Howson's property for amalgams of free groups

Question

Recall that a group has the Howson property (also known as the finitely generated intersection property) if the intersection of any two finitely generated subgroups is again finitely generated. I am interested in when the amalgam of two free groups has the Howson property.

It is a classical theorem of Burns that amalgamated products of the form $G \ast_C F$ have the Howson property so long as $G$ has the Howson property, $F$ is free, and $C$ is maximal cyclic in $F$. The proof proceeds using combinatorial group theoretic methods (i.e. normal forms for elements, Schreier transversals). On the other hand, one can inject some geometry into the picture and consider the (strictly) stronger property of local quasiconvexity in hyperbolic groups (i.e. the condition that all finitely generated subgroups are quasiconvex). Amalgams of locally quasiconvex hyperbolic groups over cyclic subgroups are again locally quasiconvex, giving many examples of cyclic amalgams with the Howson property.

Burns phrases his maximality assumption as saying $C$ is isolated in $F$, namely that if $a^n \in C$, then $a \in C$ also. It is not difficult to construct examples where the amalgam fails to have the Howson property if $C$ is fails to be isolated. Indeed, if $F, F'$ are free and $g \in F, h \in F'$, then $F \ast_{g^2 = h^3} F'$ contains a copy of $F_2 \times \mathbb{Z}$, which does not have the Howson property. Hence we may ask:

Question 1: Let $F, F'$ be finitely generated free groups, $H \leqslant F, F'$ a nontrivial isolated subgroup. When does the amalgam $G = F \ast_H F'$ have the Howson property?

As far as I am aware, there is no general result outside of the case when $H$ is cyclic. Of course, when $H$ is a free factor of both $F$ and $F'$, then $G$ is free and so has the Howson property. Outside of this trivial case, are there known examples of such $G$ with or without the property? And should one expect the answer to depend on the ranks of $F, F',$ and $H$? Moreover, when $H$ is finitely generated of infinite index and malnormal in one factor, it follows from the Bestvina-Feighn combination theorem that $G$ is hyperbolic. I would not be surprised if the following had a negative answer, though I can only imagine verifying that through their failing the Howson property.

Question 2: Suppose $H \leqslant F, F'$ is finitely generated, malnormal, of infinite index, and not cyclic. Is the amalgam $G = F \ast_H F'$ locally quasiconvex?

Answer 1

If you are willing to expand your final question slightly to include general graphs of groups then the answer to Question 2 is certainly “no”.

In their famous 2008 GAFA paper Special cube complexes, Haglund and Wise give a version of the Rips construction where the central group is the fundamental group of a thin VH complex. That is, for any finitely presented group $Q$, they construct a short exact sequence

$1\to K\to \Gamma\to Q\to 1$

where $K$ is finitely generated and $\Gamma$ is the fundamental group of a thin VH complex. I won't recall the definition of a thin VH complex, but will just note that any such group is the fundamental group of a graph of free groups with malnormal edge groups. If $Q$ is infinite then $K$ is not quasiconvex, because it has infinite index in its normaliser.

I'm sure one can also cook up an example which is an actual amalgam. For instance, if $H$ is malnormal and $b_1^{(2)}(G)=0$ then $G$ virtually fibres by Kielak's theorem, and hence is not locally quasiconvex. In fact, Wise has conjectured that local quasiconvexity fails whenever the Euler characteristic of $G$ is positive.

Not much is known about Question 1. However, one can give a conjectural answer in the framework of my paper 'Rational curvature invariants for 2-complexes', which is designed to address exactly these kinds of questions.

For a suitable graph of graphs $X$ representing $G$, the Howson property should be at least partially controlled by the invariant $\rho_+(X)$ that I define in the paper. For instance, Conjecture 12.9 says that if $\rho_+(X)<0$ then $G$ should be locally quasiconvex, and in particular have the Howson property.

It is equally reasonable to conjecture that, if $\rho_+(X)>0$, then $G$ should not have the Howson property. However, the case of $\rho_+(X)=0$ will be less definitive, since $\mathbb{Z}^2$ has the Howson property but $F\times\mathbb{Z}$ doesn't.

I won't recall the definition of $\rho_+(X)$ here, since it's a little complicated, but note that it is a rational number that can be determined by an explicit linear programming problem. My guess is that, in the case of graphs of graphs, the definition of $\rho_+(X)$ can be simplified. See §11.5 of the paper for some further discussion of $\rho_+$ for graphs of free groups. (And feel free to email me if you would like more details on any of this.)

Another reference is Wise's survey article 'An Invitation to Coherent Groups', which includes some discussion of the Howson property.