1.
Introduction
A Hom-preLie algebra was introduced by Makhlouf-Silvestrov [11]. Specifically, for a vector space L over a field K equipped with a bilinear map μ:L×L→L and a linear map α:L→L, we say that the triple (L,μ,α) is a Hom-preLie algebra if
for all x,y,z∈L. If the elements of L also satisfy the following equation
Then we call (L,μ,α) 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 (α=IdL), 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 α-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 (α)-central extension.
In [13], we study universal α-central extensions of Hom-preLie algebras. We define Hom-co-representations and low-dimensional chain complex, which derive a low-dimensional homology K-vector space of a Hom-preLie algebra. We construct a right exact covariant functor uceα of a Hom-preLie algebra which acts on a α-perfect Hom-preLie algebra L its universal α-central extension uceα(L)=αL(L)⊗αL(L)IL, where IL=⟨αL(x1)⊗x2x3−x1x2⊗αL(x3)−αL(x2)⊗x1x3+x2x1⊗αL(x3)⟩.
The purpose of this paper is to study the universal α-central extension of semi-direct product of two α-perfect Hom-preLie algebras. We introduce a Hom-action between two perfect Hom-preLie algebras (Q,IdQ) and (M,αM), giving a semi-direct product between two perfect Hom-preLie algebras. We use an associative Hom-action of (Q,IdQ) on (M,αM) to induce a Hom-action of (uce(Q),Iduce(Q)) on (uceα(M),¯αM). We obtain semi-direct product (uceα(M),¯αM)⋊(uce(Q),Iduce(Q)) and define a linear map τ⋊σ on the semi-direct product. Casas and Pacheco Rego gave the linear map τ⋊σ is a homomorphism of Hom-leibniz algebras in [5]. We add a condition that Hom-action of (Q,IdQ) on (M,αM) is uce-associative, that is, {mm′,q}={m,m′⋅q} ∀m,m′∈M,q∈Q. We also obtain a linear map τ⋊σ is a homomorphism of Hom-preLie algebras. We give a couple of necessary and sufficient conditions for the universal α-central extension of semi-direct product of two α-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 (α)-central extensions of (α)-perfect Hom-preLie algebras. In section 4 we obtain that an automorphism or a derivation can be lifted in an α-cover with certain constraints. In the final section we give some necessary and sufficient conditions about the universal α-central extension of the semi-direct product of two α-perfect Hom-preLie algebras.
Throughout this paper K denotes an arbitrary field.
2.
Hom-action
Definition 2.1. Let (M,αM) and (L,αL) be Hom-preLie algebras. A Hom-action of (L,αL) over (M,αM) consists of two bilinear maps, ρ:M⊗L→M, ρ(m⊗l)=m⋅l and λ:L⊗M→M, λ(l⊗m)=l⋅m, the following identities hold.
a) (xy)⋅αM(m)−αL(x)⋅(y⋅m)=(yx)⋅αM(m)−αL(y)⋅(x⋅m),
b) (m⋅x)⋅αL(y)−αM(m)⋅(xy)=(x⋅m)⋅αL(y)−αL(x)⋅(m⋅y),
c) (mm′)⋅αL(x)−αM(m)(m′⋅x)=(m′m)⋅αL(x)−αM(m′)(m⋅x),
d)(x⋅m)αM(m′)−αL(x)⋅(mm′)=(m⋅x)αM(m′)−αM(m)(x⋅m′),
e) αM(x⋅m)=αL(x)⋅αM(m),
f) αM(m⋅x)=αM(m)⋅αL(x),
for all x,y∈L and m,m′∈M.
If (M,αM) is an abelian Hom-preLie algebra, then the Hom-action is said to be a Hom-representation.
Example 1. a) Let (K,αK) be a subalgebra of a Hom-preLie algebra (L,αL) and (H,αH) a Hom-ideal of (L,αL). There is a Hom-action of (K,αK) over (H,αH) by the multiplication in (L,αL).
b) Let 0→(M,αM)i→(K,αK)π→(L,αL)→0 is an exact sequence of Hom-Lie algebras. If (M,αM) is an abelian Hom-preLie algebra, then we call the sequence is abelian. An abelian sequence gives a Hom-representation of (L,αL) over (M,αM) by defining ρ:M⊗L→M,ρ(m,l)=mk,π(k)=l, λ:L⊗M→M,λ(l,m)=km,π(k)=l.
Proposition 1. Let (M,αM) and (L,αL) be Hom-preLie algebras with a Hom-action of (L,αL) over (M,αM). Then (M⋊L,˜α) is a Hom-preLie algebra, where ˜α:M⋊L→M⋊L is defined by ˜α(m,l)=(αM(m),αL(l)) and multiplication
Proof. It follows by the direct computation.
Definition 2.2. [13] A short exact sequence of Hom-preLie algebras (K):0→(M,αM)i→(K,αK)π→(L,αL)→0 is said to be split if there exists a Hom-preLie algebra homomorphism σ:(L,αL)→(K,αK) such that π∘σ=IdL.
Let (M,αM) and (L,αL) be Hom-preLie algebras with a Hom-action of (L,αL) over (M,αM). We define two linear maps i:M→M⋊L,i(m)=(m,0) and π:M⋊L→L,π(m,l)=l. Then we obtain the following sequence
Furthermore, this sequence splits by σ:L→M⋊L,σ(l)=(0,l).
Definition 2.3. Let (M,αM) and (L,αL) be Hom-preLie algebras with a Hom-action of (L,αL) over (M,αM). Two extensions of (L,αL) by (M,αM), 0→(M,αM)i→(K,αK)π→(L,αL)→0 and 0→(M,αM)i′→(K′,αK′)π′→(L,αL)→0, are equivalent if there is a homomorphism of Hom-preLie algebra φ:(K,αK)→(K′,α′K) satisfies that the following commutative diagram
Lemma 2.4. Let (C,IdC) and (A,αA) be Hom-preLie algebras with a Hom-action of (C,IdC) over (A,αA). A sequence of Hom-preLie algebras 0→(A,αA)i→(B,αB)π→(C,IdC)→0 is split if and only if it is equivalent to the semi-direct sequence 0→(A,αA)j→(A⋊C,˜α)p→(C,IdC)→0.
Proof. If 0→(A,αA)i→(B,αB)π→(C,IdC)→0 is split by t:(C,IdC)→(B,αB), then the Hom-action of (C,IdC) over (A,αA) is defined by
So we obtain the following split extension:
where k:A→A⋊C,k(a)=(a,0),p:A⋊C→C,q(a,c)=c and τ:C→A⋊C,τ(c)=(0,c). Furthermore the Hom-action of (C,IdC) over (A,αA) induced by this extension coincides with the initial one:
Since φ:(A⋊C,˜α)→(B,αB),φ(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 φ:(A⋊C,˜α)→(B,αB) such that diagram (1) is commutative, then t:(C,IdC)→(B,αB) given by t(c)=φ(0,c), is a split extension.
Definition 2.5. [12] Let (M,αM) be a Hom-representation of a Hom-preLie algebra (L,αL). A derivation of (L,αL) over (M,αM) is a K-linear map d:L→M such that:
a) d(l1l2)=αL(l1)⋅d(l2)+d(l1)⋅αL(l2),
b) d∘αL=αM∘d,
for all l1,l2∈L.
Example 2. a) Let (M,αM) be a Hom-representation of (M⋊L,˜α) via π. Then the linear map θ:M⋊L→M,θ(m,l)=m, is a derivation.
b) When (M,αM)=(L,αL) is a representation follows from Example 1 a), then a derivation consists of a K-linear map d:L→L such that d(l1l2)=αL(l1)d(l2)+d(l1)αL(l2) and d∘αL=αM∘d.
Proposition 2. Let (M,αM) be a Hom-representation of a Hom-preLie algebra (L,αL). For every f-derivation d:(X,αX)→(M,αM) (d(x1x2)=d(x1)⋅αL(f(x2)) +αL(f(x1))⋅d(x2) ∀x1,x2∈X) and every homomorphism of Hom-preLie algebras f:(X,αX)→(L,αL) there is a unique homomorphism of Hom-preLie algebras h:(X,αX)→(M⋊L,˜α), such that the following diagram commute.
Conversely, every homomorphism of Hom-preLie algebras h:(X,αX)→(M⋊L,˜α), decide a homomorphism of Hom-preLie algebras f=π∘h:(X,αX)→(L,αL) and f-derivation d=θ∘h:(X,αX)→(M,αM), where θ(m,l)=x,∀m∈M,l∈L.
Proof. Let h:X→M⋊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,αL) to (M,αM) is in one-to-one correspondence with the set of Hom-preLie algebra homomorphisms h:(L,αL)→(M⋊L,˜α), such that ιπ∘h=IdL.
Proof. Take (X,αX)=(L,αL) in Proposition 2.
3.
Functorial properties
Definition 3.1. Let (L,αL) be a perfect Hom-preLie algebra. It is said to be centrally closed if its universal central extension is
i.e., HLα2(L)=0 and (uce(L),˜α)≅(L,αL).
Corollary 2. Let (L,αL) be a α-perfect Hom-preLie algebra. If 0→(Ker(Uα),αK∣) →(K,αK)Uα→(L,αL)→0 is the universal α-central extension of (L,αL), then (L,αL) is centrally closed.
Proof. HLα1(K)=HLα2(K)=0 thanks to Corollary 4.12 a) in [13]. By the proof of Corollary 4.12 b) in [13], HLα1(K)=0 if and only if (K,αK) is perfect. By Theorem 4.11 c) in [13], there exists a universal central extension 0→(HLα2(K),˜α∣)→(uce(K),˜α)uK→(K,αK)→0. Since HLα2(K)=0, uK is an isomorphism.
Definition 3.2. A Hom-preLie algebra (L,αL) is said to be simply connected if every central extension τ:(F,αF)↠(L,αL) splits uniquely as the product of Hom-preLie algebras (F,αF)=(Ker(τ),αF∣)×(L,αL).
Proposition 3. Let (L,αL) be a perfect Hom-preLie algebra. Then the following conditions are equivalent:
a) (L,αL) is simply connected.
b) (L,αL) is centrally closed.
Proof. a)⇒b) Let 0→(HLα2(L),˜α∣)→(uce(L),˜α)uL→(L,αL)→0 be the universal central extension of (L,αL), then it is split. Consequently uce(L)≅L and HLα2(L)=0.
b)⇒a) Let 0→0→(L,αL)∼→(L,αL)→0 be a universal central extension of (L,αL). So every central extension splits uniquely follows from the universal property.
Proposition 4. Let (L,αL) be a perfect Hom-preLie algebra. If μ:(L,αL)↠(M,αM) is a central extension, then the following statements hold.
a) Proposition 3 a) implies that μ:(L,αL)↠(M,αM) is a universal central extension.
b) If μ:(L,αL)↠(M,αM) is a universal α-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,αL) is perfect and every central extension splits, then μ:(L,αL)↠(M,αM) is a universal central extension. Note that every central extension of (L,αL) splits by the simply connectivity and (L,αL) is perfect by hypothesis. Then μ:(L,αL)↠(M,αM) is universal.
b) It follows from Lemma 4.10 in [13] that the composition of two central extensions is an α-central extensionon. Consider a central extension 0→(N,αN)j→(A,αA)ρ→(L,αL)→0. Let (L,αL) be perfect and μ:(L,αL)↠(M,αM) be a central extension. Note that 0→Ker(μ∘ρ)→(A,αA)μ∘ρ→(M,αM)→0 is an α-central extension. Since μ:(L,αL)↠(M,αM) is a universal α-central extension, there exists a unique homomorphism of Hom-preLie algebras φ such that μ∘ρ∘φ=μ. By Lemma 4.7 in [13], we have ρ∘φ=Id. So (L,αL) is simply connected, that is, it is centrally closed.
Let f:(L′,αL′)→(L,αL) be a homomorphism of perfect Hom-preLie algebras. It induces a linear map f⊗f:L′⊗L′→L⊗L given by (f⊗f)(x1⊗x2)=f(x1)⊗f(x2), which maps IL′ to IL. So f⊗f induces a homomorphism of Hom-preLie algebras uce(f):uce(L′)→uce(L), given by uce(f){x1,x2}={f(x1),f(x2)}. From the above conditions, the following diagram commutate.
From diagram (2), there exists a covariant right exact functor uce: Hom-preLieperf → Hom-preLieperf between the category of perfect Hom-preLie algebras. So an automorphism f of (L,αL) induces an automorphism uce(f) of (uce(L),˜α). uce(f) leaves HLα2(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 α-perfect Hom-preLie algebras. In other words, consider a homomorphism of α-perfect Hom-preLie algebras f:(L′,αL′)→(L,αL). Let IL the vector subspace of αL(L)⊗αL(L) spanned by αL(x1)⊗x2x3−x1x2⊗αL(x3)−αL(x2)⊗x1x3+x2x1⊗αL(x3),x1,x2,x3∈L, respectively IL′. f induces a linear map f⊗f:(αL′(L′)⊗αL′(L′),αL′⊗L′)→(αL(L)⊗αL(L),αL⊗L), given by f⊗f(αL′(x′1)⊗αL′(x′2))=αL(f(x′1))⊗αL(f(x′2)) such that f⊗f(IL′)⊆IL. Hence, it induces a homomorphism of Hom-preLie algebras uceα(f):(uceα′(L′),¯α′)→(uceα(L),¯α) given by uceα(f){αL′(x′1),αL′(x′2)}={αL(f(x′1)),αL(f(x′2))} such that the following diagram
is commutative.
From diagram (3), there exists a covariant right exact functor uceα: Hom-preLieα−perf → Hom-preLieα−perf between the category of α-perfect Hom-preLie algebras. So an automorphism f of (L,αL) induces an automorphism uceα(f) of (uceα(L),¯α). uceα(f) leaves Ker(Uα) invariant since diagram (3) is commutative. So we obtain the Hom-group homomorphism
Next we consider a derivation d of the α-perfect Hom-preLie algebra (L,αL). The linear map d:αL(L)⊗αL(L)→αL(L)⊗αL(L) given by d(αL(x1)⊗αL(x2))=d(αL(x1))⊗α2L(x2)+α2L(x1)⊗d(αL(x2)), keeps invariant the subspace IL of αL(L)⊗αL(L) spanned by αL(x1)⊗x2x3−x1x2⊗αL(x3)−αL(x2)⊗x1x3+x2x1⊗αL(x3),x1,x2,x3∈L. Indeed,
So it induces a linear map uceα(d):(uceα(L),¯α)→(uceα(L),¯α), given by
such that the following diagram
is commutative. Hence, a derivation d of (L,αL) induces a derivation uceα(d) of (uceα(L),¯α). uceα(d) maps Ker(Uα) on itself since diagram (4) is commutative. Consequently, we obtain the homomorphism of Hom-K-vector spaces
Lemma 3.3. Let f:(L′,αL′)→(L,αL) be a homomorphism of α-perfect Hom-preLie algebras. If d, d′∈Der(L) satisfies f∘d′=d∘f, then uceα(f)∘uceα(d′)=uceα(d)∘uceα(f).
Proof. For any x′1,x′2∈L′, we have
On the other hand
Hence we prove the lemma.
4.
Lifting automorphisms and derivations
Definition 4.1. Let (L′,αL′) be a Hom-preLie algebra. A central extension of Hom-preLie algebras f:(L′,αL′)↠(L,αL) is said to be an α-cover if (L′,αL′) is α-perfect.
Lemma 4.2. Let f:(L′,αL′)→(L,αL) be a surjective homomorphism of Hom-preLie algebras. If (L′,αL′) is α-perfect, then (L,αL) is also α-perfect.
Proof. Routine checking.
Let f:(L′,αL′)↠(L,αL) be an α-cover. By Lemma 4.2, (L,αL) is an α-perfect Hom-preLie algebra. By Theorem 4.19 in [13], it has a universal α-central extension. By means of diagram (3), we obtain the following diagram:
By Remark 4.4 in [13], Uα′:(uceα′(L′),¯α′)↠(L′,αL′) is a universal central extension. Since f:(L′,αL′)↠(L,αL) is a central extension and Uα′:(uceα′(L′),¯α′)↠(L′,αL′) is a universal central extension, by Proposition 4.15 a) in [13], the extension f∘Uα′:(uceα′(L′),¯α′)→(L,αL) is α-central which is universal in the sense of Definition 4.13 in [13].
In addition, since Uα:(uceα(L),¯α)↠(L,αL) is a universal α-central extension, there is a unique homomorphism φ:(uceα(L),¯α)↠(uceα′(L′),¯α′) satifies f∘Uα′∘φ=Uα. So we have
that is to say the following diagram
is commutative. Since f∘Uα′ is an α-central extension which is universal in the sense of Definition 4.13 in [13], we have φ∘uceα(f)=Id.
Conversely, uceα(f)∘φ=Id since the following diagram commute.
whose horizontal rows are central extensions and (uceα(L),¯α) is α-perfect, the uniqueness of the vertical homomorphism is guaranteed by Lemma 4.18 in [13]. Consequently uceα(f) is an isomorphism and we will denote the notation uceα(f)−1 by φ.
Moreover, Uα′∘uceα(f)−1:(uceα(L),¯α)↠(L′,αL′) is an α-cover. In the sequel, we will denote its kernel by
In fact, for any x∈Ker(Uα′∘uceα(f)−1), we have Uα′∘uceα(f)−1(x)=0. Hence x∈uceα(f)(Ker(Uα′)).
Conversely, for any x∈uceα(f)(Ker(Uα′)), there exists a y∈Ker(Uα′) such that x=uceα(f)(y). So Uα′∘uceα(f)−1(x)=Uα′(y)=0.
Theorem 4.3. Let f:(L′,αL′)↠(L,αL) be an α-cover. For any automorphism h on (L,αL), there is a unique θh∈Aut(L′,α′L) such that the following diagram is commutative:
if and only if the automorphism uceα(h) of (uceα(L),¯α) such that uceα(h)(C)=C. Futhermore, we obtain a group isomorphism:
Proof. Let h∈Aut(L,αL). Suppose that there is an automorphism θh on (L′,αL′) such that diagram (5) commute. Apply the functor uceα(−) to diagram (5), the following commutative diagram holds:
So uceα(h)(C)=uceα(h)∘uceα(f)(Ker(Uα′))=uceα(f)∘uceα(θh)(Ker(Uα′)) =uceα(f)(Ker(Uα′))=C.
Conversely, diagram (3) implies that Uα=f∘Uα′∘uceα(f)−1, so we have the following diagram:
If uceα(h)(C)=C, then Uα′∘uceα(f)−1∘uceα(h)(C)=Uα′∘uceα(f)−1(C)=0, so there is a unique θh:(L′,αL′)→(L′,αL′) satisfies θh∘Uα′∘uceα(f)−1=Uα′∘uceα(f)−1∘uceα(h).
On the other side, h∘f∘Uα′∘uceα(f)−1=f∘Uα′∘uceα(f)−1∘uceα(h)=f∘θh∘Uα′∘uceα(f)−1. Since (L′,αL′) is an α-perfect Hom-preLie algebra and uceα(f)−1 is an isomorphism, we have h∘f=f∘θh. Moreover, h∘f is an α-cover since Ker(h∘f)⊆Ker(f)⊆Z(L′). Hence, θh is uniquely by Lemma 4.18 in [13]. Finally, θh(Ker(f))=Ker(f). Indeed, we have f∘θh(Ker(f))=h∘f(Ker(f))=0. Conversely, for x∈Ker(f), there exists a y∈L′ such that x=θh(y). Hence, f(y)∈Ker(h)=0.
We know that Θ is well-defined, it is a monomorphism follows from the uniqueness of θh. Θ is an epimorphism, since any g∈Aut(L′,αL′) with g(Ker(f))=Ker(f), gives rise to a unique homomorphism h:(L,αL)→(L,αL) satisfies h∘f=f∘g. Consequently, g=θh and uceα(h)(C)=C.
Corollary 3. Let (L,αL) be an α-perfect Hom-preLie algebra. Then there exists a group isomorphism:
Proof. By Theorem 4.3, Uα:(uceα(L),¯α)↠(L,αL) is an α-cover. Let C=0 and uceα(f)(0)=0 in Theorem 4.3.
Theorem 4.4. Let f:(L′,αL′)↠(L,αL) be an α-cover. Denote C=uceα(f)Ker(Uα′) ⊆Ker(Uα). Then the following statements hold:
a) For any d∈Der(L,αL), there exists a δd∈Der(L′,αL′) such that the following diagram is commutative
if and only if the derivation uceα(d) of (uceα(L),¯αL) satisfies uceα(d)(C)⊆C.
b) There exists an isomorphism of Hom-vector spaces
c) Let Uα:(uceα(L),¯αL)↠(L,αL) be an α-cover. Then there exists an isomorphism of Hom-vector spaces
Proof. a) Let d∈Der(L,αL). Suppose that there exists a δd∈Der(L′,αL′), which makes diagram (6) commute. We have that the following diagram is commutative thanks to Lemma 3.3.
Hence, by diagram (4), we obtain uceα(d)(C)=uceα(d)∘uceα(f)(Ker(Uα′))=uceα(f)∘uceα(δd)(Ker(Uα′))⊆uceα(f)(Ker(Uα′))=C.
Conversely, we have that Uα=f∘Uα′∘uceα(f)−1 follows from diagram (3), hence we obtain the following diagram:
If uceα(d)(C)⊆C, then Uα′∘uceα(f)−1∘uceα(d)(C)⊆Uα′∘uceα(f)−1(C)=0, so there exists a unique δd:(L′,αL′)→(L′,αL′) such that δd∘Uα′∘uceα(f)−1=Uα′∘uceα(f)−1∘uceα(d).
On the other side, d∘f∘Uα′∘uceα(f)−1=d∘Uα=Uα∘uceα(d)=f∘Uα′∘uceα(f)−1∘uceα(d)=f∘δd∘Uα′∘uceα(f)−1. Since (L′,αL′) is an α-perfect Hom-preLie algebra and uceα(f)−1 is an isomorphism, we have d∘f=f∘δd. Moreover, d∘f is an α-cover since Ker(d∘f)⊆Ker(f)⊆Z(L′). Hence, δd is uniquely determined by Lemma 4.2. At last, we have that δd is a derivation of L′ a direct verify.
b) It well known that the map Δ is a homomorphism of Hom-vector spaces, it is a monomorphism thanks to the uniqueness of δd and it is an epimorphism, since any ρ∈Der(L′,αL′) with ρ(Ker(f))⊆Ker(f), gives rise to a unique homomorphism d:(L,αL)→(L,αL) satisfies the following diagram
is commutative, where d:(L,αL)→(L,αL) is a derivation such that uceα(d)(C)=uceα(d)∘uceα(f)(Ker(Uα′))=uceα(f)∘uceα(ρ)(Ker(Uα′))⊆uceα(f)(Ker(Uα′))=C.
c) Let C=uceα(Uα)(Ker(Uα))=0 in statement b).
5.
Universal α-central extension of a semi-direct product
Now, we give a split extension of α-perfect Hom-preLie algebras as follow
By Lemma 2.4, (G,αG)≅(M,αM)⋊(Q,IdQ), where the Hom-action of (Q,IdQ) on (M,αM) is given by q⋅m=s(q)t(m) and m⋅q=t(m)s(q), q∈Q,m∈M. Sometimes, it is necessary to assume that the previous action is associative, i.e., (mm′)⋅q=m(m′⋅q), q∈Q,m,m′∈M.
If (M,αM) is an α-perfect Hom-preLie algebra and Q is a Hom-preLie algebra (Q,IdQ). Then the direct product (G,αG)=(M,αM)×(Q,IdQ)=(M×Q,αM×IdQ) satisfies the above situation. Applying the functorial properties of uceα(−) given by diagram (3) and (Q,IdQ) is perfect, we obtain that the following diagram
is commutative. Here τ=uceα(t),π=uceα(p),σ=uceα(s). Since p∘s=IdQ, the sequence
is split. So uceα(p)∘uceα(s)=uceα(IdQ), i.e., π∘σ=Iduce(Q). Hence π is an epimorphism and there is a Hom-action of (uce(Q),Iduce(Q)) on (Ker(π),αG∣) given by:
λ:uce(Q)⊗Ker(π)→Ker(π),
and ρ:Ker(π)⊗uce(Q)→Ker(π),
By Lemma 2.4, the split sequence
is equivalent to the semi-direct product sequence, i.e.,
Let q∈Q and αM(m1),αM(m2)∈αM(M). In (uceα(G),¯αG), we have
and
These above equalities and the α-perfection of (M,αM) imply:
and
Futhermore,
and
since τ{αM(m1),αM(m2)}={t(αM(m1)),t(αM(m2))}≡{αM(m1),αM(m2)}, and σ({q1,q2})={s(q1),s(q2)}={αG(s(q1)),αG(s(q2))}.
Lemma 5.1. With the above notations, we have
Proof. For any αG(g)∈G, there exists an αM(m)∈αM(M) such that αG(g)=s(p(αG(g)))+αM(m). Hence
Conversely,
Hence, we prove the lemma.
Proposition 5. With the above notations, we have
Proof. Let {g1,g2}∈Ker(π). Then by Eq.(8), we have
Hence 0=π{g1,g2}={p(s(q1)),p(s(q2))}+{p(αM(m1)),p(αM(m2))}={q1,q2}, i.e., q1⊗q2∈IQ. So σ{q1,q2}={s(q1),s(q2)}=0. Consequently, Ker(π) has elements of the form {αM(m1),αM(m2)}. 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 α-perfect Hom-preLie algebras
Then the following statements hold
1) (uceα(G),αG)=τ(uceα(M),αM)⋊σ(uce(Q),Iduce(Q)).
2) σ(uce(Q),Iduce(Q))≅(uce(Q),Iduce(Q)).
3) (Ker(UGα),¯αG∣)≅τ(Ker(UMα),¯αM∣)⊕σ(HL2(Q),Iduce(Q)).
Proof. 1) and 2) Since π∘σ=Id, we have
Moreover, (uce(Q),Iduce(Q))≅σ(uce(Q),Iduce(Q)). By proposition 5, 1) and 2) hold.
3) Let (τ(m),σ(q))∈(uceα(G),¯αG) from 1), where m∈(uceα(M),αM) and q∈(uce(Q),Iduce(Q)). So (τ(m),σ(q))∈Ker(UGα)⇔UGα(τ(m),σ(q))=0⇔t∘UMα(m)=UGα(τ(m))=0,s∘uQ(q)=UGα(σ(q))=0⇔m∈Ker(UMα),q∈HL2(Q).
Suppose that there is an associative Hom-action of (Q,IdQ) on (M,αM), we have a Hom-action of (uce(Q),Iduce(Q)) on (uceα(M),αM) given by:
and
When it is need, the Hom-action of (Q,IdQ) on (M,αM) is uce-associative, i.e., {mm′,q}={m,m′⋅q}.
So we define the following homomorphism of Hom-preLie algebras
where the Hom-action of (Q,IdQ) on (M,αM) is uce-associative.
We obtain that τ⋊σ 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∘UMα=UGα∘τ and t is injective. Since the following diagram is commutative
UGα∘τ(Ker(UMα))=t∘UMα(Ker(UMα))=0, then τ(Ker(UMα))⊆Ker(UGα)⊆Z(uceα(G)), so
and
Consequently, uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q)⊆Ker(τ).
On the other side, we have that uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q) is an ideal of (uceα(M),αM). Then the Hom-action of (uce(Q),Iduce(Q)) on (uceα(M),αM) gives rise to a Hom-action of (uce(Q),IdQ) on
Since τ vanishes on uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q), it gives rise to ¯τ:¯uceα(M)→τ(uceα(M)). Put I=uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q), we have the following diagram
We obtain the following commutative diagram:
whose bottom row is a central extension. Since (uceα(G),¯αG) is an α-perfect Hom-preLie algebra, by Theorem 4.19 in [13], it has a universal α-central extension. So uceα(G) is centrally closed by Corollary 2, that is, uce(uceα(G))≅uceα(G). Hence, we have the following diagram
Since (uceα(G),¯αG) is centrally closed and Ψ is a central extension, Id:(uceα(G),¯αG) →(uceα(G),¯αG) is a universal central extension. Then there is a unique homomorphism of Hom-preLie algebras μ:(uceα(G),¯αG)→(¯uceα(M)⋊uce(Q),¯¯αM⋊Iduce(Q)) satisfies Ψ∘μ=Id. Since Ψ∘μ∘Ψ=Id∘Ψ=Ψ∘Id and (¯uceα(M)⋊uce(Q),¯¯αM⋊Iduce(Q)) is α-perfect, then μ∘Ψ=Id follows from Lemma 4.18 in [13]. Hence, Ψ is an isomorphism, then Ker(Ψ)=Ker(τ⋊σ)I=0. Consequently, Ker(τ⋊σ)⊆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,IdQ) on (M,αM) be uce-associative and the extension
split. Then the following conditions hold
1) The homomorphism of Hom-preLie algebras
defined by Φ({αM(m1),αM(m2)},{q1,q2})=(t(αM(m1)αM(m2)),s(q1q2)) is an epimorphism whose kernel is Ker(UMα)⊕HL2(Q). Moreover, diagram (9) is commutative.
2) Ker(τ⋊σ)≅uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q).
Theorem 5.4. The following statements are equivalent:
a) Φ:=(t∘UMα)⋊(s∘uQ):(uceα(M)⋊uce(Q),¯αM⋊Iduce(Q))→(G,αG) is a central extension. So it is an α-cover.
b) The Hom-action of (uce(Q),Iduce(Q)) on (Ker(UMα),¯αM∣) is trivial.
c) τ⋊σ is an isomorphism. Hence uceα(M)⋊uce(Q) is the universal α-central extension of (G,αG).
d) τ is injective. In particular,
Proof. a)⇔b) If Φ:=(t∘UMα)⋊(s∘uQ):(uceα(M)⋊uce(Q),¯αM⋊Iduce(Q))→(G,αG) is a central extension and Ker(Φ)=Ker(UMα)⊕HL2(Q), then the Hom-action of (uce(Q),Iduce(Q)) on (Ker(UMα),¯αM∣) is trivial. It is easy to prove the converse. Furthermore, the Hom-action is trivial, we have (uceα(M)⋊uce(Q),¯αM⋊Iduce(Q)) is α-perfect. Consequently, the extension is an α-cover.
b)⇔c) By Theorem 5.3, we have that Ker(τ⋊σ)≅uce(Q)⋅Ker(UMα)⊕Ker(UMα)⋅uce(Q), then τ∘σ is injective if and only if the Hom-action is trivial.
By diagram (9), we have that uceα(M)⋊uce(Q) is the universal α-central extension of (G,αG).
c)⇔d) It suffices to verify that τ{αM(m1),αM(m2)}=(τ∘σ)({αM(m1),αM(m2)},0), since Ker(τ)=Ker(τ∘σ). So the conclusion holds.
Since the Hom-action of (Q,IdQ) on (M,αM) is trivial, the Hom-action of (uce(Q), Iduce(Q)) on (uceα(M),¯αM) is also trivial. Consequently, (uceα(M)×uce(Q),¯αM×Iduce(Q))=(uceα(M)⋊uce(Q),¯αM×Iduce(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.