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Put $\zeta =e^{\pi i/n}$. The group representations from the preceding construction with $ϵ={\zeta }^j$, $1\le j\le n-1$, will be denoted ${\rho }_j$. Their diagonal matrices for $a$ have distinct eigenvalues. The only possible invariant lines would be these two eigenspaces, and the matrix for $b$ exchanges them. Thus each ${\rho }_j$ is irreducible. Equivalently their $a,b$ matrices do not commute, so part (a) applies. The choices $j$ and $2n-j$ give equivalent group representations by exchanging basis vectors. The displayed range contains exactly one of each pair, and distinct members have different traces at $a$ because $\cos (\pi j/n)$ is strictly decreasing there.

For a one-dimensional character of a representation, write $a\mapsto \alpha$, $b\mapsto \beta$. The conjugation relation forces ${\alpha }^2=1$, and ${\beta }^2={\alpha }^n$. Hence there are exactly four choices:For even $n$, $\beta =\pm 1$ for either $\alpha$; for odd $n$, $\alpha =1$ has $\beta =\pm 1$ and $\alpha =-1$ has $\beta =\pm i$. Finally,The sum of squares of irreducible representation dimensions proves that these four one-dimensional and $n-1$ two-dimensional group representations exhaust the irreducibles, including $n=1$ where the latter list is empty.

The Rayleigh quotient is $R[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$, interpreted through the energy quadratic form for trial states in its form domain. Expanding in normalized energy eigenstates gives $R[\psi ]=\sum E_n|c_n{|}^2/\sum |c_n{|}^2\ge E_0$. If the ground state is unique, equality requires all coefficients outside its one-dimensional eigenspace to vanish. Thus equality means the same ground-state ray, allowing normalization and an overall phase.

Use the normalized trial function $\sqrt{2/a}\cos (\pi x/a)$ on $|x|<a/2$, zero outside. Its kinetic energy is ${\pi }^2{\hslash }^2/(2m{a}^2)$ and its mean square position is $c_0{a}^2$, where $c_0=1/12-1/(2{\pi }^2)$. The energy isMinimization gives $a^4={\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_0)$ andFor the first excited well state use $\sqrt{2/a}\sin (2\pi x/a)$, zero outside. Now the kinetic energy is $2{\pi }^2{\hslash }^2/(m{a}^2)$ and the position factor is $c_1=1/12-1/(8{\pi }^2)$. Minimizing gives $a^4=4{\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_1)$ andThis particular excited-state estimate is an upper bound. The trial function is odd and the oscillator ground state is even, so it is exactly orthogonal to the ground state. Its spectral expansion therefore contains only energies at least $E_1$, giving the same variational bound on that orthogonal subspace. An arbitrary trial state without this orthogonality would bound only $E_0$, not $E_1$. The compactly supported trial functions belong to the quadratic-form domain despite derivative jumps at the well edges; their kinetic energies are correctly computed as ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$.

Write $J_r(\lambda )$ for a size-$r$ Jordan block. The characteristic polynomial fixes the total dimensions at $2$ and $-7$, while the minimal polynomial fixes the largest Jordan block at each eigenvalue. Consequently, the blocks for $2$ have total size four, largest size two, and at least one size-two block. The blocks for $-7$ both have size one. The complete list, up to similarity, isThese Jordan normal forms are inequivalent because their $2$-eigenspaces have different dimensions, two and three.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

Define the parity operator by $(P\psi )(x)=\psi (-x)$. For the usual full-line Hamiltonian operator, or any parity-invariant operator domain and boundary conditions, an even potential gives $HP=PH$: the second derivative is unchanged by reflection and $V(-x)=V(x)$. If $H\psi =E\psi$, then $H(P\psi )=E(P\psi )$. A nondegenerate energy eigenvalue has a one-dimensional eigenspace, so $P\psi =\sigma \psi$ for some scalar $\sigma$. Since $P^2=I$, ${\sigma }^2=1$. HenceThus the stationary state has even or odd parity. The symmetry must include the operator domain: an even potential alone would not imply this conclusion with asymmetric boundary conditions.

The first Young–Jucys–Murphy element is the empty sum $X_1=0$. Applying it to the nonzero simultaneous eigenvector defining a point of the joint spectrum gives $0=a_{1}v$. HenceFor the remaining spectral arguments use the following local facts, without assuming the restriction branching rule for a symmetric group: consecutive eigenvalues are distinct; a difference $a_{r+1}-a_r=\pm 1$ makes $s_r=(rr+1)$ act on the corresponding line as $\pm 1$; every other adjacent interchange gives another point of the joint spectrum in the same irreducible representation. These are the local spectral rules for Young–Jucys–Murphy elements. They follow from $s_r{X}_r{s}_r+s_r=X_{r+1}$ and the one- or two-dimensional analysis of the two neighboring eigenspaces. They are the appropriate local results used below.

Apply $V^{-1}$ to the data and phase quantum registers, leaving the flag quantum ancilla untouched. This inverse needs no extra oracle assumption: the promised dyadic eigenvalues give $U^{2^n}=I$, hence $U^{-1}=U^{2^n-1}$. The inverse of each controlled unitary gate used in $V$ can therefore be built from repeated uses of the supplied controlled-$U$, and the known Hadamard gates and quantum Fourier transform gates can be reversed. Uncomputation is necessary to erase the eigenvalue label coherently. The resulting quantum state isA quantum measurement in the computational basis of the flag followed by postselection on one yieldsThis requires $A|b⟩\ne 0$. For a normalized input and nonnegative eigenvalues, the success probability of positive quantum spectral filtering satisfies $P_{\mathrm{success}}\ge {\lambda }_{\min }^2$. In the general inequality, equality holds precisely when the input is supported on the minimum-eigenvalue eigenspace. Omitting uncomputation and discarding the phase quantum register would instead leave a mixed state with diagonal weights proportional to $|{\beta }_j{|}^2{\lambda }_j^2$, rather than the desired coherent pure state.

The printed universal nonzero-success request needs a nonkernel-input hypothesis. An $n$-qubit Hermitian operator has $2^n$ eigenvalues, counted with multiplicity. All are distinct, and the printed dyadic grid contains exactly $2^n$ possible values. They therefore occupy the entire grid, including zero: this is a multiplicity-free complete dyadic spectrum, and ${\lambda }_{\min }=0$. Taking $n=1$, $A=\mathrm{diag}(0,1/2)$, and $|b⟩=|0⟩$ satisfies every printed spectral promise but gives $A|b⟩=0$. No normalized output vector exists, so no algorithm can deliver it with nonzero probability.

For every input outside the kernel, the procedure above has $P_{\mathrm{success}}>0={\lambda }_{\min }^2$, giving the requested strict bound on the meaningful domain. The corrected result is therefore exact quantum spectral filtering conditional on $A|b⟩\ne 0$, with the boxed probability. This does not require knowing the input probability amplitudes or making extra copies of an unknown quantum state. Arbitrarily small nonzero support outside the kernel gives arbitrarily small success probability, and a failed flag measurement disturbs the input; fresh independent trials cannot be assumed when only one unknown physical input is supplied.

For homogeneous corrections at the square boundary, integration by parts gives $⟨{\psi }_0\Delta {\psi }_1⟩=⟨{\psi }_1\Delta {\psi }_0⟩$ and $⟨{\theta }_0\Delta {\theta }_1⟩=⟨{\theta }_1\Delta {\theta }_0⟩$. The first-derivative cross terms give $⟨{\psi }_0{\theta }_{1x}⟩=-⟨{\psi }_{0x}{\theta }_1⟩$ and $⟨{\theta }_0{\psi }_{1x}⟩=-⟨{\theta }_{0x}{\psi }_1⟩$. Using $\Delta {\psi }_0=q_c^2{\theta }_{0x}$ and $\Delta {\theta }_0=-{\psi }_{0x}$, these four contributions cancel in pairs. This proves exactly the supplied solvability condition, including the weight $q_c^2$ on its second row. Equivalently $({\psi }_0,q_c^2{\theta }_0)$ is the relevant adjoint eigenfunction for this coupled system.

At first order in the detuning, the inhomogeneous equations areInsert these right-hand sides into the verified solvability condition. One obtainsA further integration by parts, using the leading heat equation, givesHence the growth-rate solvability for conducting-square Darcy convection yieldsThe PDF prints the reciprocal of the required integral ratio. With its declared ${\theta }_t=ϵ\lambda \theta$, the displayed reciprocal does not follow and is false for $q_1\ne 0$. Both integrals are positive for these nonzero modes. The square's Poincare inequality gives their ratio at least $2{\pi }^2$, so it cannot equal its reciprocal. For $q_1=1$, the two explicit parities give approximately $\lambda =11.34770$ for odd $\theta$ and $\lambda =7.30164$ for even $\theta$; the printed expression instead gives about $0.00446439$ and $0.00693825$. These values provide direct mode-based counterexamples.

The corrected growth rate is positive for $q_1>0$ and negative for $q_1<0$, as expected at a convection threshold. Because the critical eigenspace is two-dimensional, an arbitrary superposition generally splits into two different first-order growth rates. The specified parity modes remain independent under the detuning: the temperature linear operator preserves reflection parity, so the cross-parity projection vanishes. This justifies applying the scalar solvability condition to either listed eigenfunction; it should not be applied as a single common eigenvalue to an arbitrary mixture.

For a $\Delta$-regular undirected graph, its Graph Laplacian is $L=\Delta I-A$, and for any complex vector $v$,If $Av=\Delta v$, the left side is zero, hence $v_i=v_j$ on every edge. In a connected graph all coordinates coincide. Thus the eigenspace is exactly the constant vectors, of dimension one. Since the adjacency matrix of a graph is symmetric and diagonalizable,

In $V_n\otimes V_n$, interchange of the two factors commutes with SU2. On the highest-weight vector $D^r{x}_1^{n-r}{x}_2^{n-r}$ it acts by $(-1{)}^r$. Each irreducible summand in the Clebsch-Gordan decomposition has multiplicity one, so this sign is its scalar action on the entire summand. The symmetric square is the $+1$ eigenspace and the exterior square the $-1$ eigenspace. ThereforeFor instance, $S^2{V}_1=V_2$ and ${\Lambda }^2{V}_1=V_0$. The dimensions agree respectively with $(n+1)(n+2)/2$ and $n(n+1)/2$.

First construct the spectral measure of a normal operator and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^{\ast }(I,T)$ generated by $T$ and $T^{\ast }$ is commutative because $T$ is a normal operator. The Gelfand transform identifies $A$ with $C({\Phi }_A)$. By spectral permanence for C-star algebras,is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal{B}(H)$. It is injective because a character is determined by its values on $T,T^{\ast }$ and $I$, and $\chi (T^{\ast })=\overline{\chi (T)}$. Thus $h$ is a homeomorphism. Apply part (a) and setThis is a projection-valued measure with $P(K)=I$ andThe associated continuous functional calculus is isometric: $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }$ for $u\in C(K)$.

If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq \mathbb{C}$. If $P(U)=0$, then $u(T)={\int }_{K}udP=0$, contradicting $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }>0$. ThereforeThis proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.

Now take an isolated point $\lambda \in K$. Its singleton is relatively open, so $P(\{\lambda \})\ne 0$. Multiplication of spectral integrals givesThus its range is contained in the eigenspace $\ker (T-\lambda I)$. Conversely, for $v$ in that kernel, the positive scalar spectral measure ${\mu }_v(B)=⟨P(B)v,v⟩$ satisfiesIt follows that ${\mu }_v(K\setminus \{\lambda \})=0$, and hence $\Vert (I-P(\{\lambda \}))v{\Vert }^2=0$. ConsequentlyThe identification of this projection with the eigenspace holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.

Let $\dim V=2n$ and let $V_{+}=\ker (S-I)$ and $V_{-}=\ker (S+I)$ be the two eigenspaces. The involution identity gives a direct sum, sinceFor $u,v$ in either one of these eigenspaces, the anti-symplectic involution identity impliesso both are isotropic subspaces of a symplectic vector space. Such a subspace has dimension at most $n$: $W\subset W^{\Omega }$ and the nondegenerate bilinear form $\Omega$ gives $\dim W^{\Omega }=2n-\dim W$, where $W^{\Omega }$ is the symplectic orthogonal complement. As their dimensions add to $2n$, both have dimension $n$. Therefore

For $p\in \mathrm{Fix}(\sigma )$, differentiation gives $(d{\sigma }_p{)}^2=I$ and $(d{\sigma }_p{)}^{\ast }{\omega }_p=-{\omega }_p$. Thus $d{\sigma }_p$ is an anti-symplectic involution of the symplectic vector space $T_{p}M$. By part (a), its $+1$ eigenspace is a Lagrangian subspace.

In the local manifold chart supplied in the question, $\sigma$ is linear, so its fixed-point set is the intersection of the chart with this $+1$ eigenspace. Therefore the fixed-point set is an embedded submanifold andEvery such tangent space has half the ambient dimension and the symplectic form vanishes on it. Hence every nonempty fixed-point set is a Lagrangian submanifold. The fixed-point set itself need not be connected.

In the matrix expression, $t^a={\lambda }^a/2$ denotes a fundamental generator, whereas the component adjoint covariant derivative uses adjoint generators. With $\mathrm{Tr}(t^d{t}^h)={\delta }^{dh}/2$,Using antisymmetry to write $f^{acd}=f^{dac}$ and $f^{bed}=f^{dbe}$ gives exactly the component mass term. This is the gauge-boson mass matrix from an adjoint scalar.

For ${\Phi }_0=v\mathrm{diag}(1,1,-2)$, the Gell-Mann matrices $t^1,t^2,t^3,t^8$ commute with ${\Phi }_0$. Their gauge bosons are therefore massless. The generators $t^4,t^5,t^6,t^7$ connect the first two eigenspaces to the third; the eigenvalue difference is $3v$. Directly taking the matrix trace giveswith all remaining entries zero. ThusThe factor $1/2$ is the canonical real-vector mass term normalization. The unbroken Lie algebra is $\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, giving the usual description SU(2) × U(1). More precisely, the centralizer of the vacuum expectation value inside $SU(3)$ isThis global quotient refines the adjoint-Higgs breaking of SU(3) to SU(2) and U(1) without changing the local gauge boson spectrum.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.