[Math] Lie algebra of the automorphism group of a Lie group

lie-algebraslie-groups

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 $\mathcal{X}(G)$.

Best Answer

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 $\mathfrak{g}$.

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