CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge 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.

Zero scalar curvature does not imply vacuum Ricci

Statement refuted

For every four-dimensional Lorentzian metric, scalar curvature $S = 0$ implies $Ric = 0$ .

Counterexample instance

Let $ℓ > 0$ , set $M = 0$ , $q_{e} = ℓ$ , $q_{m} = 0$ , and choose the RN datum on a small open interval about $r = 2 ℓ$ . Then $f = 1 + ℓ^{2} / r^{2}$ is positive there, so the metric is smooth Lorentzian and nondegenerate.

Facts & Assumptions

Given: The RN metric and curvature formulas for $r > 0$ and $f \neq 0$ .

[F1]

Exact Reissner-Nordström Ricci and curvature identities: $S = 0$ and ${Ric}_{22} = Q^{2} / r^{2}$ for the dyonic RN datum.

[A1]

$ℓ > 0$ , $M = 0$ , $q_{e} = ℓ$ , $q_{m} = 0$ , and $r = 2 ℓ$ lies in the selected positive-radius interval.

Verification

1.1A1F1algebra

At $r = 2 ℓ$ , $Q^{2} = ℓ^{2}$ and ${Ric}_{22} = Q^{2} / r^{2} = 1 / 4 \neq 0$ , while [F1] gives $S = 0$ . By continuity the same nonvanishing holds on a sufficiently small open neighborhood of that radius.

2.1F1step 1.1algebra∎

The hypotheses of the refuted statement hold on this smooth RN chart but its conclusion fails because the displayed Ricci component is nonzero. Hence the universal implication is false.

Depends on

Exact Reissner-Nordström Ricci and curvature identities