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The Beltrami-Klein model realizes the hyperbolic plane inside an open Euclidean disk, with geodesics given by straight chords. Its metric differs from the Euclidean metric, and angles are generally distorted. The straight chords make geodesic convexity especially easy to see.

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.

For a group of isometries acting properly discontinuously on the hyperbolic plane, and a point $p$ with trivial point stabilizer, the closed Dirichlet domain isIt is the intersection of the closed half-planes bounded by perpendicular bisectors between $p$ and its other group orbit points. The group orbit is locally finite and the hyperbolic plane is a proper metric space, so every point has a nearest group orbit point. This shows that the translates of $D_G(p)$ cover the plane. Distinct interiors are disjoint, since being strictly closer to two different group orbit points is impossible. Boundary ties are allowed in the closed domain. Thus it is a geometric fundamental region.

A finite group of orientation-preserving isometries of the Euclidean plane or hyperbolic plane is a cyclic group. Take one orbit and its unique minimum-radius enclosing disc. Every group element preserves the orbit and therefore the disc's centre. The orientation-preserving stabilizer of a point acts as planar rotations, so the group embeds in the circle group. In a nontrivial finite subgroup, the least positive rotation angle generates all angles by division with remainder, proving cyclicity. Orientation reversal changes the conclusion: reflection groups need not be cyclic.

A subset of a Riemannian manifold is geodesically convex when any two of its points can be joined by a minimizing geodesic segment lying in the subset. In the hyperbolic plane the joining geodesic segment is unique. Thus in the Beltrami-Klein model, geodesically convex subsets correspond to Euclidean convex sets.

A geodesic segment is the restriction of a geodesic to a compact parameter interval. In the hyperbolic plane, two points determine a unique geodesic segment, which minimizes the hyperbolic distance between them.

A six-sided polygon in the hyperbolic plane whose sides are geodesic segments. The especially useful right-angled case is a right-angled hyperbolic hexagon.

For a marked closed hyperbolic surface, the representation of its fundamental group whose image acts as the deck group on the hyperbolic plane. It is discrete and faithful, determined up to conjugation by the marking and the metric, and gives $X={\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$. For a closed orientable surface it admits determinant-one lifts. It is different from the tangent-space action of the Riemannian holonomy group.

A hyperbolic metric is a complete Riemannian metric of constant sectional curvature $-1$. On a surface it supplies local charts modeled on the hyperbolic plane.

A geodesic polygon in the hyperbolic plane is bounded by finitely many geodesic segments. For curvature $-1$, its area is its angle defect, as given by hyperbolic polygon area. Pairing equally long edges produces hyperbolic surfaces, with smoothness at glued vertices determined by their total angles.

Three corresponding side lengths determine a hyperbolic triangle up to an isometry of the hyperbolic plane. First match an ordered pair of vertices by an isometry. The third vertex lies at the intersection of two prescribed hyperbolic distance circles. There are at most two choices, interchanged by reflection in the geodesic through the matched pair; either choice can therefore be matched.

An ideal endpoint of a hyperbolic line is one of its two limiting points on the boundary at infinity. In the Poincare disc model these lie on the unit circle; in the upper half-plane model they lie on $\mathbb{R}\cup \{\infty \}$. These are boundary points, not points of the hyperbolic plane. Any two distinct ideal endpoints determine a unique hyperbolic line.

An ideal hyperbolic quadrilateral is the region bounded by four hyperbolic geodesics whose four consecutive endpoints are distinct points on the ideal boundary. The ideal vertices lie outside the hyperbolic plane. In the hyperbolic disk, the four vertices $1,i,-1,-i$ give a regular example: its sides are the interior arcs of the circles of radius one centered at $\pm 1\pm i$. Although the sides have infinite length, the quadrilateral has finite hyperbolic area $2\pi$ in curvature $-1$.

An ideal hyperbolic triangle has all three vertices on the boundary at infinity of the hyperbolic plane. All its angles are zero and its hyperbolic area in curvature $-1$ is $\pi$, by hyperbolic triangle area.

An isometry between oriented Riemannian manifolds that reverses orientation. Its differential is orthogonal with determinant minus one in oriented orthonormal frames, so it reverses the oriented volume form while preserving the associated positive volume measure. Reflection in a geodesic of the hyperbolic plane is an example. An orientation-reversing isometry still identifies the scalar Laplace spectra, even though it does not preserve the chosen complex orientation of a Riemann surface.

A nonidentity orientation-preserving isometry of the hyperbolic plane is parabolic when it has one fixed point on the ideal boundary and none in the plane. In the upper half-plane model, every parabolic isometry is conjugate to a nonzero horizontal translation.

This is one type of isometry of the hyperbolic plane, classified by its fixed points.

Two distinct hyperbolic lines are parallel, in the limiting-parallel convention, when they do not intersect in the hyperbolic plane and have exactly one common ideal endpoint. In the upper half-plane model, distinct vertical lines are parallel, sharing infinity. Disjoint lines with no common ideal endpoint are instead ultraparallel hyperbolic lines; only these have a common perpendicular of ultraparallel hyperbolic lines.

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle $2\pi$. For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle $2\pi$, so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.

The ten internal angles sum to $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

The uniformization theorem says that every simply connected Riemann surface is biholomorphic to the Riemann sphere, the complex plane, or the unit disc. Its consequence for a compact Riemann surface of genus at least two is that the universal cover is the unit disc, equivalently the hyperbolic plane.

To explain the restriction, a spherical universal cover would make the fundamental group finite. Since a finite covering map multiplies Euler characteristic, $2=m(2-2g)$ is impossible for $g\ge 2$. If the universal cover were the complex plane, its deck transformations would be biholomorphisms $z\mapsto az+b$. A nonidentity such map with $a\ne 1$ has a fixed point, impossible for a deck transformation. Thus the deck transformation group would consist of translations. A discrete subgroup of translations with compact quotient space is a rank-two Euclidean lattice, whose quotient space is a torus, again incompatible with $g\ge 2$.

On the unit disc, the Poincare metric on a Riemann surfacehas Gaussian curvature $-1$ and is invariant under every biholomorphism of the unit disc. It therefore descends through the deck transformation group to a hyperbolic metric on $X$. Its local expression is a positive multiple of $|dz{|}^2$ in every holomorphic coordinate chart, so it is compatible with the given complex structure. In fact, this hyperbolic metric is unique within that conformal class.

Let $G$ be the finite group and choose a point $p$. Its orbit $E=\{gp:g\in G\}$ is a nonempty finite set. Every group element permutes $E$ and maps a closed metric disc onto a disc of the same radius. By the permitted uniqueness of the smallest enclosing disc, it fixes that disc and therefore its centre $o$. Thus $G$ lies in the orientation-preserving stabilizer of one point.

In the Euclidean plane, translate $o$ to zero. Every isometry fixing zero is an orthogonal linear map: distance preservation gives preservation of inner products by the polarization identity, and its values on an orthonormal basis then determine the linear map. Orientation preservation gives determinant one, so it is a rotation. Consequently $G$ embeds in the circle group of rotation angles.

For the hyperbolic plane, use this precise standard fact about its isometries: after placing a point at the origin of the Poincare disc model, every orientation-preserving isometry fixing it has the form $z\mapsto e^{i\theta }z$. Equivalently, its point stabilizer is the rotation group, and its action on directions at the point is faithful. This is a general hyperbolic-plane fact allowed by the question. It again identifies $G$ with a finite subgroup of the circle group.

Finally, a nontrivial finite rotation group has a least positive angle $\theta$ in $[0,2\pi )$. Divide any other angle $\varphi$ by $\theta$: the remainder $\varphi -k\theta$ is another group angle in $[0,\theta )$, so it must be zero. Similarly the remainder of $2\pi$ is zero, and $\theta =2\pi /m$ for an integer $m$. The rotation through $\theta$ generates the group. The trivial group is cyclic as well. Therefore both finite orientation-preserving plane isometry groups are cyclic. This proves the finite orientation-preserving plane isometry groups are cyclic result using the same enclosing-disc argument in both geometries.