**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 2decomposes 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 3Let 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 4Let be subspaces of . Suppose andthen

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.*

January 30, 2010 at 10:44 am

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