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.

In fact, we have the more precise result:

Theorem 2

Suppose {\mathrm{char}(k)=0}. Let {\mathfrak{g} \subset \mathfrak{gl}(V)} be a Lie subalgebra, and let {B} be the bilinear form on {\mathfrak{gl}(V)} associated to the canonical representation on {V} (i.e. {B(x,y) = \mathrm{Tr}(xy)}). Then {\mathfrak{g}} is solvable if and only if \displaystyle B(\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]) = \{ 0\}.

This implies Cartan’s theorem. Indeed, we have a morphism of Lie algebras {\mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})} via the adjoint representation; it is not injective, but the kernel is the center {\mathfrak{z}} of {\mathfrak{g}}. In particular it is clear that {\mathfrak{g}} is solvable iff {\mathfrak{g}/\mathfrak{z}} is. Now {\mathfrak{g}/\mathfrak{z} \subset \mathfrak{gl}(\mathfrak{g})} is solvable iff the bilinear form {B} in the second theorem vanishes on {\mathfrak{g}/\mathfrak{z} \times ( [\mathfrak{g}, \mathfrak{g}] + \mathfrak{z} )/\mathfrak{z}}, according to the second result. But this bilinear form {B} on {\mathfrak{g}/\mathfrak{z}} corresponds simply to the Killing form on {\mathfrak{g}}, which factors through {\mathfrak{z}}. So it is clear that we need only prove Theorem 2, which however is still a nontrivial task

How to prove Theorem 2

One way turns out to be an easy consequence of Lie’s theorem. We already have a fairly good idea of what representations of solvable Lie algebras look like, so it is not hard to get the result. Assume that {\mathfrak{g} \subset \mathfrak{gl}(V)} is solvable. Then we can diagonalize {V} with a fixed basis so that {\mathfrak{g}} consists of upper-triangular matrices. Then {[\mathfrak{g}, \mathfrak{g}]} consists of strictly upper-triangular matrices, and it is clear that the bilinear form {B} vanishes on an upper-triangular matrix together with a strictly upper-triangular one.

The other way is harder. What we’re actually going to do is to show that any {X \in [\mathfrak{g}, \mathfrak{g}] \subset \mathfrak{gl}(V)} is nilpotent. This implies (by Engel’s theorem) that {[\mathfrak{g}, \mathfrak{g}]} is nilpotent, hence \mathfrak{g} is solvable.

I need to digress.

Jordan decompositions and replicas

Recall from linear algebra that if {V} is a vector space over an algebraically closed field {k}, then any {T \in End(V)} has a decomposition

\displaystyle T = S + N

where {S} is semisimple (i.e. diagonalizable), {N} is nilpotent, and {S,N} can be expressed as polynomials with vanishing constant-coefficients in {T}—in particular commute with each other and with {T}.

Moreover, this is unique.

Now, given a semisimple endomorphism {S} and {\phi \in \hom_{\mathbb{Q}}(k,k)}, we define the replica {\phi(S)} by applying {\phi} to each element in the diagonal realization of {S}. Note in particular that {\phi(S)} is actually a polynomial in {S}.

Lemma 3

{T \in End(V)} is nilpotent if and only if\displaystyle \mathrm{Tr}( T \phi(S)) = 0

for {S} the semisimple part of {T}, and for all {\phi}.

For the proof, assume with an appropriate choice of basis that {T} is in upper triangular form with diagonal entries {\lambda_i}; we are to prove that {\lambda_i \equiv 0}. Now in this basis, {S} is the diagonal matrix with entries {\lambda_i}, and {\phi(S)} with entries {\phi(\lambda_i)}. So, if {\phi \in \hom_{\mathbb{Q}}(k,\mathbb{Q})},

\displaystyle 0 = \mathrm{Tr}(T \phi(S)) = \sum \lambda_i \phi(\lambda_i),

which implies by another application of {\phi} that

\displaystyle \sum \phi^2(\lambda_i) = 0,

i.e. that {\phi(\lambda_i)=0} for all {\phi_i}, which implies that each {\lambda_i=0}.

We’ll finish this up in the next post.

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