For a connected semisimple Lie group $G$, $\mathrm{Inn}(G)$ is a subgroup of finite index in $\mathrm{Aut}(G)$.

Question

For a connected semisimple Lie group $G$, $\mathrm{Inn}(G)$ is a subgroup of finite index in $\mathrm{Aut}(G).$

Can anyone give reference or proof for it. I know reference for compact but in general I don't know.

Answer 1

First of all, let ${\mathfrak g}$ denote the Lie algebra of $G$. Since $G$ is connected, linearlization at $e\in G$ defines an embedding $\phi: Aut(G)\to Aut({\mathfrak g})$ (which need not be surjective, unless $G$ is simply connected, but this is not our problem). The group $Aut(G)$ is a (possibly disconnected) Lie group, the image of its Lie algebra under $d\phi$ consists of derivations of ${\mathfrak g}$. The $d\phi$-image of the Lie algebra of $Inn(G)$ (i.e. the image of adjoint representation of the Lie algebra of $G$) consists of inner derivations of ${\mathfrak g}$. Every derivation of a semisimple Lie algebra is inner. Thus, the identity component of $Aut(G)$ coincides with $Inn(G)$ (since they have the same Lie algebra). It remains to prove that the identify component of $Aut({\mathfrak g})$ is a finite index subgroup in $Aut({\mathfrak g})$. This is proven in Corollary 2 of the linked Murakami's paper "On the automorphisms of a real semisimple Lie algebra", Journal of the Mathematical Society of Japan Vol. 4, No. 2, 1952.

I am sure one can find a modern textbook reference for this; likely in Knapp's "Lie Groups Beyond an Introduction".

Answer 2

I just wanted to add a really cool fact:

Theorem: Let $G$ be a reductive group. Then, $\mathrm{Out}(G)$ is finite if and only if $Z(G)^\circ$ has rank at most $1$.

A reference is Theorem 7.1.9 (2) of this. More elementary references can be gleaned from loc. cit.