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…)