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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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A low-weight basis term does not define Upsilon

Statement refuted

Let C=CFK∞(T2,3) be the right-handed trefoil staircase on generators x0,x1,x2 with (M,A)(x0)=(0,1), (M,A)(x1)=(−1,0), (M,A)(x2)=(−2,−1) and differential ∂x1=Ux0+x2, ∂x0=∂x2=0. The claim that ΥT2,3(1) is obtained by minimizing F1(x) over the individual displayed basis terms x0,x1,x2, instead of over Maslov-zero cycles representing the normalized generator, is false. Assuming the Axiom of Countable Choice for the normalized grid/analytic comparison, under the diagonal-grid convention every grid diagram of the trefoil also has grid Upsilon value −1 at t=1.

Facts & Assumptions

Given: The explicitly displayed right-handed trefoil staircase and the definition of F1 on a finite chain.

[F1]

The Upsilon function from the filtered knot Floer complex: F1(z) is the maximum weight among terms in the support of z, and ν1 minimizes this value over Maslov-zero cycles representing the normalized generator.

[F2]

The grid Upsilon function agrees with the original Upsilon invariant now contains a proof of the grid-value comparison using the normalized grid/Heegaard supplier, whose own current Proof is present and whose [F1]–[F5] prerequisites remain explicit. MOS Proposition 2.3 supplies the chosen-variable filtered comparison and tilde-homology V-factor, and Theorem 3.3 identifies the grid filtered complex with the knot filtration; the current in-run comparison proof supplies the chain-level collapse factor and its absolute normalization argument.

Counterexample

1.1givenF1algebra

The only Maslov-zero basis terms in the displayed staircase are x0 and U−1x2; x1 has odd Maslov parity, so no integral U-translate of x1 has Maslov grading zero. Both even-degree terms are cycles, and ∂(U−1x1)=x0+U−1x2 shows they represent the same nonzero generator. Their coordinates are (0,1) and (1,0), each of F1-weight 1/2. The sum is a boundary, so every Maslov-zero cycle representing the normalized generator has filtration 1/2. Thus ν1=1/2 and Υ(1)=−2(1/2)=−1.

2.1givenF1step 1.1algebra

The displayed basis term x2 has coordinates (0,−1) and weight F1(x2)=−1/2, which is less than 1/2. It has Maslov grading −2, so it is excluded from the Maslov-zero representative set; although it is a cycle, its homology class is not the normalized degree-zero generator. The false rule would return −2(−1/2)=+1, contradicting step 1.1.

3.1givenF2step 1.1algebra∎

Under Countable Choice, [F2] gives that the diagonal grid complex has the same Upsilon value as the original invariant, so step 1.1 gives the stated grid value −1. The grid comparison uses the supplier prerequisites recorded in [F2].

Unresolved proof obligations

The finite staircase refutation is complete. Its right-handed trefoil identification is supplied by the signed three-crossing grid proof in Epsilon is stable-equivalence invariant but not a homomorphism, step 1.4, together with that item's finite staircase computation. The universal grid-value clause is supplied by [F2] under the comparison prerequisites it records; the earlier missing-comparison escalation is no longer current.

Depends on

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