# Minimal length in the free homotopy class as translation length

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.

• Why do you think your reasoning is wrong? Commented Oct 19, 2023 at 16:44
• I think that the curve could join to points that are not related by the action of $g$. In fact now that you make me think more thoroughly about it i think that thant might not be the case, but the two points are surely related by the translation by a conjugate of $g$. This is because the $\pi_1(M,p)$ and $\pi_1(M,p')$ are related by an inner automorphism, so i can change the base and lift the path using a point in it as base point. This is enough because the translation length is invariant by conjugation since the deck trasnformation act by isometries. Thank you :) Commented Oct 19, 2023 at 17:04

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.