Riemannian metric induced by a metric - MathOverflow

Riemannian metric induced by a metric

2010-11-07T10:07:13Z

Let $M$ be a smooth manifold, $\rho(p, q)$ — a differentiable metric on $M$. Can we construct Riemannian metric $g(X,Y)$ on $TM$ that induces $\rho(p, q)$? Under what conditions?

I'm sure this question has been dealt with, I just didn't find it in the quick survey of literature :)

Answer by Igor Belegradek for Riemannian metric induced by a metric

2010-11-07T12:01:15Z

There are no Riemannian manifolds with differentiable metric $\rho$. Indeed, any Riemannian metric on the real line is locally isometric to the standard Euclidean metric with $\rho(x,y)=|x-y|$, which is not differentiable. On an arbitrary Riemannian manifold $M$ if $\gamma$ is a geodesic through $p$, then the restriction of $\exp_p$ to the tangent line to $\gamma$ is a a local isometry from $\mathbb R$ to the image of $\gamma$ (by Gauss lemma). So if $\rho$ were differentiable on $M$, the metric $|x-y|$ on the tangent line would be differentiable.

The best you can hope for is that the function $x\to\rho(x,p)$ is differentiable away from $p$. This happens exactly when $\exp_p$ is a diffeomorphism, e.g. it never happens when $M$ is compact.

Answer by Dick Palais for Riemannian metric induced by a metric

2010-11-07T14:39:31Z

Here is a closely related question that may have been what the OP was driving at. Suppose that you ONLY know a Riemannian manifold as a metric space---that is you know the point set and the distance between any two points, but you do not know the metric tensor or even the differentiable structure. Can you nevertheless reconstruct these from the distance function. The answer is that you can. See:

http://www.ams.org/proc/1957-008-04/S0002-9939-1957-0088000-X/S0002-9939-1957-0088000-X.pdf