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The tau invariant and four-ball genus — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Tau of the unknot and trefoils from filtered staircases

Example

Assume the Axiom of Countable Choice. Let U be the standard unknot, let T2,3 be the closure of the positive two-strand braid with three crossings, and let T2,3† be its reverse mirror. In the normalized right-trefoil staircase over F2[U,U−1], generators a,b,c have Maslov gradings 0,−1,−2, Alexander levels 1,0,−1, and differential ∂b=Ua+c, ∂a=∂c=0; multiplication by U lowers both filtrations by one. Choose the polynomial minus lattice C−=F2[U]⟨a,b,c⟩ inside this Laurent complex, with each displayed generator at first filtration coordinate zero. Its hat complex is the quotient C−/UC−; setting U=0 refers to this lattice quotient, not to the Laurent complex in which U is invertible. The full reverse-mirror Laurent staircase is the filtered dual, with generators a∨,b∨,c∨ at Alexander levels −1,0,1 and differential ∂a∨=Ub∨, ∂c∨=b∨, ∂b∨=0. Its chosen polynomial dual lattice has hat quotient with ∂c∨=b∨ and ∂a∨=∂b∨=0. The normalized unknot complex has one generator at level zero and zero differential. These models give τ(U)=0,τ(T2,3)=1,τ(T2,3†)=−1. Both trefoils have smooth four-ball genus one.

Diagram

The braid schematic is the closure of the positive two-strand braid σ13. Both strands are oriented downward; the left-to-right strand passes over at each positive crossing, and the exterior arcs join the top and bottom endpoints. The diagram identifies the trefoil convention only; none of the filtered-complex calculations relies on the drawing.

c¾31=T2;3

Facts & Assumptions

Given: The standard unknot, the positive right-handed trefoil, its reverse mirror, and Countable Choice.

[A1]

The Axiom of Countable Choice (ACω) is the assumption retained in the Statement.

[F1]

External input, authorized by the run-specific source boundary. Hom, A survey on Heegaard Floer homology and concordance, §2.2, Figure 3(a),(c), gives the normalized CFK∞(T2,3) staircase with generators a,b,c, ∂b=Ua+c, and Maslov gradings 0,−1,−2. The displayed Alexander levels are 1,0,−1. The same section identifies CFK^(T2,3) as the i=0 filtered complex; Example 3.3 in §3.1 computes τ(T2,3)=1, and Theorem 3.1(1) states that this is the standard knot invariant. I read the figure caption, threshold definition, and examples; I do not claim to reconstruct the Heegaard-diagram computation.

[F2]

The tau level of the standard unknot is zero computes τ(U)=0 in the normalized genus-zero one-pointed model.

[F3]

External input, authorized by the run-specific source boundary. Hom, The knot Floer complex and the smooth concordance group, §2, PDF/printed p. 5, defines −K as the reverse mirror and states CFK∞(−K)≃CFK∞(K)∗, with arrows and both filtrations reversed. Its dual-basis description gives the displayed reverse-mirror staircase from [F1]. The finite filtered hat threshold sign is also proved for the supplied diagrams by The normalized filtered complex of a reverse mirror is dual. I read the duality statement and its convention; I do not claim to reconstruct its cited Heegaard-move proof.

[F4]

Euler characteristic and genus of a Seifert disk-and-band surface computes the genus of the Seifert disk-and-band surface from its circle and crossing counts; for the displayed positive three-crossing braid the counts are s=2 and c=3.

[F5]

External input, authorized by the run-specific source boundary. Hom, §3.1, Theorem 3.1(2), printed p. 8, states ∣τ(K)∣≤g4(K) over the survey's stated F2 coefficient convention (printed p. 2).

Proof

Given: The complexes and knots in the Statement.

1.1A1F1givenalgebra

By [F1], the standard right-trefoil Laurent complex has the displayed staircase. In the polynomial lattice F2[U]⟨a,b,c⟩, quotienting by U leaves ∂b=c and ∂a=∂c=0. The subcomplex at filtration at most 0 is ⟨b,c⟩, which is acyclic, while a first appears at level 1 and represents the total homology class. Thus the normalized threshold is 1, as in Hom's Example 3.3.

2.1F1F2F3step 1.1algebra

The dual differential and negated filtration from [F3] give ∂a∨=Ub∨, ∂c∨=b∨, and ∂b∨=0 for the reverse mirror. In its polynomial dual lattice, setting U=0 leaves ∂c∨=b∨ and the first nonzero total class occurs at level −1. The unknot value is [F2].

3.1F4F5step 1.1step 2.1constructalgebra∎

The oriented smoothing of the positive two-strand three-crossing braid has two Seifert circles and three bands. By [F4] it gives a genus-one spanning surface for the positive trefoil; reflecting this surface gives a genus-one spanning surface for its reverse mirror. Thus each four-ball genus is at most one. The thresholds in steps 1.1–2.1 and Hom's genus bound [F5] give the lower bound one for each; hence both smooth four-ball genera equal one.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Vanishing tau does not imply smooth sliceness

Statement refuted

Assume the Axiom of Countable Choice. The universal implication that a knot with τ(K)=0 is smoothly slice is false. Let s(K)=smax⁡(K)−1, where smax⁡(K) is the maximal quantum filtration level of a nonzero class in rational Lee homology. In the twist convention of Hedden and Ording, with two full right-handed twists, the positive-clasped Whitehead double D+(T2,3,2) has τ=0 and s=2, so its smooth four-ball genus is at least one and it is not smoothly slice.

Facts & Assumptions

Given: The Axiom of Countable Choice, the universal implication, and the specified positive-clasped, two-full-twist Whitehead double in Hedden and Ording's convention.

[A1]

The Axiom of Countable Choice (ACω) is the axiom assumption retained in the Statement.

[F1]

External input, authorized by the run-specific source boundary. Hedden and Ording, The Ozsváth–Szabó and Rasmussen concordance invariants are not equal, Theorem 1.1, printed p. 2, states τ(D+(T2,3,2))=0 and s(D+(T2,3,2))=2 for the two-twisted positive Whitehead double of the right-handed trefoil. Their Figure 1 specifies the twist box as full right-handed twists. The theorem states both values. Section 2.1, printed p. 4, reports a computer-computed Khovanov Poincaré polynomial: its homological-degree-zero support forces s∈{2,4}, and the genus-one upper bound together with Rasmussen's inequality forces s=2. The acknowledgments on printed p. 3 credit Rasmussen with the s computations. This is a published derivation using computer output, not a reproduced Lee chain-complex computation; the numerical theorem is imported here.

[F2]

External input, authorized by the run-specific source boundary. Rasmussen, Khovanov homology and the slice genus, Theorem 1 in §1, PDF p. 1, states ∣s(K)∣≤2g4(K) for every knot, where g4 is the minimum genus of a smooth properly embedded surface in B4. The source proof in §4.4 derives this from the filtered Lee cobordism map; I read that proof. This is the same smooth genus convention as Smooth and topological slice genus.

[F3]

The Alexander filtration and tau level uses the same normalized threshold convention for the standard Ozsváth–Szabó invariant in [F1]. The target total homology is one-dimensional, so nonzero image agrees with Hom's surjective-inclusion convention.

[F4]

The s in [F1] is Rasmussen's standard integer obtained from rational Lee homology with the convention s=smax⁡−1 stated in the Statement. The external Theorem 1 in [F2] uses this normalization; no local Lee-map proof is used here.

[F5]

Smooth and topological slice genus defines g4sm by minimizing the genera of smooth connected oriented properly embedded surfaces in B4 with boundary the knot.

[F6]

External input, authorized by the run-specific source boundary. Hom, A survey on Heegaard Floer homology and concordance, §3.1, Theorem 3.1(1), printed p. 8, identifies the standard τ defined by the same normalized threshold as a knot-concordance homomorphism. The survey's coefficient convention is stated on printed p. 2. I read the threshold definition and theorem statement; I do not claim to reconstruct the cited proof.

Proof

Given: Countable Choice, the proposed implication, and the Whitehead double in the Statement.

1.1A1F1F2F3F4F5F6givenalgebra

By [F1], the specified knot has τ=0 and s=2. The smooth genus bound [F2] gives 2=∣s∣≤2g4sm. Since g4sm is a nonnegative integer, it is at least one.

2.1F5step 1.1algebra∎

A smoothly slice knot bounds a smooth properly embedded disk in the four-ball and has g4sm=0. This contradicts step 1.1 for the specified knot, proving that the universal implication is false.

Sources