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In a compact hyperbolic Riemann surface without boundary tbe length minimizing geodesic between two points $p$ and $q$ is unique.

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3 Answers

up vote 7 down vote accepted

Think about a Riemann surface of genus 2 say. This is a 2-holed doughnut. It has a curve winding about one of its holes. (Or more mathematically speaking it has an embedded circle representing a nonzero homology class.) We can deform this curve so that it is length-minimizing in its homotopy class. Then the curve becomes a geodesic $C$. Then let $p$ and $q$ be "opposite" points of $C$, that is the distance from $p$ to $q$ along $C$ is a half of the length of $C$. Then $C$ splits into two equal-length geodesics between $p$ and $q$.

I suppose there may be even shorter geodesics between $p$ and $q$, but I think that choosing the homology class to minimize the length of $C$ shoudl sort that out.

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I'm not a hyperbolic geometer, but I'm going to say no. Here is my reasoning:

Let $C$ be your compact hyperbolic Riemann surface, so it is the quotient of its universal cover, the hyperbolic plane, by a cocompact Fuchsian group $\Gamma$. Let $\pi: \mathbb{H} \rightarrow \Gamma \backslash \mathbb{H} = C$ be the covering map.

We have points $P$ and $Q$ on $C$. Fix one lift $\tilde{P}$ of $P$ and consider all possible lifts $\tilde{Q}_{\gamma}$ of $Q$: this will be a full orbit of $\Gamma$. For each $\gamma \in \Gamma$, there is a unique geodesic in $\mathbb{H}$ from $P$ to $\tilde{Q}_{\gamma}$ and its image under $\pi$ is a geodesic from $P$ to $Q$.

So certainly there are lots of geodesics in question. The ones of minimal length are going to correspond to the choices of $\gamma$ such that the hyperbolic distance from $P$ to $\tilde{Q}_{\gamma}$ is minimized. But it seems pretty clear that we can choose $P$ so as to be equidistant between two such points: choose any point $Q_1$, and then choose $Q_2$ to be distinct from $Q_1$, to lie in the same $\Gamma$-orbit as $Q_1$ and to be closest (not necessarily uniquely closest!) to $Q_1$ among all such points. Let $P$ be the midpoint of the geodesic arc from $Q_1$ to $Q_2$. Then so long as $P$ is at least as far away from any other point in the $\Gamma$-orbit of $Q$ as it is from $Q_1$ and $Q_2$, on the quotient we have two length-minimizing geodesics.

I am pretty sure you can arrange for this to happen with a suitable choice of $\Gamma$. For instance, if $\Gamma$ is such that the fundamental domain is a regular hyperbolic octagon with edges identified in the usual way to get a genus $2$ quotient, then I think it works to take $Q_1$ and $Q_2$ to be adjacent vertices of the octagon. See for instance the picture towards the bottom of this page:

http://www.math.cornell.edu/~mec/Winter2009/Victor/part4.htm.

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To complement the fine answers given earlier, it could be mentioned that in the context of a hyperelliptic Riemann surface (such as any surface of genus 2, for example), the following simpler construction can be given. Among all pairs of Weierstrass points, take the pair $A,B$ of points at least distance. Then there are two minimizing segments joining $A$ and $B$ (they are switched by the hyperelliptic involution). This provides a counterexample. Note that the union of the two segments is a systolic loop.

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