If $G$ is a topological group and $G_{{e}}$ is the identity component, the it is well known that $G_{{e}}$ is a normal subgroup of $G$ and the quotient group $G/G_{{e}}$ is totally disconnected. What can we say about the subgroup $H$ (need not be normal) of $G$ such that coset space is totally disconnected? Has it been studied for some topological groups such that the coset space is totally disconnected?

  • $\begingroup$ How do you want to form the quotient when $H$ is not normal? $\endgroup$
    – Lee Mosher
    Jan 6 '18 at 16:22
  • $\begingroup$ I mean coset space. I will edit. $\endgroup$
    – Steve
    Jan 6 '18 at 16:41
  • $\begingroup$ Unlike what you suggest, the question is already interesting when $H$ is normal (see Uri's abelian example). Of course, we can restrict to the case when $G$ is totally disconnected. $\endgroup$
    – YCor
    Jan 7 '18 at 23:27

Claim 1: For any topological group $G$ and a subgroup $H$, if $G/H$ is totally disconnected (td) then $H<G$ is a closed subgroup containing $G^o$ , the identity connected component of $G$.

Let me denote by $\pi:G\to G/H$ the map given by $\pi(g)=gH$. If $H$ is not closed then $\pi(\bar{H})$ is a connected set which is not a singleton. If $H$ does not contain $G^o$ then $\pi(G^o)$ is a connected set which is not a singleton. In both cases we get that $G/H$ is not td.

For locally compact groups this is an "if and only if".

Claim 2: If $G$ is locally compact and $H<G$ is a closed subgroup containing $G^o$ then $G/H$ is td.

We may identify $G/H$ with $(G/G^o)/(H/G^o)$ thus WLOG, $G=G/G^o$ and we get that $G$ is td locally compact and $H<G$ is a closed subgroup. In this case, by van Danzig theorem, for every $x\notin H$ we can find a compact open subgroup $K<G$ such that $Kx\cap H=\emptyset$, thus $\pi(K)$ is a compact open neighborhood of $eH$ not containing $xH$.

Let me note that there exist td groups (necessarily not locally compact) which have non-trivial connected quotients.

Example: Consider the space $\ell^1$ consisting of real absolutely summable sequences as an additive group and let $G$ be its subgroup consisting of rational sequences. Endow $G$ with the subspace topology. Note that the map $G\to \mathbb{R}$, $x\mapsto \sum x_n$ is a continuous and open surjective homomorphism, thus $\mathbb{R}\simeq G/H$ where $H$ is the subgroup consisting of rational absolutely summable sequences with sum 0. Thus $G/H$ is connected. However, $G$ is totally disconnected as every two points in $G$ could be separated by disjoint open sets, using preimages of rays under the application of functionals in $\ell^\infty\simeq (\ell^1)^*$.


The answer relies crucially on the underlying topology of $G$.

For example if $G$ is a $p$-adic group (e.g. $G = \mathrm{GL}_n(\mathbb Q_p)$), then you have infinitely many open compact subgroups $H$ of infinite index (for example $H = 1 + p^N \mathrm{Mat}_n (\mathbb Z_p)$), and due to open-ness the coset space $G / H$ is totally disconnected.

If instead your $G$ is a real (or complex) Lie group, then any open subgroup $H$ has the same Lie algebra, and thus will be finite index.

I think if you do not further specify some more properties on $G$, it will be hard to give a more precise answer.

  • 2
    $\begingroup$ Isn't any discrete group a zero dimensional Lie group? Because those surely can have open subgroups of infinite index. $\endgroup$ Jan 6 '18 at 17:30
  • $\begingroup$ You are correct, I should have said classical Lie group, or positive-dimensional Lie group. $\endgroup$
    – user94041
    Jan 6 '18 at 18:40
  • $\begingroup$ Classical should be Ok, positive-dimensional is still too weak (just take the direct product of your favorite Lie group with a discrete group). $\endgroup$ Jan 6 '18 at 18:44
  • 3
    $\begingroup$ For a locally compact group, the question has an immediate answer: $G/H$ is totally disconnected iff $H$ contains $G^0$, as mentioned by Uri. No specific discussion is needed for special classes of locally compact group such as $p$-adic Lie groups. $\endgroup$
    – YCor
    Jan 7 '18 at 23:29

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.