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Let $S$ be a surface embedded in $\mathbb{R}^3$. A simple geodesic on $S$ is one that does not self-intersect. Some surfaces have simple geodesics whose length exceeds any given bound $L$. For example, a cylinder or a torus allows tight winding geodesics that are arbitrarily long before they cross themselves. But a sphere, or a Zoll surface, does not admit arbitrarily long simple geodesics, because every geodesic forms a simple closed loop.

Q. Which surfaces $S$ admit arbitrarily long simple geodesics?

To be specific: Do ellipsoids possess such geodesics?


Update (11 May 2017).

This paper settles a version of my 2-yr-old question by proving that "if the surface of a convex body $K$ contains arbitrary long closed simple geodesics, then $K$ is an isosceles tetrahedron":

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

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    $\begingroup$ If the surface is compact and oriented (and we really are using the fact that we are on a 2d surface here), then to admit an arbitrarily closed simple geodesic, the surface needs to admit a stable closed simple geodesic. This can be seeing by taking the set theoretic closure of the original geodesic, this closure has the structure of a geodesic lamination and one of the leaves should be the closed geodesic (which is also necessarily stable). $\endgroup$ – foliations Nov 12 '15 at 12:36
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    $\begingroup$ Conversely, if the surface admits a two-sided closed strictly stable simple geodeisc, then it should be possible to construct an arbitrarily long simple geodesic by minimizing in the universal cover of the tubular neighborhood of the closed geodesic . $\endgroup$ – foliations Nov 12 '15 at 12:39
  • $\begingroup$ @foliations: Thanks. Every surface has three simple, closed geodesics (Lusternik-Schnirelmann), but I am not sure what is a "stable" simple closed geodesic. $\endgroup$ – Joseph O'Rourke Nov 12 '15 at 12:53
  • $\begingroup$ Stable just means locally length minimizing. So the equator on a sphere is not stable but the neck of a hyperbola of revolution is. $\endgroup$ – foliations Nov 12 '15 at 13:02
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    $\begingroup$ @foliations the OP is asking for arbitrarily long geodesics, not infinitely long ones, so it is not clear what you mean by "the original geodesic" $\endgroup$ – Igor Rivin Nov 12 '15 at 13:23
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Elipsoid does not posess unbounded geodesics with no self-intersection.

I do not know a conceptual explanation. My explanation is that (due to integrability of the geodesic flow of ellipsoid) we know the geodesic of the ellipsoid, let me shortly describe them.

The typical geodesic viewed as a curve in the tangent bunlde lives on the Liouville torus and is a winding -- periodic or quasiperiodic -- on it. The projection of the Liouville torus to the ellipsoid is a ring (the projection is singular at two lines which project to the boundary circles of the ring and otherwise is the double cover of the interior of the ring. This implies that each such typical geodesic intersects itself.

Consider now ``untypical geodesics'', i.e., those such that their lift to the tangent bundle lies on a singular leaf of the liouville foliation or is a critical circle. The second type are already closed geodesics ( and on the ellipsoid there are at most 3 of such geodesics of the second type).

Now, the last case, i.e. the geodesic lying on a critical leaf are precisely the geodesic passing through 4 umbillic points, and we know that if a geodesic passes an umbilic point of the ellipsoid it passes through infinitely umbilic points infinitely many times which implies it has selfintersections.

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  • $\begingroup$ Thank you! I do not doubt your proof for a second. But is there any intuition that can help explain why a very nearly closed geodesic around the minor axis of the ellipsoid does not wind around near that closed geodesic many times before self-intersecting? $\endgroup$ – Joseph O'Rourke Nov 13 '15 at 5:48
  • $\begingroup$ I do not understand the question, Joseph (and would love if you give more details) but let me still try. I understood your question such that your ''going to be a counterexample'' curve stays from from one side of some closed geodesic and is closer and closer to it for big times. Then, one may think that your ''going to be a counterexample'' lies on a variation of your closed geodesic, i.e., is controlled by Jacobi vector field. By Jacobi vector field is controlled by the equation involving curvature and for positive curvatur it vanishes in finite time so something goes wrong $\endgroup$ – Vladimir S Matveev Nov 13 '15 at 8:45
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This paper

Rouyer, Joël(R-AOS); Vîlcu, Costin(R-AOS)
Simple closed geodesics on most Alexandrov surfaces. (English summary) 
Adv. Math. 278 (2015), 103–120. 
53C45 (53C22) 

Indicates that this is usually true whenever curvature is not everywhere non-negative. In genus $1,$ it is ALWAYS true when the curvature is everywhere $0.$

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Anosov constructed (Anosov, Section 8) a smooth Riemannian metric $g$ on $S^2$ such that the surface $(S^2,g)$ admits arbitrarily long non-intersecting closed geodesics. His construction starts with a standard sphere $S^2$. Split it along its equator, and insert a surface of revolution with a thinner waist. Then he deformed the metric on the upper hemisphere along with a geodesic such that the directions around the picked path get twisted faster. Then he did the same deformation on the lower hemisphere. Those long geodesics spend most of their time around the waist, while the twisting on the two hemispheres guarantees the nonintersection property. The resulting surface is very nice, and I enjoyed picturing it in my mind very much.

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  • $\begingroup$ I think the Anosov's picture is pretty similar to those described in Akopyan--Petrunin's paper (provided in the question). Beside the smoothness v.s. convexity, the main difference is that Anosov deformed the spherical cap to twist the direction, while Akopyan--Petrunin utilized the singularity to make the twist. $\endgroup$ – Pengfei May 14 '17 at 0:48

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