What can we say about $\operatorname{Aut}(G)$ for an arbitrary Lie group $G$?

Question

Let $G$ be a Lie group, $\mathfrak g$ be its Lie algebra, and $\operatorname{Aut}(G)$ be the group of its smooth automorphism. Then, my questions are:

(1) Is $\operatorname{Aut}(G)$ again a smooth manifold? And particularly a Lie group?

(2) If so, can I realize $\operatorname{End}(\mathfrak g)$ as something related to the tangent space $T_I(\operatorname{Aut}(G))$, where $I$ is the identity?

Thanks

Answer 1

This question is considered at length in Hochschild's quite lucid 1952 paper.

Answer 2

(It makes no sense to ask whether a group is Lie if no topology is specified.) I henceforth refer to (bi)continuous (equivalently, smooth) automorphisms and I endow $\DeclareMathOperator{\Aut}{\operatorname{Aut}} \Aut(G)$ with the compact-open topology.

If $G$ is connected, yes, it’s always a Lie group. It is even a closed subgroup (for the ordinary topology) of the linear algebraic group $\Aut(\mathfrak{g})$, consisting of those automorphisms whose induced automorphism of $\tilde{G}$ maps $Z$ onto itself (where $G=\tilde{G}/Z$ with $\tilde{G}$ simply connected). Beware that $\Aut(G)$ can have infinitely many components: for instance $\Aut((\mathbf{R}/\mathbf{Z})^n)\simeq\operatorname{GL}_n(\mathbf{Z})$ has infinitely components if $n\ge 2$).

An elaboration of the previous case shows that $\Aut(G)$ is a Lie group if $G$ has finitely many components; I think it remains true if $G/G^\circ$ is finitely generated (i.e., $G$ is compactly generated) but I haven’t checked.

If $G$ is arbitrary, $\Aut(G)$ is not necessarily a Lie group. Examples can be found with $G$ discrete and countable. For instance, for $p$ a prime, the group $\Aut((\mathbf{Z}[1/p]/\mathbf{Z})^n)\simeq\operatorname{GL}_n(\mathbf{Z}_p)$ is compact but not Lie if $n\ge 1$; the group $\Aut(\mathbf{Z}^{(\mathbf{N})})$ is Polish but not locally compact.