3
$\begingroup$

Consider a compact Riemann surface of genus $g\geq2$. An admissible system of Jordan curves is a finite collection of Jordan curves $\{\gamma_1,\cdots,\gamma_n\}$ such that

  1. they are nonintersecting with each other;
  2. no two are freely homotopic;
  3. none is homotopic to 0

This concept is mentioned in Masur's on a class of geodesic in teichmuller space and in this paper the author mentions that an admissible system has at most $3g-3$ Jordan curves. I am wondering why and how to prove it.

My attempt would be to consider a fundamental polygon with $4g$ sides, consecutively labeled by $$a_1,b_2,a_1^{-1},b_1^{-1},\cdots,b_g^{-1}$$ Now choose a point $p$ on the edge $a_1$, then $p$ appears on $a_1^{-1}$ as well. Connecting them gives a closed curve. We cannot do the same for $b_1$, but we can do this for $a_2,a_3,\cdots,a_g$. This gives $g$ curves. Finally an edge is also a closed curve. Now any curve not null-homotopic should cross an edge of the fundamental polygon, but I cannot seem to find another one without intersecting the existing ones. So I can find at most $g+1$ curves. Where do the $3g-3$ curves come from?

$\endgroup$
0
3
$\begingroup$

It comes from the so called "pants decomposition". A "pair of pants" (or simply pants) is a sphere with $3$ holes. Every compact surface of genus $g\geq 2$ can be decomposed into such pants. The $3g-3$ curves are the pants boundaries.

To visualize, represent your surface in the following way. Consider the sphere with $g+1$ holes, and your surface is obtained by gluing this to its double (mirror image). Now decompose the sphere with $g+1$ holes into pants. To do this you draw $g-2$ disjoint closed curves. The mirror image will contain the same number of them. So the total is $$(g+1)+2(g-2)=3g-3.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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