Minimal length in the free homotopy class as translation length

Question

Let $M$ be a compact Riemannian manifold. The fundamental group $\pi_1(M,p_0)$ is isomorphic to the group of deck transformations of the universal cover $\pi:\widetilde{M}\to M$. The translation length of an element $g\in\pi_1(M)$ is defined as $\min_{p\in\widetilde{M}}d(p,g.p)$ . Is it true that the translation length of $g$ equals the shortest length of a loop $\gamma$ in the free homotopy class of loops associated to $g$?

Here's my proof: the translation length of $g$ is realised by a curve $\gamma$ joining points $p$ and $g.p$ by compactness. Projecting $\gamma$ to $M$ we get a closed loop, and there is a path $\alpha$ joining $p_0$ and $\pi(p)$ such that $\alpha*\gamma*\alpha^{-1}$ represent $g$ in $\pi_1(M,p_0)$. So the minimal length in the free homotopy class is clearly less or equal than the translation length. On the other hand, if there was another closed loop $\gamma'$ freely homotopic to a representative of $g$ then it would lift to a curve $\widetilde{\gamma}'$ joining $p'$ to $g.p'$ for some $p'$ so the inequality is true also on the other direction.

Is it correct? I percieve the last passage as a bit sketchy, maybe because it is not clear to me why the curve would join a point with its $g$-traslate.

Answer 1

I post this solution originally appeared in the comments as an answer to close the question. Thanks to Moishe Kohan for their socratic help!

The reasoning in the question is also correct, but the curve appearing in the last line could join to points that are not related by the action of $g$, but they are for sure related by the action of a conjugate of $g$. Namely any closed loop $\gamma'$ freely homotopic to a representative of $g$ would lift to a curve $\widetilde{\gamma'}$ joining $p′$ to $\overline{g}.p$, where $\overline{g}=hgh^{-1}$.

This is because the fundamental groups $\pi_1(M,p_0)$ and $\pi_1(M,p'_0)$ are related by an inner automorphism, so we can change the base point to a point $p_0'\in Im(\gamma')$ such that $p'$ projects to $p_0'$, and lift the path $\gamma'$ with respect to this new base point. This is enough because the translation length is invariant by conjugation since the deck trasnformation act by isometries.