3
$\begingroup$

Is there a topological group which is profinite (i.e. compact and totally disconnected) and separable, but is not metrisable (equivalently is not first-countable)?

I know there are such topological space: see here. But among those listed there only the Stone-Cech compactification of $\mathbb{Z}$ has a topology which in theory can be that of a topological group (one of the other two is not Hausdorff and the second is first-countable but not metrisable, so neither can be a topological group). I don't know of any structure of a topological group on the Stone-Cech compactification of $\mathbb{Z}$, though.

If you replace "totally disconnected" with "connected" I have an example: $(\mathbb{R}/\mathbb{Z})^{2^{\aleph_0}}$. I was just wondering if the totally disconnected case had something similar. If $D$ is a finite group, is $D^{2^{\aleph_0}}$ separable?

In fact, I am even more interested in groups which are not just separable but in fact topologically finitely generated (i.e. admit a finitely generated dense subgroup).

I will appreciate all thoughts and comments on this.

$\endgroup$
  • $\begingroup$ I'm puzzled that you know how to prove that $(\mathbb{R}/\mathbb{Z})^{2^{\aleph_0}}$ is separable but not how to prove that $D^{2^{\aleph_0}}$ is separable if $D$ is finite. $\endgroup$ – Eric Wofsey Aug 18 '16 at 22:39
  • $\begingroup$ My proof is very twisted: it's by Kronecker's theorem. Using it it's clear this group contains a dense cyclic group, and hence it's separable. I'm sure there are more sensible ways to prove it, though. $\endgroup$ – Cronus Aug 18 '16 at 23:09
4
$\begingroup$

Let $F$ be any finite group and take the product $F^\mathbb{R}$. This is separable: for instance, the set of piecewise constant functions $\mathbb{R}\to F$ where the pieces are intervals with rational endpoints (and there are only finitely many pieces) is dense. As long as $F$ is nontrivial, $F^\mathbb{R}$ will not be first-countable.

On the other hand, there is no topologically finitely generated example. Indeed, a profinite group $G$ is topologically finitely generated iff there is a continuous surjective homomorphism $\widehat{F_n}\to G$, where $\widehat{F_n}$ is the profinite completion of the free group $F_n$ on $n$ generators. But $\widehat{F_n}$ is metrizable (proof: $F_n$ has only countably many different finite quotients up to isomorphism since a finite quotient is just a finite group with a collection of $n$ generators, so $\widehat{F_n}$ embeds in a countable product of finite groups). Any Hausdorff quotient of a compact metrizable space is metrizable, and hence $G$ is also metrizable.

(Note that while there is no natural group structure on $\beta\mathbb{Z}$ as you are hoping to construct, there is a natural universal way to compactify $\mathbb{Z}$ as a group, called the Bohr compactification, and this compactification is not metrizable. However, it turns out that the Bohr compactification of $\mathbb{Z}$ is not profinite! And the universal way to turn $\mathbb{Z}$ into a profinite group is just the profinite completion $\hat{\mathbb{Z}}$, which is metrizable.)

$\endgroup$
  • $\begingroup$ Very neat! You completely solved all my problems. Yes, now I remember the reason I disqualified $F^{2^{\aleph_0}}$ was because even if it is separable it's clearly not topologically finitely generated. $\endgroup$ – Cronus Aug 18 '16 at 23:12
  • $\begingroup$ The Bohr compactification is very cool, although at times the map into it is not injective, which is unfortunate. I'm not sure if this is the case with $\mathbb{Z}$; the definition of the Bohr compactification is not very constructive. $\endgroup$ – Cronus Aug 18 '16 at 23:15
  • $\begingroup$ The map into the Bohr compactification is injective for any locally compact abelian group, since by Pontryagin duality a locally compact abelian group has enough continuous homomorphisms to the compact group $\mathbb{R}/\mathbb{Z}$. $\endgroup$ – Eric Wofsey Aug 18 '16 at 23:19
  • $\begingroup$ Thank you! I'm not very fluent on these things. I really only read the Wikipedia article some years ago. $\endgroup$ – Cronus Aug 18 '16 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.