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On the universal α-central extensions of the semi-direct product of Hom-preLie algebras

  • Received: 01 October 2020 Revised: 01 November 2020 Published: 01 September 2021
  • Primary: 17A30; Secondary: 16E40

  • We study Hom-actions, semidirect product and describe the relation between semi-direct product extensions and split extensions of Hom-preLie algebras. We obtain the functorial properties of the universal α-central extensions of α-perfect Hom-preLie algebras. We give that a derivation or an automorphism can be lifted in an α-cover with certain constraints. We provide some necessary and sufficient conditions about the universal α-central extension of the semi-direct product of two α-perfect Hom-preLie algebras.

    Citation: Bing Sun, Liangyun Chen, Yan Cao. On the universal α-central extensions of the semi-direct product of Hom-preLie algebras[J]. Electronic Research Archive, 2021, 29(4): 2619-2636. doi: 10.3934/era.2021004

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  • We study Hom-actions, semidirect product and describe the relation between semi-direct product extensions and split extensions of Hom-preLie algebras. We obtain the functorial properties of the universal α-central extensions of α-perfect Hom-preLie algebras. We give that a derivation or an automorphism can be lifted in an α-cover with certain constraints. We provide some necessary and sufficient conditions about the universal α-central extension of the semi-direct product of two α-perfect Hom-preLie algebras.



    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×LL and a linear map α:LL, we say that the triple (L,μ,α) is a Hom-preLie algebra if

    α(x)(yz)(xy)α(z)=α(y)(xz)(yx)α(z).

    for all x,y,zL. If the elements of L also satisfy the following equation

    (xy)α(z)=(xz)α(y).

    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)x2x3x1x2αL(x3)αL(x2)x1x3+x2x1αL(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 τσ on the semi-direct product. Casas and Pacheco Rego gave the linear map τσ 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,mq} m,mM,qQ. We also obtain a linear map τσ 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.

    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:MLM, ρ(ml)=ml and λ:LMM, λ(lm)=lm, the following identities hold.

    a) (xy)αM(m)αL(x)(ym)=(yx)αM(m)αL(y)(xm),

    b) (mx)αL(y)αM(m)(xy)=(xm)αL(y)αL(x)(my),

    c) (mm)αL(x)αM(m)(mx)=(mm)αL(x)αM(m)(mx),

    d)(xm)αM(m)αL(x)(mm)=(mx)αM(m)αM(m)(xm),

    e) αM(xm)=αL(x)αM(m),

    f) αM(mx)=αM(m)αL(x),

    for all x,yL and m,mM.

    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:MLM,ρ(m,l)=mk,π(k)=l, λ:LMM,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 (ML,˜α) is a Hom-preLie algebra, where ˜α:MLML is defined by ˜α(m,l)=(αM(m),αL(l)) and multiplication

    (m1,l1)(m2,l2)=(m1m2+αL(l1)m2+m1αL(l2),l1l2).

    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 πσ=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:MML,i(m)=(m,0) and π:MLL,π(m,l)=l. Then we obtain the following sequence

    0(M,αM)i(ML,˜α)π(L,αL)0.

    Furthermore, this sequence splits by σ:LML,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,αK) 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(AC,˜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

    ca=t(c)i(a);   ac=i(a)t(c).

    So we obtain the following split extension:

    where k:AAC,k(a)=(a,0),p:ACC,q(a,c)=c and τ:CAC,τ(c)=(0,c). Furthermore the Hom-action of (C,IdC) over (A,αA) induced by this extension coincides with the initial one:

    ca=σ(c)j(a)=(0,c)(a,0)=(0a+IdC(c)a+00,c0)ca.

    Since φ:(AC,˜α)(B,αB),φ(a,c)=i(a)+s(c) is a homomorphism of Hom-preLie algebras such that the following diagram commutative

    (1)

    the extensions are equivalent.

    Suppose that two extensions are equivalent, that is, there is a homomorphism of Hom-preLie algebras φ:(AC,˜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:LM such that:

    a) d(l1l2)=αL(l1)d(l2)+d(l1)αL(l2),

    b) dαL=αMd,

    for all l1,l2L.

    Example 2. a) Let (M,αM) be a Hom-representation of (ML,˜α) via π. Then the linear map θ:MLM,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:LL such that d(l1l2)=αL(l1)d(l2)+d(l1)αL(l2) and dαL=αMd.

    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,x2X) and every homomorphism of Hom-preLie algebras f:(X,αX)(L,αL) there is a unique homomorphism of Hom-preLie algebras h:(X,αX)(ML,˜α), such that the following diagram commute.

    Conversely, every homomorphism of Hom-preLie algebras h:(X,αX)(ML,˜α), decide a homomorphism of Hom-preLie algebras f=πh:(X,αX)(L,αL) and f-derivation d=θh:(X,αX)(M,αM), where θ(m,l)=x,mM,lL.

    Proof. Let h:XML,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)(ML,˜α), such that ιπh=IdL.

    Proof. Take (X,αX)=(L,αL) in Proposition 2.

    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

    00(L,αL)(L,αL)0,

    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 00(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 0Ker(μρ)(A,αA)μρ(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 μρφ=μ. By Lemma 4.7 in [13], we have ρφ=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 ff:LLLL given by (ff)(x1x2)=f(x1)f(x2), which maps IL to IL. So ff 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.

    (2)

    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

    Aut(L,αL){gAut(uce(L),˜α):g(HLα2(L))=HLα2(L)}.fuce(f)

    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)x2x3x1x2αL(x3)αL(x2)x1x3+x2x1αL(x3),x1,x2,x3L, respectively IL. f induces a linear map ff:(αL(L)αL(L),αLL)(αL(L)αL(L),αLL), given by ff(αL(x1)αL(x2))=αL(f(x1))αL(f(x2)) such that ff(IL)IL. Hence, it induces a homomorphism of Hom-preLie algebras uceα(f):(uceα(L),¯α)(uceα(L),¯α) given by uceα(f){αL(x1),αL(x2)}={αL(f(x1)),αL(f(x2))} such that the following diagram

    (3)

    is commutative.

    From diagram (3), there exists a covariant right exact functor uceα: Hom-preLieαperf Hom-preLieαperf 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

    Aut(L,αL){gAut(uceα(L),¯α):g(Ker(Uα))=Ker(Uα)}fuceα(f)

    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)x2x3x1x2αL(x3)αL(x2)x1x3+x2x1αL(x3),x1,x2,x3L. Indeed,

    d(αL(x1)x2x3x1x2αL(x3)αL(x2)x1x3+x2x1αL(x3))=d(αL(x1))αL(x2x3)+α2L(x1)d(x2x3)d(x1x2)α2L(x3)αL(x1x2)d(αL(x3))d(αL(x2))αL(x1x3)α2L(x2)d(x1x3)+d(x2x1)α2L(x3)+αL(x2x1)d(αL(x3))=αL(d(x1))αL(x2)αL(x3)+α2L(x1)d(x2)αL(x3)+α2L(x1)αL(x2)d(x3)d(x1)αL(x2)α2L(x3)αL(x1)d(x2)α2L(x3)αL(x1)αL(x2)αL(d(x3))αL(d(x2))αL(x1)αL(x3)α2L(x2)d(x1)αL(x3)α2L(x2)αL(x1)d(x3)+d(x2)αL(x1)α2L(x3)+αL(x2)d(x1)α2L(x3)+αL(x2)αL(x1)αL(d(x3)) IL.

    So it induces a linear map uceα(d):(uceα(L),¯α)(uceα(L),¯α), given by

    uceα(d)({αL(x1),αL(x1)})={d(αL(x1)),α2L(x2)}+{α2L(x1),d(αL(x2))},

    such that the following diagram

    (4)

    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

    uceα:Der(L,αL){δDer(uceα(L),¯α):δ(Ker(Uα))Ker(Uα)}duceα(d),

    Lemma 3.3. Let f:(L,αL)(L,αL) be a homomorphism of α-perfect Hom-preLie algebras. If d, dDer(L) satisfies fd=df, then uceα(f)uceα(d)=uceα(d)uceα(f).

    Proof. For any x1,x2L, we have

    uceα(f)uceα(d)({αL(x1),αL(x2)})=uceα(f)({d(αL(x1)),α2L(x2)}+{α2L(x1),d(αL(x2))})={αL(d(f(x1))),α2L(f(x2))}+{α2L(f(x1)),αL(d(f(x2)))}.

    On the other hand

    uceα(d)uceα(f)({αL(x1),αL(x2)})=uceα(d)({αL(f(x1)),αL(f(x2))})={αL(d(f(x1))),α2L(f(x2))}+{α2L(f(x1)),αL(d(f(x2)))}.

    Hence we prove the lemma.

    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 fUα:(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 fUαφ=Uα. So we have

    fUαφuceα(f)=Uαuceα(f)=fUα,

    that is to say the following diagram

    is commutative. Since fUα is an α-central extension which is universal in the sense of Definition 4.13 in [13], we have φuceα(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, Uαuceα(f)1:(uceα(L),¯α)(L,αL) is an α-cover. In the sequel, we will denote its kernel by

    C:=Ker(Uαuceα(f)1)=uceα(f)(Ker(Uα)).

    In fact, for any xKer(Uαuceα(f)1), we have Uαuceα(f)1(x)=0. Hence xuceα(f)(Ker(Uα)).

    Conversely, for any xuceα(f)(Ker(Uα)), there exists a yKer(Uα) such that x=uceα(f)(y). So Uαuceα(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 θhAut(L,αL) such that the following diagram is commutative:

    (5)

    if and only if the automorphism uceα(h) of (uceα(L),¯α) such that uceα(h)(C)=C. Futhermore, we obtain a group isomorphism:

    Θ:{hAut(L,αL):uceα(h)(C)=C}{gAut(L,αL):g(Ker(f))=Ker(f)}.hθh

    Proof. Let hAut(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=fUαuceα(f)1, so we have the following diagram:

    If uceα(h)(C)=C, then Uαuceα(f)1uceα(h)(C)=Uαuceα(f)1(C)=0, so there is a unique θh:(L,αL)(L,αL) satisfies θhUαuceα(f)1=Uαuceα(f)1uceα(h).

    On the other side, hfUαuceα(f)1=fUαuceα(f)1uceα(h)=fθhUαuceα(f)1. Since (L,αL) is an α-perfect Hom-preLie algebra and uceα(f)1 is an isomorphism, we have hf=fθh. Moreover, hf is an α-cover since Ker(hf)Ker(f)Z(L). Hence, θh is uniquely by Lemma 4.18 in [13]. Finally, θh(Ker(f))=Ker(f). Indeed, we have fθh(Ker(f))=hf(Ker(f))=0. Conversely, for xKer(f), there exists a yL 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 gAut(L,αL) with g(Ker(f))=Ker(f), gives rise to a unique homomorphism h:(L,αL)(L,αL) satisfies hf=fg. 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:

    Aut(L,αL){gAut(uceα(L),¯α):g(Ker(Uα))=Ker(Uα)}.huceα(h)

    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 dDer(L,αL), there exists a δdDer(L,αL) such that the following diagram is commutative

    (6)

    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

    Δ:{dDer(L,αL):uceα(d)(C)C}{ρDer(L,αL):ρ(Ker(f))Ker(f)}.dδd

    c) Let Uα:(uceα(L),¯αL)(L,αL) be an α-cover. Then there exists an isomorphism of Hom-vector spaces

    uceα:Der(L,αL){δDer(uceα(L),¯αL):δ(Ker(Uα))Ker(Uα)}.

    Proof. a) Let dDer(L,αL). Suppose that there exists a δdDer(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=fUαuceα(f)1 follows from diagram (3), hence we obtain the following diagram:

    If uceα(d)(C)C, then Uαuceα(f)1uceα(d)(C)Uαuceα(f)1(C)=0, so there exists a unique δd:(L,αL)(L,αL) such that δdUαuceα(f)1=Uαuceα(f)1uceα(d).

    On the other side, dfUαuceα(f)1=dUα=Uαuceα(d)=fUαuceα(f)1uceα(d)=fδdUαuceα(f)1. Since (L,αL) is an α-perfect Hom-preLie algebra and uceα(f)1 is an isomorphism, we have df=fδd. Moreover, df is an α-cover since Ker(df)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 ρDer(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).

    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 qm=s(q)t(m) and mq=t(m)s(q), qQ,mM. Sometimes, it is necessary to assume that the previous action is associative, i.e., (mm)q=m(mq), qQ,m,mM.

    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 ps=IdQ, the sequence

    is split. So uceα(p)uceα(s)=uceα(IdQ), i.e., πσ=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),

    λ({q1,q2}{αG(g1),αG(g2)})={q1,q2}{αG(g1),αG(g2)}=σ({q1,q2})i({αG(g1),αG(g2)})={s(q1),s(q2)}{αG(g1),αG(g2)}={s(q1q2),αG(g1g2)}

    and ρ:Ker(π)uce(Q)Ker(π),

    ρ({αG(g1),αG(g2)}{q1,q2})={αG(g1),αG(g2)}{q1,q2}=({αG(g1),αG(g2)})σ({q1,q2})={αG(g1),αG(g2)}{s(q1),s(q2)}={αG(g1g2),s(q1q2)}.

    By Lemma 2.4, the split sequence

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

    (uceα(G),¯αG)(Ker(π),¯αG)(uce(Q),Iduce(Q)).

    Let qQ and αM(m1),αM(m2)αM(M). In (uceα(G),¯αG), we have

    {αG(s(q)),t(αM(m1))t(αM(m2))}={s(q)t(αM(m1)),αG(t(αM(m2)))}+{αG(t(αM(m1))),s(q)t(αM(m2))}{t(αM(m1))s(q),αG(t(αM(m2)))}

    and

    {t(αM(m1))t(αM(m2)),αG(s(q))}={αG(t(αM(m1))),t(αM(m2))s(q)}{αG(t(αM(m2))),t(αM(m1))s(q)}+{t(αM(m2))t(αM(m1)),αG(s(q))}.

    These above equalities and the α-perfection of (M,αM) imply:

    {s(Q),M}={sIdQ(Q),M}={αGs(Q),αM(M)αM(M)}{s(Q)αM(M),αG(αM(M))}+{αG(αM(M)),s(Q)αM(M)}{αM(M)s(Q),αG(αM(M))}{αG(M),α2G(M)}+{α2G(M),αG(M)}{αG(M),α2G(M)}{αM(M),αM(M)}

    and

    {M,s(Q)}={αM(M)αM(M),αGs(Q)}{αG(αM(M)),αM(M)s(Q)}{αM(M),αM(M)}.

    Futhermore,

    τ(uceα(M),¯αM)({αM(M),αM(M)},αG) (7)

    and

    σ(uce(Q))={s(Q),s(Q)}={αG(s(Q)),αG(s(Q))}

    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

    (uceα(G),¯αG)=({s(Q),s(Q)}+{αM(M),αM(M)},¯αG). (8)

    Proof. For any αG(g)G, there exists an αM(m)αM(M) such that αG(g)=s(p(αG(g)))+αM(m). Hence

    {αG(g1),αG(g2)}={s(p(αG(g1)))+αM(m1),s(p(αG(g2)))+αM(m2)}={s(p(αG(g1))),s(p(αG(g2)))}+{s(p(αG(g1))),αM(m2)}+{αM(m1),s(p(αG(g2)))}+{αM(m1),αM(m2)}{s(Q),s(Q)}+{s(Q),M}+{M,s(Q)}+{αM(M),αM(M)}{s(Q),s(Q)}+{αM(M),αM(M)}.

    Conversely,

    {s(Q),s(Q)}+{αM(M),αM(M)}{αG(G),αG(G)}=uceα(G).

    Hence, we prove the lemma.

    Proposition 5. With the above notations, we have

    (Ker(π),¯αG)=({αM(M),αM(M)},¯αG)=τ(uceα(M),¯αM).

    Proof. Let {g1,g2}Ker(π). Then by Eq.(8), we have

    {g1,g2}={s(q1),s(q2)}+{αM(m1),αM(m2)}uceα(G).

    Hence 0=π{g1,g2}={p(s(q1)),p(s(q2))}+{p(αM(m1)),p(αM(m2))}={q1,q2}, i.e., q1q2IQ. 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 πσ=Id, we have

    (uceα(G),αG)=(Ker(π),¯αG)σ(uce(Q),Iduce(Q)).

    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))=0tUMα(m)=UGα(τ(m))=0,suQ(q)=UGα(σ(q))=0mKer(UMα),qHL2(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:

    λ:uce(Q)uceα(M)uceα(M){q1,q2}{αM(m1),αM(m2)}{q1,q2}{αM(m1),αM(m2)}={(q1q2)αM(m1),α2M(m2)}{αM(m1)(q1q2),α2M(m2)}+{α2M(m1),(q1q2)αM(m2)}

    and

    ρ:uceα(M)uce(Q)uceα(M){αM(m1),αM(m2)}{q1,q2}{αM(m1),αM(m2)}{q1,q2}={α2M(m1),αM(m2)(q1q2)}.

    When it is need, the Hom-action of (Q,IdQ) on (M,αM) is uce-associative, i.e., {mm,q}={m,mq}.

    So we define the following homomorphism of Hom-preLie algebras

    τσ:(uceα(M),¯αM)(uce(Q),Iduce(Q))(uceα(G),¯αG)τ(uceα(M),αM)σ(uce(Q),Iduce(Q))({αM(m1),αM(m2)},{q1,q2})({t(αM(m1)),t(αM(m2))},{s(q1),s(q2)}),

    where the Hom-action of (Q,IdQ) on (M,αM) is uce-associative.

    We obtain that τσ is an epimorphism since

    (uceα(G),¯αG)τ(uceα(M),αM)σ(uce(Q),Iduce(Q)).

    Next we define a surjective homomorphism of Hom-preLie algebras

    Φ:=(tUMα)(suQ):(uceα(M)uce(Q),¯αMIduce(Q))(G,αG)({αM(m1),αM(m2)},{q1,q2})(t(αM(m1)αM(m2)),s(q1q2)),

    such that the following diagram

    (9)

    is commutative. We have that

    uce(Q)Ker(UMα)Ker(UMα)uce(Q)Ker(τ)Ker(UMα).

    Second inclusion holds since tUMα=UGατ and t is injective. Since the following diagram is commutative

    UGατ(Ker(UMα))=tUMα(Ker(UMα))=0, then τ(Ker(UMα))Ker(UGα)Z(uceα(G)), so

    τ(uce(Q)Ker(UMα))=σ(uce(Q))τ(Ker(UMα))=0

    and

    τ(Ker(UMα)uce(Q))=τ(Ker(UMα))σ(uce(Q))=0.

    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

    (¯uceα(M),¯¯αM)=(uceα(M)uce(Q)Ker(UMα)Ker(UMα)uce(Q),¯¯αM).

    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),¯¯αMIduce(Q)) satisfies Ψμ=Id. Since ΨμΨ=IdΨ=ΨId and (¯uceα(M)uce(Q),¯¯αMIduce(Q)) is α-perfect, then μΨ=Id follows from Lemma 4.18 in [13]. Hence, Ψ is an isomorphism, then Ker(Ψ)=Ker(τσ)I=0. Consequently, Ker(τσ)I.

    The above discussion can be summarized in:

    Ker(τσ)uce(Q)Ker(UMα)Ker(UMα)uce(Q).

    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

    Φ:(uceα(M)uce(Q),¯αMIduce(Q))(G,αG)

    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(τσ)uce(Q)Ker(UMα)Ker(UMα)uce(Q).

    Theorem 5.4. The following statements are equivalent:

    a) Φ:=(tUMα)(suQ):(uceα(M)uce(Q),¯αMIduce(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) τσ is an isomorphism. Hence uceα(M)uce(Q) is the universal α-central extension of (G,αG).

    d) τ is injective. In particular,

    (uceα(M×Q),¯αM×IdQ)(uceα(M)×uce(Q),¯αM×Iduce(Q)).

    Proof. a)b) If Φ:=(tUMα)(suQ):(uceα(M)uce(Q),¯αMIduce(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),¯αMIduce(Q)) is α-perfect. Consequently, the extension is an α-cover.

    b)c) By Theorem 5.3, we have that Ker(τσ)uce(Q)Ker(UMα)Ker(UMα)uce(Q), then τσ 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)}=(τσ)({αM(m1),αM(m2)},0), since Ker(τ)=Ker(τσ). 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).

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



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