Set of topologies on a group making it a compact Hausdorff topological group

Question

Maybe stupid, but from the following well known facts about compact Hausdorff (CH) spaces:

• CH topologies on a given set are pairwise incomparible (one is not finer or coarser than the other).
• There exists only one CH topology on a finite set (a direct consequence of the above)

I always wondered: are there interesting things to say about the set of CH topologies on a given set (maybe depending on its cardinal number, beginning with the first case after the finite case, namely countable)?

In particular, are there interesting things to say about the set of CH topologies on a given group that make it into a compact group?

Answer 1

The number of compact Hausdorff group topologies on a given group strongly depends on the algebraic structure of the group.

For example, any finite-dimensional torus $(\mathbb R/\mathbb Z)^n$ has $2^{\mathfrak c}$ automorphisms, among which only finitely many continuous. This implies that $(\mathbb R/\mathbb Z)^n$ has $2^{\mathfrak c}$ pairwise incomparable compact Hausdorff group topologies.

On the other hand, for any odd $n$ the group $SO(n,\mathbb R)$ has a unique compact Hausdorff topology, according to a classical result (1933) of van der Waerden.

A compact topological group $G$ is called van der Waerden (or else self-bohrifying) if every group homomorphism $G\to K$ to a compact Hausdorff topological group is continuous. It is easy to see that van der Waerden groups admit a unique compact Hausdorff topology. This paper of Hart and Kunen contains many examples of van der Waerden (=self-bohrifying) compact Hausdorff topological groups.

In particular, by Lemma 5.20 in the mentioned paper of Hart and Kunen, a countable product $\prod_{k\in\omega}G_k$ of finite groups is self-bohrifying if

1) No group occurs infinitely often in the list $(G_n)_{n\in\omega}$;

2) Each $G_k$ is either $A_{k_n}$ or $PSL(j_k, q_k)$ or $SL(j_k, q_k)$, where $\sup_k j_k=\infty$.

Answer 2

As Taras Banakh says, it really depends on the underlying group. Some comments in the direction of having a unique CH group topology (which of course is not the case in general):

Profinite groups are residually finite, whereas connected compact Hausdorff groups are divisible. So if a group $G$ admits a CH group topology, then $G$ has a largest divisible normal subgroup $N$, which is the connected component of the identity in any CH group topology on $G$, and then $G/N$ is residually finite and admits a profinite topology. So the problem effectively reduces to looking at residually finite groups (where CH = profinite) and divisible groups (where CH => connected).

A difficult theorem of Nikolov and Segal says the following: if $G$ is a profinite group with a finitely generated dense subgroup, then every finite index subgroup of $G$ is open. In other words, the profinite topology of $G$ as an abstract group is compact, and it is the only CH group topology on $G$.

Another common situation where you get a unique CH topology is when there is a CH topology generated as group topology by normalizers of centralizers. This is the case, for instance, for the automorphism group of a locally finite rooted tree (this group does not necessarily have a dense finitely generated subgroup). A sufficient condition for this kind of argument to work is as follows: suppose your group $G$ acts faithfully on the Cantor set, such that the pointwise fixator of any proper clopen subset is infinite. Give $G$ the coarsest group topology such that the action is continuous. If that topology is compact, then it is the unique CH group topology for $G$.