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

$\begingroup$ Expand for $\vec{q}=\vec{p}+\vec{\epsilon}$. $\endgroup$ – AHusain Oct 19 '18 at 9:37

$\begingroup$ A related question: mathoverflow.net/questions/37651/… $\endgroup$ – Ben McKay Oct 19 '18 at 9:44

3$\begingroup$ 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. $\endgroup$ – Ben McKay Oct 19 '18 at 9:45

3$\begingroup$ There's an answer at math.SE that answers this question: math.stackexchange.com/a/198721/18934 $\endgroup$ – Suvrit Oct 19 '18 at 11:52

$\begingroup$ 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. $\endgroup$ – Matt F. Nov 14 '18 at 9:19