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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.

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.

Take $N$ to be a closed compact Riemannian manifold and use the positive Laplace-Beltrami operator. Its Riemannian heat kernel $K_N(t,x,y)$ is the integral kernel of $e^{-t{\Delta }_N}$:It solves $({\partial }_t+{\Delta }_x)K_N=0$ for $t>0$ and tends to the identity kernel as $t\downarrow 0$. For an orthonormal eigenbasis, the spectral expansion of the Riemannian heat kernel isThe bar is required for a complex basis. With boundary, a specified invariant boundary condition must also be included in this definition.

For the quotient formula use the covering-space interpretation: $U$ acts freely and properly discontinuously by Riemannian isometries. On compact $N$ such a discrete group is finite. Freeness alone, without this covering hypothesis, does not justify the image sum: an infinite dense subgroup of circle rotations, for example, acts freely but has no manifold quotient. Positive-dimensional group actions require a different quotient analysis.

For lifts $x,y$ of $\bar{x},\bar{y}\in M$, the heat kernel on a finite isometric quotient isInvariance of $K_N$ under simultaneous Riemannian isometries and reindexing the sum show that this is independent of both lifts. A local isometry commutes with the Laplacian, so the sum satisfies the quotient heat equation. For its initial condition, integrate over a fundamental domain $F\subset N$ against a lifted function. The terms combine into the integral over all of $N$, whose initial limit is $f(x)$. Uniqueness of the heat evolution proves the formula. There is no factor $1/|U|$ in this kernel formula.

On the diagonal the sum is $U$-invariant. Its integral over $F$ is therefore $1/|U|$ times its integral over $N$, giving the heat traceThus the normalization factor appears in the trace, not the kernel. Put $F_t(g)={\int }_N{K}_N(t,x,gx)dV(x)$. For any finite isometry group $T$, changing variables $x=hy$ proves $F_t(hg{h}^{-1})=F_t(g)$: it is a class function on $T$.

For Gassmann equivalent subgroups $U_1,U_2\le T$, their intersections with each conjugacy class have equal size, and their orders are equal. If both act freely, the heat traces of their quotients satisfyso they agree for every $t>0$. Since $Z_M(t)={\sum }_{\lambda }m(\lambda )e^{-t\lambda }$, equality determines the spectrum with multiplicities: take $t\to \infty$ to recover the smallest eigenvalue and its multiplicity, subtract that term, and repeat. This proves Sunada theorem. Equivalently, projection onto $U_i$-invariant functions averages the group action, and the quotient eigenvalue multiplicity is $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, with ${\chi }_{\lambda }$ the eigenspace character of a representation. Almost conjugacy equalizes these averages.

For a finite subgroup $U$ of a group $T$ of Riemannian isometries of a closed manifold $N$, assume $U$ acts freely and define $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$. Invariance of the Riemannian heat kernel and change of variables show that $I_t$ is a class function on $T$. For $U\le T$, the heat kernel on a finite isometric quotient and integration over a fundamental domain give the displayed heat trace formula. Grouping by conjugacy classes proves equality for Gassmann equivalent subgroups. More generally the same argument allows fixed points of $T$ outside the freely acting subgroups; only the quotient under consideration must be smooth.