Orthogonal subspace to ideal for Killing form is ideal

Statement
Suppose $$L$$ is a finite-dimensional Lie algebra over a field $$F$$, and $$\kappa$$ is the fact about::Killing form on $$L$$.

Suppose $$I$$ is an ideal of $$L$$. Define:

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

Then, $$I^\perp$$ is also an ideal of $$L$$.

About orthogonal subspaces

 * Orthogonal subspace to derivation-invariant subalgebra for Killing form is derivation-invariant
 * Orthogonal subspace to subalgebra for Killing form not is subalgebra

About ideals and Killing forms

 * Killing form on ideal equals restriction of Killing form

Facts used

 * 1) uses::Associativity-like relation between Killing form and Lie bracket: This states that for all $$x,y,z$$ in the Lie algebra, $$\kappa([x,y],z) = \kappa(x,[y,z])$$.

Proof
Given: A finite-dimensional Lie algebra $$L$$ over a field $$F$$. $$\kappa$$ is the Killing form on $$L$$, and $$I$$ is an ideal of $$L$$. $$I^\perp = \{ x \mid \kappa(x,y) = 0 \ \forall \ y \in I\}$$.

To prove: $$I^\perp$$ is an ideal.

Proof:


 * 1) $$I^\perp$$ is a linear subspace: This follows from the fact that $$\kappa$$ is $$F$$-bilinear.
 * 2) For $$x \in I^\perp$$ and $$y \in L$$, $$[x,y] \in I^\perp$$: To prove this, it suffices to show that for any $$z \in I$$, $$\! \kappa([x,y],z) = 0$$. But by fact (1), $$\! \kappa([x,y],z) = \kappa(x,[y,z])$$. Since $$I$$ is an ideal and $$z \in I$$, $$[y,z] \in I$$, so $$\! \kappa(x,[y,z]) = 0$$. This completes the proof.