Determine the orbits of \[ \operatorname{SL}_2(\mathbb Z) \] acting by fractional linear transformations on \[ \mathbb Q\cup\{\infty\}, \] and determine the orbits of the principal congruence subgroup \(\Gamma(2)\) on the same set.
Solution
There is one \(\operatorname{SL}_2(\mathbb Z)\)-orbit and three \(\Gamma(2)\)-orbits: \[ \boxed{ \begin{aligned} \mathcal O_\infty &=\left\{\frac pq:\gcd(p,q)=1,\ p\text{ odd},\ q\text{ even}\right\},\\ \mathcal O_0 &=\left\{\frac pq:\gcd(p,q)=1,\ p\text{ even},\ q\text{ odd}\right\},\\ \mathcal O_1 &=\left\{\frac pq:\gcd(p,q)=1,\ p\text{ odd},\ q\text{ odd}\right\}. \end{aligned} } \] Here \(\infty=1/0\) belongs to \(\mathcal O_\infty\); the representatives of the three orbits are \(\infty,0,1\).
1. The action of \(\operatorname{SL}_2(\mathbb Z)\) on \(\mathbb Q\cup\{\infty\}\) is transitive.
Let \(p/q\in\mathbb Q\) be in lowest terms. By Bezout’s identity there are integers \(r,s\) such that \[ ps-qr=1. \] Then \[ g= \begin{pmatrix} p&r\\ q&s \end{pmatrix} \in\operatorname{SL}_2(\mathbb Z), \] and \[ g\cdot\infty=\frac pq. \] Thus every rational boundary point lies in the orbit of \(\infty\), and so does \(\infty\) itself. Hence there is exactly one orbit.
2. Reduction modulo \(2\) gives three invariants for the \(\Gamma(2)\) action.
Represent a boundary point by a primitive column vector \[ \binom pq, \qquad \gcd(p,q)=1, \] with \(p/q\) as its projective class. Since the vector is primitive, its reduction modulo \(2\) is one of the three nonzero vectors \[ \binom10, \qquad \binom01, \qquad \binom11 \] in \((\mathbb F_2)^2\).
Every matrix in \[ \Gamma(2)=\{g\in\operatorname{SL}_2(\mathbb Z):g\equiv I\pmod2\} \] acts trivially on \((\mathbb F_2)^2\). Therefore the parity type of \((p,q)\) is constant on each \(\Gamma(2)\)-orbit. The three displayed parity classes are thus distinct orbits or unions of orbits.
3. Every primitive pair \((p,q)\) with \(p\) odd and \(q\) even lies in the orbit of \(\infty\).
Choose integers \(b,d\) with \[ pd-qb=1. \] Because \(p\) is odd and \(q\) is even, reduction modulo \(2\) forces \(d\) to be odd. All solutions are \[ (b+kp,d+kq), \qquad k\in\mathbb Z. \] Since \(p\) is odd, choose the parity of \(k\) so that \(b+kp\) is even; the parity of \(d+kq\) remains odd. Renaming this solution \((b,d)\) gives \[ g= \begin{pmatrix} p&b\\ q&d \end{pmatrix} \equiv \begin{pmatrix}1&0\\0&1\end{pmatrix} \pmod2. \] Thus \(g\in\Gamma(2)\) and \[ g\cdot\infty=\frac pq. \]
4. Every primitive pair \((p,q)\) with \(p\) even and \(q\) odd lies in the orbit of \(0\).
Choose integers \(a,c\) with \[ aq-cp=1. \] Modulo \(2\), this forces \(a\) to be odd. Replacing a solution by \[ (a+kp,c+kq) \] preserves the equation; because \(q\) is odd and \(p\) is even, choose \(k\) so that \(c+kq\) is even while \(a+kp\) remains odd. Then \[ g= \begin{pmatrix} a&p\\ c&q \end{pmatrix} \in\Gamma(2), \] and \[ g\cdot0=\frac pq. \]
5. Every primitive pair \((p,q)\) with both \(p\) and \(q\) odd lies in the orbit of \(1\).
Choose integers \(a,c\) with \[ aq-cp=1. \] Modulo \(2\), since \(p\) and \(q\) are both odd, \[ a-c\equiv1\pmod2, \] so \(a\) and \(c\) have opposite parity. Replacing \((a,c)\) by \[ (a+p,c+q) \] flips both parities without changing \(aq-cp\). Hence we may arrange \[ a\text{ odd}, \qquad c\text{ even}. \] Set \[ b=p-a, \qquad d=q-c. \] Then \(b\) is even, \(d\) is odd, and \[ ad-bc =a(q-c)-(p-a)c =aq-cp =1. \] Thus \[ g= \begin{pmatrix} a&b\\ c&d \end{pmatrix} \in\Gamma(2). \] Finally, \[ g\cdot1 =\frac{a+b}{c+d} =\frac pq. \]
6. Hence these are exactly the three \(\Gamma(2)\)-orbits.
By 2 no two different parity classes can lie in the same \(\Gamma(2)\)-orbit. By <1>3–<1>5 each parity class is itself one orbit, represented respectively by \[ \infty=\frac10, \qquad 0=\frac01, \qquad 1=\frac11. \] This proves the classification.