Geodesics for a Cone Metric - MathOverflow

Geodesics for a Cone Metric

2012-03-10T21:42:48Z

Here is a question that I hope/suspect is elementary but cannot find a reference for. Suppose we are given a surface, S, with a conformally Euclidean metric, |f(z)||dz|, where f(z) is meromorphic. Puncture the surface S at all of the zeroes and poles of f(z) (at the "cone points"), and denote the resulting surface S*.

Is it true that there exists a unique geodesic in every homotopy class of curves in S*?

Thanks!

Answer by Robert Kucharczyk for Geodesics for a Cone Metric

2012-03-10T22:28:04Z

The answer to the question as stated is "no": take for $S$ the plane $\mathbb{C}$ with the metric induced by the differential $z\mathrm{d}z$; then $S^{\ast }=\mathbb{C}\smallsetminus {0}$ and there is simply no geodesic from $1$ and $\mathrm{i}$. Or, if you want closed loops and free homotopy, there is no geodesic in the free homotopy class of the loop around $0$. Intuitively, it is clear what happens: if you draw a curve and try to pull it straight, you are forced to go through the cone point at the origin.

If, however, you consider the surface $S$ itself and extend the definition of a geodesic so as to work on Euclidean metrics with cone points as well, the answer is "yes" if $S$ is complete as a metric space. The usual definition of geodesics in this context is one which works for all metric spaces: locally isometric maps from an interval. The statement you asked for is then a consequence of the Arzelà-Ascoli theorem.

These things are treated in detail in the textbook Quadratic differentials by Kurt Strebel.

Answer by Misha for Geodesics for a Cone Metric

2012-03-10T23:39:06Z

Maybe you are asking for uniqueness, not existence? Then the answer is yes because your metric admits a locally CAT(0) completion in the answer below. This uniqueness result, originally, I think, due to Teichmuller, should be also in Strebel's book and is based on a Gauss-Bonnet calculation.