Consider a discrete Gromov-hyperbolic group $\Gamma$ (and its Cayley graph $\mathcal{G}$ w.r.t. some generating set). The notion of Gromov-boundary, indicated with $\partial\Gamma$, is naturally associated with the group.

It is well known that two geodesics starting at the same point (assume a vertex) in the graph $\mathcal{G}$ and having the same limit point at infinity (i.e., on the boundary) have to stay at a bounded distance: if $\gamma_1$, $\gamma_2$ are s.t. $\gamma_1(0)=\gamma_2(0)$ and $\gamma_1(\infty)=\gamma_2(\infty)$, then, there exists $K>0$ s.t. $\sup_{t\in [0,\infty)}\mathrm{dist}(\gamma_1(t),\gamma_2(t))<K$, where "$\mathrm{dist}$" denotes the path-distance in the graph.

Let us now specialize to the case where $\Gamma$ has the presentation $\langle a_1,b_1,a_2,b_2\mid [a_1,b_1][a_2,b_2]\rangle$. This is the fundamental group of an orientable, connected and compact surface of genus $2$, for example, a connected sum of two tori.

The Cayley graph of this group is (dual to) a tessellation of the hyperbolic disc by (hyperbolic) octagons, obtained joining at each vertex eight different octagons.

It can be proved that two geodesics $\gamma_1,\gamma_2$ s.t. $\gamma_1(0)=\gamma_2(0)$ and $\gamma_1(n)=\gamma_2(n)$ for some $n\in\mathbb{N}$ satisfy $\sup_{t\in[0,n]}\mathrm{dist}(\gamma_1(t),\gamma_2(t))\leq 4$, meaning in particular that two vertices on the geodesics corresponding to the same parameter $t\in\mathbb{N}$ are at most four edges apart.

My claim is that the same bound holds for the case where $\gamma_1(\infty)=\gamma_2(\infty)$, but $\gamma_1(t)\neq \gamma_2(t)$ for $0<t<\infty$.

Frist of all, I wonder if this claim is actually true. I ask if someone is aware of a counterexample. Then I ask if someone is already in possess of a proof of this claim. I strike that I am referring to a very specific case: I'm interested in the particular group presented above.


1 Answer 1


This is a consequence of Strebel's classification of geodesic triangles in C'(1/6) polygonal complexes; see Ghys and de la Harpe's book, Sur les groupes hyperboliques d'après Mikhael Gromov. enter image description here

Notice that the cases IV and V are not possible in the hyperbolic plane because the tessellation is T(4).

Let $P_1, \ldots, P_n$ be a sequence of polygons which intersect both $\gamma_1$ and $\gamma_2$, where $\gamma_1(0)=\gamma_2(0)=o \in P_1$, and suppose that further $P_n$, no polygon intersect both $\gamma_1$ and $\gamma_2$. Next, take $k$ large enough and consider a geodesic triangle $[o, \gamma_1(k), \gamma_2(k)]$. Following the classification above, we are in case II or III, where the central polygon is $P_n$. Finally, conclude that it is possible to find a $k$ so that $d(\gamma_1(k),\gamma_2(k))$ is arbitrarily large. The situation is roughly the following:

enter image description here

As a consequence, $\gamma_1(+ \infty) \neq \gamma_2(+ \infty)$.

Consequently, if you suppose that $\gamma_1(+ \infty)= \gamma_2(+ \infty)$, then the classification implies that any geodesic triangle $[o,\gamma_1(k),\gamma_2(k)]$ must have the form I, so that $\gamma_1(k)$ and $\gamma_2(k)$ belong to a common polygon. Hence $d(\gamma_1(k), \gamma_2(k)) \leq 4$.

There are probably some missing cases in the previous argument, but I think a rigourous proof can be extracted from it. (Because you are working in the hyperbolic plane, Strebel's classification can probably be proved directly and elementarily.)

  • $\begingroup$ Thanks for your reference, I actually knew about Strebel's classification. Your idea is clear and straightforward. The fact is that I miss one point: "Finally, conclude that it is possible to find a $k$ so that $d(\gamma_1(k),\gamma_2(k))$ is arbitrarily large." Why is that? I don't see this. Maybe is elementary, but I can't get it at first sight. Could you please clearify? I apologize for my further question. $\endgroup$
    – EM90
    Oct 22, 2016 at 10:57
  • $\begingroup$ I added a picture to make the idea clearer. $\endgroup$
    – Seirios
    Oct 22, 2016 at 16:14
  • $\begingroup$ Yeah, the picture actually helps. I guess that the point is that the segment joining $\gamma_1(k),\gamma_2(k)$ is a geodesic, too, and it "walks along" the same polygons, so that we can estimate quite accurately its length. Thus, choosing the right amount of polygons "after" $P_n$, we get a length as large as we want. Thanks a lot! $\endgroup$
    – EM90
    Oct 24, 2016 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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