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Let $(M,g)$ be a riemannian Manifold, we can use the metric $g$ to obtain a metric $d_g:M\times M\to \mathbb{R}$.

I ask for a kind of converse, we can start with a metric $d:M\times M\to \mathbb{R}$ and ask when we can recover a metric $g$ such that $d=d_g$. What obstructions need to be required on $d$ so the answer is positive, at least necessary?

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    You'll want $M$ to be at least path connected, otherwise $d_g$ is not well-defined. Note that $(d_g)^2$ is always smooth when defined, so if you remember the smooth structure on $M$ that's one necessary condition on $d$. Another obvious one is that $d$ must induce the right topology on $M$ – so this rules out e.g. the discrete metric.2012-09-16
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    I think for any two points $x$, $y$ there must exist a rectifiable path from $x$ to $y$ whose length (determined using $d$) is exactly $d(x,y)$. What I'm sure is that at least the infimum of all path lengths from $x$ to $y$ must be $d(x,y)$.2012-09-16
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    @celtschk The second part is correct: every Riemannian manifold is a *length space*: $d$ is the infimum of lengths of connecting curves. If $M$ is also complete, then the first property holds as well ($M$ is a *geodesic space*), by the [Hopf-Rinow theorem](http://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem). Without completeness it's not true: consider the plane minus a point.2012-09-17

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