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Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?

I know it's true for $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?

Dai
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  • First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $\mathbb R^n$? – Jack Lee Jan 13 '19 at 22:38
  • Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature). – Jack Lee Jan 13 '19 at 22:39
  • In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $\mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them. – Jack Lee Jan 13 '19 at 22:41
  • Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them." – Jack Lee Jan 13 '19 at 23:18
  • I notice completeness is needed for exsistense of geodesic. I added it to assumption of question. Assume simple connectedness, then can't I apply Gauss Bonnet to the disk enclosed by two geodesics connecting A and B(assume there are two for the sake of contradiction)? I edit the question: Is it true for complete simply connected Riemannian manifolds, negative curved implies unique geodesic between two points? – Dai Jan 14 '19 at 17:08
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    I think https://math.stackexchange.com/questions/170964/the-number-of-geodesics-of-a-complete-riemann-manifold-with-non-positive-section?rq=1 answers you question affirmatively. – Jason DeVito - on hiatus Jan 14 '19 at 18:00
  • The problem with that Gauss-Bonnet argument is this: how do you show that two geodesics joining the same points must necessarily enclose a disk? – Jack Lee Jan 14 '19 at 18:04
  • In the case of orientable surfaces, these are topologically sphere or disk. So loop should enclosed a disk. In higher dimension, the extended disk from loop may not be embedding into manifold, so is problematic. – Dai Jan 14 '19 at 20:56

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This is a direct consequence of the Cartan-Hadamard theorem. Pretty much every Riemannian Geometry textbook will have it. Note that you only need the sectional curvature to be $\le 0$ for the exponential map $\exp_p$ to be a diffeomorphism (for each $p\in M$).

Moishe Kohan
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