It is well-known that hyperbolic space is delta-hyperbolic, but what is delta? - MathOverflow

It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-04-30T02:18:31Z

Recall that a space is $\delta$-hyperbolic if there is some number $\delta$ with the property that every point on an edge of a geodesic triangle lies within $\delta$ of another edge. For example a tree is $0$-hyperbolic. One of the basic facts about standard hyperbolic space is that it is $\delta$-hyperbolic for some $\delta$, and I am looking for the smallest delta which makes this true.

Full disclosure: I stole this question from Dima Burago, who brought it up as an example of of a useless problem about which he is nevertheless a little curious. I haven't exactly burned the midnight oil, but I can't solve it.

Answer by Matthew Stover for It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-04-30T02:38:49Z

Let $T$ be a triangle in $\mathbb{H}^2$. Its area is $\pi - \alpha - \beta - \gamma$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles. You can find how slim this triangle is by considering an inscribed circle in $T$. The radius of this triangle, thus $\delta$, are bounded above by the area, so to find the $\delta$ that works for all triangles, you take the limit and consider an ideal triangle $T_\infty$. You can explicitly compute that the inscribed circle minimizing distance between the sides has length $4 \log \phi$, where $\phi$ is the golden ratio. (See here and here.)

Answer by S. Carnahan for It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-04-30T03:17:54Z

We can use the isometry group of $H^n$ to reduce to the case of an ideal triangle in the upper half plane, with vertices at -1, 1, and infinity. We want to find the distance between $i$ and the vertical geodesic with real part 1. To find the shortest geodesic, we reflect $i$ in the vertical line, and take half the distance between $i$ and $i+2$. The distance formula yields $\tanh^{-1}\left(\frac{|(i+2)-i|}{|(i+2) + i|} \right) = \tanh^{-1}(1/\sqrt2)$. This is about 0.8813735.

Answer by Sam Nead for It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-04-30T03:41:26Z

Consider the ideal triangle with vertices at infinity, zero, and one. Let $C$ be the semicircle perpendicular to the vertical line $[0, \infty]$ and meeting $1/2 + i/2$ (ie the midpoint of the semicircle $[0,1]$). So $C$ meets $[0, \infty]$ at the point $i \cdot \sqrt{2}/2$. Scale down by a factor of $\sqrt{2}/2$ to get the point $1/\sqrt{2} + i/\sqrt{2}$. Use a Mobius transformation to rotate this by $\pi/2$ about $i$ to get $i(1+\sqrt{2})$. Now integrate $dy/y$ to get $\log(1 + \sqrt{2})$. This is approximately 0.88137358702.

Answer by Junyan Xu for It is well-known that hyperbolic space is delta-hyperbolic, but what is delta?

2010-06-13T09:09:31Z

See http://www.math.umn.edu/~am/book/outercircles.pdf, p.14-15 for detailed treatment.