So, we now know that in any representation of a semisimple Lie algebra {\mathfrak{g}}, the semisimple elements act by diagonalizable operators. Now if we were given an abelian subalgebra {\mathfrak{h} \subset \mathfrak{g}} whose elements were semisimple, then every simple representation {V} of {\mathfrak{g}} would split into simple {\mathfrak{h}}-modules {W}, and perhaps we could study {V} through the various {W}. After all, each {W} is easy to describe: it is one-dimensional, and the action of {\mathfrak{h}} can be described by a linear functional on it (i.e., a weight).

The {\mathfrak{h}} in question that will be chosen are Cartan subalgebras.

More precisely, a Cartan subalgebra of the semisimple Lie algebra {\mathfrak{g}} is a maximal abelian subalgebra that consists only of semisimple elements. The special linear algebra {\mathfrak{sl}_n} of matrices of trace zero has the diagonal matrices of trace zero as a Cartan subalgebra, for instance.

In this post, I shall prove the existence of Cartan subalgebras. We will always be working over an algebraically closed field of characteristic zero.

Given {H \in \mathfrak{g}} and {\lambda \in k}, we write:

\displaystyle \mathfrak{g}_{\lambda}(H) := \{ v \in \mathfrak{g}: (\mathrm{ad} H - \lambda)^k v = 0 \ \mathrm{for} \ k \ \mathrm{large} \}

i.e. one of the Jordan “weight spaces” for {H}. This has an interesting property:

\displaystyle \boxed{ [\mathfrak{g}_{\lambda}(H), \mathfrak{g}_{\mu}(H)] \subset \mathfrak{g}_{\lambda + \mu}(H) .}

The property is proved using the identity for derivations {\delta} of an algebra:

So, in particular, {\mathfrak{g}_0(H)} is a subalgebra, containing {H}. This will be a candidate for the Cartan subalgebra. When {H} minimizes {\dim(\mathfrak{g}_0(H))}, we say that {H} is regular. Our main goal is to prove

Theorem 1 {\mathfrak{g}_0(H)} is a Cartan subalgebra when {H} is regular.

 

We already know it is a Lie subalgebra. First, let’s prove it is nilpotent.

Proposition 2 When {\mathfrak{g}} is any Lie algebra and {H} is regular, then {\mathfrak{g}_0(H)} is nilpotent.

 

We have a decomposition of vector spaces {\mathfrak{g} = \mathfrak{g}_0(H) \oplus \mathfrak{g}_{*}(H)}, where {\mathfrak{g}_*(H) = \bigoplus_{\lambda \neq 0} \mathfrak{g}_{\lambda}(H)}. Then {[ \mathfrak{g}_0(H), \mathfrak{g}_*(H)] \subset \mathfrak{g}_*(H)} by the boxed equality. Now choose {G \in \mathfrak{g}_0(H) }. We will show that {\mathrm{ad} G} is nilpotent on {\mathfrak{g}_0(H)}. By Engel’s theorem, this will imply that {\mathfrak{g}_0(H)} is nilpotent.

Now {\mathrm{ad} H} has no nonzero eigenvalues on {\mathfrak{g}_{*}(H)} by definition, and in particular is invertible. Thus for all {G} in a Zariski dense subset of {\mathfrak{g}_0(H)}, we have {\mathrm{ad} G} invertible on {\mathfrak{g}_{*}(H)}. But then {\mathrm{ad} G} preserves the decomposition {\mathfrak{g}_0(H) + \mathfrak{g}_*(H)}, and the maximal subspace on which it acts nilpotently must be contained within {\mathfrak{g}_0(H)}. In particular, the rank of this subspace is at most {\dim(\mathfrak{g}_0(H))}—but by regularity of {H}, we find that in fact {\mathrm{ad} G} is nilpotent on {\mathfrak{g}_0(H)} for all {G} in that Zariski dense subset just mentioned. Nilpotence is a closed condition, so we can drop the Zariski dense clause though. As a result, we are done.

Now, let’s get abelianness.

Proposition 3 (Step Two) If {\mathfrak{g}} is semisimple and {H} regular, then {\mathfrak{g}_0(H)} is abelian.

 

First, we make an important remark. The subspaces {\mathfrak{g}_{\lambda}(H), \mathfrak{g}_{\mu}(H)} are orthogonal under the Killing form if {\lambda \neq - \mu}. This is because if { x \in \mathfrak{g}_{\lambda}(H), y \in \mathfrak{g}_{\mu}(H)}, then {\mathrm{ad} x \mathrm{ad} y} deals with the decomposition

\displaystyle \mathfrak{g} = \bigoplus_{w} \mathfrak{g}_w(H)

by sending each {\mathfrak{g}_w(H)} into {\mathfrak{g}_{w+\lambda+\mu}(H)}, by the boxed statement. When {\lambda + \mu \neq 0}, this operation must have trace zero.

In particular, we find that the Killing form {B} restricted to {\mathfrak{g}_0(H)} is necessarily nondegenerate. So, it is sufficient to prove that {B([x,y],z)=0} if {x,y,z \in \mathfrak{g}_0(H)}. But, we can always choose the basis of {\mathfrak{g}} so that {\mathfrak{g}_0(H)} acts by upper-triangular matrices—Lie’s theorem. But for any upper-traingular matrices {A,B,C}, we have that {[A,B]C} is strictly upper-triangular and thus of trace zero, whence the claim. This proves the proposition.

So we see that {\mathfrak{g}_0(H)} is abelian; it is in fact maximal abelian, because it contains everything commuting with {H}.

Proposition 4 (Final Step) Hypotheses as above, {\mathfrak{g}_0(H)} consists of semisimple elements.

 

Indeed, let {x \in \mathfrak{g}_0(H)} split as {x = s+ n}. Then {s} commutes with everything that commutes with {x}, in particular with all of {\mathfrak{g}_0(H)}. Therefore {s \in \mathfrak{g}_0(H)}, and same for {n}. So {\mathfrak{g}_0(H)} contains semisimple and nilpotent parts of its elements. But, if we choose a basis for {\mathfrak{g}} such that {\mathfrak{g}_0(H)} acts by upper-triangular matrices on {\mathfrak{g}} (by Lie’s theorem), then it follows that if {n \in \mathfrak{g}_0(H)} is nilpotent and {y \in \mathfrak{g}_0(H)} is arbitrary,

\displaystyle B(y,n) = \mathrm{Tr}( (\mathrm{ad} y)( \mathrm{ad} n)) = 0,

because {n} must act by a strictly upper-triangular matrix. By nondegeneracy of the Killing form, we find {n=0}.

It turns out that any two Cartan subalgebras are conjugate under a special type of automorphism. Maybe I’ll get to this someday, but for now, finding the straight path up Mount Bourbaki to the representation theory of semisimple Lie algebras is the paramount goal.   Also, another side remark: it’s possible to talk about Cartan subalgebras of any Lie algebra (the definition is slightly different), and many of the proofs here generalize.  Cf. Serre’s Complex Semisimple Lie Algebras or Sternberg’s book on the subject, for instance.

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