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,

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

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

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

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

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 $L=H\oplus H^\perp$ .

$\qedsymbol$