TITLE: Comparing two fundamental domains for $\Gamma(2)$
QUESTION [9 upvotes]: (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.

REPLY [2 votes]: 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.
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