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The Killing form is invariant and nondegenerate on a complex semisimple Lie algebra

Statement

Let $g$ be a complex semisimple Lie algebra, and let $B$ be its Killing form from The Killing form of a semisimple Lie algebra. Then

$B ([x, y], z) = B (x, [y, z]) (x, y, z \in g),$

and $B$ is nondegenerate.

Facts & Assumptions

Given: A complex semisimple Lie algebra $g$ and its Killing form $B$ .

Proof

technique · direct

1.1

Using ${ad}_{[x, y]} = [{ad}_{x}, {ad}_{y}]$ and cyclicity of trace, one gets $B ([x, y], z) = tr ([{ad}_{x}, {ad}_{y}] {ad}_{z}) = tr ({ad}_{x} [{ad}_{y}, {ad}_{z}]) = B (x, [y, z])$ .

givenalgebra

2.1

The radical ${x \in g : B (x, g) = 0}$ is an ideal because step 1.1 makes it stable under brackets. Cartan's semisimplicity criterion states that a finite-dimensional Lie algebra over characteristic $0$ is semisimple if and only if its Killing form is nondegenerate. Since $g$ is complex semisimple, this criterion makes the radical zero.

givenstep 1.1

3.1

Hence $B$ is invariant and has zero radical, so it is nondegenerate.

step 1.1step 2.1∎

Depends on

The Killing form of a semisimple Lie algebra

Used by

The quadratic Casimir element Definition
The Killing form pairs only opposite root spaces Proposition
Root-space decomposition relative to a Cartan subalgebra Theorem

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Sources

Pavel Etingof, Lie Groups and Lie Algebras I (standard reference, not scraped)