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? 



Hyperbolicity on Riemann Surfaces
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, $D$. 
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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.
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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 $X$.


