15
$\begingroup$

Let $\Sigma$ be a closed hyperbolic surface. Is it true that for any finite collection of points $x_1,\ldots,x_n\in\Sigma$ there exists a closed geodesic $\gamma$ containing none of them?

Remark: It is possible for a point $x$ to lie in infinitely many closed geodesics (e.g. if $x$ is a fixed point for an orientation preserving isometry of order 2). Still, I wonder if any such point need to be a fixed point for some isometry $\phi\in$ Isom$^+(\Sigma)$ ?

$\endgroup$
7
  • 1
    $\begingroup$ Do you know for a single point? $\endgroup$ – YCor Jan 28 '15 at 23:07
  • $\begingroup$ Yes, I managed to prove that it is possible to find three simple geodesic $\alpha,\beta$ and $\gamma$ that pairwise intersect only once and that cannot all intersect in the same point. $\endgroup$ – Federico Vigolo Jan 29 '15 at 10:49
  • $\begingroup$ I just realized that the answer to the question in the remark should be negative: let $p\colon\Sigma'\to\Sigma$ be a Riemannian cover; if $x\in\Sigma$ is a point that is contained in infinitely many geodesic then so are all its pre-images $p^{-1}(x)$. It should then be possible to find a cover and a point in $p^{-1}(x)$ that is not fixed by any isometry. (Maybe the answer is still positive if $\Sigma$ has genus 2?) $\endgroup$ – Federico Vigolo Jan 29 '15 at 11:02
  • $\begingroup$ Suppose $\Sigma$ has genus $g$. Then one can easily find a set of $3g - 3$ disjoint closed geodesics on $\Sigma$. So the statement must be true when $n < 3g-3$. $\endgroup$ – Jeremy Kahn Dec 17 '15 at 0:05
  • $\begingroup$ The preimages of these points is a countable set in the hyperbolic plane. There are an uncountable number of geodesics originating from some basepoint. Isn't that enough? $\endgroup$ – Steve D Mar 6 '16 at 15:45
1
$\begingroup$

I'm not an expert on this stuff so some of these calculations could be wrong:

The number of geodesics of length $L$ in a Riemann surface is asymptotic to $e^L/L$. If we show that the number of geodesics of length $L$ in a Riemann surface passing through a point $x$ is $o(e^L/L)$, we have shown that the answer is yes.

We may as well take $x$ to be $i$ the upper half plane. Then an element of the fundamental group is a geodesic if and only if the corresponding automorphism of the upper half plane takes $i$ straight to that point without turning, which means it consists of a rotation, then multiplication by $y$, then an inverse multiplication, which corresponds to a symmetric matrix.

As $\pi_1(\Sigma)$ is a subgroup of $SL_2(\mathbb R)$, its transpose $\pi_1(\Sigma)^T$ is as well. Let $\Sigma'$ be the Riemann surface, of possibly infinite genus, whose fundamental group is the intersection of those two groups. Then the inverse-transpose automorphism of the fundamental group gives a map from $\Sigma'$ to itself which, because it corresponds to conjugation by the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, is geodesic inversion around the point $x$. An element of the fundamental group corresponds to a geodesic through $x$ if and only if it is fixed through inversion.

Assume $\Sigma'$ has finite genus. Given any geodesic of length $L$ through $x$, the halfway point of geodesic is also a fixed point of that involution. Consider the halfway points of all geodesics of length $<L$ in the upper half plane - they lie inside the set of inverse images of fixed points, which is discrete and periodic, hence has bounded density. They are also all in a ball of radius $L/2$, which has volume $e^{L/2}$, so there are at most $\approx e^{L/2}$ of them.

Intuitively there should be even fewer through a given point in the infinite genus case than in the finite case but I don't see how to establish this. The problem is that the density of the fixed points could increase as they approach infinity on the surface - though they would have to increase very rapidly to beat this estimate. I would expect that the number of geodesics on a surface that just lift to an infinite degree cover is already o(e^L/L), though, but I don't know a proof of this either.

$\endgroup$

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.