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Suppose a fixed pair of Gassmann equivalent finite covers of a compact two-dimensional orbifold has isometric limiting hyperbolic metrics. Choose nearby orbifold-smooth nowhere locally homogeneous metrics and rescale them to the displayed distinct areas. Scaling by a positive constant multiplies area by that constant in dimension two, preserves local rigidity, and preserves convergence when the constants tend to one. Sunada theorem supplies equal spectra within each pair. Local rigidity forces any quotient isometry to descend to the identity on the base; nonconjugate covering subgroups therefore exclude within-pair isometry. Distinct areas exclude all cross-pair isometries and spectra, because the leading heat trace coefficient is area divided by $4\pi t$. Fixed markings of the two isometric limiting covers give convergence to the same metric. The existence of those isometric limits is an additional geometric input, not a consequence of Gassmann equivalence alone.

Choose distinct odd primes and take products of the nonisomorphic Gassmann equivalent regular subgroups, selecting one of the two groups at each prime. Product conjugacy-class counts give $2^n$ Gassmann equivalent subgroups of the same finite product of symmetric groups. Their fundamental groups are distinguished by whether the unique Sylow subgroup at each chosen prime is abelian. The closed-manifold realization of a finitely presented fundamental group and Sunada theorem therefore produce $2^n$ pairwise nonhomeomorphic, and hence nonisometric, isospectral closed four-manifolds. Gassmann equivalence alone, without a topology distinction, does not imply nonhomeomorphism.

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.

Subgroups $H_1,H_2$ of a finite group $G$ are Gassmann equivalent if $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$. Equivalently their coset permutation representations have the same character of a representation. They need not be conjugate subgroups. The Sunada theorem turns this equality into equality of Laplacian eigenfunction multiplicities.

For the positive Laplace-Beltrami operator on a closed compact Riemannian manifold, the heat trace is $Z_M(t)={\sum }_j{e}^{-t{\lambda }_j}={\int }_M{K}_M(t,x,x)dV(x)$. It records every eigenvalue with multiplicity. In a finite free isometric quotient it equals $|U{|}^{-1}{\sum }_{u\in U}{\int }_N{K}_N(t,x,ux)dV(x)$. The orbital integrals form a class function of the ambient finite isometry group, so Gassmann equivalence gives equal quotient heat traces and hence Sunada theorem.

Two Riemannian manifolds are isospectral for a specified differential operator if their eigenvalues, counted with multiplicities and with the same boundary conditions, agree. Riemannian isometries preserve the Laplace-Beltrami operator spectrum, but the converse can fail. The transplantation theorem and the Sunada theorem give systematic constructions.

For an odd prime $p$, embed the elementary abelian group $(C_p{)}^3$ and the Heisenberg group over a prime field regularly in $S_{p^3}$. Every nonidentity element has order $p$ and hence regular permutation cycle type $p^{p^2}$. The two subgroups have equal intersection counts with every symmetric-group conjugacy class, so they are Gassmann equivalent, but they are not isomorphic because one is abelian and the other is not. Using a compact simply connected isometric cover with ambient deck group $S_{p^3}$ gives quotients with these nonisomorphic fundamental groups; Sunada theorem makes them isospectral while their topology differs.

Embed $C_4\times C_4$ and $Q_8\times C_2$ in $S_{16}$ by their regular representations. Both have one element of order one, three of order two, and twelve of order four. An order-$r$ element in a regular action has cycle type $r^{16/r}$. Their intersections with every conjugacy class of $S_{16}$ consequently have the same size, so they are Gassmann equivalent. The first group is abelian and the second is not, so their underlying groups are not isomorphic. Finite-cover constructions from the Sunada theorem therefore give isospectral manifolds with different fundamental groups.

The Sunada theorem is the following. Let $X\to M_0$ be a finite normal Riemannian covering of a connected closed Riemannian manifold, with finite deck transformation group $T$. If $U_1,U_2\le T$ satisfy Gassmann equivalence, meaningthen $U_1\backslash X$ and $U_2\backslash X$ have the same scalar Laplacian spectrum, with multiplicities. Equivalently, an epimorphism ${\pi }_1(M_0)\to T$ yields isospectral covers corresponding to the inverse images of $U_1,U_2$.

Here is the representation-theoretic proof. Every finite-dimensional Laplacian eigenspace $E_{\lambda }$ of $X$ is a group representation of $T$, since isometries commute with the Laplace-Beltrami operator. Eigenfunctions on $U_i\backslash X$ identify with $E_{\lambda }^{U_i}$. Averaging gives the projection $P_i=|U_i{|}^{-1}{\sum }_{u\in U_i}u$, soThe two subgroup orders agree, and the character of a representation ${\chi }_{\lambda }$ is constant on conjugacy classes. The displayed multiplicities are therefore equal for every $\lambda$. This proof also applies when $T$ has fixed points on $X$, provided the two subgroups $U_i$ act freely, so their quotients are smooth manifolds. We shall use this extension for an orbifold base in Question 4.

The precise geometric test for an isometry uses the full isometry group of a common universal cover. Equip the simply connected cover $\tilde{X}$ of $M_0$ with the lifted metric and let ${\Gamma }_i$ be the subgroups of its deck transformation group corresponding to the two covers. ThenIndeed, such an $F$ descends to an isometry of the quotients. Conversely an isometry of the quotients lifts to their simply connected covers; lift its inverse with compatible basepoints to see that the lift is bijective. Since it preserves the lifted metrics locally, it is a global isometry, and equivariance of the covering maps shows that it conjugates the deck subgroups. This proves the criterion, rather than assuming that an isometry must preserve the map to $M_0$. In particular, nonconjugacy inside $T$ alone does not establish nonisometry; the full covering space may have additional isometries. Nonisomorphic fundamental groups are a sufficient topological obstruction, and length-identified geometric configurations provide another.

For a concrete nonisomorphic Gassmann equivalent regular subgroups example, fix an odd prime $p$ and take $T=S_{p^3}$. Embed $E=(C_p{)}^3$ and $B={\mathrm{UT}}_3({\mathbb{F}}_p)$ through their regular permutation actions. Both have order $p^3$. Every nonidentity element of $E$ has order $p$. For $B$, write its elements as $I+N$ with $N^3=0$; thenover ${\mathbb{F}}_p$, including $p=3$, so its nonidentity elements also have order $p$. In a regular action, each nonidentity order-$p$ element has cycle type $p^{p^2}$. Thus each subgroup meets the identity class in one element, the class of that cycle type in $p^3-1$ elements, and every other conjugacy class in zero elements. They are Gassmann equivalent. They are not isomorphic: $E$ is abelian, whereas the Heisenberg group $B$ has $[I+E_{12},I+E_{23}]=I+E_{13}\ne I$.

To obtain arbitrarily many examples in dimension five, choose $r$ distinct odd primes $p_1,\dots ,p_r$ with $2^r\ge n$. For each prime use the preceding pair $E_j,B_j\le S_{p_j^3}$. SetA conjugacy class in a direct product is a product of conjugacy classes, and its intersection count with $U_{\epsilon }$ is the product of the individual counts. Thus all $2^r$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{\epsilon }$ is abelian precisely when ${\epsilon }_j=0$, and an abstract group isomorphism preserves this property at each prime.

We use the closed-manifold realization of a finitely presented fundamental group in dimension five. Its construction here is short. From a finite presentation of $T$ with $s$ generators, start with the connected sum of $s$ copies of $S^1\times S^4$. Its fundamental group is free on those generators. Represent the finitely many relators by disjoint embedded circles, which is possible by general position for curves in a manifold in dimension five. Their oriented normal bundles are trivial. Perform framed circle surgery, removing $S^1\times D^4$ and inserting $D^2\times S^3$. The Seifert-van Kampen theorem shows that each surgery kills exactly the normal closure of the corresponding relator: the attaching circle bounds the new disk, and the new $S^3$ factor contributes no fundamental group. The resulting closed connected five-manifold $M_0$ has ${\pi }_1(M_0)=T$.

Choose any smooth Riemannian metric on $M_0$ and lift it to its universal cover $X$. Since $T$ is finite, $X$ is a compact simply connected five-manifold on which $T$ acts freely and isometrically. The Sunada theorem makes all $U_{\epsilon }\backslash X$ isospectral, whileshows that they are pairwise nonhomeomorphic, hence nonisometric. Taking any $n$ members proves for every positive integer $n$, there are $n$ pairwise nonisometric closed five-manifolds with the same Laplacian spectrum.

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.

Let $p:N\to M=U\backslash N$ be a finite normal Riemannian covering, so $U$ acts freely by deck isometries. With $x,y$ any lifts of $\bar{x},\bar{y}$, the heat kernel on a finite isometric quotient satisfiesThere is no factor $1/|U|$ in this kernel formula. Simultaneous invariance of $H_N$ under isometries proves independence of the lifts, by reindexing $U$. Each term solves the lifted heat equation. If $F_U$ is a fundamental domain and $f$ a function on $M$, the integral of the right side against $f(\bar{y})$ over $F_U$ equalsbecause the translates $u{F}_U$ tile $N$ up to sets of measure zero. This converges to $f(p(x))$ as $t\downarrow 0$, proving the initial condition. Uniqueness of the heat equation solution on the closed manifold gives the formula. For a noncompact finite cover, use the minimal heat kernels, or compatible Friedrichs heat realizations. The normalized pullback $f\mapsto |U{|}^{-1/2}f\circ p$ is unitary onto the deck-invariant $L^2$ subspace and preserves the Dirichlet energy, so it intertwines the heat semigroups and yields the same kernel formula. No uniqueness of unrestricted noncompact heat-equation solutions is needed.

Set $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$ for an isometry $g$. Diagonal integration and the covering degree giveUnlike the kernel formula, the trace formula has an averaging factor: integration of a quotient function over $N$ is $|U|$ times integration over $U\backslash N$.

Suppose now $U\le T$, with $T$ any larger group of isometries of the closed manifold $N$. Change variables $x=hy$ and use simultaneous heat-kernel invariance to obtainThus $I_t$ is a class function, and the Sunada orbital heat-trace formula can be grouped by conjugacy classes $C$ of $T$:Only classes meeting the finite subgroup $U$ contribute, so this class sum is finite even if the larger isometry group $T$ is infinite.

For finite $T$, Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every conjugacy class. Summing these counts gives $|U_1|=|U_2|$, so the displayed formula gives identical heat traces. Their spectra, with multiplicities, agree: as $t\to \infty$, the constant limit identifies the multiplicity of zero; after subtracting it, the slowest-decaying exponential identifies the least positive eigenvalue and its multiplicity. Repeating determines every subsequent eigenvalue. This proves the Sunada theorem in the stated covering situation.

There is an important convention when the universal cover $N$ has an infinite deck group $T$. For compact quotient manifolds, the subgroups must have finite index; the appropriate generalized Gassmann condition is equality of the characters of the finite coset permutation representations, not equality of possibly infinite cardinalities of conjugacy-class intersections. Let $K$ be the intersection of the two subgroup cores. Each core is the kernel of the corresponding finite coset action, so $K$ is normal and of finite index in $T$, and $K\subseteq U_1\cap U_2$. The manifold $X=K\backslash N$ is then a compact finite normal cover of $M_0$. The finite group $G=T/K$ acts on $X$, and $H_i=U_i/K$ have equal coset permutation characters in $G$. For a finite group, the fixed-coset count isEquality at the identity gives equal subgroup orders, and then equality for all $g$ gives almost-conjugacy. Applying the proved finite theorem to $H_i\backslash X=U_i\backslash N$ establishes isospectrality also in this finite-index interpretation. It avoids taking the divergent heat trace of a noncompact universal cover.

To realize a finitely presented group $T=⟨a_1,\dots ,a_r\mid R_1,\dots ,R_s⟩$, begin with the closed smooth four-manifold $B={\#}_{j=1}^r(S^1\times S^3)$, whose fundamental group is free on the $a_j$. Represent the relator words by disjoint smoothly embedded circles; general position permits this in dimension four, including separating self-intersections of representatives. Their oriented normal rank-three bundles are trivial, so perform framed surgeries replacing $S^1\times D^3$ by $D^2\times S^2$. Removing the circles' tubular neighborhoods leaves the fundamental group unchanged, by codimension-three general position. The Seifert-van Kampen theorem shows that each replacement kills exactly the normal closure of its relator. The resulting closed smooth $M_0$ has ${\pi }_1(M_0)\cong T$. This gives a concrete closed-manifold realization of a finitely presented fundamental group. Equip it with any smooth Riemannian metric and lift that metric to the universal and intermediate covers; all covering maps become local isometries.

To ensure nonisometry, one must choose nonconjugate subgroups. If $U_1=U_2$, or they are conjugate, the quotients are isometric for every lifted metric, so Gassmann equivalence alone cannot imply nonisometry. For nonconjugate subgroups, choose the base metric to have no nonidentity local isometry between open sets, using the Sunada local isometry lemma in dimension at least two. If $F:M_1\to M_2$ were an isometry, a local inverse of the first covering followed by $F$ and the second covering would be a local isometry of $M_0$, hence the identity. Thus the covering maps satisfy $p_{2}F=p_1$ everywhere. Classification of connected covers then forces $U_1,U_2$ to be conjugate in ${\pi }_1(M_0)=T$, a contradiction. Therefore nonconjugate almost-conjugate subgroups and a locally rigid base metric give isospectral nonisometric quotients.

Take the nonnegative scalar Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$ on a closed compact Riemannian manifold, and include every eigenvalue with its multiplicity. The spectrum contains substantial global geometric information, but generally does not determine the Riemannian metric up to isometry. Specifying the operator and its domain matters: the spectrum on functions is different information from the spectra of the Hodge Laplacian on differential forms, and a boundary requires fixed boundary conditions.

The Weyl law makes dimension and volume audible. If $N(\lambda )$ counts eigenvalues at most $\lambda$, thenwhere ${\omega }_d$ is the volume of the Euclidean unit ball. Thus the growth exponent recovers $d$, and its leading coefficient recovers $\mathrm{Vol}(M)$. The multiplicity of zero gives the number of connected components: a function in the kernel has $\int |\nabla f{|}^2=0$ and hence is constant on each component.

More detailed invariants come from the heat trace, which is determined directly by the spectrum:The first heat invariants for a closed manifold areHere $R$ is scalar curvature, $\mathrm{Ric}$ the Ricci curvature, and $\mathrm{Rm}$ the Riemann curvature tensor. A divergence term in the local coefficient integrates to zero on a closed manifold. The expansion supplies further integrated curvature expressions, rather than a pointwise curvature map.

On a closed surface, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$ and $|\mathrm{Rm}{|}^2=4K^2$, so the same coefficients becomeThe Gauss-Bonnet theorem gives $\int KdA=2\pi \chi (M)$; hence the constant term of the heat trace is $\chi (M)/6$. In particular the Euler characteristic, and for a connected orientable surface its genus, are spectrally determined. There is even spectral determination of hyperbolic curvature on a surface: if this spectrum agrees with that of a curvature-$-1$ surface, the coefficients give $\int K=-A$ and $\int K^2=A$, whenceThus $K=-1$ everywhere. This determines local curvature, while leaving room for globally nonisometric hyperbolic surfaces.

Geodesic information is also accessible in important settings. For closed hyperbolic surfaces, the Selberg trace formula makes the scalar spectrum equivalent to the unmarked length spectrum with its appropriate multiplicities. This does not supply a marked length spectrum: it does not identify which group element or intersecting curve has a given length. For general manifolds, one must not simply assert this equivalence; the singularities of the wave trace can be affected by degeneracies and cancellations.

The historic development separates these positive results from the inverse problem. Weyl's 1911 eigenvalue asymptotics established the leading geometric information. The Minakshisundaram-Pleijel heat expansion of 1949 brought local curvature into the short-time analysis. In 1964 Milnor exhibited nonisometric flat tori in dimension sixteen with identical scalar spectra, showing that all eigenvalues together need not recover the metric. Kac's 1966 question about a vibrating drum popularized the corresponding inverse problem for planar domains. Vignéras's 1980 hyperbolic examples and the Sunada theorem of 1985 extended the supply of counterexamples; the latter turns Gassmann equivalence of finite subgroups into a geometric construction. The four-dimensional tetracode length-isospectral lattice construction and the planar examples of Gordon, Webb and Wolpert in 1992 made nonuniqueness explicit in smaller dimensions and in the original drum setting.

Nor is scalar isospectrality a universal topological invariant strong enough to determine the manifold. For example, Tetra and Didi are scalar-isospectral closed orientable flat three-manifolds with first Betti numbers one and zero. Their topology therefore differs. The spectra of all Hodge Laplacians would detect these Betti numbers through the dimensions of their zero eigenspaces, illustrating the extra information obtained by changing the operator. Dimension, volume and many integrated invariants are determined; the full metric and, in general, the topology are not.

Use the canonical nonnegative Laplace-Beltrami operator and its Riemannian heat kernel for $t>0$. For the spectral conclusion we use closed compact manifolds, so the operator has a discrete spectrum and a finite heat trace. The kernel identity itself can also be formulated for the canonical Friedrichs extension on a noncompact manifold; this specifies the heat operator rather than assuming uniqueness among unrestricted solutions of the heat equation.

An isometry $h$ preserves the Laplacian, its defining quadratic form and therefore its heat semigroup. ConsequentlySet $H(x,y,t)={\sum }_{g\in U}{K}_N(\tilde{x},g\tilde{y},t)$. A different choice of lifts has the form $h\tilde{x},k\tilde{y}$ with $h,k\in U$, because the covering is normal. The resulting sum issince $g\mapsto h^{-1}gk$ permutes $U$. This proves independence of both lifts. Local smooth sections of the covering map then show that $H$ is smooth for positive time. There are finitely many summands, so spatial and time differentiation commute with the sum, and local isometry gives $({\partial }_t+{\Delta }_x)H=0$.

For the integral and initial condition, use the sheet-counting formulaIt follows by dividing the base into measurable pieces lying in evenly covered charts and using the locally isometric inverse branches. For a smooth test function $\varphi$ on the base, this formula givesOn a closed manifold these integrals are finite, and the finite sheet sum entails no convergence exchange. The right side tends to $\varphi (p\tilde{x})=\varphi (x)$ as $t\downarrow 0$ by the initial property of the Riemannian heat kernel. Thus $H$ has the correct delta initial condition.

One can identify the operator exactly, also avoiding a noncompact uniqueness issue, by heat semigroup descent through a finite normal covering. If $d=|U|$, the map $I\varphi =d^{-1/2}\varphi \circ p$ is unitary onto the $U$-invariant subspace of $L^2(N)$, becauseLocally isometric pullback proves the second equality. Smooth compactly supported invariant functions descend, and averaging over the finite group gives the corresponding equality of the closed quadratic-form domains. Hence the Friedrichs extensions and their heat semigroups intertwine. The preceding integral is precisely that descended semigroup, provingFor the noncompact version, finite covering maps are proper, so compactly supported test functions have compactly supported lifts; the same form argument is valid. The canonical heat kernels are sub-Markovian, which bounds integrals against bounded test functions. All operations used above are therefore defined. In particular there is no factor $1/|U|$ in the kernel formula: integrating over the sheets already accounts for its normalization.

Now take a finite ambient group $T$ of isometries of a closed $N$, and free Gassmann equivalent subgroups $U_1,U_2$. Define the orbital integralIt is finite, and the conjugacy-invariance identity makes it a class function on $T$. The heat kernel on a finite isometric quotient and the sheet-counting formula give the quotient heat tracesThe averaging factor belongs here, since the cover has $|U_i|$ sheets. By Gassmann equivalence, the two subgroups meet each conjugacy class in the same number of elements, and summing these numbers also gives $|U_1|=|U_2|$. Since ${\Phi }_t$ is constant on each class, the two heat traces are equal for every $t>0$.

Finally, a discrete nonnegative eigenvalue multiset is determined by its heat trace. Its zero multiplicity is ${\lim }_{t\to \infty }Z(t)$. After subtracting the zero terms, the smallest remaining eigenvalue and its multiplicity are recovered from the leading exponential decay; subtract those terms and repeat. Compactness ensures only finitely many eigenvalues in any bounded interval, so this procedure recovers the entire multiset. We have proved the Sunada theorem:The closedness assumption is used for this heat-trace deduction; on a general noncompact manifold one cannot replace it by a possibly infinite trace without further analysis.

If the initial manifold is not compact, the same isospectral conclusion can be interpreted for the canonical self-adjoint operators without taking an infinite trace. To see this, decompose the unitary representation of the finite group $T$ as $L^2(N)={⨁}_{\rho }{V}_{\rho }\otimes {\mathcal{H}}_{\rho }$, with $V_{\rho }$ irreducible and ${\mathcal{H}}_{\rho }$ its multiplicity space. Commutation with the group makes the Laplacian act on each summand as $I\otimes D_{\rho }$, by the Schur lemma applied to its bounded resolvents. The projection onto the $U_i$-invariant factor is $|U_i{|}^{-1}{\sum }_{u\in U_i}\rho (u)$, and its trace is therefore $\dim V_{\rho }^{U_i}=|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\rho }(u)$. Gassmann equivalence makes these dimensions equal. Unitary identifications $V_{\rho }^{U_1}\to V_{\rho }^{U_2}$, tensored with the identity on each multiplicity space, intertwine the restricted Laplacians. The normalized pullbacks identify those restrictions with the two quotient operators. This Sunada unitary equivalence without a finite heat trace proves equality of their full spectra, including continuous spectrum when present.

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.

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.

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 $\ell (a)$.