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…)