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For the nonnegative Hodge Laplacian convention $\Delta f=-\mathrm{div}\mathrm{grad}f$, a smooth function on a closed manifold with $\Delta f\ge 0$ has $\int \Delta f=0$ and hence $\Delta f=0$. Integration by parts gives $\int |df{|}^2=0$, so it is constant on each connected component. Global constancy requires connectedness; boundary conditions are needed if a boundary is allowed.

On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

For a smooth map $f:M\to N$ between oriented connected closed manifolds of the same dimension, a normalized volume form on the target gives $\mathrm{deg}f={\int }_M{f}^{\ast }\omega$. The formula agrees with the degree as a sum of local degrees. A change of normalized top-form adds an exact form by top-dimensional de Rham cohomology, whose pullback has zero integral by Stokes theorem.

For the positive Laplace-Beltrami operator on a closed manifold, the singular support of the wave trace lies in zero together with signed lengths of closed geodesics. Under the clean fixed-set hypotheses, its singularity at a periodic-geodesic length has an expansion determined by the geodesic flow and its linearized return map. In particular, when the relevant leading coefficient does not vanish through cancellation, that length is determined by the spectrum. The inclusion of singular supports alone does not justify an unconditional recovery of every geodesic length.

A perpendicular Jacobi field on a Riemannian surface has scalar equation $\ddot{y}+Ky=0$. For strictly negative curvature on a closed manifold, the displayed derivative is uniformly positive definite, so the quadratic-form criterion for an Anosov flow makes the geodesic flow Anosov. When $K=-1$, the stable and unstable lines in the canonical frame are respectively $\mathbb{R}(H-V)$ and $\mathbb{R}(H+V)$, with factors $e^{-t}$ and $e^t$.

A harmonic one-form is a differential one-form $\alpha$ satisfying $\Delta \alpha =0$ for the Hodge Laplacian. On a closed manifold, the Hodge decomposition theorem identifies these forms with degree-one de Rham cohomology. The Bochner-Weitzenbock formula for one-forms implies that harmonic one-forms are parallel under nonnegative Ricci curvature.

On a closed manifold with a Riemannian metric, a smooth solution of ${\partial }_{t}u+\Delta u=0$ for the positive Laplace-Beltrami operator satisfies the displayed identity by integration by parts. Zero initial data force the solution to vanish, proving uniqueness and the semigroup property for heat evolution.

On a closed manifold with a Riemannian metric, the minimum of a smooth heat solution cannot fall below its initial minimum. At a spatial minimum, $\Delta u\le 0$ for the positive Laplace-Beltrami operator. Adding a small increasing function of time and considering the first crossing gives the assertion. Applying the result to nonnegative initial functions shows that the Riemannian heat kernel is nonnegative.

On a closed manifold, the positive Laplace-Beltrami operator has a smooth heat kernel and its heat semigroup is a trace-class operator for positive time. Expanding in a smooth orthonormal eigenbasis gives $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{\varphi }_j(x)\overline{{\varphi }_j(y)}$. Elliptic regularity bounds each derivative of ${\varphi }_j$ polynomially in ${\lambda }_j$, while the exponential dominates those powers, so the expansion converges with all derivatives. Integrating the diagonal and using the unit norm of each eigenfunction proves the formula. For a bundle operator, insert the fibre matrix trace inside the integral.

Glue the boundary of a Möbius band to the projective line in the real projective plane by a mapping degree $n$. The band is the mapping cylinder of the degree-two boundary covering of its core. Collapsing the connecting tree in the resulting double mapping cylinder gives a CW complex with one vertex, edges $a,b$, and two attaching words $a^2$ and $a^{-n}{b}^2$. Its cellular chain complex has degree-two boundary matrix $\left(\begin{array}{cc}2& -n\\ 0& 2\end{array}\right)$. Its determinant is four, so $H_2=0$. The Smith normal form gives $H_1\cong \mathbb{Z}/4$ for odd $n$ and $(\mathbb{Z}/2{)}^2$ for even $n$; $H_0=\mathbb{Z}$ and higher homology vanishes. The space cannot be homotopy equivalent to a closed manifold: mod-two homology forces a potential manifold to be a surface, and its Euler characteristic is one, whereas the only such closed connected surface is the projective plane, with different integral $H_1$.

For $n\ge 2$, the Wang sequence and the mapping degree $(-1{)}^{n+1}$ of the antipodal map give $H_0=H_1=\mathbb{Z}$. If $n$ is odd, $H_n=H_{n+1}=\mathbb{Z}$; if $n$ is even, $H_n=\mathbb{Z}/2$ and $H_{n+1}=0$. All other groups vanish. At $n=1$ the mapping torus is a torus with $H_1={\mathbb{Z}}^2$. At $n=0$, the two-point swapping mapping torus is a circle. Every example is already a closed manifold; for $n\ge 1$ it is orientable exactly when $n$ is odd.

For maps between oriented connected closed manifolds of the same dimension, a homotopy induces equal maps on top-dimensional homology, hence equal coefficients on the fundamental class. For smooth maps and a smooth homotopy, the degree by integration of a pullback volume form gives another proof: the Generalized Stokes theorem on the homotopy cylinder makes the difference of endpoint pullback integrals zero because the target top form is closed. This is an obstruction to extending a degree-one sphere identity over a ball.

For the positive scalar Laplace-Beltrami operator on a closed manifold with a Riemannian metric, the coefficient of $t^2$ in the normalized heat trace is the displayed integral. The local coefficient additionally has a scalar-curvature divergence term whose integral vanishes. In dimension two, $R=2K$, $|\mathrm{Ric}{|}^2=2K^2$, $|\mathrm{Rm}{|}^2=4K^2$, giving $a_2=\frac{1}{15}{\int }_M{K}^{2}dA$. Together with volume and total Gaussian curvature, it determines the variance of curvature and detects constant curvature.

The vector space of functions satisfying $\Delta f=\lambda f$ for a specified Laplace-Beltrami operator. On a closed manifold its dimension is finite and is the spectral multiplicity of $\lambda$. A finite isometric group action preserves this space; invariant elements are the Laplacian eigenfunctions on the corresponding smooth quotient.

The spectrum of a specified self-adjoint Laplace-Beltrami operator. On a closed manifold of positive dimension, it is a discrete sequence of nonnegative eigenvalues tending to infinity, counted with the dimensions of their Laplacian eigenspaces. On a noncompact manifold or a manifold with boundary, the domain and boundary conditions are part of the operator definition and continuous spectrum can occur.

For maps of oriented connected closed manifolds of the same dimension, apply functoriality of homology to $f_{\ast }[M]=(\mathrm{deg}f)[N]$ and $g_{\ast }[N]=(\mathrm{deg}g)[L]$. The composite coefficient is their product. This recovers the multiplication of signed sheet counts for compositions of finite covering maps.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.

Choose a smooth cutoff $\eta (x,y)$ that is one near the diagonal and supported where these normal coordinates exist, and form the heat parametrixIt tends to the identity kernel as $t\downarrow 0$: in normal coordinates the Gaussian is an approximate identity, and $u_{0}J=J^{1/2}\to 1$ at the centre. Its residual $R_N=({\partial }_t+{\Delta }_x)P_N$ is $\eta g_t{t}^N{\Delta }_x{u}_N$ plus cutoff-derivative terms supported a positive distance from the diagonal. Those latter terms are smaller than every power of $t$. Hence $R_N=O(t^{N-m/2})$ uniformly. Every prescribed finite number of derivatives has the same kind of estimate after increasing $N$; differentiating a Gaussian costs only finitely many powers of $t$.

For $N$ large enough that the residual is continuous and bounded up to time zero, use the Volterra parametrix correction. For kernels defineSet $Q=-R_N+R_N\ast R_N-R_N\ast R_N\ast R_N+\cdots$. A bounded residual with bound $C$ givesso this series converges and $Q+R_N+R_N\ast Q=0$. Because $P_N$ has identity initial kernel, differentiating the convolution gives $({\partial }_t+{\Delta }_x)(P_N\ast Q)=Q+R_N\ast Q$. Thusis an exact kernel with the correct initial condition. Taking $N$ sufficiently large gives any required finite regularity; uniqueness of the heat equation makes the kernels obtained from different $N$ equal. For uniqueness, the difference applied to smooth initial data has zero initial value, and the energy identity $\frac{d}{dt}\Vert v{\Vert }_2^2=-2\Vert \nabla v{\Vert }_2^2$ forces $v=0$. Parabolic regularity then gives smoothness for positive time.

The Gaussian part of $P_N$ has uniformly bounded spatial $L^1$ norm, so the correction is $O(t^{N+1-m/2})$. This proves, uniformly near the diagonal,with corresponding derivative estimates on increasing the truncation order. In particular on every parabolic neighbourhood $r\le B\sqrt{t}$ it is the stated Gaussian expansion with relative remainder $O(t^{N+1})$, and on the diagonal it yields the complete heat trace expansion. The standard off-diagonal Gaussian version on compact subsets of a sufficiently small convex normal neighbourhood follows from the same convolution estimates with the Gaussian factor retained: the inequalitylocalizes the convolution to the unique joining geodesic; its transverse Hessian is positive, and Gaussian integration gives the successive transport orders. The additive estimate above already specifies an appropriate uniform near-diagonal remainder without asserting a relative estimate far from the diagonal on the basis of a crude bound.

If $M$ is noncompact, use the minimal Friedrichs extension heat semigroup; if there is a boundary, specify boundary conditions. The same construction is local in the interior. One may extend a relatively compact coordinate neighbourhood to a closed auxiliary manifold; localization errors are supported a positive distance away and are exponentially small near the smaller diagonal neighbourhood. Thus the local coefficients remain the same. Finite integrated heat invariants and a discrete spectrum require the global hypotheses used above.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

For a smooth function $f$ and the Levi-Civita connection $\nabla$, the Riemannian Hessian is the covariant two-tensorThe connection rules show that this is linear over smooth functions in both vector-field arguments, so it is a tensor. Torsion-freeness gives its symmetry:Its metric trace defines the paper's Laplace-Beltrami operator, $\Delta ={\mathrm{div}}_g{\mathrm{grad}}_g$; its eigenvalues on a closed manifold are nonpositive. We use $P=-\Delta$ when listing nonnegative eigenvalues or writing $e^{-tP}$.

At $p$, choose an orthonormal basis $e_1,\dots ,e_d$ of the tangent space and unit-speed geodesics ${\gamma }_i(t)={\exp }_p(t{e}_i)$. Differentiating twice along a curve givesThe covariant acceleration vanishes on a geodesic. Taking the trace at $t=0$ proves the geodesic trace formula for the Laplace-Beltrami operator:The last equality is Taylor's formula along the geodesics, and expresses the Laplacian entirely through nearby function values.

For stabilization, let $X_1,X_2$ be the assumed isospectral nonisometric four-dimensional tori, and let $k=n-4>0$. Take the same small flat torus $F_{\epsilon }={\mathbb{R}}^k/\epsilon {\mathbb{Z}}^k$ as an extra factor. On a Riemannian product,If ${\varphi }_j$ and ${\psi }_{\ell }$ are eigenfunctions with eigenvalues ${\lambda }_j$ and ${\mu }_{\ell }$, then ${\varphi }_j{\psi }_{\ell }$ has eigenvalue ${\lambda }_j+{\mu }_{\ell }$. Products of orthonormal eigenbases are complete in the product $L^2$ space, so this lists all eigenvalues with multiplicities, including coincidences between different sums. Consequently the two products are isospectral.

One must also prove that multiplying by a common factor has not accidentally made the two manifolds isometric. ChooseThe injectivity radii are positive because the factors are compact. A closed product geodesic of length less than $\delta$ projects to a closed geodesic of length less than $\delta$ in $X_i$. Such a projected geodesic must be constant: otherwise its initial vector over one period would be a nonzero vector of norm below the injectivity radius mapping back to its starting point under the exponential map. Thus all sufficiently short closed geodesics lie in the $F_{\epsilon }$ directions. Conversely the $k$ coordinate circles, of length $\epsilon$, pass through every point and span those directions.

It follows that the smooth distribution tangent to the small flat-torus factor is characterized intrinsically by the span of tangent vectors to closed geodesics of length below $\delta$. A hypothetical product isometry would preserve this smooth distribution and its orthogonal complement. The latter has leaves $X_i\times \{q\}$; the isometry and its inverse take whole leaves to whole leaves, thereby giving an isometry $X_1\cong X_2$, a contradiction. This proves isospectral stabilization by a small flat torus:This argument works even without assuming that the original four-dimensional torus metrics are flat.

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.