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Take a hyperbolic torus with a single cone point of angle $2\pi /7$. Its orbifold fundamental group is $⟨a,d\mid [d,a{]}^7=1⟩$. An epimorphism to $\mathrm{PSL}(3,2)$ whose commutator has exact order $7$ has torsion-free kernel. The point and line stabilizers have index $7$ and order $24$, hence avoid all nontrivial cone stabilizers. Their quotient covers are smooth surfaces with Euler characteristic $7(-6/7)=-6$, so have genus $4$. Their equal coset permutation characters give equal heat traces by the Sunada theorem. Nonisometry still requires a geometric distinction, such as the intersection test for nonisometric finite covers.

The orbifold fundamental group records loops together with their local isotropy relations, so a loop around a cone point of order $m$ has its $m$th power trivial rather than being trivial itself. For an oriented closed surface of genus $g$ with cone orders $m_1,\dots ,m_r$, its presentation is $⟨a_1,b_1,\dots ,a_g,b_g,c_1,\dots ,c_r\mid c_i^{m_i}=1,{\prod }_j[a_j,b_j]c_1\cdots c_r=1⟩$. A torsion-free finite-index subgroup of a hyperbolic surface orbifold gives a smooth locally isometric surface cover. The absence of torsion means that the subgroup meets no nonidentity local cone stabilizer.

Put $T=\mathrm{PSL}(3,2)=\mathrm{GL}(3,{\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. It acts on the seven points and seven lines of the Fano plane. Take $U_1$ to be a point stabilizer and $U_2$ to be a line stabilizer; both have index seven and order twenty-four. The fact that point and hyperplane stabilizers are almost conjugate follows by comparing fixed-point counts: for a matrix $g$, the numbers of fixed nonzero vectors and fixed nonzero covectors are bothbecause $g-I$ and its transpose have the same rank, and the dual action $g^{-T}$ has the same fixed space as $g^T$. Over ${\mathbb{F}}_2$, nonzero vectors represent points and nonzero covectors represent lines without any further scalar quotient. The two permutation characters therefore agree. The coset-character formulathen proves Gassmann equivalence. They are not conjugate: a point stabilizer has orbits of sizes one and six on points and hence preserves no three-point line. Its conjugates are again point stabilizers, whereas $U_2$ preserves its line. The point and line stabilizers form the required nonconjugate Gassmann pair.

We construct a family using the Lambert quadrilateral construction of a cone torus. Let $\alpha =\pi /14$. We use the elementary hyperbolic tri-rectangle fact that a quadrilateral with three right angles, remaining angle $\alpha$, and side lengths $x,y$ adjacent to the right-angle vertex opposite $\alpha$ exists uniquely forHere $x,y$ are the sides at the opposite right-angle corner, not the sides incident to $\alpha$. Given any $x>0$, choose $y=\mathrm{arsinh}(\cos \alpha /\sinh x)$. Reflect four copies around the opposite right-angle vertex. The result is a symmetric hyperbolic quadrilateral with all four corner angles $\alpha$ and opposite sides of equal lengths. Pair opposite sides by the axis translations. The quotient is an oriented torus with one cone point, whose total angle is $4\alpha =2\pi /7$. Its two central axis loops $a,d$ are embedded closed geodesics, of lengths $2x,2y$, intersecting once at right angles.

Cut this cone torus along $a$. The result is a hyperbolic pair of pants with two geodesic boundaries of length $\ell =2x$ and the cone point. Reglue those equal boundaries with a Fenchel–Nielsen twist $\tau \in \mathbb{R}$. This gives a two-parameter family $B_{\ell ,\tau }$, with the same cone angle, and marked simple classes $a,d$ meeting once. Near any zero-twist member their geodesic representatives remain embedded, transverse and disjoint from the cone point. The length of $a$ remains $\ell$, and the right-angle matrix product givesFor example, the zero-twist holonomies are $\mathrm{diag}(e^x,e^{-x})$ and $\left(\begin{array}{cc}\cosh y& \sinh y\\ \sinh y& \cosh y\end{array}\right)$; multiply the latter by $\mathrm{diag}(e^{\tau /2},e^{-\tau /2})$ to obtain the twist formula. Their commutator has trace $2-4{\sinh }^{2}x{\sinh }^{2}y=-2\cos (\pi /7)$, confirming exact elliptic order seven. The side-pairing construction supplies discreteness; the trace computation alone would not do so.

The orbifold fundamental group isSend $a$ to the supplied generator $A$ and $d$ to $D$. Their commutator $C=[D,A]$ has order seven, so this defines an epimorphism $\rho :\Gamma \to T$. Its kernel avoids all nontrivial cone stabilizers, which are conjugates of powers of $[d,a]$; therefore it is torsion-free. It defines a smooth closed regular hyperbolic cover $X_{\ell ,\tau }\to B_{\ell ,\tau }$ with deck group $T$. The subgroups $U_i$, of order twenty-four, also avoid every order-seven cone stabilizer. Thusare smooth closed seven-sheeted covers of the cone torus. The orbifold Euler characteristic is ${\chi }_{\mathrm{orb}}(B)=-(1-1/7)=-6/7$, givingEquivalently there is one preimage of the cone point, with local degree seven, and its lifted angle is $2\pi$. The Sunada theorem, in the free-subgroup-action form proved above, makes $S_1(\ell ,\tau )$ and $S_2(\ell ,\tau )$ isospectral for every parameter.

Nonisometry requires further work. A cycle of length $r$ in the monodromy permutation of a simple base geodesic gives a primitive simple lifted geodesic of length $r$ times the base length. From the supplied $A$ action, each cover has one lift of $a$ of length $\ell$ and two of length $3\ell$. From the supplied $D$ action, each has one lift of $d$ of each length ${\ell }_d,2{\ell }_d,4{\ell }_d$. Let ${\beta }_i$ be the unique two-cycle lift of $d$, and let ${\alpha }_{i,1},{\alpha }_{i,2}$ be the two three-cycle lifts of $a$. The fact that cycle intersections count intersections of lifted curves gives their intersection numbers: the seven crossings above the one base crossing are labelled by the sheets, so a crossing belongs to both lifted components exactly when its sheet lies in both cycles. ConsequentlyIndeed, in the first cover the two three-cycles have supports $\{1,2,5\},\{3,6,4\}$ and the two-cycle has support $\{0,3\}$; in the second they have supports $\{1,4,3\},\{2,5,6\}$ and $\{2,5\}$. The lifted curves intersect only over the single base crossing, so these are actual intersection counts of the geodesics, not merely abstract sheet-incidence numbers.

To make this a test against every isometry, we must identify these curves intrinsically, rather than require the isometry to respect the covering map. We use generic isolation of lifted simple geodesics, and give the needed genericity argument. Lengths of closed geodesics are real-analytic functions of the cone-torus length and twist parameters. For a primitive hyperbolic class $w$ and an integer $1\le r\le 7$, consider the equalitiesApart from $w=a,r=3$ or $w=d,r=2$, none is an identity on the full two-parameter family. To see this for the first equality, pinch the simple curve $a$ while keeping the cone angle fixed. The collar lemma makes every class crossing $a$ long. A class disjoint from $a$ lies in the cut pair of pants; its limiting holonomy has positive translation length unless it is a power of one of the new cusp loops, which are conjugate to $a$ in the torus group. Cone-peripheral classes are elliptic and do not represent closed geodesics. Thus the only primitive class with length tending to zero in this degeneration is $a$, up to inversion and conjugacy. This excludes an identity with a positive multiple of $\ell$. Pinching $d$ proves the corresponding assertion for ${\ell }_d$. The argument uses the usual hyperbolic collar and cusp degeneration facts, obtainable by the right-angle quadrilateral formulas; the cone angle $2\pi /7<\pi$ remains fixed. Analytic continuation on the connected length-twist parameter space shows that the same identities cannot hold on a small open neighbourhood of a zero-twist member.

There are countably many class equations, and each nontrivial real-analytic equation has a closed zero set with empty interior. The Baire category theorem therefore gives a parameter $({\ell }_0,{\tau }_0)$ in that small neighbourhood at which both target lengths are attained only by the prescribed lift components. This isolation persists on a smaller open neighbourhood. Indeed, near a fixed compact hyperbolic metric there are uniformly only finitely many primitive geodesic classes below a fixed length bound, by discreteness of the covering group and a compact fundamental region; nearby metrics are uniformly bilipschitz. Only finitely many classes can therefore approach the two fixed finite target lengths. Their strict inequalities persist under a sufficiently small parameter change.

Now keep $\tau ={\tau }_0$ and let $\ell$ vary in a sufficiently small interval $I$ around ${\ell }_0$. For each $\ell \in I$, the only primitive geodesics of length $3\ell$ in either cover are the two $\alpha$ components, and the only primitive geodesic of length $2{\ell }_d$ is $\beta$. An isometry would have to preserve these length classes and their intersection-count multiset, contradicting $\{0,1\}\ne \{0,2\}$. Henceis an interval-parametrized family of isospectral nonisometric hyperbolic Riemann surfaces of genus four. Shrink $I$ again so a small length window around $3{\ell }_0$ contains only those two components for every parameter. Their common length $3\ell$ varies strictly, so members at distinct parameters cannot all be the same surface in different markings. The spectra agree within each pair; this construction does not assert a nontrivial continuously isospectral deformation of one hyperbolic surface.

Take $T$ to be a finite group acting by orientation-preserving isometries. We give both an unbranched construction for every such abstract group and the orbifold construction needed for low-genus examples. The inherited quotient metric and the existence of some unrelated smooth hyperbolic metric on the underlying topological surface are different questions.

For a free action, choose generators $t_1,\dots ,t_s$ of $T$ and a closed orientable surface of genus $g_0\ge \max (2,s)$. Its surface group has generators $a_j,b_j$ and relation ${\prod }_j[a_j,b_j]=1$. Sending $a_j$ to the chosen generators, the remaining $a_j$ to one, and all $b_j$ to one defines a surjective homomorphism to $T$. Its kernel determines a connected regular finite cover $M$. Give the base a Riemannian metric of constant curvature $-1$ by the uniformization theorem, and pull it back to $M$. Its deck group is $T$, acting freely and isometrically. All subgroup quotients are then smooth covers as well.

For the more general construction, prescribe an oriented compact hyperbolic orbifold of signature $(g_0;m_1,\dots ,m_r)$, withIts orbifold fundamental group has presentationChoose an epimorphism $\theta :\Gamma \to T$ for which every $\theta (c_i)$ has exact order $m_i$. All finite-order elements of the uniformizing group are conjugate to powers of cone generators, so exact orders make $K=\ker \theta$ torsion-free. Realize $\Gamma$ as a cocompact Fuchsian group on the hyperbolic plane. ThenThe source notation $T/M$ is interpreted as the quotient of $M$ by $T$, as in the consistent subgroup notation $U\backslash M$. The manifold $M$ is smooth; $M_0$ has cone angles $2\pi /m_i$ and may have singularities.

The smoothness criterion from cone-monodromy cycles is explicit. On the $n=[T:U]$ cosets, let ${\sigma }_i$ be the permutation induced by $\theta (c_i)$. A cycle of length $\ell$ has local degree $\ell$ over the cone point. The resulting total angle is $2\pi \ell /m_i$, so its residual cone order is $m_i/\ell$. Therefore the inherited metric on $M_1$ is smooth precisely when every cycle of every ${\sigma }_i$ has length $m_i$. Equivalently,Indeed a nonidentity cone-generator power fixes a coset exactly when its conjugate lies in $U$. The inherited metric on $M_0$ itself is smooth precisely when there are no nontrivial cone stabilizers, equivalently when the $T$ action on $M$ is free.

For the Euler characteristics, first distinguish the ordinary characteristic of the underlying base from its orbifold Euler characteristic:To derive the covering formulas, triangulate the underlying base with all cone points as vertices. Ordinary vertices, open edges and faces lift with their full degree. A cone point whose monodromy has $c_i$ cycles has $c_i$ preimage vertices rather than $n$; thus the lifted Euler characteristic loses $n-c_i$ there. This is the Euler characteristic from coset cycle counts:For the full regular cover $M$, each cone generator acts on $T$ in cycles of length $m_i$, so it has $|T|/m_i$ cycles. ConsequentlyWhen $M_1$ is smooth, all its cone cycles likewise have full length and $c_i=n/m_i$. HenceThese are also the Riemann-Hurwitz formula with the ramification deficits made explicit. The Gauss-Bonnet theorem gives area $-2\pi \chi$ for each smooth closed hyperbolic surface, consistently multiplying with covering degree.

Finally, if one asks whether the underlying surface admits any smooth metric of constant curvature $-1$, rather than whether this quotient metric is smooth, the criterion is ordinary Euler characteristic less than zero. Necessity follows from the Gauss-Bonnet theorem; sufficiency is the uniformization theorem for genus at least two. Thus an orbifold quotient with cone singularities can sometimes be given a different smooth hyperbolic metric, but this change does not preserve its role as the locally isometric quotient in the spectral construction. A sphere or torus underlying $M_0$ cannot be made a closed smooth curvature-$-1$ surface at all.

For the constant-curvature constructions, the three data sets lead respectively to genera two, three and four. The first two pairs are isometric through a lifted reflection; the third gives nonisometric pairs under the length-identification hypothesis stated below. Nonconjugacy of the two subgroups inside the finite group is not by itself a test for nonisometry.

Replace each $T_i$, if necessary, by the joint image of its two coset actions, generated by the corresponding pair of generator permutations. The joint kernel lies in both subgroups, so passing to this finite quotient leaves the two coset actions unchanged. We use these faithful joint images in the constructions below.

First verify the Gassmann equivalence needed for the Sunada theorem for all group elements, not just the displayed generators. Write $P_{A,U},P_{B,U},P_{A,V},P_{B,V}$ for their permutation matrices. Invertible matrices $C_i$ satisfyingintertwine every word in the generators, hence give equal permutation characters. Here are explicit certificates calculated from the coset actions. Define $C_{kj}=1$ when column label $j$ belongs to the $k$th support below, and zero otherwise. For $i=1$, labels run from one to twelve; the row supports in their order areFor $i=2$ and $i=3$, labels run from zero to six, and the row supports are respectivelyDirect substitution gives the two intertwining identities, and elimination gives $\det C_1=-1024$, $\det C_2=\det C_3=32$. These nonzero determinants certify equivalent linear permutation representations. Equal characters imply almost conjugacy: for a subgroup $U$,because a fixed coset corresponds to $t^{-1}gt\in U$, and each specified conjugate has $|C_T(g)|$ conjugating elements. The subgroup orders agree since their indices agree, so equality of these characters gives equal intersection counts with every conjugacy class.

Nonconjugacy can also be checked directly: a bijective equivariant sheet relabelling would be determined by the image of one label, since the actions are transitive. Propagating each of the twelve or seven possible images by the two generators gives no consistent bijection in any of the three cases.

For genus two, use $i=1$. The generator orders are $3,3$, and their product has order six; on both twelve-sheet coset spaces their cycles are respectively $3^4,3^4,6^2$. Map the hyperbolic triangle group $\Delta (3,3,6)$ onto the joint finite permutation image $T_1$. Its kernel is torsion-free because these orders are exact, giving a smooth normal cover $M$ of the sphere orbifold with cone orders $3,3,6$. The subgroups act freely by the smoothness criterion from cone-monodromy cycles, since all coset cycles have their full orders. The two quotient surfaces therefore haveTheir equal permutation characters give isospectrality by the Sunada theorem.

There is nevertheless an isometry. The sheet relabellingfixes labels three and five and satisfiesTo see the geometric significance, write the triangle-group generators as $a=s_1{s}_2$, $b=s_2{s}_3$, where $s_j$ reflect in triangle sides. Conjugation by $s_2$ sends both $a,b$ to their inverses. The two displayed permutation identities say exactly that reflecting the triangles and relabelling the sheets preserves their gluings. This reflection intertwining of triangle-cover coset actions lifts the base reflection to an orientation-reversing isometry of the quotient surfaces. Thus the genus-two pair from these hyperbolic triangles is isometric, despite the subgroups being nonconjugate in the orientation-preserving finite group.

For genus three, use $i=2$. Both generators and their product act in seven-cycles, so the same construction with $\Delta (7,7,7)$ gives smooth seven-sheet quotients andTheir isospectrality follows from the certificate $C_2$. The relabelling $R_2(j)=4-j(\mathrm{mod}7)$, namely $(04)(13)(56)$ with label two fixed, again satisfies $R_2{A}_U{R}_2^{-1}=A_V^{-1}$ and $R_2{B}_U{R}_2^{-1}=B_V^{-1}$. Therefore this constant-curvature genus-three pair is also isometric by the same lifted triangle reflection. This assertion concerns the constructed triangle metrics, rather than arbitrary metrics placed independently on the two topological surfaces.

For genus four, use $i=3$ and a different base: a hyperbolic torus with one cone point of order seven. Its orbifold fundamental group has presentationMap $a,b$ to $A_3,B_3$. Multiplying the given permutations shows that their commutator is a seven-cycle in both coset actions, so this map has torsion-free kernel and both subgroup quotients are smooth. The base orbifold Euler characteristic is $-(1-1/7)=-6/7$, givingThe certificate $C_3$ and the Sunada theorem give an isospectral pair for each base metric. This is the cone-torus construction of genus-four Sunada surfaces.

For nonisometry, choose simple closed base geodesics $\alpha ,\beta$ representing $a,b$, avoiding the cone point and meeting once. We use the following additional geometric hypothesis, as permitted in this question: choose a cone-torus metric such that, in each cover, the length $4\ell (\alpha )$ identifies its unique degree-four lift, and the length $3\ell (\beta )$ identifies precisely its two degree-three lifts, among all primitive closed geodesics. This is a generic length-separation choice; it excludes coincidences with other geodesic lengths. No claim about an arbitrary symmetric choice of base metric is needed.

A lift component of a base geodesic corresponds to a cycle of its monodromy permutation. The degree-four $A_3$ cycle is $\{2,6,4,5\}$ on the $U_3$ sheets and $\{0,1,6,3\}$ on the $V_3$ sheets. The degree-three $B_3$ cycles are $\{1,2,5\},\{3,6,4\}$ on the first cover and $\{1,4,3\},\{2,5,6\}$ on the second. Since the base curves meet once, their lifted geometric intersection numbers are the sizes of the intersections of these sheet-label sets. The cycle intersections count intersections of lifted curves therefore gives the two unordered multisetsNegative curvature ensures that the geodesic representatives realize geometric intersection number. An isometry would preserve the length-identified degree-four lift and the set of the two length-identified degree-three lifts, even if it exchanged the latter, so it would preserve this unordered multiset. The multisets differ. The intersection test for nonisometric finite covers thus proves the genus-four pair is nonisometric for this choice of constant-curvature metric. Without a condition excluding relevant accidental length coincidences, subgroup nonconjugacy alone would not establish that conclusion.

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.