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Two horocycles have a unique common orthogonal hyperbolic line when their ideal centres of horocycles differ: it is the line with those two ideal endpoints. If the centres agree, an isometry puts both horocycles on horizontal lines; every vertical hyperbolic line meets both orthogonally, so there are infinitely many. This classification remains valid for intersecting or tangent horocycles.

The ideal centre of a horocycle is its point of tangency to the boundary at infinity; a horizontal horocycle in the upper half-plane model has centre infinity. Sending the centre to infinity by an isometry makes the horocycle horizontal. Orthogonal hyperbolic lines then become vertical, proving that a hyperbolic line meets a horocycle orthogonally exactly when it has the centre as an ideal endpoint.

Use curvature $-1$. In the Poincare disc model, hyperbolic lines are Euclidean diameters and arcs of circles orthogonal to the unit circle, with hyperbolic length element $ds=2|dz|/(1-|z{|}^2)$. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis, with $ds=|dz|/\mathrm{Im}z$. Both displayed hyperbolic metrics are positive scalar multiples of the Euclidean metric, so their angles agree with Euclidean angles. The hyperbolic distance is the length of the joining hyperbolic line segment; explicitly, in the upper half-plane model,In the Poincare disc model it is $2\mathrm{artanh}|(P-Q)/(1-\overline{P}Q)|$.

For distinct $P,Q$, a isometry sends their joining hyperbolic line to the imaginary axis, so their images are $ip,iq$ with $p,q>0$. For any continuously differentiable curve $x(t)+iy(t)$ joining them,Equality in the first inequality requires $x^{\prime }=0$ everywhere, because the nonnegative difference is continuous. Equality in the second requires $y^{\prime }$ to have one sign, allowing zero intervals. Thus equality holds precisely for a monotone reparametrisation of the joining hyperbolic segment. Conversely every such reparametrisation gives equality. For $P=Q$, equality means length zero and the constant curve; monotonicity is understood non-strictly.

Two distinct hyperbolic lines are parallel hyperbolic lines if they are disjoint in the plane and have exactly one common ideal endpoint; they are ultraparallel hyperbolic lines if they are disjoint and have no common ideal endpoint. To prove the common perpendicular of ultraparallel hyperbolic lines theorem, send one line to the imaginary axis. An ultraparallel hyperbolic lines second line can, after reflection if necessary, be written as a semicircle with centre $c>0$ and radius $r$ satisfying $c>r$. A hyperbolic line perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $R>0$. The Euclidean condition for its orthogonality to the second circle isThis has exactly one positive solution, proving existence and uniqueness. Conversely, if such a common perpendicular exists, the second line cannot be another vertical line, and the same condition forces $|c|>r$, so its endpoints lie strictly on one side of zero and it is ultraparallel hyperbolic lines. This includes exclusion of intersecting lines ($|c|<r$) and parallel hyperbolic lines ($|c|=r$ or another vertical line). The statement concerns distinct lines: a line coincident with itself would have many perpendiculars.

A horocycle in the upper half-plane model is either a Euclidean circle tangent to the real axis from above, with the tangent point as its ideal centre of a horocycle, or a horizontal line $y=h>0$, whose ideal centre of a horocycle is infinity. A isometry sending this centre to infinity sends the horocycle to a horizontal line. The hyperbolic lines meeting that horizontal line orthogonally are exactly the vertical lines; therefore the hyperbolic lines orthogonal to a horocycle are exactly those with its ideal centre as an endpoint.

If two horocycles have distinct horocycle centres, the unique hyperbolic line with those two ideal endpoints meets both orthogonally. If they have the same horocycle centre, send it to infinity: both become horizontal lines and every vertical hyperbolic line meets both orthogonally. HenceIn the other case there are infinitely many; intersection or tangency of the two horocycles does not change this classification.

Interpret the printed congruence entrywise in the integer lattice: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 $\pm 1$. Here the trace is even, excluding $\pm 1$; trace zero would give $d=-a$ and $1=-a^2-bc\equiv -1(\mathrm{mod}4)$, 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 towhich 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 $e^{\pi iz}$ 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