# Optimal $\delta$ for Gromov's $\delta$-hyperbolicity of the hyperbolic plane

What is the minimal $$\delta$$ such that the hyperbolic plane is $$\delta$$-hyperbolic, in the sense of the four point definition of Gromov?

Four point definition of Gromov: A metric space $$(X, d)$$ is $$\delta$$-hyperbolic if, for all $$w, x, y, z \in X$$, $$d(w, x) + d(y, z) \leq \text{max}\{d(x, y) + d(w, z), d(x, z) + d(w, y) \} +2\delta.$$

Empirically, the minimal value seems to be approximately $$0.693$$.

There is a related question, but this concerns the optimal $$\delta$$ in the $$\delta$$-slim definition. While this implies a bound on the $$\delta$$ of the four point definition, it hasn't yet helped me to derive the minimal value.

Any help (or a reference) would be greatly appreciated!

• Take 4 points far apart and compute the difference. Dec 29, 2020 at 6:40
• If you take 4 points at infinity, their only invariant is the cross ratio. So it should be possible to compute $\delta$ as a function of the cross ratio and find its minimum. Dec 29, 2020 at 7:18
• @ThiKu: You do not want any of the four distances to be $\infty$. Dec 29, 2020 at 7:36
• @dodd: ThiKu certainly meant to consider the limit as point go to infinity of the difference $d(w,z)+d(y,z)-\max\{d(x,y)+d(w,z),d(x,z)+d(w,y)\}$. Dec 29, 2020 at 10:37
• For this, you could get points to go to infinity one at a time, using Buseman functions (which are limits of différence between distance functions). Dec 29, 2020 at 10:38

Indeed, the hyperbolic plane is $$\log(2)$$-hyperbolic (with the 4-point definition of hyperbolicity) and this is the optimal constant. The result is nontrivial and first appeared as Corollary 5.4 in
The answer is $$\delta = \ln(2) \approx 0.693147181$$.
With the claim in hand, we can compute $$\delta$$ in the upper half plane model. We place the points at $$0, 1, \infty, -1$$. We place identical horocircles at each of these points. These are cyclically tangent, and all have the same minimal distance $$\delta/2$$ from the point $$i$$. The points of tangency are cyclically permuted by the order four rotation about $$i$$. If we take boundary of the horosphere about $$\infty$$ to be the line $$y = H$$ then we discover that the order four element (fixing $$i$$) sends $$1 + iH$$ to $$-1 + 2i/H = -1 + iH$$. Thus $$H = \sqrt{2}$$. So $$\delta$$ is twice the distance from $$i$$ to $$i\sqrt{2}$$ and we are done.
The proof of the claim appears to be difficult. We have to prove that, given four material points, we can increase $$\delta$$ by first moving them "outward" to lie on a circle (tricky), then to lie symmetrically on the circle (medium), and then increase the radius of the circle to infinity (easy).