Killing Form on $\mathfrak{su}(n)$

Question

I am a little confused as to what is the Killing Form on $\mathfrak{su}(n)$. Since it is only a real Lie-Algebra, one needs the Killing-Form to assume real values. But what is the Killing form exactly on those matrices? On Wikipedia, it sais $B(X,Y) = 2n\cdot \text{tr}(XY)$ but I do not see why this should always be a real number. Some sources use $B(X,Y) = 2n \text{Re}(\text{tr}(XY))$ and now I am not sure what to believe. If anyone could resolve my confusion, i would be very happy.

Answer 1

We don't need to do an explicit computation here. $\mathfrak{su}(n)$, by definition, consists of skew-hermitian matrices, so they satisfy $X^{\dagger} = -X$. This gives $(XY)^{\dagger} = Y^{\dagger} X^{\dagger} = (-Y)(-X) = YX$. And this gives, by the cyclicity of the trace,

$$2 \text{ tr}(XY) = \text{tr}(XY) + \text{tr}(YX) = \text{tr} \left( XY + (XY)^{\dagger} \right) \in \mathbb{R}$$

because $XY + (XY)^{\dagger}$ is hermitian.