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

I am gathering material for an exposition on Lie theory and I am after proofs that a Lie group with finite centre is compact if and only if its Killing form is negative definite.

I know of one, sketched as follows: for the "if" direction, the group's adjoint representation image must be a subgroup of $O\left(\mathbf{g},\,-B\right)$ (where $\mathbf{g}$ is the group's Lie algebra and $B$ the Killing form), since the Killing form is Ad-invariant. Thus the original group must be a finite cover (owing to the finite centre) of this compact beast. In the "only if" direction, one uses the compactness to build an Ad-invariant average over the group $G$ of an arbitrary pair of elements $x$ and $y$ the Lie algebra $B\left(x,y\right)=\int_{\gamma\in G} \left<{\rm Ad}_\gamma x, {\rm Ad}_\gamma y\right> d\mu(G)$, with $\mu$ the Haar measure. Some simple gymnastics on this integral then shows it is negative definite.

I would like a citation for the above proof if you know one.

Does any one know of other proofs different from this one? With citations to a primary source, if you can, please.

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 $\mathcal{G}$.