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Which smooth, closed surfaces $S \subset \mathbb{R}^3$ have no single geodesic $\gamma$ that fills $S$ densely?

Say a geodesic $\gamma$ "fills $S$ densely" if the closure of the set of points through which $\gamma$ passes equals $S$. Some examples:

My assumption is that almost all surfaces have geodesics that fill them. Is this known, under any interpretation of "almost all"? I would also be interested in extending the list of exceptional surfaces beyond {sphere, Zoll, ellipsoid}. Thanks for pointers!


Answers Summary (18Apr2013):

  • (Robert Bryant, Mikhail Katz) Any surface of revolution with poles has no dense geodesic. This holds for convex or nonconvex surfaces of revolution.
  • (Robert Bryant) There are generalizations of Liouville surfaces (due to Goryachev-Chaplygin and to Dullin-Matveev) that have no dense geodesic.
  • (Misha Kapovich) Every surface may be perturbed by gluing on "focusing caps" so that it has dense geodesics.
  • (Keith Burns) Guess: There is always a dense geodesic on a closed Riemannian surface of genus $\ge 2$.
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Joseph: You meant "Bryant", not "Grant". –  Misha Apr 18 '13 at 13:21
    
@Misha: Whoops! Thanks; corrected. –  Joseph O'Rourke Apr 18 '13 at 13:58
    
@Joseph: I'm not sure why you write "Certain Zoll surfaces...". I thought that the definition of 'Zoll surface' was that each of the geodesics closes, i.e., is periodic (not necessarily with the same period for all geodesics), so that no geodesic is dense in the surface. At what caveat are you hinting by using the word 'certain'? –  Robert Bryant Apr 20 '13 at 13:09
    
@Robert: You are right, Robert. Now corrected. Thanks! –  Joseph O'Rourke Apr 20 '13 at 14:12
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4 Answers

up vote 26 down vote accepted
  1. Any surface of revolution in $3$-space with poles will have this property. The reason is that, in this case, any geodesic either goes through a pole (i.e., a point where the axis of revolution meets the surface) and is a profile curve that lies in a plane or else, because of the Clairaut integral, it avoids that pole by some positive distance. Thus, no geodesic on the surface is dense in the surface.

  2. You mention ellipsoids, which furnish examples of these special surfaces. These are examples of so-called 'Liouville surfaces', i.e., Riemannian surfaces $(S,g)$ for which there exist two independent first integrals of the geodesic flow on $T^\ast S$ that are quadratic functions on the fibers of $T^\ast S\to S$, one of which is the co-metric associated to $g$ and the other of which is an independent first integral. As you probably know, surfaces of revolution are surfaces for which there exist a first integral of the geodesic flow that is linear on the fibers of $T^\ast S\to S$, namely the Clairaut integral. It has been known for some time that there are metrics on the $2$-sphere that don't possess any 'extra' first integrals that are linear or quadratic functions on the fibers of $T^\ast S\to S$, but do possess first integrals that are cubic or quartic functions on the fibers of $T^\ast S\to S$. These are due to Goryachev-Chaplygin (early 20th century) and Dullin-Matveev (2004). These are also examples for which no geodesic winds densely over the surface. All of these work because there are 'conservation laws' for the geodesic flow of a particular kind, and they properly generalize the Liouville surfaces (which includes the famous case of ellipsoids).

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Thank you, Robert! I didn't realize this was implied by the Clairaut integral. –  Joseph O'Rourke Apr 14 '13 at 14:22
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@Joseph: Yes, the point is that, in polar coordinates, the metric is $ds^2 = dr^2 + f(r)^2\ d\theta^2$, where $f(0)=0$ and $f'(0)=1$. The energy integral gives $\dot r^2 + f(r)^2\ \dot\theta^2 = 1$ on a unit speed geodesic, and the Clairaut integral, $f(r)^2\dot\theta = C$, is constant on a geodesic. When $C=0$, the geodesic goes through the pole at $r=0$, but when $C\not=0$, the above two equations show that $f(r)^2\ge C^2$, so $r$ has a positive minimum on the geodesic. –  Robert Bryant Apr 14 '13 at 14:33
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Donnay and Pugh proved that every embedded surface $S\subset R^3$ can be $C^0$-perturbed so that the new metric has ergodic geodesic flow, see here. In particular, the new metric will have dense geodesics (moreover, "generic" geodesics will be dense in the unit tangent bundle).

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From the Abstract: "In this paper we show that any surface in $\mathbb{R}^3$ can be modified by gluing on small ‘focusing caps’ so that its geodesic flow becomes ergodic." Thanks, Misha! –  Joseph O'Rourke Apr 14 '13 at 19:55
    
Nice answer. It would be interesting to know whether there could be metrics on aspherical surfaces where no geodesic is dense. –  katz Apr 15 '13 at 7:49
    
@katz: Misha, I do not know, but you should ask Keith Burns, he would know the answer or if this is an open problem. –  Misha Apr 15 '13 at 13:02
    
Good idea. I haven't spoken to him in years. Could you ask? –  katz Apr 15 '13 at 17:12
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Update: Keith Burns does not know an answer (which means this should be treated as an open problem); his guess is that there is always a dense geodesic on a closed Riemannian surface of genus $\ge 2$. –  Misha Apr 16 '13 at 19:35
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Clairaut's relation shows that any simply connected surface of revolution has this property, whether convex or not (for the reason stated by Robert).

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@katz: Thanks. I, too, realized, shortly after I posted, that 'convex' wasn't necessary, so I edited my answer and re-posted it just before I saw yours. –  Robert Bryant Apr 14 '13 at 13:52
    
For arbitrary metrics on hyperbolic surfaces, will ergodicity guarantee that any metric will have a dense geodesic? For arbitrary merics on tori I am not sure. –  katz Apr 14 '13 at 14:52
    
A trivial remark is that there are many Riemannian metrics on closed hyperbolic surfaces, so that geodesic flow is not ergodic and there are no dense geodesics in the unit tangent bundle. However these are local constructions which do not prevent existence of dense geodesics on the surface itself. –  Misha Apr 15 '13 at 16:59
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A billiard with no dense orbits as in the question dense orbits in billiards may be the "flattened" (doubly covered) limit of such surfaces.

[Edit in response to Misha:] Examples of billiards with no dense orbits: Circle or ellipse billiards that are limits of ellipsoids. The Penrose solution to the illumination problem: see

N. Chernov and G. Galperin "Search light in billiard tables" Regular and Chaotic Dynamics 8, 225-241 (2003).

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@Carl: Which billiard do you think has no dense orbits? –  Misha Apr 19 '13 at 0:40
    
@Misha: Thanks - I have added some examples. –  Carl Apr 19 '13 at 8:28
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To summarize: 1. It is unknown if there is a polygonal billiard (convex or not) which does not admit a dense billiard trajectory. 2. It is unknown if there is a region with smooth boundary (apart from ellipses) which does not admit a dense billiard trajectory. 3. There is example of Rauch (1978) for (2), which has one non-smooth boundary point. –  Misha Apr 19 '13 at 13:13
    
@Misha: A useful summary of the boundary between the known & unknown! –  Joseph O'Rourke Apr 20 '13 at 23:45
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