6
$\begingroup$

Can you define a group topology on a group by specifying which subgroups should be discrete with respect to that topology (where a subgroup $S$ of $G$ is discrete if each $s\in S$ has an open neighborhood $U_s$ such that if $s_1\neq s_2$, then $U_{s_1}$ and $U_{s_2}$ are disjoint)? If not, to what extent is the topology on a group defined if one specifies that a certain collection of subgroups should be discrete? For example, taking the example of the additive group $\mathbb Q$, can you define the usual topology inherited from the reals by specifying that the only discrete subgroups should be of the form $q\mathbb{Z}$ for $q\in \mathbb{Q}$?

Thanks

$\endgroup$
  • $\begingroup$ I posted (as an answer) and deleted the following example: Define a topology on $\mathbb Q$ by declaring the open sets to be of the form $r\mathbb Z+U$ for any non-zero rational $r$, and any $U$ which is open in the usual topology. I believe it is a $T_2$ topological group, but problem with it is that the subgroup $\mathbb Z$ is not discrete. Your question might be related to so-called discretely generated topological spaces (perhaps providing a positive answer for such topological groups, that would include first-countable, monotonically normal), see e.g. $\endgroup$ – Mirko Mar 27 '14 at 15:38
  • $\begingroup$ In general it's sort or hopeless. For instance, the topological additive group of $p$-adics $\mathbf{Z}_p$ has only the trivial subgroup as a discrete subgroup. The same holds for the infinite product $G$ of infinitely many copies of $\mathbf{Z}_p$. Now consider a non-continuous homomorphism $f:G\to C_p$, where $C_p=\mathbf{Z}/p\mathbf{Z}$, and endow $G$ with a new finer topology, namely the topology induced by the diagonal embedding into $G\times C_p$. Then the new topology also has no nontrivial discrete subgroup. Both are Hausdorff. $\endgroup$ – YCor Mar 28 '14 at 13:08
  • $\begingroup$ I have realised that my question is false for a different reason to that given by user46855 - for if a certain subgroup is discrete under some topology, then it will most certainly be discrete under any finer topology. So then a different question arises - is there always a finest topology under which a certain collection of subgroups in discrete, i.e. is the set of all topologies making each of those subgroups discrete closed under arbitrary meets? $\endgroup$ – Liam Baker Apr 8 '14 at 7:30
  • $\begingroup$ Since, as you say, your condition is preserved by refining, so that the discrete topology is the finest topology satisfying the desired condition, surely you mean to ask whether there is a coarsest topology making exactly the desired subgroups discrete? $\endgroup$ – LSpice Oct 20 '19 at 21:05
  • 1
    $\begingroup$ @LSpice yes indeed; thank you for the correction! $\endgroup$ – Liam Baker Oct 22 '19 at 8:37
4
$\begingroup$

For a negative answer to the first question: In a topological group discrete (and more generally locally compact, or even locally closed) subgroups are closed. So in a compact group the discrete subgroups are exactly the finite subgroups. Taking a compact group with a discontinuous group automorphism (example: the unit circle, using choice to decompose it in indecomposable components as divisible abelian group) transporting the topology with the automorphism gives a different compact group topology.

For more general related results, see Comfort's chapter about topological groups in the handbook of set theoretical topology, specifically 11.3 pag. 1248 and 11.1 pag. 1247. Two samples from that source:

Ross [1965] [...] raised the following question: If $T_1$ and $T_2$ are locally compact topological group topologies on the Abelian group $G$ with the same closed subgroups, must $(G, T_1)$ and $(G, T_2)$ be topologically isomorphic? RICKERT [1967] and RAJAGOPALAN [1968] showed that under various mild and natural additional hypotheses the answer is "Yes"; recently MOSKALENKO [1981] has constructed an example showing that the answer to Ross' question is "No".

MOSKALENKO, Z.I. [1981] Existence of nonisomorphic locally compact topologizations of an abelian group with identical sets of closed subgroups, Ukrainski{ Mat. Z., 33, 820-823. [In Russian. English translation: Ukranian Math. J., 33 (1982), 620-622.]

RAJAGOPALAN, M. and H. SUBRAHMANIAN [1976] Dense subgroups of locally compact groups, Colloq. Math., 35, 289-292.

RICKERT, N.W. [1967] Locally compact topologies for groups, Trans. Amer. Math. Soc., 126, 225-235.

JANAKIRAMAN and SOUNDARARAJAN [1982] have given several characterizations of those compact, totally disconnected Abelian topological groups which admit strictly finer totally bounded topological group topologies with the same closed subgroups, showing that such groups exist in profusion

The above results are not directly applicable to the additive group of rational numbers with the usual (locally pre-compact hence not locally bounded but not locally compact) topology. One knows a great profusion of non locally bounded ring topologies on the rational numbers [see the books about topological fields that I have already cited in another answer], but since their definition is not plain I have not checked what are their discrete subgroups.

$\endgroup$

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.