Take the 2-minute tour ×

Could somebody teach me about the relationship, if any, between hyperbolicity in groups (in Gromov sense) and hyperbolicity in 3-dimensional orbifolds? To be more specific, let Q be a 3-dimensional orbifold. If the orbifold fundamental group of Q is a hyperbolic group (in Gromov sense), then can we say that Q is a hyperbolic orbifold?

share|improve this question
Finite groups are Gromov hyperbolic, so $3$-manifolds like the $3$-sphere and lens spaces have hyperbolic fundamental groups but are not actually hyperbolic. If you want examples with infinite fundamental groups, take the connect sum of two lens spaces. –  Andy Putman Feb 5 '12 at 6:20
If you want someone to teach you about the relationship, read Gromov's foundational paper. –  Igor Rivin Feb 5 '12 at 19:56
To add to Andy's answer, you need to assume that the orbifold is irreducible and "good", so that there are no essential suborbifolds of Euler characteristic $>0$. –  Ian Agol Feb 5 '12 at 20:41
The orbifold theorem was proved before geometrization, but a unified proof is now available: front.math.ucdavis.edu/1101.3733 –  Ian Agol Feb 6 '12 at 5:49
Agol - I was playing it safe and allowing $Q$ to have empty orbifold locus. –  Sam Nead Feb 6 '12 at 10:20

1 Answer 1

up vote 8 down vote accepted

[See Peter Scott's Bulletin article for more information.] Typically, we say an orbifold $Q$ is hyperbolic if it comes to us as a quotient of hyperbolic space $H^n$ by the action of a discrete group $G$ of isometries. If the action $G$ is cocompact then $G$ will be a Gromov hyperbolic group. This is the "easy direction".

On the other hand, if $Q$ is an orbifold with enough topological hypotheses (for example, dimension three, irreducible, "good" as Agol says, perhaps more...) then, if the orbifold fundamental group of $Q$ is Gromov hyperbolic it follows from the geometrization theorem (Perelman and so on) that $Q$ is orbifold homeomorphic to a quotient as in the first paragraph.

So, roughly, the two notions are equivalent. However one direction is easy -- it follows from basic definitions in the field of coarse geometry -- and the other direction is one of the most famous recent results in mathematics.

share|improve this answer
Many many thanks, Sam!!! –  scarlet Feb 6 '12 at 1:08

Your Answer


