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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.

Branched-cover correction terms and sliceness — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

L(3,1) correction terms in the stated orientation convention

Example

Use the orientation and cyclic labels for which d(L(p,1),i)=((2i−p)2−p)/(4p). For p=3, the correction terms are d(L(3,1),0)=1/2, d(L(3,1),1)=−1/6, and d(L(3,1),2)=−1/6. Reversing orientation negates all three values.

Facts & Assumptions

Given: The stated orientation convention and the three labels i=0,1,2.

[F1]

For every p≥1 and canonical label i, the stated L(p,1) correction-term formula holds; reversing orientation negates the value at the conjugate label (Lens-space correction-term formula for L(p,1)).

Proof

technique · substitute each label into the formula and simplify

Given: The data above.

1.1F1givenalgebra

For p=3,i=0, (2i−p)2−p=9−3=6 and 4p=12, so [F1] gives d(L(3,1),0)=6/12=1/2.

2.1F1step 1.1algebra∎

For i=1, the numerator is (−1)2−3=−2; for i=2, it is 12−3=−2. Both denominators are 12, so the other two values are −2/12=−1/6. Orientation reversal negates each by [F1].

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A nonextendible Spin-c structure with nonzero d on a rational-ball boundary

Statement refuted

The false claim is: if a closed oriented three-manifold Y with H∗(Y;Q)≅H∗(S3;Q) bounds a smooth compact four-manifold B with H∗(B;Q)≅H∗(point;Q), then a nonzero correction term d(Y,t) for any boundary Spinc structure t obstructs the existence of B.

Example

Let B2 be the rational homology ball bounded by L(4,1) constructed as the complement of a tubular neighborhood of a conic in CP2. The canonical Spinc label i=0 has d(L(4,1),0)=3/4. The extension set has exactly two elements and every extending structure has correction term zero. Therefore label i=0 is nonextendible, even though this rational-ball filling exists.

Facts & Assumptions

Given: AC, the rational ball B2 with boundary L(4,1), and the lens-space label i=0.

[A1]

AC is assumed by the construction and the extension-set supplier (The Axiom of Choice).

[F1]

The conic complement is a rational homology four-ball with boundary L(4,1) in the selected orientation (A rational homology ball bounded by L(4,1)).

[F2]

The surgery matrix for L(4,1) is nonsingular and its first homology has order 4 (Inverse-matrix formula for a surgery linking form).

[F3]

Over a rational-ball boundary, the extending Spinc structures form a coset of order ∣H1(Y)∣, and every extending structure has zero correction term (Spin-c extensions over a rational ball form a metabolizer coset, Correction terms vanish on extending Spin-c structures over rational homology balls).

[F4]

For the selected orientation, d(L(4,1),i)=((2i−4)2−4)/16 (Lens-space correction-term formula for L(p,1)).

Proof

technique · compute the extension-set size and the correction term at a label outside that set

Given: The data above.

1.1A1F1F2F3givenalgebra

By [F1] and [F2], B2 is a rational homology ball with ∣H1(L(4,1))∣=4. Therefore [F3] gives exactly 4=2 extending Spinc structures.

2.1A1F3F4step 1.1algebra∎

At label i=0, [F4] gives d(L(4,1),0)=(16−4)/16=3/4≠0. Every extending structure has d=0 by [F3], so label 0 is not in the two-element extension set. Thus a nonextending structure has nonzero correction term while the rational-ball filling exists, refuting the stated claim.

Sources