5
$\begingroup$

An $\epsilon$-net of a closed hyperbolic surface $X$ is a finite set of points $p_i$ such that the family of balls centered at $p_i$ with radius $\epsilon$ is a cover of $X$, and the family of balls centered at $p_i$ with radius $\epsilon/2$ are distinct pair by pair.

My question is that if there is a closed geodesic goes through all $B(p_i,\epsilon)$ of some $\epsilon$-net, and if not, under which conditions of X and $\epsilon$-net we can find that geodesic?

Thank you so much!

$\endgroup$
5
$\begingroup$

Yes, such a closed geodesic always exists. See Theorem 1.1 of this paper by Basmajian, Parlier, and Souto (which I found by searching under the term "density of closed geodesics on a hyperbolic surface").

Now, to be honest, what you want is simpler than what is proved in that paper. Namely, it is well known that the geodesic flow has a dense (non-closed) geodesic $\gamma : \mathbb{R} \to X$. Take a subsegment $\gamma | [-M,+M]$, close it off with a uniformly short segment --- namely one of length at most the diameter of $X$ --- and straighten. If $M$ is sufficiently large then the result of straightening is a closed geodesic $\gamma_M$. Furthermore, for each $\epsilon>0$, if $M$ is sufficiently large then $\gamma_M$ is $\epsilon$-dense meaning that it hits every $\epsilon$-ball in $X$.

$\endgroup$
  • $\begingroup$ Thank you for the reference. Actually, I am reading that article and try to construct a closed geodesic which is shorter than one in the article. Following your answer, maybe we can not find an upper bound of the closed geodesic $\gamma_M$. Also, I don't understand why M is sufficiently large then the result of straightening is a closed geodesic $\gamma_M$. Could you please explain a little bit more or give me any reference? Thank you so much. $\endgroup$ – Markiff Sep 5 '17 at 7:09
  • 1
    $\begingroup$ The proof uses quasi geodesics. If you know what those are, and if you know the behavior of quasi geodesics in the hyperbolic plane, then the proof should be an exercise. $\endgroup$ – Lee Mosher Sep 5 '17 at 13:45

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.