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In the curvature-minus-one Poincare 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.

A real Möbius transformation represented by $A\in {\mathrm{SL}}_2(\mathbb{R})$ with $|\mathrm{tr}A|<2$. It fixes a point of the Poincare 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.

The metric area of a hyperbolic surface. In the Poincare 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 Poincare 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 Poincare 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 Poincare 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.

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 Poincare half-plane model.

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 Poincare 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 Poincare 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 Poincare 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 Poincare 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

The groupacts on the Poincare half-plane model byA direct calculation givesThereforeso every element of PSL2(R) is an isometry of the hyperbolic metric.

For a continuously differentiable curve $\gamma (t)=x(t)+iy(t)$ in the Poincare half-plane model, its hyperbolic length isThe hyperbolic lines are the geodesics: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.

The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane Möbius transformation to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $i{y}_0$ and $i{y}_1$. Every joining curve satisfiesEquality holds for the monotone vertical parametrizationThus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho (z,w)$.

Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the formThis is the normal form underlying the common perpendicular of ultraparallel hyperbolic lines. Put $z=e^{u+i\theta }$. Sincethe metric becomesEvery path from $l$ to $m$ therefore hasEquality requires $\theta =\pi /2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance

For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.

The Poincare half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

Every orientation-preserving isometry of the Poincare half-plane model has the formwith matrices identified up to multiplication by $-I$. Thus $G\cong PSL(2,\mathbb{R})$.

For $z=x+iy\in \mathbb{H}$, the affine map $(z-x)/y$ sends $z$ to $i$, proving transitivity on points. Real translations and positive dilations act transitively on finite boundary points, while inversion $z\mapsto -1/z$ exchanges $0$ and $\infty$; hence the action on $\mathbb{R}\cup \{\infty \}$ is transitive. A hyperbolic line is determined by its unordered pair of boundary endpoints, and a real Möbius transformation can send any such pair to $\{0,\infty \}$. Therefore $G$ is also transitive on hyperbolic lines.

In the Poincare half-plane model, the Riemannian metricassigns to a smooth curve $\gamma (t)=(x(t),y(t))$ the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is

The two geodesics from $e^{i\alpha }$ and $e^{i\beta }$ to infinity are the vertical lines $x=\cos \alpha$ and $x=\cos \beta$, while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are $\alpha$, $\pi -\beta$, and zero at the ideal vertex, and henceTriangulating a geodesic polygon with $n$ sides into $n-2$ triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.