When Killing form equals a constant times the trace

Question

How to find an element $0\not =a\in \mathbb C$ such that $\kappa_L (x,y)=aTr(xy)$ for all $x,y\in L$. Where $L$ is:

1. $A_l$
2. $B_l$
3. $C_l$
4. $D_l,\ \ l>2$.

Why such an $a$ is unique?

Answer 1

Because $L$ is simple, any two non-degenerate symmetric invariant bilinear forms are scalar multiples of each other, over an algebraically closed field of characteristic zero. The proof follows from Schur's Lemma (compare with this question). Since for simple Lie algebras the Killing form and the trace form are both non-degenerate, there is a unique $a\neq 0$ such that $ \kappa(x,y)=a\cdot tr(xy)$ for all $x,y\in L$. Actually, the unique value $a$ for each classical Lie algebra $L$ is explicitly known, and listed here. One can also directly prove these formulas, e.g., that $\kappa(x,y)=2ntr(xy)$ for all $x,y \in \mathfrak{sl}_n(\mathbb{C})$, thereby proving uniqueness of the scalar again. Reference: This homework, exercise $4$.