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

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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.

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Failure of a full-rank standard lattice embedding obstructs smooth sliceness under the gluing hypotheses

Statement

Assume the Axiom of Choice. Let $K \subset S^{3}$ be a smooth oriented knot whose double branched cover $M = Σ_{2} (K)$ is a rational homology sphere. Suppose $M$ bounds a compact connected smooth simply connected positive-definite four-manifold $W$ with boundary intersection lattice $Q_{W}$ of rank $n$ . Assume $Q_{W}$ has no isometric full-rank embedding into the standard dot-product lattice $Z^{n}$ . If every smooth proper slice disk $D$ for $K$ has a branched-cover boundary inclusion $π_{1} (M) \to π_{1} (B_{D})$ whose image normally generates $π_{1} (B_{D})$ , then $K$ is not smoothly slice. This conditional obstruction gives no numerical lower bound on the smooth four-ball genus beyond excluding genus zero.

Statement references

Conditional branched-cover gluing and lattice embedding

Donaldson diagonalization for smooth simply connected definite four-manifolds

Facts & Assumptions

Given: Full AC; the knot, filling, nonembedding condition, and universal normal-generation condition in the statement.

[F1]

A smooth slice disk and the normal-generation condition produce a closed smooth simply connected positive-definite gluing with a full-rank isometric embedding of the filling lattice. A slice-branched-cover gluing produces a closed definite four-manifold

[F2]

Every closed smooth simply connected positive-definite oriented four-manifold has standard diagonal integral intersection lattice. Donaldson diagonalization theorem for positive-definite four-manifolds

[A1]

The boundary-normal-generation assumption holds for every possible smooth slice disk of $K$ ; this assumption is retained because [F1] does not prove it universally.

[A2]

Full AC is assumed by the branched-cover and intersection-form supplier chain.

Proof

Given: The filling, nonembedding condition, and all hypotheses in the statement.

1.1givenA1A2F1construct

Suppose a smooth slice disk $D$ existed. The universal assumption [A1] supplies the normal-generation hypothesis for this actual $D$ , so the proved conditional gluing implication [F1] gives a closed smooth simply connected positive-definite manifold $X$ and an isometric full-rank embedding $Q_{W} ↪ Q_{X}$ . Its rank is $n$ by the same gluing result. This uses the explicit hypothesis and does not infer normal generation from rational homology.

2.1F2step 1.1algebra∎

By [F2], $Q_{X}$ is isometric to $Z^{n}$ with its dot product. Composing that isometry with the embedding in step 1.1 contradicts the assumed absence of an isometric full-rank embedding $Q_{W} ↪ Z^{n}$ . Thus no smooth slice disk exists, proving precisely the conditional nonsliceness conclusion. No larger lower bound on the smooth four-ball genus is deduced.

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