Are hyperbolic triangle groups hyperbolic?

Question

This might be a silly question, but are hyperbolic triangle groups hyperbolic, in the sense of Gromov?

By a hyperbolic triangle group, I mean a group given by a presentation, $$\langle a, b, c; a^p, b^q, c^r, abc\rangle$$ where $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$.

I think they are, and it seems to be implied in some places, but nowhere seems to state it explicitly (apart from cough wikipedia cough cough).

These groups act on they Hyperbolic plane in some way (they correspond to tilings of the plane with triangles which preserve the orientation of the triangles), so it is natural to generalise my question: is there some criteria $\mathcal{C}$ (faithfully, say) we can place on a group such that,

$G$ is Hyperbolic in the sense of Gromov if and only if $G$ acts on some hyperbolic plane in a $\mathcal{C}$ way.

(I should say that I understand that the "hyperbolic" which Gromov talks about is really talking about the Cayley graph having some hyperbolic properties, such as linear area and the $\delta$-thin triangle condition, and so on. However, hyperbolic groups can be defined in so many different-but-equivalent way and this one seems, well, a natural one to think about, even if it isn't necessarily easy to work with!)

Answer 1

Here is one way to prove it: 1) If the triangle group is uniform, then Milnor-Schwarz shows that it is word-hyperbolic (since it acts properly discontinuously and cocompacty by isometries on the hyperbolic plane).

2) If the triangle group is nonuniform, then we can argue as follows: a) There is a single quasi-isometry class of nonuniform lattices in $SL_2(\mathbb R)$ (see, eg., Farb: "The quasi-isometry classification of lattices in semisimple Lie groups"). b) $SL_2(\mathbb Z)$ is word-hyperbolic (e.g. as being isomorphic to $\mathbb Z/4\mathbb Z \ast_{\mathbb Z/2\mathbb Z} \mathbb Z/6\mathbb Z$). c) Therefore the index 2 subgroup in the triangle group consisting of the orientation-preserving isometries is word-hyperbolic. d) Thus, the triangle group itself is word-hyperbolic.

However, the part 2) feels a bit overkill.

Answer 2

I ran across this old question, and thought I would augment the answer of @JonasSchulz with a more straightforward proof in the nonuniform case, where one or more of $p,q,r$ is $+\infty$. And let me work with the corresponding reflection group $$\langle a,b,c : a^2=b^2=c^2=(ab)^p = (bc)^q = (ca)^r = 1\rangle $$ in which the given group sits with index 2; whichever of $p,q,r$ equals $\infty$, simply remove the corresponding relator from the presentation.

In all cases, uniform or nonuniform, one applies the Poincare polygon theorem to conclude that this group acts on the hyperbolic plane properly discontinuously with a fundamental domain being a triangle $\tau$ with angles $\pi/p$, $\pi/q$, $\pi/r$, and with generators being reflections in the sides. Whichever of $p,q,r$ is infinite, the corresponding vertex is on the circle at infinity.

If one or more vertex is infinite then one can embed a tree $T$ into $\mathbb{H}^2$ which is invariant under the group, which is an equivariant deformation retraction, and on which the group acts properly discontinuously and cocompactly. Since trees are hyperbolic metric spaces, the group is Gromov hyperbolic. Alternatively, the group has a free subgroup of finite index, by the theorem of Karass-Pietrowski-Solitar.

The tree $T$ is constructed by specifying $T \cap \tau$, and that specification depends on how many of $p,q,r$ equal $+\infty$. If just one of $p,q,r$ equals $\infty$, $T \cap \tau$ equals the side opposite the infinite vertex. If two of $p,q,r$ equal $\infty$ then $T \cap \tau$ is the perpendicular segment from the finite vertex to the opposite side. And if all three of $p,q,r$ equal infinity then $T \cap \tau$ is the union of the three perpendiculars from the "geometric barycenter" of $\tau$ to the three sides; in this case evidently the group is $\mathbb{Z}/2 * \mathbb{Z}/2 * \mathbb{Z}/2$ which contains a rank 2 free group with index 2.