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Is there an example of a nontrivial discrete amenable group with vanishing integral homology?

To put the question in contrapositive. Given arbitrary acyclic group $Q$, is there some reason for the Johnson cocycle in $Z^1_b(Q, (\ell^{\infty}(Q)/\mathbb R)^*)$ (universal cocycle obstructing amenability, which sends pair of elements $(g_1, g_2)$ to the difference of delta measures supported on them) to be nontrivial?

It is well known that every perfect group is a homomorphic image of some acyclic group. If the answer to the question is negative, can we say something about homology of the kernels of those acyclic covers of perfect amenable groups?

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Feb 29 at 10:17
  • $\begingroup$ @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. $\endgroup$
    – Denis T
    Commented Feb 29 at 11:26
  • $\begingroup$ Is anything known about acyclicity of groups that are virtually solvable? elementary amenable? $\endgroup$
    – Igor Belegradek
    Commented Feb 29 at 12:34
  • $\begingroup$ 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?...) $\endgroup$
    – Matt Zaremsky
    Commented Feb 29 at 14:15
  • $\begingroup$ @MattZaremsky these map onto the 2-element cyclic group (signature on $S_n$) $\endgroup$
    – YCor
    Commented Mar 2 at 12:52

1 Answer 1

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(1) Acyclic amenable groups do exist, because binate amenable groups exists: for instance, Philipp Hall's "universal locally finite group", which is by definition the Fraïssé limit of all finite groups, is such an example.

A reference is § 3.1.3 in A.J. Berrrick, "A topologist’s view of perfect and acyclic groups" in "Invitations to Geometry and Topology", Oxford University Press 2002.

(2) On the other hand, your "contrapositive question" is in fact not equivalent because the coefficients are non-trivial.

On the contrary, it is true that the Johnson cocycle is non-trivial for all non-amenable groups, even in "usual" cohomology:

Indeed the non-triviality in H^1 (bounded or not!) is simply a consequence of the long exact sequence of coefficients in either bounded or ordinary cohomology.

Put simply: the absence of an invariant mean gives a non-trivial class in H^1 (with non-trivial coefficients, not Z...), and this class happens to be bounded.

Nicolas Monod

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