Consider an orientable surface $S$ of arbitrary genus, possibly with boundaries, and with marked points and/or punctures. I will assume that every boundary has at least one marked point so that the surface can be triangulated by triangles with vertices at the marked points/punctures.
The mapping class group $MCG(S)$ acts transitively on the set of triangulations, the action can be represented by a sequence of flips (a.k.a. Whitehead moves), the action does not change the "combinatorics" of the triangulation (there are some subtle details in how one defines what I mean by "combinatorics", depending on whether one decides to give names to the punctures or not, but I my question doesn't really depend too much on these details).
It is natural to consider a fundamental domain $\mathcal{D}$ for this action, i.e. a set of triangulations such that every combinatorial type of triangulation is contained precisely once, or equivalently a set of triangulations such that any other triangulation is MCG-equivalent to a triangulation in $\mathcal{D}$.
It seems to me that there is no canonical way of choosing such a set $\mathcal{D}$ (although I would be interested in hearing any comments about this). Also, there are many ways to generate it, perhaps the simplest is just to start from a triangulation $T$ and flip it in all possible ways until a set $\mathcal{D}$ is generated.
I would like to know if there is a "dual" way to generate $\mathcal{D}$ in the following sense: give a collection of arcs (i.e. curves with endpoints at the marked points/punctures) $\mathcal{A}$ such that all possible triangulations built out of arcs in $\mathcal{A}$ produces a set $\mathcal{D}$.
It is not difficult to see why such a set $\mathcal{A}$ always exists. Perhaps it is a bit surprising that $\mathcal{A}$ itself contains pairs of arcs which are related by the action of the MCG (it is not itself a "fundamental domain" of arcs). Potentially, this could be a much more efficient way to describe the set of all triangulations. After all, even in the case of a disk with $n$ marked points one has a Catalan number of triangulations but only $n(n-2)/2$ different arcs.
To sum up: is there a general description for how to generate sets of arcs $\mathcal{A}$ on a surface $S$ such the set of triangulations $\mathcal{D}$ one can build out of arcs in this $\mathcal{A}$ contain every triangulation up to $MCG(S)$ precisely once?
