Why does the Killing form act on the Cartan subalgebra as $\kappa(h,k)=\sum_\alpha \alpha(h)\alpha(k)$? Chose a finite dimensional semi-simple Lie algebra $L$ and a Cartan subalgebra $H$. Why do we have $$\kappa_L (h,k)=\sum_{\alpha\in\Phi (L,H)}\alpha(h)\alpha(k),$$ where $\kappa$ is the Killing form and $\Phi(L,H)$ is the set of roots?
I tried to work something out via the root space decomposition and Lie's theorem but I didn't get that far. Any thoughts?
 A: Take $h\in H$. Remember that the action of $\operatorname{ad}(h)$ on $L$ is diagonalisable and that the eigenvalues (other than $0$) are precisely the numbers $\alpha(h)$, with $\alpha\in\Phi(L,H)$. Therefore, with respect to some basis of $L$, the matrix of $\operatorname{ad}(h)$ is a diagonal matrix such that the entries of the main diagonal are the numbers $\alpha(h)$ ($\alpha\in\Phi(L,H)$) and $0$. If $k$ is another element of $H$, the same thing will be true with respect to the same basis. Therefore, the matrix of $\operatorname{ad}(h)\circ\operatorname{ad}(k)$ will be a diagonal matrix whose entries within the main diagonal will be the numbers $\alpha(h)\alpha(k)$ ($\alpha\in\Phi(L,H)$) and $0$. So, its trace is the sum that you have mentioned.
A: The Cartan decomposition says that
$$
L=H\oplus \sum_{\alpha\in \Phi}L_{\alpha}.
$$
The matrices of the restrictions of $ad(h)$ to $L_{\alpha}$ for $h\in H$ can be taken simultaneously in lower-triangular form with diagonal $(\alpha(h),\ldots ,\alpha(h))$ by Lie's Theorem. With $\dim L_{\alpha}=1$ and taking the trace we obtain that
$$
\kappa(h,k)=tr(ad(h)\cdot ad(k))=\sum_{\alpha\in \Phi}\alpha(h)\alpha(k)
$$
for $h,k\in H$.
