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Let $M$ be a closed Riemannian manifold. What are the necessary and sufficient conditions on $M$ to ensure that for every point $p \in M$, and every geodesic $\gamma: [0, \infty) \to M$ with $\gamma(0) = p$, we have that $\liminf_{t \to \infty} d(\gamma(t), p) = 0$?

Note: Here d denotes the usual Riemannian distance.

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    $\begingroup$ A theorem of Anosov shows that the geodesic flow on the unit sphere bundle of a compact manifold with negative curvature is ergodic. so for most $p$, the geodesic that started at $p$ will return infinitely often to any neighborhood of $p$. $\endgroup$
    – Liviu Nicolaescu
    Apr 11, 2021 at 11:49
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    $\begingroup$ @LiviuNicolaescu that's definitely an answer... unless the OP really wants both necessary and sufficient, in which case I can't really see any way to formulate a simple answer that includes both Anosov's theorem and Zoll manifolds. $\endgroup$
    – Willie Wong
    Apr 11, 2021 at 11:54
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    $\begingroup$ I wasn't finished with my coffee when I wrote the comment. The geodesic flow on any compact Riemann manifold is volume preserving and the Poincare recurrence theorem shows that any trajectory of the flow will intersect any open set infinitely often. $\endgroup$
    – Liviu Nicolaescu
    Apr 11, 2021 at 12:34
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    $\begingroup$ Hm, from what I understand the Poincare recurrence theorem ensures recurrence for the trajectory of almost every point. Does this easily extend to every point? $\endgroup$
    – Nate River
    Apr 11, 2021 at 12:48
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    $\begingroup$ Certainly geodesic flows on manifolds of negative curvature have some points that are not recurrent. $\endgroup$
    – Anthony Quas
    Apr 11, 2021 at 16:54

2 Answers 2

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This is a comment that is meant to show that one should not look for such Riemannian manifolds among negatively curved ones. Indeed, let $\Sigma$ be a compact surface with a metric of negative curvature. Then for any point $p$ there is a geodesic $\gamma: [0,\infty)$ such that $\lim \inf_{t\to \infty}(p,\gamma(t))>0$.

Proof. Take any simple closed geodesic $\eta$ on $\Sigma$ that doesn't contain $p$. Parametrise $\eta$ by $S^1=\mathbb R/\mathbb Z$. Consider a geodesic segment $\gamma_s$ that joins $p$ with a point $\eta(s)$ and let $s\to \infty$. You will be able to vary $\gamma_s$ continuously changing $s$ continuously (because curvature is negative). Then segments $\gamma_s$ will converge in the limit to a geodesic ray that accumulates to $\eta$.

This construction generalises to any dimension.

One might wonder about positive curvature. Of course, if we take a round sphere, this is a good example, all trajectories are periodic. However, I am not quite sure if a generic $2$-sphere of positive curvature has the property you are asking for. Here a toy model are billiards (https://en.wikipedia.org/wiki/Dynamical_billiards) in convex domains in $\mathbb R^2$. This is a heavily studied subject. However, if one takes the second simplest billiard - an ellipse, then the property you are looking for doesn't hold for a subset of trajectories of codimension $1$. Namely, if you take a trajectory that passes through a focus and continue it to infinity, it will converge to the large axes of the ellipse. For all other trajectories, that don't pass through a focus indeed, there is recurrence - they come back to $p$ as close as you want.

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  • $\begingroup$ How can $s$ go to infinity in $S^1$? $\endgroup$
    – R W
    Jun 15, 2021 at 13:52
  • $\begingroup$ $S^1=\mathbb R/\mathbb Z$ you go to infinity in $\mathbb R$. So $\gamma_s\ne \gamma_{s+1}$. For example if $\Sigma=\mathbb R^2/\mathbb Z^2=T^2$, $p=(0,0)$ and the geodesic is the vertical circle $x_1=1/2$, then in the limit this construction will give you a vertical geodesic in through $(0,0)$ $\endgroup$
    – Dmitri Panov
    Jun 15, 2021 at 14:01
  • $\begingroup$ OK - so you are saying the following. Take a closed geodesic not passing through $p$, then any geodesic ray asymptotic to it (which can be chosen to be issued from $p$) has the required property. I agree. It is actually best seen by looking at the universal covering space. Still, it is a bit misleading to say that $s\in S^1$ goes to infinity in $\mathbb R$ :) $\endgroup$
    – R W
    Jun 15, 2021 at 15:29
  • $\begingroup$ You are right, I was sloppy. To spell out this procedure one should indeed better go to the universal cover. $\endgroup$
    – Dmitri Panov
    Jun 15, 2021 at 15:58
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    $\begingroup$ The connection is as follows. Suppose you what to study the billiard $ax^2+by^2=1$. Then you consider the geodesic flow on the ellipsoid $ax^2+by^2+\varepsilon z^2=1$ and let $\varepsilon \to 0$. Then the geodesic flow "converges" to the billiard, this a classical idea. By the way, the geodesic flow on ellipsoids is very well understood. It is worth checking if such accumulating trajectories always exist (I forgot, but suspect they do). I guess the moral of this billiard example is that on a positively curved surface an infinite geodesic can accumulate to a closed one (I believe it's true). $\endgroup$
    – Dmitri Panov
    Jun 17, 2021 at 7:33
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Just so that it doesn't get lost in the comments: in this paper Nadirashvili shows that a a $C^2$ Riemannian metric in the 2-torus for which the geodesic flow is recurrent (all points are Poisson stable) must be flat.

N. S. Nadirashvili, “Conditions of stability in the sense of Poisson of a geodesic flow on a torus”, Mat. Zametki, 44:1 (1988), 147–149

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