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I'm following a book in Humphrey's Introdutction to Lie Algebra's and Representation Theory. I'm reading the proof that a semisimple Lie algebra is the direct sum of simple modules.

It uses the following observation without proof: If $L$ is a semisimple Lie algebra and $I \subset L$ is a subspace (it's actually an ideal here), then $\dim I + \dim I^\perp = \dim L$.

Why is that true?

This is ofcourse familiar if the Killing from is replaced by an inner product.

What exactly is needed from a bilinear form for this to be true? (Nondegenericity I guess, being symmetric too maybe)

Actually, in this proof it's enough to show that $\dim I + \dim I^\perp \geq \dim L$ so

If there's an easy proof of the $\geq$ direction it would be interseting too.

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What exactly is needed from a bilinear form for this to be true? (Nondegenericity I guess, being symmetric too maybe)

Try to show that if $V$ is a finite dimensional vector space equipped with a non-degenerate bilinear form $\langle -, - \rangle$, and $W$ is a subspace, then $W^{\perp} \cong (V/W)^*$ which comes from the restriction of the isomorphism $V \to V^*$ coming from the bilinear form.

In this situation, semisimplicity implies that the Killing form is non-degenerate, thus what I said applies.

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  • $\begingroup$ This works. Thanks. The hard part for me was to prove the the map $W^\perp \rightarrow (V/W)^*$ is onto. When I thought of a non-degenerate bilinear form as one represented by an invertible matrix, this part cleared out too. $\endgroup$ – Gils Jan 26 '13 at 12:24

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