# How to prove that the hyperbolic space is $\delta$-hyperbolic

How can I prove that the hyperbolic space $$\mathbb{H}^n$$ (in any of its realization as hyperboloid, Poincaré disc or Poincaré half-space) is $$\delta$$-hyperbolic? For the moment I am not interested in the optimal $$\delta$$.

Almost every source I looked upon states it as well-known, but I couldn't find a thorough proof. I tried to follow the hints on "Embeddings of Gromov hyperbolic spaces" by Bonk and Schramm, but I don't see how to connect the Bounded metric space case to the hyperbolic space. I also tried to prove the 4 points inequality explicitly for the hyperboloid in $$\mathbb{R}^{n+1}$$ with distance $$d(A,B)=\mathrm{arcosh}(-\langle A\mid B\rangle),\text{ where }\langle A\mid B\rangle=\sum_{i=1}^na_ib_i-a_{n+1}b_{n+1},$$ but I got stuck.

Is there any trick I can use? Or is there some easy argument that I'm missing?

Many thanks to anyone who could help me!

• The correct terminology is "Gromov-hyperbolic". The wording "$\delta$-hyperbolic" is correct only if the value of $\delta$ matters.
– YCor
Commented Oct 18, 2022 at 13:35
• @YCor: I wouldn’t be quite so categorical. I have seen the two terms used interchangeably in the literature, whether or not the precise constant matters. Commented Oct 18, 2022 at 13:44
• @AndyPutman I'm aware of its quite common misuse in the literature.
– YCor
Commented Oct 18, 2022 at 13:50
• @YCor: That's an amusing consequence of taking the terminology very literally! :) Commented Oct 18, 2022 at 16:21
• @YCor: Alternatively one can just assume, quite reasonably, that the OP meant "How to prove that the hyperbolic space is $\delta$-hyperbolic [for some $\delta$]". This is perfectly reasonable English. Let's not accuse new users of mistakes they haven't even made.
– HJRW
Commented Oct 18, 2022 at 16:42

A standard reference for this kind of questions is "Metric Spaces of Non-Positive Curvature" by Bridson and Haefliger. Here is their proof.

It is enough to do it for $$\mathbb{H}^2$$. If the side of a triangle is far from the other, it means that a large half-disk is contained in the triangle. As the area of a triangle is bounded by $$\pi$$, the radius of a such a half-disk is bounded (as the area of a disk goes to infinity with the radius).

For the optimal $$\delta$$, see this answser.

• Thank you very much! To complete the proof I just need to motivate why it suffices to show it for n=2 and why exactly half of the disc is contained in the triangle. $\bullet$ For the latter problem, my idea is to argue as follows: the part of the disc inside the triangle is sent to the part outside via a reflection with respect to the geodesic line on which the side I'm considering lies. This reflection should be an isometry, hence the volume of the ball inside the triangle is half the volume of the ball. Is the argument right? What about extending to n>2? Thank you in advance Commented Oct 18, 2022 at 14:51
• @GiacomoGavelli: for the first part, just notice that any three points in $\mathbb{H}^n$ are contained in an isometrically embedded copy of $\mathbb{H}^2$.
– HJRW
Commented Oct 18, 2022 at 16:38
• @GiacomoGavelli yes, your argument shows that the area of the half disk is half the area of the disk.
– FMB
Commented Oct 20, 2022 at 8:00