Associativity-like relation between Killing form and Lie bracket

Statement
Suppose $$F$$ is a field, $$L$$ is a Lie algebra over $$F$$, and $$\kappa$$ is the fact about::Killing form on $$L$$. Then, for any $$x,y,z \in L$$, we have:

$$\! \kappa([x,y],z) = \kappa(x,[y,z])$$.

Related facts

 * Orthogonal subspace to ideal for Killing form is ideal
 * Killing form is symmetric

Facts used

 * 1) uses::Trace of product of two linear transformations is independent of their order (in the form uses::trace of product of linear transformations is invariant under cyclic permutations)

Proof
Given: A Lie algebra $$L$$ over a field $$F$$ with Killing form $$\kappa$$.

To prove: $$\! \kappa([x,y],z) = \kappa(x,[y,z])$$ for all $$x,y,z \in L$$.

Proof: Using the definition of $$\kappa$$, we get:

$$\kappa([x,y],z) - \kappa(x,[y,z]) = \operatorname{tr}\left(\operatorname{ad}[x,y] \circ \operatorname{ad}(z) - \operatorname{ad}(x) \circ \operatorname{ad}[y,z]\right)$$.

Using the fact that Lie ring acts as derivations by adjoint action, in particular, that $$\operatorname{ad}[x,y] = [\operatorname{ad}(x),\operatorname{ad}(y)]$$, we get:

$$\kappa([x,y],z) - \kappa(x,[y,z]) = \operatorname{tr}\left(\operatorname{ad}(x) \circ \operatorname{ad}(y) \circ \operatorname{ad}(z) - \operatorname{ad}(y) \circ \operatorname{ad}(x) \circ \operatorname{ad}(z) - \operatorname{ad}(x) \circ \operatorname{ad}(y) \circ \operatorname{ad}(z) + \operatorname{ad}(x) \circ \operatorname{ad}(z) \circ \operatorname{ad}(y)\right)$$.

Canceling out, we get:

$$\kappa([x,y],z) - \kappa(x,[y,z]) = \operatorname{tr}\left(-\operatorname{ad}(y) \circ \operatorname{ad}(x) \circ \operatorname{ad}(z)+ \operatorname{ad}(x) \circ \operatorname{ad}(z) \circ \operatorname{ad}(y)\right)$$.

Let $$A = \operatorname{ad}(y)$$ and $$B = \operatorname{ad}(x) \circ \operatorname{ad}(z)$$. Then:

$$\kappa([x,y],z) - \kappa(x,[y,z]) = \operatorname{tr}(-AB + BA)$$.

Since $$\operatorname{tr}(AB) = \operatorname{tr}(BA)$$ we get the required result.