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Given a Riemannian manifold $(M,g)$ is it possible to calculate the distance between two points on this manifold. Is it possible the inverse? That means: given a formula of the distance, for example: $$d(p,q)=\dfrac{1}{\sqrt2}\left(\sqrt{p}-\sqrt{q}\right)^2$$ is it possible to calculate the metric tensor and so define a manifold in which this formula is valid? Thanks

YCor
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Riccardo.Alestra
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  • Expand for $\vec{q}=\vec{p}+\vec{\epsilon}$. – AHusain Oct 19 '18 at 09:37
  • A related question: https://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function – Ben McKay Oct 19 '18 at 09:44
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    It is not difficult to prove that the distance function determines the Riemannian metric, should a Riemannian metric exist with the given distance function. Determining when such a Riemannian metric exists has been considered in the literature, but I can't remember a reference at the moment. – Ben McKay Oct 19 '18 at 09:45
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    There's an answer at math.SE that answers this question: https://math.stackexchange.com/a/198721/18934 – Suvrit Oct 19 '18 at 11:52
  • It's actually hard to determine if a metric is Riemannian. But this $d$ violates the triangle inequality with $d(1,4)=d(4,9)=1/\sqrt{2}$ and $d(1,9)=4/\sqrt{2}$, so it's not a metric at all. –  Nov 14 '18 at 09:19
  • A more general version: https://mathoverflow.net/questions/45154/riemannian-metric-induced-by-a-metric – Paul Reynolds Feb 24 '20 at 19:24

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