Timeline for Example of a non-normal infinite index subgroup of a non-amenable group with certain properties.
Current License: CC BY-SA 2.5
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 5, 2011 at 21:39 | comment | added | Łukasz Grabowski | @Alain: What about the assumption of having no normal subgroups of finite index? Anyway, I admit that after Andreas Thom gave his answer here mathoverflow.net/questions/59317/… I gave up thinking about it, since my only motivation was to construct the equivalence relation as described above | |
| May 4, 2011 at 14:52 | comment | added | Alain Valette | The question has been answered by Kate J in this question: mathoverflow.net/questions/59166/… | |
| Apr 19, 2011 at 19:11 | answer | added | Alain Valette | timeline score: 4 | |
| Mar 26, 2011 at 0:22 | comment | added | Jesse Peterson | Sorry, I think now that essential freeness is actually equivalent to any non-identity element having infinitely many non-fixed points, and so is not equivalent to faithfulness, as I put in my last comment. But I still think that the invariant mean should give you only non-strong ergodicity, and not in general give hyperfinite. | |
| Mar 25, 2011 at 5:08 | comment | added | Jesse Peterson |
If $G/H$ has a $G$ invariant mean then this will imply that the action of $G$ on $\{ 0, 1 \}^{G/H}$ will not be strongly ergodic (in fact this is if and only if). But this actions will be essentially free as soon as it is faithful and so in this case should not give an amenable equivalence relation unless $G$ itself is amenable.
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| Mar 24, 2011 at 14:04 | history | edited | Łukasz Grabowski | CC BY-SA 2.5 |
added 1417 characters in body; added 16 characters in body
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| Mar 23, 2011 at 18:31 | comment | added | Jesse Peterson | How do you produce an amenable equivalence relation from such a subgroup? | |
| Mar 23, 2011 at 16:16 | comment | added | Łukasz Grabowski | @Kate: this time not, because I write "no finite-index subgroup of H is normal in G", so it's automatic :-) | |
| Mar 23, 2011 at 16:13 | comment | added | Kate Juschenko | I guess you need index strictly greater than 1 again? | |
| Mar 23, 2011 at 15:54 | history | asked | Łukasz Grabowski | CC BY-SA 2.5 |