5
$\begingroup$

This is a follow-up question to a question from math.stackexchange: https://math.stackexchange.com/q/1436253/67563 Had it not been for the exchange there between myself and @Lee_Mosher in the comments I would not have thought the question that follows research-level, but apparently the details I'm concerned with have not been worked out in the literature.

Consider the Fuchsian group $\Gamma:=\Big\langle\begin{pmatrix}1&1\\1&2\end{pmatrix}, \begin{pmatrix}1&-1\\-1&2\end{pmatrix}\Big\rangle$. In the literature it is stated (with varying justifications) that $T:=\mathbb{H}^2/\Gamma$ is a punctured torus and that the Dirichlet domain $D$ for $T$ centered at $i\in\mathbb{H^2}$ is the region colored in yellow in the figure below. I am interested in verifying this explicitly via the following method of constructing $D$.

A Dirichlet domain can be formed by taking the images under each $\gamma\in\Gamma$ of its center $c$, then looking at the geodesic $g_{\gamma}$ from $c$ to $\gamma(c)$, then taking the perpendicular bisector $b_{\gamma}$ of $g_{\gamma}$, then taking the intersection of all the half-spaces that are on the same side of $b_{\gamma}$ as $c$ (varying over all $\gamma\in\Gamma$).

enter image description here

Let $c=i$ and let $\gamma=\begin{pmatrix}1&1\\1&2\end{pmatrix}$. $\gamma(i)=\dfrac{i+1}{i+2}=\frac{3}{5}+\frac{1}{5}i$. The geodesic $g_{\gamma}$ from $i$ to $\gamma(i)$ is a segment of the circle centered at $-\frac{1}{2}$ of radius $\frac{5}{4}$, shown in blue in the figure. This can be verified by calculating that the Euclidean distance between $-\frac{1}{2}$ and $i$, as well as the Euclidean distance between $-\frac{1}{2}$ and $\frac{3}{5}+\frac{1}{5}i$, is $\frac{5}{4}$.

The problem arises when we consider $b_{\gamma}$. The distance between $-\frac{1}{2}$ and $p:=\frac{1}{2}+\frac{1}{2}i$ is also $\frac{5}{4}$, therefore $g_{\gamma}$ intersects the edge of the Dirichlet domain that connects $0$ and $1$ at $p$, as shown in the figure. But that edge is not $b_{\gamma}$ because the geodesic perpendicular to it at $p$ is the vertical line rising out of $\frac{1}{2}$ (the dotted line in the figure). Even worse, the hyperbolic distance between $i$ and $\frac{1}{2}+\frac{1}{2}i$ is equal to the hyperbolic distance between $\frac{1}2+\frac{1}2i$ and $\frac{3}{5}+\frac{1}{5}i$, as follows. \begin{align} d_{\mathbb{H}^2}(x_1+iy_1,x_2+iy_2)&=\mathrm{arcosh}\Big(1+\frac{(x_2-x_1)^2+(y_2-y_1)^2}{2y_1y_2}\Big)\\ \Longrightarrow d_{\mathbb{H}^2}(i,\tfrac{1}{2}+\tfrac{1}{2}i)&= \mathrm{arcosh}\Big(1+\frac{(\tfrac{1}{2}-0)^2+(\tfrac{1}{2}-1)^2}{2\cdot 1\cdot\tfrac{1}{2}}\Big)=\mathrm{arcosh}(\tfrac{3}{2}),\\ \text{and }d_{\mathbb{H}^2}(\tfrac{1}{2}+\tfrac{1}{2}i,\tfrac{3}{5}+\tfrac{1}{5}i)&= \mathrm{arcosh}\Big(1+\frac{(\tfrac{3}{5}-\tfrac{1}{2})^2+(\tfrac{1}{5}-\tfrac{1}{2})^2}{2\cdot\tfrac{1}{2}\cdot\tfrac{1}{5}}\Big)=\mathrm{arcosh}(\tfrac{3}{2}). \end{align} And so $b_{\gamma}$ must be a segment of a half-circle which traverses the Dirichlet domain like the red curve in the figure, but this is not possible.

Where is the mistake?

$\endgroup$
  • $\begingroup$ Where in the literature have you seen this precise claim? $\endgroup$ – j.c. Sep 24 '15 at 21:47
  • $\begingroup$ @j.c. For one I thought this was done in Bonahon's Low-Dimensional Geometry, but looking for the exact statement now I'm not finding it. He discusses this fundamental domain in sections 5.5 and 8.2. Exercise 5.13 is to show $D$ has the symmetry group of the square, from which I thought this followed. I could have sworn I saw some online references stating this too but I'm not finding those either! I would be very happy to learn that the assumption is false and that my computations are correct. $\endgroup$ – j0equ1nn Sep 24 '15 at 22:40
  • $\begingroup$ I do not have time to do computations, but you can try to check if you got the right domain by verifying conditions of Poincare's fundamental polygon theorem. First, check which elements (if any!) of $\Gamma$ pair the edges of $D$. Then check if these elements generate $\Gamma$: This should be easy since $\Gamma$ is free. Then check ideal vertex cycle conditions (namely that the vertex cycles yield parabolic elements). If these conditions hold, you got your domain, if not, then not. $\endgroup$ – Moishe Kohan Sep 25 '15 at 0:12
  • 2
    $\begingroup$ I don't think $D$ is the Dirichlet domain centered at $i$. I think your computation is correct. Also note that the vertical sides imply that if $D$ were the Dirichlet domain, then there would be points in the $\Gamma$-orbit of $i$ located at $2+i$ and $-2+i$, which I don't think is the case here. I also don't think it's the Ford domain, as the geodesics are not isometric circles, and with a Ford domain for a group with cusps such as this you would expect the two vertical sides to be identified by a parabolic element. $\endgroup$ – Grant Lakeland Sep 25 '15 at 22:31
  • 3
    $\begingroup$ I want to add that $D$ is both a Ford domain and the Dirichlet domain centered at $i$ for the principal congruence subgroup $\Gamma(2)$ of the modular group. In that case, the quotient space is a thrice-punctured sphere - the vertical sides are identified by $\left( \begin{smallmatrix}1 & 2 \\ 0 & 1\end{smallmatrix} \right) $ and the nonvertical by $\left( \begin{smallmatrix}1 & 0 \\ 2 & 1\end{smallmatrix} \right)$. This fact may be the cause of some of the confusion here. $\endgroup$ – Grant Lakeland Sep 25 '15 at 22:42
0
$\begingroup$

As discussed in the comments, the contradiction is that $D$ isn't really the Dirichlet domain for $G$ centered at $i$. The red edge is in fact an edge of the domain.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.