𝔤 ⊗ 𝔤 AS A LEIBNIZ ALGEBRA

R. KURDIANI AND T. PIRASHVILI

0. Introduction

Let 𝔤 be a Lie algebra. We define the following bracket

[x ⊗ y, a ⊗ b] := [x, [a, b]] ⊗ y + x ⊗ [y, [a, b]]  (1)

on 𝔤⊗𝔤. It turns out that 𝔤⊗𝔤 equipped with this bracket in general is not a Lie algebra but only a Leibniz algebra. Let us recall that Leibniz algebras are non-anti-commutative generalization of Lie algebras [8], [9]. More precisely a Leibniz algebra 𝔤 is a vector space equipped with a bracket

[−, −] : 𝔤 ⊗ 𝔤 → 𝔤,

satisfying the Leibniz identity:

[x, [y, z]] = [[x, y], z] − [[x, z], y].  (2)

Clearly any Lie algebra is a Leibniz algebra, and conversely any Leibniz algebra 𝔤 with property [x, x] = 0, x ∈ 𝔤 is a Lie algebra.

Any Leibniz algebra 𝔥 gives rise to a Lie algebra 𝔥_{Lie}, which is obtained as the quotient of 𝔥 by the relation [x, x] = 0. For a Lie algebra 𝔤 we put

𝔤\underline{⊗}𝔤 := (𝔤 ⊗ 𝔤)_{Lie}.

We prove that there exist a central extension of Lie algebras

0 → γ(𝔤) → 𝔤\underline{⊗}𝔤 → 𝔤 ⊠ 𝔤 → 0,

where 𝔤 ⊠ 𝔤 is the Lie algebra version of the nonabelian tensor product of Brown and Loday [1] constructed by Ellis [2], [3].

Among other things, we prove that the dimension of the abelian Lie algebra γ(𝔤) equals to dim_{U𝔤}(𝔤, 𝔤) provided 𝔤 is a finite dimensional semi-simple Lie algebra over a field of characteristic zero. In this case we also compute the Leibniz homology of 𝔤 ⊗ 𝔤.

1. Preliminaries on Leibniz algebras

It follows from Leibniz identity (2) that in any Leibniz algebra one has

[x, [y, y]] = 0, [x, [y, z]] + [x, [z, y]] = 0.  (3)

Let 𝔤 be a Leibniz algebra. A subspace 𝔥 ⊂ 𝔤 is called left (resp. right) ideal if for any a ∈ 𝔥 and x ∈ 𝔤 one has [x, a] ∈ 𝔥 (resp. [a, x] ∈ 𝔥). If 𝔥 is both left and right ideal, then 𝔥 is called two-sided ideal. For a Leibniz algebra 𝔤 one puts

Z^{r}(𝔤) = {a ∈ 𝔤 | [x, a] = 0, x ∈ 𝔤}.

Moreover, we let 𝔤^{ann} be the subspace of 𝔤 spanned on elements of the form [x, x], x ∈ 𝔤. Clearly for any x, y ∈ 𝔤 one has

[x, y] + [y, x] ∈ 𝔤^{ann}.

By (3) one has also

𝔤^{ann} ⊂ Z^{r}(𝔤).

Lemma 1.1. Z^{r}(𝔤) and 𝔤^{ann} are two sided ideals of 𝔤. Moreover

[Z^{r}(𝔤), 𝔤] ⊂ 𝔤^{ann}.

Proof. Since 𝔤^{ann} ⊂ Z^{r}(𝔤) and [𝔤, Z^{r}(𝔤)] = 0 it suffices to show only the last inclusion. To prove this one observes that for any u ∈ Z^{r}(𝔤) and x ∈ 𝔤 by definition one has [x, u] = 0 and so

[u, x] = [u, x] + [x, u] ∈ 𝔤^{ann}.

It is clear that the quotient

𝔤_{Lie} := 𝔤/𝔤^{ann}

is a Lie algebra, which satisfies the following universal property: any Leibniz homomorphism 𝔤 → 𝔥 into a Lie algebra 𝔥 factors trough 𝔤_{Lie}. Since 𝔤^{ann} ⊂ Z^{r}(𝔤) we see that

𝔤^{Lie} := 𝔤/Z^{r}(𝔤)

is also a Lie algebra. Thus by definition one has a central extension of Lie algebras

0 → 𝔤^{a} → 𝔤_{Lie} → 𝔤^{Lie} → 0,

where 𝔤^{a} = Z^{r}(𝔤)/𝔤^{ann}.

The lower central series or descending central series of 𝔤 is the sequence

⋯ ⊂ C^{n}𝔤 ⊂ ⋯ ⊂ C^{2}𝔤 ⊂ C^{1}𝔤

of two sided ideals of 𝔤 defined inductively as follows

C^{1}𝔤 = 𝔤 and C^{n+1}𝔤 = [C^{n}𝔤 , 𝔤], n > 0.

A Leibniz algebra 𝔤 is called nilpotent if C^{n}𝔤 = 0 for some n. Let n be the smallest integer such that C^{n}𝔤 = 0. Then n − 1 is called class of nilpotency of 𝔤. It is clear that 𝔤 is nilpotent of class n − 1 iff for any x_{1}, ⋯, x_{n} ∈ 𝔤 one has

[x_{1}, [x_{2}, ⋯, [x_{n−1}, x_{n}] ⋯] = 0.

In this case [⋯ [x_{1}, x_{2}], x_{3}], ⋯, x_{n}] = 0 as well.

The derived series of a Leibniz algebra 𝔤 is the sequence

⋯ ⊂ D^{n}𝔤 ⊂ ⋯ ⊂ D^{2}𝔤 ⊂ D^{1}𝔤

of two sided ideals of 𝔤 defined inductively as follows

D^{1}𝔤 = 𝔤 and D^{n+1}𝔤 = [D^{n}𝔤 , D^{n}𝔤].

It is clear that, then D^{m}𝔤 ⊂ C^{m}(𝔤) for any m. A Lie algebra 𝔤 is called solvable if D^{n}𝔤 = 0 for some n.

Examples 1.2. As was observed in [10] if 𝔤 is a Lie algebra, M is a (right) 𝔤-module and

f : M → 𝔤

is a 𝔤-homomorphism from M to the adjoint representation of 𝔤, then

[m_{1}, m_{2}] := [m_{1}, f(m_{2})], m_{1}, m_{2} ∈ M  (4)

defines a Leibniz algebra structure on M. We let Le(f) be this particular Leibniz algebra.

We recall also (see [10]) that in this way one gets any Leibniz algebra. More precisely, by (3) the bracket

[−, −] : 𝔤 ⊗ 𝔤 → 𝔤,

factors through the map

[−, −] : 𝔤 ⊗ 𝔤_{Lie} → 𝔤,

and hence yields a 𝔤_{Lie}-module structure on 𝔤. Thus the projection p : 𝔤 → 𝔤_{Lie} is a 𝔤_{Lie}-homomorphism and it follows that one has an isomorphism of Leibniz algebras 𝔤 ≅ Le(p). We also recall that (see [9]) the class ch(𝔤) of the extension

0 → 𝔤^{ann} → 𝔤 → 𝔤_{Lie} → 0

in Ext_{U(𝔤_{Lie})}^{1}(𝔤_{Lie}, 𝔤^{ann}) = H^{1}(𝔤_{Lie}, Hom(𝔤_{Lie}, 𝔤^{ann})) is called the characteristic element of 𝔤. Here U(𝔤_{Lie}) is the classical universal enveloping algebra of the Lie algebra 𝔤_{Lie} and H^{∗} denotes the Lie algebra cohomology. It follows that the triple (𝔤_{Lie}, 𝔤^{ann}, ch(𝔤)) characterize 𝔤 completely.

Let 𝔤 be a Leibniz algebra. We let δ_{n} : 𝔤^{⊗n} → 𝔤^{⊗n−1} be a linear map given by

δ_{n}(x_{1} ⊗ ⋯ ⊗ x_{n}) =

∑_{i<j}(−1)^{n−j}x_{1} ⊗ ⋯ ⊗ [x_{i}, x_{j}] ⊗ ⋯ ⊗ x̂_{j} ⊗ ⋯ ⊗ x_{n}.

For any n ≥ 0 the vector space 𝔤^{n} has a natural 𝔤_{Lie}-module structure, which is defined by

[x_{1} ⊗ ⋯ ⊗ x_{n}, x] = ∑_{i} x_{1} ⊗ ⋯ ⊗ [x_{i}, x] ⊗ ⋯ ⊗ x_{n}.

One observes that δ_{n} is a 𝔤_{Lie}-homomorphism (see also [8]). Note that δ_{3} : 𝔤 ⊗ 𝔤 ⊗ 𝔤 → 𝔤 ⊗ 𝔤 is given by

δ_{3}(x ⊗ y ⊗ z) = −[x, y] ⊗ z + [x, z] ⊗ y + x ⊗ [y, z]

and δ_{2} = [−, −] : 𝔤⊗𝔤 → 𝔤. By [8] δ_{n−1}δ_{n} = 0. So one has the complex

⋯ → 𝔤^{⊗n} ⟶^{δ_{n}} 𝔤^{⊗n−1} → ⋯ → 𝔤 ⊗ 𝔤 → 𝔤 → k.

Homology of this complex are called Leibniz homology of 𝔤 and it is denoted by HL_{∗}𝔤. If 𝔤 is a Lie algebra, then clearly HL_{1}𝔤 = H_{1}𝔤.

Let us return to the homomorphism f : M → 𝔤 of 𝔤-modules. Then one has the inclusions:

Le(f)^{ann} ⊂ Ker(f) ⊂ Z^{r}(Le(f)).

Thus

0 → Ker(f)/Le(f)^{ann} → Le(f)_{Lie} → Im(f) → 0

and

0 → Z^{r}(Le(f))/Ker(f) → Im(f) → Le(f)^{Lie} → 0

are central extensions of Lie algebras. One observes that

Z^{r}(Le(f)) = {a ∈ M | [x, fa] = 0, x ∈ M}.

Let us also note that

0 → Ker(f) → Le(f) → Im(f) → 0

is an abelian extension of Leibniz algebras. Moreover [Le(f), Ker(f)] = 0, in other words Ker(f) is an antisymmetric representation of Im(f) in the sense of [9].

2. Elementary properties of 𝔤 ⊗ 𝔤

Let 𝔤 be a Lie algebra. As was mentioned in the previous section 𝔤 ⊗ 𝔤 is a 𝔤-module via

[a ⊗ b, x] = [a, x] ⊗ b + a ⊗ [b, x].

Then one easily observes that the commutator map

[−, −] : 𝔤 ⊗ 𝔤 → 𝔤

is 𝔤-linear. Thus by the previous section the tensor product 𝔤 ⊗ 𝔤 carries the Leibniz algebra structure. One observes that this structure is nothing else but one given by (1).

Lemma 2.1. The map given by

x ⊗ y ↦ −y ⊗ x

is an automorphism of the Leibniz algebra 𝔤 ⊗ 𝔤.

The proof is obvious.

Lemma 2.2. Let 𝔤 be a nilpotent Lie algebra, then 𝔤 ⊗ 𝔤 is a nilpotent Leibniz algebra. More precisely, if the class of nilpotency of 𝔤 is ≤ 2n, then the class of nilpotency of 𝔤 ⊗ 𝔤 is ≤ 2n − 1.

Proof. Clearly

[C^{i}𝔤 ⊗ C^{j}𝔤, C^{k}𝔤 ⊗ C^{l}𝔤] ⊂ C^{i+k+l}𝔤 ⊗ C^{j}𝔤 + C^{i}𝔤 ⊗ C^{j+k+l}𝔤.

It follows that [C^{i}𝔤⊗C^{j}𝔤, 𝔤⊗𝔤] ⊂ C^{i+2}𝔤⊗C^{j}𝔤+C^{i}𝔤⊗C^{j+2}𝔤. By induction we get

C^{m}(𝔤 ⊗ 𝔤) ⊂ ∑_{i+j=m−1} C^{2i+1}𝔤 ⊗ C^{2j+1}𝔤.

If C^{2n+1}𝔤 = 0 and i+j = 2n−1, then 2i+1 ≥ 2n+1 or 2j +1 ≥ 2n+1 and therefore C^{2n}(𝔤 ⊗ 𝔤) = 0.

Lemma 2.3. Let 𝔤 be a solvable Lie algebra of class m, then 𝔤 ⊗ 𝔤 is a solvable Leibniz algebra of class m. More generally, if f : M → 𝔤 is a 𝔤-homomorphism into the adjoint representation of a Lie algebra 𝔤 with property f(M) ⊂ D^{k}𝔤 and D^{n}𝔤 = 0. Then D^{n+2−k}(Le(f)) = 0.

Proof. The first part indeed follows from the second one. One needs to take M = 𝔤 ⊗ 𝔤, k = 2 and f to be the commutator map. Since f : Le(f) → 𝔤 is a homomorphism of Leibniz algebras and f(Le(f)) ⊂ D^{k}𝔤 we have f(D^{m}(Le(f)) ⊂ D^{k+m−1}𝔤. Hence

f(D^{n−k+1}Le(f)) ⊂ D^{n}𝔤 = 0.

By definition we have

D^{n+2−k}(Le(f)) = [D^{n+1−k}(Le(f)), D^{n+1−k}(Le(f))] =
= [D^{n−k+1}(Le(f)), f(D^{n−k+1}Le(f))] = 0.

Remark 2.4. We leave as an exercise to the interested reader to show that, if 𝔤 is a Leibniz algebra, then the bracket (1) still makes 𝔤 ⊗ 𝔤 into Leibniz algebra, as well as the facts that Lemma 2.1, Lemma 3.1, Lemma 3.2, Corollary 3.3 and the fact that sequence (5) is exact are still true in this generality.

3. Relation with nonabelian tensor product

Lemma 3.1. The image of the map δ_{3} : 𝔤⊗𝔤⊗𝔤 → 𝔤⊗𝔤 is an abelian two-sided ideal in 𝔤 ⊗ 𝔤. Moreover Im(δ_{3}) ⊂ Z^{r}(𝔤 ⊗ 𝔤).

Proof. Im(δ_{3}) is a right ideal because δ_{3} is a 𝔤-homomorphism. The inclusion Im(δ_{3}) ⊂ Z^{r}(𝔤 ⊗ 𝔤) follows from the fact that [−, −] ∘ δ_{3} = 0. This equality implies that Im(δ_{3}) is also a left ideal.

Thanks to 3.1 the vector space

𝔤 ∗ 𝔤 = Coker(δ_{3} : 𝔤 ⊗ 𝔤 ⊗ 𝔤 → 𝔤 ⊗ 𝔤)

has the Leibniz algebra structure. We let x ∗ y be the image of x ⊗ y ∈ 𝔤 ⊗ 𝔤 into 𝔤 ∗ 𝔤. Since [x, y] ∗ z = [x, z] ∗ y + x ∗ [y, z], we see that

[x ∗ y, a ∗ b] = [x, y] ∗ [a, b].

Therefore 𝔤∗𝔤 is a variation of the ”non-abelian Leibniz tensor product” given in [5].

By the definition of Leibniz homology one has an exact sequence of Leibniz algebras

0 → HL_{2}𝔤 → 𝔤 ∗ 𝔤 ⟶^{[−,−]} 𝔤 → HL_{1}𝔤 → 0.  (5)

Here HL_{2}𝔤 and HL_{1}𝔤 are abelian Lie algebras. Moreover one can show that HL_{2}𝔤 is a central subalgebra of 𝔤 ∗ 𝔤.

As usual we let Γ(𝔤) be the subspace of 𝔤 ⊗ 𝔤 spanned by symmetric tensors x ⊗ x, x ∈ 𝔤. Clearly x ⊗ y + y ⊗ x ∈ Γ(𝔤), x, y ∈ 𝔤.

Lemma 3.2. Γ(𝔤) is an abelian two-sided ideal of 𝔤 ⊗ 𝔤. Moreover Γ(𝔤) ⊂ Z^{r}(𝔤 ⊗ 𝔤).

Proof. Take b = a in (1) and use (3) to get [x ⊗ y, a ⊗ a] = 0. This shows Γ(𝔤) ⊂ Z^{r}(𝔤 ⊗ 𝔤) and hence Γ(𝔤) is a left ideal. If one puts y = x in (1) one gets that [x ⊗ x, a ⊗ b] = z ⊗ x + x ⊗ z ∈ Γ(𝔤), where z = [x, [a, b]].

Corollary 3.3. One has an abelian extension of Leibniz algebras

0 → Γ(𝔤) → 𝔤 ⊗ 𝔤 → 𝔤 ∧ 𝔤 → 0,

where 𝔤 ∧ 𝔤 is a Leibniz algebra under the bracket

[x ∧ y, a ∧ b] := [x, [a, b]] ∧ y − [y, [a, b]] ∧ x.

Let Γ̃(𝔤) be the subspace of 𝔤 ∗ 𝔤 spanned by the images of the elements

x ⊗ [y, z] + [y, z] ⊗ x ∈ Γ(𝔤) ⊂ 𝔤 ⊗ 𝔤, x, y, z ∈ 𝔤

and [u, v] ⊗ [u, v], u, v ∈ 𝔤 under the homomorphism 𝔤 ⊗ 𝔤 → 𝔤 ∗ 𝔤.

Lemma 3.4. One has inclusions (𝔤 ∗ 𝔤)^{ann} ⊂ Γ̃(𝔤) ⊂ Z^{r}(𝔤 ∗ 𝔤). Hence Γ̃(𝔤) is a two-sided ideal of 𝔤 ∗ 𝔤.

Proof. In 𝔤 ∗ 𝔤 we have [x ∗ y, x ∗ y] = [x, y] ∗ [x, y], which implies the first inclusion. The second one follows from the previous lemma.

Corollary 3.5. The quotient 𝔤 ⊠ 𝔤 := (𝔤 ∗ 𝔤)/Γ̃(𝔤) is a Lie algebra.

Remark 3.6. The Lie algebra 𝔤 ⊠ 𝔤 is nothing else but the ”nonabelian tensor product of Lie algebras” as it is defined in [3], while the quotient 𝔤 ∗ 𝔤 by the image of Γ(𝔤) under the projection 𝔤 ⊗ 𝔤 → 𝔤 ∗ 𝔤 is nothing else but the ”second non-abelian exterior power of Lie algebras” as it is defined in [2]. We let 𝔤 ⋏ 𝔤 be this quotient.

Summary above statements we have the following

Proposition 3.7. The underling vector space of the non-abelian tensor product 𝔤 ⊠ 𝔤 is isomorphic to the quotient of the tensor product 𝔤 ⊗ 𝔤 by the subspace generated by the elements

x ⊗ [y, z] + [y, z] ⊗ x,
[x, y] ⊗ [x, y],
[x, y] ⊗ z + [y, z] ⊗ x + [z, x] ⊗ y,

where x, y, z ∈ 𝔤.

Comparing the definitions one sees that there is a commutative diagram of Leibniz algebras

𝔤 ⊗ 𝔤 ⟶ 𝔤 ∗ 𝔤 ⟶ 0
↓  ↓
𝔤 ∧ 𝔤 ⟶ 𝔤 ⋏ 𝔤 ⟶ 0,

which implies that the sequence

𝔤 ∧ 𝔤 ∧ 𝔤 ⟶^{δ_{3}} 𝔤 ∧ 𝔤 → 𝔤 ⋏ 𝔤 → 0

is also exact. Thus one has also the following diagram

0 ⟶ HL_{2}𝔤 ⟶ 𝔤 ∗ 𝔤 ⟶^{[−,−]} 𝔤 ⟶ HL_{1}𝔤 ⟶ 0
↓  ↓  ∥  ∥
0 ⟶ H_{2}𝔤 ⟶ 𝔤 ⋏ 𝔤 ⟶^{[−,−]} 𝔤 ⟶ H_{1}𝔤 ⟶ 0,

compare also with [2] and [3].

Let us also recall that there are exact sequences

0 → Γ(H_{1}𝔤) → 𝔤 ⊠ 𝔤 → 𝔤 ⋏ 𝔤 → 0,  (6)

HL_{3}𝔤 → H_{3}𝔤 ⟶^{α} H_{0}(𝔤, Γ(𝔤)) → HL_{2}𝔤 → H_{2}𝔤 → 0.

The first one is constructed in [3], while the second one in [11]. Let us recall that the homomorphism α is the dual of the classical homomorphism of Koszul (see [6])

{Invariant quadratic forms on 𝔤} → H_{3}𝔤.

For a Lie algebra 𝔤 we let 𝔤\underline{⊗}𝔤 be the Liezation of 𝔤 ⊗ 𝔤, that is

𝔤\underline{⊗}𝔤 := (𝔤 ⊗ 𝔤)_{Lie}.

Since 𝔤⊠𝔤 is also a Lie quotient of 𝔤⊗𝔤, we have a canonical surjective homomorphism

p̄: 𝔤\underline{⊗}𝔤 → 𝔤 ⊠ 𝔤.

we let γ(𝔤) be the kernel of p̄. Since γ(𝔤) is contained in the image of

Ker([−, −] : 𝔤 ⊗ 𝔤 → 𝔤)

under the canonical projection 𝔤 ⊗ 𝔤 → 𝔤\underline{⊗}𝔤, we see that

0 → γ(𝔤) → 𝔤\underline{⊗}𝔤 ⟶^{p̄}𝔤 ⊠ 𝔤 → 0  (7)

is a central extension of Lie algebras. Here p is the canonical projection 𝔤 ⊗ 𝔤 → 𝔤 ⊠ 𝔤.

Lemma 3.8. If 𝔤 is a perfect Lie algebra, then p̄ has a section, thus 𝔤\underline{⊗}𝔤 is a product of 𝔤 ⊠ 𝔤 and an abelian Lie algebra γ(𝔤).

Proof. If H_{1}𝔤 = 0, then (6) shows that 𝔤 ⊠ 𝔤 ≅ 𝔤 ⋏ 𝔤. Thus one has an universal central extension (compare with [2] and [3])

0 → H_{2}𝔤 → 𝔤 ⊠ 𝔤 → 𝔤 → 0

and therefore 𝔤 ⊠ 𝔤 is a super perfect Lie algebra (that is H_{1}(𝔤 ⊠ 𝔤) = H_{2}(𝔤 ⊠ 𝔤) = 0 see [4]) and p̄ has a section.

Let us recall that D^{2}𝔤 denotes the commutator subalgebra [𝔤.𝔤] of 𝔤.

Lemma 3.9. [Γ𝔤, D^{2}𝔤] ⊂ (𝔤 ⊗ 𝔤)^{ann}

Proof. Take a ∈ 𝔤 and x = [b, c] ∈ D^{2}𝔤. Then

[a ⊗ a, x] = [a ⊗ a, b ⊗ c] =
= [a ⊗ a, b ⊗ c] + [b ⊗ c, a ⊗ a] ∈ (𝔤 ⊗ 𝔤)^{ann}.

Lemma 3.10. There is a well-defined homomorphism

H_{0}(D^{2}𝔤, Γ𝔤) → 𝔤\underline{⊗}𝔤.

Moreover, the image of the composite H_{0}(D^{2}𝔤, ΓD^{2}𝔤) → H_{0}(D^{2}𝔤, Γ𝔤) → 𝔤\underline{⊗}𝔤 lies in γ(𝔤).

Proof. Since H_{0}(D^{2}𝔤, Γ𝔤) ≅ (Γ𝔤)/[Γ𝔤, D^{2}𝔤] Lemma 3.9 shows that the composite

Γ𝔤 → 𝔤 ⊗ 𝔤\underline{⊗}

factors trough H_{0}(D^{2}𝔤, Γ𝔤) and the expected homomorphism indeed exists. Thanks to Proposition 3.7 the image of [x, y] ⊗ [x, y] into 𝔤 ⊠ 𝔤 is zero and we obtain the second part of Lemma.

Thus we obtain the homomorphism

β : H_{0}(D^{2}𝔤, ΓD^{2}𝔤) → γ(𝔤).

We let

ᾱ : H_{3}(D^{2}𝔤) → γ(𝔤)

be the composite of α_{D^{2}𝔤} with β. We recall that

α_{𝔤} : H_{3}(𝔤) → H_{0}(𝔤, Γ𝔤)

was defined in (6).

4. 𝔤 ⊗ 𝔤 for a semisimple Lie algebra 𝔤

Lemma 4.1. Let 𝔤 be a perfect Lie algebra. Then (𝔤 ⊗ 𝔤)^{ann} is a 𝔤submodule of 𝔤 ⊗ 𝔤.

Proof. Let us recall that for any Leibniz algebra 𝔥 the vector space 𝔥^{ann} is a 𝔥_{Lie}-submodule of 𝔥. We take h = 𝔤 ⊗ 𝔤. Then (1) shows that the action of 𝔥_{Lie} = 𝔤\underline{⊗}𝔤 on 𝔤 ⊗ 𝔤 factors trough the commutator map 𝔤\underline{⊗}𝔤 → 𝔤. By the assertion the last map is an epimorphism and hence the result.

In the rest of this section we assume that 𝔤 is a finite dimensional semisimple Lie algebra over a field of characteristic zero. By the classical results of Whitehead 𝔤 is a superperfect Lie algebra, that is H_{1}𝔤 = H_{2}𝔤 = 0. Thanks to [11] in fact HL_{i}𝔤 = 0 for all i > 0. It is a well known fact that the category of finite dimensional 𝔤-modules is a semi-simple abelian category. As a consequence any such module M has a functorial decomposition M = M^{t}⊕M^{nt}, where M^{t} is a 𝔤-module with trivial 𝔤-action, while M^{nt} is a 𝔤-submodule with H^{0}(𝔤, M^{nt}) = 0. Clearly M^{t} = H^{0}(𝔤, M).

Lemma 4.2. Let f : M → 𝔤 be a 𝔤-homomorphism from a finite dimensional 𝔤-module to the adjoint representation of a finite dimensional semisimple Lie algebra 𝔤. Then (Le(f))^{ann} ⊂ M^{nt}. Moreover, if f^{t} is the restriction of f on M^{t}, then the composite

M^{t} ⊂ M → Le(f)_{Lie}

yields an isomorphism from Le(f^{t}) to a central subalgebra of Le(f)_{Lie}.

Proof. Since 𝔤^{t} = 0 and f is a 𝔤-homomorphism it follows that f(M^{t}) = 0. Thus M^{t} ⊂ Z^{r}(Le(f)). Take a, b ∈ M and write a = a^{t} + a^{nt}, b = b^{t} + b^{nt} according to the decomposition M = M^{t} ⊕ M^{nt}. Then f(a) = f(a^{nt}) and f(b) = f(b^{nt}). Hence

[a, b] + [b, a] = [a, f(b)] + [b, f(a)] = [a, f(b^{nt})] + [b, f(a^{nt})] =
= [a^{nt}, f(b^{nt})] + [b^{nt}, f(a^{nt})] ∈ M^{nt}.

Since (Le(f))^{ann} is spanned by elements of the form [a, b] + [b, a] we see that (Le(f))^{ann} ⊂ M^{nt}. Therefore (Le(f))^{ann} ∩ M^{t} = 0 and hence the restriction of the canonical projection Le(f) → Le(f)_{Lie} to M^{t} is a monomorphism and the result follows.

For any finite dimensional simisimple Lie algebra 𝔤 we put

r(𝔤) := dimHom_{U𝔤}(𝔤, 𝔤).

The following is well-known

Lemma 4.3. Let 𝔤 be a finite dimensional semisimple Lie algebra. Then dimH^{0}(𝔤, 𝔤 ⊗ 𝔤) = r(𝔤). Moreover, if 𝔤 is a simple Lie algebra over algebraicaly closed field, then r(𝔤) = 1.

Proof. The Killing form yields an isomorphism 𝔤^{∗} ≅ 𝔤 of U𝔤-modules and hence 𝔤 ⊗ 𝔤 ≅ Hom(𝔤, 𝔤). It follows that

H^{0}(𝔤, 𝔤 ⊗ 𝔤) ≅ Hom_{U𝔤}(𝔤, 𝔤).

The last assertion follows from the Schur lemma, because the adjoint representation of a simple Lie algebra is a simple 𝔤-module.

Lemma 4.4. Let f : M → 𝔤 be a 𝔤-homomorphism to the adjoint representation of a Lie algebra 𝔤. Assume M ≅ M_{1} ⊕M_{2} as 𝔤-modules, f(M_{1}) = 0 and the action of 𝔤 on M_{1} is trivial. Then Le(f) as a Leibniz algebra is isomorphic to the product of the abelian Leibniz algebra M_{1} and Le(f_{2}), where f_{2} is the restriction of f on M_{2}.

The proof is obvious.

Theorem 4.5. Let 𝔤 be a finite dimensional semisimple Lie algebra. Then 𝔤 ⊗ 𝔤 as a Leibniz algebra is isomorphic to the direct product of an abelian Lie algebra of dimension r(𝔤) and a Leibniz algebra 𝔥. Moreover 𝔥_{Lie} ≅ 𝔤. Thus one has an isomorphism of Lie algebras

𝔤\underline{⊗}𝔤 ≅ k^{r(𝔤)} × 𝔤.

Proof. The first part is a consequence of Lemma 4.4. Indeed we take f = [−, −] : 𝔤 ⊗ 𝔤 → 𝔤 to be the commutator map and M_{1} = (𝔤 ⊗ 𝔤)^{t}, M_{2} = (𝔤 ⊗ 𝔤)^{nt}. Since 𝔤^{t} = 0, we have f(M_{1}) = 0. Therefore the conditions of Lemma 4.4 hold and hence

𝔤 ⊗ 𝔤 ≅ (𝔤 ⊗ 𝔤)^{t} × 𝔥,

where (𝔤 ⊗ 𝔤)^{t} is an abelian Lie algebra and h is a Leibniz algebra which is isomorphic to Le((𝔤 ⊗ 𝔤)^{nt} → 𝔤). By Lemma 4.3 we know that (𝔤 ⊗ 𝔤)^{t} ≅ k^{r(𝔤)}. Thus we proved the first part of the theorem. Since 𝔤 is superperfect, we have 𝔤 ⊠ 𝔤 ≅ 𝔤 and by Lemma 3.8 𝔤\underline{⊗}𝔤 is a product of the Lie algebra 𝔤 and an abelian Lie algebra 𝔞. We have to prove that 𝔞 ≅ k^{r(𝔤)}. By Lemma 4.1 𝔤\underline{⊗}𝔤 is a 𝔤-module and the quotient map 𝔤 ⊗ 𝔤 → 𝔤\underline{⊗}𝔤 is a 𝔤-homomorphism. Hence it yields an epimorphism

(𝔤 ⊗ 𝔤)^{t} → (𝔤\underline{⊗}𝔤)^{t},

which is also a monomorphism thanks to Lemma 4.2. It follows from Lemma 4.3 that (𝔤 ⊗ 𝔤)^{t} ≅ k^{r(𝔤)} and hence the result.

Remark. The same proof shows that if 𝔤 is a semisimple, then

(𝔤 ∧ 𝔤)_{Lie} ≅ 𝔤.

Lemma 4.6. Let 𝔤 be a finite dimensional semisimple Lie algebra. Then ch(𝔤 ⊗ 𝔤) = 0 and the map

ᾱ : H_{3}(D^{2}𝔤) → γ(𝔤)

is an isomorphism.

Proof. We know that 𝔤 ⊗ 𝔤 ≅ k^{r(𝔤)} × 𝔥, with 𝔥_{Lie} ≅ 𝔤. Thus ch(𝔤 ⊗ 𝔤) is the pullback of ch(𝔥) along the projection 𝔤 ⊗ 𝔤 → 𝔥. Since ch(𝔥) ∈ H^{1}(𝔤, 𝔥^{ann}) = 0 we obtain ch(𝔤 ⊗ 𝔤) = 0 as well. To proof the second assertion, one observes that for a semisimple Lie algebra 𝔤, we have D^{2}𝔤 = 𝔤 and the map α is an isomorphism thanks to [11], thus the result follows from 4.5.

Corollary 4.7. Let 𝔤 be a finite dimensional semisimple Lie algebra. Then

HL_{n}(𝔤 ⊗ 𝔤) ≅ k^{r(𝔤)×n}, n > 0.

Proof. We know that 𝔤⊗𝔤 ≅ k^{r(𝔤)}×𝔥. By Proposition 4.3 [11] we know that HL_{n}𝔥 = 0 for n > 0 and HL_{0}𝔥 ≅ k. Thus the result follows from the Loday-Künneth theorem for Leibniz homology [7].

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