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A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner-product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

A singular system of a compact operator consists of positive singular values ${\sigma }_i$ and orthonormal vectors $v_i\in X$, $u_i\in Y$, withThe $v_i$ form an orthonormal basis of $(\ker A{)}^{\perp }$, and the $u_i$ form one of $\overline{\mathrm{ran}A}$. There are finitely many positive singular values for finite-rank $A$; otherwise ${\sigma }_i\to 0$. Obtain the system by applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$, choosing $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$, and setting $u_i=A{v}_i/{\sigma }_i$. The inner product identities verify orthonormality and the displayed relations.

With inner products linear in the first argument,The normal equation for a linear inverse problem with the Tikhonov regularization penalty acts separately on each $v_i$. Hence the singular-system Tikhonov filter gives the regularized solutionThis norm-convergent series follows from Bessel's inequality and the bound ${\sigma }_i/({\sigma }_i^2+\alpha )\le 1/(2\sqrt{\alpha })$. Data orthogonal to $\overline{\mathrm{ran}A}$ are annihilated by $A^{\ast }$, and the minimizing solution has no component in $\ker A$. The same formula with $y^{\delta }$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by ${\sigma }_i$, with convergence governed by the Picard criterion.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,

Assume $n\ge 2$. The place-permutation action on ${\mathbb{C}}^n$ preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of $S_n$ is two-transitive, so the irreducible augmentation criterion for a transitive group action makes $W$ irreducible. The two-row Young permutation module decomposition of $M^{(n-1,1)}\cong {\mathbb{C}}^n$ identifies its nontrivial summand as $V^{(n-1,1)}$. Hence $W\cong V^{(n-1,1)}$.

All standard Young tableaux of shape $(n-1,1)$ are $T_j$, $2\le j\le n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of $w_j$ with $w_l$, $j<l$, is zero because the coefficients of $w_j$ sum to zero. Directly summing the action of $(ar)$ gives $X_r{w}_j=c_{T_j}(r)w_j$.

For $s_i=(ii+1)$, $s_1{w}_2=-w_2$. For $2\le i\le n-1$, its only nontrivial two-dimensional block isEvery other $w_j$ is fixed, including all $w_j$ with $j>2$ when $i=1$. These formulas follow by swapping coordinates $i,i+1$ in the displayed vectors, and are the Young orthogonal form with axial distance $i$ for $T_i$.

Put $m=n-k$ and take the subgroup $S_m\times S_k$ permuting the first $m$ and last $k$ letters. Define the skew representation of a symmetric group as the multiplicity spaceThe last-letter copy of $S_k$ commutes with $S_m$, so it acts on a map $f$ by $g\cdot f={\rho }_{\lambda }(g)\circ f$. This gives a genuine $\mathbb{C}{S}_k$-module, without claiming that the whole restricted module is itself the skew representation. Equivalently,

Fix a prefix tableau of shape $\mu$ and complete it to shape $\lambda$. Iterated restriction branching rule for a symmetric group identifies the multiplicity-space orthonormal basis with standard skew Young tableaux, using labels $1,\dots ,k$ for the last $k$ cells. The Young orthogonal form restricts to this basis. When $R=s_{i}T$ is standard and $d=c_T(i+1)-c_T(i)$,An admissible interchange has $|d|\ge 2$, so its off-diagonal coefficient is nonzero. Consequently $w_R$ belongs to the group algebra span of $w_T$. The preceding reduced-path argument reaches every standard skew tableau, so this span is the whole module. Every $w_T$ is a cyclic vector for a group representation. The inherited invariant inner product makes this a unitary representation.

Use the sign conventionFor the Levi-Civita connection this is the Riemannian curvature two-form, an element of ${\Omega }^2(M;\mathrm{End}(TM))$: $R(X,Y)$ is an endomorphism of the tangent bundle, alternating in $X,Y$, and the expression has tensoriality in all arguments. It is the curvature form of a connection for the tangent-bundle connection.

For the first Bianchi identity, take the cyclic sum in $X,Y,Z$. Torsion-freeness says ${\nabla }_{Y}Z-{\nabla }_{Z}Y=[Y,Z]$, so the double-derivative terms combine to ${\sum }_{\mathrm{cyc}}{\nabla }_X[Y,Z]$. The remaining terms may be cyclically relabelled as $-{\sum }_{\mathrm{cyc}}{\nabla }_{[Y,Z]}X$. Using torsion-freeness once more, followed by the Jacobi identity for vector fields, givesThe Ricci curvature is the tracefor any orthonormal basis. The sectional curvature of the two-plane spanned by independent $u,v$ isThese conventions give positive curvature on a round sphere. In dimension three, put $K_{ij}=K(e_i,e_j)$ and $r_i=\mathrm{Ric}(e_i,e_i)$. Curvature symmetries giveSolving this linear system yields the sectional curvatures from Ricci curvature in dimension three formulaHere $\mathrm{Scal}=r_1+r_2+r_3$ is the scalar curvature. The chosen orthonormal basis need not diagonalize Ricci; its diagonal evaluations already determine these three sectional curvatures.

At each point, use the inverse metric on the cotangent space. It induces an inner product on $p$-forms byEquivalently, wedges of distinct members of an orthonormal coframe form an orthonormal basis. Choose an orientation to define the ordinary global Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩d{\mathrm{vol}}_g$. It maps $p$-forms to $(n-p)$-forms and obeys ${\ast }^2=(-1{)}^{p(n-p)}$. We use real forms; complex forms are obtained by complex-linear extension. Without an orientation an ordinary global star is not available without twisting the target, so orientation is implicit in this first part.

With the codifferential ${\delta }_p=(-1{)}^{n(p+1)+1}\ast d\ast$ for $p>0$ and ${\delta }_0=0$, define the Hodge Laplacian $\Delta =d\delta +\delta d$. On functions it is the nonnegative-sign Laplace-Beltrami operator, $\Delta f=-\mathrm{div}\mathrm{grad}f$. A harmonic differential form satisfies $\Delta \alpha =0$.

For $\alpha$ of degree $p$, the displayed sign formula and star square giveApplying the identities again, at degrees $p-1$ and $p+1$, proves the Hodge star commutes with the Hodge Laplacian relationSince star is invertible, $\alpha$ is harmonic if and only if $\ast \alpha$ is harmonic. This part does not need compactness; zero-degree and top-degree terms are interpreted as zero when their degrees are outside the range.

For the remaining cohomological statements assume a compact oriented manifold without boundary. The Hodge decomposition theorem gives the $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^p=\ker \Delta$ is finite-dimensional and consists of smooth forms. Formal adjointness gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic differential forms are closed differential forms and coclosed differential forms.

If $\Delta f\ge 0$, formal adjointness and $\Delta 1=0$ give ${\int }_M\Delta f=0$. A continuous nonnegative function with zero integral vanishes, so $\Delta f=0$. Then $\Vert df{\Vert }_{L^2}^2=⟨\Delta f,f⟩=0$. Thus the a function with nonnegative Laplacian on a closed manifold is locally constant assertion isIt is globally constant when $M$ is connected. The PDF does not explicitly include connectedness here; on two disjoint circles one may take different constants, so literal global constancy needs that qualification. Boundaryless is the standard manifold convention in this argument; otherwise boundary conditions are necessary.

For a closed differential form $\alpha$, write its Hodge decomposition as $\alpha =h+d\beta +\delta \gamma$. Since $d\alpha =dh=0$, we have $d\delta \gamma =0$. Formal adjointness yields $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$, so $\alpha =h+d\beta$. Thus $h$ represents its de Rham cohomology class. If two harmonic representatives differ by $d\beta$, their difference $h$ has $\Vert h{\Vert }^2=⟨d\beta ,h⟩=⟨\beta ,\delta h⟩=0$. Therefore

Finally let $h\in G$. A connected Lie group admits a smooth path $h_t$ from the identity to $h$. The maps $R_{h_t}$ give a smooth homotopy between the identity and $R_h$. The homotopy invariance of de Rham cohomology says their pullbacks agree on cohomology. The maps $R_{h_t}$ preserve orientation, since their Jacobian signs cannot change from the identity along the path. As an orientation-preserving isometry, $R_h$ commutes on forms with star, exterior differentiation, codifferentiation and consequently the Hodge Laplacian. Hence $R_h^{\ast }\alpha$ is harmonic and represents the same cohomology class as a harmonic $\alpha$. Uniqueness now proves the harmonic forms are fixed by a connected isometric group action conclusionThe argument also applies componentwise when $M$ is disconnected.

For each subset $J\subseteq \{1,\dots ,n\}$, take the Walsh character $f_J(x)={\prod }_{j\in J}{x}_j$, including $f_{\varnothing }\equiv 1$. Under the uniform distribution on a finite set these functions are an orthonormal basis: $f_J{f}_K=f_{J△K}$ has mean zero unless $J=K$, since the coordinates are independent fair signs. There are $2^n$ such functions, the full dimension of the function space.

Flipping a coordinate in $J$ changes the sign, while flipping one outside $J$ leaves the value unchanged. The total probability of a sign change is $|J|/(2n)$, soThe spectrum of lazy random walk on the hypercube is thereforeAll eigenvalues are nonnegative and the second largest is $1-1/n$. Hence both the ordinary spectral gap and the absolute spectral gap equal $\boxed{\gamma =1/n}$. The requested bound in part (c) is smaller than the exact gap by a factor $2n$; even the refined increasing-coordinate canonical paths Poincare bound loses a factor $(n+1)/2$. The true optimal Poincare inequality for a reversible Markov chain constant is $n$.

Use a real orthonormal basis of Fourier modes on the time circle:The eigenvalue of $L_{\omega }$ on the constant mode is ${\omega }^2$, and each sine/cosine pair has eigenvalue ${\omega }^2+{\nu }_r^2$. Retaining these modes giveswhere $C_N$ can depend on $N,\beta$ but is independent of $\omega$. Each real Gaussian integral contributes $\sqrt{2\pi /{\lambda }_r}$. Consequently the Fourier-cutoff oscillator functional determinant isA spectral cutoff between successive eigenvalue pairs selects precisely this finite subspace. In comparing two frequencies, the displayed regularization keeps the same number of modes $N$; holding an absolute spectral threshold fixed need not give the same $N$ at finite cutoff.

Assume first $0<k<N$. Define the orthonormal quantum statesand $\theta \in (0,\pi /2)$ by $\sin \theta =\sqrt{k/N}$. The relevant invariant subspace is the two-dimensional span of these quantum states, not the entire Hilbert space. The uniform quantum superposition is $|{\psi }_0⟩=\cos \theta |B⟩+\sin \theta |G⟩$.

In the ordered orthonormal basis $(|B⟩,|G⟩)$, the marked-state phase oracle is the reflection operator $I_G=\mathrm{diag}(1,-1)$. Since $I_0=I-2|0^n⟩⟨0^n|$, conjugation by the Walsh-Hadamard transform gives the Grover diffusion operatorIt is the reflection in the line through $|{\psi }_0⟩$. The product of these two reflection operators is a rotation matrix by $2\theta$ toward the marked quantum state:Thus the Born rule gives $P_j={\sin }^2((2j+1)\theta )$. Choose a nonnegative integer $j_{\ast }$ nearest to $\pi /(4\theta )-1/2$. Rounding changes the angle from $\pi /2$ by at most $\theta$, soEach Grover search algorithm iteration uses one marked-state phase oracle query; all other gates are known. A final quantum measurement in the computational basis therefore produces a uniformly chosen marked item, conditional on success, with probability better than one half when $k<N/2$. This uses the supplied marked count $k$ to select the stopping time; an unknown $k$ needs an additional search schedule or counting procedure. If $k=N$, no oracle query is needed; if $k=0$, no marked item exists, so a successful search is impossible.

Let $V$ denote the coherent exact quantum phase estimation unitary operator from the preceding part, with the data quantum register written first. For every eigenvector of $A$,because $e^{2\pi iA}|u_j⟩=e^{2\pi i{\lambda }_j}|u_j⟩$ and ${\lambda }_j=c_j/2^n$ lies in $[0,1)$, with no phase-aliasing ambiguity. Apply $V$ coherently to the input quantum state and a clean quantum register; do not measure the eigenvalue label. By linearity it produces ${\sum }_j{\beta }_j|u_j⟩|c_j⟩$.

Append a quantum ancilla in $|0⟩$, and use the provided quantum variable rotation with ${\theta }_{c_j}=\arcsin {\lambda }_j$, choosing the nonnegative cosine. The resulting normalized quantum state isThe sum covers both terms; this fixes the potentially ambiguous sum placement in the abbreviated display. The orthonormal basis of eigenvectors and the unit length of each rotated quantum ancilla show that its norm is one. This is quantum spectral filtering by multiplication, not the reciprocal rotation used in the HHL algorithm.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

Let $W_j=V_{j+1}\ominus V_j$ denote the orthogonal complement of $V_j$ inside $V_{j+1}$. The Meyer-Mallat theorem constructs a wavelet whose integer function translations form an orthonormal basis of $W_0$. Dilation then gives the orthogonal decomposition into orthogonal complements $L^2(\mathbb{R})={⨁}_{j\in \mathbb{Z}}{W}_j$, so its translates and dilates form an orthonormal wavelet basis.

Using the MRA low-pass filter, choose the high-pass symbol $n(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$. The quadrature mirror filter identity makes the two analysis channels orthonormal. The resulting Fourier transform formula isEquivalently, with the refinement coefficients from the scaling refinement equation, put $g_k=(-1{)}^{k-1}\overline{h_{1-k}}$ and $\psi (x)=\sqrt{2}{\sum }_k{g}_k\phi (2x-k)$. This index choice gives precisely the displayed high-pass symbol. Changing all $g_k$ by one constant unit phase gives the same orthonormal wavelet construction. In particular the often-used $(-1{)}^k\overline{h_{1-k}}$ convention differs only by a global minus sign. The basis is ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ for $j,k\in \mathbb{Z}$.