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The Zoll surfaces have the property that all of their geodesics are closed. If one futher stipulates that all geodesics are also simple, i.e., non-self-intersecting, does this leave only the sphere?

Apologies for the simplicity of this question, but I am not finding an answer in the literature, and I suspect many just know this off the top of their head. Thanks!

flag	asked Jun 18 2010 at 12:23 Joseph O'Rourke 44.1k●1●50●216

What wrong with a torus (or indeed, any constant curvature Riemann surface) ? – David Lehavi Jun 18 2010 at 12:48

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Most geodesics on a torus are non-closed. – Torsten Ekedahl Jun 18 2010 at 12:58

Anton Petrunin · Accepted Answer · 2010-06-18 13:09:16Z UTC

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From Guillemin's "The Radon transform on Zoll surfaces", it follows that there are deformations of $S^2$ which keep all geodesics closed AND simple.

answered Jun 18 2010 at 13:09

Anton Petrunin
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Thanks so much! V. Guillemin, The Radon transform on Zoll surfaces. Advan. Math. 22 (1976), pp. 85–119. I will have to retrieve that paper! – Joseph O'Rourke Jun 18 2010 at 13:17

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Yes, many metrics have this property. There is a very detailled reference by Arthur Besse: Manifolds all of whose Geodesic are closed. – Benoît Kloeckner Jun 18 2010 at 13:18

@Benoit: Thanks! I wonder from the title of that book (which I just ordered through Interlibrary Loan) if this class of manifolds is of interest: every closed geodesic is simple (but not necessarily every geodesic is closed). – Joseph O'Rourke Jun 18 2010 at 14:07

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According to Berger's "Panoramic view of Riemannian geometry", p. 439, Gromoll and Grove proved in 1981 that the hypothesis that all geodesics are closed implies that they have the same length and are simple. – BS Feb 29 2012 at 18:34

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@BS: you should assume something about the space, otherwise the space forms would give a counterexample. – Anton Petrunin Mar 3 2012 at 2:57

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Surfaces all of whose geodesics are both closed and simple

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