CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Auxiliary timelike vector has negative form

Statement refuted

Assume Countable Choice and the indefinite auxiliary form and subsidiary quotient of Lorenz auxiliary subsidiary null quotient. Every vector in the auxiliary space $K$ has nonnegative quadratic form $[ϵ, ϵ]_{η}$ .

Construction

Assume Countable Choice. Let $E = (5 / 4, 3 / 2) \times (- 1 / 8, 1 / 8)^{2}$ . Then $E \subset {k : 1 < ∣ k ∣ < 2}$ because $∣ k ∣ > 5 / 4$ and $∣ k ∣^{2} < (3 / 2)^{2} + 2 (1 / 8)^{2} < 4$ . Its Lebesgue measure is $∣ E ∣ = (1 / 4)^{3} = 1 / 64$ by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included. Set $ϵ (k) = 1_{E} (k) (1, 0, 0, 0) \in K$ . Then $[ϵ, ϵ]_{η} = - 1 / 64 < 0,$ but $ϵ \notin K_{0}$ because $- ϵ^{0} + n \cdot ϵ = - 1$ on $E$ . Therefore this negative-form auxiliary vector is not a physical state in the subsidiary quotient.

Facts & Assumptions

Given: Countable Choice, the indefinite auxiliary form and subsidiary space, and the null-quotient interface.

[A1]

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

[F1]

The physical subsidiary space requires $- ϵ^{0} + n \cdot ϵ = 0$ almost everywhere; only its null quotient is positive. Lorenz auxiliary subsidiary null quotient

[F2]

The Lebesgue measure of a rectangular box is the product of its side lengths. A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

Counterexample

Given: [A1], [F1], and the box $E$ in the statement.

1.1A1F2givenalgebra

The box $E$ has measure $1 / 64$ by [F2], so $ϵ \in L^{2} (R^{3}; C^{4})$ . Its spatial components vanish and its time component equals one on $E$ , hence $[ϵ, ϵ]_{η} = \int_{E} (- 1) d^{3} k = - 1 / 64 < 0.$

2.1F1F2step 1.1given∎

On $E$ , $- ϵ^{0} + n \cdot ϵ = - 1$ , so the subsidiary condition in [F1] fails on a set of measure $1 / 64 > 0$ by [F2]. The quotient in [F1] is formed only from $K_{0}$ ; this vector is outside that space and cannot represent a negative-norm physical quotient class.

Auxiliary timelike vector has negative form

Statement refuted

Construction

Facts & Assumptions

Counterexample

Depends on

Dependency tree · 2 levels

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