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Smooth sliceness and four-ball genus obstructions — Examples and Counterexamples
1 · Prerequisites
- Abelian Categories
- Absolute and Conditional Convergence; Rearrangement; Products
- Absolute Continuity and the Sharp Fundamental Theorem of Calculus
- Analyticity of Holomorphic Functions; Liouville and Morera
- Applications of the Fundamental Group
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- The de Rham Theorem and Degree
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- The Determinant of a Linear Operator, Cofactors and Cramer's Rule
- The Diagram Lemmas in an Abelian Category
- The Divergence Theorem and Classical Stokes
- The Dolbeault Complex and Integral Solutions
- The Duality of Lᵖ and L^q
- The Exponential Function
- The Exterior Derivative and Cartan Calculus
- The Formal Laurent Series Field ℝ((t⁻¹)): Cauchy Complete, Non-Archimedean, Not Complete
- The Fundamental Group
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- The Fundamental Theorems of Calculus
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2 · Logical flowchart
3 · Definitions, theorems and proofs
None yet.
4 · Examples, counterexamples and false statements
ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →
E8 as the lattice input to a conditional sliceness obstruction
Example
Let be a smooth oriented knot and suppose its double branched cover bounds a compact simply connected positive-definite smooth four-manifold whose boundary intersection lattice is isometric to . Then this particular filling lattice has rank eight and has no isometric full-rank embedding into the standard lattice . 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
Facts & Assumptions
Given: A knot and a positive-definite filling with boundary intersection lattice isometric to .
[F1]
The E8 lattice is rank eight, positive definite, has determinant one, and admits no isometric full-rank embedding into . The E8 lattice is even unimodular and is not diagonal
Verification
1.1givenF1construct
An isometry identifies the rank of with eight and transports any proposed isometric full-rank embedding to one from .
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.
CounterexampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →
The topological E8 manifold does not satisfy the smooth diagonalization conclusion
Statement refuted
Under full AC, every closed connected simply connected oriented topological four-manifold with positive-definite integral intersection form has that form isometric over the integers to the standard diagonal lattice.
Statement references
Facts & Assumptions
Given: Full AC and the topological-category assertion in the Statement refuted.
[F1]
There is a closed connected simply connected oriented topological four-manifold with intersection form . The supplier explicitly applies Freedman's external realization theorem under the run-specific TOP exception; no local reconstruction of that existence theorem is claimed. Topological realization of the E8 intersection form
[F2]
The form is positive definite and unimodular but not isometric over the integers to the standard diagonal rank-eight lattice. The E8 lattice is even unimodular and is not diagonal
Counterexample
Given: Full AC, the asserted topological extension and the realization/lattice facts [F1]–[F2].
1.1F1
Take the manifold furnished by [F1]. It is closed, connected, simply connected, oriented and topological, so it meets the asserted manifold hypotheses.
2.1F1F2step 1.1∎
Its intersection form is positive definite by [F2], but the same local lattice result proves it is not isometric to the standard diagonal form. Thus satisfies every hypothesis of the refuted assertion and violates its conclusion.
Sources
Remarks
Donaldson's diagonalization theorem assumes a smooth four-manifold. This topological realization does not meet that hypothesis, and the lattice calculation alone does not give a knot nonsliceness certificate.