On the universal $\alpha$-central extensions of the semi-direct product of Hom-preLie algebras

1.   Introduction

A Hom-preLie algebra was introduced by Makhlouf-Silvestrov [11]. Specifically, for a vector space $L$ over a field $\mathbb{K}$ equipped with a bilinear map $\mu :L\times L\rightarrow L$ and a linear map $\alpha:L\rightarrow L$, we say that the triple $(L,\mu,\alpha)$ is a Hom-preLie algebra if

for all $x,y,z\in L$. If the elements of $L$ also satisfy the following equation

Then we call $(L,\mu,\alpha)$ is a Hom-Novikov algebra. Clearly, a Hom-Novikov algebra is a Hom-preLie algebra. Moreover, Hom-preLie algebras generalizes the notation of pre-Lie algebras ($\alpha = {{\rm{Id}}}_{L}$), which has been extensively studied in the construction and Classification of Hom-Novikov algebras (Yau [14])etc. Since Hom-preLie algebras are a kind of Hom-Lie admissible algebras, there are some close connections between Hom-preLie algebra and Hom-Lie algebra theories. For example, a derivation of a Hom-preLie algebra with respect to a Hom-representation is an $\alpha$-derivation which is introduced in[12].

In recent year, the universal central extension of a perfect Leibniz algebra was studied in several articles[2,6,3,1,8,9,10]. In [4,5,7], authors study universal ($\alpha$)-central extension.

In [13], we study universal $\alpha$-central extensions of Hom-preLie algebras. We define Hom-co-representations and low-dimensional chain complex, which derive a low-dimensional homology $\mathbb{K}$-vector space of a Hom-preLie algebra. We construct a right exact covariant functor ${\frak{uce}}_{\alpha}$ of a Hom-preLie algebra which acts on a $\alpha$-perfect Hom-preLie algebra $L$ its universal $\alpha$-central extension ${\frak{uce}}_\alpha(L) = \frac{\alpha_L(L)\otimes \alpha_L(L)}{I_L}$, where $I_L = \langle\alpha_L(x_1)\otimes x_2x_3- x_1x_2\otimes \alpha_L(x_3)-\alpha_L(x_2)\otimes x_1x_3+ x_2x_1\otimes \alpha_L(x_3)\rangle.$

The purpose of this paper is to study the universal $\alpha$-central extension of semi-direct product of two $\alpha$-perfect Hom-preLie algebras. We introduce a Hom-action between two perfect Hom-preLie algebras $(Q, {{\rm{Id}}}_Q)$ and $(M,\alpha_M)$, giving a semi-direct product between two perfect Hom-preLie algebras. We use an associative Hom-action of $(Q, {{\rm{Id}}}_Q)$ on $(M,\alpha_M)$ to induce a Hom-action of $({\frak{uce}}(Q), {{\rm{Id}}}_{{\frak{uce}}(Q)})$ on $({\frak{uce}}_{\alpha}(M),\overline{\alpha_M})$. We obtain semi-direct product $({\frak{uce}}_{\alpha}(M),\overline{\alpha_M})\rtimes ({\frak{uce}}(Q), {{\rm{Id}}}_{{\frak{uce}}(Q)})$ and define a linear map $\tau\rtimes \sigma$ on the semi-direct product. Casas and Pacheco Rego gave the linear map $\tau\rtimes \sigma$ is a homomorphism of Hom-leibniz algebras in [5]. We add a condition that Hom-action of $(Q, {{\rm{Id}}} _Q)$ on $(M,\alpha_M)$ is ${\frak{uce}}$-associative, that is, $\{mm',q\} = \{m,m'\centerdot q\}$ $\forall m,m'\in M, q\in Q$. We also obtain a linear map $\tau\rtimes \sigma$ is a homomorphism of Hom-preLie algebras. We give a couple of necessary and sufficient conditions for the universal $\alpha$-central extension of semi-direct product of two $\alpha$-perfect Hom-preLie algebras by the above results.

The paper is organized as follows. Section 2 a preliminary section which contains Hom-actions and semidirect product of Hom-preLie algebras. We describe the relation between the semi-direct product extension and split extensions of Hom-preLie algebras. In section 3 we analyzing the functorial properties of the universal ($\alpha$)-central extensions of ($\alpha$)-perfect Hom-preLie algebras. In section 4 we obtain that an automorphism or a derivation can be lifted in an $\alpha$-cover with certain constraints. In the final section we give some necessary and sufficient conditions about the universal $\alpha$-central extension of the semi-direct product of two $\alpha$-perfect Hom-preLie algebras.

Throughout this paper $\mathbb{K}$ denotes an arbitrary field.

2.   Hom-action

Definition 2.1. Let $(M, \alpha_M)$ and $(L,\alpha_L)$ be Hom-preLie algebras. A Hom-action of $(L,\alpha_L)$ over $(M, \alpha_M)$ consists of two bilinear maps, $\rho: M \otimes L \to M,$ $\rho(m\otimes l) = m\centerdot l$ and $\lambda: L \otimes M \to M,$ $\lambda(l\otimes m) = l\centerdot m$, the following identities hold.

$\rm a)$ $(xy) \centerdot \alpha_M(m) - \alpha_L(x) \centerdot (y \centerdot m) = (yx) \centerdot \alpha_M(m)-\alpha_L(y) \centerdot (x \centerdot m),$

$\rm b)$ $(m\centerdot x)\centerdot \alpha_L(y)-\alpha_M(m)\centerdot (xy) = (x\centerdot m)\centerdot \alpha_L(y)-\alpha_L(x)\centerdot (m\centerdot y),$

$\rm c)$ $(mm')\centerdot \alpha_L(x)-\alpha_M(m)(m'\centerdot x) = (m'm)\centerdot \alpha_L(x)-\alpha_M(m')(m\centerdot x),$

$\rm d)$$(x\centerdot m)\alpha_M(m')-\alpha_L(x)\centerdot (mm') = (m\centerdot x)\alpha_M(m')-\alpha_M(m) (x\centerdot m'),$

$\rm e)$ $\alpha_M(x\centerdot m) = \alpha_L(x) \centerdot \alpha_M(m),$

$\rm f)$ $\alpha_M(m\centerdot x) = \alpha_M(m) \centerdot \alpha_L(x),$

for all $x, y \in L$ and $m, m' \in M$.

If $(M, \alpha_M)$ is an abelian Hom-preLie algebra, then the Hom-action is said to be a Hom-representation.

Example 1. a) Let $(K, \alpha_K)$ be a subalgebra of a Hom-preLie algebra $(L, \alpha_L)$ and $(H, \alpha_H)$ a Hom-ideal of $(L, \alpha_L)$. There is a Hom-action of $(K, \alpha_K)$ over $(H, \alpha_H)$ by the multiplication in $(L, \alpha_L)$.

b) Let $0 \to (M,\alpha_M) \stackrel{i}\to (K,\alpha_K) \stackrel{\pi}\to (L,\alpha_L) \to 0$ is an exact sequence of Hom-Lie algebras. If $(M,\alpha_M)$ is an abelian Hom-preLie algebra, then we call the sequence is abelian. An abelian sequence gives a Hom-representation of $(L, \alpha_L)$ over $(M, \alpha_M)$ by defining $\rho : M \otimes L \to M, \rho(m,l) = mk, \pi(k) = l$, $\lambda : L \otimes M \to M, \lambda(l,m) = km, \pi(k) = l$.

Proposition 1. Let $(M, \alpha_M)$ and $(L,\alpha_L)$ be Hom-preLie algebras with a Hom-action of $(L,\alpha_L)$ over $(M, \alpha_M)$. Then $(M\rtimes L,\tilde{\alpha})$ is a Hom-preLie algebra, where $\tilde{\alpha}:M\rtimes L \to M\rtimes L$ is defined by $\tilde{\alpha}(m,l) = (\alpha_M(m),\alpha_L(l))$ and multiplication

Proof. It follows by the direct computation.

Definition 2.2. [13] A short exact sequence of Hom-preLie algebras $(K) : 0 \to (M, \alpha_M) \stackrel{i} \to (K,\alpha_K) \stackrel{\pi} \to (L, \alpha_L) \to 0$ is said to be split if there exists a Hom-preLie algebra homomorphism $\sigma : (L, \alpha_L) \to (K, \alpha_K)$ such that $\pi \circ \sigma = { {{\rm{Id}}}}_L$.

Let $(M, \alpha_M)$ and $(L,\alpha_L)$ be Hom-preLie algebras with a Hom-action of $(L,\alpha_L)$ over $(M, \alpha_M)$. We define two linear maps $i:M \to M\rtimes L,i(m) = (m,0)$ and $\pi:M\rtimes L \to L, \pi(m,l) = l$. Then we obtain the following sequence

Furthermore, this sequence splits by $\sigma: L \to M\rtimes L, \sigma(l) = (0,l)$.

Definition 2.3. Let $(M, \alpha_M)$ and $(L,\alpha_L)$ be Hom-preLie algebras with a Hom-action of $(L,\alpha_L)$ over $(M, \alpha_M)$. Two extensions of $(L,\alpha_L)$ by $(M, \alpha_M)$, $0 \to (M,\alpha_M) \stackrel{i}\to (K,\alpha_K) \stackrel{\pi}\to (L,\alpha_L) \to 0$ and $0 \to (M,\alpha_M) \stackrel{i'}\to (K',\alpha_{K'}) \stackrel{\pi'}\to (L,\alpha_L) \to 0$, are equivalent if there is a homomorphism of Hom-preLie algebra $\varphi: (K,\alpha_K)\to (K',\alpha_K')$ satisfies that the following commutative diagram

Lemma 2.4. Let $(C,{ {{\rm{Id}}}}_C)$ and $(A, \alpha_A)$ be Hom-preLie algebras with a Hom-action of $(C,{ {{\rm{Id}}}}_C)$ over $(A, \alpha_A)$. A sequence of Hom-preLie algebras $0 \to (A,\alpha_A) \stackrel{i}\to (B,\alpha_B) \stackrel{\pi}\to (C,{ {{\rm{Id}}}}_C) \to 0$ is split if and only if it is equivalent to the semi-direct sequence $0 \to (A,\alpha_A) \stackrel{j}\to (A\rtimes C,\tilde{\alpha}) \stackrel{p}\to (C,{ {{\rm{Id}}}}_C) \to 0$.

Proof. If $0 \to (A,\alpha_A) \stackrel{i}\to (B,\alpha_B) \stackrel{\pi}\to (C,{ {{\rm{Id}}}}_C) \to 0$ is split by $t:(C,{ {{\rm{Id}}}}_C)\to (B,\alpha_B),$ then the Hom-action of $(C,{ {{\rm{Id}}}}_C)$ over $(A,\alpha_A)$ is defined by

So we obtain the following split extension:

where $k:A\to A\rtimes C,k(a) = (a,0),p:A\rtimes C \to C,q(a,c) = c$ and $\tau:C\to A\rtimes C,\tau(c) = (0,c)$. Furthermore the Hom-action of $(C,{ {{\rm{Id}}}}_C)$ over $(A,\alpha_A)$ induced by this extension coincides with the initial one:

Since $\varphi: (A\rtimes C,\tilde{\alpha}) \to (B,\alpha_B),\varphi(a,c) = i(a)+s(c)$ is a homomorphism of Hom-preLie algebras such that the following diagram commutative

the extensions are equivalent.

Suppose that two extensions are equivalent, that is, there is a homomorphism of Hom-preLie algebras $\varphi: (A\rtimes C,\tilde{\alpha}) \to (B,\alpha_B)$ such that diagram (1) is commutative, then $t: (C, { {{\rm{Id}}}}_C) \to (B, \alpha_B)$ given by $t(c) = \varphi(0, c)$, is a split extension.

Definition 2.5. [12] Let $(M, \alpha_M)$ be a Hom-representation of a Hom-preLie algebra $(L, \alpha_L)$. A derivation of $(L, \alpha_L)$ over $(M, \alpha_M)$ is a $\mathbb{K}$-linear map $d : L \to M$ such that:

$\rm a)$ $d(l_1l_2) = \alpha_L(l_1)\centerdot d(l_2)+d(l_1)\centerdot \alpha_L(l_2),$

$\rm b)$ $d\circ \alpha_L = \alpha_M\circ d,$

for all $l_1,l_2\in L$.

Example 2. a) Let $(M, \alpha_M)$ be a Hom-representation of $(M \rtimes L,\tilde{ \alpha})$ via $\pi$. Then the linear map $\theta: M \rtimes L \to M, \theta(m, l) = m$, is a derivation.

b) When $(M, \alpha_M) = (L, \alpha_L)$ is a representation follows from Example 1 $a)$, then a derivation consists of a $\mathbb{K}$-linear map $d : L\to L$ such that $d(l_1l_2) = \alpha_L(l_1) d(l_2)+d(l_1) \alpha_L(l_2)$ and $d\circ \alpha_L = \alpha_M\circ d$.

Proposition 2. Let $(M, \alpha_M)$ be a Hom-representation of a Hom-preLie algebra $(L, \alpha_L)$. For every $f$-derivation $d : (X, \alpha_X) \to (M, \alpha_M)$ $(d(x_1x_2) = d(x_1)\centerdot\alpha_L(f(x_2))$ $+\alpha_L(f(x_1))\centerdot d(x_2)\ \forall x_1,x_2\in X)$ and every homomorphism of Hom-preLie algebras $f : (X, \alpha_X) \to (L, \alpha_L)$ there is a unique homomorphism of Hom-preLie algebras $h : (X, \alpha_X) \to (M \rtimes L, \tilde{\alpha})$, such that the following diagram commute.

Conversely, every homomorphism of Hom-preLie algebras $h : (X, \alpha_X) \to (M \rtimes L, \tilde{\alpha})$, decide a homomorphism of Hom-preLie algebras $f = \pi \circ h : (X, \alpha_X) \to (L, \alpha_L)$ and $f$-derivation $d = \theta \circ h : (X, \alpha_X) \to (M, \alpha_M)$, where $\theta(m,l) = x,\forall m\in M,l \in L.$

Proof. Let $h : X \to M \rtimes L, h(x) = (d (x), f (x))$ be a homomorphism. Then the homomorphism $h$ satisfies all the conditions.

Corollary 1. The set of all derivations from $(L, \alpha_L)$ to $(M, \alpha_M)$ is in one-to-one correspondence with the set of Hom-preLie algebra homomorphisms $h : (L, \alpha_L) \to (M \rtimes L, \tilde{\alpha})$, such that $\iota\pi \circ h = { {{\rm{Id}}}}_L$.

Proof. Take $(X, \alpha_X) = (L, \alpha_L)$ in Proposition 2.

3.   Functorial properties

Definition 3.1. Let $(L, \alpha_L)$ be a perfect Hom-preLie algebra. It is said to be centrally closed if its universal central extension is

i.e., $HL_2^\alpha(L) = 0$ and $({\frak{uce}}(L),\tilde{\alpha})\cong (L,\alpha_L).$

Corollary 2. Let $(L, \alpha_L)$ be a $\alpha$-perfect Hom-preLie algebra. If $0 \to ({ {\rm{Ker}}}(U_\alpha), \alpha_{K_{\mid}})$ $\to (K, \alpha_K) \stackrel{U_\alpha}\to (L,\alpha_L) \to 0$ is the universal $\alpha$-central extension of $(L,\alpha_L)$, then $(L,\alpha_L)$ is centrally closed.

Proof. $HL_1^\alpha(K) = HL_2^\alpha(K) = 0$ thanks to Corollary 4.12 a) in [13]. By the proof of Corollary 4.12 b) in [13], $HL_1^\alpha(K) = 0$ if and only if $(K,\alpha_K)$ is perfect. By Theorem 4.11 c) in [13], there exists a universal central extension $0 \to (HL_2^{\alpha}(K), \widetilde{\alpha}_{\mid}) \to ({\frak{uce}}(K), \widetilde{\alpha}) \stackrel{u_K}\to (K,\alpha_K) \to 0$. Since $HL_2^\alpha(K) = 0$, $u_K$ is an isomorphism.

Definition 3.2. A Hom-preLie algebra $(L,\alpha_L)$ is said to be simply connected if every central extension $\tau:(F,\alpha_F)\twoheadrightarrow (L,\alpha_L)$ splits uniquely as the product of Hom-preLie algebras $(F,\alpha_F) = ({ {\rm{Ker}} (\tau)}, \alpha_{F\mid})\times(L, \alpha_L).$

Proposition 3. Let $(L,\alpha_L)$ be a perfect Hom-preLie algebra. Then the following conditions are equivalent:

$a)$ $(L,\alpha_L)$ is simply connected.

$b)$ $(L,\alpha_L)$ is centrally closed.

Proof. $a)\Rightarrow b)$ Let $0 \to (HL_2^{\alpha}(L), \widetilde{\alpha}_{\mid}) \to ({\frak{uce}}(L), \widetilde{\alpha}) \stackrel{u_L}\to (L,\alpha_L) \to 0$ be the universal central extension of $(L,\alpha_L),$ then it is split. Consequently ${\frak{uce}}(L)\cong L$ and $HL_2^\alpha(L) = 0.$

$b)\Rightarrow a)$ Let $0 \to 0 \to (L,\alpha_L) \stackrel{\sim}\to (L,\alpha_L) \to 0$ be a universal central extension of $(L,\alpha_L)$. So every central extension splits uniquely follows from the universal property.

Proposition 4. Let $(L,\alpha_L)$ be a perfect Hom-preLie algebra. If $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is a central extension, then the following statements hold.

$a)$ Proposition 3 a) implies that $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is a universal central extension.

$b)$ If $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is a universal $\alpha$-central extension, then statements a) and b) hold in Proposition 3.

Proof. $a)$ It follows from Theorem 4.11 b) in [13] that if $(L,\alpha_L)$ is perfect and every central extension splits, then $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is a universal central extension. Note that every central extension of $(L,\alpha_L)$ splits by the simply connectivity and $(L,\alpha_L)$ is perfect by hypothesis. Then $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is universal.

$b)$ It follows from Lemma 4.10 in [13] that the composition of two central extensions is an $\alpha$-central extensionon. Consider a central extension $0 \to (N, \alpha_N) \stackrel{j} \to (A,\alpha_A) \stackrel{\rho} \to (L, \alpha_L) \to 0$. Let $(L,\alpha_L)$ be perfect and $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ be a central extension. Note that $0 \to { {\rm{Ker}}}(\mu\circ \rho) \to (A,\alpha_A) \stackrel{\mu\circ \rho} \to (M, \alpha_M) \to 0$ is an $\alpha$-central extension. Since $\mu:(L,\alpha_L)\twoheadrightarrow (M,\alpha_M)$ is a universal $\alpha$-central extension, there exists a unique homomorphism of Hom-preLie algebras $\varphi$ such that $\mu\circ\rho\circ\varphi = \mu$. By Lemma 4.7 in [13], we have $\rho\circ\varphi = {{\rm{Id}}}$. So $(L,\alpha_L)$ is simply connected, that is, it is centrally closed.

Let $f : (L', \alpha_{L'})\to (L, \alpha_L)$ be a homomorphism of perfect Hom-preLie algebras. It induces a linear map $f\otimes f : L'\otimes L'\to L\otimes L$ given by $(f\otimes f)(x_1\otimes x_2) = f (x_1)\otimes f (x_2)$, which maps $I_{L'}$ to $I_L$. So $f\otimes f$ induces a homomorphism of Hom-preLie algebras ${\frak{uce}}(f) : {\frak{uce}}(L') \to {\frak{uce}}(L)$, given by ${\frak{uce}}(f) \{x_1, x_2\} = \{f(x_1), f(x_2)\}$. From the above conditions, the following diagram commutate.

From diagram (2), there exists a covariant right exact functor ${\frak{uce}}$: Hom-preLie$^{\rm{perf}}$ $\to$ Hom-preLie$^{\rm{perf}}$ between the category of perfect Hom-preLie algebras. So an automorphism $f$ of $(L, \alpha_L)$ induces an automorphism ${\frak{uce}}(f)$ of $({\frak{uce}}(L), \tilde{\alpha})$. ${\frak{uce}}(f)$ leaves $HL_2^\alpha(L)$ invariant since diagram (2) is commutative. Consequently, we obtain the Hom-group homomorphism

Similar to the above discussion, we also obtain the functorial properties of $\alpha$-perfect Hom-preLie algebras. In other words, consider a homomorphism of $\alpha$-perfect Hom-preLie algebras $f : (L', \alpha_{L'})\to (L, \alpha_L).$ Let $I_L$ the vector subspace of $\alpha_L(L)\otimes \alpha_L(L)$ spanned by $\alpha_L(x_1)\otimes x_2x_3- x_1x_2\otimes \alpha_L(x_3)-\alpha_L(x_2)\otimes x_1x_3+ x_2x_1\otimes \alpha_L(x_3),x_1, x_2, x_3 \in L$, respectively $I_{L'}$. $f$ induces a linear map $f\otimes f: (\alpha_{L'} (L') \otimes \alpha_{L'} (L') , \alpha_{L'\otimes L'}) \to (\alpha_{L} (L) \otimes \alpha_{L} (L), \alpha_{L\otimes L})$, given by $f\otimes f(\alpha_{L'} (x'_1) \otimes \alpha_{L'} (x'_2)) = \alpha_{L} (f(x'_1))\otimes \alpha_{L} (f(x'_2))$ such that $f\otimes f(I_{L'})\subseteq I_L$. Hence, it induces a homomorphism of Hom-preLie algebras ${\frak{uce}}_\alpha(f) : ({\frak{uce}}_{\alpha'}(L'),\overline{\alpha'}) \to ({\frak{uce}}_\alpha(L),\overline{\alpha})$ given by ${\frak{uce}}_\alpha(f) \{\alpha_{L'} (x'_1), \alpha_{L'}(x'_2)\} = \{\alpha_{L} (f(x'_1)), \alpha_{L} (f(x'_2))\}$ such that the following diagram

is commutative.

From diagram (3), there exists a covariant right exact functor ${\frak{uce}}_\alpha$: Hom-preLie$^{\alpha-\rm{perf}}$ $\to$ Hom-preLie$^{\alpha-\rm{perf}}$ between the category of $\alpha$-perfect Hom-preLie algebras. So an automorphism $f$ of $(L, \alpha_L)$ induces an automorphism ${\frak{uce}}_\alpha(f)$ of $({\frak{uce}}_\alpha(L),\overline{\alpha})$. ${\frak{uce}}_\alpha(f)$ leaves ${ {\rm{Ker}}}(U_{\alpha})$ invariant since diagram (3) is commutative. So we obtain the Hom-group homomorphism

Next we consider a derivation $d$ of the $\alpha$-perfect Hom-preLie algebra $(L, \alpha_L)$. The linear map $d: \alpha_L(L)\otimes\alpha_L(L) \to \alpha_L(L)\otimes\alpha_L(L)$ given by $d(\alpha_L(x_1)\otimes\alpha_L(x_2)) = d(\alpha_L(x_1))\otimes\alpha_L^2(x_2) + \alpha_L^2(x_1) \otimes d(\alpha_L(x_2))$, keeps invariant the subspace $I_L$ of $\alpha_L(L)\otimes \alpha_L(L)$ spanned by $\alpha_L(x_1)\otimes x_2x_3- x_1x_2\otimes \alpha_L(x_3)-\alpha_L(x_2)\otimes x_1x_3+ x_2x_1\otimes \alpha_L(x_3),x_1, x_2, x_3 \in L$. Indeed,

So it induces a linear map ${\frak{uce}}_\alpha(d) : ({\frak{uce}}_\alpha(L) , \overline{\alpha}) \to ({\frak{uce}}_\alpha(L) , \overline{\alpha})$, given by

such that the following diagram

is commutative. Hence, a derivation $d$ of $(L,\alpha_L)$ induces a derivation ${\frak{uce}}_{\alpha}(d)$ of $({\frak{uce}}_{\alpha}(L),\overline{\alpha})$. ${\frak{uce}}_{\alpha}(d)$ maps ${ {\rm{Ker}}}(U_{\alpha})$ on itself since diagram (4) is commutative. Consequently, we obtain the homomorphism of Hom-$\mathbb{K}$-vector spaces

Lemma 3.3. Let $f : (L', \alpha_{L'}) \to (L, \alpha_L)$ be a homomorphism of $\alpha$-perfect Hom-preLie algebras. If $d,\ d' \in { {{\rm{Der}}}}(L)$ satisfies $f \circ d' = d\circ f$, then ${\frak{uce}}_\alpha(f)\circ {\frak{uce}}_\alpha(d') = {\frak{uce}}_\alpha(d)\circ {\frak{uce}}_\alpha(f).$

Proof. For any $x'_1,x'_2\in L'$, we have

On the other hand

Hence we prove the lemma.

4.   Lifting automorphisms and derivations

Definition 4.1. Let $(L', \alpha_{L'})$ be a Hom-preLie algebra. A central extension of Hom-preLie algebras $f : (L', \alpha_{L'}) \twoheadrightarrow(L, \alpha_L)$ is said to be an $\alpha$-cover if $(L', \alpha_{L'})$ is $\alpha$-perfect.

Lemma 4.2. Let $f : (L', \alpha_{L'}) \to (L, \alpha_L)$ be a surjective homomorphism of Hom-preLie algebras. If $(L', \alpha_{L'})$ is $\alpha$-perfect, then $(L, \alpha_L)$ is also $\alpha$-perfect.

Proof. Routine checking.

Let $f : (L', \alpha_{L'}) \twoheadrightarrow (L, \alpha_L)$ be an $\alpha$-cover. By Lemma 4.2, $(L, \alpha_L)$ is an $\alpha$-perfect Hom-preLie algebra. By Theorem 4.19 in [13], it has a universal $\alpha$-central extension. By means of diagram (3), we obtain the following diagram:

By Remark 4.4 in [13], $U_{\alpha'} : ({\frak{uce}}_{\alpha'}(L'),\overline{\alpha'})\twoheadrightarrow (L',\alpha_{L'})$ is a universal central extension. Since $f : (L', \alpha_{L'}) \twoheadrightarrow (L, \alpha_L)$ is a central extension and $U_{\alpha'} : ({\frak{uce}}_{\alpha'}(L'),\overline{\alpha'})\twoheadrightarrow (L',\alpha_{L'})$ is a universal central extension, by Proposition 4.15 a) in [13], the extension $f \circ U_{\alpha'} : ({\frak{uce}}_{\alpha'}(L'),\overline{\alpha'}) \to (L, \alpha_L)$ is $\alpha$-central which is universal in the sense of Definition 4.13 in [13].

In addition, since $U_{\alpha} : ({\frak{uce}}_{\alpha}(L),\overline{\alpha})\twoheadrightarrow (L,\alpha_{L})$ is a universal $\alpha$-central extension, there is a unique homomorphism $\varphi : ({\frak{uce}}_{\alpha}(L),\overline{\alpha}) \twoheadrightarrow ({\frak{uce}}_{\alpha'}(L'),\overline{\alpha'})$ satifies $f\circ U_{\alpha'}\circ \varphi = U_{\alpha}$. So we have

that is to say the following diagram

is commutative. Since $f\circ U_{\alpha'}$ is an $\alpha$-central extension which is universal in the sense of Definition 4.13 in [13], we have $\varphi\circ {\frak{uce}}_{\alpha}(f) = {{\rm{Id}}}$.

Conversely, ${\frak{uce}}_{\alpha}(f) \circ\varphi = {{\rm{Id}}}$ since the following diagram commute.

whose horizontal rows are central extensions and $({\frak{uce}}_{\alpha}(L),\overline{\alpha})$ is $\alpha$-perfect, the uniqueness of the vertical homomorphism is guaranteed by Lemma 4.18 in [13]. Consequently ${\frak{uce}}_{\alpha}(f)$ is an isomorphism and we will denote the notation ${\frak{uce}}_{\alpha}(f)^{-1}$ by $\varphi$.

Moreover, $U_{\alpha'}\circ {\frak{uce}}_{\alpha}(f)^{-1}:({\frak{uce}}_{\alpha}(L),\overline{\alpha})\twoheadrightarrow (L',\alpha_{L'})$ is an $\alpha$-cover. In the sequel, we will denote its kernel by

In fact, for any $x\in { {\rm{Ker}}}(U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1})$, we have $U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}(x) = 0$. Hence $x\in {\frak{uce}}_{\alpha}(f)({ {\rm{Ker}}}(U_{\alpha'}))$.

Conversely, for any $x\in {\frak{uce}}_{\alpha}(f)({ {\rm{Ker}}}(U_{\alpha'}))$, there exists a $y\in{ {\rm{Ker}}}(U_{\alpha'})$ such that $x = {\frak{uce}}_{\alpha}(f)(y)$. So $U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}(x) = U_{\alpha'}(y) = 0$.

Theorem 4.3. Let $f : (L', \alpha_{L'}) \twoheadrightarrow (L, \alpha_L)$ be an $\alpha$-cover. For any automorphism $h$ on $(L, \alpha_{L})$, there is a unique $\theta_h \in Aut(L', \alpha_L')$ such that the following diagram is commutative:

if and only if the automorphism ${\frak{uce}}_{\alpha}(h)$ of $({\frak{uce}}_{\alpha}(L),\overline{\alpha})$ such that ${\frak{uce}}_{\alpha}(h)(C) = C$. Futhermore, we obtain a group isomorphism:

Proof. Let $h \in Aut(L, \alpha_L)$. Suppose that there is an automorphism $\theta_h$ on $(L', \alpha_{L'})$ such that diagram (5) commute. Apply the functor ${\frak{uce}}_{\alpha}(-)$ to diagram (5), the following commutative diagram holds:

So ${\frak{uce}}_{\alpha}(h)(C) = {\frak{uce}}_{\alpha}(h)\circ {\frak{uce}}_{\alpha}(f)(Ker (U_{\alpha'})) = {\frak{uce}}_{\alpha}(f)\circ {\frak{uce}}_{\alpha}(\theta_h)( { {\rm{Ker}}} (U_{\alpha'}))$ $= {\frak{uce}}_{\alpha}(f)({ {\rm{Ker}}} (U_{\alpha'})) = C$.

Conversely, diagram (3) implies that $U_{\alpha} = f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}$, so we have the following diagram:

If ${\frak{uce}}_{\alpha}(h) (C) = C$, then $U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} \circ {\frak{uce}}_{\alpha}(h) (C) = U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}(C) = 0$, so there is a unique $\theta_h : (L', \alpha_{L'})\to (L', \alpha_{L'})$ satisfies $\theta_h \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} = U_{\alpha'}\circ {\frak{uce}}_{\alpha}(f)^{-1}\circ {\frak{uce}}_{\alpha}(h)$.

On the other side, $h \circ f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} = f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}\circ {\frak{uce}}_{\alpha}(h) = f \circ\theta_h \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}$. Since $(L', \alpha_{L'})$ is an $\alpha$-perfect Hom-preLie algebra and ${\frak{uce}}_{\alpha}(f)^{-1}$ is an isomorphism, we have $h \circ f = f \circ \theta_h.$ Moreover, $h\circ f$ is an $\alpha$-cover since ${ {\rm{Ker}}}(h\circ f)\subseteq { {\rm{Ker}}}(f)\subseteq Z(L')$. Hence, $\theta_h$ is uniquely by Lemma 4.18 in [13]. Finally, $\theta_h ({ {\rm{Ker}}}(f)) = { {\rm{Ker}}}(f)$. Indeed, we have $f\circ \theta_h({ {\rm{Ker}}}(f)) = h \circ f({ {\rm{Ker}}}(f)) = 0.$ Conversely, for $x\in { {\rm{Ker}}}(f)$, there exists a $y\in L'$ such that $x = \theta_h(y)$. Hence, $f(y)\in { {\rm{Ker}}}(h) = 0$.

We know that $\Theta$ is well-defined, it is a monomorphism follows from the uniqueness of $\theta_h$. $\Theta$ is an epimorphism, since any $g\in Aut(L',\alpha_{L'})$ with $g ({ {\rm{Ker}}}(f)) = { {\rm{Ker}}}(f)$, gives rise to a unique homomorphism $h : (L, \alpha_L) \to(L, \alpha_L)$ satisfies $h \circ f = f \circ g$. Consequently, $g = \theta_h$ and ${\frak{uce}}_{\alpha}(h) (C) = C$.

Corollary 3. Let $(L, \alpha_L)$ be an $\alpha$-perfect Hom-preLie algebra. Then there exists a group isomorphism:

Proof. By Theorem 4.3, $U_\alpha:({\frak{uce}}_{\alpha}(L),\overline{\alpha})\twoheadrightarrow (L,\alpha_L)$ is an $\alpha$-cover. Let $C = 0$ and ${\frak{uce}}_{\alpha}(f)(0) = 0$ in Theorem 4.3.

Theorem 4.4. Let $f : (L', \alpha_{L'}) \!\twoheadrightarrow\! (L, \alpha_L)$ be an $\alpha$-cover. Denote $C\! = \!{\frak{uce}}_{\alpha}(f) { {\rm{Ker}}}(U_{\alpha'})$ $\subseteq { {\rm{Ker}}}(U_\alpha).$ Then the following statements hold:

a) For any $d \in { {{\rm{Der}}}} (L, \alpha_L)$, there exists a $\delta_d \in { {{\rm{Der}}}}(L', \alpha_{L'})$ such that the following diagram is commutative

if and only if the derivation ${\frak{uce}}_{\alpha}(d)$ of $({\frak{uce}}_{\alpha}(L) , \overline{\alpha_L})$ satisfies ${\frak{uce}}_{\alpha}(d)(C) \subseteq C$.

b) There exists an isomorphism of Hom-vector spaces

c) Let $U_\alpha : ({\frak{uce}}_{\alpha}(L),\overline{\alpha_L}) \twoheadrightarrow (L, \alpha_L)$ be an $\alpha$-cover. Then there exists an isomorphism of Hom-vector spaces

Proof. a) Let $d \in { {{\rm{Der}}}}(L, \alpha_L)$. Suppose that there exists a $\delta_d\in { {{\rm{Der}}}}(L', \alpha_{L'})$, which makes diagram (6) commute. We have that the following diagram is commutative thanks to Lemma 3.3.

Hence, by diagram (4), we obtain ${\frak{uce}}_{\alpha}(d)(C) = {\frak{uce}}_{\alpha}(d)\circ {\frak{uce}}_{\alpha}(f)(Ker (U_{\alpha'})) = {\frak{uce}}_{\alpha}(f)\circ {\frak{uce}}_{\alpha}(\delta_d)( { {\rm{Ker}}} (U_{\alpha'})) \subseteq {\frak{uce}}_{\alpha}(f)({ {\rm{Ker}}} (U_{\alpha'})) = C$.

Conversely, we have that $U_{\alpha} = f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}$ follows from diagram (3), hence we obtain the following diagram:

If ${\frak{uce}}_{\alpha}(d) (C) \subseteq C$, then $U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} \circ {\frak{uce}}_{\alpha}(d) (C) \subseteq U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}(C) = 0$, so there exists a unique $\delta_d : (L', \alpha_{L'})\to (L', \alpha_{L'})$ such that $\delta_d \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} = U_{\alpha'}\circ {\frak{uce}}_{\alpha}(f)^{-1}\circ {\frak{uce}}_{\alpha}(d)$.

On the other side, $d \circ f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1} = d \circ U_{\alpha} = U_{\alpha}\circ {\frak{uce}}_{\alpha}(d) = f \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}\circ {\frak{uce}}_{\alpha}(d) = f\circ \delta_d \circ U_{\alpha'} \circ {\frak{uce}}_{\alpha}(f)^{-1}$. Since $(L', \alpha_{L'})$ is an $\alpha$-perfect Hom-preLie algebra and ${\frak{uce}}_{\alpha}(f)^{-1}$ is an isomorphism, we have $d \circ f = f \circ \delta_d.$ Moreover, $d\circ f$ is an $\alpha$-cover since ${ {\rm{Ker}}}(d\circ f)\subseteq { {\rm{Ker}}}(f)\subseteq Z(L')$. Hence, $\delta_d$ is uniquely determined by Lemma 4.2. At last, we have that $\delta_d$ is a derivation of $L'$ a direct verify.

b) It well known that the map $\Delta$ is a homomorphism of Hom-vector spaces, it is a monomorphism thanks to the uniqueness of $\delta_d$ and it is an epimorphism, since any $\rho\in { {{\rm{Der}}}}(L',\alpha_{L'})$ with $\rho ({ {\rm{Ker}}}(f)) \subseteq{ {\rm{Ker}}}(f)$, gives rise to a unique homomorphism $d : (L, \alpha_L) \to(L, \alpha_L)$ satisfies the following diagram

is commutative, where $d : (L, \alpha_L) \to (L, \alpha_L)$ is a derivation such that ${\frak{uce}}_{\alpha}(d) (C) = {\frak{uce}}_{\alpha} (d) \circ {\frak{uce}}_{\alpha}(f) ({ {\rm{Ker}}}(U_{\alpha'})) = {\frak{uce}}_{\alpha}(f) \circ {\frak{uce}}_{\alpha}(\rho) ({ {\rm{Ker}}}(U_{\alpha'}))\subseteq {\frak{uce}}_{\alpha}(f) ({ {\rm{Ker}}}(U_{\alpha'})) = C$.

c) Let $C = {\frak{uce}}_{\alpha}(U_\alpha) ({ {\rm{Ker}}} (U_\alpha)) = 0$ in statement b).

5.   Universal $\alpha$-central extension of a semi-direct product

Now, we give a split extension of $\alpha$-perfect Hom-preLie algebras as follow

By Lemma 2.4, $(G,\alpha_G)\cong (M,\alpha_M)\rtimes(Q,{ {{\rm{Id}}}}_Q)$, where the Hom-action of $(Q,{ {{\rm{Id}}}}_Q)$ on $(M,\alpha_M)$ is given by $q \centerdot m = s(q)t(m)$ and $m \centerdot q = t(m)s(q)$, $q \in Q, m \in M$. Sometimes, it is necessary to assume that the previous action is associative, i.e., $(mm')\centerdot q = m(m'\centerdot q)$, $q \in Q, m,m' \in M$.

If $(M,\alpha_M)$ is an $\alpha$-perfect Hom-preLie algebra and $Q$ is a Hom-preLie algebra $(Q, { {{\rm{Id}}}}_Q)$. Then the direct product $(G,\alpha_G) = (M,\alpha_M)\times (Q, { {{\rm{Id}}}}_Q) = (M \times Q, \alpha_M \times { {{\rm{Id}}}}_Q)$ satisfies the above situation. Applying the functorial properties of ${\frak{uce}}_{\alpha}(-)$ given by diagram (3) and $(Q, { {{\rm{Id}}}}_Q)$ is perfect, we obtain that the following diagram

is commutative. Here $\tau = {\frak{uce}}_{\alpha}(t),\pi = {\frak{uce}}_{\alpha}(p),\sigma = {\frak{uce}}_{\alpha}(s)$. Since $p \circ s = { {{\rm{Id}}}}_Q$, the sequence

is split. So ${\frak{uce}}_{\alpha}(p) \circ {\frak{uce}}_{\alpha}(s) = {\frak{uce}}_{\alpha}({ {{\rm{Id}}}}_Q)$, i.e., $\pi\circ \sigma = { {{\rm{Id}}}}_{{\frak{uce}}(Q)}.$ Hence $\pi$ is an epimorphism and there is a Hom-action of $({\frak{uce}}(Q),{ {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ on $({ {\rm{Ker}}}(\pi), \alpha_{G\mid})$ given by:

$\lambda: {\frak{uce}}(Q) \otimes { {\rm{Ker}}}(\pi)\to { {\rm{Ker}}}(\pi)$,

and $\rho: { {\rm{Ker}}}(\pi)\otimes{\frak{uce}}(Q) \to { {\rm{Ker}}}(\pi)$,

By Lemma 2.4, the split sequence

is equivalent to the semi-direct product sequence, i.e.,

Let $q \in Q$ and $\alpha_M(m_1), \alpha_M(m_2) \in \alpha_M(M)$. In $({\frak{uce}}_{\alpha}(G),\overline{\alpha_G})$, we have

and

These above equalities and the $\alpha$-perfection of $(M, \alpha_M)$ imply:

and

Futhermore,

and

since $\tau\{\alpha_M(m_1), \alpha_M(m_2)\} = \{t (\alpha_M(m_1)) , t (\alpha_M(m_2))\} \equiv \{\alpha_M(m_1), \alpha_M(m_2)\}$, and $\sigma(\{q_1, q_2\}) = \{s(q_1), s(q_2)\} = \{\alpha_G(s(q_1)), \alpha_G(s(q_2))\}.$

Lemma 5.1. With the above notations, we have

Proof. For any $\alpha_G(g) \in G$, there exists an $\alpha_M(m) \in \alpha_M(M)$ such that $\alpha_G(g) = s(p(\alpha_G(g))) + \alpha_M(m)$. Hence

Conversely,

Hence, we prove the lemma.

Proposition 5. With the above notations, we have

Proof. Let $\{g_1, g_2\} \in { {\rm{Ker}}}(\pi)$. Then by Eq.(8), we have

Hence $0 = \pi\{g_1, g_2\} = \{p(s(q_1)) , p(s(q_2))\} + \{p(\alpha_M(m_1)) , p(\alpha_M(m_2))\} = \{q_1, q_2\}$, i.e., $q_1\otimes q_2 \in I_Q$. So $\sigma\{q_1, q_2\} = \{s(q_1) , s(q_2)\} = 0$. Consequently, ${ {\rm{Ker}}}(\pi)$ has elements of the form $\{\alpha_M(m_1), \alpha_M(m_2)\}$. It is easy to prove the reverse inclusion. Eq.(7) gives a proof of the second equality.

Theorem 5.2. Consider a split extension of $\alpha$-perfect Hom-preLie algebras

Then the following statements hold

1) $({\frak{uce}}_\alpha(G), \alpha_G) = \tau ({\frak{uce}}_\alpha(M), \alpha_M)\rtimes \sigma({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$.

2) $\sigma({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})\cong ({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)}).$

3) $({ {\rm{Ker}}}( U_\alpha^G), \overline{\alpha_{G\mid}})\cong \tau({ {\rm{Ker}}}( U_\alpha^M), \overline{\alpha_{M\mid}})\oplus \sigma(HL_2(Q),{ {{\rm{Id}}}}_{{\frak{uce}}(Q)}).$

Proof. 1) and 2) Since $\pi\circ\sigma = { {{\rm{Id}}}}$, we have

Moreover, $({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})\cong \sigma({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$. By proposition 5, 1) and 2) hold.

3) Let $(\tau (m) , \sigma(q))\in ({\frak{uce}}_\alpha(G), \overline{\alpha_G})$ from 1), where $m \in ({\frak{uce}}_\alpha(M) , \alpha_M)$ and $q \in({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$. So $(\tau (m) , \sigma(q))\in { {\rm{Ker}}}(U_\alpha^G)\Leftrightarrow U_\alpha^G(\tau (m) , \sigma(q)) = 0\Leftrightarrow t\circ U_\alpha^M(m) = U_\alpha^G(\tau (m)) = 0,s\circ u_Q(q) = U_\alpha^G(\sigma (q)) = 0\Leftrightarrow m\in { {\rm{Ker}}}(U_\alpha^M), q\in HL_2(Q).$

Suppose that there is an associative Hom-action of $(Q, { {{\rm{Id}}}}_Q)$ on $(M, \alpha_M)$, we have a Hom-action of $({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ on $({\frak{uce}}_\alpha(M), \alpha_M)$ given by:

and

When it is need, the Hom-action of $(Q , { {{\rm{Id}}}}_Q$) on $(M, \alpha_M)$ is ${\frak{uce}}$-associative, i.e., $\{mm',q\} = \{m,m'\centerdot q\}$.

So we define the following homomorphism of Hom-preLie algebras

where the Hom-action of $(Q , { {{\rm{Id}}}}_Q$) on $(M, \alpha_M)$ is ${\frak{uce}}$-associative.

We obtain that $\tau\rtimes\sigma$ is an epimorphism since

Next we define a surjective homomorphism of Hom-preLie algebras

such that the following diagram

is commutative. We have that

Second inclusion holds since $t\circ U_\alpha^M = U_\alpha^G\circ\tau$ and $t$ is injective. Since the following diagram is commutative

$U_{\alpha}^G \circ\tau({ {\rm{Ker}}}(U_\alpha^M)) = t\circ U_\alpha^M({ {\rm{Ker}}}(U_\alpha^M)) = 0$, then $\tau({ {\rm{Ker}}}(U_\alpha^M))\subseteq { {\rm{Ker}}}(U_\alpha^G)\subseteq Z({\frak{uce}}_{\alpha}(G))$, so

and

Consequently, ${\frak{uce}}(Q)\centerdot{ {\rm{Ker}}}(U_\alpha^M)\oplus{ {\rm{Ker}}}(U_\alpha^M)\centerdot{\frak{uce}}(Q)\subseteq { {\rm{Ker}}}(\tau).$

On the other side, we have that ${\frak{uce}}(Q)\centerdot{ {\rm{Ker}}}(U_\alpha^M)\oplus{ {\rm{Ker}}}(U_\alpha^M)\centerdot{\frak{uce}}(Q)$ is an ideal of $({\frak{uce}}_{\alpha}(M) , \alpha_M)$. Then the Hom-action of $({\frak{uce}}(Q) , { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ on $({\frak{uce}}_\alpha(M), \alpha_M)$ gives rise to a Hom-action of $({\frak{uce}}(Q), { {{\rm{Id}}}}_Q)$ on

Since $\tau$ vanishes on ${\frak{uce}}(Q)\centerdot{ {\rm{Ker}}}(U_\alpha^M)\oplus{ {\rm{Ker}}}(U_\alpha^M)\centerdot{\frak{uce}}(Q)$, it gives rise to $\overline{\tau} : \overline{{\frak{uce}}_\alpha(M)}\to \tau({\frak{uce}}_\alpha(M))$. Put $I = {\frak{uce}}(Q)\centerdot{ {\rm{Ker}}}(U_\alpha^M)\oplus{ {\rm{Ker}}}(U_\alpha^M)\centerdot{\frak{uce}}(Q)$, we have the following diagram

We obtain the following commutative diagram:

whose bottom row is a central extension. Since $({\frak{uce}}_\alpha(G),\overline{\alpha_G})$ is an $\alpha$-perfect Hom-preLie algebra, by Theorem 4.19 in [13], it has a universal $\alpha$-central extension. So ${\frak{uce}}_\alpha(G)$ is centrally closed by Corollary 2, that is, ${\frak{uce}}({\frak{uce}}_\alpha(G))\cong {\frak{uce}}_\alpha(G)$. Hence, we have the following diagram

Since $({\frak{uce}}_\alpha(G),\overline{\alpha_G})$ is centrally closed and $\Psi$ is a central extension, ${ {{\rm{Id}}}}: ({\frak{uce}}_\alpha(G),\overline{\alpha_G})$ $\to ({\frak{uce}}_\alpha(G),\overline{\alpha_G})$ is a universal central extension. Then there is a unique homomorphism of Hom-preLie algebras $\mu : ({\frak{uce}}_\alpha(G),\!\overline{\alpha_G})\! \to\! \left(\overline{{\frak{uce}}_{\alpha}(M)}\rtimes{\frak{uce}}(Q),\!\overline{\overline{\alpha_M}}\rtimes{ {{\rm{Id}}}}_{{\frak{uce}}(Q)}\right)$ satisfies $\Psi \circ \mu = { {{\rm{Id}}}}$. Since $\Psi \circ \mu \circ \Psi \!\! = \!\!{ {{\rm{Id}}}} \circ\Psi \!\! = \!\!\Psi\circ{ {{\rm{Id}}}}$ and $\left(\!\overline{{\frak{uce}}_{\alpha}(M)}\rtimes{\frak{uce}}(Q),\!\overline{\overline{\alpha_M}}\rtimes{ {{\rm{Id}}}}_{{\frak{uce}}(Q)}\!\right)$ is $\alpha$-perfect, then $\mu \circ \Psi = { {{\rm{Id}}}}$ follows from Lemma 4.18 in [13]. Hence, $\Psi$ is an isomorphism, then ${ {\rm{Ker}}}(\Psi) = \frac{{ {\rm{Ker}}}(\tau\rtimes\sigma)}{I} = 0$. Consequently, ${ {\rm{Ker}}}(\tau\rtimes\sigma)\subseteq I$.

The above discussion can be summarized in:

We can obtain the following theorem from the above results.

Theorem 5.3. Let the Hom-action of $(Q, { {{\rm{Id}}}}_Q)$ on $(M, \alpha_M)$ be ${\frak{uce}}$-associative and the extension

split. Then the following conditions hold

1) The homomorphism of Hom-preLie algebras

defined by $\Phi(\{\alpha_M(m_1), \alpha_M(m_2)\} , \{q_1, q_2\}) = (t(\alpha_M(m_1)\alpha_M(m_2)) , s(q_1q_2))$ is an epimorphism whose kernel is ${ {\rm{Ker}}}(U_\alpha^ M) \oplus HL_2(Q)$. Moreover, diagram (9) is commutative.

2) ${ {\rm{Ker}}}(\tau\rtimes\sigma)\cong {\frak{uce}}(Q)\centerdot { {\rm{Ker}}}(U_\alpha^ M)\oplus { {\rm{Ker}}}(U_\alpha^ M)\centerdot {\frak{uce}}(Q).$

Theorem 5.4. The following statements are equivalent:

a) $\Phi: = (t\circ U_\alpha^M)\rtimes(s\circ u_Q):({\frak{uce}}_\alpha(M)\rtimes {\frak{uce}}(Q), \overline{\alpha_M}\rtimes{ {{\rm{Id}}}}_{{\frak{uce}}(Q)})\rightarrow (G, \alpha_G)$ is a central extension. So it is an $\alpha$-cover.

b) The Hom-action of $({\frak{uce}}(Q), { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ on $({ {\rm{Ker}}}(U_\alpha^ M), \overline{\alpha_{M\mid}})$ is trivial.

c) $\tau\rtimes \sigma$ is an isomorphism. Hence ${\frak{uce}}_\alpha(M) \rtimes{\frak{uce}}(Q)$ is the universal $\alpha$-central extension of $(G, \alpha_G)$.

d) $\tau$ is injective. In particular,

Proof. $a)\Leftrightarrow b)$ If $\Phi: = (t\circ U_\alpha^M)\rtimes(s\circ u_Q):({\frak{uce}}_\alpha(M)\rtimes {\frak{uce}}(Q), \overline{\alpha_M}\rtimes{ {{\rm{Id}}}}_{{\frak{uce}}(Q)})\rightarrow (G, \alpha_G)$ is a central extension and ${ {\rm{Ker}}}(\Phi) = { {\rm{Ker}}}(U^M_\alpha)\oplus HL_2(Q)$, then the Hom-action of $({\frak{uce}}(Q), { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ on $({ {\rm{Ker}}}(U_\alpha^ M), \overline{\alpha_{M\mid}})$ is trivial. It is easy to prove the converse. Furthermore, the Hom-action is trivial, we have $({\frak{uce}}_\alpha(M)\rtimes {\frak{uce}}(Q),\overline{\alpha_M}\rtimes { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$ is $\alpha$-perfect. Consequently, the extension is an $\alpha$-cover.

$b)\Leftrightarrow c)$ By Theorem 5.3, we have that ${ {\rm{Ker}}}(\tau\rtimes\sigma)\cong {\frak{uce}}(Q)\centerdot { {\rm{Ker}}}(U_\alpha^ M)\oplus { {\rm{Ker}}}(U_\alpha^ M)\centerdot {\frak{uce}}(Q)$, then $\tau\circ \sigma$ is injective if and only if the Hom-action is trivial.

By diagram (9), we have that ${\frak{uce}}_\alpha(M)\rtimes {\frak{uce}}(Q)$ is the universal $\alpha$-central extension of $(G, \alpha_G)$.

$c)\Leftrightarrow d)$ It suffices to verify that $\tau \{\!\alpha_M \!(m_1) , \alpha_M \!(m_2)\!\} \!\! = \!\!(\tau\circ\sigma) ( \{\alpha_M \!(m_1) , \alpha_M \!(m_2)\} \! , 0\!)$, since ${ {\rm{Ker}}}(\tau ) \!\! = \!\!{ {\rm{Ker}}}(\tau\circ \sigma )$. So the conclusion holds.

Since the Hom-action of $(Q, { {{\rm{Id}}}}_Q)$ on $(M, \alpha_M)$ is trivial, the Hom-action of $( {\frak{uce}}(Q),$ ${ {{\rm{Id}}}}_{ {\frak{uce}}(Q)})$ on $({\frak{uce}}_\alpha(M), \overline{\alpha_M})$ is also trivial. Consequently, $({\frak{uce}}_\alpha(M)\times {\frak{uce}}(Q),\overline{\alpha_M}\times { {{\rm{Id}}}}_{{\frak{uce}}(Q)}) = ({\frak{uce}}_\alpha(M)\rtimes {\frak{uce}}(Q),\overline{\alpha_M}\times { {{\rm{Id}}}}_{{\frak{uce}}(Q)})$. The proof of the particular case is completed by statement c).

Acknowledgments

The authors would like to thank the referee for valuable comments and suggestions on this article.