It is not difficult to see that the curves of constant geodesic curvature on a geometric sphere are all circles: simple, closed curves that are geometric circles lying in a plane:
     Sphere curves
My question is:

Q. What are the curves of constant (positive) geodesic curvature on an ellipsoid?

Earlier Dmitri Panov asked a more general MO question, "Curves of constant curvature on $S^2$." My question is more specific. I would be interested to know how large is the class of simple (non-self-intersecting), closed, constant-curvature curves on an ellipsoid, whether there are nonsimple closed curves, whether there are infinitely long curves, etc. Dmitri's question revealed that many general questions are open, but perhaps there has been a special study made of the ellipsoid? I have so far not found any literature specifically on this. Thanks for pointers!

Addendum. I've made a few experiments which suggest that simple, closed curves of constant curvature might not be uncommon. My calculations were incorrect---Sorry to mislead!

Further edit. I've now rewritten the calculations, which (I think!) are now correct within numerical accuracy. Here is a portion of a (non-simple) constant-curvature curve on the ellipsoid $\frac{x^2}{2^2}+y^2+z^2=1$:
          Ellipsoid curve

  • $\begingroup$ do you want a geometric picture or a mathematical formula? $\endgroup$ – Will Sawin Jul 20 '12 at 14:58
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    $\begingroup$ @Will: A geometric picture! $\endgroup$ – Joseph O'Rourke Jul 20 '12 at 17:59
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    $\begingroup$ In the degenerate case of a cylinder, you can clearly get non-simple closed curves curves. Draw a circle on a piece of paper and roll the paper up into a tube so that it self-intersects. For a prolate ellipsoid whose long axis is long compared to the scale set by the curve's curvature, you should get the same behavior. $\endgroup$ – Ben Crowell Jul 22 '12 at 20:19
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    $\begingroup$ It's strongly counterintuitive to me that randomly chosen initial conditions would lead with nonzero probability to a simple closed curve. For the curves you showed, did you choose initial conditions that had special symmetry? Did you verify with high precision, or only visually, that they returned to the same point with the same tangent vector? $\endgroup$ – Ben Crowell Jul 23 '12 at 5:07
  • $\begingroup$ @Ben: You were right to be suspicious! My method of computing the angular turn at each point of the curve was incorrect. I agree that closure should not be so prevalent. $\endgroup$ – Joseph O'Rourke Jul 23 '12 at 11:57

You may want to have a look at the article Foliation by constant mean curvature spheres, by Rugang Ye, Pacific Journal of Mathematics, 147 (1991), 381–396.

In this article, the author shows, given a Riemannian surface $M$, that every nondegenerate critical point $p$ of the Gauss curvature has a punctured neighborhood that is foliated by closed curves of constant geodesic curvature. (Think of the rings of a bulls-eye centered on $p$.)

In particular, for an ellipsoid with three distinct axes, the six critical points of the Gauss curvature (where the axes meet the surface) are nondegenerate, and hence there exists a foliation by closed, constant geodesic curvature curves centered on each one. These must be real-analytic, so I imagine that the closed constant geodesic curvature foliations around each axis extend to foliate the entire ellipse away from the corresponding axis points. This would give three distinct $1$-parameter families of closed constant geodesic curvature curves on the ellipsoid with three distinct axis lengths. I don't know whether the classic differential geometers were able to compute these curves explicitly or not.

Of course, there will be many other closed curves with constant geodesic curvature on the ellipsoid that don't fall into these families. (Just think of the ones with geodesic curvature $0$, i.e., the closed geodesics, for example.) However, the 'generic' curve of constant geodesic curvature on the ellipsoid does not close.

On the other hand, it was asserted (without explicit proof) by Darboux (see Livre VI, Chapitre VII, Paragraph 622, footnote 1 of his Leçons sur la Théorie Générale des Surfaces) that the only surfaces for which all of the curves of constant geodesic curvature are closed are the surfaces of constant Gauss curvature. I don't know where a proof might first have appeared in the literature, but I once asked Victor Guillemin about this, and he produced a proof. In Ye's article quoted above, he shows that, if one has a sequence of closed curves of constant geodesic curvature whose curvatures go to infinity and that converge to a point $p$ in the surface, then $p$ must be a critical point of the Gauss curvature, and this provides an alternative proof of Darboux' claim.

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    $\begingroup$ "Foliation by constant mean curvature spheres": PDF download. $\endgroup$ – Joseph O'Rourke Jul 24 '17 at 13:16
  • $\begingroup$ I don't quite get the last deduction of Darboux' claim from Rick's paper. Maybe all constant curvature curves close up, but they might spin around many times before closing up (so are immersed, rather than embedded). I didn't look at Darboux' book, so maybe he meant simple closed curves. Of course, it seems implausible, but I didn't get the logic. $\endgroup$ – Ian Agol Jul 24 '17 at 18:04
  • $\begingroup$ @IanAgol: You are right that there is more to say, but I think it can be completed as follows: Take the tautological forms $(\omega_1,\omega_2,\omega_{12})$ on the unit circle bundle $B$ of the surface, and on $B\times\mathbb{R}$, consider the integral curves of the system $dt=\omega_2=\omega_1-t\omega_{12}=0$. By hypothesis, these are all closed, with the $t=0$ curves being the unit circles themselves. You can now compute the derivative of the Poincaré return map around the $t=0$ curves and find that it is the identity. Thus, the curves close up as simple curves on the surface for $t$ small. $\endgroup$ – Robert Bryant Jul 25 '17 at 8:20

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