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!)

  • 3
    $\begingroup$ When $G$ acts on a CAT(-1) space (such as the hyperbolic plane), it is indeed hyperbolic. Here, I say "acts" when I mean properly and cocompactly (as the triangle groups do). It is an open question if all (Gromov) hyperbolic groups are CAT(-1). That being said, not all hyperbolic groups can "act" on the hyperbolic plane, simply because the quotient is a a very special type of space, and your group has to be the fundamental group of that space, and there are restrictions there. $\endgroup$
    – user641
    Feb 2, 2012 at 15:50
  • 2
    $\begingroup$ Gromov hyperbolic groups are often refereed to as it word-hyperbolic. Just to add to your general confusion, there is also a concept of hyperbolic Coxeter groups. The hyperbolic triangle groups have index 2 in corresponding hyperbolic Coxeter groups. It turns out that all hyperbolic Coxeter groups are word-hyperbolic, but there are word-hyperbolic Coxeter groups that are not hyperbolic Coxeter groups. I I think it has been proved that a Coxeter group is word-hyperbolic if and only if it doesn't contain a free abelian group of rank2 , and it is easy to decide that from the Coxeter diagram. $\endgroup$
    – Derek Holt
    Feb 2, 2012 at 20:35
  • $\begingroup$ @DerekHolt: I know hyperbolic and word hyperbolic do not coincide for semigroups - is this perhaps similar? I mean, every presentation of a coxeter group defines a monoid, and there is something about embedding these monoids in the groups (was this perhaps a paper of yours?). Do you perhaps have a reference for this non-coinciding? $\endgroup$
    – user1729
    Feb 2, 2012 at 21:07
  • $\begingroup$ No, I am afraid I can't help much here. I know that Duncan and Gilman gave a language-theoretical definition of word-hyperbolic semigroups that turns out to be equivalent to word-hyperbolic for groups, butI have never come across any other definition of hyperbolic semigroups. $\endgroup$
    – Derek Holt
    Feb 2, 2012 at 22:28
  • $\begingroup$ @DerekHolt: I have dug into my notes...so, embeddable monoids was in a paper of Dehornoy and Godelle, and was for Artin groups, not Coxeter ones. The hyperbolic semigroups I saw in a talk of Alan Cain's. The slides for the talk can be found here, www-history.mcs.st-and.ac.uk/~alanc/talks/conf.html. He defines a semi-group to be hyperbolic if the Cayley graph is hyperbolic, as in group theory, but crucially this does not imply a solution to the word problem (while being word-hyperbolic does). $\endgroup$
    – user1729
    Feb 3, 2012 at 10:41

2 Answers 2


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.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .