MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been looking at the numerical behavior of a particular quantity (of no direct importance here, though if you must know the gory details start with figure 17 here) associated to the geodesic flow on a surface of constant negative curvature and genus $g$. The behavior is quantitatively similar for $g = 2,3,4$ and physical intuition based on this quantity suggests that some "intrinsic" timescale--prime candidates are the mixing or relaxation time--should therefore depend on $g$ only weakly or even not at all.

So: what is known about the behavior of the mixing and relaxation (or similar) times associated to these flows as $g$ varies?

share|cite|improve this question
There are many different metrics of curvature -1 on a surface of genus g. Do you have particular metrics in mind? For surfaces of genus 2, the mixing time can go to infinity as you vary the metric: if there are two genus-one subsurfaces separated by a long thin tube, the portion of the unit tangent bundle on one side takes a long time to mix with the portion on the other side. – Bill Thurston Dec 8 '10 at 23:58
Let's say for concreteness that the surface of constant negative curvature is given by a identifying appropriate edges of a regular $8g-4$-gon in the disk model with metric $ds^2 = dz d\bar{z}/(1-|z|^2)^2$. – Steve Huntsman Dec 9 '10 at 0:59
@Steve -- do you mean the regular $4g$-gon? The rate of mixing has to depend on $g$, because the surface that the $4g$-gon gives has injectivity radius an increasing function of $g$. (It grows like $\log(g)$.) – Sam Nead Dec 9 '10 at 10:30
@Sam--thanks. I mean $8g-4$: see for the edge identification. – Steve Huntsman Dec 9 '10 at 13:45
@Steve - Thank you for the figure. So you are using the regular, right angled, $8g−4$-gon. Let's call it $P$. If you draw $P$ in the most symmetric fashion in the disc model of $H^2$ then the center $O \in P$ will be distance $\log(g)$ (more or less) from the boundary of $P$. So for subsets of the unit tangent "close to" the point $O$ you'll have to flow for at least that long before any mixing at all happens. – Sam Nead Dec 9 '10 at 18:13
up vote 4 down vote accepted

Here is another thought that struck me on the way home, that I should have realized earlier. Suppose that $S$ is a closed hyperbolic surface, of genus $g$. Then the area of $S$ is $-2\pi\chi(S) = 2\pi(2g - 2)$. Since the area of a disk in the hyperbolic plane is exponential in its radius, it follows that the diameter of $S$ is at least logarithmic in $g$. The mixing time of a space has to be at least the diameter, right? So this gives a uniform lower bound on the mixing time.

Thurston's comment is pointing out that there is no uniform upper bound. To see this: The injectivity radius is one-half the systole (the length of the shortest closed geodesic). For a hyperbolic surface, the collar lemma implies that as the injectivity radius goes to zero the diameter goes to infinity (this is the previously mentioned "long thin tube"). Thus the mixing time also has to grow, by the previous paragraph.

I roughly expect the mixing time can be estimated from the logarithm of the genus and the inverse of the injectivity radius. One reference for the geometric facts above is Peter Buser's book "Geometry and spectra of compact Riemann surfaces".

share|cite|improve this answer

OK, so I wanted to elaborate here on Sam's helpful comments.

Set $N = 8g-4$. An explicit description of the $N$-gon $F$ that I have in mind is \begin{equation} F = D \ \backslash \ \bigcup_{j=1}^{N} \left(\sqrt{a-1} \cdot D + \sqrt{a} e^{2\pi i(j-2g)/N}\right) \end{equation} with $a = \sec \frac{2\pi}{N}$. In particular, the nearest point to the origin is at a Euclidean distance $u := \sqrt{a} - \sqrt{a-1}$, so the hyperbolic distance is $d = \int_0^u \frac{dr}{1-r^2} = \frac{1}{2}\log\frac{1+u}{1-u}$, which evidently grows as $\log g$.

A bit more context also: I expect that $t_g f(g) \approx const$, where $t_g$ is whatever timescale and $f(g)$ is the quantity mentioned in the question. If $t_g \sim \log g$ then I'd expect that $f(g) \sim 1/\log g$, which is actually a weak enough dependence on $g$ to not be surprising based on the numerics alluded to in the question.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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