On complex nilpotent Leibniz superalgebras of nilindex n+m

ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M L.M. CAMACHO, J. R. G´ OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV Abstract. We present the description up to isomorphism of Leibniz superalgebras with characteristic sequence (n|m1, . . . , mk) and nilindex n+m, where m=m1+· · · +mk, n and m(m6= 0) are dimensions of even and odd parts, respectively. Mathematics Subject Classification 2000: 17A32, 17B30. Key Words and Phrases: Lie superalgebras, Leibniz superalgebras, nilindex, characteristic sequence. 1. Introduction Extensive investigations of Lie algebras theory have lead to appearance of more general algebraic objects, such as Mal’cev algebras, Lie superalgebras, Leibniz algebras, and others. The well-known Lie superalgebras are generalizations of Lie algebras and for many years they attract the attention of both the mathematicians and physicists. The systematical exposition of basic Lie superalgebras theory can be found in the monograph [12] and the papers related with the nilpotent Lie superalgebras are [5], [8], [10], and [11]. Leibniz superalgebras are the generalization of Leibniz algebras and, on the other hand, they naturally generalize Lie superalgebras. Recall that Leibniz algebras are a ”non antisymmetric” generalization of Lie algebras [13]. The study of nilpotent Leibniz algebras [1]–[3] shows that many nilpotent properties of Lie algebras can be extended for nilpotent Leibniz algebras. The results of the nilpotent Leibniz algebras may help us to investigate the nilpotent Leibniz superalgebras. In the description of Leibniz superalgebras structure the crucial task is to prove the existence of a suitable basis (so-called the adapted basis) in which the multiplication of the superalgebra has the most convenient form. In contrast to Lie superalgebras for which problem of the description of superalgebras with the maximal nilindex is difficult [10], for nilpotent Leibniz superalgebras it turns out to be comparatively easy and was solved in [1]. The distinctive property of such Leibniz superalgebras is that they are single-generated. The next step - the description of Leibniz superalgebras with the dimensions of even and odd parts, respectively equal to nand m, and with nilindex n+mat this moment seems to be very complicated. Therefore, such Leibniz superalgebras can be studied by applying restrictions on their characteristic sequences [4]–[9]. Following this Partially supported by the PAICYT, FQM143 of the Junta de Andaluc´ıa (Spain). The third author was supported by grant NATO-Reintegration ref. CBP.EAP.RIG.983169. 1 2 L.M. CAMACHO, J. R. G ´ OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV approach we investigate such Leibniz superalgebras with characteristic sequence C(L) = (n|m1, . . . , mk), where m1+···+mk=m. Taking into account the results of the papers [6] and [9], where some cases of nilpotent Leibniz superalgebras were described, in this work we investigate the rest cases. Thus, we complete the description of Leibniz superalgebras with characteristic sequence C(L) = (n|m1,...,mk) and nilindex n+m. All over the work we consider spaces and algebras over the field of complex numbers. By asterisks (∗) we denote the appropriate coefficients at the basic elements of superalgebra. 2. Preliminaries AZ2-graded vector space G=G0⊕G1is called a Lie superalgebra if it is equipped with a product [−,−] which satisfies the following conditions: 1. [Gα,Gβ]⊆ Gα+β(mod 2), 2. [x, y] = −(−1)αβ[y, x], 3. (−1)αγ [x, [y, z]]+(−1)αβ[y, [z, x]]+(−1)βγ [z, [x, y]] = 0 – Jacobi superidentity for any x∈ Gα, y ∈ Gβ, z ∈ Gγand α, β, γ ∈Z2. AZ2-graded vector space L=L0⊕ L1is called a Leibniz superalgebra if it is equipped with a product [−,−] which satisfies the following conditions: 1. [Lα,Lβ]⊆ Lα+β(mod 2), 2. [x, [y, z]] = [[x, y], z]−(−1)αβ[[x, z], y]−Leibniz superidentity, for any x∈ L, y ∈ Lα, z ∈ Lβand α, β ∈Z2. The vector spaces L0and L1are said to be even and odd parts of the superalgebra L, respectively. Note that if in Lthe graded identity [x, y] = −(−1)αβ[y, x] holds, then the Leibniz superidentity and Jacobi superidentity coincide. Thus, Leibniz superalgebras are a generalization of Lie superalgebras. For examples of Leibniz superalgebras we refer to [1]. Denote by Leibn,m the set of Leibniz superalgebras with dimensions of the even part and the odd part equal to nand m, respectively. Let V=V0⊕V1, W =W0⊕W1be two Z2-graded spaces. A linear map f:V→Wis called of degree α(denoted as deg(f) = α), if f(Vβ)⊆Wα+βfor all β∈Z2. Let Land L′be Leibniz superalgebras. A linear map f:L → L′is called a homomorphism of Leibniz superalgebras if 1. f(L0)⊆ L′ 0and f(L1)⊆ L′ 1,i.e. deg(f) = 0; 2. f([x, y]) = [f(x), f(y)] for all x, y ∈ L. Moreover, if fis bijection then it is called an isomorphism of Leibniz superalgebras Land L′. For a given Leibniz superalgebra Lwe define a descending central sequence in the following way: L1=L,Lk+1 = [Lk,L1], k ≥1. Definition 2.1. A Leibniz superalgebra Lis called nilpotent, if there exists s∈N such that Ls= 0.The minimal number swith this property is called nilindex of the superalgebra L. The sets R(L) = {z∈ L | [L, z] = 0}and Z(L) = {z∈ L | [L, z] = [z, L] = 0} ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 3 are called the right annihilator and the center of a superalgebra L,respectively. Using the Leibniz superidentity it is not difficult to see that R(L) is an ideal of the superalgebra L. Moreover, the elements of the form [a, b] + (−1)αβ[b, a], (a∈ Lα, b ∈ Lβ) belong to R(L). The following theorem describes the nilpotent Leibniz superalgebras with maximal nilindex. Theorem 2.2. [1] Let Lbe a Leibniz superalgebra of the variety Leibn,m with nilindex equal to n+m+ 1.Then Lis isomorphic to one of the following nonisomorphic superalgebras: [ei, e1] = ei+1,1≤i≤n−1; [ei, e1] = ei+1,1≤i≤n+m−1, [ei, e2] = 2ei+2,1≤i≤n+m−2, (omitted products are equal to zero). It should be noted that for the second superalgebra we have m=nwhen n+m is even and m=n+ 1 if n+mis odd. Moreover, it is clear that the Leibniz superalgebra has the maximal nilindex if and only if it is single-generated. Let L=L0⊕ L1be a nilpotent Leibniz superalgebra. For an arbitrary element x∈L0,the operator of right multiplication Rxis a nilpotent endomorphism of the space Li,where i∈ {0,1}.Denote by Ci(x) (i∈ {0,1}) the descending sequence of the dimensions of Jordan blocks of the operator Rx.Consider the lexicographical order on the set Ci(L0). Definition 2.3. A sequence C(L) = max x∈L0\[L0,L0]C0(x)max ex∈L0\[L0,L0]C1(ex) is said to be the characteristic sequence of the Leibniz superalgebra L. Similar as in case of Lie superalgebras [7] (Corollary 3.0.1) it can be proved that the characteristic sequence is an invariant under isomorphisms. For Leibniz superalgebras we introduce the analogue of the zero-filiform Leibniz algebras. Definition 2.4. A Leibniz superalgebra L ∈ Leibn,m is called zero-filiform if C(L) = (n|m) Denote by ZFn,m the set of all zero-filiform Leibniz superalgebras from Leibn,m. From [2] it can be concluded that the even part of a zero-filiform Leibniz superalgebra is a zero-filiform Leibniz algebra, therefore a zero-filiform superalgebra is not a Lie superalgebra. Further, we need the result on existence of an adapted basis for zero-filiform Leibniz superalgebras. Theorem 2.5. [9] In an arbitrary superalgebra from ZFn,m there exists a basis {x1, x2,...,xn, y1, y2,...,ym}which satisfies the following conditions:                [xi, x1] = xi+1,1≤i≤n−1, [xn, x1] = 0, [xi, xk] = 0,1≤i≤n, 2≤k≤n, [yj, x1] = yj+1,1≤j≤m−1, [ym, x1] = 0, [yj, xk] = 0,1≤j≤m, 2≤k≤n. 4 L.M. CAMACHO, J. R. G ´ OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV 3. Zero-filiform Leibniz superalgebra with nilindex equal to n+m This section is devoted to the description of zero-filiform Leibniz superalgebras with nilindex equal to n+m. Let L ∈ ZFn,m with nilindex equal to n+m. Evidently, Lhas two generators. Moreover, from Theorem 2.5 it follows that one generator lies in L0and the second generator lies in L1. Without loss of generality it can be assumed that in an adapted basis the generators are x1and y1. In the adapted basis of Lwe introduce the notations. [xi, y1] = m X j=2 αi,jyj,1≤i≤n, [yi, y1] = n X j=2 βi,jxj,1≤i≤m. In the above notation the following lemma holds. Lemma 3.1. [yi, yj] = min{i+j−1,m}−i X s=0 (−1)sCs j−1 n−j+s+1 X t=2 βi+s,txt+j−s−1,(1) where 1≤i, j ≤m. Proof. The proof is deduced by induction on jat any value of i. Since in the work [9] the set ZFn,2was already described, we consider the set ZFn,m(m≥3). Case ZF2,m (m≥3). Theorem 3.2. Let Lbe a Leibniz superalgebra with the nilindex m+ 2 from ZF2,m (m≥3).Then mis odd and Lis isomorphic to the following superalgebra: [x1, x1] = x2, [yi, x1] = yi+1,1≤i≤m−1, [x1, yi] = −yi+1,1≤i≤m−1, [yi, ym+1−i] = (−1)j+1x2,1≤i≤m−1. Proof. From (1) we easily obtain [yi, yj] = (−1)j−1βi+j−1,2x2,2≤i+j≤m+ 1, [yi, yj] = 0, m + 2 ≤i+j≤2m. (2) It should be noted that βm,26= 0.Indeed, if βm,2= 0,then Lm−1={x2, ym−1, ym}, Lm={x2, ym}and Lm+1 ={ym}which imply that [ym−1, y1] = ax2and [x2, y1] = bym,where ab 6= 0. The chain of the equalities abym= [ax2, y1] = [[ym−1, y1], y1] = 1 2[ym−1,[y1, y1]] = 0 implies a contradiction to the property ab 6= 0.Therefore, βm,26= 0. The simple analysis of the products leads to x2∈ Z(L) (since x2∈ Lm+1 ⊆ Z(L)). Using the Leibniz superidentity we have [x1, yi] = α1,2yi+1 +···+α1,m−i+1ym,1≤i≤m−1. ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 5 The expression [y1, x1] + [x1, y1] lies in R(L).Hence (1 + α1,2)y2+α1,3y3+···+α1,mym(3) belongs to R(L),as well. If either α1,26=−1 or there exists i(3 ≤i≤m) such that α1,i 6= 0,then multiplying the linear combination (3) from the right side required times to x1 we deduce ym∈ R(L).However, by (2) we have [y1, ym] = (−1)m−1βm,2x2which implies that βm,2= 0 and we get a contradiction with condition βm,26= 0. If α1,2=−1 and α1,i = 0 (3 ≤i≤m), then by applying the Leibniz superidentity for the basic elements {x1, yi, yi}we obtain β2i,2= 0 for 1 ≤i≤[m 2]. Note that in case mis even we obtain βm,2= 0 which is a contradiction. Therefore, mis odd. Let us introduce new notations γs=β2s−1,2,1≤s≤m+ 1 2. Then we obtain the family L(γ1, γ2,...,γm+1 2) :        [x1, x1] = x2, [yi, x1] = yi+1,1≤i≤m−1, [x1, yi] = −yi+1,1≤i≤m−1, [yi, yj] = (−1)j−1γi+j 2,2x2, i +jis even,2≤i+j≤m+ 1, m is odd. Make the following general transformation of the generator basic elements: x′ 1=b1x1, y′ 1= m+1 2 X s=1 a2s−1y2s−1. Then x′ 2=b2 1x2and y′ 2i−1=b2(i−1) 1 m−2(i−1)+1 2 X s=1 a2s−1y2s+2i−3,1≤i≤m+ 1 2, y′ 2i=b2i−1 1 m−2(i−1)−1 2 X s=1 a2s−1y2s+2i−2,1≤i≤m−1 2. Choosing the parameters aias follows a1=s1 bm−3 1γm+1 2 , a3=−a1γm−1 2 2γm+1 2 , ai=−a2 1γm−i+2 2+ 2a1a3γm−i+4 2+···+ (2a1ai−2+···+ 2ai−3 2ai+1 2)γm−1 2 2a1γm+1 2 − − (2a3ai−2+···+ 2ai−3 2ai+5 2+a2 i+1 2 )γm+1 2 2a1γm+1 2 ,for i+ 1 2odd. ai=− a2 1γm−i+2 2+ 2a1a3γm−i+4 2+···+ (2a1ai−2+···+ 2ai−5 2ai+3 2+a2 i−1 2 )γm−1 2 2a1γm+1 2 − −(2a3ai−2+···+ 2ai−1 2ai+3 2)γm+1 2 2a1γm+1 2 ,for i+ 1 2even. 6 L.M. CAMACHO, J. R. G ´ OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV when 4 ≤i≤m, we obtain [y′ m, y′ 1] = x′ 2,[y′ i, y′ 1] = 0 for 1 ≤i≤m−1. Then applying Leibniz superidentity get the rest brackets [y′ i, y′ j] = 0,1≤i, j ≤m, i +j6=m+ 1, [y′ i, y′ j] = (−1)j−1x′ 2,1≤i, j ≤m, i +j=m+ 1.. Thus, we obtain the superalgebra of the theorem.  Case ZFn,m (n≥3, m ≥3). Lemma 3.3. Any Leibniz superalgebra from ZFn,m (n≥3, m ≥3) has nilindex less that n+m. Proof. Let us assume the contrary, i.e. Lis a Leibniz superalgebra from ZFn,m (n≥ 3, m ≥3) and Lhas the nilindex equal to n+m. Then in the adapted basis we have L={x1, x2,...,xn, y1, y2,...,ym}, L2={x2,...,xn, y2,...,ym}, L3⊃ {x3,...,xn, y3,...,ym}. Let us suppose that L3={x3,...,xn, y2,...,ym},i.e. x2/∈ L3and y2∈ L3. Then there exits i0(2 ≤i0≤n) such that [xi0, y1] = αi0,2y2+···+αi0,mymwith αi0,26= 0.Since xi∈ R(L) for 2 ≤i≤nand R(L) is an ideal, then αi0,2y2+···+ αi−0,mym∈ R(L). Multiplying the product [xi0, y1] on the right side consequently to the basic element x1(m−1)−times we easily obtain that y2, y3,...,ym∈ R(L),that is L2=R(L). By induction one can prove the following [xi, y1] =      m+1−i P j=2 α1,jyj+i−1,if i+ 1 ≤m, 0,if i+ 1 > m. (4) Since L2=R(L) then y2can appear only in the products [xi, y1] for 2 ≤i≤n) or [yj, x1] for 2 ≤j≤m−1).However, in the first case from (4) we conclude that y2does not lie in L3and in the second case the element y2can not be obtained, i.e. in both cases we have a contradiction with the assumption L3={x3,...,xn, y2,...,ym}. Thus, L3={x2,...,xn, y3,...,ym}.Let sbe a natural number such that x2∈ Ls\ Ls+1. Suppose s≤m. Then we have Li={x2,...,xn, yi,...,ym},2≤i≤s, Ls+1 ={x3,...,xn, ys,...,ym} and in the equality [ys−1, y1] = n P j=2 βs−1,jxjthe coefficient βs−1,2is not zero. From Lemma 3.1 we have [y1, ys] = s−1 X i=0 (−1)iCi s−1 n−s+i+1 X t=2 β1+i,txt+s−i−1, ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 7 in which the coefficient βs−1,2occurs. Taking into account the equality [ys, y1] = n X j=2 βs,jxjwe conclude that x3∈lin < [y1, ys],[ys, y1], x4,...,xn>. Therefore Ls+2 =< x3,...,xn, ys+1,...,ym>, i.e. ys∈ Ls+1 \ Ls+2 and α2,s 6= 0. Consider the equalities [ys−1,[y1, y1]] = 2[[ys−1, y1], y1] = 2 "n X t=2 βs−1,t[xt, y1]#= 2βs−1,2[x2, y1]+ X i≥s+1 (∗)yi. On the other hand [ys−1,[y1, y1]] = 0,because [y1, y1]∈ R(L). The basic element ysappears only in the product [x2, y1].Hence we have that βs−1,2α2,sys+P i≥s+1 (∗)yi= 0 which implies βs−1,2α2,s = 0.This contradicts to the assumption s≤m. Let us consider now the case s=m+ 1.Then we have L={x1, x2,...,xn, y1, y2,...,ym}, Li={x2, x3,...,xn, yi, yi+1 ...,ym},2≤i≤m, Lm+i−1={xi, xi+1 ...,xn},2≤i≤n. Since x2∈ Lm+1 we have [ym, y1] = n P i=2 βm,ixiwith βm,26= 0. The sum [y1, x1] + [x1, y1] lies in R(L) since [y1, x1] + [x1, y1] = (1 + α1,2)y2+ α1,3y3+···+α1,mym∈ R(L). If [y1, x1] + [x1, y1] = 0,then using the Leibniz superidentity we have [x1,[ym, y1]] = [[x1, ym], y1] + [[x1, y1], ym] = −[y2, ym] = βm,2x3+X i≥4 (∗)xi. On the other hand [x1,[ym, y1]] = n X i=2 βm,i[x1, xi] = 0. Hence, βm,2= 0 which is a contradiction. Thus, [y1, x1] + [x1, y1]6= 0.Continuing the same argumentation as in the proof of Theorem 3.2 we obtain ym∈ R(L). Therefore [y1, ym] = m−1 X i=0 (−1)iCi m−1 n−m+i+1 X t=2 β1+i,txt+m−i−1= 0. The minimal value of the expression t+m−i−1 is reached when i=m−1 and t= 2.Thus, [y1, ym] = (−1)m−1Cm−1 m−1βm,2x2+P i≥4 (∗)xiwhich implies that βm,2= 0.That is a contradiction with the assumption that nilindex of Lis equal to n+m.  4. Leibniz superalgebras with the characteristic sequence (n|m1, m2,...,mk)and nilindex n+m Leibniz superalgebras with the characteristic sequence equal to (n|m−1,1) and with the nilindex n+mwere examined in [6]. Therefore, in this section we shall consider the Leibniz superalgebras Lof nilindex n+mwith the characteristic sequence equal to (n|m1, m2,...,mk) with conditions m1≤m−2. 8 L.M. CAMACHO, J. R. G ´ OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV From the definition of characteristic sequence there exits a basis {x1, x2,...xn, y1, y2,...ym}in which the operator Rx1|L1has the following form: Rx|L1=    Jmj10··· 0 0Jmj2··· 0 ··· ··· ··· ··· 0 0 ··· Jmjk    , where (mj1, mj2, . . . , mjk) is a permutation of (m1, m2,...,mk).Without loss of generality, by a shifting of the basic elements we can assume that operator Rx1|L1 has the following form Rx1|L1=    Jm10··· 0 0Jm2··· 0 ··· ··· ··· ··· 0 0 ··· Jmk    . It means that the basis {x1, x2,...xn, y1, y2,...ym}satisfies the following conditions: [xi, x1] = xi+1,1≤i≤n−1, [yj, x1] = yj+1,for j /∈ {m1, m1+m2,...,m1+m2+···+mk}, [yj, x1] = 0,for j∈ {m1, m1+m2,...,m1+m2+···+mk}. (5) It is clear that two generators can not lie in L0.In fact, in [2] the table of multiplication of the Leibniz algebra L0is presented and it has only one generator. Theorem 4.1. Let Lbe a Leibniz superalgebra of nilindex n+mwith characteristic sequence (n|m1, m2,...,mk),where m1≤m−2.Then both generators can not belong to L1at the same time. Proof. Let Leibniz superalgebra L=L0⊕L1has nilindex n+mand let {x1, x2,...,xn} be a basis of L0and {y1, y2,...,ym}a basis of L1. Suppose that two generators lie in L1.Then they should be from the set {y1, ym1+1, ym1+m2+1,...,ym1+m2+···+mk−2+1, ym1+m2+···+mk−1+1} Without loss of generality, the generators can be chosen as {y1, ym1+1}. Consider the following cases: Case 1. Let [y1, y1]∈ L0\L2 0.Then consider the Leibniz superalgebra generated by the element < y1> . Since [y1, y1]∈ L0\L2 0we can assume [y1, y1] = x1.Then from the products in (5) we deduce {x1, x2,...,xn, y2, y3,...,ym1} ⊆< y1> . It it easy to see that ym1+1 /∈< y1> . Indeed, if ym1+1 ∈< y1>, then {ym1+2,...,ym1+m2} ⊆< y1>which implies C(L)≥(n|m1+m2, m3,...,mk).That is a contradiction to the condition of characteristic sequence of L, because C(L) = (n|m1, m2,...,mk). Thus, the Leibniz superalgebra generated by the basic element y1consist of {x1, x2,...,xn, y1, y2,...,ym1}. Since the superalgebra < y1>is single-generated then from Theorem 2.2 we have that either m1=nor m1=n+1 and the multiplication in < y1>has the following form: [xi, x1] = xi+1,1≤i≤n−1, [yj, x1] = yj+1,1≤j≤m1−1, [xi, y1] = 1 2yi+1,1≤i≤m1−1, [yj, y1] = xj,1≤j≤n. ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 9 Case m1=n. Since y1and yn+1 are generators we have L={x1, x2,...,xn, y1,...,yn, yn+1,...,ym}, L2={x1, x2,...,xn, y2,...,yn, yn+2,...,ym}. Besides, x1/∈ L3.Otherwise, if x1∈ L3,then there exists z∈ L1such that z∈ L2/L3.Thereby z∈lin < [y1, y1],[y1, yn+1],[yn+1, y1],[yn+1, yn+1]>and taking into account that [yi, yj]∈ L0we obtain z∈ L0which is a contradiction. Thus, L3={x2,...,xn, y2,...,yn, yn+2,...,ym}. If L2k={xi, xi+1,...,xn, yj, . . . , yn, yn+2,...,ym}, then by a similar way one can prove that L2k+1 ={xi+1,...,xn, yj,...,yn, yn+2,...,ym}. In fact, if z∈ L2k/L2k+1,then zhas to be generated by 2kproducts of the generators (but they are from L1). Hence this products belong to L0,and we have z∈ L0. Applying the similar argumentation we get L2k+2 ={xi+1,...,xn, yj+1,...,yn, yn+2,...,ym}. Continuing with the process, we obtain that L2n+1 ={yi1, yi2,...,yik}and L2n+2 = 0.Since dim(L2n+1/L2n+2) = 1 then L2n+1 ={yn+2}and nilindex should be equal to 2n+ 2.Thus, m=n+ 2 and we have L={x1, x2,...,xn, y1,...,yn, yn+1, yn+2}, L2k={xk,...,xn, yk+1,...,yn, yn+2},1≤k≤n−1, L2k+1 ={xk+1,...,xn, yk+1,...,yn, yn+2},1≤k≤n−1, L2n={xn, yn+2},L2n+1 ={yn+2},L2n+2 ={0}. Furthermore, L2n= [L2n−1,L] =<[xn, y1],[xn, yn+1],[yn, y1],[yn, yn+1],[yn+2, y1], [yn+2, yn+1]> . Note that the element yn+2 can be obtained only from product [xn, yn+1] (because [xn, y1] = 0, otherwise we get a contradiction with the property of characteristic sequence). However, [xn, yn+1] = [[xn−1, x1], yn+1] = [xn−1,[x1, yn+1]] + [[xn−1, yn+1], x1] = 0, which deduce L2n={xn},that is a contradiction with the condition of nilindex. Case m1=n+ 1. In this case, similar to the previous case, we get a contradiction. Case 2. Let [y1, y1]/∈ L0\ L2 0and [ym1+1, ym1+1]∈ L0\ L2 0.Then applying the same arguments for ym1+1 as for y1in Case 1, we obtain a contradiction with the fact that both generators lie in L1,as well. Case 3. Let [y1, y1]/∈ L0\ L2 0and [ym1+1, ym1+1]/∈ L0\ L2 0.Then, without loss of generality, we can assume that [y1, ym1+1] = x1, [ym1+1, y1] = n P i=1 bixi. If b1= 1,then making the change of basis y′ 1=y1+ym1+1 we obtain [y′ 1, y′ 1]∈ L0\ L2 0.Therefore this case can be reduced to Case 1. If b16= 1,then [y1, ym1+1]−[ym1+1, y1] = (1 −b1)x1+b2x2+···+bnxn∈ R(L) and since xi∈ R(L) (2 ≤i≤n) we get x1∈ R(L). From the Leibniz superidentity