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. (more…)