Finitely presented non-residually amenable groups without free subgroups

Question

Does there exist a finitely presented group that does not contain a nonabelian free group and is not residually amenable?

Answer 1

Let $G$ be the group of piecewise homographic self-homeomorphisms of the real projective line $\mathbb{P}=\mathbb{P}^1(\mathbf{R})$, and $G_\infty$ the stabilizer of $\infty$. So $G_\infty$ can be viewed as the group of self-homeomorphisms of $\mathbf{R}$, piecewise homographic (i.e. piecewise of the form $x\mapsto\frac{ax+b}{cx+d}$) with finitely many breakpoints.

The group $G$, or rather its $C^1$-subgroup was studied a long time ago as automorphism group of a Moulton plane (a fake projective plane); it was also considered by Greenberg.

It is not hard to show that $G_\infty$ has no nonabelian free subgroup (it's an immediate adaptation of the analogous result in the piecewise affine case, e.g., in Thompson's group $F$; see [M, Theorem 14] for a short sketch including all ideas).

Monod discovered that $G_\infty$ is non-amenable. This is quite involved but intuitively, the idea is as follows. Let $A$ be a countable dense unital subring of $\mathbf{R}$, and $G(A)$ the elements of $G$ with breakpoints in $A$ and acting on pieces as $\mathrm{PGL}_2(A)$. Then $G(A)$ induces on $\mathbb{P}$ the same equivalence relation as its subgroup $\mathrm{PGL}_2(A)$. Carrière and Ghys proved that this equivalence relation is non-amenable (in a suitable sense, pertaining to topological dynamics of groups). Then $G_\infty(A)$ also induces the same equivalence, the only difference being $\{\infty\}$ being apart; this tiny difference doesn't affect non-amenability of the action. Now a non-amenable equivalence relation can't be induced by an amenable group. So $G_\infty(A)$ is non-amenable. In this way Monod produced finitely generated subgroups of $G_\infty(A)$ that are not amenable (while without non-abelian free subgroup).

Next Lodha and Moore produced such subgroup that are in addition finitely presented, using explicit generators. I claim their group $\Gamma$ has its third derived subgroup $\Gamma'''$ containing every nontrivial normal subgroup.

Let $\Gamma_{0}$ be the subgroup of compactly supported elements in $\Gamma$ (the support being viewed in $\mathbf{R}$): this is the kernel of the homomorphism consisting in taking germs at $\pm\infty$ (which is valued in a metabelian group). Hence $\Gamma''\subset\Gamma_{0}\subset$ and in particular $\Gamma_0$ is non-trivial.

Their group $\Gamma$ contains a fixed-point-free self-homeomorphism $\rho$, so for every compact subset $K$ there exists $n$ such that $\rho^n(K)\cap K$ is empty. This implies, by a simple commutator trick (see for instance Lemma 3.3.4 in Burillo's book) that every normal subgroup of $\Gamma$ contains $[\Gamma_0,\Gamma_0]$. In particular, $\Gamma'''=[\Gamma_0,\Gamma_0]$.

Thus the intersection of nontrivial normal subgroups of $\Gamma$ is nontrivial. So if $\Gamma$ is not P (P any property passing to subgroups), then it's not residually P. So $\Gamma$ is not residually amenable.

Probably a closer look shows that $[\Gamma_0,\Gamma_0]=\Gamma'''$ is simple, but it would need some more details: the above is enough (it implies that $[\Gamma_0,\Gamma_0]$ has no nontrivial proper conjugacy $\Gamma$-invariant subgroup and in particular is characteristically simple). Maybe $\Gamma''$ itself is simple but this might be significantly more technical.

[CG] Y Carrière, E. Ghys. Relations d'équivalence moyennables sur les groupes de Lie. Comptes Rendus Acad. Sci. t.300 Sér.I no.19, 1985, 677–680. (French) link at Ghys' webpage

[LM] Y. Lodha, J.T. Moore. A nonamenable finitely presented group of piecewise projective homeomorphisms. Groups Geom. Dyn. 10 (2016), no. 1, 177–200. arXiv link