ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Flat torus harmonic forms are constant-coefficient forms

Example

Assume the Axiom of Choice, $n \geq 2$ and an integer $k \geq 0$ , with $(\binom{n}{k}) = 0$ when $k > n$ . On $T^{n} = R^{n} / Z^{n}$ with its flat metric, the forms $d x^{i_{1}} \land \dots \land d x^{i_{k}}$ , $i_{1} < \dots < i_{k}$ , form a basis of $H^{k}$ . Every real de Rham class therefore has a unique constant-coefficient representative, and $\dim H_{d R}^{k} (T^{n}) = (\binom{n}{k})$ .

Facts & Assumptions

Given: full AC, $n \geq 2$ , an integer $k \geq 0$ and the flat torus.

[F1]

The compact Hodge theorem supplies a unique harmonic representative of each de Rham class (Harmonic representatives for compact de Rham and Dolbeault complexes).

[F2]

The quotient $T^{n} = R^{n} / Z^{n}$ is a boundaryless smooth manifold, and $q$ is a surjective local isometry with $q^{*} g = \sum_{i} d x^{i} \otimes d x^{i}$ . Its quotient charts have locally translational transition maps (Flat torus model geometry, Proof, quotient-chart and metric construction). Smooth $k$ -forms are exterior-bundle sections (A smooth differential $k$ -form), and $ι_{X}$ is the contraction of Interior product of a form by a vector field.

[F3]

The cube $[0, 1]^{n}$ is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, assertion 1). For continuous functions on a cube, iterated Riemann integrals equal the multiple integral, and integrating a continuous derivative gives its endpoint difference (A continuous function on a closed rectangle has repeated Riemann integrals in every coordinate order, all equal to its multiple integral, Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

Verification

1.1F2F3given

Each class has a representative in $[0, 1]^{n}$ , by subtracting the integer parts of its coordinates. Thus $q ([0, 1]^{n}) = T^{n}$ ; pulling an open cover back to this compact cube and projecting a finite subcover proves compactness. The local coframes $d x^{j}$ agree across the translational changes of [F2], so they descend to a global orthonormal coframe, and its dual vector fields are the descended $\partial_{j}$ . For a form $u$ , its pullback is invariant under integer translations because $q (x + m) = q (x)$ ; therefore its coefficients in this coframe are smooth periodic functions.

2.1F1F2F3step 1.1algebra

Write $u = \sum_{I} u_{I} d x^{I}$ in that coframe. The coordinate formula and [F1] give $d = \sum_{j} ε (d x^{j}) \partial_{j}$ . Integrating each coefficient derivative over $[0, 1]^{n}$ by [F3] cancels opposite-face terms by periodicity, so the formal adjoint identity gives $d^{*} = - \sum_{j} ι_{\partial_{j}} \partial_{j}$ . Here the density integral equals the cube integral: subdivide the cube into finitely many small boxes lying in quotient charts and use their Euclidean density; the shared faces have zero Riemann content. The wedge/contraction identity proved in [F1], $ε (d x^{j}) ι_{\partial_{l}} + ι_{\partial_{l}} ε (d x^{j}) = δ_{j l} I$ , now gives $Δ u = - \sum_{I, j} \partial_{j}^{2} u_{I} d x^{I}$ . One more integration by parts using [F3] gives $(Δ u, u)_{2} = \sum_{I, j} \int_{[0, 1]^{n}} ∣ \partial_{j} u_{I} ∣^{2}$ .

3.1F1step 2.1algebra∎

If $u$ is harmonic, the sum of nonnegative integrals in step 2.1 is zero. Each derivative is continuous, so a nonzero value would give a neighbourhood with positive integral; hence every $\partial_{j} u_{I}$ vanishes everywhere. Integration along line segments in $R^{n}$ gives constant coefficients. Conversely constant coefficients give $Δ u = 0$ by step 2.1. The increasing wedge basis has $(\binom{n}{k})$ members, and [F1] identifies this harmonic space with de Rham cohomology, proving the asserted representative and dimension claims.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

83 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.