Hom-Novikov Algebras and Hom-Novikov-Poisson Algebras  111Corresponding
author<EMAIL_ADDRESS>
Lamei Yuan${}^{\,{\ddagger}}$, Hong You${}^{\,{\ddagger}\,{\dagger}}$

${\ddagger}$Academy of Fundamental and Interdisciplinary Sciences,

Harbin Institute of Technology, Harbin 150080, China

†Department of Mathematics, Suzhou University, Suzhou 200092, China

E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract. The purpose of this paper is to study Hom-Novikov algebras and Hom-
Novikov-Poisson algebras, both of which were defined by Yau. In the paper, we
give several constructions leading us to some interesting examples of Hom-
Novikov algebras and Hom-Novikov-Poisson algebras. Also, we introduce the
notion of quadratic Hom-Novikov algebras and provide some properties.

Key words: Novikov algebras, Hom-Novikov algebras, quadratic Hom-Novikov
algebras, Novikov-Poisson algebras, Hom-Novikov-Poisson algebras.
2000 Mathematics Subject Classification: 17A30, 17B45, 17B81

§1. Introduction

Novikov algebras were introduced in connection with the Poisson brackets of
hydrodynamic type [3, 8, 9] and Hamiltonian operators in the formal
variational calculus [12, 21]. The theoretical study of Novikov algebras was
started by Zel’manov [31] and Filipov [11]. But the term “Novikov algebra” was
first used by Osborn [19]. The left multiplication operators of a Novikov
algebra form a Lie algebra. Thus, it is effective to relate the study of
Novikov algebras to the theory of Lie algebras [4]. Novikov algebras are a
special class of left-symmetric algebras (or under other names such as pre-Lie
algebras, quasi-associative algebras and Vinberg algebras), arising from the
study of affine manifolds, affine structures and convex homogeneous cones [2,
15, 20]. Left-symmetric algebras have close relations with many important
fields in mathematics and mathematical physics, such as infinite-dimensional
Lie algebras [3, 9], classical and quantum Yang-Baxter equation [10, 13],
quantum field theory [7] and so on.

Novikov-Poisson algebras were originally introduced by Xu [21, 22] motivated
by studying simple Novikov algebras and irreducible modules. Every Novikov-
Poisson algebra with associative commutative unity can be constructed from an
associative commutative derivation algebra. That there exists a relationship
between associative commutative derivation algebras and Novikov algebras was
also pointed out in [12]. Infinite-dimensional Novikov algebras over an
algebraically closed field of characteristic $0$ were classified in [23] under
some restrictions on the structure of right and left multiplication operators.
It turned out that such Novikov algebras are Novikov-Poisson algebras and are
obtained from associative commutative differentially simple algebras with
unity. While it was proved in [32] that a Novikov algebra is simple over a
field of characteristic not $2$ if and only if its associative commutative
derivation algebra is differentially simple.

A quadratic Novikov algebra, introduced in [6, 33], is a Novikov algebra with
a symmetric nondegenerate invariant bilinear form. The motivation for studying
quadratic Novikov algebras came from the fact that Lie algebras or associative
algebras with symmetric nondegenerate invariant bilinear forms have important
applications in several areas of mathematics and physics, such as the
structure theory of finite-dimensional semisimple Lie algebras, the theory of
complete integrable Hamiltonian systems and the classification of statistical
models over two-dimensional graphs.

Hom-Lie algebras were initially introduced by Hartwig, Larsson and Silvestrov
in [14] motivated by examples of deformed Lie algebras coming from twisted
discretizations of vector fields. This kind of algebras includes Lie algebras
as a subclass in which the twisting map is the identity map. As
generalizations of Hom-Lie algebras, quasi-hom-Lie algebras [16], Hom-Lie
superalgebras [1] and Hom-Lie color algebras [30] were introduced,
respectively. Both the Hom-Lie superalgebras and Hom-Lie color algebras are
special cases of the quasi-hom-Lie algebras. There has been a lot of progress
in the study of Hom-Lie algebras (see e.g., [24, 25, 26, 27]).

Hom-associative algebras generalizing associative algebras to a situation
where associativity law is twisted by a linear map were introduced in [18]. It
is well known that there is always a Lie algebra associated to an associative
algebra via the commutator bracket. An analogous result holds for Hom-
associative algebras, i.e., the commutator bracket multiplication defined by
the multiplication in a Hom-associative algebra leads naturally to Hom-Lie
algebras. This result was also extended to the Hom-Lie superalgebra [1] and
the Hom-Lie color algebra [30] cases.

Following the patterns of Hom-Lie algebras and Hom-associative algebras, Yau
in [28] introduced Hom-Novikov algebras, in which the two defining identities
are twisted by a linear map. It turned out that Hom-Novikov algebras can be
constructed from Novikov algebras, commutative Hom-associative algebras and
Hom-Lie algebras along with some suitable linear maps. Later, Yau in [29]
defined a Hom-Novikov-Poisson algebra as a twisted generalization of Novikov-
Poisson algebras.

In the present paper, we consider both Hom-Novikov algebras and Hom-Novikov-
Poisson algebras. In Section 2, we summarize some definitions including
Novikov algebras, Novikov-Poisson algebras, Hom-associative algebras, Hom-Lie
algebras, Hom-Novikov algebras and Hom-Novikov-Poisson algebras and related
results.

In Section 3, we focus on Hom-Novikov algebras. We establish the relationships
between Hom-Novikov algebras and Hom-Lie algebras as well as the relations
between Hom-Novikov algebras and Novikov algebras. Also, we provide a
construction of Hom-Novikov algebras from a commutative Hom-associative
algebra together with a suitable linear map but not a derivation.

In Section 4, we extend the notion of quadratic Novikov algebras to quadratic
Hom-Novikov algebras, which are Hom-Novikov algebras with a symmetric
nondegenerate bilinear form satisfying equation (4.3). We reduce the case
where the twisting map is an automorphism or an involution to the study of the
relationships between quadratic Hom-Novikov algebras and quadratic Novikov
algebras as well as the relations between quadratic Hom-Novikov algebras and
quadratic Hom-Lie algebras.

In Section 5, we study Hom-Novikov-Poisson algebras. We prove that the tensor
of two Hom-Novikov-Poisson algebras is still a Hom-Novikov-Poisson algebra. In
addition, we give a way to construct Hom-Novikov-Poisson algebras by using a
commutative Hom-associative algebra along with a derivation, generalizing a
construction in Novikov algebra case due to Dorfman and Gel’fand, in Novikov-
Poisson algebra case due to Xu and in Hom-Novikov algebra case due to Yau.
Finally, we provide some interesting examples.

Throughout this paper, $\mathbb{F}$ denotes a field of characteristic zero.
All vector spaces and tensor products are assumed to be over $\mathbb{F}$,
except where otherwise indicated.

§2. Preliminaries

In this section, we summarize some definitions concerning Novikov algebras and
Hom-algebras, and related results. For detailed discussions and examples we
refer the reader to the literatures, for instance, [14, 18, 21, 22, 28, 29]
and references therein.

###### Definition 2.1

A Novikov algebra $(\mathcal{A},\mu)$ is a vector space with a bilinear
product $\mu:\mathcal{A}\times\mathcal{A}\longrightarrow\mathcal{A}$
satisfying

$\displaystyle(xy)z$ $\displaystyle=$ $\displaystyle(xz)y,$ (2.1)
$\displaystyle(xy)z-x(yz)$ $\displaystyle=$ $\displaystyle(yx)z-y(xz),$ (2.2)

for all $x,y,z\in\mathcal{A}$, and where $\mu(x,y)=xy$.

In the sequel, for simplicity, we often write $xy$ instead of $\mu(x,y)$ since
confusion rarely occurs. Note that Novikov algebras are a special class of
left-symmetric algebras with only equation (2.2) satisfied. Define an
associator on $\mathcal{A}$ by

$\displaystyle a(x,y,z)=(xy)z-x(yz),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{A}.$

Then equation (2.2) is equivalent to

$\displaystyle a(x,y,z)=a(y,x,z),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{A}.$

In other words, the associator is symmetric in $x,y$, from which the left-
symmetric algebra takes its name.

The commutator of a Novikov algebra (or a left-symmetric algebra)
$\mathcal{A}$

$[x,y]=xy-yx,\hskip 8.5359pt\mbox{for all}\ x,y\in\mathcal{A},$

defines a Lie algebra $Lie(\mathcal{A})$, which is called the sub-adjacent Lie
algebra of $\mathcal{A}$. Hence both Novikov algebras and left-symmetric
algebras are Lie-admissible algebras.

In order to establish tensor theory of Novikov algebras, Xu in [22] introduced
Novikov-Poisson algebras, which play important roles in the study of simple
Novikov algebras.

###### Definition 2.2

A Novikov-Poisson algebra is a vector space $\mathcal{A}$ with two operations
$\cdot$ and $\ast$ such that $(\mathcal{A},\cdot)$ forms a commutative
associative algebra and $(\mathcal{A},\ast)$ forms a Novikov algebra for which

$\displaystyle(x\cdot y)\ast z$ $\displaystyle=$ $\displaystyle x\cdot(y\ast
z),$ (2.3) $\displaystyle(x\ast y)\cdot z-x\ast(y\cdot z)$ $\displaystyle=$
$\displaystyle(y\ast x)\cdot z-y\ast(x\cdot z),$ (2.4)

for all $x,y,z\in\mathcal{A}$.

For an endomorphism of a Novikov-Poisson algebra $(\mathcal{A},\cdot,\ast)$,
we mean a linear map $\alpha:\mathcal{A}\longrightarrow\mathcal{A}$ preserving
the two operations, i.e.,

$\displaystyle\alpha(x\cdot y)=\alpha(x)\cdot\alpha(y),\hskip
8.5359pt\alpha(x\ast y)=\alpha(x)\ast\alpha(y),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$

The notion of Hom-Lie algebras was initially introduced by Hartwig, Larsson
and Silvestrov in[14] motivated by examples of deformed Lie algebras coming
from twisted discretizations of vector fields. A slightly more general
definition of Hom-Lie algebras was given by Makhlouf and Silvestrov in [18],
where the Hom-associative algebras were also introduced.

###### Definition 2.3

A Hom-Lie algebra is a triple $(L,[\cdot,\cdot],\alpha)$ consisting of a
vector space $L$, a bilinear map $[\cdot,\cdot]:L\times L\longrightarrow L$
and a linear map $\alpha:L\longrightarrow L$ satisfying the following two
conditions

$\displaystyle[x,y]=-[y,x],\hskip 14.22636pt\mbox{(skew-symmetry)}$
$\displaystyle[[x,y],\alpha(z)]+[[z,x],\alpha(y)]+[[y,z],\alpha(x)]=0,\hskip
14.22636pt\mbox{(Hom-Jacobi \ identity)}$

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

Clearly, Lie algebras are spacial classes of Hom-Lie algebras when $\alpha$ is
the identity map. If the linear map $\alpha$ is an algebra homomorphism with
respect to the bracket $[\cdot,\cdot]$, then the Hom-Lie algebras are a
spacial case of quasi-hom-Lie algebras [16]. Moreover, all these classes are
special cases of the more general quasi-Lie algebras [17], in which Lie color
algebras and in particular Lie superalgebras are also included.

###### Definition 2.4

A Hom-associative algebra is a triple $(V,\mu,\alpha)$ consisting of a linear
space $V$, a bilinear map $\mu:V\times V\longrightarrow V$ and a linear map
$\alpha:V\longrightarrow V$ satisfying

$\alpha(x)(yz)=(xy)\alpha(z),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{A}.$

We say that a Hom-associative algebra $(\mathcal{A},\mu,\alpha)$ is
commutative if $xy=yx$ holds for all $x,y\in\mathcal{A}$. The following result
provides a construction of Hom-Lie algebras from Hom-associative algebras,
extending the fundamental construction of Lie algebras from associative
algebras via commutator bracket multiplication.

###### Proposition 2.5

(see [References, Proposition 1.6]) To any Hom-associative algebra defined by
a multiplication $\mu$ and a linear map $\alpha$ over an $\mathbb{F}$-linear
space $V$, one may associate a Hom-Lie algebra defined by the bracket

$\displaystyle[x,y]=xy-yx,\hskip 8.5359pt\mbox{for all}\ x,y\in V.$

Following the patterns of Hom-Lie algebras and Hom-associative algebras, Yau
in[28] introduced a twisted generalization of Novikov algebras, called Hom-
Novikov algebras.

###### Definition 2.6

A Hom-Novikov algebra is a triple $(\mathcal{A},\mu,\alpha)$ consisting of a
vector space $\mathcal{A}$, a bilinear map
$\mu:\mathcal{A}\times\mathcal{A}\longrightarrow\mathcal{A}$ and an algebra
homomorphism $\alpha:\mathcal{A}\longrightarrow\mathcal{A}$ satisfying

$\displaystyle(xy)\alpha(z)$ $\displaystyle=$ $\displaystyle(xz)\alpha(y),$
(2.5) $\displaystyle(xy)\alpha(z)-\alpha(x)(yz)$ $\displaystyle=$
$\displaystyle(yx)\alpha(z)-\alpha(y)(xz),$ (2.6)

for all $x,y,z\in\mathcal{A}.$

Clearly, Novikov algebras are examples of Hom-Novikov algebras by setting
$\alpha=\rm{id}$, where $\rm{id}$ is the identity map. It was shown in [28]
that any Novikov algebra can be deformed into a Hom-Novikov algebra along with
an algebra endomorphism.

###### Proposition 2.7

(see [References, Theorem 1.1] ) Let $(\mathcal{A},\mu)$ be a Novikov algebra
and $\alpha:\mathcal{A}\longrightarrow\mathcal{A}$ be an algebra homomorphism.
Then $(\mathcal{A},\alpha\circ\mu,\alpha)$ is a Hom-Novikov algebra, where
$\circ$ denotes composition of functions.

Similar to Hom-Novikov algebras, Yau in [29] introduced Hom-Novikov-Poisson
algebras.

###### Definition 2.8

A Hom-Novikov-Poisson algebra is a quadruple
$(\mathcal{A},\cdot,\ast,\alpha)$, where $\mathcal{A}$ is a vector space,
$\cdot$ and $\ast$ are bilinear maps and $\alpha$ is an algebra homomorphism
of $\mathcal{A}$, such that $(\mathcal{A},\ast,\alpha)$ forms a Hom-Novikov
algebra, $(\mathcal{A},\cdot,\alpha)$ forms a commutative Hom-associative
algebra and the following two compatible conditions hold:

$\displaystyle(x\cdot y)\ast\alpha(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot(y\ast z),$ (2.7) $\displaystyle(x\ast
y)\cdot\alpha(z)-\alpha(x)\ast(y\cdot z)$ $\displaystyle=$
$\displaystyle(y\ast x)\cdot\alpha(z)-\alpha(y)\ast(x\cdot z),$ (2.8)

for all $x,y,z\in\mathcal{A}$.

We can recover Novikov-Poisson algebras from Hom-Novikov-Poisson algebras when
$\alpha={\rm id}$. Some more interesting examples of Hom-Novikov-Poisson
algebras will be constructed in Section 5.

§3. Hom-Novikov algebras

In this section, we establish the relationships between Hom-Novikov algebras
and Novikov algebras as well as the relations between Hom-Novikov algebras and
Hom-Lie algebras. Also, we provide a construction of Hom-Novikov algebras from
a commutative Hom-associative algebra along with a suitable linear map but not
a derivation. Thus, it is different from that given in [28].

Just as a Novikov algebra can be deformed into a Lie algebra via the
commutator bracket, we have the similar result for Hom-Novikov algebras and
Hom-Lie algebras, because in Hom-Novikov algebras hold Jacobi-like identities.

###### Proposition 3.1

Let $(\mathcal{A},\mu,\alpha)$ be a Hom-Novikov algebra. For all
$x,y,z\in\mathcal{A}$, one has

$\displaystyle[x,y]\alpha(z)+[y,z]\alpha(x)+[z,x]\alpha(y)$ $\displaystyle=$
$\displaystyle 0,$ (3.1)
$\displaystyle\alpha(x)[y,z]+\alpha(y)[z,x]+\alpha(z)[x,y]$ $\displaystyle=$
$\displaystyle 0,$ (3.2)

where $\mu(x,y)=xy$ and $[x,y]=xy-yx$.

Proof. For all $x,y,z\in\mathcal{A}$, we have

$\displaystyle[x,y]\alpha(z)$ $\displaystyle=$
$\displaystyle(xy)\alpha(z)-(yx)\alpha(z),$ $\displaystyle{[y,z]}\alpha(x)$
$\displaystyle=$ $\displaystyle(yz)\alpha(x)-(zy)\alpha(x),$
$\displaystyle{[z,x]}\alpha(y)$ $\displaystyle=$
$\displaystyle(zx)\alpha(y)-(xz)\alpha(y).$

Since equation (2.5) holds for all $x,y,z\in\mathcal{A}$, we have

$\displaystyle[x,y]\alpha(z)+[y,z]\alpha(x)+[z,x]\alpha(y)$ $\displaystyle=$
$\displaystyle\big{(}(xy)\alpha(z)-(xz)\alpha(y)\big{)}+\big{(}(yz)\alpha(x)-(yx)\alpha(z)\big{)}+\big{(}(zx)\alpha(y)-(zy)\alpha(x)\big{)}$
$\displaystyle=$ $\displaystyle 0,$

which proves equation (3.1). Now using equations (2.6) and (3.1), we have

$\displaystyle\alpha(x)[y,z]+\alpha(y)[z,x]+\alpha(z)[x,y]$ $\displaystyle=$
$\displaystyle\alpha(x)(yz)-\alpha(x)(zy)+\alpha(y)(zx)-\alpha(y)(xz)+\alpha(z)(xy)-\alpha(z)(yx)$
$\displaystyle=$
$\displaystyle\big{(}\alpha(x)(yz)-\alpha(y)(xz)\big{)}+\big{(}\alpha(y)(zx)-\alpha(z)(yx)\big{)}+\big{(}\alpha(z)(xy)-\alpha(x)(zy)\big{)}$
$\displaystyle=$
$\displaystyle\big{(}(xy)\alpha(z)-(yx)\alpha(z)\big{)}+\big{(}(yz)\alpha(x)-(zy)\alpha(x)\big{)}+\big{(}(zx)\alpha(y)-(xz)\alpha(y)\big{)}$
$\displaystyle=$ $\displaystyle[x,y]\alpha(z)+[y,z]\alpha(x)+[z,x]\alpha(y)$
$\displaystyle=$ $\displaystyle 0,$

which proves equation (3.2) and the proposition. $\Box$

###### Corollary 3.2

Let $(\mathcal{A},\mu,\alpha)$ be a Hom-Novikov algebra. Define a bilinear map
$[\cdot,\cdot]:\mathcal{A}\times\mathcal{A}\longrightarrow\mathcal{A}$ by

$[x,y]=\mu(x,y)-\mu(y,x),\hskip 8.5359pt\mbox{for all}\ x,y\in\mathcal{A}.$

Then $HLie(\mathcal{A})=(\mathcal{A},[\cdot,\cdot],\alpha)$ is a Hom-Lie
algebra with the same twisting map $\alpha$.

The Lie admissible algebras were introduced by A.A. Albert in 1948. Makhlouf
and Silvestrov in [18] extended the notions and results about Lie admissible
algebras to Hom-algebras, called Hom-Lie admissible algebras. According to
Corollary 3.2, any Hom-Novikov algeba is Hom-Lie admissible with the same
twisting map. We call such a Hom-Lie algebra $HLie(\mathcal{A})$ the sub-
adjacent Hom-Lie algebra of the Hom-Novikov algebra
$(\mathcal{A},\mu,\alpha)$.

###### Definition 3.3

Let $(\mathcal{A},\mu,\alpha)$ be a Hom-Novikov algebra, which is called

  * (i)

regular if $\alpha$ is an algebra automorphism;

  * (ii)

involutive if $\alpha$ is an involution, i.e., $\alpha^{2}=\rm{id}.$

Yau in [28] gave a way to construct Hom-Novikov algebras, starting from a
Novikov algebra and an algebra endomorphism. In the following, we provide a
construction of Novikov algebras from Hom-Novikov algebras along with an
algebra automorphism.

###### Proposition 3.4

If $(\mathcal{A},\mu,\alpha)$ is an involutive Hom-Novikov algebra, then
$(\mathcal{A},\alpha\circ\mu)$ is a Novikov algebra.

Proof.  For convenience, we write $\mu(x,y)=xy$ and $x\star y=\alpha(xy),$ for
all $x,y\in\mathcal{A}$. Hence, it needs to show

$\displaystyle(x\star y)\star z$ $\displaystyle=$ $\displaystyle(x\star
z)\star y,$ (3.3) $\displaystyle(x\star y)\star z-x\star(y\star z)$
$\displaystyle=$ $\displaystyle(y\star x)\star z-y\star(x\star z),$ (3.4)

for all $x,y,z\in\mathcal{A}$. Since $\alpha$ is an involution, we have

$\displaystyle(x\star y)\star z$ $\displaystyle=$
$\displaystyle\alpha\big{(}\alpha(xy)z\big{)}=\big{(}\alpha^{2}(xy)\big{)}\alpha(z)=(xy)\alpha(z).$

Similarly, we have $(x\star z)\star y=(xz)\alpha(y)$, from which equation
(3.3) follows since $(\mathcal{A},\mu,\alpha)$ is a Hom-Novikov algebra.
Furthermore, using equation (2.6), we have

$\displaystyle(x\star y)\star z-x\star(y\star z)$ $\displaystyle=$
$\displaystyle\alpha\big{(}\alpha(xy)z\big{)}-\alpha\big{(}x\alpha(yz)\big{)}$
$\displaystyle=$ $\displaystyle(xy)\alpha(z)-\alpha(x)(yz)$ $\displaystyle=$
$\displaystyle(yx)\alpha(z)-\alpha(y)(xz)$ $\displaystyle=$
$\displaystyle\alpha\big{(}\alpha(yx)z\big{)}-\alpha\big{(}y\alpha(xz)\big{)}$
$\displaystyle=$ $\displaystyle(y\star x)\star z-y\star(x\star z),$

which proves equation (3.4) and the proposition. $\Box$

###### Proposition 3.5

If $(\mathcal{A},\mu,\alpha)$ is a regular Hom-Novikov algebra, then
$(\mathcal{A},[\cdot,\cdot]_{\alpha^{-1}}=\alpha^{-1}\circ[\cdot,\cdot])$ is a
Lie algebra, where $[x,y]=\mu(x,y)-\mu(y,x)=xy-yx,$ for all
$x,y\in\mathcal{A}.$ In particular, if $\alpha$ is an involution, then
$(\mathcal{A},[\cdot,\cdot]_{\alpha}=\alpha\circ[\cdot,\cdot])$ is a Lie
algebra.

Proof.  For any $x,y,z\in\mathcal{A}$, we have

$\displaystyle[[x,y]_{\alpha^{-1}},z]_{\alpha^{-1}}$ $\displaystyle=$
$\displaystyle[\alpha^{-1}(xy-yx),z]_{\alpha^{-1}}$ $\displaystyle=$
$\displaystyle\alpha^{-1}\big{(}\alpha^{-1}(xy-yx)z-z\alpha^{-1}(xy-
yx)\big{)}$ $\displaystyle=$ $\displaystyle\alpha^{-2}\big{(}(xy-
yx)\alpha(z)-\alpha(z)(xy-yx)\big{)}$ $\displaystyle=$
$\displaystyle\alpha^{-2}\big{(}[x,y]\alpha(z)-\alpha(z)[x,y]\big{)}.$

Similarly, we have

$\displaystyle[[y,z]_{\alpha^{-1}},x]_{\alpha^{-1}}$ $\displaystyle=$
$\displaystyle\alpha^{-2}\big{(}[y,z]\alpha(x)-\alpha(x)[y,z]\big{)},$
$\displaystyle{[[z,x]_{\alpha^{-1}},y]_{\alpha^{-1}}}$ $\displaystyle=$
$\displaystyle\alpha^{-2}\big{(}[z,x]\alpha(y)-\alpha(y)[z,x]\big{)}.$

Then it follows from equations (3.1) and (3.2) that

$[[x,y]_{\alpha^{-1}},z]_{\alpha^{-1}}+[[y,z]_{\alpha^{-1}},x]_{\alpha^{-1}}+{[[z,x]_{\alpha^{-1}},y]_{\alpha^{-1}}}=0.$

Clearly, $[x,y]=-[y,x].$ Since $\alpha^{-1}$ is an automorphism, we have

$[x,y]_{\alpha^{-1}}=-[y,x]_{\alpha^{-1}},$

which proves that $(\mathcal{A},[\cdot,\cdot]_{\alpha^{-1}})$ is a Lie
algebra. It follows immediately that $(\mathcal{A},[\cdot,\cdot]_{\alpha})$ is
also a Lie algebra since $\alpha=\alpha^{-1}$ when $\alpha$ is an involution.
$\Box$

Let $(\mathcal{A},\mu)$ be a commutative associative algebra and $D$ be a
derivation of $(\mathcal{A},\mu)$. Then the new product

$\displaystyle x\star_{\lambda}y=xD(y)+\lambda xy,\hskip 8.5359pt\mbox{for
all}\ x,y\in\mathcal{A},$ (3.5)

makes $(\mathcal{A},\star_{\lambda})$ become a Novikov algebra for $\lambda=0$
by Gel’fand and Dorfman [12], for $\lambda\in\mathbb{F}$ by Filipov [11], and
for a fixed element $\lambda\in\mathcal{A}$ by Xu [21]. Yau [28] generalized
this construction to Hom-Novikov algebras $(\mathcal{A},\mu,\alpha)$ in
$\lambda=0$ setting together with an additional condition that $D$ commutes
with $\alpha$ . It is natural to consider other type of linear maps to replace
the derivation $D$ in (3.5), such that $(\mathcal{A},\star_{\lambda})$ forms a
Novikov algebra or a Hom-Novikov algebra.

Let $(\mathcal{A},\mu,\alpha)$ be a commutative Hom-associative algebra with a
linear selfmap $\partial$ commuting with $\alpha$. Consider the following
operation on $\mathcal{A}$:

$\displaystyle x\star y=x\partial(y),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$ (3.6)

For all $x,y,z\in\mathcal{A}$, by straightforward calculation, we have

$\displaystyle(x\star y)\star\alpha(z)$ $\displaystyle=$
$\displaystyle(x\partial(y))\star\alpha(z)$ $\displaystyle=$
$\displaystyle(x\partial(y))\partial(\alpha(z))$ $\displaystyle=$
$\displaystyle(x\partial(y))\alpha(\partial(z))$ $\displaystyle=$
$\displaystyle\alpha(x)(\partial(y)\partial(z)),$

and

$\displaystyle(x\star z)\star\alpha(y)$ $\displaystyle=$
$\displaystyle(x\partial(z))\star\alpha(y)$ $\displaystyle=$
$\displaystyle(x\partial(z))\partial(\alpha(y))$ $\displaystyle=$
$\displaystyle(x\partial(z))\alpha(\partial(y))$ $\displaystyle=$
$\displaystyle\alpha(x)(\partial(z)\partial(y))$ $\displaystyle=$
$\displaystyle\alpha(x)(\partial(y)\partial(z)),$

where both commutativity and Hom-associativity of $(\mathcal{A},\mu,\alpha)$
are used. Hence, we have

$\displaystyle(x\star y)\star\alpha(z)=(x\star z)\star\alpha(y).$

Similarly, we have

$\displaystyle(x\star y)\star\alpha(z)-\alpha(x)\star(y\star
z)=\alpha(x)(\partial(y)\partial(z))-\alpha(x)\partial(y\partial(z)),$ (3.7)

and

$\displaystyle(y\star x)\star\alpha(z)-\alpha(y)\star(x\star
z)=\alpha(y)(\partial(x)\partial(z))-\alpha(y)\partial(x\partial(z)).$ (3.8)

If, in addition, $\partial$ satisfies the following condition

$\displaystyle\partial(x\partial(y))=\partial(x)\partial(y),\hskip
8.5359pt\mbox{for all}\ x,y\in\mathcal{A}.$ (3.9)

Then it follows from equations (3.7) and (3.8) that

$\displaystyle(x\star y)\star\alpha(z)-\alpha(x)\star(y\star z)=(y\star
x)\star\alpha(z)-\alpha(y)\star(x\star z)=0.$

Now from the discussions above, we obtain

###### Theorem 3.6

Let $(\mathcal{A},\mu,\alpha)$ be a commutative Hom-associative algebra with a
linear selfmap $\partial$ satisfying equation (3.9) and commuting with
$\alpha$. Then $(\mathcal{A},\star,\alpha)$ is a Hom-Novikov algebra, where
$\star$ is defined by (3.6).

We have the following corollary by setting $\alpha={\rm id}$ in the theorem.

###### Corollary 3.7

Let $(\mathcal{A},\mu)$ be a commutative associative algebra with a linear
selfmap $\partial$ satisfying condition (3.9). Then $(\mathcal{A},\star)$ is a
Novikov algebra in which $\star$ is defined by (3.6).

###### Example 3.8

Let $(\mathcal{A}=\mathcal{A}_{0}\oplus\mathcal{A}_{1},\mu)$ be a complex
superalgebra, where $\mathcal{A}_{0}=\mathbb{C}[t,t^{-1}]$ is the Laurent
polynomials in one variable and $\mathcal{A}_{1}=\theta\mathbb{C}[t,t^{-1}]$
in which $\theta$ is the Grassman variable satisfying $\theta^{2}=0$. We
assume that $t$ commutes with $\theta$. The generators of $\mathcal{A}$ are of
the form $t^{n}$ and $\theta t^{n}$ for all $n\in\mathbb{Z}$. Then
$\mathcal{A}$ is a commutative associative superalgebra under the usual
multiplication.

Define a linear map $\partial:\mathcal{A}\longrightarrow\mathcal{A}$ by

$\displaystyle\partial(t^{n}+\theta t^{m})=t^{n},\hskip 8.5359pt\mbox{for
all}\ m,n\in\mathbb{Z}.$ (3.10)

It is easy to check that $\partial$ satisfies condition (3.9). According to
Corollary 3.7, $(\mathcal{A},\star_{1})$ is a Novikov algebra, where the
operation $\star_{1}$ is defined by

$\displaystyle(t^{n_{1}}+\theta t^{m_{1}})\star_{1}(t^{n_{2}}+\theta
t^{m_{2}})=(t^{n_{1}}+\theta t^{m_{1}})\partial(t^{n_{2}}+\theta
t^{m_{2}})=t^{n_{1}+n_{2}}+\theta t^{m_{1}+n_{2}},$

for all $n_{1},n_{2},m_{1},m_{2}\in\mathbb{Z}$.

Furthermore, define an algebra endomorphism $\alpha$ on $\mathcal{A}$ by

$\alpha(t^{n}+\theta t^{m})=(t+c)^{n},\hskip 8.5359pt\mbox{for all}\
m,n\in\mathbb{Z},$

where $c$ is a fixed element in $\mathbb{C}$. Then
$(\mathcal{A},\alpha\circ\mu,\alpha)$ is a commutative Hom-associative
algebra. For convenience, we write $\cdot$ instead of $\alpha\circ\mu$, then
we have

$\displaystyle(t^{n_{1}}+\theta t^{m_{1}})\cdot(t^{n_{2}}+\theta
t^{m_{2}})=(t+c)^{n_{1}+n_{2}},$

for all $n_{1},n_{2},m_{1},m_{2}\in\mathbb{Z}$. Let $\partial$ be as defined
by (3.10). Clearly, $\partial$ commutes with $\alpha$. Moreover, we have

$\displaystyle\partial\big{(}(t^{n_{1}}+\theta
t^{m_{1}})\cdot\partial(t^{n_{2}}+\theta
t^{m_{2}})\big{)}=(t+c)^{n_{1}+n_{2}},$

and

$\displaystyle\partial(t^{n_{1}}+\theta
t^{m_{1}})\cdot\partial(t^{n_{2}}+\theta t^{m_{2}})=(t+c)^{n_{1}+n_{2}}.$

Hence, $\partial$ satisfies condition (3.9). Define a new bilinear operation
on $\mathcal{A}$ by

$\displaystyle(t^{n_{1}}+\theta t^{m_{1}})\star_{2}(t^{n_{2}}+\theta
t^{m_{2}})=(t^{n_{1}}+\theta t^{m_{1}})\cdot\partial(t^{n_{2}}+\theta
t^{m_{2}})=(t+c)^{n_{1}+n_{2}},$

for all $n_{1},n_{2},m_{1},m_{2}\in\mathbb{Z}$. Then
$(\mathcal{A},\star_{2},\alpha)$ forms a Hom-Novikov algebra by Theorem 3.6.

§4. Quadratic Hom-Novikov Algebras

In this section, we extend the notion of quadratic Novikov algebra to
quadratic Hom-Novikov algebras and provide some properties.

Recall that a quadratic Lie algebra is a triple
$(\mathcal{G},[\cdot,\cdot],B)$, where $(\mathcal{G},[\cdot,\cdot])$ is a Lie
algebra and $B:\mathcal{G}\times\mathcal{G}\longrightarrow\mathbb{F}$ is a
symmetric nondegenerate bilinear form satisfying

$\displaystyle B([x,y],z)=B(x,[y,z]),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{G},$ (4.1)

which is called the invariance of $B$. It is easy to see that identity (4.1)
is equivalent to

$\displaystyle B([x,y],z)=-B(y,[x,z]),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{G}.$

Assume that $(\mathcal{A},\mu)$ is a Novikov algebra. A bilinear form
$B:\mathcal{A}\times\mathcal{A}\longrightarrow\mathbb{F}$ is said to be
invariant or associative if and only if

$B(xy,z)=B(x,yz),\hskip 8.5359pt\mbox{for all}\ x,y,z\in\mathcal{A}.$

A Novikov algebra $(\mathcal{A},\mu)$ with a symmetric nondegenerate invariant
bilinear form is called a quadratic Novikov algebra and denoted by
$(\mathcal{A},\mu,B)$.

Benayadi and Makhlouf in [5] extended the notion of quadratic Lie algebra to
Hom-Lie algebras and obtained quadratic Hom-Lie algebras.

###### Definition 4.1

A quadratic Hom-Lie algebra is a quadruple
$(\mathcal{A},[\cdot,\cdot],\alpha,B)$ such that
$(\mathcal{A},[\cdot,\cdot],\alpha)$ is a Hom-Lie algebra with a symmetric
invariant nondegenerate bilinear form $B$ satisfying

$\displaystyle B(\alpha(x),y)=B(x,\alpha(y)),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$ (4.2)

We can define quadratic Hom-Novikov algebras as follows.

###### Definition 4.2

A quadratic Hom-Novikov algebra $(\mathcal{A},\mu,\alpha,B)$ is a Hom-Novikov
algebra $(\mathcal{A},\mu,\alpha)$ with a symmetric nondegenerate bilinear
form satisfying

$\displaystyle B(xy,\alpha(z))=B(\alpha(x),yz),\hskip 8.5359pt\mbox{for all}\
x,y,z\in\mathcal{A}.$ (4.3)

We recover quadratic Novikov algebras when $\alpha={\rm id}$. The following
result says the sub-adjacent Hom-Lie algebra of a quadratic Hom-Novikov
algebra is also quadratic.

###### Proposition 4.3

Let $(\mathcal{A},\mu,\alpha,B)$ be a quadratic Hom-Novikov algebra and
$HLie(\mathcal{A})=(\mathcal{A},[\cdot,\cdot],\alpha)$ be the sub-adjacent
Hom-Lie algebra of $\mathcal{A}$. If $\alpha$ is an automorphism satisfying

$\displaystyle B(\alpha(x),y)=B(x,\alpha(y)),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$ (4.4)

Then $(\mathcal{A},[\cdot,\cdot],\alpha,B_{\alpha})$ is a quadratic Hom-Lie
algebra, where $B_{\alpha}(x,y)=B(\alpha(x),y)$, for all $x,y\in\mathcal{A}$.

Proof.  Since $B$ is a nondegenerate bilinear form and $\alpha$ is an
automorphism, $B_{\alpha}$ is a nondegenerate bilinear form on $\mathcal{A}$.
For all $x,y,z\in\mathcal{A}$, using the properties of $B$, we have

$\displaystyle B_{\alpha}([x,y],z)$ $\displaystyle=$ $\displaystyle
B\big{(}\alpha([x,y]),z\big{)}$ $\displaystyle=$ $\displaystyle
B([x,y],\alpha(z))$ $\displaystyle=$ $\displaystyle
B(xy,\alpha(z))-B(yx,\alpha(z))$ $\displaystyle=$ $\displaystyle
B(\alpha(x),yz)-B(\alpha(x),zy)$ $\displaystyle=$ $\displaystyle
B(\alpha(x),[y,z])$ $\displaystyle=$ $\displaystyle B_{\alpha}(x,[y,z]).$

Hence $B_{\alpha}$ is invariant. Using symmetry of $B$ and equation (4.4), we
have

$\displaystyle
B_{\alpha}(x,y)=B(\alpha(x),y)=B(y,\alpha(x))=B(\alpha(y),x)=B_{\alpha}(y,x),$

which proves $B_{\alpha}$ is symmetric. Using equation (4.4) again, we have

$B_{\alpha}(\alpha(x),y)=B(\alpha(\alpha(x)),y)=B(\alpha(x),\alpha(y))=B_{\alpha}(x,\alpha(y)),$

which completes the proof. $\Box$

###### Corollary 4.4

Let $(\mathcal{A},\mu,B)$ be a quadratic Novikov algebra with an automorphism
$\alpha$ satisfying equation (4.4) and $(\mathcal{A},[\cdot,\cdot])$ be the
sub-adjacent Lie algebra. Then
$(\mathcal{A},[\cdot,\cdot]_{\alpha}=\alpha\circ[\cdot,\cdot],\alpha,B_{\alpha})$
forms a quadratic Hom-Lie algebra, where $B_{\alpha}(x,y)=B(\alpha(x),y)$, for
all $x,y\in\mathcal{A}$.

Proof.  It follows from [References, Corollary 2.6] that
$(\mathcal{A},[\cdot,\cdot]_{\alpha},\alpha)$ is a Hom-Lie algebra. Using the
similar arguments as those in the proof of Proposition 4.3, we get
$B_{\alpha}$ is a symmetric nondegenerate bilinear form with equation (4.2)
satisfied. It remains to show that $B_{\alpha}$ is invariant. For all
$x,y,z\in\mathcal{A}$, using invariance and symmetry of $B$, we have

$\displaystyle B_{\alpha}([x,y]_{\alpha},z)$ $\displaystyle=$ $\displaystyle
B\big{(}\alpha([x,y]_{\alpha}),z\big{)}$ $\displaystyle=$ $\displaystyle
B\big{(}[x,y]_{\alpha},\alpha(z)\big{)}$ $\displaystyle=$ $\displaystyle
B(\alpha(x)\alpha(y),\alpha(z))-B(\alpha(y)\alpha(x),\alpha(z))$
$\displaystyle=$ $\displaystyle
B(\alpha(x),\alpha(y)\alpha(z))-B(\alpha(z)\alpha(y),\alpha(x))$
$\displaystyle=$ $\displaystyle B(\alpha(x),[y,z]_{\alpha})$ $\displaystyle=$
$\displaystyle B_{\alpha}(x,[y,z]_{\alpha}),$

which proves the invariance of $B_{\alpha}$ and the result. $\Box$

###### Proposition 4.5

Let $(\mathcal{A},\mu,\alpha,B)$ be a quadratic Hom-Novikov algebra, where
$\alpha$ is an involution satisfying equation (4.4). Then
$(\mathcal{A},\star=\alpha\circ\mu,B)$ is a quadratic Novikov algebra.

Proof.  $({A},\star=\alpha\circ\mu,\alpha)$ is a Hom-Novikov algebra by
Proposition 3.4. It suffices to show that $B$ is invariant under the operation
$\star$. For all $x,y,z\in\mathcal{A}$, we have

$\displaystyle B(x,y\star
z)=B(x,\alpha(y)\alpha(z))=B(\alpha(x),yz)=B(xy,\alpha(z))=B(\alpha(x)\alpha(y),z)=B(x\star
y,z),$

which completes the proof. $\Box$

###### Proposition 4.6

Let $(\mathcal{A},\mu,\alpha,B)$ be a quadratic Hom-Novikov algebra, where
$\alpha$ is an automorphism satisfying equation (4.4). Then,
$(\mathcal{A},\star=\alpha\circ\mu,\alpha^{2},B_{\alpha^{2}})$ is a quadratic
Hom-Novikov algebra, where $B_{\alpha^{2}}(x,y)=B(\alpha^{2}(x),y)$, for all
$x,y\in\mathcal{A}$.

Proof. It follows from [References, Corollary 2.12]) that
$(\mathcal{A},\alpha\circ\mu,\alpha^{2})$ forms a Hom-Novikov algebra. Since
$B$ is a nondegenerate bilinear form on $\mathcal{A}$ and $\alpha$ is an
automorphism, $B_{\alpha^{2}}$ is a nondegenerate bilinear form. For all
$x,y,z\in\mathcal{A}$, using the hypothesis, we have

$B_{\alpha^{2}}(x,y)=B(\alpha^{2}(x),y)=B(\alpha(x),\alpha(y))=B(x,\alpha^{2}(y))=B(\alpha^{2}(y),x)=B_{\alpha^{2}}(y,x).$

Thus, $B_{\alpha^{2}}$ is symmetric. Moreover, we have

$\displaystyle B_{\alpha^{2}}\big{(}\alpha^{2}(x),y\star z\big{)}$
$\displaystyle=$ $\displaystyle
B\big{(}\alpha^{4}(x),\alpha(y)\alpha(z)\big{)}$ $\displaystyle=$
$\displaystyle B\big{(}\alpha^{3}(x),\alpha^{2}(y)\alpha^{2}(z)\big{)}$
$\displaystyle=$ $\displaystyle
B\big{(}\alpha^{2}(x)\alpha^{2}(y),\alpha^{3}(z)\big{)}$ $\displaystyle=$
$\displaystyle B\big{(}\alpha(x)\alpha(y),\alpha^{4}(z)\big{)}$
$\displaystyle=$ $\displaystyle B_{\alpha^{2}}\big{(}x\star
y,\alpha^{2}(z)\big{)},$

which proves the invariance of $B_{\alpha^{2}}$ and the proposition. $\Box$

###### Corollary 4.7

Let $(\mathcal{A},\mu,\alpha,B)$ be a quadratic Hom-Novikov algebra and
$\alpha$ be an automorphism satisfying equation (4.4). Then
$(\mathcal{A},\alpha^{n}\circ\mu,\alpha^{n+1},B_{\alpha^{n+1}})$ is a
quadratic Hom-Novikov algebra, where
$B_{\alpha^{n+1}}(x,y)=B(\alpha^{n+1}(x),y)$, for all $n>0$ and
$x,y\in\mathcal{A}$.

Let $(\mathcal{A},\mu,\alpha)$ be a Hom-Novikov algebra, whose center is
denoted by $\mathcal{Z}(\mathcal{A})$ and defined by

$\displaystyle\mathcal{Z}(\mathcal{A})=\\{x\in\mathcal{A}|xy=yx=0,\ \mbox{for
all}\ y\in\mathcal{A}\\}.$

Let $(\mathcal{G},[\cdot,\cdot],\beta)$ be a Hom-Lie algebra. The lower
central series of $\mathcal{G}$ is defined as usual, i.e.,
$\mathcal{G}^{1}=\mathcal{G}$,
$\mathcal{G}^{i+1}=[\mathcal{G},\mathcal{G}^{i}]$ for all $i\geq 1$. We call
$\mathcal{G}$ is $i$-step nilpotent if $\mathcal{G}^{i+1}=0$. The center of
the Hom-Lie algebra is denoted by $\mathcal{C}(\mathcal{G})$ and defined by

$\displaystyle\mathcal{C}(\mathcal{G})=\\{x\in\mathcal{G}|[x,y]=0,\ \mbox{for
all}\ y\in\mathcal{G}\\}.$

###### Proposition 4.8

Let $(\mathcal{A},\mu,\alpha,B)$ be a quadratic Hom-Novikov algebra and
$HLie(\mathcal{A})$ be the sub-adjacent Hom-Lie algebra. If $\alpha$ is an
automorphism, then
$[HLie(\mathcal{A}),HLie(\mathcal{A})]\subseteq\mathcal{Z}(\mathcal{A}).$ As a
consequence, $HLie(\mathcal{A})$ is $2$-step nilpotent.

Proof.  For all $x,y,z,w\in\mathcal{A}$, using equations (2.5), (2.6) and
(4.3), we have

$\displaystyle B(\alpha(x)[y,z],\alpha^{2}(w))$ $\displaystyle=$
$\displaystyle B(\alpha^{2}(x),[y,z]\alpha(w))$ $\displaystyle=$
$\displaystyle B(\alpha^{2}(x),(yz)\alpha(w)-(zy)\alpha(w))$ $\displaystyle=$
$\displaystyle B(\alpha^{2}(x),\alpha(y)(zw)-\alpha(z)(yw))$ $\displaystyle=$
$\displaystyle
B(\alpha(x)\alpha(y),\alpha(zw))-B(\alpha(x)\alpha(z),\alpha(yw))$
$\displaystyle=$ $\displaystyle B((xy)\alpha(z)-(xz)\alpha(y),\alpha^{2}(w))$
$\displaystyle=$ $\displaystyle 0.$

Using the symmetry of $B$ and equation (4.3), we have

$B([y,z]\alpha(w),\alpha^{2}(x))=B(\alpha(x)[y,z],\alpha^{2}(w))=0,$

which implies $[y,z]\subseteq\mathcal{Z}(\mathcal{A})$ since $\alpha$ is an
automorphism and $B$ is nondegenerate. Hence, we have
$[HLie(\mathcal{A}),HLie(\mathcal{A})]\subseteq\mathcal{Z}(\mathcal{A}).$
Obviously, $\mathcal{Z}(\mathcal{A})\subseteq\mathcal{C}(HLie(\mathcal{A}))$.
Then it follows that $HLie(\mathcal{A})$ is $2$-step nilpotent. $\Box$

§5. Hom-Novikov-Poisson Algebras

In this section, we discuss Hom-Novikov-Poisson algebras. We give different
ways to construct Hom-Novikov-Poisson algebras and provide some interesting
examples.

The following result (see also [References, Corollary 2.15])says that any Hom-
Novikov-Poisson algebra can be constructed from a Novikov-Poisson algebra and
an algebra endomorphism.

###### Proposition 5.1

Let $\big{(}\mathcal{A},\mu,\nu\big{)}$ be a Novikov-Poisson algebra such that
$\big{(}\mathcal{A},\nu\big{)}$ is a commutative associative algebra and
$\big{(}\mathcal{A},\mu\big{)}$ is a Novikov algebra. If
$\alpha:\mathcal{A}\longrightarrow\mathcal{A}$ is an algebra homomorphism.
Then
$\mathcal{A}_{\alpha}=\big{(}\mathcal{A},\mu_{\alpha}=\alpha\circ\mu,\nu_{\alpha}=\alpha\circ\nu,\alpha\big{)}$
forms a Hom-Novikov-Poisson algebra.

Proof. It follows that $\big{(}\mathcal{A},\mu_{\alpha},\alpha\big{)}$ is a
Hom-Novikov algebra from Proposition 2.7 and
$\big{(}\mathcal{A},\nu_{\alpha},\alpha\big{)}$ is a Hom-associative algebra
from [References, Corollary 2.6]. Clearly,
$\big{(}\mathcal{A},\nu_{\alpha},\alpha\big{)}$ is commutative since $\alpha$
is an algebra homomorphism and $(\mathcal{A},\nu)$ is commutative. By
Definition 2.8, it suffices to check conditions (2.7) and (2.8). For all
$x,y,z\in\mathcal{A}$, Using equation (2.4), we have

$\displaystyle\nu_{\alpha}\big{(}\mu_{\alpha}(x,y),\alpha(z)\big{)}-\mu_{\alpha}\big{(}\alpha(x),\nu_{\alpha}(y,z)\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\nu\big{(}\alpha\circ\mu(x,y),\alpha(z)\big{)}-\alpha\circ\mu\big{(}\alpha(x),\alpha\circ\nu(y,z)\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\alpha\circ\nu\big{(}\mu(x,y),z\big{)}-\alpha\circ\alpha\circ\mu\big{(}x,\nu(y,z)\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\alpha\circ\Big{(}\nu\big{(}\mu(x,y),z\big{)}-\mu\big{(}x,\nu(y,z)\big{)}\Big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\alpha\circ\Big{(}\nu\big{(}\mu(y,x),z\big{)}-\mu\big{(}y,\nu(x,z)\big{)}\Big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\nu\big{(}\alpha\circ\mu(y,x),\alpha(z)\big{)}-\alpha\circ\mu\big{(}\alpha(y),\alpha\circ\nu(x,z)\big{)}$
$\displaystyle=$
$\displaystyle\nu_{\alpha}\big{(}\mu_{\alpha}(y,x),\alpha(z)\big{)}-\mu_{\alpha}\big{(}\alpha(y),\nu_{\alpha}(x,z)\big{)},$

which proves equation (2.8). Using equation (2.3), we have

$\displaystyle\mu_{\alpha}\big{(}\nu_{\alpha}(x,y),\alpha(z)\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\mu\big{(}\alpha\circ\nu(x,y),\alpha(z)\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\alpha\circ\mu\big{(}\nu(x,y),z\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\alpha\circ\nu\big{(}x,\mu(y,z)\big{)}$
$\displaystyle=$
$\displaystyle\nu_{\alpha}\big{(}\alpha(x),\mu_{\alpha}(y,z)\big{)},$

which proves equation (2.7) and the result. $\Box$

Suppose that $\big{(}\mathcal{A}_{1},\mu^{1},\nu^{1}\big{)}$ and
$\big{(}\mathcal{A}_{2},\mu^{2},\nu^{2}\big{)}$ are Novikov-Poisson algebras.
Define two operations $\nu^{1}\otimes\nu^{2}$ and $\mu^{1}\otimes\mu^{2}$ on
$\mathcal{A}_{1}\otimes\mathcal{A}_{2}$ by

$\displaystyle(\nu^{1}\otimes\nu^{2})(x_{1}\otimes x_{2},y_{1}\otimes y_{2})$
$\displaystyle=$
$\displaystyle\nu^{1}(x_{1},y_{1})\otimes\nu^{2}(x_{2},y_{2}),$ (5.1)
$\displaystyle(\mu^{1}\otimes\mu^{2})(x_{1}\otimes x_{2},y_{1}\otimes y_{2})$
$\displaystyle=$
$\displaystyle\mu^{1}(x_{1},y_{1})\otimes\nu^{2}(x_{2},y_{2})+\nu^{1}(x_{1},y_{1})\otimes\mu^{2}(x_{2},y_{2}),$
(5.2)

for all $x_{1},\ y_{1}\in\mathcal{A}_{1}$ and $x_{2},\
y_{2}\in\mathcal{A}_{2}$. Then
$\big{(}\mathcal{A}_{1}\otimes\mathcal{A}_{2},\mu^{1}\otimes\mu^{2},\nu^{1}\otimes\nu^{2}\big{)}$
forms a Novikov-Poisson algebra (cf.[21, 22]). Additionally, assume that
$\alpha$ and $\beta$ are algebra homomorphisms of
$\big{(}\mathcal{A}_{1},\mu^{1},\nu^{1}\big{)}$ and
$\big{(}\mathcal{A}_{2},\mu^{2},\nu^{2}\big{)}$, respectively. It follows from
Proposition 5.1 that
$\big{(}\mathcal{A}_{1},\mu^{1}_{\alpha},\nu^{1}_{\alpha},\alpha\big{)}$ and
$\big{(}\mathcal{A}_{2},\mu^{2}_{\beta},\nu^{2}_{\beta},\beta\big{)}$ are Hom-
Novikov-Poisson algebras. Define a linear map $\alpha\otimes\beta$ on
$\mathcal{A}_{1}\otimes\mathcal{A}_{2}$ by

$\displaystyle(\alpha\otimes\beta)(x\otimes y)=\alpha(x)\otimes\beta(y),\hskip
8.5359pt\mbox{for all}\ x\in\mathcal{A}_{1},\ y\in\mathcal{A}_{2}.$ (5.3)

For all $x_{1},y_{1}\in\mathcal{A}_{1}$ and $x_{2},y_{2}\in\mathcal{A}_{2}$,
we have

$\displaystyle(\alpha\otimes\beta)\big{(}\nu^{1}\otimes\nu^{2}(x_{1}\otimes
x_{2},y_{1}\otimes y_{2})\big{)}$ $\displaystyle=$
$\displaystyle(\alpha\otimes\beta)\big{(}\nu^{1}(x_{1},y_{1})\otimes\nu^{2}(x_{2},y_{2})\big{)}$
$\displaystyle=$
$\displaystyle\alpha\circ\nu^{1}(x_{1},y_{1})\otimes\beta\circ\nu^{2}(x_{2},y_{2})$
$\displaystyle=$
$\displaystyle\nu^{1}(\alpha(x_{1}),\alpha(y_{1}))\otimes\nu^{2}(\beta(x_{2}),\beta(y_{2}))$
$\displaystyle=$
$\displaystyle(\nu^{1}\otimes\nu^{2})\big{(}\alpha(x_{1})\otimes\beta(x_{2}),\alpha(y_{1})\otimes\beta(y_{2})\big{)}$
$\displaystyle=$
$\displaystyle(\nu^{1}\otimes\nu^{2})\big{(}\alpha\otimes\beta(x_{1}\otimes
x_{2}),\alpha\otimes\beta(y_{1}\otimes y_{2})\big{)}.$

Similarly, we have

$\displaystyle(\alpha\otimes\beta)\big{(}\mu^{1}\otimes\mu^{2}(x_{1}\otimes
x_{2},y_{1}\otimes
y_{2})\big{)}=(\mu^{1}\otimes\mu^{2})\big{(}\alpha\otimes\beta(x_{1}\otimes
x_{2}),\alpha\otimes\beta(y_{1}\otimes y_{2})\big{)}.$

Thus $\alpha\otimes\beta$ is an algebra homomorphism of
$\big{(}\mathcal{A}_{1}\otimes\mathcal{A}_{2},\mu^{1}\otimes\mu^{2},\nu^{1}\otimes\nu^{2}\big{)}$.

On the other hand, consider the following two operations on
$\mathcal{A}_{1}\otimes\mathcal{A}_{2}$ :

$\displaystyle(\nu^{1}_{\alpha}\otimes\nu^{2}_{\beta})(x_{1}\otimes
x_{2},y_{1}\otimes y_{2})$ $\displaystyle=$
$\displaystyle\nu^{1}_{\alpha}(x_{1},y_{1})\otimes\nu^{2}_{\beta}(x_{2},y_{2}),$
$\displaystyle(\mu^{1}_{\alpha}\otimes\mu^{2}_{\beta})(x_{1}\otimes
x_{2},y_{1}\otimes y_{2})$ $\displaystyle=$
$\displaystyle\mu^{1}_{\alpha}(x_{1},y_{1})\otimes\nu^{2}_{\beta}(x_{2},y_{2})+\nu^{1}_{\alpha}(x_{1},y_{1})\otimes\mu^{2}_{\beta}(x_{2},y_{2}),$

for all $x_{1},y_{1}\in\mathcal{A}_{1}$, $x_{2},y_{2}\in\mathcal{A}_{2}$.
Then, we have

$\displaystyle(\nu^{1}_{\alpha}\otimes\nu^{2}_{\beta})(x_{1}\otimes
x_{2},y_{1}\otimes y_{2})$ $\displaystyle=$
$\displaystyle\nu^{1}_{\alpha}(x_{1},y_{1})\otimes\nu^{2}_{\beta}(x_{2},y_{2})$
$\displaystyle=$
$\displaystyle\alpha\circ\nu^{1}(x_{1},y_{1})\otimes\beta\circ\nu^{2}(y_{1},y_{2})$
$\displaystyle=$
$\displaystyle\nu^{1}(\alpha(x_{1}),\alpha(y_{1}))\otimes\nu^{2}(\beta(x_{2}),\beta(y_{2}))$
$\displaystyle=$
$\displaystyle(\nu^{1}\otimes\nu^{2})\big{(}\alpha(x_{1})\otimes\beta(x_{2}),\alpha(y_{1})\otimes\beta(y_{2})\big{)}$
$\displaystyle=$
$\displaystyle(\nu^{1}\otimes\nu^{2})\big{(}\alpha\otimes\beta(x_{1}\otimes
x_{2}),\alpha\otimes\beta(y_{1}\otimes y_{2})\big{)}.$

Similarly, we have

$\displaystyle(\mu^{1}_{\alpha}\otimes\mu^{2}_{\beta})(x_{1}\otimes
x_{2},y_{1}\otimes
y_{2})=(\mu^{1}\otimes\mu^{2})\big{(}\alpha\otimes\beta(x_{1}\otimes
x_{2}),\alpha\otimes\beta(y_{1}\otimes y_{2})\big{)}.$

Now from the discussions above, we obtain

$\displaystyle\mu^{1}_{\alpha}\otimes\mu^{2}_{\beta}$ $\displaystyle=$
$\displaystyle(\alpha\otimes\beta)\circ(\mu^{1}\otimes\mu^{2})=(\mu^{1}\otimes\mu^{2})_{\alpha\otimes\beta},$
$\displaystyle\nu^{1}_{\alpha}\otimes\nu^{2}_{\beta}$ $\displaystyle=$
$\displaystyle(\alpha\otimes\beta)\circ(\nu^{1}\otimes\nu^{2})=(\nu^{1}\otimes\nu^{2})_{\alpha\otimes\beta}.$

Then, according to Proposition 5.1, we have the following result (see also
[References, Corollary 3.6]):

###### Theorem 5.2

With notations above. Let $\big{(}\mathcal{A}_{1},\mu^{1},\nu^{1}\big{)}$ and
$\big{(}\mathcal{A}_{2},\mu^{2},\nu^{2}\big{)}$ be two Novikov-Poisson
algebras. If $\alpha$ and $\beta$ are algebra homomorphisms of
$\big{(}\mathcal{A}_{1},\mu^{1},\nu^{1}\big{)}$ and
$\big{(}\mathcal{A}_{2},\mu^{2},\nu^{2}\big{)}$, respectively. Then
$\big{(}\mathcal{A}_{1}\otimes\mathcal{A}_{2},\mu^{1}_{\alpha}\otimes\mu^{2}_{\beta},\nu^{1}_{\alpha}\otimes\nu^{2}_{\beta},\alpha\otimes\beta\big{)}$
forms a Hom-Novikov-Poisson algebra.

We extend in the following the Yau’s Theorem (see [28]) to the Hom-Novikov-
Poisson algebra case. The following theorem gives a way to construct Hom-
Novikov-Poisson algebras from a commutative Hom-associative algebra along with
a derivation.

###### Theorem 5.3

Let $(\mathcal{A},\cdot,\alpha)$ be a commutative Hom-associative algebra and
$\partial$ be a derivation such that $\partial\alpha=\alpha\partial$. Define a
new operation on $\mathcal{A}$ by

$\displaystyle x\star y=x\cdot\partial(y),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$ (5.4)

Then $(\mathcal{A},\cdot,\star,\alpha)$ forms a Hom-Novikov-Poisson algebra.

Proof. It follows from [References, Theorem 1.2] that
$(\mathcal{A},\star,\alpha)$ is a Hom-Novikov algebra. We need to check the
two compatible conditions (2.7) and (2.8). For any $x,y,z\in\mathcal{A}$, we
have

$\displaystyle(x\cdot y)\star\alpha(z)$ $\displaystyle=$ $\displaystyle(x\cdot
y)\cdot\partial(\alpha(z))$ $\displaystyle=$ $\displaystyle(x\cdot
y)\cdot\alpha(\partial(z))$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot(y\cdot\partial(z))$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot(y\star z),$

which proves equation (2.7). Furthermore, we have

$\displaystyle(x\star y)\cdot\alpha(z)-\alpha(x)\star(y\cdot z)$
$\displaystyle=$
$\displaystyle(x\cdot\partial(y))\cdot\alpha(z)-\alpha(x)\cdot\partial(y\cdot
z)$ $\displaystyle=$
$\displaystyle(x\cdot\partial(y))\cdot\alpha(z)-\alpha(x)\cdot(\partial(y)\cdot
z+y\cdot\partial(z))$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot(\partial(y)\cdot
z)-\alpha(x)\cdot\big{(}\partial(y)\cdot z+y\cdot\partial(z)\big{)}$
$\displaystyle=$ $\displaystyle-\alpha(x)\cdot\big{(}y\cdot\partial(z)\big{)}$
$\displaystyle=$ $\displaystyle-(x\cdot y)\cdot\alpha(\partial(z)).$

Similarly, we have

$\displaystyle(y\star x)\cdot\alpha(z)-\alpha(y)\star(x\cdot z)=-(y\cdot
x)\cdot\alpha(\partial(z)).$

Since $(\mathcal{A},\cdot,\alpha)$ is commutative, we get

$\displaystyle(x\star y)\cdot\alpha(z)-\alpha(x)\star(y\cdot z)=(y\star
x)\cdot\alpha(z)-\alpha(y)\star(x\cdot z),$

which completes the proof. $\Box$

The following result is due to Xu [22].

###### Proposition 5.4

Let $(\mathcal{A},\cdot,\ast)$ be a Novikov-Poisson algebra, such that
$(\mathcal{A},\cdot)$ is a commutative associative algebra and
$(\mathcal{A},\ast)$ is a Novikov algebra. Suppose that $(\mathcal{A},\cdot)$
contains an identity element $1$, that is, $1\cdot x=x\cdot 1=x$ for all
$x\in\mathcal{A}$. Define a map
$\partial:\mathcal{A}\longrightarrow\mathcal{A}$ by

$\displaystyle\partial(x)=1\ast x-(1\ast 1)\cdot x,\hskip 8.5359pt\mbox{for
all}\ x\in\mathcal{A}.$ (5.5)

Then $\partial$ is a derivation of $(\mathcal{A},\cdot)$.

In the following, we give some examples of Hom-Novikov-Poisson algebras by
using Theorem 5.3 and Proposition 5.4.

###### Example 5.5

Assume that $(\mathcal{A},\cdot,\ast)$ is a Novikov-Poisson algebra, where
$(\mathcal{A},\cdot)$ is a commutative associative algebra with unity $1$ and
$(\mathcal{A},\ast)$ is a Novikov algebra. Equation (2.3) implies that

$x\cdot(1\ast 1)=x\ast 1,\hskip 8.5359pt\mbox{for all}\ x\in\mathcal{A}.$

Since $(\mathcal{A},\cdot)$ is commutative, we have

$\displaystyle\partial(x)=1\ast x-x\ast 1,\hskip 8.5359pt\mbox{for all}\
x\in\mathcal{A},$ (5.6)

in which $\partial$ is defined by (5.5). Let $\alpha$ be an algebra
homomorphism of $(\mathcal{A},\cdot,\ast)$. Define a new multiplication on
$\mathcal{A}$ by

$\displaystyle x\bullet y=\alpha(x\cdot y),\hskip 8.5359pt\mbox{ for all}\
x,y\in\mathcal{A}.$

Then $(\mathcal{A},\bullet,\alpha)$ is a commutative Hom-associative algebra.
Obviously, $\alpha(1)=1$. As a consequence, $\alpha$ commutes with $\partial$.
Furthermore, $\partial$ is a derivation of $(\mathcal{A},\bullet)$ since it is
a derivation of $(\mathcal{A},\cdot)$ by Proposition 5.4. Now define a new
operation on $\mathcal{A}$ by

$x\star y=x\bullet\partial(y),\hskip 8.5359pt\mbox{for all}\
x,y\in\mathcal{A}.$

Then, thanks to Theorem 5.3, $(\mathcal{A},\bullet,\star,\alpha)$ is a Hom-
Novikov-Poisson algebra.

###### Example 5.6

Let the ground field be the complex field $\mathbb{C}$ and $\Delta$ be a
nonzero abelian subgroup of $\mathbb{C}$. Suppose that $f$ is a nontrivial
homomorphism of $\Delta$ into the additive group of $\mathbb{C}$. Consider a
vector space $\mathcal{A}$ with basis $\\{x_{a}|a\in\Delta\\}$. Fix an element
$q$ in $\Delta$. Define a multiplication on $\mathcal{A}$ by

$\displaystyle x_{a}\cdot x_{b}=x_{a+b+q},\hskip 8.5359pt\mbox{for all}\
a,b\in\Delta.$

Then $(\mathcal{A},\cdot)$ is a commutative associative algebra. Define a
linear map $\partial_{1}:\mathcal{A}\longrightarrow\mathcal{A}$ by

$\displaystyle\partial_{1}(x_{a})=f(a+q)x_{a},\hskip 8.5359pt\mbox{for all}\
a\in\Delta.$

It is easy to check that $\partial_{1}$ is a derivation of
$(\mathcal{A},\cdot)$. Then we obtain a Novikov-Poisson algebra
$(\mathcal{A},\cdot,\ast)$ with the operation $\ast$ defined by

$\displaystyle x_{a}\ast
x_{b}=x_{a}\cdot\partial_{1}(x_{b})=f(b+q)x_{a+b+q},\hskip 8.5359pt\mbox{for
all}\ a,b\in\Delta.$

Note that $x_{-q}$ is an identity element of $(\mathcal{A},\cdot)$. Define
another linear map $\partial_{2}:\mathcal{A}\longrightarrow\mathcal{A}$ by

$\displaystyle\partial_{2}(x_{a})=x_{-q}\ast x_{a}-x_{a}\ast x_{-q},\hskip
8.5359pt\mbox{for all}\ a\in\Delta.$

Then it follows from Proposition 5.4 and the discussions presented in Example
5.5 that $\partial_{2}$ is a derivation of $(\mathcal{A},\cdot)$.

Let $\alpha:\mathcal{A}\longrightarrow\mathcal{A}$ be a linear map defined by

$\displaystyle\alpha(x_{a})=e^{a+q}x_{a},\hskip 8.5359pt\mbox{for all}\
a\in\Delta.$

Then $\alpha$ is an algebra homomorphism of $(\mathcal{A},\cdot,\ast)$ and
$\alpha(x_{-q})=x_{-q}$. Moreover, $\alpha$ commutes with $\partial_{2}$.
Define a new multiplication on $\mathcal{A}$ by

$\displaystyle x_{a}\bullet x_{b}=\alpha(x_{a}\cdot x_{b}),\hskip
8.5359pt\mbox{for all}\ a,b\in\Delta,$

which makes $(\mathcal{A},\bullet,\alpha)$ form a commutative Hom-associative
algebra. Also, $\partial_{2}$ is a derivation of $(\mathcal{A},\bullet)$.
Define a bilinear map $\star$ on $\mathcal{A}$ by

$\displaystyle x_{a}\star x_{b}=x_{a}\bullet\partial_{2}(x_{b}),\hskip
8.5359pt\mbox{for all}\ a,b\in\Delta,$

To be more precise, we have

$\displaystyle x_{a}\star x_{b}=e^{a+b+2q}f(b+q)x_{a+b+q},\hskip
8.5359pt\mbox{for all}\ a,b\in\Delta.$

According to Theorem 5.3, $(\mathcal{A},\bullet,\star,\alpha)$ is a Hom-
Novikov-Poisson algebra.

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