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Let $M^{n}$ a Riemannian manifold orientable with positive curvature and even dimension. Let $\gamma$ a closed geodesic. Prove that $\gamma$ is homotopic to a closed curve whose lenght is strictly less than $\gamma$. I would any tips for solve this problem, because , i honestly i do know which tool use.

Thanks.

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    $\begingroup$ You probably need some assumptions about compactness or completeness of $M$. The key will be to produce a deformation of $\gamma$ that, due to some fact about positive curvature, has shorter length. What tools describe deformations of geodesics? (Hint: starts with "J" and ends with "acobi field") $\endgroup$
    – Neal
    Commented Dec 8, 2016 at 1:27
  • $\begingroup$ @Neal i will think according to your tips. Thanks. $\endgroup$ Commented Dec 8, 2016 at 1:45
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    $\begingroup$ @Neal: No need for compactness/completeness. This is just a local issue in a small neighborhood of $\gamma$. $\endgroup$ Commented Dec 8, 2016 at 5:55
  • $\begingroup$ @MoisheCohen Good call. $\endgroup$
    – Neal
    Commented Dec 8, 2016 at 18:32

2 Answers 2

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Consider a parallel translation $P : T_pM\rightarrow T_pM$ along a closed geodesic $\gamma$ where $\gamma (0)=\gamma (l)=p$

Since $M$ is orientable, then $P$ is orientation preserving isometry on $T_pM$ so that it has a fixed point $v$. When $\gamma(t,s)=\exp_{\gamma (t)}\ sV(t)$ where $V$ is a parallel field s.t. $V(0)=v$, then ${\rm length}\ \gamma(t,s)<{\rm length}\ \gamma(t,0)$ for $s>0$, since $M$ has a positive curvature.

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Either I didn't understand you, or the statement is false. There exist non- simply connected positively curved manifolds of even dimension (e.g. product of two lens spaces). So choose any non-trivial class in fundamental group and consider the shortest (smooth) loop in this class by minimizing the length functional. It will be geodesic.

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  • $\begingroup$ Your example does not have strictly positive curvature. It is a standard result that found e.g. in do Carmo's book that oriented even dimensional positively curved manifolds are simply connected. Your example has nonnegative curvature only. $\endgroup$ Commented Dec 25, 2016 at 19:25

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