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}}.

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