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.