Cartan's second criterion

Statement
The following are equivalent for a finite-dimensional Lie algebra over an algebraically closed field $$F$$ of characteristic zero:


 * 1) $$L$$ is a fact about::semisimple Lie algebra over $$F$$: in other words, $$L$$ has no nonzero  abelian ideals, or equivalently, $$L$$ has no nonzero  solvable ideals.
 * 2) The fact about::Killing form on $$L$$ is nondegenerate.

Related facts

 * Cartan's first criterion
 * Lie's theorem: This forms the basis of the proof of Cartan's first criterion.

Breakdown for non-algebraically closed fields

 * Cartan's second criterion fails for non-algebraically closed fields
 * Cartan's second criterion fails for prime characteristic

Facts used

 * 1) uses::Orthogonal subspace to ideal for Killing form is ideal
 * 2) uses::Killing form on ideal equals restriction of Killing form
 * 3) uses::Cartan's first criterion: This states that a Lie algebra $$A$$ is solvable if and only if $$\kappa_A(x,y) = 0$$ for all $$x \in A, y \in [A,A]$$, where $$\kappa_A$$ denotes the Killing form on $$A$$.
 * 4) uses::Abelian ideal is in nullspace for Killing form

Semisimple implies nondegenerate Killing form
Given: $$L$$ is a semisimple Lie algebra. Define:

$$L^\perp = \{ x \in L \mid \kappa(x,y) = 0 \ \forall \ y \in L \}$$.

To prove: $$L^\perp = 0$$.

Proof:


 * 1) By fact (1), $$L^\perp$$ is an ideal.
 * 2) By fact (2), the restriction of $$\kappa$$ to $$L^\perp$$ equals the Killing form on $$L^\perp$$.
 * 3) By definition, $$\kappa(x,y) = 0$$ for all $$x,y \in L^\perp$$. Combining this with the previous step, we see that the Killing form on $$L^\perp$$ is identically zero.
 * 4) By fact (3), we obtain that $$L^\perp$$ is solvable.
 * 5) Finally, since $$L$$ has no nonzero solvable ideal, $$L^\perp = 0$$.

Nondegenerate Killing form implies semisimple
We prove the contrapositive: if the Lie algebra is not semisimple, the Killing form is nondegenerate.

Given: A Lie algebra $$L$$ with a nonzero abelian ideal $$A$$ and Killing form $$\kappa$$.

To prove: $$\kappa$$ is degenerate.

Proof: By fact (4), $$A$$ is in the nullspace for $$\kappa$$, i.e.,

$$\kappa(x,y) = 0 \ \forall x \in A, y \in L$$.

Since $$A$$ is nonzero, we obtain that $$\kappa$$ is degenerate.