Journal of Lie Theory Volume 8 (1998) 429–431 C ◯1998 Heldermann Verlag

The Euler-Poincaré characteristic of a Lie algebra

Teimuraz Pirashvili ^{∗}

Communicated by J. Faraut

Abstract. We will show that the Euler-Poincaré characteristic of a finite dimensional Lie algebra is zero if the ground field has characteristic zero or the Lie algebra is not perfect.

Let V^{∗} = ⨁_{i≥0} V^{i} be a graded vector space over a field k. We assume that each V^{i} has finite dimension. We recall that Euler-Poincaré characteristic of V^{∗} is given by

χ(V^{∗}) := ∑_{i≥0}(−1)^{i} dim V^{i}.

Let g be a Lie algebra over k and M be a g-module. Assume dim g < ∞ and dim M < ∞. We let χ(g, M) denote the Euler-Poincaré characteristic of H^{∗}(g, M). When M = k with trivial action, we write χ(g) instead of χ(g, k). The goal of this paper is to prove the following result.

Theorem 1. Let g be a finite dimensional Lie algebra and M be a finite dimensional g-module. Assume that one of the following conditions hold

i) H_{1}(g) ≠ 0.

ii) char k = 0 and g ≠ 0.

Then χ(g, M) = 0.

For solvable Lie algebras this was proved before in [1]. In order to prove the theorem we need to make some simple observations.

Lemma 2. Let V^{∗} be a finite dimensional graded vector space, whose nonzero components are concentrated in odd degrees. Then

χ(Λ^{∗}(V^{∗})) = 0,

ISSN 0949–5932 / $2.50 ^{_{C} ◯} Heldermann Verlag where Λ^{∗} denotes the exterior algebra.

∗ Supported by GAS Grant 1.5.

Proof. Since Λ^{∗}(V^{∗} ⊕ W^{∗}) ≅ Λ^{∗}(V^{∗}) ⊗ Λ^{∗}(W^{∗}) and the Euler-Poincaré characteristic has multiplicative property, one can reduce the statement to the case where V^{∗} is concentrated in degree 2k + 1 and is one-dimensional. Then one has χ(Λ^{∗}(V^{∗})) = 1 + (−1)^{2k+1} = 0.

Lemma 3. If g is one dimensional Lie algebra and M is a finite dimensional g-module, then χ(g, M) = 0.

Proof. By definition of Lie algebra cohomology, one has an exact sequence

0 → H^{0}(g, M) → M → M → H^{1}(g, M) → 0

and H^{i}(g, M) = 0 for i > 1. Thus χ(g, M) = 0.

Lemma 4. If k has characteristic zero, g is a finite dimensional semi-simple Lie algebra and M is a finite dimensional g-module, then χ(g, M) = 0.

Proof. It is well known that H^{∗}(g) is an exterior algebra on odd degree generators, thus Lemma 2 gives that χ(g) = 0. Moreover, assume that M is a finite dimensional g-module. Then one has an isomorphism H^{∗}(g, M) ≅ H^{∗}(g) ⊗ H^{0}(g, M). Therefore

χ(g, M) = χ(g) (dim H^{0}(g, M)) = 0.

Lemma 5. Let

0 → a → g → h → 0

be a short exact sequence of finite dimensional Lie algebras and M be a finite dimensional g-module. Then

χ(g, M) = ∑_{i≥0}(−1)^{i}χ(h, H^{i}(a, M)).

Moreover, if the action of h on the cohomology H^{∗}(a, M) is trivial, then we have

χ(g, M) = χ(a, M)χ(h).

Proof. Since the Euler-Poincaré characteristic does not change after taking homology, the Hochschild-Serre spectral sequence gives:

χ(g, M) = χ(E_{2}^{∗∗}) = ∑_{p,q≥0}(−1)^{p+q} dim E_{2}^{pq}.

To finish the proof one remarks that

∑_{q≥0}(−1)^{q} ∑_{p≥0}(−1)^{p} dim H^{p}(h, H^{q}(a, M)) = ∑_{i≥0}(−1)^{i}χ(h, H^{i}(a, M)).

Proof of the Theorem. i) By assumption there exists an epimorphism from g into a nonzero abelian Lie algebra. On the other hand, from any nonzero abelian Lie algebra there exists an epimorphism onto a one dimensional Lie algebra. Hence there exists an exact sequence of Lie algebras

0 → a → g → h → 0

with dim h = 1. Now we can use Lemma 5 and Lemma 3 to finish the proof.

ii) By i) one can assume that g is not solvable. Let r be the radical of g and s be the factor-algebra g/r. By Lemma 5 one has χ(g, M) = ∑_{i≥0}(−1)^{i}χ(s, H^{i}(r, M)). Since s ≠ 0 is a semi-simple Lie algebra, it has zero Euler-Poincaré characteristic (by Lemma 4) and the result follows.

References

[1] Malliavin-Brameret, M.-P., Caractéristiques d’Euler-Poincaré d’algèbres de Lie rèsolubles, C. R. Acad. Sci. Paris Sér. A-B 284 (1977), A1487– A1488.

A. Razmadze Math. Inst. Alexidze str. 1, Tbilisi, 380093. Republic of Georgia. pira@mathematik.uni-bielefeld.de

Received February 23, 1998