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Every finitely generated group acts freely and properly discontinuously by Riemannian isometries on a connected manifold in any dimension $d\ge 3$. Choose a surjection $F_r\to T$ and the regular covering of ${\#}_r(S^1\times S^{d-1})$ associated with its kernel. Pulling back a metric makes the deck transformation group isometric. Uniqueness of lifts makes its action free. The cover can be noncompact and need not be simply connected; no claim about realizing every group as a closed three-manifold fundamental group is involved.

For a finite free action by Riemannian isometries, the Riemannian heat kernel on the quotient is the image sum shown above. Integrating over a fundamental domain combines all images into the integral on the covering manifold, proving the initial condition and showing that there is no averaging factor in the kernel. The heat trace does have an averaging factor $1/|U|$, because the integral of a quotient function over the cover is $|U|$ times its quotient integral. Mere freeness of a general nondiscrete group action is not enough for this covering formula.

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 covers branched over three real points with a real base point, choose the two generating loops so complex conjugation reverses each. A sheet permutation satisfying the displayed identities lifts that conjugation to an anticonformal equivalence of the covers, including the filled branch points. For closed hyperbolic Riemann surfaces this is a Riemannian isometry. Such an isometry need not come from conjugating the subgroups inside the original deck group, so a Gassmann equivalent pair can be isometric despite nonconjugacy there.

Use the positive Laplace-Beltrami operator, $\Delta =-\mathrm{div}\nabla$. A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice and the metric descends from the Euclidean metric. Its dual lattice isIf $V$ is the covolume of $\Lambda$, the functions ${\varphi }_{\xi }(x)=V^{-1/2}{e}^{2\pi i⟨\xi ,x⟩}$ descend to the torus. Direct differentiation gives the spectrum of a flat torus:The multiset includes one entry for each dual vector. In particular zero has multiplicity one. Writing $\Lambda =B{\mathbb{Z}}^d$ identifies the dual vectors with $B^{-T}m$, $m\in {\mathbb{Z}}^d$. Standard Fourier series on the unit cube then prove that these functions are an orthonormal basis of $L^2$ on the torus. If $\Delta f=\lambda f$, its Fourier coefficients satisfy $(4{\pi }^2|\xi {|}^2-\lambda )\hat{f}(\xi )=0$. Thus every eigenfunction is a linear combination of the indicated shell of frequencies, with no additional eigenvalues. Compactness gives the discrete self-adjoint realization, whose eigenfunctions are smooth. For real functions the two characters at $\xi ,-\xi$ become cosine and sine, giving the same real multiplicities. The opposite Laplacian sign would negate all displayed eigenvalues without changing the rigidity conclusion.

To prove two-dimensional spectral rigidity, let $L={\Lambda }^{\ast }$ and recover its Gram matrix from its vector-length multiset. Let $a$ be the shortest nonzero length and choose $v\in L$ with $|v|=a$. This vector is primitive: a proper integer multiple would have a shorter lattice vector. Hence $L\cap \mathbb{R}v=\mathbb{Z}v$. From the spectral multiset subtract exactly two occurrences of each length $ka$, $k=1,2,\dots$. The remaining multiset consists precisely of the vectors not on this line, including correct multiplicities at coincident lengths. Its least length $b$ is the length of a shortest vector $w$ independent of $v$.

The pair $(v,w)$ is a basis of $L$. Otherwise take a lattice point in a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ and reduce its two coefficients into $[-1/2,1/2]$. The resulting nonzero lattice vector $z=\alpha v+\beta w$ is not on $\mathbb{R}v$, by primitivity of $v$. Butwith strict inequality: when both coefficients are nonzero the triangle inequality is strict for independent vectors, and when one vanishes its length is at most $b/2$. This contradicts the definition of $b$. Subtract an integer multiple of $v$ from $w$ and, if necessary, change its sign to arrange $0\le c=⟨v,w⟩\le a^2/2$; minimality of $b$ makes this reduction possible without decreasing its length.

The covolume $A$ of $L$ is also spectral. The count of lattice vectors with length at most $R$ obeysFor example, translate a bounded fundamental parallelogram of diameter bound $D$: the union of tiles with centres in $B_R$ contains $B_{R-D}$ and lies inside $B_{R+D}$. Comparing their areas proves the leading coefficient. The spectrum supplies this count. Since $(v,w)$ is a lattice basis, its parallelogram area satisfies $A^2=a^2{b}^2-c^2$. ThereforeAll three quantities are determined by the spectrum. Equal Gram matrices give an orthogonal map between the dual lattices; taking duals gives an orthogonal map between the original lattices, descending to a Riemannian isometry of the tori. Isospectral flat two-tori are isometric. This is the two-dimensional lattice reconstruction from vector lengths; it uses lengths with multiplicities, not merely the set of distinct lengths.

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.

First realize the free isometric action. Choose generators $t_1,\dots ,t_r$ of the finitely presented group $T$. In every dimension $d\ge 3$, the connected sumhas fundamental group the free group $F_r$. Map its generators onto the $t_j$ and take the connected regular covering $N\to B_d$ corresponding to the kernel. Its deck transformation group is $F_r/\ker \pi \cong T$. Lift any Riemannian metric on $B_d$ to $N$. Deck transformations are then Riemannian isometries, and a deck transformation fixing one point is the identity by uniqueness of lifts. This gives the free isometric realization of a finitely generated group. For the trivial group use $S^d$. If $T$ is infinite the covering manifold need not be compact, which is allowed in this first construction. Simple connectivity of $N$ is not being asserted here.

The requested nonhomeomorphism conclusion needs an extra hypothesis. For a counterexample even with distinct subgroups, take $T=S_3$, $U_1=⟨(12)⟩$ and $U_2=⟨(23)⟩$. They are conjugate and therefore Gassman equivalent. For every free isometric action of $T$, the conjugating element induces a Riemannian isometry $U_1\backslash N\to U_2\backslash N$. These quotients cannot be nonhomeomorphic. Identical subgroups give an even simpler obstruction. Thus the stated equivalence condition alone cannot imply the claimed topology.

A sufficient version is: for a finite $T$ with Gassmann equivalent, nonisomorphic subgroups $U_1,U_2$, there are closed isospectral manifolds with fundamental groups $U_1,U_2$, and hence they are not homeomorphic. To prove it, realize $T$ as the fundamental group of a closed smooth manifold $B$ of dimension four. Here is the needed closed-manifold realization of a finitely presented fundamental group. Start with a four-dimensional zero-handle and one one-handle for each generator. Attach two-handles along disjoint embedded boundary circles representing the relators, with arbitrary framings. The boundary is three-dimensional, so finitely many such loops can be chosen disjoint. The resulting compact handle manifold $W$ has ${\pi }_1(W)=T$ by the Seifert-van Kampen theorem.

The inclusion $\partial W\to W$ is surjective on fundamental groups: relative to the boundary the dual handle decomposition uses only handles of indices two, three and four, none of which introduces a fundamental-group generator. Double $W$ along its boundary. In the amalgamated product for the double, both boundary maps are the same surjection onto $T$, so the two copies of $T$ are identified completely and ${\pi }_1(B)=T$. The double is closed, connected and smooth after smoothing its collar. This construction works in any dimension at least four; it does not claim that every finitely presented group is a closed three-manifold group.

Let $N$ now be the universal cover of $B$ with the lifted metric. Since $T$ is finite, $N$ is compact and simply connected, with a free isometric deck action of $T$. By Sunada theorem, $M_i=U_i\backslash N$ are isospectral. Because $N$ is their universal cover, ${\pi }_1(M_i)\cong U_i$. Nonisomorphic fundamental groups rule out a homeomorphism. For an infinite $T$ the same qualified argument applies to finite-index subgroups with equal coset characters: divide first by the intersection $K={\mathrm{Core}}_T(U_1)\cap {\mathrm{Core}}_T(U_2)$ of their finite-index subgroup cores, producing a compact intermediate cover and a finite isometry group to which Sunada applies.

For explicit nonhomeomorphic examples, let $p$ be an odd prime. Let $H_p=(\mathbb{Z}/p\mathbb{Z}{)}^3$, and let $K_p$ be the Heisenberg group over a prime field, whose elements are triples with multiplicationBoth have order $p^3$, and every nonidentity element has order $p$. For $K_p$, induction givesso the assertion follows at $k=p$, since $p$ is odd. But $H_p$ is abelian and $K_p$ is not: $(1,0,0)$ and $(0,1,0)$ do not commute.

Embed both groups regularly in $S_{p^3}$. Every nonidentity element in either regular representation has cycle type $p^{p^2}$. Hence both embedded subgroups meet the identity class once, the class of that cycle type $p^3-1$ times, and all other classes zero times. They are nonisomorphic Gassmann equivalent regular subgroups. Taking $p=3$ and using the closed four-manifold construction with $T=S_{27}$ gives an explicit pair of isospectral nonhomeomorphic quotients, completing the intended construction under a sufficient hypothesis.

Finally choose distinct odd primes $p_1,\dots ,p_n$ and takeConjugacy classes in a direct product are products of classes, so their intersection counts factor. Thus all $2^n$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{ϵ}$ is its $j$th factor, and whether it is abelian records ${ϵ}_j$. Realize $T_n$ as the fundamental group of a closed four-manifold and quotient its universal cover by these subgroups. The binary family of nonhomeomorphic Sunada quotients givesThis existence statement is valid; the earlier universal nonhomeomorphism assertion from Gassmann equivalence alone is not.

Let $m=\mathrm{ord}(A)$, $n=\mathrm{ord}(B)$ and $r=\mathrm{ord}(AB)$. Cut the Riemann sphere along arcs joining three marked points, say $0,1,\infty$, so the complement is simply connected. Take copies labeled by the elements of $T$ and glue the slit banks with monodromies given by left multiplication by $A,B,(AB{)}^{-1}$. The product relation ensures consistency. The surface is connected because $A,B$ generate $T$. At a puncture with monodromy order $m$, fill each cycle using the local complex coordinate $z=w^m$; use the analogous charts at the other two punctures. This constructs the three-branch-point regular surface cover $N\to {\mathbb{P}}^1$.

Right multiplication by inverses defines a left action of $T$, commutes with the monodromy gluing and extends holomorphically over the filled points. Average a compatible metric over this finite group to obtain a metric for which $T$ acts by Riemannian isometries. When the genus is at least two, the unique compatible curvature-minus-one metric from the uniformization theorem is an invariant choice. Away from the three marked fibers the action is free. The nontrivial stabilizer subgroups are the conjugates of $⟨A⟩$, $⟨B⟩$ and $⟨AB⟩$, of orders $m,n,r$, omitting order-one groups.

There are $|T|/m$, $|T|/n$, $|T|/r$ points over the corresponding branch points. The Riemann-Hurwitz formula givesThis construction also covers the spherical and toroidal cases. If one monodromy has order one it is simply an unbranched marked fiber, with no nontrivial stabilizer.

For the supplied coset actions, faithfulness gives $m=n=3$. Compose permutations with the rightmost factor acting first. The two products $AB$ have cycle decompositionsThus $r=6$. The index is twelve and each subgroup has order eight, so $|T|=96$. ConsequentlyAll cycles of $A,B,AB$ on each coset set have the full orders $3,3,6$. No nontrivial power of any of these branch monodromies fixes a coset. The fixed-coset criterion means that each $U_i$ has trivial intersection with every conjugate of each cyclic branch stabilizer. Since these are all stabilizers on $N$, both $U_i$ act freely. Their quotient surfaces are therefore smooth closed surfaces, not cone-point orbifolds. Sunada theorem supplies isospectrality, and the unbranched Euler characteristic formula givesChoose the hyperbolic metric for these surfaces and their common genus-nine cover.

In this natural construction the two genus-two surfaces are isometric, by an orientation-reversing map. Define the sheet permutationDirect composition verifiesComplex conjugation of the base sphere fixes its three real branch points and reverses the two chosen generating loops, so their monodromies become the inverses. These identities supply a lift, after relabeling sheets by $P$, to an anticonformal map between the two branched covers. It extends across the branch points by the local power charts. An anticonformal map between closed hyperbolic Riemann surfaces is a Riemannian isometry, by uniqueness of the compatible hyperbolic metric.

Equivalently, take the hyperbolic triangle group of signature $(3,3,6)$. Reflection in the side joining the two order-three vertices conjugates its two vertex rotations to their inverses. The displayed sheet permutation identifies the reflected subgroup for the first covering with the subgroup for the second, up to an irrelevant change of base sheet. This is the orientation-reversing equivalence of branched-cover monodromy. The exhibited isometry is anticonformal, so the conclusion is about Riemannian isometry, without asserting that this map is holomorphic. Merely knowing the subgroups were almost conjugate would not have settled isometry; the explicit inverse-monodromy symmetry does. Nor does the data force this symmetry for every arbitrarily chosen invariant deformation of the cover's metric: it is present for the constructed hyperbolic metric.

A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.

A useful precise form of the transplantation theorem is as follows. Take $N$ congruent copies of a Euclidean tile with labelled flat faces. For each face colour $s$, encode the gluings by a symmetric signed permutation matrix $M_s$: a glued pair gives off-diagonal entries $+1$ in the two corresponding positions; an unglued face gives a diagonal entry $-1$ for a Dirichlet boundary condition, or $+1$ for a Neumann boundary condition. Corresponding face identifications must use the same maps in the tile coordinates. Suppose two assemblies have matrices $M_s$ and $M_s^{\prime }$ and there is an invertible matrix $C$ withThen, for every eigenvalue $\lambda$, $f=(f_1,\dots ,f_N)\mapsto Cf$ is an isomorphism between their Laplacian eigenfunction spaces. The two assemblies are isospectral, including multiplicities.

To see why, pull every tile restriction back to the same Euclidean tile. A boundary value vector $u$ and an outward normal derivative vector $v$ satisfyAt an internal face these are continuity of the function and equality of oppositely directed derivatives. At a diagonal $-1$ they impose $u_i=0$; at a diagonal $+1$ they impose $v_i=0$. The intertwining identities preserve both equations, and the Euclidean Laplacian commutes with constant linear combinations of tile functions. The resulting piecewise function is a weak eigenfunction, with the prescribed boundary conditions, and interior elliptic regularity makes it smooth wherever the assembled manifold is smooth. At polygonal corners one uses the usual weak domain of the Dirichlet Laplacian or Neumann Laplacian, rather than assuming smoothness through a corner. The inverse matrix gives the inverse transplantation.

Here is a completely specified propeller pair. Label the seven triangles $0,\dots ,6$, with central triangle $0$, and colour the three sides $a,b,c$. In the left and right assemblies the internal-edge pairings areEvery pair means reflection of one triangle across the indicated side; the other faces are external boundary. Each assembly has a central triangle and three two-triangle arms. These are the propeller domains. The drawing uses one acute scalene triangle, and each colour always denotes the side opposite the same marked vertex of the starting triangle.

Let $P_s,Q_s$ be the ordinary permutation matrices obtained by adding fixed points to the pairings. The following matrix supplies the Neumann boundary condition transplantation:There is a short way to construct and check it. Start with the support $\{1,2,4\}$ in row $0$, and require the support in row $Q_s(i)$ to be $P_s$ applied to the support in row $i$. Traversing the six edges of the right-hand tree gives exactly the displayed rows; checking the remaining fixed-point relations verifies $N{P}_s=Q_{s}N$ for all three colours. Every row has three ones and every two different rows meet in exactly one column, soThe right-hand side has eigenvalues $9$ once and $2$ six times, proving $N$ is invertible, with $|\det N|=24$. Its incidence pattern is that of the Fano plane, which explains the uniform intersections.

For the Dirichlet boundary condition, put $D=\mathrm{diag}(1,-1,-1,1,-1,1,1)$. In both assemblies this gives opposite signs to adjacent triangles. Thus their signed face matrices are $M_s=-D{P}_{s}D$ and $M_s^{\prime }=-D{Q}_{s}D$: the off-diagonal gluing entries are positive and the unpaired diagonal entries are negative. The matrixthen satisfies $C{M}_s=M_s^{\prime }C$ and is invertible. The transplantation theorem proves equality of the Dirichlet Laplacian spectra; $N$ proves equality of the Neumann Laplacian spectra as well.

Finally vary the three labelled side lengths $(a,b,c)$ of the initial triangle in an open region of strict triangle inequalities, with an acute scalene shape sufficiently close to an equilateral triangle. Reflection constructs both domains continuously from these three independent real parameters. At the equilateral configuration the tiles have disjoint interiors. Choose a sufficiently small neighbourhood of that configuration with every triangle angle ${\alpha }_i$ in $(\pi /4,\pi /2)$. At each central vertex exactly four tile angles meet, forming a sector of angle $4{\alpha }_i<2\pi$, while nonincident pieces stay separated. Thus there is an open three-dimensional parameter region in which both assemblies are embedded planar domains.

For a scalene triangle in this region, the three central vertices are exactly the three reflex boundary vertices: their interior angles are $4{\alpha }_i$, while the other boundary vertices have angles ${\alpha }_i$ or $2{\alpha }_i<\pi$. Hence a planar isometry of the domains must map the central triangle onto the central triangle. Put those two central triangles in the same position. Their unequal side lengths force any such isometry to fix the central triangle pointwise. Yet the first arm across, for example, side $a$ attaches its outer triangle across colour $c$ in $L$, and across colour $b$ in $R$. These occupy different positions, so the identity does not identify the domains. An intrinsic Riemannian isometry between connected planar domains also restricts locally to a rigid Euclidean motion and hence is one globally. The two propellers are nonisometric throughout this open three-parameter family. Scaling is included as one of the parameters; modulo scaling the family has two parameters.

The helicoid has no angular identification: its height coordinate is $v$. ConsequentlyThe catenoid has total curvature $-2\pi$. A global Riemannian isometry preserves both the Gaussian curvature and the area element, hence preserves total curvature. ThereforeThe same obstruction is visible in the parameter map: increasing $v$ by $2\pi$ gives a different point of the helicoid but the same point of the catenoid.

Suppose a finite group $G$ acts by Riemannian isometries on a compact Riemannian manifold $M$, and Gassmann equivalent subgroups $H_1,H_2$ act freely. The quotient manifolds $H_i\backslash M$ have equal Laplace-Beltrami operator spectra. For an eigenfunction space $E$, the quotient multiplicity is $|H_i{|}^{-1}{\sum }_{h\in H_i}\mathrm{tr}(h|E)$, and the equality follows by summing its character of a representation over conjugacy classes. Nonconjugacy inside $G$ alone does not guarantee that the quotients are nonisometric.