Is automorphism group of lie group a lie group?

Question

Let $G$ be a lie group. Is its automorphism group $\text{Aut}(G)$ of smooth automorphisms then a lie group?

According to G. Hochschild is $\text{Aut}(G)$ when equipped with the compact-open topology a lie group if the component group $G/G_0$ is finitely generated. Here is link https://www.jstor.org/stable/1990752?seq=1#metadata_info_tab_contents

What if $G/G_0$ is not finitely generated? I have tried to find an example of lie group with non-finitely generated component group whose automorphism group is not a lie group when equipped with the compact-open topology, but haven't found one. I suspect a discrete countable group whose automorphism group is uncountable can be such an example.

Answer 1

In practice I think it's very rare for people to consider Lie groups $G$ where $\pi_0(G)$ isn't just finite, but technically any countable discrete group is a (zero-dimensional) Lie group. So it suffices to find a (necessarily infinitely generated) countable discrete group $G$ whose automorphism group is not discrete when equipped with the compact-open topology. (You already know this but I'm clarifying for any other readers.)

We can take $G = \mathbb{Q}/\mathbb{Z}$ (with the discrete topology). It's a nice exercise to show that $\text{End}(G)$ is the profinite integers $\widehat{\mathbb{Z}}$ and so $\text{Aut}(G)$ is the group of units of the profinite integers, which is in particular uncountable. The compact-open topology here is the product topology on $G^G$; I don't understand this topology very well but I'd guess it doesn't have any uncountable discrete subspaces and hence that $\text{Aut}(G)$ is not discrete in it. The nicest possibility is that $\text{Aut}(G)$ would just have the profinite topology (which is certainly not discrete) but I don't know if that's true.

Answer 2

The compact-open topology is the coarsest topology on $\text{Aut}(G)$ so the action $\text{Aut}(G)\times G\longrightarrow G$ is continuous. With that topology the answer is NO.

Take $G$ to be any countable discrete group with uncountable automorphism group $\text{Aut}(G)$, for example $G=\mathbb{Q}/\mathbb{Z}$. If $X$ and $Y$ are topological spaces and $Y$ discrete, then $\mathcal{C}(X,Y)$ is totally disconnected wrt. the compact-open topology. To see this, let $f\in\mathcal{C}(X,Y)$ and $A\subseteq \mathcal{C}(X,Y)$ be a connected subset so $f\in A$. Then $\forall x \in X \quad B_x\equiv \{h\in A \mid h(x)=f(x)\}$ is non-empty and clopen in $A$. So $A=B_x \quad\forall x\in X$. $A=\bigcap_{x\in X}B_x=\{f\}$. That means $\mathcal{C}(X,Y)$ is totally disconnected.

Since $G$ is discrete, $\text{Aut}(G)$ is totally disconnected wrt. the compact-open topology. Since $\text{Aut}(G)$ is also uncountable, it is not a manifold.