ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)

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.

A finite Q-exact torus integral with constant value

Physical scope

A dimensionless finite-dimensional graded integral used only as an illustration of the Ward identity; it is not identified with the S4 field configuration space and gives no continuum or empirical prediction.

Example

Let $M = R^{2} / (2 π Z)^{2}$ be the two-torus with dimensionless angular coordinates $x, y$ modulo $2 π$ ; $d x, d y$ denote the induced global angular one-forms. Orient $M$ by the volume form $μ = d x \land d y$ and define the graded integral $I (F) = \int_{M} F_{[2]}$ .

The complex differential-form algebra is $Ω^{∙} (M; C) = Ω^{0} \oplus Ω^{1} \oplus Ω^{2}$ , where $C^{\infty} (M; C)$ is the space of smooth functions periodic in both coordinates, $Ω^{0} = C^{\infty} (M; C)$ , $Ω^{1} = {a d x + b d y : a, b \in C^{\infty} (M; C)}$ , $Ω^{2} = {c d x \land d y : c \in C^{\infty} (M; C)}$ , and $Ω^{k} = 0$ for $k > 2$ . The product is the exterior product, determined by $d x \land d x = d y \land d y = 0$ and $d y \land d x = - d x \land d y$ . The exterior derivative is the degree-one linear map $d$ defined by $d f = (\partial_{x} f) d x + (\partial_{y} f) d y$ , $d (a d x + b d y) = (\partial_{x} b - \partial_{y} a) d x \land d y$ , and $d α = 0$ for $α \in Ω^{2}$ ; these formulas give the graded Leibniz rule and $d^{2} = 0$ .

Let the odd derivation be $Q = d$ , set the dimensionless action $S = 0$ , choose the even insertion $O = 1 + λ d x \land d y$ for $λ \in R$ , and choose the odd deformation $V = \sin (x) d y$ . For every dimensionless real $t$ , the integral $Z (t) = I (O e^{- S - t Q V})$ equals $4 π^{2} λ$ . This is a finite-dimensional regulator toy integral only; it is not a model of the S4 field configuration space or a continuum physical prediction.

Example references

Finite-dimensional Q-exact deformation theorem

Facts & Assumptions

Given: The closed oriented torus $M$ , its dimensionless coordinates and volume form, $Q = d$ , $S = 0$ , $O = 1 + λ d x \land d y$ , and $V = \sin (x) d y$ as specified in the Example.

[F1]

Finite-Dimensional Q-Exact Deformation Invariance: if $Q$ is an odd derivation, $Q S = Q O = 0$ , $(Q^{2} V) e^{- S - t Q V} = 0$ , the Q-exact boundary integral vanishes, and the integrand and its derivative have the stated compact-parameter L1 majorants, then the finite-dimensional graded integral is constant in $t$ .

Verification

1.1givenalgebra

The coefficient product rule in the displayed coordinate formulas gives $d (f g) = d f g + f d g$ for $f, g \in Ω^{0}$ and $d (f α) = d f \land α + f d α$ , $d (α f) = d α f - α \land d f$ for $α \in Ω^{1}$ ; for $α, β \in Ω^{1}$ , both sides of the graded-Leibniz identity have degree three and vanish because $Ω^{3} = 0$ . Thus $d$ is an odd derivation. For $f \in Ω^{0}$ , $d^{2} f = (\partial_{x} \partial_{y} f - \partial_{y} \partial_{x} f) d x \land d y = 0$ by equality of mixed partial derivatives, and $d^{2} = 0$ on $Ω^{1} \oplus Ω^{2}$ because $d Ω^{2} = 0$ . Hence $Q = d$ has $Q^{2} = 0$ . The torus has no boundary, and for every smooth periodic one-form $α = a (x, y) d x + b (x, y) d y$ , $\int_{M} d α = \int_{0}^{2 π} \int_{0}^{2 π} (\partial_{x} b - \partial_{y} a) d x d y = 0$ by periodicity. Thus $I (Q F) = 0$ for each admissible odd form $F$ . Also $Q S = Q O = 0$ and $Q V = d V = \cos (x) d x \land d y$ .

2.1givenF1step 1.1

The top-degree coefficient of $O e^{- t Q V}$ and its $t$ -derivative are smooth on compact $M$ and bounded uniformly when $t$ ranges over any compact interval; a constant bound is integrable on $M$ . Therefore all finite-dimensional hypotheses of [F1] hold.

3.1givenF1step 2.1algebra∎

Since $(d x \land d y)^{2} = 0$ , $e^{- t Q V} = 1 - t \cos (x) d x \land d y$ . Multiplying by $O$ gives top-degree part $(λ - t \cos (x)) d x \land d y$ , whose integral is $4 π^{2} λ - t \int_{0}^{2 π} \int_{0}^{2 π} \cos (x) d x d y = 4 π^{2} λ$ . This agrees with the constant value supplied by [F1].

Depends on

Finite-Dimensional Q-Exact Deformation Invariance