most recent 30 from http://mathoverflow.net 2013-05-20T22:00:14Z http://mathoverflow.net/feeds/question/28622 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28622/surfaces-all-of-whose-geodesics-are-both-closed-and-simple Joseph O'Rourke 2010-06-18T12:23:33Z 2010-06-18T13:09:16Z

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

http://mathoverflow.net/questions/28622/surfaces-all-of-whose-geodesics-are-both-closed-and-simple/28627#28627 Anton Petrunin 2010-06-18T13:09:16Z 2010-06-18T13:09:16Z

<p>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.</p>