- In this problem, consider the group \(G\) of matrices \(\left( \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right)\) with integer entries, determinant 1, and such that a and d have the same parity, b and c have the same parity, and c and d have opposite parity. This group also acts on the upper half-plane by fractional linear transformations. To the group \(G\) corresponds the fundamental domain \(\mathcal { F }\) defined by \(| \tau | \geq 1 , | \mathrm { R e } ( \tau ) | \leq 1\) , and Im \(. ( \tau ) \geq 0\) (see Figure 1). Also, let
\[ S (\tau) = - 1 / \tau \leftrightarrow \left( \begin{array}{c c} 0 & - 1 \\ 1 & 0 \end{array} \right) \quad \text { and } \quad T _ {2} (\tau) = \tau + 2 \leftrightarrow \left( \begin{array}{c c} 1 & 2 \\ 0 & 1 \end{array} \right). \]
Prove that every fractional linear transformation corresponding to \(g \in G\) is a composition of finitely many S, \(T _ { 2 }\) and their inverses, in analogy with the previous problem.
Solution
Let \(G_0 = \langle S, T_2 \rangle\), and recall that for \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\operatorname{SL}_2(\ZZ)\) and \(\tau\in\mathbb H\), \[\Im(M\tau) = \frac{\Im\tau}{\abs{c\tau + d}^2}.\] Put \(\mathcal{F} = \{\tau \in \mathbb{H} : |\tau| \ge 1,\ |\Re\tau| \le 1\}\).
1. \(G_0 \subseteq G\).
For \(S\): \(a=d=0\) have the same parity, \(b=-1\) and \(c=1\) have the same parity, and \(c=1\), \(d=0\) have opposite parity. For \(T_2\): \(a=d=1\), \(b=2\) and \(c=0\) are even, and \(c=0\), \(d=1\) have opposite parity. Since \(G\) is a group, it contains \(\langle S,T_2\rangle\).
2. Every \(G_0\)-orbit in \(\mathbb H\) meets \(\mathcal F\).
Fix \(\tau\in\mathbb H\). The numbers \(c\tau+d\), for bottom rows \((c,d)\) of matrices in \(G_0\), lie in the lattice \(\ZZ\tau+\ZZ\) and are nonzero, so \(\abs{c\tau+d}\) attains a positive minimum; hence some \(\tau'\) in the orbit maximizes \(\Im\) over the orbit. Choose \(k\in\ZZ\) with \(\tau'' = \tau' + 2k\) satisfying \(|\Re\tau''| \le 1\); then \(\Im\tau''=\Im\tau'\) is still maximal. If \(|\tau''| < 1\), then \(\Im(S\tau'') = \Im\tau''/|\tau''|^2 > \Im\tau''\), contradicting maximality. So \(\tau''\in\mathcal F\).
3. If \(h\in G\) and \(h(2i)\in\mathcal F\), then \(h(2i)=2i\).
For \(M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in G\), the entries \(c\) and \(d\) have opposite parity, so \(\abs c\ne\abs d\); the same holds for the bottom row \((-c,a)\) of \(M^{-1}\), since \(a\equiv d\). For \(\sigma=x+iy\in\mathcal F\) and such a bottom row with \(c\ne0\), \[\abs{c\sigma+d}^2=c^2\abs\sigma^2+2cdx+d^2\ge c^2-2\abs{cd}+d^2=(\abs c-\abs d)^2\ge1,\tag{$*$}\] and the first inequality is strict when \(\abs\sigma>1\) and \(\abs x<1\).
Let \(h=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in G\), \(\tau=2i\), and \(\tau'=h\tau\in\mathcal F\). If \(\Im\tau'<\Im\tau\), then \(\abs{-c\tau'+a}<1\) for the bottom row of \(h^{-1}\); by \((*)\) this forces \(c=0\), and then \(a=\pm1\), a contradiction. So \(\Im\tau'\ge\Im\tau\) and \(\abs{c\tau+d}\le1\). Since \(\abs\tau=2>1\) and \(\Re\tau=0\), the strict form of \((*)\) excludes \(c\ne0\). Hence \(c=0\), \(a=d=\pm1\), \(b\) is even, and \(\tau'=\tau\pm b\). As \(\abs{\Re\tau'}\le1\) and \(\Re\tau=0\), \(\abs b\le1\), so \(b=0\) and \(\tau'=\tau\).
4. The only elements of \(\operatorname{SL}_2(\ZZ)\) fixing \(2i\) are \(\pm I\).
If \(\frac{2ia+b}{2ic+d}=2i\), then \(2ia+b=-4c+2id\), so \(d = a\) and \(b = -4c\). Then \(ad - bc = a^2 + 4c^2 = 1\) forces \(c=0\), \(b=0\), \(a=d=\pm1\).
5. Q.E.D.
Let \(g\in G\). By step 2 there is \(\gamma\in G_0\) with \(\gamma g(2i)\in\mathcal F\). By step 1, \(\gamma g\in G\), so step 3 gives \(\gamma g(2i)=2i\) and step 4 gives \(\gamma g=\pm I\). Hence the transformation \(g\) equals \(\gamma^{-1}\), a finite composition of \(S\), \(T_2\), and their inverses.