Just a silly little question which arose in connection with infinite Galois groups and their Krull topology:- can a given abstract group be endowed with distinct, non-homeomorphic, profinite topologies ? (I asked this question several years ago on the Topology Q+A and was told the question is undecidable and has something to do with supercompact cardinals). As I'm not that well-versed as concerns large cardinals etc., could someone verify/elucidate this please ? Thank you in advance ! Stephan.
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5$\begingroup$ Dear Stephan: 1) Good question: +1. 2) If you are willing to include (part of) your name in the question anyway, you should at least consider including it in your username (e.g. something like Stephan [first letter of your surname]). This is a professional forum and -- while it's certainly not required -- a lot of us much prefer when people use (some appropriate portion of) their actual names, rather than epithets, nicknames or pseudonyms. $\endgroup$– Pete L. ClarkCommented Apr 23, 2011 at 19:24
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$\begingroup$ "non-homeomorphic"?? All (separable) profinite topologies make your group homeomorphic to a Cantor set... or do you really care about non-separable topologies? A perhaps more interesting question is to find two topologies on the same abstract group that make them non-isomorphic in the category of topological groups. $\endgroup$– André HenriquesCommented Apr 23, 2011 at 21:07
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$\begingroup$ @ Pete L. Clark: Dear Pete, first of all thanks for your comments ! Secondly, as I'm quite new to this forum, I'm probably not aware of all the customs here - I simply gave my "generic" username when I "enrolled" here. (In particular, I wouldn't know how to "roll" the matter back ?!). At the time I didn't feel that I was doing any harm. (I stand corrected.) Thanks also for the point ! Kind regards, Stephan F. Kroneck. $\endgroup$– Stephan F. KroneckCommented Apr 23, 2011 at 21:51
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$\begingroup$ @ André Henriques: thanks for pointing out a possible source of misunderstanding; indeed, (as in the context of Galois groups with their Krull topology) the question was intended that way, i.e. that the given group structure be compatible with the profinite topology and that one has a topological group. Kind regards, Stephan. $\endgroup$– Stephan F. KroneckCommented Apr 23, 2011 at 21:51
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5$\begingroup$ For finitely generated profinite groups, the uniqueness of the topology follows from a result of Nikolov and Segal (see mathoverflow.net/questions/34290/…). $\endgroup$– Ian AgolCommented Apr 23, 2011 at 23:59
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2 Answers
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Yes.
I have classified some abelian examples: there are uncountably many pairwise non-homeomorphic pro-$p$ topologies that can be placed on the (unrestricted) product of any countable collection of cyclic $p$-groups of unbounded exponent.
The results are presented here, but I am in the process of redrafting http://arxiv.org/abs/1101.3005
answered Apr 24, 2011 at 13:47
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As Agol said in a finitely generated profinite group every subgroup of finite index is open. Therefore, the topology is unique and detremined by the algebra. This was first proved by Serre for pro-$p$ groups and eventually Nikolov and Segal proved it for any profinite groups.
Now, take $\mathbb{F}_p[[t]]$ formal power series over a field of $p$-elements and take their its abelian group. Then it is a metric pro-$p$ group which is the same as being countably based at $1$. On the other hand, take a vector space over $\mathbb{F}_p$ of a countable dimension and take its pro-$p$ complition. I am almost sure (so you might want to check the details) that in both cases you have a vector space of dimesnion $2^{\aleph_0}$ so the groups are isomorphic abstractly. But in the first case the topology is countably based at $1$ (and therefore in any points) while it is not countably based at $1$ in the second case.
You can read more about similar situation in Wilson's book on profinite groups on the chapter on free group.
answered Apr 24, 2011 at 9:19
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$\begingroup$ @ Agol & Yiftach Barnea: first of all thank you for your comments ! @ Agol: I suspect that in the case of say the absolute Galois group of the algebraic closure of $\Bbb Q$ the result of Segal will not apply; at present I am trying to think of non-trivial cases when it does. @ Yiftach Barnea: Unfortunately I do not know the book by Wilson (googled it, OUP, obviously) and would not know how to get hold of a copy (without purchasing it). Will have to check your remarks, but it would seem that the final answer to my question is a resounding "yes" ! ... $\endgroup$ Commented Apr 24, 2011 at 10:31
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$\begingroup$ @ Yiftach Barnea: (ctd.) ... That just leaves me to wonder why the answer on the Topology Q + A mentioned supercompact cardinals (maybe when comparing the dimensions you mentioned the details are not that simple after all ?) Kind regards, Stephan. @ Agol: By "non-trivial", I only meant the case of Galois groups of separable algebraic closures - what sources did you have in mind of fg. profinite groups ? Kind regards, Stephan. $\endgroup$ Commented Apr 24, 2011 at 10:39
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$\begingroup$ Wilson's book is pretty standard you should be able to find a copy in some library. Sorry, I have no idea what supercompact cardinals are and why and if they are related to any of this. $\endgroup$ Commented Apr 24, 2011 at 11:19
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$\begingroup$ An uncountable cardinal $\kappa$ is supercompact if there is a normal fine measure on $P_\kappa\theta$ for every ordinal $\theta$, which is equivalent to the existence of an elementary embedding of the universe into a transitive class $j:V\to M$ with critical point $\kappa$, such that $M^\theta\subset M$. Such cardinals are the best-known upper bound for the consistency of the Proper Forcing Axiom and other forcing axioms, such as Martin's Maximum, which are sometimes used in consistency arguments to construct such kind of examples, although I don't know the connection in this case. $\endgroup$ Commented Apr 24, 2011 at 12:25
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$\begingroup$ @ all: Thanks to all of you for your comments and insights; I would have liked to have been able to award several of you the reputations points for an answer ! In particular, I'm just glad that I can lay the matter to rest (it's been nagging at me for years !). So, I guess I would vote to close. Kind regards, Stephan. $\endgroup$ Commented Apr 26, 2011 at 12:45