TITLE: Constraints on the homology of amenable groups
QUESTION [11 upvotes]: 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 $d$.)

REPLY [11 votes]: 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.

REPLY [10 votes]: 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 $n \geq 3$.