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For Riemann surfaces there are at least to possible notions of hyperbolicity. The classical one given by the Uniformization Theorem, or equivalently the type problem, which essentially says that a simply connected Riemann surfaces is conformally equivalent to one of the following:

  • Riemann Sphere $\mathbb{C}\cup\{\infty\}$ (elliptic type).
  • Complex plane (parabolic type).
  • Open unit disk (hyperbolic type).

On the other hand, given a Riemann surface one can asks if it is hyperbolic in the Gromov's sense. In other words, does there exists $\delta>0$ such that all the geodesic triangles in the surface are $\delta$-thin?

It seems to me that this two notions of hyperbolicity are not equivalent and one can have counterexamples in both directions. For instance, the two dimensional torus $\mathbb{T}^2$ is hyperbolic in Gromov's sense (since it is compact), but it's also a quotient of the Euclidean plane by a free action of a discrete group of isometries and therefore, of parabolic type.

My questions are: what is a sufficient condition for a surface of hyperbolic type to be Gromov's hyperbolic? what is known about the relation of these two notions?

Related Question: Let $G$ be an infinite planar graph with uniformly bounded degree and assume that the simple random walk is transient. Is the graph necessarily Gromov's hyperbolic?

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  • $\begingroup$ "necessary condition to guarantee" makes no sense. Do you mean "sufficient condition"? Please edit the question. $\endgroup$
    – Sam Nead
    Commented Aug 19, 2011 at 21:19
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    $\begingroup$ In order to talk about hyperbolicity in Gromov's sense, you need to specify a specific metric structure on the Riemann surface, as they only come equipped with a conformal structure (pointed out in R W's answer). A natural choice is the Riemannian metric with canonical constant Gaussian curvature, but do you want to restrict to this case? I take it from some comments below that you don't, and would like to consider Riemann surfaces with some arbitrary choice of Riemannian metric conformally equivalent to this one. It might be worth editing something like this into your question. $\endgroup$
    – j.c.
    Commented Aug 24, 2011 at 16:37

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NEW ANSWER:

As there has been much confusion on this point (some of it mine...):

Definition: A Riemannian 2-manifold $S$ is of hyperbolic type if the universal cover of $S$ is conformally equivalent to the open unit disk, $D$.

On the other hand we have

Definition: A hyperbolic surface $S$ is a surface equipped with a complete Riemannian metric of constant curvature minus one.

It is an exercise to show that all hyperbolic surfaces are surfaces of hyperbolic type. On the other hand, a surface of hyperbolic type need not be hyperbolic. As an easy example of this, choose your favorite positive function $f$ on the disk $D$ and use $f$ to scale the Poincare metric. This new metric is (almost surely) not constant curvature but is conformally equivalent to the Poincare metric.

With these definitions in place: the original question is ill-posed. Knowing that a surface $S$ is of hyperbolic type does not suffice to tell us the metric. To be precise, there are conformally equivalent metrics $\rho_0$ and $\rho_1$ on the open disk $D$ so that the first is Gromov hyperbolic and the second is not. (Eg, let $\rho_0$ be the Poincare metric while $\rho_1$ has larger and larger "mushrooms" as you walk to infinity.)

OLD ANSWER (written in terms of the above definitions):

I'll assume that you are asking for a sufficient condition to ensure that a hyperbolic surface $S$ is Gromov hyperbolic. One condition is that $S$ has finite area. In this case $S$ has a compact core (which is of no interest in this setting) and a finite number of cusps. A cusp is obtained by modding out a horodisk by a parabolic isometry. All cusps are quasi-isometric to rays. Thus $S$ is quasi-isometric to a tree having one vertex and one ray per cusp.

A simpler condition is that $\pi_1(S)$ is finitely generated. The allowed surfaces are now somewhat more complicated: in addition to cusps there can be funnels in the complement of the compact core. A funnel is obtained by modding out a half-plane by a hyperbolic isometry. All funnels are quasi-isometric to the hyperbolic plane. (This is a nice exercise!) So, here, $S$ is quasi-isometric to the one-point union of a collection of hyperbolic planes and rays.

As for the "opposite direction": When the group is infinitely generated things can be very strange. For example, consider any cubic, connected graph $X$, of infinite diameter, where all edges have length one. (This is a very large class of metric spaces, even after passing to quasi-isometric equivalence classes.) Then, for any such graph $X$ there is a hyperbolic surface $S_X$ quasi-isometric to $X$.

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    $\begingroup$ Thanks Sam! This is exactly what I meant. Can you please explain me a little bit more why this condition is sufficient? $\endgroup$
    – ght
    Commented Aug 20, 2011 at 0:17
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    $\begingroup$ This might be a stupid question but how do you prove that a simply connected hyperbolic surface (i.e. conformally equivalent to the unit disk) is Gromov's hyperbolic? $\endgroup$
    – Val
    Commented Aug 20, 2011 at 0:56
  • $\begingroup$ I think you also need the surface to have finite area to be q.i. to a tree, so that there are rank 1 cusps. Otherwise, the surface (with finitely generated fundamental group) will be geometrically finite, and have finitely many ends which are each q.i. to the hyperbolic plane or cusps. $\endgroup$
    – Ian Agol
    Commented Aug 21, 2011 at 18:37
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    $\begingroup$ @ght I think there is some mismatch between your definitions of "surface" here. In his answer and comments, Sam Nead is working with hyperbolic surfaces in the sense of hyperbolic geometry, i.e. those with Riemannian metrics of constant negative Gaussian curvature. You are more concerned with Riemann surfaces of hyperbolic type, which only carry a conformal structure (a whole equivalence class of Riemannian metrics!). His last comment thus points out that a "Riemannian manifold of hyperbolic type" and a "hyperbolic surface" thus are not the same. $\endgroup$
    – j.c.
    Commented Aug 24, 2011 at 16:27
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    $\begingroup$ @All - Reading Val's question more carefully, I see that Val is not asking about hyperbolic surfaces (as he/she wrote) but rather asking about surfaces of hyperbolic type (as he/she also wrote!). So, Val and ght are asking the same question "Is every surface of hyperbolic type in fact Gromov hyperbolic?" The answer to this question is "No". This does not contradict anything I said about hyperbolic surfaces. I'll edit my answer to make this clear. $\endgroup$
    – Sam Nead
    Commented Aug 24, 2011 at 19:48
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I don't quite understand your question about surfaces as the notions of hyperbolicity you are talking about deal with two different structures: a Riemann surface is endowed with a conformal structure, whereas Gromov's hyperbolicity is a property of metric spaces (in particular, of Riemannian manifolds).

There is also some confusion in the torus example: when saying that it is of parabolic type you are talking about its universal cover, whereas when claiming that it is Gromov hyperbolic you are talking about the torus itself (actually it is not really fair to say that a compact metric space is Gromov hyperbolic).

Finally, concerning graphs the answer is no, because you can always attach to your favorite transient planar graph a sequence of circles with increasing radii - it won't change transience and planarity, but will prevent the resulting graph from being hyperbolic.

EDIT: I would still strongly advise against basing any examples or counterexamples on "compact hyperbolic spaces". Although formally they do have the $\delta$-hyperbolicity property, the whole point of developing this theory was to look at the large scale geometry of such spaces, and in particular at their behavior at infinity. If you wish, the notion of a compact hyperbolic space is as rich as that of a compact vector space. This is what I meant by saying that "it is not really fair to say that a compact metric space is Gromov hyperbolic".

As for your revised question, one can formulate it in a more general way: when is a quotient of a Gromov hyperbolic space also Gromov hyperbolic? In order to see its scope, you may first look at the discrete case, where any regular graph is a quotient of the corresponding homogeneous tree.

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    $\begingroup$ Seeing that compact metric spaces are Gromov hyperbolic, I think it's perfectly fair to call them Gromov hyperbolic. $\endgroup$
    – Autumn Kent
    Commented Aug 19, 2011 at 14:50
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    $\begingroup$ Given a Riemann surface you can always put a metric on it and consider it as a metric space. Therefore, both notions of hyperbolicity make sense. A connected metric surface is a quotient of one of the following - the sphere (curvature +1) - the Euclidean plane (curvature 0) - the hyperbolic plane (curvature −1). by a free action of a discrete subgroup of isometries. Hence, with this notion the torus is a Riemann surface and manifold who is of parabolic type. Other thing is that ALL compact metric spaces are indeed Gromov hyperbolic so the torus is Gromov hyperbolic. $\endgroup$
    – ght
    Commented Aug 19, 2011 at 15:47
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About the related question: it is a result of Babai that a (connected, locally finite) vertex-transitive, planar graph is isomorphic to the 1-skeleton of an Archimedean tiling of the sphere, or the Euclidean plane, or the hyperbolic plane. So assuming transience singles out the hyperbolic plane, and implies Gromov-hyperbolicity for the graph. See http://www.cs.uchicago.edu/files/tr_authentic/TR-2001-04.ps

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  • $\begingroup$ Is this Theorem 3.1 of that paper? If so, you need to add the "zero or one endedness" hypothesis. In particular, as far as I understand the definition of Archimedean tilings, some hypothesis is needed to exclude regular trees. $\endgroup$
    – Sam Nead
    Commented Aug 24, 2011 at 19:58
  • $\begingroup$ @ Sam: Correct. I quoted in a rush. $\endgroup$
    – Alain Valette
    Commented Aug 24, 2011 at 22:19

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