I am trying to prove that the Poincaré disk, $\mathbb{D}$, is $\delta$-hyperbolic with respect to the slim triangle definition for hyperbolicity. I have been stuck for a while on where to begin, particularly with respect to ideal triangles.
I am trying to do this from a Geometrical Group Theory point of view, as I am not familiar enough with topology to use those tools.
Many thanks
On
One way to do it is to start by proving that the area of any geodesic triangle is less than $\pi$ (actually the area is exactly $\pi-\alpha-\beta-\gamma$, where $\alpha$, $\beta$ and $\gamma$ are the angles at the vertexes).
Then you can conclude by contradiction as follows: suppose no $\delta$ works. Then for each positive real number $R$ you can find a geodesic triangle $T_R$ which is not $R$-thin. This implies the existence of a point on one edge such that the disc of radius $R$ does not meet the two other sides. Hence half of this disc is in the interior of the triangle. From this you can easily deduce that the volume of your triangle must have an area bigger than half the volume of the disc. Hence for $R$ large enough we would get a geodesic triangle with area bigger than $\pi$, which is absurd.
The orientation-preserving isometry group is $PSL(2,R)$ and it acts transitively on triples of pairwise distinct points on the ideal boundary, i.e. on the set of non-degenerate ideal triangles. In other words: all ideal triangles are isometric. Thus it suffices to compute $\delta$ for one ideal triangle. You will get $\delta=\log(\sqrt{2}+1)$.
Then you still have to prove that this implies the same estimate for non-ideal triangles.