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Looking at the qeustion(Existence of a Riemannian metric inducing a given distance.), it answered that not all distance functions will have a corresponding Riemannian metric. My question is then, what are the conditions for some metric function to have a corresponding metric? An example in the linked question's answer is using a consequence of Gauss's lemma, but this does not seem to be sufficient in and of itself.

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    This is addressed by Moishe Kohan's answer in the linked question. – Kajelad May 13 '21 at 02:46
  • Those are sufficent but not neccessary conditions. – Jean Du Plessis May 13 '21 at 09:50
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    @JeanDuPlessis: These are both sufficient and necessary as explained in my answer. – Moishe Kohan May 13 '21 at 13:57
  • @MoisheKohan Thanks for the response, my intuition tells me that a unit disk with a Euclidean metric would have a corresponding Riemannian metric, but its geodesics aren't extendable, so where is my intuition failing me here? Also it seems like the euclidean plane with the manhattan metric would satisfy the requirements – Jean Du Plessis May 14 '21 at 20:20
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    Closed unit disk is not a Riemannian manifold; open unit disk has extendible geodesics. As for the Manhattan metric, "it seems" is utterly false. If you are interested in this subject, my suggestion is start by reading the book by Burago, Burago and Ivanov "A course in metric geometry." – Moishe Kohan May 14 '21 at 20:53
  • @MoisheKohan Thanks, I'll check out the book. For the disk example, both the open and closed disk have the same distance function essentially, so why would one have a corresponding Riemannian metric and the other not (outside just the underlying space's properties) – Jean Du Plessis May 15 '21 at 22:15
  • The closed disk does not satisfy the geodesic extension property (as any manifold with nonempty boundary) while the open disk does. (The extension property of course does not mean that geodesics extend indefinitely, only a tiny bit, depending on the geodesic.) – Moishe Kohan May 16 '21 at 03:23

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