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

**Invariant bilinear forms **

Let be a Lie algebra over the field and representations. Recall the following.

1. is *invariant* under if for all .

2. is a representation of : define . This is isomorphic as a -module to the tensor product , where is regarded as a -module. We can think of as a -module because the enveloping algebra is a Hopf algebra under the homomorphism given by for , and extended further.

3. Let be a bilinear form on , i.e. a linear map . Then is said to be *invariant* under if for all

if we treat as an element of , this is the same as saying it is invariant in the sense of 1 above.

Ok, all good. Given a representation as above, we have a particular example of an invariant and symmetric bilinear form on 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

where are the corresponding endomorphisms of corresponding to . An important special case of this is when we are considering the adjoint representation of on itself; then this is called the **Killing form**.

What we shall prove is the following:

Theorem 1 (Cartan)Let the ground field be of characteristic zero. The Lie algebra is solvable if and only if

for the Killing form of the adjoint representation.

In fact, we have the more precise result:

Theorem 2Suppose . Let be a Lie subalgebra, and let be the bilinear form on associated to the canonical representation on (i.e. ). Then is solvable if and only if

This implies Cartan’s theorem. Indeed, we have a morphism of Lie algebras via the adjoint representation; it is not injective, but the kernel is the center of . In particular it is clear that is solvable iff is. Now is solvable iff the bilinear form in the second theorem vanishes on , according to the second result. But this bilinear form on corresponds simply to the Killing form on , which factors through . 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 is solvable. Then we can diagonalize with a fixed basis so that consists of upper-triangular matrices. Then consists of strictly upper-triangular matrices, and it is clear that the bilinear form 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 is nilpotent. This implies (by Engel’s theorem) that is nilpotent, hence is solvable.

I need to digress.

**Jordan decompositions and replicas **

Recall from linear algebra that if is a vector space over an algebraically closed field , then any has a decomposition

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

Moreover, this is *unique*.

Now, given a semisimple endomorphism and , we define the **replica** by applying to each element in the diagonal realization of . Note in particular that is actually a polynomial in .

Lemma 3is nilpotent if and only if

for the semisimple part of , and for all .

For the proof, assume with an appropriate choice of basis that is in upper triangular form with diagonal entries ; we are to prove that . Now in this basis, is the diagonal matrix with entries , and with entries . So, if ,

which implies by another application of that

i.e. that for all , which implies that each .

We’ll finish this up in the next post.

January 29, 2010 at 8:46 pm

[…] Climbing Mount Bourbaki Thoughts on mathematics « Cartan’s solvability criterion for Lie algebras […]

January 10, 2011 at 6:20 pm

Great job again. I am not done reading this, but please fix a few things with potential for confusing readers:

– In the definition of “invariant”, $x\in V$ should be $v\in V$.

– The definition of the $\mathfrak g$-module structure on $\operatorname*{hom}_k (V,W)$ should be $(Xf)(v) = X(fv) – f(Xv)$, not $(Xf)(v) = f(Xv) – X(fv)$.

– In Theorem 1, please say that you mean the Killing form of the adjoint representation, otherwise there is no difference to Theorem 2.

– Theorem 2: “and” misspelt “adn”.

– $\mathfrak{gl}(g$ one line below Theorem 2.

January 10, 2011 at 9:52 pm

Thanks for the corrections!

January 18, 2011 at 6:40 pm

Sorry, I don’t get your logic in the proof of Theorem 2. You say that if B(g, [g, g]) = 0, then g is nilpotent, hence solvable. But, using the other direction of Theorem 2, wouldn’t this yield that any solvable Lie algebra is nilpotent?

Of course, you are meaning that [g, g] is nilpotent, not g, but I am not sure about this either. The mistake might be a confusion between “X is nilpotent (as a matrix)” and “X is ad-nilpotent”.

Also, two more typos: “upper-traingular” and $x -> 1 (x) x + x (x) 1″ (the -> arrow should be |->). Besides, in the definition of an invariant bilinear form, you forgot “for all X in g”.

January 19, 2011 at 8:27 am

Corrected, thanks!

You are right that the correct statement is that [g,g] is nilpotent. I realize I never stated this in the proof on Engel’s theorem, but a Lie algebra of nilpotent (so automatically Ad-nilpotent) matrices is automatically nilpotent (as a Lie algebra).