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…)

For the next few weeks, I’m probably going to be doing primarily algebra posts.

Invariant bilinear forms

Let ${\mathfrak{g}}$ be a Lie algebra over the field ${k}$ and ${V,W}$ representations. Recall the following.

1. ${v \in V}$ is invariant under ${\mathfrak{g}}$ if ${Xv = 0}$ for all ${X \in \mathfrak{g}}$.

2. ${\hom_k(V,W)}$ is a representation of ${\mathfrak{g}}$: define $(Xf)v = X(fv) - f(Xv)$. This is isomorphic as a ${\mathfrak{g}}$-module to the tensor product ${W \otimes V^{\vee}}$, where ${V^{\vee}}$ is regarded as a ${\mathfrak{g}}$-module. We can think of ${W \otimes V^{\vee}}$ as a ${\mathfrak{g}}$-module because the enveloping algebra ${U\mathfrak{g}}$ is a Hopf algebra under the homomorphism ${U\mathfrak{g} \rightarrow U \mathfrak{g} \otimes U \mathfrak{g}}$ given by ${x \mapsto 1 \otimes x + x \otimes 1}$ for ${x \in \mathfrak{g}}$, and extended further.

3. Let ${B}$ be a bilinear form on ${V}$, i.e. a linear map ${B: V \otimes V \rightarrow k}$. Then ${B}$ is said to be invariant under ${\mathfrak{g}}$ if for all ${v,v' \in V, X \in \mathfrak{g}}$

$\displaystyle B(Xv, v') + B(v, Xv') = 0;$

if we treat ${B}$ as an element of ${(V \otimes V)^{\vee}}$, this is the same as saying it is invariant in the sense of 1 above.

Ok, all good. Given a representation ${V}$ as above, we have a particular example ${B_V}$ of an invariant and symmetric bilinear form on ${\mathfrak{g}}$ in this post (which in particular shows why some of what I just posted here is redundant; I hadn’t looked back when I started writing it) given by

$\displaystyle B_V(x,y) := \mathrm{Tr}( x_V y_V),$

where ${x_V,y_V}$ are the corresponding endomorphisms of ${V}$ corresponding to ${x,y \in \mathfrak{g}}$. An important special case of this is when we are considering the adjoint representation of ${\mathfrak{g}}$ on itself; then this is called the Killing form.

What we shall prove is the following:

Theorem 1 (Cartan)

Let the ground field ${k}$ be of characteristic zero. The Lie algebra ${\mathfrak{g}}$ is solvable if and only if$\displaystyle B(\mathfrak{g}, [\mathfrak{g},\mathfrak{g}]) = \{ 0 \},$

for ${B}$ the Killing form of the adjoint representation. (more…)