On HL2 (L, L) for semisimple Leibniz algebras

On HL2(L, L)for semisimple Leibniz algebras L M Camacho1, M Ladra2and R M Turdibaev3 1Dpto. Matem´atica Aplicada I. University of Sevilla, 41012, Spain, 2,3Department of Algebra, University of Santiago de Compostela, 15782, Spain. E-mail: [email protected], [email protected], [email protected] Abstract. In this paper we present a decomposition of HLn(L, L) into a direct sum of some subspaces for a finite dimensional complex semisimple Leibniz algebra L. Furthermore, we provide a more specific decomposition in case n= 2 into two subspaces. We verify that one of those subspaces annihilates for specific Leibniz algebras with liezation sl2and some others. 1. Introduction Leibniz algebras were rediscovered by A. Bloh [3] and later were given another impulse of investigation due to works of J.–L. Loday [7] who discovered that the classical Chevalley–Eilenberg boundary map in the exterior module of a Lie algebra can be lifted to the tensor module. To prove that it becomes a chain complex the only identity needed is a so called Leibniz identity which yields to definition of Leibniz algebra and Leibniz (co)homology. In this work we study Leibnz algebra cohomology and using an idea from [1] we decompose the cochain space CLn(L, L) into a direct sub of special subspaces that we call level spaces. Further we concentrate on so called semisimple Leibniz algebras. They are Leibniz algebras with corresponding Lie algebras being semisimple. Semisimple Leibniz algebra Lcan be naturally consider as a Z2- graded algebra with L0being its liezation and L1=I. We notice that due to structural properties of semisimple Leibniz algebras, coboundary operator preserves the level spaces. Thus we are able to establish in Theorem 4.3 a decomposition of HLn(L, L) into direct sum of nsubspaces, that we label from −n+ 1 to 0. Our goal is to check proposed conjecture in [2] that HL2(L, L) = 0 for any finite-dimensional complex semisimple Leibniz algebra L. Authors in [2] validate the claim for simple Leibniz algebra with liezation L/I ∼ =sl2over C. In order to conjecture to hold, we elaborate on each of the subspaces in the doi:10.1088/1742-6596/697/1/012008 decomposition of HL2(L, L) to be zero and present equivalent conditions to the conjecture to be true in Proposition 5.6 and Proposition 5.7 in terms of Lie algebra and their representations. In the last section, we establish that space of level −1 is zero for all finitedimensional Leibniz algebras with liezation sl2in Theorem 6.1 and we check analogous statement for another Leibniz algebra with liezation sl2⊕sl2in Theorem 6.2. 2. Preliminaries Given a Lie algebra and its non-trivial module a different algebraic structure arises on the direct sum. Example 2.1. ([5]) Let gbe a Lie algebra and Mbe a g-module. Consider L=g⊕Mwith a bracket [(g1, m1),(g2, m2)] := ([g1, g2],−g2.m1).Then [[x, y], z] = [[x, z], y]+[x, [y, z]] (1) holds for any x, y, z ∈Land this algebra is not a Lie algebra if the action of g on Mis not trivial. Looking to the above identity (1) one notices that the right multiplication operator [−, z] by any element zis a derivation (see [6]) and satisfies a so called Leibniz rule. Although defined earlier by A. Bloh [3] these objects attracted more attention after series of J.-L. Loday and his collaborators’ works. Definition 2.2. An algebra (L, [−,−]) over a field Kis called a Leibniz algebra if identity (1) holds. Leibniz algebra of Example 2.1 is denoted as guMand is called hemisemidirect product Leibniz algebra in [5]. Any Lie algebra is a Leibniz algebra but not the converse. Given a Leibniz algebra Lits two sided ideal generated by elements [x, x] for all x∈Lis very important. In this work we denote this ideal by I. The quotient Leibniz algebra L/I is easily seen to be a Lie algebra, called the liezation of Leibniz algebra L. Definition 2.3. The set AnnR(L) = {x∈L|[L, x] = 0}of a Leibniz algebra L is called the right annihilator of L. One can show that AnnR(L) is an ideal of L. Note that due to Leibniz identity it follows that [L, I] = 0. Thus Iis a subset of a right annihilator AnnR(L) = {x∈L|[L, x] = 0}. The center is defined as Z(L) = {x∈L| [x, L]=[L, x]=0}. If I=Lthen Leibniz algebra is a trivial algebra with all products being zero. If I= 0 then Lis a Lie algebra. In this work we assume Ito be non trivial, eliminating Lie and trivial Leibniz algebras from the study. One has a short exact sequence 0 →I→Lf −→ g→0. Note that projection f is universal in the sense that a Leibniz map from Lto any Lie algebra factors through f. Arepresentation Mof a Leibniz algebra Lis introduced in [6]. doi:10.1088/1742-6596/697/1/012008 2 Definition 2.4. A vector space Mis called a representation or bimodule over a Leibniz algebra Lif there are two bilinear maps: [−,−]: L×M→Mand [−,−]: M×L→M satisfying the following three axioms [m, [x, y]] = [[m, x], y]−[[m, y], x], [x, [m, y]] = [[x, m], y]−[[x, y], m], [x, [y, m]] = [[x, y], m]−[[x, m], y], for any m∈M,x, y ∈L. A Leibniz bimodule is called anti-symmetric when [x, m] = 0 for x∈L, m ∈ M. Provided that, the last two required identities vanish. Since ideal Iis in the right-annihilator, if x∈Ior y∈Ithe first identity vanishes as well. Therefore, the action of Lon anti-symmetric L-bimodule Mis determined by the action of complement of Ithat is required to satisfy only the first axiom. Now for g∈gdefine g◦m:= −[m, ϕ(g)],where ϕis an inverse map of f. These are well-defined, since if g=g′then ϕ(g−g′)∈Iand Iis in the right annihilator. Moreover, the first identity turns into [g1, g2]◦m=−[m, ϕ([g1, g2])] = −[m, [ϕ(g1), ϕ(g2)])] = −[[m, ϕ(g1)], ϕ(g2)] + [[m, ϕ(g2)], ϕ(g1)] = g2◦[m, ϕ(g1)] −g1◦[m, ϕ(g2)] =g1◦(g2◦m)−g2◦(g1◦m), i.e. Mbecomes a left g-module (or a right g-module with action m◦g= [m, ϕ(g)]). Conversely, if Mis a left g-module with an action ◦:g×M→M, then by defining [m, l] := −f(l)◦mfor l∈Limplies by the arguments above that M is an anti-symmetric L-bimodule. Thus, anti-symmetric Leibniz L-bimodule is equivalent to a Lie algebra gmodule of it liezation g.For instance, an ideal generated by squares Ican be considered as a Lie algebra module over liezation g. Call Leibniz algebra Lsemisimple if gis semisimple. Lis called simple if the only non-trivial ideal of Lis I= [L, L].These agrees with suggested definition in [1]. T. Pirashvili proved the following statement. Proposition 2.5. ([10]) Let f:L→gbe an epimorphism from an arbitrary finite dimensional Leibniz algebra Lto semisimple Lie algebra g.Then fadmits a section. Considering a semisimple Leibniz algebra Lits liezation by definition is a semisimple Lie algebra. The fundamental projection onto its liezation admits a section due to the above proposition. This leads to a key fact that is the base of the study on decomposition of cohomology of semisimple Leibniz algebras in the following sections. doi:10.1088/1742-6596/697/1/012008 3 Corollary 2.6. Let Lbe a finite dimensional semisimple Leibniz algebra with liezation g.Then L∼ =guI. Acohomology of a Leibniz algebra Lwith coefficients in the representation Mare defined in [8] as follows. Define the space of n-cochains CLn(L, M) = HomK(L⊗n, M) for n≥0 and aK-homomorphism ∂n:CLn(L, M)→CLn+1(L, M) by (∂nf)(x1, . . . , xn+1) := [x1, f(x2, . . . , xn+1)]+ + n+1 ∑ i=2 (−1)i[f(x1, . . . , bxi, . . . , xn+1), xi]+ +∑ 1≤i<j≤n+1 (−1)j+1f(x1, . . . , xi−1,[xi, xj], xi+1, . . . , bxj, . . . , xn+1). This (CL∗(L, M), ∂) is a cochain complex. Its n-th cohomology group is well defined by HLn(L, M) := ZLn(L, M)/BLn(L, M),where the elements ZLn(L, M) := ker ∂nand BLn(L, M) := im ∂n−1are called n-cocycles and n-coboundaries, respectively. Recall that it is noted in [8] that ML:= {m∈M|[l, m] = 0,∀l∈L}=HL0(L, M) which is called bisubmodule of left invariants of M. In Leibniz algebras a derivation is defined as usual. Definition 2.7. A linear map d:L→Mis called a derivation of Lin Mif d([x, y]) = [d(x), y]+[x, d(y)] for any x, y ∈L. Space of all derivations from Lto Mis denoted by Der(L, M). Moreover, for a given m∈Ma map Rm:L→Mdefined by Rm(l) = [l, m] is a derivation. It is called an inner derivation and running through all of the elements of the bimodule we obtain space of inner derivation Innder(L, M). It is known that HL1(L, M) = Der(L, M)/Innder(L, M). If Mis an anti-symmetric L-bimodule then inner derivations are zero and HL1(L, M) = Der(L, M) = {f:L→M|f([l1, l2]) = [f(l1), l2]}. Further, in [8] considering Laas the antisymmetric representation whose underlying Kmodule is Land action L×La→Lis Leibniz bracket of L, it is claimed that HL1(L, La)= 0.Indeed, taking an identity map id|Lwe see that id ∈HL1(L, La).With similar arguments once can obtain that HL1(L, I)= 0 where Iis an L-module by the algebra bracket action. Note that when Mis symmetric L-bimodule [8] establishes HL1(L, M) = HL1(g, M) = H1(g, M) for liezation gof Leibniz algebra L. However, similar statement for anti-symmetric Mdoes not hold. One of the main tools in this work relies on T. Pirashvili’s result [10] on Leibniz cohomology of semisimple Lie algebras. doi:10.1088/1742-6596/697/1/012008 4 Theorem 2.8. [10] Let gbe a finite dimensional semisimple Lie algebra and Mbe a Leibniz representation of g. Then HLn(g, M) = 0 for all n≥2. It was proved by constructing some spectral sequences. The same result was proved by P. Ntolo [9] using Casimir element and constructing the explicit homotopy. 3. Decomposition of CLn(L, L)for a Leibniz algebra Let Vbe a vector space over a field Kand V=V0⊕V1be a non-trivial decomposition. The goal of this section is to present the decomposition of HomK(V⊗n, V ) that will be heavily used in the next sections. For a natural number nand an integer kdefine I(n, k) to be a set of bijections from {1,2, . . . , n}to a set of kones and n−kzeros for 0 ≤k≤nand I(n, k) = ∅ otherwise. For a bijection π∈ I(n, k) and a number jequal to 0 or 1 let us denote for short Hj π(1)...π(n):= HomK(Vπ(1) ⊗ · · · ⊗ Vπ(n), Vj). Denote by CLn(V, V ) = HomK(V⊗n, V ) and define for −n≤k≤1 a subspace CLn(V, V )(−k)=(⊕ π∈I(n,k) H0 π(1)...π(n))⊕(⊕ π∈I(n,k+1) H1 π(1)...π(n)). Note that there are (n k)+(n k+1)=(n+1 k+1)summands in the above decomposition. Moreover, for any Hj π(1)...π(n)in this decomposition one has k=j−(π(1) + · · · +π(n)). Let us call subspace CLn(V, V )(k)of level k. The following statement presents a decomposition isomorphic to CLn(V, V ) into subspaces of all possible levels −n, . . . , 0,1. Proposition 3.1. CLn(V, V )∼ =⊕1 k=−nCLn(V, V )(−k). Proof. We have CLn(V, V ) = Hom(V⊗n, V ) = Hom((V0⊕V1)⊗n, V0⊕V1) ∼ =Hom((V0⊕V1)⊗n, V0)⊕Hom((V0⊕V1)⊗n, V1) ∼ =Hom(V1⊗V1⊗ · · · ⊗ V1, V0)⊕ ⊕(⊕ i∈I(n,n−1) H0 i(1)...i(n)⊕ ⊕ i∈I(n,n) H1 i(1)...i(n))⊕ . . . ⊕(⊕ i∈I(n,0) H0 i(1)...i(n)⊕ ⊕ i∈I(n,1) H1 i(1)...i(n))⊕ ⊕Hom(V0⊗V0⊗ · · · ⊗ V0, V1), or in more compact form CLn(V, V )∼ = n ⊕ k=−1(⊕ i∈I(n,k) H0 i(1)...i(n))⊕(⊕ i∈I(n,k+1) H1 i(1)...i(n)). Hence, the claim is proved. doi:10.1088/1742-6596/697/1/012008 5 4. Decomposition of HLn(L, L)for semisimple Leibniz algebra Now consider a semi-simple Leibniz algebra Las an adjoint representation of L, i.e. Lacting on itself by Leibniz algebra bracket. Due to Corollary 2.6 we have L∼ =Iug. Setting L0=gand L1=Icochain spaces CLn(L, L) admits the decomposition of Proposition 3.1. However, due to [L, I] = 0 one can obtain more properties of the coboundary operator. Proposition 4.1. Coboundary operator ∂preserves the level spaces. Proof. For a non-zero φ∈CLn(L, L)(k)consider ∂φ(x1, . . . , xn, xn+1), where exactly mof x1, . . . , xn+1 belong to I. It is evident that if m < k or m > k + 1 then ∂φ(x1, . . . , xn, xn+1) = 0. We will consider the other possible cases. Let φ∈H0 i(1)...i(n)for some i∈ I(n, k). If m=kthen due to im φ∈git follows that ∂φ(x1, . . . , xn, xn+1)∈g. If m=k+ 1 then ∂φ(x1, . . . , xn, xn+1) belongs to Iif x1∈Iand zero otherwise. Hence, ∂(H0 i(1)...i(n))⊆(⊕ j∈I(n+1,k)H0 j(1)...j(n+1))⊕H1 1i(1)...i(n). Now consider φ∈H1 i(1)...i(n)for some i∈ I(n, k+1). It is obvious that m=k implies ∂φ(x1, . . . , xn, xn+1) = 0. One verifies the last possibility m=k+ 1 to yield ∂(H1 i(1)...i(n))⊆ ⊕ j∈I(n+1,k+1) H1 j(1)...j(n+1). Thus ∂(CLn(L, L)(−k))⊆CLn+1(L, L)(−k)for all −1≤k≤n. Let us denote by ZLn(L, L)(k)the kernel of restriction of ∂on CLn(L, L)(n) and by BLn(L, L)(k)the image of restriction of ∂on CLn−1(L, L)(k)for all −n≤k≤1. Proposition 4.2. Let Lbe a finite-dimensional semisimple Leibniz algebra. Then ZLn(L, L)(−n)= 0 for all positive integers n. Proof. Let φ∈ZLn(L, L)(−n). Then φ:⊗nI→gand ∂φ(g, i1, . . . , in) = [g, φ(i1, . . . , in)] = 0 for all g∈g, i1, . . . , in∈I. This yields im φ∈Z(g) which is zero since gis semi-simple. Therefore, φ= 0 and we are done. Provided with Proposition 4.1 one has BLn(L, L)(k)⊆ZLn(L, L)(k) and consider subspace ZLn(L, L)(k)/BLn(L, L)(k). Let us denote it by HLn(L, L)(k), for all −n≤k≤1. As a consequence on Propositions 3.1 and 4.1 we have the following Theorem 4.3. Let Lbe a finite-dimensional semisimple Leibniz algebra. Then HLn(L, L)∼ =HLn(L, L)(−n+1) ⊕ · · · ⊕ HLn(L, L)(−1) ⊕HLn(L, L)(0) for n≥2and HL1(L, L)∼ =HL1(L, L)(0) ⊕HL1(g, I). doi:10.1088/1742-6596/697/1/012008 6 Proof. Clearly, we have BLn(L, L)∼ =⊕1 k=−nBLn(L, L)(−k), ZLn(L, L)∼ =⊕1 k=−nZLn(L, L)(−k). Note that by definition BLn(L, L)(−n)= 0 and together with Proposition 4.2 we obtain HLn(L, L)(−n)= 0. Observe that HLn(L, L)(1) =HLn(g, I). If n > 1 by Theorem 2.8 it is zero. Hence, the result follows. 5. HL2(L, L)for semisimple Leibniz algebra Let us describe BL2(L, L) = BL2(L, L)(−1) ⊕BL2(L, L)(0) ⊕BL2(L, L)(1) first. Proposition 5.1. A coboundary operator ∂acts on CL1(L, L)as follows: H0 1,→H0 10 ⊕H0 01 ⊕H1 11, H1 1⊕H0 0→H0 10 ⊕H0 00, H1 0→H1 00. Proof. Consider ∂d for d∈CL1(L, L). We list the only non-zero actions of ∂d on an element (x, y)∈(Li(1), Li(2)) depending on d. 1. Let d∈CL1(L, L)(−1) =H0 1. Then ∂(H0 1)⊆H1 11 ⊕H0 01 ⊕H0 10. Indeed, ∂d(x0, y1) = [x0, d(y1)] ∈L0, ∂d(x1, y0) = [d(x1), y0]−d([x1, y0]) ∈L0, ∂d(x1, y1) = [x1, d(y1)] ∈L1. Moreover, if ∂d = 0 then from [x0, d(y1)] = 0 we obtain d= 0, i.e. ∂sends H0 1to H1 11⊕H0 01⊕H0 10 injectively. In particular, this implies dim BL2 (−1)(L, L) = dim H0 1= dim Hom(L1, L0) = dim L1·dim L0. 2. Let d∈CL1(L, L)(0) =H1 1⊕H0 0. Then we analyse separate cases. 2.1 Let d1∈H1 1. Then ∂(H1 1)⊆H1 10. Indeed, ∂d1(x1, y0) = [d1(x1), y0]−d1([x1, y0]) ∈L1. 2.2 Let d0∈H0 0. Then ∂(H0 0)⊆H0 00 ⊕H1 10. Indeed, ∂d0(x0, y0) = [d0(x0), y0]+[x0, d0(y0)] −d0([x0, y0]) ∈L0, ∂d0(x1, y0) = [x1, d0(y0)] ∈L1. 3. Let d∈CL1(L, L)(1) =H1 0. Then ∂(H1 0)⊆H1 00. Indeed, ∂d(x0, y0) = [d(x0), y0]−d([x0, y0]) ⊆L1. Now we will concentrate on ZL2(L, L). For a φ∈CL2(L, L) we have ∂φ(x, y, z) = [x, φ(y, z)] −[φ(x, y), z]+[φ(x, z), y] +φ(x, [y, z]) −φ([x, y], z) + φ([x, z], y) (2) By Proposition 4.2 there is a decomposition ZL2(L, L) = ZL2(L, L)(−1) ⊕ ZL2(L, L)(0) ⊕ZL2(L, L)(1). We have ZL2(L, L)(−1) = ker ∂|H0 10⊕H0 01⊕H1 11 and proposition below shows that coboundary operator takes every component of CL2(L, L)(−1) injectively. doi:10.1088/1742-6596/697/1/012008 7 Proposition 5.2. A coboundary operator ∂acts on CL2(L, L)(−1) as follows H0 01 ,→H1 101 ⊕H0 010 ⊕H0 001, H0 10 ,→H1 110 ⊕H0 100 ⊕H0 010, H1 11 ,→H1 101 ⊕H1 110. Proof. We have CL2(L, L)(−1) =H0 10 ⊕H0 01 ⊕H1 11. Now let us examine how ∂ acts on each of the subspaces. Case 1. Let φ∈H0 01. Then in the system (2) the only non-zero equalities are ∂φ(x1, y0, z1) = [x1, φ(y0, z1)] ∈L1, ∂φ(x0, y0, z1) = [x0, φ(y0, z1)] + [φ(x0, z1), y0]−φ([x0, y0], z1)∈L0, ∂φ(x0, y1, z0) = −[φ(x0, y1), z0] + φ(x0,[y1, z0]) + φ([x0, z0], y1)∈L0. Hence, H0 01 →H1 101 ⊕H0 010 ⊕H0 001. Now if φ∈ker ∂|H0 01 then [x1, φ(y0, z1)] = 0 implies φ(y0, z1)⊆I=L1while im φ⊆L0. Thus φ= 0 and H0 01 ,→H1 101 ⊕H0 010 ⊕H0 001. Case 2. Let φ∈H0 10. Similarly, we have the following equalities from system (2): ∂φ(x0, y1, z0) = [x0, φ(y1, z0)] ∈L0 ∂φ(x1, y0, z0) = −[φ(x1, y0), z0]+[φ(x1, z0), y0] +φ(x1,[y0, z0]) −φ([x1, y0], z0) + φ([x1, z0], y0)∈L0 ∂φ(x1, y1, z0) = [x1, φ(y1, z0)] ∈L1 Hence, H0 10 →H1 110 ⊕H0 100 ⊕H0 010. Analogously as above, if φ∈ker ∂|H0 10 then equality [x1, φ(y1, z0)] = 0 implies φ= 0. Therefore, H0 10 ,→H1 110 ⊕H0 100 ⊕ H0 010. Case 3. Let φ∈H1 11, then system (2) implies ∂φ(x1, y0, z1) = [φ(x1, z1), y0]−φ([x1, y0], z1)∈L1, ∂φ(x1, y1, z0) = −[φ(x1, y1), z0] + φ(x1,[y1, z0]) + φ([x1, z0], y1)∈L1. Assuming φ∈ker ∂|H1 11 from the equalities ∂φ(x1, z0, y1) = ∂φ(x1, y1, z0)=0 we obtain φ(x1,[y1, z0]) = 0. Due to [I, g] = Iit implies φ= 0. Hence, H1 11 ,→H1 101 ⊕H1 110. We have ∂(H0 00 ⊕H1 01 ⊕H1 10)⊆H0 000 ⊕H1 100 ⊕H1 010 ⊕H1 001. Now let us examine how ∂acts on each of the subspaces. Proposition 5.3. Coboundary operator acts on CL2(L, L)(0) as follows ∂:H0 00 ,→H1 100 ⊕H0 000 ∂:H1 10 →H1 100 ∂:H1 01 ,→H1 010 ⊕H1 001 and ZL2(L, L)(0) = ker ∂|H0 00⊕H1 10 . doi:10.1088/1742-6596/697/1/012008 8 Proof. Case 1. Assume φ∈H0 00, by system (2) we have ∂φ(x1, y0, z0) = [x1, φ(y0, z0)] ∈L1, ∂φ(x0, y0, z0) = [x0, φ(y0, z0)] −[φ(x0, y0), z0] + [φ(x0, z0), y0] +φ(x0,[y0, z0]) −φ([x0, y0], z0) + φ([x0, z0], y0)∈L0. Given that φ∈ker ∂|H0 00 then [x1, φ(y0, z0)] = 0 which implies φ= 0. Thus, ∂:H0 00 ,→H1 100 ⊕H0 000. Case 2. Assuming φ∈H1 10 non-zero equalities of system (2) yields ∂φ(x1, y0, z0) = −[φ(x1, y0), z0]+[φ(x1, z0), y0] +φ(x1,[y0, z0]) −φ([x1, y0], z0) + φ([x1, z0], y0)⊆L1. Hence, ∂:H1 10 →H1 100. Case 3. Let φ∈H1 01. System (2) provides ∂φ(x0, y1, z0) = −[φ(x0, y1), z0] + φ(x0,[y1, z0]) + φ([x0, z0], y1)⊆L1; ∂φ(x0, y0, z1) = [φ(x0, z1), y0]−φ([x0, y0], z1)∈L1. Then ∂:H1 01 ,→H1 010 ⊕H1 001 since assuming φ∈ker ∂|H1 01 the last two equalities yield [x1, φ(y0, z0)] = 0 which implies φ= 0. From above it follows that ZL2(L, L)(0) = ker ∂|H0 00⊕H1 01⊕H1 10 = ker ∂|H0 00⊕H1 10 ⊕ker ∂|H1 01 = ker ∂|H0 00⊕H1 10 , which finishes the proof. As a consequence of the last two propositions we obtain Theorem 5.4. For a finite dimensional semisimple Leibniz algebra Lover C we have an isomorphism HL2(L, L)∼ =HL2(L, L)(−1) ⊕HL2(L, L)(0), where HL2(L, L)(−1) = ker ∂|H0 10⊕H0 01⊕H1 11 /∂(H0 1)and HL2(L, L)(0) = ker ∂|H0 00⊕H1 10 /∂(H1 1⊕H0 0). From this point we will concentrate our study on each of the subspaces of level −1 and 0 separately. Proposition 5.5. Any 2-cocycle φ=φ0 10 +φ0 01 +φ1 11 ∈ker ∂|H0 10⊕H0 01⊕H1 11 is uniquely determined by a map ϕ∈Hom(I⊗g)→gthat satisfies ϕ(x1,[y0, z0]) = ϕ([x1, y0], z0)−ϕ([x1, z0], y0). Moreover, φis determined by the following equalities            φ0 01(x0,[z1, y0]) = −[x0, ϕ(z1, y0)] φ1 11(x1,[z1, y0]) = −[x1, ϕ(z1, y0)] φ0 10([x1, y0], z0) = ϕ([x1, y0], z0)−[ϕ(x1, y0), z0] (3) doi:10.1088/1742-6596/697/1/012008 9 Using Φϕ(xk, e, f) = 0 for 1 ≤k≤m−1 we have the following chain of equalities (m−2k)d(xk) = d(xk.h) = ϕ(xk, h) = ϕ(xk,[e, f]) =ϕ(xk.e, f)−ϕ(xk.f, e) = =−k(m+ 1 −k)ϕ(xk−1, f)−ϕ(xk+1, e) =−k(m+ 1 −k)d(xk)−ϕ(xk+1, e). This results in ϕ(xk+1, e) = −(k+1)(m−k)d(xk) = d(xk+1.e) for 1 ≤k≤m−1. Now Φϕ(x0, e, f) = 0 yields ϕ(x1, e) = −ϕ(x0, h) = −md(x0) = d(x1.e) and Φϕ(x0, e, h) = 0 results in ϕ(x0, e) = 0 which is the same as d(x0.e) = 0. Thus we have ϕ(xk, e) = d(xk.e) for all 0 ≤k≤m. Thus ϕ(i, g) = d(i.g) for all i∈I, g ∈gand by Theorem 5.6 in this case we obtain HL2(L, L)(−1) = 0. Now assume that Iis a finite dimensional sl2-module. Then it is completely reducible. Let Jbe an irreducible submodule of I. Then for restriction of ϕ on Hom(J⊗g,g) by the previous construction we have a map dJ∈Hom(J, g) such that ϕ(j, g) = d(j.g). Defining d∈Hom(I, g) as a direct sum of dJof all irreducible submodules we obtain the desired map. Consider a Lie algebra g=sl2⊕sl2and let I=I1⊕I2be a sum of two irreducible sl2-modules of the same dimension (hence, isomorphic). As shown in work [4] the action of gon Iis as follows: xk.e1=−k(m+ 1 −k)xk−1, k = 1, . . . , m. xk.f1=yk+1, k = 0, . . . , m −1, xk.h1= (m−2k)yk, k = 0, . . . , m, yk.e1=−k(m+ 1 −k)yk−1, k = 1, . . . , m. yk.f1=yk+1, k = 0, . . . , m −1, yk.h1= (m−2k)yk, k = 0, . . . , m, xj.e2=yj.h2=yjj= 0, . . . , m xj.h2=yj.f2=−xjj= 0, . . . , m where I1= Span{x0, x1, . . . , xm}, I2= Span{y0, y1, . . . , ym}and sli 2= ⟨ei, fi, hi⟩for i= 1,2. Let L1=I1⊕I2⊕sl1 2⊕sl2 2be a Leibniz algebra with table of multiplication as above. Proposition 6.2. For Leibniz algebra L1we have HL2(L1, L1)(−1) = 0. Proof. Define a map d:I→sl1 2⊕sl2 2by d(xk) = ϕ(xk−1, f1), d(yk) = ϕ(yk−1.f1) for all 1 ≤k≤mand d(x0) = 1 mϕ(x0, h1), d(y0) = 1 mϕ(y0, h1). Then by the proof of Theorem 6.1 we have ϕ(i, g) = d(i.g) for all i∈I1⊕I2and g∈sl1 2. Hence, we only need to show the desired equality when g∈sl2 2. Conditions Φϕ(xk, f2, h2) = 0 and Φϕ(yk, e2, h2) = 0 yield ϕ(xk, f2) = 0 = d(xk.f2), ϕ(yk, e2) = 0 = d(yk.e2), doi:10.1088/1742-6596/697/1/012008 16 respectively, for 0 ≤k≤m. Next conditions Φϕ(xk, e2, f2) = 0 and Φϕ(xk, e2, h2) = 0 gives us ϕ(xk, h2) = ϕ(yk, f2) and ϕ(xk, e2) = ϕ(yk, h2) for 0 ≤k≤m, respectively. Recall xk.h2=yk.f2and xk.e2=yk.h2. Using Φϕ(xk, e2, f1) = 0 one obtain ϕ(xk+1, e2) = ϕ(yk, f1) = d(yk.f1) = d(yk+1) = d(xk+1.e2) for 0 ≤k≤m−1. One derives equality ϕ(x0, e2) = 1 mϕ(y0, h1) = d(y0) = d(x0.e2) using Φϕ(x0, e2, h1) = 0. Similarly, using Φϕ(xk, f1, h2) = 0 one obtain ϕ(xk+1, h2) = −ϕ(xk, f1) = −d(xk.f1) = −d(xk+1) = d(xk+1.h2) for 0 ≤k≤m−1. As for the only undefined missing part one verifies ϕ(x0, h2) = −1 mϕ(x0, h1) = −d(x0) = d(x0.h2) using Φϕ(x0, h1, h2) = 0. Hence, we have proved that ϕ(i, g) = d(i.g) for all i∈I1⊕I2, g ∈sl1 2⊕sl2 2 and by Theorem 5.6 in this case we obtain HL2(L1, L1)(−1) = 0. Acknowledgments Authors were supported by Ministerio de Economa y Competitividad (Spain), grant MTM2013-43687-P (European FEDER support included). The third author was also supported by Xunta de Galicia, grant GRC2013-045 (European FEDER support included). References [1] Abdukassymova A S, Dzhumadil’daev A S 2000 Simple Leibniz algebras of rank 1 Abstract presented to the IX International Conference of the Representation Theory of Algebras pp 17–18 [2] Adashev J Q, Ladra M, Omirov B A 2015 On the second cohomology group of simple Leibniz algebras arXiv:1502.00609 pp 1–11 [3] Bloh A 1965 On a generalization of the concept of Lie algebra Dokl. Akad. 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