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Worked cases for metabolic Seifert forms

Example

Assume full AC. The disk Seifert surface of the oriented unknot has the rank-zero Seifert lattice and is metabolic. In the positive-push convention, the standard positive Hopf-band plumbing for the right-handed trefoil has matrix $V = (\begin{array}{cc} - 1 & 0 \\ - 1 & - 1 \end{array})$ . Since $V$ is invertible, its rational isometric structure has symmetric form $B = V + V^{T} = (\begin{array}{cc} - 2 & - 1 \\ - 1 & - 2 \end{array})$ , which is negative definite; the integral Seifert lattice is not metabolic.

Example references

The positive-push Seifert matrix convention

The definition of a metabolic Seifert lattice

The positive Hopf-plumbing matrix calculation

Facts & Assumptions

Given: Full AC, the disk Seifert surface of the oriented unknot, and the standard positive Hopf-plumbing Seifert surface of the right-handed trefoil.

[A1]

Full AC is the choice-function axiom used by the knot Seifert-pairing interface. The Axiom of Choice

[F1]

The rank-zero lattice is allowed; its zero subgroup is a primitive rank-zero metabolizer. Metabolic Seifert forms and algebraic knot concordance

[F2]

The positive Hopf-plumbing basis has matrix $V = (\begin{array}{cc} - 1 & 0 \\ - 1 & - 1 \end{array})$ . A positive Hopf plumbing gives the trefoil Seifert matrix

[F3]

Since $V$ is invertible, $(V + V^{T}, V^{- 1} V^{T})$ satisfies the rational isometric-structure definition. Integral Witt classes and rational Seifert isometric structures, An invertible Seifert matrix gives a rational isometric structure

[F4]

A metabolic lattice must have a rank-half direct summand on which the Seifert pairing vanishes. Metabolic Seifert forms and algebraic knot concordance

Proof

1.1A1F1givenalgebra

The disk has $H_{1} = 0$ , so its Seifert matrix is the empty matrix. The subgroup $M = 0$ has rank zero, is a direct summand, and the pairing vanishes on it. By [F1], this is metabolic and represents the identity class.

2.1F2F3F4algebra∎

By [F2] and [F3], the trefoil's rational structure has symmetric form $B = (\begin{array}{cc} - 2 & - 1 \\ - 1 & - 2 \end{array})$ . For a vector $(m, n) \in Q^{2}$ , $B ((m, n), (m, n)) = - 2 (m^{2} + m n + n^{2}) = - 2 (m + n / 2)^{2} - 3 n^{2} / 2$ , which is negative whenever $(m, n) \neq (0, 0)$ . Thus the rational structure has no nonzero isotropic line. For an integral vector the Seifert self-pairing is $- (m^{2} + m n + n^{2}) < 0$ , so no rank-one subgroup has identically zero Seifert pairing, and [F4] rules out an integral metabolizer.

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Sources

Dale Rolfsen, Knots and Links (standard reference, not scraped)
Charles Livingston, A Survey of Classical Knot Concordance (standard reference, not scraped)