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Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

The natural expectation is orbital stability, since a spatial translation of a travelling wave is another travelling wave. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert asymptotic stability for arbitrary disturbances without a norm and a class of initial data. The requested essential spectrum calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.

In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term givesThe first equation to study is its linearization,and the associated eigenvalue equation $\mathcal{L}\varphi =\lambda \varphi$. Differentiating the front equation gives $\mathcal{L}{w}_c^{\prime }=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted L2 space the limiting dispersion relation is $\lambda =-{\xi }^2+ic\xi +k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.

There is a small functional-analytic correction to the printed definition. Use the complete exponentially weighted L2 spaceThe additional requirement of unweighted L2 space membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=1_{[-N,0]}$ belongs to the intersection and is a weighted Cauchy sequence, but its weighted limit $1_{(-\infty ,0]}$ does not belong to unweighted L2 space. The usual Hilbert space for this spectral argument is the completion $X_{\gamma }$. We use that completion explicitly, rather than treating the intersection as a complete space.

Multiplication $T_{\gamma }v=e^{\gamma z}v$ is a unitary map from $X_{\gamma }$ onto L2 space. Take the domain $T_{\gamma }^{-1}{H}^2(\mathbb{R})$ for $\mathcal{L}$. The transformed closed operator isThis follows from $T_{\gamma }{\partial }_z{T}_{\gamma }^{-1}={\partial }_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the Fourier transform, areIn particular the necessary strict leading-edge condition is ${\gamma }^2-c\gamma +k<0$. For $c>2\sqrt{k}$ this meansA particularly convenient choice isWith this choice there is no first derivative, and ${\mathcal{L}}_{\gamma }={\partial }_z^2+k-c^2/4-2w_c(z)$ is a self-adjoint differential operator on $H^2(\mathbb{R})$. This also avoids possible ambiguities between definitions of essential spectrum for nonnormal operators.

The relevant perturbation theorem is essential-spectrum invariance under relatively compact perturbations: if $A_0$ is a closed operator, $V:D(A_0)\to H$ is compact with respect to the graph norm, and $A_0+V$ is closed on the same domain, their Fredholm essential spectrum is the same. For self-adjoint operators this is the usual essential spectrum after removing isolated finite-multiplicity eigenvalues. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the graph norm domain to $H$, and use compact perturbation invariance of Fredholm operators for each $\lambda$.

Define a step-potential comparison operatorBoth $A_0$ and ${\mathcal{L}}_{\gamma }$ are self-adjoint on $H^2(\mathbb{R})$. Their difference is multiplication byIt is relatively compact, not necessarily compact as a multiplication map on all of L2 space. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb{R})$. On each bounded interval the Rellich-Kondrachov compactness theorem gives a subsequence converging in L2 space. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\Vert V{\Vert }_{\infty }$ is arbitrarily small and the sequence has a uniform L2 norm bound, so those tails are uniformly small. Thus the products converge globally in L2 space, proving graph-domain compactness.

Finally,Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\Vert h{\Vert }_2^2$, so no spectrum lies above this endpoint. For every $\lambda \le k-c^2/4$, choose $\xi$ with $\lambda =k-c^2/4-{\xi }^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\Vert (A_0-\lambda )h_N{\Vert }_2\to 0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the Weyl criterion, proving that every such $\lambda$ belongs to the essential spectrum. The relatively compact perturbation theorem therefore givesThis proves the weighted essential spectrum of a Fisher travelling front is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:Since $w_c\ge 0$, the full transformed spectrum lies below $-(c^2/4-k)$, and the evolution given by its linearization decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero eigenfunction, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted L2 norm alone control the nonlinear remainder at $z\to -\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.

At the minimal speed $c_{\ast }=2\sqrt{k}$, the best choice is $\gamma =\sqrt{k}$. Now ${\gamma }^2-c\gamma +k=(\gamma -\sqrt{k}{)}^2$, andThe essential spectrum has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.

Let $X=ϵx$ and suppose $k^2(X)$ stays positive and sufficiently smooth in the region considered. Write $y=\exp [{ϵ}^{-1}{S}_0(X)+S_1(X)+\cdots ]$ in ${ϵ}^2{y}_{XX}+k^2(X)y=0$. The leading and first transport equations of the WKB approximation areWith $K=\sqrt{k^2}>0$, they give $S_0^{\prime }=\pm iK$ and $S_1=-\frac{1}{2}\log K+\text{constant}$. Thus the leading WKB approximation for a slowly varying oscillator isThe amplitude is fixed by the transport equation, not just by replacing the constant wave number in a sinusoid. The local validity condition is that variation of $K$ over a wavelength is small, for example $|dK/dx|/K^2\ll 1$.

If $k^2=-{\kappa }^2<0$, the same calculation gives the evanescent formAt a classical turning point where $k^2$ vanishes, these amplitudes diverge and the separated WKB descriptions cease to be uniform. For a simple zero, an Airy turning-point connection formula replaces them in an $O({ϵ}^{2/3})$ interval in $X$; it connects oscillation to exponential behavior with the usual quarter-phase shift. A higher-order zero needs a different local canonical problem, so the Airy statement assumes a simple turning point.

For a smooth positive weight $r$ on the closed interval, let $T={\int }_0^{\pi }\sqrt{r(x)}dx$. The local wave number in the Sturm-Liouville problem is $\sqrt{\lambda }\sqrt{r}$, and the condition at zero selects a sine. Imposing the condition at $\pi$ gives the WKB quantization conditionFor weighted normalization ${\int }_0^{\pi }r|y_n{|}^{2}dx=1$, rapidly oscillating averaging gives $C_n\sim \sqrt{2/T}$. There is no turning-point phase correction here because $r$ stays positive at both endpoints with Dirichlet boundary conditions.

In the specified example $r(x)=(x+1{)}^{-2}$, so $T=\log (1+\pi )=L$ and ${\lambda }_n^{\mathrm{WKB}}=(n\pi /L{)}^2$. To solve exactly, set $t=\log (x+1)$ and $y=\sqrt{x+1}w(t)$. Direct differentiation reduces the Euler-Cauchy equation to $w^{\prime \prime }+(\lambda -1/4)w=0$ on $0<t<L$, with zero endpoint values. Hence the inverse-square weighted Dirichlet spectrum isValues $\lambda \le 1/4$ give no nonzero solution satisfying both endpoint conditions. The WKB eigenfunction shape after leading quantization is actually identical to this exact shape; the leading eigenvalue omits $1/4$.

Using the exact eigenvalue as the reference, its relative error isThus $n\ge 3$ is sufficient and necessary among positive integers. Referencing the approximate value instead gives $n>5L/\pi$, and the same integer answer.

Let $U$ be real and twice continuously differentiable, take a nonzero real wave number, and write $c=c_r+i{c}_i$. Exponential instability requires $k{c}_i>0$, in particular $c_i\ne 0$, so $U-c$ has no zero on the channel. Divide the Rayleigh equation for inviscid shear flow by $U-c$, multiply by ${\psi }^{\ast }$ and integrate. The wall conditions remove the boundary term, givingIts imaginary part isThe weight is nonnegative and positive wherever the nontrivial eigenfunction does not vanish. A nontrivial solution of this regular second-order equation cannot vanish on an interval. If $U^{\prime \prime }$ had one strict sign without changing sign, the integral could not vanish. If $U^{\prime \prime }$ were identically zero, the real identity would instead give the contradiction $I=0$. Thus $U^{\prime \prime }$ must take both signs, and continuity gives an interior zero. This proves Rayleigh's inflection-point theorem: an unstable profile must have an inflection point, in fact a sign change of $U^{\prime \prime }$ rather than merely an isolated zero without sign change.

Adding $i$ times the equation for $B$ to the equation for $A$ combines the two real amplitudes into $P=A+iB$. With the fluctuating coupling absent, the derivative terms satisfy $-r{B}_x+ir{A}_x=ir{P}_x$, so the complex evolution isBoth real amplitudes satisfy the zero Dirichlet boundary condition, hence $P(0)=P(\ell )=0$. A unit-modulus change of dependent variable removes the first derivative: put $P=e^{-irx/2}F$. Direct differentiation givesThe eigenfunctions satisfying the Dirichlet boundary condition are therefore $e^{-irx/2}\sin (n\pi x/\ell )$, with temporal eigenvaluesThe first nonzero stationary mode appears when ${\sigma }_1=0$, givingThe complex constant $C$ supplies the phase freedom of this unperturbed mean-field dynamo model.

To display the dipole and quadrupole parity in a mean-field dynamo, center the coordinate at the midpoint: $y=x-\ell /2$, $F(y)=\cos (\pi y/\ell )$, $h=r_0/2$. Choose a real amplitude $C_0$. The dipole choice is $P_D=C_{0}F(y)e^{-ihy}$, soHere $A_D$ is an even function and $B_D$ an odd function of $y$, and the cosine envelope vanishes at both boundaries. The quadrupole choice is $P_Q=i{P}_D$, which givesNow $A_Q$ is an odd function and $B_Q$ an even function. Both choices have the same stationary threshold because the unperturbed complex equation is invariant under a constant phase rotation.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.

We use a closed manifold of positive dimension $n$. The nonnegative positive Laplace-Beltrami operator on functions isIt satisfies $⟨\Delta f,f⟩={\int }_M|df{|}_g^{2}d{\mathrm{vol}}_g$. Its domain as a self-adjoint operator on $L^2(M)$ is $H^2(M)$. The analytic argument in Question 4(d), with $q=0$, gives a smooth orthonormal eigenbasis and ordered eigenvalues $0={\lambda }_0\le {\lambda }_1\le \cdots \to \infty$.

We prove a quantitative counting bound from local Fourier series, rather than assume a spectral asymptotic formula. Choose a finite coordinate cover and smooth functions ${\chi }_{\alpha }$ compactly supported in its charts with ${\sum }_{\alpha }{\chi }_{\alpha }^2=1$. This can be obtained by normalizing an ordinary nonnegative partition of unity by the square root of the sum of its squares. Embed each chart support in a fixed coordinate torus, and extend $v_{\alpha }={\chi }_{\alpha }f$ by zero there. Uniform comparison of the coordinate and metric norms, and differentiation of the cutoffs, give constants independent of $f$ such thatLet $P_R$ retain the Fourier modes $|k|\le R$. Parseval identity givesThe map $f\mapsto (P_R{v}_{\alpha }{)}_{\alpha }$ has target dimension at most $C_0{R}^n$, since there are at most $C_n{R}^n$ lattice points in each cutoff ball.

Let $E_{\Lambda }$ span all eigenfunctions with eigenvalues at most $\Lambda$. For $f\in E_{\Lambda }$, $\Vert df{\Vert }_2^2\le \Lambda \Vert f{\Vert }_2^2$. If every retained local coefficient vanishes, the preceding estimates implyChoose $R=b\sqrt{1+\Lambda }$ with fixed $b$ sufficiently large. Then this is impossible unless $f=0$, so the finite-dimensional coefficient map is injective on $E_{\Lambda }$. Therefore the eigenvalue counting function satisfiesTaking $\Lambda ={\lambda }_k$ gives $k+1\le C_1(1+{\lambda }_k{)}^{n/2}$, so ${\lambda }_k\ge c_1(k+1{)}^{2/n}-1$. For all sufficiently large $k$, this implies ${\lambda }_k\ge A_0{k}^{2/n}$.

If $M$ is connected, the zero eigenvalue is simple: $\Delta f=0$ implies $df=0$, hence $f$ is constant. Thus every ${\lambda }_k$ for $k\ge 1$ is positive. Taking the minimum of $A_0$ and the finitely many positive ratios ${\lambda }_k/k^{2/n}$ below the large-$k$ threshold provesConnectedness is implicit in the printed claim. If $M$ has $b>1$ connected components, zero has multiplicity $b$, and the claim fails already at $k=1$. The same proof gives ${\lambda }_k\ge A\max \{k-b+1,0{\}}^{2/n}$ for all $k$, as well as the universal counting bound above. These finite initial zero modes do not affect part (d).

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

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

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

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

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.

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.