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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 :)

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I have a question of my own: does the Nash embedding theorem help here? – Suvrit Nov 7 '10 at 10:14
Let $M:=\mathbb R^d$. Then a metric $\rho$ on $M$ amounts to a function of $2d$ variables that satifies certain inequalities whereas a Riemannian metric $g$ on $M$ amounts to a collection of $d(d+1)/2$ functions of $d$ variables satisfying certain inequalities. This simple count of degrees of freedom shows that there must be strong conditions on $\rho$ to admit a corresponding $g$. – Christian Blatter Nov 7 '10 at 10:30
@Suvrit I doubt it. @Christian Good point. I feel a bit fishy about it, though, I'll dig into it deeper :) @Paolo Yeah, I was thinking in this line, too. – Alexei Averchenko Nov 7 '10 at 10:51
up vote 13 down vote accepted

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:

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And there is a more recent work on the subject by Lytchak-Yaman, see – Igor Belegradek Nov 7 '10 at 14:47
I was interested in finding the conditions, under which for $\rho(p,q)$ there is a $g(X,Y)$ such that $\rho_g(p,q) \equiv \rho(p,q)$, where $\rho_g(p,q)$ is constructed from $g$ as usual, and how can such $g$ be constructed from $\rho$. Thanks for the links :) – Alexei Averchenko Nov 7 '10 at 15:26
Upd: I think the problem in Igor's comment and the one in my previous comment are equivalent, because $g \mapsto \rho_g$ is injective, and so it is bijective to it's total image. – Alexei Averchenko Nov 7 '10 at 15:30

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.

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And an interesting question for the OP to think about is, what if it is assumed that $\rho^k$ is differentiable for some integer power $k$ of the distance function. (For $k = 2$, the Hessian of the squared distance function gives you a positive definite symmetric bilinear form on the tangent space, i.e. a metric tensor.) – Willie Wong Nov 7 '10 at 14:18
I think, the whole guestion is misguided, and a better question is under what conditions $\rho$ comes from a Riemannian metric, and what is the regularity of the metric. This has been studied by various authors, and e.g. Nikolaev proved that if $\rho$ is a length metric on a closed topological manifold $X$ of curvature bounded above and below in comparison sense, then $X$ has a $C^{1,\alpha}$ Riemannian metric that induces $\rho$. – Igor Belegradek Nov 7 '10 at 14:38
This is the first time that passwords are needed to open academic pdf-files that I downloaded from <br><br>;<br>; Did I miss something there? – Pengfei Nov 29 '10 at 11:48

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