CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Small invariants do not alone control derivative corrections

Statement refuted

In coordinates $(t, x, y, z)$ on Minkowski $R^{4}$ with $η = diag (- 1, 1, 1, 1)$ and orientation $d t \land d x \land d y \land d z$ , define $X_{ϵ} = - (F_{ϵ})_{a b} (F_{ϵ})^{a b} / 4$ and $Y_{ϵ} = - (F_{ϵ})_{a b} (⋆_{η} F_{ϵ})^{a b} / 4$ . The refuted assertion is: for every fixed finite $α, β \in R$ and every family $(F_{ϵ})_{0 < ϵ < 1}$ of smooth exact real two-forms on $R^{4}$ (that is, $F_{ϵ} = d A_{ϵ}$ for a smooth real one-form $A_{ϵ}$ ), if $\sup_{R^{4}} (∣ α X_{ϵ} ∣ + ∣ β Y_{ϵ} ∣) \to 0$ as $ϵ ↓ 0$ , then there exist $C \geq 0$ and $ϵ_{0} \in (0, 1)$ such that $\sup_{R^{4}} ∣ \partial_{x} (F_{ϵ})_{t x} ∣ \leq C$ for every $0 < ϵ < ϵ_{0}$ . The assertion says that small invariant corrections alone give a local uniform bound on this derivative scale.

Statement references

Nonlinear electromagnetic constitutive model with a $C^{2}$ Lagrangian

Lorentzian exterior calculus and Hodge star

Facts & Assumptions

Given: Minkowski $R^{4}$ with length coordinates $(t, x, y, z)$ , metric $η = diag (- 1, 1, 1, 1)$ , orientation $d t \land d x \land d y \land d z$ , fixed positive scales $F_{0}, L_{0}, k_{*}$ , fixed finite $α, β$ , and $0 < ϵ < 1$ .

[F1]

Nonlinear electromagnetic constitutive model with a $C^{2}$ Lagrangian defines $X = - F_{a b} F^{a b} / 4$ and $Y = - F_{a b} (⋆_{g} F)^{a b} / 4$ and allows arbitrary fixed coefficients $α, β$ in $L = X + α X^{2} + β Y^{2}$ . If $F$ and the fixed scale $F_{0}$ have unit $F$ , then $[X] = [Y] = F^{2}$ and $[α] = [β] = F^{- 2} = [F_{0}]^{- 2}$ ; the definition makes no claim of derivative control from these invariants.

[F2]

Lorentzian exterior calculus and Hodge star gives the Lorentzian Hodge component formula. For the simple two-form $d t \land d x$ , its dual has only the complementary $d y \land d z$ component, so its contraction with $d t \land d x$ is zero.

[A1]

$F_{0}$ is a fixed field-strength scale with unit $F$ , $L_{0}$ is a fixed length scale, and $k_{*} > 0$ is a fixed inverse-length scale; thus $F_{0} L_{0}$ has the local-potential component units in length coordinates. The fixed finite coefficients $α, β$ have units $F^{- 2}$ , and $ϵ \in (0, 1)$ is dimensionless. No coefficient is assigned a QED value.

Counterexample

Define the smooth one-form $A_{ϵ} = F_{0} L_{0} ϵ^{3} \cos ⁣ (\frac{x}{L_{0} ϵ^{2}}) d t$ and its curvature $F_{ϵ} = d A_{ϵ}$ .

1.1givenF1A1algebra

Differentiating the coefficient gives $F_{ϵ} = F_{0} ϵ \sin ⁣ (\frac{x}{L_{0} ϵ^{2}}) d t \land d x .$ It is exact, smooth for every $ϵ > 0$ , and $d F_{ϵ} = d^{2} A_{ϵ} = 0$ . It is a valid curvature configuration; no field equation is claimed.

1.2givenF1F2A1algebra

Put $E_{ϵ} = F_{0} ϵ \sin (x / (L_{0} ϵ^{2}))$ . The only nonzero independent component is $(F_{ϵ})_{t x} = E_{ϵ}$ , and raising indices with $η$ gives $(F_{ϵ})^{t x} = - E_{ϵ}$ . Hence $F_{a b} F^{a b} = - 2 E_{ϵ}^{2}$ and $X_{ϵ} = E_{ϵ}^{2} / 2$ . By [F2], $Y_{ϵ} = 0$ .

2.1givenA1step 1.1step 1.2algebra

Since $∣ \sin ∣ \leq 1$ , $∣ α X_{ϵ} ∣ + ∣ β Y_{ϵ} ∣ \leq ∣ α ∣ F_{0}^{2} ϵ^{2} / 2$ , which tends to zero. Direct differentiation gives $\partial_{x} (F_{ϵ})_{t x} = \frac{F_{0}}{L_{0} ϵ} \cos ⁣ (\frac{x}{L_{0} ϵ^{2}}),$ whose supremum is $F_{0} / (L_{0} ϵ)$ ; the phase wave number is $k_{ϵ} = (L_{0} ϵ^{2})^{- 1}$ . Thus both the derivative supremum and $k_{ϵ} / k_{*}$ diverge.

3.1step 2.1algebra∎

The family has invariant corrections tending uniformly to zero while $\sup ∣ \partial_{x} (F_{ϵ})_{t x} ∣$ diverges by step 2.1. It therefore contradicts the refuted uniform derivative-bound implication. A derivative-truncation criterion must state its own derivative hypotheses and cutoff; the invariant inequalities alone do not provide them.

Remarks

This is a smooth field configuration with a global potential and no singularities. It is not asserted to solve the nonlinear constitutive equation or represent an observation. The example neither evaluates an effective-action coefficient nor supplies a QED threshold.

Small invariants do not alone control derivative corrections

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Remarks

Depends on

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