# Homotopy between a closed geodesic and a closed curve.

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.

• 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")
– Neal
Commented Dec 8, 2016 at 1:27
• @Neal i will think according to your tips. Thanks. Commented Dec 8, 2016 at 1:45
• @Neal: No need for compactness/completeness. This is just a local issue in a small neighborhood of $\gamma$. Commented Dec 8, 2016 at 5:55
• @MoisheCohen Good call.
– Neal
Commented Dec 8, 2016 at 18:32

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.