# What can we say about $Aut(G)$ for an arbitrary Lie group $G$?

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

(1) Is $Aut(G)$ again a smooth manifold? and particularly a Lie group?

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

Thanks

• By "automorphism" do you mean "smooth automorphism"? – Igor Rivin Mar 3 '16 at 14:22

(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 $\mathrm{Aut}(G)$ 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 $\mathrm{Aut}(\mathfrak{g})$, 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 $\mathrm{Aut}(G)$ can have infinitely many components: for instance $\mathrm{Aut}((\mathbf{R}/\mathbf{Z})^n)\simeq\mathrm{GL}_n(\mathbf{Z})$ has infinitely components if $n\ge 2$).
An elaboration of the previous case shows that $\mathrm{Aut}(G)$ 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, $\mathrm{Aut}(G)$ is not necessarily a Lie group. Examples can be found with $G$ discrete and countable. For instance, for $p$ a prime, the group $\mathrm{Aut}((\mathbf{Z}[1/p]/\mathbf{Z})^n)\simeq\mathrm{GL}_n(\mathbf{Z}_p)$ is compact but not Lie if $n\ge 1$; the group $\mathrm{Aut}(\mathbf{Z}^{(\mathbf{N})})$ is Polish but not locally compact.