It is known that for any (possibly infinite) Galois extension $L/K$, the Galois group is profinite, when endowed with the Krull topology (inherited from the inverse limit).
Can $\mathrm{Aut}(\Bbb C)$ be endowed with a topology with respect to which it is a profinite group (i.e. compact, Hausdorff, totally disconnected)?
Some thoughts:
According to the answer here, one might define a basis of open neighborhoods of the identity via $\{ \mathrm{Aut}(\Bbb C/K) \mid K \text{ finitely generated} \}$, and we obtain a topological group which is Hausdorff and totally disconnected. Is it correct? (The answer doesn't give any proof). If moreover it is locally compact, we could conclude that $\mathrm{Aut}(\Bbb C)$ is a locally profinite group.
But the answer above tells that this topological group is compact iff every $z \in \Bbb C$ has finite orbit. This is not the case, since $ \Bbb C \cong \overline{\Bbb Q(X_a : a \in \Bbb R)}$ and $X_a \mapsto X_{a+1}$ (for all $a \in \Bbb R$) extends to an automorphism of $\Bbb C$ with infinite orbit. However, the answer doesn't give any references, and this might not be the only topology that we may consider on $\mathrm{Aut}(\Bbb C)$ (of course, the discrete topology is totally disconnected and Hausdorff but not compact, while the trivial topology is compact but not Hausdorff).
Thank you!
