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
then
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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