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.