Hom-algebras were firstly studied by Hartwig, Larsson and Silvestrov in [4], where they introduced the structure of Hom-Lie algebras in the context of the deformations of Witt and Virasoro algebras. Determination of derivation algebras is an important task in Lie algbra, see [11]. Later, Larsson and Silvestrov extended the notion of Hom-Lie algebras to quasi-Hom Lie algebras and quasi-Lie algebras, see [5] and [6]. Wang et al. (see [10]) studied the structure of the generalized Hom-Lie algebras (i.e., the Hom-Lie algebras in Yetter-Drinfeld category $^{H}_{H}\mathcal{YD}$).

Let $H$ be a Hopf algebra, and $A$ be an $H$-module algebra. Cohen and Westreich [2] showed that if $H$ is quasitriangular and $A$ is quantum commutative with respect to $H$, then $A_0\subseteq Z(A)$. It is now a naive but natural question to ask whether we can obtain same results for the generalized Hom-Lie algebras that are analogous to [2]. This becomes our main motivation of the paper.

To give a positive answer to the question above, we organize this paper as follows.

In Section 2, we recall some basic definitions about Yetter-Drinfeld modules, (monoidal) Hom-Lie algebras and generalized Hom-Lie algebras. In Section 3, we discuss the central invariant of generalized Hom-Lie algebras (see Theorem 3.8).

In this section we recall some basic definitions and results related to our paper. Throughout the paper, all algebraic systems are supposed to be over a field ${k}$. The reader is referred to [1] as general references about monoidal Hom-algebras and monoidal Hom-Lie algebras, to [7] and [9] about Hopf algebras, and [3] about Yetter-Drinfeld categories. If $C$ is a coalgebra, we use the Sweedler-type notation for the comultiplication: $\Delta(c)=c_{1}\otimes c_{2}$ for all $c\in C.$

From now on we always assume that $H$ is a Hopf algebra with a bijective antipode $S$. The Yetter-Drinfeld category $^{H}_{H}\mathcal{YD}$ is a braided monoidal category whose objects $M$ are both left $H$-modules and left $H$-comodules, morphisms are both left $H$-linear and $H$-colinear maps and satisfy the compatibility condition

$ {{h}_{1}}{{m}_{(-1)}}\otimes {{\mathit{h}}_{2}}\cdot {{m}_{0}}={{({{h}_{1}}\cdot m)}_{(-1)}}{{h}_{2}}\otimes {{({{h}_{1}}\cdot m)}_{0}} $

or equivalently $ \rho(h\cdot m)=h_{1}m_{(-1)}S(h_{3})\otimes (h_{2}\cdot m_{0}), $ where the $H$-module action is denoted $h\cdot m$ and the $H$-comodule structure map is denoted by $\rho_{M}:M\rightarrow H\otimes M, \rho(m)=m_{(-1)}\otimes m_{0}$ for all $h\in H, m\in M$. The braiding $\tau$ is given by $ \tau(m\otimes n) = m_{(-1)}\cdot n \otimes m_{0} $ for all $m \in M, n \in N$, $M, N$ are objects in ${}^{H}_{H}\mathcal{YD}$.

Let $A$ be an object in $^{H}_{H}\mathcal{YD}$, the braiding $\tau$ is called symmetric on $A$ if the following condition holds, for any $a, b\in A, $ $ ((a_{(-1)}\cdot b)_{(-1)}\cdot a_{0})\otimes(a_{(-1)}\cdot b)_{0}=a\otimes b, $ which is equivalent to the following condition

$ \begin{eqnarray} a_{(-1)}\cdot b\otimes a_{0}=b_{0}\otimes S^{-1}(b_{(-1)})\cdot a. \end{eqnarray} $

(2.1)

Recall from [1] that a monoidal Hom-algebra is a triple $(A, \mu, \alpha)$ consisting of a linear space $A$, a $k$-linear map $\mu :A\otimes \mathit{A}\to A$ and a homomorphism $\alpha:A\rightarrow A$ for all $a, b, c\in A$, such that

$ \begin{eqnarray*} \alpha(ab)=\alpha(a)\alpha(b), ~\alpha(1_A)=1_A, ~\alpha(a)(bc)=(ab)\alpha(c), ~1_{A}a=a1_{A}=\alpha(a). \end{eqnarray*} $

Recall from [1] that a monoidal Hom-Lie algebra is a triple $(L, [, ], \alpha)$ consisting of a linear space $L$, a bilinear map $[, ]:L\otimes L\rightarrow L$ and a homomorphism $\alpha:L\rightarrow L$ satisfying

$ \begin{eqnarray*} && \alpha[l, l']=[\alpha(l), \alpha(l')], \\ &&[l, l']=-[l', l]~~({\rm Skew-symmetry}), \\ &&\circlearrowleft_{l, l', l''}[\alpha(l), [l', l'']]=0~~({\rm Hom-Jacobi~identity}) \end{eqnarray*} $

for all $l, l', l''\in L$, where $\circlearrowleft$ denotes the summation over the cyclic permutation on $l, l', l''.$

Let $H$ be a Hopf algebra. Recall from [10] that a generalized Hom-Lie algebra is a triple $(L, [, ], \alpha)$, which is a monoidal Hom-Lie algebra in a Yetter-Drinfeld category $^{H}_{H}\mathcal{YD}$, where $L$ is an object in $^{H}_{H}\mathcal{YD}$, $\alpha:L\rightarrow L$ is a homomorphism in $^{H}_{H}\mathcal{YD}$ and $[, ]:L\otimes L\rightarrow L$ is a morphism in $^{H}_{H}\mathcal{YD}$ satisfying

(1) $H$-Hom-skew-symmetry

$ \begin{eqnarray} [l, l']=-[l_{(-1)}\cdot l', l_{0}], ~l, l'\in L. \end{eqnarray} $

(2.2)

(2) $H$-Hom-Jacobi identity

$ \begin{eqnarray} \{l\otimes l'\otimes l''\}+\{(\tau\otimes 1)(1\otimes\tau)(l\otimes l'\otimes l'')\}+\{(1\otimes\tau)(\tau\otimes 1)(l\otimes l'\otimes l'')\}=0 \end{eqnarray} $

(2.3)

for all $l, l', l''\in L$, where $\{l\otimes l'\otimes l''\}$ denotes $[\alpha(l), [l', l'']]$ and $\tau$ the braiding for $L$.

In this section we always assume that the braiding $\tau$ is symmetric. We consider some $H$-analogous of classical concepts of ring theory and of Lie theory as follows.

Let $A$ be a monoidal Hom-algebra in $^{H}_{H}\mathcal{YD}$. An $H$-Hom-ideal $U$ of $A$ is not only $H$-stable but also $H$-costable such that $\alpha(U)\subseteq U$ and $(AU)A=A(UA)\subseteq U.$

Let $L$ be a generalized Hom-Lie algebra. An $H$-Hom-Lie ideal $U$ of $L$ is not only $H$-stable but also $H$-costable such that $\alpha(U)\subseteq U$ and $[U, L]\subseteq U.$

Define the center of $L$ to be $Z_{H}(L)=\{l\in L|[l, L]_{H}=0\}.$ It is easy to see that $Z_{H}(L)$ is not only $H$-stable but also $H$-costable.

$L$ is called $H$-prime if the product of any two non-zero $H$-Hom-ideals of $L$ is non-zero. It is called $H$-semiprime if it has no non-zero nilpotent $H$-Hom-ideals, and is called $H$-simple if it has no nontrivial $H$-Hom-ideals.

Definition 3.1  If $A$ is a monoidal Hom-algebra in $^{H}_{H}\mathcal{YD}$, the monoidal Hom-subalgebra of $H$-invariant is the set

$ A_0=\{a\in A|h\cdot a=\varepsilon(h)a, \alpha(a)=a\}. $

Example 3.2  Let $\{x_{1}, x_{2}, x_{3}\}$ be a basis of a 3-dimensional linear space $A$. The following multiplication $m$ and linear map $\alpha$ on $A$ define a monoidal Hom-algebra (see [8]):

$ \begin{eqnarray*} &&m(x_{1}, x_{1})=x_{1}, m(x_{1}, x_{2})=x_{2}, m(x_{1}, x_{3})=bx_{3}, \\ &&m(x_{2}, x_{1})=x_{2}, m(x_{2}, x_{2})=x_{2}, m(x_{2}, x_{3})=bx_{3}, \\ &&m(x_{3}, x_{1})=bx_{3}, m(x_{3}, x_{2})=0, m(x_{3}, x_{3})=0, \\ &&\alpha(x_{1})=x_{1}, \alpha(x_{2})=x_{2}, \alpha(x_{3})=bx_{3}, \end{eqnarray*} $

where $b$ is a parameter in $k$. Let $G$ be the cyclic group of order $2$ generated by $g$. The group algebra $H=kG$ is a Hopf algebra in the usual way,

$ \begin{eqnarray*} &&\rho(x_{1})=e\otimes x_{1}, \rho(x_{2})=e\otimes x_{2}, \rho(x_{3})=g\otimes x_{3}, \\ &&e\cdot x_{i}=x_{i}, g\cdot x_{1}=x_{1}, g\cdot x_{2}=x_{2}, g\cdot x_{3}=-x_{3}, i=1, 2, 3, \end{eqnarray*} $

where $e$ is the unit of the group $G$. It is not hard to check that $A$ is a monoidal Hom-algebra in $^{H}_{H}\mathcal{YD}$.

We assume that $A_0$ is a linear space spanned by $\{x_{1}, x_{2}\}$, then $A_0$ is the monoidal Hom-subalgebra of $H$-invariant.

From a monoidal Hom-algebra $(L, \alpha)$ in $^{H}_{H}\mathcal{YD}$, Wang et al. [10] gave a derived monoidal Hom-Lie algebra $(L, [, ], \alpha)$ in $^{H}_{H}\mathcal{YD}$ (that is a generalized Hom-Lie algebra) as follows for all $a, b\in L$,

$ \begin{eqnarray} [, ]:L\otimes L\rightarrow L, \ \ [a, b]=ab-(a_{(-1)}\cdot b)a_{0}. \end{eqnarray} $

(3.1)

In what follows, we always assume that the generalized Hom-Lie algebra means the above generalized Hom-Lie algebra.

The following lemma is referred to [10].

Lemma 3.3  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra. Then

(1) $[\alpha(a), bc]=[a, b]\alpha(c)+(a_{-1}\cdot\alpha(b))[a_{0}, c]$;

(2) $[ab, \alpha(c)]=\alpha(a)[b, c]+[a, b_{(-1)}\cdot c]\alpha(b_{0})$;

(3) $[ab, \alpha(c)]=[\alpha(a), bc]+[a_{(-1)}\cdot\alpha(b), (a_{0(-1)}\cdot c)a_{00}]$ for all $a, b, c\in L.$

Define $ad_x(l)=[x, l]$ for all $x, l\in L$. By Lemma 3.3 (1), we have

$ ad_{\alpha(x)}(lm)=ad_x(l)\alpha(m)+(x_{(-1)}\cdot \alpha(l))ad_{x_0}(m). $

Lemma 3.4  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra, and let $x\in L_0$. Then

(1) $\tau_{L, L}(x\otimes y)=y\otimes x$, $\tau_{L, L}(y\otimes x)=x\otimes y$;

(2) $ad_x(y)=xy-yx$;

(3) $ad_x(yz)=ad_x(y)\alpha(z)+\alpha(y)ad_x(z)$;

(4) $ad^2_x(yz)=ad^2_x(y)\alpha^2(z)+2\alpha(ad_x(y)ad_x(z))+\alpha^2(y)ad^2_x(z)$, for all $y, z\in L$.

Lemma 3.5  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra. Assume that $L$ is $H$-simple. Then $Z_H(L)_0$ is a field.

Proof  Note that ${{\mathit{Z}}_{\mathit{H}}}{{\rm{(}\mathit{L}\rm{)}}_{\rm{0}}}\rm{=}{{\mathit{Z}}_{\mathit{H}}}\rm{(}\mathit{L}\rm{)}\cap {{\mathit{L}}_{\rm{0}}}\rm{=}\mathit{Z}\rm{(}\mathit{L}\rm{)}\cap {{\mathit{L}}_{\rm{0}}}\rm{=}\mathit{Z}{{\rm{(}\mathit{L}\rm{)}}_{\rm{0}}}$, where $Z(L)$ is the usual center of $L$. Taking $0\neq x\in Z_H(L)_0$, we have that $Lx=I\neq 0$ is an $H$-Hom-ideal, thus $I=L$. That is to say that for some $y\in L$, we obtain $xy=yx=1$. Since

$ \begin{eqnarray*} \alpha^2(h\cdot y)&=&\alpha(h\cdot y)1=\alpha(h\cdot y)(xy)= \alpha(h_1\cdot y)(\varepsilon(h_2)xy)\\ &=& \alpha(h_1\cdot y)((h_2\cdot x)y)=(h\cdot (yx))\alpha(y)\\ &=&(h\cdot 1)\alpha(y)=\varepsilon(h)1\alpha(y)=\varepsilon(h)\alpha^2(y). \end{eqnarray*} $

We can get $h\cdot y=\varepsilon(h)y$, that is, $y\in L_0$.

We need to show $y\in Z_H(L)$. For any $z\in L$, by Lemma 3.4 (1), $[z, x]=zx-xz=0$. Thus $yz-zy=0$, i.e., $[y, z]=yz-zy=0$ by Lemma 3.4 (2). This shows that $y\in Z_H(L)$.

Lemma 3.6  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra, and let $x\in L_0$, $l, m\in L$. Then

(1) $ad^2_x(xl)=\alpha^2(x)ad^2_x(l)$;

(2) if $ad^2_x(L)=0$ and char $(k)\neq 2$, then $ad_x(l)(Lad_x(m))=0$.

Proof  (1) It is easy to show that (1) holds by Lemma 3.4 (4).

(2) For all $l, m\in L$, we have

$ \begin{eqnarray*} 0&=&ad^2_x(lm)=ad^2_x(l)\alpha^2(m)+2\alpha(ad_x(l)ad_x(m))+\alpha^2(l)ad^2_x(m)\\ &=&2ad_x(\alpha(l))ad_x(\alpha(m)), \end{eqnarray*} $

and so $ad_x(l)ad_x(m)=0$ since char $(k)\neq 0$. Thus by Lemma 3.4 (3), one gets

$ ad_x(l)(Lad_x(m))=0 $

for all $l, m\in L$.

Lemma 3.7  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra. If $L$ is $H$-simple with char $(k)\neq 2$, assume that $I$ is an $H$-Hom-Lie ideal of $[L, L]$. Let $x\in I_0$ satisfying

(ⅰ) $ad_x(I)=0$;

(ⅱ) $ad^2_x([L, L])=0$.

Thus $x\in Z_H(L)$.

Proof  Let $x\in I_0$. For any $m\in L$, $l\in [L, L]$ and $y\in I$. By Lemma 3.3 (1),

$ \begin{eqnarray}\label{a} 0=ad^2_x([\alpha(l), my])=ad^2_x([l, m]\alpha(y))+ad^2_x((l_{(-1)}\cdot\alpha(m))[l_0, y]). \end{eqnarray} $

(3.2)

First, we have

$ \begin{eqnarray*} && ad^2_x([l, m]\alpha(y))\\ &=& ad^2_x([l, m])\alpha^3(y)+2\alpha(ad_x([l, m])ad_x(\alpha(y)))+\alpha^2([l, m])ad^2_x(\alpha(y))\\ &\stackrel{(i)}{=}& ad^2_x([l, m])\alpha^3(y) \stackrel{(ii)}{=}0. \end{eqnarray*} $

Hence

$ \begin{eqnarray}\label{b} ad^2_x([l, m]\alpha(y))=0. \end{eqnarray} $

(3.3)

Similarly,

$ \begin{eqnarray*} && ad^2_x((l_{(-1)}\cdot\alpha(m))[l_0, y])\\ & =&ad^2_x(l_{(-1)}\cdot\alpha(m))\alpha^2([l_0, y]) + 2\alpha(ad_x(l_{(-1)}\cdot\alpha(m))ad_x([l_0, y]))\\ &&+ \alpha^2(l_{(-1)}\cdot\alpha(m))ad^2_x([l_0, y]). \end{eqnarray*} $

Since $l\in [L, L]$ and $[, ]$ is $H$-colinear, $l_0\in [L, L]$, $ad_x([l_0, y])\stackrel{(i)}{=}0$ and $ad^2_x([l_0, y])\stackrel{(ii)}{=}0$. Hence

$ \begin{eqnarray}\label{c} ad^2_x((l_{(-1)}\cdot\alpha(m))[l_0, y])= ad^2_x(l_{(-1)}\cdot\alpha(m))\alpha^2([l_0, y]). \end{eqnarray} $

(3.4)

Substituting (3.3) and (3.4) into (3.2), we obtain

$ \begin{eqnarray}\label{d} ad^2_x(l_{(-1)}\cdot\alpha(m))\alpha^2([l_0, y])=0 \end{eqnarray} $

(3.5)

for all $y\in I$, $l\in [L, L]$, $m\in L$. We now consider two cases.

(1) If $[I, [L, L]]=0$, then we have $ad^2_x(L)=0$. By Lemma 3.6 (2), $ad_x(l)(Lad_x(m))=0$. Since $L$ is $H$-simple, we get $ad_x(L)=0$, $\forall\ l\in L$, and hence $x\in Z_H(L)$.

(2) Now assume $U=[I, [L, L]]\neq 0$. It is easy to see that $U$ is an $H$-Hom-Lie ideal of $[L, L]$. By (3.5) we have $ad^2_x(L)U=0$. Let $Q=\{y\in L|yU=0\}$, then $Q$ is an $H$-stable $H$-costable left ideal of $L$, we claim $Q=0$. If not, then $L=QL$ since $L$ is $H$-simple. By (2.1) we have

$ QL\subseteq [Q, L]+LQ\subseteq [Q, L]+Q. $

Thus $L=Q+[Q, L]$. Let $y\in Q$, $l\in [L, L]$ and $u\in U$. Then

$ [y, l] u=ylu=y[l, u]. $

Since $[l, u]\in U$, $y[l, u]=0$, and thus $[Q, [L, L]]\subseteq Q$ and $Q[L, L]\subseteq Q$. Hence

$ L=QL=Q(Q+[Q, L])\subseteq Q. $

This implies $LU=0$, which contradicts the assumption $U\neq 0$. Hence, $Q=0$, and so $ad^2_x(L)=0$. Similarly to case (1), one gets $x\in Z_H(L)$.

Theorem 3.8  Let $(L, [, ], \alpha)$ be a generalized Hom-Lie algebra. Let $L$ be $H$-simple with char $(k)\neq 2$, and assume that $V$ is an $H$-Hom-Lie ideal of $[L, L]$ such that $[V_0, V]\subseteq Z_H(L)_0$. Then $V_0\subseteq Z_H(L)_0$.

Proof  Let $V$ be an $H$-Hom-Lie ideal of $[L, L]$ such that $[V_0, V]\subseteq Z_H(L)_0$. Let $x\in V_0$. We consider the following two cases:

(1) $ad_x(V)=0$, which implies that $x\in Z_H(L)_0$ by Lemma 3.7.

(2) $ad_x(V)\neq 0$. For $v\in V$, we have

$ \begin{eqnarray*} &&[[x, [x, L]], \alpha(v)]\\ & \stackrel{(2.2)}{=}&-[[x, [x, L]]_{(-1)}\cdot \alpha(v), [x, [x, L]]_0]\\ & \stackrel{(2.1)}{=}&-[\alpha(v_0), [x, [x, S^{-1}(v_{(-1)})\cdot L]]]\\ & \stackrel{(2.3)}{=}&[[x, v_{0(-1)}S^{-1}(v_{(-1)})\cdot \alpha(L)], [\alpha(v_{00}), x]]+[x, [[x, L], v]]\\ &=& [[x, \alpha(L)], [\alpha(v), x]]+[x, [[x, L], v]]\\ &\subseteq&[[x, L], [V, x]]+[x, [[x, L], v]]\\ &\subseteq&0+[x, [[L, L], v]]\subseteq [x, V]\subseteq Z_H(L)_0. \end{eqnarray*} $

We obtain $[ad^2_x(L), V]\subseteq Z_H(L)_0$. By Lemma 3.6 (1), we have $ad^2_x(xl)=\alpha^2(x)ad^2_x(l)$. If $ad^2_x(l)\neq 0$ for some $l\in L$, then $(ad^2_x(l))^{-1}\in Z_H(L)_0$ by Lemma 3.5. In this case, it is easy to see that $x\in Z_H(L)_0$. Now we assume $ad_{x}^{2}(L)\varsubsetneq {{Z}_{H}}{{(L)}_{0}}$. Let $y\in L$ with $ad^2_x(y)\notin Z_H(L)_0$. Then we choose $z\in V$ such that $0 \neq ad_z(x)=u\in Z_H(L)_0$. Thus there exist $v_1, v_2, v_3\in Z_H(L)_0$ such that $[z, ad^2_x(y)]=v_1$, $[\alpha(z), ad^2_x(xy)]=v_2$ and $[\alpha^2(z), ad^2_x(x^2y)]=v_3$. Now we have

$ \begin{eqnarray*} v_2 &=& [\alpha(z), ad^2_x(xy)]=[\alpha(z), xad^2_x(y)]\\ &=& [z, x]\alpha(ad^2_x(y))+x[z, ad^2_x(y)]=u\alpha(ad^2_x(y))+xv_1. \end{eqnarray*} $

By Lemma 3.5, $u$ is invertible. Thus $ ad^2_x(y)=\alpha^{-1}(u^{-1}v_2-u^{-1}(xv_1)). $ However, by $v_1\in Z_H(L)$, $x\in V_0$ and Lemma 3.4 (1), we have $ xv_1=v_1x, $ and so $ ad^2_x(y)=\alpha^{-1}(u^{-1}v_2-u^{-1}(v_1x)). $ Similarly, we have

$ \begin{eqnarray*} v_3 &=& [\alpha^2(z), ad^2_x(x^2y)]=[\alpha(\alpha(z)), xad^2_x(xy)]\\ &=& [\alpha(z), x]\alpha(ad^2_x(xy))+x[\alpha(z), \alpha(ad^2_x(xy))]\\ &=& [\alpha(z), \alpha(x)]\alpha(ad^2_x(xy))+x[\alpha(z), \alpha(ad^2_x(xy))]\\ &=& \alpha(u)\alpha(ad^2_x(xy))+xv_2\\ &=& u\alpha(ad^2_x(xy))+xv_2, \end{eqnarray*} $

and thus $ ad^2_x(xy)=\alpha^{-1}(u^{-1}v_3-u^{-1}(v_2x)). $ Using Lemma 3.6 (1), we have

$ \begin{eqnarray*} ad^2_x(xy) &=& xad^2_x(y)=\alpha^{-1}(\alpha(x)(u^{-1}v_2)-\alpha(x)(u^{-1}(v_1x)))\\ &=& \alpha^{-1}((xu^{-1})\alpha(v_2)-(xu^{-1})\alpha(v_1x))=\alpha^{-1}(u^{-1}(v_2x)-u^{-1}(v_1x^2)). \end{eqnarray*} $

Hence $v_1x^2-2v_2x+v_3=0$, that is, $x^2+\theta^1 x+ \theta^0=0$, where $\theta^1=-2v_2/v_1$, $\theta^0=v_3/v_1$, and $\theta^1, \theta^0\in Z_H(L)$. And so by Lemma 3.3 (2) and Lemma 3.4 (1) we have

$ \begin{eqnarray*} 0 &=&[-\theta^0, \alpha(z)]=[x^2, \alpha(z)]+[\theta^1 x, \alpha(z)]\\ &=& \alpha([x^2, z])+\alpha(\theta^1)[x, z]+[\theta^1, x_{(-1)}\cdot z]\alpha(x_0)\\ &=& \alpha([x^2, z])+\alpha(\theta^1)[x, z]. \end{eqnarray*} $

By Lemma 3.4 (1), one has $\alpha([x^2, z])=-\alpha(\theta^1)[x, z]=\alpha(\theta^1)u, $ and similarly we have

$ \alpha([x^2, z])=\alpha(x[x, z]+[x, z]x)=2\alpha([x, z]x)=-2\alpha(ux)=-2ux. $

Since $u\in Z_H(L)_0$, $\alpha(\theta^1)=-2x$. Since char $(k)\neq 2$, $x=-(1/2)\theta^1\in Z_H(L)$.