Whitehead lemmas in Lie algebra cohomology for non-algebraically closed fields

Question

I read in Weibel's homological algebra that Whitehead's first and second lemmas are true for any characteristic 0 field. I mean the following:

Whitehead Lemma(s): Let g be a semisimple Lie algebra over a characteristic 0 field, V any finite-dimensional representation of g, then H^1(g, V) = H^2(g, V) = 0.

The proof uses

Theorem 7.8.9 : H^i(g, M) = 0 for any nontrivial irreducible representation M of g if g is semisimple over char 0 field.

The proof of this uses the Casimir operator. The Casimir operator is said to act by a scalar, but doesn't this assertion use Schur's lemma? Schur's lemma requires the field to be algebraically closed.

Whitehead's first lemma implies the complete reducibility of finite-dimensional g-modules, but isn't complete reducibility only true for semisimple Lie algebras over an algebraically closed field of char 0?

Whitehead's second lemma implies the Levi decomposition. Does Levi decomposition require the field to be algebraically closed?

Does anyone know where I can find a proof of the Whitehead Lemma(s) (assuming it is true) when the field is not assumed to be algebraically closed?

Answer 1

There is no problem about the Whitehead lemmas over an arbitrary field of characteristic 0 in the context of (1) complete reducibility of finite dimensional representations of a semisimple Lie algebra, (2) existence of a Levi decomposition of a finite dimensional Lie algebra. Besides the modern treatment by Weibel, there are various older treatments such as Bourbaki, Chapter 1 of Groupes et algebres de Lie (1960) and Jacobson, Sections III.7, III.9 of Lie Algebras (1962) followed by III.10 on Lie algebra cohomology. Along the way, one can make use of a Casimir element in the universal enveloping algebra: this is a commuting operator in the representations studied, which is what counts. (Of course, when the field is large enough and the representation is irreducible, Schur's Lemma provides a scalar, but this is a corollary of the general Schur formulation.)

For a treatment emphasizing highest weight representations, such as I gave in my short graduate text, it's not satisfactory to work over such an arbitrary field. At the least, one needs a splitting field for a semisimple Lie algebra; my convention, not needed for some of the generalities, was that the field be algebraically closed. So I may be partly responsible for confusing the issue of just how large the field needs to be at each step of the theory.

In particular, complete reducibility does not require that a Casimir element operate on a representation space by a scalar. However, irreducible representations over arbitrary fields of characteristic 0 might be larger than those over a splitting field. The Cartan-Weyl theory is best understood over a large enough field, where eigenvalues are available when needed.

P.S. I've just tracked down my copy of Weibel. Concerning his treatment of these topics, I've never gone through this in detail and find it difficult to sort out his proof of the Whitehead lemmas without going back into the earlier part of his book. He does want the Casimir element to act by scalars when needed even though the field is arbitrary of characteristic 0, but this argument doesn't seem to enter into the other more self-contained treatments mentioned above. Anyway, it's worth looking at Jacobson's Chapter III.

Answer 2

Cohomology does not change under field extensions, so just extend everything to the algebraic closure to prove your result in char 0 in general. Of course, field extensions do not preserve irreducibility, but they preserve semi-simplicity, and preserve the property of having a trivial factor in the decomposition into irreducibles. (That works if you allow to use the complete reducibility in the algebraically closed case, not if you want to prove it using Whitehead's lemma.)