10
$\begingroup$

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!

$\endgroup$
5
  • 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
  • 6
    $\begingroup$ @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

6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.