Does there exist a compact Riemannan manifold $M^n$ and an $L > 0$ such that the number of homotopy classes of simple closed curves $\gamma$ on $M^n$ whose shortest representatives have length at most $L$ is infinite? For surfaces ($n=2$) with constant curvature metrics, this is impossible. Thanks!
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1$\begingroup$ There is no canonical geodesic representative of a homotopy class of curves in a Riemannian manifold in general (there is one if the curvature is negative). Moreover the notion of homotopy class of simple closed curves does not make sense in dimension larger than 2. $\endgroup$– Jean-Marc SchlenkerCommented Aug 31, 2011 at 15:13
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$\begingroup$ I really meant shortest representative. I changed the question to reflect that. And I put simple in there because I'm also interested in other metrics on surfaces -- in higher dimensions, the condition is vacuous. $\endgroup$– Julia ECommented Aug 31, 2011 at 15:22
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This cardinality is always finite, for any compact locally simply connected metric space. If there were infinitely many non-homotopic curves of length $\le L$, they would have a converging subsequence (by Arzela-Ascoli). In a locally simply connected space, any two sufficiently close curves are homotopic, so curves in the sequence are eventually homotopic to their limit, a contradiction.
answered Aug 31, 2011 at 16:55
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$\begingroup$ +1 @Sergei. That's very elegant. $\endgroup$ Commented Aug 31, 2011 at 17:07