= Solution
Interpret the printed congruence entrywise in the integer lattice:
$$
a,d\in1+2\mathbb Z,\qquad b,c\in2\mathbb Z.
$$
It then defines the usual level-two <principal congruence subgroup> of $\mathrm{SL}_2(\mathbb Z)$, despite the PDF's ambient $\mathrm{SL}_2(\mathbb R)$. A literal ideal congruence inside $\mathbb R$ would be vacuous because $2\mathbb R=\mathbb R$, and would make the asserted conclusion false. The integer-lattice interpretation is essential.
Pass to $\overline{\Gamma(2)}=\Gamma(2)/\{\pm I\}$, which has exactly the same action. Reduction modulo two maps the <modular group> onto $\mathrm{SL}_2(\mathbb F_2)$, a group of order six: the reductions of $S$ and $T$ generate it. Its kernel is $\overline{\Gamma(2)}$, so the <index of a subgroup> is six. The <standard fundamental domain of the modular group> has <hyperbolic area> $\pi/3$, and hence the quotient has <hyperbolic area> $2\pi$.
There are no nonidentity <elliptic Möbius transformations> in $\overline{\Gamma(2)}$. An integral matrix representing an <elliptic Möbius transformation> has <trace> $0$ or $\pm1$. Here the <trace> is even, excluding $\pm1$; <trace> zero would give $d=-a$ and $1=-a^2-bc\equiv-1\pmod4$, impossible. Thus the effective action is a free <properly discontinuous group action>, and the quotient is a <Riemann surface>.
A <cusp of a modular group> is represented by a rational boundary point. Their orbits correspond to
$$
\mathrm{SL}_2(\mathbb F_2)/\langle\overline T\rangle,
$$
which has $6/2=3$ elements. They are represented by $\infty,0,1$, or by the three nonzero parity vectors of a primitive numerator-denominator pair. Each <width of a cusp> is two. A union of six copies of the <standard fundamental domain of the modular group> gives a fundamental region for this subgroup. Removing small <horocycle> neighbourhoods of its cusps leaves a <compact> core. Adding one point at each <cusp of a modular group>, using the local parameter $e^{\pi i z}$ after moving that cusp to infinity, gives a <compact> <Riemann surface> $\overline X$.
For a finite-area <hyperbolic surface> of <genus> $g$ with $r$ cusps, the <Gauss-Bonnet theorem> gives area $2\pi(2g-2+r)$. Thus $2\pi=2\pi(2g-2+3)$ and $g=0$. A <compact> <genus>-zero <Riemann surface> is the <Riemann sphere>, by the <uniformization theorem>. A <Möbius transformation> sends the three added points to $0,1,\infty$. Restricting it gives
$$
\boxed{\mathbb H/\Gamma(2)\cong\mathbb P^1\setminus\{0,1,\infty\}
=\mathbb C\setminus\{0,1\}.}
$$
Back to article page
