Definitions: A geodesic space is a metric space $(X,d)$ such that every two points $x,y \in X$ can be joined by a geodesic. (A path $[0,1] \to X$ is a rectifiable curve if and only if it has a parametrization for which it is Lipschitz continuous. A rectifiable curve is a geodesic if and only if the inequality in the definition of Lipschitz continuous is an equality, according to this document.)

Context: Consider the second part of Lemma 2.2 in the same aforementioned document:

Let $X$ be a complete metric space... $X$ is a geodesic space if and only if, for all $x, y \in X$ there is a point $z \in X$ such that $d(x,y)= d(y,z)=\frac{1}{2}d(x,y)$.

In particular, such metric spaces are specific examples of convex metric spaces.

Now compare this with the following equivalent formulation of the parallel postulate:

Let a line segment join the midpoint of two sides of a given triangle. That line segment will be half as long as the third side.

Interpretation: Triangles are defined in terms of geodesics (or at least paths), thus a triangle can be defined in any geodesic space, even if it's not a vector space. Thus we can test in any geodesic space whether or not the parallel postulate holds (it seems, since the above appears to imply that the midpoint of a line segment/path/geodesic always exists for geodesic spaces). Now, for Riemannian manifolds, I think I recall that it is a corollary of Gauss-Bonnet that the parallel postulate holds if and only if the space is locally Riemannian isometric to Euclidean space.

Question: Using the above facts, can we conclude that, for a general geodesic space, that it is locally topologically isometric to Euclidean space if and only if the parallel postulate holds?

Note: I am not quite sure what to tag this question, since it is an attempt to use metric geometry (which assumes only the existence of a topological metric, i.e. not a Riemannian one), to understand Euclidean geometry in the context of general Riemannian geometry.

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    $\begingroup$ It is not true that triangles can be defined in every geodesic space. I believe that this may be done only in those spaces where every two points are joined by a unique geodesic. I am afraid that this is a very strong condition, extremely few spaces satisfying it. $\endgroup$ – Alex M. Apr 1 '17 at 11:38
  • $\begingroup$ @AlexM. Why does one need a unique geodesic? I don't see yet how that's necessary. $\endgroup$ – Chill2Macht Apr 1 '17 at 13:18
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    $\begingroup$ If there are several geodesics between $A$ and $B$, and a single one between $B$ and $C$ and, respectively, between $A$ and $C$, which of those geodesics between $A$ and $B$ will be "the" edge of $\triangle ABC$? Now imagine that there are several geodesics between $B$ and $C$ and, respectively, between $A$ and $C$ - this becomes a nightmare, doesn't it? In particular, let $A$ and $B$ be opposed points on the sphere (say, North and South poles), and let $C$ be some arbitrary other point (say, on the Equator). Try to imagine what $\triangle ABC$ will be... $\endgroup$ – Alex M. Apr 1 '17 at 14:20
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    $\begingroup$ Could you expand on your question? First of all, what does it mean to be "topologically isometric"? Do you simply mean "isometric"? Secondly, in your question what exactly do you mean by the parallel postulate? Is this the property of triangles that you stated? Thirdly, what do you mean by "Euclidean space"? For instance, does a (possibly incomplete and nonseparable) inner product vector space qualify? $\endgroup$ – Moishe Kohan Jan 29 '18 at 18:13
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    $\begingroup$ Then $\ell_2$ would be a counter example. However, if you also require connected, complete, local compact and finite dimensional (maybe locally compact can be dropped) then indeed your space is locally isometric to $E^n$ for some $n$. This is a nontrivial theorem. $\endgroup$ – Moishe Kohan Jan 31 '18 at 14:40

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