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Suppose $S$ is a surface of finite type with nonempty boundary. Now consider the arc complex $\mathcal{A}$. The action of Mod(S)(mapping class group) on the set of all vertices has finitely many orbits (see this post). My question is

What are the orbits of higher cells? Are there finitely many orbits? In other words how the quotient looks like?

From the above post we know it will consists of finitely many vertices. I think a variation of the above proof is also gives the answer for 1-cells i.e. edges. But what about cells of higher dimension.

PS: Any link , reference, paper will be extremely helpful. Thanks in advance.

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  • $\begingroup$ @YvesCornulier thanks for pointing out the mistake. You are absolutely right. $\endgroup$
    – Cusp
    Commented Dec 26, 2013 at 17:34
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    $\begingroup$ The same argument as in the above post shows that there are finitely many orbits of $k$-cells for all $k$. To describe them, you need to keep track of the homeomorphism type of the complement together with the order that the arcs come out of the base-point and the way that the arcs make up the boundary components of the complement. I recommend meditating on the section of Farb-Margalit's primer called "The change of coordinates principal" if you find this kind of thing tricky. $\endgroup$
    – Andy Putman
    Commented Dec 26, 2013 at 17:58
  • $\begingroup$ @Andy I was reading that book. But as I was trying with examples with three or four arcs and I was not convinced whether the general result is true or not as the pictures are becoming complicated. After reading the comment I tried to build up an inductive argument on the number of arcs and it worked. Thanks for the comment. $\endgroup$
    – Cusp
    Commented Dec 27, 2013 at 5:26

2 Answers 2

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The quotient complex was studied in a paper by Penner, "The structure and singularities of quotient arc complexes", Journal of Topology 1 (2008), 527-550. An earlier version of the paper is available on the arXiv. He enumerated all cases when the quotient is a sphere (this only happens when the genus and number of boundary components are small) and he found there are exactly four cases when the quotient is a manifold other than a sphere:

(1) annulus with two punctures.

(2) pair of pants with one puncture.

(3) genus 0 with 4 boundary circles and no punctures.

(4) genus 1 with 2 boundary circles and no punctures.

In each case one is considering arcs with endpoints either at punctures or at one distinguished point in each boundary circle. As a fun exercise I once tried to figure out what the manifold was in these four cases. According to some very sketchy notes I still have, I got answers in the first three cases:

(1) $S^2\times S^3$.

(2) twisted product of $S^2$ and $S^5$.

(3) $S^2\times S^7$.

In case (4) Penner showed the quotient is a closed 7-manifold but I didn't determine what it was. Someone should figure this out and also check the answers in (1-3). As I recall there were some rather pretty pictures that came up.

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  • $\begingroup$ Thanks for the reference. I will try the above exercises. $\endgroup$
    – Cusp
    Commented Dec 27, 2013 at 5:28
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Here is a closely related mathoverflow question.

How many triangulations of the genus $g$ surface on $n$ vertices?

Suppose that $S$ has genus $g$ and $n$ boundary components. Then all maximal simplices in the arc complex $A(S)$ have the same dimension, and correspond to hexagonal decompositions of $S$. Alternatively, if you crush the boundary components to points, they correspond to triangulations of the surface with marked points. Two maximal simplices are adjacent if and only if the corresponding triangulations differ by a diagonal flip. This gives a way to talk about the top dimensional simplices.

As $g$ and $n$ increase, this "graph of triangulations" has very rich combinatorics and is a subject of current research. As just a hint of the difficulties involved, it is very difficult to even count the number of vertices of the graph of triangulations (the number of top dimensional simplices of $A(S)$). Even getting good asymptotics is a challenge, as the above linked mathoverflow question indicates.

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