Symmetry, Integrability and Geometry: Methods and Applications Leibniz Algebras and Lie Algebras ⋆

This paper concerns the algebraic structure of finite-dimensional complex Leibniz algebras. In particular, we introduce left central and symmetric Leibniz algebras, and study the poset of Lie subalgebras using an associative bilinear pairing taking values in the Leibniz kernel.


Introduction
Throughout the present paper, a left Leibniz algebra means a nonassociative C-algebra M with a product (or bracket) [ ] satisfying the following identity for all a, b, c ∈ M : Leibniz algebras were first studied for their own sake by Loday [5] (see also [6,Section 10.6]). The rationale for the present work is partially motivated by the triangular decomposition of a vertex operator algebra (VOA) V . In the most widely studied case when V is of CFT-type, i.e. V n = 0 for n < 0 and V 0 = C1 is spanned by the vacuum vector, the summands in (2) are Lie algebras. In the general case (2) satisfies only the weaker condition of being a decomposition into Z-graded left Leibniz algebras with respect to the 0 th operation in V (cf. [4]). This decomposition plays a rôle in our work [7] where, among other things, we are interested in the Lie subalgebras of V 1 and the vertex subalgebras of V that they generate. This leads directly to the main theme of the present paper, which is the study of the poset of Lie subalgebras of certain kinds of Leibniz algebras. Readers uninured to VOA theory need not be concerned -the present paper deals solely with Leibniz algebras, and no further mention of VOAs will be made. The definition of Leibniz algebra is not left-right symmetric; a right Leibniz algebra, in which the map b → [ba] for fixed a is a derivation, is not necessarily a left Leibniz algebra. We call This paper is a contribution to the Special Issue on New Directions in Lie Theory. The full collection is available at http://www.emis.de/journals/SIGMA/LieTheory2014.html an algebra that is both a left and right Leibniz algebra a symmetric Leibniz algebra. Notice from (1) that a left Leibniz algebra M satisfies and dually, a right Leibniz algebra satisfies We call M a left central Leibniz algebra if it is a left Leibniz algebra that also satisfies (4). Equivalently, M is both the left and right centralizer of C(M ). There is a hierarchy of algebras and in fact each containment is strict.
We now describe the contents of the present paper, which deals solely with finite-dimensional, complex, left Leibniz algebras M . We tacitly assume this in everything that follows. In Section 2 we discuss some basic facts, in particular concerning Levi subalgebras and Levi decompositions of Leibniz algebras. Levi subalgebras (i.e., semisimple Lie subalgebras that complement the solvable radical) are readily seen to exist in M (see [3] and Section 2 below). Malcev's theorem for Lie algebras does not extend to left Leibniz algebras, though it does for various special classes, including left central Leibniz algebras. Section 3 is devoted to these issues.
The remainder of the paper revolves about the symmetric bilinear pairing ψ : . This is a feature of all Leibniz algebras, but for left central Leibniz algebras it is particularly efficacious, because in this case it is associative. Then the radical R of ψ is a 2-sided ideal of M , and the poset L of Lie subalgebras of M coincides with the set of Leibniz subalgebras which are also isotropic subspaces. General properties of this set-up are developed in Section 4, including (Proposition 1) the fact that R coincides with the intersection of the maximal elements of L. We also give (Lemma 7) a general construction of a class of left central Leibniz algebras based solely on Lie-theoretic data. In Sections 5 and 6 we consider left central Leibniz algebras with dim C(M ) = 1 (so that ψ is a trace form in the usual sense). Such algebras arise, for example, as quotients M/U whenever U is a hyperplane of C(M ). We prove (Theorem 3) that there is a Lie subalgebra L ⊆ M that is also a maximal isotropic subspace of M (and therefore also a maximal element of L) and such that L/R is nilpotent. In Section 6 we assume that M is also symmetric, a property that holds precisely when M ⊆ R (Lemma 8). We prove (Theorem 4) that a symmetric Leibniz algebra with dim C(M ) = 1 has a 2-sided ideal of codimension at most 1 that arises from the construction of Lemma 7. In this way, we more-or-less obtain a characterization of such Leibniz algebras in terms of Lie-theoretic data.

Basic facts
We call a Lie subalgebra S of M that complements B as in (6)

On Malcev's theorem for Leibniz algebras
Malcev's theorem for finite-dimensional complex Lie algebras includes the statements that all Levi subalgebras are conjugate by the exponential of an inner derivation, and every semisimple Lie subalgebra is contained in a Levi subalgebra. We shall see that both of these assertions are generally false for Leibniz algebras. On the other hand, versions of the Malcev theorem can be proved for certain classes of Leibniz algebras. This section is concerned with these questions. We begin with a general construction. Let S be a finite-dimensional complex Lie algebra Here, of course, [ab] denotes the bracket in S.
Suppose that S is simple and N a simple S-module that is not the adjoint module. Then Hom S (S, N ) = 0, so that S is the unique Levi subalgebra of M . In the case of Lie algebras, if there is a unique Levi subalgebra then it is an ideal. However, as our construction shows, this is not necessarily true for Leibniz algebras.
Suppose that T ⊆ S is a simple Lie subalgebra. Replacing S by T in the previous paragraphs, we see that if d ∈ Hom T (T, N ) then In order for T 1 to be contained in a Levi subalgebra, it is necessary and sufficient that d is the restriction d = Res S T (c) for some c ∈ Hom S (S, N ). Arguing in the same way as the discussion preceding Example 1, we know that S 2 = {(x, c(x)) | x ∈ S 1 } for some c ∈ Hom S 1 (S 1 , N ). We will show that the linear map f : M → M defined by These definitions apply to all Leibniz algebras. ψ is particularly useful in the study of central Leibniz algebras because of the next result.  (b) L is a totally isotropic subspace; (c) L + R ∈ L.
Proof . The equivalence of (a) and (b) follows directly from the definitions. Next, if L + R is a Lie algebra then L is necessarily a Lie subalgebra, so that (c) ⇒ (a). Conversely, if (a) holds then L is a Lie subalgebra of M . In this case, L + R is itself a Leibniz subalgebra because R is a 2-sided ideal. Moreover, if a ∈ R, b ∈ L, then ψ(a + b, a + b) = ψ(a, a) + 2[bb] + 2ψ(a, b) = 0. Therefore, R + L is totally isotropic, and the equivalence of (a) and (b) applied to R + L shows that (c) holds. This completes the proof of the lemma.  Proof . L is clearly a Leibniz subalgebra of M . Let S ⊆ L be a Levi subalgebra. Because L/R is semisimple we have L = R + S, and this is a Lie subalgebra thanks to the equivalence of (a) and (c) in Lemma 2.
Lemma 5. M/R has no nonzero semisimple ideals. In particular, every minimal ideal of M/R is abelian.
Proof . Suppose that U/R is an ideal of M/R that is also a semisimple Lie algebra. By Lemma 4, U ∈ L, and in particular it is totally isotropic. Then U ⊆ R by Lemma 3. Therefore U/R is both abelian and semisimple, whence it reduces to 0. This proves the first assertion of the lemma. Because every minimal ideal of the Lie algebra M/R is either abelian or semisimple, the second assertion follows. This completes the proof of the lemma. Proof . Let L ∈ L * . Then R + L is also a Lie subalgebra by Lemma 2, so that R ⊆ L because L is maximal. This shows that R is contained in every L ∈ L * , and hence also in their intersection.
To prove the opposite containment, we use induction on dim N . There is nothing to prove if M is a Lie algebra, so we may, and shall, assume that this is not the case. Therefore, N = 0. Suppose that dim N ≥ 2. Then every hyperplane N 0 ⊆ N is a 2-sided ideal in M , and is itself contained in every maximal Lie subalgebra L. Then by induction we obtain L∈L * L/N 0 = R(M/N 0 ), which says that if a ∈ ∩L then ψ(a, x) ∈ N 0 for all x ∈ M . Then ψ(a, x) ∈ ∩N 0 = 0 (the last intersection ranging over hyperplanes of N ), and thus a ∈ R, as required This reduces us to the case when dim N = 1, so that we can think of ψ as a C-valued bilinear form. We assume this for the remainder of the proof.
Because M is not a Lie algebra, R = M . Suppose that R has codimension 1. Then every nonzero element of M/R is anisotropic, so that if a ∈ M \ R then a cannot be contained in a Lie subalgebra of M . Thus in this case, R is the unique element in L * , and the desired result is clear.
Finally, suppose that M/R has dimension at least 2. Because ψ is a nondegenerate C-valued bilinear form on M/R, M/R is spanned by isotropic vectors, i.e., elements a + R with [aa] = 0. Such elements a are contained in some maximal Lie subalgebra; therefore, if b ∈ ∩ L∈L * L, a and b generate a Lie subalgebra of M , so that [ab] + [ba] = 0. Because the isotropic vectors a + R span, we can conclude that [ab] + [ba] = 0 for all a ∈ M , whence b ∈ R. This completes the proof of the proposition.
We will need the next lemma in Section 5. Proof . Write B ⊥ = S 0 ⊕ B 0 , where S 0 and B 0 are a Levi factor and the solvable radical respectively for B ⊥ . Because B is an ideal in M then so too is B 0 . Therefore, B 0 ⊆ B ∩ B ⊥ . Thus B 0 = rad(B), and in particular B 0 is totally isotropic. So we see that there is an orthogonal decomposition B ⊥ = S 0 ⊥ B 0 , and since both summands are totally isotropic then B ⊥ is a totally isotropic ideal of M . Now we can apply Lemma 3, with B ⊥ playing the rôle of U , to conclude that S 0 ⊆ R. Therefore, B ⊥ ⊆ B, and the lemma is proved.
We now describe the construction of a class of left central Leibniz algebras that depends only on Lie-theoretic data (L 1 , L 2 , R, α, π) satisfying (a)-(c) below. The set-up is as follows: (a) a pair of Lie algebras L 1 , L 2 with a common ideal R and L 2 /R abelian: for x 1 , x 2 ∈ L 1 , y 1 , y 2 ∈ L 2 , and where [ ] is the bracket in L 1 or L 2 .
Theorem 3. Suppose that M is a left central Leibniz algebra of rank 1, and let B be as in Lemma 6. Then there is at least one maximal Lie algebra L ∈ L * with the following properties: (a) L is a maximal isotropic subspace of M ; Proof . If L ∈ L * then R ⊆ L (cf. Proposition 1). So part (b) makes sense. To prove the theorem, we use induction on dim M . If every nonzero element of M/R is anisotropic, then R is both the unique maximal Lie subalgebra of M and the unique maximal totally isotropic subspace, in which case all parts of the theorem are obvious. So we may, and shall, assume that M/R contains nonzero isotropic elements. As in the proof of Proposition 1, this implies that dim(M/R) ≥ 2. We assume first that there is a (nonzero) ideal U/R in M/R which is totally isotropic. If M/R is solvable, we choose any such ideal U/R; if M/R is nonsolvable we take U = B ⊥ . Note that in the second case, B is a proper ideal of M , so that B ⊥ /R is nonzero thanks to the assumption that M has rank 1. Moreover, B ⊥ /R is totally isotropic in the second case because of Lemma 6.
With U/R chosen in this way, U ⊥ is a proper ideal of M, U ⊥ ⊆ B, and U = rad(U ⊥ ). By induction there is a Lie subalgebra L ⊆ U ⊥ which is a maximal isotropic subspace of U ⊥ and such that L/U is nilpotent. If L 1 is a maximal isotropic subspace of M that contains L, then L 1 ⊆ U ⊥ , and this implies that L = L 1 since L is maximal isotropic in U ⊥ . So L is maximal isotropic in M .
It remains to show that L/R is nilpotent. Let a ∈ M, u ∈ U, x ∈ L. Then ψ(a, [ux]) = ψ([au], x) = 0, which shows that [LU ] ⊆ R. Another way to say this is U/R ⊆ Z(L/R). Since we already know that L/U is nilpotent, we can conclude that L/R is too. This completes the proof of the theorem in this case. Thus we may, and now shall, suppose that no nonzero ideal U/R of M/R is totally isotropic. Now take any minimal nonzero ideal U/R ⊆ M/R. Then U/R is nondegenerate, and because M has rank 1 there is an orthogonal decomposition M/R = U/R ⊥ U ⊥ /R. Continuing in this way, we obtain an orthogonal decomposition with each U j /R a minimal, nondegenerate ideal of M/R. By Lemma 5, each U j /R is abelian. Then M/R is itself an abelian Lie algebra, and every nonzero element of M/R generates a 1dimensional ideal. In particular, since M/R contains nonzero isotropic elements, it also has a nonzero isotropic ideal. This contradiction completes the proof of the theorem. Consider the left central Leibniz algebra M = M (L 1 , L 2 , R, α, π) with product (9) and its quotient algebraM = M/D (cf. Lemma 7) introduced at the end of Section 4. Since D is contained in the radical of M , it follows from Lemma 8 thatM is symmetric if, and only if, M is symmetric. From (9) and Lemma 8 once more, this holds if, and only if, L 1 /R is abelian (recall that L 2 /R is abelian by construction) and x 1 .y 2 − x 2 .y 1 ∈ R (x 1 , x 2 ∈ L 1 , y 1 , y 2 ∈ L 2 ). We assert that the second condition is a consequence of the first. Indeed, because π : L 1 /R → Hom C (L 2 /R, Z) is an injection of L 1 -modules then L 1 acts trivially on im π, i.e., im π annihilates each element x 1 .y 1 . Since im π has no nonzero invariants in its action on L 2 /R, it follows that x 1 .y 1 ∈ R, and since this holds for any x 1 ∈ L 1 , y 1 ∈ L 2 then the second condition indeed follows, as asserted. Thus we have proved Now suppose that M 0 is a symmetric Leibniz algebra of rank 1. Thus N = C(M 0 ) has dimension 1 and ψ defines a trace form on M 0 with radical R. Consequently, we can find a pair of maximal, totally isotropic subspaces L 1 , L 2 ⊆ M 0 such that L 1 ∩ L 2 = R and L 1 + L 2 has codimension at most 1 in M 0 . (L 1 + L 2 = M 0 if, and only if, dim(M 0 /R) is even.) Because M 0 /R is an abelian Lie algebra by Lemma 8, it follows that H := L 1 + L 2 is a 2-sided ideal with Lie subalgebras L 1 , L 2 ; indeed, L 1 , L 2 ∈ L * . We will show that H ∼ =M (L 1 , L 2 , R, α, π) for suitably defined maps α, π. Both L 1 , L 2 are ideals of M 0 , in particular the left adjoint defines a morphism of Lie algebras α : L 1 → Der(L 2 ) satisfying the assumptions of (b) at the end of Section 4. Similarly, R = rad(M ) and we define the morphism of L 1 -modules π : L 1 /R → Hom C (L 2 /R, N ) as π(x 1 , y 1 ) := ψ(x 1 , y 1 ) (x 1 ∈ L 1 , y 1 ∈ L 2 ). It is easily seen that the assumptions of (c) at the end of Section 4 also hold. (The main point is the associativity of ψ.) Thus the set-up discussed in Section 4 holds, and we can apply Lemma 7 to obtain the left central Leibniz algebras M (L 1 , L 2 , R, α, π) andM . We assert that H ∼ =M . To see this, one checks that the map ϕ : M (L 1 , L 2 , R, α, π) → H, (x 1 , y 1 ) → x 1 + y 1 , is a surjective morphism of Leibniz algebras with kernel {(x 1 , y 1 ) | x 1 + y 1 = 0} = {(a, −a) | a ∈ R}. Now our assertion follows from the very construction ofM . We have proved Theorem 4. A symmetric Leibniz algebra of rank 1 has an ideal of codimension at most 1 isomorphic toM (L 1 , L 2 , R, α, π).