CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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The s-invariant is not twice the knot Floer tau invariant

Statement refuted

Assume the Axiom of Countable Choice. The universal identity $s (K) = 2 τ (K)$ for every oriented knot $K$ is false. The positive-clasped Whitehead double of the right-handed trefoil with Hedden–Ording twist parameter $t = 2$ has $s = 2$ and $τ = 0$ , so its two sides are unequal.

Statement references

Rasmussen's s-invariant

Knot Floer tau invariant

Hedden–Ording, The Ozsváth–Szabó and Rasmussen concordance invariants are not equal (2005), Theorem 1.1

Rasmussen, Khovanov homology and the slice genus (2004), §1

Facts & Assumptions

Given: ${A C}_{ω}$ , the universal identity and the positive-clasped Whitehead double $D_{+} (T_{2, 3}, 2)$ with Hedden–Ording twist parameter $t = 2$ .

[F1]

The worked example proves the values $s (D_{+} (T_{2, 3}, 2)) = 2$ and $τ (D_{+} (T_{2, 3}, 2)) = 0$ in the Hedden–Ording twist convention (Unknot, trefoils, and the 2-twisted positive Whitehead double, The Hedden–Ording t=2 positive Whitehead double has tau zero and s two).

[A1]

${A C}_{ω}$ is inherited from the diagram-independent $s$ definition and the joint Whitehead-value computation; the finite filtered threshold defining $τ$ introduces no choice assumption (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Rasmussen’s s-invariant, The Hedden–Ording t=2 positive Whitehead double has tau zero and s two).

[L1]

The joint numerical lemma proves both the exact Lee value and the normalized full filtered knot Floer threshold for this actual Whitehead double. The Hedden–Ording t=2 positive Whitehead double has tau zero and s two

Proof

Given: The proposed universal identity and the specified positive-clasped Whitehead double with Hedden–Ording twist parameter $t = 2$ .

1.1A1F1L1givenalgebra

For this actual knot, [F1] and [L1] give $s (D_{+} (T_{2, 3}, 2)) = 2$ and $τ (D_{+} (T_{2, 3}, 2)) = 0$ . Hence twice its normalized tau value is $2 \cdot 0 = 0$ .

2.1step 1.1algebra∎

If the asserted identity held for every oriented knot, substituting this knot would give $2 = 0$ . Since these integers are unequal, the displayed knot is an actual counterexample to the universal identity.

Depends on

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Sources

Matthew Hedden and Philip Ording, The Ozsváth–Szabó and Rasmussen concordance invariants are not equal, arXiv:math/0512348v1 (standard reference, not scraped)
Jacob Rasmussen, Khovanov homology and the slice genus, arXiv:math/0402131v1 (standard reference, not scraped)