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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.