Timeline for Do acyclic amenable groups exist?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 2 at 20:12 | comment | added | Matt Zaremsky | @YCor Oh duh, right, use $A_n$ for twisting instead, that still successfully kills the abelianization of $H_n$ but now doesn't create any new stuff. (Still no idea about higher homology.) | |
| Mar 2 at 12:52 | comment | added | YCor | @MattZaremsky these map onto the 2-element cyclic group (signature on $S_n$) | |
| Mar 2 at 1:14 | vote | accept | Denis T | ||
| Mar 1 at 13:10 | answer | added | Nicolas Monod | timeline score: 12 | |
| Feb 29 at 14:15 | comment | added | Matt Zaremsky | I feel like maybe twisted Houghton groups are a candidate? By twisted I mean take the usual Houghton group $H_n$ and semidirect it with $S_n$ (so, elementary amenable). These groups are perfect, and homologically stable (arxiv.org/abs/1509.07639), though I can't find anything about whether the stable homology is (non-)trivial. By analogy to Higman-Thompson groups, the stability feels like it could lead to them being acyclic, as in arxiv.org/abs/1411.5035. (Or, maybe this doesn't work, and there's some obvious reason twisted $H_n$ has non-trivial second homology?...) | |
| Feb 29 at 12:34 | comment | added | Igor Belegradek | Is anything known about acyclicity of groups that are virtually solvable? elementary amenable? | |
| Feb 29 at 11:26 | comment | added | Denis T | @YCor Owen Tanner calculated homology of interval exchange groups in many cases, and results seem to imply that they are never acyclic (neither their derived subgroups, by looking at spectral sequences). I think I will add a list of non-examples which at first glance may look promising. | |
| Feb 29 at 10:21 | history | edited | YCor | CC BY-SA 4.0 |
removed trivial example
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| Feb 29 at 10:17 | comment | added | YCor | Let $\Lambda$ be a subgroup of $\mathbf{R}$ containing $\mathbf{Q}$ such that $\Lambda/\mathbf{Q}$ is also isomorphic to $\mathbf{Q}$. Let $G_\Lambda$ be the set of interval exchanges such that $f(x)-x\in\Lambda$. (Interval exchange: right continuous permutation of $[0,1[$ that is piecewise a translation with finitely many breakpoints) Then $G_\Lambda$ is known to be amenable (this is essentially due to Juschenko-Monod), and its derived subgroup $G'_\Lambda$ is a simple group (essentially due to Sah). I do not know whether $G'_\Lambda$ is acyclic. | |
| Feb 29 at 10:04 | history | edited | Denis T | CC BY-SA 4.0 |
added 98 characters in body
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| Feb 29 at 0:06 | history | asked | Denis T | CC BY-SA 4.0 |