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Improvement after J-M Schlenker's comment below :
This post has been divided into two parts, the second part is here.

Question : Is a finite dimensional metric space, uniquely geodesic if and only if it is CAT(0) ?

In the case of a negative answer :
- Is CAT(0) assumption necessary ? Is it sufficient ?
- What are the classical counter-examples ?
- Is there a slight additive assumption for having a positive answer ?

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    $\begingroup$ The answer to the main question is no. It is easy to produce a counter-example taking for instance the graph of a rotationally invariant function $u$ in the plane, with a "bump" at the origin, with the induced metric. With a proper choice of $u$ there are two minimizing geodesics between say $(1,0)$ and $(-1,0)$. However I think you can get a positive answer if you add an hypothesis of non-positive curvature. $\endgroup$ – Jean-Marc Schlenker Aug 21 '13 at 13:43
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    $\begingroup$ A simple example in the same spirit is the sphere minus some pole caps. $\endgroup$ – HenrikRüping Aug 21 '13 at 13:52
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    $\begingroup$ And here comes the negative answer: If you just take a small pole cap it is still uniquely geodesic, but not CAT(0). Only antipodal points have nonunique geodesics. So if the cap does not contain a pair of antipodal points, its uniquely geodesic. $\endgroup$ – HenrikRüping Aug 21 '13 at 21:09
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    $\begingroup$ Please try to avoid reformulating a question as dramatically as you did: one cannot understand the comments after such an edit. Ask another question instead. $\endgroup$ – Benoît Kloeckner Aug 22 '13 at 12:02
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    $\begingroup$ @Sébastien Palcoux: I know, all I say is that your edit make the first comments here irrelevant. This kind of edit is really a pain for people who try to get interested in your questions. $\endgroup$ – Benoît Kloeckner Aug 22 '13 at 12:12
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The CAT(0) asumption is sufficient: it implies that any two points are connected by a unique geodesic segment. This is well-known and follows from the definitions.

However, as pointed out by HenrikRüping, the CAT(0) asumption is not necessary, you can for instance perturb the hyperbolic plane by putting a small lump of positive curvature near a point and the resulting metric will not be CAT(0) but will still be uniquely geodesic.

You might however get a positive result in this direction if you add some smoothness and a (fairly strong) topological asumption, namely, that your space is the universal cover of a torus. Then you can use the result by Burago and Ivanov here that any Riemannian metric without conjugate point on a torus is flat. If a metric is uniquely geodesic then it is very close to having no conjugate points, and by their result it has to be flat, hence CAT(0). So as far as I can see it's conceivable that a uniquely geodesic distance on the universal cover of an $n$-dimensional torus is flat, hence CAT(0) (?)

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  • $\begingroup$ Perhpas the well-known result you say is the Cartan-Hadamard theorem. $\endgroup$ – Sebastien Palcoux Aug 22 '13 at 11:22
  • $\begingroup$ Is CAT(0) assumption sufficient alone (it follows from the definition, as you say) ? Or do you need to add (without "finite dimensional" assumption) "complete", "simply-connected" assumptions ? $\endgroup$ – Sebastien Palcoux Aug 22 '13 at 11:29
  • $\begingroup$ In fact, I'm looking for a (as general as possible) characterization of "uniquely geodesic" spaces, that's why I have improve the post like this, and I have opened this new post. $\endgroup$ – Sebastien Palcoux Aug 22 '13 at 11:34
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    $\begingroup$ I think that the definition of CAT(0) here clearly implies that two points are connected by at most one geodesic -- otherwise just consider the triangle made of those two geodesics (with one degenerate edge), then apply the CAT(0) condition, and conclude that they are at distance $0$. Actually no topological assumption is necessary, so to have meaningful statement you should say that locally CAT(0) and simply connected imply globally CAT(0) and thus uniqueness of geodesics, or something like that. $\endgroup$ – Jean-Marc Schlenker Aug 22 '13 at 11:48
  • $\begingroup$ I'll try to look at your new question as soon as I have some time (this evening or tomorrow). $\endgroup$ – Jean-Marc Schlenker Aug 22 '13 at 11:51
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A good counterexample is the Teichmuller space of a closed oriented surface $S$. It is uniquely geodesic by Teichmuller's theorem, but it is not $CAT(0)$.

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Here is another counterexample. Let us endow the Euclidean space $V=\mathbb{R}^n$ with the distance induced by the usual $\ell^p$-norm. Then, if $1<p< \infty$, then $V$ is uniquely geodesic. However, $V$ is CAT(0) if and only if $p=2$. (The proofs of these statements are very easy; as far as I remember, they may be found in the book by Bridson and Haefliger on metric spaces of non-positive curvature).

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