Consider the Fuchsian group $\Gamma:=\Big\langle\begin{pmatrix}1&1\\1&2\end{pmatrix}, \begin{pmatrix}1&-1\\-1&2\end{pmatrix}\Big\rangle$. A commonly studied fact is that when this acts upon the upper half-plane $\mathbb{H}^2$, the quotient $T:=\mathbb{H}^2/\Gamma$ is a hyperbolic once-punctured torus. For instance taking the Ford domain of $\Gamma$, we get as a fundamental region the hyperbolic quadrilateral in the center of the following tessellation. enter image description here

I am interested in getting similar pictures using Dirichlet domains instead. For $p\in\mathbb{H}^3$ let $D(p)$ denote the Dirichlet domain for $\Gamma$ centered at $p$. Note that since $\Gamma$ is torsion-free, there are no restrictions on which $p$ can be used.

What is the effect on $D(p)$ as we vary $p$? Is $D(p)$ always a four-sided polygon with ideal vertices? Also, for a given $p$ what elements of $\Gamma$ contribute sides to $D(p)$, and which ones occur in gluing the sides?

UPDATE: Even an investigation of the simplest scenario here lead me to complications, see https://mathoverflow.net/q/219139/14835 for more on that.


1 Answer 1


Here is a partial answer, which may point the way to a reasonable conjecture. Along the way I do answer one of your several questions.

First a correction: the Ford domain is independent of $p$ (up to covering transformation). It depends on the Fuchsian group alone. The Dirichlet domain $D(p)$ depends, of course, on $p$.

Next some setup, designed to learn as much about $D(p)$ as possible without doing any actual work.

The boundary $\partial D(p)$, when projected to the quotient punctured torus $T$, becomes an embedded 1-complex $S(p)\subset T$. $S(p)$ is a closed subset of $T$, and it is not compact, indeed $S(p)$ accumulates on the puncture of $T$.

Let $\infty$ denote the "puncture" of $T$, aka the end of $T$. Denote $\overline T = T \cup \infty$ to be the one-point compactification, aka the end compactification in this case, so $\overline T$ is an ordinary, unpunctured, compact torus. Denote the closure of $S(p)$ in $\overline T$ as $\overline S(p) = S(p) \cup \infty$. Then $\overline S(p)$ is itself a compact 1-complex. The sides of $\partial D(p)$ glue up in pairs to form the 1-cells of $\overline S(p)$, and so the number of $1$-cells of $\overline S(p)$ is always half the number of sides of $D(p)$.

The topology of $\overline S(p)$ has the following restrictions. First, $\overline S(p)$ is a spine of the once-punctured torus $\overline T - \{p\}$, and so $\overline S(p)$ is a graph of rank $2$ and Euler characteristic $-1$. Its vertex $\infty$ has valence $\ge 1$. All finite vertices have valence $\ge 3$. So by an Euler characteristic argument on the finite graph $\overline S(p)$ the possibilities for $\overline S(p)$ can be classified into the following seven types, in order of "decreasing genericity":

  1. Valence 1 at $\infty$, three finite vertices of valence $3$, and five 1-cells.
  2. Valence 1 at $\infty$, one finite vertex of valence $4$, one finite vertex of valence $3$, and four 1-cells (this comes in two sub-types: the 1-cell with endpoint at $\infty$ can have opposite endpoint either of valence $3$ or $4$).
  3. Valence 1 at $\infty$, one finite vertex of valence $5$, and three 1-cells.
  4. Valence 2 at $\infty$, two finite vertices of valence $3$, and three 1-cells.
  5. Valence 2 at $\infty$, one finite vertex of valence $4$, and two 1-cells.
  6. Valence 3 at $\infty$, one finite vertex of valence $3$, two 1-cells.
  7. Valence 4 at $\infty$, no finite vertices, and two 1-cells.

With that, you have asked:

  • Is $D(p)$ always a four-sided polygon with ideal vertices (in other words, an ideal quadrilateral)?

which is equivalent to asking

  • Does $\overline S(p)$ always have type 7?

For example, in your picture if you take $p=i$ (not $p=2i$) then $D(i)$ is the ideal quadrilateral with endpoints $-1,0,1,\infty$ as your picture shows, and this is an an "ideal square" meaning that its isometric symmetry group is the order 8 dihedral group. The "center" of this ideal square is $p=i$. So, with the choice $p=i$, type 7 indeed occurs.

However, the answer to your question is "no". For example, if $p$ is near the puncture, e.g. if $p=\lambda i$ for very large $\lambda$, then it's not hard to see that the valence of $\infty$ equals $1$ and so either type 1, 2, or 3 occurs.

Perhaps a more natural conjecture is that on an open dense set of values of $p$, what occurs is type 1, which is the most generic type. An alternative conjecture is that, since the shape of $T$ itself is rather nongeneric in the Teichmuller space, on an open dense set of values of $p$ what occurs is type 2 or 3.

  • $\begingroup$ This is fantastic! I'm going to work on this some more and probably comment again. In response to the correction: I am aware that the Ford domain doesn't depend on a $p$, I didn't mean to suggest that it does. I like to think of the Ford domain like $\underset{p\rightarrow\infty}{\lim}D(p)$. $\endgroup$
    – j0equ1nn
    Sep 15, 2015 at 20:47
  • $\begingroup$ @j0equ1nn: Glad to help. Keep me posted, I find this stuff interesting. $\endgroup$
    – Lee Mosher
    Sep 15, 2015 at 20:56
  • $\begingroup$ I made a careless error in computing $D(i)$, in particular in computing $\gamma^{-1}(i)$ and $\delta^{-1}(i)$, which makes everything I say after that nonsense so I'm modifying the part of the question about that example. $\endgroup$
    – j0equ1nn
    Sep 17, 2015 at 9:42
  • $\begingroup$ @Lee_Mosher So now I'm wondering how to go about determining which case we're in for a given point. Even for the simplest example, $p=i$, I'm not seeing how we know the edges are as you said. I understand why the toroidal edge pairings are given by the generators, and I understand how to compute from $D(p)$+pairings, the generators. But how do we know which $\gamma\in\Gamma$ contribute edges to $D(p)$? $\endgroup$
    – j0equ1nn
    Sep 17, 2015 at 10:30
  • 1
    $\begingroup$ In that case $p=i$ it's simply an issue of symmetry. Ihe ideal quadrilateral with vertices $-1,0,1,\infty$ is an "ideal square" by which I mean that it's symmetry group is the maximal possible symmetry group, namely the dihedral group of order 8. The point $i$ is the unique point fixed by that entire symmetry group. You can write out the eight fractional linear transformations in this group completely explicitly. So, the Dirichlet domain $D(i)$ must exhibit the same symmetry group. $\endgroup$
    – Lee Mosher
    Sep 17, 2015 at 12:56

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