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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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  • $\begingroup$ Joseph: You meant "Bryant", not "Grant". $\endgroup$
    – Misha
    Commented Apr 18, 2013 at 13:21
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    $\begingroup$ @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'? $\endgroup$
    – Robert Bryant
    Commented Apr 20, 2013 at 13:09
  • $\begingroup$ @Robert: You are right, Robert. Now corrected. Thanks! $\endgroup$
    – Joseph O'Rourke
    Commented Apr 20, 2013 at 14:12

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  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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  • $\begingroup$ Thank you, Robert! I didn't realize this was implied by the Clairaut integral. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 14, 2013 at 14:22
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    $\begingroup$ @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. $\endgroup$
    – Robert Bryant
    Commented Apr 14, 2013 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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  • $\begingroup$ 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! $\endgroup$
    – Joseph O'Rourke
    Commented Apr 14, 2013 at 19:55
  • $\begingroup$ Nice answer. It would be interesting to know whether there could be metrics on aspherical surfaces where no geodesic is dense. $\endgroup$
    – Mikhail Katz
    Commented Apr 15, 2013 at 7:49
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    $\begingroup$ 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 $\ge 2$. $\endgroup$
    – Misha
    Commented Apr 16, 2013 at 19:35
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    $\begingroup$ I don't believe this. I think you can `plumb in' a closed geodesic whose linearization is elliptic a la KAM. Then, by KAM the geod. flow will have ``elliptic islands' which block the density of any geodesic. $\endgroup$
    – Richard Montgomery
    Commented Oct 14, 2013 at 2:31
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    $\begingroup$ @Richard: I think you mean that KAM blocks density in the unit tangent bundle, but the question is about density in the surface. A dense geodesic could (in principle) keep intersecting the KAM geodesic at angle bounded away from zero. $\endgroup$
    – Misha
    Commented Oct 14, 2013 at 5:06
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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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  • $\begingroup$ @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. $\endgroup$
    – Robert Bryant
    Commented Apr 14, 2013 at 13:52
  • $\begingroup$ 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. $\endgroup$
    – Mikhail Katz
    Commented Apr 14, 2013 at 14:52
  • $\begingroup$ 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. $\endgroup$
    – Misha
    Commented Apr 15, 2013 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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  • $\begingroup$ @Carl: Which billiard do you think has no dense orbits? $\endgroup$
    – Misha
    Commented Apr 19, 2013 at 0:40
  • $\begingroup$ @Misha: Thanks - I have added some examples. $\endgroup$
    – user25199
    Commented Apr 19, 2013 at 8:28
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    $\begingroup$ 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. $\endgroup$
    – Misha
    Commented Apr 19, 2013 at 13:13
  • $\begingroup$ @Misha: A useful summary of the boundary between the known & unknown! $\endgroup$
    – Joseph O'Rourke
    Commented Apr 20, 2013 at 23:45
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I feel there is some implicit idea here but nobody formulated it, so let me try:

a smooth closed surface has no filling geodesic if and only if its geodesic flow is integrable

What do you think about this conjecture ? For me it seems probable since integrable flow have their Arnold tori that don't fill up all the space: for example, for the ellipsoid, the picture of the space that a geodesic fills in is a projection of an Arnold tori (an annulus).

All the examples stated above are integrable, and Burns guess is nice because surfaces of big genus can't have an integrable flow (Kozlov theorem, if I am not wrong...)

I am really excited about this conjecture.

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    $\begingroup$ False: the flat torus has integrable geodesic flow, but some geodesics are dense. $\endgroup$
    – Ben McKay
    Commented Oct 18, 2017 at 11:16

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