Another piece of the proof
Ok, so we should try to apply the nilpotence criterion to the proof of Theorem 2 yesterday, and thus finish up the proof of Cartan’s criterion. So, we’re writing as , and we are going to show that
for all . (This is the notation about replicas.) Now for , and we have
So if we succeed in proving the following lemma, we will be done!
Lemma 1 (Key Lemma) Let be a subalgebra of . Let and let be the semisimple part of some . Then .
Once we prove this, we will be able to apply the nilpotence criterion together with our assumption about and conclude is nilpotent.
But now, we need more machinery.
Tensor products and the Jordan decomposition
Given a vector space , consider . Given , we define in a way to mimic the constructions on Lie representations. Define
So if is a representation of and acts by on , then acts by on considereed as a -module.
Lemma 2 decomposes into semisimple and nilpotent parts via , where form the decomposition of .
Indeed, it is easy to see that and commute with each other. is easily seen to be nilpotent. We have decompositions of into decompositions of one-dimensional vector spaces invariant under . This gives a decomposition of into one-dimensional spaces which are seen to be -invariant by the definition. So are at least semisimple and nilpotent, and also commute with . If the official Jordan decomposition is , then the equation
the fact that these four transformations commute with each other, and the fact that a transformation simultaneously semisimple and nilpotent is zero, show that our putative splitting claimed in the lemma is in fact not an impostor to the one true Jordan decomposition.
Let’s record a corollary:
Corollary 3 Let be subspaces of . Suppose and .Then and .
Indeed, and are, as elements in the Jordan decomposition of , polynomials in without constant term, so this is clear.
Corollary 4 Let be subspaces of . Suppose and
where is the semisimple part.
This now follows because ; it is easily checked that then corresponds to .
Finally, we can now prove the key lemma. Indeed, let , and apply the above corollary.
It’s all done now. Source: Serre, Lie Algebras and Lie Groups.