left BiHom-Poisson modules from a given one.
\betaegin{prop}\lambdaabel{HJB-HJB}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ be a BiHom-Poisson algebra and  $V_{\phi,\psi}=(V,\phi, \psi)$  be a left BiHom-Poisson $A$-module with the structure maps $\lambdaambda$ and $\muathrm{h}o$.  Then for each $n, m\in\muathbb{N},$ the maps
\betaegin{eqnarray}
\lambdaambda^{(n,m)}=\lambdaambda\circ(\alphalpha^n\betaeta^m\otimes Id_V)\lambdaabel{nmj1}\\
\muathrm{h}o^{(n,m)}=\muathrm{h}o\circ(\alphalpha^n\betaeta^m\otimes Id_V)\lambdaabel{nmj2}
\end{eqnarray}
give the BiHom-module $(V,\phi, \psi)$ the structure of a left BiHom-Poisson $A$-module that we denote
by $V_{\phi,\psi}^{(n,m)}.$
\end{prop}
\betaegin{proof}
Since the structure map $\lambdaambda$ is a morphism of BiHom-modules, we get:
\betaegin{eqnarray}
&&\phi\lambdaambda^{(n,m)}=\phi\muathrm{h}o_l\circ(\alphalpha^n\betaeta^m\otimes Id_V)
\mubox{ \textsl{( by (\rhoef{nmj1}) )}}\nonumber\\
&=&\lambdaambda\circ(\alphalpha^{n+1}\betaeta^m\otimes\phi)
=\lambdaambda\circ(\alphalpha^n\betaeta^m\otimes Id_V)\circ(\alphalpha\otimes\phi)
=\lambdaambda^{(n,m)}\circ(\alphalpha\otimes\phi)\nonumber
\end{eqnarray}
 Similarly, we get that $\psi\lambdaambda^{(n,m)}=\lambdaambda^{(n,m)}\circ(\betaeta\otimes\psi),\
 \phi\muathrm{h}o^{(n,m)}=\muathrm{h}o^{(n,m)}\circ(\phi\otimes\alphalpha)$ and $\psi\muathrm{h}o^{(n,m)}=\muathrm{h}o^{(n,m)}\circ(\psi\otimes\betaeta).$  Thus $\muathrm{h}o_l^{(n)}$ and $\muathrm{h}o_r^{(n)}$ are morphisms of BiHom-modules. First, pick $(x,y)\in A^{\times 2}$ and $v\in V,$ then using (\rhoef{lbhm1}) in the second line for $V_{\phi,\psi},$ we get
 \betaegin{eqnarray}
&& \lambdaambda^{(n,m)}(\alphalpha(x),\lambdaambda^{(n,m)}(y,v))=\lambdaambda(\alphalpha^{n+1}\betaeta^m(x),
 \lambdaambda(\alphalpha^n\betaeta^m(y),v))\nonumber\\
 &&=\lambdaambda(\muu(\alphalpha^n\betaeta^m(x),\alphalpha^n\betaeta^m(y)),\psi(v))=\lambdaambda^{(n,m)}
 (\muu(x,y),\psi(v))\nonumber
 \end{eqnarray}
 Secondly, we compute
 \betaegin{eqnarray}
 &&\muathrm{h}o^{(n,m)}(\{\betaeta(x),y\},\psi(v))=\muathrm{h}o(\{\alphalpha^n\betaeta^{m+1}(x),\betaeta(y)\},
 \psi(v))\nonumber\\
 &&=\muathrm{h}o(\alphalpha\betaeta\alphalpha^n\betaeta^m(x),\muathrm{h}o(\alphalpha^n\betaeta^m(y),v))-\muathrm{h}o(\betaeta\alphalpha^n\betaeta^m(y),\muathrm{h}o(\alphalpha\alphalpha^n\betaeta^m(x),v)) \mubox{ ( by (\rhoef{lbhm2}) in  $V_{\phi,\psi}$ )} \nonumber\\
 &&=\lambdaambda^{(n,m)}(\alphalpha\betaeta(x),\lambdaambda^{(n,m)}(y,v))-\lambdaambda^{(n,m)}(\betaeta(y),\lambdaambda^{(n,m)}(\alphalpha(y),v))\nonumber
 \end{eqnarray}
 Next, we obtain
 \betaegin{eqnarray}
&&\muathrm{h}o^{(n,m)}(\betaeta\alphalpha(x),\lambdaambda^{(n,m)}(y,v))=\muathrm{h}o(\alphalpha\betaeta\alphalpha^n\betaeta^m(x),\lambdaambda(\alphalpha^n\betaeta^m(y),v))\nonumber\\
&&=\lambdaambda\{\betaeta\alphalpha^n\betaeta^m(x),\alphalpha^n\betaeta^m(y)\},\psi(v))+
\lambdaambda(\betaeta\alphalpha^n\betaeta^(y),\muathrm{h}o(\alphalpha\alphalpha^n\betaeta^m(x),v)) \mubox{ ( by (\rhoef{lbhm3}) in  $V_{\phi,\psi}$ )} \nonumber\\
&&=\lambdaambda^{(n,m)}(\{\betaeta(x),y\},\psi(v))+\lambdaambda^{(n,m)}(\betaeta(y),
\muathrm{h}o^{(n,m)}(\alphalpha(x),v))\nonumber
 \end{eqnarray}
 Finally, we compute
 \betaegin{eqnarray}
 &&\muathrm{h}o^{(n,m)}(\muu(\betaeta(x),y),\psi(v))=\muathrm{h}o(\muu(\betaeta\alphalpha^n\betaeta^m(x),\alphalpha^n\betaeta^m(y)),\psi(v))\nonumber\\
 &&=\lambdaambda(\alphalpha\betaeta\alphalpha^n\betaeta^m(x),\muathrm{h}o(\alphalpha^n\betaeta^m(y),v))+
 \lambdaambda(\betaeta\alphalpha^n\betaeta^m(y),\muathrm{h}o(\alphalpha\alphalpha^n\betaeta^m(x),v))
 \mubox{ ( by (\rhoef{lbhm3}) in  $V_{\phi,\psi}$ )} \nonumber\\
 &&=\lambdaambda^{(n,m)}(\alphalpha\betaeta(x),\muathrm{h}o^{(n,m)}(y,v))+\lambdaambda^{(n,m)}(
 \betaeta(y), \muathrm{h}o^{(n,m)}(\alphalpha(x),v)).\nonumber
 \end{eqnarray}
 Hence, $V_{\phi,\psi}^{(n,m)}$ is a left BiHom-Poisson $A$-module.
\end{proof}
 Proposition \rhoef{HJB-HJB} reads for the case of right BiHom-Poisson module as:
\betaegin{prop}\lambdaabel{HJB-HJBr}
Let $(A,\{\cdot,\cdot\},\muu,\alphalpha,\betaeta)$ 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 $\deltaelta$.  Then for each $n, m\in\muathbb{N},$ the maps
\betaegin{eqnarray}
\wedge^{(n,m)}=\wedge\circ(Id_V\otimes\alphalpha^n\betaeta^m)\lambdaabel{nmj1}\\
\deltaelta^{(n,m)}=\deltaelta\circ(Id_V\otimes\alphalpha^n\betaeta^m)\lambdaabel{nmj2}
\end{eqnarray}
give the BiHom-module $(V,\phi, \psi)$ the structure of a right BiHom-Poisson $A$-module that we denote
by $V_{\phi,\psi}^{(n,m)}.$
\end{prop}
\betaegin{thm}\lambdaabel{HJB-JB}
Let $(A,\{\cdot,\cdot\}, \muu)$ be a Poisson algebra, $V$ be a left Poisson $A$-module with the structure maps
$\lambdaambda$, $\muathrm{h}o$ and $\alphalpha,\betaeta$ be  endomorphisms of the Jordan algebra $A$
and $\phi, \psi$ be  linear self-maps of $V$ such that $\phi\circ\lambdaambda=\lambdaambda\circ(\alphalpha\otimes\phi),$
$\phi\circ\muathrm{h}o=\muathrm{h}o\circ(\alphalpha\otimes\phi),$
$\psi\circ\lambdaambda=\lambdaambda\circ(\betaeta\otimes\psi)$ and
$\psi\circ\muathrm{h}o=\muathrm{h}o\circ(\betaeta\otimes\psi)$.
Write $A_{\alphalpha,\betaeta}$ for the BiHom-Poisson algebra $(A,\{\cdot,\cdot\}_{\alphalpha,\betaeta}=\{\cdot,\cdot\}(\alphalpha\otimes\betaeta),\muu_{\alphalpha,\betaeta}=\muu(\alphalpha\otimes\betaeta), \alphalpha, \betaeta)$ and
$V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:
\betaegin{eqnarray}
\tilde{\lambdaambda}=\lambdaambda\circ(\alphalpha\betaeta\otimes\psi) \mubox{ and }
\tilde{\muathrm{h}o}=\muathrm{h}o\circ(\alphalpha\betaeta\otimes\psi)
\end{eqnarray}
give the BiHom-module $V_{\phi,\psi}$ the structure of a left BiHom-Poisson $A_{\alphalpha,\betaeta}$-module.
\end{thm}
\betaegin{proof}
It is clear that $\tilde{\lambdaambda}$ and $\tilde{\muathrm{h}o}$ are morphisms of
BiHom-modules. Netx, pick $(x,y)\in A^{\times 2}$ and $v\in V,$ then
using (\rhoef{lbhm1}) for  $V_{\phi,\psi}$ in the second line  we get:
 \betaegin{eqnarray}
&& \tilde{\lambdaambda}(\alphalpha(x),\tilde{\lambdaambda}(y,v))=\lambdaambda(\alphalpha^2\betaeta(x),
 \lambdaambda(\alphalpha\betaeta^2(y),\psi^2(v)))\nonumber\\
 &&=\lambdaambda(\muu(\alphalpha^2\betaeta(x),\alphalpha\betaeta^2(y)),\psi^2(v))=\lambdaambda(\alphalpha\betaeta\muu(\alphalpha(x),\betaeta(y)),\psi^2(v))=\tilde{\lambdaambda}
 (\muu_{\alphalpha,\betaeta}(x,y),\psi(v))\nonumber
 \end{eqnarray}
 Secondly, we compute
 \betaegin{eqnarray}
 &&\tilde{\muathrm{h}o}(\{\betaeta(x),y\}_{\alphalpha,\betaeta},\psi(v))=\muathrm{h}o(\{\alphalpha^2\betaeta^2(x),\alphalpha\betaeta^2(y)\},
 \psi^2(v))\nonumber\\
 &&=\muathrm{h}o(\alphalpha^2\betaeta^2(x),\muathrm{h}o(\alphalpha\betaeta^2(y),\psi^2(v)))-\muathrm{h}o(\alphalpha\betaeta^2(y),\muathrm{h}o(\alphalpha^2\betaeta^2(x),\psi^2(v)) \mubox{ ( by (\rhoef{lbhm2}) in  $V_{\phi,\psi}$ )} \nonumber\\
 &&=\tilde{\lambdaambda}(\alphalpha\betaeta(x),\tilde{\lambdaambda}(y,v))-\tilde{\lambdaambda}(\betaeta(y),\tilde{\lambdaambda}(\alphalpha(y),v))\nonumber
 \end{eqnarray}
 Next, we obtain
 \betaegin{eqnarray}
&& \tilde{\muathrm{h}o}(\betaeta\alphalpha(x), \tilde{\lambdaambda}(y,v))=\muathrm{h}o(\alphalpha^2\betaeta(x),\lambdaambda(\alphalpha\betaeta^2(y),\psi^2(v)))\nonumber\\
&&=\lambdaambda\{\alphalpha^2\betaeta(x),\alphalpha\betaeta^2(y)\},\psi^2(v))+
\lambdaambda(\alphalpha\betaeta^2(y),\muathrm{h}o(\alphalpha^2\betaeta(x),\psi^2(v))) \mubox{ ( by (\rhoef{lbhm3}) in  $V_{\phi,\psi}$ )} \nonumber\\
&&= \tilde{\lambdaambda}(\{\betaeta(x),y\}_{\alphalpha,\betaeta},\psi(v))+ \tilde{\lambdaambda}(\betaeta(y),
 \tilde{\muathrm{h}o}(\alphalpha(x),v))\nonumber
 \end{eqnarray}
 Finally, we compute
 \betaegin{eqnarray}
 && \tilde{\muathrm{h}o}(\muu_{\alphalpha,\betaeta}(\betaeta(x),y),\psi(v))=\muathrm{h}o(\muu(\alphalpha^2\betaeta^2(x),\alphalpha\betaeta^2(y)),\psi^2(v))\nonumber\\
 &&=\lambdaambda(\alphalpha^2\betaeta^2(x),\muathrm{h}o(\alphalpha\betaeta^2(y),\psi^2(v)))+
 \lambdaambda(\alphalpha\betaeta^2(y),\muathrm{h}o(\alphalpha^2\betaeta^2(x),\psi^2(v)))
 \mubox{ ( by (\rhoef{lbhm3}) in  $V_{\phi,\psi}$ )} \nonumber\\
 &&= \tilde{\lambdaambda}(\alphalpha\betaeta(x), \tilde{\muathrm{h}o}(y,v))+ \tilde{\lambdaambda}
 (\betaeta(y),  \tilde{\muathrm{h}o}(\alphalpha(x),v)))\nonumber
 \end{eqnarray}
 Hence, $V_{\phi,\psi}^{(n,m)}$ is a left BiHom-Poisson $A$-module.
\end{proof}
\betaegin{cor}
Let $(A,\{\cdot,\cdot\}, \muu)$ be a Poisson algebra, $V$ be a left Poisson $A$-module with the structure maps
$\lambdaambda$ and $\muathrm{h}o$, $\alphalpha,\ \betaeta$ be  endomorphisms of the Poisson algebra $A$ and $\phi, \psi$ be  linear self-maps of $V$ such that $\phi\circ\lambdaambda=\lambdaambda\circ(\alphalpha\otimes\phi),$
$\phi\circ\muathrm{h}o=\muathrm{h}o\circ(\alphalpha\otimes\phi),$
$\psi\circ\lambdaambda=\lambdaambda\circ(\betaeta\otimes\psi)$ and
$\psi\circ\muathrm{h}o=\muathrm{h}o\circ(\alphalpha\otimes\psi)$.\\
Write $A_{\alphalpha,\betaeta}$ for the BiHom-Poisson algebra $(A,\{\cdot,\cdot\}_{\alphalpha,\betaeta}=\{\alphalpha\otimes\betaeta\}, \muu_{\alphalpha,\betaeta}=\muu(\alphalpha\otimes\betaeta), \alphalpha, \betaeta)$ and
$V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:
\betaegin{eqnarray}
\tilde{\lambdaambda}^{(n,m)}=\lambdaambda\circ(\alphalpha^{n+1}\betaeta^{m+1}\otimes\psi) \mubox{ and }
\tilde{\muathrm{h}o}^{(n,m)}=\muathrm{h}o\circ(\alphalpha^{n+1}\betaeta^{m+1}\otimes\psi)
\end{eqnarray}
give the BiHom-module $V_{\phi,\psi}$ the structure of a left BiHom-Poisson $A_{\alphalpha,\betaeta}$-module for all
$n, m\in\muathbb{N}$.
\end{cor}
\betaegin{proof} The proof follows from Proposition \rhoef{HJB-HJB} and Theorem \rhoef{HJB-JB}.
\end{proof}
Similarly, we can prove the following result which is the analogous of
Theorem \rhoef{HJB-JB} for right BiHom-Poisson modules.
\betaegin{thm}\lambdaabel{HJB-JB}
Let $(A,\{\cdot,\cdot\}, \muu)$ be a Poisson algebra, $V$ be a right Poisson $A$-module with the structure maps
$\lambdaambda$, $\muathrm{h}o$ and $\alphalpha,\betaeta$ be  endomorphisms of the Jordan algebra $A$
and $\phi, \psi$ be  linear self-maps of $V$ such that $\phi\circ\lambdaambda=\lambdaambda\circ(\alphalpha\otimes\phi),$
$\phi\circ\muathrm{h}o=\muathrm{h}o\circ(\alphalpha\otimes\phi),$
$\psi\circ\lambdaambda=\lambdaambda\circ(\betaeta\otimes\psi)$ and
$\psi\circ\muathrm{h}o=\muathrm{h}o\circ(\betaeta\otimes\psi)$.
Write $A_{\alphalpha,\betaeta}$ for the BiHom-Poisson algebra $(A,\{\cdot,\cdot\}_{\alphalpha,\betaeta}=\{\cdot,\cdot\}(\alphalpha\otimes\betaeta), \muu_{\alphalpha,\betaeta}=\muu(\alphalpha\otimes\betaeta), \alphalpha, \betaeta)$ and
$V_{\phi,\psi}$ for the BiHom-module $(V,\phi,\psi).$ Then the maps:
\betaegin{eqnarray}
\tilde{\wedge}=\wedge\circ(\phi\otimes \alphalpha\betaeta) \mubox{ and }
\tilde{\deltaelta}=\muathrm{h}o\circ(\phi\otimes\alphalpha\betaeta)
\end{eqnarray}
give the BiHom-module $V_{\phi,\psi}$ the structure of a right BiHom-Poisson $A_{\alphalpha,\betaeta}$-module.
\end{thm}
In the case of Poisson algebras, we can form semidirect products when given a left (or a right) module. Similarly, we have
\betaegin{thm}
Let $(A, \{\cdot,\cdot\}, \muu, \alphalpha, \betaeta)$ be a BiHom-Poisson algebra and  $(V,\phi, \psi)$  be a left $A$-module with the structure maps $\lambdaambda$ and $\muathrm{h}o$. Then $(A\oplus V,[\deltaot,\deltaot],\alphast, \tilde{\alphalpha}, \tilde{\betaeta})$ is
a BiHom-Poisson algebra where
$\alphast,[\cdot,\cdot]: (A\oplus V)^{\otimes 2}\lambdaongrightarrow A\oplus V,~$
$(a+u)\alphast(b+v):=\muu(a,b)+\lambdaambda(a,v)+\lambdaambda(\alphalpha^{-1}\betaeta(b),\psi^{-1}\phi(u)),$  $[a+u,b+v]:=\{a,b\}+\muathrm{h}o(a,v)-\muathrm{h}o(\alphalpha^{-1}\betaeta(b),\psi^{-1}\phi(u))$  and
$\tilde{\alphalpha}, \tilde{\betaeta}: A\oplus V\lambdaongrightarrow A\oplus V,$ $\tilde{\alphalpha}(a+u):=\alphalpha(a)+\phi(u)$  and $\tilde{\betaeta}(a+u):=\betaeta(a)+\psi(u)$ called the
semidirect product of the BiHom-Poisson $(A,\muu, \{\cdot,\cdot\},\alphalpha, \betaeta)$ and  $(V,\phi, \psi).$
\end{thm}
\betaegin{proof} Clearly, $\tilde{\alphalpha}$ and $\tilde{\betaeta}$ are multiplicative
with respect to $\alphast$ and $[\cdot,\cdot].$ Next
\betaegin{eqnarray}
&&\tilde{\betaeta}(a+u)\alphast\tilde{\alphalpha}(b+v)=(\betaeta(a)+\psi(u))\alphast(\alphalpha(b)+\phi(v))\nonumber\\
&&=\muu(\betaeta(a),\alphalpha(b))+\lambdaambda(\betaeta(a),\phi(v))+\lambdaambda(\betaeta(b),\phi(u)\nonumber\\
&&=\muu(\betaeta(b),\alphalpha(a))+\lambdaambda(\betaeta(b),\phi(u)+\lambdaambda(\betaeta(a),\phi(v)) \mubox{ ( by the BiHom-commutativity of $\muu$ )}\nonumber\\
&&=\tilde{\betaeta}(b+v)\alphast\tilde{\alphalpha}(u+v).
\end{eqnarray}
Next, pick  $(a,b,c)\in A^{\times 2}$ and $(u,v,w)\in V^{\times 2},$ then
\betaegin{eqnarray}
&&\Big((a+u)\alphast(b+v)\Big)\alphast\betaar{\betaeta}(c+w)=
\Big(\muu(a,b)+\lambdaambda(a,v)+\lambdaambda(\alphalpha^{-1}\betaeta(b),\psi^{-1}\phi(u))\Big)
\alphast(\betaeta(c)+\psi(w))\nonumber\\
&&=\muu(\muu(a,b),\betaeta(c))+\lambdaambda(\muu(a,b),\psi(w))+\lambdaambda(\alphalpha^{-1}\betaeta^2(c),
\lambdaambda(\betaeta^{-1}\alphalpha(a),\phi\psi^{-1}(v))\nonumber\\
&&+\lambdaambda(\alphalpha^{-1}\betaeta^2(c),\lambdaambda(b,(\psi^{-1}\phi)^2(u)))
\mubox{  ( using $\lambdaambda$ is a morphism )}\nonumber\\
&&=\muu(\alphalpha(a),\muu(b,c))+\lambdaambda(\muu(a,b),\psi(w))+\lambdaambda(\muu((\alphalpha^{-1}\betaeta)^2(c),\betaeta^{-1}\alphalpha(a)),\phi(v))
\nonumber\\
&&+\lambdaambda(\muu((\alphalpha^{-1}\betaeta)^2(c),b),\psi^{-1}\phi^2(u))
\mubox{  ( using  BiHom-associativity  and (\rhoef{lbhm1}) )}\\
&&=\muu(\alphalpha(a),\muu(b,c))+\lambdaambda(\muu(a,b),\psi(w))+\lambdaambda(\muu(a,\betaeta\alphalpha^{-1}(c)),\phi(v))
\nonumber\\
&&+\lambdaambda(\muu(\alphalpha^{-1}\betaeta(b),\alphalpha^{-1}\betaeta(c),\psi^{-1}\phi^2(u))
\mubox{  ( using  BiHom-commutativity )}\nonumber
\end{eqnarray}
Similarly, we prove that
\betaegin{eqnarray}
&&\betaar{\alphalpha}(a+u)\alphast\Big((b+v)\alphast(c+w)\Big)
=\muu(\alphalpha(a),\muu(b,c))+\lambdaambda(\muu(a,b),\psi(w))+\lambdaambda(\muu(a,\betaeta\alphalpha^{-1}(c)),\phi(v))\nonumber\\
&&+\lambdaambda(\muu(\alphalpha^{-1}\betaeta(b),\alphalpha^{-1}\betaeta(c),\psi^{-1}\phi^2(u))
\nonumber
\end{eqnarray}
Hence $(A\oplus V, \alphast,\betaar{\alphalpha}, \betaar{\betaeta})$ is a BiHom-commutative BiHom-associative algebra. Now, observe that $(A\oplus V,[\cdot,\cdot],\betaar{\alphalpha},\betaar{\betaeta})$ is a BiHom-Lie algebra (Proposition 4.9, \cite{GRAZIANI} ).
 Finally, let $(a,b,c)\in A^{\times 2}$ and $(u,v,w)\in V^{\times 2}.$ Then
 \betaegin{eqnarray}
 &&[\betaar{\alphalpha}\betaar{\betaeta}(a+u),(b+v)\alphast (c+w)]=[\alphalpha\betaeta(a)+\phi\psi(u),
 \muu(b,c)+\lambdaambda(b,w)+\lambdaambda(\alphalpha^{-1}\betaeta(),\psi^{-1}\phi(v)]\nonumber\\
 &&=\{\alphalpha\betaeta(a),\muu(b,c)\}+\muathrm{h}o(\alphalpha\betaeta(a),\lambdaambda(b,w))+\muathrm{h}o(\alphalpha\betaeta(a),\lambdaambda(\alphalpha^{-1}\betaeta(c),\psi^{-1}\phi(v)))
 \nonumber\\
 &&-\muathrm{h}o(\muu(\betaeta\alphalpha^{-1}(b),\betaeta\alphalpha^{-1}(c)),\psi\psi^{-1}\phi^2(u))
 =\muu(\{\betaeta(a),b\},\betaeta(c))\nonumber\\
 &&+\muu(\betaeta(c),\{\alphalpha(a),c\})+\lambdaambda(\{\betaeta(a),b\},\psi(w))
 +\lambdaambda(\betaeta(b),\muathrm{h}o(\alphalpha(a),w))+\lambdaambda(\{\betaeta(a),\alphalpha^{-1}\betaeta(c)\},\phi(v))\nonumber\\
 &&+\lambdaambda(\betaeta^2\alphalpha^{-1}(c),\muathrm{h}o(\alphalpha(a),\psi^{-1}\phi(v)))
 -\lambdaambda(\betaeta(b),\muathrm{h}o(\betaeta\alphalpha^{-1}(c),\psi^{-1}\phi^2(u)))\nonumber\\
 &&-\lambdaambda(\betaeta^2\alphalpha^{-1}(c),\muathrm{h}o(b,\psi^{-1}\phi^2(u)))
 \mubox{ ( by (\rhoef{BHL}}), (\rhoef{lbhm3}), (\rhoef{lbhm4}) )\nonumber\\
&& =\Big(\muu(\{\betaeta(a),b\},\betaeta(c))+ \lambdaambda(\{\betaeta(a),b\},\psi(w))+
 \lambdaambda(\betaeta^2\alphalpha^{-1}(c),\muathrm{h}o(\alphalpha(a),\psi^{-1}\phi(v)))\nonumber\\
 &&-\lambdaambda(\betaeta^2\alphalpha^{-1}(c),\muathrm{h}o(b,\psi^{-1}\phi^2(u)))  \Big)
  +\Big(\muu(\betaeta(c),\{\alphalpha(a),c\})+ \lambdaambda(\betaeta(b),\muathrm{h}o(\alphalpha(a),w))
   \nonumber\\
 &&  -\lambdaambda(\betaeta(b),\muathrm{h}o(\betaeta\alphalpha^{-1}(c),\psi^{-1}\phi^2(u)))+\lambdaambda(\{\betaeta(a),\alphalpha^{-1}\betaeta(c)\},\phi(v)) \Big)
 \mubox{ ( rearranging terms )}\nonumber\\
 &&=[\betaar{\betaeta}(a+u),(b+v)]\alphast\betaar{\betaeta}(c+w)+\betaar{\betaeta}(b+v)\alphast[\betaar{\alphalpha}(a+u),(c+w)]\nonumber
 \end{eqnarray}
 Hence, the conclusion follows.
\end{proof}
\betaegin{rmk} Consider the  split null extension $A\oplus V$ determined by the left BiHom-Poisson module $(V,\phi,\psi)$ for the BiHom-Poisson algebra
$(A, \{\cdot,\cdot\}, \muu, \alphalpha,\betaeta)$ 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\rhoightarrow A\oplus V $ given by $i(v)=(0,v)$ and a surjective homomorphism of BiHom-modules $\pi : A\oplus V\rhoightarrow A$ 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
$\sigmaigma: A\rhoightarrow A\oplus V$ given by $\sigmaigma(a)=(a,0)$ 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 $\pi.$
 \end{rmk}

\betaegin{defn}
Let $(A, \{\cdot,\cdot\}, \muu, \alphalpha,\betaeta)$ be a BiHom Poisson algebra. A skew-symmetric $n$-linear map $f: \underbrace{A\times \cdots \times A}_{n ~times}\rhoightarrow A$ that is a
derivation in each argument is called an $n$-BiHom-cochain, if it satisfies
$$\betaegin{array}{llllllllllll}
f(\alphalpha(x_1), \cdots, \alphalpha(x_n))&=&\alphalpha\circ f(x_{1}, \cdots, x_{n}),\lambdaabel{cochain2}\\
f(\betaeta(x_1), \cdots, \betaeta(x_n))&=&\betaeta\circ f(x_{1}, \cdots, x_{n}).\lambdaabel{cochain3}
\end{array}$$
The set of $n$-Hom-cochains is denoted by $C_{\alphalpha,\betaeta}^n(A,A)$, for $n\geq1$.
\end{defn}

\betaegin{defn}
Let $(A, \{\cdot,\cdot\}, \muu, \alphalpha,\betaeta)$ be a regular BiHom Poisson algebra. For $n=1,2,$ the coboundary operator $\deltaelta^{n}:C_{\alphalpha,\betaeta}^n(A,A)\rhoightarrow C_{\alphalpha,\betaeta}^{n+1}(A,A)$ is defined as follows:
\betaegin{eqnarray}
&&\deltaelta^{1}f(x,y)=\{\alphalpha(x),f(y)\}-\{f(x),\alphalpha(y)\}-f(\{\alphalpha^{-1}\betaeta(x),y\})\\
&&\deltaelta^{2}f(x,y,z)=\{\alphalpha\betaeta(x),f(y,z)\}-\{\alphalpha\betaeta(y),f(x,z)\}+
\{\alphalpha\betaeta(y),f(x,z)\}
\nonumber\\
&&-f(\{\alphalpha^{-1}\betaeta(x),y\},\betaeta(z))+f(\{\alphalpha^{-1}\betaeta(x),z\},\betaeta(y))-f(\{\alphalpha^{-1}\betaeta(y),z\},\betaeta(x))
\end{eqnarray}
\end{defn}

\betaegin{lem}
The coboundary operators $\deltaelta^i$ are well defined, for $i=1, 2$.
\end{lem}
\betaegin{proof}
For any $x,y,z\in A$ we have:
\betaegin{eqnarray}
&&\deltaelta^{1}f(\alphalpha(x),\alphalpha(y))=\{\alphalpha^2(x),f\alphalpha(y)\}-\{f\alphalpha(x),\alphalpha^2(y)\}-f(\{\alphalpha^{-1}\betaeta\alphalpha(x),\alphalpha y)\})\nonumber\\
&&=\{\alphalpha^2(x),\alphalpha f(y)\}-\{\alphalpha f(x),\alphalpha^2(y)\}-f(\{\alphalpha\alphalpha^{-1}\betaeta(x),\alphalpha(y)\})=\alphalpha\circ\deltaelta^1 f(x, y)\nonumber
\end{eqnarray}
and
\betaegin{eqnarray}
&&\deltaelta^{2}f(\alphalpha(x),\alphalpha(y),\alphalpha(z))=\{\alphalpha\betaeta\alphalpha(x),f(\alphalpha(y),\alphalpha(z))\}-\{\alphalpha\betaeta\alphalpha(y),f(\alphalpha(x),\alphalpha(z))\}
\nonumber\\
&&+
\{\alphalpha\betaeta\alphalpha(y),f(\alphalpha(x),\alphalpha(z))\}-f(\{\alphalpha^{-1}\betaeta\alphalpha(x),\alphalpha(y)\},\betaeta\alphalpha(z))+f(\{\alphalpha^{-1}\betaeta\alphalpha(x),\alphalpha(z)\},\betaeta\alphalpha(y))\nonumber\\
&&-f(\{\alphalpha^{-1}\betaeta\alphalpha(y),\alphalpha(z)\},\betaeta\alphalpha(x))=
\{\alphalpha\betaeta\alphalpha(x),\alphalpha f(y,z)\}-\{\alphalpha\betaeta\alphalpha(y),\alphalpha f(x,z)\}
\nonumber\\
&&+
\{\alphalpha\betaeta\alphalpha(y),\alphalpha f(x,z)\}-f(\alphalpha(\{\alphalpha^{-1}\betaeta(x),y\}),\alphalpha\betaeta(z))+f(\alphalpha(\{\alphalpha^{-1}\betaeta()x,z\}),\alphalpha\betaeta(y))\nonumber\\
&&-f(\alphalpha(\{\alphalpha^{-1}\betaeta(y),z\}),\alphalpha\betaeta(x))=\alphalpha\circ\deltaelta^2 f(x,y,z)\nonumber
\end{eqnarray}
In the same way, we obtain $\deltaelta^{1}f\circ\betaeta=\betaeta\circ\deltaelta^1f$ and $\deltaelta^{2}f\circ\betaeta=\betaeta\circ\deltaelta^2 f.$

Then $\deltaelta^i$ are well defined, for $i=1,2$.
\end{proof}
\betaegin{thm}
With notations as above, we have
$$\deltaelta^{2}\circ\deltaelta^{1}=0.$$
\end{thm}
\betaegin{proof} Let $f\in C_{\alphalpha,\betaeta}^1(A,A)$ and $(x,y,z)\in A^{\times 3}$ then, we have:
\betaegin{eqnarray}
&&\deltaelta^{2}\circ\deltaelta^{1} f(x,y,z)=
\{\alphalpha\betaeta(x),\deltaelta^{1} f(y,z)\}-\{\alphalpha\betaeta(y),\deltaelta^{1} f(x,z)\}+
\{\alphalpha\betaeta(y),\deltaelta^{1} f(x,z)\}
\nonumber\\
&&-\deltaelta^{1} f(\{\alphalpha^{-1}\betaeta(x),y\},\betaeta(z))+\deltaelta^{1} f(\{\alphalpha^{-1}\betaeta(x),z\},\betaeta(y))-\deltaelta^{1} f(\{\alphalpha^{-1}\betaeta(y),z\},\betaeta(x))\nonumber\\
&=&\{\alphalpha\betaeta(x),\{\alphalpha(y),f(z)\}\}-\{\alphalpha\betaeta(x),\{\alphalpha(z),f(y)\}\}
-\{\alphalpha\betaeta(x),f(\{\alphalpha^{-1}\betaeta(y),z\})\}\nonumber\\
&&-\{\alphalpha\betaeta(y),\{\alphalpha(x),f(z)\}\}+\{\alphalpha\betaeta(y),\{\alphalpha(z),f(x)\}\}
+\{\alphalpha\betaeta(y),f(\{\alphalpha^{-1}\betaeta(x),z\})\}\nonumber\\
&&+\{\alphalpha\betaeta(z),\{\alphalpha(x),f(y)\}\}-\{\alphalpha\betaeta(z),\{\alphalpha(y),f(x)\}\}
-\{\alphalpha\betaeta(z),f(\{\alphalpha^{-1}\betaeta(x),y\})\}\nonumber\\
&&-\{\{\betaeta(x),\alphalpha(y)\}, f\betaeta(z)\}+\{\alphalpha\betaeta(z),f(\{\alphalpha^{-1}\betaeta(x),y\}),\}+f(\{\{\alphalpha^{-2}\betaeta^2(x),\alphalpha^{-1}\betaeta(y)\},\betaeta(z)\})\nonumber\\
&&+\{\{\betaeta(x),\alphalpha(z)\}, f\betaeta(y)\}-\{\alphalpha\betaeta(y),f(\{\alphalpha^{-1}\betaeta(x),z\}),\}-f(\{\{\alphalpha^{-2}\betaeta^2(x),\alphalpha^{-1}\betaeta(z)\},\betaeta(y)\})\nonumber\\
&&-\{\{\betaeta(y),\alphalpha(z)\}, f\betaeta(x)\}+\{\alphalpha\betaeta(x),f(\{\alphalpha^{-1}\betaeta(y),z\}),\}+f(\{\{\alphalpha^{-2}\betaeta^2(y),\alphalpha^{-1}\betaeta(z)\},\betaeta(x)\})\nonumber\\
&=&\{\betaeta^2(\alphalpha\betaeta^{-1}(x)),\{\betaeta(\alphalpha\betaeta^{-1}(y)),\alphalpha(\alphalpha^{-1}f(z))\}\}+\{\betaeta^2(\alphalpha\betaeta^{-1}(x)),\{\betaeta(\alphalpha^{-1} f(y)),\alphalpha(\betaeta\alphalpha^{-1}(z))\}\}\nonumber\\
&&-\{\alphalpha\betaeta(x),f(\{\alphalpha^{-1}\betaeta(y),z\})\}-\{\betaeta^2(\alphalpha\betaeta^{-1}(y)),\{\betaeta(\alphalpha^{-1}f(z)),\alphalpha(\betaeta\alphalpha^{-1}(x))\}\}\nonumber\\
&&+\{\betaeta^2(\alphalpha\betaeta^{-1}(y)),\{\betaeta(\betaeta^{-1}\alphalpha(z)),\alphalpha(\alphalpha^ {-1}f(x))\}\}
+\{\alphalpha\betaeta(y),f(\{\alphalpha^{-1}\betaeta(x),z\})\}\nonumber\\
&&+\{\betaeta^2(\alphalpha\betaeta^{-1}(z)),\{\betaeta(\betaeta^{-1}\alphalpha(x)),\alphalpha(\alphalpha^{-1}f(y))\}\}+\{\betaeta^2(\alphalpha\betaeta^{-1}(z)),\{\betaeta(\alphalpha^{-1} f(x)),\alphalpha(\alphalpha\betaeta^{-1}(y))\}\}\nonumber\\
&&-\{\alphalpha\betaeta(z),f(\{\alphalpha^{-1}\betaeta(x),y\})\}
+\{\betaeta^2(\alphalpha^{-1}f(z)),\{\betaeta(\alphalpha\betaeta^{-1}(x)),\alphalpha(\betaeta^{-1}\alphalpha(y))\}\}
\nonumber\\
&&+\{\alphalpha\betaeta(z),f(\{\alphalpha^{-1}\betaeta(x),y\}),\}-f(\{\betaeta^2(\alphalpha^{-1}(z)),\{\betaeta(\alphalpha^{-1}(x)),\alphalpha(\alphalpha^{-1}(y))\}\})\nonumber\\
&&+\{\betaeta^2(\alphalpha^{-1}f(y)),\{\betaeta(\betaeta^{-1}\alphalpha(z)),\alphalpha(\betaeta^{-1}\alphalpha(x))\}\}-\{\alphalpha\betaeta(y),f(\{\alphalpha^{-1}\betaeta(x),z\}),\}
\nonumber\\
&&-f(\{\betaeta^2(\alphalpha^{-1}(y)),\{\betaeta(\alphalpha^{-1}(z)),\alphalpha(\alphalpha^{-1}(x))\}\})+\{\betaeta^2(\alphalpha^{-1}f(x)),\{\betaeta(\alphalpha\betaeta^{-1}(y)),\alphalpha(\alphalpha\betaeta^{-1}(z))\}\}\nonumber\\
&&+\{\alphalpha\betaeta(x),f(\{\alphalpha^{-1}\betaeta(y),z\}),\}-f(\{\betaeta^2(\alphalpha^{-1}(x)),\{\betaeta(\alphalpha^{-1}(y)),\alphalpha(\alphalpha^{-1}(z))\}\})\nonumber\\
&& \mubox{ ( since  $\{u,v\}=-\{\betaeta\alphalpha^{-1}(v),\alphalpha\betaeta^{-1}(u)\}$ $\forall u,v\in A$)}\nonumber\\
&&=0 \mubox{ ( by  the BiHom-Jaobi identity) }. \nonumber
\end{eqnarray}
\end{proof}
For $n=1,2,$ the map $f\in C_{\alphalpha,\betaeta}^n(A,A)$ is called an $n$-BiHom-cocycle $\deltaelta^n f=0.$ We denote the subspace spanned by $n$-Bihom-cocycles by
 $Z_{\alphalpha,\betaeta}^n(A,A)$ and $B_{\alphalpha,\betaeta}^n(A,A)=\deltaelta^{n-1}C_{\alphalpha,\betaeta}^{n-1}(A,A).$ Since  $\deltaelta^2\circ\deltaelta^1=0,$ $B_{\alphalpha,\betaeta}^2(A,A)$
 is a subspace of $Z_{\alphalpha,\betaeta}^2(A,A).$ Hence we can define a cohomology space $H_{\alphalpha,\betaeta}^2(A,A)$ of as the factor space
 $Z_{\alphalpha,\betaeta}^2(A,A)/B_{\alphalpha,\betaeta}^2(A,A).$

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\end{document}