Cup-product in Hom-Leibniz cohomology and Hom-Zinbiel algebras

TL;DR

This paper introduces a cup-product in Hom-Leibniz cohomology and demonstrates that it satisfies the graded Hom-Zinbiel relation, expanding the algebraic structures in Hom-cohomology theories.

Contribution

It defines a new cup-product in Hom-Leibniz cohomology and proves its relation to Hom-Zinbiel algebra structures, which is a novel theoretical development.

Findings

Cup-product in Hom-Leibniz cohomology is defined.

The cup-product satisfies the graded Hom-Zinbiel relation.

Establishes a new connection between Hom-Leibniz and Hom-Zinbiel algebraic structures.

Abstract

We define a cup-product in Hom-Leibniz cohomology and show that the cup-product satisfies the graded Hom-Zinbiel relation.

Equations

$$\cup :H{L}_{\alpha ,{\alpha }_1}^n(L;A)\times H{L}_{\alpha ,{\alpha }_1}^m(L;A)\to H{L}_{\alpha ,{\alpha }_1}^{n+m}(L;A),$$

$${\alpha }_1(a\cup b)\cup c=a\cup {\alpha }_1(b\cup c)+(-1{)}^{|b||c|}a\cup {\alpha }_1(c\cup b),$$

$$\mu ({\alpha }_1(x),\mu (y,z))=\mu (\mu (x,y),{\alpha }_1(z)),\text{ for all }x,y,z\in A.$$

$$\mu (a_i,a_j)=\{\begin{array}{ll}{a}_1& \text{mbox}if(i,j)=(1,1)\\ a_2& \text{mbox}if(i,j)\ne (1,1).\end{array}$$

$${\circlearrowleft }_{x,y,z}\mu (\mu (x,y),\alpha (z))=0,$$

$$[x,y{]}_{d,\alpha }:=[\alpha (x),d\alpha (y)].$$

$$\alpha =\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right].$$

$$():R\otimes R\to R\text{and}{\alpha }_1:R\to R,$$

$$((xy){\alpha }_1(z))=({\alpha }_1(x)(yz))+({\alpha }_1(x)(zy)),$$

$$((xy){\alpha }_1(z))=({\alpha }_1(x)(yz))+(-1{)}^{|y||z|}({\alpha }_1(x)(zy)).$$

$$\sigma (1)<\cdots <\sigma (n)\text{mbox}and\sigma (n+1)<\cdots <\sigma (n+m).$$

$$s{h}_{n,m}:=\sum _{\sigma }\sigma ,$$

$$\sigma (v_1\dots v_p)=(v_{{\sigma }^{-1}(1)}\dots v_{{\sigma }^{-1}(p)}),$$

$${\rho }_{n,m}:=1_1\otimes {\widetilde{sh}}_{n-1,m}=L^{\otimes n+m}\to L^{\otimes n+m},$$

$${\rho }_{n,m}(x_1,\dots ,x_{n+m})=\sum _{\sigma }\text{mbox}sgn(\sigma )(x_1,x_{\sigma (2)},\dots ,x_{\sigma (n+m)}),$$

$${\tau }_{r,m}(cb)=bc.$$

$$({\rho }_{n,m}\otimes 1_r)\circ {\rho }_{n+m,r}=(1_n\otimes {\rho }_{m,r}+(-1{)}^{rm}\circ {\tau }_{r,m}\circ {\rho }_{r,m})\circ {\rho }_{n,m+r}$$

$$[x\otimes r,y\otimes s]=[x,y]\otimes (rs)-[y,x]\otimes (sr),$$

$$\alpha \otimes {\alpha }_1:L\otimes R\to L\otimes R,(\alpha \otimes {\alpha }_1)(x\otimes r)=\alpha (x)\otimes {\alpha }_1(r),$$

$$\sum _{\sigma \in S_3}sgn(\sigma )\sigma [\alpha (x)\otimes {\alpha }_1(r),[y\otimes s,z\otimes t]]=0,$$

$$\sum _{\sigma \in S_3}sgn(\sigma )\sigma$$

$$\displaystyle-[\alpha(x),[z,y]]\otimes(\alpha_{1}(r)(st))+[[z,y],\alpha(x)]\otimes((ts)\alpha_{1}(r))\big{)}=0.$$

$$\sum _{\sigma \in S_3}sgn(\sigma )\sigma$$

$$\displaystyle-([[x,z],\alpha(y)]-[[x,y],\alpha(z)])\otimes(\alpha_{1}(r)(st))+[[z,y],\alpha(x)]\otimes((ts)\alpha_{1}(r))\big{)}=0.$$

$$({\alpha }_1(r)(st))+({\alpha }_1(r)(ts))-((rs){\alpha }_1(t))+({\alpha }_1(r)(st))+({\alpha }_1(r)(ts))-((rs){\alpha }_1(t))$$

$$2(({\alpha }_1(r)(st))+({\alpha }_1(r)(ts))-((rs){\alpha }_1(t)))=0$$

$$\ast :R\otimes R\to R,x\ast y=xy+yx.$$

$${\alpha }_1(x)\ast (y\ast z)$$

$$\displaystyle=\alpha_{1}(x)\big{(}(yz)+(zy)\big{)}+\big{(}(yz)+(zy)\big{)}\alpha_{1}(x)$$

$$\displaystyle=\alpha_{1}(x)\big{(}(yz)+(zy)\big{)}+(yz)\alpha_{1}(x)+(zy)\alpha_{1}(x)$$

$$\displaystyle=\alpha_{1}(x)\big{(}(yz)+(zy)\big{)}+\alpha_{1}(y)\big{(}(zx)+(xz)\big{)}+\alpha_{1}(z)\big{(}(yx)+(xy)\big{)}.$$

$$(x\ast y)\ast {\alpha }_1(z)$$

$$\displaystyle=\big{(}(xy)+(yx)\big{)}\alpha_{1}(z)+\alpha_{1}(z)\big{(}(xy)+(yx)\big{)}$$

$$\displaystyle=(xy)\alpha_{1}(z)+(yx)\alpha_{1}(z)+\alpha_{1}(z)\big{(}(xy)+(yx)\big{)}$$

$$\displaystyle=\alpha_{1}(x)\big{(}(yz)+(zy)\big{)}+\alpha_{1}(y)\big{(}(xz)+(zx)\big{)}+\alpha_{1}(z)\big{(}(xy)+(yx)\big{)}.$$

Full text

1. Introduction

The concept of Hom-Lie algebras was introduced by Hartwig, Larsson and Silvestrov in the study of algebraic structures describing some q-deformations of the Witt and the Virasoro algebras [4]. Hom-Leibniz algebras was introduced by Makhlouf and Silvestrov generalizing the notion of Leibniz algebras [7]. Hom-Leibniz algebra is obtained by twisting the Leibniz identity by a self linear map $\alpha$. A Hom-Leibniz algebra is same as a Leibniz algebra when $\alpha$ is the identity map.

There is a cup-product on the graded Hochschild cohomology $H^{\ast}(A,~{}A)$ of an associative algebra $A$ and with this cup-product $H^{\ast}(A,~{}A)$ carries a Gerstenhaber algebra structure [3]. A cup-product and Gerstenhaber algebra on Hom-associative algebras have been studied in [2]. Loday defined a cup-product on the graded cohomology groups of the Leibniz algebras [6]. The cup-product on the cohomology satisfies the graded Zinbiel algebra relation. A cup-product on the equivariant cohomology of Leibniz algebra equipped with an action of a finite group has been studied in [9].

Exploring the similarities and dissimilarities between the category of Leibniz algebras and the larger category of Hom-Leibniz algebras is an interesting theme in the literature. This paper contributes to the development of the non-associative algebra theory in this direction. The main purpose of this paper is to extend Loday’s result to Hom-Leibniz algebra case. We introduce cohomology for Hom-Leibniz algebra $(L,[.,.],\alpha)$ with coefficients in a Hom-associative, commutative algebra $(A,\mu,\alpha_{1})$ and define a cup-product on the graded cohomology $HL^{\ast}_{\alpha,\alpha_{1}}(L;~{}A)$ of the Hom-Leibniz algebra $L$

[TABLE]

which satisfies the following graded algebra relation:

[TABLE]

where $a\in HL^{n}_{\alpha,\alpha_{1}}(L;~{}A),~{}b\in HL^{m}_{\alpha,\alpha_{1}}(L;~{}A),~{}c\in HL^{r}_{\alpha,\alpha_{1}}(L;~{}A)$.

2. Preliminaries

In this section, we recall some definitions and basic facts of Hom-algebras and shuffle algebras. In this paper, $k$ always denotes any field.

Definition 2.1**.**

A Hom-associative algebra over $k$ is a triple $(A,\mu,\alpha_{1})$ consists of a $k$-vector space $A$ together with a $k$-bilinear map $\mu:A\times A\rightarrow A$ and a $k$-linear map $\alpha_{1}:A\rightarrow A$ satisfying $\alpha_{1}(\mu(x,y))=\mu(\alpha_{1}(x),\alpha_{1}(y))$ and

[TABLE]

Example 2.2**.**

Let $A$ be a two dimensional vector space with basis $\{a_{1},a_{2}\}$. Define a multiplication $\mu:A\times A\rightarrow A$ by

[TABLE]

A linear map $\alpha_{1}:A\rightarrow A$ is defined by $\alpha_{1}(a_{1})=a_{1}-a_{2}$ and $\alpha_{1}(a_{2})=0$. Then the triple $(A,\mu,\alpha_{1})$ defines a Hom-associative algebra structure.

Definition 2.3**.**

A Hom-Lie algebra is a triple $(A,\mu,\alpha)$ consisting of a linear space $A$, a skew-bilinear map $\mu:V\times V\rightarrow V$ and a linear space homomorphism $\alpha:A\rightarrow A$ satisfying the Hom-Jacobi identity

[TABLE]

for all $x,y,z$ in $A$, where $\circlearrowleft_{x,y,z}$ denotes summation over the cyclic permutations on $x,y,z$.

A Hom-Lie algebra whose endomorphism $\alpha$ is the identity is a Lie algebra.

Definition 2.4**.**

A Hom-Leibniz algebra is a $k$-linear vector space $L$ equipped with a $k$-bilinear map $[.,.]:L\times L\to L$ and a $k$-linear map $\alpha:L\to L$ satisfying the identity:

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

We denote a Hom-Leibniz algebra as $(L,[.,.],\alpha)$. A Hom-Leibniz algebra $(L,[.,.],\alpha)$ is called multiplicative if it satisfies $[\alpha(x),\alpha(y)]=\alpha([x,y])$.

A homomorphism between Hom-Leibniz algebras $(L_{1},[.,.]_{1},\alpha_{1})$ and $(L_{2},[.,.]_{2},\alpha_{2})$ is a $k$-linear map $\phi:L_{1}\to L_{2}$ which satisfies $\phi([x,y]_{1})=[\phi(x),\phi(y)]_{2}$ and $\phi\circ\alpha_{1}=\alpha_{2}\circ\phi$.

Example 2.5**.**

Any Hom-Lie algebra is a Hom-Leibniz algebra as in the presence of skew-symmetry Hom-Jacobi identity is same as Hom-Leibniz identity.

Example 2.6**.**

Given a Leibniz algebra $(L,[.,.])$ and a Leibniz algebra homomorphism $\alpha:L\to L$, one always get a Hom-Leibniz algebra $(L,[.,.]_{\alpha},\alpha)$, where $[x,y]_{\alpha}=[\alpha(x),\alpha(y)]$.

Example 2.7**.**

Let $(L,d)$ be a differential Lie algebra with the Lie bracket $[.,.]$ and $\alpha:L\to L$ is an endomorphism such that

1. (1)

$\alpha[x,y]=[\alpha(x),\alpha(y)],$** 2. (2)

$\alpha\circ d=d\circ\alpha.$**

Then $L$ is a Hom-Leibniz algebra with the new bracket defined as

[TABLE]

Example 2.8**.**

Let $L$ is a two-dimensional $\mathbb{C}$-vector space with basis $\{e_{1},e_{2}\}$. We define a bracket as $[e_{2},e_{2}]=e_{1}$ and zero else where and the endomorphism is given by the matrix

[TABLE]

It is easy to check that $(L,[.,.],\alpha)$ is a Hom-Leibniz algebra which is not Hom-Lie.

Definition 2.9**.**

A Hom-Zinbiel algebra is a triple $(R,(),\alpha_{1})$ consisting of a vector space $R$ together with two $k$-linear maps

[TABLE]

satisfying the relation

[TABLE]

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

For a graded vector space $R$, we can define a notion of a graded Hom-Zinbiel algebra, satisfying the following graded relation

[TABLE]

2.1. Shuffle algebra

Let $S_{p}$ be the permutation group of $p$ elements $1,\ldots,p.$ A permutation $\sigma\in S_{p}$ is called a $(n,m)$-shuffle if $n+m=p$ and

[TABLE]

In the group algebra $k[S_{p}],$ let $sh_{n,m}$ be the element

[TABLE]

where the summation is over all $(n,m)\mbox{-shuffles}.$

For any vector space $V$ we let $\sigma\in S_{p}$ act on $V^{\otimes p}$ by

[TABLE]

where the generator $v_{1}\otimes\cdots\otimes v_{p}$ of $V^{\otimes p}$ is denoted by $v_{1}\ldots v_{p}.$

Note that the linear map from $k[S_{p}]$ to itself induced by $\sigma\mapsto\mbox{sgn}(\sigma)\sigma^{-1}$ for $\sigma\in S_{p}$ is an anti-homomorphism. Let us denote the image of $\alpha\in k[S_{p}]$ under this map by $\tilde{\alpha}.$ For any non-negative integers $n$ and $m$ we define

[TABLE]

is given by

[TABLE]

where the above sum is over all $(n-1,m)$-shuffles $\sigma$.

For the generators $b=b_{1}\ldots b_{m},~{}c=c_{1}\ldots c_{r}$. Suppose $\tau_{m,r}$ is a permutation such that

[TABLE]

Remark 2.10**.**

For non-negative integers $n,~{}~{}m,~{}~{}r,$ under the anti-homomorphism $\alpha\to\tilde{\alpha}$ we have the following equality

[TABLE]

(See Proposition (1.8) of [6])

3. Some results on Hom-Leibniz and Hom-Zinbiel algebras

In this section, we prove some results regarding Hom-Leibniz and Hom-Zinbiel algebras along the same line as Leibniz and Zinbiel algebras.

Proposition 3.1**.**

Suppose $(L,[.,.],\alpha)$ is a Hom-Leibniz algebra and $(R,(),\alpha_{1})$ is a Hom-Zinbiel algebra. Then the tensor product $L\otimes R$ together with the bracket and $k$-linear map

[TABLE]

is a Hom-Lie algebra.

Proof.

Skew-symmetry part is trivial. Hom-Jacobi identity is same as the following relation

[TABLE]

here permutation group $S_{3}$ is acting same way on both sets $x,y,z$ and $r,s,t$. We have,

[TABLE]

Using Hom-Leibniz relation on $L$ this is same as

[TABLE]

Note that $S_{3}$ acting symmetrically on the sum. Thus, it is enough to prove that the coefficient of $([x,y],\alpha(z)]]$ is [math].

[TABLE]

using the Hom-Zinbiel relation. ∎

Proposition 3.2**.**

Let $(R,\alpha_{1})$ be a Hom-Zinbiel algebra. Then, $(R,\alpha_{1},\ast)$ is a Hom-associative and commutative algebra, where

[TABLE]

Proof.

Commutativity is trivial from the definition. We only need to check the Hom-associativity.

[TABLE]

[TABLE]

Thus, $(R,\alpha_{1},\ast)$ is a Hom-associative, commutative algebra. ∎

4. Cohomology of Hom-Leibniz algebras

In this section, we introduce a cohomology for Hom-Leibniz algebras with coefficients in Hom-associative and commutative algebras. To define cochain complex suitable for our work, we need to consider multiplicative Hom-Leibniz algebras. From now on by Hom-Leibniz algebra we always mean a multiplicative Hom-Leibniz algebra.

Suppose $(L,[.,.],\alpha)$ is a Hom-Leibniz algebra. Denote $CL_{n}(L)=L^{\otimes n}$ and define

[TABLE]

where

[TABLE]

Theorem 4.1**.**

For all $n\geq 1$, $d_{n-1}d_{n}=0$.

Proof.

Proof follows from the Theorem (3.1) of [1]. ∎

Thus, we have a chain complex

[TABLE]

The homology groups of this chain complex are called homology groups of the Hom-Leibniz algebra $(L,[.,.],\alpha)$ and $n$th homology group is denoted by $HL^{\alpha}_{n}(L)$. Note that for $\alpha=id$, $HL^{\alpha}_{n}(L)$ is same as the Leibniz homology of $L$ defined in [5]. Suppose $(L,[.,.],\alpha)$ is a Hom-Leibniz algebra and let $(A,\mu,\alpha_{1})$ be a Hom-associative and commutative algebra over $k$.

Set $CL_{\alpha,\alpha_{1}}^{n}(L;~{}A)=\{f:L^{\otimes n}\to A\mid\alpha_{1}\circ f=f\circ\alpha^{\otimes n}\}$, for all $n\geq 1$. We define

[TABLE]

by $\delta^{n}_{\alpha}(c)=c\circ d_{n+1},$ $c\in CL_{\alpha,\alpha_{1}}^{n}(L;~{}A),$ where $d_{n+1}:L^{\otimes(n+1)}\rightarrow L^{\otimes n}$ is the boundary map (7). Explicitly, for any $c\in CL_{\alpha,\alpha_{1}}^{n}(L;~{}A)$ and $(x_{1},\ldots,x_{n+1})\in L^{\otimes(n+1)},$ $\delta^{n}_{\alpha}(c)(x_{1},\ldots,x_{n+1})$ is given by the expression

[TABLE]

Note that $\delta^{n}_{\alpha}(c)$ need to satisfy the condition $\alpha_{1}\circ\delta^{n}_{\alpha}(c)=\delta^{n}_{\alpha}(c)\circ\alpha^{\otimes n}$ to be a member of $CL_{\alpha,\alpha_{1}}^{n+1}(L;~{}A)$. This is why we need to consider multiplicative Hom-Leibniz algebras to define cochain complex in this way.

Clearly, $\delta_{\alpha}^{2}=0$ as $d^{2}=0,$ and therefore, $(CL_{\alpha,\alpha_{1}}^{\sharp}(L;~{}A),\delta)$ is a cochain complex. Its homology groups are called Hom-Leibniz cohomology groups of $L$ with coefficients in $A$ and $n$th cohomology group is denoted by $HL_{\alpha,\alpha_{1}}^{n}(L;~{}A).$

5. Cup-product in Hom-Leibniz cohomology

Main aim of this section is to define a cup-product operation on the graded cohomology of the Hom-Leibniz algebra and show that the cup-product satisfies the graded Hom-Zinbiel relation.

Proposition 5.1**.**

Given a Hom-associative and commutative algebra $(A,\mu,\alpha_{1})$ there is a induced map

[TABLE]

for all $n\geq 1$.

Proof.

Using the twisting map $\alpha_{1}$ of $A$ we define a map

[TABLE]

Suppose $f\in CL_{\alpha,\alpha_{1}}^{n}(L;~{}A)$ is a cocycle. This implies $\delta^{n}_{\alpha}f=0$. We need to show $\delta^{n}_{\alpha}(\alpha_{1}\circ f)=0$. For all $(x_{1}\ldots,x_{n+1})$, we have,

[TABLE]

Now suppose $f=\delta^{n-1}_{\alpha}g$ for some $g\in CL_{\alpha,\alpha_{1}}^{n-1}(L;~{}A)$. Using the linearity of $\alpha_{1}$ we have $\alpha_{1}\circ f=\alpha_{1}\circ(\delta^{n-1}_{\alpha}g)=\delta^{n-1}_{\alpha}(\alpha_{1}\circ g)$.

Thus, under the image of $\underline{\alpha_{1}}$ cocycles goes to cocycles and coboundaries goes to coboundaries. So $\underline{\alpha_{1}}$ induces a map on the cohomology level

[TABLE]

Note that we have denoted the induced map on the cohomology level by the same notation as for the twisting map $\alpha_{1}$ of $A$. ∎

Lemma 5.2**.**

Given a Hom-Leibniz algebra $(L,[.,.],\alpha)$ and a Hom-associative, commutative algebra $(A,\mu,\alpha_{1})$, there is a bilinear operation on the graded space $CL_{\alpha,\alpha_{1}}^{\ast}(L;~{}A)$,

[TABLE]

defined as

[TABLE]

for $f\in CL_{\alpha}^{n}(L;~{}A)$ and $g\in CL_{\alpha}^{m}(L;~{}A)$ and the operation $\cup$ satisfies

[TABLE]

Proof.

To prove

[TABLE]

It is enough to show the following

[TABLE]

We define

[TABLE]

as \delta^{i,j}_{\alpha}(x_{1},\ldots,x_{p}):=\big{(}\alpha(x_{1}),\ldots,\alpha(x_{i-1}),[x_{i},x_{j}],\alpha(x_{i+1}),\ldots,\widehat{\alpha(x_{j})},\ldots,\alpha(x_{n})\big{)}. Note that, $d_{p}=\sum_{1\leq i<j\leq p}(-1)^{j}\delta^{i,j}_{\alpha}$.

Consider

[TABLE]

where $1\leq k<l\leq{n+1}$ and $\sigma$ is a $(n,m)$-shuffle acting on $\{2,\ldots,n+m+1\}$. This operator is a part of \big{(}(d_{n+1}\circ(\alpha^{m-1})^{\otimes{n+1}})\otimes(\alpha^{n})^{\otimes m}\big{)}\circ\rho_{n+1,m}.

As $\sigma$ is a $(n,m)$-shuffle, $\sigma(k)<\sigma(l)$. The equation (14) is same as the following equation

[TABLE]

for some permutation $\omega$. As $l\leq n+1$ and $\sigma$ is a $(n,m)$-shuffle, $\omega$ is a $(n-1,m)$-shuffle.

Thus, equation (15) is a part of \big{(}(\alpha^{m-1})^{\otimes n}\otimes(\alpha^{n-1})^{\otimes m}\big{)}\circ\rho_{n,m}\circ d_{n+m+1}.

Now consider,

[TABLE]

where $n+1\leq k<l\leq n+m+1$ and $\sigma$ is a $(n-1,m+1)$-shuffle acting on $\{2,\ldots,n+m+1\}$. This operator is a part of \big{(}(\alpha^{m})^{\otimes n}\otimes(d_{m+1}\circ(\alpha^{n-1})^{\otimes m+1})\big{)}\circ\rho_{n,m+1}. As $\sigma$ is a $(n-1,m+1)$-shuffle, $\sigma(k)<\sigma(l).$ The equation (16) is same as the following equation

[TABLE]

for some permutation $\omega$. As $k\geq n+1$ and $\sigma$ is a $(n,m)$-shuffle, $\omega$ is a $(n-1,m)$-shuffle.

Thus, equation (17) is a part of \big{(}(\alpha^{m-1})^{\otimes n}\otimes(\alpha^{n-1})^{\otimes m}\big{)}\circ\rho_{n,m}\circ d_{n+m+1}.

Using the same argument from [6], it is easy to check that both sides of (5) contains the same number of elements. We just showed that terms in the right hand side belongs to terms in the left hand side. This prove the validity of the equation (5). Therefore, the cup product is well-defined. ∎

Theorem 5.3**.**

Let $(L_{1},[.,.],\alpha)$ and $(L_{2},[.,.],\bar{\alpha})$ be two Hom-Leibniz algebras and $(A,\mu,\alpha_{1})$ be any Hom-associative, commutative algebra. The cup-product is functorial, that is, if $\phi:L_{1}\to L_{2}$ is a homomorphism then the induced map

[TABLE]

commutes with $\cup$, that is,

[TABLE]

Proof.

Let $f\in CL^{n}_{\bar{\alpha},\alpha_{1}}(L_{2};~{}A)$ and $g\in CL^{m}_{\bar{\alpha},\alpha_{1}}(L_{2};~{}A)$. To prove the theorem, it is sufficient to show

[TABLE]

on cochains.

[TABLE]

∎

Corollary 5.4**.**

If $\phi:L_{1}\to L_{2}$ is an isomorphism then the induced map on the cohomology level

[TABLE]

is also an isomorphism.

Remark 5.5**.**

If $f_{2}\cup f_{2}=0$, that is, $f$ is a square zero element for some $f_{2}\in HL^{n}_{\bar{\alpha},\alpha_{1}}(L_{2};~{}A)$ and $f_{1}\cup f_{1}\neq 0$ for any $f_{1}\in HL^{n}_{\alpha,\alpha_{1}}(L_{1};~{}A)$. Then we can conclude that $L_{1}$ and $L_{2}$ can not be ismorphic as Hom-Leibniz algebras. In this way the cup-product can be used to distinguish Hom-Leibniz algebras.

Now we show the cup-product satisfes the graded Hom-Zinbiel relation.

Theorem 5.6**.**

The operation $\cup$ induces a well defined bilinear product on the graded cohomology, called the cup-product,

[TABLE]

and the cup product $\cup$ together with the linear map

[TABLE]

satisfies the following graded Hom-Zinbiel formula

[TABLE]

where $a\in HL^{n}_{\alpha,\alpha_{1}}(L;~{}A),~{}b\in HL^{m}_{\alpha,\alpha_{1}}(L;A),~{}c\in HL^{r}_{\alpha,\alpha_{1}}(L;~{}A)$.

Proof.

Suppose $a,b,c$ is represented by cocycles $f,g,h$ respectively. From the formula (11) of Lemma 5.2 it is clear that if $f,g$ are cocycles then $f\cup g$ is also a cocycle and if either $f$ or $g$ is a coboundary then $f\cup g$ is a coboundary. Thus, the cup-product defined on the cochain level induces a cup-product on the cohomology level.

Since $A$ is commutative, we have,

[TABLE]

Where $\tau_{r,m}$ is a permutaion defined in (5). Now, pre-composing both sides of (6) by $\mu\circ((f(\alpha^{m+r-1})^{\otimes n}\otimes g(\alpha^{n+r-1})^{\otimes m})\otimes h(\alpha^{m+n-1})^{\otimes r})$

[TABLE]

This is same as

[TABLE]

This implies

[TABLE]

Thus,

[TABLE]

∎

6. A simple computation

We illustrate our theory by giving a simple computation which is based on the Example 2.8 with coefficients in a Hom-associative and commutative algebra given in the Example 2.2. Here we first recall Examples 2.8 and 2.2.

Let $L$ be a two dimensional $\mathbb{C}$-vector space with basis $\{e_{1},e_{2}\}$. Then $L$ is a Hom-Leibniz algebra with respect to the following bracket and endomorphisms:

[TABLE]

The endomorphism $\alpha$ is given by

[TABLE]

Let $A$ be a two dimensional vector space with basis $\{a_{1},a_{2}\}$. The multiplication $\mu$ of the Hom-associative and commutative algebra $A$ is defined as:

[TABLE]

The endomorphism $\alpha_{1}:A\rightarrow A$ is defined by $\alpha_{1}(a_{1})=a_{1}-a_{2}$ and $\alpha_{1}(a_{2})=0$.

We have the following chain complex:

[TABLE]

We write boundary maps $d_{n}$ for $n\leq 3$,

[TABLE]

To understand the boundary maps $d_{2},d_{3}$, it is enough to write down boundary maps on the basis elements. Here we write only non-zero cases.

[TABLE]

We define a map $f:L\to A$ as $f(e_{2})=a_{2}-a_{1},~{}~{}~{}f(e_{1})=0.$

We need to check that $\alpha_{1}\circ f=f\circ\alpha$.

[TABLE]

Thus, $f\in CL^{1}_{\alpha,\alpha_{1}}(L,~{}A)$. Now,

[TABLE]

Note that

[TABLE]

One can check easily from the definition of the cup-product that

[TABLE]

Thus,

[TABLE]

Remark 6.1**.**

It may be remarked that although cohomology operations are well understood in topology and these operations are useful in distingushing spaces, however these operations have not been explored well in the context of cohomology of algebras. Having a cup-product is the first step towards it. In a future project we intend to address this problem in general, and hence, cup-product in cohomology of Leibniz algebras or Hom Leibniz algebras are being studied at present. Further, any new algebraic structure on cohomology of a type of algebras helps in the classification problem of the algebras of a given type. This is one of the motivation behind introducing cup-product.

Acknowledgements: The author express his gratitute to the esteemed referee for her/his useful comments and suggestions on the earlier version of the manuscript that have improved the exposition. The author would like to thank Prof. Goutam Mukherjee of Indian Statistical Institute, Kolkata, for his guidance and reading a draft version of this article. The author is also thankful to Apurba Das and Surojit Ghosh for their help and useful comments on this article.