(0). My question concerns the relation between two different fundamental domains for the group
$$
\Gamma(2)= \left\{
\begin{bmatrix} a & b \\
c & d
\end{bmatrix}\in SL_2(\mathbb Z) \; \big\lvert \; a-1\equiv b \equiv c \equiv d-1\equiv 0 \, \mbox{ mod }\, 2\;
\right\}\subset SL_2(\mathbb Z).
$$
(1). Since $\Gamma(2)$ has index 6 in $SL_2(\mathbb Z)$, one can find a fundamental domain $D$ for $\Gamma(2)$ formed by six copy of the standard fundamental domain for the modular group $SL_2(\mathbb Z)$. Below is a picture of such a $D$: it is the union of the triangles 1 to 6. (This picture is extracted from the book "Fuchsian groups" by S. Katok, where the construction of $D$ is detailed in Example F p. 141).
(2). On the other hand, let $\Delta(\infty,\infty,\infty)$ be the triangle group of type $[\infty,\infty,\infty]$. If $T$ stands for an ideal hyperbolic triangle in the hyperbolic plane $\mathbb H$ (to fix things, one can take $0$, $1$ and $i\infty$ as the ideal vertices of $T$ in the upper half-plane model of $\mathbb H$) then by definition $\Delta(\infty,\infty,\infty)$ is the subgroup (of index 2) of orientation-preserving isometries of the subgroup of ${\rm Isom}(\mathbb H)$ spanned by the three reflections with respect to the sides of $T$.
(3). It is well known that $\Gamma(2)$ can be described as $\Delta(\infty,\infty,\infty)$ (see for example p. 442 of the book "Lectures on the theory of functions of a complex variable II" by G. Sasone or p. 94 of the paper "Arithmetic triangle groups" by K. Takeuchi).
From the second description of $\Gamma(2)$ as $\Delta(\infty,\infty,\infty)$, it comes that as a fundamental domain $D'$ for $\Gamma(2)$, one can take the union of $T$ (with vertices $0,1$ and $i\infty$ as in (2).) with its translate $T+1$ (the ideal hyperbolic triangle with vertices $1,2$ and $i\infty$).
Question: how to obtain $D$ from $D'$ (or the converse) by cutting and pasting?
I guess that this is well-known.
A reference would be welcome.
Best Answer
I start from the hexagon $D$ and use the same notations as in the figure in the question. Take the matrices $S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ as usual. The boundaries of D have the following correspondences:
\begin{eqnarray*} [p+2,\infty] &=& T^2([p,\infty]),\\ [0,v] &=& ST^{-2}S([0,p]),\\ [1,p+2] &=& TST^{-2}ST^{-1}([1,v]). \end{eqnarray*}
Now, cut off the hyperbolic triangle $(0,v,1)$ and glue it back to the other side. The resulting domain after translating 1/2 units is an ideal quadrilateral, as desired.