Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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
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 https://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.
