# Transposed Hom-Poisson and Hom-pre-Lie Poisson algebras and bialgebras

Ismail Laraiedh  Departement of Mathematics, Faculty of Sciences,
Sfax University, Box 1171, 3000 Sfax, Tunisia
<EMAIL_ADDRESS>Departement of Mathematics,
College of Sciences and Humanities Al Quwaiiyah,
Shaqra University, Kingdom of Saudi Arabia
<EMAIL_ADDRESS>Sergei Silvestrov  Division of Mathematics and
Physics,
School of Education, Culture and Communication,
Mälardalen University, Box 883, 72123 Västerås, Sweden
<EMAIL_ADDRESS>
###### Abstract

The notions of transposed Hom-Poisson and Hom-pre-Lie Poisson algebras are
introduced. Their bimodules and matched pairs are defined and the relevant
properties and theorems are given. The notion of Manin triple of transposed
Hom-Poisson algebras is introduced, and its equivalence to the transposed Hom-
Poisson bialgebras is investigated. The notion of $\mathcal{O}$-operator is
exploited to illustrate the relations existing between transposed Hom-Poisson
and Hom-pre-Lie Poisson algebras.

##  1 Introduction

A Poisson algebra $(A,\mu,\\{\cdot,\cdot\\})$ consists of a commutative
associative algebra $(A,\mu)$ together with a Lie algebra structure
$\\{\cdot,\cdot\\}$, that satisfies, for all $x,y,z\in A$, the Leibniz
identity

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

Poisson algebras appear in many areas of mathematics and physics being
important for integrable systems, Poisson geometry, Poisson brackets and
Poisson manifolds, geometry of smooth loops, vertex algebras, quantization
theory, deformation quantization, classical mechanics, Hamiltonian mechanics,
quantum mechanics, quantum field theory, Lie theory and operads [2, 3, 13, 17,
21, 25, 26, 27, 29, 30, 33, 34, 35, 36, 38, 44, 45, 48, 57, 58, 74, 75, 76,
80, 92, 91, 90]. Poisson algebras led to development of other fundamental
structures such as noncommutative Poisson algebras [92], Jacobi algebras
(generalized Poisson algebras) [2, 20, 27, 46, 58], Gerstenhaber algebras and
Lie-Rinehart algebras [32, 49, 60, 79] and Novikov-Poisson algebras [18, 93]
arising from Novikov algebra in connection with the Poisson brackets of
hydrodynamic type [12] and Hamiltonian operators in the formal variational
calculus [31].

A transposed Poisson algebra [6] is a triple $(A,\mu,\\{\cdot,\cdot\\})$
consists of a commutative associative algebra $(A,\mu)$ and a Lie algebra
$(A,\\{\cdot,\cdot\\})$, satisfying, for all $x,y,z\in A$, the transposed
Leibniz identity

$2\mu(z,\\{x,y\\})=\\{\mu(z,x),y\\}+\\{x,\mu(z\cdot y)\\}.$ (1.1)

While both the Poisson algebra and transposed Poisson algebra are built from a
commutative associative operation and a Lie operation, the intersection of the
compatibility conditions of the Poisson algebra and of the transposed Poisson
algebra is trivial, meaning that the structures of transposed Poisson algebras
and Poisson algebras are incompatible in this sense [6].

The theory of Hom-algebras has been initiated in [40, 52, 53] motivated by
quasi-deformations of Lie algebras of vector fields, in particular
q-deformations of Witt and Virasoro algebras. Hom-Lie algebras and more
general quasi-Hom-Lie algebras were introduced first by Hartwig, Larsson and
Silvestrov in [40] where a general approach to discretization of Lie algebras
of vector fields using general twisted derivations ($\sigma$-derivations) and
a general method for construction of deformations of Witt and Virasoro type
algebras based on twisted derivations have been developed. The general quasi-
Lie algebras, containing the quasi-Hom-Lie algebras and Hom-Lie algebras as
subclasses, as well their graded color generalization, the color quasi-Lie
algebras including color quasi-hom-Lie algebras, color hom-Lie algebras and
their special subclasses the quasi-Hom-Lie superalgebras and hom-Lie
superalgebras, have been first introduced in [40, 52, 53, 54, 55, 84].
Subsequently, various classes of Hom-Lie admissible algebras have been
considered in [69]. In particular, in [69], the Hom-associative algebras have
been introduced and shown to be Hom-Lie admissible, that is leading to Hom-Lie
algebras using commutator map as new product, and in this sense constituting a
natural generalization of associative algebras as Lie admissible algebras
leading to Lie algebras using commutator map. Furthermore, in [69], more
general $G$-Hom-associative algebras including Hom-associative algebras, Hom-
Vinberg algebras (Hom-left symmetric algebras), Hom-pre-Lie algebras (Hom-
right symmetric algebras), and some other Hom-algebra structures, generalizing
$G$-associative algebras, Vinberg and pre-Lie algebras respectively, have been
introduced and shown to be Hom-Lie admissible, meaning that for these classes
of Hom-algebras, the operation of taking commutator leads to Hom-Lie algebras
as well. Also, flexible Hom-algebras have been introduced, connections to Hom-
algebra generalizations of derivations and of adjoint maps have been noticed,
and some low-dimensional Hom-Lie algebras have been described. In Hom-algebra
structures, defining algebra identities are twisted by linear maps. Since the
pioneering works [40, 52, 53, 54, 55, 69], Hom-algebra structures have
developed in a popular broad area with increasing number of publications in
various directions. Hom-algebra structures include their classical
counterparts and open new broad possibilities for deformations, extensions to
Hom-algebra structures of representations, homology, cohomology and formal
deformations, Hom-modules and hom-bimodules, Hom-Lie admissible Hom-
coalgebras, Hom-coalgebras, Hom-bialgebras, Hom-Hopf algebras, $L$-modules,
$L$-comodules and Hom-Lie quasi-bialgebras, $n$-ary generalizations of BiHom-
Lie algebras and BiHom-associative algebras and generalized derivations, Rota-
Baxter operators, Hom-dendriform color algebras, Rota-Baxter bisystems and
covariant bialgebras, Rota-Baxter cosystems, coquasitriangular mixed
bialgebras, coassociative Yang-Baxter pairs, coassociative Yang-Baxter
equation and generalizations of Rota-Baxter systems and algebras, curved
$\mathcal{O}$-operator systems and their connections with tridendriform
systems and pre-Lie algebras, BiHom-algebras, BiHom-Frobenius algebras and
double constructions, infinitesimal BiHom-bialgebras and Hom-dendriform
$D$-bialgebras, Hom-algebras have been considered [1, 4, 5, 7, 8, 9, 11, 10,
16, 14, 15, 19, 22, 23, 24, 28, 37, 39, 41, 42, 43, 47, 50, 51, 52, 55, 56,
59, 61, 62, 63, 64, 65, 67, 68, 66, 70, 72, 71, 73, 77, 78, 81, 83, 82, 85,
86, 87, 88, 95, 94, 97, 98, 96, 99].

A Hom-type generalization of Poisson algebras, called Hom-Poisson algebras,
was introduced in [72]. We aim in this paper to introduce the notions of
transposed Hom-Poisson and Hom-pre-Lie Poisson algebras. The concept of
bimodules and matched pairs are constructed, and we give some related results.
Next, we introduce notion of Manin triple of transposed Hom-Poisson algebras
and transposed Hom-Poisson bialgebras and then give the equivalence between
them. Also, we introduce and study the notion of an $\mathcal{O}$-operator of
transposed Hom-Poisson and Hom-pre-Lie Poisson algebras.

The paper is organized as follows. In Section 2, we introduce the notion of
transposed Hom-Poisson algebra, and explain connections between transposed
Hom-Poisson algebras and Hom- Poisson algebras. Moreover, we define the notion
of a bimodule of a transposed Hom-Poisson algebra and provide some properties.
In Section 3, we introduce the notion of Manin triples for transposed Hom-
Poisson algebras and transposed Hom-Poisson bialgebras. Also, the equivalences
between their Manin triples are established. In Section 4, we establish
definition of Hom-pre-Lie Poisson algebra and we develop some construction
theorems. Their bimodule and matched pair are defined and their related
relevant properties are also given. In Section 5, we introduce and study the
notion of an $\mathcal{O}$-operator of transposed Hom-Poisson algebras
generalizing the notion of Rota-Baxter operators. Their relationship with Hom-
pre-Lie Poisson algebras is also described.

##  2 Transposed Hom-Poisson algebras

Throughout this paper $\mathbb{K}$ is an algebraically closed field of
characteristic $0$ and $A$ is a $\mathbb{K}$-vector space. We refer to the
standard one-to-one correspondence between linear maps
$\mu:V_{1}\otimes\dots\otimes V_{n}\rightarrow W$ and multilinear maps
$\mu:V_{1}\times\dots\times V_{n}\rightarrow W$ given by
$\mu(v_{1},\dots,v_{n})=\mu(v_{1}\otimes\dots\otimes v_{n})$, whenever the
same notation is used for these maps.

In this section, we introduce the notion of transposed Hom-Poisson algebra
with a connection to commutative Hom-associative algebra and Hom-Lie algebra.
Next, we study the notion of bimodule and matched pair of transposed Hom-
Poisson algebras.

######  Definition 2.1.

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

######  Definition 2.2.

A Hom-algebra is a triple $(A,\mu,\alpha)$ where $(A,\alpha)$ is a Hom-module
and $\mu:A^{\otimes 2}\rightarrow A$ is a linear map.

######  Definition 2.3.

A Hom-algebra $(A,\mu,\alpha)$ is called multiplicative, if the linear map
$\alpha:A\to A$ is moreover an algebra homomorphism of the algebra $(A,\mu)$,
that is,

$\alpha\circ\mu=\mu\circ(\alpha\otimes\alpha),$

or equivalently, in multilinear maps notation,
$\alpha\circ\mu(x,y)=\mu(\alpha(x),\alpha(y)),$ for $x,y\in A.$

In any Hom-algebra $(A,\mu,\alpha)$, Hom-associator ($\alpha$-associator) is
defined by

$as_{A}(x,y,z)=\mu(\mu(x,y),\alpha(z))-\mu(\alpha(x),\mu(y,z))$

for all $x,y,z\in A$. Hom-associators are generalizations to Hom-algebras of
the associators in algebras (when $\alpha=id$).

######  Definition 2.4.

A Hom-associative algebra is a Hom-algebra $(A,\mu,\alpha)$ satisfying

$\displaystyle as_{A}(x,y,z)=0,$ (Hom-associativity) (2.1)

or equivalently, $\mu(\alpha(x),\mu(y,z))=\mu(\mu(x,y),\alpha(z)),$ for
$x,y,z\in A$.

######  Definition 2.5.

A commutative Hom-associative algebra is a Hom-algebra $(A,\mu,\alpha)$
satisfying, for all $x,y,z\in A$,

$\displaystyle\mu(x,y)=\mu(y,x),$ (Commutativity) (2.2) $\displaystyle
as_{A}(x,y,z)=0.$ (Hom-associativity) (2.3)

######  Example 2.6.

Consider a two-dimensional $\mathbb{K}$-linear space $A$ with basis
$\\{e_{1},e_{2}\\}$.

  1. 1)

The triple $(A,\mu,\alpha)$ with the bilinear product and linear map defined
on $A$ by

$\displaystyle\mu(e_{1},e_{1})=-e_{1},\quad$
$\displaystyle\mu(e_{1},e_{2})=\mu(e_{2},e_{1})=e_{2},\quad\mu(e_{2},e_{2})=e_{1},$
$\displaystyle\alpha(e_{1})=e_{1},\quad\alpha(e_{2})=-e_{2},$

is a commutative Hom-associative algebra.

  2. 2)

The triple $(A,\mu,\alpha)$ with the bilinear product and linear map defined
on $A$ by

$\displaystyle\mu(e_{1},e_{1})=e_{1},$
$\displaystyle\quad\mu(e_{1},e_{2})=\mu(e_{2},e_{1})=0,\quad\mu(e_{2},e_{2})=e_{2},$
$\displaystyle\quad\alpha(e_{1})=e_{1},\quad\alpha(e_{2})=0.$

is a commutative Hom-associative algebra.

######  Example 2.7.

Consider a three-dimensional $\mathbb{K}$-linear space $A$ with basis
$\\{e_{1},e_{2},e_{3}\\}$.

  1. 1)

The triple $(A,\mu,\alpha)$ with the bilinear product and linear map defined
on $A$ by

$\displaystyle e_{1}\cdot e_{1}$ $\displaystyle=e_{1},\quad e_{2}\cdot
e_{2}=e_{2}+e_{3},\quad e_{3}\cdot e_{3}=e_{2}+e_{3}$ $\displaystyle
e_{2}\cdot e_{3}$ $\displaystyle=e_{2}+e_{3},\quad e_{3}\cdot
e_{2}=e_{2}+e_{3},$ $\displaystyle\alpha(e_{1})$ $\displaystyle=e_{1},$

where the non written products and images of $\alpha$ are zero, is a
commutative Hom-associative algebra.

  2. 2)

The triple $(A,\mu,\alpha)$ with the bilinear product and linear map defined
on $A$ by

$\displaystyle e_{1}\cdot e_{1}$ $\displaystyle=p_{1}e_{1},\quad e_{2}\cdot
e_{2}=p_{2}e_{2},\quad e_{3}\cdot e_{3}=p_{3}e_{3},$
$\displaystyle\alpha(e_{1})$ $\displaystyle=e_{1},\quad\alpha(e_{2})=e_{2},$

where the non written products and images of $\alpha$ are zero, is a
commutative Hom-associative algebra.

######  Definition 2.8.

A derivation of a commutative Hom-associative algebra $(A,\mu,\alpha)$ is a
linear map $D:A\rightarrow A$ satisfying for all $x,y\in A$,

$\displaystyle\alpha\circ D$ $\displaystyle=D\circ\alpha,$ (2.4)
$\displaystyle D(x\cdot y)$ $\displaystyle=D(x)\cdot y+x\cdot D(y).$ (2.5)

We will use in this article a definition of bimodule of a commutative Hom-
associative algebras including Hom-module map condition (2.7), while we note
that there are also other definitions of Hom-modules and Hom-bimodules of Hom-
associative algebras, for example the more general notions requiring only
(2.6), [8, 9, 70, 72, 39, 95, 98].

######  Definition 2.9.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra and let
$(V,\beta)$ be a Hom-module. Let $s:A\rightarrow gl(V)$ be a linear map. The
triple $(s,\beta,V)$ is called a bimodule of $(A,\cdot,\alpha)$ if for all
$x,y\in A,v\in V$,

$\displaystyle s(x\cdot y)\beta(v)=s(\alpha(x))s(y)v,$ (2.6)
$\displaystyle\beta(s(x)v)=s(\alpha(x))\beta(v).$ (2.7)

######  Proposition 2.10.

Let $(s,\beta,V)$ be a bimodule of a commutative Hom-associative algebra
$(A,\cdot,\alpha)$. Then, the direct sum $A\oplus V$ of vector spaces is a
commutative Hom-associative algebra with multiplication defined for all
$x_{1},x_{2}\in A,v_{1},v_{2}\in V$ by

$\displaystyle(x_{1}+v_{1})\cdot^{\prime}(x_{2}+v_{2})$
$\displaystyle=x_{1}\cdot x_{2}+(s(x_{1})v_{2}+s(x_{2})v_{1}),$
$\displaystyle(\alpha\oplus\beta)(x_{1}+v_{1})$
$\displaystyle=\alpha(x_{1})+\beta(v_{1}).$

###### Proof.

For all $x_{1},x_{2}\in A,~{}v_{1},v_{2}\in V$,

$\displaystyle(x_{1}+v_{1})\cdot^{\prime}(x_{2}+v_{2})=x_{1}\cdot
x_{2}+s(x_{1})v_{2}+s(x_{2})v_{1}$ $\displaystyle\quad=x_{2}\cdot
x_{1}+s(x_{2})v_{1}+s(x_{1})v_{2}=(x_{2}+v_{2})\cdot^{\prime}(x_{1}+v_{1}),$
$\displaystyle\Big{[}(x_{1}+v_{1})\cdot^{\prime}(x_{2}+v_{2})\Big{]}\cdot^{\prime}(\alpha+\beta)(x_{3}+v_{3})$
$\displaystyle\quad=\Big{[}x_{1}\cdot
x_{2}+s(x_{1})v_{2}+s(x_{2})v_{1}\Big{]}\cdot^{\prime}(\alpha(x_{3})+\beta(v_{3}))$
$\displaystyle\quad=(x_{1}\cdot x_{2})\cdot\alpha(x_{3})+s(x_{1}\cdot
x_{2})\beta(v_{3})+s(\alpha(x_{3}))(s(x_{1})v_{2}+s(x_{2})v_{1})$
$\displaystyle\quad=\alpha(x_{1})\cdot(x_{2}\cdot
x_{3})+s(\alpha(x_{1}))s(x_{2})\beta(v_{3})+s(x_{3}\cdot
x_{1})v_{2}+s(x_{3}\cdot x_{2})v_{1}~{}~{}(by\eqref{Cond1})$
$\displaystyle\quad=(\alpha+\beta)(x_{1}+v_{1})\cdot^{\prime}\Big{[}(x_{2}+v_{2})\cdot^{\prime}(x_{3}+v_{3})\Big{]}.$

Then, $(A\oplus V,\cdot^{\prime},\alpha+\beta)$ is a commutative Hom-
associative algebra. ∎

The commutative Hom-associative algebra constructed in Proposition 2.10 is
denoted by $(A\oplus V,\cdot^{\prime},\alpha+\beta)$ or
$A\times_{s,\alpha,\beta}V.$

######  Theorem 2.11.

Let $\mathcal{A}=(A,\cdot_{A},\alpha)$ and $\mathcal{B}=(B,\cdot_{B},\beta)$
be two commutative Hom-associative algebras. Suppose that there are linear
maps $s_{A}:A\rightarrow gl(B)$ and $s_{B}:B\rightarrow gl(A)$ such that
$(s_{A},\beta,B)$ is a bimodule of $\mathcal{A}$, and $(s_{B},\alpha,A)$ is a
bimodule of $\mathcal{B}$ satisfying, for any $x,y\in A,~{}a,b\in B$,

$\displaystyle s_{A}(\alpha(x))(a\cdot_{B}b)$
$\displaystyle=\beta(a)\cdot_{B}(s_{A}(x)b)+s_{A}(s_{B}(b)x)\beta(x),$ (2.8)
$\displaystyle(s_{A}(x)a)\cdot_{B}\beta(b)+s_{A}(s_{B}(a)x)\beta(b)$
$\displaystyle=\beta(a)\cdot_{B}(s_{A}(x)b)+s_{A}(s_{B}(b)x)\beta(a),$ (2.9)
$\displaystyle s_{B}(\beta(a))(x\cdot_{A}y)$
$\displaystyle=\alpha(x)\cdot_{A}(s_{B}(a)y)+s_{B}(s_{A}(y)a)\alpha(a),$
(2.10) $\displaystyle(s_{B}(a)x)\cdot_{A}\alpha(y)+s_{B}(s_{A}(x)a)\alpha(y)$
$\displaystyle=\alpha(x)\cdot_{A}(s_{B}(a)y)+s_{B}(s_{A}(y)a)\alpha(x).$
(2.11)

Then, $(A,B,s_{A},\beta,s_{B},\alpha)$ is called a matched pair of commutative
Hom-associative algebras. In this case, there is a commutative Hom-associative
algebra structure on the direct sum $A\oplus B$ of the underlying vector
spaces of $A$ and $B$ given by

$\begin{array}[]{llllll}(x+a)\cdot(y+b)&=&x\cdot_{A}y+(s_{A}(x)b+s_{A}(y)a)+a\cdot_{B}b+(s_{B}(a)y+s_{B}(b)x),\cr(\alpha\oplus\beta)(x+a)&=&\alpha(x)+\beta(a).\end{array}$

###### Proof.

For any $x,y,z\in A$ and $a,b,c\in B$,

$\displaystyle(\alpha+\beta)(x+a)\cdot((y+b)\cdot(z+c))$
$\displaystyle\quad=(\alpha(x)+\beta(a))(y\cdot_{A}z+s_{B}(b)z+s_{B}(c)y+b\cdot
c+s_{A}(y)c+s_{A}(z)b)$
$\displaystyle\quad=\alpha(x)\cdot_{A}(y\cdot_{A}z)+\alpha(x)\cdot_{A}s_{B}(b)z+\alpha(x)\cdot_{A}s_{B}(c)y+s_{B}(\beta(a))(y\cdot_{A}z)$
$\displaystyle\quad\quad+s_{B}(\beta(a))s_{B}(b)z+s_{B}(\beta(a))s_{B}(c)y+s_{B}(b\cdot_{B}c)\alpha(x)+s_{B}(s_{A}(y)c)\alpha(x)$
$\displaystyle\quad\quad+s_{B}(s_{A}(z)b)\alpha(x)+\beta(a)\cdot_{B}(b\cdot_{B}c)+\beta(a)\cdot_{B}s_{A}(y)c+\beta_{A}(\alpha(x))s_{A}(y)c$
$\displaystyle\quad\quad+s_{A}(\alpha(x))s_{A}(z)b+s_{A}(y\cdot_{A}z)\beta(a)+s_{A}(s_{A}(b)z)\beta(a)+s_{A}(s_{B}(c)y)\beta(a),$
$\displaystyle((x+a)\cdot(y+b))\cdot(\alpha+\beta)(z+c)$
$\displaystyle\quad=(x\cdot_{A}y+s_{B}(a)y+s_{B}(b)x+a\cdot_{B}b+s_{A}(x)b+s_{A}(y)a)\cdot(\alpha(z)+\beta(c))$
$\displaystyle\quad=(x\cdot_{A}y)\cdot_{A}\alpha(z)+s_{B}(a)y\cdot_{A}\alpha(z)+s_{B}(b)x\cdot_{A}\alpha(z)+s_{B}(a\cdot_{B}b)\alpha(z)$
$\displaystyle\quad\quad+s_{B}(s_{A}(x)b)\alpha(z)+s_{B}(s_{A}(y)a)\alpha(z)+s_{B}(\beta(c))(x\cdot_{A}y)+s_{A}(\beta(c))s_{B}(a)y$
$\displaystyle\quad\quad+s_{B}(\beta(c))s_{B}(b)x+(a\cdot_{B}b)\cdot_{B}\beta(c)+(s_{A}(x)b)\cdot_{B}\beta(c)+(s_{A}(y)a)\cdot_{B}\beta(c)$
$\displaystyle\quad\quad+r_{A}(\alpha(z))(a\cdot_{B}b)+s_{A}(\alpha(z))(s_{A}(x)b)+s_{A}(\alpha(z))(s_{A}(y)a)+s_{A}(x\cdot_{A}y)\beta(c)$
$\displaystyle\quad\quad+s_{A}(s_{B}(a)y)\beta(c)+(s_{B}(b)x)\beta(c).$

Then, by (2.2), (2.1) and (2.8)-(2.11), we deduce that
$(\alpha+\beta)(x+a)\cdot((y+b)\cdot(z+c))=((x+a)\cdot(y+b))\cdot(\alpha+\beta)(z+c)$.
∎

######  Definition 2.12 ([40]).

A Hom-Lie algebra is a triple $(A,[\cdot,\cdot],\alpha)$ consisting of a
linear space $A$, bilinear map (bilinear product) $[\cdot,\cdot]:A\otimes
A\rightarrow A$, and a linear map $\alpha:A\rightarrow A$ satisfying, for all
$x,y,z\in A$,

$\displaystyle[x,y]=-[y,x],$ (Skew-symmetry) (2.12)
$\displaystyle[\alpha(x),[y,z]]+[\alpha(y),[z,x]]+[\alpha(z),[x,y]]=0.$ (Hom-
Jacobi identity) (2.13)

######  Example 2.13.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra and
$D_{1},D_{2}$ be commuting derivations (that is, $D_{1}D_{2}=D_{2}D_{1}$).
Then, there is a Hom-Lie algebra $(A,[\cdot,\cdot],\alpha)$, where for all
$x,y\in A$,

$[x,y]=D_{1}(x)\cdot D_{2}(y)-D_{1}(y)\cdot D_{2}(x).$ (2.14)

######  Definition 2.14 ([4, 82]).

Let $(A,[\cdot,\cdot],\alpha)$ be a Hom-Lie algebra and $(V,\beta)$ be a Hom-
module. Let $\rho:A\rightarrow gl(V)$ be a linear map. The triple
$(\rho,\beta,V)$ is called a representation of $A$ if for all $x,y\in
A,~{}v\in V$,

$\displaystyle\rho([x,y])\beta(v)$ $\displaystyle=$
$\displaystyle\rho(\alpha(x))\circ\rho(y)v-\rho(\alpha(y))\circ\rho(x)v,$
(2.15) $\displaystyle\beta(\rho(x)v)$ $\displaystyle=$
$\displaystyle\rho(\alpha(x))\beta(v).$ (2.16)

######  Proposition 2.15.

Let $(\rho,\beta,V)$ be a representation of a Hom-Lie algebra
$(A,\\{\cdot,\cdot\\},\alpha)$. Then, the direct sum $A\oplus V$ of vector
spaces is turned into a Hom-Lie algebra by defining the multiplication in
$A\oplus V$ by

$\displaystyle[x_{1}+v_{1},x_{2}+v_{2}]$ $\displaystyle=$
$\displaystyle\\{x_{1},x_{2}\\}+\rho(x_{1})v_{2}-\rho(x_{2})v_{1},$
$\displaystyle(\alpha\oplus\beta)(x_{1}+v_{1})$ $\displaystyle=$
$\displaystyle\alpha(x_{1})+\beta(v_{1}).$

We denote such a Hom-Lie algebra by $(A\oplus V,[\cdot,\cdot],\alpha+\beta),$
or $A\times_{\rho,\alpha,\beta}V.$

Now, we introduce the notion of matched pair of Hom-Lie algebra

######  Theorem 2.16 ([83]).

Let $(A,\\{\cdot,\cdot\\}_{A},\alpha)$ and $(B,\\{\cdot,\cdot\\}_{B},\beta)$
be two Hom-Lie algebras. Suppose that there are linear maps
$\rho_{A}:A\rightarrow gl(B)$ and $\rho_{B}:B\rightarrow gl(A)$ such that
$(\rho_{A},\beta,B)$ is a representation of $A$ and $(\rho_{B},\alpha,A)$ is a
representation of $B$ satisfying for any $x,y\in A,~{}a,b\in B$,

$\displaystyle\begin{array}[]{lll}\rho_{B}(\beta(a))\\{x,y\\}_{A}&=&\\{\rho_{B}(a)x,\alpha(y)\\}_{A}+\\{\alpha(x),\rho_{B}(a)(y)\\}_{A}\\\
&+&\rho_{B}(\rho_{A}(y)(a))\alpha(x)-\rho_{B}(\rho_{A}(x)(a))\alpha(y),\end{array}$
(2.19)
$\displaystyle\begin{array}[]{lll}\rho_{A}(\alpha(x))\\{a,b\\}_{B}&=&\\{\rho_{A}(x)a,\beta(b)\\}_{B}+\\{\beta(a),\rho_{A}(x)(b)\\}_{B}\\\
&+&\rho_{A}(\rho_{B}(b)(x))\beta(a)-\rho_{A}(\rho_{B}(a)(x))\beta(b).\end{array}$
(2.22)

Then, $(A,B,\rho_{A},\beta,\rho_{B},\alpha)$ is called a matched pair of Hom-
Lie algebras. In this case, there is a Hom-Lie algebra structure on the vector
space $A\oplus B$ of the underlying vector spaces of $A$ and $B$ given by

$\displaystyle[x+a,y+b]$
$\displaystyle=\\{x,y\\}_{A}+\rho_{A}(x)b-\rho_{A}(y)a$ (2.23)
$\displaystyle\quad\quad+\\{a,b\\}_{B}+\rho_{B}(a)y-\rho_{B}(b)x,$ (2.24)
$\displaystyle(\alpha\oplus\beta)(x+a)$ $\displaystyle=\alpha(x)+\beta(a).$
(2.25)

Hom-Poisson algebras were introduced in [72].

######  Definition 2.17 ([72]).

A _Hom-Poisson algebra_ is a quadruple $(V,\mu,\\{\cdot,\cdot\\},\alpha)$
consisting of a vector space $V$, bilinear maps $\mu:V\times V\rightarrow V$
and $\\{\cdot,\cdot\\}:V\times V\rightarrow V$, and a linear map
$\alpha:V\rightarrow V$ satisfying,

  1. (i)

$(V,\mu,\alpha)$ is a commutative Hom-associative algebra,

  2. (ii)

$(V,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Lie algebra,

  3. (iii)

for all $x,y,z$ in $V$,

$\\{\alpha(x),\mu(y,z)\\}=\mu(\alpha(y),\\{x,z\\})+\mu(\alpha(z),\\{x,y\\}).$
(2.26)

Condition (2.26), expressing the compatibility between the multiplication and
the Poisson bracket, can be reformulated equivalently as

$\\{\mu(x,y),\alpha(z)\\}=\mu(\\{x,z\\},\alpha(y))+\mu(\alpha(x),\\{y,z\\}).$
(2.27)

######  Example 2.18 ([72]).

Let $\\{e_{1},e_{2},e_{3}\\}$ be a basis of a $3$-dimensional vector space $A$
over $\mathbb{K}$. The following multiplication $\cdot$, skew-symmetric
bracket $\\{\cdot,\cdot\\}$ and linear map $\alpha$ on $V$ define a Hom-
Poisson algebra:

$\displaystyle e_{1}\cdot e_{1}=e_{1},\quad e_{1}\cdot e_{2}=e_{3},\quad
e_{2}\cdot e_{1}=e_{3}$
$\displaystyle\\{e_{1},e_{2}\\}=ae_{2}+be_{3},\quad\\{e_{1},e_{3}\\}=ce_{2}+de_{3},$
$\displaystyle\alpha(e_{1})=\lambda_{1}e_{2}+\lambda_{2}e_{3},\quad\alpha(e_{2})=\lambda_{3}e_{2}+\lambda_{4}e_{3},\quad\alpha(e_{3})=\lambda_{5}e_{2}+\lambda_{6}e_{3},$

where
$a,b,c,d,\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}$
are parameters in $\mathbb{K}.$

######  Proposition 2.19.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra, and let
$D_{1}$ and $D_{2}$ be commuting derivations, $D_{1}D_{2}=D_{2}D_{1}$. For all
$x,y\in A$, let

$\\{x,y\\}=D_{1}(x)\cdot D_{2}(y)-D_{1}(y)\cdot D_{2}(x).$ (2.28)

Then, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Poisson algebra.

###### Proof.

First, we prove that $(A,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Lie algebra. For
all $x,y,z\in A$,

$\displaystyle\\{x,y\\}=D_{1}(x)\cdot D_{2}(y)-D_{1}(y)\cdot
D_{2}(x)=-(D_{1}(y)\cdot D_{2}(x)-D_{1}(x)\cdot D_{2}(y))$
$\displaystyle\quad\quad\quad=-\\{y,x\\},$
$\displaystyle\circlearrowleft_{x,y,z}\\{\alpha(x),\\{y,z\\}\\}=\circlearrowleft_{x,y,z}\Big{(}D_{1}(\alpha(x))\cdot
D_{2}(\\{y,z\\})-D_{1}(\\{y,z\\})\cdot D_{2}(\alpha(x))\Big{)}$
$\displaystyle\quad=\circlearrowleft_{x,y,z}\Big{(}D_{1}(\alpha(x))\cdot
D_{2}(D_{1}(y)\cdot D_{2}(z)-D_{1}(z)\cdot D_{2}(y))$
$\displaystyle\quad\quad\quad-D_{1}(D_{1}(y)\cdot D_{2}(z)-D_{1}(z)\cdot
D_{2}(y))\cdot D_{2}(\alpha(x))\Big{)}$
$\displaystyle\quad=\circlearrowleft_{x,y,z}\Big{(}\underbrace{D_{1}(\alpha(x))\cdot(D_{2}D_{1}(y)\cdot
D_{2}(z))}_{a_{1}(x,y,z)}+\underbrace{D_{1}(\alpha(x))\cdot(D_{1}(y)\cdot
D_{2}^{2}(z))}_{b_{1}(x,y,z)}$
$\displaystyle\quad\quad\quad-\underbrace{D_{1}(\alpha(x))\cdot(D_{2}D_{1}(z)\cdot
D_{2}(y))}_{c_{1}(x,y,z)}-\underbrace{D_{1}(\alpha(x))\cdot(D_{1}(z)\cdot
D_{2}^{2}(y))}_{d_{1}(x,y,z)}$
$\displaystyle\quad\quad\quad-\underbrace{(D_{1}^{2}(y)\cdot D_{2}(z))\cdot
D_{2}(\alpha(x))}_{e_{1}(x,y,z)}-\underbrace{(D_{1}(y)\cdot
D_{1}D_{2}(z))\cdot D_{2}(\alpha(x))}_{f_{1}(x,y,z)}$
$\displaystyle\quad\quad\quad+\underbrace{(D_{1}^{2}(z)\cdot D_{2}(y))\cdot
D_{2}(\alpha(x))}_{g_{1}(x,y,z)}+\underbrace{(D_{1}(z)\cdot
D_{1}D_{2}(y))\cdot D_{2}(\alpha(x))}_{h_{1}(x,y,z)}\Big{)}.$

Using (2.1)-(2.4), observe that

$\displaystyle\circlearrowleft_{x,y,z}(b_{1}(x,y,z)-d_{1}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D_{1}(x))\cdot(D_{1}(y)\cdot
D_{2}^{2}(z))-\alpha(D_{1}(x))\cdot(D_{1}(z)\cdot D_{2}^{2}(y))\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D_{1}(x))\cdot(D_{1}(y)\cdot
D_{2}^{2}(z))-\alpha(D_{1}(z))\cdot(D_{1}(x)\cdot D_{2}^{2}(y))\Big{)}=0,$
$\displaystyle\circlearrowleft_{x,y,z}(g_{1}(x,y,z)-e_{1}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}(D_{1}^{2}(z)\cdot
D_{2}(y))\cdot\alpha(D_{2}(x))-(D_{1}^{2}(y)\cdot
D_{2}(z))\cdot\alpha(D_{2}(x))\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}(D_{1}^{2}(z)\cdot
D_{2}(y))\cdot\alpha(D_{2}(x))-(D_{1}^{2}(y)\cdot
D_{2}(x))\cdot\alpha(D_{2}(z))\Big{)}=0,$
$\displaystyle\circlearrowleft_{x,y,z}(a_{1}(x,y,z)-f_{1}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D_{1}(x))\cdot(D_{2}D_{1}(y)\cdot
D_{2}(z))-(D_{1}(y)\cdot D_{1}D_{2}(z))\cdot\alpha(D_{2}(x))\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D_{1}(x))\cdot(D_{1}D_{2}(y)\cdot
D_{2}(z))-\alpha(D_{1}(y))\cdot(D_{1}D_{2}(z)\cdot D_{2}(x))\Big{)}$
$\displaystyle\quad\quad=0,\quad\mbox{(\hbox{using} $D_{1}D_{2}=D_{2}D_{1}$)}$
$\displaystyle\circlearrowleft_{x,y,z}(h_{1}(x,y,z)-c_{1}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}(D_{1}(z)\cdot
D_{1}D_{2}(y))\cdot\alpha(D_{2}(x))-\alpha(D_{1}(x))\cdot(D_{2}D_{1}(z)\cdot
D_{2}(y))\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}(D_{1}(z)\cdot
D_{1}D_{2}(y))\cdot\alpha(D_{2}(x))-(D_{1}(x)\cdot
D_{1}D_{2}(z))\cdot\alpha(D_{2}(y))\Big{)}$
$\displaystyle\quad\quad=0.\quad\mbox{(\hbox{using}~{}$D_{1}D_{2}=D_{2}D_{1}$)}$

This shows that $(A,\\{\cdot,\cdot\\},\alpha)$ satisfies the Hom-Jacobi
identity $\circlearrowleft_{x,y,z}\\{\alpha(x),\\{y,z\\}\\}=0$.

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

$\displaystyle\\{x,z\\}\cdot\alpha(y)+\alpha(x)\cdot\\{y,z\\}$
$\displaystyle=\underbrace{(D_{1}(x)\cdot
D_{2}(z))\cdot\alpha(y)}_{i_{1}(x,y,z)}-\underbrace{(D_{1}(z)\cdot
D_{2}(x))\cdot\alpha(y)}_{j_{1}(x,y,z)}$
$\displaystyle\quad+\underbrace{\alpha(x)\cdot(D_{1}(y)\cdot
D_{2}(z))}_{k_{1}(x,y,z)}-\underbrace{\alpha(x)\cdot(D_{1}(z)\cdot
D_{2}(y))}_{l_{1}(x,y,z)}.$

Using (2.1)-(2.4),

$\displaystyle i_{1}(x,y,z)+k_{1}(x,y,z)$
$\displaystyle\quad\quad=(D_{1}(x)\cdot
D_{2}(z))\cdot\alpha(y)+\alpha(x)\cdot(D_{1}(y)\cdot D_{2}(z))$
$\displaystyle\quad\quad=(D_{1}(x)\cdot y)\cdot\alpha(D_{2}(z))+(x\cdot
D_{1}(y))\cdot\alpha(D_{2}(z))$ $\displaystyle\quad\quad=D_{1}(x\cdot
y)\cdot\alpha(D_{2}(z))$ $\displaystyle\quad\quad=D_{1}(x\cdot y)\cdot
D_{2}(\alpha(z)),$ $\displaystyle j_{1}(x,y,z)+l_{1}(x,y,z)$
$\displaystyle\quad\quad=(D_{1}(z)\cdot
D_{2}(x))\cdot\alpha(y)+\alpha(x)\cdot(D_{1}(z)\cdot D_{2}(y))$
$\displaystyle\quad\quad=\alpha(D_{1}(z))\cdot(D_{2}(x)\cdot
y)+\alpha(D_{1}(z))\cdot(x\cdot D_{2}(y))$
$\displaystyle\quad\quad=\alpha(D_{1}(z))\cdot D_{2}(x\cdot y)$
$\displaystyle\quad\quad=D_{1}(\alpha(z))\cdot D_{2}(x\cdot y).$

This means that

$\displaystyle\\{x,z\\}\cdot\alpha(y)+\alpha(x)\cdot\\{y,z\\}$
$\displaystyle=D_{1}(x\cdot y)\cdot D_{2}(\alpha(z))-D_{1}(\alpha(z))\cdot
D_{2}(x\cdot y)$ $\displaystyle=\\{x\cdot y,\alpha(z)\\}.$

Therefore, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ satisfies the Hom-Leibniz
identity. ∎

######  Definition 2.20.

A transposed Hom-Poisson algebra is a quadruple
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$, where $(A,\cdot,\alpha)$ is a
commutative Hom-associative algebra and $(A,\\{\cdot,\cdot\\},\alpha)$ is a
Hom-Lie algebra, satisfying for all $x,y,z\in A$ the transposed Hom-Leibniz
identity,

$2\alpha(z)\cdot\\{x,y\\}=\\{z\cdot x,\alpha(y)\\}+\\{\alpha(x),z\cdot y\\}.$
(2.29)

######  Definition 2.21.

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

  1. (i)

Hom-subalgebra of $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ if $\alpha(H)\subseteq
H,~{}H\cdot H\subseteq H~{}\textrm{and}~{}\\{H,H\\}\subseteq H.$

  2. (ii)

Hom-ideal of $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ if $\alpha(H)\subseteq
H,~{}A\cdot H\subseteq H~{}\textrm{and}~{}\\{A,H\\}\subseteq H.$

######  Proposition 2.22.

Let $(A_{1},\cdot_{1},\\{\cdot,\cdot\\}_{1},\alpha_{1})$ and
$(A_{2},\cdot_{2},\\{\cdot,\cdot\\}_{2},\alpha_{2})$ be two transposed Hom-
Poisson algebras. Define two operations $\cdot$ and $[\cdot,\cdot]$ on
$A_{1}\otimes A_{2}$ for all $x_{1},y_{1}\in A_{1}$, $x_{2},y_{2}\in A_{2}$ by

$\displaystyle(x_{1}\otimes x_{2})\cdot(y_{1}\otimes y_{2})$
$\displaystyle=x_{1}\cdot_{1}y_{1}\otimes x_{2}\cdot_{2}y_{2},$ (2.30)
$\displaystyle[x_{1}\otimes x_{2},y_{1}\otimes y_{2}]$
$\displaystyle=\\{x_{1},y_{1}\\}_{1}\otimes
x_{2}\cdot_{2}y_{2}+x_{1}\cdot_{1}y_{1}\otimes\\{x_{2},y_{2}\\}_{2},$ (2.31)
$\displaystyle(\alpha_{1}\otimes\alpha_{2})(x_{1}\otimes x_{2})$
$\displaystyle=\alpha_{1}(x_{1})\otimes\alpha_{2}(x_{2}),$ (2.32)

Then, $(A_{1}\otimes A_{2},\cdot,[\cdot,\cdot],\alpha_{1}\otimes\alpha_{2})$
is a transposed Hom-Poisson algebra.

######  Proposition 2.23.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra, and let $D$
be a derivation. Define the multiplication, for all $x,y\in A$, by

$\\{x,y\\}=x\cdot D(y)-D(x)\cdot y.$ (2.33)

Then, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a transposed Hom-Poisson
algebra.

###### Proof.

First, we prove that $(A,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Lie algebra. For
all $x,y,z\in A$,

$\displaystyle\\{x,y\\}$ $\displaystyle=x\cdot D(y)-D(x)\cdot y=-(y\cdot
D(x)-D(y).x)=-\\{y,x\\},$ $\displaystyle\circlearrowleft_{x,y,z}$
$\displaystyle\\{\alpha(x),\\{y,z\\}\\}=\circlearrowleft_{x,y,z}\\{\alpha(x),y\cdot
D(z)-D(y)\cdot z\\}$
$\displaystyle=\circlearrowleft_{x,y,z}\Big{(}\alpha(x)\cdot D(y\cdot
D(z)-D(y)\cdot z)-D(\alpha(x))\cdot(y\cdot D(z)-D(y)\cdot z)\Big{)}$
$\displaystyle=\circlearrowleft_{x,y,z}\Big{(}\underbrace{\alpha(x)\cdot(D(y)\cdot
D(z))}_{a_{2}(x,y,z)}+\underbrace{\alpha(x)\cdot(y\cdot
D^{2}(z))}_{b_{2}(x,y,z)}-\underbrace{\alpha(x)\cdot(D^{2}(y)\cdot
z)}_{c_{2}(x,y,z)}$
$\displaystyle\quad\quad\quad-\underbrace{\alpha(x)\cdot(D(y)\cdot
D(z))}_{d_{2}(x,y,z)}-\underbrace{D(\alpha(x))\cdot(y\cdot
D(z))}_{e_{2}(x,y,z)}+\underbrace{D(\alpha(x))\cdot(D(y)\cdot
z)}_{f_{2}(x,y,z)}\Big{)}.$

Using (2.1)-(2.4),

$\displaystyle\circlearrowleft_{x,y,z}(a_{2}(x,y,z)-d_{2}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(x)\cdot(D(y)\cdot
D(z))-\alpha(x)\cdot(D(y)\cdot D(z))\Big{)}=0,$
$\displaystyle\circlearrowleft_{x,y,z}(b_{2}(x,y,z)-c_{2}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(x)\cdot(y\cdot
D^{2}(z))-\alpha(x)\cdot(D^{2}(y)\cdot z)\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(x)\cdot(y\cdot
D^{2}(z))-\alpha(z)\cdot(x\cdot D^{2}(y))\Big{)}=0,$
$\displaystyle\circlearrowleft_{x,y,z}(f_{2}(x,y,z)-e_{2}(x,y,z))$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D(x))\cdot(D(y)\cdot
z)-\alpha(D(x))\cdot(y\cdot D(z))\Big{)}$
$\displaystyle\quad\quad=\circlearrowleft_{x,y,z}\Big{(}\alpha(D(x))\cdot(D(y)\cdot
z)-\alpha(D(z))\cdot(D(x)\cdot y)\Big{)}=0.$

This proves that $(A,\\{\cdot,\cdot\\},\alpha)$ satisfies the Hom-Jacobi
identity. Next, for all $x,y,z\in A$,

$\displaystyle\\{z\cdot x,\alpha(y)\\}+\\{\alpha(x),z\cdot y\\}$
$\displaystyle\quad=(z\cdot x)\cdot D(\alpha(y))-D(z\cdot
x)\cdot\alpha(y)+\alpha(x)\cdot D(z\cdot y)-D(\alpha(x))\cdot(z\cdot y)$
$\displaystyle\quad=\underbrace{(z\cdot x)\cdot
D(\alpha(y))}_{i_{2}(x,y,z)}-\underbrace{(D(z)\cdot
x)\cdot\alpha(y)}_{j_{2}(x,y,z)}-\underbrace{(z\cdot
D(x))\cdot\alpha(y)}_{k_{2}(x,y,z)}$
$\displaystyle\quad\quad+\underbrace{\alpha(x)\cdot(D(z)\cdot
y)}_{l_{2}(x,y,z)}+\underbrace{\alpha(x)\cdot(z\cdot
D(y))}_{m_{2}(x,y,z)}-\underbrace{D(\alpha(x))\cdot(z\cdot
y)}_{n_{2}(x,y,z)}.$

Using (2.1)-(2.4),

$\displaystyle l_{2}(x,y,z)-j_{2}(x,y,z)$
$\displaystyle\quad\quad=\alpha(x)\cdot(D(z)\cdot y)-(D(z)\cdot
x)\cdot\alpha(y)$ $\displaystyle\quad\quad=\alpha(x)\cdot(D(z)\cdot
y)-\alpha(x)\cdot(D(z)\cdot y)=0,$ $\displaystyle i_{2}(x,y,z)+m_{2}(x,y,z)$
$\displaystyle\quad\quad=(z\cdot x)\cdot\alpha(D(y))+\alpha(x)\cdot(z\cdot
D(y))$ $\displaystyle\quad\quad=\alpha(z)\cdot(x\cdot
D(y))+\alpha(z)\cdot(x\cdot D(y))=2\alpha(z)\cdot(x\cdot D(y)),$
$\displaystyle k_{2}(x,y,z)+n_{2}(x,y,z)$ $\displaystyle\quad\quad=(z\cdot
D(x))\cdot\alpha(y)+\alpha(D(x))\cdot(z\cdot y)$
$\displaystyle\quad\quad=\alpha(z)\cdot(D(x)\cdot y)+\alpha(z)\cdot(D(x)\cdot
y)=2\alpha(z)\cdot(D(x)\cdot y).$

This means that

$\\{z\cdot x,\alpha(y)\\}+\\{\alpha(x),z\cdot y\\}=2\alpha(z)\cdot(x\cdot
D(y))-2\alpha(z)\cdot(D(x)\cdot y)=2\alpha(z)\\{x,y\\}.$

Therefore, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ satisfies the transposed Hom-
Leibniz identity. ∎

######  Example 2.24.

Consider a two dimensional $\mathbb{K}$-linear space $A$ with basis
$\\{e_{1},e_{2}\\}$. Consider the commutative Hom-associative algebra
$(A,\cdot,\alpha)$ defined on $A$ by

$e_{1}\cdot e_{1}=-e_{1},~{}e_{1}\cdot e_{2}=e_{2}\cdot
e_{1}=e_{2},~{}\alpha(e_{1})=-e_{1},~{}\alpha(e_{2})=-e_{2},$

and the derivation $D$ on $A$ defined by $D(e_{1})=0,~{}D(e_{2})=\lambda
e_{2},~{}~{}\lambda\in\mathbb{K}.$ Then, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$
is a transposed Hom-Poisson algebra, where for all $(i,j)\neq(1,2)$,

$\displaystyle\\{e_{1},e_{2}\\}$ $\displaystyle=e_{1}\cdot
D(e_{2})-D(e_{1})\cdot e_{2}=\lambda e_{2},$ (2.34)
$\displaystyle\\{e_{i},e_{j}\\}$ $\displaystyle=0.$ (2.35)

######  Lemma 2.25 ([6]).

Let $(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra. Then, for
all $x,y,z,t\in A,$

$\\{xz,yt\\}+\\{xt,yz\\}=2zt\\{x,y\\}.$ (2.36)

###### Proof.

Using (1.1),

$\displaystyle\\{xz,yt\\}+\\{x,ytz\\}=2z\\{x,yt\\},$
$\displaystyle\\{xt,yz\\}+\\{xtz,y\\}=2z\\{xt,y\\}.$

Taking the sum of the above two identities gives

$(\\{xz,yt\\}+\\{xt,yz\\})+(\\{x,ytz\\}+\\{xtz,y\\})=2z(\\{x,yt\\}+\\{xt,y\\}).$

This means that

$\displaystyle\\{x,ytz\\}+\\{xtz,y\\}=2tz\\{x,y\\},$ $\displaystyle
2z(\\{x,yt\\}+\\{xt,y\\})=4zt\\{x,y\\}.$

Therefore, (2.36) is satisfied. ∎

######  Lemma 2.26 ([6]).

Let $(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra. For any
$h\in A$, define a linear map $\alpha_{h}:A\rightarrow A$ by

$\alpha_{h}(x)=h\cdot x,~{}~{}\hbox{for~{}all}~{}x\in A.$ (2.37)

Then, $(A,\\{\cdot,\cdot\\},\alpha_{h})$ is a Hom-Lie algebra.

######  Lemma 2.27.

Let $(A,\cdot)$ be a commutative associative algebra and for all $h\in A$
$\alpha_{h}$ the linear map define in (2.37). Then, $(A,\cdot,\alpha_{h})$ is
a commutative Hom-associative algebra.

###### Proof.

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

$(x\cdot y)\cdot\alpha_{h}(z)=(x\cdot y)\cdot h\cdot z=h\cdot(x\cdot y)\cdot
z=h\cdot x\cdot(y\cdot z)=\alpha_{h}(x)\cdot(y\cdot z).$

This means that $(A,\cdot,\alpha_{h})$ is a commutative Hom-associative
algebra. ∎

######  Theorem 2.28.

Let $(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra. For each
$h\in A$, let $\alpha_{h}$ be the linear map defined in (2.37). Then,
$(A,\cdot,\\{\cdot,\cdot\\},\alpha_{h})$ is a transposed Hom-Poisson algebra.

###### Proof.

Let $(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra. By Lemma
2.26 and Lemma 2.27, $(A,\\{\cdot,\cdot\\},\alpha_{h})$ is a Hom-Lie algebra
and $(A,\cdot,\alpha_{h})$ is a commutative Hom-associative algebra
respectively. Now, using Lemma 2.25, we show that the transposed Hom-Leibniz
identity is satisfied:

$\displaystyle\\{z\cdot x,\alpha_{h}(y)\\}+\\{\alpha_{h}(x),z\cdot y\\}$
$\displaystyle=\\{z\cdot x,h\cdot y\\}+\\{h\cdot x,z\cdot y\\}$
$\displaystyle=2z\cdot h\\{x,y\\}\quad\quad\mbox{(by \eqref{lemmaidentity2})}$
$\displaystyle=2\alpha_{h}(z)\\{x,y\\}\qed$

######  Example 2.29.

Consider a two-dimensional $\mathbb{K}$-linear space $A$ with basis
$\\{e_{1},e_{2}\\}$, and consider the transposed Poisson algebra
$(A,\cdot,\\{\cdot,\cdot\\})$ defined on $A$ by

$e_{1}\cdot e_{2}=e_{2}\cdot e_{1}=e_{1},~{}e_{2}\cdot
e_{2}=e_{2},~{}\\{e_{1},e_{2}\\}=e_{2}.$

Let $\alpha_{e_{1}}:A\rightarrow A$ be the linear map defined by
$\alpha_{e_{1}}(e_{1})=0,\alpha_{e_{1}}(e_{2})=e_{1}.$ Then, by Theorem 2.28,
$(A,\cdot,\\{\cdot,\cdot\\},\alpha_{e_{1}})$ is a transposed Hom-Poisson
algebra.

######  Lemma 2.30.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
Then, for all $x,y,z\in A$,

$\alpha(x)\cdot\\{y,z\\}+\alpha(y)\cdot\\{z,x\\}+\alpha(z)\cdot\\{x,y\\}=0.$
(2.38)

###### Proof.

Let $x,y,z\in A$. By (2.29),

$\displaystyle\\{x\cdot z,\alpha(y)\\}+\\{\alpha(x),y\cdot
z\\}=2\alpha(z)\cdot\\{x,y\\},$ $\displaystyle\\{y\cdot
x,\alpha(z)\\}+\\{\alpha(y),z\cdot x\\}=2\alpha(x)\cdot\\{y,z\\},$
$\displaystyle\\{z\cdot y,\alpha(x)\\}+\\{\alpha(z),x\cdot
y\\}=2\alpha(y)\cdot\\{z,x\\}.$

Taking the sum of the three identities above gives (2.38). ∎

######  Proposition 2.31.

For a linear space $A$ with a commutative Hom-associative multiplication
$\cdot$ and a Hom-Lie bracket $\\{\cdot,\cdot\\}$.

  1. (i)

If either $\cdot$ or $\\{\cdot,\cdot\\}$ is zero, then
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a transposed Hom-Poisson algebra, as
well as a Hom-Poisson algebra.

  2. (ii)

$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is both a Hom-Poisson algebra and a
transposed Hom-Poisson algebra if and only if
$\alpha(x)\cdot\\{y,z\\}=\\{x\cdot y,\alpha(z)\\}=0$ for all $x,y,z\in A.$

###### Proof.

The statement (i) holds, as all axioms are trivially satisfied. Let us prove
(ii). It is easy to check that if $\alpha(x)\cdot\\{y,z\\}=\\{x\cdot
y,\alpha(z)\\}=0$, for all $x,y\in A$, then
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is both a Hom-Poisson algebra and a
transposed Hom-Poisson algebra. Conversely, let $x,y,z\in A$. By (2.26),

$\\{z\cdot
x,\alpha(y)\\}=-\alpha(x)\cdot\\{y,z\\}-\alpha(z)\cdot\\{y,x\\},~{}~{}\\{\alpha(x),z\cdot
y\\}=\alpha(y)\cdot\\{x,z\\}+\alpha(z)\cdot\\{x,y\\}.$

Next, by (2.29),

$0=\\{z\cdot x,\alpha(y)\\}+\\{\alpha(x),z\cdot
y\\}-2\alpha(z)\cdot\\{x,y\\}=\\{x,z\\}\cdot\alpha(y)-\\{y,z\\}\cdot\alpha(x).$

Then, by (2.38),
$\\{x,y\\}\cdot\alpha(z)=\\{x,z\\}\cdot\alpha(y)-\\{y,z\\}\cdot\alpha(x)=0.$
By (2.26) again, $\\{x\cdot y,\alpha(z)\\}=0.$ ∎

######  Example 2.32.

By Proposition 2.31, the transposed Hom-Poisson algebra
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ in Example 2.24 is Hom-Poisson algebra if
and only if $\lambda=0$.

######  Remark 2.33.

The above Proposition means that, with the relations of the commutative Hom-
associative product and the Hom-Lie bracket, the intersection of the Hom-
Leibniz identity and transposed Hom-Leibniz identity is trivial. That is, the
structures of transposed Hom-Poisson algebras and Hom-Poisson algebras are
incompatible.

######  Definition 2.34.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha),$ and
$(A^{\prime},\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime},\alpha^{\prime})$ be
transposed Hom-Poisson algebras. A morphism
$f:(A,\cdot,\\{\cdot,\cdot\\},\alpha)\rightarrow(A^{\prime},\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime},\alpha^{\prime})$
is a linear map $f:A\rightarrow A^{\prime}$ satisfying for all $x,y\in A$,

$\displaystyle f(x\cdot y)=f(x)\cdot^{\prime}f(y),\quad
f(\\{x,y\\})=\\{f(x),f(y)\\}^{\prime},\quad f\circ\alpha=\alpha^{\prime}\circ
f.$

The following theorem provides a procedure for construction of the transposed
Hom-Poisson algebras from transposed Poisson algebras and morphisms.

######  Theorem 2.35.

Let $\mathcal{A}=(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra
and $\alpha:\mathcal{A}\rightarrow\mathcal{A}$ be a transposed Poisson
algebras morphism. Define $\cdot_{\alpha},\ \\{\cdot,\cdot\\}_{\alpha}:A\times
A\rightarrow A$ for all $x,y\in A$, by $x\cdot_{\alpha}y=\alpha(x\cdot y)$ and
$\\{x,y\\}_{\alpha}=\alpha(\\{x,y\\})$. Then,
$\mathcal{A}_{\alpha}=(A_{\alpha}=A,\cdot_{\alpha},\\{\cdot,\cdot\\}_{\alpha},\alpha)$
is a transposed Hom-Poisson algebra called the $\alpha$-twist or Yau twist of
$(A,\cdot,\\{\cdot,\cdot\\})$. Moreover, assume that
$\mathcal{A}^{\prime}=(A^{\prime},\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime})$
is another transposed Poisson algebra and
$\alpha^{\prime}:\mathcal{A}^{\prime}\rightarrow\mathcal{A}^{\prime}$ is a
transposed Poisson algebras morphism. Let
$f:\mathcal{A}\rightarrow\mathcal{A}^{\prime}$ be a transposed Poisson
algebras morphism satisfying $f\circ\alpha=\alpha^{\prime}\circ f$. Then,
$f:\mathcal{A}_{\alpha}\rightarrow\mathcal{A}^{\prime}_{\alpha}$ is a
transposed Hom-Poisson algebras morphism.

######  Definition 2.36.

Let $\mathcal{A}=(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra
and $\alpha:\mathcal{A}\to\mathcal{A}$ be a transposed Poisson algebras
morphism.

  1. (i)

$\mathcal{A}_{\alpha}=(A_{\alpha}=A,\cdot_{\alpha}=\alpha\circ\cdot,\\{\cdot,\cdot\\}_{\alpha}=\alpha\circ\\{\cdot,\cdot\\})$
is called the $\alpha$-twisting of $\mathcal{A}$.

  2. (ii)

The $\alpha$-twisting $\mathcal{A}^{\prime}_{\alpha}$ of $\mathcal{A}$ is
called trivial if $\cdot_{\alpha}=0=\\{\cdot,\cdot\\}_{\alpha}$, and non-
trivial if either $\cdot_{\alpha}\neq 0$ or $\\{\cdot,\cdot\\}_{\alpha}\neq
0$.

  3. (iii)

$\mathcal{A}$ is called rigid if every twisting of $\mathcal{A}$ is either
trivial or isomorphic to $\mathcal{A}$.

######  Proposition 2.37.

Let $\mathcal{A}=(A,\cdot,\\{\cdot,\cdot\\})$ be a transposed Poisson algebra.
Suppose there exist a morphism $\alpha\colon\mathcal{A}\to\mathcal{A}$ such
that either $\cdot_{\alpha}=\alpha\circ\cdot$ is not associative, or
$\\{\cdot,\cdot\\}_{\alpha}=\alpha\circ\\{\cdot,\cdot\\}$ does not satisfy the
Jacobi identity. Then, $\mathcal{A}$ is not rigid.

###### Proof.

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

######  Definition 2.38.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
A representation of $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a quadruple
$(s,\rho,\beta,V)$ such that $(s,\beta,V)$ is a bimodule of the Hom-
commutative-associative algebra $(A,\cdot,\alpha)$ and $(\rho,\beta,V)$ is a
representation of the Hom-Lie algebra $(A,\\{\cdot,\cdot\\},\alpha)$
satisfying, for all $x,y\in A,~{}v\in V,$

$\displaystyle
2s(\\{x,y\\})\beta(v)=\rho(\alpha(x))s(y)v-\rho(\alpha(y))s(x)v,$ (2.39)
$\displaystyle 2s(\alpha(x))\rho(y)v=\rho(x\cdot
y)\beta(v)+\rho(\alpha(y))s(x)v,$ (2.40)
$\displaystyle\beta(s(x)v)=s(\alpha(x))\beta(v),~{}~{}~{}~{}\beta(\rho(x)v)=\rho(\alpha(x))\beta(v).$
(2.41)

######  Proposition 2.39.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra,
and let $(s,\rho,\beta,V)$ be its representation. Then, $(A\oplus
V,\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime},\alpha+\beta)$ is a transposed
Hom-Poisson algebra, where $(A\oplus V,\cdot^{\prime},\alpha+\beta)$ is the
semi-direct product commutative Hom-associative algebra
$A\ltimes_{s,\alpha,\beta}V$, and $(A\oplus
V,\\{\cdot,\cdot\\}^{\prime},\alpha+\beta)$ is the semi-direct product Hom-Lie
algebra $A\ltimes_{\rho,\alpha,\beta}V$.

###### Proof.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra,
and let $(s,\rho,\beta,V)$ be its representation. By Proposition 2.10 and
Proposition 2.15, $(A\oplus V,\cdot^{\prime},\alpha+\beta)$ is a commutative
Hom-associative algebra, and $(A\oplus
V,\\{\cdot,\cdot\\}^{\prime},\alpha+\beta)$ is a Hom-Lie algebra respectively.
Now, we show that the transposed Hom-Leibniz identity is satisfied. For all
$x_{1},x_{2},x_{3}\in A$ and $v_{1},v_{2},v_{3}\in V,$

$\displaystyle
2(\alpha+\beta)(x_{3}+v_{3})\cdot^{\prime}\\{x_{1}+v_{1},x_{2}+v_{2}\\}^{\prime}-\\{(x_{3}+v_{3})\cdot^{\prime}(x_{1}+v_{1}),(\alpha+\beta)(x_{2}+v_{2})\\}^{\prime}$
$\displaystyle\quad\quad-\\{(\alpha+\beta)(x_{1}+v_{1}),(x_{3}+v_{3})\cdot^{\prime}(x_{2}+v_{2})\\}^{\prime}$
$\displaystyle=2(\alpha+\beta)(x_{3}+v_{3})\cdot^{\prime}(\\{x_{1},x_{2}\\}+\rho(x_{1})v_{2}-\rho(x_{2})v_{1})$
$\displaystyle\quad\quad-\\{x_{3}\cdot
x_{1}+s(x_{3})v_{1}+s(x_{1})v_{3},(\alpha+\beta)(x_{2}+v_{2})\\}^{\prime}$
$\displaystyle\quad\quad-\\{(\alpha+\beta)(x_{1}+v_{1}),x_{3}\cdot
x_{2}+s(x_{3})v_{2}+s(x_{2})v_{3}\\}^{\prime}$
$\displaystyle=2\Big{(}\alpha(x_{3})\cdot\\{x_{1},x_{2}\\}+s(\alpha(x_{3}))\rho(x_{1})v_{2}-s(\alpha(x_{3}))\rho(x_{2})v_{1}+s((\\{x_{1},x_{2}\\})\beta(v_{3})\Big{)}$
$\displaystyle\quad\quad-\Big{(}\\{x_{3}\cdot
x_{1},\alpha(x_{2})\\}+\rho(x_{3}\cdot
x_{1})\beta(v_{2})-\rho(\alpha(x_{2}))s(x_{3})v_{1}-\rho(\alpha(x_{2}))s(x_{1})v_{3}\Big{)}$
$\displaystyle\quad\quad-\Big{(}\\{\alpha(x_{1}),x_{3}\cdot
x_{2}\\}+\rho(\alpha(x_{1}))s(x_{3})v_{2}+\rho(\alpha(x_{1}))s(x_{2})v_{3}-\rho(x_{3}\cdot
x_{2})\beta(v_{1})\Big{)}$
$\displaystyle=\Big{(}2\alpha(x_{3})\cdot\\{x_{1},x_{2}\\}-\\{x_{3}\cdot
x_{1},\alpha(x_{2})\\}-\\{\alpha(x_{1}),x_{3}\cdot x_{2}\\}\Big{)}$
$\displaystyle\quad\quad+\Big{(}2s(\alpha(x_{3}))\rho(x_{1})v_{2}-\rho(x_{3}\cdot
x_{1})\beta(v_{2})-\rho(\alpha(x_{1}))s(x_{3})v_{2}\Big{)}$
$\displaystyle\quad\quad-\Big{(}2s(\alpha(x_{3}))\rho(x_{2})v_{1}-\rho(x_{3}\cdot
x_{2})\beta(v_{1})-\rho(\alpha(x_{2}))s(x_{3})v_{1}\Big{)}$
$\displaystyle\quad\quad+\Big{(}2s(\\{x_{1},x_{2}\\})\beta(v_{3})+\rho(\alpha(x_{2})s(x_{1})v_{3}-\rho(\alpha(x_{1}))s(x_{2})v_{3}\Big{)}=0+0+0+0=0.$

Then, $(A\oplus V,\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime},\alpha+\beta)$ is
a transposed Hom-Poisson algebra. ∎

The transposed Hom-Poisson algebra $(A\oplus
V,\cdot^{\prime},\\{\cdot,\cdot\\}^{\prime},\alpha+\beta)$ in Proposition 2.39
is denoted by $A\times_{s,\rho,\alpha,\beta}V.$

######  Example 2.40.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra,
and let $S_{\cdot}(a)b=a\cdot b=b\cdot a$ and $ad(a)b=\\{a,b\\}$, for all
$a,b\in A$. Then, $(S_{\cdot},ad,\alpha,A)$ is a regular representation of
$A$.

######  Theorem 2.41.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra
and $(s,\rho,\beta,V)$ a representation. Define
$s^{\ast},\rho^{\ast}:A\rightarrow End(V^{\ast})$ by

$\langle s^{\ast}(x)u^{\ast},v\rangle=-\langle
s(x)v,u^{\ast}\rangle,~{}~{}\langle\rho^{\ast}(x)u^{\ast},v\rangle=-\langle\rho(x)v,u^{\ast}\rangle.$
(2.42)

Let $\alpha^{\ast}:A^{\ast}\rightarrow
A^{\ast},~{}\beta^{\ast}:V^{\ast}\rightarrow V^{\ast}$ be the dual maps of
$\alpha$ and $\beta$ respectively, that is,

$\alpha^{\ast}(x^{\ast})(y)=x^{\ast}(\alpha(y)),~{}~{}\beta^{\ast}(u^{\ast})(v)=u^{\ast}(\beta(v)).$
(2.43)

If, in addition, for any $x,y\in A,~{}x^{\ast}\in A^{\ast},~{}v\in
V,~{}v^{\ast}\in V^{\ast},$

$\displaystyle
2s(\\{x,y\\})\beta(v)=s(y)\rho(\alpha(x))v-s(x)\rho(\alpha(y))v,$ (2.44)
$\displaystyle 2\rho(y)s(\alpha(x))v=\rho(x\cdot
y)\beta(v)+s(x)\rho(\alpha(y))v,$ (2.45)
$\displaystyle\beta(s(x)v)=s(x)\beta(v),~{}~{}\beta(\rho(\alpha(x))v)=\rho(x)\beta(v),$
(2.46)

then $(-s^{\ast},\rho^{\ast},\beta^{\ast},V^{\ast})$ is a representation of
$A$. Moreover,
$A\times_{-s^{\ast},\rho^{\ast},\alpha^{\ast},\beta^{\ast}}V^{\ast}$ is also
transposed Hom-Poisson algebra.

###### Proof.

For any $x,y\in A,~{}x^{\ast}\in A^{\ast},~{}v\in V,~{}u^{\ast}\in V^{\ast}$,
according to (2.42) and (2.43),

$\displaystyle\langle\beta^{\ast}(\rho^{\ast}(x)u^{\ast}),v\rangle$
$\displaystyle=-\langle\rho^{\ast}(x)u^{\ast},\beta(v)\rangle=-\langle\rho(x)\beta(v),u^{\ast}\rangle$
$\displaystyle=-\langle\beta(\rho(\alpha(x))v),u^{\ast}\rangle=-\langle\rho(\alpha(x))v,\beta^{\ast}(u^{\ast})\rangle=\langle\rho^{\ast}(\alpha(x))\beta^{\ast}(u^{\ast}),v\rangle.$

So, (2.41) holds for $\rho^{\ast}$. Similarly, (2.41) holds for $-s^{\ast}$.
According to (2.42)-(2.45),

$\displaystyle\langle-2s^{\ast}(\\{x,y\\})\beta^{\ast}(u^{\ast})+\rho^{\ast}(\alpha(x))s^{\ast}(y)u^{\ast}+\rho^{\ast}(\alpha(y))s^{\ast}(x)u^{\ast},v\rangle$
$\displaystyle\quad\quad=\langle-2\beta(s(\\{x,y\\})v)+s(y)\rho(\alpha(x))v-s(x)\rho(\alpha(y))v,u^{\ast}\rangle=0,$
$\displaystyle\langle-2s^{\ast}(\alpha(x))\rho^{\ast}(y)v-\rho(x\cdot
y)\beta^{\ast}(u^{\ast})+\rho^{\ast}(\alpha(y))s^{\ast}(x)u^{\ast},v\rangle$
$\displaystyle\quad\quad=\langle\beta(\rho(x\cdot
y)v)+s(x)\rho(\alpha(y))v-2\rho(y)s(\alpha(x)),u^{\ast}\rangle=0.$

Therefore, (2.39) and (2.40) hold for
$(-s^{\ast},\rho^{\ast},\beta^{\ast},V^{\ast})$. Then,
$(-s^{\ast},\rho^{\ast},\beta^{\ast},V^{\ast})$ is a representation of
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$. ∎

######  Example 2.42.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
The quadruples $(A,S,ad,\alpha)$ and
$(A^{\ast},-S^{\ast},ad^{\ast},\alpha^{\ast})$ are representations of
transposed Hom-Poisson algebra $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$.

######  Theorem 2.43.

Let $\mathcal{A}=(A,\cdot_{A},\\{\cdot,\cdot\\}_{A},\alpha)$ and
$\mathcal{B}=(B,\cdot_{B},\\{\cdot,\cdot\\}_{B},\beta)$ be two transposed Hom-
Poisson algebras. Suppose that there are linear maps
$s_{A},\rho_{A}:A\rightarrow gl(B)$ and $s_{B},\rho_{B}:B\rightarrow gl(A)$
such that $A\bowtie^{\rho_{A},\beta}_{\rho_{B},\alpha}B$ is a matched pair of
Hom-Lie algebras, and $A\bowtie^{s_{A},\beta}_{s_{B},\alpha}B$ is a matched
pair of commutative Hom-associative algebras, and for all $x,y\in A,~{}a,b\in
B$, the following equalities hold:

$\displaystyle\begin{split}2s_{A}(\alpha(x))\\{a,b\\}_{B}&=\rho_{B}(s_{B}(a)x)\beta(b)+\\{s_{A}(x)a,\beta(b)\\}_{B}\\\
&\quad-\rho_{A}(s_{B}(b)x)\beta(a)+\\{\beta(a),s_{A}(x)b\\}_{B},\end{split}$
(2.47)
$\displaystyle\begin{split}2\beta(a)\cdot_{B}\rho_{A}(x)b&-2s_{A}(\rho_{B}(b)x)\beta(a)\\\
&=\\{s_{A}(x)a,\beta(b)\\}_{B}+\rho_{A}(s_{B}(a)x)\beta(b)+\rho_{A}(\alpha(x))(a\cdot_{B}b),\end{split}$
(2.48)
$\displaystyle\begin{split}2s_{B}(\beta(a))\\{x,y\\}_{A}&=\rho_{A}(s_{A}(x)a)\alpha(a)+\\{s_{B}(a)x,\alpha(y)\\}_{A}\\\
&\quad-\rho_{B}(s_{A}(y)a)\alpha(x)+\\{\alpha(y),s_{B}(a)y\\}_{A},\end{split}$
(2.49)
$\displaystyle\begin{split}2\alpha(x)\cdot_{A}\rho_{B}(a)y&-2s_{B}(\rho_{A}(y)a)\alpha(x)\\\
&=\\{s_{B}(a)x,\alpha(y)\\}_{A}+\rho_{B}(s_{A}(x)a)\alpha(y)+\rho_{B}(\beta(a))(x\cdot_{A}y).\end{split}$
(2.50)

Then, $(A,B,s_{A},\rho_{A},\beta,s_{B},\rho_{B},\alpha)$ is called a matched
pair of transposed Hom-Poisson algebras. In this case, on the direct sum
$A\oplus B$ of the underlying vector spaces of $\mathcal{A}$ and
$\mathcal{B}$, there is a transposed Hom-Poisson algebra structure which is
given for any $x,y\in A,~{}a,b\in B$ by

$\displaystyle(x+a)\cdot(y+b)$
$\displaystyle=x\cdot_{A}y+(l_{A}(x)b+r_{A}(y)a)+a\cdot_{B}b+(l_{B}(a)y+r_{B}(b)x),$
$\displaystyle[x+a,y+b]$
$\displaystyle=\\{x,y\\}_{A}+\rho_{A}(x)b-\rho_{A}(y)a+\\{a,b\\}_{B}+\rho_{B}(a)y-\rho_{B}(b)x,$
$\displaystyle(\alpha\oplus\beta)(x+a)$ $\displaystyle=\alpha(x)+\beta(a).$

###### Proof.

By Theorem 2.11 and Theorem 2.16, we deduce that $(A\oplus
B,\cdot,\alpha+\beta)$ is a commutative Hom-associative algebra and $(A\oplus
B,[\cdot,\cdot],\alpha+\beta)$ is a Hom-Lie algebra. Now, the rest, it is easy
(in a similar way as for Proposition 2.11) to verify the transposed Hom-
Leibniz identity satisfied. ∎

##  3 Manin triples and bialgebras of transposed Hom-Poisson algebras

In this section, we introduce the notion of Manin triples for transposed Hom-
Poisson algebras and transposed Hom-Poisson bialgebras and establish the
equivalences between their Manin triples.

######  Definition 3.1.

A bilinear form $\mathfrak{B}$ on a transposed Hom-Poisson algebra
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is called invariant if for all $x,y,z\in
A$,

$\mathfrak{B}(x\cdot y,\alpha(z))=\mathfrak{B}(\alpha(x),y\cdot
z),\quad\mathfrak{B}(\\{x,y\\},\alpha(z))=\mathfrak{B}(\alpha(x),\\{y,z\\}).$

######  Definition 3.2.

A Manin triple of transposed Hom-Poisson algebras is a triple of transposed
Hom-Poisson algebras $(A,A_{1},A_{2})$ together with a nondegenerate symmetric
invariant bilinear form $\mathfrak{B}$ on $A$ satisfying the following
conditions:

  1. (i)

$A_{1}$ and $A_{2}$ are transposed Hom-Poisson subalgebras of $A$;

  2. (ii)

as linear spaces, $A=A_{1}\oplus A_{2}$;

  3. (iii)

$A_{1}$ and $A_{2}$ are isotropic with respect to $\mathfrak{B}$, that is, for
any $x_{1},y_{1}\in A_{1}$ and any $x_{2},y_{2}\in A_{2}$,
$\mathfrak{B}(x_{1},y_{1})=0=\mathfrak{B}(x_{2},y_{2}).$

######  Definition 3.3.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
Suppose there is a transposed Hom-Poisson algebra structure
$(A^{*},\circ,[\cdot,\cdot],\alpha^{\ast})$ on the dual space of $A$, and
there is a transposed Hom-Poisson algebra structure on the direct sum $A\oplus
A^{*}$ of the underlying linear spaces of $A$ and $A^{*}$, such that
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ and
$(A^{*},\circ,[\cdot,\cdot],\alpha^{\ast})$ are subalgebras, and the natural
symmetric bilinear form on $A\oplus A^{*}$ given, for all $x,y\in
A,a^{*},b^{*}\in A^{*}$, by

$\displaystyle\mathfrak{B}_{d}(x+a^{*},y+b^{*})$ $\displaystyle=\langle
a^{*},y\rangle+\langle x,b^{*}\rangle,\;$ (3.1)
$\displaystyle(\alpha+\alpha^{\ast})(x+a^{\ast})$
$\displaystyle=\alpha(x)+\alpha^{\ast}(a^{\ast}),$ (3.2)

is invariant, then $(A\oplus A^{*},A,A^{*},\alpha+\alpha^{\ast})$ is called a
standard Manin triple of transposed Hom-Poisson algebras associated to
$\mathfrak{B}_{d}$.

######  Remark 3.4.

A standard Manin triple of transposed Hom-Poisson algebras is a Manin triple
of transposed Hom-Poisson algebras.

######  Theorem 3.5.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ and
$(A^{\ast},\circ,[\cdot,\cdot],\alpha^{\ast})$ be two transposed Hom-Poisson
algebras. Then, $(A\oplus A^{\ast},A,A^{\ast})$ is a standard Manin triple of
transposed Poisson algebras if and only if
$(A,A^{\ast},-S^{\ast}_{\cdot},ad^{\ast}_{\\{\cdot,\cdot\\}},\alpha^{\ast},-S^{\ast}_{\circ},ad^{\ast}_{[\cdot,\cdot]},\alpha)$
is a matched pair of transposed Hom-Poisson algebras.

###### Proof.

Consider the following four maps, given for all $x,v,u\in A$,
$x^{\ast},v^{\ast},u^{\ast}\in A^{\ast}$, by

$\displaystyle S^{\ast}_{\cdot}:A\rightarrow gl(A^{\ast}),\langle
S^{\ast}_{\cdot}(x)u^{\ast},v\rangle=-\langle u^{\ast},S_{\cdot}(x)v\rangle,$
$\displaystyle ad^{\ast}_{\\{\cdot,\cdot\\}}:A\rightarrow gl(A^{\ast}),\langle
ad^{\ast}_{\\{\cdot,\cdot\\}}(x)u^{\ast},v\rangle=\langle
u^{\ast},ad_{\\{\cdot,\cdot\\}}(x)v\rangle,$ $\displaystyle
S^{\ast}_{\circ}:A^{\ast}\rightarrow gl(A),\langle
S^{\ast}_{\circ}(x^{\ast})u,v^{\ast}\rangle=-\langle
u,S_{\circ}(x^{\ast})v^{\ast}\rangle,$ $\displaystyle
ad^{\ast}_{[\cdot,\cdot]}:A^{\ast}\rightarrow gl(A),\langle
ad^{\ast}_{[\cdot,\cdot]}(x^{\ast})u,v^{\ast}\rangle=\langle
u,ad_{[\cdot,\cdot]}(x^{\ast})v^{\ast}\rangle.$

If
$(A,A^{\ast},-S^{\ast}_{\cdot},ad^{\ast}_{\\{\cdot,\cdot\\}},\alpha^{\ast},-S^{\ast}_{\circ},ad^{\ast}_{[\cdot,\cdot]},\alpha)$
is a matched pair of transposed Hom-Poisson algebras, then
$A\bowtie^{-S^{\ast}_{\cdot},ad^{\ast}_{\\{\cdot,\cdot\\}},\alpha^{\ast}}_{-S^{\ast}_{\circ},ad^{\ast}_{[\cdot,\cdot]},\alpha}A^{\ast}$
is a transposed Hom-Poisson algebra with the product $\ast$ and
$[\cdot,\cdot]_{\rho}$ given by

$\displaystyle(x+a)\ast(y+b)=x\cdot
y-(S^{\ast}_{\cdot}(x)b+S^{\ast}_{\cdot}(y)a)+a\circ
b-(S^{\ast}_{\circ}(a)y+S^{\ast}_{\circ}(b)x),$ (3.3)
$\displaystyle[[x+a,y+b]]=\\{x,y\\}+ad^{\ast}_{\\{\cdot,\cdot\\}}(x)b-ad^{\ast}_{\\{\cdot,\cdot\\}}(y)a+[a,b]+ad^{\ast}_{[\cdot,\cdot]}(a)y-ad^{\ast}_{[\cdot,\cdot]}(b)x.$

Let us compute and compare
$\displaystyle\mathfrak{B}_{d}((x+a)\ast(y+b),(z+c))$ and
$\mathfrak{B}_{d}((x+a),(y+b)\ast(z+c))$, for all $x,y,z\in A,a,b,c\in A^{*}:$

$\displaystyle\mathfrak{B}_{d}((x+a)\ast(y+b),z+c)$
$\displaystyle\quad=B_{d}(x\cdot
y-S^{\ast}_{\cdot}(x)b-S^{\ast}_{\cdot}(y)a+a\circ
b-S^{\ast}_{\circ}(a)y-S^{\ast}_{circ}(b)x,z+c)$ $\displaystyle\quad=\langle
x\cdot y-S^{\ast}_{\circ}(a)y-S^{\ast}_{\circ}(b)x,c\rangle+\langle z,a\circ
b-S^{\ast}_{\cdot}(x)b-S^{\ast}_{\cdot}(y)a\rangle$
$\displaystyle\quad=\langle x\cdot y,c\rangle-\langle
S^{\ast}_{\circ}(a)y,c\rangle-\langle S^{\ast}_{\circ}(b)x,c\rangle+\langle
z,a\circ b\rangle-\langle z,S^{\ast}_{\cdot}(x)b\rangle-\langle
z,S^{\ast}_{\cdot}(y)a\rangle$ $\displaystyle\quad=\langle x\cdot
y,c\rangle+\langle y,S_{\circ}(a)c\rangle+\langle
x,S_{\circ}(b)c\rangle+\langle z,a\circ b\rangle+\langle
S_{\cdot}(x)z,b\rangle+\langle S_{\cdot}(y)z,a\rangle$
$\displaystyle\quad=\langle x\cdot y,c\rangle+\langle y,a\circ
c\rangle+\langle x,b\circ c\rangle+\langle z,a\circ b\rangle+\langle x\cdot
z,b\rangle+\langle y\cdot z,a\rangle,$
$\displaystyle\mathfrak{B}_{d}((x+a),(y+b)\ast(z+c))$
$\displaystyle\quad=B_{d}((x+a),y\cdot
z-S^{\ast}_{\cdot}(y)c-S^{\ast}_{\cdot}(z)b-S^{\ast}_{\circ}(b)z-S^{\ast}_{\circ}(c)y)$
$\displaystyle\quad=\langle x,b\circ
c-S^{\ast}_{\cdot}(y)c-S^{\ast}_{\cdot}(z)b\rangle+\langle y\cdot
z-S^{\ast}_{\circ}(b)z-S^{\ast}_{\circ}(c)y,a\rangle$
$\displaystyle\quad=\langle x,b\circ c\rangle-\langle
x,S^{\ast}_{\cdot}(y)c\rangle-\langle x,S^{\ast}_{\cdot}(z)b\rangle+\langle
y\cdot z,a\rangle-\langle S^{\ast}_{\circ}(b)z,a\rangle-\langle
S^{\ast}_{\circ}(c)y,a\rangle$ $\displaystyle\quad=\langle x,b\circ
c\rangle+\langle S_{\cdot}(y)x,c\rangle+\langle S_{\cdot}(z)x,b\rangle+\langle
y\cdot z,a\rangle+\langle z,S_{\circ}(b)a\rangle+\langle
y,S_{\circ}(c)a\rangle$ $\displaystyle\quad=\langle x,b\circ c\rangle+\langle
y\cdot x,c\rangle+\langle z\cdot x,b\rangle+\langle y\cdot z,a\rangle+\langle
z,b\circ a\rangle+\langle y,c\circ a\rangle.$

It follows that
$\mathfrak{B}_{d}((x+a)\ast(y+b),z+c)=\mathfrak{B}_{d}((x+a),(y+b)\ast(z+c)).$
Similarly,
$\mathfrak{B}_{d}([[x+a,y+b]],z+c)=\mathfrak{B}_{d}((x+a),[[y+b,z+c]]).$ We
deduce the invariance of the standard bilinear form $\mathfrak{B}_{d}$ on
$A\oplus A^{\ast}$. Therefore, $(A\oplus A^{\ast},A,A^{\ast})$ is a standard
Manin triple of transposed Poisson algebras $A$ and $A^{\ast}.$ ∎

######  Definition 3.6.

A Lie bialgebra is a quadruple $(A,\\{\cdot,\cdot\\},\delta,\alpha)$ where
$(A,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Lie algebra and
$\delta:A\rightarrow\wedge^{2}A$ is a linear map such that $(A,\delta,\alpha)$
is a Hom-Lie coalgebra and $\delta$ is a $1$-cocycle on $A$ with coefficients
in $A\otimes A$, that is, it satisfies the following equation for all $x,y\in
A$,

$\delta(\\{x,y\\})=(ad_{x}\otimes\alpha+\alpha\otimes
ad_{x})\delta(y)-(ad_{y}\otimes\alpha+\alpha\otimes ad_{y})\delta(x).$ (3.4)

A Hom-Lie bialgebra $(A,\\{\cdot,\cdot\\},\delta,\alpha)$ is called coboundary
if $\delta$ is a $1$-coboundary, that is, there exists an $r\in A\otimes A$
with $\alpha^{\otimes 2}(r)=r$, such that for all $x\in A$,

$\delta(\\{x,y\\})=(ad_{x}\otimes\alpha+\alpha\otimes ad_{x})\delta(y))r.$
(3.5)

In this case, a coboundary Lie bialgebra is usually denoted by
$(A,\\{\cdot,\cdot\\},r,\alpha).$

######  Definition 3.7.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra. An
infinitesimal Hom-bialgebra structure on $A$ is a linear map
$\Delta:A\rightarrow A\otimes A$ such that $(A,\Delta)$ is a Hom-coassociative
coalgebra and $\Delta$ is a 1-cocycle, that is, for all $a,b\in A$,

$\displaystyle\Delta(a\cdot
b)=(S(\alpha(a))\otimes\alpha)\Delta(b)+(\alpha\otimes
S(\alpha(b)))\Delta(a).$ (3.6)

An infinitesimal Hom-bialgebra $(A,\cdot,\Delta,\alpha)$ is called coboundary
if $\Delta$ 1-coboundary, that is, there exists an $r\in A\otimes A$ with
$\alpha^{\otimes 2}(r)=r$, such that for all $a\in A$,

$\displaystyle\Delta(a)=(S(a)\otimes\alpha-\alpha\otimes S(a))r.$ (3.7)

In this case, a coboundary infinitesimal Hom-bialgebra is denoted by
$(A,\cdot,r,\alpha)$.

######  Definition 3.8.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
Suppose that it is equipped with two comultiplications
$\delta,\Delta:A\rightarrow A\otimes A$ such that $(A,\Delta,\delta,\alpha)$
is a transposed Hom-Poisson coalgebra, that is, $(A,\delta,\alpha)$ is a Hom-
Lie coalgebra and $(A,\Delta,\alpha)$ is a Hom-cocommutative coassociative
coalgebra, and they satisfy the following compatibility condition for all
$x\in A$:

$\displaystyle(\alpha\otimes\Delta)\delta(x)=(\delta\otimes\alpha)\Delta(x)+(\tau\otimes
id)(\alpha\otimes\delta)\Delta(x).$

If in addition, $(A,\\{\cdot,\cdot\\},\delta,\alpha)$ is a Hom-Lie bialgebra,
$(A,\cdot,\Delta,\alpha)$ is a commutative and associative infinitesimal Hom-
bialgebra, and $\delta$ and $\Delta$ are compatible in the sense that

$\displaystyle\delta(x\circ y)$ $\displaystyle=$
$\displaystyle(S(\alpha(y))\otimes\alpha)\delta(x)+(S(\alpha(x))\otimes\alpha)\delta(y)$
$\displaystyle-(\alpha\otimes ad_{x})\Delta(y)-(\alpha\otimes
ad_{y})\Delta(x),$ $\displaystyle\Delta(\\{x,y\\})$ $\displaystyle=$
$\displaystyle(ad_{\alpha(x)}\otimes\alpha+\alpha\otimes
ad_{\alpha(x)}\Delta(y)$
$\displaystyle+(S(\alpha(y))\otimes\alpha-\alpha\otimes
S(\alpha(y)))\delta(x),$

then $(A,\cdot,\\{\cdot,\cdot\\},\delta,\Delta,\alpha)$ is called a transposed
Hom-Poisson bialgebra.

######  Theorem 3.9.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra
equipped with two comultiplications $\delta,\Delta:A\rightarrow A\otimes A$.
Suppose that the mappings $\delta^{\ast},\Delta^{\ast}:A^{\ast}\otimes
A^{\ast}\subset(A\otimes A)^{\ast}\rightarrow A^{\ast}$ induce a transposed
Hom-Poisson algebra structure on $(A^{\ast},\alpha^{\ast})$, where
$\delta^{\ast}$ and $\Delta^{\ast}$ correspond to the Hom-Lie bracket and the
product of the commutative Hom-associative algebra respectively. Set
$[\cdot,\cdot]=\delta^{\ast}$, $\circ=\Delta^{\ast}$. Then, the following
conditions are equivalent:

  1. (i)

$(A,\\{\cdot,\cdot\\},\cdot,\delta,\Delta,\alpha)$ is a transposed Hom-Poisson
bialgebra.

  2. (ii)

$(A,A^{\ast},ad^{\ast}_{\\{\cdot,\cdot\\}},-S^{\ast}_{\cdot},\alpha^{\ast}ad^{\ast}_{[\cdot,\cdot]},-S^{\ast}_{\circ},\alpha)$
is a matched pair of the transposed Hom-Poisson algebras.

  3. (iii)

$(A\oplus A^{\ast},A,A^{\ast},\alpha+\alpha^{\ast})$ is a standard Manin
triple of the transposed Hom-Poisson algebras with the bilinear form defined
in (3.1), and the isotropic subalgebras are $A$ and $A^{\ast}$.

##  4 Hom-pre-Lie Poisson algebras

In this section, we provide definition of Hom-pre-Lie Poisson algebra, develop
some constructions theorems and define their bimodule and matched pair and
describe some relevant properties.

######  Definition 4.1 ([69]).

A Hom-pre-Lie algebra is a triple $(A,\ast,\alpha)$ consisting of a vector
space $A$, a bilinear product $\ast:A\otimes A\rightarrow A,$ and a linear map
$\alpha:A\rightarrow A$ satisfying, for all $x,y,z\in A$,

$(x\ast y)\ast\alpha(z)-\alpha(x)\ast(y\ast z)=(y\ast
x)\ast\alpha(z)-\alpha(y)\ast(x\ast z).$ (4.1)

The equation (4.1) is called Hom-pre-Lie identity.

######  Lemma 4.2.

Let $(A,\ast,\alpha)$ be a Hom-pre-Lie algebra. Then,
$(A,[\cdot,\cdot],\alpha)$ is a Hom-Lie algebra with $[x,y]=x\ast y-y\ast x,$
for any $x,y\in A$. We say that $(A,[\cdot,\cdot],\alpha)$ is the sub-adjacent
Lie algebra of $(A,\ast,\alpha)$ and denoted by $A^{c}$.

###### Proof.

First, we prove $[\cdot,\cdot]$ satisfies Hom-skewsymmetry,

$[x,y]=x\ast y-y\ast x=-(y\ast x-x\ast y)=-[y,x].$

Now, we prove the Hom-Jacobi-identity,

$\displaystyle[\alpha(x),[y,z]]+[\alpha(y),[z,x]]+[\alpha(z),[x,y]]$
$\displaystyle\quad=[\alpha(x),y\ast z-z\ast y]+[\alpha(y),z\ast x-x\ast
z]+[\alpha(z),x\ast y-y\ast x]$ $\displaystyle\quad=\alpha(x)\ast(y\ast
z)-\alpha(x)\ast(z\ast y)-(y\ast z)\ast\alpha(x)$
$\displaystyle\quad\quad+(z\ast y)\ast\alpha(x)+\alpha(y)\ast(z\ast
x)-\alpha(y)\ast(x\ast z)$ $\displaystyle\quad\quad-(z\ast
x)\ast\alpha(y)+(x\ast z)\ast\alpha(y)+\alpha(z)\ast(x\ast y)$
$\displaystyle\quad\quad-\alpha(z)\ast(y\ast x)-(x\ast y)\ast\alpha(z)+(y\ast
x)\ast\alpha(z)$ $\displaystyle\quad=(y\ast
x)\ast\alpha(z)-\alpha(y)\ast(x\ast z)-((x\ast
y)\ast\alpha(z)-\alpha(x)\ast(y\ast z))$ $\displaystyle\quad\quad+(z\ast
y)\ast\alpha(x)-\alpha(z)\ast(y\ast x)-((y\ast
z)\ast\alpha(x)-\alpha(z)\ast(y\ast x))$ $\displaystyle\quad\quad+(x\ast
z)\ast\alpha(y)-\alpha(x)\ast(z\ast y)-((z\ast
x)\ast\alpha(y)-\alpha(z)\ast(x\ast y))$ $\displaystyle\quad=0+0+0=0.\qed$

Let us recall now the notion of bimodule and matched pair of a Hom-pre-Lie
algebra.

######  Definition 4.3 ([89]).

Let $(A,\ast,\alpha)$ be a Hom-pre-Lie algebra, and $(V,\beta)$ be a Hom-
module. Let $l_{\ast},r_{\ast}:A\rightarrow gl(V)$ be two linear maps. The
quadruple $(l_{\ast},r_{\ast},\beta,V)$ is called a bimodule of $A$ if for all
$x,y\in A,v\in V$,

$\displaystyle l_{\ast}(\\{x,y\\})\beta(v)$ $\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))l_{\ast}(y)v-l_{\ast}(\alpha(y))l_{\ast}(x)v$ (4.2)
$\displaystyle r_{\ast}(\alpha(y))\rho(x)v$ $\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))r_{\ast}(y)-r_{\ast}(x\ast y)\beta(v)$ (4.3)
$\displaystyle\beta(l_{\ast}(x)v)$ $\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))\beta(v),$ (4.4) $\displaystyle\beta(r_{\ast}(x)v)$
$\displaystyle=$ $\displaystyle r_{\ast}(\alpha(x))\beta(v),$ (4.5)

where $\\{x,y\\}=x\ast y-y\ast x$ and $\rho=l_{\ast}-r_{\ast}.$

######  Proposition 4.4 ([89]).

Let $(l_{\ast},r_{\ast},\beta,V)$ be a bimodule of a Hom-pre-Lie algebra
$(A,\ast,\alpha)$. Then, the direct sum $A\oplus V$ of vector spaces is turned
into a Hom-pre-Lie algebra by defining multiplication in $A\oplus V$ for all
$x_{1},x_{2}\in A,v_{1},v_{2}\in V$ by

$\displaystyle(x_{1}+v_{1})\ast^{\prime}(x_{2}+v_{2})$ $\displaystyle=$
$\displaystyle x_{1}\ast x_{2}+(l_{\ast}(x_{1})v_{2}+r_{\ast}(x_{2})v_{1}),$
$\displaystyle(\alpha\oplus\beta)(x_{1}+v_{1})$ $\displaystyle=$
$\displaystyle\alpha(x_{1})+\beta(v_{1}).$

We denote such a Hom-pre-Lie algebra by $(A\oplus
V,\ast^{\prime},\alpha+\beta),$ or
$A\times_{l_{\ast},r_{\ast},\alpha,\beta}V.$

######  Proposition 4.5.

Let $(l_{\ast},r_{\ast},\beta,V)$ be a bimodule of a Hom-pre-Lie algebra
$(A,\ast,\alpha)$. Let $(A,\\{\cdot,\cdot\\},\alpha)$ be the subadjacent Hom-
Lie algebra of $(A,\ast,\alpha)$. Then, $(l_{\ast}-r_{\ast},\beta,V)$ is a
representation of Hom-Lie algebra $(A,\\{\cdot,\cdot\\},\alpha)$.

###### Proof.

For all $x,y\in A,~{}v\in V$,

$\displaystyle\beta((l_{\ast}-r_{\ast})(x)v)=$
$\displaystyle\beta(l(x)v)-\beta(r(x)v)$ $\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))\beta(v)-r_{\ast}(\alpha(x))\beta(v)=(l_{\ast}-r_{\ast})(\alpha(x))\beta(v),$

and also

$\displaystyle(l_{\ast}-r_{\ast})(\alpha(x))\circ(l_{\ast}-r_{\ast})(y)v-(l_{\ast}-r_{\ast})(\alpha(y))\circ(l_{\ast}-r_{\ast})(x)v$
$\displaystyle\quad=l_{\ast}(\alpha(x))l_{\ast}(y)v-r_{\ast}(\alpha(x))l_{\ast}(y)v-l_{\ast}(\alpha(x))r_{\ast}(y)v+r_{\ast}(\alpha(x))r_{\ast}(y)v$
$\displaystyle\quad\quad-
l_{\ast}(\alpha(y))l_{\ast}(x)v+l_{\ast}(\alpha(y))r_{\ast}(x)v+r_{\ast}(\alpha(y))l_{\ast}(x)v-r_{\ast}(\alpha(y))r_{\ast}(x)v$
$\displaystyle\quad=\Big{(}l_{\ast}(\alpha(x))l_{\ast}(y)v-l_{\ast}(\alpha(y))l_{\ast}(x)v\Big{)}$
$\displaystyle\quad\quad+\Big{(}r_{\ast}(\alpha(y))l_{\ast}(x)v-l_{\ast}(\alpha(x))r_{\ast}(y)v-r_{\ast}(\alpha(y))r_{\ast}(x)v\Big{)}$
$\displaystyle\quad\quad-\Big{(}r_{\ast}(\alpha(x))l_{\ast}(y)v-r_{\ast}(\alpha(x))r_{\ast}(y)v-l_{\ast}(\alpha(y))r_{\ast}(x)v\Big{)}$
$\displaystyle\quad=l_{\ast}(\\{x,y\\})\beta(v)-r_{\ast}(x\ast
y)\beta(v)+r_{\ast}(y\ast x)\beta(v)$
$\displaystyle\quad=l_{\ast}(\\{x,y\\})\beta(v)-r_{\ast}(\\{x,y\\})\beta(v)=\rho(\\{x,y\\})\beta(v).$

Therefore, (2.15) and (2.16) are satisfied. ∎

######  Theorem 4.6 ([89]).

Let $(A,\ast_{A},\alpha)$ and $(B,\ast_{B},\beta)$ be two Hom-pre-Lie
algebras. Suppose that there are linear maps
$l_{\ast_{A}},r_{\ast_{A}}:A\rightarrow gl(B)$ and
$l_{\ast_{B}},r_{\ast_{B}}:B\rightarrow gl(A)$ such that
$(l_{\ast_{A}},r_{\ast_{A}},\beta,B)$ is a bimodule of $A$ and
$(l_{\ast_{B}},r_{\ast_{B}},\alpha,A)$ is a bimodule of $B$ satisfying for any
$x,y\in A,~{}a,b\in B$ and for

$\displaystyle\\{x,y\\}_{A}=x\ast y-y\ast
x,\quad\rho_{A}=l_{\ast_{A}}-r_{\ast_{A}},$ $\displaystyle\\{a,b\\}_{B}=a\ast
b-b\ast a,\quad\rho_{B}=l_{\ast_{B}}-r_{\ast_{B}},$

the following properties:

$\displaystyle\begin{split}r_{\ast_{A}}(\alpha(x))\\{a,b\\}_{B}&=r_{\ast_{A}}(r_{\ast_{B}}(b)x)\beta(a)-r_{\ast_{A}}(l_{\ast_{B}}(a)x)\beta(b)\\\
&+\beta(a)\ast_{B}r_{\ast_{A}}(x)b)-\beta(b)\ast_{B}(r_{\ast_{A}}(x)a),\end{split}$
(4.6)
$\displaystyle\begin{split}l_{\ast_{A}}(\alpha(x))(a\ast_{B}b)&=(\rho_{A}(x)a)\ast_{B}\beta(b)-l_{\ast_{A}}(\rho_{B}(a)x)\beta(b)\\\
&+\beta(a)\ast_{B}(l_{\ast_{A}}(x)b)+r_{\ast_{A}}(r_{\ast_{B}}(b)x)\beta(a),\end{split}$
(4.7)
$\displaystyle\begin{split}r_{\ast_{B}}(\beta(a))\\{x,y\\}_{B}&=r_{\ast_{B}}(r_{\ast_{A}}(y)xa)\alpha(x)-r_{\ast_{B}}(l_{\ast_{A}}(x)a)\alpha(y)\\\
&+\alpha(x)\ast_{A}r_{\ast_{B}}(a)y)-\alpha(y)\ast_{A}(r_{\ast_{B}}(a)x),\end{split}$
(4.8)
$\displaystyle\begin{split}l_{\ast_{B}}(\beta(a))(x\ast_{A}y)&=(\rho_{B}(a)x)\ast_{A}\alpha(y)-l_{\ast_{B}}(\rho_{A}(x)a)\alpha(y)\\\
&+\alpha(x)\ast_{A}(l_{\ast_{B}}(a)y)+r_{\ast_{B}}(r_{\ast_{A}}(y)a)\alpha(x).\end{split}$
(4.9)

Then, $(A,B,l_{\ast_{A}},r_{\ast_{A}},\beta,l_{\ast_{B}},r_{\ast_{B}},\alpha)$
is called a matched pair of Hom-pre-Lie algebras. In this case, there exists a
Hom-pre-Lie algebra structure on the vector space $A\oplus B$ of the
underlying vector spaces of $A$ and $B$ given by

$\displaystyle(x+a)\ast(y+b)$
$\displaystyle=x\ast_{A}y+(l_{\ast_{A}}(x)b+r_{\ast_{A}}(y)a)+a\ast_{B}b+(l_{\ast_{B}}(a)y+r_{\ast_{B}}(b)x),$
$\displaystyle(\alpha\oplus\beta)(x+a)$ $\displaystyle=\alpha(x)+\beta(a).$

We denote this Hom-pre-Lie algebra by
$A\bowtie^{l_{\ast_{A}},r_{\ast_{A}},\beta}_{l_{\ast_{B}},r_{\ast_{B}},\alpha}B$.

######  Proposition 4.7.

Let $(A,B,l_{\ast_{A}},r_{\ast_{A}},\beta,l_{\ast_{B}},r_{\ast_{B}},\alpha)$
be a matched pair of Hom-pre-Lie algebras $(A,\ast_{A},\alpha)$ and
$(B,\ast_{B},\beta)$. Then,
$(A,B,l_{\ast_{A}}-r_{\ast_{A}},\beta,l_{\ast_{B}}-r_{\ast_{B}},\alpha)$ is a
matched pair of the associated Hom-Lie algebras
$(A,\\{\cdot,\cdot\\}_{A},\alpha)$ and $(B,\\{\cdot,\cdot\\}_{B},\beta).$

###### Proof.

Let $(A,B,l_{\ast_{A}},r_{\ast_{A}},\beta,l_{\ast_{B}},r_{\ast_{B}},\alpha)$
be a matched pair of Hom-pre-Lie algebras $(A,\ast_{A},\alpha)$ and
$(B,\ast_{B},\beta)$. By Proposition 4.5, the linear maps
$l_{\ast_{A}}-r_{\ast_{A}}:A\rightarrow gl(B)$ and
$l_{\ast_{B}}-r_{\ast_{B}}:B\rightarrow gl(A)$ are representations of the
underlying Hom-Lie algebras $(A,\\{\cdot,\cdot\\}_{A},\alpha)$ and
$(B,\\{\cdot,\cdot\\}_{B},\beta)$, respectively. Therefore, (2.19) is
equivalent to (4.6)-(4.7), and (2.22) is equivalent to (4.8)-(4.9). ∎

Next, we define Hom-pre-Lie Poisson algebras and study their properties.

######  Definition 4.8.

A Hom-pre-Lie Poisson algebra is a quadruple $(A,\cdot,\ast,\alpha)$ such that
$(A,\cdot,\alpha)$ is a commutative Hom-associative algebra and
$(A,\ast,\alpha)$ is a Hom-pre-Lie algebra satisfying for all $x,y,z\in A$,
the following compatibility conditions:

$\displaystyle(x\cdot y)\ast\alpha(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot(y\ast z),$ (4.10) $\displaystyle(x\ast
y)\cdot\alpha(z)-(y\ast x)\cdot\alpha(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\ast(y\cdot z)-\alpha(y)\ast(x\cdot z).$ (4.11)

######  Example 4.9.

Consider a two-dimensional $\mathbb{K}$-linear space $A$ with basis
$\\{e_{1},e_{2}\\}$. Then, $(A,\cdot,\ast,\alpha)$ is a Hom-pre-Lie Poisson
algebra, where

$e_{1}\cdot e_{1}=ae_{2},~{}e_{1}\ast e_{1}=e_{1},~{}e_{1}\ast
e_{2}=e_{2},~{}\alpha(e_{1})=2e_{1},~{}\alpha(e_{2})=2e_{2},~{}~{}a\in\mathbb{K}.$

######  Definition 4.10.

Morphism
$f:(A,\cdot,\ast,\alpha)\rightarrow(A^{\prime},\cdot^{\prime},\ast^{\prime},\alpha^{\prime})$
of Hom-pre-Lie Poisson algebras are linear maps $f:A\rightarrow A^{\prime}$
satisfying for all $x,y\in A$,

$\displaystyle f(x\cdot y)=f(x)\cdot^{\prime}f(y),\quad f(x\ast
y)=f(x)\ast^{\prime}f(y),\quad f\circ\alpha=\alpha^{\prime}\circ f,$

which can be illustrated by the following standard commutative diagrams:

$\textstyle{A\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f\times
f}$$\scriptstyle{\cdot}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{A^{\prime}\otimes
A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cdot^{\prime}}$$\textstyle{A^{\prime}}$
$\textstyle{A\otimes
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f\times
f}$$\scriptstyle{\ast}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{A^{\prime}\otimes
A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\ast^{\prime}}$$\textstyle{A^{\prime}}$
$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\alpha}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha^{\prime}}$$\textstyle{A^{\prime}}$

######  Theorem 4.11.

Let $\mathcal{A}=(A,\cdot,\ast)$ be a pre-Lie Poisson algebra and
$\alpha:\mathcal{A}\rightarrow\mathcal{A}$ be a pre-Lie Poisson algebras
morphism. Define $\cdot_{\alpha}:A\times A\rightarrow A$ and
$\ast_{\alpha}:A\times A\rightarrow A$ for all $x,y\in A$ by
$x\cdot_{\alpha}y=\alpha(x\cdot y)$ and $x\ast_{\alpha}y=\alpha(x\ast y).$
Then,
$\mathcal{A}_{\alpha}=(A_{\alpha}=A,\cdot_{\alpha},\ast_{\alpha},\alpha)$ is a
Hom-pre-Lie Poisson algebra, called $\alpha$-twist or Yau twist of
$\mathcal{A}$. Moreover, assume that
$\mathcal{A}^{\prime}=(A^{\prime},\cdot^{\prime},\ast^{\prime})$ is another
pre-Lie Poisson algebra, and
$\alpha^{\prime}:\mathcal{A}^{\prime}\rightarrow\mathcal{A}^{\prime}$ is a
pre-Lie Poisson algebras morphism. Let
$f:\mathcal{A}\rightarrow\mathcal{A}^{\prime}$ be a pre-Lie Poisson algebras
morphism satisfying $f\circ\alpha=\alpha^{\prime}\circ f$. Then,
$f:\mathcal{A}_{\alpha}\rightarrow\mathcal{A}^{\prime}_{\alpha}$ is a Hom-pre-
Lie Poisson algebras morphism.

###### Proof.

We shall only prove (4.10) in $\mathcal{A}_{\alpha}$, as the other relation is
proved analogously. For any $x,y,z\in A$,

$\displaystyle(x\cdot_{\alpha}y)\ast_{\alpha}\alpha(z)=$
$\displaystyle\alpha(\alpha(x)\cdot\alpha(y))\ast\alpha^{2}(z)$
$\displaystyle=$
$\displaystyle(\alpha^{2}(x)\cdot\alpha^{2}(y))\ast\alpha^{2}(z)$
$\displaystyle=$
$\displaystyle\alpha^{3}(x)\cdot(\alpha^{2}(y)\ast\alpha^{2}(z))\quad\mbox{(by
\eqref{eq:pre-Poisson 1}~{}in~{}$\mathcal{A}$)}$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot_{\alpha}(y\ast_{\alpha}z).$

The second assertion follows from

$\displaystyle f(x\cdot_{\alpha}y)=$ $\displaystyle
f(\alpha(x)\cdot\alpha(y))=f(\alpha(x))\cdot^{\prime}f(\alpha(y))=\alpha^{\prime}f(x)\cdot^{\prime}\alpha^{\prime}f(y)=f(x)\cdot^{\prime}_{\alpha^{\prime}}f(y).$

Similarly, $f(x\ast_{\alpha}y)=f(x)\ast^{\prime}_{\alpha^{\prime}}f(y)$. ∎

######  Proposition 4.12.

Let $\mathcal{A}=(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra and
$\alpha^{\prime}:\mathcal{A}\rightarrow\mathcal{A}$ be a morphism. Then,
$(A,\alpha^{\prime}\circ\cdot,\alpha^{\prime}\circ\ast,\alpha\circ\alpha^{\prime})$
is a Hom-pre-Lie Poisson algebra.

######  Corollary 4.13.

If $\mathcal{A}=(A,\cdot,\ast,\alpha)$ is a multiplicative Hom-pre-Lie Poisson
algebra, then for any $n\in\mathbb{N}^{\ast}$,

  1. (i)

The $n{\rm th}$ derived Hom-pre-Lie Poisson algebra of type $1$ of
$\mathcal{A}$ is defined by

$\mathcal{A}_{1}^{n}=(A,\cdot^{(n)}=\alpha^{n}\circ\cdot,\ast^{(n)}=\alpha^{n}\circ\ast,\alpha^{n+1}).$

  2. (ii)

The $n{\rm th}$ derived Hom-pre-Lie Poisson algebra of type $2$ of $A$ is
defined by

$\mathcal{A}_{2}^{n}=(A,\cdot^{(2^{n}-1)}=\alpha^{2^{n}-1}\circ\cdot,\ast^{(2^{n}-1)}=\alpha^{2^{n}-1}\circ\ast,\alpha^{2^{n}}).$

###### Proof.

Apply Theorem 4.12 with $\alpha^{\prime}=\alpha^{n}$ and
$\alpha^{\prime}=\alpha^{2^{n}-1}$ respectively. ∎

######  Proposition 4.14.

Let $(A_{1},\cdot_{1},\ast_{1},\alpha_{1})$ and
$(A_{2},\cdot_{2},\ast_{2},\alpha_{2})$ be two Hom-pre-Lie Poisson algebras.
Define two operations $\cdot$ and $\ast$ on $A_{1}\otimes A_{2}$ for all
$x_{1},y_{1}\in A_{1},x_{2},y_{2}\in A_{2}$ by

$\displaystyle(x_{1}\otimes x_{2})\cdot(y_{1}\otimes
y_{2})=x_{1}\cdot_{1}y_{1}\otimes x_{2}\cdot_{2}y_{2},$ (4.12)
$\displaystyle(x_{1}\otimes x_{2})\ast(y_{1}\otimes
y_{2})=(x_{1}\ast_{1}y_{1})\otimes(x_{2}\cdot_{2}y_{2})+(x_{1}\cdot_{1}y_{1})\otimes(x_{2}\ast_{2}y_{2}),$
(4.13) $\displaystyle(\alpha_{1}\otimes\alpha_{2})(x_{1}\otimes
x_{2})=\alpha_{1}(x_{1})\otimes\alpha_{2}(x_{2}).$ (4.14)

Then, $(A_{1}\otimes A_{2},\cdot,\ast,\alpha_{1}\otimes\alpha_{2})$ is a Hom-
pre-Lie Poisson algebra.

######  Theorem 4.15.

Let $\mathcal{A}=(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra. For
any $x,y\in A$, let $\\{x,y\\}=x\ast y-y\ast x$. Then,
$\mathcal{A}^{c}=(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a transposed Hom-
Poisson algebra, called the sub-adjacent transposed Hom-Poisson algebra of
$(A,\cdot,\ast,\alpha)$.

###### Proof.

By Lemma 4.2, $(A,\\{\cdot,\cdot\\},\alpha)$ is a Hom-Lie algebra. The
transposed Hom-Leibniz identity is proved as follows:

$\displaystyle 2\alpha(z)\cdot(x\ast y-y\ast x)-((z\cdot x)\ast\alpha(y))$
$\displaystyle\quad-\alpha(y)\ast(z\cdot x))-(\alpha(x)\ast(z\cdot y)-(z\cdot
y)\ast\alpha(x))$ $\displaystyle=2\alpha(z)\cdot(x\ast
y)-2\alpha(z)\cdot(y\ast x)-(z\cdot x)\ast\alpha(y)$
$\displaystyle\quad+\alpha(y)\ast(z\cdot x)-\alpha(x)\ast(z\cdot y)+(z\cdot
y)\ast\alpha(x)$ $\displaystyle=(\alpha(z)\cdot(x\ast y)-(z\cdot
x)\ast\alpha(y))+((z\cdot y)\ast\alpha(x)$
$\displaystyle\quad-\alpha(z)\cdot(y\ast x))+(\alpha(z)\cdot(x\ast
y)-\alpha(z)\cdot(y\ast x)$ $\displaystyle\quad-\alpha(y)\ast(z\cdot
x)+\alpha(x)\ast(z\cdot y))$ $\displaystyle=0+0+0=0.\quad\mbox{(by
\eqref{eq:pre-Poisson 1}, \eqref{eq:pre-Poisson 2} and by commuatativity)}$

Thus, $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a Hom-transposd Poisson algebra.
∎

The relation existing between a transposed Hom-Poisson algebra and Hom-pre-Lie
Poisson algebra, as illustrated by the following diagram:

Hom-comm ass alg+ Hom-pre-Lie alg$\scriptstyle{[x,y]=x\ast y-y\ast x}$Hom-pre-
Lie Poisson alg$\scriptstyle{[x,y]=}$$\scriptstyle{x\ast y-y\ast x}$Hom-comm
ass alg+Hom-Lie algtransposed Hom-Poisson alg.

In the following we introduce the notions of bimodule and matched pair of Hom-
pre-Lie Poisson algebras and related relevant properties are also given

######  Definition 4.16.

Let $\mathcal{A}=(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra. A
bimodule of $\mathcal{A}$ is a $5$-tuple $(s,l_{\ast},r_{\ast},\beta,V)$ such
that $(l_{\ast},r_{\ast},\beta,V)$ is a bimodule of the Hom-pre-Lie algebra
$(A,\ast,\alpha)$ and $(s,\beta,V)$ is a bimodule of the commutative Hom-
associative algebra $(A,\cdot,\alpha)$ satisfying for all $x,y\in A$, $v\in V$
and $\\{x,y\\}=x\ast y-y\ast x,~{}\rho=l_{\ast}-r_{\ast}$,

$\displaystyle l_{\ast}(x\cdot y)\beta(v)$ $\displaystyle=$ $\displaystyle
s(\alpha(x))l_{\ast}(y)v,$ (4.15) $\displaystyle r_{\ast}(\alpha(y))s(x)v$
$\displaystyle=$ $\displaystyle s(x\ast y)\beta(v),$ (4.16) $\displaystyle
r_{\ast}(\alpha(y))s(x)v$ $\displaystyle=$ $\displaystyle
s(\alpha(x))r_{\ast}(y)v,$ (4.17) $\displaystyle s(\\{x,y\\})\beta(v)$
$\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))s(y)v-l_{\ast}(\alpha(y))s(x)v,$ (4.18) $\displaystyle
s(\alpha(y))\rho(x)v$ $\displaystyle=$ $\displaystyle
l_{\ast}(\alpha(x))s(y)v-r_{\ast}(x\cdot y)\beta(v).$ (4.19)

######  Proposition 4.17.

If $(s,l_{\ast},r_{\ast},\beta,V)$ is a bimodule of a Hom-pre-Lie Poisson
algebra $(A,\cdot,\ast,\alpha),$ then, there is a Hom-pre-Lie Poisson algebra
structure on $(A\oplus V,\cdot^{\prime},\ast^{\prime},\alpha+\beta),$ where
$(A\oplus V,\cdot^{\prime},\alpha+\beta)$ is the semi-direct product
commutative Hom-associative algebra $A\ltimes_{s,\alpha,\beta}V$, and
$(A\oplus V,\ast^{\prime},\alpha+\beta)$ is the semi-direct product Hom-Lie
algebra $A\ltimes_{l_{\ast},r_{\ast},\alpha,\beta}V$.

###### Proof.

Let $(s,l_{\ast},r_{\ast},\beta,V)$ be a bimodule of a Hom-pre-Lie Poisson
algebra $(A,\cdot,\ast,\alpha).$ By Proposition 2.10 and Proposition 4.4,
$(A\oplus V,\cdot^{\prime},\alpha+\beta)$ is a commutative Hom-associative
algebra and $(A\oplus V,\ast^{\prime},\alpha+\beta)$ is a Hom-pre-Lie algebra
respectively. Now, we prove the axioms (4.10) and (4.11) in $A\oplus V$. For
any $x_{1},x_{2},x_{3}\in A$ and $v_{1},v_{2},v_{3}\in V$,

$\displaystyle((x_{1}+v_{1})\cdot^{\prime}(x_{2}+v_{2}))\ast^{\prime}(\alpha+\beta)(x_{3}+v_{3})$
$\displaystyle\quad=(x_{1}\cdot
x_{2}+s(x_{1})v_{2}+s(x_{2})v_{1})\ast^{\prime}(\alpha(x_{3})+\beta(v_{3}))$
$\displaystyle\quad=(x_{1}\cdot x_{2})\ast\alpha(x_{3})+l_{\ast}(x_{1}\cdot
x_{2})\beta(v_{3})+r_{\ast}(\alpha(x_{3}))s(x_{1})v_{2}+r_{\ast}(\alpha(x_{3}))s(x_{2})v_{1},$
$\displaystyle(\alpha+\beta)(x_{1}+v_{1})\cdot^{\prime}((x_{2}+v_{2})\ast^{\prime}(x_{3}+v_{3}))$
$\displaystyle\quad=(\alpha(x_{1})+\beta(v_{1}))\cdot^{\prime}(x_{2}\ast
x_{3}+l_{\ast}(x_{2})v_{3}+r_{\ast}(x_{3})v)$
$\displaystyle\quad=\alpha(x_{1})\cdot(x_{2}\ast
x_{3})+s(\alpha(x_{1}))l_{\ast}(x_{2})v_{3}+s(\alpha(x_{1}))r_{\ast}(x_{3})v+s(x_{2}\ast
x_{3})\beta(v_{1}).$

By (4.15)-(4.17), and (4.10) in $A$,

$\displaystyle((x_{1}+v_{1})\cdot^{\prime}(x_{2}+v_{2}))\ast^{\prime}(\alpha+\beta)(x_{3}+v_{3})$
$\displaystyle\quad\quad\quad\quad\quad\quad=(\alpha+\beta)(x_{1}+v_{1})\cdot^{\prime}((x_{2}+v_{2})\ast^{\prime}(x_{3}+v_{3})),$
$\displaystyle((x_{1}+v_{1})\ast^{\prime}(x_{2}+v_{2}))\cdot^{\prime}(\alpha+\beta)(x_{3}+v_{3})-((x_{2}+v_{2})\ast^{\prime}(x_{1}+v_{1}))\cdot^{\prime}(\alpha+\beta)(x_{3}+v_{3})$
$\displaystyle\quad=(x_{1}\ast
x_{2}+l_{\ast}(x_{1})v_{2}+r_{\ast}(x_{2})v_{1})\cdot^{\prime}(\alpha(x_{1})+\beta(v_{2}))$
$\displaystyle\quad\quad-(x_{2}\ast
x_{1}+l_{\ast}(x_{2})v_{1}+r_{\ast}(x_{1})v_{2})\cdot^{\prime}(\alpha(x_{3})+\beta(v_{3}))$
$\displaystyle\quad=(x_{1}\ast x_{2})\cdot\alpha(x_{3})+s(x_{1}\ast
x_{2})\beta(v_{3})+s(\alpha(x_{3}))l_{\ast}(x_{1})v_{2}+s(\alpha(x_{3}))r_{\ast}(x_{2})v_{1}$
$\displaystyle\quad\quad-(x_{2}\ast x_{1})\cdot\alpha(x_{3})-s(x_{2}\ast
x_{1})\beta(v_{3})-s(\alpha(x_{3}))l_{\ast}(x_{2})v_{1}-s(\alpha(x_{3}))r_{\ast}(x_{1})v_{2},$
$\displaystyle(\alpha+\beta)(x_{1}+v_{1})\ast^{\prime}((x_{2}+v_{2})\cdot^{\prime}(x_{3}+v_{3}))-(\alpha+\beta)(x_{2}+v_{2})\ast^{\prime}((x_{1}+v_{1})\cdot^{\prime}(x_{3}+v_{3}))$
$\displaystyle\quad=(\alpha(x_{1})+\beta(v_{1}))\ast^{\prime}(x_{2}\ast
x_{3}+s(x_{2})v_{3}+s(x_{3})v_{2})$
$\displaystyle\quad\quad-(\alpha(x_{2})+\beta(v_{2}))\ast^{\prime}(x_{1}\cdot
x_{3}+s((x_{1})v_{3}+s(x_{3})v_{1})$
$\displaystyle\quad=\alpha(x_{1})\ast(x_{2}\cdot
x_{3})+s(\alpha(x_{1}))s(x_{2})v_{3}+l_{\ast}(\alpha(x_{1}))s(x_{3})v_{2}+r_{\ast}(x_{1}\cdot
x_{3})\beta(v_{2})$ $\displaystyle\quad\quad-\alpha(x_{2})\ast(x_{1}\cdot
x_{3})-l_{\ast}(\alpha(x_{2}))s(x_{1})v_{3}-l_{\ast}(\alpha(x_{2}))s(x_{3})v_{1}-r_{\ast}(x_{1}\cdot
x_{3})\beta(v_{2}).$

By (4.18)-(4.19), and (4.11) in $A$,

$\displaystyle((x_{1}+v_{1})\ast^{\prime}(x_{2}+v_{2}))\cdot^{\prime}(\alpha+\beta)(x_{3}+v_{3})-((x_{2}+v_{2})\ast^{\prime}(x_{1}+v_{1}))\cdot^{\prime}(\alpha+\beta)(x_{3}+v_{3})$
$\displaystyle=(\alpha+\beta)(x_{1}+v_{1})\ast^{\prime}((x_{2}+v_{2})\cdot^{\prime}(x_{3}+v_{3}))$
$\displaystyle\quad\quad\quad\quad\quad\quad-(\alpha+\beta)(x_{2}+v_{2})\ast^{\prime}((x_{1}+v_{1})\cdot^{\prime}(x_{3}+v_{3})).\qed$

We denote such Hom-pre-Lie Poisson algebra by
$A\times_{s,l_{\ast},r_{\ast},\alpha,\beta}V$.

######  Example 4.18.

Let $(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra. Then,
$(S,L_{\ast},R_{\ast},\alpha,A)$ is called a regular bimodule of
$(A,\cdot,\ast,\alpha)$, where for all $(x,y)\in A\times A$,

$S(x)y=x\cdot y,\quad L_{\ast}(x)y=x\ast y,\quad R_{\ast}(x)y=y\ast x.$

######  Proposition 4.19.

If
$f:\mathcal{A}=(A,\cdot_{1},\ast_{1},\alpha)\longrightarrow(A^{\prime},\cdot_{2},\ast_{2},\beta)$
is a Hom-pre-Lie Poisson algebras morphism, then
$(s_{1},l_{\ast_{1}},r_{\ast_{1}},\beta,A^{\prime})$ becomes a bimodule of
$\mathcal{A}$ via $f$, that is, for all $(x,y)\in A\times A^{\prime}$,
$s_{1}(x)y=f(x)\cdot_{2}y,$ $l_{\ast_{1}}(x)y=f(x)\ast_{2}y,$
$r_{\ast_{1}}(x)y=y\ast_{2}f(x).$

###### Proof.

We prove the axioms (4.15)-(4.19), as the others being proved similarly. For
any $x,y\in A$ and $z\in A^{\prime}$,

$\displaystyle l_{\ast_{1}}(x\cdot_{1}y)\beta(z)$
$\displaystyle=f(x\cdot_{1}y)\ast_{2}\beta(z)$
$\displaystyle=(f(x)\cdot_{2}f(y))\ast_{2}\beta(z)$ $\displaystyle=\beta
f(x)\cdot_{2}(f(y)\ast_{2}z)$
$\displaystyle=f(\alpha(x))\cdot_{2}(f(y)\ast_{2}z)$
$\displaystyle=s_{1}(\alpha(x))(f(y)\ast_{2}z)$
$\displaystyle=s_{1}(\alpha(x))l_{\ast_{1}}(y)z,$ $\displaystyle
r_{\ast_{1}}(\alpha(y))s_{1}(x)(z)$
$\displaystyle=r_{\ast_{1}}(\alpha(y))(z\cdot_{2}f(x))$
$\displaystyle=(z\cdot_{2}f(x))\ast_{2}f(\alpha(y))$
$\displaystyle=(z\cdot_{2}f(x))\ast_{2}\beta(f(y))$
$\displaystyle=(f(x)\ast_{2}f(y))\cdot_{2}\beta(z)$
$\displaystyle=f(x\ast_{1}y)\cdot_{2}\beta(z)$
$\displaystyle=s_{1}(x\ast_{1}y)\beta(z),$ $\displaystyle
r_{\ast_{1}}(\alpha(y))s_{1}(x)z$
$\displaystyle=s_{1}(x)z\ast_{2}f(\alpha(y))$
$\displaystyle=s_{1}(x)z\ast_{2}\beta(f(y))$
$\displaystyle=(f(x)\cdot_{2}z)\ast_{2}\beta(f(y))$
$\displaystyle=\beta(f(x))\cdot_{2}(z\ast_{2}f(y))$
$\displaystyle=f(\alpha(x))\cdot_{2}(z\ast_{2}f(y))$
$\displaystyle=s_{1}(\alpha(x))(z\ast_{2}f(y))$
$\displaystyle=s_{1}(\alpha(x))r_{\ast_{1}}(y)z,$ $\displaystyle
s_{\ast_{1}}(\\{x,y\\}_{1})\beta(z)$
$\displaystyle=s_{1}(x\ast_{1}y-y\ast_{1}x)\beta(z)$
$\displaystyle=s_{1}(x\ast_{1}y)\beta(z)-s_{1}(y\ast_{1}x)\beta(z)$
$\displaystyle=f(x\ast_{1}y)\cdot_{2}\beta(z)-f(y\ast_{1}x)\cdot_{2}\beta(z)$
$\displaystyle=(f(x)\ast_{2}f(y))\cdot_{2}\beta(z)-(f(y)\ast_{2}f(x))\cdot_{2}\beta(z)$
$\displaystyle=\beta(f(x))\ast_{2}(f(y)\cdot_{2}z)-\beta(f(y))\ast_{2}(f(x)\cdot_{2}z)$
$\displaystyle=f(\alpha(x))\ast_{2}(f(y)\cdot_{2}z)-f(\alpha(y))\ast_{2}(f(x)\cdot_{2}z)$
$\displaystyle=l_{\ast_{1}}(\alpha(x))(f(y)\cdot_{2}z)-l_{\ast_{1}}(\alpha(y))(f(x)\cdot_{2}z)$
$\displaystyle=l_{\ast_{1}}(\alpha(x))s_{1}(y)z-l_{\ast_{1}}(\alpha(y))s_{1}(x)z,$
$\displaystyle s_{1}(\alpha(y))\rho_{1}(x)v$
$\displaystyle=s_{1}(\alpha(y))(l_{\ast_{1}}-r_{\ast_{1}})(x)z$
$\displaystyle=s_{1}(\alpha(y))l_{\ast_{1}}(x)z-s_{1}(\alpha(y))r_{\ast_{1}}(x)z$
$\displaystyle=f(\alpha(y))\cdot_{2}l_{\ast_{1}}(x)z-f(\alpha(y))\cdot_{2}r_{\ast_{1}}(x)z$
$\displaystyle=\beta(f(y))\cdot_{2}(f(x)\ast_{2}z)-\beta
f(y)\cdot_{2}(z\ast_{2}f(x))$
$\displaystyle=\beta(f(x))\cdot_{2}(z\cdot_{2}f(y))-\beta(z)\ast_{2}(f(x)\cdot_{2}f(y))$
$\displaystyle=f(\alpha(x))\ast_{2}(z\cdot_{2}f(y))-\beta(z)\ast_{2}(f(x)\cdot_{1}y)$
$\displaystyle=l_{\ast_{1}}(\alpha(x))s_{1}(y)z-r_{\ast_{1}}(x\cdot_{1}y)\beta(z).\qed$

######  Corollary 4.20.

Let $(s,l_{\ast},r_{\ast},\beta,V)$ be a bimodule of a Hom-pre-Lie Poisson
algebra $(A,\cdot,\ast,\alpha)$. Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be
the subadjacent of $(A,\cdot,\ast,\alpha)$. Then,
$(s,l_{\ast}-r_{\ast},\beta,V)$ is a representation of
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$.

###### Proof.

It follows from the relation between the Hom-pre-Lie Poisson algebra and the
associated transposed Hom-Poisson algebra. More precisely by Poposition 4.5,
we deduce that $(l_{\ast}-r_{\ast},\beta,V)$ is a representation of
$(A,\\{\cdot,\cdot\\},\alpha)$. Now, the rest, it is easy (in a similar way as
for Poposition 4.5) to verify the axioms (2.39)-(2.40) ∎

######  Theorem 4.21.

Let $\mathcal{A}=(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra, and
let $V_{\beta}=(s,l_{\ast},r_{\ast},\beta,V)$ be a bimodule of $\mathcal{A}$.
Let $\alpha^{\prime}$ be a morphism of $\mathcal{A}$ such that the maps
$\alpha$ and $\alpha^{\prime}$ commute, and let $\beta^{\prime}$ be a linear
map of $V$ such that the maps $\beta$ and $\beta^{\prime}$ commute.
Furthermore, suppose that

$\left\\{\begin{array}[]{lllllll}\beta^{\prime}\circ
s=(s\circ\alpha^{\prime})\beta^{\prime},\\\ \beta^{\prime}\circ
l_{\ast}=(l_{\ast}\circ\alpha^{\prime})\beta^{\prime},\\\ \beta^{\prime}\circ
r_{\ast}=(r_{\ast}\circ\alpha^{\prime})\beta^{\prime},\end{array}\right.$

$\mathcal{A}_{\alpha^{\prime}}$ is the Hom-pre-Lie Poisson algebra
$(A,\cdot_{\alpha^{\prime}},\ast_{\alpha^{\prime}},\alpha\alpha^{\prime})$,
and
$V_{\beta^{\prime}}=(\widetilde{s},\widetilde{l}_{\ast},\widetilde{r}_{\ast},\beta\beta^{\prime},V)$,
where
$\widetilde{s}=(s\circ\alpha^{\prime})\beta^{\prime},~{}\widetilde{l}_{\ast}=(l_{\ast}\circ\alpha^{\prime})\beta^{\prime},~{}\widetilde{r}_{\ast}=(r_{\ast}\circ\alpha^{\prime})\beta^{\prime}.$
Then, $V_{\beta^{\prime}}$ is a bimodule of $\mathcal{A}_{\alpha^{\prime}}$.

###### Proof.

We prove only the axioms (4.15)-(4.19), as the others being proved similarly.
For any $x,y\in A$ and $v\in V$,

$\displaystyle\widetilde{l}_{\ast}(x\ast_{\alpha}^{\prime}y)\beta\beta^{\prime}(v)$
$\displaystyle=l_{\ast}(\alpha^{\prime 2}(x)\ast\alpha^{\prime
2}(y))\beta\beta^{\prime 2}(v)$ $\displaystyle=s(\alpha\alpha^{\prime
2}(x))l_{\ast}(\alpha^{\prime 2}(y))\beta^{\prime 2}(v)\quad\mbox{(by
\eqref{3,4} in $\mathcal{A}$)}$
$\displaystyle=\widetilde{s}(\alpha\alpha^{\prime}(x))\widetilde{l}_{\ast}(y)v,$
$\displaystyle\widetilde{r_{\ast}}(\alpha\alpha^{\prime}(y))\widetilde{s}(x)v$
$\displaystyle=\widetilde{r_{\ast}}(\alpha\alpha^{\prime}(y))s(\alpha^{\prime}(x))\beta^{\prime}(v)$
$\displaystyle=r_{\ast}(\alpha\alpha^{\prime
2}(y))\beta^{\prime}s((\alpha^{\prime}(x))\beta^{\prime}(v)$
$\displaystyle=r_{\ast}(\alpha\alpha^{\prime 2}(y))s((\alpha^{\prime
2}(x))\beta^{\prime 2}(v)$ $\displaystyle=s(\alpha^{\prime
2}(x)\ast\alpha^{\prime 2}(y))\beta\beta^{\prime 2}(v)$
$\displaystyle=s(\alpha^{\prime}(x)\ast_{\alpha^{\prime}}\alpha^{\prime}(y))\beta\beta^{\prime
2}(v)$
$\displaystyle=s(\alpha^{\prime}(x)\ast_{\alpha^{\prime}}\alpha^{\prime}(y))\beta\beta^{\prime
2}(v)$
$\displaystyle=\widetilde{s}(x\ast_{\alpha^{\prime}}y)\beta\beta^{\prime}(v),$
$\displaystyle\widetilde{r_{\ast}}(\alpha\alpha^{\prime}(y))\widetilde{s}(x)v$
$\displaystyle=\widetilde{r_{\ast}}(\alpha\alpha^{\prime}(y))s(\alpha^{\prime}(x))\beta^{\prime}(v)$
$\displaystyle=r_{\ast}(\alpha\alpha^{\prime 2}(y))s(\alpha^{\prime
2}(x))\beta^{\prime 2}(v)$ $\displaystyle=s(\alpha\alpha^{\prime
2}(x))r_{\ast}(\alpha^{\prime 2}(y))\beta^{\prime 2}(v)$
$\displaystyle=s(\alpha\alpha^{\prime
2}(x))\widetilde{r_{\ast}}(\alpha^{\prime}(y))\beta^{\prime}(v)$
$\displaystyle=\widetilde{s}(\alpha\alpha^{\prime}(x))\widetilde{r_{\ast}}(y)v,$
$\displaystyle\widetilde{s}(\\{x,y\\}_{\alpha^{\prime}})\beta\beta^{\prime}(v)$
$\displaystyle=\widetilde{s}(\\{\alpha^{\prime}(x),\alpha^{\prime}(y)\\})\beta\beta^{\prime}(v)$
$\displaystyle=s(\\{\alpha^{\prime 2}(x),\alpha^{\prime
2}(y)\\})\beta\beta^{\prime 2}(v)$
$\displaystyle=l_{\ast}(\alpha\alpha^{\prime 2}(x))s(\alpha^{\prime
2}(y))\beta^{\prime 2}(v)$ $\displaystyle-l_{\ast}(\alpha\alpha^{\prime
2}(y))s(\alpha^{\prime 2}(x))\beta^{\prime 2}(v)$
$\displaystyle=l_{\ast}(\alpha\alpha^{\prime
2}(x))\widetilde{s}(\alpha^{\prime}(y))\beta^{\prime}(v)$ $\displaystyle-
l_{\ast}(\alpha\alpha^{\prime
2}(y))\widetilde{s}(\alpha^{\prime}(x))\beta^{\prime}(v)$
$\displaystyle=\widetilde{l_{\ast}}(\alpha\alpha^{\prime}(x))\widetilde{s}(y)v$
$\displaystyle-\widetilde{l_{\ast}}(\alpha\alpha^{\prime}(y))\widetilde{s}(x)v,$
$\displaystyle\widetilde{s}(\alpha\alpha^{\prime}(y))\widetilde{\rho}(x)v$
$\displaystyle=\widetilde{s}(\alpha\alpha^{\prime}(y))\rho(\alpha^{\prime}(x))\beta^{\prime}(v)$
$\displaystyle=\widetilde{s}(\alpha\alpha^{\prime 2}(y))\rho(\alpha^{\prime
2}(x))\beta^{\prime 2}(v)$ $\displaystyle=l_{\ast}(\alpha\alpha^{\prime
2}(y))\rho(\alpha^{\prime 2}(x))s(\alpha^{\prime 2}(y))\beta^{\prime 2}(v)$
$\displaystyle-r_{\ast}(\alpha^{\prime 2}(x)\cdot\alpha^{\prime
2}(y))\beta\beta^{\prime 2}(v)$ $\displaystyle=l_{\ast}(\alpha\alpha^{\prime
2}(y))\rho(\alpha^{\prime
2}(x))\widetilde{s}(\alpha^{\prime}(y))\beta^{\prime}(v)$
$\displaystyle-\widetilde{r_{\ast}}(\alpha^{\prime}(x)\cdot\alpha^{\prime}(y))\beta\beta^{\prime}(v)$
$\displaystyle=\widetilde{l_{\ast}}(\alpha\alpha^{\prime}(x))\widetilde{s}(y)v$
$\displaystyle-\widetilde{r_{\ast}}(x\cdot_{\alpha}^{\prime}y)\beta\beta^{\prime}(v).\qed$

Taking $\alpha^{\prime}=\alpha^{p}$ and $\beta^{\prime}=\beta^{q}$ leads to
the following statement.

######  Corollary 4.22.

Let $\mathcal{A}=(A,\cdot,\ast,\alpha)$ be a multiplicative Hom-pre-Lie
Poisson algebra, and $(s,l_{\ast},r_{\ast},\beta,V)$ a bimodule of
$\mathcal{A}$. Then, $V_{\beta^{q}}$ is a bimodule of $A_{\alpha^{p}}$ for any
nonnegative integers $p$ and $q$.

######  Theorem 4.23.

Let $(A,\cdot_{A},\ast_{A},\alpha)$ and $(B,\cdot_{B},\ast_{B},\beta)$ be two
Hom–pre-Lie Poisson algebras. Suppose that there are linear maps
$s_{A},l_{\ast_{A}},r_{\ast_{A}}:A\rightarrow gl(B),$ and
$s_{B},l_{\ast_{B}},r_{\ast_{B}}:B\rightarrow gl(A)$ such that
$A\bowtie^{l_{\ast_{A}},r_{\ast_{A}},\beta}_{l_{\ast_{B}},r_{\ast_{B}},\alpha}B$
is a matched pair of Hom-pre-Lie algebras and
$A\bowtie^{s_{A},\beta}_{s_{B},\alpha}B$ is a matched pair of commutative Hom-
associative algebra, and for all $x,y\in A,~{}a,b\in B$, and for

$\\{x,y\\}_{A}=x\ast_{A}y-y\ast_{A}x,~{}\\{a,b\\}_{B}=a\ast_{B}-b\ast_{B}a,~{}\rho_{A}=l_{\ast_{A}}-r_{\ast_{A}},~{}\rho_{B}=l_{\ast_{B}}-r_{\ast_{B}},$

the following equalities hold:

$\displaystyle r_{\ast_{A}}(\alpha(x))(a\cdot_{B}b)$
$\displaystyle=\beta(a)\cdot_{B}(r_{\ast_{A}}(x)b)+s_{A}(l_{\ast_{B}}(b)x)\beta(a),$
(4.20)
$\displaystyle(s_{A}(x)a)\ast_{B}\beta(b)+l_{\ast_{A}}(s_{B}(a)x)\beta(b)$
$\displaystyle=s_{A}(\alpha(x))(a\ast_{B}b),$ (4.21) $\displaystyle
l_{\ast_{A}}(s_{B}(a)x)\beta(b)+(s_{A}(x)a)\ast_{B}\beta(b)$
$\displaystyle=\beta(a)\cdot_{B}(l_{\ast_{A}}(x)b)+s_{A}(r_{B}(b)x)\beta(a),$
(4.22)
$\displaystyle\begin{split}s_{A}(\alpha(x))(\\{a\ast_{B}b\\}_{B})&=\beta(a)\ast_{B}(s_{A}(x)b)+r_{\ast_{A}}(s_{B}(b)x)\beta(a)\\\
&-\beta(b)\ast_{B}(s_{A}(x)a)-r_{A}(s_{B}(a)x)\beta(b),\end{split}$ (4.23)
$\displaystyle\begin{split}(\rho_{A}(x)a)\cdot_{B}\beta(b)-s_{A}(\rho_{B}(a)x)\beta(b)&=l_{\ast_{A}}(\alpha(x))(a\cdot_{B}b)\\\
&-\beta(a)\ast_{B}(l_{\ast_{A}}(x)b)-r_{\ast_{A}}(s_{B}(b)x)\beta(a),\end{split}$
(4.24) $\displaystyle r_{\ast_{B}}(\beta(a))(ax\cdot_{A}y)$
$\displaystyle=\alpha(x)\cdot_{A}(r_{\ast_{B}}(a)y)+s_{B}(l_{\ast_{A}}(y)a)\alpha(x),$
(4.25)
$\displaystyle(s_{B}(a)x)\ast_{A}\alpha(y)+l_{\ast_{B}}(s_{A}(x)a)\alpha(y)$
$\displaystyle=s_{B}(\beta(a))(x\ast_{A}y),$ (4.26) $\displaystyle
l_{\ast_{B}}(s_{A}(x)a)\alpha(y)+(s_{B}(a)x)\ast_{A}\alpha(y)$
$\displaystyle=\alpha(x)\cdot_{A}(l_{\ast_{B}}(a)y)+s_{B}(r_{A}(y)a)\alpha(x),$
(4.27)
$\displaystyle\begin{split}s_{B}(\beta(a))(\\{x\ast_{A}y\\}_{A})&=\alpha(x)\ast_{A}(s_{B}(a)y)+r_{\ast_{B}}(s_{A}(y)a)\alpha(x)\\\
&-\alpha(y)\ast_{A}(s_{B}(a)x)-r_{B}(s_{A}(x)a)\alpha(y),\end{split}$ (4.28)
$\displaystyle\begin{split}(\rho_{B}(a)x)\cdot_{A}\alpha(x)-s_{B}(\rho_{A}(x)a)\alpha(y)&=l_{\ast_{B}}(\beta(a))(x\cdot_{A}y)\\\
&-\alpha(x)\ast_{A}(l_{\ast_{B}}(a)y)-r_{\ast_{B}}(s_{A}(y)a)\alpha(x).\end{split}$
(4.29)

Then,
$(A,B,s_{A},l_{\ast_{A}},r_{\ast_{A}},\beta,s_{B},l_{\ast_{B}},r_{\ast_{B}},\alpha)$
is called a matched pair of Hom-pre-Lie Poisson algebras. In this case, there
exists a Hom-pre-Lie Poisson algebra structure on the direct sum $A\oplus B$

###### Proof.

The proof is obtained in a similar way as for Theorem 2.11. ∎

Let
$A\bowtie^{s_{A},l_{\ast_{A}},r_{\ast_{A}},\beta}_{s_{B},l_{\ast_{B}},r_{\ast_{B}},\alpha}B$
denote this Hom-pre-Lie Poisson algebra.

######  Corollary 4.24.

Let
$(A,B,s_{A},l_{\ast_{A}},r_{\ast_{A}},\beta,s_{B},l_{\ast_{B}},r_{\ast_{B}},\alpha)$
be a matched pair of Hom-pre-Lie Poisson algebras
$(A,\cdot_{A},\ast_{A},\alpha)$ and $(B,\cdot_{B},\ast_{B},\beta)$.

Then,
$(A,B,s_{A},l_{\ast_{A}}-r_{\ast_{A}},\beta,s_{B},l_{\ast_{B}}-r_{\ast_{B}},\alpha)$
is a matched pair of the associated transposed Hom-Poisson algebras
$(A,\cdot_{A},\\{\cdot,\cdot\\}_{A},\alpha)$ and
$(B,\cdot_{B},\\{\cdot,\cdot\\}_{B},\beta)$.

###### Proof.

Let
$(A,B,s_{A},l_{\ast_{A}},r_{\ast_{A}},\beta,s_{B},l_{\ast_{B}},r_{\ast_{B}},\alpha)$
be a matched pair of Hom-pre-Lie Poisson algebras
$(A,\cdot_{A},\ast_{A},\alpha)$ and $(B,\cdot_{B},\ast_{B},\beta)$. Then,
$(A,B,s_{A},\beta,s_{B},\alpha)$ is a matched pair of commutative Hom-
associative algebras $(A,\cdot_{A},\alpha)$ and $(B,\cdot_{B},\beta)$, and by
Proposition 4.7,
$(A,B,l_{\ast_{A}}-r_{\ast_{A}},\beta,l_{\ast_{B}}-r_{\ast_{B}},\alpha)$ is a
matched pair of the associated Hom-Lie algebras
$(A,\\{\cdot,\cdot\\}_{A},\alpha)$ and $(B,\\{\cdot,\cdot\\}_{B},\beta)$.
Besides, by Corollary 4.20, the linear maps
$s_{A},l_{\ast_{A}},r_{\ast_{A}}:A\rightarrow gl(B)$ and
$s_{B},l_{\ast_{B}},r_{\ast_{B}}:B\rightarrow gl(A)$ are representations of
the underlying transposed Hom-Poisson algebras
$(B,\cdot_{B},\\{\cdot,\cdot\\}_{B},\beta)$ and
$(A,\cdot_{A},\\{\cdot,\cdot\\}_{A},\alpha)$ respectively. Therefore,
(2.47)-(2.48) are equivalent to (4.20)-(4.24), and (2.49)-(2.50) are
equivalent to (4.25)-(4.29). ∎

##  5 $\mathcal{O}$-operators of transposed Hom-Poisson algebras

In this section, we introduce and study the notion of an
$\mathcal{O}$-operator of transposed Hom-Poisson algebras generalizing the
notion of Rota-Baxter operators. Their properties and relationship with Hom-
pre-Lie Poisson algebras is also described.

######  Definition 5.1.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra and
$(s,\beta,V)$ be a bimodule. Then, a linear map $T:V\rightarrow A$ is called
an $\mathcal{O}$-operator associated to $(s,\beta,V)$, if $T$ satisfies for
all $u,v\in V,$

$\displaystyle\alpha T=T\beta,$ (5.1) $\displaystyle T(u)\cdot
T(v)=T(s(T(u))v+s(T(v))u).$ (5.2)

######  Lemma 5.2.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra, and
$(l,r,\beta,V)$ be a bimodule. Let $T:V\rightarrow A$ be an
$\mathcal{O}$-operator associated to $(s,\beta,V)$. Then, there exists a
commutative Hom-associative algebra structure on $V$ given for all $u,v\in V$
by

$\displaystyle u\diamond v=s(T(u))v+s(T(v))u.$

###### Proof.

First, we prove the commutativity condition. For all $u,v\in V$ ,

$u\diamond v=s(T(u))v+s(T(v))u=s(T(v))u+s(T(u))v=v\diamond u.$

Now, we prove the Hom-associativity condition. For all $u,v,w\in V$,

$\displaystyle(u\diamond
v)\diamond\beta(w)=(s(T(u))v+s(T(v))u)\diamond\beta(w)$
$\displaystyle\quad=s\Big{[}T(s(T(u))v+s(T(v))u)\Big{]}\beta(w)+s(T(\beta(w)))(s(T(u))v+s(T(v))u)$
$\displaystyle\quad=s(T(u)\cdot T(v))\beta(w)+s(\alpha T(w))s(T(u))v+s(\alpha
T(w))s(T(v))u$ $\displaystyle\quad=s(T(u)\cdot T(v))\beta(w)+s(T(w)\cdot
T(u))\beta(v)+s(T(w)\cdot T(v))\beta(u),~{}~{}(\textsl{by}~{}\eqref{Cond1})$
$\displaystyle\beta(u)\diamond(v\diamond
w)=\beta(u)\diamond(s(T(v))w+s(T(w))v)$
$\displaystyle\quad=s(\alpha(T(u))(s(T(v))w+s(T(w))v)+s\Big{[}T(s(T(v))w+s(T(w))v)\Big{]}\beta(u)$
$\displaystyle\quad=s(\alpha T(u))s(T(v))w+s(\alpha T(u))s(T(w))v+s(T(v)\cdot
T(w))\beta(u)$ $\displaystyle\quad=s(T(u)\cdot T(v))\beta(w)+s(T(u)\cdot
T(w))\beta(v)+s(T(v)\cdot T(w))\beta(u).\quad\mbox{(by \eqref{Cond1})}$

Then, $(u\diamond v)\diamond\beta(w)=\beta(u)\diamond(v\diamond w)$. ∎

######  Corollary 5.3.

Let $(A,\cdot,\alpha)$ be a commutative Hom-associative algebra, and
$(s,\beta,V)$ be a bimodule. Let $T:V\rightarrow A$ be an
$\mathcal{O}$-operator associated to $(s,\beta,V)$. Then, $T$ is a morphism
from the commutative Hom-associative algebra $(V,\diamond,\beta)$ to
$(A,\cdot,\alpha)$.

######  Definition 5.4 ([83]).

Let $\mathcal{A}=(A,[\cdot,\cdot],\alpha)$ be a Hom-Lie algebra, and
$(\rho,\beta,V)$ be a representation of $\mathcal{A}$. Then, a linear map
$T:V\rightarrow A$ is called an $\mathcal{O}$-operator associated to
$(\rho,\beta,V)$, if $T$ satisfies for all $u,v\in V$,

$\displaystyle\alpha T$ $\displaystyle=T\beta,$ (5.3)
$\displaystyle[T(u),T(v)]$ $\displaystyle=T(\rho(T(u))v-\rho(T(v))u).$ (5.4)

######  Example 5.5.

An $\mathcal{O}$-operator on a Hom-Lie algebra
$\mathcal{A}=(A,[\cdot,\cdot],\alpha)$ with respect to the adjoint
representation is called a Rota-Baxter operator on $\mathcal{A}$.

######  Lemma 5.6 ([83]).

Let $T:V\rightarrow A$ be an $\mathcal{O}$-operator on a Hom-Lie algebra
$(A,[\cdot,\cdot],\alpha)$ with respect to a representation $(\rho,\beta,V)$.
Define a multiplication $\ast$ on $V$ by

$u\ast v=\rho(T(u)v),\text{ for all }u,v\in V.$ (5.5)

Then, $(V,\ast,\alpha)$ is a Hom-pre-Lie algebra.

######  Corollary 5.7.

Let $T:V\rightarrow A$ be an $\mathcal{O}$-operator on a Hom-Lie algebra
$(A,[\cdot,\cdot],\alpha)$ with respect to a representation $(\rho,\beta,V)$.
Then, $T$ is a morphism from the Hom-Lie algebra $(V,[\cdot,\cdot]_{C},\beta)$
to $(A,[\cdot,\cdot],\alpha)$, where $[u,v]_{C}=u\ast v-v\ast u,~{}$ for all
$u,v\in V.$

######  Definition 5.8.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra,
and let $(s,\rho,\beta,V)$ be a representation of
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$. A linear operator $T:V\rightarrow A$ is
called an $\mathcal{O}$-operator on $A$ if $T$ is both an
$\mathcal{O}$-operator on the commutative Hom-associative algebra
$(A,\cdot,\alpha)$ and an $\mathcal{O}$-operator on the Hom-Lie algebra
$(A,\\{\cdot,\cdot\\},\alpha)$.

######  Example 5.9.

An $\mathcal{O}$-operator on a transposed Hom-Poisson algebra
$\mathcal{A}=(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ with respect the regular
bimodule and the adjoint representation is called a Rota-Baxter operator on
$\mathcal{A}$.

######  Theorem 5.10.

Assume that $\mathcal{A}=(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ is a transposed
Hom-Poisson algebra, and $T:V\rightarrow A$ is an $\mathcal{O}$-operator on
$\mathcal{A}$ with respect to the representation $(s,\rho,\beta,V)$. Define
new operations $\diamond$ and $\ast$ on $V$ by

$u\diamond v=s(T(u))v+s(T(v))u,~{}u\ast v=\rho(T(u))v.$ (5.6)

Then, $(V,\diamond,\ast,\alpha)$ is a Hom-pre-Lie Poisson algebra. Therefore
$V$ is a transposed Hom-Poisson algebra as the sub-adjacent transposed Hom-
Poisson algebra of this Hom-pre-Lie Poisson algebra and $T$ is a homomorphism
of transposed Hom-Poisson algebras. Furthermore, $T(V)=\\{T(v)\colon v\in
V\\}\subset A$ is a subalgebra of $\mathcal{A}$ and there is an induced Hom-
pre-Lie Poisson algebra structure on $T(V)$ given for all $u,v\in V$ by

$T(u)\diamond^{\prime}T(v)=T(u\diamond v),~{}~{}T(u)\ast^{\prime}T(v)=T(u\ast
v).$ (5.7)

Moreover, $T$ is a homomorphism of Hom-pre-Lie Poisson algebras.

###### Proof.

By Lemma 5.2 and Lemma 5.6, we deduce that $(A,\cdot,\alpha)$ is a commutative
Hom-associative algebra and $(A,\ast,\alpha)$ is a Hom-pre-Lie algebra. Next,
we prove the axiom (4.10). For any $u,v,w\in V$,

$\displaystyle(u\diamond v)\ast\beta(w)=$
$\displaystyle(s(T(u))v+s(T(v))u)\ast\beta(w)$ (5.8) $\displaystyle=$
$\displaystyle\rho\Big{(}T(s(T(u))v+s(T(v))u)\Big{)}\beta(w)$ (5.9)
$\displaystyle=$ $\displaystyle\rho(T(u)\cdot T(v))\beta(w),$ (5.10)
$\displaystyle\beta(u)\diamond(v\ast w)=$
$\displaystyle\beta(u)\diamond(\rho(T(v))w)$ (5.11) $\displaystyle=$
$\displaystyle s(T(\beta(u)))\rho(T(v))w+s(T(\rho(T(v))w))\beta(u),$

Hence, we get (4.10) by (5.1), (2.39) and (2.40). Now, we prove the axiom
(4.11),

$\displaystyle(u\ast v)\diamond\beta(w)-(v\ast u)\diamond\beta(w)$
$\displaystyle=(\rho(T(u))v)\diamond\beta(w)-(\rho(T(v))u)\diamond\beta(w)$
$\displaystyle=s(T(\rho(T(u))v)\beta(w)+s(T(\beta(w)))\rho(T(u))v$
$\displaystyle-s(T(\rho(T(v))u)\beta(w)+s(T(\beta(w)))\rho(T(v))u$
$\displaystyle=s(T(\rho(T(u))v-T(\rho(T(v))u)\beta(w)$
$\displaystyle+s(T(\beta(w))\rho(T(u))v-s(T(\beta(w))\rho(T(v))u$
$\displaystyle=s(\\{T(u),T(v)\\})\beta(w)+s(T(\beta(w))\rho(T(u))v$
$\displaystyle-s(T(\beta(w))\rho(T(v))u,$ $\displaystyle\beta(u)\ast(v\diamond
w)-\beta(v)\ast(u\diamond w)$ $\displaystyle=\beta(u)\ast(s(T(v))w+s(T(w))v)$
$\displaystyle-\beta(v)\ast(s(T(u))w+s(T(w))u)$
$\displaystyle=\rho(T(\beta(u)))(s(T(v))w+s(T(w))v)$
$\displaystyle-\rho(T(\beta(v)))(s(T(u))w+s(T(w))u)$
$\displaystyle=\rho(T(\beta(u)))s(T(v))w+\rho(T(\beta(u)))s(T(w))v$
$\displaystyle-\rho(T(\beta(v)))s(T(u))w+\rho(T(\beta(v)))s(T(w))u.$

Hence, we get (4.11) by (5.1), (2.39) and (2.40).

On the other hand,

$\displaystyle T(\\{u,v\\}_{C})=$ $\displaystyle T(u\ast v-v\ast
u)=T(\rho(T(u))v-\rho(T(v))u)=\\{T(u),T(v)\\},$ $\displaystyle T(u\diamond
v)=$ $\displaystyle T(s(T(u))v+s(T(v))u)=T(u)\cdot T(v),$

which implies that $T$ is a homomorphism of transposed Hom-Poisson algebras.
The other conclusions follow immediately. ∎

######  Corollary 5.11.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra.
There is a Hom-pre-Lie Poisson algebra structure on
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ such that its sub-adjacent transposed
Hom-Poisson algebra is exactly $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ if and
only if there exists an invertible $\mathcal{O}$-operator on
$(A,\cdot,\\{\cdot,\cdot\\},\alpha).$

###### Proof.

Suppose that there exists an invertible $\mathcal{O}$-operator $T:V\rightarrow
A$ associated to the representation $(l,r,\rho,V)$, then the compatible Hom-
pre-Lie Poisson algebra structure on $A$, for all $x,y\in A$ is given by

$x\cdot y=T(s(x)T^{-1}(y)+s(y)T^{-1}(x)),\;\;x\ast y=T(\rho(x)T^{-1}(y)).$

Conversely, let $(A,\cdot,\ast,\alpha)$ be a Hom-pre-Lie Poisson algebra, and
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be the sub-adjacent transposed Hom-
Poisson algebra. Then, the identity map $id$ is an $\mathcal{O}$-operator on
$(A,\cdot,\ast,\alpha)$ with respect to the regular representation
$(S,ad,\alpha,A)$. ∎

######  Example 5.12.

Let $(A,\cdot,\\{\cdot,\cdot\\},\alpha)$ be a transposed Hom-Poisson algebra
and $R:A\longrightarrow A$ a Rota-Baxter operator. Define new operations on
$A$ by

$x\diamond y=R(x)\cdot y+x\cdot R(y),\quad x\ast y=\\{R(x),y\\}.$

Then, $(A,\diamond,\ast,\alpha)$ is a Hom-pre-Lie Poisson algebra and $R$ is a
homomorphism from the sub-adjacent transposed Hom-Poisson algebra
$(A,\diamond,\\{\cdot,\cdot\\}_{C},\alpha)$ to
$(A,\cdot,\\{\cdot,\cdot\\},\alpha)$, where $\\{x,y\\}_{C}=x\ast y-y\ast x$.

##  6 Acknowledgement

Support from The Royal Swedish Academy of Sciences Foundations is gratefully
acknowledged.

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