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](https://mathoverflow.net/questions/49963/is-the-action-of-the-mapping-class-group-transitive-on-embedded-arcs)). 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.



Action of Mapping Class Group on Arc complex
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.


