# How does the mixing time of a geodesic flow on a surface vary with the genus?

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?

• 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 imgur.com/XDADm.png 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

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.
Set $N = 8g-4$. An explicit description of the $N$-gon $F$ that I have in mind is $$F = D \ \backslash \ \bigcup_{j=1}^{N} \left(\sqrt{a-1} \cdot D + \sqrt{a} e^{2\pi i(j-2g)/N}\right)$$ 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.