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We can join the north pole and the south pole of an sphere by an unlimited number of geodesics.

1: Is this property still valid if we take any manifold that is diffeomorphic to the sphere, i.e. are there any two points in this manifold that are connected by an infinite number of geodesic?

2: If the answer of question 1 is no, is it possible to find a manifold diffeomorphic to the sphere, such that, for all two distinct points there is only one geodesic passing through this points?

Thanks

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For (1), are you asking about any metric? For instance, if you've chosen the pullback of the spherical metric, then every two points are connected by uncountably many geodesics. –  Neal Oct 27 '12 at 13:09

1 Answer 1

up vote 3 down vote accepted

For $2$ the answer is no. In fact, more generally, on any closed manifold there are always at least two points with two geodesics between them.

The reason is that every closed Riemannian manifold has at least one closed geodesic. I thought this result was due to Birhoff, but according to http://www.encyclopediaofmath.org/index.php/Closed_geodesic it's due to Lyusternik and Fet. (In the non-simply connected case, it's not too hard to prove and is due to Cartan).

Now, let $\gamma:[0,L]$ be a unit speed closed geodesic. We may assume wlog that $\gamma$ is minimal in the sense that if $\gamma$ is restricted to any subinterval of $[0,L]$, the resulting geodesic is not a closed geodesic.

Now, consider the points $\gamma(0)$ and $\gamma(L/2)$. If these are not the same point, then the geodesic $\gamma$ and "follow $\gamma$ backwards from $\gamma(0)$" are two geodesics between $2$ different points.

If, on the other hand, $\gamma(0) = \gamma(L/2)$ (but $\gamma'(0)\neq \gamma'(L/2)$, since otherwise we'd contradict minimality of $\gamma$), then we repeat the argument with $\gamma(0)$ and $\gamma(L/4)$. Eventually the sequence $\gamma(L/2), \gamma(L/4),...,\gamma(L/2^k)$ gets within the injectivity radius at $\gamma(0)$, and then the argument stops.

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Very interesting @JasonDeVito. Thank you. Any idead for 1? –  Tomás Oct 27 '12 at 13:22
    
The two geodesics produced by this proof are always continuations of each other. Is is also possible to find two points with geodesics between them whose maximal extensions are different subsets of the manifold? (This would necessarily require stronger conditions than closedness because it's not true for the real projective plane). –  Henning Makholm Oct 27 '12 at 13:23
    
@Tomás, no, just take an ellipsoid with three different axis lengths, so it is not a surface of revolution. There was a fair amount of published work either side of 1900 about this sort of thing. The dissertation of Bliss characterizing all geodesics of a torus of revolution ("anchor ring") was published in the Annals about 1901. –  Will Jagy Oct 27 '12 at 15:27
    
Thanks you , @WillJagy. –  Tomás Oct 28 '12 at 11:12

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