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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?

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  • $\begingroup$ Hi, the two statements in the bullits are unclear, at least for me. Cannot be compared in what sense? In the second- do you mean that only one exists? $\endgroup$ – Amir Sagiv Feb 9 at 23:39
  • $\begingroup$ @AmirSagiv in the sense that the relation of inclusion between compact Hausdorff topologies on a set, is discrete. $\endgroup$ – YCor Feb 10 at 1:13
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    $\begingroup$ The first question seems reasonable, but the second one seems oddly phrased (and the title of your question is similarly a bit confusing to me). Do you want these CH topologies to be group topologies on the given group? $\endgroup$ – Yemon Choi Feb 10 at 3:18
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    $\begingroup$ For instance, there was an old MO question, the answers to which pointed out that if you equip a countably infinite group with a LCH group topology then this topology must be the usual disctrete topology (this follows from a Baire category argument). On the other hand, there are many non-homeomorphic countable compact sets. $\endgroup$ – Yemon Choi Feb 10 at 3:20
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    $\begingroup$ The two questions are somewhat unrelated, I'm not sure it's a good idea to keep them separately. Anyway, there's a reasonable classification of countable Hausdorff countable spaces up to homeomorphism (there are $\aleph_1$ many, classified by a countable ordinal and a nonzero finite number) and hence this provides some information on the set of compact Hausdorff topologies on a given countable set (there are $\aleph_1$ many modulo the group of permutations). $\endgroup$ – YCor Feb 10 at 4:57
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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 $\sup_k j_k=\infty$.

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    $\begingroup$ Ok thanks, that's exactly what I was looking for, I will have a look at this paper. Do you know where the Van Der Waerden result of 1933 appears ? Is the result you mention on the tori is a consequence of the various complex structures on tori via change of lattices ? Does the terminology "self-bohrifying" refer to Bohr ? $\endgroup$ – huurd Feb 10 at 8:27
  • $\begingroup$ @huurd The coordinates of the paper of van der Waerden can be found in this paper of Comfort (topology.auburn.edu/tp/reprints/v28/tp28205.pdf) who uses the terminology of van der Waerden groups. $\endgroup$ – Taras Banakh Feb 10 at 21:06
  • $\begingroup$ @huurd The result on tori implies from the algebraic description of the circle group $\mathbb R/\mathbb Z$ - it is the dirsct sum of the quasi-cyclic group (that is the group of roots of the unit) and continuum many copies of the group $\mathbb Q$ of rationals. There are $2^{\mathfrak c}$ many permutations of these copies of $\mathbb Q$, which yields $2^{\mathfrak c}$ automorphisms of $\mathbb R/\mathbb Z$. $\endgroup$ – Taras Banakh Feb 10 at 21:08
  • $\begingroup$ @huurd Yes, self-bohrifying means that the identity map G→bG from a group G to its Bohr compactification bG is bijective $\endgroup$ – Taras Banakh Feb 10 at 22:30
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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 $G$.

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