Which surfaces admit unbounded-length simple geodesics?

Question

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?

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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).

Answer 1

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.

Answer 2

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 $0.$

Answer 3

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.