Proof that Lie group with finite centre is compact if and only if its Killing form is negative definite

Solution 1:

If $G$ is a compact Lie group with discrete centre then its Lie algebra $\mathcal{G}$ has a trivial centre $Z(\mathcal{G})=\{0\}$. Since $G$ is compact, it admits a bi-invariant metric (as you mention above) and $\mathrm{ad}_x$ is antisymmetric for all $x\in\mathcal{G}$. Moreover, $\mathrm{ad}_x$ is diagonalizable and has all his eigenvalues of the forme $i\lambda$, $(\lambda>0)$ then its Killing form is negative definite: $$K(x,x)=\mathrm{tr}(\mathrm{ad}_x\mathrm{ad}_x)=-\sum_{k=1}^n\lambda_k^2<0$$ In the other direction, one may use the Myers theorem. But first, we must prove that the Ricci curvature of $\mathcal{G}$ is strictly positive, for some bi-invariant scalar product on $\mathcal{G}$.