The Radical of the Killing Form

The first and most important structural result using the Killing form regards its “radical”. We never really defined this before, but it’s not hard: the radical of a binary form $B$ on a vector space $V$ is the subspace consisting of all $v\in V$ such that $B(v,w)=0$ for all $w\in V$ . That is, if we regard $B$ as a linear map $v\mapsto B(v,\underline{\hphantom{X}})$ , the radical is the kernel of this map. Thus we see that $B$ is nondegenerate if and only if its radical is zero; we’ve only ever dealt much with nondegenerate bilinear forms, so we’ve never really had to consider the radical.

Now, the radical of the Killing form $\kappa$ is more than just a subspace of $L$ ; the associative property tells us that it’s an ideal. Indeed, if $s$ is in the radical and $x,y\in L$ are any other two Lie algebra elements, then we find that

$\displaystyle\kappa([s,x],y)=\kappa(s,[x,y])=0$

thus $[s,x]$ is in the radical as well.

We recall that there was another “radical” we’ve mentioned: the radical of a Lie algebra is its maximal solvable ideal. This is not necessarily the same as the radical of the Killing form, but we can see that the radical of the form is contained in the radical of the algebra. By definition, if $x$ is in the radical of $\kappa$ and $y\in L$ is any other Lie algebra element we have

$\displaystyle\kappa(x,y)=\mathrm{Tr}(\mathrm{ad}(x),\mathrm{ad}(y))=0$

Cartan’s criterion then tells us that the radical of $\kappa$ is solvable, and is thus contained in $\mathrm{Rad}(L)$ , the radical of the algebra. Immediately we conclude that if $L$ is semisimple — if $\mathrm{Rad}(L)=0$ — then the Killing form must be nondegenerate.

It turns out that the converse is also true. In fact, the radical of $\kappa$ contains all abelian ideals $I\subseteq L$ . Indeed, if $x\in I$ and $y\in L$ then $\mathrm{ad}(x)\mathrm{ad}(y):L\to I$ , and the square of this map sends $L$ into $[I,I]=0$ . Thus $\mathrm{ad}(x)\mathrm{ad}(y)$ is nilpotent, and thus has trace zero, proving that $\kappa(x,y)=0$ , and that $x$ is contained in the radical of $\kappa$ . So if the Killing form is nondegenerate its radical is zero, and there can be no abelian ideals of $L$ . But the derived series of $\mathrm{Rad}(L)$ eventually hits zero, and its last nonzero term is an abelian ideal of $L$ . This can only work out if $\mathrm{Rad}(L)$ is already zero, and thus $L$ is semisimple.

So we have a nice condition for semisimplicity: calculate the Killing form and check that it’s nondegenerate.

September 6, 2012 - Posted by John Armstrong | Algebra, Lie Algebras

3 Comments »

[…] taking its determinant to find . Since this is nonzero, we conclude that is nondegenerate, which we know means that is semisimple — at least in fields where . Share this:StumbleUponDiggRedditLike […]

Pingback by Back to the Example « The Unapologetic Mathematician | September 7, 2012 | Reply
[…] to the Killing form . The associativity of shows that is also an ideal, just as we saw for the radical. Indeed, the radical of is just . Anyhow, Cartan’s criterion again shows that the […]

Pingback by Decomposition of Semisimple Lie Algebras « The Unapologetic Mathematician | September 8, 2012 | Reply
[…] of . Then we can define to be the subspace orthogonal (with respect to ) to , and the fact that the Killing form is nondegenerate tells us that , and thus […]

Pingback by All Derivations of Semisimple Lie Algebras are Inner « The Unapologetic Mathematician | September 11, 2012 | Reply

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