A semisimple Lie algebra is one that has no nonzero abelian ideals. This is equivalent to the absence of solvable ideals. Indeed, if
had a solvable ideal
, we could consider the derived series of
, i.e.
. These are ideals because, by the Jacobi identity, the Lie product of two ideals is an ideal. These
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 representations for
semisimple
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 is not generally semisimple; we are restricting ourselves to finite-dimensional
-modules.
Cartan’s criterion
Before getting there, we will prove a basic criterion for semisimplicity.
Theorem 1 (Cartan) The Lie algebra
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 two–post spiel on.
First, let’s suppose is not semisimple, i.e. has a nonzero abelian ideal
. This part of the proof is straightforward and elementary. Then given
, we have
by the hypothesis on . So
is nilpotent, of trace zero, and thus
belongs to the kernel of the Killing form.
Now, assume is semisimple. This implies that the center of
is zero, so the
-homomorphism
is an injection. Let
be the kernel of the Killing form
, i.e.
if and only if for all
,
By invariance of the Killing form, this is a Lie ideal in fact. Now by the above identification, and if
is the trace form on
, we have
By Cartan’s criterion (or, rather, Theorem 2 here), 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 Lie algebra. (We can always get a noncanonical embedding of any Lie algebra (always assumed finite-dimensional) in some
by a hard theorem of Ado though.) Second, we have a very convenient nondegenerate and invariant bilinear form.
Simple decomposition
Semisimplicity also allows 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
satisfies
.
Theorem 2 A semisimple Lie algebra
is isomorphic to a direct sum
where the
are simple Lie algebras, determined uniquely.
First, we prove existence. The basic lemma is that if is an ideal in
, then we can choose an appropriate ideal
and write
where the direct sum is in the sense of Lie algebras. Indeed, we take to be the orthogonal complement of
under the Killing form. By invariance this is also an ideal, and we have a direct sum decomposition as vector spaces. Since
is an ideal contained in both
, it is zero and we have proved our “basic lemma.”
It is now clear how to prove existence of the direct sum decomposition of into simple components. If
has no nontrivial ideals, we are done; if not we make a splitting decomposition as above and work inductively on the dimension of
.
For uniqueness, suppose we have collections of simple Lie algebras and
I claim that the can be obtained up to isomorphism by renumbering the
. We prove this by induction on the number of
. There is a surjective homomorphism obtained by quotienting:
i.e. a family of homomorphisms . At least one of these is nonzero, and if the restriction
is nonzero, then it is in fact an isomorphism because the kernel is an ideal, and so is the image. (Proof of the latter:
)
So suppose is nonzero. I claim that there is only one such index
, because if there were
with
isomorphisms, then
contradiction. Given this, we can remove to get an isomorphism
and proceed inductively to get uniqueness.
As a corollary of this result and the corresponding fact about simple Lie algebras, we find:
for any semisimple Lie algebra .
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