# Canonical fundamental domain for a discrete subgroup Γ of SL₂(R) acting on hyperbolic plane

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?

• 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.

• No point in the upper half-plane is "canonical" is this context since $\mathrm{SL}_2(\mathbb{R})$ acts transitively on it. – Robin Chapman May 23 '10 at 16:12
• Well, the imaginary axis is kind of special since it's fixed by the action of complex conjugation. And $SO_2(\mathbb R)$ fixes $i$. – Akela May 23 '10 at 16:26
• @Akela: complex conjugation rests on specified coordinate. No natural complex conjugation acting on complex manifold in general. If $G$ is non-compact 3-dim. conn'd ss Lie group with center of order 2 (so isom. to ${\rm{SL}}_ 2(\mathbf{R})$, but avoided coordinates in favor of structure) then its quotient by a maximal compact subgroup $K$ has canonical complex structure that makes left $G$-action on $G/K$ be through holomorphic automorphisms. But no complex conjugation. Also, all choices of $K$ are conjugate, and choice of $K$ "is" choice of $i$ in your case, but no canonical choice. – BCnrd May 23 '10 at 17:05
• @Akela: You've asked 26 questions so far, of which only four have accepted answers. Let me suggest accepting more answers than that, because it's helpful to readers. I also don't think that you should delete the other question. The question is reasonable enough, and stability of the list of questions is also helpful to readers. – Greg Kuperberg May 23 '10 at 17:26
• To those voting up the previous: Akela went ahead and accepted many more answers, in fact plenty enough for the time being. So that's cool. I'll just leave up the previous comment as general advice. – Greg Kuperberg May 23 '10 at 18:28

As I alluded to in a comment to one of your previous questions, many people have thought about how to explicitly compute fundamental domains, homology, etc., for congruence subgroups of $SL_2(\mathbb Z)$: this leads to the theory of modular symbols, which is not only a theory, but serves as the foundation for all computational work on modular forms too. (See papers of Mazur and Manin from the 1970s, and for the computational aspects, more recent writings of Cremona, William Stein, and others.)

On a related note, let me remark that it it actually quite feasible, and reasonable, to generate a fundamental domain for a finite index subgroup by translating the fundamental domain for $SL_2(\mathbb Z)$.
For a carefully worked example, see here (especially the discussion on page 6).

• dear Prof. Emerton: I had seen "Parabolic points and zeta functions of modular curves" the triangulation done in it. However I asked a more naive question withholding that information. This is because I had seen that the triangulation in that paper was especially suited for the purposes of computing special values of $L$-functions. I wanted to know whether it is possible to get "nicer" domains once we let go of the number-theoretic goals in our mind. For this purpose I also dropped the requirement that $\Gamma$ is a subgroup of $SL_2 (\mathbb Z)$. – Akela May 23 '10 at 23:53
• Also in a version of this question I edited out, I had acknowledged your helpful comment to the previous question, in addition to sigfpe's. – Akela May 23 '10 at 23:54

EDITED: One "canonical" choice for the basepoint for the Voronoi decomposition, also called the Dirichlet decomposition, is a point of of maximal injectivity radius. For example, when displaying the Dirichlet domain, SnapPea searches for a point of maximum injectivity radius. If your hyperbolic manifold has a single cusp (as the modular curve $H^2/\rm{SL}(2,Z)$ does) then any limit of the Voronoi domain, as the basepoint exits the cusp, is called a Ford domain.

To get a truly canonical choice, in the once-cusped surface case we can turn to the Bowditch-Epstein construction (in dimension three this is due to Epstein-Penner - see the comments below). Roughly this corresponds to taking a Dirichlet domain about the parabolic point at infinity. A more precise way to say this: chop off the cusp using a horotorus of very small area. Let the EP domain be the set of points in M with a unique shortest path to the horotorus.

This is the domain that SnapPea's "cusp window" displays automatically. When the manifold has several cusps building the EP domain involves a choice of relative volume of the cusps. Again, one could invent a "canonical" choice by taking all of the cusps to have the same volume...

I suppose that "canonical" here means "one of finitely many". For a more detailed discussion of the two and three dimensional cases please see the books "Indra's Pearls" or the somewhat more advanced "Outer circles".

Just to clear up a possible point of confusion in the above -- the EP domain is not an $n$-cell. For a once-cusped surface (eg the modular surface) the EP domain is an annulus with one end at the cusp and the other end glued together to give the surface. For a once-cusped three-manifold (eg the complement of the figure eight knot) the EP domain is a two torus crossed with $R_{\geq 0}$. Again one end goes to the cusp and the other is glued to give the three-manifold.

Finally, the Voronoi domain does not limit to the EP domain as the basepoint exits the cusp. As stated above, as the basepoint exits the cusp, the result of "gluing the faces of the Voronoi domain corresponding to parabolic isometries" limits to the EP domain.

• I didn't know the term "Ford domain", and it seems that the standard domain for $SL(2,\mathbb{Z})$ is its Ford domain. – Greg Kuperberg May 23 '10 at 17:45
• That is correct. One could argue that the standard domain is also a Dirichlet domain, using the orbifold point of order two as a basepoint. – Sam Nead May 23 '10 at 22:47
• @Sam: Firstly, in this case the Ford domain is not quite "canonical" in your sense, because we need to choose a fundamental region for the action of the stabilizer of $\infty$. Obviously, each finitely generated group will indeed have finitely many nice choices. – Grant Lakeland May 28 '10 at 17:47
• (sorry, I had two comments and they were too long to fit together) Secondly, if we use coudy's definition of Dirichlet domain, I believe it's usually required that the basepoint for a Dirichlet domain is not fixed by any non-trivial element of the group, because the half-spaces we intersect must be well-defined. However, taking any point on the imaginary axis above $i$ will lead to the standard fundamental domain for $PSL_2(\mathbb{Z})$. – Grant Lakeland May 28 '10 at 17:49
• @Grant - you only need to choose a fundamental region if you insist on cutting the ford domain open to get an $n$-cell. If you do so insist then there are uncountably many possibilities... (Even for the modular surface!) – Sam Nead May 28 '10 at 18:05

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.

I know of two ways to make a canonical or somewhat canonical fundamental domain of a geometric group action.

1) A discrete reflection group of hyperbolic isometries (or a discrete reflection group in any isotropic geometry in any dimension) has a canonical fundamental domain bounded by its reflection walls. If you pass to a subgroup of a reflection group, then as you say you can stitch together copies of the small fundamental domain to make a larger one. $SL(2,\mathbb{Z})$ is half of a reflection group $\Gamma$, and this could be the main motivation for standard choices for its fundamental domain. The same fundamental domain is also the triangle of another reflection group, one which is a different index 2 subgroup of this $\Gamma$.

2) The Voronoi tiling of a free orbit of a group $\Gamma$ always gives you a Voronoi fundamental domain, as you say. Even if your favorite orbit is not free, you can interpret it as a limit of free orbits and take the limit of the Voronoi tilings. Even though the standard fundamental domain of $SL(2,\mathbb{Z})$ might not be a Voronoi domain, I think that it is a limiting Voronoi domain, obtained by taking $z_0 \to \infty$ or $z_0 \to 0$ along the imaginary axis.