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For a paired $2s$-gon with $r$ base vertex cycles of a paired polygon, there are one face, $s$ edges and $r$ vertices on the base, so $\chi (M_0)=1-s+r$. The full finite-group construction has $|T|$ faces, $s|T|$ edges and ${\sum }_j|T|/q_j$ vertices, giving the displayed ordinary Euler characteristic. If all lifted angles are $2\pi$, the orbifold Euler characteristic formula gives equivalently $\chi (M)=|T|{\chi }_{\mathrm{orb}}(M_0)$.

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.

Use an oriented geodesic hyperbolic $2s$-gon $D$ with paired edges of equal length. Identify each pair by a length-preserving map reversing the boundary directions, to make an oriented base surface $M_0$ with possibly singular vertices. Assign to the $s$ edge pairs generators $A_1,\dots ,A_s$ of $T$, with inverse labels for reversed crossings. For each $g\in T$, take a copy $D_g$ and glue its positively labelled edge of type $k$ to the corresponding mate in $D_{g{A}_k}$. The generators make the glued surface $M$ connected. Left multiplication $D_g\mapsto D_{hg}$ commutes with these right-labelled gluings and gives the $T$ action with quotient $M_0$.

Existence for an arbitrary finite group can even be arranged with no cone singularities. If $T$ has $d$ generators, take $h\ge \max \{2,d\}$ and a regular hyperbolic polygon with $4h$ edges, all angles $\pi /(2h)$, paired with side word ${\prod }_{i=1}^h{a}_i{b}_i{a}_i^{-1}{b}_i^{-1}$. Its vertices form one cycle of angle $2\pi$, giving a smooth base of genus $h$. Label the $a_i$ by the generators of $T$, padded by identities, and every $b_i$ by the identity. The single vertex word is ${\prod }_i[A_i,1]=1$, so every lifted vertex also has angle $2\pi$. The resulting connected ordinary normal covering map has degree $|T|$ and deck group $T$.

This is the paired-polygon construction of finite surface covers. Interior and open-edge points are regular hyperbolic points; possible singularities occur only at vertices. When the base has cone points, the map is an orbifold covering and may be branched as a map between the underlying topological surfaces. It should not be called an unbranched manifold covering at those branch points.

A vertex cycle of a paired polygon is the collection of vertices identified to one point on $M_0$. Let $r$ be the number of cycles, ${\theta }_j$ the sum of the polygon angles in cycle $j$, and $C_j\in T$ the gluing word recorded on going around it. If $q_j=|C_j|$ is the order of that element, a lift closes after $q_j$ turns. It has angle $q_j{\theta }_j$, and there are $|T|/q_j$ such lifted vertices. The base and lifted cone orders, when they are integer orbifold orders, areAn orbifold homomorphism requires $q_j\mid m_j$. The lifted point is smooth exactly when $q_j{\theta }_j=2\pi$, equivalently $q_j=m_j$ in that case. For arbitrary cone metrics the angle formulas remain valid even when an angle does not correspond to an integer orbifold order.

Counting faces, edges and vertices gives the Euler characteristic of a paired-polygon surface cover:These are ordinary topological Euler characteristics, rather than orbifold Euler characteristics. If the lifted surface is smooth, the equivalent orbifold formula isFor a closed oriented smooth surface, $\chi =2-2g$ defines its genus. These formulas also follow from the Riemann-Hurwitz formula by subtracting the ramification deficits $|T|-|T|/q_j$ over each base vertex.

For $U\le T$, the quotient map $M\to M_1=U\backslash M$ is an ordinary normal covering if and only if $U$ acts freely. All possible point stabilizers in this construction are conjugates of $⟨C_j⟩$, so the concrete necessary and sufficient condition isThis is the freeness criterion for a subgroup of a polygon-cover deck group. Under it the deck group is $U$, the degree is $|U|$, andThere is no requirement $U◃T$ for this map. That different condition characterizes normality of the intermediate cover $M_1\to M_0$. If fixed points are retained and one instead speaks of an orbifold covering, the quotient is regular as an orbifold cover, but ordinary Euler characteristic need not divide by $|U|$; orbifold Euler characteristic does.

For the low-genus examples takeLet $U_1$ be a nonzero-vector stabilizer and $U_2$ a plane stabilizer. Their actions have seven points, so $|U_1|=|U_2|=24$ and $[T:U_i]=7$. They are almost conjugate subgroups: the numbers of fixed nonzero vectors for a matrix $A$ and its dual action $A^{-T}$ are both $2^{\dim \ker (A-I)}-1$. Thus the two coset permutation characters agree, which is equivalent to Gassmann equivalence. They are not conjugate: a point stabilizer has a common fixed nonzero vector, whereas a plane stabilizer fixes no nonzero vector globally. In particular, since $7\nmid 24$, neither subgroup contains any nonidentity element of a cyclic subgroup of order seven. All cone monodromies of exact order seven therefore act freely on the intermediate covers.

For genus three, start with a hyperbolic triangle whose three angles are $\pi /7$ and double it across one edge. The resulting quadrilateral $D$ has anglesThe side pairings produce three base vertex cycles, each of total angle $2\pi /7$, and an underlying sphere. Choose generators $A,B$ with $|A|=|B|=|AB|=7$; the three vertex words are $A,B,(AB{)}^{-1}$. Hence $s=2$, $r=3$ and all $q_j=7$. The full cover is smooth, andThe hyperbolic polygon area of $D$ is $8\pi /7$, so the seven-sheeted surfaces have area $8\pi$, agreeing with Gauss-Bonnet. Their equal spectra follow from the Sunada theorem. For complete numerical generator data, one possible pair isThese generate all of $T$ and have the required orders. This construction establishes the requested isospectrality; it does not establish nonisometry. In the constant-curvature triangle example a lifted reflection can identify the point and plane covers, as in reflection intertwining of triangle-cover coset actions. Breaking that symmetry with a generic metric gives nonisometric genus-three surfaces, but then the metric need not have constant curvature. Those are distinct assertions.

For genus four, use the cone-torus construction of genus-four Sunada surfaces. Take a regular hyperbolic quadrilateral with four anglesand pair opposite sides in the usual torus pattern $a,b,a^{-1},b^{-1}$. All four vertices belong to one cycle, whose total angle is $2\pi /7$; the underlying base is a torus with one cone point. Choose $A,B$ of order four generating $T$, with commutator $[A,B]$ of order seven. For example,Only the commutator is a cone monodromy, so the order-four side labels do not create cone points of order four. Here $s=2$, $r=1$, $q_1=7$, givingThe quadrilateral has area $12\pi /7$, and each seven-sheeted smooth quotient has area $12\pi$. Again the order-seven cone stabilizer meets neither $U_i$ nontrivially, so the metrics are smooth and the Sunada theorem gives identical spectra. The required group and subgroup orders are therefore 168 and 24 in both constructions; the genus-three side-generator orders are 7 and 7, and the displayed genus-four side-generator orders are 4 and 4, with cone-word order 7 in both cases.

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.

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.

The fiber monodromy is periodic when isotopy is allowed to move the boundary, in the sense of a periodic surface homeomorphism. Here is a justification from the Seifert fibered space structure, rather than a special formula about torus knots. Its base is a disk with cone-point orders $p,q$. The fiber Seifert surface is an incompressible surface because its inclusion in the mapping torus is injective on fundamental groups. It is also a boundary-incompressible surface: lift a proposed boundary-compressing disk to the infinite cyclic cover $\Sigma \times \mathbb{R}$ and project to $\Sigma$. This would homotope an essential arc of $\Sigma$ into $\partial \Sigma$, contradicting its essentiality. The classification of incompressible surfaces in Seifert fibered spaces therefore applies. A vertical surface in a Seifert fibered space has zero intersection with a regular Seifert fiber, whereas the fibration class evaluates on $f=pqm$ as $pq$. Thus the Seifert surface is isotopic to a horizontal surface in a Seifert fibered space, with $pq$ intersections per regular orbit.

Following the oriented Seifert fibers from one intersection to the next gives a first-return homeomorphism of the horizontal topological surface. It is a representative of the fiber monodromy, and its $pq$-th power is the identity. It has order $pq$: a regular orbit meets the connected horizontal surface in a Seifert fibered space in $pq$ cyclically ordered points. This realizes the periodic case of the Nielsen–Thurston classification theorem. It is neither an Anosov homeomorphism nor a pseudo-Anosov homeomorphism.

There is a boundary convention here. A representative fixed pointwise on the boundary is not periodic relative to the boundary: its $pq$-th power is a boundary Dehn twist (with sign determined by the return-map convention). Capping the boundary, or allowing it to rotate during isotopy, gives the finite-order representative intended by “periodic”. This distinguishes periodic surface type from literal finite order in the mapping class group relative to the boundary.

For any fibered knot, its Alexander module is $H_1(\Sigma ;\mathbb{Z})$ with the deck transformation given by homological monodromy, so its Alexander polynomial of a knot is $\det (tI-{\varphi }_{\ast })$ up to a unit. The general Turaev torsion relation $\Delta (t)\doteq (1-t)\tau (X)$ and part (b) therefore giveThis rational expression is a polynomial: in its factorization into cyclotomic polynomials, $\gcd (p,q)=1$ cancels all denominator factors. It is monic, has constant coefficient one, and has polynomial degree $pq+1-p-q=(p-1)(q-1)$. The characteristic polynomial has polynomial degree $2g$, is monic, and has constant coefficient $\det {\varphi }_{\ast }=1$, because homological monodromy preserves the intersection form. These facts remove the unit ambiguity and giveThe sign change between $\det ({\varphi }_{\ast }-tI)$ and $\det (tI-{\varphi }_{\ast })$ is trivial because $\mathrm{rank}{H}_1(\Sigma )=2g$ is even. As a separate geometric check, the orbifold Euler characteristic of the base is $1/p+1/q-1$, and its $pq$-sheeted horizontal surface in a Seifert fibered space has Euler characteristic $\chi (\Sigma )=p+q-pq=1-2g$.