Leibniz Algebras Associated to Extensions of sl 2

LEIBNIZ ALGEBRAS ASSOCIATED TO EXTENSIONS OF sl2 L. M. Camacho1, S. Gómez-Vidal1, and B. A. Omirov2 1Departamento Matemática Aplicada I, Universidad de Sevilla, Sevilla, Spain 2Institute of Mathematics, National University of Uzbekistan, Tashkent, Uzbekistan In this work, we investigate the structure of Leibniz algebras whose associated Lie algebra is a direct sum of sl2 and the solvable radical. In particular, we obtain the description of such algebras when: the ideal generated by the squares of elements of a Leibniz algebra is irreducible over sl2 and when the dimension of the radical is equal to two. Key Words: Irreducible module; Lie algebra; Leibniz algebra; Simple algebra; Solvability. 2010 Mathematics Subject Classification: 17A32; 17A60; 17B30; 17B60. 1. INTRODUCTION The notion of Leibniz algebra was introduced in 1993 by J.-L. Loday [6] as a generalization of Lie algebras. In the last 20 years, the theory of Leibniz algebras has been actively studied, and many results of the theory of Lie algebras have been extended to Leibniz algebras. From the theory of Lie algebras it is known that every finite dimensional Lie algebra is decomposed into a semidirect sum of a semisimple subalgebra and solvable radical (Levi’s Theorem) [5]. Moreover, thanks to Mal’cev’s results [7] the study of solvable Lie algebras is reduced to the study of nilpotent algebras. Recently, Barnes proved an analogue of Levi’s Theorem for Leibniz algebras [4]. Namely, a Leibniz algebra is decomposed into a semidirect sum of its solvable radical and a semisimple Lie algebra. Recall, an algebra L over a field F is called Leibniz algebra if for any elements x y z ∈ L the Leibniz identity holds: x y z = x y z − x z y where − − is multiplication of L. Let L be a Leibniz algebra and I = ideal < x x  x ∈ L> be the ideal of L generated by all squares. Then I is the minimal ideal with respect to the property that L/I is a Lie algebra. The natural epimorphism L→L/I determines the associated Lie algebra L/I of the Leibniz algebra L. The inherent properties of non-Lie Leibniz algebras imply that the subspace spanned by the squares of elements of the algebra is a nontrivial ideal (further denoted by I). Moreover, the ideal Iis abelian and hence, it belongs to the solvable radical. Due to the existence of the non-trivial ideal Iin non-Lie Leibniz algebra we can not consider the notion of simplicity in ordinary sense. In [1], the adaptation of simplicity for Leibniz algebras was proposed. Namely, a Leibniz algebra Lis said to be simple if it only has the following ideals: 0,I,Land L L = I. Obviously, this definition agrees with the definition of a simple Lie algebra. In fact, Abdykassymova and Dzhumadil’daev in [2] suggested the following construction of Leibniz algebras. Let Gbe a simple Lie algebra and Mbe an irreducible skew-symmetric G- module (i.e., x m =0 for all x∈G m ∈M). Then the vector space Q=G+M equipped with the multiplication x +m y +n =x y +m y where m n ∈M x y ∈Gis a simple Leibniz algebra. From the analogue of Levi’s decomposition for Leibniz algebras (see [4]), we conclude that the above construction is universal for a simple Leibniz algebra assuming G=L/I M =I Q =L, that is, any simple Leibniz algebra is realized by this construction. According to [5], in any three-dimensional simple Lie algebra there exits a basis e f h such that the table of multiplication has the form e h =2e f h =−2f e f =h h e =−2e h f =2f f e =−h This Lie algebra is denoted by sl2, and the basis e f h is called the canonical basis. Theorem 1.1 ([5]).For each integer m=012, there exists one, up to isomorphism, irreducible sl2-module Mof dimension m+1The module Mhas a basis x0x 1  xmsuch that the representing transformations E,F, and Hcorresponding to the canonical basis e f h are given by Hxk=m −2kxk0≤k≤m Fxm=0 Fxk=xk+10≤k≤m−1 Ex0=0 Exk=−km +1−kxk−11≤k≤m In [8], the authors described the complex finite-dimensional Leibniz algebras whose associated Lie algebra is isomorphic to sl2The crucial role in that classification is played by Theorem 1.1. Taking a similar approach, a property of Leibniz algebras with an associated Lie algebra sl2˙ +R where Ris a solvable radical, is obtained. In addition, in this article, we present the classification of such algebras for a two-dimensional radical R. Throughout the article, we consider finite-dimensional algebras over the field of complex numbers. Moreover, in the multiplication table of an algebra we omit the null products. 2. COMPLEX LEIBNIZ ALGEBRAS WHOSE ASSOCIATED LIE ALGEBRAS ARE ISOMORPHIC TO sl2˙ +R. In this section, we consider the Leibniz algebra Lfor which its corresponding Lie algebra is a semidirect sum of sl2and a two-dimensional solvable ideal R In addition, we shall assume that Iis a right irreducible module over sl2. Let Lbe a Leibniz algebra such that L/I ≃sl2⊕R where Ris a solvable Lie algebra and ¯e ¯ h ¯ f x0x 1x m,y1 y2ynare the bases of sl2I, and R, respectively. Let e h f x0x 1x my 1y 2y nbe a basis of the algebra Lsuch that e =¯e h =¯ h f =¯ f yi=yi1≤i≤n Then taking into account the Theorem 1.1, we have e h =2e+ m  j=0 aj ehxj h f =2f+ m  j=0 aj hf xj e f =h+ m  j=0 aj ef xj h e =−2e+ m  j=0 aj hexjf h =−2f+ m  j=0 aj fhxj f e =−h+ m  j=0 aj fexj e yi= m  j=0 ijxjf yi= m  j=0 ijxj h yi= m  j=0 ijxj xkh=m −2kxk0≤k≤m xkf=xk+10≤k≤m−1 xke=−km +1−kxk−11≤k≤m where 1 ≤i≤n. Similar as in the proof of Proposition 3.1 of the [8], one can get e h =2e h f =2f e f =h h e =−2e f h =−2f f e =−h e e =0f f =0 h h =0 Let us denote the following vector spaces: sl−1 2=<ehf> R −1=<y 1y 2y n> Proposition 2.1. Let Lbe a Leibniz algebra whose quotient L/I sl2⊕R, where R is a solvable ideal and Iis a right irreducible module over sl2with dimI = 3Then sl−1 2R −1=0 Proof. It is clearly sufficient to prove the equality for the basic elements of sl−1 2 and R−1. Consider the Leibniz identity e e yi =e e yi−e yi e =−e yi e =− m  j=0 ijxje= m  j=1 −mj +jj −1ijxj−11≤i≤n On the other hand, we have that e e yi =e m j=0ijxj=0 for 1 ≤i≤n. Comparing the coefficients of the basic elements, we obtain ij =0 for 1 ≤j≤ m Thus, e yi=i0x0with 1 ≤i≤n Consider the equalities 0=e m  j=0 ijxj=e f yi =e f yi−e yi f =h yi−i0x0f=h yi−i0x1 Then we have that h yi=i0x1with 1 ≤i≤n. From the equalities 0=e h yi =e h yi−e yi h =2e yi−i0x0h =2i0x0−mi0x0=i02−mx0 it follows that i0=0 for 1 ≤i≤n Taking into account that m= 2 (because of dimI = 3), we get e yi=h yi=0 with 1 ≤i≤n From the equalities 0=f e yi =f e yi−f yi e =h yi−f yi e =−f yi e =− m  j=0 ijxje=− m  j=0 −mj +jj −1ijxj−1 we derive ij =0 for 1 ≤j≤m. Consequently, f yi=i0x0for all 1 ≤i≤n. Similarly, from 0=f f yi =f f yi−f yi f =f yi f =i0x0f=i0x1 we obtain f yi=0 for all 1 ≤i≤n Thus, we obtain e yi=f yi=h yi=0 with 1 ≤i≤n We complete the proof of the proposition.  3. COMPLEX LEIBNIZ ALGEBRAS WHOSE ASSOCIATED LIE ALGEBRA IS ISOMORPHIC TO sl2˙ +R,dim R =2 In this section, we consider the case when Ris a two-dimensional solvable radical. From the classification of two-dimensional Lie algebras (see [5]) we know that in Rthere exists a basis y1 y2in which the following table of multiplication of Rhas this form: y1 y2=y1y2 y1=−y1 In the case when dimI = 3 and Ia right irreducible module over sl2, summarizing the results of the Proposition 2.1, we get the following table of multiplication: e h =2e h f =2f e f =h h e =−2e f h =−2f f e =−h xkh=m −2kxk0≤k≤m xkf=xk+10≤k≤m−1 xke=−km +1−kxk−11≤k≤m yie= m  j=0 aj iexj1≤i≤2 yif= m  j=0 aj if xj1≤i≤2 yih= m  j=0 aj ihxj1≤i≤2 xky i= m  j=0 ak ijxj0≤k≤m 1≤i≤2 y1y 2=y1+ m  j=0 bj 12xjy 2y 1=−y1 y1y 1= m  j=0 bj 1xjy 2y 2= m  j=0 bj 2xj (1) where e h f x0x 1x my 1y 2is a basis of L Let us present the following theorem which describes the Leibniz algebras with the condition L/I sl2⊕R, where dimI = 3 dimR =2, and Ia right irreducible module over sl2. Theorem 3.1. Let Lbe a Leibniz algebra whose quotient L/I sl2⊕R, where Ris a two-dimensional solvable radical and Iis a right irreducible module over sl2dimI = 3. Then there exists a basis e h f x0x 1x my 1y 2of Lsuch that the table of multiplication in Lhas the following form: e h =2e h f =2f e f =h h e =−2e f h =−2f f e =−h xkh=m −2kxk0≤k≤m xkf=xk+10≤k≤m−1 xke=−km +1−kxk−11≤k≤m y1y 2=y1y 2y 1=−y1 xky 2=axk0≤k≤m a ∈ Proof. Let Lbe an algebra satisfying the conditions of the theorem, then we get the table of multiplication 1. We consider the chain of the equalities 0=xi h y1 =xi h y1−xiy 1 h =m −2ixiy 1− m  k=0 ai 1kxkh =m −2i m  k=0 ai 1kxk− m  k=0 ai 1km −2kxk= m  k=0 ai 1km −2i−m −2kxk = m  k=0 2ai 1kk −ixk from which we have ai 1k=0 with 0 ≤i≤mand i= k Thus, xiy 1=ai 1ixi=a1ixiwith 0 ≤i≤m Similarly, 0=xi h y2 =xi h y2−xiy 2 h =m −2ixiy 2− m  k=0 ai 2kxih =m −2i m  k=0 ai 2kxi− m  k=0 ai 2km −2kxk= m  k=0 ai 2km −2i−m −2kxk = m  k=0 2ai 2kk −ixk we get xiy 2=ai 2ixiwith 0 ≤i≤m From the identity xiy 1y 2 =xiy 1 y2−xiy 2 y1, we deduce xiy 1+ m  k=0 bk 12xk=ai 1ixiy 2−ai 2ixiy 1 ⇒xiy 1=ai 1iai 2ixi−ai 2iai 1ixi=0 and we have xiy 1=0 with 0 ≤i≤m that is, I y1=0. We consider the Leibniz identity xiy 2 e =xiy 2 e −xi e y2 for 0 ≤i≤m Then, 0=ai 2ixie−−mi +ii −1xi−1y 2 =ai 2i−mi +ii −1xi−1−ai 2i−1−mi +ii −1xi−1 =−−mi +ii −1ai 2i−ai 2i−1xi−1 which leads to ai 2i=ai 2i−1, that is, we get xiy 2=axiwith 0 ≤i≤mand some a∈. Thus, we have obtained the product I R−1. Verifying that 0=y1 m  j=0 aj 1fxj=y1y 1 f =y1y 1 f −y1 f y1=y1y 1 f = m  j=0 bj 1xjf= m−1  j=0 bj 1xj+1 we obtain bj 1=0 for 0 ≤j≤m−1, i.e., y1y 1=b1 mxm The equalities 0=y1y 1 h =y1y 1 h −y1 h y1=y1y 1 h =am 1xmh=−mam 1xm deduce bm 1=0, hence y1y 1=0 From the identities 0=y2y 1 h =y2y 1 h −y2 h y1=−y1 h 0=y2y 1 f =y2y 1 f −y2 f y1=−y1 f 0=y2y 1 e =y2y 1 e −y2 e y1=−y1 e we obtain y1h=y1f=y1e=0 Using the above obtained products and 0=y1y 2 f =y1y 2 f −y1 f y2=y1y 2 f =y1+ m  k=0 bk 12xkf= m  k=0 bk 12xkf= m−1  k=0 bk 12xk+1 we get bi 12 =0 with 0 ≤i≤m−1 Now from 0=y1y 2 h =y1y 2 h −y1 h y2=y1y 2 h =y1+bm 12xmh=−mbm 12xm we get bm 12 =0Consequently, bm 12 =0 for all 0 ≤i≤m, i.e., y1y 2=y1. Thus, we obtain the following table of multiplication: e h =2e h f =2f e f =h h e =−2e f h =−2f f e =−h xkh=m −2kxk0≤k≤m xkf=xk+10≤k≤m−1 xke=−km +1−kxk−11≤k≤m y2e= m  j=0 aj 2exjy 2f= m  j=0 aj 2fxjy 2h= m  j=0 aj 2hxj y1y 2=y1y 2y 1=−y1y 2y 2= m  j=0 bj 2xj xky 2=axk0≤k≤m In order to complete the proof of the theorem, we need to prove that y2y 2= 0and R−1sl −1 2=0. Consider two cases. Case 1. Let a= 0 be. Then taking the change of the basic element y2 =y2− m  j=0 bj 2 axj we get y2 y 2 =y2− m  j=0 bj 2 axjy 2− m  j=0 bj 2 axj =y2y 2−m  j=0 bj 2 axjy 2= m  j=0 bj 2xj− m  j=0 bj 2xj=0 which leads to y2y 2=0 Consider 0=y2y 2 h =y2y 2 h −y2 h y2=−y2 h y2 =− m  j=0 aj 2hxjy 2=− m  j=0 aj 2haxj which gives aj 2h=0 for 0 ≤j≤m. Similarly, from the equalities 0=y2y 2 f =y2y 2 f −y2 f y2=− m  j=0 aj 2fxjy 2=− m  j=0 aj 2faxj 0=y2y 2 e =y2y 2 e −y2 e y2=− m  j=0 aj 2exjy 2=− m  j=0 aj 2eaxj we get aj 2f=aj 2e=0 for 0 ≤j≤m. Hence, R−1sl −1 2=0. Thus, we proved the theorem for the case of a= 0. Case 2. Let a=0 be. Then, we consider the identity y2y 2 f =y2y 2 f −y2 f y2 and we derive 0= m  j=0 bi 2xif= m−1  j=0 bi 2xi+1 ⇒bi 2=00≤i≤m−1 ie y2y 2=bm 2xm From the chain of the equalities 0=y2y 2 h =y2y 2 h −y2 h y2=bm 2xmh=−mbm 2xm we obtain bm 2=0, that is, y2y 2=0 Let us take the change of the basic element in the form y2 =y2− m  j=1 aj−1 2e −mj +jj −1xj Then y2 e=y2e− m  j=1 aj−1 2e −mj +jj −1xje =y2e− m  j=1 aj−1 2e −mj +jj −1−mj +jj −1xj−1 = m  j=0 aj 2exj− m  j=1 aj−1 2exj−1 = m  j=0 aj 2exj− m−1  j=0 aj 2exj=am 2exm Thus, we can assume that y2e=am 2exmy 2h= m  j=0 aj 2hxjy 2f= m  j=0 aj 2fxj We have y2 e h =y2 e h −y2 h e =am 2exmh− m  j=0 aj 2hxje =−mam 2exm− m  j=0 aj 2h−mj +jj −1xj−1