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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.

Here “bumpy” means a nowhere locally homogeneous metric, not the different convention concerning nondegenerate closed geodesics. A smooth Riemannian metric has this property if every local isometry between connected open subsets is the identity on its domain; in particular, no two distinct open subsets are isometric. The Sunada local isometry lemma states that, on a compact smooth manifold without boundary of dimension at least two, such Riemannian metrics form a residual set and are dense in the $C^{\infty }$ topology. The dimension restriction matters: in dimension one, arclength coordinates always give nontrivial local isometries.

An element $j_x^{k}f$ of the jet bundle of maps $J^k(M,N)$ records the source point $x$, target value $f(x)$ and all derivatives through order $k$. If $\dim M=m$ and $\dim N=n$, there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivative multi-indices of order $r$, and $n$ components for each one. ThusFor the projection $j_x^{k}f\mapsto (x,f(x))$ onto $M\times N$, the fiber has dimension $n((\genfrac{}{}{0ex}{}{m+k}{k})-1)$. In source and target coordinates its Taylor coefficients identify it withThis is a coordinate-dependent identification for $k>1$, since a change of coordinates mixes different derivative orders. For $k=1$ the fiber is canonically $\mathrm{Hom}(T_{x}M,T_{y}N)$; for higher orders, truncation to the previous jet bundle of maps is an affine bundle modeled on ${\mathrm{Sym}}^k(T_x^{\ast }M)\otimes T_{y}N$. For $k=0$ the fiber is a point.

The density assertion in the printed question requires distinct closed domains. Literally, for $i=k$, the identity is an isometry for every Riemannian metric, so ${\mathcal{S}}_{ii}$ is the whole metric space and its complement is empty. Merely indexing the same closed ball twice has the same defect. The intended assertion is for distinct embedded closed balls $K_i={\overline{U}}_i$ and $K_k={\overline{U}}_k$, with an isometry meaning an onto diffeomorphism preserving the metric. The proof below establishes precisely this corrected assertion.

Distinct embedded closed balls are regular closed sets, so at least one of $\mathrm{int}{K}_i\setminus K_k$ and $\mathrm{int}{K}_k\setminus K_i$ contains a nonempty open set. Indeed, if both were empty, taking closures would give both inclusions and hence equality. Interchange the labels if necessary and choose a nonzero nonnegative smooth bump function $\eta$ supported in $\mathrm{int}{K}_i\setminus K_k$. For any given Riemannian metric $g$, make the localized volume perturbationThe Riemannian volume form satisfies $d{V}_{g_t}=e^{mt\eta }d{V}_g$. ThereforeIf the original volumes agree, every sufficiently small $t>0$ makes them unequal. If they already disagree, continuity preserves that inequality for sufficiently small $t$. An onto isometry must preserve volume, so each of these perturbed Riemannian metrics belongs to $\mathcal{C}{\mathcal{S}}_{ik}$. Finally $g_t\to g$ in $C^{\infty }$: all derivatives of $e^{2t\eta }-1$ tend to zero uniformly on compact sets. This proves density of $\mathcal{C}{\mathcal{S}}_{ik}$ for distinct closed balls, in fact in every positive dimension. The localized volume perturbation settles this particular density claim directly. It also does not imply the stronger Sunada local isometry lemma by itself, since a general local isometry need not send one preselected basis ball onto another.

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.