Complex cyclic Leibniz superalgebras

Complex cyclic Leibniz superalgebras L. M. Camacho1 · R. M. Navarro2 · B. A. Omirov3 Abstract Since Loday introduction of Leibniz algebras as a generalisation of Lie algebras, many results of the theory of Lie algebras have been extended to Leibniz algebras. Cyclic Leibniz algebras, which are generated by one element, have no equivalent into Lie algebras, though. This fact provides cyclic Leibniz algebras with important properties. Throughout the present paper we extend the concept of cyclic to Leibniz superalgebras, obtaining then the definition, as long as the description and classification of finite-dimensional complex cyclic Leibniz superalgebras. Furthermore, we prove that any cyclic Leibniz superalgebra can be obtained by means of infinitesimal deformations of the null-filiform Leibniz superalgebra. We also obtain a description of irreducible components in the variety of Leibniz algebras and superalgebras. Keywords Leibniz superalgebra · Cyclic Leibniz superalgebra · Null-filiform superalgebra · Infinitesimal deformation · Irreducible component Mathematics Subject Classification 17A32 · 17A70 · 17B30 · 13D10 This work has been supported by Agencia Estatal de Investigación (Spain), Grant MTM2016-79661-P (European FEDER support included, EU) and by the European Union under the 2014-2020 ERDF Operational Programme and by the Department of Economy, Knowledge, Business and University of the Regional Government of Andalusia. Project Reference : FEDER-UCA18-107643. BL. M. Camacho [email protected] R. M. Navarro rnav[email protected] B. A. Omirov [email protected] 1Dpto. Matemática Aplicada I, Universidad de Sevilla, Sevilla, Spain 2Dpto. de Matemáticas, Universidad de Extremadura, Cáceres, Spain 3National University of Uzbekistan, Tashkent, Uzbekistan 1 Introduction Leibniz algebras were introduced in the 90’s by Loday as a generalization of Lie algebras and since then many results of the theory of Lie algebras have been extended to Leibniz algebras [3,4,7,10,14,15]. The notion of Leibniz superalgebras was introduced in [1] (they were also considered under the name of graded Leibniz algebras in [12]). Furthermore, Leibniz superalgebras are generalizations of Leibniz algebras and, on the other hand, they naturally generalize well-known Lie superalgebras. It should be noted that the study of Leibniz superalgebras is much more complicated since the graded anti-symmetric identity does not hold in Leibniz superalgebras. The most interesting examples of non-Lie Leibniz algebras are cyclic Leibniz algebras, those Leibniz algebras generated by one element and do not appear in Lie algebras, except one-dimensional abelian algebra. In nilpotent algebras case such algebras are called null-filiform Leibniz algebras [2], while solvable one-generated Leibniz algebras are called cyclic algebras [16]. To the study of Leibniz superalgebras have been devoted a few works, we just mention the papers [6,9] (and reference therein), where the description of nilpotent Leibniz superalgebras of nilindex equal to the dimension of the superalgebra is obtained. In this paper we extend the results on one-generated Leibniz superalgebras (which are always solvable). More precisely, we extend the concept of cyclic Leibniz algebras to Leibniz superalgebras, obtaining then the definition, as long as the description and classification of finite-dimensional complex cyclic Leibniz superalgebras. In fact, we showed that any cyclic Leibniz superalgebra is an infinitesimal deformation of null-filiform superalgebra. On the other hand, let us consider V an (n + m)-dimensional vector space over a field F. Then, the bilinear maps V × V → V form an (n + m)3-dimensional affine space. Recall that a set is called irreducible if it cannot be represented as a union of two nontrivial closed subsets (in the sense of Zariski topology), otherwise it is called reducible. The maximal irreducible closed subset of a variety is called an irreducible component. Moreover, from algebraic geometry we know that an algebraic variety is the union of its irreducible components and that closures of open irreducible subsets produce irreducible components. Therefore, for the description of a variety it is very important to find all open sets. Thus, throughout this paper we find an irreducible component of the variety of (n + m)-dimensional Leibniz superalgebras, which is a closure of union of orbits of cyclic Leibniz superalgebras. Henceforth, we will consider vector spaces and algebras over the field of complex numbers C. 2 Basic notions and preliminaries results Let us start with some basics regarding Leibniz superalgebras. We suppose that the reader is familiarised with the basics of Leibniz algebras. Definition 2.1 [1]. A Z2-graded vector space L = L0¯ ⊕ L ¯1 over a field F is called a Leibniz superalgebra if it is equipped with a product [−, −] which satisfies the following conditions: [Lα,Lβ]⊆Lα+β(mod 2)for allα, β ∈Z2 [x,[y,z]] = [[x,y],z]−(−1)αβ [[x,z],y]− graded Leibniz identity for all x∈L,y∈Lα,z∈Lβ,α, β ∈Z2. Note that if a Leibniz superalgebra L satisfies the identity [x,y]=−(−1)αβ [y,x] for all x∈Lα,y∈Lβ, then the graded Leibniz identity is reduced to the graded Lie identity. Therefore Leibniz superalgebras are a generalization of Lie superalgebras. Note that as usual L=L0is a Leibniz algebra. Hence, if we replace L1by {0}in the next results and definitions we obtain the equivalent results for Leibniz algebras. Recall that isomorphisms between Leibniz superalgebras are assumed to be consistent with the Z2-graduation. We call a Z2-graded vector space M=M0⊕M1a module over a Leibniz superalgebra Lif there are two bilinear maps: [−,−] : L×M→Mand [−,−] : M×L→M satisfying the following three axioms [m,[x,y]] = [[m,x],y]−(−1)|x||y|[[m,y],x], [x,[m,y]] = [[x,m],y]−(−1)|y||m|[[x,y],m], [x,[y,m]] = [[x,y],m]−(−1)|m||y|[[x,m],y], for any m∈M|m|,x∈L|x|,y∈L|y|. Given a Leibniz superalgebra L,letCn(L,M)be the space of all F-linear homogeneous mappings L⊗n→M,n≥0 and C0(L,M)=M. This space is graded by Cn(L,M)=Cn 0(L,M)⊕Cn 1(L,M)with Cn p(L,M)= n0+n1=n n1+r≡p(mod 2) Hom(L⊗n0 0⊗L⊗n1 1,Mr) Let dn:Cn(L,M)→Cn+1(L,M)be an F-homomorphism defined by (dnf)(x1,...,xn+1):= [x1,f(x2,...,xn+1)] + n+1  i=2 (−1)i+|xi|(|f|+|xi+1|+···+|xn+1|)[f(x1,...,xi,...,xn+1), xi] + 1≤i<j≤n+1 (−1)j+1+|xj|(|xi+1|+···+|xj−1|)f(x1,...,xi−1,[xi,xj], xi+1,...,xj,...,xn+1), where f∈Cn(L,M)and xi∈L. Since the differential operator d = i≥0 di satisfies the property d ◦ d = 0, the cohomology group is well defined and HLn p(L, M) = ZLn p(L, M)/BLn p(L, M), p ∈{0, 1} where the elements ZLn 0(L, M) (BLn 0(L, M)) and ZLn 1(L, M) (BLn 1(L, M))are called even n-cocycles (even n-coboundaries) and odd n-cocycles (odd n-coboundaries), respectively. Let us remark that formula for dn can be obtained from the differential operator for color Leibniz algebras [8]. Note that the space ZL1(L, L) consists of derivations of the superalgebra L, which are defined by the condition: d([x, y]) = (−1)|d||y|[d(x), y]+[x, d(y)]. If we denote by Rx the right multiplication operator, i.e., Rx : L → L, then due to graded Leibniz identity it can be seen that Rx is a derivation. Note that the graded Leibniz identity can be expressed in the following form: R[x,y] = Ry Rx − (−1)αβ Rx Ry where x ∈ Lα, y ∈ Lβ . We denote by R(L) the set of all right multiplication operators. It is not difficult to check that R(L) with the multiplication defined as follows < Ra, Rb >:= Ra Rb − (−1)αβ Rb Ra for Ra ∈ R(L)α, Rb ∈ R(L)β , becomes a Lie superalgebra. For a Leibniz superalgebra L = L0 ⊕ L1 we consider the set Annr (L), Annr (L) = {X ∈ L :[L, X]= 0} which is called the right annihilator of L. It is easy to see that Annr (L) is a two-sided ideal of L and [X, X]∈ Annr (L) for any X ∈ L0, this notion is consistent with the right annihilator in Leibniz algebras. If we consider I = ideal < [X, Y]+(−1)αβ [Y , X]: X ∈ Lα, Y ∈ Lβ >, then I ⊆ Annr (L). A deformation of a Leibniz superalgebra L is a one-parameter family Lt of Leibniz superalgebras with the bracket μt = μ0 + tϕ1 + t2ϕ2 + ··· , where ϕi are L-valued even 2-cochains, i.e., elements of Hom(L ⊗ L, L)0 = C2(L, L)0. Two deformations Lt , Lt with corresponding laws μt,μt are equivalent if there exists a linear automorphism ft = id + f1t + f2t2 + ··· of L, where fi are elements of C1(L, L)0 such that the following equation holds μ t (x, y) = ft −1(μt ( ft (x), ft (y))) for any x, y ∈ L. Applying the graded Leibniz identity for the superalgebras Lt implies that the 2- cochain ϕ1 is an even 2-cocycle, i.e. d2ϕ1 = 0. If ϕ1 vanishes identically, the first non vanishing ϕi will be a 2-cocycle. If μt is an equivalent deformation with cochains ϕi , then ϕ 1 − ϕ1 = d1 f1, hence every equivalence class of deformations defines uniquely an element of HL2(L, L)0. Note that the linear integrable deformation ϕsatisfies the condition ϕ(x,ϕ(y,z)) −ϕ(ϕ(x,y), z)+(−1)|y||z|ϕ(ϕ(x,z), y)=0. It should be noted that a Leibniz algebra is a superalgebra with trivial odd part and the definition of cohomology groups of Leibniz superalgebras extends the definition of cohomology groups of Leibniz algebras given in [13]. For a Leibniz superalgebra Lconsider the following central lower and derived series L1=L,Lk+1=[Lk,L1],k≥1, L[1]=1,L[s+1]=[L[s],L[s]],s≥1. Definition 2.2 A Leibniz superalgebra Lis said to be nilpotent, (respectively, solvable) if there exists p∈Nsuch that Lp=0 (respectively, L[q]=0). Now we give the notion of null-filiform Leibniz superalgebra. Definition 2.3 An n-dimensional Leibniz superalgebra is said to be null-filiform if dim Li=n+1−i,1≤i≤n+1. Similarly to the case of nilpotent Leibniz algebras [2] it is easy to check that a Leibniz superalgebra is null-filiform if and only if it is single-generated. Moreover, a null-filiform superalgebra has the maximal nilindex. Theorem 2.1 [1]LetLbe a null-filiform Leibniz superalgebra of the variety Leibn,m. Then Lis isomorphic to one of the following non-isomorphic superalgebras: NFn:[xi,x1] =xi+1,1≤i≤n−1;NFn,m:⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ [yi,y1]=xi,1≤i≤n, [xi,y1]=1 2yi+1,1≤i≤m−1, [yj,x1]=yj+1,1≤j≤m−1, [xi,x1]=xi+1,1≤i≤n−1. where {x1,x2,...,xn}and {y1,y2,...,ym}are bases of the even and odd parts, respectively. Remark 2.1 Note that the first superalgebra is a null-filiform Leibniz algebra [2] and from the assertion of Theorem 2.1 we conclude that in the case of non-trivial odd part of the null-filiform Leibniz superalgebra NFn,mthere are two possibilities for m, namely, m=nor m=n+1. Next, we recall some results about cyclic Leibniz algebras. We transfrom the results in [16] regarding cyclic left-Leibniz algebras to the cyclic right-Leibniz algebras. Thus, let Abe an n-dimensional vector space over Ccontaining a nonzero element a. Suppose we have a linear operator T:A→Asuch that ais a cyclic vector for T, i.e., such that B={a,T(a),...,Tn−1(a)}is a basis of A. Therefore Tn(a)= α1a+α2T(a)+···+αnTn−1(a)for some α1,...,α n∈C. Let us define a product A×Ca→Aby v(ca)=cT(v) for all v∈Aand c∈C, therefore Tcan be seen as the right multiplication operator by a(that is T(a)=Ra). For simplicity we denote ak:= Rk−1 a(a). Let us recall now the Leibniz identity [[x,y],z]=[[x,z],y]+[x,[y,z]] which can be rewritten as Rz([x,y])=[Rz(x), y]+[x,Rz(y)]. This latter condition is nothing but the condition for Rzto be a derivation. Therefore we have a general equivalence, i.e. Rzis a derivation for all z∈Aiff Ais a Leibniz algebra Additionally, in our case (one-generated algebras) we have that right multiplication operator is a derivation (that is, Ais a Leibniz algebra) if and only if Raj=0for all j≥2. In fact if Ais a Leibniz algebra is a well-known fact that Ra2=0 and then by induction it can be easily check that Raj=0 for all j≥2. For the converse, suppose Raj=0 for all j≥2. Then, we consider an arbitray x∈A, then xcan be expressed as a linear combination of the basis vectors {a,a2,...,an}, that is: x=c1a+c2a2+···+an. Since Raj=0 for all j≥2, then Rx=c1Ra,soit suffices to prove that Rais a derivation. The condition for being a derivation over the basis vectors remains: Ra([ai,aj])=[Ra(ai), aj]+[ai,Ra(aj)] this equation trivially holds for all j≥2asRaj=0. For j=1 and since [ai,a2]=0 we get ai+2=ai+2which trivially holds. Let us now remark that the authors in [16] use a different characterization for (left) Leibniz algebras to have (left) multiplication operator as a derivation. Nevertheless, the classificaton Theorem remains valid and adapting the result to right Leibniz algebras we obtain Theorem 2.2 [16]letAbe an n-dimensional cyclic Leibniz algebra over C. Then Ais isomorphic to a Leibniz algebra spanned by {a,a2,...,an}and with bracket products one of the following: (1) NFn:[ai,a]=ai+1,1≤i≤n−1 (2) CNFn:[ai,a]=ai+1,1≤i≤n−1 [an,a]=an (3) CNFn(0,...,1,α k+1,...,α n): ⎧ ⎨ ⎩ [ai,a]=ai+1,1≤i≤n−1 [an,a]=ak+αk+1ak+1 +···+αnan,2≤k≤n−1,(α k+1,··· ,α n)∈Cn−k/∼ With the equivalence relation ∼defined over Cd, with d=n−k, as follows: (γ1,γ 2,...,γ d)∼(γ  1,γ 2,...,γ d)if (γ1,γ 2,...,γ d)=(ωdγ 1,ω d−1γ 2, ...,ωγ d) for some (d+1)-th root of unity ω. 3 Complex cyclic Leibniz superalgebras Let A=A0⊕A1be a (n+m)-dimensional Z2-graded vector space, i.e. dim(A0)=n and dim(A1)=m. Suppose we have a non-zero element aand a linear operator T: A→Asuch that B={a,T(a), . . . , Tn+m−1(a)}is a homogeneous basis for A. Then Tn+m(a)=α1a+α2T(a)+···+αn+m−1Tn+m−1(a)for some α1,...,α n+m−1∈C. We define the following product: A×Ca→Aby v(ca)=cT(v) for all v∈Aand c∈C, therefore Tcan be seen as the right multiplication by a, i.e. Ra. We denote Rk−1 a(a)by ak. We intend to extend this product linearly such that right multiplication is a derivation, i.e. A=A0⊕A1is a Leibniz superalgebra. Note that Rais a (super) derivation if and only if A=A0⊕A1is a Leibniz superalgebra. In fact, the condition for being a derivation of a Leibniz superalgebra (for more details see [11]) is d([x,y])=(−1)|d||y|[d(x), y]+[x,d(y)]. Since the degree of Rzas homomorphism between Z2-graded vector spaces is the same as the degree of the homogeneous element z, that is |Rz|=|z|, then the condition for Rz to be a derivation is exactly Rz([x,y])=(−1)|z||y|[Rz(x), y]+[x,Rz(y)].This last condition can be rewritten [[x,y],z]=(−1)|z||y|[[x,z],y]+[x,[y,z]] which is nothing but the graded Leibniz identity. Remark 3.1 In the setting defined above, a∈A1in order not to have a degenerate case (Leibniz algebra case). In fact, if |a|= 0 then |aj|= 0 for all 2 ≤j≤n+m, which means that all the basis vectors are even and m=0, being A=A0a Leibniz algebra. Note that cyclic Leibniz algebras have been already studied in [16]. Hereafter we consider aan odd element, and consequently ajis even for jeven and odd otherwise. Remark 3.2 In the setting defined above, we have only two possibilites for m, that is, either m=nor m=n+1. Proposition 3.1 Let A=A0⊕A1be a (n+m)-dimensional Z2-graded vector space such that we have a non-zero odd element a and a linear operator T :A→A verifying that B ={a,T(a), ..., Tn+m−1(a)}is a homogeneous basis for A. The right multiplication by a, i.e. Rais defined as the linear operator T together with the following product: A×Ca→Awith v(ca)=cT(v) for all v∈Aand c ∈C. Under the above conditions, right multiplication operator is a derivation (Ais a Leibniz superalgebra) if and only if Raj=0for all j ≥3and Ra2=2RaRa. Proof Suppose right multiplication operator is a derivation, then R(A)-thesetof all right multiplication operators - has Lie superalgebra structure with the following multiplication: <Ra,Rb>:= RaRb−(−1)¯ i¯ jRbRa,Ra∈R(A)¯ i,Rb∈R(A)¯ j a a it can be seen then that Ra2 =< Ra, Ra >= 2Ra Ra. Since R j (x) = Ra(Ra j−1(x)) for all j ≥ 2 and for all x ∈ A and a j := Ra j−1(a), then a j+1 = R j (a) = Ra(Ra j−1(a)) = Ra(a j ) =[a j , a]. Therefore, we have Ra3 = R[a2,a] =< Ra2 , Ra >= Ra2 Ra − Ra Ra2 = 2Ra Ra Ra − 2Ra Ra Ra = 0. Since Ra j+1 = R[a j ,a] =< Ra j , Ra >= Ra j Ra − (−1) j Ra Ra j , j ≥ 3 then by induction we have Ra j = 0 for all j ≥ 3. On the other hand, any x ∈ A can be expressed as a linear combination of the basis vectors, i.e., x = α1a + α2a2 + ··· + αn+man+m. This latter fact leads to Rx = α1 Ra +α2 Ra2 on account of Ra j = 0 for all j ≥ 3. Since it is enough to consider x an homogeneous element, then either α1 = 0 or α2 = 0 and therefore it suffices to show that Ra and Ra2 are derivations over A, Ra an odd derivation and Ra2 an even one. By linearity we only have to check the condition of being a derivation over the basis vectors {a, a2,...,an+m }. On account of | a j | is either 0 or 1 depending on if j is even or odd respectively, then the equation for Ra to be an odd derivation remains Ra([ai , a j ]) = (−1) j [Ra(ai ), a j ]+[ai , Ra(a j )] this equation trivially holds for all j ≥ 3as Ra j = 0. For j = 2 and since [ai , a3]= 0 it remains Ra([ai , a2]) =[Ra(ai ), a2] replacing Ra2 by 2Ra Ra the above equation trivially holds, i.e. 2ai+3 = 2ai+3.For j = 1 we get Ra(ai+1) =−[ai+1, a]+[ai , Ra(a)] but as Ra2 = 2Ra Ra, then we have from the above equation ai+2 =−ai+2 +2ai+2 = ai+2. Finally, the condition for Ra2 to be an even derivation over the basis vectors is Ra2 ([ai , a j ]) =[Ra2 (ai ), a j ]+[ai , Ra2 (a j )] this equation trivially holds for j ≥ 3, and also holds for j = 2 and j = 1 taking into account [ai , a4]= 0 and [ai , a3]= 0, respectively.  Corollary 3.1 If A is Leibniz superalgebra with m = n, then an+m+1 is an odd element and can be expressed as a linear combination of the odd basis vectors, i.e., an+m+1 = α1a + α3a3 + ...αn+m−1an+m−1, then since Ra j = 0 for all j ≥ 3 we have 0 =[a, an+m+1]=[a,(α1a + α3a3 + ...αn+m−1an+m−1)]= α1a2, which implies that α1=0. Analogously, if m =n+1, then an+m+1is and even element and can be expressed as a linear combination of the even basis vectors, i.e. an+m+1=α2a2+α4a4+...α n+m−1an+m−1, then 0=[a,an+m+1]=[a,(α 2a2+α4a4+...α n+m−1an+m−1)]=α2a3 obtaining then α2=0. 4 Classification of cyclic Leibniz superalgebras over C Since throughout this section we consider Leibniz superalgebras, we denote by [,] the standard bracket product as usual. Next, we start with the nilpotent case. Proposition 4.1 The only one, up to isomorphism, non-degenerate nilpotent cyclic Leibniz superalgebra is exactly N Fn,mwith m =norm=n+1. Proof Note that the only one, up to isomorphism, null-filiform Leibniz superalgebra (non Leibniz algebra) is NFn,m. Moreover, in order to have a non-trivial odd part we have only two possibilites for m(m=nor m=n+1). For more details it can be consulted [1]. After a very carefully chosen isomorphism, change of scale, we have a new law for NFn,mverifying in particular [y i,y 1]=x i,1≤i≤n,and [x i,y 1]=y i+1,1≤i≤m−1 and with some non-null scalars in the other two set of brackets. Replacing then y 1by a(and then x 1=a2,y 2=a3,x 2=a4,and so on) we obtain [ak,a]=Ra(ak)=ak+1which concludes the proof.  Thus, hereafter we consider the non-nilpotent case, i.e. [an+m,a] = 0 Proposition 4.2 Any finite dimensional non-nilpotent and non-degenerate cyclic Leibniz superalgebra, L =L¯ 0⊕L¯ 1, with dim(L¯ 0)=n and dim(L¯ 0)=m, can be expressed in a homogeneous basis {a,a2,...an+m}by one of the following laws •If m=n: [ai,a]=ai+1,1≤i≤n+m−1, [an+m,a]=αkak,3≤k≤n+m−1, [ai,a2]=2ai+2,1≤i≤n+m−2, [an+m−1,a2]=2αkak,3≤k≤n+m−1, [an+m,a2]=2αkak+1,3≤k≤n+m−1, with some αk= 0and k odd, •If m=n+1: [ai,a]=ai+1,1≤i≤n+m−1, [an+m,a]=αkak,4≤k≤n+m−1, [ai,a2]=2ai+2,1≤i≤n+m−2, [an+m−1,a2]=2αkak,4≤k≤n+m−1, [an+m,a2]=2αkak+1,4≤k≤n+m−1, with some αk= 0and k even. where the omitted products are equal to zero. Moreover, aifor all i odd is the basis of the odd part and aifor all i even is the basis of the even part of the Leibniz superalgebra. References 1. 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