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.