I want to prove that on a semisimple Lie algebra $\mathfrak{g}$ over ${\bf R}$:

$\mathfrak{g}$ is compact if and only if the Killing form is strictly negative definite.

Here the Lie algebra is compact if Int$(\mathfrak{g})$, which is a Lie group of $\{ ad_X | X\in \mathfrak{g}\}$, is compact.

I am reading a Helgason's book : The proof is short and I cannot catch the strategy.

The proof : If Killing form $B$ is strictly negative definitie then $O(B)$ is a set of all linear transformations which leave $B$ invariant. Then $O(B)$ is compact. I cannot understand.

Thank you in advance.


1 Answer 1


The key is to look at just the vector space structure of $\mathfrak g$ and then use some nice topological properties.

So for now, think of $\mathfrak g$ as just a real vector space, say $\mathbb R^n$, so that the Killing form is essentially an inner product. Since the Killing form is negative definite, there is a an element $A \in GL_n(\mathbb R)$ with all negative eigenvalues such that $B(x,y) = x^T A y$. The collection of all linear transformations which preserve B will be this more generalized orthogonal group $O(B)$, which is closed and bounded by precisely the same arguments as those showing that $O(n)$ is compact (just no longer using the Euclidean inner product).

Now $B$ is adjoint invariant and all Lie algebra automorphisms of $\mathfrak g$ must preserve the commutator (and hence the adjoints) we get that $\operatorname{Aut}(\mathfrak g) \subseteq O(B)$. Furthermore, $\operatorname{Int}(\mathfrak g)$ is a closed subgroup of $\operatorname{Aut}(\mathfrak g)$ and so is also a closed subspace of $O(B)$ and hence compact. Thus the Lie algebra is also compact as required.

  • $\begingroup$ We do not need the negativity in the proof right? $\endgroup$
    – S. D. Z
    Feb 23, 2020 at 4:26

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