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A flat torus is ${\mathbb{R}}^d/\Lambda$ with its quotient Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its volume is the lattice covolume. The dual lattice indexes its complete Fourier series of Laplacian eigenfunctions.

Subgroups $H_1,H_2$ of a finite group $G$ are Gassmann equivalent if $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$. Equivalently their coset permutation representations have the same character of a representation. They need not be conjugate subgroups. The Sunada theorem turns this equality into equality of Laplacian eigenfunction multiplicities.

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.

A scalar $\lambda$ for which a nonzero Laplacian eigenfunction satisfies this equation, with specified boundary conditions or on a closed manifold, is an eigenvalue of the positive Laplacian. A Dirichlet Laplacian eigenvalue imposes zero boundary values; other boundary conditions generally change the spectrum.

Use the Galerkin method: substitute the retained trigonometric modes and project onto them with the $L^2$ inner product over the rectangle. Each mode satisfies the printed boundary conditions. Write $X=\alpha x$, $Z=\pi z$, and abbreviate the four mode prefactors by $C=2\sqrt{2}\beta /\alpha$, $D=2\sqrt{2}/\beta$, $V_1=2\sqrt{2}\sigma \pi R_{\Omega }/(\alpha \beta )$ and $V_2=\sigma R_{\Omega }/\alpha$. The Jacobian determinant in the advection terms is $J(f,g)=f_x{g}_z-f_z{g}_x$.

The stream function is a single Laplacian eigenfunction, so ${\nabla }^2\psi =-{\beta }^2\psi$ and $J(\psi ,{\nabla }^2\psi )=0$ exactly. For the temperature modes, direct multiplication givesFor the transverse-velocity modes, the retained products areFor example, $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$ and $\sin X\cos 2X=(\sin 3X-\sin X)/2$ supply the signs of the two retained $ac$ and $ae$ couplings. The third harmonics are orthogonal to the retained modes and are discarded by this Galerkin method; they are not claimed to vanish in the full partial differential equations.

Before changing time, the five coefficient equations areDividing by ${\beta }^2$ gives the stated five-mode rotating-convection truncation, withThe squared rotation parameter is $q=r_{\Omega }^2\ge 0$. The printed $R_{\Omega }$ is proportional to rotation rate, so it is a square-root-type rotation parameter rather than the more usual Taylor number proportional to squared rotation rate; the algebra uses the coefficient as printed. At zero rotation the physical transverse velocity modes vanish; the normalized $d,e$ equations can still be understood through their continuous zero-rotation limit.

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.

Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.

Only the Laplacian eigenfunctions whose eigenvalues lie in the intersection of the unstable band $(q_{-},q_{+})$ with the allowed spectrum can grow. A lower bound $k^2>k_0^2$ removes all unstable modes if $k_0^2\ge q_{+}$, even when the unrestricted threshold in (iii) is exceeded.

If all real wavenumbers above the cutoff are possible, an equivalent strict threshold isIt is unchanged when $k_0^2\le \sqrt{A\gamma /D}$, and becomes $AD+\gamma +D{k}_0^2+A\gamma /k_0^2$ when $k_0^2>\sqrt{A\gamma /D}$. A bounded geometry generally gives a discrete spectrum, so the stronger requirement is that an actual allowed eigenvalue falls in the band.

Consider two finite assemblies of the same number $n$ of congruent bounded Euclidean tiles. Label their matching faces so the same prescribed face identifications are used in both assemblies. For each face type $c$, encode the first assembly by a symmetric involution $A_c$: a paired face exchanges its two tile indices, an exterior Dirichlet boundary condition contributes diagonal $-1$, and an exterior Neumann boundary condition contributes diagonal $+1$. Let $B_c$ encode the second assembly. Face coordinates are pulled back to the same reference face; if identifications have additional face isometries, these pullbacks must be included in the intertwining operators.

The transplantation theorem says that a constant matrix $C$ with $C{A}_c=B_{c}C$ for every face type sends the vector of tile restrictions of a Laplacian eigenfunction on the first assembly to one on the second, by $g=Cf$. If $C$ is invertible, it is a bijection of each eigenspace and the assemblies are isospectral with multiplicities.

To prove it, each component solves the same interior equation $\Delta f_j=\lambda f_j$, and a constant linear combination solves that equation too. On a face, let $u$ be the vector of boundary values and $v$ the vector of outward normal derivatives. All matching and exterior conditions are exactlyFor an internal face these equations say that values agree and the two outward derivatives sum to zero. For diagonal $-1$ they impose zero value, and for diagonal $+1$ zero normal derivative. The intertwining identity gives $B_c(Cu)=Cu$ and $B_c(Cv)=-Cv$, so the transplanted functions match and satisfy the correct boundary conditions. Applying $C^{-1}$ proves the eigenspace bijection. At corners the same reasoning is interpreted in the finite-energy weak domain: no value jump and cancellation of normal flux prevent an extra distributional source. It gives the usual self-adjoint Dirichlet/mixed Laplacians on polygonal assemblies.

Here is an explicit transplantation by reflection parity. Take a rectangle of width $2a$ and height $b$, with the Dirichlet boundary condition on every exterior side. Cut it at its vertical midline into two congruent half-rectangles, pulling the right-hand half back by reflection. The midline gluing matrix isReplace the connected assembly by the disjoint union of two width-$a$, height-$b$ rectangles: the first has Dirichlet conditions on all sides; the second has Dirichlet on three sides and Neumann on the side representing the cut. Its corresponding matrix is $D=\mathrm{diag}(-1,1)$. The invertible, indeed orthogonal, matrixintertwines the midline conditions. Every other face has matrix $-I$ on both assemblies, which automatically intertwines. The transplantation theorem proves equality of the complete spectra, although one surface is connected and the other has two components with the required uniform and mixed conditions.

Equivalently, odd reflection eigenfunctions vanish on the cut, while even reflection eigenfunctions have zero normal derivative there. A direct separation-of-variables check gives the connected rectangle's eigenvaluesEven $m=2j$ gives the fully Dirichlet half-rectangle spectrum, and odd $m=2j+1$, $j\ge 0$, gives the mixed half-rectangle spectrum. The split preserves every multiplicity, including coincidences between different pairs of indices.

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.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.

Linearization about $w=0$ removes the quadratic term and givesThe three homogeneous boundary conditions select the vertical normal modesTake a horizontal Laplacian eigenfunction satisfyingOn $W_{j}f$, the full Laplacian has eigenvalue $-(a^2+j^2{\pi }^2)$. The growth rate is consequentlyNeutrality occurs atFor fixed nonzero $a$, this increases strictly with $j$, so the first unstable vertical mode is $j=1$ and

Take a nonzero separated solution $\phi =X(x)Y(y)$. Dividing by $XY$ separates the Laplacian eigenfunction equation into $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(\lambda -\mu )Y$. The zero-endpoint Dirichlet boundary conditions admit no nonzero linear or hyperbolic factors, so both separation constants must be positive squares. HenceAll separated nonzero solutions have this form. At a repeated eigenvalue, linear combinations of the corresponding modes are also solutions, though they need not remain separable.

Reflection across a cut splits Laplacian eigenfunctions into odd and even parts. The odd restriction has a Dirichlet boundary condition on the cut; the even restriction has a Neumann boundary condition there. The displayed matrix intertwines the two-tile swap with $\mathrm{diag}(-1,1)$. A connected uniform-Dirichlet rectangle of width $2a$ is therefore isospectral to the disjoint union of width-$a$ rectangles with all-Dirichlet and three-Dirichlet/one-Neumann conditions. This illustrates why changing boundary conditions can conceal connectedness in the spectrum.

For two assemblies of congruent Euclidean tiles, let $M_s,N_s$ encode the gluing or boundary reflection at each labelled face. An invertible constant matrix $C$ satisfying $C{M}_s=N_{s}C$ carries tile restrictions of Laplacian eigenfunctions bijectively to those on the second assembly. Boundary values satisfy $M_{s}u=u$ and outward normal derivatives satisfy $M_{s}v=-v$, so the intertwining identities preserve matching and boundary conditions. Use diagonal $-1$ for a Dirichlet boundary condition and $+1$ for a Neumann boundary condition.