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Let $M$ be a closed Riemannian manifold of dimension $n \geq 2$. I am looking at sufficient conditions for the following two properties:

  1. existence of closed geodesics of arbitrarily long length on $M$. Of course, we assume that a closed geodesic must be traversed only once to find its length

  2. existence of arbitrarily long geodesic segments. By geodesic segment, we mean geodesics with domains restricted to closed intervals, and whose start and endpoints are not necessarily the same.

One can see that on round spheres, neither of the above is possible. On the other hand, on negatively curved manifolds, both are true (due to ergodicity of the geodesic flow, I presume). They are true also on tori. Also, the first condition implies the second condition.

Being a beginner, I am not sure what is known in the literature, or how to go about thinking about such questions. Any advice is highly appreciated!

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If you restrict yourself to simple (non-self-intersecting) closed geodesics on convex surfaces, then a complete characterization is known, as described in these two papers:

(1) Protasov, Vladimir Yu. "Simple Closed Geodesics on a Polyhedron." The Mathematical Intelligencer (2024): 1-8.

"the disphenoid [isosceles tetrahedron] admits arbitrarily long geodesics. Then we show that no other convex polyhedra possess this property."

(2) Akopyan, Arseniy, and Anton Petrunin. "Long geodesics on convex surfaces." arXiv preprint arXiv:1702.05172 (2017).

"If the surface of a convex body $K$ contains arbitrary long closed simple geodesics, then $K$ is an isosceles tetrahedron." ... "Note that the main theorem implies in particular that a smooth convex surface does not have arbitrarily long simple closed geodesics."

Figure from Akopyan-Petrunin.

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    $\begingroup$ I think that is a complete characterization for polyhedra with flat faces, not for smooth compact surfaces with arbitrary curvature. $\endgroup$
    – Ben McKay
    Commented Aug 3 at 15:09
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    $\begingroup$ @BenMcKay: The second paper does not assume $K$ is a polyhedron. I edited to make that clearer. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 3 at 15:16
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If I understood the question correctly the second condition is just a question if your manifold has at least one non-closed geodesic. It is a rather generic property.

Any metric on a sphere (except possibly in dimension 3) with all geodesics closed is necessarily Zoll (i.e. the lengths of all geodesics are the same), due to results of Grove-Gromoll and Radeschi-Wilking. In dimension $\ge 3$ Zoll manifolds are known to be symmetric, and in dimension 2 they are quite well studied.

There is no chance for a surface other than $S^2$ and $\mathbb RP^2$ to have a metric with all geodesics closed, so any surface of higher genus with any metric satisfy your property 2.

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  • $\begingroup$ Unless I am missing something, what you are saying is that a topological sphere (or maybe a simply connected manifold) with all geodesics closed is characterized. But that does not necessarily say anything about other manifolds, right? Why can't some other topological manifold have a metric with all geodesics closed? $\endgroup$
    – H. Saito
    Commented Aug 3 at 18:49
  • $\begingroup$ @H.Saito "Why can't some other topological manifold have a metric with all geodesics closed? " I only claimed it for surfaces! Considering the exponential map at some point of a surface with all geodesics closed, one can see that the surface can be covered by a sphere, hence its genus is 0. $\endgroup$
    – Dmitrii Korshunov
    Commented Aug 3 at 20:34

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