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 $1$.

The natural candidates are, of course, closed geodesics of minimal length...

  • $\begingroup$ The only isometric invariant of "closed connected Riemannian manifolds of dimension 1" (a.k.a. circles) is their length. You must mean something else. $\endgroup$ – alvarezpaiva Jun 6 '12 at 21:27
  • $\begingroup$ I think its about closed geodesics without self-intersection. $\endgroup$ – user1688 Jun 6 '12 at 21:54
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    $\begingroup$ Do you mean 'isometric' in the strict sense that the restriction of the ambient distance function to the product of the geodesic with itself yields a distance function that is the same as the distance function for a circle in the Euclidean plane? $\endgroup$ – Robert Bryant Jun 6 '12 at 22:06
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    $\begingroup$ If it were about closed geodesics without self-intersection, then slope $1/n$ would work for any $n$. Thus, I think it's what Robert Bryant said. $\endgroup$ – Will Sawin Jun 6 '12 at 22:46

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.

Balacheff, Croke and Katz in "A Zoll counterexample to a geodesic length conjecture" 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.


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