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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.

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.

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.

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.

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.

Use the upper-half-plane model of the hyperbolic plane, whose geodesics are vertical lines and semicircles perpendicular to the real boundary. Hyperbolic and Euclidean angles agree. An isometry takes one of the ultraparallel hyperbolic lines to the imaginary axis. The other then has two finite endpoints on the same side; reflecting if necessary, write them as $0<a<b$. Its Euclidean centre and radius are $c=(a+b)/2$ and $r=(b-a)/2$.

A geodesic perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $q$. Two intersecting Euclidean circles meet orthogonally exactly when their squared centre distance equals the sum of their squared radii. Orthogonality to the second line therefore requiresSince $a<\sqrt{ab}<b$, the two semicircles intersect at one point in the upper half-plane. This constructs the common perpendicular of ultraparallel hyperbolic lines and proves uniqueness, because its radius is forced.

For the three-line continuation, the three common perpendiculars can be pairwise disjoint, but need not be. Explicit configurations of three ultraparallel lines make both claims transparent. For the positive example, take the three semicircles of radius $1/2$ centred at $-5/2,0,5/2$. Their endpoint intervals are disjoint. Their common perpendiculars are the semicircles with centres $-5/4,5/4,0$ and radii $\sqrt{21}/4,\sqrt{21}/4,\sqrt{6}$, respectively. The two smaller semicircles are disjoint because $2(\sqrt{21}/4)<5/2$; each is nested inside the largest because $5/4+\sqrt{21}/4<\sqrt{6}$. Hence their full geodesics are disjoint. For the negative example, take three concentric semicircles of radii $1,2,3$. They are pairwise ultraparallel, but every common perpendicular is the imaginary axis, so the perpendiculars coincide.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.

Work initially with $\mathrm{Re}s>1$, where the nonholomorphic Eisenstein series and its differentiated series converge locally absolutely. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$, the identitiesshow that each summand at $\gamma \tau$ becomes the summand indexed by $(ma+nc,mb+nd)$ at $\tau$. Reindexing the nonzero integer pairs therefore provesHere $y^s$ and the denominator powers use logarithms of positive real numbers, so there is no ambiguity for complex $s$.

For the nonnegative Laplace-Beltrami operator of the hyperbolic plane, $\Delta y^s=-y^2{\partial }_y^2{y}^s=s(1-s)y^s$. If $(m,n)=r(c,d)$ with $(c,d)$ primitive and $r>0$, choose a matrix in $S{L}_2(\mathbb{Z})$ with bottom row $(c,d)$. The corresponding summand isThe Laplace-Beltrami operator commutes with hyperbolic isometries, so this summand has eigenvalue $s(1-s)$. This also covers $m=0$, or follows there directly. Termwise differentiation now givesThe equation describes an automorphic eigenfunction; it does not assert that this noncuspidal function belongs to the square-integrable discrete spectrum.