# Is a $0$-hyperbolic group free?

In his article, Abderezak Ould Houcine asks the following question: If $G$ is a hyperbolic group, let $\delta_0(G)$ denote the infinimum of $\delta$ for which $G$ is $\delta$-hyperbolic. When $\delta_0(G)=0$?

Of course, free groups are such groups. But are they other ones?

If $\Gamma$ is a (Cayley) graph and $C$ a cycle of minimum length $\ell$, then taking three points $x,y,z \in C$ satisfying $d(x,y)=d(y,z)=d(z,x)= \ell/3$, the triangle induced by $C$ is geodesic and it is clearly not $0$-thin. So it seems that the only $0$-hyperbolic graphs are trees, and consequently, the only $0$-hyperbolic groups would be free.

Am I missing something? Do you have an example of non-free $0$-hyperbolic group?

• The only comment I have is that the minimum length of a cycle need not be equal to $1$; but your argument should work with any cycle length. Otherwise, looks good to me. – Lee Mosher Aug 1 '14 at 22:43
• In fact, it was not $1$ but $\mathrm{l}$. It is clearer if I write $\ell$, sorry. – Seirios Aug 2 '14 at 6:21

## 2 Answers

If a graph is not a tree, then it contains a shortest contactible loop, which is an isometric image of a circle. A circle of length $L$ has $\delta\ge L/4$ (using the $4$-point form of hyperbolicity, with $x,z,y,v$ placed at equal distances around the circle). So yes, a $0$-hyperbolic graph is a tree. And if it's the Cayley graph of a group, the group is free.

It's possible that the remark was really intended for the metric $\ell_S$ considered in the paper later; this metric is defined only on the set of vertices, which seems to allow non-free groups to have $0$-hyperbolicity in this metric.

As noted by just a brick in the wall, the answer depends on the definition of $\delta$-hyperbolicity: if it is defined with respect to the geometric realization of a Cayley graph (ie. a 1-complex), then free groups are the only $0$-hyperbolic groups; if it is defined with respect to the set of vertices of a Cayley graph, the answer does not seem to be so trivial.

To complete the previous answer, I just mention that $\mathbb{Z}_2 \ast \mathbb{Z}_3$ is a $0$-hyperbolic group for the second definition above. Its Cayley graph with respect to the usual generating set is roughly: Let $x,y,z,w$ be three vertices. If the union of geodesics between each pair of points does not contain a triangle, then the $4$-point form of hyperbolicity holds since we are essentially in a subtree. Otherwise, we are in the situation as illustrated below: Now, we easily check that $$d(y,w)+d(x,z) =d(x,a)+d(a,z)+d(y,b)+d(b,w)= d(y,z)+d(x,w),$$ so that the $4$-point form of hyperbolicity holds too.

EDIT: Another argument is to isometrically embed the 0-squeleton of the Cayley graph above into the following tree: (Blue edges have length 1/2.)