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!

  • 2
    $\begingroup$ Take 4 points far apart and compute the difference. $\endgroup$
    – markvs
    Dec 29, 2020 at 6:40
  • 4
    $\begingroup$ 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. $\endgroup$
    – ThiKu
    Dec 29, 2020 at 7:18
  • $\begingroup$ @ThiKu: You do not want any of the four distances to be $\infty$. $\endgroup$
    – markvs
    Dec 29, 2020 at 7:36
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    $\begingroup$ @dodd: ThiKu certainly meant to consider the limit as point go to infinity of the difference $d(w,z)+d(y,z)-\max\{d(x,y)+d(w,z),d(x,z)+d(w,y)\}$. $\endgroup$ Dec 29, 2020 at 10:37
  • 2
    $\begingroup$ 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). $\endgroup$ Dec 29, 2020 at 10:38

2 Answers 2


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

Nica, Bogdan; Špakula, Ján, Strong hyperbolicity, Groups Geom. Dyn. 10, No. 3, 951-964 (2016). ZBL1368.20057.


The answer is $\delta = \ln(2) \approx 0.693147181$.

Claim: The correct placement of the four points at infinity is at the corners of an ideal square.

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).


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