This is probably an insanely hard question, but given an abstract metric space, is there some way to determine whether it's a manifold with a Riemannian, or more generally a Finslerian, metric? If that's too hard, one could start off by assuming that the underlying space is a manifold. The example that got me thinking about this was the induced metric on the 2sphere embedded in $R^3$...the underlying space is obviously a smooth manifold, and the metric should be smooth(the geodesics would even be great circles, as they are for the standard metric on $S^2$,) but I don't see how you could prove the triangle inequality pointwise, as you'd have to to show that it's Finslerian, let alone Riemannian. This example is already incredibly simple, since it's a homogeneous space with a the metric induced by being a subspace of another homogeneous space.

$\begingroup$ Every smooth manifold has lots of Riemannian metrics. Locally it's clear, and then you patch the local metrics together using a partition of unity. I don't know anything about Finsler manifolds, but wikipedia seems to suggest that every Riemannian manifold is Finsler. $\endgroup$ – Kevin H. Lin Dec 11 '09 at 2:05

6$\begingroup$ Kevin Lin, I think you misread the question. $\endgroup$ – Harry Gindi Dec 11 '09 at 2:20
If $X$ is a metric space and $x$, $y\in X$, a segment from $x$ to $y$ is a subset $S\subseteq X$ such that $x$, $y\in S$ and $S$ is isometric to $[0,d(x,y)]$.
Let now $n\in\{1,2,3\}$ and let $X$ be a metric space which is locally compact, $n$dimensional and such that (i) every two points are the endpoints of a unique segment, (ii) if two segments have an endpoint and one other point in common, then one is contained in the other, and (iii) every segment can be extended, at either end, to a larger segment. Then $X$ is homeomorphic to $\mathbb R^n$.
This is a metric characterization of $\mathbb R^n$ for small $n$. Using it locally, you get a characterization of topological manifolds of small dimension. I imagine higher dimensions or smoothness are harder to come by...
The theorem above, along with quite a few other nice results, is proved in [Berg, Gordon O. Metric characterizations of Euclidean spaces. Pacific J. Math. 48 (1973), 1128.]
LATER...
Of course, the paper [Nikolaev, I. G. A metric characterization of Riemannian spaces. Siberian Adv. Math. 9 (1999), no. 4, 158.] is relevant here! The paper gives purely metric characterizations of Riemannian manifolds of all smoothnesses up to $C^\infty$. The MR review gives a small history of the problem and references to earlier work.

5$\begingroup$ By the way... what an amazing feat to be able to get such a characterization, isn't it? $\endgroup$ – Mariano SuárezÁlvarez Dec 11 '09 at 6:28

$\begingroup$ It is! I'll look up the paper from work on Monday, but I wasn't really expecting that there would be an answer to my question. Thanks! $\endgroup$ – Gordon Craig Dec 13 '09 at 2:18

1$\begingroup$ Mariano, what fails in dimension n > 3? Is it just that there are other spaces which satisfy these characteristics? $\endgroup$ – Tom LaGatta Dec 29 '09 at 8:17
I think that if the underlying space is a smooth manifold $M$, a metric $d$ (in the "metric space" sense) on it will give rise to a Riemannian metric if and only if it satisfies a smoothness condition (I assume we want the Riemannian metric to be smooth). Fix a point $p \in M$. Since our manifold is smooth, there is a natural identification between a neighborhood of the origin of the vector space $TM_p$ and a neighborhood of $p$ in $M$. So, we can assume that our metric $d$ is defined on this neghborhood of origin in $TM_p$. For a vector $v \in TM_p$, we define its length to be $\vec v=\lim_{t\to 0} (d(t\vec v,0)/t)$. If this limit converges, then $\vec v \mapsto \vec v^2$ should be a quadratic form on $TM_p$ and the polarization identity should give us a positive definite bilinear form $g$ on $TM_p$ such that $g(\vec v,\vec v)=\vec v^2$ for all $\vec v$. The smoothness condition on $d$ should, I guess, be that the function $d(t\vec v,0)$ is always smooth (so the limit always exists), and that the resulting $g$ should be smooth as a function of $p$. In this case, we have our Riemannian metric.
So, the problem seems to reduce to the question of when does the topological manifold $M$ have a smooth structure such that the metric $d$ satisfies the above smoothness condition. It would also be good to find a simpler way to state the smoothness condition.
I'm also thinking that people might know a way to define Riemannian metrics in categories other than that of smooth manifolds. This might be more appropriate for this question, but I have no idea how to deal with that.
(I would've put this in a comment, but I don't have enough karma for that. I'm a bit intimidated that nobody has posted something like this yet  I hope I haven't missed something stupid.)

3$\begingroup$ I do not see how you get a a natural identification between a neighborhood of the origin in the tangent space and a neighborhood of the point. An exponential map would do it, but you need a riemannian metric to get one. Moreover, it seems to me that the chordal distance on the unit sphere of Euclidean space satisfies your smoothness assumption, however it does not derive from a riemannian metric since the distance between two points is strictly smaller than the length of the shortest curve between them (in other words, it is not even a length space). $\endgroup$ – Benoît Kloeckner Apr 30 '10 at 16:15