## Can Lie algebra cohomology prove Cartan’s Semisimplicity Criterion?

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Here is what I mean by "Cartan's semisimplicity criterion":

Let $\mathfrak g$ be a finite-dimensional Lie algebra over a field of characteristic $0$. Assume that the center of $\mathfrak g$ is trivial. Then, the following three assertions are equivalent:

(1) The Killing form on $\mathfrak g\times \mathfrak g$ is nondegenerate.

(2) Every short exact sequence of finite-dimensional representations of $\mathfrak g$ splits.

(3) Every subrepresentation of the adjoint representation of $\mathfrak g$ has a complementary subrepresentation.

What I am looking for is a slick proof for this equivalence (although the only thing I really need is a proof of (1) $\Longrightarrow$ (3)). I am aware of the proof in Fulton-Harris Appendix C, but this could fill an hour of talking and seems to involve many unmotivated ideas. Is there something more explanatory? Using cohomology perhaps? Is the whole thing obvious from an advanced viewpoint? (I don't mean using the classification of simple Lie algebras, of course...) Maybe newer ideas such as Lie algebroids, algebraic groups etc. can help?

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Presumably you've looked at the treaments in Weibel chapter 6 or 7(the one on Lie algebra cohomology), or the proof in Serre's book "Lie algebras and Lie groups". – Daniel Pomerleano Jan 10 2011 at 0:27

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Throughout you mean finite dimensional reps. – Mariano Suárez-Alvarez Jan 10 2011 at 2:19

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By the way, this is false as stated. Abelian lie algebras satisfy (3). – Ben Webster♦ Jan 10 2011 at 2:31

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Yes, it seems like there is the implicit assumption that $\mathfrak{g}$ has trivial center. – Keerthi Madapusi Pera Jan 10 2011 at 3:41

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Cohomology is implicit in the algebraic version of Weyl's complete reducibility theorem (see Jacobson's 1962 book or Weibel), but not strictly needed. The quadratic Casimir element is crucial, though you can avoid universal enveloping algebras here. The efficient algebraic argument by Brauer is used in Bourbaki Groupes et algebres de Lie (Chap. 1), Serre's lectures, my book, Fulton & Harris, etc. What is "simplest" depends a lot on what you already know. Weyl's original proof for compact groups is the most transparent step beyond Maschke's theorem for finite groups. – Jim Humphreys Jan 10 2011 at 14:51
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## 2 Answers

My favorite proof is showing that the trivial module is projective in the category $\mathcal R$ of finite dimensional reps; you can do this using cohomology and the quadratic Casimir (so here the Killing form comes in). After that, showing that all fin. dim. reps. are semisimple amounts to showing that they are all projective in $\mathcal R$, and this follows from the identity $\hom_\mathfrak g(M,N)=\hom_\mathfrak g(k,\hom(M,N))$ and a little cohomological yoga.

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The fact that the trivial module is the maximally complicated one has always marveled me. – Mariano Suárez-Alvarez Jan 10 2011 at 3:39

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I think once you internalize the translation principle, it suddenly becomes the most natural thing in the world. – Ben Webster♦ Jan 10 2011 at 4:54

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But it is a more general phenomenon than that: the same thing happens for local algebras, where the residue ring contains all the complexity, for example, or non-negatively graded ones. – Mariano Suárez-Alvarez Jan 10 2011 at 4:58

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I would argue that that's completely different; that's just because it's the only simple module. In the Lie algebra situation, being able to tensor gets much more out of one simple than you have any right to. – Ben Webster♦ Jan 10 2011 at 5:45

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My favorite proof is the Harish-Chandra isomorphism.

Choose the PBW isomorphism $U(\mathfrak g)\cong U(\mathfrak n_-)\otimes U(\mathfrak h)\otimes U(\mathfrak n_+)$ and let $p: U(\mathfrak g)\to U(\mathfrak h)\cong \mathbb k[\mathfrak h^*]$ be the projection killing all higher order terms in the other factors. Then

Theorem. The restriction of $p$ to the center $Z(U(\mathfrak g))$ is an injection whose image is the invariant functions for the $\rho$-shifted Weyl group action $w\bullet \lambda =w(\lambda+\rho)-\rho$ on $\mathfrak h^*$.

This means that the center acts in the same way on two different highest weight representations if and only if $\lambda+\rho$ and $\mu+\rho$ are in the same orbit of the Weyl group (consider the action on the highest weight vector). This can't happen for two different dominant weight vectors ($\lambda+\rho$ is in the interior of the dominant Weyl chamber), so the action of the center distinguishes different simple representations.

Now we need to show that these can't glue to each other; you can easily reduce to the case of an extension $V\to W \to V$ for a simple $V$. In this case, just pick a splitting on the highest weight space, and let $V'\subset W$ be the sub-rep generated by the preimage of the highest weight in the quotient copy of $V$. By PBW, the intersection of this with the highest weight space is 1-dimensional, so it cannot have any intersection with the sub-module, but it maps surjectively to the quotient. Thus it is a complement and the sequence splits.

I think this shows that there really is something delicate about this semi-simplicity; this isn't the lowest tech proof, but they're always a little subtle. You can also see this from how wildly semi-simplicity fails in the infinite-dimensional case; look at some literature on Verma modules and category $\mathcal O$ to see how complicated this can get.

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