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Cartan's criterion for semisimplicity

DEFINITION 5.42   We call a Lie algebra $ L$ over $ k$ non degenerate if the Killing form $ \kappa_L$ of $ L$ is non degenerate.

LEMMA 5.43   Every non degenerate Lie algebra $ L$ over a field $ k$ is semisimple.

PROOF.. Assume that there exists a non trivial abelian ideal $ A$ of $ L$ . Let $ y_0$ be a non zero element of $ A$ . Then for any $ x\in L$ , $ z=\operatorname{ad}(x)\operatorname{ad}(y_0)$ is nilpotent. Indeed,

$\displaystyle z(L)=\operatorname{ad}(x)\operatorname{ad}(y_0) (L) = \operatorname{ad}(x)([y_0,L])\subset \operatorname{ad}(x)(A)\subset A,
$

$\displaystyle z^2(L)=\operatorname{ad}(x)\operatorname{ad}(y_0) (z(L))\subset \operatorname{ad}(x) \operatorname{ad}(y_0) (A)=\operatorname{ad}(x)[y_0,A]=0.
$

Thus $ \kappa(x,y_0)=\operatorname{tr}(\operatorname{ad}(x)\operatorname{ad}(y_0))=0$ for for any $ x\in L$ . That means, $ y_0\in L^\perp$ . This is contrary to the assumption on $ L$ . $ \qedsymbol$

PROPOSITION 5.44   Let $ n$ be a positive integer. Let $ k$ be a field of characteristic $ p \in \operatorname{Ccs}(n)$ . Let $ L$ be a Lie algebra of dimension $ n$ . Then the following conditions are equivalent:
  1. $ L$ is semisimple.
  2. $ L$ is non degenerate.
  3. $ L$ is a direct sum of simple ideals.

PROOF.. ( $ (1) \implies (2)$ ): Assume $ L$ is semisimple. Let us take an ideal $ I=L^\perp$ of $ L$ . Then the Killing form on $ I$ is identically equal to zero. since $ \dim(I)\leq n$ , $ I$ is a solvable algebra. Since $ L$ is semisimple, this implies $ I=0$ .

( $ (2) \implies (1)$ ): holds (regardless of the base field) in view of the previous lemma.

( $ (3)\implies (2)$ ): We see that simple algebras are non degenerate in view of the argument above. Thus $ L$ is also non degenerate.

( $ (2)\implies (3)$ ): Let $ H$ be a nontrivial ideal of $ L$ . Then $ H \cap H^\perp$ is an abelian ideal of $ L$ . Indeed, for any $ y, z \in H\cap H^{\perp}$ and for any $ x\in L$ , we have

$\displaystyle \kappa(x,[y,z])=\kappa([x,y],z)\in \kappa(H,H^\perp)=0
$

So that $ [y,x]\in L^\perp=0$ . On the other hand, by the previous lemma we see that $ H \cap H^\perp$ is semisimple and so we have $ H\cap H^\perp=0$ . Accordingly we have $ L=H\oplus H^\perp$ .

$ \qedsymbol$


next up previous
Next: examples Up: generalities in finite dimensional Previous: Cartan's criterion for solvability(Ccs)
2007-12-19