aa r X i v : . [ m a t h . R A ] O c t LEIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS
J ¨ORG FELDVOSS
Abstract.
In this paper we define the basic concepts for left or right Leibnizalgebras and prove some of the main results. Our proofs are often variationsof the known proofs but several results seem to be new.
Introduction
This paper is mainly a survey on (left or right) Leibniz algebras from the pointof view of the theory of non-associative (i.e., not necessarily associative) algebras,but there are also several new results. Leibniz algebras were first introduced byBloh in the mid sixties of the last century (see [11, 12, 13]) and then forgottenfor nearly thirty years. In the early 1990’s they were rediscovered by Loday whotogether with his students and collaborators developed much of the theory of Leibnizalgebras, Leibniz bimodules, and Leibniz cohomology (see, for example, [35, 36,20]). A left (resp. right) Leibniz algebra is a vector space with a multiplicationfor which every left (resp. right) multiplication operator is a derivation (i.e., alinear operator satisfying the usual Leibniz product rule). As such Leibniz algebrasare non-anticommutative versions of Lie algebras. In particular, Leibniz algebrasare examples of non-associative algebras (see [44]). In contrast to other paperson this topic, we study Leibniz algebras exclusively from this point of view. Wehave tried to make the paper sufficiently self-contained so that it could serve asa first introduction to Leibniz algebras, their modules (or representations), andtheir cohomology. Leibniz algebras play an important role in different areas ofmathematics and physics (see [35]). In the last three decades numerous papers onLeibniz algebras appeared and many results have been duplicated. In this paperwe develop the basics of the theory of Leibniz algebras in a systematic way byconsidering them as a special class of non-associative algebras. In the following wewill describe the contents of the paper in more detail.The first section is devoted to some background material on non-associativealgebras which will be useful for the rest of the paper. In particular, we introducethe concept of a radical of an arbitrary algebra and prove three of its properties
Date : October 17, 2018.2010
Mathematics Subject Classification.
Primary 17A32; Secondary 17A05, 17A20, 17A60,17A65, 17A99.
Key words and phrases.
Non-associative algebra, derived series, solvable algebra, radical, Liemultiplication algebra, nilpotent algebra, flexible algebra, power-associative algebra, nil algebra,opposite algebra, Leibniz algebra, center, Leibniz kernel, Leibniz module, Engel’s theorem, Lie’stheorem, trace form, invariant symmetric bilinear form, Killing form, minimally degenerate, Car-tan’s solvability criterion, extension of Leibniz algebras, Leibniz bimodule, Leibniz cohomology,decendent central series, simple Leibniz algebra, Lie-simple Leibniz algebra, semisimple Leibnizalgebra, first Whitehead lemma, second Whitehead lemma. that can be used as axioms for such a concept. We refer the reader to [44] for moredetails and most of the proofs.In the second section we give the definition of left resp. right Leibniz algebrasand prove some basic results. We use several low-dimensional Leibniz algebrasto illustrate the concepts to be introduced and the results to be proved. Amongother notions, we define the (left/right) center and the Leibniz kernel of a Leibnizalgebra. The latter measures how much a Leibniz algebra deviates from being aLie algebra. Moreover, we associate several Lie algebras to a Leibniz algebra anddiscuss how these are related to each other. We include an example showing that leftor right Leibniz algebras are not necessarily power-associative. This falsifies a claimby Barnes in [6]. On the other hand, we prove that symmetric Leibniz algebras(i.e., algebras satisfying the left and the right Leibniz identity) are flexible, power-associative, and nil. The terms of the derived series of an arbitrary algebra areusually only subalgebras. At the end of the second section we show that for left orright Leibniz algebras each term of their derived series is an ideal.Section 3 contains definitions of left Leibniz modules and Leibniz bimodulesof a left Leibniz algebra. In particular, following Eilenberg [28] we motivate thedefining identities of a Leibniz bimodule by considering abelian extensions of a leftLeibniz algebra. We prove some basic properties of Leibniz bimodules followingmainly Loday [35, 36] who introduced and investigated Leibniz bimodules for aright Leibniz algebra. In addition, we also briefly discuss trace forms associatedto finite-dimensional left Leibniz modules (see also [2] and [22]). Similarly to theLeibniz kernel of a Leibniz algebra, we introduce the anti-symmetric kernel of aLeibniz bimodule. Using this concept, we give a very short “Schur’s lemma type”proof of the fact that irreducible Leibniz bimodules are either symmetric or anti-symmetric (see also the proof of Theorem 3.1 in [29]). Note that this proof neitherneeds to assume that the Leibniz algebra is finite dimensional nor that the Leibnizbimodule is finite dimensional as in [7, Theorem 1.4]). In the fourth section wedefine the cohomology of a left Leibniz algebra in analogy to the cohomology ofa right Leibniz algebra in [35, 36], and describe the Leibniz cohomology spaces indegree 0 and 1. We also show explicitly how Leibniz 2-cocycles give rise to abelianextensions of left Leibniz algebras.The remaining three sections of the paper are devoted to several results for nilpo-tent, solvable, and semisimple Leibniz algebras, respectively. We give variants ofthe known proofs of Engel’s and Lie’s theorem for Leibniz algebras as well as derivesome of their applications. Furthermore, we prove Cartan’s solvability criterionfor Leibniz algebras. We characterize the nilpotency and solvability of a Leibnizalgebra in terms of the nilpotency and solvability of their associated Lie algebras,respectively. These results seem to be new. In the last section we derive somestructural properties of (semi)simple Leibniz algebras and explain that the firstWhitehead lemma does not hold for Leibniz algebras. In a previous version of thispaper we derived the second Whitehead lemma for Leibniz algebras from Levi’stheorem for Leibniz algebras along the lines of the proof of [44, Proposition 3.22].Unfortunately, our proof was not correct. We are very grateful to Bakhrom Omirovfor bringing this to our attention. Recently, in joint work with Friedrich Wagemann we found a proof of the second Whiteheadlemma for Leibniz algebras by using spectral sequences.
EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 3
In this paper all algebras are defined over a field. For a subset X of a vectorspace V over a field F we let h X i F be the subspace of V that is spanned by X . Weuse [ − , − ] to denote the commutator of linear operators or matrices. The identityfunction on a set S will be denoted by id S , the set of non-negative integers will bedenoted by N , and the set of positive integers will be denoted by N .1. Non-associative algebras
In this section we briefly recall some of the definitions and results on non-associative algebras that we will need in the remainder of the paper. For moredetails and most of the proofs we refer the reader to [44].An algebra A is a vector space over a field with a bilinear mapping A × A → A ,( x, y ) xy , the multiplication of A . The usual definitions of the concepts of subalgebra , left or right ideal , ideal (= left and right ideal), homomorphism , iso-morphism , etc., are the same as for associative algebras since they do not use theassociativity of the multiplication. Moreover, the fundamental homomorphism the-orem , the isomorphism theorems , and the correspondence theorems for subalgebras and for one-sided or two-sided ideals , respectively, continue to hold with the sameproofs as in the associative case.Let S and T be two non-empty subsets of an algebra A over a field F . Then ST := h st | s ∈ S , t ∈ T i F is the F -subspace of A spanned by the products st . In particular, S := SS (see[44, p. 9]).The derived series of subalgebras A (0) ⊇ A (1) ⊇ A (2) ⊇ · · · of A is defined recursively by A (0) := A and A ( n +1) := ( A ( n ) ) for every non-negativeinteger n .Note that A ( m + n ) = ( A ( m ) ) ( n ) for all non-negative integers m and n . Moreover,if φ : A → B is a homomorphism of algebras, then φ ( A ( n ) ) = φ ( A ) ( n ) for everynon-negative integer n .An algebra A is called solvable if A ( r ) = 0 for some non-negative integer r . Analgebra A is called abelian if AA = 0 (i.e., if any product of elements in A is zero).The next result is an immediate consequence of the compatibility of homomor-phisms of algebras with the derived series. Proposition 1.1.
Subalgebras and homomorphic images of solvable algebras aresolvable.
The following results are well-known (see [44, Proposition 2.2 and 2.3]).
Proposition 1.2.
Extensions of solvable algebras by solvable algebras are solvable.
Proposition 1.3.
The sum of two solvable ideals of an algebra is solvable.
Let A be a finite-dimensional algebra, and let R be a solvable ideal of maximaldimension. If I is any solvable ideal of A , then it follows from Proposition 1.3 that R + I is a solvable ideal of A . Since R ⊆ R + I , for dimension reasons we have that R = R + I , and thus I ⊆ R . This shows that R is the largest solvable ideal of A .The ideal R is called the radical of the algebra A and will be denoted by Rad( A ). J ¨ORG FELDVOSS
Proposition 1.4.
Every finite-dimensional algebra A contains a largest solvableideal Rad( A ) satisfying the following properties: (a) Rad(Rad( A )) = Rad( A ) . (b) Rad( A / Rad( A )) = 0 . (c) If φ : A → B is a homomorphism of algebras, then φ (Rad( A )) ⊆ Rad( φ ( A )) .Proof. (a) follows from the solvability of Rad( A ), and (c) is an immediate conse-quence of Proposition 1.1.(b): According to the correspondence theorem for ideals, there exists an ideal I of A such that Rad( A ) ⊆ I and Rad( A / Rad( A )) = I / Rad( A ). This implies that I / Rad( A ) is solvable, and therefore Proposition 1.2 yields that I is solvable. Hence I ⊆ Rad( A ), and so Rad( A / Rad( A )) = 0. (cid:3) Let a ∈ A be an arbitrary element. Then the left multiplication operator L a : A → A , x ax is linear and L ( A ) := { L a | a ∈ A } is a subspace of the associative algebra End( A ) of linear operators on A . Similarly,the right multiplication operator R a : A → A , x xa is linear and R ( A ) := { R a | a ∈ A } is a subspace of End( A ). Let Mult( A ) denote the subalgebra of End( A ) that isgenerated by L ( A ) ∪ R ( A ), the associative multiplication algebra of A (see [44,Section 2 in Chapter II]). Let gl ( A ) denote the general linear Lie algebra on theunderlying vector space of A with the commutator [ X, Y ] := X ◦ Y − Y ◦ X as Liebracket. Moreover, let Lie( A ) denote the subalgebra of gl ( A ) that is generated by L ( A ) ∪ R ( A ), the Lie multiplication algebra of A (see [44, Section 3 in Chapter II]).Finally, letDer( A ) := { D ∈ End( A ) | ∀ x, y ∈ A : D ( xy ) = D ( x ) y + xD ( y ) } denote the derivation algebra of A . Note that Der( A ) is a subalgebra of gl ( A ), andtherefore Der( A ) is another Lie algebra associated to A .An algebra A is called nilpotent if there exists a positive integer n such that anyproduct of n elements in A , no matter how associated, is zero. This generalizes theconcept of nilpotency for associative algebras. Note that every nilpotent algebrais solvable (see [44, p. 18]). For any subset S of an algebra A let S ∗ denote thesubalgebra of Mult( A ) generated by { L s : A → A | s ∈ S } ∪ { R s : A → A | s ∈ S } . Then an ideal I of A is nilpotent if, and only if, I ∗ is nilpotent (see[44, Theorem 2.4]). In particular, A is nilpotent if, and only if, Mult( A ) = A ∗ isnilpotent.An algebra A is called flexible if L x ◦ R x = R x ◦ L x holds for every element x ∈ A ,or equivalently, if the identity x ( yx ) = ( xy ) x is satisfied for all elements x, y ∈ A (see [44, p. 28]). An algebra A is called power-associative if any subalgebra of A generated by one element is associative (see [44, p. 30]). In this case one can definepowers of an element x ∈ A recursively by x := x and x n +1 := xx n for everypositive integer n . These powers then satisfy the usual power laws x m + n = x m x n and ( x m ) n = x mn (see [44, p. 30]). Alternative algebras, Jordan algebras, and Liealgebras are flexible and power-associative (see [44, pp. 28, 30, and 92]).An element x of a power-associative algebra is called nilpotent if x n = 0 forsome positive integer n . A subset of a power-associative algebra consisting only of EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 5 nilpotent elements is called nil (see [44, p. 30]). Note that every solvable power-associative algebra is nil (see [44, p. 31]).The opposite algebra A op of an algebra A with multiplication ( x, y ) xy hasthe same underlying vector space structure and the opposite multiplication ( x, y ) x · y := yx . Since the derived series of A op coincides with the derived series of A ,the opposite algebra of a solvable algebra is solvable. Moreover, it is clear from thedefinition of nilpotency, that the opposite algebra of a nilpotent algebra is nilpotent.2. Leibniz algebras – Definition and Examples A left Leibniz algebra is an algebra L such that every left multiplication operator L x : L → L , y xy is a derivation. This is equivalent to the identity(2.1) x ( yz ) = ( xy ) z + y ( xz )for all x, y, z ∈ L , the left Leibniz identity , which in turn is equivalent to the identity(2.2) ( xy ) z = x ( yz ) − y ( xz )for all x, y, z ∈ L .Similarly, one defines a right Leibniz algebra to be an algebra L such that everyright multiplication operator R y : L → L , x xy is a derivation. This is equivalentto the identity(2.3) ( xy ) z = ( xz ) y + x ( yz )for all x, y, z ∈ L , the right Leibniz identity , which in turn is equivalent to theidentity(2.4) x ( yz ) = ( xy ) z − ( xz ) y for all x, y, z ∈ L .Following Mason and Yamskulna [38] we call an algebra a symmetric Leibnizalgebra if it is at the same time a left and a right Leibniz algebra. Note that everyLie algebra is a symmetric Leibniz algebra.It is clear that the opposite algebra of a left Leibniz algebra is a right Leibnizalgebra and that the opposite algebra of a right Leibniz algebra is a left Leibnizalgebra. Consequently, the opposite algebra of a symmetric Leibniz algebra is againa symmetric Leibniz algebra. Therefore, in most situations it is enough to consideronly left or right Leibniz algebras.The following results are direct consequences of the left and right Leibniz identity,respectively. Lemma 2.1. If L is a left Leibniz algebra, then L x = 0 for every element x ∈ L . Lemma 2.2. If L is a right Leibniz algebra, then R x = 0 for every element x ∈ L .Proof. We only prove Lemma 2.1 as this yields Lemma 2.2 by considering theopposite algebra. Let x, y ∈ L be arbitrary elements. Then we obtain from identity(2.2) that L x ( y ) = x y = x ( xy ) − x ( xy ) = 0 , which shows that L x = 0. (cid:3) J ¨ORG FELDVOSS
Every abelian (left or right) Leibniz algebra is a Lie algebra, but there are manyLeibniz algebras that are not Lie algebras (see, for example, [20, 4, 5, 27, 1, 2, 3,39, 18, 31, 30, 34, 16, 15, 17, 22, 21, 23, 24, 25]). We will use the following threeexamples to illustrate the concepts introduced in this section.
Examples. (1) Let A ℓ := F e ⊕ F f be a two-dimensional vector space with multiplication ee = f e = f f = 0, and ef = f . Then A ℓ is a left Leibniz algebra, but nota right Leibniz algebra. We have that A (1) ℓ = F f and A (2) ℓ = 0. Hence A ℓ is solvable.(2) Let N := F e ⊕ F f be a two-dimensional vector space with multiplication ee = ef = f e = 0, and f f = e . Then N is a symmetric Leibniz algebra.We will see in Section 5 that N is nilpotent.(3) Let S ℓ := sl ( C ) × L (1), where sl ( C ) is the Lie algebra of traceless complex2 × L (1) is the two-dimensional left sl ( C )-module (see [32,Lemma 7.2]). Then S ℓ with multiplication ( X, a )( Y, b ) := ([
X, Y ] , X · b )for any X, Y ∈ sl ( C ) and any a, b ∈ L (1) is a left Leibniz algebra (seeSection 3 and Lemma 4.7) . Moreover, it can be shown that S ℓ is simple(see Section 7 for the definition of the simplicity of Leibniz algebras). Remark.
One can prove that up to isomorphism A ℓ , A op ℓ , and N are the onlytwo-dimensional left or right non-Lie Leibniz algebras (see [22, pp. 11/12]).Let L be a left or right Leibniz algebra. Then C ℓ ( L ) := { c ∈ L | L c = 0 } = { c ∈ L | ∀ x ∈ L : cx = 0 } is called the left center of L , C r ( L ) := { c ∈ L | R c = 0 } = { c ∈ L | ∀ x ∈ L : xc = 0 } is called the right center of L , and C ( L ) := C ℓ ( L ) ∩ C r ( L )is called the center of L . Remark.
For a Lie algebra the left center, the right center, and the center are allthe same.It is an immediate consequence of the definitions that the left center is a rightideal and the right center is a left ideal. More precisely, we have the followingresults.
Proposition 2.3.
Let L be a left Leibniz algebra. Then L C ℓ ( L ) ⊆ C ℓ ( L ) and C ℓ ( L ) L = 0 . In particular, C ℓ ( L ) is an abelian ideal of L . Proposition 2.4.
Let L be a right Leibniz algebra. Then C r ( L ) L ⊆ C r ( L ) and L C r ( L ) = 0 . In particular, C r ( L ) is an abelian ideal of L . This is known as the hemi-semidirect product of sl ( C ) and L (1) (see [40, Definition 1.5]). Note that, as for Lie algebras, the given definition of the center of a left or right Leibnizalgebra is not the one used for other non-associative algebras (see [44, p. 14]). The reason for thisis that, in general, Leibniz algebras, contrary to alternative algebras or Jordan algebras, are farfrom being associative (see Proposition 2.15 and Proposition 2.16).
EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 7
Proof.
We only prove Proposition 2.3 as this yields Proposition 2.4 by consideringthe opposite algebra. Let c ∈ C ℓ ( L ) and x, y ∈ L be arbitrary elements. Then weobtain from identity (2.2) that L xc ( y ) = ( xc ) y = x ( cy ) − c ( xy ) = xL c ( y ) − L c ( xy ) = 0 , which shows that L xc = 0, i.e., xc ∈ C ℓ ( L ). This proves the first statement,and the second statement is an immediate consequence of the definition of the leftcenter. (cid:3) Examples. (1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1we have that C ℓ ( A ℓ ) = F f and C r ( A ℓ ) = F e . Hence C ( A ℓ ) = 0.(2) For the two-dimensional nilpotent symmetric Leibniz algebra N from Ex-ample 2 we have that C ( N ) = C ℓ ( N ) = C r ( N ) = F e .(3) For the five-dimensional simple left Leibniz algebra S ℓ from Example 3 wehave that C ℓ ( S ℓ ) = L (1) and C r ( S ℓ ) = 0. Hence C ( S ℓ ) = 0.Let L be a left or right Leibniz algebra over a field F . ThenLeib( L ) := h x | x ∈ L i F is called the Leibniz kernel of L . The Leibniz kernel measures how much a left orright Leibniz algebra deviates from being a Lie algebra. In particular, a left or rightLeibniz algebra is a Lie algebra if, and only if, its Leibniz kernel vanishes.The following results are well-known (see [20, Proposition 1 (a)], [19, Proposi-tion 1.1.4], and [7, Lemma 1.3 and its proof]). Proposition 2.5.
Let L be a left Leibniz algebra. Then L Leib( L ) ⊆ Leib( L ) and Leib( L ) ⊆ C ℓ ( L ) . Proposition 2.6.
Let L be a right Leibniz algebra. Then Leib( L ) L ⊆ Leib( L ) and Leib( L ) ⊆ C r ( L ) . Corollary 2.7. If L is a symmetric Leibniz algebra, then Leib( L ) ⊆ C ( L ) .Proof. We only prove Proposition 2.6 as this yields Proposition 2.5 by consideringthe opposite algebra. As C r ( L ) is a subspace of L , the second statement is animmediate consequence of Lemma 2.2.Since Leib( L ) is a subspace of L , it is enough for the proof of the first statementto show that x y ∈ Leib( L ) for any elements x, y ∈ L . By using again Lemma 2.2,we obtain that( x + y ) = ( x + y )( x + y ) = x x + x y + yx + y = x y + y , and therefore x y = ( x + y ) − y ∈ Leib( L ). (cid:3) Examples. (1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1we have that Leib( A ℓ ) = F f = C ℓ ( A ℓ ). Moreover, Leib( A ℓ ) ∩ C r ( A ℓ ) = 0.(2) For the two-dimensional nilpotent symmetric Leibniz algebra N from Ex-ample 2 we have that Leib( N ) = F e = C ℓ ( N ) = C r ( N ) = C ( N ).(3) For the five-dimensional simple left Leibniz algebra S ℓ from Example 3 wehave that Leib( S ℓ ) = L (1) = C ℓ ( S ℓ ). Moreover, Leib( S ℓ ) ∩ C r ( S ℓ ) = 0. J ¨ORG FELDVOSS
Remark.
Let L be the two-dimensional solvable left Leibniz algebra from Exam-ple 1 or the five-dimensional simple left Leibniz algebra from Example 3. We havethat L Leib( L ) = Leib( L ). This shows that for a left Leibniz algebra L , L Leib( L )can be non-zero and L Leib( L ) is not necessarily properly contained in Leib( L ). Ofcourse, a similar statement holds for right Leibniz algebras. But for a symmetricLeibniz algebra L we have that L Leib( L ) = 0 = Leib( L ) L as Leib( L ) ⊆ C ( L ).The following result is a consequence of Proposition 2.5 or Proposition 2.6 (see[22, p. 11] and [29, p. 479]). Proposition 2.8.
Let L be a left or right Leibniz algebra. Then Leib( L ) is anabelian ideal of L . Moreover, if L = 0 , then Leib( L ) = L .Proof. It follows from Proposition 2.5 or Proposition 2.6 that in either case Leib( L )is an ideal of L such that Leib( L )Leib( L ) = 0.Suppose now that Leib( L ) = L . Then we have that LL = 0. In particular, everysquare of L is zero. Consequently, we obtain that L = Leib( L ) = 0. (cid:3) Remark:
It is an immediate consequence of Proposition 2.8 that every one-dimensional left or right Leibniz algebra is a Lie algebra. Moreover, with a littlemore work one can use Proposition 2.8 to classify the two-dimensional left or rightLeibniz algebras (see [22, pp. 11/12]).Let L be a left or right Leibniz algebra. Then by definition of the Leibniz kernel, L Lie := L / Leib( L ) is a Lie algebra. We call L Lie the canonical Lie algebra associatedto L . In fact, the Leibniz kernel is the smallest ideal such that the correspondingfactor algebra is a Lie algebra (see [29, Theorem 2.5]). Proposition 2.9.
Let L be a left or right Leibniz algebra. Then Leib( L ) is thesmallest ideal I of L such that L / I is a Lie algebra.Proof. Let I be an ideal of L such that L / I is a Lie algebra. Then it follows fromthe anti-commutativity of L / I that x ∈ I for every element x ∈ L . Since I is asubspace of L , we conclude that Leib( L ) ⊆ I . (cid:3) Proposition 2.10.
Let L be a left Leibniz algebra. Then the set L ( L ) of leftmultiplication operators of L is an ideal of the derivation algebra Der( L ) of L , and L /C ℓ ( L ) is a Lie algebra that is isomorphic to L ( L ) .Proof. By definition of a left Leibniz algebra, L ( L ) ⊆ Der( L ). Since the multipli-cation of L is bilinear, L ( L ) is a subspace of Der( L ).Let x, y ∈ L and D ∈ Der( L ) be arbitrary elements. Then the computation[ D, L x ]( y ) = ( D ◦ L x − L x ◦ D )( y ) = D ( xy ) − xD ( y ) = D ( x ) y = L D ( x ) ( y )shows that [ D, L x ] = L D ( x ) ∈ Der( L ) for any element x ∈ L and any derivation D ∈ Der( L ). This completes the proof that L ( L ) is an ideal of Der( L ).Now consider the linear transformation L : L → Der( L ) defined by x L x . Bydefinition, Ker( L ) = C ℓ ( L ) and Im( L ) = L ( L ). Finally, it follows from the identity[ D, L y ] = L D ( y ) that L xy = L L x ( y ) = [ L x , L y ] for any elements x, y ∈ L , and thus L is a homomorphism of Leibniz algebras. Then the fundamental homomorphismtheorem shows that L induces a Leibniz algebra isomorphism from L/C ℓ ( L ) onto L ( L ). (cid:3) EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 9
Remark.
I am very grateful to Friedrich Wagemann for pointing out to me thefour-term exact sequence of left Leibniz algebras0 → C ℓ ( L ) → L L → Der( L ) → Out ℓ ( L ) → , where Out ℓ ( L ) := Der( L ) /L ( L ) is the Lie algebra of outer derivations of the leftLeibniz algebra L , and in which only L is not necessarily a Lie algebra (see [26,Proposition 1.8]).By combining Proposition 2.5 and Proposition 2.10 with the fundamental homo-morphism theorem we obtain the following result. Corollary 2.11.
Let L be a left Leibniz algebra. Then the set L ( L ) of left multi-plication operators of L is a homomorphic image of the canonical Lie algebra L Lie associated to L . The first part of the next result is [29, Theorem 2.3].
Proposition 2.12.
Let L be a right Leibniz algebra. Then the set R ( L ) of rightmultiplication operators of L is an ideal of the derivation algebra Der( L ) of L , and L /C r ( L ) is a Lie algebra that is isomorphic to R ( L ) op .Proof. The proof of Proposition 2.12 is very similar to the proof of Proposition 2.10.Nevertheless, we include the whole argument in this case as well, since it shows whyleft multiplication operators are preferable when one writes functions on the left oftheir arguments.By definition of a right Leibniz algebra, R ( L ) ⊆ Der( L ). Since the multiplicationof L is bilinear, R ( L ) is a subspace of Der( L ).Let x, y ∈ L and D ∈ Der( L ) be arbitrary elements. Then the computation[ D, R x ]( y ) = ( D ◦ R x − R x ◦ D )( y ) = D ( yx ) − D ( y ) x = yD ( x ) = R D ( x ) ( y )shows that [ D, R x ] = R D ( x ) ∈ Der( L ) for any element x ∈ L and any derivation D ∈ Der( L ). This completes the proof that R ( L ) is an ideal of Der( L ).Now consider the linear transformation R : L → Der( L ) defined by x R x . Bydefinition, Ker( R ) = C r ( L ) and Im( R ) = R ( L ). Finally, it follows from the identity[ D, R x ] = R D ( x ) that R xy = R R y ( x ) = [ R y , R x ] for any elements x, y ∈ L , and thus R : L → Der( L ) op is a homomorphism of Leibniz algebras. Then the fundamentalhomomorphism theorem shows that R induces a Leibniz algebra isomorphism from L/C r ( L ) onto R ( L ) op . (cid:3) Remark.
Similarly to [26, Proposition 1.8], we obtain from the proof of Proposi-tion 2.12 the four-term exact sequence of right Leibniz algebras0 → C r ( L ) → L R → Der( L ) op → Out r ( L ) op → , where Out r ( L ) := Der( L ) /R ( L ) is the Lie algebra of outer derivations of the rightLeibniz algebra L , and in which only L is not necessarily a Lie algebra.By combining Proposition 2.6 and Proposition 2.12 with the fundamental homo-morphism theorem we obtain the following result. Corollary 2.13.
Let L be a right Leibniz algebra. Then the set R ( L ) of right mul-tiplication operators of L is a homomorphic image of the opposite of the canonicalLie algebra L opLie associated to L . The Lie multiplication algebra of a symmetric Leibniz algebra can be described asfollows. This generalizes the corresponding result for Lie algebras (see [44, p. 21]).
Theorem 2.14.
The Lie multiplication algebra
Lie( L ) = L ( L ) + R ( L ) of a sym-metric Leibniz algebra L is an ideal of the derivation algebra Der( L ) of L .Proof. Recall that the Lie multiplication algebra Lie( L ) of the symmetric Leibnizalgebra L is the smallest subalgebra of the general linear Lie algebra gl ( L ) thatcontains L ( L ) ∪ R ( L ). It follows from Proposition 2.10 and Proposition 2.12 that L ( L ) and R ( L ) are ideals of Der( L ). But as a sum of ideals, L ( L ) + R ( L ) is anideal of Der( L ), and thus a subalgebra of Der( L ) which in turn is a subalgebra of gl ( L ). This shows that Lie( L ) = L ( L ) + R ( L ). (cid:3) Example.
Let N be the two-dimensional nilpotent symmetric Leibniz algebra fromExample 2. Then L ( N ) = F f = R ( N ), and therefore Lie( N ) = F f . On the otherhand, Leib( N ) = F e , and thus N Lie ∼ = F f ∼ = Lie( N ). Question.
What is the relationship between the canonical Lie algebra L Lie asso-ciated to a symmetric Leibniz algebra L and the Lie multiplication algebra Lie( L )of L ? Note that if L is a Lie algebra, then Lie( L ) ∼ = L /C ( L ) (see [46, Theo-rem 1.1.2 (4)] or Proposition 2.10). So in this case L Lie = Lie( L ) if C ( L ) = 0. Butas Leib( N ) = C ( N ), the isomorphism Lie( L ) ∼ = L /C ( L ) is compatible with theprevious example.The following results generalize [14, Exercise I.1.6] from Lie algebras to leftand right Leibniz algebras. As in the case of Lie algebras they are an immediateconsequence of the left and right Leibniz identity, respectively. Note the mixture ofleft and right in both propositions which we will see again in Proposition 5.9 andProposition 5.10. Proposition 2.15.
A left Leibniz algebra L is associative if, and only if, L ⊆ C r ( L ) . Proposition 2.16.
A right Leibniz algebra L is associative if, and only if, L ⊆ C ℓ ( L ) . These results show that, in general, left and right Leibniz algebras are far frombeing associative. In fact, it follows from Proposition 5.9 and Proposition 5.10below that associative left/right Leibniz algebras are nilpotent. Examples. (1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1we have that A ℓ = F f and C r ( A ℓ ) = F e . Hence A ℓ is not associative.(2) For the two-dimensional nilpotent symmetric Leibniz algebra N from Ex-ample 2 we have that N = F e = C ℓ ( N ) = C r ( N ). Hence N is associative.(3) For the five-dimensional simple left Leibniz algebra S ℓ from Example 3 wehave that S ℓ = S ℓ and C r ( S ℓ ) = 0. Hence S ℓ is not associative. Note that in dimension 3 there are five isomorphism classes of nilpotent non-Lie Leibniz alge-bras of which four (including one 1-parameter family) are associative (see [22, Theorem 6.4]). Indimension 4 there are seventeen isomorphism classes of indecomposable nilpotent non-Lie Leibnizalgebras of which eleven (including three 1-parameter families) are associative (see [3, Theo-rem 3.2], but compare this with [24] for the correct total number of isomorphism classes).
EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 11
In the sentence after the proof of [6, Corollary 1.3] Barnes claims that left Leib-niz algebras are power-associative. But the following example shows that this isnot always the case. Let A ℓ denote the two-dimensional left Leibniz algebra fromExample 1. Then ( e + f ) ( e + f ) = 0 = f = ( e + f )( e + f ) , which yields that A ℓ is not power-associative. The following result shows that Barnes’ claim holds for symmetric Leibniz alge-bras. In fact, like Lie algebras, such algebras are flexible.
Proposition 2.17.
Every symmetric Leibniz algebra is flexible and power-associative.Moreover, x = 0 for every element x .Proof. Let L be a symmetric Leibniz algebra. It follows from Lemma 2.2 that yx = 0 for any elements x and y of a right Leibniz algebra. Hence the left Leibnizidentity for L yields x ( yx ) = ( xy ) x + yx = ( xy ) x for any elements x and y of L ,and thus L is flexible.According to Lemma 2.1, we have that x y = 0 for any elements x and y of aleft Leibniz algebra. This implies that x x = 0 = xx for any element x ∈ L . Inparticular, third powers are well-defined and are equal to zero. Consequently, weobtain by induction that symmetric Leibniz algebras are power-associative. (cid:3) So in this respect symmetric Leibniz algebras are only slightly more general thanLie algebras; instead of squares of arbitrary elements of a Lie algebra being zero,cubes of arbitrary elements of a symmetric Leibniz algebra are zero.In general, the terms of the derived series of an algebra are only subalgebras.We finish this section by proving that any term in the derived series of a left or aright Leibniz algebra is indeed an ideal.
Proposition 2.18.
Let L be a left or right Leibniz algebra. Then L ( n ) is an idealof L for every non-negative integer n .Proof. We only prove the result for left Leibniz algebras as this yields the resultfor right Leibniz algebras by considering the opposite algebra. We first prove that L ( n ) is a left ideal of L for every non-negative integer n . We proceed by inductionon n . The base step n = 0 is clear as LL (0) = LL ⊆ L = L (0) . For the inductionstep let n > n −
1. Thenwe obtain from the left Leibniz identity and the induction hypothesis that LL ( n ) = L [ L ( n − L ( n − ] ⊆ [ LL ( n − ] L ( n − + L ( n − [ LL ( n − ] ⊆ L ( n − L ( n − = L ( n ) . Next, we prove that L ( n ) is a right ideal of L for every non-negative integer n .We again proceed by induction on n . The base step n = 0 is clear as L (0) L = LL ⊆ L = L (0) . For the induction step let n > This example also shows that left or right Leibniz algebras are not necessarily flexible. The proof of Proposition 2.17 shows that this result holds more generally for left centralLeibniz algebras or right central Leibniz algebras. The latter notions were introduced by Masonand Yamskulna in [38]: A left central Leibniz algebra is a left Leibniz algebra L such that Leib( L ) ⊆ C r ( L ) and a right central Leibniz algebra is a right Leibniz algebra L such that Leib( L ) ⊆ C ℓ ( L ).It follows from Proposition 2.5 and Proposition 2.6 that a left (resp. right) Leibniz algebra is left(resp. right) central if, and only if, Leib( L ) ⊆ C ( L ). statement is true for n −
1. Then we obtain from identity (2.2) and the inductionhypothesis that L ( n ) L = [ L ( n − L ( n − ] L ⊆ L ( n − [ L ( n − L ] ⊆ L ( n − L ( n − = L ( n ) . This completes the proof. (cid:3) Leibniz modules
In this and in the next section we consider only left Leibniz algebras. We leaveit to the interested reader to formulate the corresponding definitions and results forright Leibniz algebras (see [35, 36]).Let L be a left Leibniz algebra over a field F . A left L -module is a vector space M over F with an F -bilinear left L -action L × M → M , ( x, m ) x · m such that( xy ) · m = x · ( y · m ) − y · ( x · m )is satisfied for every m ∈ M and all x, y ∈ L .The usual definitions of the notions of submodule , irreducibility , complete re-ducibility , composition series , homomorphism , isomorphism , etc., hold for left Leib-niz modules.Moreover, every left L -module M gives rise to a homomorphism λ : L → gl ( M )of left Leibniz algebras, defined by λ x ( m ) := x · m , and vice versa. We call λ the left representation of L associated to M . We call the kernel of λ the annihilator of M and denote it by Ann L ( M ). Examples. (1) Every left Leibniz algebra is a left module over itself via the Leibniz multi-plication. This module is called the left adjoint module and will be denotedby L ad ,ℓ . The associated representation L : L → gl ( L ) is called the leftadjoint representation of L . Note that Ann L ( L ad ,ℓ ) = C ℓ ( L ).(2) The ground field F of any left Leibniz algebra is a left L -module via x · α := 0for every element x ∈ L and every scalar α ∈ F . This module is called the trivial left module of L .The next result generalizes the second part of Proposition 2.5 and reduces thestudy of left Leibniz modules to Lie modules. Lemma 3.1.
Let L be a left Leibniz algebra, and let M be a left L -module. Then Leib( L ) ⊆ Ann L ( M ) .Proof. Since Leib( L ) is a subspace of L , it is enough to show that x ∈ Ann L ( M )for any element x ∈ L . We have that λ x = λ x ◦ λ x − λ x ◦ λ x = 0, and therefore itfollows that x ∈ Ker( λ ) = Ann L ( M ). (cid:3) By virtue of Lemma 3.1, every left L -module is an L Lie -module, and vice versa.This is the reason that in [9, D´efinition 1.1.14] left Leibniz modules are calledLie modules. Consequently, many properties of left L -modules follow from thecorresponding properties of L Lie -modules. As one application of this point of view,we discuss trace forms associated to finite-dimensional left Leibniz modules (seealso [2, 22]).For any finite-dimensional left representation λ : L → gl ( M ) of a left Leibnizalgebra L we define its trace form κ λ : L × L → F by κ λ ( x, y ) := tr( λ x ◦ λ y ). EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 13
Every trace form is an invariant symmetric bilinear form. As usual, a bilinear form β : L × L → F is called invariant if β ( xy, z ) = β ( x, yz ) for any elements x, y, z ∈ L .The subspace L λ := { x ∈ L | ∀ y ∈ L : κ λ ( x, y ) = 0 } of L is called the radical of κ λ . Example.
As for Lie algebras, the trace form κ := κ L associated to the left adjointrepresentation L of a finite-dimensional left Leibniz algebra L is called the Killingform of L . We will denote its radical by L ⊥ .The next result is a consequence of Lemma 3.1 and the invariance as well as thesymmetry of κ λ . Lemma 3.2.
Let L be a left Leibniz algebra, and let λ : L → gl ( M ) be a finite-dimensional left representation of L . Then L λ is an ideal of L with Leib( L ) ⊆ L λ .Proof. The inclusion Leib( L ) ⊆ L λ follows immediately from Lemma 3.1.Let x, y ∈ L and z ∈ L λ be arbitrary elements. Then we obtain from theinvariance of κ λ that κ λ ( zx, y ) = κ λ ( z, xy ) = 0, i.e., zx ∈ L λ . This shows that L λ is a right ideal of L .Moreover, using the symmetry in conjunction with the invariance of κ λ , weconclude that κ λ ( xz, y ) = κ λ ( y, xz ) = κ λ ( yx, z ) = κ λ ( z, yx ) = 0, i.e., xz ∈ L λ .This proves that L λ is a left ideal of L . (cid:3) Lemma 3.2 implies that a trace form on a left non-Lie Leibniz algebra is nevernon-degenerate. We call a trace form associated to a finite-dimensional left rep-resentation λ of a left Leibniz algebra L minimally degenerate if L λ = Leib( L ). Example.
Let A ℓ be the two-dimensional solvable left Leibniz algebra from Ex-ample 1 in Section 2. Recall that Leib( A ℓ ) = F f = 0. Since κ ( e, e ) = 1 = 0 yields A ⊥ ℓ = A ℓ , we conclude from Lemma 3.2 that A ⊥ ℓ = Leib( A ℓ ). Hence the Killingform of A ℓ is minimally degenerate.The correct concept of a module for left Leibniz algebras is the notion of a Leibnizbimodule. In order to motivate the appropriate definition of a bimodule for a leftLeibniz algebra, we follow the approach that Eilenberg proposed for any given classof non-associative algebras (see [28] and [44, pp. 25/26]).Let L be a left Leibniz algebra over a field F , and let M be a vector space overthe same ground field. Then the Cartesian product L × M with componentwiseaddition and componentwise scalar multiplication is a vector space over F . Supposethat L acts on M from the left via the F -bilinear map L × M → M , ( x, m ) x · m and L acts on M from the right via the F -bilinear map M × L → M , ( m, x ) m · x .Then we define a multiplication on L × M by( x , m )( x , m ) := ( x x , m · x + x · m ) In [22, Definition 5.6] a minimally degenerate Killing form is called non-degenerate , but thiscontradicts the usual definition of a non-degenerate bilinear form. Note that non-Lie Leibnizalgebras can admit non-degenerate invariant symmetric bilinear forms, which, of course, cannotbe trace forms. For example, this is the case for the two-dimensional nilpotent symmetric Leibnizalgebra N from Example 2 in Section 2 (see [10, Example 2.2]). for any elements x , x ∈ L and m , m ∈ M . To ensure that L × M satisfies theleft Leibniz identity, we compute( x , m )[( x , m )( x , m )] = ( x , m )( x x , m · x + x · m )= ( x ( x x ) , m · ( x x ) + x · ( m · x + x · m ))= ( x ( x x ) , m · ( x x ) + x · ( m · x ) + x · ( x · m )) , [( x , m )( x , m )]( x , m ) = ( x x , m · x + x · m )( x , m )= (( x x ) x , ( m · x + x · m ) · x + ( x x ) · m ))= (( x x ) x , ( m · x ) · x + ( x · m ) · x + ( x x ) · m )) , ( x , m )[( x , m )( x , m )] = ( x , m )( x x , m · x + x · m )= ( x ( x x ) , m · ( x x ) + x · ( m · x + x · m ))= ( x ( x x ) , m · ( x x ) + x · ( m · x ) + x · ( x · m ))for any elements x , x , x ∈ L and m , m , m ∈ M . The vector space L × M satisfies the left Leibniz identity if, and only if, the left-hand side of the first identityequals the sum of the left-hand sides of the second and third identities. Hence theright-hand side of the first identity must equal the sum of the right-hand sides ofthe second and third identities. For the first components of the right-hand sidesthis is just the left Leibniz identity for L . From the desired equality for the secondcomponents of the right-hand sides one can read off the following three identitieswhich are sufficient for L × M to satisfy the left Leibniz identity.(3.1) x · ( y · m ) = ( xy ) · m + y · ( x · m )(3.2) x · ( m · y ) = ( x · m ) · y + m · ( xy )(3.3) m · ( xy ) = ( m · x ) · y + x · ( m · y )for every m ∈ M and all x, y ∈ L .This motivates the following definition. An L -bimodule is a vector space M with a bilinear left L -action and a bilinear right L -action such that the followingcompatibility conditions are satisfied:(LLM) ( xy ) · m = x · ( y · m ) − y · ( x · m )(LML) ( x · m ) · y = x · ( m · y ) − m · ( xy )(MLL) ( m · x ) · y = m · ( xy ) − x · ( m · y )for every m ∈ M and all x, y ∈ L .It is an immediate consequence of (LLM) that every Leibniz bimodule is a leftLeibniz module. Moreover, by combining (LML) and (MLL) we obtain the followingresult. Lemma 3.3.
Let L be a left Leibniz algebra, and let M be an L -bimodule. Then ( x · m ) · y + ( m · x ) · y = 0 holds for every m ∈ M and all x, y ∈ L . As for left Leibniz modules, the usual definitions of the notions of subbimodule , irreducibility , complete reducibility , composition series , homomorphism , isomor-phism , etc., hold for Leibniz bimodules. EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 15
Let L be a left Leibniz algebra over a field F , and let V be a vector space over F .Then a pair ( λ, ρ ) of linear transformations λ : L → End F ( V ) and ρ : L → End F ( V )is called a representation of L on V if the following conditions are satisfied:(3.4) λ xy = λ x ◦ λ y − λ y ◦ λ x (3.5) ρ xy = λ x ◦ ρ y − ρ y ◦ λ x (3.6) ρ xy = ρ y ◦ ρ x + λ x ◦ ρ y for every m ∈ M and all x, y ∈ L .Then every L -bimodule M gives rise to a representation ( λ, ρ ) of L on M via λ x ( m ) := x · m and ρ x ( m ) := m · x . Conversely, every representation ( λ, ρ ) of L on M defines an L -bimodule structure on M via x · m := λ x ( m ) and m · x := ρ x ( m ).In order to avoid confusion, the annihilator of an L -bimodule M will be denotedby Ann bi L ( M ), and it is defined as Ann bi L ( M ) := Ker( λ ) ∩ Ker( ρ ).An L -bimodule M is called symmetric if m · x = − x · m for every x ∈ L and every m ∈ M . An L -bimodule M is called anti-symmetric if m · x = 0 for every x ∈ L and every m ∈ M . Moreover, an L -bimodule M is called trivial if x · m = 0 = m · x for every x ∈ L and every m ∈ M . Note that an L -bimodule M is trivial if, andonly if, M is symmetric and anti-symmetric. Moreover, Ann bi L ( M ) = Ann L ( M ) if M is symmetric or anti-symmetric and Ann bi L ( M ) = L if, and only if, M is trivial. Examples. (1) Every left Leibniz algebra L is a bimodule over itself via the Leibniz multi-plication, the so-called adjoint bimodule L ad of L . The associated represen-tation is ( L, R ), where L denotes the left multiplication operator of L and R denotes the right multiplication operator of L . This representation is calledthe adjoint representation of L . Note that Ann bi L ( L ad ) = C ( L ). Moreover,unless the ground field has characteristic two, the adjoint bimodule L ad issymmetric if, and only if, L is a Lie algebra. (2) The ground field F of any left Leibniz algebra L is a trivial L -bimodule via x · α := 0 =: α · x for every element x ∈ L and every scalar α ∈ F . Lemma 3.4.
Let L be a left Leibniz algebra, and let M be an anti-symmetric or asymmetric L -bimodule. Then Leib( L ) ⊆ Ann bi L ( M ) .Proof. Since every L -bimodule is a left L -module, it follows from Lemma 3.1 thatLeib( L ) ⊆ Ker( λ ). If M is anti-symmetric, then ρ = 0, and the assertion is clear.On the other hand, if M is symmetric, then λ x = − ρ x for every element x ∈ L .Hence Ker( ρ ) = Ker( λ ), and therefore Leib( L ) ⊆ Ker( ρ ) follows from the alreadyestablished inclusion Leib( L ) ⊆ Ker( λ ). (cid:3) Let L be a left Leibniz algebra over a field F , and let M be an L -bimodule. Wecall M := h x · m + m · x | x ∈ L , m ∈ M i F the anti-symmetric kernel of M (see [36, p. 145]). In particular, the adjoint bimodule of a symmetric Leibniz algebra is not always symmetricas the name might suggest.
Example (cf. also [29, p. 479]) . Let L be a left Leibniz algebra. Then the identity xy + yx = ( x + y ) − x − y shows that ( L ad ) ⊆ Leib( L ). Moreover, if the ground field of L has characteristic = 2, then the identity x = ( xx + xx ) shows that ( L ad ) = Leib( L ).The next result generalizes Proposition 2.5 to arbitrary Leibniz bimodules (see[36, p. 145] and also the proof of Theorem 3.1 in [29]). Proposition 3.5.
Let L be a left Leibniz algebra, and let M be an L -bimodule.Then M is an anti-symmetric L -subbimodule of M .Proof. Let x, y ∈ L and m ∈ M be arbitrary elements. Then we obtain from (3.1)and (3.2) that x · ( y · m + m · y ) = x · ( y · m ) + x · ( m · y )= [( xy ) · m + m · ( xy )] + [ y · ( x · m ) + ( x · m ) · y ] ∈ M . Since M is a subspace of M , we conclude that L M ⊆ M . Moreover, it is animmediate consequence of Lemma 3.3 that M L = 0. (cid:3) By definition of the anti-symmetric kernel, M sym := M/M is a symmetric L -bimodule. We call M sym the symmetrization of M . In fact, the anti-symmetrickernel is the smallest subbimodule such that the corresponding factor module issymmetric. (This should be compared with the analogous statement for the adjointLeibniz bimodule in Proposition 2.9.) Proposition 3.6.
Let L be a left Leibniz algebra, and let M be an L -bimodule.Then M is the smallest L -subbimodule N of M such that M/N is a symmetric L -bimodule.Proof. Let N be a L -subbimodule of M such that M/N is a symmetric L -bimodule.Then it follows from the symmetry of M/N that x · ( m + N ) = − ( m + N ) · x , andconsequently, x · m + m · x ∈ N for every element x ∈ L and every element m ∈ M .Since N is a subspace of M , we conclude that M ⊆ N . (cid:3) The next result was first established for finite-dimensional irreducible Leibniz bi-modules over finite-dimensional right or left Leibniz algebras (see [37, Theorem 3.1]and [7, Theorem 1.4]). Then in [29, Theorem 3.1] this was generalized to arbitraryirreducible Leibniz bimodules over any right Leibniz algebra. At the same time theproof is simplified considerably.
Theorem 3.7.
Let L be a left Leibniz algebra. Then every irreducible L -bimoduleis symmetric or anti-symmetric.Proof. Let M be an irreducible L -bimodule. According to Proposition 3.5, we havethat M = 0 or M = M . In the former case M is symmetric and in the latter case M is anti-symmetric. (cid:3) The final result of this section discusses how a left Leibniz module can be madeinto an anti-symmetric Leibniz bimodule or into a symmetric Leibniz bimodule.
Proposition 3.8.
Let L be a left Leibniz algebra. Then the following statementshold: (a) Every left L -module with a trivial right action is an anti-symmetric L -bimodule. EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 17 (b)
Every left L -module determines a unique symmetric L -bimodule.Proof. (a): Since (LLM) holds for every left L -module, and each term in (LML) aswell as in (MLL) contains at least one trivial right action, the assertion follows.(b): In order for M to be a symmetric L -bimodule, the right action of L on M has to be defined by m · x := − x · m for every element m ∈ M and every element x ∈ L . Since (LLM) is automatically satisfied, one only needs to verify (LML)and (MLL). Let x, y ∈ L and m ∈ M be arbitrary elements. Then it follows from(LLM) that ( x · m ) · y = − y · ( x · m )= ( xy ) · m − x · ( y · m )= x · ( m · y ) − m · ( xy )and ( m · x ) · y = y · ( x · m )= x · ( y · m ) − ( xy ) · m = m · ( xy ) − x · ( m · y ) . This completes the proof. (cid:3)
Note that for the trivial left Leibniz module the Leibniz bimodule structuresobtained from parts (a) and (b) of Proposition 3.8 both give rise to the trivialLeibniz bimodule.
Examples.
Let L be a left Leibniz algebra and consider the left adjoint L -module L ad ,ℓ .(1) According to Proposition 3.8 (a), L ad ,ℓ with a trivial right action is ananti-symmetric L -bimodule, which we call the anti-symmetric adjoint bi-module L a of L . The associated representation is ( L, anti-symmetric adjoint representation of L , where L denotes the left multipli-cation operator of L .(2) By virtue of Proposition 3.8 (b), L ad ,ℓ has a unique symmetric L -bimodulestructure, which we call the symmetric adjoint bimodule L s of L . Theassociated representation is ( L, − L ), the so-called symmetric adjoint rep-resentation of L . 4. Leibniz cohomology
Let L be a left Leibniz algebra over a field F , and let M be an L -bimodule. Forany non-negative integer n set C n ( L , M ) := Hom F ( L ⊗ n , M ) and consider the lineartransformation d n : C n ( L , M ) → C n +1 ( L , M ) defined by( d n f )( x ⊗ · · · ⊗ x n +1 ) := n X i =1 ( − i +1 x i · f ( x ⊗ · · · ⊗ ˆ x i ⊗ · · · ⊗ x n +1 )+ ( − n +1 f ( x ⊗ · · · ⊗ x n ) · x n +1 + X ≤ i Let L be a left Leibniz algebra, and let M be an L -bimodule.Then HL ( L , M ) = M L .Proof. We have that ( d m )( x ) = − m · x for any m ∈ M and any x ∈ L . HenceHL ( L , M ) = Ker( d ) = M L . (cid:3) In particular, we obtain from Lemma 3.4 and the definition of M L the followingresult. Corollary 4.2. Let L be a left Leibniz algebra, and let M be an L -bimodule. Thenthe following statements hold: (a) If M is symmetric, then HL ( L , M ) ∼ = M L Lie . (b) If M is anti-symmetric, then HL ( L , M ) = M . The subspaceDer( L , M ) := { f ∈ Hom F ( L , M ) | ∀ x, y ∈ L : f ( xy ) = f ( x ) · y + x · f ( y ) } of Hom F ( L , M ) is called the space of derivations from L to M . The subspaceIder( L , M ) := { g ∈ Hom F ( L , M ) | ∃ m ∈ M : g ( x ) = m · x } of Der( L , M ) is called the space of inner derivations from L to M . Proposition 4.3. Let L be a left Leibniz algebra, and let M be an L -bimodule.Then HL ( L , M ) = Der( L , M ) / Ider( L , M ) .Proof. It follows from ( d m )( x ) = − m · x that Ider( L , M ) = Im( d ). More-over, we have that ( d f )( x ⊗ y ) = x · f ( y ) + f ( x ) · y − f ( xy ) for any m ∈ M , f ∈ Hom F ( L , M ), and x, y ∈ L . Hence Ker( d ) = Der( L , M ), and thereforeHL ( L , M ) = Ker( d ) / Im( d ) = Der( L , M ) / Ider( L , M ). (cid:3) Corollary 4.4. Let L be a left Leibniz algebra over a field F , and let M be an L -bimodule. Then the following statements hold: (a) If M is symmetric, then HL ( L , M ) ∼ = HL ( L Lie , M ) = H ( L Lie , M ) . (b) If M is anti-symmetric, then HL ( L , M ) = Hom L ( L , M ) , where the latterdenotes the vector space of homomorphisms of left L -modules.Proof. (a): Let f ∈ Der( L , M ) and let x ∈ L be arbitrary. Since M is symmetric,we have that f ( x ) = f ( x ) · x + x · f ( x ) = − x · f ( x ) + x · f ( x ) = 0. Since f is linear,it follows that f (Leib( L )) = 0. Hence f induces a derivation from L Lie to M . Thisyields the isomorphism HL ( L , M ) ∼ = HL ( L Lie , M ). EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 19 Let d n Lie denote the coboundary map for the Chevalley-Eilenberg cohomology ofLie algebras (see [14, Exercise I.3.12]). Then we have that( d f )( x ∧ y ) = x · f ( y ) − y · f ( x ) − f ( xy )for any m ∈ M , any f ∈ Hom F ( L Lie , M ), and any x, y ∈ L Lie . Since M is symmet-ric, we obtain that Ker( d ) = Der( L Lie , M ). Similarly, ( d m )( x ) = x · m for any m ∈ M and any x ∈ L Lie . Hence Im( d ) = Ider( L Lie , M ), and thus we have thatHL ( L Lie , M ) = H ( L Lie , M ).(b): Since M is anti-symmetric, we have thatDer( L , M ) = { f ∈ Hom F ( L , M ) | ∀ x, y ∈ L : f ( xy ) = x · f ( y ) } as well as Ider( L , M ) = { } , and therefore it follows from Proposition 4.3 thatHL ( L , M ) = Hom L ( L , M ). (cid:3) In particular, we obtain from Corollary 4.4 the following result. Corollary 4.5. If L is a left Leibniz algebra over a field F , then HL ( L , F ) ∼ = L / L ∼ = L Lie / L . The next result is reminiscent of Seligman’s non-vanishing theorem for modularLie algebras (see [45, p. 102]). Corollary 4.6. If L = 0 is a left Leibniz algebra, then HL ( L , L a ) = 0 .Proof. It follows from Corollary 4.4 (b) that 0 = id L ∈ Hom L ( L , L a ) ∼ = HL ( L , L a ). (cid:3) Finally, we briefly discuss abelian extensions of left Leibniz algebras. Let L and K be left Leibniz algebras. Then any short exact sequence 0 → K → E → L → extension of L by K , and K is called the kernel of the extension. Twoextensions 0 → K ι → E π → L → → K ι ′ → E ′ π ′ → L → L by K arecalled equivalent if there is a homomorphism φ : E → E ′ such that φ ◦ ι = ι ′ and π ′ ◦ φ = π . Every extension 0 → K → E → L → K givesrise to an L -bimodule structure on K (see [19, Proposition 1.3.6]). It can be shownthat the set of equivalence classes of all extensions of L by a given L -bimodule M (considered as an abelian Leibniz algebra) are in natural bijection to HL ( L , M )(see [19, Theorem 1.3.13] for left Leibniz algebras and [36, Proposition 1.9] for rightLeibniz algebras). We will only prove the following ingredient of establishing sucha bijection. Lemma 4.7. Let L be a left Leibniz algebra, let M be an L -bimodule, and let f ∈ C ( L , M ) be a Leibniz -cocycle. Then E := L × M with Leibniz productdefined by ( x , m )( x , m ) := ( x x , m · x + x · m + f ( x ⊗ x )) for any x , x ∈ L and any m , m ∈ M is a left Leibniz algebra.Proof. Since the Cartesian product L × M with componentwise addition and com-ponentwise scalar multiplication is a vector space, we only need to verify the leftLeibniz identity. We have that( d f )( x ⊗ x ⊗ x ) = x · f ( x ⊗ x ) − x · f ( x ⊗ x ) − f ( x ⊗ x ) · x − f ( x x ⊗ x ) + f ( x ⊗ x x ) − f ( x ⊗ x x ) for any x , x , x ∈ L . Since f is a 2-cocycle, we have that x · f ( x ⊗ x )+ f ( x ⊗ x x ) = x · f ( x ⊗ x )+ f ( x ⊗ x ) · x + f ( x x ⊗ x )+ f ( x ⊗ x x ) . We compute( x , m )[( x , m )( x , m )] = ( x , m )( x x , m · x + x · m + f ( x ⊗ x ))= ( x ( x x ) , m · ( x x ) + x · ( m · x ) + x · ( x · m )+ x · f ( x ⊗ x ) + f ( x ⊗ x x )) , [( x , m )( x , m )]( x , m ) = ( x x , m · x + x · m + f ( x ⊗ x ))( x , m )= (( x x ) x , ( m · x ) · x + ( x · m ) · x + f ( x ⊗ x ) · x + ( x x ) · m + f ( x x ⊗ x )) , ( x , m )[( x , m )( x , m )] = ( x , m )( x x , m · x + x · m + f ( x ⊗ x ))= ( x ( x x ) , m · ( x x ) + x · ( m · x ) + x · ( x · m )+ x · f ( x ⊗ x ) + f ( x ⊗ x x ))for any elements x , x , x ∈ L and m , m , m ∈ M .It follows from the left Leibniz identity, (3.1), (3.2), (3.3), and the 2-cocycleidentity for f that the right–hand side of the first identity equals the sum of theright-hand sides of the second and third identities. Hence the left-hand side of thefirst identity also equals the sum of the left-hand sides of the second and thirdidentities, and thus L × M satisfies the left Leibniz identity. (cid:3) Nilpotent Leibniz algebras Let L be a left or right Leibniz algebra. Then the left descending central series L ⊇ L ⊇ L ⊇ · · · of L is defined recursively by L := L and n +1 L := L n L for every positive integer n . Similarly, the right descending central series L ⊇ L ⊇ L ⊇ · · · of L is defined recursively by L := L and L n +1 := L n L for every positive integer n . Note that Proposition 5.2 is just [4, Lemma 1]. For the convenience of thereader we include its proof. Proposition 5.1. Let L be a left Leibniz algebra. Then ( n L ) n ∈ N is a descendingfiltration of L , i.e., n L ⊇ n +1 L and m L n L ⊆ m + n L for all positive integers m and n . In particular, n L is an ideal of L for every positive integer n . Proposition 5.2. Let L be a right Leibniz algebra. Then ( L n ) n ∈ N is a descendingfiltration of L , i.e., L n ⊇ L n +1 and L m L n ⊆ L m + n for all positive integers m and n . In particular, L n is an ideal of L for every positive integer n .Proof. We only prove Proposition 5.2 as this yields Proposition 5.1 by consideringthe opposite algebra. Firstly, we show that the right descending central series isindeed descending, i.e., L n +1 ⊆ L n for every positive integer n . We proceed by This definition is consistent with the definition of L given in Section 1. EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 21 induction on n . The base step n = 1 is clear. For the induction step let n > n − 1. Then L n +1 = L n L ⊆ L n − L = L n . Next, we prove L m L n ⊆ L m + n by induction on n . The base step n = 1 is animmediate consequence of the definition: L n L = L n L = L n +1 . For the induction step let n > n − 1. It follows from identity (2.4) and by applying the induction hypothesistwice that L m L n = L m ( L n − L ) ⊆ ( L m L n − ) L + ( L m L ) L n − ⊆ L m + n − L + L m +1 L n − ⊆ L m + n . This completes the proof. (cid:3) Following [4, p. 2] and [41, p. 3829] we define the descending central series L ⊇ L ⊇ L ⊇ · · · of a left or right Leibniz algebra L recursively by L := L and L n := n − P k =1 L k L n − k for every positive integer n . Note that Proposition 5.4 is just [4, Lemma 2]. Forthe convenience of the reader we include its proof. Proposition 5.3. Let L be a left Leibniz algebra. Then L n = n L for every positiveinteger n . In particular, ( L n ) n ∈ N is a descending filtration of L , i.e., L n ⊇ L n +1 and L m L n ⊆ L m + n for all positive integers m and n . Proposition 5.4. Let L be a right Leibniz algebra. Then L n = L n for every positiveinteger n . In particular, ( L n ) n ∈ N is a descending filtration of L , i.e., L n ⊇ L n +1 and L m L n ⊆ L m + n for all positive integers m and n .Proof. We only prove the first statement in Proposition 5.4 as this in conjunctionwith Proposition 5.2 implies the remaining statements. We proceed by inductionon n . The base step n = 1 is an immediate consequence of the definitions. For theinduction step let n > n . Then by applying the induction hypothesis and Proposition 5.2we obtain that L n = n − X k =1 L k L n − k = n − X k =1 L k L n − k ⊆ L n . On the other hand, it follows from the induction hypothesis that L n = L n − L = L n − L ⊆ n − X k =1 L k L n − k = L n . This completes the proof. (cid:3) By combining Proposition 5.3 and Proposition 5.4 we obtain the following result(cf. also [10, Proposition 2.13]). Corollary 5.5. If L is a symmetric Leibniz algebra, then L n = n L = L n for everypositive integer n . Recall from Section 1 that an algebra L is called nilpotent if there exists a positiveinteger n such that any product of n elements in L , no matter how associated, iszero (see [44, p. 18]). The following observation is clear. Lemma 5.6. A left or right Leibniz algebra L is nilpotent if, and only if, thereexists a positive integer n such that L n = 0 . Examples. (1) The two-dimensional solvable left Leibniz algebra A ℓ from Example 1 inSection 2 is not nilpotent as ( A ℓ ) n = n A ℓ = F f = 0 for every integer n ≥ A nr = 0 for every integer n ≥ N from Example 2 in Sec-tion 2 is nilpotent as N n = n N = N n = 0 for every integer n ≥ S ℓ from Example 3 inSection 2 is not nilpotent as ( S ℓ ) n = n S ℓ = S nℓ = S ℓ = 0 for everypositive integer n .In the proof of Proposition 5.8 we will need the following result. Lemma 5.7. If φ : L → K is a homomorphism of left or right Leibniz algebras,then φ ( L n ) = φ ( L ) n for every positive integer n .Proof. We proceed by induction on n . The base step n = 1 follows from φ ( L ) = φ ( L ) = φ ( L ) . For the induction step let n > n . Then we obtain from the induction hypothesisthat φ ( L n ) = φ ( n − X k =1 L k L n − k ) = n − X k =1 φ ( L k ) φ ( L n − k )= n − X k =1 φ ( L ) k φ ( L ) n − k = φ ( L ) n . This completes the proof. (cid:3) The next result is an immediate consequence of Lemma 5.7. Proposition 5.8. Subalgebras and homomorphic images of nilpotent left or rightLeibniz algebras are nilpotent. Proposition 5.9. If I is an ideal of a left Leibniz algebra L such that I ⊆ C r ( L ) and L / I is nilpotent, then L is nilpotent. Proposition 5.10. If I is an ideal of a right Leibniz algebra L such that I ⊆ C ℓ ( L ) and L / I is nilpotent, then L is nilpotent.Proof. We only prove Proposition 5.10 as this yields Proposition 5.9 by consideringthe opposite algebra.Since L / I is nilpotent, there exists a positive integer r such that L r = L r ⊆ I .But by hypothesis, we have that I ⊆ C ℓ ( L ), i.e., IL =0. Hence L r +1 = L r +1 = L r L ⊆ IL = 0, and therefore L is nilpotent. (cid:3) Proposition 5.11. The sum of two nilpotent ideals of a left or right Leibniz algebrais nilpotent. EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 23 Proof. Let L be a right Leibniz algebra. We begin by proving that I n is a rightideal of L for any right ideal I of L and for any positive integer n . We will proceedby induction on n . The base step I L = IL ⊆ I = I follows from the fact that I is a right ideal of L . For the induction step let n > n − 1. Then we obtain from the right Leibniz identityand the induction hypothesis that I n L = ( I n − I ) L ⊆ ( I n − L ) I + I n − ( IL ) ⊆ I n − I = I n . Now let J and K be two nilpotent ideals of L . We will prove that( J + K ) n ⊆ n − X r =1 ( J r ∩ K n − r )for every positive integer n . We will again proceed by induction on n . The basestep ( J + K ) = J + K = J + K is an immediate consequence of the definition ofthe right descending central series. For the induction step let n ≥ n . Then we obtain from the inductionhypothesis and the fact that J r and K n − r are right ideals of L :( J + K ) n +1 = ( J + K ) n ( J + K ) ⊆ n − X r =1 ( J r ∩ K n − r )( J + K ) ⊆ n − X r =1 ( J r ∩ K n − r ) J + n − X r =1 ( J r ∩ K n − r ) K ⊆ n − X r =1 ( J r +1 ∩ K n − r ) + n − X r =1 ( J r ∩ K n +1 − r )= n X s =2 ( J s ∩ K n +1 − s ) + n − X r =1 ( J r ∩ K n +1 − r )= J n ∩ K + n − X r =2 ( J r ∩ K n +1 − r ) + J ∩ K n = n X r =1 ( J r ∩ K n +1 − r ) . Finally, as J and K are nilpotent, there exist positive integers s and t such that J s = 0 and K t = 0. Hence we obtain that ( J + K ) s + t − = 0 which shows that J + K is a nilpotent ideal of L . (cid:3) Using Proposition 5.11 one can proceed similar to the solvable case to establishthe existence of a largest nilpotent ideal (see Section 1). We leave the details tothe interested reader (see also [31, Proposition 1]).Now we prove the Lie analogue of a consequence of [44, Theorem 2.4] for sym-metric Leibniz algebras. Theorem 5.12. For every symmetric Leibniz algebra L the following statementsare equivalent: (i) L is nilpotent. (ii) L Lie is nilpotent. (iii) L /C ( L ) is nilpotent. (iv) Lie( L ) is nilpotent.Proof. The implication (i) ⇒ (ii) is an immediate consequence of Proposition 5.8. Itfollows from Corollary 2.7 that there is a natural epimorphism L Lie → L /C ( L ) ofLie algebras. Hence another application of Proposition 5.8 yields the implication(ii) ⇒ (iii). Moreover, the implication (iii) ⇒ (i) follows from Proposition 5.9.Suppose now that L /C ( L ) is nilpotent. Then we obtain from Proposition 2.10that there is a natural epimorphism L /C ( L ) → L /C ℓ ( L ) ∼ = L ( L ), and therefore L ( L ) is nilpotent. Similarly, we obtain from Proposition 2.12 that there is a natu-ral epimorphism L /C ( L ) → L /C r ( L ) ∼ = R ( L ) op , and thus R ( L ) is nilpotent. Henceit follows from Proposition 5.11 and Theorem 2.14 that Lie( L ) is nilpotent. Thisestablishes the implication (iii) ⇒ (iv). Finally, the implication (iv) ⇒ (i) can beobtained from Proposition 5.8 in conjunction with Proposition 2.10 and Proposi-tion 5.10. (cid:3) Example. Let A ℓ denote the two-dimensional non-nilpotent solvable left Leib-niz algebra A ℓ from Example 1 in Section 2. As has been observed above, wehave that C ℓ ( A ℓ ) = F f = Leib( A ℓ ) and C r ( A ℓ ) = F e . Hence L Lie = L / Leib( L ), L ( L ) ∼ = L /C ℓ ( L ), and R ( L ) ∼ = [ L /C r ( L )] op are one-dimensional. So each of theseLie algebras is abelian, and thus nilpotent, but A ℓ is not nilpotent. This showsthat the implication (ii) ⇒ (i) in Theorem 5.12 neither holds for left nor for rightLeibniz algebras. Moreover, neither of the implications (iii) ⇒ (i) and (iv) ⇒ (i) inTheorem 5.12 holds for left (resp. right) Leibniz algebras if one replaces C ( L ) by C ℓ ( L ) (resp. C r ( L )) and Lie( L ) by L ( L ) (resp. R ( L )).The Leibniz analogue of Engel’s theorem for Lie algebras of linear transforma-tions was first proved by Patsourakos [41, Theorem 7] and later by Barnes [7, The-orem 1.2] who used the corresponding result for Lie algebras in his proof. Note thatPatsourakos does not have to assume that the representation is finite-dimensional.For the convenience of the reader we include a variation of Barnes’ proof. Theorem 5.13. Let L be a finite-dimensional left Leibniz algebra, and let ( λ, ρ ) bea representation of L on a non-zero finite-dimensional vector space M such that λ x is nilpotent for every element x ∈ L . Then ρ x is nilpotent for every element x ∈ L ,and there exists a non-zero vector m ∈ M such that λ x ( m ) = 0 = ρ x ( m ) for everyelement x ∈ L .Proof. We first prove that for any element x ∈ L the nilpotency of λ x implies thenilpotency of ρ x (see also [41, Lemma 6]). Namely, we have ρ nx = ( − n − ρ x ◦ λ n − x for every element x ∈ L and every positive integer n . This can be shown byinduction on n . The base step n = 1 is trivial. For the induction step let n ≥ n . Then it follows from the inductionhypothesis and Lemma 3.3 that ρ n +1 x = ρ x ◦ ρ nx = ( − n − ρ x ◦ ρ x ◦ λ n − x = − ( − n − ρ x ◦ λ x ◦ λ n − x = ( − n ρ x ◦ λ nx . Since the L -bimodule M is finite-dimensional, it has an irreducible L -subbi-module N . We obtain from Theorem 3.7 that N is symmetric or anti-symmetric.In the former case we have that ρ x = − λ x , and in the latter case we have that ρ x = 0 for every x ∈ L . It follows from the linearity of λ and (3.4) that λ ( L ) is aLie subalgebra of gl ( M ). So the existence of a non-zero vector m ∈ N such that λ x ( m ) = 0 for every x ∈ L can be obtained from Engel’s theorem for Lie algebras EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 25 of linear transformations (see [32, Theorem 3.3]). Finally, this and the symmetryor anti-symmetry of N yield ρ x ( m ) = 0 for every x ∈ L . (cid:3) We conclude this section by several applications of Theorem 5.13. The first resultis just a reformulation of Theorem 5.13 in terms of a composition series of a Leibnizbimodule (see also [41, Corollary 9]). Corollary 5.14. Let L be a finite-dimensional left Leibniz algebra over a field F ,and let ( λ, ρ ) be a representation of L on a non-zero finite-dimensional vector space M such that λ x is nilpotent for every element x ∈ L . Then the following statementshold: (a) If M is irreducible, then M is the one-dimensional trivial L -bimodule. (b) If M is finite-dimensional, then every composition series M $ M $ · · · $ M n = M of M satisfies dim F M j = j , λ x ( M j ) ⊆ M j − , and ρ x ( M j ) ⊆ M j − forevery integer j ∈ { , . . . , n } and every element x ∈ L .Proof. (a) is an immediate consequence of the proof of Theorem 5.13 as we did notuse the finite dimension of M in the irreducible case.(b): We apply part (a) to each composition factor M j /M j − (1 ≤ j ≤ n ). (cid:3) Next, we specialize Corollary 5.14 and Theorem 5.13 to the adjoint Leibnizbimodule. Corollary 5.15. Let L be a finite-dimensional nilpotent left Leibniz algebra overa field F , and let I be a d -dimensional ideal of L . Then the following statementshold: (a) There exists an ascending chain L $ L $ · · · $ L n = L of ideals of L such that I = L d , dim F L j = j , LL j ⊆ L j − , and L j L ⊆ L j − for every integer j ∈ { , . . . , n } . (b) If I = 0 , then LI $ I and IL $ I (c) If I = 0 , then I ∩ C ( L ) = 0 .Proof. Since L is nilpotent, we have that L x is nilpotent for every element x ∈ L .(a): Choose a composition series of the adjoint L -bimodule that contains I (see[46, Proposition 1.1.1]) and apply Corollary 5.14 to the adjoint representation ( L, R )of L .(b) is an immediate consequence of part (a).(c): Note that I is an L -subbimodule of the adjoint L -bimodule. It follows fromTheorem 5.13 that there exists a non-zero element y ∈ I such that L x ( y ) = 0 = R x ( y ), i.e., xy = 0 = yx for every element x ∈ L . Hence 0 = y ∈ C r ( L ) ∩ C ℓ ( L ) = C ( L ). (cid:3) Finally, we prove Engel’s theorem for left Leibniz algebras (see [4, Theorem 2]and [41, Corollary 10]). Corollary 5.16. If L is a finite-dimensional left Leibniz algebra such that L x isnilpotent for every element x ∈ L , then L is nilpotent. Proof. In the notation of Corollary 5.15, we have that k L ⊆ L n − k +1 for everypositive integer k , and therefore n +1 L = 0. We proceed by induction on k to provethe former statement. The base step k = 1 is just L = L = L n . For the inductionstep let k > k − L ⊆ L n − k +2 is true. Then weobtain from the induction hypothesis in conjunction with Corollary 5.15 (a) that k L = L k − L ⊆ LL n − k +2 ⊆ L n − k +1 . (cid:3) Solvable Leibniz algebras In the case of solvable Leibniz algebras we have one-sided analogues of Theo-rem 5.12. These show that the solvability of a left or right Leibniz algebra L isequivalent to the solvability of several Lie algebras associated to L . Proposition 6.1. For every left Leibniz algebra L the following statements areequivalent: (i) L is solvable. (ii) L Lie is solvable. (iii) L /C ℓ ( L ) is solvable. (iv) L ( L ) is solvable. Proposition 6.2. For every right Leibniz algebra L the following statements areequivalent: (i) L is solvable. (ii) L Lie is solvable. (iii) L /C r ( L ) is solvable. (iv) R ( L ) is solvable.Proof. We only prove Proposition 6.2 as this yields Proposition 6.1 by consideringthe opposite algebra.The implication (i) ⇒ (ii) is an immediate consequence of Proposition 1.1. It fol-lows from Proposition 2.6 that there is a natural epimorphism L Lie = L / Leib( L ) → L /C r ( L ) of Lie algebras. Hence another application of Proposition 1.1 yields theimplication (ii) ⇒ (iii). Moreover, the implication (iii) ⇒ (i) follows from Proposi-tion 1.2. Finally, the remaining equivalence of (iii) and (iv) is an immediate conse-quence of Proposition 2.12. (cid:3) Theorem 6.3. For every symmetric Leibniz algebra L the following statements areequivalent: (i) L is solvable. (ii) L Lie is solvable. (iii) L /C ( L ) is solvable. (iv) L ( L ) is solvable. (v) R ( L ) is solvable. (vi) Lie( L ) is solvableProof. The equivalence of (i) and (ii) follows from Proposition 6.1 or Proposition 6.2and the implication (i) ⇒ (iii) is an immediate consequence of Proposition 1.1.Suppose now that L /C ( L ) is solvable. Then we obtain from Proposition 2.10that there is a natural epimorphism L /C ( L ) → L /C ℓ ( L ) ∼ = L ( L ), and therefore L ( L ) is solvable. Similarly, we obtain from Proposition 2.12 that there is a natural EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 27 epimorphism L /C ( L ) → L /C r ( L ) ∼ = R ( L ) op , and thus R ( L ) is solvable. This estab-lishes the implications (iii) ⇒ (iv) and (iii) ⇒ (v). Each of the implications (iv) ⇒ (i)and (v) ⇒ (i) follows from Proposition 6.1 and Proposition 6.2, respectively.The implication (i) ⇒ (vi) is a consequence of Theorem 2.14 and Proposition 1.3in conjunction with the already established implications. Finally, the implication(vi) ⇒ (iv) can be obtained from Theorem 2.14 and Proposition 1.1. (cid:3) The Leibniz analogue of Lie’s theorem for solvable Lie algebras of linear trans-formations was proved by Patsourakos [42, Theorem 2]. For the convenience of thereader we include a proof that follows very closely the proof of Theorem 5.13 anduses the corresponding result for Lie algebras. Theorem 6.4. Let L be a finite-dimensional solvable left Leibniz algebra over analgebraically closed field of characteristic zero, and let ( λ, ρ ) be a representation of L on a non-zero finite-dimensional vector space M . Then M contains a commoneigenvector for the linear transformations in λ ( L ) ∪ ρ ( L ) .Proof. Since the L -bimodule M is finite-dimensional, it has an irreducible L -sub-module N . We obtain from Theorem 3.7 that N is symmetric or anti-symmetric.In the former case we have that ρ x = − λ x , and in the latter case we have that ρ x = 0 for every x ∈ L . Similarly to the proof of Proposition 2.10, one can showthat λ ( L ) is a Lie subalgebra of gl ( M ) such that λ ( L ) ∼ = L / Ann L ( M ). The latterisomorphism in conjunction with Proposition 1.1 yields that λ ( L ) is solvable. Sothe existence of a common eigenvector m ∈ N for λ ( L ) can be obtained from Lie’stheorem for solvable Lie algebras of linear transformations (see [32, Theorem 4.1]).Finally, this and the symmetry or anti-symmetry of N imply that m is also acommon eigenvector for ρ ( L ). (cid:3) We conclude this section by several applications of Theorem 6.4. The first resultis just a reformulation of Theorem 6.4 in terms of a composition series of a Leibnizbimodule (see also [42, Corollary 2] and [22, Theorem 3.2]). Corollary 6.5. Let L be a finite-dimensional solvable left Leibniz algebra over analgebraically closed field F of characteristic zero, and let M be a non-zero finite-dimensional L -bimodule. Then the following statements hold: (a) If M is irreducible, then dim F M = 1 . (b) Every composition series M $ M $ · · · $ M n = M of M satisfies dim F M j = j for any ≤ j ≤ n .Proof. (a) is an immediate consequence of the proof of Theorem 6.4.(b): We apply part (a) to each composition factor M j /M j − (1 ≤ j ≤ n ). (cid:3) Remark. Corollary 6.5 (and thus also Theorem 6.4) is not true for ground fieldsof prime characteristic as already can be seen for the non-abelian two-dimensionalLie algebra (see [46, Example 5.9.1]). Moreover, the one-dimensional Lie algebraspanned by any proper rotation of a two-dimensional real vector space shows thatCorollary 6.5 and Theorem 6.4 in general hold only over algebraically closed fields.Next, we specialize Corollary 6.5 to the adjoint Leibniz bimodule (see also [22,Corollary 3.3]). Corollary 6.6. Let L be a finite-dimensional solvable left Leibniz algebra over analgebraically closed field F of characteristic zero, and let I be a d -dimensional idealof L . Then there exists an ascending chain L $ L $ · · · $ L n = L of ideals of L such that I = L d and dim F L j = j for every ≤ j ≤ n .Proof. Choose a composition series of the adjoint L -bimodule that contains I (see[46, Proposition 1.1.1]) and apply Corollary 6.5 to the adjoint L -bimodule. (cid:3) We can also employ Engel’s theorem for left Leibniz algebras to prove the fol-lowing result (see [4, Theorem 4], [42, Corollary 3], and [31, Corollary 6]). Corollary 6.7. Let L be a finite-dimensional solvable left Leibniz algebra over analgebraically closed field F of characteristic zero. Then L x is nilpotent for everyelement x ∈ L . In particular, L is nilpotent.Proof. According to Corollary 6.6, there exists an ascending chain0 = L $ L $ · · · $ L n = L of ideals of L such that dim F L j = j for every 1 ≤ j ≤ n . Hence one can choosesuccessively a basis { x , . . . , x j } of L j (1 ≤ j ≤ n ) such that the correspondingmatrices of L ( L ) are upper triangular. By virtue of the proof of Proposition 2.10,we have that L ( L ) = [ L ( L ) , L ( L )], and thus the matrices of L ( L ) are strictlyupper triangular. Hence L x is nilpotent for every element x ∈ L . In particular,we obtain from Corollary 5.16 that L is nilpotent. (cid:3) Finally, we give a proof of Cartan’s solvability criterion for left Leibniz algebras(see also [2, Theorem 3.7] and [22, Theorem 3.5]). Theorem 6.8. Let L be a finite-dimensional left Leibniz algebra over a field ofcharacteristic zero. Then L is solvable if, and only if, κ ( x, y ) = 0 for every element x ∈ L and every element y ∈ L .Proof. Suppose first that L is solvable and the ground field of L is algebraicallyclosed. It follows from Corollary 6.6 that L ( L ) can be simultaneously representatedby upper triangular matrices. Then the proof of Proposition 2.10 shows that thecorresponding matrix of L yz = [ L y , L z ] = L y ◦ L z − L z ◦ L y is a strictly uppertriangular matrix for any y, z ∈ L . Hence κ ( x, yz ) = tr( L x ◦ L yz ) = 0 for any x, y, z ∈ L .Suppose now that κ ( x, y ) = 0 for every element x ∈ L and every element y ∈ L ,and the ground field of L is again algebraically closed. It follows from Proposi-tion 2.10 that L ( L ) is a Lie subalgebra of gl ( L ). In particular, we obtain as beforethat tr( L x ◦ [ L y , L z ]) = tr( L x ◦ L yz ) = κ ( x, yz ) = 0 for any x, y, z ∈ L . So theother implication is a consequence of [32, Theorem 4.3] in conjunction with Propo-sition 6.1.Finally, in case the ground field F of L is not algebraically closed, a base fieldextension will show the assertion. Namely, let F be an algebraic closure of F , set L := L ⊗ F F , and let L a ⊗ α ( b ⊗ β ) := ab ⊗ αβ for any a, b ∈ L and any α, β ∈ F denote the left multiplication operator of L . Since L = L ⊗ F F , we obtain byinduction that L ( n ) = L ( n ) ⊗ F F for every non-negative integer n . Consequently, L is solvable if, and only if, L is solvable. Moreover, we obtain for the Killing form κ of L that κ ( x ⊗ , y ⊗ 1) = tr( L x ⊗ ◦ L y ⊗ ) = tr( L x ◦ L y ) = κ ( x, y ) for all x, y ∈ L EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 29 as L a ⊗ α ( b ⊗ β ) = L a ( b ) ⊗ αβ for any a, b ∈ L and any α, β ∈ F . This can be usedto show that κ ( x, y ) = 0 for every element x ∈ L and every element y ∈ L if, andonly if, κ ( x, y ) = 0 for every element x ∈ L and every element y ∈ L . (cid:3) Semisimple Leibniz algebras A left or right Leibniz algebra L is called simple if 0, Leib( L ), L are the only idealsof L , and Leib( L ) $ L (see [27, Definition 1] for the first part of this definitionor [39, Definition 2.2], [22, Definition 5.1], [29, Definition 2.6]). Note that there isalso another definition of simplicity for Leibniz algebras in the literature, namely,requiring that 0 and L are the only ideals of L . Then it follows from Proposition 2.8that every simple left or right Leibniz algebra has to be a Lie algebra.Exactly as for Lie algebras, we call a left or right Leibniz algebra L perfect incase L = L holds. Then the first result in this section is an immediate consequenceof these definitions. Proposition 7.1. Every simple left or right Leibniz algebra is perfect. The next result follows from the correspondence theorem for ideals. Note alsothat the condition Leib( L ) $ L implies that the canonical Lie algebra L Lie associ-ated to a simple left or right Leibniz algebra L is not abelian. Proposition 7.2. If L is a simple left or right Leibniz algebra, then L Lie is a simpleLie algebra. We will see in the example after Theorem 7.11 that the converse of Proposition 7.2does not always hold.In analogy with the above definition of simplicity, we call a left or right Leibnizalgebra L semisimple when Leib( L ) contains every solvable ideal of L . Proposition 7.3. Every simple left or right Leibniz algebra is semisimple.Proof. Let I be any solvable ideal of the simple left or right Leibniz algebra L .Then either I = 0, I = Leib( L ), or I = L . In the first two cases we have that I ⊆ Leib( L ), and we are done. So suppose that I = L . From Proposition 7.1 weobtain by induction that L = L ( n ) holds for every non-negative integer n . Since byhypothesis I is solvable, there exists a non-negative integer r such that I ( r ) = 0.Hence L = L ( r ) = I ( r ) = 0 which contradicts the requirement Leib( L ) $ L . (cid:3) In the finite-dimensional case we have the following result which often is takenas the definition of semisimplicity for Leibniz algebras (see [22, Definition 5.2] or[29, Definition 2.12]). Proposition 7.4. A finite-dimensional left or right Leibniz algebra L is semisimpleif, and only if, Leib( L ) = Rad( L ) .Proof. It follows from Proposition 2.8 and Proposition 1.4 that Leib( L ) ⊆ Rad( L ).This in conjunction with the semisimplicity of L proves the “only if”-part of theassertion, and the converse follows from Proposition 1.4 which says that Rad( L ) isthe largest solvable ideal of L . (cid:3) Proposition 7.5. If L is a semisimple left Leibniz algebra, then Leib( L ) = C ℓ ( L ) . Proposition 7.6. If L is a semisimple right Leibniz algebra, then Leib( L ) = C r ( L ) . Corollary 7.7. If L is a semisimple symmetric Leibniz algebra, then Leib( L ) = C ( L ) .Proof. We only prove Proposition 7.6 as this yields Proposition 7.5 by consideringthe opposite algebra.It follows from Proposition 2.6 that Leib( L ) ⊆ C r ( L ). According to Proposi-tion 2.4, C r ( L ) is an abelian ideal of L , and thus the semisimplicity of L yields that C r ( L ) ⊆ Leib( L ). (cid:3) Proposition 7.8. A left or right Leibniz algebra L is semisimple if, and only if, L Lie is semisimple.Proof. Suppose that L is semisimple, and let i be any solvable ideal of the Lie alge-bra L Lie = L / Leib( L ). Let π : L → L Lie denote the natural epimorphism of Leibnizalgebras. Then I := π − ( i ) is an ideal of L , and it follows from Proposition 1.2applied to i = I + Leib( L ) / Leib( L ) ∼ = I / I ∩ Leib( L ) that I is solvable. Hencethe semisimplicity of L yields that I ⊆ Leib( L ), i.e., i = 0. Consequently, L Lie issemisimple.In order to prove the converse, suppose that L Lie = L / Leib( L ) is semisimple,and let I be any solvable ideal of the left or right Leibniz algebra L . Then byProposition 1.1 we obtain that I + Leib( L ) / Leib( L ) is a solvable ideal of L Lie .Since the latter Lie algebra is semisimple, we have that I + Leib( L ) / Leib( L ) = 0,and thus I ⊆ Leib( L ). Hence L is semisimple. (cid:3) Since finite-dimensional semisimple Lie algebras over a field of characteristic zeroare perfect, it follows from Proposition 7.8 in conjunction with Corollary 4.5 thatthe same is true for Leibniz algebras (see [22, Corollary 5.5]). Corollary 7.9. Every finite-dimensional semisimple left Leibniz algebra over afield of characteristic zero is perfect. Similar to the Killing form of a finite-dimensional semisimple Lie algebra overa field of characteristic zero being non-degenerate, the Killing form of a finite-dimensional semisimple Leibniz algebra over a field of characteristic zero is mini-mally degenerate (see [22, Theorem 5.8]). As for Lie algebras, this can be obtainedfrom Cartan’s solvability criterion for Leibniz algebras (see Theorem 6.8). Proposition 7.10. The Killing form of every finite-dimensional semisimple leftLeibniz algebra over a field of characteristic zero is minimally degenerate.Proof. Let L be a finite-dimensional semisimple left Leibniz algebra over a field ofcharacteristic 0, and let I := L ⊥ denote the radical of the Killing form κ of L .Recall that I is an ideal of L . The proof of [32, Lemma 5.1] shows that κ I = κ | I × I ,where κ I denotes the Killing form of I . Then we have that κ I ( x, y ) = κ ( x, y ) = 0for every x ∈ I and every y ∈ I . Hence Theorem 6.8 shows that I is solvable.Since L is semisimple, we obtain that L ⊥ = I ⊆ Leib( L ), and thus it follows fromLemma 3.2 that L ⊥ = Leib( L ), i.e., κ is minimally degenerate. (cid:3) Remark. The example after Lemma 3.2 shows that contrary to Lie algebras,where the non-degeneracy of the Killing form implies semisimplicity, the converseof Proposition 7.10 is not true.Next, we give some insight into the structure of a finite-dimensional semisimpleLeibniz algebra in characteristic zero that can be obtained from the analogue of EIBNIZ ALGEBRAS AS NON-ASSOCIATIVE ALGEBRAS 31 Levi’s theorem for Leibniz algebras and is due to Pirashvili [43, Proposition 2.4]and Barnes [8, Theorem 1]. The first part of Theorem 7.11 was already observed byFialowski and Mih´alka [29, Corollary 2.14] and the third part is [30, Theorem 3.1]. Theorem 7.11. If L is a finite-dimensional semisimple left Leibniz algebra overa field of characteristic zero, then there exists a semisimple Lie subalgebra s of L such that L = s ⊕ Leib( L ) and Leib( L ) is an anti-symmetric completely reducible s -bimodule. Moreover, if L is simple, then s is simple and Leib( L ) is irreducible.Proof. The existence of the semisimple Lie subalgebra s with L = s ⊕ Leib( L ) is animmediate consequence of Levi’s theorem for Leibniz algebras and Proposition 7.4.It follows from Proposition 2.5 that Leib( L ) is an ideal of L such that Leib( L ) L = 0.Hence Leib( L ) is an anti-symmetric L -bimodule. In particular, Leib( L ) is also ananti-symmetric s -bimodule. Finally, Weyl’s theorem (see [32, Theorem 6.3]) yieldsthat Leib( L ) is a completely reducible s -bimodule.Suppose now that L is simple. Then we obtain from Proposition 7.2 that s ∼ = L / Leib( L ) = L Lie is also simple. Let M be a non-zero proper L -subbimodule ofLeib( L ). Then M is an ideal of L that is different from 0, Leib( L ), and L , whichcontradicts the simplicity of L . Hence Leib( L ) is an irreducible L -bimodule. Byvirtue of Proposition 2.8, Leib( L ) is also an irreducible s -bimodule. (cid:3) A left or right Leibniz algebra L is called Lie-simple if L Lie is simple. Accordingto Proposition 7.3 and Proposition 7.8, Lie-simple left or right Leibniz algebras aresemisimple, but they are not always simple. Example (see also [22, Example 5.3]) . Let s be any simple Lie algebra, and let M and M ′ be two non-trivial irreducible left s -modules. Then consider the leftLeibniz algebra L := s × [ M ⊕ M ′ ] with multiplication ( x, a )( y, b ) := ( xy, x · b ) forany x, y ∈ s and any a, b ∈ M ⊕ M ′ (see also Example 3 in Section 2). The identity( x, a )( x, a ) = (0 , x · a ) for any x ∈ s and any a ∈ M ⊕ M ′ shows that Leib( L ) ⊆ M ⊕ M ′ . Since M and M ′ are non-trivial irreducible left s -modules, we have that s M = M and s M ′ = M ′ . This in conjunction with the above identity shows theother inclusion, and therefore Leib( L ) = M ⊕ M ′ . Hence L Lie = L / Leib( L ) ∼ = s issimple, and thus L is Lie-simple. But as M and M ′ are ideals of L that are differentfrom 0, Leib( L ), and L , we conclude that L is not simple. Remark. The same argument as in the previous example in conjunction withProposition 7.8 proves that L = s × M with multiplication ( x, a )( y, b ) := ( xy, x · b )for any x, y ∈ s and any a, b ∈ M , where s is a semisimple Lie algebra and M isa direct sum of non-trivial irreducible left s -modules, is a semisimple left Leibnizalgebra. Note also that there exist semisimple left Leibniz algebras that cannotbe decomposed as a direct product of (Lie-)simple left Leibniz algebras (see [30,Examples 2 and 3]).We obtain as an immediate consequence of Proposition 7.8 in conjunction withCorollary 4.4 (a) and the first Whitehead lemma for Lie algebras (see [33, The-orem 13 in Chapter III]) the corresponding result for symmetric bimodules oversemisimple Leibniz algebras. More precisely, L is the hemi-semidirect product of s and the s -bimodule Leib( L ). Theorem 7.12. If L is a finite-dimensional semisimple left Leibniz algebra over afield of characteristic zero, then HL ( L , M ) = 0 holds for every symmetric finite-dimensional L -bimodule M . Remark. Note that it follows from Corollary 4.6 that Theorem 7.12 does not holdfor anti-symmetric bimodules over semisimple Leibniz algebras. Acknowledgments. 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Farnsteiner: Modular Lie Algebras and Their Representations , Monographsand Textbooks in Pure and Applied Mathematics, vol. , Marcel Dekker, New York/Basel,1988. A version of this paper is also available on the arXiv as “Some basic properties of Leibnizalgebras” (arXiv:1302.3345v2 [math.RA], March 1, 2013). Department of Mathematics and Statistics, University of South Alabama, Mobile,AL 36688-0002, USA E-mail address ::