A semisimple Lie algebra ${\mathfrak{g}}$ is one that has no nonzero abelian ideals. This is equivalent to the absence of solvable ideals. Indeed, if ${\mathfrak{g}}$ had a solvable ideal ${\mathfrak{z}}$, we could consider the derived series of ${\mathfrak{z}}$, i.e. ${D^1\mathfrak{z} = [\mathfrak{z},\mathfrak{z}], D^n\mathfrak{z} = [D^{n-1}\mathfrak{z}, D^{n-1}\mathfrak{z}]}$. These are ideals because, by the Jacobi identity, the Lie product of two ideals is an ideal. These ${D^n \mathfrak{z}}$ eventually become zero by the hypothesis of solvability, and the last nonzero one is abelian.

One justification for the epithet “semisimple” is that the category of finite-dimensional representations is in fact semisimple, i.e. that any exact sequence of ${\mathfrak{g}}$ representations for ${\mathfrak{g}}$ semisimple $\displaystyle 0 \rightarrow V' \rightarrow V \rightarrow V'' \rightarrow 0$

splits. This is what happens for finite groups, because by Maschke’s theorem the group algebra of a finite group is semisimple. Nevertheless, the enveloping algebra ${U\mathfrak{g}}$ is not generally semisimple; we are restricting ourselves to finite-dimensional ${U \mathfrak{g}}$-modules.

Cartan’s criterion

Before getting there, we will prove a basic criterion for semisimplicity.

Theorem 1 (Cartan) The Lie algebra ${\mathfrak{g}}$ is semisimple if and only if its Killing form is nondegenerate.

The proof turns out to be a relatively easy consequence of Cartan’s criterion for solvability, which I’ve already given a twopost spiel on. (more…)