Profinite groups as Galois groups over $\Bbb C(X)$

Question

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\ab}{\mathrm{ab}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\prolim}{\varprojlim} $ It is known that every finite group is isomorphic to the Galois group of some finite extension $K / \C(T)$, using Riemann existence theorem (Remark 2.14 a) in Völklein, Groups as Galois groups, an introduction).

Is every profinite group $G$ of cardinality $2^{\aleph_0}$ isomorphic (as topological group) to the Galois group of some extension of $\C(X)$ ?

If not, what if we require these profinite groups to be abelian, or/and torsion-free?

Ideas :

Notice that not all profinite groups are Galois groups over $\C(X)$, since $\Gal(L/K) \subset L^L$ has cardinality $2^{|L|}$ (when $L$ is infinite), which is at most $2^{| \overline{\C(X)}|} = 2^{2^{\aleph_0}}$. By a work of Haran and Jarden (2000), one knows that the absolute Galois group of $\C(X)$ is the profinite completion of the (discrete) free group $F_{2^{\aleph_0}}$, which has cardinality $2^{2^{\aleph_0}}$ (I think). So I could also ask about profinite groups of such cardinality, but I'm sticking to ${2^{\aleph_0}}$ for now.

Every group of cardinality ${2^{\aleph_0}}$ is a quotient of $F_{2^{\aleph_0}}$, but I'm not sure if every profinite group of such cardinality is a topological quotient of $\widehat{F_{2^{\aleph_0}}}$ (i.e. quotient by a closed subgroup).

(An analoguous question for $\Q$ was asked here).

Answer 1

The answer is yes. It is a theorem of Douady (see e.g. Theorem 3.4.8 in Szamuely's "Galois Groups and Fundamental Groups") that the absolute Galois group of $\mathbb C(X)$ is isomorphic to the free profinite group on the set $\mathbb C$. This can be deduced in purely group-theoretic way from the results you already know thanks to Riemann existence theorem, namely that for any finite set $S\subseteq\mathbb C$, the quotient parametrizing covers unramified outside $S\cup\{\infty\}$ is isomorphic to free profinite group on $S$. Analogous results hold for any algebraically closed field of characteristic zero.

Now, any profinite group with at most $2^{\aleph_0}$ elements is a quotient of a free profinite group on $2^{\aleph_0}$ generators, meaning it appears as a Galois group of an extension of $\mathbb C(X)$.

We can in fact give a precise characterization of groups which appear this way: these are those profinite groups which have a dense subset of size $2^{\aleph_0}$. I would expect not every profinite group of cardinality $2^{2^{\aleph_0}}$ to have such a subset, but I don't have a counterexample off the top of my head.