# Lie algebra of the automorphism group of a Lie group?

Let $G$ be a Lie group and $\text{Aut}(G)$ the group of all Lie group automorphisms of $G$. If $\text{Aut}(G)$ can be interpreted to be a Lie group (for example, in the context of synthetic differential geometry), is there a nice characterization of it's Lie algebra?

The Lie algebra of $\text{Diff}(G)$ is $\mathcal{X}(G)$, the space of all vector fields on $G$, so the Lie algebra of $\text{Aut}(G)$ should be a Lie subalgebra of $\mathcal{X}(G)$.

Edit, 12/26/15: Following YCor's comments below, let's restrict to the case that $G$ is connected for simplicity. Here $\text{Aut}(G)$ is always a Lie group (in just plain old differential geometry), because it's always a closed subgroup of $\text{Aut}(\mathfrak{g})$, which is in turn a closed subgroup of $GL(\mathfrak{g})$.
The Lie algebra of $\text{Aut}(\mathfrak{g})$ is quite nice: it's the Lie algebra $\text{Der}(\mathfrak{g})$ of derivations on $\mathfrak{g}$. In general (for example, when $G = S^1$), $\text{Aut}(G)$ will be a proper subgroup of $\text{Aut}(\mathfrak{g})$, but the two agree if $G$ is simply connected. $\text{Aut}(G)$ always has a subgroup $\text{Inn}(G)$ of inner automorphisms, and correspondingly its Lie algebra always contains the subalgebra $\text{Inn}(\mathfrak{g})$ of inner derivations of $\mathfrak{g}$.
• The topology on $\mathrm{Aut}(G)$ has to be specified: it's usually the topology of uniform convergence on compact subset, which is also the compact-open topology. When $G/G_0$ is finite, or more generally is finitely generated, this makes $\mathrm{Aut}(G)$ a Lie group. In more generality this fails, e.g. when $G=(\mathbf{Z}[1/p]/\mathbf{Z})$, then $\mathrm{Aut}(G)$ is topologically isomorphic to $\mathbf{Z}_p$ (the $p$-adics); when $G$ is a vector space of infinite countable dimension over $\mathbf{F}_p$ or $\mathbf{Q}$ it is not even locally compact. – YCor Dec 26 '15 at 23:27
• Also, in an example such as $G=\mathbf{R}\times\mathbf{Z}$, the group $\mathrm{Aut}(G)$ is a Lie group but $\mathrm{Aut}(G)_0$ has larger dimension than $\mathrm{Der}(\mathfrak{g})$, since the kernel of the restriction map $\mathrm{Aut}(G)\to\mathrm{Aut}(G_0)$ has positive dimension. – YCor Dec 26 '15 at 23:35
• @YCor: thanks for the correction. I had in mind the case that $G/G_0$ is finite, and for $\text{Aut}(G)$ to have the topology induced from the obvious topology on $GL(\mathfrak{g})$. I'll just edit to restrict to the connected case for simplicity. – Qiaochu Yuan Dec 26 '15 at 23:47
• Even when $G/G_0$ is finite, the action of $\mathrm{Aut}(G)$ on $G_0$ is not necessarily faithful (e.g., when $G=A\times B$ when $A$ is the circle group and $B$ is the 2-element group), so the topology induced by the topology of $\mathrm{GL}(\mathfrak{g)}$ is not Hausdorff. But if $G$ is connected it's fine (it coincides with the compact-open topology). – YCor Dec 27 '15 at 0:14
The Lie algebra of $Aut(G)$ is the space of derivations of the Lie algebra of $G$
• This is not true without some hypotheses on $G$ (simply connected suffices). For example, it's not true when $G = S^1$. – Qiaochu Yuan Dec 26 '15 at 9:29