Does there exist a complete, non-compact Riemannian manifold $(M^n, g)$, which has only one ray starting from one fixed point $p\in M^n$?
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2$\begingroup$ Yes, a capped round cylinder. From any point away from the cap there is only one ray going along the axis of the cylinder. $\endgroup$– Igor BelegradekCommented Jul 28, 2017 at 1:46
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$\begingroup$ @ Igor Belegradek, From any point away from the cap, except one direction(which tangent to the circle), there is a ray starting from any other directions, so your example seems to be not right. $\endgroup$– mmaatthhCommented Jul 28, 2017 at 2:04
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1$\begingroup$ The problem could be stated in a clearer way. $\endgroup$– Wlod AACommented Jul 28, 2017 at 6:48
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3$\begingroup$ Igor's example is correct (with the standard definition of a ray). What definition are you using? $\endgroup$– MishaCommented Jul 28, 2017 at 8:02
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2$\begingroup$ @mmaatthh: you seem to be confused about the definition of a ray. In general, deciding when a geodesic is a ray is nontrivial. A study of rays on certain surfaces of revolution can be found in arxiv.org/abs/1108.1515. $\endgroup$– Igor BelegradekCommented Jul 28, 2017 at 11:55
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1 Answer
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Recall that a ray is a geodesic defined on $[0,\infty)$ that minimizes the distance between any of its points.
A capped round cylinder has required properties: From any point sufficiently far away from the cap there is only one ray going along the axis of the cylinder. (This is because no ray passed through the cap, and rays on a cylinder are easy to analyse). It is also easy to get the same conclusion at each point not on the cap.
In general, deciding when a geodesic is a ray is nontrivial even for surfaces of revolution. These matters are discussed at length in http://arxiv.org/abs/1108.1515.
answered Aug 6, 2017 at 20:36