Alphabeta Math
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Trefoil and figure-eight Levine–Tristram and Arf calculations

Example

Assume full AC. Use the positive-push Seifert matrices VT=(−10−1−1),VF=(−10−11) for the positive trefoil and the figure-eight knot, respectively. At ω=−1, their Levine–Tristram signatures are −2 and 0. In the displayed symplectic bases both quadratic forms have q(a)=q(b)=1, so both Arf invariants are 1. Their normalized Conway polynomials are 1+z2 and 1−z2, respectively.

Facts & Assumptions

Given: Full AC and the two displayed oriented knot Seifert matrices in the positive-push convention.

[A1]

Full AC is the ambient hypothesis for the Seifert pairing. The Axiom of Choice

[F1]

The trefoil matrix VT is the positive-push matrix for the positive trefoil. Trefoil and figure-eight Alexander polynomials from Seifert matrices

[F2]

The figure-eight matrix VF is a positive-push Seifert matrix; its skew part is the standard symplectic block. Seifert pairing and matrix of an oriented knot, The skew part of a Seifert matrix is the surface intersection form

[F3]

The Arf invariant is q(a)q(b) for a genus-one symplectic basis and equals a2 modulo two. Knot Arf invariant from the Seifert form, The Arf invariant is the second Conway coefficient modulo two

[F4]

For a Seifert matrix V, the Seifert determinant ΘK(u)=det⁡(u−1V−uVT) gives the normalized Conway polynomial ΘK(u)=∇K(u−u−1). Seifert determinant constructs the normalized Conway polynomial

[F5]

The Levine–Tristram signature of the knot is independent of the chosen surface and basis, including the one-sided-average value at an Alexander root. Seifert stabilization preserves the Levine–Tristram signature

Proof

1.1A1F1F2F5givenalgebra

Since 1−(−1)=2, H−1(V)=2(V+VT). For the trefoil, VT+VTT=(−2−1−1−2) has eigenvalues −1,−3, so the signature is −2. For the figure-eight, VF+VFT=(−2−1−12) has determinant −5<0, hence one positive and one negative eigenvalue, so the signature is 0.

2.1F1F2F3F4step 1.1algebra∎

The standard intersection block is (01−10) for both matrices. The quadratic values on the basis vectors are their diagonal entries modulo two, namely (1,1) for each matrix. Therefore Arf⁡=1⋅1=1 for each knot. Their Alexander determinants are det⁡(VT−tVTT)=1−t+t2 and det⁡(VF−tVFT)=−1+3t−t2. Substituting t1/2−t−1/2=z with the symmetric normalization gives ∇T=1+z2 and ∇F=1−z2; both z2 coefficients are odd, agreeing with the Arf values.

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