9
$\begingroup$

A well-known result of Hedlund and Morse states that if a Riemannian metric on a closed surface of genus $g > 1$ has no conjugate points, then it carries transitive geodesics (i.e., geodesics whose velocity vectors are dense in the unit tangent bundle).

Are there Riemannian metrics on closed surfaces of genus $g > 1$ that do not carry a transitive geodesic and if so what is the weakest condition known under which the existence of transitive geodesics has been proved?

Addendum. As Misha remarks in his answer, it is easy to construct to metrics on any closed surface that do not carry a transitive geodesic. However, the metrics I'm interested in have the additional property that the lifted metric on the universal cover has no trapped geodesics. In other words, no geodesic stays forever in a compact subset of the open unit disc.

$\endgroup$
3
  • 1
    $\begingroup$ Take a round sphere, drill a hole in it from the north pole to the south pole. This is a torus without transitive geodesics and with no trapped geodesics. You can attach some handles on periphery if you'd like to. $\endgroup$ Aug 6, 2013 at 16:17
  • $\begingroup$ Thanks Andrey !! Could you please write this as an answer ? $\endgroup$ Aug 6, 2013 at 16:36
  • $\begingroup$ This is not much different from katz's answer, so I'd rather not. Here is a possible continuation: assume additionally that the set of non-periodic geodesics is dense. $\endgroup$ Aug 6, 2013 at 17:24

2 Answers 2

4
$\begingroup$

This is both: answer on the new version of the question and comment on comment of Andrey Gogolev, who asked whether one can make the question more complicated assuming additionally that the set of non-periodic geodesics is dense.

Hier is an example that answers both: it is not much different from the answer of Andrey from his comment (and also from katz's answer).

Take a torus with periodic coordinates $x,y$ and a liouville metric $(X(x) -Y(y)) (dx^2 + dy^2)$ on it (whose geodesic flow is integrable). Make two small holes on the torus near the points where $X$ has minimum and $Y$ has maximum and connect by a neck. For generic $X$ and $Y$ and generic neck the set of nonperiodic geodesic is dence but still it has a 'regular' region in the tangent space where no geodesic passing through at least one point of neck can come. The closure of a typical geodesic from the regular region is a torus in the tangent space which is projected to an annulus on the torus and after passing to the universal cover this annulus becomes an infinite unboundend band.

One more remark is that by KAM theory, this examples survives if you slightly perturbe the metric.

$\endgroup$
10
  • $\begingroup$ Thanks Vladimir. It does seems then that the Morse-Hedlund theorem is as good as it gets. By the way that theorem has now been extended to Finsler metrics by Gomes and Ruggiero (for reversible metrics it is just the same techniques, but the extension is not trivial for non-reversible metrics). $\endgroup$ Aug 7, 2013 at 17:16
  • $\begingroup$ What is Morse-Hedlund theorem and why one can not make it better? $\endgroup$ Aug 7, 2013 at 17:30
  • $\begingroup$ @Matveev: ?? that's the question posted. $\endgroup$ Aug 7, 2013 at 18:00
  • $\begingroup$ Of course. Sorry. As always I skipped the introduction and went to the question immediately. By the way, Victor Bangert claimed that he could proved the existence of a finsler metric on a closed surface of genus >1 whose geodesic flow is integrable -- he did not publish the proof as far as I know though $\endgroup$ Aug 7, 2013 at 18:12
  • $\begingroup$ I think Gabriel Paternain proved that a real analytic Finsler metric on a surface of genus g > 1 cannot be integrable. That makes me quite curious about this example of Bangert !! $\endgroup$ Aug 7, 2013 at 19:44
4
$\begingroup$

A connected sum of a hyperbolic surface with the round sphere along a thin neck near the south pole will not be transitive in this sense, because any geodesic arriving from the thin neck will never reach vectors close to the tangent vectors of the equator, because any great circle near the equator stays away from the south pole. Related thread Surfaces filled densely by a geodesic

$\endgroup$
1
  • $\begingroup$ Thanks for the remark. I was really thinking of metrics that do not contain trapped geodesics in their universal cover. I'll change the question to reflect this. $\endgroup$ Aug 6, 2013 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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