TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Fourier–Galerkin assembly correctness and termination

Statement

Assume the Axiom of Choice and Countable Choice. Let $d \geq 1$ and $b, N \geq 0$ be integers, let $ℏ > 0$ and $ω > 0$ be real, and set $T = 2 π / ω$ . Let $H_{k} \in M_{d} (C)$ for $∣ k ∣ \leq b$ satisfy $H_{- k} = H_{k}^{*}$ , set $H_{k} = 0$ for $∣ k ∣ > b$ , and put $H (t) = \sum_{∣ k ∣ \leq b} H_{k} e^{i k ω t}$ . Set $q = (2 N + 1) d$ . Fix $τ > 0$ and a total routine $S_{τ}$ on ${M \in M_{q} (C) : M = M^{*}}$ with the standard inner product on $C^{q}$ , that for each input $M$ returns real numbers ${\tilde{λ}}_{1}, \dots, {\tilde{λ}}_{q}$ , an orthonormal basis $x_{1}, \dots, x_{q}$ , and residuals $r_{j} = ∥ M x_{j} - {\tilde{λ}}_{j} x_{j} ∥ \leq τ$ as in the algorithm Definition. Let $H_{T} = L^{2} ((0, T), d t; C^{d})$ with its componentwise $L^{2}$ inner product, and let $K f = H (t) f - i ℏ f^{'} (t)$ on the domain of vectors whose components lie in $H^{1} (0, T)$ and have equal endpoint traces. Let $S_{N}$ be the span of $T^{- 1 / 2} e^{i n ω t} v$ with $- N \leq n \leq N$ and $v \in C^{d}$ , and let $Π_{N}$ be its orthogonal projection. The matrix with blocks $(K_{N})_{n m} = H_{n - m} + n ℏ ω δ_{n m} I_{d}$ is exactly the matrix of $Π_{N} K ∣_{S_{N}}$ in the ordered Fourier basis and is Hermitian. The block-assembly portion of the algorithm terminates after exactly $(2 N + 1)^{2}$ block assignments. The complete procedure is well-defined and terminates, because it then calls $S_{τ}$ once on the Hermitian matrix $K_{N}$ . This is an exact compression result only; it asserts no convergence or spectral-accuracy guarantee for finite-section eigenvalues and no residual guarantee beyond the stipulated solver interface.

Statement references

The Axiom of Choice

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Floquet Fourier–Galerkin assembly

Finite-level Sambe operator with Fourier domain

The twisted-periodic momentum operator on an interval

Facts & Assumptions

Given: AC and Countable Choice, the Fourier coefficients, bandwidth $b$ , cutoff $N$ , $d \geq 1$ , $ω > 0$ , $ℏ > 0$ , $T = 2 π / ω$ , $τ > 0$ , and the total routine $S_{τ}$ with the declared interface from the Statement.

[A1]

AC and Countable Choice are carried from the self-adjoint Sambe operator and periodic Fourier supplier (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Finite-level Sambe operator with Fourier domain defines $K = H (t) - i ℏ d / d t$ on componentwise periodic $H^{1} (0, T)$ and states that its domain contains every smooth periodic vector-valued function.

[F2]

The twisted-periodic momentum operator on an interval at twist $θ = 0$ gives the complete orthonormal Fourier family $e_{n} (t) = T^{- 1 / 2} e^{i n ω t}$ in scalar $L^{2} (0, T)$ .

[F3]

Floquet Fourier–Galerkin assembly assigns $(K_{N})_{n m} = H_{n - m} + n ℏ ω δ_{n m} I_{d}$ for each ordered pair $n, m \in {- N, \dots, N}$ , with $H_{k} = 0$ for $∣ k ∣ > b$ and $H_{- k} = H_{k}^{*}$ . The finite Fourier polynomial $H (t) = \sum_{k} H_{k} e^{i k ω t}$ is smooth and $T$ -periodic; it is Hermitian because $H (t)^{*} = \sum_{k} H_{k}^{*} e^{- i k ω t} = \sum_{k} H_{k} e^{i k ω t} = H (t)$ after reindexing.

[F4]

Under the algorithm Definition, $S_{τ}$ is total on finite Hermitian matrices and is called exactly once after the finite block assembly; it returns the declared finite output record unchanged.

Proof

Given: [A1], [F1]–[F3], and the finite Fourier polynomial in the Statement.

1.1A1F1F2givenalgebra

For $- N \leq m \leq N$ and $v \in C^{d}$ , $e_{m} (t) v = T^{- 1 / 2} e^{i m ω t} v$ is smooth and periodic, so it lies in $D (K)$ by [F1]. Multiplication by $H (t)$ gives $\sum_{k} T^{- 1 / 2} e^{i (m + k) ω t} H_{k} v$ ; its coefficient in harmonic $n$ is $H_{n - m} v$ . The derivative term is $- i ℏ \partial_{t} (e_{m} v) = m ℏ ω e_{m} v$ , which contributes only when $n = m$ . Thus $Π_{N} K (e_{m} v)$ has $n$ th coefficient $(H_{n - m} + n ℏ ω δ_{n m} I_{d}) v$ .

1.2F3algebra

For every pair $n, m$ , [F3] gives $(K_{N})_{n m}^{*} = H_{n - m}^{*} + n ℏ ω δ_{n m} I_{d} = H_{m - n} + m ℏ ω δ_{m n} I_{d} = (K_{N})_{m n}$ , since $H_{- k} = H_{k}^{*}$ and the diagonal term is present only when $n = m$ . Hence $K_{N}$ is Hermitian.

2.1F2F3step 1.1

The coefficient just computed is precisely the block assigned by [F3] in row $n$ , column $m$ . Since the ordered Fourier basis spans $S_{N}$ , equality of all its matrix blocks proves that $K_{N}$ is exactly the matrix of $Π_{N} K ∣_{S_{N}}$ .

3.1F3F4step 1.2given∎

There are $(2 N + 1)^{2}$ ordered pairs $(n, m)$ , each assignment fills one finite $d$ by $d$ block, so the block assembly terminates after exactly that many assignments. By step 1.2, $K_{N}$ is Hermitian, and it has dimension $q = (2 N + 1) d$ ; hence it is in the domain of the total routine $S_{τ}$ in [F4]. The algorithm makes one call $R : = S_{τ} (K_{N})$ after assembly, and totality gives a returned finite record. Thus the complete procedure is well-defined and terminates. This exact compression and termination argument proves no convergence as $N \to \infty$ , spectral accuracy, or residual claim beyond the routine's stipulated interface.

Fourier–Galerkin assembly correctness and termination

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources