# On a Riemannian manifold, calculate the metric from the distance [closed]

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

• Expand for $\vec{q}=\vec{p}+\vec{\epsilon}$. – AHusain Oct 19 '18 at 9:37
• A related question: mathoverflow.net/questions/37651/… – Ben McKay Oct 19 '18 at 9:44
• 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 9:45
• There's an answer at math.SE that answers this question: 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. – Matt F. Nov 14 '18 at 9:19