- Let \(\operatorname { S L _ { 2 } } ( \mathbb { Z } )\) denote the set of \(2 \times 2\) matrices with integer entries and determinant 1, that is,
\[ \operatorname{SL} _ {2} (\mathbb {Z}) = \left\{g = \left( \begin{array}{c c} a & b \\ c & d \end{array} \right): a, b, c, d \in \mathbb {Z} \text { and } a d - b c = 1 \right\}. \]
This group acts on the upper half-plane by the fractional linear transformation \(g ( \tau ) = ( a \tau + b ) / ( c \tau + d )\) . Together with this action comes the so-called fundamental domain \({ \mathcal { F } } _ { 1 }\) in the complex plane defined by
\[ \mathcal {F} _ {1} = \{\tau \in \mathbb {C}: | \tau | \geq 1, | \operatorname{Re} (\tau) | \leq 1 / 2 \text { and } \operatorname{Im} (\tau) > 0 \}. \]
It is illustrated in Figure 3.
Consider the two elements in \(\operatorname { S L _ { 2 } } ( \mathbb { Z } )\) defined by \(S ( \tau ) = - 1 / \tau\) and \(T _ { 1 } ( \tau ) = \tau + 1\) These correspond (for example) to the matrices
\[ \left( \begin{array}{c c} 0 & - 1 \\ 1 & 0 \end{array} \right) \quad \text {and} \quad \left( \begin{array}{c c} 1 & 1 \\ 0 & 1 \end{array} \right), \]
respectively. Let g be the subgroup of \(\operatorname { S L _ { 2 } } ( \mathbb { Z } )\) generated by S and \(T _ { 1 }\) .
Show that for every \(\tau \in\) H there exists \(g \in { \mathfrak { g } }\) such that \(g ( \tau ) \in \mathcal { F } _ { 1 }\)
We say that two points \(\tau\) and \(\tau ^ { \prime }\) are congruent if there exists \(g \in \operatorname{SL}_2(\mathbb Z)\) such that \(g(\tau)=\tau^\prime\). Prove that if \(\tau,\tau^\prime \in \mathcal F_1\) are distinct and congruent, then either \(\operatorname{Re}(\tau)=\pm1/2\) and \(\tau^\prime=\tau\mp1\), or \(|\tau|=1\) and \(\tau^\prime=-1/\tau\). [Hint: Say \(\tau ^ { \prime } = g ( \tau )\) . Why can one assume that Im \(( \tau ^ { \prime } ) \geq \operatorname { I m } ( \tau )\) , and therefore \(| c \tau + d | \le 1 ?\) Now consider separately the possibilities \(c = - 1\) \(c = 0\) , or \(c = 1 .\) ]
Prove that S and \(T _ { 1 }\) generate the modular group in the sense that every fractional linear transformation corresponding to \(g \in \mathrm { S L } _ { 2 } ( \mathbb { Z } )\) is a composition of finitely many \(S \mathrm { { s } }\) and \(T _ { \mathrm { 1 } } \mathrm { ^ { , } s }\) , and their inverses. Strictly speaking, the matrices associated to \(S\) and \(T _ { 1 }\) generate the projective special linear group \(\mathrm { P S L _ { 2 } ( Z ) }\) , which equals \(\operatorname{SL}_2(\mathbb Z)\) modulo \(\{\pm I\}\). [Hint: Observe that 2i is in the interior of \(\mathcal { F } _ { 1 }\) . Now map \(g ( 2 i )\) back into \(\mathcal { F } _ { 1 }\) by using part (a). Use part (b) to conclude.]
Solution
Let \(\Gamma=\langle S,T_1\rangle\), where \[ S(\tau)=-\frac1\tau, \qquad T_1(\tau)=\tau+1. \] For \[ g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}_2(\mathbb Z), \] a direct calculation gives \[ \Im(g\tau)=\frac{\Im\tau}{|c\tau+d|^2}. \tag{1} \]
For part (a), start with \(z=\tau\in\mathbb H\). Translate by an integral power of \(T_1\) so that \[ |\Re z|\le\frac12. \] If \(|z|\ge1\), then \(z\in\mathcal F_1\). If \(|z|<1\), apply \(S\). Since \[ \Im(Sz)=\frac{\Im z}{|z|^2}>\Im z, \] this strictly increases the imaginary part; then translate again into the strip \(|\Re z|\le1/2\), and repeat.
The process must terminate. Indeed every intermediate point has the form \[ z=\gamma\tau, \qquad \gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma, \] and its imaginary part never decreases. Thus by (1), for every stage after the first, \[ |c\tau+d|^2\le1. \] There are only finitely many integer pairs \((c,d)\) satisfying this inequality: if \(\tau=x+iy\), then \(|c|y\le1\), so only finitely many \(c\) occur, and then \(|cx+d|\le1\) leaves only finitely many \(d\). Since every application of \(S\) strictly increases the imaginary part, infinitely many such steps are impossible. Hence the algorithm stops at a point of \(\mathcal F_1\).
For part (b), suppose \(\tau,\tau'\in\mathcal F_1\) and \(\tau'=g\tau\). Since congruence is symmetric, interchange \(\tau\) and \(\tau'\) if necessary so that \[ \Im\tau'\ge\Im\tau. \] Equation (1) then gives \[ |c\tau+d|\le1. \tag{2} \] Write \(\tau=x+iy\). From \(|x|\le1/2\) and \(|\tau|\ge1\), \[ y\ge\frac{\sqrt3}{2}. \] Thus (2) implies \(|c|\le1\).
If \(c=0\), the determinant condition gives \(a=d=\pm1\), so \[ \tau'=\tau+n \] for some \(n\in\mathbb Z\). Since both real parts lie in \([-1/2,1/2]\), either \(n=0\), giving \(\tau'=\tau\), or the points lie on the vertical boundary and \[ \Re\tau=\frac12,\quad \tau'=\tau-1, \] or \[ \Re\tau=-\frac12,\quad \tau'=\tau+1. \]
Now suppose \(|c|=1\). Since \(g\) and \(-g\) act identically, take \(c=1\); then \(ad-b=1\), and \[ \tau'=\frac{a\tau+b}{\tau+d}=a-\frac{1}{\tau+d}, \] while (2) says \(|\tau+d|\le1\). If \(|d|\ge2\), then \(|\tau+d|\ge|d|-|x|\ge\frac32\), which is impossible. If \(d=\pm1\), then \[ |\tau\pm1|^2=|\tau|^2\pm2x+1\ge2\pm2x\ge1, \] with equality only for \(|\tau|=1\) and \(x=\mp\frac12\). If \(d=0\), then \(|\tau|\le1\), so \(|\tau|=1\). In every case \(|\tau|=1\), so \(1/\tau=\bar\tau\).
If \(d=0\), then \(\tau'=a-\bar\tau=a-x+iy\). This lies in \(\mathcal F_1\) only if \(|a-x|\le\frac12\): either \(a=0\) and \(\tau'=-1/\tau\), or \(x=\pm\frac12\), \(a=\pm1\), and \(\tau'=\tau\).
If \(d=1\), then \(\tau=e^{2\pi i/3}\), \(\tau+1=e^{i\pi/3}\), and \(\tau'=a-e^{-i\pi/3}=a-\frac12+\frac{\sqrt3}2i\). This lies in \(\mathcal F_1\) only for \(a=0\), where \(\tau'=\tau\), or \(a=1\), where \(\tau'=\tau+1=-1/\tau\). The case \(d=-1\) is the reflection \(\tau\mapsto-\bar\tau\) of this one: \(\tau=e^{i\pi/3}\) and \(\tau'\in\{\tau,\tau-1\}\), with \(\tau-1=-1/\tau\).
Thus a point \(\tau'\ne\tau\) with \(|c|=1\) has \(|\tau|=1\) and \(\tau'=-1/\tau\), and distinct congruent points of \(\mathcal F_1\) occur exactly as stated.
For part (c), let \(G\in\operatorname{SL}_2(\mathbb Z)\). By part (a), there is \(h\in\Gamma\) such that \[ hG(2i)\in\mathcal F_1. \] But \(2i\) is an interior point of \(\mathcal F_1\), and \(hG(2i)\) is congruent to \(2i\). Part (b) therefore forces \[ hG(2i)=2i. \] If \[ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}_2(\mathbb Z) \] fixes \(2i\), then \[ \frac{2ia+b}{2ic+d}=2i. \] Comparing real and imaginary parts gives \[ b=-4c, \qquad a=d. \] The determinant condition becomes \[ a^2+4c^2=1, \] so \(c=0\), \(a=d=\pm1\), and \(b=0\). Hence \(A=\pm I\). Therefore \[ hG=\pm I, \] so in \(\operatorname{PSL}_2(\mathbb Z)\), \[ G=h^{-1}\in\Gamma. \] Thus \(S\) and \(T_1\) generate \(\operatorname{PSL}_2(\mathbb Z)=\operatorname{SL}_2(\mathbb Z)/\{\pm I\}\), equivalently every associated fractional linear transformation is a finite composition of \(S,T_1\), and their inverses.