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.

First, let’s suppose ${\mathfrak{g}}$ is not semisimple, i.e. has a nonzero abelian ideal ${\mathfrak{h}}$. This part of the proof is straightforward and elementary. Then given ${x \in \mathfrak{g}, y \in \mathfrak{h}}$, we have

$\displaystyle (\mathrm{ad} x)(\mathrm{ad} y) \mathfrak{g} \subset \mathfrak{h}, \quad (\mathrm{ad} x)(\mathrm{ad} y) \mathfrak{h} = \{ 0 \}$

by the hypothesis on ${\mathfrak{h}}$. So ${ (\mathrm{ad} x)(\mathrm{ad} y) }$ is nilpotent, of trace zero, and thus ${\mathfrak{h}}$ belongs to the kernel of the Killing form.

Now, assume ${\mathfrak{g}}$ is semisimple. This implies that the center of ${\mathfrak{g}}$ is zero, so the ${\mathrm{ad}}$-homomorphism ${\mathfrak{g} \rightarrow gl(\mathfrak{g})}$ is an injection. Let ${\mathfrak{z}}$ be the kernel of the Killing form ${B}$, i.e. ${z \in \mathfrak{z}}$ if and only if for all ${x \in \mathfrak{g}}$,

$\displaystyle B(x,z) = 0.$

By invariance of the Killing form, this is a Lie ideal in fact. Now ${\mathfrak{z} \subset gl(\mathfrak{g})}$ by the above identification, and if ${B_{\mathfrak{g}}}$ is the trace form on ${gl(\mathfrak{g})}$, we have

$\displaystyle B_{\mathfrak{g}}(\mathfrak{z}, \mathfrak{z}) = 0.$

By Cartan’s criterion (or, rather, Theorem 2 here), ${\mathfrak{z}}$ is solvable, so it must be zero, and the Killing form is thus nondegenerate.

The preceding discussion highlighted two reasons why semisimplicity is convenient: first, we get a canonical embedding in a ${gl_n}$ Lie algebra. (We can always get a noncanonical embedding of any Lie algebra (always assumed finite-dimensional) in some ${gl_n}$ by a hard theorem of Ado though.) Second, we have a very convenient nondegenerate and invariant bilinear form.

Simple decomposition

Semisimplicity also allows ${\mathfrak{g}}$ to be built up from simple pieces, where we call a Lie algebra simple if it is nonabelian and has no nontrivial ideals (except zero and itself). In particular, a simple Lie algebra ${\mathfrak{s}}$ satisfies ${\mathfrak{s} = [\mathfrak{s},\mathfrak{s}]}$.

Theorem 2 A semisimple Lie algebra ${\mathfrak{g}}$ is isomorphic to a direct sum ${\bigoplus_i \mathfrak{g}_i}$ where the ${\mathfrak{g}_i}$ are simple Lie algebras, determined uniquely.

First, we prove existence. The basic lemma is that if ${\mathfrak{h}}$ is an ideal in ${\mathfrak{g}}$, then we can choose an appropriate ideal ${\mathfrak{h}'}$ and write

$\displaystyle \mathfrak{g} = \mathfrak{h} \oplus \mathfrak{h}',$

where the direct sum is in the sense of Lie algebras. Indeed, we take ${\mathfrak{h}'}$ to be the orthogonal complement of ${\mathfrak{h}}$ under the Killing form. By invariance this is also an ideal, and we have a direct sum decomposition as vector spaces. Since ${[\mathfrak{h}, \mathfrak{h}']}$ is an ideal contained in both ${\mathfrak{h}, \mathfrak{h}'}$, it is zero and we have proved our “basic lemma.”

It is now clear how to prove existence of the direct sum decomposition of ${\mathfrak{g}}$ into simple components. If ${\mathfrak{g}}$ has no nontrivial ideals, we are done; if not we make a splitting decomposition as above and work inductively on the dimension of ${\mathfrak{g}}$.

For uniqueness, suppose we have collections of simple Lie algebras ${\mathfrak{g}_i, \mathfrak{g}_j'}$ and

$\displaystyle \mathfrak{g} = \bigoplus_i \mathfrak{g}_i \simeq \bigoplus_j \mathfrak{g}'_j .$

I claim that the ${\mathfrak{g}_j'}$ can be obtained up to isomorphism by renumbering the ${\mathfrak{g}_i}$. We prove this by induction on the number of ${j}$. There is a surjective homomorphism obtained by quotienting:

$\displaystyle f: \bigoplus_i \mathfrak{g}_i \rightarrow \mathfrak{g}_1',$

i.e. a family of homomorphisms ${f_i: \mathfrak{g}_i \rightarrow \mathfrak{g}_1'}$. At least one of these is nonzero, and if the restriction ${f_i: \mathfrak{g}_i \rightarrow \mathfrak{g}_1'}$ is nonzero, then it is in fact an isomorphism because the kernel is an ideal, and so is the image. (Proof of the latter: ${[\mathfrak{g}_1', f(\mathfrak{g}_i)] = \left[ f\left( \mathfrak{g} \right), f(\mathfrak{g}_i) \right] = f([\mathfrak{g}, \mathfrak{g}_i]) \subset f(\mathfrak{g_i}).}$)

So suppose ${f_i: \mathfrak{g}_i \rightarrow \mathfrak{g}_1'}$ is nonzero. I claim that there is only one such index ${i}$, because if there were ${i,i'}$ with ${f_i, f_{i'}}$ isomorphisms, then

$\displaystyle \mathfrak{g}_1' = [\mathfrak{g}_1', \mathfrak{g}_1'] = [f(\mathfrak{g}_i), f(\mathfrak{g}_{i'})] = f([\mathfrak{g}_i, \mathfrak{g}_{i'}]) = f(\{0\}) = \{0\},$

contradiction. Given this, we can remove ${\mathfrak{g}_i, \mathfrak{g}_1'}$ to get an isomorphism

$\displaystyle \bigoplus_{i' \neq i} \mathfrak{g}_{i'} \simeq \bigoplus_{j \neq 1} \mathfrak{g}_j'$

and proceed inductively to get uniqueness.

As a corollary of this result and the corresponding fact about simple Lie algebras, we find:

$\displaystyle \mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$

for any semisimple Lie algebra ${\mathfrak{g}}$.