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Let $M$ be a complete connected Riemannian manifold. Fix $p \in M$.

Assume every point in $M$ has a unique minimizing geodesic connecting it to $p$. Is it true that for every point, the only geodesic connecting it to $p$ is the minimizing one?

(Does the answer change if we count periodic geodesic as one?)

Remarks:

(1) In all the examples I have checked so far this holds. Of course, if for some $q \in M$ there is a unique minimizing geodesic, it does not imply there is a unique geodesic from $p$ to $q$ (think of a circle).

For spheres,tori,cylinders - the claim vacuously holds, since there is no point satisfying the hypothesis (every point has an "antipodal" point where there is more than one minimizing geodesic).

(2) If the assertion is true, it implies something quite strong on manifolds where minimizing geodesics are always unique; By this answer such a manifold must be diffeomorphic to $\mathbb{R}^n$ (via at the exponential map).

Actually, one can adapt the argument of that answer to see that if there is one such point $p$ (that minimizing geodesics from it to all other points are unique), then the manifold will be diffeomorphic to $\mathbb{R}^n$.

Thus, to refute the conjecture, it is enough to find a manifold with one such point which is not diffeomorphic to $\mathbb{R}^n$.

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Asaf Shachar
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1 Answers1

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According to section 4 of this paper:

Stephanie B. Alexander, I. David Berg, and Richard L. Bishop, Cut loci, minimizers, and wavefronts in Riemannian manifolds with boundary, Michigan Math Journal, Volume 40, Issue 2 (1993), 229-237

for a connected complete Riemannian manifold $(M,g)$ without boundary the cut-locus $C(p)$ of a point $p\in M$ equals the closure of the set $N(p)$ consisting of points $q\in M$ such that there is more than one minimizing geodesic between $p$ and $q$.

Now, your assumption is that $N(p)=\emptyset$ for some $p\in M$. Hence, $C(p)=\emptyset$ for this $p\in M$. Thus, according to the answer to your earlier question (I think you asked this question several times, I remember writing an answer), it follows that any point in $M$ is connected to $p$ by a unique geodesic as $\exp_p$ is a diffeomorphism.

Moishe Kohan
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  • Thanks. I am looking into the paper now. By the way, this is the question you have answered before: http://math.stackexchange.com/questions/1612453/geodesics-are-minimizing-in-a-simply-connected-manifold-without-conjugate-points (It is different from the current question and from the one on manifolds with globally minimizing geodesics). Also, note that here I only assumed $N(p)=\emptyset$ for one fixed point $p$, not for every $p \in M$. However, it does not change the conclusion, since in this case, for this specific $p$, $\text{exp}_p$ will be a diffeomorphism... – Asaf Shachar Jul 25 '16 at 06:33
  • (This essentially follows from the answer to the global minimizers question). – Asaf Shachar Jul 25 '16 at 06:33
  • No, the OP's assumption is not that $N(p) = \emptyset$ for every $p$. The OP's assumption is that there exist $p \in M$ such that $N(p) = \emptyset$. From this, indeed, one may use the result that you cite to conclude that $C(p) = emptyset$. Since, in general, $\exp_p$ gives a diffeomorphism between $M \setminus C(p)$ and some open subset of $T_p M$, it follows here that $M$ is diffeomorphic to a star-shaped open subset of $\Bbb R^n$, hence to $\Bbb R^n$. – Alex M. Jul 25 '16 at 10:18