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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.

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.