Timeline for Distinct, non-homeomorphic, profinite topologies on a given abstract group ?
Current License: CC BY-SA 3.0
7 events
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| Apr 26, 2011 at 12:45 | comment | added | Stephan F. Kroneck | @ 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. | |
| Apr 24, 2011 at 12:25 | comment | added | Joel David Hamkins | 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. | |
| Apr 24, 2011 at 11:19 | comment | added | Yiftach Barnea | 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. | |
| Apr 24, 2011 at 11:15 | history | edited | Yiftach Barnea | CC BY-SA 3.0 |
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| Apr 24, 2011 at 10:39 | comment | added | Stephan F. Kroneck | @ 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. | |
| Apr 24, 2011 at 10:31 | comment | added | Stephan F. Kroneck | @ 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" ! ... | |
| Apr 24, 2011 at 9:19 | history | answered | Yiftach Barnea | CC BY-SA 3.0 |