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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.

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.

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 a finite presentation $T=⟨x_1,\dots ,x_r\mid R_1,\dots ,R_s⟩$. For a closed-manifold realization of a finitely presented fundamental group, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with fundamental group the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ hasby the Seifert-van Kampen theorem. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this handle decomposition upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the fundamental group: there are no new one-handles or two-handle relations. Thus ${\pi }_1(B)\to {\pi }_1(W)$ is an isomorphism andThese are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed fundamental group.

Now assume $T$ finite. Choose any smooth Riemannian metric on $B$ and lift it to its universal cover $\tilde{B}$. This cover is compact, since its degree is $|T|$, and simply connected. Its deck transformations give a free isometric $T$-action. For the two Gassmann equivalent subgroups putEach is a closed Riemannian manifold, and $\tilde{B}$ is its simply connected universal cover, with deck transformation group $U_i$. Consequently ${\pi }_1(M_i)\cong U_i$.

Here is the complete spectral argument behind the Sunada theorem in this situation. Each finite-dimensional upstairs eigenspace $E_{\lambda }$ is a representation of $T$, because the action is isometric and commutes with the Laplace-Beltrami operator. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_{\lambda }^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its trace givesThe character ${\chi }_{\lambda }$ is constant on each conjugacy class of $T$. Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. The quotients are isospectral, with ${\pi }_1(M_i)\cong U_i$.

If $U_2=t{U}_1{t}^{-1}$, the map $[x{]}_{U_1}\mapsto [tx{]}_{U_2}$ is well defined, invertible and isometric because $t$ acts by an isometry upstairs. Conjugate subgroups give isometric quotients.

For an explicit example with different fundamental groups, use the order-sixteen Gassmann pairEmbed each in $T=S_{16}$ by left multiplication on its own sixteen elements, identified with the same sixteen-point set. These regular representations are faithful. Both groups have one identity element, three elements of order two, and twelve of order four: in $C_4^2$ the elements of order dividing two are precisely its four even-coordinate elements; in $Q_8\times C_2$ they are the four elements in $\{1,-1\}\times C_2$. Every remaining element has order four.

In a regular action, an order-$k$ element is a product of $16/k$ disjoint $k$-cycles. Thus each subgroup intersects the identity conjugacy class of $S_{16}$ in one element, the class of type $2^8$ in three, the class of type $4^4$ in twelve, and every other class in zero. They are Gassmann equivalent. However $H_1$ is abelian and $H_2$ is nonabelian, so they are not isomorphic. Apply the preceding construction with fundamental group $S_{16}$ and these subgroups. It producesIn particular they cannot even be homeomorphic, and therefore cannot be isometric.

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.