4
$\begingroup$

This question already has an answer here:

Let $G$ be a locally compact totally disconnected group and let $\phi$ be a surjective homomorphism from $G\to H$ (added later: where $H$ has the topology coinduced by $\phi$). Is H also locally compact and totally disconnected? If not is there a homomorphism from such a group to a Lie group?

More generally what happens if we remove local compactness? Is total disconnectedness still preserved

$\endgroup$

marked as duplicate by Francois Ziegler, Jan-Christoph Schlage-Puchta, Gregory Arone, Community Sep 11 '18 at 3:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 5
    $\begingroup$ I think you are missing some conditions to pin down the examples/scenario you have in mind. For otherwise, I can cook up silly counterexamples as follows: take $H$ to be your favourite topological group, and then take $G$ to be $H$ equipped with the discrete topology. $\endgroup$ – Yemon Choi Aug 30 '18 at 2:02
  • $\begingroup$ Or you can take $H$ to be a single point and $\phi$ trivial... $\endgroup$ – Arturo Magidin Aug 30 '18 at 2:26
  • 2
    $\begingroup$ @ArturoMagidin using trivial $H$ is not a counterexample, since the trivial group with its only possible topology is totally disconnected (its only connected subsets are points). $\endgroup$ – KConrad Aug 30 '18 at 3:12
  • 2
    $\begingroup$ Downvoting until the OP clarifies what was actually intended, given that the stated question has a trivial/silly counterexample $\endgroup$ – Yemon Choi Aug 30 '18 at 11:42
  • 2
    $\begingroup$ This question is fully answered here: mathoverflow.net/a/290157/89334 where in particular an example of a (Hausdorff) totally disconnected group with a (Hausdorff) connected quotient is given. $\endgroup$ – Uri Bader Sep 1 '18 at 16:05
8
$\begingroup$

The quotient group $H$ will be totally disconnected and locally compact if it is Hausdorff (or at least $T_1$) and $H$ inherits the quotient topology.

Let $\phi: G\to H$ be a surjective homomorphism of Hausdorff topological groups, $G$ locally compact and totally disconnected. Let $1, h \in H$, and $1\in U, h\notin U$, $U \subset H$ an open subset. Then $\phi^{-1}(U)$ is an open subset of $G$, saturated by cosets of the closed normal subgroup $\phi^{-1}(1)= N \lhd G$. Then $1\in G$ has a basis of clopen subgroup neighborhoods, so we can find $C < G$ clopen with $C \subset \phi^{-1}(U)$. Then $C\cdot N \subset \phi^{-1}(U)$ is an open subgroup, hence $C\cdot N$ is closed (the union of its non-trivial cosets is an open complement). Moreover $\phi$ is an open map, and $1 \in \phi(C), h\in \phi(G-C\cdot N)$ are clopen sets separating $1$ from $h$. Hence $H$ is totally disconnected.

On the other hand, if $H$ is not Hausdorff, then it might be connected. For example, $\mathbb{Z}_p/\mathbb{Z}$ has the trivial quotient topology, since $ \mathbb{Z}$ is dense. Usually topological groups are required to be Hausdorff, so I'm not sure if you were allowing this pathology.

$\endgroup$
6
$\begingroup$

No [Update: well, the answer was no to the original question, but the OP later without comment edited the question in a way that invalidates this answer.] . Let $\mu_{p^\infty}$ be the group of $p$th-power roots of unity in $\mathbf C^\times$, with the topology it gets as a subset of $\mathbf C^\times$. This group is not locally compact, since there is no compact neighborhood of 1 in $\mu_{p^\infty}$: a nonempty compact Hausdorff space without isolated points is uncountable, while the full group $\mu_{p^\infty}$ is countable. We'll show $\mu_{p^\infty}$ with its topology inside $\mathbf C^\times$ is the image of $\mathbf Q_p$ under a continuous homomorphism.

Set $\chi \colon \mathbf Q_p \rightarrow \mu_{p^\infty}$ by $\chi(x) = e^{2\pi i\{x\}_p}$, where $\{x\}_p$ is the $p$-adic fractional part of the $p$-adic number $x$. This $\chi$ is a homomorphism with kernel $\mathbf Z_p$, which is open in $\mathbf Q_p$, so $\chi$ is locally constant on $\mathbf Q_p$ and thus is continuous. It is surjective since for $0 \leq a < p^n$ we have $e^{2\pi i a/p^n} = \chi(a/p^n)$.

Remark: as abstract groups $\mu_{p^\infty} \cong \mathbf Q_p/\mathbf Z_p$, but as topological groups they are not the same since $\mathbf Q_p/\mathbf Z_p$ (with its quotient topology) has the discrete topology. The group $\mathbf Q_p/\mathbf Z_p$ with its discrete topology is locally compact and totally disconnected, so that would not provide a counterexample for your question.

$\endgroup$
  • $\begingroup$ This is a special instance of the general principle commented above by @Yemon Choi. $\endgroup$ – Uri Bader Sep 1 '18 at 16:09
  • $\begingroup$ After some fiddling I agree this does fit Yemon's scenario: replacing the domain of $\chi$ with its quotient by the kernel equipped with the quotient topology, the domain becomes $\mathbf Q_p/\mathbf Z_p$ as a discrete group. $\endgroup$ – KConrad Sep 2 '18 at 1:38
4
$\begingroup$

A rather trivial counterexample: Take $H$ to be your favourite topological group that fails to be locally compact and/or totally disconnected; take $G$ to be $H$ with the discrete topology. Then $G$ is locally compact and totally disconnected (any discrete space is both), and the identity function $G \to H$ is a continuous surjective homomorphism.

So in sum: Neither local compactness nor total disconnectedness is preserved under images, either separately or together.

Edit: I notice after posting that Yemon Choi already posted this in comments; but since it answers the question, I’ll leave this up unless Yemon would prefer to make it an answer himself.

$\endgroup$
  • 2
    $\begingroup$ Thanks Peter. TBH I was hoping that the OP would amend and am a bit surprised that several users have seen fit to give elaborate answers when there was a trivial counterexample. $\endgroup$ – Yemon Choi Aug 30 '18 at 11:41
  • $\begingroup$ @YemonChoi and my elaborate answer wasn't even right ... $\endgroup$ – Nik Weaver Aug 30 '18 at 12:52

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