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$^+(\Sigma)$ ?

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    $\begingroup$ 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

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.


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