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In the curvature-minus-one Poincaré half-plane model, a geodesic polygon with $n$ sides and interior angles ${\alpha }_1,\dots ,{\alpha }_n$ has areaThis is the polygonal Gauss-Bonnet theorem.

The Cayley transform gives a conformal bijection between an unbounded half-plane and the unit disc. Two conventions are useful. For the right half-plane, useFor the upper half-plane, useRotating the upper half-plane onto the right half-plane by $z\mapsto -iz$ explains the relationship $C(z)=R(-iz)$. The factors of $i$ distinguish which half-plane is mapped onto the unit disc. ^[ref]

In particular, $C(i)=0$, $C(2i)=1/3$ and $C(1+i)=(1-2i)/5$. The whole positive imaginary axis becomes the real diameter: $C(iy)=(y-1)/(y+1)$ for $y>0$. Horizontal grid lines become circles tangent to the unit circle at $1$ when completed by their endpoint at infinity; they are not hyperbolic geodesics.

On the Riemann sphere, $C$ maps the extended real line onto the unit circle:Every finite real $x$ has $|C(x)|=1$; the remaining point $w=1$ corresponds to infinity. The pole $z=-i$ is outside the upper half-plane, and the pole of $C^{-1}$ at $w=1$ is on the boundary of the unit disc. Neither pole obstructs the interior conformal bijection.

For a short verification, subtract the squared moduli:These identities put each interior image in the required domain; solving $w(z+i)=z-i$ gives the stated inverse function. Both maps are holomorphic, with $C^{\prime }(z)=2i/(z+i{)}^2\ne 0$, so the conformal bijection preserves Euclidean angles between tangent directions.

Euclidean distance is not preserved: $i$ and $2i$ are one unit apart, whereas $0$ and $1/3$ are only one third of a unit apart. The striking feature is that $C$ does preserve length for the hyperbolic metrics of the Poincaré half-plane model and Poincaré disk model:Consequently it is an isometry for these hyperbolic metrics, including their induced hyperbolic distances. For the same example, the corresponding hyperbolic geodesic segments have equal length:Thus the Cayley transform lets us choose the most convenient picture: a straight boundary for calculations, or a bounded unit disc for visualization. It also transports automorphisms of the unit disk into upper half-plane automorphisms by function composition $C^{-1}\circ g\circ C$.

The composition of hyperbolic reflections in two ultraparallel hyperbolic lines has infinite order. Map their common perpendicular to the imaginary axis in the Poincaré half-plane model. The two reflecting hyperbolic lines become semicircles centred at zero, with distinct radii $r_1,r_2$. Reflection in the radius-$r$ semicircle is $z\mapsto r^2/\overline{z}$, so their composition is $z\mapsto (r_2/r_1{)}^{2}z$. Every positive power remains nonidentity, because $(r_2/r_1{)}^{2k}\ne 1$ for $k\ge 1$.

Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

A real Möbius transformation represented by $A\in {\mathrm{SL}}_2(\mathbb{R})$ with $|\mathrm{tr}A|<2$. It fixes a point of the Poincaré half-plane model; an integral matrix of this type has trace $0$ or $\pm 1$. In the Poincare disc model, moving its fixed point to zero makes it a rotation about zero.

For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

The metric area of a hyperbolic surface. In the Poincaré half-plane model the density is $dxdy/y^2$. A finite-area torsion-free quotient of genus $g$ with $r$ cusps has area $2\pi (2g-2+r)$ by the Gauss-Bonnet theorem.

In the Poincaré half-plane model, a hyperbolic circle of radius $R$ centred at $u+iv$ is the Euclidean circle $(x-u{)}^2+(y-v\cosh R{)}^2=v^2{\sinh }^{2}R$. Its Euclidean centre is shifted upward from its hyperbolic centre, and its Euclidean radius is $v\sinh R$. The lower height $v{e}^{-R}$ is positive. The formula follows by mapping the centre to zero in the Poincare disc model and using radial distance $2\mathrm{artanh}\rho$.

A surface with a complete Riemannian metric of constant Gaussian curvature $-1$. Its local model is the Poincaré half-plane model; the orientable case is a quotient by a free properly discontinuous group action of orientation-preserving isometries.

Every orientation-preserving isometry of the Poincaré half-plane model lies in $PS{L}_2(\mathbb{R})$. Every orientation-reversing isometry is the composition of one of these with the hyperbolic reflection $z\mapsto -\overline{z}$ in the imaginary axis. Equivalently, the full isometry group consists of real Möbius and anti-Möbius transformations, modulo nonzero scalar matrices.

For the refractive index $\mu =1/y$ on the upper half-plane, Fermat principle and the Beltrami identity give $1/[y\sqrt{1+(dy/dx{)}^2}]=1/R$. The nonvertical rays are arcs of circles centered on $y=0$; vertical lines are the remaining rays. These are also geodesics of the Poincaré half-plane model. A ray entering the strip $0<y\le 1$ toward decreasing $y$ at angle $\theta$ to the vertical has limiting horizontal displacement $\tan (\theta /2)$, at most one. The line $y=0$ is approached only after infinite optical path length because the refractive index diverges there.

A nonidentity real Möbius transformation represented by $A\in {\mathrm{SL}}_2(\mathbb{R})$ with $|\mathrm{tr}A|=2$. It has one ideal boundary fixed point and is conjugate to a nonzero translation on the Poincaré half-plane model.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

The Poincaré half-plane model has Riemannian metric $d{s}^2=(d{x}^2+d{y}^2)/y^2=|dz{|}^2/(\mathrm{Im}z{)}^2$. Every Möbius transformation preserving this half-plane can be represented as $f(z)=(Az+B)/(Cz+D)$ with real coefficients and $\Delta =AD-BC>0$. Direct calculation givesTherefore $|df|/\mathrm{Im}f=|dz|/\mathrm{Im}z$, proving it is an isometry.

The imaginary axis is a hyperbolic geodesic. Its distance from $ib$ to $ic$ is $|{\int }_b^{c}dy/y|=|\log (c/b)|$. This is also the minimum over all paths, since path length is at least $|\int dy/y|$. The two vertical points at distance $r$ from $ib$ are $ib{e}^{-r}$ and $ib{e}^r$. Reflection in the imaginary axis preserves the hyperbolic circle and its centre. Since it is assumed to be a Euclidean circle, its Euclidean centre lies on that axis and its two vertical intersections determine the Euclidean centre and radius:This proves the hyperbolic circle in the upper half-plane formula for the specified centre.

For the horizontal-distance calculation, the geodesic joining $ib$ to $a+ib$ is the semicircle of centre $a/2$ and radius $R=\sqrt{b^2+a^2/4}$. Parameterize it as $z=a/2+R{e}^{i\theta }$. Its metric length is $|d\theta |/\sin \theta$. If the right endpoint angle is ${\theta }_0$, then $\sin {\theta }_0=b/R$, $\cos {\theta }_0=a/(2R)$, and the other endpoint angle is $\pi -{\theta }_0$. HenceThe equality of horizontal hyperbolic and Euclidean distances therefore requiresLet $t=a/2>0$. The derivative of $t/\sinh t$ has the sign of $\sinh t-t\cosh t$. This quantity vanishes at zero and has derivative $-t\sinh t<0$, so the height function is strictly decreasing. Its limits are one at $a\downarrow 0$ and zero at $a\to \infty$. Continuity and the intermediate value theorem prove that each $0<b<1$ has exactly one positive matching separation $a$.

Use the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ with metric $d{s}^2=(d{x}^2+d{y}^2)/y^2$. Its hyperbolic lines are vertical lines and semicircles centred on the real boundary, hence orthogonal to that boundary. Each has two ideal endpoints in $\mathbb{R}\cup \{\infty \}$. Distinct parallel hyperbolic lines are disjoint inside $\mathbb{H}$ and share one ideal endpoint; distinct ultraparallel hyperbolic lines are disjoint and share none. Intersecting lines are neither. Because the metric is conformal to the Euclidean metric, the two notions of angle agree.

Apply a hyperbolic isometry taking one line $l$ to the imaginary axis, whose ideal endpoints are $0,\infty$. If the other line $l^{\prime }$ is ultraparallel, its endpoints are two finite nonzero real numbers of the same sign. Reflect if necessary to write them as $0<a<b$. Then $l^{\prime }$ is the semicircle with centre $c=(a+b)/2$ and radius $r=(b-a)/2$.

Every hyperbolic line perpendicular to the imaginary axis must be a semicircle centred at zero: at the intersection its tangent has to be horizontal, forcing its real centre to coincide with zero. Let its radius be $R$. Orthogonality to the circle representing $l^{\prime }$ requires $c^2=R^2+r^2$, since the radii to the intersection form a right triangle. ConsequentlyThis has exactly one positive solution. The circles intersect above the real axis: $c-r=a<\sqrt{ab}<b=c+r$. Hence the required common perpendicular of ultraparallel hyperbolic lines exists and is unique.

Conversely, if $l^{\prime }$ intersects the imaginary axis, its endpoints straddle zero and the same equation would require $R^2=ab<0$, which is impossible. If $l^{\prime }$ has endpoint zero, it gives $R^2=0$, not a line. If its shared endpoint is infinity, $l^{\prime }$ is another vertical line; no circle centred at zero meets a distinct vertical line with horizontal tangent. Thus parallel lines have no common perpendicular either. These cases exhaust distinct lines, proving a unique common perpendicular exists exactly for ultraparallel lines.

In the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ has Riemannian metricIts orientation-preserving isometries are the Möbius transformations $z\mapsto (az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full isometry group also contains their compositions with $z\mapsto -\overline{z}$, which reverse orientation. The hyperbolic lines are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.

A hyperbolic line with finite endpoints $p<q$ is mapped to the imaginary axis byIts coefficient determinant is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving isometry. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis hyperbolic line. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the imaginary axis, $T_2^{-1}{T}_1$ sends the first to the second. This proves transitivity on hyperbolic lines.

For two ultraparallel hyperbolic lines, choose their common perpendicular and map it to the imaginary axis. A hyperbolic line perpendicular to that axis is a semicircle centred at zero: orthogonality at $iy$ forces the circle centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The hyperbolic reflection in the radius-$r$ semicircle is $S_r(z)=r^2/\overline{z}$; it fixes that semicircle pointwise and reverses orientation. Their composition isIts $k$th power is $z\mapsto {\lambda }^{k}z$, and no positive power is the identity since $\lambda >0$ and $\lambda \ne 1$. Conjugation by the chosen isometry preserves order of a group element, so the original composition has infinite order. Reversing the order of the reflections replaces $\lambda$ by ${\lambda }^{-1}$ and gives the same conclusion.

First obtain the hyperbolic circle in the upper half-plane. For centre $w=u+iv$, the fractional-linear map $\zeta =(z-w)/(z-\overline{w})$ takes the Poincaré half-plane model isometrically to the Poincare disc model, sending $w$ to zero. Direct differentiation transforms the metric to $4|d\zeta {|}^2/(1-|\zeta {|}^2{)}^2$. A radial segment from zero to modulus $\rho$ has length $2\mathrm{artanh}\rho$. Any path to that modulus has at least this length, by $|d\zeta |\ge |d\rho |$; thus it is the hyperbolic distance.

A circle of hyperbolic radius $R$ therefore satisfiesSquaring and rearranging givesHence it is a circle in Euclidean geometry, with Euclidean centre $u+iv\cosh R$ and radius $v\sinh R$. Its lowest point has height $v{e}^{-R}>0$, so it stays entirely in the upper half-plane. The centres differ whenever $R>0$: for example, the hyperbolic circle centred at $i$ has Euclidean centre $i\cosh R$, rather than $i$.

For hyperbolic side-side-side congruence, use the standard fact that hyperbolic isometries act transitively on ordered point pairs with the same positive separation, and that reflection in a hyperbolic line is an isometry. Map $A,B$ to $A^{\prime },B^{\prime }$. The image of $C$ has prescribed distances $b$ from $A^{\prime }$ and $a$ from $B^{\prime }$, just as $C^{\prime }$ does. There are at most two such points, interchanged by reflection in the line $A^{\prime }{B}^{\prime }$.

For completeness, normalize $A^{\prime },B^{\prime }$ to points on the positive imaginary axis. Their distance circles are Euclidean circles with centres on that axis by the formula above. Subtracting their equations determines one height $y$, and then the abscissa is determined up to its sign. Reflection $z\mapsto -\overline{z}$ interchanges these choices. Therefore either the chosen isometry or its composition with that reflection maps all three labelled vertices correctly. Equal three side lengths imply a hyperbolic isometry between the triangles.

Two sides and an unspecified equal angle do not suffice. Here is an explicit hyperbolic side-side-angle ambiguity. Fix the angle at $A$ to be $\pi /6$, let $\cosh a=3/2$ and $\cosh b=2$, and chooseBoth are positive and distinct. The hyperbolic cosine rule at $A$ becomesSetting $u=e^c$, the condition is $u^2-6u+7=0$, whose two roots are $3\pm \sqrt{2}$. Construct each triangle from sides $b,c_{\pm }$ meeting at angle $\pi /6$; its opposite side is the same $a$ by the cosine rule. The triangles therefore share $a,b,A$ but have different third sides and cannot be isometric. If the common angle is specifically the included angle between the two given sides, the cosine rule determines the third side and side-side-side congruence does apply.

Use the Gaussian curvature $-1$ normalizationof the Poincaré half-plane model. This is a complete conformal metric, invariant under real Möbius transformations. Its quotient by the free effective principal congruence subgroup is again complete: a geodesic lifts to the complete covering plane and extends there for all time. Transporting the quotient metric through the biholomorphism supplies the required complete conformal metric.

A nonconstant closed geodesic corresponds to a hyperbolic Möbius transformation $A$ of $\Gamma (2)$, and its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Indeed its eigenvalues have magnitudes $\lambda ,{\lambda }^{-1}$ with $\lambda >1$, and its axis quotient has length $2\log \lambda$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $\Gamma (2)$, $bc\equiv 0(\mathrm{mod}4)$ implies $ad\equiv 1(\mathrm{mod}4)$. The odd numbers $a,d$ therefore have the same residue modulo four, andA hyperbolic Möbius transformation must have $|\mathrm{tr}A|>2$, so its smallest possible absolute trace is six. It is attained byConsequentlyThis element is primitive: a proper power would have a shorter root represented by a hyperbolic Möbius transformation, contradicting the trace bound. No essential simple closed geodesic exists on the three-punctured sphere, since every essential simple loop is peripheral and corresponds to a parabolic Möbius transformation. The minimizing closed geodesic is therefore nonsimple. Finally, the PDF leaves the numerical Gaussian curvature unspecified: the displayed length uses Gaussian curvature $-1$; for Gaussian curvature $-{\kappa }^2$, it is divided by $\kappa$.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincaré half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

The hyperbolic lines $AB$ and $CD$ are both perpendicular to $BC$. They are therefore ultraparallel hyperbolic lines, and $BC$ is their unique common perpendicular of ultraparallel hyperbolic lines.

The common perpendicular realizes the minimum hyperbolic distance between two ultraparallel lines. This follows directly in the Poincaré half-plane model after a Möbius transformation sends the two lines to concentric semicircles: the supplied distance formula is minimized along their common radial geodesic. The segment $AD$ also joins the two lines, but it is different from $BC$ in a nondegenerate quadrilateral. Hence