ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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.

E8 as the lattice input to a conditional sliceness obstruction

Example

Let $K \subset S^{3}$ be a smooth oriented knot and suppose its double branched cover bounds a compact simply connected positive-definite smooth four-manifold $W$ whose boundary intersection lattice is isometric to $L_{E_{8}}$ . Then this particular filling lattice has rank eight and has no isometric full-rank embedding into the standard lattice $Z^{8}$ . This verifies the algebraic nonembedding input for the conditional KT-26 sliceness obstruction. It does not assert that a specified knot has such a filling, that every slice-disk branched cover satisfies boundary normal generation, or that the conditional obstruction proves any further numerical four-ball-genus bound.

Example references

The explicit E8 Gram matrix

E8 determinant and nonembedding calculation

Facts & Assumptions

Given: A knot $K$ and a positive-definite filling $W$ with boundary intersection lattice isometric to $L_{E_{8}}$ .

[F1]

The E8 lattice is rank eight, positive definite, has determinant one, and admits no isometric full-rank embedding into $Z^{8}$ . The E8 lattice is even unimodular and is not diagonal

Verification

1.1givenF1construct

An isometry $Q_{W} ≅ L_{E_{8}}$ identifies the rank of $Q_{W}$ with eight and transports any proposed isometric full-rank embedding $Q_{W} ↪ Z^{8}$ to one from $L_{E_{8}}$ .

2.1F1step 1.1∎

Such an embedding is excluded by [F1]. Thus the algebraic nonembedding premise holds for this filling. No knot-level conclusion follows without the separate branched-cover gluing and gauge-theory hypotheses.

Depends on

Used by

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Dependency tree · two levels

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