Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let S be a surface of genus g with some parked points (n of them). Assume $n \geq 2$ and fix two of the marked points. Consider the set of embedded arcs going between these two special points. The group of diffeomorphisms preserving the marked points acts on this set. What are the orbits? How many are there? Equivalently, we can instead consider embedded arcs up to isotopy and consider the action of the mapping class group of the surface.

I am specifically interested in two cases: genus zero with four marked points, and genus 1 with two marked points. However I think the general question is also interesting and it seems like the sort of thing that has been studied before. I just no idea where to look.

share|improve this question
1  
I love the "parked points" :-D –  Bruno Martelli Dec 20 '10 at 13:52
1  
As long as the two points are distinct, there is only one orbit. If the two marked points coincide then there are finitely many orbits. –  Sam Nead Dec 20 '10 at 14:24
    
@Sam Nead: Does that also mean that if I consider closed arcs (that doesn't intersect any of the points) then there are finitely many orbits? Where is a good reference for this kind of stuff? –  Chris Schommer-Pries Dec 20 '10 at 14:57
1  
Yes, it is true also for closed arcs. As a nice reference I would suggest the book of Farb Margalit www.math.uchicago.edu/~margalit/mcg/mcgv406.pdf –  Bruno Martelli Dec 20 '10 at 16:15
    
The newest version of that book is apparently 4.08 math.utah.edu/~margalit/primer –  j.c. Dec 20 '10 at 19:02

1 Answer 1

up vote 6 down vote accepted

There is only one orbit.

Suppose you have two such arcs $\lambda, \lambda'$. Let $S_\lambda$ be obtained from $S$ by removing a regular neighborhood of $\lambda$: the regular neighborhood is a disc, containing the two marked points in its interior. The two surfaces $S_\lambda$ and $S_{\lambda'}$ so obtained have the same Euler characteristic and the same number (one) of boundary components, hence they are diffeomorphic.

You can arrange the diffeomorphism so that it preverves the other marked points (simply move them by isotopy). You can then extend it to the removed discs and get a diffeomorphism of $S$ preserving the punctures and sending $\lambda$ to $\lambda'$.

share|improve this answer
    
Thanks! This is the answer I was hoping for. Another great example of how MO is the fastest substance known to mathematics. –  Chris Schommer-Pries Dec 20 '10 at 14:52
4  
The same argument shows that there are finitely many orbits in the general case, btw. Classification of surfaces is a wonderful thing. –  Igor Rivin Dec 20 '10 at 15:14

Your Answer

 
discard

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.