TITLE: Canonical fundamental domain for a discrete subgroup Γ of SL₂(R) acting on hyperbolic plane
QUESTION [8 upvotes]: Let a discrete subgroup $\Gamma$ of $SL_2(\mathbb R)$ act on the hyperbolic plane by Möbius transformations. Is there a "best" or "most canonical" fundamental domain for this action? 
Some (mostly unhelpful) observations:

For the action of $SL_2(\mathbb Z)$ the usually taken fundamental domain is this one. Any particular reason?
Unfortunately the hyperbolic triangle with vertices $0, 1, \infty$ also won't do for $SL_2(\mathbb Z)$, as it is not (quite) a fundamental domain.
If we are considering a finite-index sugroup of $SL_2(\mathbb Z)$, then we can form the fundamental domain as a finite union of the fundamental domain of  $SL_2(\mathbb Z)$. This is not very nice-looking, but it can be determined effectively given a description of $\Gamma$.
Let the action be specified for $\Gamma$. Then, for any point $z_0 \in \mathbb H$, there is the standard polygon around it. This was pointed out to me by sigfpe. Thanks a lot, sigfpe. This is a very nice and very canonical construction, once you have a starting point $z_0$. But this is very theoretical. We do not know what are the vertices, in a computable way. We just know that they exist. And then again, what would be a canonical choice for the point $z_0$? If any point inside the hyperbolic plane is canonical, it is $i$. But the two examples considered above for  $SL_2(\mathbb Z)$ does not arise as a fundamental polygon or standard polygon for $i$. In fact $i$ is on the edge of both of these fundamental domains.

REPLY [3 votes]: I found interesting the question of why we normally use "that" fundamental domain for the action of $PSL_2(\mathbb{Z})$. I thought it worthy of its own answer, though I make no claims that the reasons I will list are exhaustive (and I'd be interested to see others!).

Once we have a fundamental domain, and as long as we keep track of the ways the sides get identified by the group, the Poincaré Polyhedron Theorem allows us to recover a presentation for the group, where side-pairings correspond to generators, and vertex cycles to relations. If, say, we were interested in finding minimal generating sets, that would correspond to finding fundamental domains with minimal numbers of sides. Since $PSL_2(\mathbb{Z})$ has rank 2, the "cleanest" fundamental domains we can find (in this sense) are those with 4 sides. (Here we regard the two halves of the bottom circle arc as distinct sides, since they are identified by the linear fractional transformation $z \rightarrow \frac{-1}{z}$, and we regard $i$ as a vertex.) Further, the two matrices which correspond to the side-pairings of the usual domain are $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right)$, which are frequently taken as generators for $PSL_2(\mathbb{Z})$. So, in this regard, the domain "fits in" with other existing conventions.
As Greg Kuperberg mentions, $PSL_2(\mathbb{Z})$ is a subgroup of index 2 in the $(2,3,\infty)$ triangle reflection group, and it's easy to see how our domain is just one copy of this triangle unioned with its reflection along a vertical side.
As coudy mentions, our domain is a Dirichlet domain, and the proof of this fact is fairly straightforward (and elegant, in my opinion). Certainly, it's easier to see with this particular domain than with other domains one might try to construct.
I suspect that one reason this domain is used so frequently is that short, elementary proofs exist that it is indeed a fundamental domain, which avoid any of the discussion listed above. In what little I have glanced at on automorphic forms, this type of proof is given at an introductory stage, so that one can move on to more content. I like to see the simplest arguments as being tied up with the previously mentioned Ford domains, though I'm not sure how useful others might find it. Essentially, the unit circle is the isometric circle of the map $z \rightarrow \frac{-1}{z}$, which (because this particular element has order 2, or is its own inverse) means the circle is sent to itself, and its exterior and interior are interchanged. This observation is part of showing that no two points of our domain are equivalent under the action of the group. Using other domains involves using more than one (isometric) circle, so our domain is simplest in this sense.