3
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ I love the "parked points" :-D $\endgroup$
    – Bruno Martelli
    Commented Dec 20, 2010 at 13:52
  • 1
    $\begingroup$ 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. $\endgroup$
    – Sam Nead
    Commented Dec 20, 2010 at 14:24
  • $\begingroup$ @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? $\endgroup$
    – Chris Schommer-Pries
    Commented Dec 20, 2010 at 14:57
  • 1
    $\begingroup$ 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 $\endgroup$
    – Bruno Martelli
    Commented Dec 20, 2010 at 16:15
  • $\begingroup$ The newest version of that book is apparently 4.08 math.utah.edu/~margalit/primer $\endgroup$
    – j.c.
    Commented Dec 20, 2010 at 19:02

1 Answer 1

Reset to default
7
$\begingroup$

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 $\lambda'$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! This is the answer I was hoping for. Another great example of how MO is the fastest substance known to mathematics. $\endgroup$
    – Chris Schommer-Pries
    Commented Dec 20, 2010 at 14:52
  • 4
    $\begingroup$ The same argument shows that there are finitely many orbits in the general case, btw. Classification of surfaces is a wonderful thing. $\endgroup$
    – Igor Rivin
    Commented Dec 20, 2010 at 15:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .