Show that the Cartan-Killing form of $sl_n\mathbb{C}$ is $\langle{\rm diag}(a),{\rm diag}(b)\rangle=2n\sum_i a_i b_i$ Consider $sl_n\mathbb{C}$ as aLie-algebra, and choose h the CSA formed by diagonal matrixes. I can i demonstrate that the Cartan-Killing form in $sl_n\mathbb{C}$ is $<diag(a_i),diag(b_i)>=2n \sum_{i=1}^n a_ib_i$?
 A: Granting that $sl(n)$ is simple, one might first show that, up to scalars, there is a unique $sl(n)$-invariant non-degenerate symmetric bilinear form on it. To see this, note that a non-degenerate bilinear form gives an $sl(n)$-module map $f$ of $sl(n)$ to its dual $sl(n)^*$, by $f(x)(y)=\langle x,y\rangle$. The non-degeneracy is equivalent to trivial kernel, so, by finite-dimensionality, equivalent to $f$ being an isomorphism. Given two such pairings, and corresponding maps $f,g$, the composite $f^{-1}\circ g$ is an $sl(n)$-isomorphism of $sl(n)$ to itself. By Schur's Lemma, this is a scalar. [done]
Thus, it suffices to evaluate Casimir and the "trace pairing" on a single pair of elements of the Cartan subalgebra, to evaluate the constant. For example, $x=y=(1,-1,0,...,0)$ (diagonal) has eigenvalues $+2$ (once), $-2$ (once), $+1$ ($n-2$ times), and $-1$ ($n-2$ times). Applying it twice and adding up gives $2n$.
(To show from scratch that $sl(n)$ is simple is not too hard, if one wants...)
