Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
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. –  Jean-Marc Schlenker Aug 31 '11 at 15:13
    
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. –  Julia E Aug 31 '11 at 15:22
add comment

1 Answer 1

up vote 18 down vote accepted

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.

share|improve this answer
    
+1 @Sergei. That's very elegant. –  Tom LaGatta Aug 31 '11 at 17:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.