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Let $(M,g)$ be a $2$-dimensional oriented complete Riemannian manifold without boundary. Let $f:\big(\Bbb S^1,1\big)\to (M,x_0)$ be a smooth loop such that $f$ has minimum length among all admissible loops in the given free homotopy class of loops. That is, if $f':\big(\Bbb S^1,1\big)\to (M,x_0')$ is another admissible loop such that there is a continuous map $H:\Bbb S^1\times [0,1]\to M$ with $H(-,0)=f, H(-,1)=f'$, then length of $f$ is no more than length of $f'$.

Consider the following commutative diagram obtained from a smooth lifting. $\require{AMScd}$ \begin{CD} (\Bbb R,0) @>\displaystyle\ell>> (\widetilde M,\widetilde {x_0})\\ @V\text{Universal Cover} V V @VV \text{Universal Cover}V\\ (\Bbb S^1,1) @>>\displaystyle f> (M,x_0)\\ \end{CD}

Let $p:=\ell(t_0)$ and $q:=\ell(t_1)$ be two distinct points, and $\lambda:=\ell\big|[t_0,t_1]$. Is it true that $\text{Length}_{\widetilde g}(\lambda)\leq \text{Length}_{\widetilde g}(\lambda')$ for all admissible curves $\lambda'$ with the same endpoints $p$ and $q$?

$\bullet$ Always consider the pull-back metric on every covering.

$\bullet$ In a Riemannian manifold, every minimizing curve is a geodesic when it is given a unit-speed parameterization, i.e. $\text{im}(\lambda)$ is the image of some geodesic.

$\bullet$ If needed, one may assume $f$ is not null-homotopic.

$\textbf{Definition:}$ On a smooth manifold $M$ a curve $\gamma:[a,b]\to M$ is said to be admissible if there is a partition $a=a_0<a_1<...<a_{n-1}<a_n=b$ of $[a,b]$ such that $\gamma\big|[a_{k-1},a_k]$ is smooth with non-vanishing velocity for each $k=1,...,n$.

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I just saw the "orientability" hypothesis on your question, which kills this answer. But I'm going to leave the answer anyway, because there are orientable generalizations in higher dimensions.


For a (nonorientable) counterexample, take $\widetilde M \to M$ to be the universal covering of the projective plane $$f : S^2 \mapsto \mathbb R P^2 $$ where $S^2$ has its usual metric and the (unique) nontrivial deck transformation is the antipodal map of $S^2$. The group $\mathbb R P^2$ is cyclic of order $2$. There is a constant speed, smooth map $f : S^1 \to \mathbb R P^2$ which represents that element and has minimal length, that minimum being equal to $\pi$ (so $f$ has speed $1/2$). The corresponding lift $\ell : \mathbb R \to S^2$ is a constant speed universal covering map of a great circle in $S^2$; in fact, the covering map restricted to that great circle gives a 2-to-1 covering of the circle $\text{image}(f)$. But $p = \ell(0)=\ell(2\pi) = q$ and so the length of the restriction $\lambda = \ell \mid [0,2\pi]$ is strictly greater than the constant path $\lambda'$ with the same endpoints as $\ell$, namely the point $p = q$.


But there is a restricted situation in which the statement you want is true, namely that the Gaussian curvature of $M$ is non-positive at each point. There is a theorem in differential geometry which says that on any nonpositively curved complete simply connected Riemannian manifold, every locally geodesic path is globally geodesic (this is not the same as, but is closely connected to, the Cartan-Hadamard Theorem). Since each of your maps $f$ is locally geodesic, it follows that each of their lifts $\lambda$ is locally geodesic, and by applying that theorem it follows that $\lambda$ is globally geodesic.

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  • $\begingroup$ Nice, I want a reference for this: any nonpositively curved complete simply connected Riemannian manifold, every locally geodesic path is globally geodesic. So, the last paragraph does hold if $M$ covers a hyperbolic closed surface and $g$ is just the pull-back metric on $M$ as $\text{Sectional curvature=Gaussian curvature=}-1$ is preserved under local isometry. Am I right? Thank you for your help. $\endgroup$
    – Someone
    Commented Feb 27, 2021 at 16:32
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    $\begingroup$ Yes, that's correct. It holds on flat surfaces as well e.g. tori and Klein bottles with locally Euclidean metrics. $\endgroup$
    – Lee Mosher
    Commented Feb 27, 2021 at 16:46
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    $\begingroup$ As for references, I don't know the differential geometry textbooks well enough, but as said I suspect this fact can be found in conjunction with the Cartan-Hadamard theorem. Another very closely related fact is that in non-positive curvature, geodesics have no conjugate points. $\endgroup$
    – Lee Mosher
    Commented Feb 27, 2021 at 16:49
  • $\begingroup$ Thank you very much, I will try. $\endgroup$
    – Someone
    Commented Feb 27, 2021 at 16:50

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