CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Equal zero-traces do not force smooth concordance

Statement refuted

Assume the Axiom of Choice. The proposed statement is that for every pair of smooth oriented knots $K_{0}, K_{1} \subset S^{3}$ , a smooth diffeomorphism $X_{0} (K_{0}) ≅ X_{0} (K_{1})$ implies that $K_{0}$ and $K_{1}$ are smoothly concordant.

Statement references

Piccirillo, Introduction and §4

Hom, §3

Facts & Assumptions

Given: Full AC, the Conway knot $C$ and its Figure 2 trace partner $K_{P}$ .

[A1]

The Axiom of Choice, written AC, is assumed. The Axiom of Choice

[A2]

AC implies dependent choice and hence countable choice ${A C}_{ω}$ ; this supplies every retained local countable-choice hypothesis. AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

Smooth knot concordance: a smooth concordance is a properly embedded smooth annulus in $S^{3} \times [0, 1]$ with product collars.

[F2]

Lee computations for the Conway knot and its Piccirillo partner gives $s (C) = 0$ by its direct Conway Lee computation. Only its Conway conclusion is used here.

[F3]

Lee cobordism maps give the genus filtration bound gives $∣ s (K_{1}) - s (K_{0}) ∣ \leq 2 g$ for an actual smoothly properly embedded oriented genus- $g$ cobordism with product collars.

[F4]

Piccirillo’s partner has the same zero-trace and Rasmussen invariant two proves under full AC that $X_{0} (C) ≅ X_{0} (K_{P})$ and $s (K_{P}) = 2$ , using the retained local framed witness, the explicitly imported published zero-trace equality for the named knots, and the direct rational Lee calculation.

Counterexample

1.1A1A2F1F4givencontradiction

The oriented knots $C, K_{P}$ of [F4] have diffeomorphic zero-traces. Suppose they were smoothly concordant. By [F1] the concordance is a properly embedded product-collared annulus, hence a genus-zero cobordism.

2.1A1A2F3step 1.1algebra

Apply [F3] with $g = 0$ . It gives $∣ s (C) - s (K_{P}) ∣ \leq 0$ , so the two invariants must agree.

3.1A2F2F4step 1.1step 2.1algebracontradiction∎

But [F2] and [F4] give $s (C) = 0$ , $s (K_{P}) = 2$ . Their difference has absolute value two, contradicting step 2.1. Thus these actual equal-zero-trace knots are not smoothly concordant, refuting the unchanged universal Statement.

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Sources

Lisa Piccirillo, The Conway knot is not slice (standard reference, not scraped)
Jennifer Hom, Getting a handle on the Conway knot (standard reference, not scraped)