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Q1. For a smooth, closed (compact) surface $S$ embedded in $\mathbb{R}^3$, under which conditions is it true that, for every pair of points $a,b \in S$, there are an infinite number of geometrically distinct geodesics connecting $a$ to $b$?

By "geometrically distinct" (my terminology) I mean that there is a point on one geodesic not on the other. So, if $S$ is the geometric unit-radius sphere $\mathbb{S}^2 \subset \mathbb{R}^3$, there are only two geodesics connecting $a$ to $b$, the two subarcs of the great circle through $\{a,b\}$.

I cannot think of another $S$ that has only a finite number of distinct geodesics connecting each pair of points. For example, any two points on a torus are connected by winding geodesics [left, below], as well as other less obvious geodesics [right below].


      TorusGeodesics
      Images from: Irons, Mark L. "The curvature and geodesics of the torus." 2008. (PDF Download.)
An infinite cylinder would be a counterexample, but it is not closed & compact. A narrow version of Q1 is:

Q2. Could it be that only $\mathbb{S}^2$ has a finite number of connecting geodesics? That every other $S$ has an $\infty$# between every point pair?

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    $\begingroup$ I'm not sure what you mean by the infinite (circular) cylinder being a counterexample. After all, any two points on such a cylinder that don't lie on a common geodesic circle are joined by infinitely many distinct geodesics. $\endgroup$ – Robert Bryant Jan 10 '15 at 3:28
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    $\begingroup$ Also, a smooth compact surface smoothly embedded in $\mathbb{R}^3$ that is not the $2$-sphere must have an infinite fundamental group and hence must have infinitely many distinct (in your sense) geodesics joining any two distinct points. $\endgroup$ – Robert Bryant Jan 10 '15 at 3:30
  • $\begingroup$ @RobertBryant: I meant: On the $\infty$-cylinder, there are two points that only have two connecting geodesics (because they lie on a circle wrapping the cylinder). If the cylinder is finitely capped, even these points are connected by an $\infty$-# of geodesics. $\endgroup$ – Joseph O'Rourke Jan 10 '15 at 3:32
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    $\begingroup$ Well, this is a consequence of Morse Theory: If $S$ is the surface and $a$ and $b$ are points on it, then each fixed-endpoint homotopy class of curves $c:[0,1]\to S$ with $c(0)=a$ and $c(1)=b$ contains at least one constant speed (minimal) geodesic. When the compact orientable surface is not the $2$-sphere, the set of such fixed-endpoint homotopy classes is infinite because it is isomorphic (though not canonically) to the fundamental group of the surface, which is infinite. Now, it is easy to see that these cannot fall into a finite number of geodesics that are distinct in your sense. $\endgroup$ – Robert Bryant Jan 10 '15 at 3:40
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    $\begingroup$ for the sphere a and b must not be antipodal to have only finitely many connecting geodesics. But you have implicitely addressed that by demanding finiteness for every pair. $\endgroup$ – Manfred Weis Jan 10 '15 at 7:11
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Any smooth compact surface smoothly embedded in $\mathbb{R}^3$ that is not the $2$-sphere must have an infinite fundamental group and hence must have infinitely many distinct (in your sense) geodesics joining any two distinct points.

This result follows from Morse theory: If $S$ is the surface and $a$ and $b$ are points on it, then each fixed-endpoint homotopy class of curves $c:[0,1]\to S$ with $c(0)=a$ and $c(1)=b$ contains at least one constant speed (minimal) geodesic. When $S$ is not the $2$-sphere, the set of such fixed-endpoint homotopy classes is infinite because it is isomorphic (though not canonically) to the fundamental group of the surface, which is infinite. Now, it is easy to see that these minimal geodesics in each fixed-endpoint homotopy class cannot fall into a finite number of geodesics that are distinct in your sense.

Note that this does not depend on the surface being embedded isometrically into $\mathbb{R}^3$, just on having the fundamental group be sufficiently 'large', thus, the only potential compact counterexamples are $S^2$ and $\mathbb{RP}^2$, and then probably only for very special metrics on these.

Added remark: It is, perhaps, an interesting question as to which Riemannian metrics $g$ on the $2$-sphere have the property that, for the generic pair of points $a$ and $b$, there are only a finite number of distinct geodesics joining $a$ to $b$. (Note that even the standard unit sphere in $3$-space does not have the property that each pair of distinct points lie on only a finite number of distinct geodesics, since antipodal points lie on a whole circle of geodesics in this case. This is why I'm considering the weaker condition of asking finiteness only for generic pairs.) One general class in which this is true is that of Zoll metrics, i.e., when all of the geodesics are closed. In that case, an application of Sard's Theorem shows that the generic pair of points on the surface lie on only a finite number of geodesics, in the sense that the set of pairs that lie on an infinite number of geodesics is a set of measure zero in the $4$-manifold that consists of (ordered) pairs of points on the surface.

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  • $\begingroup$ So it seems that, for both the 2-sphere and Zoll metrics, the key to generic finiteness is that "all of the geodesics are closed." $\endgroup$ – Joseph O'Rourke Jan 12 '15 at 1:00

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