Condition for Riemannian distance to be equal to metric distance

If $$M$$ is a metric space than it is a topological space and if it is locally homeomorphic to to $$R^n$$ we say that it is a manifold and if we equipped this manifold with a inner product $$g_p$$ on tangent space $$T_pM$$ we can define a "Riemannian distance" $$L=\int \sqrt {g(v(t),v(t))}dt$$.

In $$R^n$$ the metric distance and this "Riemannian distance" have the same value.

Are there others examples of this ? What is the conditions for this to be true?

• The Riemannian distance does not depend on the initial metric. It only depends on topology/differential structure. So this makes this question a bit invalid. There is only one metric equal to the Riemannian distance on $M$ and it is the Riemannian distance itself. And lots of equivalent metrics are different. What kind of condition you are looking for? – freakish Apr 18 at 12:45
• So it is coincidence that in $R^n$ they are equal . Are there others coincidence ? – amilton moreira Apr 18 at 12:48