Cartan’s Criterion

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

If $L\subseteq\mathfrak{gl}(V)$ is a linear Lie algebra and $\mathrm{Tr}(xy)=0$ for all $x\in[L,L]$ and $y\in L$ , then $L$ 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 $M$ consists of all the $x\in\mathfrak{gl}(V)$ such that $\mathrm{ad}(x)$ sends $L$ to $[L,L]$ . Clearly $L\subseteq M$ , but it may not be all of $M$ ; our hypothesis states that $\mathrm{Tr}(xy)=0$ for all $y\in L$ , but the criterion needs it to hold for all $y\in M$ .

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

$\displaystyle\begin{aligned}\mathrm{Tr}([x,y]z)&=\mathrm{Tr}(xyz-yxz)\\&=\mathrm{Tr}(xyz)-\mathrm{Tr}(yxz)\\&=\mathrm{Tr}(xyz)-\mathrm{Tr}(xzy)\\&=\mathrm{Tr}(xyz-xzy)\\&=\mathrm{Tr}(x[y,z])\end{aligned}$

Now, if $x,y\in L$ — so $[x,y]\in[L,L]$ — and $z\in M$ then

$\displaystyle\mathrm{Tr}([x,y]z)=\mathrm{Tr}(x[y,z])=\mathrm{Tr}([y,z]x)$

But since $z\in M$ we know that $[y,z]\in[L,L]\subseteq L$ , which means that the hypothesis kicks in: $[y,z]\in[L,L]$ and $x\in L$ so $\mathrm{Tr}([y,z]x)=0$ .

Then we know that all $x\in[L,L]$ are nilpotent endomorphisms, which makes them ad-nilpotent. Engel’s theorem tells us that $[L,L]$ is nilpotent, which means $L$ is solvable.

We can also extend this out to abstract Lie algebras: if $L$ is any Lie algebra such that $\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}(y))=0$ for all $x\in[L,L]$ and $y\in L$ , then $L$ is solvable. Indeed, we can apply the linear version to the image $\mathrm{ad}(L)\subseteq\mathfrak{gl}(L)$ to see that this algebra is solvable. The kernel $\mathrm{Ker}(\mathrm{ad})$ is just the center $Z(L)$ , which is abelian and thus automatically solvable. The image $\mathrm{ad}(L)$ is thus the solvable quotient of $L$ by a solvable kernel, so we know that $L$ itself is solvable.

September 1, 2012 - Posted by John Armstrong | Algebra, Lie Algebras

3 Comments »

[…] Where we have used the trace identity from last time. […]

Pingback by The Killing Form « The Unapologetic Mathematician | September 3, 2012 | Reply
[…] Cartan’s criterion then tells us that the radical of is solvable, and is thus contained in , the radical of the algebra. Immediately we conclude that if is semisimple — if — then the Killing form must be nondegenerate. […]

Pingback by The Radical of the Killing Form « The Unapologetic Mathematician | September 6, 2012 | Reply
[…] 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 intersection is solvable, but since is semisimple this means , and we can […]

Pingback by Decomposition of Semisimple Lie Algebras « The Unapologetic Mathematician | September 8, 2012 | Reply

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