Semisimple implies complete reducibility

Question

Why does a semisimple Lie algebra imply complete reducibility?

I have that a semisimple Lie algebra is a Lie algebra with no non-zero solvable ideals. Complete reducibility means that every invariant subspace of a representation has a complement.

But I do not understand why the Lie algebra must be semisimple to have complete reducibility.

Answer 1

As Qiaochu points out, this only holds in the finite dimensional case and the proof is somewhat involved. I think it's better to give a reference here than try and repeat a long argument you could find in any text on the subject:

• See Chapter II.6 of Humphrey's book Introduction to Lie algebras and representation theory. In particular, Theorem 6.3 of that section is what you're looking for.

Answer 2

The main algebraic reason behind Weyl's theorem on complete reducibility is Whitehead's first lemma, i.e., that the first Lie algebra cohomology with finite-dimensional modules as coefficients vanishes for finite-dimensional semisimple Lie algebras of characteristic zero.

It is easy to see that we need semisimple for this result. A first step is to see that a semisimple Lie algebra has only inner derivations, under the above assumptions.

The property of being semisimple is very important also for the classification theory and for the representation theory of Lie algebras.