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Take an orientable surface $S_g^s$ of genus $g$ with no boundaries and $s$ points removed and fix a complete hyperbolic metric of finite area (assuming that the Euler characteristic allows an hyperbolic structure on $S$). We know from hyperbolic geometry that there exists a unique geodesic in every non trivial free-homotopy class of curves. I know for sure that there is at least one point on $S_g^s$ through which there is no simple geodesic, but I have no idea about how to prove it. Could you help me with that?

Thank you!

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1 Answer 1

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Let $C$ be a horodisk neighborhood of the puncture $p$ so that the horocyclic boundary has length less than $2$. Then no simple, closed geodesic $g$ in $S$ intersects $C$.

Proof: Lift $g$ and $C$ to the upper half plane so that the $p$ is taken to the point at infinity and $C = \{ z \in \mathbb{C} \mid \rm{Im}(z) \geq 1 \}$. Suppose that $g$ meets $C$. Thus each lift of $g$ has horizontal width at least $2$. Thus adjacent lifts of $g$ intersect, a contradiction.

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thanks a lot: you've really helped me! –  fatoddsun May 23 '11 at 14:30
You are welcome! If I could ask - where does this problem come from? –  Sam Nead May 23 '11 at 16:12
Sure you can ask! It comes from a proof included in Harer's paper "Virtual Cohomological dimension of the mapping class group of an orientable surface" –  fatoddsun May 27 '11 at 21:21
@fatoddsun - Ok, here is something cool but harder than what I said: my proof needs $s > 0$. But the statement is still true for closed surfaces! This follows from a famous paper of Birman and Series. I don't know a proof off the top of my head. –  Sam Nead Sep 15 '11 at 22:00

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