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Suppose that $M$ is a riemannian manifold, such that for any two distinct points there is a unique geodesic line connecting them.

Does that imply that these geodesics are length-minimizing? That is to say, is the length of a geodesic between two points equal to the distance between the points?

My effores led to being able to prove it, given the "geodesic triangle inequality" - the length of a geodesic between two points is $\leq$ the sum of distances of two geodesics connecting the two points to a third point. However I can't prove it.

35T41
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2 Answers2

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No.

Think of the sphere.

For any two distinct points on the sphere, there is a single great circle connecting them.

Assuming they are not antipodal, you can go around the "long way" or the "short way".

Fly by Night
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  • OK, thanks. What if I assume that all geodesics can be continued forever without self intersecting? – 35T41 Jan 22 '18 at 18:44
  • Wikipedia answers that: "The distance $d(p,q)$ between two points $p$ and $q$ of a Riemannian manifold $M$ is defined as the infimum of the length taken over all continuous, piecewise continuously differentiable curves $\gamma : [a,b] \to M$ such that $\gamma(a)=p$ and $\gamma(b)=q$. With this definition of distance, geodesics in a Riemannian manifold are then the locally distance-minimizing paths." – Fly by Night Jan 22 '18 at 18:49
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Part of your question is unclear -- your hypothesis said that "for any two distinct points there is a unique geodesic line connecting them," and then you asked if these geodesic lines are length-minimizing. Ordinarily, I would think a "geodesic line" would mean a geodesic defined for all time and without self-intersections; but it could also mean simply a geodesic defined for all time (such as great circles on the sphere), or even a maximal geodesic (such as the maximal geodesics in the unit disk, which are just portions of lines). On the other hand, the term "length-minimizing" usually applies to a geodesic segment, that is, a geodesic defined on a closed, bounded interval.

For definiteness, I'll interpret your question to mean the following:

Suppose $(M,g)$ is a Riemannian manifold in which

  • all geodesics can be continued for all time, and
  • every pair of points can be connected by a unique geodesic segment.

Does this imply that each geodesic segment is length-minimizing?

With this interpretation, the answer is yes. The fact that all geodesics can be continued for all time means that $M$ is geodesically complete. It's then a consequence of the Hopf-Rinow theorem that every pair of points can be connected by a length-minimizing geodesic segment. Since you are assuming that every pair of points can be connected by a unique geodesic segment, that segment must be length-minimizing.

Jack Lee
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