The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

3 Comments »

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

    Pingback by The Killing Form « The Unapologetic Mathematician | September 3, 2012 | Reply

  2. […] 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

  3. […] 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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