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Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

Use unnormalized parabolic induction, the convention fixed by the supplied sequence and by the occurrence of constants in $V(1)$. Its functions satisfy $f(bg)=\chi (b)f(g)$, and $G$ acts by right translation. Characters in this smooth category are smooth characters. Put $\lambda ={\chi }^w{\delta }^{-1}$, retaining the question's modulus $\delta (\mathrm{diag}(a,d))=|d/a|$.

First derive the evaluation adjunction for unnormalized induction. A $G$-intertwiner $A:V(\chi )\to V(\xi )$ gives a functional $\ell (v)=A(v)(1)$. Since $A$ intertwines and induced functions have left covariance, $\ell (\pi (b)v)=\xi (b)\ell (v)$. Conversely such a functional definesThis function has the required left covariance. Any compact-open subgroup fixing $v$ fixes it on the right, so it is a smooth induced function; compactness of $B\backslash G$ removes any additional support issue. The construction is $G$-equivariant, and the two operations are inverse. Since $\xi$ is trivial on $U$, these functionals factor precisely through the Jacquet module, givingNow use the given exact sequence $0\to \lambda \to J\to \chi \to 0$, where $J=V(\chi {)}_U$ has dimension two.

If $\lambda \ne \chi$, choose $t_0\in T$ on which their values differ. Its operator on $J$ is upper triangular with distinct eigenvalues, so has two one-dimensional eigenspaces. Since $T$ is abelian, all its operators commute with this operator and preserve the eigenspaces. They are the linear character lines $\lambda$ and $\chi$. This proves that distinct-character extensions of an abelian group split, so $J\cong \lambda \oplus \chi$. A map between one-dimensional linear character spaces is zero unless the linear characters agree, and otherwise is one-dimensional. Therefore the required dimension is one for $\xi =\lambda$ or $\xi =\chi$, and zero for any other linear character.

The case $\lambda =\chi$ needs an extra argument: the exact sequence alone would also permit a split self-extension with two quotient functionals. Write $\chi (\mathrm{diag}(a,d))={\chi }_1(a){\chi }_2(d)$. Equality with $\lambda$ forces ${\chi }_1/{\chi }_2=|\cdot |$, so $\chi =(\eta \circ \det ){\chi }_0$, where $\eta ={\chi }_2$ and ${\chi }_0(\mathrm{diag}(a,d))=|a|$. Multiplication of induced functions by $\eta (\det g)$ identifies this with a determinant-character twist of $V({\chi }_0)$. Twisting is invertible, so it is enough to prove nonsplitting for ${\chi }_0$.

Use the open Bruhat decomposition coordinateFor $x\ne 0$ the matrix factorizationshows that $g_f(x)=|x{|}^{-1}f(1)$ for all sufficiently large $|x|$: the lower-unipotent matrix approaches the identity, where $f$ is locally constant. The coordinate map is injective, since this open cell is dense in the projective-line quotient and induced functions are locally constant sections.

For $t=\mathrm{diag}(a,d)$, direct multiplication givesThere exists a section $f_0$ with $f_0(1)=1$ and coordinate functionIndeed the displayed factorization changes to the lower chart near infinity, where this tail corresponds to the constant function one. Thus it extends smoothly across the missing point. With $t=\mathrm{diag}(\varpi ,1)$ and $|\varpi |=q^{-1}$, the difference is the compactly supported functionNormalize additive Haar measure by $\mathrm{vol}(\mathfrak{o})=1$. ThenThis difference survives in the Jacquet module. To prove that assertion, the unipotent action translates coordinates: $g_{\pi (n(y))f}(x)=g_f(x+y)$. Any such translation difference is compactly supported because the tail is $f(1)|x{|}^{-1}$ and $|x+y|=|x|$ when $|x|$ is sufficiently large. Its integral is zero: integrate over a sufficiently large additive ball, invariant under translation by $y$, and change variables. A finite sum of unipotent translation differences also has integral zero. Hence $h$, whose integral is nonzero, cannot be a coinvariant relation. In particular, on $J$ the operator $\pi (t)-{\chi }_0(t)I$ is nonzero. This proves the equal-character Jacquet self-extension for GL2 is nonsplit.

Any $T$-equivariant functional $J\to {\chi }_0$ must kill the nonzero image of that operator. Since $J$ has dimension two, its space of such functionals has dimension at most one; the quotient in the supplied exact sequence already provides one. Twisting back gives the same conclusion for the common linear character $\chi =\lambda$. For any different $\xi$, a surjection $J\to \xi$ is impossible because both composition factors of $J$ are $\chi$. Combining the two cases givesWhen the two listed linear characters coincide, the dimension remains one; they are not two independent contributions.

We construct the two domains explicitly from seven congruent triangles. Let a scalene reference triangle have sides $a,b,c$ opposite its three labelled vertices. Choose its three side lengths in a small open neighbourhood of $(1,1,1)$, with a fixed strict ordering. Thus the three angles are distinct and lie between $\pi /4$ and $\pi /2$. Each copy keeps these side labels.

For the first assembly glue the following tile pairs, reflecting across the indicated side:Each coloured adjacency graph is a tree with central tile $7$ and three two-tile arms. Place tile $7$ in the plane and obtain every other tile by the prescribed reflections. At the equilateral reference shape the copies have disjoint interiors and form a polygonal disk; the same holds throughout a sufficiently small neighbourhood, since the only contacts between tiles are the prescribed edges and their incident vertices. In particular each assembly is a bounded connected planar domain.

To prove equality of spectra, index tiles by the seven nonzero vectors $x\in {\mathbb{F}}_2^3$, using $x=(x_1,x_2,x_3)$ with label $x_1+2x_2+4x_3$. LetThese involutions act on points by $x\mapsto s_{s}x$ and on nonzero covectors by $y\mapsto s_s^{-T}y=s_s^{T}y$. Their swaps are exactly the two columns of the table. Write $P_s,Q_s$ for the corresponding permutation matrices. The Fano plane incidence matrixsatisfies $B{P}_s=Q_{s}B$, since incidence is preserved by a linear transformation and its inverse transpose. Each point lies on three lines, while two different points have one common line. Thereforewhere $J$ is the all-ones matrix, proving that $B$ is invertible.

For the Dirichlet boundary condition, a glued side has a positive swap and an exposed side has diagonal entry $-1$. Both trees have the same bipartition, with sign matrixTheir signed reflection matrices are $M_s=-D{P}_{s}D$ and $N_s=-D{Q}_{s}D$. On a glued edge its endpoint signs are opposite, so the off-diagonal entries are $+1$; on an exposed edge the entry is $-1$. Consequently the invertible matrix $C=DBD$ satisfies

We spell out why this gives isospectrality. Pull each function back from the seven tiles to the reference triangle. Its seven boundary values along side $s$ form a vector $u_s$ satisfying $M_s{u}_s=u_s$: this imposes equal values at glued edges and zero values at exposed edges. Outward normal derivatives of a smooth eigenfunction satisfy $M_s{v}_s=-v_s$, giving opposite normal derivatives on glued edges. The identities above preserve both conditions under $u\mapsto Cu$. Since $C$ is constant, it also commutes with the scalar Laplacian on each triangle.

One can include all weak eigenfunctions without any corner regularity issue by orthogonalization of a transplantation matrix. The matrix $C^{T}C$ commutes with the symmetric $M_s$, so $O=C(C^{T}C{)}^{-1/2}$ is orthogonal and still intertwines $M_s,N_s$. On the disjoint tiles, $O$ preserves the $L^2$ norm and the sum of the Dirichlet energies; it maps the trace-matching form domain $H_0^1({\Omega }_P)$ bijectively to $H_0^1({\Omega }_L)$. Thus it is a unitary equivalence of the Dirichlet Laplacians, proving equality of every eigenvalue with its multiplicity.

Finally, the only reentrant vertices of either polygon are the three vertices of tile $7$, with interior angles four times the corresponding reference-triangle angles. All other boundary angles are less than $\pi$. These three distinct angles force any hypothetical isometry to map each labelled central vertex to its counterpart. An isometry between connected open planar domains is a restriction of a rigid motion: its derivative is orthogonal, and preservation of the flat Levi-Civita connection makes that derivative constant. After aligning the two central triangles, this rigid motion fixes three noncollinear points and is the identity. But the two domains do not coincide: the tile beyond side $a$ of the central triangle has its next attachment along $b$ in the point assembly and along $c$ in the line assembly. These occupy different open regions. This contradiction proves nonisometry.

Therefore the three freely varying side lengths give a three-parameter family of nonisometric Dirichlet-isospectral bounded planar domains. The parameters include scale; no congruence normalization removes it.

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.

Throughout the hyperbolic discussion take curvature $-1$ and $g\ge 2$. A Y-piece is a pair of pants with a hyperbolic metric and three geodesic boundary components. Three labelled positive boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3$ determine it up to isometry. To see existence and uniqueness, cut along the three shortest perpendicular seams. The two resulting right-angled hyperbolic hexagons have alternating side lengths ${\ell }_i/2$, which determine each hexagon. Conversely, any three positive alternating lengths produce such a hexagon, and doubling along the other sides produces the desired Y-piece. In particular the seam opposite ${\ell }_i$ satisfies

An X-piece is a four-holed hyperbolic sphere with geodesic boundaries, obtained by joining two Y-pieces along one pair of equal-length boundaries. For a chosen separating curve, it is determined bywhere $s$ is the common gluing length and $\tau$ is the relative arclength twist, measured between chosen seam endpoints. The identification reverses boundary orientation so that the surface is oriented. For an unmarked gluing, $\tau$ is periodic modulo $s$; a marking retains $\tau \in \mathbb{R}$, because a full Dehn twist changes the marking. These six parameters describe the X-piece with its chosen separating curve; different curve choices or boundary symmetries can describe the same unmarked surface.

Choose a pants decomposition of a closed oriented genus-$g$ surface. Each Y-piece has Euler characteristic $-1$, and gluing boundary circles does not change the total Euler characteristic. Thus there are $2g-2$ pieces. Their $3(2g-2)$ boundary circles are paired, giving $3g-3$ gluing curves. Choose their disjoint geodesic representatives, then prescribe each cuff length and its relative twist. The Fenchel–Nielsen coordinates areThey determine the marked hyperbolic surface and hence its unmarked isometry class, with $6g-6$ real parameters. One must also fix the combinatorial gluing pattern of the chosen pants decomposition. Full twists and other changes of marking identify some parameter sets in unmarked moduli space of Riemann surfaces. The genus restriction matters: a sphere and a torus do not admit a closed curvature-$-1$ metric or such a decomposition into geodesic Y-pieces.

For a fixed oriented topological surface $S_g$, Teichmüller space ${\mathcal{T}}_g$ consists of marked Riemann surfaces $(X,f:S_g\to X)$. Two markings represent the same point when a conformal isomorphism between the surfaces carries one marking to the other up to homotopy. By the uniformization theorem, for $g\ge 2$ these are equivalently marked curvature-$-1$ metrics. The displayed Fenchel–Nielsen coordinates identify ${\mathcal{T}}_g$ with a cell of real dimension $6g-6$; the mapping class group quotient gives the unmarked moduli space of Riemann surfaces.

The spectral version of the Wolpert generic spectral rigidity theorem says that outside a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$, the Laplace spectrum determines the hyperbolic surface up to isometry. The exceptional subset has positive codimension and is invariant under changes of marking. Orientation reversal must be allowed, since the Laplacian cannot detect orientation. Equivalently, a generic surface is determined by its unmarked length spectrum, counted with multiplicities. This is a generic uniqueness statement, not a claim that every isospectral pair is isometric; the construction in Question 4 lies in the exceptional locus.

Three major ingredients and their roles are as follows.

For completeness, a related finiteness statement sometimes grouped with this theorem is that a fixed closed hyperbolic surface has only finitely many isospectral isometry classes. Its short proof uses three particularly concrete tools: a common discrete length spectrum; Mumford's compactness theorem, applied to the common positive systole; and finitely many determining marked lengths. Compactness would give an accumulation point if an isospectral family were infinite. Nearby markings make each determining length converge, while discreteness of the fixed spectrum makes it eventually constant. The determining lengths then force eventual equality of the marked surfaces, a contradiction. This finiteness of isospectral hyperbolic surfaces statement does not replace the stronger generic uniqueness assertion above.

An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.