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Let $M$ be a manifold (finite dimension) and a $d:M\times M\mapsto \mathbb{R}^+$ metric on $M$.

Does this define a Riemannian manifold $(M, g)$? Is it possible to derive a metric tensor $g$ from the metric $d$?

I don't care about the pathologies, let's say that you have geodesic completeness or any kind of regularities you want on $M$ and $d$.

Just for motivational background: it is well known how to get a metric on a Riemannian manifold. My question is the other way around.

BGabor
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  • I feel there are some metric $d$ can't induce a metric tensor . For example, $d(x,x)=0, d(x,y)=1 ,\forall x\in M, \forall y\ne x$. – Enhao Lan May 31 '17 at 12:58
  • You might look into metric derivatives https://en.wikipedia.org/wiki/Metric_derivative. This does not give you directly the metric tensor, but gives you a derivative of functions and in particular lets you define tangent spaces – wonko May 31 '17 at 13:00

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