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.
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 ${\rm length}\ \gamma(t,s)&lt;{\rm length}\
\gamma(t,0)$ for $s&gt;0$, since $M$ has a positive curvature.
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.