TITLE: Killing form: clean proof that two spaces are orthogonal?
QUESTION [10 upvotes]: Let $G$ be a linear algebraic group whose Lie algebra $\mathfrak{g}$ is semisimple. Let $x$ be a regular semisimple element of $G$. Write $\mathfrak{t}$ for the Lie algebra of the maximal torus $T = C(x)$ centralizing $x$, and let $\mathfrak{h}$ be the image of $\mathfrak{g}$ under the map $g\mapsto \textrm{Ad}_x(g)-g$.
Since $g$ is semisimple, we may diagonalize it; then all the elements of $\mathfrak{h}$ have only zeroes in the diagonal. It is not hard to see that, for $G$ a classical group ($\textrm{SL}_n$, $\textrm{SO}_n$, $\textrm{Sp}_{2 n}$, over an arbitrary field), the spaces $\mathfrak{t}$ and $\mathfrak{h}$ are orthogonal under the Killing form: the Killing form is then a multiple of $(X,Y)\mapsto \textrm{tr}(X Y)$, and, for $X$ diagonal and $Y$ having only zeroes in the diagonal, $X Y$ clearly has only zeroes in the diagonal, and thus has trace $0$.
Is there a clean proof that works for all linear algebraic groups $G$ with $\mathfrak{g}$ semisimple -- preferably one with as little computation or casework as possible?

REPLY [12 votes]: $$(X, \mathrm{Ad}_x(Y) - Y) = (X,\mathrm{Ad}_x(Y)) - (X,Y) = (\mathrm{Ad}_{x^{-1}}(X),Y) - (X,Y) = (X,Y) - (X,Y) = 0$$
Using the fact that the Killing form is invariant under the adjoint action and that $\mathrm{Ad}_x(X) = X$ for $X \in \mathfrak{t}$. I can't think of a reason why this wouldn't work in any characteristic although I'm much more familiar with characteristic 0.