The Killing Form

We can now define a symmetric bilinear form on our Lie algebra by the formula

It’s symmetric because the cyclic property of the trace lets us swap and and get the same value. It also satisfies another identity which is referred to as “associativity”, though it may not appear like the familiar version of that property at first:

Where we have used the trace identity from last time.

This is called the Killing form, named for Wilhelm Killing and not nearly so coincidentally as the Poynting vector. It will be very useful to study the structures of Lie algebras.

First, though, we want to show that the definition is well-behaved. Specifically, if is an ideal, then we can define to be the Killing form of . It turns out that is just the same as , but restricted to take its arguments in instead of all of .

A lemma: if is any subspace of a vector space and has its image contained in , then the trace of over is the same as its trace over . Indeed, take any basis of and extend it to one of ; the matrix of with respect to this basis has zeroes for all the rows that do not correspond to the basis of , so the trace may as well just be taken over .

Now the fact that is an ideal means that for any the mapping is an endomorphism of sending all of into . Thus its trace over is the same as its trace over all of , and the Killing form on applied to is the same as the Killing form on applied to the same two elements.

Cartan’s Criterion

It’s obvious that if is nilpotent then will be solvable. And Engel’s theorem tells us that if each is ad-nilpotent, then is itself nilpotent. We can now combine this with our trace criterion to get a convenient way of identifying solvable Lie algebras.

If is a linear Lie algebra and for all and , then is solvable. We’d obviously like to use the trace criterion to show this, but we need a little massaging first.

The catch is that our consists of all the such that sends to . Clearly , but it may not be all of ; our hypothesis states that for all , but the criterion needs it to hold for all .

To get there, we use the following calculation, which is a useful lemma in its own right:

Now, if — so — and then

But since we know that , which means that the hypothesis kicks in: and so .

Then we know that all are nilpotent endomorphisms, which makes them ad-nilpotent. Engel’s theorem tells us that is nilpotent, which means is solvable.

We can also extend this out to abstract Lie algebras: if is any Lie algebra such that for all and , then is solvable. Indeed, we can apply the linear version to the image to see that this algebra is solvable. The kernel is just the center , which is abelian and thus automatically solvable. The image is thus the solvable quotient of by a solvable kernel, so we know that itself is solvable.