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A while ago I asked this question in Math Stackexchange. Since I didn't receive an answer so far, I thought I'd ask it here.

Suppose $Y$ is a proper length space, where every pair of points $x,y\in Y$ can be joined by a unique length minimizing geodesic (i.e. global geodesic). Can there still be more than one local geodesic joining some two points, or is it necessarily so that the space is also uniquely locally geodesic?

If a uniquely geodesic space which is not uniquely locally geodesic exists, can one even cook up Riemannian manifold as an example?

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up vote 5 down vote accepted

Without further hypotheses it is easy to cook up a counterexample. Start with a standard torus of revolution, choose a point $p$ on the "outer" circle $\alpha$ and follow this circle until you reach the first conjugate point $q$ from $p$. It is well known that for points $x$ past $q$, the geodesic is no longer minimizing, and in fact there is a pair of geodesics $\beta$ and $\gamma$ joining $p$ to $x$. Now simply "trim" your manifold by removing everything outside the digonal region bounded by $\alpha$ and $\beta$. Here every point is joined to $p$ by a unique geodesic but for points $x$ on $\alpha$ beyond $q$, the geodesic $\alpha$ is no longer minimizing.

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Aren't $p$ and $q$ joined by two minimizing geodesics? – Pablo Lessa Aug 12 '14 at 13:15
No, $q$ is the "last" point with a unique minimizer. – Mikhail Katz Aug 12 '14 at 13:20
I don't believe you :). If $\beta_x$ is the minimizing geodesic joining each $x$ on $\alpha$ to $p$. What's the limit of $\beta_x$ when $x$ converges to $q$? Are you saying this goes to $\alpha$? Proof or reference? – Pablo Lessa Aug 12 '14 at 13:42… – Mikhail Katz Aug 12 '14 at 14:04
Thanks! After looking at chapter 8 I believe you now :). – Pablo Lessa Aug 12 '14 at 14:56

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