Given a manifold and its metric tensor, how can I compute the distance between two points on the manifold?

What are the high level steps?

Edit: In particular, suppose the manifold is an open unit ball in $R^d$ $$B = \{ x \in \mathbb R ^d: |x| < 1 \},$$ and the metric tensor is $$\frac{2}{(1-\|x\|^2)^2} g_E$$ where $x \in B$ and $g_E$ is the Euclidean metric tensor. How should one compute the distance between two points on the the manifold?

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    $\begingroup$ The metric $d(p, q)$ is just the infimum of the set of lengths of paths connecting $p, q$---it's an instructive exercise to verify that this really defines a metric. Of course, it's not practical to consider all possible paths, so to compute a distance one needs to use additional observations, but what facts are available depends very much on the setting. For that reason, I suggest including a concrete example. $\endgroup$ – Travis Willse Mar 21 at 21:34
  • $\begingroup$ @Travis Thanks! I edited the question with a concrete example. $\endgroup$ – user25004 Mar 21 at 23:51
  • $\begingroup$ Possible duplicate of Distance in the Poincare Disk model of hyperbolic geometry $\endgroup$ – Arctic Char Mar 22 at 5:21

If you have a curve on the manifold (say two-dimensional) $u_i=u_i(t)$, between $t=t_1$ and $t=t_2$, and your metric tensor is $g_{ij}$ (covariant components), then the length of the curve is given by $$ \int\limits_{t_1}^{t_2}\sqrt{g_{ij}(u(t))\partial_tu^i\partial_tu^j}\,dt. $$ If you choose your curve to be a geodesic, then you get the distance.

  • $\begingroup$ So the geodesics should be derived first? $\endgroup$ – user25004 Mar 22 at 13:17
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    $\begingroup$ @user25004 : as Travis pointed out, a metric can be induced by considering the infimum of the length of the paths connectig A and B, that is, the length of the curve that minimize lenght. This curve is a geodesic. So yes, you have to derive the geodesic first $\endgroup$ – Gabriele Cassese Mar 22 at 14:22
  • $\begingroup$ Yes you have to solve the second order ODEs for them. Given your metric you find the Christoffel symbols etc. Your concrete metric differs by a factor from the Euclidean one so many computations will be simple. $\endgroup$ – GReyes Mar 22 at 19:06

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