Geodesic circles on riemannian manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:46:23Z http://mathoverflow.net/feeds/question/98990 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98990/geodesic-circles-on-riemannian-manifolds Geodesic circles on riemannian manifolds Huliber 2012-06-06T20:55:20Z 2012-06-07T01:49:14Z <p>Can one always find, in a compact riemannian manifold, a closed geodesic isometric to a usual circle when endowed with the ambient distance ? For instance, in the usual flat torus, the only geodesics verifying this property are those of slope \$0\$, \$1/2\$ and \$1\$. </p> <p>The natural candidates are, of course, closed geodesics of minimal length...</p> http://mathoverflow.net/questions/98990/geodesic-circles-on-riemannian-manifolds/99002#99002 Answer by Vitali Kapovitch for Geodesic circles on riemannian manifolds Vitali Kapovitch 2012-06-07T01:49:14Z 2012-06-07T01:49:14Z <p>I'm not sure but I believe you are asking if there always exists a closed geodesic such that it gives a distance preserving embedding of \$S^1\$ with respect to the length metric on the circle and the ambient metric on the target manifold. If this is the question then the answer is no.</p> <p>Balacheff, Croke and Katz in <a href="http://www.springerlink.com/content/ax605170431104q2/" rel="nofollow">"A Zoll counterexample to a geodesic length conjecture"</a> have an example of a metric on \$S^2\$ where the length of the shortest closed geodesic is strictly bigger than twice the diameter of \$S^2\$. This of course gives a counterexample to the above question too.</p>