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A geodesic metric space can locally be approximated by a vector space. This approximation provides it with a natural manifold structure. It means that geodesic metric space is more fundamental concept and to be a manifold is just a quality of it.
Talking about a Finsler space for example, which is a metric space by construction, we would not better call it a manifold upfront, but instead check the conditions for it to be a manifold.
Can we say what those conditions are for a Finsler space to be a manifold? Or is it so that basically any Finsler space is always a manifold and that is the reason why they are so often called simply Finsler manifolds?

MirOdin
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    A geodesic metric space can locally be approximated by a vector space. This approximation provides it with a natural manifold structure.

    That is a rather bold statement, which I am inclined to view as false. The Sierpinski carpet, when endowed with path-length metric, is a geodesic space -- what vector space approximates it? What kind of a manifold is it? If your concept of manifold is somehow more general than a topological manifold, it would help to know what it is.

     –  Dec 18 '14 at 18:37
  • The context of my initial statement is Finsler spaces, not fractals at all. Lets call it smooth geodesic metric space or tell me please how else to specify the expectation of smoothness of geodesics to be able to reasonably formulate my main question: Is basically any Finsler space always a manifold or what are the additional requirements, say over $\mathbb R^n$? – MirOdin Dec 18 '14 at 19:56
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    Step 1: Define "Finsler space". Step 2: Define "Manifold". Step 3: compare the definitions. –  Dec 18 '14 at 19:57
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    A manifold is a topological space that is locally Euclidean (i.e., around every point, there is a neighborhood that is topologically the same as the open unit ball in $\mathbb R^n$).


    > A Finsler space is a smooth manifold possessing a Finsler metric.


    Looks like a vicious circle to me.

     – MirOdin Dec 18 '14 at 20:21
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    So, you answered your own question: every Finsler space is a manifold, by definition. –  Dec 18 '14 at 20:23
  • I am not satisfied with this definition of Finsler space which from the very beginning starts with a manifold. Can we define it constructively in more general terms starting with some underlying topological space than adding some measure which would provide it with smooth geodesic structure. I expect that there is a way to do it in such a way that without directly referencing any manifold attributes from the beginning, as a result we would be able to construct a topological space with manifold properties. – MirOdin Dec 18 '14 at 22:51
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    @Vladimir: You are asking "Alexandrov's question" in the context of Finsler's metrics. In the Riemannian case, work of various people culminated in Nikolaev's solution of Alexandrov's problem on characterizing Riemannian manifolds as locally compact metric spaces satisfying some curvature restrictions (roughly, upper and lower curvature bounds). As far as I know, the problem for Finsler manifolds is wide-open. Check references here http://www.math.illinois.edu/~inik/papers.htm – Moishe Kohan Dec 18 '14 at 23:56
  • See also here: http://mathoverflow.net/questions/8513/characterization-of-riemannian-metrics – Moishe Kohan Dec 18 '14 at 23:59
  • @studiosus - Thank you, this gives an answer to my question and the solution of Nikolaev you referenced gives a positive answer: In pure geometric terms (synthetic coordinateless description of geometry) it is shown that (Alexandov's) metric spaces with intrinsic metric having bounded curvature $\leqslant K$ and $\geqslant K'$, though strictly saying are not Riemannian spaces and in general case are not manifolds, nevertheless they resemble Riemannian manifolds and may be regarded as in some sense generalized Riemannian manifolds (non-negativity of curvature is not assumed there). – MirOdin Dec 19 '14 at 21:04
  • and I tend to see it as a confirmation that a Riemannian manifold can be constructed as a type of a metric space! – MirOdin Dec 19 '14 at 21:12
  • @peth: In order to make them $C^1$-smooth manifolds with $C^1$-smooth metric you also need to add local compactness. To get $C^\infty$ you need a much more technical condition that Nikolaev also worked out. – Moishe Kohan Dec 19 '14 at 21:20
  • You never answered the question of user147623 -- where did you ever read the statement that every geodesic metric space is a topological manifold? – Chill2Macht Jul 15 '17 at 10:57

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In the Riemannian setting, this question is answered in detail in my answer here. The problem for Finsler manifolds is wide-open (to the best of my knowledge).

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