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Leibniz algebras as non-associative algebras

In this paper we define the basic concepts for left or right Leibniz algebras and prove some of the main results. Our proofs are often variations of the known proofs and several results seem to be new.


Introduction
This paper is mainly a survey on (left or right) Leibniz algebras from the point of 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 by Bloh in the mid sixties of the last century (see [11,12,13]) and then forgotten for nearly thirty years. In the early 1990's they were rediscovered by Loday who together with his students and collaborators developed much of the theory of Leibniz algebras, Leibniz bimodules, and Leibniz cohomology (see, for example, [35,36,20]). A left (resp. right) Leibniz algebra is a vector space with a multiplication for which every left (resp. right) multiplication operator is a derivation (i.e., a linear operator satisfying the usual Leibniz product rule). As such Leibniz algebras are non-anticommutative versions of Lie algebras. In particular, Leibniz algebras are examples of non-associative algebras (see [44]). In contrast to other papers on this topic, we study Leibniz algebras exclusively from this point of view. We have tried to make the paper sufficiently self-contained so that it could serve as a first introduction to Leibniz algebras, their modules (or representations), and their cohomology. Leibniz algebras play an important role in different areas of mathematics and physics (see [35]). In the last three decades numerous papers on Leibniz algebras appeared and many results have been duplicated. In this paper we develop the basics of the theory of Leibniz algebras in a systematic way by considering them as a special class of non-associative algebras. In the following we will describe the contents of the paper in more detail.
The first section is devoted to some background material on non-associative algebras which will be useful for the rest of the paper. In particular, we introduce the concept of a radical of an arbitrary algebra and prove three of its properties that can be used as axioms for such a concept. We refer the reader to [44] for more details and most of the proofs.
In the second section we give the definition of left resp. right Leibniz algebras and prove some basic results. We use several low-dimensional Leibniz algebras to illustrate the concepts to be introduced and the results to be proved. Among other notions, we define the (left/right) center and the Leibniz kernel of a Leibniz algebra. The latter measures how much a Leibniz algebra deviates from being a Lie algebra. Moreover, we associate several Lie algebras to a Leibniz algebra and discuss how these are related to each other. We include an example showing that left or right Leibniz algebras are not necessarily power-associative. This falsifies a claim by 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, powerassociative, and nil. The terms of the derived series of an arbitrary algebra are usually only subalgebras. At the end of the second section we show that for left or right Leibniz algebras each term of their derived series is an ideal.
Section 3 contains definitions of left Leibniz modules and Leibniz bimodules of a left Leibniz algebra. In particular, following Eilenberg [28] we motivate the defining identities of a Leibniz bimodule by considering abelian extensions of a left Leibniz algebra. We prove some basic properties of Leibniz bimodules following mainly Loday [35,36] who introduced and investigated Leibniz bimodules for a right Leibniz algebra. In addition, we also briefly discuss trace forms associated to finite-dimensional left Leibniz modules (see also [2] and [22]). Similarly to the Leibniz kernel of a Leibniz algebra, we introduce the anti-symmetric kernel of a Leibniz 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 antisymmetric (see also the proof of Theorem 3.1 in [29]). Note that this proof neither needs to assume that the Leibniz algebra is finite dimensional nor that the Leibniz bimodule is finite dimensional as in [7,Theorem 1.4]). In the fourth section we define the cohomology of a left Leibniz algebra in analogy to the cohomology of a right Leibniz algebra in [35,36], and describe the Leibniz cohomology spaces in degree 0 and 1. We also show explicitly how Leibniz 2-cocycles give rise to abelian extensions of left Leibniz algebras.
The remaining three sections of the paper are devoted to several results for nilpotent, solvable, and semisimple Leibniz algebras, respectively. We give variants of the known proofs of Engel's and Lie's theorem for Leibniz algebras as well as derive some of their applications. Furthermore, we prove Cartan's solvability criterion for Leibniz algebras. We characterize the nilpotency and solvability of a Leibniz algebra 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 some structural properties of (semi)simple Leibniz algebras and explain that the first Whitehead lemma does not hold for Leibniz algebras. In a previous version of this paper we derived the second Whitehead lemma for Leibniz algebras from Levi's theorem 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 Omirov for bringing this to our attention. 1 In this paper all algebras are defined over a field. For a subset X of a vector space V over a field F we let X F be the subspace of V that is spanned by X. We use [−, −] to denote the commutator of linear operators or matrices. The identity function on a set S will be denoted by id S , the set of non-negative integers will be denoted by N 0 , and the set of positive integers will be denoted by N.

Non-associative algebras
In this section we briefly recall some of the definitions and results on nonassociative algebras that we will need in the remainder of the paper. For more details 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, isomorphism, etc., are the same as for associative algebras since they do not use the associativity of the multiplication. Moreover, the fundamental homomorphism theorem, the isomorphism theorems, and the correspondence theorems for subalgebras and for one-sided or two-sided ideals, respectively, continue to hold with the same proofs as in the associative case.
Let S and T be two non-empty subsets of an algebra A over a field F. Then is the F-subspace of A spanned by the products st. In particular, S 2 := SS (see [44, p. 9]). The derived series of subalgebras of A is defined recursively by A (0) := A and A (n+1) := (A (n) ) 2 for every non-negative integer 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 every non-negative integer n.
An algebra A is called solvable if A (r) = 0 for some non-negative integer r. An algebra 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 homomorphisms of algebras with the derived series. Let A be a finite-dimensional algebra, and let R be a solvable ideal of maximal dimension. 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). Let a ∈ A be an arbitrary element. Then the left multiplication operator L a : A → A, x → ax is linear and 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 An algebra A is called nilpotent if there exists a positive integer n such that any product of n elements in A, no matter how associated, is zero. This generalizes the concept of nilpotency for associative algebras. Note that every nilpotent algebra is solvable (see [44, p. 18]). For any subset S of an algebra A let S * denote the subalgebra of Mult(A) generated by {L 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, 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 define powers of an element x ∈ A recursively by x 1 := x and x n+1 := xx n for every positive 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 Lie algebras 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 for some positive integer n. A subset of a power-associative algebra consisting only of nilpotent elements is called nil (see [44, p. 30]). Note that every solvable powerassociative algebra is nil (see [44, p. 31]).
The opposite algebra A op of an algebra A with multiplication (x, y) → xy has the 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 the definition of nilpotency, that the opposite algebra of a nilpotent algebra is nilpotent.

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 for all x, y, z ∈ L, the left Leibniz identity, which in turn is equivalent to the identity for all x, y, z ∈ L.
Similarly, one defines a right Leibniz algebra to be an algebra L such that every right multiplication operator R y : L → L, x → xy is a derivation. This is equivalent to the identity for all x, y, z ∈ L, the right Leibniz identity, which in turn is equivalent to the identity for all x, y, z ∈ L. Following Mason and Yamskulna [38] we call an algebra a symmetric Leibniz algebra if it is at the same time a left and a right Leibniz algebra. Note that every Lie algebra is a symmetric Leibniz algebra.
It is clear that the opposite algebra of a left Leibniz algebra is a right Leibniz algebra and that the opposite algebra of a right Leibniz algebra is a left Leibniz algebra. Consequently, the opposite algebra of a symmetric Leibniz algebra is again a symmetric Leibniz algebra. Therefore, in most situations it is enough to consider only 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 2 = 0 for every element x ∈ L.
Lemma 2.2. If L is a right Leibniz algebra, then R x 2 = 0 for every element x ∈ L.

Examples.
(1) Let A ℓ := Fe ⊕ Ff 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 not a right Leibniz algebra. We have that A (2) Let N := Fe ⊕ Ff 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. for any X, Y ∈ sl 2 (C) and any a, b ∈ L(1) is a left Leibniz algebra (see Section 3 and Lemma 4.7) 2 . 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 only two-dimensional left or right non-Lie Leibniz algebras (see [22, pp. 11/12]).
Let L be a left or right Leibniz algebra. Then Remark. For a Lie algebra the left center, the right center, and the center are all the same.
It is an immediate consequence of the definitions that the left center is a right ideal and the right center is a left ideal. More precisely, we have the following results.  2 This is known as the hemi-semidirect product of sl 2 (C) and L(1) (see [40,Definition 1.5]). 3 Note that, as for Lie algebras, the given definition of the center of a left or right Leibniz algebra is not the one used for other non-associative algebras (see [44, p. 14]). The reason for this is that, in general, Leibniz algebras, contrary to alternative algebras or Jordan algebras, are far from being associative (see Proposition 2.15 and Proposition 2.16).
Proof. We only prove Proposition 2.3 as this yields Proposition 2.4 by considering the opposite algebra. Let c ∈ C ℓ (L) and x, y ∈ L be arbitrary elements. Then we obtain 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 left center.

Examples.
(1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1 we have that C ℓ (A ℓ ) = Ff and C r (A ℓ ) = Fe. Hence C(A ℓ ) = 0.    Proof. We only prove Proposition 2.6 as this yields Proposition 2.5 by considering the opposite algebra. As C r (L) is a subspace of L, the second statement is an immediate consequence of Lemma 2.2.
Since Leib(L) is a subspace of L, it is enough for the proof of the first statement to show that x 2 y ∈ Leib(L) for any elements x, y ∈ L. By using again Lemma 2.2, we obtain that and therefore x 2 y = (x 2 + y) 2 − y 2 ∈ Leib(L).

Examples.
(1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1 we have that Leib(A ℓ ) = Ff = C ℓ (A ℓ ). Moreover, Leib(A ℓ ) ∩ C r (A ℓ ) = 0. Remark. Let L be the two-dimensional solvable left Leibniz algebra from Example 1 or the five-dimensional simple left Leibniz algebra from Example 3. We have that LLeib(L) = Leib(L). This shows that for a left Leibniz algebra L, LLeib(L) can be non-zero and LLeib(L) is not necessarily properly contained in Leib(L). Of course, a similar statement holds for right Leibniz algebras. But for a symmetric Leibniz algebra L we have that LLeib(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]). 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, every square of L is zero. Consequently, we obtain that L = Leib(L) = 0.
Remark: It is an immediate consequence of Proposition 2.8 that every onedimensional left or right Leibniz algebra is a Lie algebra. Moreover, with a little more work one can use Proposition 2.8 to classify the two-dimensional left or right Leibniz 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 associated to L. In fact, the Leibniz kernel is the smallest ideal such that the corresponding factor 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 the smallest 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 from the anti-commutativity of L/I that x 2 ∈ I for every element x ∈ L. Since I is a subspace of L, we conclude that Leib(L) ⊆ I. Proposition 2.10. Let L be a left Leibniz algebra. Then the set L(L) of left multiplication 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 multiplication 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 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 . By definition, 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 Lx(y) = [L x , L y ] for any elements x, y ∈ L, and thus L is a homomorphism of Leibniz algebras. Then the fundamental homomorphism theorem shows that L induces a Leibniz algebra isomorphism from L/C ℓ (L) onto L(L).
Remark. I am very grateful to Friedrich Wagemann for pointing out to me the four-term exact sequence of left Leibniz algebras where Out ℓ (L) := Der(L)/L(L) is the Lie algebra of outer derivations of the left Leibniz 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 homomorphism theorem we obtain the following result. 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 why left multiplication operators are preferable when one writes functions on the left of their arguments.
By definition of a right Leibniz algebra, R(L) ⊆ Der(L). Since the multiplication of 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 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 . By definition, Ker(R) = C r (L) and Im(R) = R(L). Finally, it follows from the identity for any elements x, y ∈ L, and thus R : L → Der(L) op is a homomorphism of Leibniz algebras. Then the fundamental homomorphism theorem shows that R induces a Leibniz algebra isomorphism from Remark. Similarly to [26, Proposition 1.8], we obtain from the proof of Proposition 2.12 the four-term exact sequence of right Leibniz algebras where Out r (L) := Der(L)/R(L) is the Lie algebra of outer derivations of the right Leibniz 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 homomorphism theorem we obtain the following result.
Corollary 2.13. Let L be a right Leibniz algebra. Then the set R(L) of right multiplication operators of L is a homomorphic image of the opposite of the canonical Lie algebra L op Lie associated to L.
The Lie multiplication algebra of a symmetric Leibniz algebra can be described as follows. This generalizes the corresponding result for Lie algebras (see [44, p. 21]). These results show that, in general, left and right Leibniz algebras are far from being associative. In fact, it follows from Proposition 5.9 and Proposition 5.10 below that associative left/right Leibniz algebras are nilpotent. 4

Examples.
(1) For the two-dimensional solvable left Leibniz algebra A ℓ from Example 1 we have that For the five-dimensional simple left Leibniz algebra S ℓ from Example 3 we have that S 2 ℓ = S ℓ and C r (S ℓ ) = 0. Hence S ℓ is not associative.
In the sentence after the proof of [6, Corollary 1.3] Barnes claims that left Leibniz algebras are power-associative. But the following example shows that this is not always the case. Let A ℓ denote the two-dimensional left Leibniz algebra from Example 1. Then which yields that A ℓ is not power-associative. 5 The following result shows that Barnes' claim holds for symmetric Leibniz algebras. In fact, like Lie algebras, such algebras are flexible. Proof. Let L be a symmetric Leibniz algebra. It follows from Lemma 2.2 that yx 2 = 0 for any elements x and y of a right Leibniz algebra. Hence the left Leibniz identity for L yields x(yx) = (xy)x + yx 2 = (xy)x for any elements x and y of L, and thus L is flexible.
According to Lemma 2.1, we have that x 2 y = 0 for any elements x and y of a left Leibniz algebra. This implies that x 2 x = 0 = xx 2 for any element x ∈ L. In particular, third powers are well-defined and are equal to zero. Consequently, we obtain by induction that symmetric Leibniz algebras are power-associative. 6 So in this respect symmetric Leibniz algebras are only slightly more general than Lie 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 a right Leibniz algebra is indeed an ideal.
Proposition 2.18. Let L be a left or right Leibniz algebra. Then L (n) is an ideal of L for every non-negative integer n.
Proof. We only prove the result for left Leibniz algebras as this yields the result for 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 induction on n. The base step n = 0 is clear as LL (0) = LL ⊆ L = L (0) . For the induction step let n > 0 be an integer and assume that the statement is true for n − 1. Then we obtain from the left Leibniz identity and the induction hypothesis that 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 > 0 be an integer and assume that the 5 This example also shows that left or right Leibniz algebras are not necessarily flexible. 6 The proof of Proposition 2.17 shows that this result holds more generally for left central Leibniz algebras or right central Leibniz algebras. The latter notions were introduced by Mason and Yamskulna in [38]: A left central Leibniz algebra is a left Leibniz algebra L such that Leib(L) ⊆ Cr(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 induction hypothesis that This completes the proof.

Leibniz modules
In this and in the next section we consider only left Leibniz algebras. We leave it to the interested reader to formulate the corresponding definitions and results for right Leibniz algebras (see [35,36]).
Let L be a left Leibniz algebra over a field F.
is satisfied for every m ∈ M and all x, y ∈ L.
The usual definitions of the notions of submodule, irreducibility, complete reducibility, composition series, homomorphism, isomorphism, etc., hold for left Leibniz 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 multiplication. This module is called the left adjoint module and will be denoted by L ad,ℓ . The associated representation L : L → gl(L) is called the left adjoint 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 := 0 for 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 the study of left Leibniz modules to Lie modules.
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, Définition 1.1.14] left Leibniz modules are called Lie modules. Consequently, many properties of left L-modules follow from the corresponding properties of L Lie -modules. As one application of this point of view, we discuss trace forms associated to finite-dimensional left Leibniz modules (see also [2,22]).
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 Example. As for Lie algebras, the trace form κ := κ L associated to the left adjoint representation L of a finite-dimensional left Leibniz algebra L is called the Killing form 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 the symmetry of κ 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 the invariance 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.
The correct concept of a module for left Leibniz algebras is the notion of a Leibniz bimodule. In order to motivate the appropriate definition of a bimodule for a left Leibniz algebra, we follow the approach that Eilenberg proposed for any given class of 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 over the same ground field. Then the Cartesian product L × M with componentwise addition and componentwise scalar multiplication is a vector space over F. for any elements x 1 , x 2 ∈ L and m 1 , m 2 ∈ M . To ensure that L × M satisfies the left Leibniz identity, we compute for any elements x 1 , x 2 , x 3 ∈ L and m 1 , m 2 , m 3 ∈ M . The vector space L × M satisfies the left Leibniz identity if, and only if, the left-hand side of the first identity equals the sum of the left-hand sides of the second and third identities. Hence the right-hand side of the first identity must equal the sum of the right-hand sides of the second and third identities. For the first components of the right-hand sides this is just the left Leibniz identity for L. From the desired equality for the second components of the right-hand sides one can read off the following three identities which are sufficient for L × M to satisfy the left Leibniz identity. (3.1) x · (y · m) = (xy) · m + y · (x · m) 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 following compatibility conditions are satisfied: for every m ∈ M and all x, y ∈ L.
It is an immediate consequence of (LLM) that every Leibniz bimodule is a left Leibniz module. Moreover, by combining (LML) and (MLL) we obtain the following result. 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, isomorphism, etc., hold for Leibniz bimodules.
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: (1) Every left Leibniz algebra L is a bimodule over itself via the Leibniz multiplication, the so-called adjoint bimodule L ad of L. The associated representation is (L, R), where L denotes the left multiplication operator of L and R denotes the right multiplication operator of L. This representation is called the 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 is symmetric if, and only if, L is a Lie algebra. 8 (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. Let L be a left Leibniz algebra over a field F, and let M be an L-bimodule. We call the anti-symmetric kernel of M (see [36, p. 145]). 8 In particular, the adjoint bimodule of a symmetric Leibniz algebra is not always symmetric as the name might suggest.
Example (cf. also [29, p. 479]). Let L be a left Leibniz algebra. Then the identity shows that (L ad ) 0 ⊆ Leib(L). Moreover, if the ground field of L has characteristic = 2, then the identity x 2 = 1 2 (xx + xx) shows that (L ad ) 0 = 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]). Proof. Let x, y ∈ L and m ∈ M be arbitrary elements. Then we obtain from (3.1) and (3.2) that   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 This completes the proof.
Note that for the trivial left Leibniz module the Leibniz bimodule structures obtained from parts (a) and (b) of Proposition 3.8 both give rise to the trivial Leibniz 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 an anti-symmetric L-bimodule, which we call the anti-symmetric adjoint bimodule L a of L. The associated representation is (L, 0), the so-called antisymmetric adjoint representation of L, where L denotes the left multiplication operator of L. (2) By virtue of Proposition 3.8 (b), L ad,ℓ has a unique symmetric L-bimodule structure, which we call the symmetric adjoint bimodule L s of L. The associated representation is (L, −L), the so-called symmetric adjoint representation of L.

Leibniz cohomology
Let L be a left Leibniz algebra over a field F, and let M be an L-bimodule. For any non-negative integer n set C n (L, M ) := Hom F (L ⊗n , M ) and consider the linear transformation d n : C n (L, M ) → C n+1 (L, M ) defined by for any f ∈ C n (L, M ) and all elements x 1 , . . . , x n+1 ∈ L.
It is proved in [19, Lemma 1.3.1] that (C n (L, M ), d n ) n∈N0 is a cochain complex, i.e., d n+1 • d n = 0 for every non-negative integer n. Of course, the original idea of defining Leibniz cohomology as the cohomology of such a cochain complex for right Leibniz algebras is due to Loday [36, (1.8)] (see also [12]). Hence one can define the cohomology of L with coefficients in M by     Let d n Lie denote the coboundary map for the Chevalley-Eilenberg cohomology of Lie algebras (see [14,Exercise I.3.12]). Then we have that In particular, we obtain from Corollary 4.4 the following result.  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 → 0 is called an extension of L by K, and K is called the kernel of the extension. Two Every extension 0 → K → E → L → 0 with abelian kernel K gives rise to an L-bimodule structure on K (see [19,Proposition 1.3.6]). It can be shown that 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 2 (L, M ) (see [19,Theorem 1.3.13] for left Leibniz algebras and [36, Proposition 1.9] for right Leibniz algebras). We will only prove the following ingredient of establishing such a bijection.
for any x 1 , x 2 ∈ L and any m 1 , m 2 ∈ M is a left Leibniz algebra.
Proof. Since the Cartesian product L × M with componentwise addition and componentwise scalar multiplication is a vector space, we only need to verify the left Leibniz identity. We have that for any x 1 , x 2 , x 3 ∈ L. Since f is a 2-cocycle, we have that We compute for any elements x 1 , x 2 , x 3 ∈ L and m 1 , m 2 , m 3 ∈ M . It follows from the left Leibniz identity, (3.1), (3.2), (3.3), and the 2-cocycle identity for f that the right-hand side of the first identity equals the sum of the right-hand sides of the second and third identities. Hence the left-hand side of the first identity also equals the sum of the left-hand sides of the second and third identities, and thus L × M satisfies the left Leibniz identity.

Nilpotent Leibniz algebras
Let L be a left or right Leibniz algebra. Then the left descending central series of L is defined recursively by 1 L := L and n+1 L := L n L for every positive integer n. Similarly, the right descending central series of L is defined recursively by L 1 := L and L n+1 := L n L for every positive integer n. 9 Note that Proposition 5.2 is just [4, Lemma 1]. For the convenience of the reader we include its proof.
Proposition 5.1. Let L be a left Leibniz algebra. Then ( n L) n∈N is a descending filtration 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 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. 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 considering the opposite algebra. Firstly, we show that the right descending central series is indeed descending, i.e., L n+1 ⊆ L n for every positive integer n. We proceed by induction on n. The base step n = 1 is clear. For the induction step let n > 1 be an integer and assume that the statement is true for n − 1. Then Next, we prove L m L n ⊆ L m+n by induction on n. The base step n = 1 is an immediate consequence of the definition: For the induction step let n > 1 be an integer and assume that the statement is true for n − 1. It follows from identity (2.4) and by applying the induction hypothesis twice that This completes the proof. Proof. We only prove the first statement in Proposition 5.4 as this in conjunction with Proposition 5.2 implies the remaining statements. We proceed by induction on n. The base step n = 1 is an immediate consequence of the definitions. For the induction step let n > 1 be an integer and assume that the statement is true for any integer less than n. Then by applying the induction hypothesis and Proposition 5.2 we obtain that On the other hand, it follows from the induction hypothesis that This completes the proof.
Corollary 5.5. If L is a symmetric Leibniz algebra, then L n = n L = L n for every positive integer n.
Recall from Section 1 that an algebra L is called nilpotent if there exists a positive integer n such that any product of n elements in L, no matter how associated, is zero (see [44, p. 18]). The following observation is clear. Examples.
(1) The two-dimensional solvable left Leibniz algebra A ℓ from Example 1 in Section 2 is not nilpotent as (A ℓ ) n = n A ℓ = Ff = 0 for every integer n ≥ 3. But note that A n r = 0 for every integer n ≥ 3. (2) The two-dimensional symmetric Leibniz algebra N from Example 2 in Section 2 is nilpotent as N n = n N = N n = 0 for every integer n ≥ 3. (3) The five-dimensional simple left Leibniz algebra S ℓ from Example 3 in Section 2 is not nilpotent as (S ℓ ) n = n S ℓ = S n ℓ = S ℓ = 0 for every positive 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 For the induction step let n > 1 be an integer and assume that the statement is true for any integer less than n. Then we obtain from the induction hypothesis that This completes the proof.
The next result is an immediate consequence of Lemma 5.7.
Proposition 5.8. Subalgebras and homomorphic images of nilpotent left or right Leibniz 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 considering the 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.
Proposition 5.11. The sum of two nilpotent ideals of a left or right Leibniz algebra is nilpotent.
Proof. Let L be a right Leibniz algebra. We begin by proving that I n is a right ideal of L for any right ideal I of L and for any positive integer n. We will proceed by induction on n. The base step I 1 L = IL ⊆ I = I 1 follows from the fact that I is a right ideal of L. For the induction step let n > 1 be an integer and assume that the statement is true for n − 1. Then we obtain from the right Leibniz identity and the induction hypothesis that I n L = (I n−1 I)L ⊆ (I n−1 L)I + I n−1 (IL) ⊆ I n−1 I = I n .
Now let J and K be two nilpotent ideals of L. We will prove that for every positive integer n. We will again proceed by induction on n. The base step (J + K) 1 = J + K = J 1 + K 1 is an immediate consequence of the definition of the right descending central series. For the induction step let n ≥ 1 be an integer and assume that the statement is true for n. Then we obtain from the induction hypothesis and the fact that J r and K n−r are right ideals of L: 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−1 = 0 which shows that J + K is a nilpotent ideal of L.
Using Proposition 5.11 one can proceed similar to the solvable case to establish the existence of a largest nilpotent ideal (see Section 1). We leave the details to the interested reader (see also [31,Proposition 1]  (iv) Lie(L) is nilpotent.
Proof. The implication (i)⇒(ii) is an immediate consequence of Proposition 5.8. It follows from Corollary 2.7 that there is a natural epimorphism L Lie → L/C(L) of Lie 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 2.10 that 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 natural epimorphism L/C(L) → L/C r (L) ∼ = R(L) op , and thus R(L) is nilpotent. Hence it follows from Proposition 5.11 and Theorem 2.14 that Lie(L) is nilpotent. This establishes the implication (iii)⇒(iv). Finally, the implication (iv)⇒(i) can be obtained from Proposition 5.8 in conjunction with Proposition 2.10 and Proposition 5.10.
Example. Let A ℓ denote the two-dimensional non-nilpotent solvable left Leibniz algebra A ℓ from Example 1 in Section 2. As has been observed above, we have that C ℓ (A ℓ ) = Ff = Leib(A ℓ ) and C r (A ℓ ) = Fe. 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 these Lie algebras is abelian, and thus nilpotent, but A ℓ is not nilpotent. This shows that the implication (ii)⇒(i) in Theorem 5.12 neither holds for left nor for right Leibniz algebras. Moreover, neither of the implications (iii)⇒(i) and (iv)⇒(i) in Theorem 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 transformations was first proved by Patsourakos [41,Theorem 7] and later by Barnes [7, Theorem 1.2] who used the corresponding result for Lie algebras in his proof. Note that Patsourakos 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) 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 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 every element x ∈ L.
Proof. We first prove that for any element x ∈ L the nilpotency of λ x implies the nilpotency of ρ x (see also [41,Lemma 6]). Namely, we have ρ n x = (−1) n−1 ρ x • λ n−1 x for every element x ∈ L and every positive integer n. This can be shown by induction on n. The base step n = 1 is trivial. For the induction step let n ≥ 1 and assume that the statement is true for n. Then it follows from the induction hypothesis and Lemma 3.3 that Since the L-bimodule M is finite-dimensional, it has an irreducible L-subbimodule 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 a Lie 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 of linear transformations (see [32,Theorem 3.3]). Finally, this and the symmetry or anti-symmetry of N yield ρ x (m) = 0 for every x ∈ L.
We conclude this section by several applications of Theorem 5.13. The first result is just a reformulation of Theorem 5.13 in terms of a composition series of a Leibniz bimodule (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 statements hold: (a) If M is irreducible, then M is the one-dimensional trivial L-bimodule. Next, we specialize Corollary 5.14 and Theorem 5.13 to the adjoint Leibniz bimodule. 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 from Theorem 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).
Corollary 5.16. If L is a finite-dimensional left Leibniz algebra such that L x is nilpotent 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 every positive integer k, and therefore n+1 L = 0. We proceed by induction on k to prove the former statement. The base step k = 1 is just 1 L = L = L n . For the induction step let k > 1 be an integer and assume that k−1 L ⊆ L n−k+2 is true. Then we obtain from the induction hypothesis in conjunction with Corollary 5.15 (a) that k L = L k−1 L ⊆ LL n−k+2 ⊆ L n−k+1 .

Solvable Leibniz algebras
In the case of solvable Leibniz algebras we have one-sided analogues of Theorem 5.12. These show that the solvability of a left or right Leibniz algebra L is equivalent to the solvability of several Lie algebras associated to L. Proposition 6.1. For every left Leibniz algebra L the following statements are equivalent: Proposition 6.2. For every right Leibniz algebra L the following statements are equivalent: Proof. We only prove Proposition 6.2 as this yields Proposition 6.1 by considering the opposite algebra. The implication (i)⇒(ii) is an immediate consequence of Proposition 1.1. It follows 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 the implication (ii)⇒(iii). Moreover, the implication (iii)⇒(i) follows from Proposition 1.2. Finally, the remaining equivalence of (iii) and (iv) is an immediate consequence of Proposition 2.12. Theorem 6.3. For every symmetric Leibniz algebra L the following statements are equivalent: Proof. The equivalence of (i) and (ii) follows from Proposition 6.1 or Proposition 6.2 and 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 2.10 that 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 epimorphism L/C(L) → L/C r (L) ∼ = R(L) op , and thus R(L) is solvable. This establishes 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 1.3 in conjunction with the already established implications. Finally, the implication (vi)⇒(iv) can be obtained from Theorem 2.14 and Proposition 1.1.
The Leibniz analogue of Lie's theorem for solvable Lie algebras of linear transformations was proved by Patsourakos [42,Theorem 2]. For the convenience of the reader we include a proof that follows very closely the proof of Theorem 5.13 and uses the corresponding result for Lie algebras. Proof. Since the L-bimodule M is finite-dimensional, it has an irreducible L-submodule 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 show that λ(L) is a Lie subalgebra of gl(M ) such that λ(L) ∼ = L/Ann L (M ). The latter isomorphism in conjunction with Proposition 1.1 yields that λ(L) is solvable. So the existence of a common eigenvector m ∈ N for λ(L) can be obtained from Lie's theorem 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 a common eigenvector for ρ(L).  Remark. Corollary 6.5 (and thus also Theorem 6.4) is not true for ground fields of prime characteristic as already can be seen for the non-abelian two-dimensional Lie algebra (see [46,Example 5.9.1]). Moreover, the one-dimensional Lie algebra spanned by any proper rotation of a two-dimensional real vector space shows that Corollary 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 an algebraically closed field F of characteristic zero, and let I be a d-dimensional ideal of L. Then there exists an ascending chain 0 = L 0 L 1 · · · L n = L of ideals of L such that I = L d and dim F L j = j for every 1 ≤ 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.
We can also employ Engel's theorem for left Leibniz algebras to prove the following 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 an algebraically closed field F of characteristic zero. Then L x is nilpotent for every element x ∈ L 2 . In particular, L 2 is nilpotent.
Proof. According to Corollary 6.6, there exists an ascending chain 0 = L 0 L 1 · · · L n = L of ideals of L such that dim F L j = j for every 1 ≤ j ≤ n. Hence one can choose successively a basis {x 1 , . . . , x j } of L j (1 ≤ j ≤ n) such that the corresponding matrices of L(L) are upper triangular. By virtue of the proof of Proposition 2.10, we have that L(L 2 ) = [L(L), L(L)], and thus the matrices of L(L 2 ) are strictly upper triangular. Hence L x is nilpotent for every element x ∈ L 2 . In particular, we obtain from Corollary 5.16 that L 2 is nilpotent.
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 of characteristic zero. Then L is solvable if, and only if, κ(x, y) = 0 for every element x ∈ L and every element y ∈ L 2 .
Proof. Suppose first that L is solvable and the ground field of L is algebraically closed. It follows from Corollary 6.6 that L(L) can be simultaneously representated by upper triangular matrices. Then the proof of Proposition 2.10 shows that the corresponding matrix of L yz = [L y , L z ] = L y • L z − L z • L y is a strictly upper triangular 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 2 , and the ground field of L is again algebraically closed. It follows from Proposition 2.10 that L(L) is a Lie subalgebra of gl(L). In particular, we obtain as before that tr(L x • [L y , L z ]) = tr(L x • L yz ) = κ(x, yz) = 0 for any x, y, z ∈ L. So the other implication is a consequence of [32,Theorem 4.3] in conjunction with Proposition 6.1.
Finally, in case the ground field F of L is not algebraically closed, a base field extension 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 2 = L 2 ⊗ F F, we obtain by induction 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 ⊗ 1, y ⊗ 1) = tr(L x⊗1 • L y⊗1 ) = tr(L x • L y ) = κ(x, y) for all x, y ∈ L as L a⊗α (b ⊗ β) = L a (b) ⊗ αβ for any a, b ∈ L and any α, β ∈ F. This can be used to show that κ(x, y) = 0 for every element x ∈ L and every element y ∈ L 2 if, and only if, κ(x, y) = 0 for every element x ∈ L and every element y ∈ L 2 . . Note that there is also 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 2.8 that every simple left or right Leibniz algebra has to be a Lie algebra.

Semisimple Leibniz algebras
Exactly as for Lie algebras, we call a left or right Leibniz algebra L perfect in case L = L 2 holds. Then the first result in this section is an immediate consequence of these definitions. The next result follows from the correspondence theorem for ideals. Note also that the condition Leib(L) L 2 implies that the canonical Lie algebra L Lie associated 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 simple Lie algebra.
We will see in the example after Theorem 7.11 that the converse of Proposition 7.2 does not always hold.
In analogy with the above definition of simplicity, we call a left or right Leibniz algebra L semisimple when Leib(L) contains every solvable ideal of L. 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 we obtain by induction that L = L (n) holds for every non-negative integer n. Since by hypothesis 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 2 .
In the finite-dimensional case we have the following result which often is taken as the definition of semisimplicity for Leibniz algebras (see [22,Definition 5.2] or [29, Definition 2.12]).   Theorem 7.11. If L is a finite-dimensional semisimple left Leibniz algebra over a 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. 10 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 an immediate 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 an anti-symmetric s-bimodule. Finally, Weyl's theorem (see [32,Theorem 6.3]) yields that 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 of Leib(L). Then M is an ideal of L that is different from 0, Leib(L), and L, which contradicts the simplicity of L. Hence Leib(L) is an irreducible L-bimodule. By virtue of Proposition 2.8, Leib(L) is also an irreducible s-bimodule.
A left or right Leibniz algebra L is called Lie-simple if L Lie is simple. According to Proposition 7.3 and Proposition 7.8, Lie-simple left or right Leibniz algebras are semisimple, but they are not always simple.
Example (see also [22 Remark. The same argument as in the previous example in conjunction with Proposition 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 is a direct sum of non-trivial irreducible left s-modules, is a semisimple left Leibniz algebra. Note also that there exist semisimple left Leibniz algebras that cannot be 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 with Corollary 4.4 (a) and the first Whitehead lemma for Lie algebras (see [33, Theorem 13 in Chapter III]) the corresponding result for symmetric bimodules over semisimple Leibniz algebras.
Theorem 7.12. If L is a finite-dimensional semisimple left Leibniz algebra over a field of characteristic zero, then HL 1 (L, M ) = 0 holds for every symmetric finitedimensional L-bimodule M .
Remark. Note that it follows from Corollary 4.6 that Theorem 7.12 does not hold for anti-symmetric bimodules over semisimple Leibniz algebras.