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Write $Y={\mathbb{P}}^1\setminus \{0,1,\infty \}$. One concrete uniformization uses the image in ${\mathrm{PSL}}_2(\mathbb{R})$ of the level-two principal congruence subgroupIt acts on the complex upper half-plane by Möbius transformations. Integer matrix entries make the group discrete, and the action is properly discontinuous. There are no nonidentity elliptic stabilizers: an elliptic element would have integer even trace with absolute value less than two, hence trace zero. But then $d=-a$ gives $bc=-(a^2+1)\equiv 2(\mathrm{mod}4)$, contradicting the evenness of $b,c$. Thus the quotient is an ordinary Riemann surface, with no orbifold points.

A fundamental region is the ideal quadrilateralIts vertical sides are identified by $T(\tau )=\tau +2$, and its semicircular sides by $U(\tau )=\tau /(1-2\tau )$. The ideal vertices are $-1,0,1,\infty$; $-1$ and $1$ represent one puncture, while zero and infinity give two others. Filling these three punctures gives genus zero: the identified quadrilateral has three vertices, two paired edges and one face, so its compactification has Euler characteristic $3-2+1=2$. Its complex compactification is a sphere, and a Möbius transformation puts the punctures at $0,1,\infty$. HenceThis is the thrice-punctured sphere as a modular quotient. Equivalently, the uniformization theorem applied to its universal covering map produces the upper-half-plane model: the sphere cannot cover this noncompact surface, and an entire covering from $\mathbb{C}$ would contradict Little Picard theorem by omitting both zero and one.

The curvature-minus-one metric $d{s}_{\mathbb{H}}=|d\tau |/\mathrm{Im}\tau$ is invariant under every real Möbius transformation preserving $\mathbb{H}$. Indeed, for determinant-one coefficients,Thus it descends to the hyperbolic metric on $Y$. This defines its distance by the infimum of path lengths; a choice of local lift does not affect lengths.

For the Schwarz-Pick lemma, first prove the Schwarz lemma. If $h:\mathbb{D}\to \mathbb{D}$ is a holomorphic function and $h(0)=0$, the function $h(z)/z$ extends as a holomorphic function across zero. On $|z|=r$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ yield $|h(z)|\le |z|$ and $|h^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative makes $h(z)/z$ a constant of modulus one, again by the maximum modulus principle, so $h$ is a rotation.

For a general holomorphic function $F:\mathbb{D}\to \mathbb{D}$, conjugate by the automorphisms of the unit disk $A_a(z)=(z-a)/(1-\overline{a}z)$, putting $a$ and $F(a)$ at zero. Applying the proved Schwarz lemma givesThe disk hyperbolic metric is $2|dz|/(1-|z{|}^2)$, and its distance is twice the inverse hyperbolic tangent of the displayed pseudohyperbolic distance. Both forms of Schwarz-Pick lemma therefore express metric contraction. Equality in the derivative at one point makes $F$ a disk automorphism by the equality case already proved.

For $X=\mathbb{D}\setminus \{0\}$ and a holomorphic function $f:X\to Y$, compose with a universal covering map of $X$ and lift to a universal covering map of $Y$. The simply connected source of the lift is a disk or half-plane. Schwarz-Pick lemma there bounds the lifted metric derivative, and the covering maps are local isometries, soThe second inequality follows by applying the length inequality to every joining path and taking the infimum. The hyperbolic metric on the punctured disk is its own intrinsic metric,obtained from the covering $z=e^{i\tau }$. It is not merely the ordinary disk metric restricted to the punctured disk.

Yes, equality can occur for distinct points. The cyclic subgroup $⟨T⟩$ lies in $\Gamma (2)$. Its quotient is conformally the punctured disk, using $z=e^{\pi i\tau }$, because this identifies precisely $\tau \sim \tau +2$. The subgroup inclusion induces a covering map that is a holomorphic functionIt is a local isometry. Choose a sufficiently small geodesically convex ball around any target point that is evenly covered. Two distinct sufficiently close points in one lift of that ball are joined by the lift of its minimizing target geodesic. The lift has the same length, giving $d_X\le d_Y$ for this pair, while contraction gives the reverse inequality. Thus equality holds.

For context, if the lifted map is not an automorphism, the derivative inequality is strict at every point; integrating along a minimizing geodesic in the complete punctured-disk metric gives a strict distance inequality for every distinct pair. Equality forces the covering-map case, but even a covering map need not preserve all distances globally because shorter paths can pass through other sheets. This is the distinction captured by equality in hyperbolic Schwarz-Pick contraction.

In curvature-minus-one polar coordinates the hyperbolic metric is $d{s}^2=d{r}^2+{\sinh }^{2}rd{\theta }^2$. Thus the area element is $\sinh rdrd\theta$, givingIt is asymptotic to $\pi e^r$.

For a usual locally finite tessellation with compact congruent tiles of nonempty interior, let their diameter be at most $D$ and choose in each a congruently placed interior ball of radius $r_0>0$. These balls have disjoint interiors. The number of tiles meeting a geodesic segment of length one is uniformly bounded: their ball centres lie in a ball of radius $D+1$, and the disjoint $r_0$-balls lie in the larger ball of radius $D+1+r_0$. Comparing areas supplies a constant $K$. A path of length $R$, divided into at most $\lceil R\rceil$ such segments, therefore crosses at most $K(\lceil R\rceil +1)$ tiles. A small generic perturbation avoids vertices if steps are defined by crossing sides.

Consequently the union of tiles at most $n$ steps from the initial tile contains a metric ball of radius $cn-C$ for fixed $c>0,C$. If $A$ is tile area, their number $N(n)$ is at least $\mathrm{Area}B(cn-C)/A$, an exponential lower bound. Conversely any chain of $n$ adjacent tiles lies in $B(D(n+1))$ about a point in the initial tile; the interiors are disjoint, so $AN(n)\le \mathrm{Area}B(D(n+1))$. Thus $N(n)$ is bounded above and below by positive exponential functions of $n$. No precise common growth exponent is claimed.

An explicit example is the regular right-angled pentagon tiling $\{5,4\}$: reflect a regular hyperbolic pentagon with all angles $\pi /2$ in its sides. Four pentagons meet at each vertex, and their interiors tile the plane. Its tiles have area $3\pi -5\pi /2=\pi$ by the hyperbolic polygon area formula. The compactness and ordinary local finiteness assumptions exclude pathological decompositions with degenerate tiles.

Use the curvature $-1$ hyperbolic metric $ds=2|dz|/(1-|z{|}^2)$. For the disk Möbius transformation ${\varphi }_w(z)=(z-w)/(1-\overline{w}z)$, set $r_w(z)=|{\varphi }_w(z)|$. The hyperbolic distance satisfiesThe triangle inequality givesThus convergence of the exponential sum is independent of the base point. Taking the base point zero, it is equivalent to the Blaschke condition ${\sum }_n(1-|z_n|)<\infty$. This condition implies that only finitely many terms lie in any compact subdisk, including finitely many repetitions of any point.

To construct the required holomorphic function, for $a\ne 0$ define the Blaschke factorFor $|z|\le R<1$, writing $a=|a|e^{i\alpha }$ givesConsequently, after the finitely many factors corresponding to small $|a|$ are removed, the sum of $|1-b_a|$ converges uniformly on each compact subdisk. Those factors are uniformly close to one, so their logarithms form an absolutely convergent series there. Their product tail is holomorphic and nonvanishing. Including the finitely many removed factors gives the Blaschke productwith exactly the requested zero multiplicities. Each finite product has modulus at most one in the disk, so the limit does too. Hence $f=B/2$ maps into the open disk and has exactly the prescribed zeros, proving the implication from the summability condition.

Let $f$ be the nonzero bounded holomorphic function in the zero-set condition. Choose $v$ with $f(v)\ne 0$ and compose with a disk Möbius transformation carrying zero to $v$. Call the resulting function $g$ and its zeros $a_n$, counted with multiplicity. Then $g(0)\ne 0$ and $|g|\le 1$.

For a radius $r$ with no zero on its circle, Jensen's formula givesTo see the formula directly, divide $g$ by its finitely many interior zero factors. The logarithm of the modulus of the resulting zero-free holomorphic function is harmonic near the closed disk and satisfies the mean value property. The circle average of $\log |r{e}^{i\theta }-a|$ for $|a|<r$ is $\log r$, obtained by expanding $\log (1-(a/r)e^{-i\theta })$. Subtracting its value at zero gives exactly the displayed zero sum.

Let $r\uparrow 1$ through zero-free circles. Increasing limits of the nonnegative summands giveSince $1-|a_n|\le -\log |a_n|$, the Blaschke condition follows for the transformed sequence. The hyperbolic metric identity and the base-point comparison proved above giveThis proves the reverse implication and completes the equivalence.

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

Let $V={\mathbb{F}}_2^3$. Since ${\mathbb{F}}_2^{\times }=\{1\}$ and the scalar centre is trivial,Choose $U_1$ to stabilize a nonzero vector and $U_2$ to stabilize a two-dimensional subspace. There are seven choices of either object, so both subgroups have index $7$ and order $24$.

Their coset permutation characters agree. The number of nonzero vectors fixed by $g$ is $2^{\dim \ker (g-I)}-1$. A two-dimensional subspace is the kernel of a unique nonzero covector, so its stabilizer action is the dual action $g^{-T}$. Its fixed-point count isusing equality of the matrix ranks of a matrix and its transpose. For a subgroup $H$, counting the representatives of fixed cosets givesThus the equal characters and subgroup orders give equal intersection counts with every conjugacy class. This proves $U_1,U_2$ are Gassmann equivalent.

They are not conjugate. For example, the stabilizer of $e_1$ contains every matrix $\left(\begin{array}{cc}1& b\\ 0& B\end{array}\right)$ with $b\in {\mathbb{F}}_2^{1\times 2}$ and $B\in \mathrm{GL}(2,2)$. An invariant plane containing $e_1$ would give an invariant line in $V/⟨e_1⟩$, impossible for the full $\mathrm{GL}(2,2)$ action. An invariant plane not containing $e_1$ would be a complement to that line, but the arbitrary shears $b$ do not preserve any such complement. Hence a point stabilizer preserves no plane, whereas every conjugate of $U_2$ does. The two subgroups are nonconjugate.

We now use a cone-torus construction of genus-four Sunada surfaces. Let $B$ be a hyperbolic torus with one cone point of angle $2\pi /7$. Its orbifold fundamental group has presentationMap $a,d$ to the supplied generators $A,D$ of $G$. Their commutator has exact order $7$, so this gives a surjection $\Gamma \to G$ with torsion-free kernel $K$: every finite-order element of $\Gamma$ is conjugate to a power of the cone generator, and no nonidentity such power lies in $K$. Thus $N={\mathbb{H}}^2/K$ is a smooth closed hyperbolic surface with a $G$-action. Its quotients by $U_1,U_2$ are smooth too, since order-$24$ subgroups contain no element of order $7$ and therefore meet no cone stabilizer.

The orbifold Euler characteristic of $B$ is $-(1-1/7)=-6/7$. Each quotient cover has degree $7$, soThe Sunada theorem proved in Question 3 makes $S_1,S_2$ isospectral for every choice of the base hyperbolic metric.

Here is an explicit supply of base metrics. A hyperbolic trirectangle has three right angles and a fourth angle $\alpha =\pi /14$. If its sides adjacent to the opposite right-angle vertex have lengths $r,s$, the trirectangle identity is $\sinh r\sinh s=\cos \alpha$. Thus $r>0$ varies freely, with $s=\mathrm{arsinh}(\cos \alpha /\sinh r)$. Reflecting four copies about their two perpendicular centre lines gives a quadrilateral with four angles $\pi /14$ and opposite sides of equal length. Identifying opposite sides produces a cone torus: all four corners give total angle $2\pi /7$. Its two centre-line geodesics intersect once and have lengths $2r,2s$. Cut along the shorter chosen geodesic $d$ and reglue with a small twist $\tau$; this retains the cone angle and allows an additional metric parameter. The remaining cut surface has two equal-length geodesic boundaries and the cone point.

We give a concrete intersection test for nonisometric finite covers, rather than relying on subgroup nonconjugacy alone. Choose $\ell (d)=\epsilon$ sufficiently small. The trirectangle geometry gives an embedded collar of width $\log (1/\epsilon )+O(1)$ about $d$. The shortest perpendicular joining the two boundary components after cutting has length $\delta =2\log (1/\epsilon )+O(1)$. At zero twist it closes to $a$, the unique shortest closed geodesic crossing $d$ once. Uniqueness follows from uniqueness of the shortest perpendicular; all other once-crossing classes have a positive length gap, since only finitely many closed geodesics have bounded length. For sufficiently small twists, the same marked class $a$ remains the unique shortest once-crossing geodesic. A geodesic crossing $d$ at least twice traverses the collar at least twice, so for small $\epsilon$ its length is greater than $\ell (a)$.

Geodesics disjoint from $d$ remain entirely in the cut surface and have twist-independent lengths. In a bounded window around $\ell (a)$ they supply only finitely many possible length values, including lengths of iterates. Meanwhile $\ell (a)$ varies nontrivially with $\tau$ by the hyperbolic seam identity: its hyperbolic cosine has a positive multiple of $\cosh (\tau /2)$. Choose a small twist away from those finitely many coincidences. Then the only base geodesic of length $\ell (a)$ is $a$, up to orientation. By taking $\epsilon$ smaller if necessary, the only base geodesics of length at most $2\epsilon$ are powers of $d$: those crossing $d$ are long, and the nonperipheral geodesics of the limiting two-cusped cone pair of pants have lengths bounded away from zero.

In each cover, primitive lift components of a base geodesic are indexed by cycles of its monodromy permutation, with length equal to the cycle size times its base length. The permutation of $A$ has a unique singleton $\{0\}$ in each action, so each $S_i$ has a uniquely length-identified primitive lift ${\alpha }_i$ of length $\ell (a)$. The permutation of $D$ has one two-cycle: it is $\{0,3\}$ for $U_1$ and $\{2,5\}$ for $U_2$. Hence each $S_i$ has a unique primitive lift ${\beta }_i$ of length $2\epsilon$; the iterate of the singleton lift is not primitive. At the single base intersection of $a,d$, intersections of their lifted components are exactly their common sheet labels. ThereforeAn isometry would have to preserve the uniquely identified primitive lengths and their geometric intersection numbers. This contradiction proves that $S_1$ and $S_2$ are not isometric, even allowing orientation reversal.

Finally, let $\epsilon$ range over a sufficiently small positive interval, choosing a permissible twist for each value. The singleton cycle of $D$ makes the hyperbolic systole of each cover equal to $\epsilon$. Distinct values therefore give distinct isometry classes. We obtain uncountably many pairs of nonisometric, isospectral genus-four Riemann surfaces.

Throughout the hyperbolic discussion take curvature $-1$ and $g\ge 2$. A Y-piece is a pair of pants with a hyperbolic metric and three geodesic boundary components. Three labelled positive boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3$ determine it up to isometry. To see existence and uniqueness, cut along the three shortest perpendicular seams. The two resulting right-angled hyperbolic hexagons have alternating side lengths ${\ell }_i/2$, which determine each hexagon. Conversely, any three positive alternating lengths produce such a hexagon, and doubling along the other sides produces the desired Y-piece. In particular the seam opposite ${\ell }_i$ satisfies

An X-piece is a four-holed hyperbolic sphere with geodesic boundaries, obtained by joining two Y-pieces along one pair of equal-length boundaries. For a chosen separating curve, it is determined bywhere $s$ is the common gluing length and $\tau$ is the relative arclength twist, measured between chosen seam endpoints. The identification reverses boundary orientation so that the surface is oriented. For an unmarked gluing, $\tau$ is periodic modulo $s$; a marking retains $\tau \in \mathbb{R}$, because a full Dehn twist changes the marking. These six parameters describe the X-piece with its chosen separating curve; different curve choices or boundary symmetries can describe the same unmarked surface.

Choose a pants decomposition of a closed oriented genus-$g$ surface. Each Y-piece has Euler characteristic $-1$, and gluing boundary circles does not change the total Euler characteristic. Thus there are $2g-2$ pieces. Their $3(2g-2)$ boundary circles are paired, giving $3g-3$ gluing curves. Choose their disjoint geodesic representatives, then prescribe each cuff length and its relative twist. The Fenchel–Nielsen coordinates areThey determine the marked hyperbolic surface and hence its unmarked isometry class, with $6g-6$ real parameters. One must also fix the combinatorial gluing pattern of the chosen pants decomposition. Full twists and other changes of marking identify some parameter sets in unmarked moduli space of Riemann surfaces. The genus restriction matters: a sphere and a torus do not admit a closed curvature-$-1$ metric or such a decomposition into geodesic Y-pieces.

For a fixed oriented topological surface $S_g$, Teichmüller space ${\mathcal{T}}_g$ consists of marked Riemann surfaces $(X,f:S_g\to X)$. Two markings represent the same point when a conformal isomorphism between the surfaces carries one marking to the other up to homotopy. By the uniformization theorem, for $g\ge 2$ these are equivalently marked curvature-$-1$ metrics. The displayed Fenchel–Nielsen coordinates identify ${\mathcal{T}}_g$ with a cell of real dimension $6g-6$; the mapping class group quotient gives the unmarked moduli space of Riemann surfaces.

The spectral version of the Wolpert generic spectral rigidity theorem says that outside a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$, the Laplace spectrum determines the hyperbolic surface up to isometry. The exceptional subset has positive codimension and is invariant under changes of marking. Orientation reversal must be allowed, since the Laplacian cannot detect orientation. Equivalently, a generic surface is determined by its unmarked length spectrum, counted with multiplicities. This is a generic uniqueness statement, not a claim that every isospectral pair is isometric; the construction in Question 4 lies in the exceptional locus.

Three major ingredients and their roles are as follows.

For completeness, a related finiteness statement sometimes grouped with this theorem is that a fixed closed hyperbolic surface has only finitely many isospectral isometry classes. Its short proof uses three particularly concrete tools: a common discrete length spectrum; Mumford's compactness theorem, applied to the common positive systole; and finitely many determining marked lengths. Compactness would give an accumulation point if an isospectral family were infinite. Nearby markings make each determining length converge, while discreteness of the fixed spectrum makes it eventually constant. The determining lengths then force eventual equality of the marked surfaces, a contradiction. This finiteness of isospectral hyperbolic surfaces statement does not replace the stronger generic uniqueness assertion above.

A Riemann surface is a connected one-complex-dimensional complex manifold: it has local complex coordinates with holomorphic transition maps. These coordinates induce an orientation and a conformal structure. Fix a closed oriented topological surface $S_g$ of genus $g$. A point of Teichmüller space is a marked Riemann surface $(X,f)$, where $f:S_g\to X$ is an orientation-preserving marking. Two pairs are equivalent if there is a biholomorphism $h:X\to X^{\prime }$ with $h\circ f$ isotopic to $f^{\prime }$. Thus the marking remembers how curves on the reference surface sit in $X$; forgetting it gives the moduli space of Riemann surfaces, obtained by quotienting by the mapping class group.

For $g\ge 2$, the uniformization theorem supplies a unique curvature-$-1$ hyperbolic metric in each conformal class. A pants decomposition has $3g-3$ cuffs, and their positive lengths together with their real twists give Fenchel–Nielsen coordinates. HenceFor completeness, ${\mathcal{T}}_0$ is a point, whereas ${\mathcal{T}}_1$ is the upper half-plane, parametrizing marked complex tori $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$.

The Wolpert generic spectral rigidity theorem says that, for fixed $g\ge 2$, there is a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$ such that a hyperbolic surface outside that subset is determined up to isometry by its unmarked length spectrum, equivalently its Laplacian spectrum. This is an unoriented isometry statement: reversing orientation cannot be detected by either spectrum.

The Huber spectral equivalence theorem says that two closed curvature-$-1$ hyperbolic surfaces have the same Laplacian spectrum, with multiplicities, if and only if they have the same unmarked length spectrum, with multiplicities. One may consistently count unoriented closed geodesics including their iterates; alternatively, the primitive multiset determines the multiset with iterates and conversely. This equivalence comes from the Selberg trace formula.

The Buser finite length spectrum theorem is the finite-determination ingredient: for each $g\ge 2$ and $\epsilon >0$ there is a finite number $L(g,\epsilon )$ such that, for any two closed hyperbolic surfaces of genus $g$ with hyperbolic systole at least $\epsilon$, equality of their length multisets through $L(g,\epsilon )$ implies equality of their entire length spectra. The finite-length statement is used in Buser's chapter on Wolpert's theorem. It determines the full spectrum, not necessarily the isometry class.

Here is a proof of the Buser finite length spectrum theorem through finite spectral matching by polynomial invariants. We use the following subsidiary results, stating the required content. First, Mumford's compactness theorem makes the part of the genus-$g$ moduli space of Riemann surfaces with hyperbolic systole at least $\epsilon$ compact. Second, marked hyperbolic holonomy representations of a closed oriented surface lift from ${\mathrm{PSL}}_2(\mathbb{R})$ to ${\mathrm{SL}}_2(\mathbb{R})$, and normalized representatives can be chosen continuously on sufficiently small parameter charts. Their hyperbolic translation length satisfiesThird, the length spectrum of a compact hyperbolic surface is locally finite, and metrics on compact marked parameter charts are uniformly bilipschitz equivalent. Finally, the Hilbert basis theorem implies that a descending sequence of algebraic sets in a finite-dimensional affine space stabilizes. These are compactness, holonomy, elementary geometric finiteness and algebraic finiteness results; none is the finite unmarked-spectrum assertion we are proving.

By compactness, finitely many relatively compact marked charts cover the thick moduli space of Riemann surfaces. Choose normalized lifted hyperbolic holonomy representations on these charts and take their compact closures. Their finite union $K$ is a compact set of determinant-one matrix tuples representing every surface in the thick part, possibly more than once. Each tuple records the images of the $2g$ generators of the fundamental group. Thus $K$ sits in a fixed finite-dimensional real matrix space. Every group word is a polynomial in these matrix entries: multiplication is polynomial, and for a determinant-one two-by-two matrix its inverse is its adjugate, also polynomial. Consequently each $q_{\gamma }$ is a polynomial on the ambient space. Squared traces remove the sign ambiguity of the lift, and $4{\cosh }^2(\ell /2)$ is strictly increasing for positive $\ell$.

Enumerate the unoriented nontrivial free homotopy classes as ${\gamma }_1,{\gamma }_2,\dots$, including the iterates, and write ${\ell }_j(x)$ for their geodesic lengths at $x\in K$. The uniform finiteness of short geodesic classes follows here directly: on each of the finitely many compact marked charts, choose a fixed reference metric $h$ and a uniform bilipschitz constant $C$. A class with ${\ell }_j(x)\le R$ somewhere on that chart has reference length at most $CR$. Only finitely many classes have that reference length, and a finite union of these finite collections is finite. In particular, for every $j$ the candidate setis finite. If two parameters in $K$ have equal lengths for classes $j,k$, then $k\in F_j$. These candidate sets depend on the fixed compact set $K$ and the source label $j$, not on the number of labels currently being matched.

For every $N$, define $E_N$ in the full ambient space of pairs of matrix tuples $(x,y)$ as follows. Require an injective choice $\sigma :\{1,\dots ,N\}\to \mathbb{N}$ with $\sigma (j)\in F_j$ andand independently require an injective choice $\tau :\{1,\dots ,N\}\to \mathbb{N}$ with $\tau (j)\in F_j$ andThere are finitely many possible pairs of choices $(\sigma ,\tau )$ because every $F_j$ is finite. For each choice these are finitely many polynomial equations. Therefore $E_N$ is a finite union of algebraic sets, hence an algebraic set. Restricting an injective matching to its first $N$ entries provesThe Hilbert basis theorem now supplies $N_0$ such that $E_N=E_{N_0}$ for all $N\ge N_0$. This step takes place in a fixed affine space and uses polynomial equations, so it gives stabilization of the actual sets, not just stabilization of germs at individual parameters.

For $(x,y)\in E_{N_0}\cap (K\times K)$, stabilization gives injective partial length matchings in both directions for every $N$. Fix any length $a>0$. Only finitely many labels have length $a$ at $x$ or at $y$, by local finiteness. Take $N$ large enough to contain every such label at $x$. Its injective matching into the labels at $y$ preserves $q_j$, and hence preserves length, so the multiplicity of $a$ at $x$ is at most its multiplicity at $y$. The reverse matching gives the opposite inequality. Thus the two multiplicities are equal for every $a$, proving equality of the entire unmarked length spectra. No assumed global matching or unjustified exchange of infinitely many neighborhoods is needed.

Finally choose a numberSuppose the two length multisets agree through $L$. For each of the first $N_0$ labels at $x$, select a distinct equal-length label at $y$, which is possible precisely because the truncated multisets agree with multiplicity. Each selected label belongs to the corresponding $F_j$. Do the same in the reverse direction. Then $(x,y)\in E_{N_0}$, and the previous paragraph proves equality of the full length spectra. Since $K$ was selected using only the fixed thick genus-$g$ parameter space, the resulting finite $L$ depends only on $g,\epsilon$. This proveswith each intersection interpreted as a multiset. By the Huber spectral equivalence theorem, the corresponding entire Laplacian spectra also agree. This is the finite matching reduction needed before the analytic and geometric rigidity steps in the Wolpert generic spectral rigidity theorem.

On the unit three-ball $B^3$, the distance in the Poincare ball model is the infimum of the lengthsover piecewise smooth paths joining the two points. Equivalently the hyperbolic metric as a metric-space distance isIt has curvature-minus-one normalization; in particular $d(0,x)=2\mathrm{artanh}|x|$.

For a nonempty finite set $p_1,\dots ,p_m$, define $R(x)={\max }_{j}d(x,p_j)$. The triangle inequality gives $|R(x)-R(y)|\le d(x,y)$, so $R$ is continuous. Take $R_0=R(p_1)$. Any improving center satisfies $d(x,p_1)\le R(x)\le R_0$, and thus lies in the closed hyperbolic ball $K=\overline{B}(p_1,R_0)$.

This ball is compact: an isometry sending $p_1$ to the origin identifies it with the Euclidean closed ball of radius $\tanh (R_0/2)<1$, and the Euclidean and hyperbolic topologies agree inside $B^3$. The continuous function $R$ attains a minimum on $K$. Centers outside $K$ have $R>R_0$, so this is also the global minimum. Its associated closed ball proves a minimum enclosing ball in hyperbolic space exists. For the empty finite set, any center and radius zero suffice.

Zeros are counted with their prescribed finite multiplicities, and the function is not identically zero. The criterion for zero sets in the unit disc is local finiteness: no point of the disc is an accumulation point of the zero sequence, and no point has infinite multiplicity. Necessity follows from the identity theorem for holomorphic functions. For sufficiency, factor a finite number of zeros at zero as $z^m$. For an infinite sequence of remaining zeros $a_n$, local finiteness implies $|a_n|\to 1$. With Weierstrass elementary factorsuse $f(z)=z^m{\prod }_n{E}_n(z/a_n)$. On each compact subdisc, eventually $|z/a_n|\le q<1$ and the logarithm of the tail factor is $-{\sum }_{j>n}(z/a_n{)}^j/j$, bounded in modulus by $q^{n+1}/((n+1)(1-q))$. These bounds are summable, so infinite product convergence from logarithmic tails gives locally uniform convergence and nonvanishing away from the listed zeros. Its finite initial factors give exactly their multiplicities. For a finite sequence an ordinary polynomial suffices.

Use the curvature-minus-one hyperbolic metric, with length element $2|dz|/(1-|z{|}^2)$. Integrating radially givesTherefore the printed sum condition is equivalent to the Blaschke condition ${\sum }_n(1-|a_n|)<\infty$.

To prove necessity for bounded $f$, first suppose $f(0)\ne 0$ and choose a radius $r<1$ with no zero on its circle. Remove the finitely many factors $(z-a_n)$ for $|a_n|<r$. The logarithm of the modulus of the remaining zero-free function is harmonic on the closed disc. Its mean value property, together with the mean of $\log |r{e}^{i\theta }-a|$ being $\log r$ for $|a|<r$ (expand the logarithm in powers of $a/(r{e}^{i\theta })$), proves Jensen's formulaLet $r\uparrow 1$ through such radii. Each summand increases to $-\log |a_n|$, and $-\log t\ge 1-t$ for $0<t\le 1$ by integration of $1/t\ge 1$. Consequently ${\sum }_n(1-|a_n|)<\infty$. If $f$ has order $m$ at zero, apply the argument to $f/z^m$, which is still bounded: outside $|z|=1/2$ it is bounded by $2^m\Vert f{\Vert }_{\infty }$, and inside use the maximum modulus principle. Add the finite contribution of the zeros at zero.

For sufficiency use a Blaschke product. For $a\ne 0$ its normalized Blaschke factor isOn $|z|\le R<1$, $|1-b_a(z)|\le (1-|a|)(1+R)/(1-R)$. The Blaschke condition thus makes the product converge locally uniformly; outside the specified zeros, the tail logarithms converge absolutely and the product is nonzero. Each factor has modulus at most one in the disc, so the limit is bounded by one. A factor $z^m$ supplies any finite multiplicity at zero. This proves

For the elementary inequality, put $H(t)=\log ((1+t)/(1-t))-2t$. Then $H(0)=0$ andso $\boxed{\log ((1+t)/(1-t))\ge 2t}$ for $0\le t<1$.

For any point $w$ not among the zeros, the pseudohyperbolic distance formula givesApply the proved inequality with $t=e^{-\rho (w,a)}$ to obtain $\log |b_a(w)|\le -2e^{-\rho (w,a)}$. Sum over finite products and pass to the limit, including factors at zero. This givesThe sum is finite: the triangle inequality gives $e^{-\rho (w,a_n)}\le e^{\rho (0,w)}{e}^{-\rho (0,a_n)}$. If $w$ is itself a zero, $B(w)=0$ and the same bound is immediate, so no substitution at the excluded value $t=1$ is needed. A unimodular constant in $B$ has no effect.

If “zeros at the points” is interpreted as a set without assigned multiplicities, first delete repeated entries throughout. Otherwise infinitely repeating a single zero would invalidate the claimed necessity of the series condition; the standard zero-sequence convention counts multiplicities.

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.

Sunada's theorem states that if a finite group $T$ acts isometrically on a closed Riemannian manifold $X$, and two Gassmann equivalent subgroups $H_1,H_2$ act freely, then $H_1\backslash X$ and $H_2\backslash X$ are isospectral for the function Laplace-Beltrami operator. Gassmann equivalence means $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$ of $T$. For each Laplace-Beltrami operator eigenspace $E_{\lambda }$, its invariant subspace has dimensionThe character of a representation is a class function, so the two multiplicities agree. A necessary condition for nonisometry is that the subgroups are not conjugate in $T$: conjugation by an ambient deck transformation would induce an isometry. This is not sufficient when the common covering has additional geometric symmetries.

Use the oriented hyperbolic triangle groupand the epimorphism $\rho :\Gamma \to T$ given by $a\mapsto s$, $b\mapsto d$. Its base orbifold is the doubled hyperbolic triangle with angles $\pi /3,\pi /3,\pi /6$, having signature $(0;3,3,6)$. Let $K=\ker \rho$ and $K_i={\rho }^{-1}(H_i)$.

The groups $K_i$ are torsion free. Indeed, torsion elements in an oriented triangle group are conjugate to nontrivial powers of its three vertex rotations. Such powers map either to elements of order three, which cannot lie in a subgroup of order $96/12=8$, or to the order-two cube $z=(sd{)}^3$. The latter is central and absent from both $H_i$, so it is absent from every conjugate as well. Thus the degree-twelve orbifold coversare smooth, closed, connected orientable surfaces. The common cover $X=K\backslash {\mathbb{H}}^2$ is smooth too; $H_i$ act freely on it. Orbifold Euler characteristic givesFor example, this formula follows by triangulating the base and weighting a vertex of order $r$ by $1/r$ before multiplying by the cover degree. Equivalently, each doubled triangle has hyperbolic area $\pi /3$, so $N_i$ has area $4\pi$ and Gauss-Bonnet theorem gives the same genus.

There is one extra issue in obtaining the same hyperbolic limit on a single marked surface: the two limiting hyperbolic covers must themselves be isometric. Nonconjugacy and the displayed element orders alone do not prove this. An explicit realization of the supplied example resolves the issue. The group seed is recorded in the original genus-two construction; the following coset actions and reflection identities give a direct verification. Label the twelve sheets so that the two right-coset actions of $s,d$ areThese are concrete permutation data for an order-$96$ realization with the required two subgroups. The permutationsatisfies, by direct substitution,Reflection of the doubled triangle reverses both vertex rotations. Apply this reflection on every fundamental domain and relabel sheets by $P$. The two identities ensure that identified edges go to identified edges, so this descends to a hyperbolic isometry $F_0:N_1\to N_2$. Fix $N=N_1$, let $g$ be its hyperbolic metric, and use $F_0$ to mark $N_2$ by the same smooth surface. Thus both pulled-back limiting Riemannian metrics are exactly $g$. The explicit reversed-action check is what makes the common limit justified, rather than treating it as a consequence of Sunada's theorem.

Choose smooth orbifold Riemannian metrics $h_k$ on the base converging to its hyperbolic metric $h_0$, with no nonidentity local isometry on the regular part. The density assertion in Sunada's local isometry lemma applies in orbifold charts as well: use equivariant perturbations at the finitely many cone points and ordinary local perturbations elsewhere. Smoothness here is orbifold smoothness, so the lifted Riemannian metrics on $X$ and the two smooth covers are ordinary smooth Riemannian metrics. Such choices can be made in successively smaller smooth neighbourhoods of $h_0$; cone orders and the topological coverings stay fixed.

For area separation of convergent Sunada families, adjust each $h_k$ by a constant positive factor so thatExplicitly, multiply the initially chosen Riemannian metric by the ratio of the desired area to its actual area. In dimension two this ratio is exactly the area multiplier. The ratios tend to one, so smooth convergence persists, and constant scaling preserves the absence of local isometries. Lift these adjusted Riemannian metrics to $N_i$ and pull them back by the fixed markings to obtain $g_{ik}$, for $i=1,2$. The three required properties follow below.