Theorem: If $G$ is a compact semisimple Lie group, then its Killing form is negative definite.
In its proof: Since $G$ is compact there is an Ad-invariant inner product on $\mathfrak g$.
Since each $ad_X$ is skew-symmetric, let $ad_X=(a_{ij})$ relative to an orthonormal basis of $\mathfrak g$. Then $$B(X,X)=tr(ad_X\circ ad_X)=\sum_i\sum_ja_{ij}a_{ji}=-\sum_{i,j}a^2_{ij}\leq0$$ Since $G$ is semisimple, $B$ is nondegenerate, so the above sum is strictly less than zero. $ \diamond $
I have a question here: Where the Ad-invariant inner product is used in the proof above?
Best Answer
The fact that $B$ is $Ad$-invariant implies that the matrix of $ad_x$ is skew symmetric in a orthogonal basis, the result follows from the fact that if $X$ is skew symmetric, $trace(X^2)\leq 0$ if $X$ is skew-symmetric and $B$ is non degenerated.
N.B: if $X$ is skew-syymetric, $X+X^T=0$ implies that $X(X+X^T)=X^2+XX^T=0$, thus $trace(X^2)=-trace(XX^T)$. We know that $trace(XX^T)\geq 0$.
If $B$ is $Ad$-invariant, $B(Ad(exp(tz))x,Ad(exp(tz))y)=B(x,y)$. If you calculate the differential at $t=0$, you obtain $B([z,x],y)+B(x,[z,y])=0$