MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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*?


share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.