Identification of $H$ with $H^{*}$ relativ the Killing-form Let $H$ be a maximal toral subalgebra of a semisimple Lie Algebra $L$.
The identification of $H^{*}$ and H relativ the Killing-form says, that to $\phi\in H^{*}$ corresponds the unique element $t_{\phi}\in H$ satisfying $\phi(h)=\kappa(t_{\phi}, h)$ for all $h\in H$.
Since the restriction of $\kappa$ to $H$ is nondegenerate, this scertainly makes sense for a single $h$, but I don´t understand, why the unique $t_{\phi}$ is the same for all $h\in H$.
I guess the main argument is, that $ad_{L}H$ is simultaneously diagnalisable, but I don´t know what exactly about simultaneosly diagonalisability it is that yields this point.
Thank you for helping me.
 A: For the purposes of this, it is easier to forget that there is any sort of Lie algebra in the picture. The important thing is that we have a finite dimensional vector space $H$ with a non-degenerate bilinear form $\kappa: H\otimes H\to \mathbb{C}$ (it is also not important that the field is $\mathbb{C}$ actually).
Given this, we define a map from $H^*$ to $H$ by $\phi\mapsto t_{\phi}$ where $t_{\phi}$ is defined by the requirement that for all $h\in H$ we have $\phi(h) = \kappa(t_{\phi},h)$. So we need to see that there is such a $t_{\phi}$ and that it is uniquely determined by the above property.
But in fact, it is probably easier to do this the other way around, by defining a bijective linear map $\Phi$ from $H$ to $H^*$ by $h\mapsto \phi_h$ where we define $\phi_h(h') = \kappa(h,h')$. This is clearly a linear map, and it is injective precisely because $\kappa$ is non-degenerate. So it must be surjective and thus bijective. Now it is easy to see that if we define $t_{\phi} = \Phi^{-1}(\phi)$ then this satisfies the above conditions, and that any other element satisfying the conditions will map to $\phi$ under $\Phi$, which means that it is uniquely determined.
