I'm reading the proof of the Preissman Theorem, in Do Carmo's book of Riemannian Geometry. A crucial step in this demonstration is the following lema,
Lema: Let $M$ be a compact riemannian manifold, and $\alpha$ a non trivial deck transformation of the universal covering $\widetilde{M}$, where we are considering $\widetilde{M}$ with a covering metric. So the statement is that $\alpha$ leaves invariant a geodesic $\widetilde{\gamma}$ of $\widetilde{M}$, in this sense
$$\alpha(\widetilde{\gamma}(-\infty,\infty))=\widetilde{\gamma}(-\infty,\infty).$$
Sketch of proof: Let $\pi:\widetilde{M}\to M$ be the covering transformation. Let $\widetilde{p}\in \widetilde{M}$ and $p=\pi(\widetilde{p}).$ Let $g\in \pi_1(M,p)$ be the element corresponding to $\alpha$ by the known isomorphism $\pi_1(M,p)\simeq Aut(\widetilde{M}).$ By the Cartan Theorem, there is a closed geodesic $\gamma$ in the class of free homotopy $M$ given by $g.$
The main idea now is to show that, $\alpha$ fixes the extension of a lifting of $\gamma.$ For this, we obtain a deck tranformation that clearly fix the lifting of $\gamma$ (just take a deck transformation $\beta$ associated to the class of homotopy of $\gamma$ with a base point $q\in \gamma$). And then show that they coincide in one point and therefore must be the same, $\alpha=\beta$.
My Question: Is there any reason to believe that the geodesic wich will be fixed by $\alpha$ is precisely the lifting of a geodesic given by the Cartan Theorem? Or was that just an insight wich the person who'd demonstrated the theorem have?
For those who do not remember this is the statement of the theorem cartan
Cartan Theorem: Let $M$ be a compact riemannian manifold. Let $\pi_1(M)$ be the set of all the classes of free homotopy of $M.$ Then in each non trival class there is a closed geodesic. (i.e a closed curve which is geodesic in all of its points.)
