13
$\begingroup$

Let $X$ be an infinite set, and let $G$ be the symmetric group on $X$. I want to understand $G$ by putting a topology on it, without imposing any more structure on $X$. What 'interesting' possibilities are there and what is known about them?

In particular, I have heard of the pointwise convergence topology (an open neighbourhood of g is a set of permutations that agree on some specified finite set of points), and found some papers on this, but are there any other topologies that have been studied?

What if I take the coarsest topology compatible with the group operations such that either a) the stabiliser of any subset is closed, b) the stabiliser of any partition is closed, or c) both are closed? I think these will be coarser than the pointwise convergence topology, because in the pointwise convergence topology the stabiliser of any first-order structure is closed. Is there a useful characterisation of the open subgroups and/or closed subgroups?

Improve this question
$\endgroup$
3
  • $\begingroup$ I just want to "second" that the topology you mention first -- I think of it as the natural [i.e., compact-open] function space topology on Sym(X) induced from the discrete topology on X -- has led to some interesting results. E.g. when rearranging an infinite series, it allows you to make statements like "almost every rearrangement [in the sense of Baire category] satisfies...." Could you provide some motivation for considering these other topologies? $\endgroup$
    – Pete L. Clark
    Commented Feb 17, 2010 at 16:50
  • 1
    $\begingroup$ I suppose it's inspired by the Zariski topology, which is quite coarse but still makes many of the 'natural' subgroups closed (eg the classical groups as subgroups of GL_n). In a permutation group context the most obvious kinds of proper subgroup are the intransitive ones (preserving a subset) and the imprimitive ones (preserving a partition). I also want to avoid giving finite subsets any special status a priori, so the topology behaves well under, for instance, dividing out by an equivalence relation with infinite parts. $\endgroup$
    – Colin Reid
    Commented Feb 17, 2010 at 17:36
  • 2
    $\begingroup$ I think it is known that if $X$ is countable, $G$ is a Polish group and $p:G\to Sym(X)$ is a universally measurable homomorphism (where $Sym(X)$ has its usual - i.e. pointwise - topology), then $p$ is automatically continuous. I don't know if it helps, but it does show that the usual topology on $Sym(X)$ is the minimal canonical one. $\endgroup$
    – Pandelis Dodos
    Commented Feb 17, 2010 at 17:58

1 Answer 1

Reset to default
9
$\begingroup$

It is known that there are exactly two separable group topologies on $S_\infty$ (i.e., on the group of permutations of a countable set): one is antidiscrete and the other one is topology of pointwise convergence. This is a statement of Theorem 6.26 in here. Hence Polish topology on $S_\infty$ is unique.

There are some abstract characterizations of closed subgroups, for example they are exactly the Polish groups with a countable basis of identity consisting of open subgroups. Another characterization is: a Polish group G is homeomorphic to a closed subgroup of $S_\infty$ if and only if it admits a compatible left invariant ultrametric. You may want to look at Section 1.5 of Becker and Kechris "The Descriptive Set Theory of Polish Group Action".

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks - it looks like the topologies I suggested are all actually the same as the pointwise convergence topology if X is countable. Is this true more generally? $\endgroup$
    – Colin Reid
    Commented Feb 17, 2010 at 20:54
  • $\begingroup$ I don't know what is the general situation. I added a remark to the post on characterization of closed subgroups. $\endgroup$
    – Konstantin Slutsky
    Commented Feb 17, 2010 at 22:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .