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I've learned about the notion of topologizability from "On topologizable and non-topologizable groups" by Klyachko, Olshanskii and Osin (http://arxiv.org/abs/1210.7895) - a discrete group $G$ is topologizable iff there exists a topology on $G$ which makes it into a Hausdorff non-discrete topological group.

Main question: Is every infinite amenable group topologizable?


Main motivation for this question is that perhaps naively non-topologizability seemed to me such a strange property that I hoped it could be used to show existence of non-sofic groups.

Question: Is there an infinite non-topologizable sofic group?

After failing to prove that sofic groups are topologizable I thought it would be still interesting to prove that infinite "elementary sofic" groups are topologizable. Elementary sofic groups are for the purpose of this discussion the class of "groups which are provably sofic by current methods", i.e. it contains all amenable groups, is closed under taking free products amalgamated over amenable groups, extensions with sofic kernel and amenable quotient, /any other results which are in the literature/, and with the property that if G is residually elementary sofic then G is elementary sofic.

Question: Is there an infinite non-topologizable elementary sofic group?

Unfortunately my plan to answer the above question negatively failed at step 1, and hence the Main question.

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Are the integers topologizable? What does that topology look like? –  Ben Willson Nov 27 '12 at 22:08
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Wouldn't the evenly spaced integer topology work? –  Nathan Nov 27 '12 at 23:24
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@Ben: you can use profinite topology, e.g. fix a prime number $p$; then integer $n$ is "near" to $0$ if large power of $p$ divides $n$. Other way to topologize integers is - take the action of integers $Z$ on the circle such that the generator $t$ of $Z$ acts by irrational rotation, and define $t^k$ to be "close" to the identity element iff $t^k$ is a rotation by a "small" angle. This is special case of "find an infinite cyclic subggroup of a compact group and induce the topology" –  Łukasz Grabowski Nov 28 '12 at 0:19
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@Simone: AFAI understand, Bohr compactif'n B(G) is in particular compact, so by Peter-Weyl thm if G embeds in B(G) then G is res. linear, and so by Malcev thm if it is fin. generated then it is res. finite. So to produce example of an amenable group G which doesn't embed into B(G) take a fin. generated simple amenable group, or easier take a fin.gen. solvable non-res. finite group (e.g. BS(1,n)). Taking a simple group is more convincing though, because the image in B(G) is trivial, whereas if the image in B(G) is infinite one could still hope for inducing a (Hausdorff) topology on G somehow... –  Łukasz Grabowski Dec 3 '12 at 18:12
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I essentially stopped thinking about this, because I don't know how to proceed, but I thought I'd share one idea which at one point I thought was hopeful: G acts on a certain metric space X by isometries - namely fix a mean m on G and define X to be the set of subsets of G up to sets of mean 0. The metric on X is d(A,B) = m(A-B \cup B-A). There are various topologies on Isom(X) but I failed to prove any of them gives a Hausdorff non-discrete topology on G. –  Łukasz Grabowski Dec 9 '12 at 22:45

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