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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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A positive Hopf plumbing gives the trefoil Seifert matrix

Statement

Assume the full Axiom of Choice. A positive Hopf band is an oriented annulus with unknotted core whose two boundary components, with their induced orientations, have linking number +1. Let F be the genus-one Seifert surface obtained by plumbing two positive Hopf bands in a square, with oriented core curves a,b forming a positive handle basis, and choose the plumbing side so that a+ bounds a disk in S3 disjoint from b. The oriented boundary of this standard positive two-band plumbing is the right-handed trefoil. In the positive-push convention Vij=lk⁡(ai+,aj) with ordered basis (a,b), the Seifert matrix is

V=(−10−1−1).

More generally, the same matrix conclusion holds for any genus-one Seifert surface obtained by plumbing two oriented annuli with induced boundary linking number +1, provided their core curves form the stated positive handle basis and the stated disk for a+ is disjoint from b. This broader matrix assertion does not identify the boundary knot.

Facts & Assumptions

Given: The stated positive Hopf plumbing, its oriented core curves a,b, the positive-push convention, and full AC for the knot Seifert-pairing interface.

[A1]

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

[F1]

The Seifert matrix entry is Vij=lk⁡(ai+,aj), with the stated positive normal. Seifert pairing and matrix of an oriented knot

[F2]

The stated plumbing convention gives a⋅Fb=1 and the specified bounding disk for a+ is disjoint from b; each Hopf annulus has boundary linking +1 by the definition in the Statement.

[F4]

The canonical three-positive-crossing closure has a genus-one disk-and-band surface, whose three pair-of-band Hopf annuli all have core self-pairing −1. The canonical positive three-crossing braid has the negative trefoil form

[F3]

The skew matrix of the positive-push form is the oriented intersection matrix in the same ordered basis. The skew part of a Seifert matrix is the surface intersection form

Proof

1.1A1F2F4givenalgebra

For the positive Hopf bands with unknotted cores, the standard plumbing has the three crossing bands of the positive two-strand three-crossing closure: its two annular strips use successive pairs of the three bands, and their common disk is the plumbing square with its four alternating attaching edges. In oriented braid coordinates (x,y,t) use (X,Y,Z)=(x,−t,y), of determinant +1, to draw the closure on the (X,Y) page with viewer +Z. The overpass runs down-left and the underpass down-right at each of the three crossings, while the two outside closure arcs are crossing-free. This is precisely the right-hand trefoil diagram of Rolfsen printed146, identified as right-hand on printed201; the comparison fixes the handedness using actual crossing data. The two core classes have intersection number a⋅Fb=1 by [F2]. Since F has genus one and one boundary component, its first homology has rank two; the intersection matrix in the core classes has determinant 1. Therefore a,b form an integral basis.

2.1A1F1F2F3step 1.1algebra∎

The following calculation also proves the broader matrix assertion: it uses only the stated annulus boundary linkings, positive handle basis, and disjoint disk, so unknotted cores are unnecessary here. By [F1], each diagonal is the self-linking of an annulus core. Equally oriented parallel cores transport to the two boundary components with one orientation reversed, so their linking is minus the boundary linking, namely −1. A small positive normal push returns to the first core through curves disjoint from the second. Thus V11=V22=−1. The disk specified in [F2] bounds a+ in the complement of b, hence lk⁡(a+,b)=0 and V12=0. By [F3], V12−V21=a⋅Fb=1, so V21=−1. These four entries give the displayed matrix.

Diagram

This schematic shows the plumbing square and the oriented core classes. Red and blue identify the two Hopf bands; the horizontal core is a and the vertical core is b. The orientation of the core tangents and the oriented surface gives a⋅Fb=+1. The diagram records the local basis convention only; the matrix proof uses the linking and intersection calculations above.

\begin{tikzpicture}[scale=0.9,>=stealth]
  \fill[red!14] (-2,-0.35) rectangle (2,0.35);
  \fill[blue!14] (-0.35,-2) rectangle (0.35,2);
  \draw[thick,red,->] (-2,0) -- (2,0);
  \draw[thick,blue,->] (0,-2) -- (0,2);
  \draw[dashed] (-0.35,-0.35) rectangle (0.35,0.35);
  \node[red] at (-1.3,0.55) {$a$};
  \node[blue] at (0.55,1.3) {$b$};
  \node at (0,-0.62) {plumbing square};
\end{tikzpicture}

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