Edit (March 24): My first question has been answered nicely, but I am still looking for an answer to the second one.

Due to the Kan–Thurston theorem, the homology of an arbitrary group can be anything you want.

Using the Rips complex, we can see that hyperbolic groups are $F_{\infty}$ and, if torsion-free, even of type $F$, i.e., there exists a model of $BG$ that is a finite CW-complex.

The other side of Bridson's universe of finitely presented groups is where amenable groups live. I wondered about constraints that amenablility imposes on the cohomology of a group. More precisely, let me ask the following two separate questions.

Let $G$ denote a finitely presented torsion-free amenable group.

  • Is $G$ always of type $F$?

(True for nilpotent groups, see Brown's book. In general, this is probably false or open: Wikipedia taught me that it is an unresolved conjecture to prove that Thompson's group $F$ is not amenable.)

  • Can it happen that $H_{j}(G;Z)$ is non-trivial in a single degree $d$?

(For $d = 1$, we can obviously choose $G = \mathbf{Z}$, but I do not know any examples for bigger $d$.)


Is every finitely presented torsion-free amenable group of homotopical type F?

It's false even for metabelian groups.

For $i=1,2,3$, let $G_i$ be a copy of a solvable Baumslag-Solitar group, say $\langle t_i,x_i|t_ix_it_i^{-1}=x_i^2\rangle$. Start with $H=G_1\times G_2\times G_3$. It has a homomorphism $\phi$ onto $\mathbf{Z}$ mapping each $t_i$ to $1$ and each $x_i$ to $0$. Let $G$ be its kernel. Then $G$ is finitely presented, but not of type FP$_3$ / F$_3$.

This follows from the "FP$_m$-conjecture" for metabelian groups. Namely $H$ is not finitely presented by the Bieri-Strebel criterion [3], and $H$ is not FP$_3$ / F$_3$, as the "FP$_m$-conjecture" is known for $m=3$ in a semidirect product of two abelian groups [2], and also known in general when the Prüfer rank is finite [1].

Note: the exponent $2$ could be replaced with any number $\ge 2$, possibly depending on $i$. Also, $\phi_i(t_i)$ could be defined as any positive integer $n_i$, with the same conclusion (while if $\phi(t_1),\phi(t_2)>0$ and $\phi(t_3)<0$ the resulting kernel has type F).

[1] H. Åberg, Bieri–Strebel valuations (of finite rank), Proc. London Math. Soc. (3) 52 (1986) 269–304.

[2] R. Bieri, J. Harlander, On the FP3-conjecture for metabelian groups, J. London Math. Soc. (2) 64 (2001) 595–610.

[3] R. Bieri, R. Strebel, Valuations and finitely presented metabelian groups, Proc. London Math. Soc. (3) 41 (1980) 439–464.

  • $\begingroup$ (Note: taking $n$ copies of the Baumslag-Solitar and defining $H_n$ similarly, we get a metabelian group that is FP$_{n-1}$ and not FP$_n$. By the way, F$_n$ and FP$_n$ are equivalent for metabelian groups.) $\endgroup$
    – YCor
    Mar 15 at 10:14
  • 2
    $\begingroup$ "$F_n$ and $FP_n$ are equivalent for metabelian groups." - for the record, they're also equivalent for finitely presented groups. $\endgroup$ Mar 15 at 11:10
  • $\begingroup$ @MattZaremsky could you give a reference? $\endgroup$
    – ARG
    Mar 16 at 9:54
  • 2
    $\begingroup$ @ARG this is certainly in Brown's book "cohomology of groups". (Homotopical property F$_n$ is equivalent to [FP$_n$ + finitely presented]) $\endgroup$
    – YCor
    Mar 16 at 10:12
  • 1
    $\begingroup$ Yeah, it should be in Brown's book, and also in Geoghegan's book "Topological Methods in Group Theory". My favorite "reason" for (fin. pres. + FP_n)=(F_n) is that it's a group theoretic version of Hurewicz: (simply connected + homology vanishes)=(homotopy vanishes). $\endgroup$ Mar 16 at 10:59

Abels' groups provide simple examples of solvable groups that are finitely presented but not of type $F$. Given a ring $R$, define the group $$A_n(R):= \left\{ \left( \begin{array}{ccccc} 1 &&&& \\ & d_1 && \ast & \\ && \ddots && \\ & 0 && d_{n-1} & \\ &&&& 1 \end{array} \right) \in \mathrm{GL}(n+1,R), \ d_1,\ldots, d_{n-2} \in R^\times \right\}.$$ Extending Abels' result, Brown proved that, for every prime $p$ and every $n \geq 1$, the (virtually torsion-free) group $A_n(\mathbb{Z}[1/p])$ is of type $FP_{n-1}$ but not of type $FP_n$, and it is finitely presented whenever $n \geq 3$.

  • 1
    $\begingroup$ Note: $A_n(\mathbf{Z}[1/p])$ is not torsion-free, but its normal subgroup of index $2^{n-1}$ of elements with positive diagonal is torsion-free. $\endgroup$
    – YCor
    Mar 15 at 10:16
  • $\begingroup$ @JensReinhold: I added a remark in my answer about your second question. $\endgroup$
    – AGenevois
    Mar 24 at 10:21
  • $\begingroup$ @AGenevois: About this thing for the second question, I think the Euler characteristic being zero just means that if the homology is concentrated in a single positive dimension, then the non-zero thing can only possibly be a Betti number of 1 in an odd dimension. (But OK, I see the question is phrased a little wrong, it should have said non-trivial homology in a single positive degree.) (Also, for the record, I have no idea whether an example exists with, e.g., $d=3$.) $\endgroup$ Mar 24 at 18:09
  • $\begingroup$ @MattZaremsky: You're right, my comment is not so interesting... $\endgroup$
    – AGenevois
    Mar 26 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.