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 {X \in [\mathfrak{g},\mathfrak{g}] \subset \mathfrak{gl}(V)} as {X = S + N}, and we are going to show that

\displaystyle \mathrm{Tr}( X \phi(S)) = 0

for all {\phi \in \hom_{\mathbb{Q}}(k,k)}. (This is the notation about replicas.) Now {X = \sum [A_i, B_i]} for {A_i,B_i \in \mathfrak{g}}, and we have

\displaystyle \mathrm{Tr}(X \phi(S)) = \sum \mathrm{Tr}( [A_i, B_i] \phi(S) ) = \sum \mathrm{Tr}( B_i [ A_i, \phi(S)]).

So if we succeed in proving the following lemma, we will be done!

Lemma 1 (Key Lemma) Let {\mathfrak{g}} be a subalgebra of {\mathfrak{gl}(V)}. Let {A \in \mathfrak{g}} and let {S} be the semisimple part of some {X \in \mathfrak{g} \subset \mathfrak{gl}(V)}. Then {[A, S] \in [\mathfrak{g},\mathfrak{g}]}.

 

Once we prove this, we will be able to apply the nilpotence criterion together with our assumption about {B} and conclude {X} is nilpotent.

But now, we need more machinery.

Tensor products and the Jordan decomposition

Given a vector space {V}, consider {V_{p,q} := V^{\otimes p} \otimes V^{\vee \otimes q}}. Given {T \in End(V)}, we define {T_{p,q} \in End(V_{p,q})} in a way to mimic the constructions on Lie representations. Define

\displaystyle T_{p,q} := \sum_{i=1}^p (1_V \otimes \dots T \dots 1_V \otimes 1_{V^{\vee \otimes q}}) - \sum_{j=1}^q (1_{V^{\otimes p}} \otimes 1_{V^{\vee}} \otimes \dots T^* \dots \otimes 1)

So if {V} is a representation of {\mathfrak{g}} and {X} acts by {T} on {V_{p,q}}, then {X} acts by {T_{p,q}} on {V_{p,q}} considereed as a {\mathfrak{g}}-module.

Lemma 2 {T_{p,q}} decomposes into semisimple and nilpotent parts via {T_{p,q} = S_{p,q} + N_{p,q}}, where {S,N} form the decomposition of {T}.

Indeed, it is easy to see that {S_{p,q}} and {N_{p,q}} commute with each other. {N_{p,q}} is easily seen to be nilpotent. We have decompositions of {V, V^{\vee}} into decompositions of one-dimensional vector spaces invariant under {S}. This gives a decomposition of {V_{p,q}} into one-dimensional spaces which are seen to be {S_{p,q}}-invariant by the definition. So {S_{p,q}, N_{p,q}} are at least semisimple and nilpotent, and also commute with {T_{p,q}}. If the official Jordan decomposition is {T_{p,q} = S_{(p,q)} + N_{(p,q)}}, then the equation

\displaystyle S_{p,q} + N_{p,q} = S_{(p,q)} + N_{(p,q)},

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 {W \subset W'} be subspaces of {V_{p,q}}. Suppose {T \in End(V)} and {T_{p,q} (W') \subset W}.Then {S_{p,q}(W') \subset W} and {N_{p,q} (W') \subset W}.

 

Indeed, {S_{p,q}} and {N_{p,q}} are, as elements in the Jordan decomposition of {T_{p,q}}, polynomials in {T_{p,q}} without constant term, so this is clear.

Corollary 4 Let {W \subset W'} be subspaces of {\mathfrak{gl}(V)}. Suppose {T \in End(V)} and\displaystyle (\mathrm{ad} T) (W') \subset W;

then\displaystyle (\mathrm{ad} S) (W') \subset W

where {S} is the semisimple part. 

 

This now follows because {\mathfrak{gl}(V) = End(V) \simeq V_{1,1} = V \otimes V^{\vee}}; it is easily checked that {\mathrm{ad} T} then corresponds to {T_{1,1}}.

Finally, we can now prove the key lemma. Indeed, let {W = [\mathfrak{g},\mathfrak{g}], W' = \mathfrak{g}}, and apply the above corollary.

 It’s all done now.  Source: Serre, Lie Algebras and Lie Groups.

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