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Let $M$ be a compact manifold (w/o boundary). Suppose that there is no closed geodesics on $M$ of length precisely $C$. I am trying to prove that there is an open cover $\{U_j\}$ of $M$ and $\epsilon >0$ with the following property: if $\gamma$ is a unit speed geodesic in $M$, $C-\epsilon < t < C+\epsilon$, then $\gamma(0)$ and $\gamma(t)$ do not lie in the same $U_j$.

It seems like this would be amenable to a compactness argument of some sort, perhaps using an exponential neighborhood at each point. But, I am not seeing a good way to do it. Any suggestions?

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Do you want to prove this for all j? – Tomás Oct 7 '12 at 1:16
up vote 4 down vote accepted

The exponential map considered as a map on the tangent bundle $\exp: TM \to M$ is continuous. Let $S_C := \{(p,v)\in TM\mid |v|_p = C\}$.

By assumption, we have $d(p,\exp_p(v)) > 0$ for all $(p,v)\in S_C$. So by compactness of $S_C$ and continuity of $(p,v)\mapsto d(p,\exp_p(v))$, there exists $\delta > 0$ such that $d(p,\exp_p(v)) \ge \delta$ for all $(p,v)\in S_C$. If you now choose a small $\epsilon > 0$, then (again using compactness) you will still have $d(p,\exp_p(v)) \ge \delta/2$ for all $(p,v)\in TM $ such that $C-\epsilon< |v| < C+\epsilon$. So this $\epsilon > 0$ and any covering of diameter $<\delta/2$ will do the job.

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Hey Sam. I have two questions. What assumption are you assumig when you say $d(p,exp_{p}(v))>0$. I think this is always true when $|v|_{p}\neq 0$ or no? When you conclude $d(p,exp_{p}(v))>\delta$, are you assuming $(p,v)\in S_{C}$? – Tomás Oct 7 '12 at 1:28
No, $d(p, \exp_p(v))$ could be zero if there was a closed geodesic. For example on the sphere you have $$\exp_p(v) = p\cos(|v|) + \frac{v}{|v|}\sin(|v|)$$ so $\exp_p(v) = p$ if $|v|$ is an integer multiple of $2\pi$. To the second question, yes. I have forgotten to mention that, thank you. – Sam Oct 7 '12 at 1:34
ok you are right. Thanks by the patient. – Tomás Oct 7 '12 at 1:39
@Tomás: that there is no closed curve of length exactly $C$ implies that $p\notin \{\exp_pv|v\in T_pM, |v|=C\}$. – Berci Oct 7 '12 at 1:41

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