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The normalized pullback is a unitary map from $L^2$ of the quotient onto the deck-invariant subspace of $L^2$ of the cover. Local isometry and the sheet-counting integration formula preserve the Dirichlet quadratic form. Therefore the Friedrichs extension and its heat semigroup intertwine with this map. A finite sum of lifted Riemannian heat kernels gives the quotient kernel; independence of lifts follows by reindexing the deck group. This formulation avoids a uniqueness claim for unrestricted heat-equation solutions on a noncompact or incomplete manifold.

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 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.

Fix a bounded tile in Euclidean space $\Omega$ with labelled flat boundary faces. For each face label $s$, encode an assembly of $n$ copies by an involutive real symmetric signed permutation matrix $M_s$: a paired interface contributes a two-by-two swap, an unglued face with a Neumann boundary condition contributes $+1$ on the diagonal, and an unglued face with a Dirichlet boundary condition contributes $-1$. Use the same face-identification convention on each copy. If a second assembly has matrices $N_s$ and there is an invertible constant matrix $C$ satisfyingthen the map $u=(u_1,\dots ,u_n)\mapsto Cu$ is an isomorphism between corresponding Laplace-Beltrami operator eigenspaces. This is the transplantation theorem.

To prove it, restrict an eigenfunction to the tile copies and identify each copy with $\Omega$. Along a face, let $a$ be its vector of boundary values and $b$ its vector of outward normal derivatives. At a glued interface the values agree and the outward derivatives are opposite. At an unglued face with a Neumann boundary condition the derivative vanishes, and at a face with a Dirichlet boundary condition the value vanishes. All three cases are encoded byThe intertwining identities give $N_s(Ca)=Ca$ and $N_s(Cb)=-Cb$, so the transplanted function satisfies the second assembly's interface and boundary conditions. On each tile its Laplace-Beltrami operator is the same constant linear combination of the original Laplacians, hence it has the same eigenvalue. Across an interface, matching values and opposite normal derivatives make it a weak solution; elliptic regularity, or the allowed reflection principle, makes it a genuine smooth solution there. At corners the operator is understood by its finite-energy, Friedrichs extension; the same matching argument applies to its quadratic form domain. Applying $C^{-1}$ reverses the construction. Thus dimensions of all eigenspaces agree, including multiplicities, and the two specified boundary-value spectra coincide.

The requested pair exists with explicitly specified mixed boundary conditions and polygonal boundary. Take four copies $Q_0,Q_1,Q_2,Q_3$ of an octagonal regular polygon in Euclidean space. Label four alternate sides $a,b,c,d$ in cyclic order; the four intervening sides always have a Neumann boundary condition. The labelled sides are disjoint even at their endpoints. Parameterize every side in the positive boundary direction of the reference octagon. When pairing same-labelled sides, identify equal parameters. This produces a locally flat surface with boundary: a neighbourhood of an interior point of an interface consists of two half-disks glued along their diameters. All tile vertices remain boundary points, with boundary angles $3\pi /4$ or $3\pi /2$; there are no interior cone points.

Here are the complete gluing and boundary conditions. An entry $ij$ pairs the corresponding sides of $Q_i,Q_j$. The indicated unpaired sides have a Dirichlet boundary condition; every other unpaired side has a Neumann boundary condition.Both assemblies are compact and connected. The Laplace-Beltrami operator is the nonnegative operator $P=-\Delta$ associated with the finite-energy quadratic form $\int |\nabla u{|}^2$, with zero Sobolev trace on the specified Dirichlet arcs. Thus the mixed boundary condition is defined without requiring classical differentiability at a corner or at a Dirichlet–Neumann junction.

The four signed permutation matrices for $D$ areFor $B$ they areUse the Hadamard matrixThe intertwining relations can be checked on all columns at once. Writing the columns of $C$ as $h_0,h_1,h_2,h_3$, direct application of the displayed $N_s$ givesFor the four intervening faces both assemblies have the identity face matrix, so these intertwining relations hold as well. Also $C^{T}C=C^2=4I$. This four-tile mixed-boundary transplantation between a disk and a nonorientable surface gives a bijection of every pair of eigenspaces, preserving eigenvalues and multiplicities.

There is also a direct argument on the quadratic form domains that handles all boundary corners. The orthogonal matrix $U=C/2$ acts on tile restrictions, preserves their L2 inner product and the sum of their Dirichlet energies, and commutes with taking a Sobolev trace. The condition $M_{s}u=u$ on each face encodes equality of the two traces at a paired interface and vanishing trace at an unpaired Dirichlet face. Intertwining takes these conditions exactly to $N_s(Uu)=Uu$. Consequently $U$ maps the whole form domain onto the other form domain and preserves its quadratic form. The associated self-adjoint operators are unitarily equivalent, proving isospectrality for the stated mixed boundary conditions without any assumption of smoothness at a Dirichlet–Neumann junction.

We now verify the actual topology using the topology of disk tiles glued along disjoint boundary arcs. Because the gluing arcs in each tile are disjoint, the resulting surface with boundary is a thickening of its gluing graph, with one vertex for each closed disc tile and one edge for each paired arc. One can construct this homotopy equivalence by taking a small central closed disc in each tile and narrow strips to its paired arcs. What remains is a collar neighbourhood of the boundary, which retracts onto this connected system of closed discs and strips. Shrinking each central closed disc and each strip then gives the gluing graph. Equivalently, begin with a spanning tree of the gluing graph: its closed disc tiles joined along boundary arcs form one closed disc, and every remaining pair of boundary arcs adds one band. Thus for $n$ tiles and $e$ paired arcs the fundamental group is a free group of rank $e-n+1$, and the Euler characteristic is $n-e$.

For $D$, the gluing graph is the pathGluing a new closed disc to a closed disc along one boundary arc again gives a closed disc; applying this three times proves that $D$ is homeomorphic to a closed disc. In particularFor $B$, the five paired arcs give the graph edges $13,12,02,01,23$. It is connected, and henceHere $F_2$ is the free group on two generators, so $B$ is not simply connected. To decide orientation of a surface, give every reference tile its counterclockwise orientation. Our same-parameter gluing preserves boundary directions, so a global orientation requires opposite orientation signs on paired tiles: the two induced boundary orientations at an interior interface must be opposite. Such signs exist exactly when the gluing graph is bipartite. In $B$ the cyclehas length three. Following it forces the sign to reverse three times and return to its starting tile with the opposite sign. This proves that $B$ is a nonorientable surface. It has nonempty boundary because every intervening octagon side remains unpaired. Thus $D$ is a simply-connected orientable surface with boundary, whereas its isospectral partner $B$ is neither simply connected nor orientable, exactly as required, for the specified mixed boundary problem.

The boundary qualification is essential. With smooth boundary and the same pure Dirichlet boundary condition or pure Neumann boundary condition on both surfaces, the following existing obstruction applies. Polygonal corners and the mixed boundary conditions in the construction above place it outside that obstruction.

With the interpretation of compact smooth Riemannian surfaces with the same pure Dirichlet boundary condition or pure Neumann boundary condition, the printed simply-connected/nonorientable pair cannot exist. The two-dimensional heat trace has the following expansion (the smooth-boundary coefficients are given in the heat-coefficient formulas, Theorems 5.1–5.2):where the boundary sign is minus for Dirichlet and plus for Neumann. The Gauss-Bonnet theorem identifies the constant term with $\chi (M)/6$. Isospectrality therefore forces equal Euler characteristics. A compact connected simply-connected surface with boundary is a disk and has $\chi =1$. A connected nonorientable surface with $c\ge 1$ crosscaps and $b\ge 1$ boundary components has $\chi =2-c-b\le 0$. Thus no such smooth pure-Dirichlet or pure-Neumann pair exists. This is the Euler characteristic from the heat trace of a bordered surface.

Piecewise-smooth boundaries introduce corner terms, and mixed boundary conditions change the heat coefficients, as the explicit construction above demonstrates. Moreover, in the usual pure-Neumann polygonal reflection-transplantation setup, the pure Neumann reflection transplantation preserves Euler characteristic argument applies even with corners. The numbers of tiles agree. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$, so these agree too. For each tile-vertex label, vertex classes are orbits of the subgroup generated by incident-face involutions. Their number is the dimension of invariant vectors in that permutation representation, which is unchanged by the intertwiner. Hence the numbers of vertices agree, proving equal $V-E+F$. Thus this standard tiled construction also cannot supply a disk on one side and a nonorientable surface on the other.

For comparison, the following existing construction gives an orientable and a nonorientable pair with the same spectrum for the pure Neumann boundary condition. Both have nontrivial fundamental groups; this is a separate example, not the construction used to meet the simply-connected requirement above. This eight-tile Neumann transplantation across orientability makes the comparison explicit. The finite seed is the affine modulo-eight example in the original orientability construction; the intertwiner and the topology checks are derived below. Take eight copies of a regular Euclidean hexagon and label three pairwise nonadjacent sides $\sigma ,t,u$. Every remaining side is a side with a Neumann boundary condition. Index the copies by $j\in \mathbb{Z}/8\mathbb{Z}$. For the first assembly pair same-labelled faces according tofor the second, usewith all arithmetic modulo eight. A fixed point means an unglued side with a Neumann boundary condition. Pair faces by the reflection convention, matching the same endpoint parameter in the reference tiles. The distinguished sides are disjoint, so no interior vertex singularities are created; the resulting surfaces have polygonal boundary corners.

Let $R{e}_j=e_{j+1}$ on the eight-dimensional tile-index space and setIt commutes with the permutation matrices for $t$ and $\sigma$, because its coefficients are invariant under reversing cyclic order. For $u_1,u_2$, write $C_{ij}=c_{i-j}$, where $c_0=c_4=1$, $c_1=c_7=2$, and all other $c_r$ vanish. The required relation $C{M}_{u_1}=M_{u_2}C$ is exactly $c_r=c_{3r+4}$, verified from these four nonzero entries. Thus all three intertwining relations hold.

The cyclic Fourier vectors diagonalize $C$. Its eigenvalues arenamely $6,2\sqrt{2},2,-2\sqrt{2},-2,-2\sqrt{2},2,2\sqrt{2}$. None is zero, so $C$ is invertible and the two assemblies are isospectral manifolds for the Neumann boundary condition.

We prove their topology as well. Their gluing graphs are connected: the first contains the spanning path $0,1,3,6,2,7,5,4$, and the second contains the spanning cycle $0,1,7,2,6,3,5,4,0$. Each has eight tile vertices and ten paired-face edges. Since tiles are disks and gluing arcs are disjoint, each surface retracts up to homotopy to this multigraph; thus its Euler characteristic is $8-10=-2$ and its fundamental group is a free group of rank three.

An orientation assigns a sign to each reference tile, and a reflection gluing requires opposite signs at the ends of every graph edge. The second graph admits this assignment: take sign $+$ on $\{0,5,6,7\}$ and sign $-$ on $\{1,2,3,4\}$. Every paired face joins opposite signs, proving orientability. In the first graph the cyclehas odd length. Transporting orientation around it reverses the initial sign, proving nonorientability. Hence the first surface is also not simply connected. The second surface is orientable but likewise not simply connected; the disk required in the question is supplied instead by the four-tile mixed-boundary construction above.

The four-tile construction resolves the printed simply-connected requirement under explicit mixed boundary conditions. The eight-tile comparison only concerns orientation of a surface under a pure Neumann boundary condition.