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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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The solid-torus skein module has a core-multicurve basis

Statement

Assume the Axiom of Choice and let R=Z[A±1]. Identify the solid torus with A×[−1,1], where A is a planar annulus. A simple closed curve is called essential if it does not bound a disk in A. Every framed link in this product admits a regular annular diagram and an integer framing correction from its blackboard framing. Let z be one blackboard-framed annular core curve, and for n≥1 let zn denote n parallel copies of that curve, with z0 the empty link. In the polynomial ring R[z], z is also the formal variable sent to the core class [z]. Then the classes [zn], n≥0, form an R-basis of KA(A×[−1,1]). Equivalently, the map J:R[z]→KA(A×[−1,1]) sending zn to [zn] is an R-module isomorphism.

Facts & Assumptions

Given: The Axiom of Choice, the annulus-product model of the solid torus, and its framed-link skein module.

[A1]

The Axiom of Choice, written AC (The Axiom of Choice).

[A2]

Countable choice, written ACω, means every countable family of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)).

[A3]

In ZF, AC implies ACω (AC implies DC implies countable choice).

[F1]

The module quotient is generated by framed links and imposes the crossing and zero-framed trivial-circle relations (Kauffman bracket skein module of a three-manifold).

[F2]

Every framed link in A×(−1,1) admits a regular annular diagram, with its framing correction recorded by signed full-twist marks (Framed link diagrams and local Reidemeister moves in plane products).

[F3]

The state expansion Ψ is well-defined on the skein module and sends a diagram with twist correction m to (−A3)m∑SAa(S)−b(S)δu(S)ze(S) (Annular state expansion for the Kauffman bracket skein module).

[F4]

A positive or negative full framing twist on a component multiplies its skein class by −A3 or −A−3, respectively (Full framing twists in the Kauffman bracket skein module).

Proof

1.1A1F3givenconstructalgebra

Define J by the stated R-linear assignment. On zn take the crossingless annular diagram of n parallel essential circles with blackboard framing. It has m=0, a=b=u=0, and e=n, so [F3] gives Ψ(J(zn))=zn. By linearity, Ψ∘J is the identity on R[z], and hence J is injective.

2.1A1A2A3F1F2F3F4step 1.1inductioncasesalgebra∎

Let L be any framed link in the solid torus. By [F2], choose a regular annular diagram D and write mj for the framing correction on its jth component, with m=∑jmj. Apply [F4] componentwise: each positive twist contributes −A3, each negative twist contributes −A−3=(−A3)−1, so [L]=(−A3)m[D], where D carries its blackboard framing. Induct on the number of crossings. At one crossing, [F1] replaces D by its two smoothing diagrams with coefficients A and A−1, and each smoothing has one fewer crossing. Applying the induction hypothesis to each term yields [D]=∑SAa(S)−b(S)[DS], where each DS is a crossingless annular state with blackboard framing. Every embedded state circle in the annulus either bounds a disk or is essential and cuts the annulus into two annuli, so it is isotopic to a core circle. For finitely many disjoint essential circles, cutting along them gives annular regions between consecutive circles; thus they are parallel copies of the core. Among the contractible circles choose one whose disk contains no other state circle; its disk has a small thickening disjoint from the remaining link, so the trivial-circle relation in [F1] removes it with factor δ. Repeating removes all contractible circles. Therefore [L]=(−A3)m∑SAa(S)−b(S)δu(S)[ze(S)]=J(Ψ([L])), where the last equality is [F3]. Thus J∘Ψ is the identity on every link generator, and by R-linearity on the whole quotient. Together with step 1.1 this proves J and Ψ are inverse isomorphisms, so the displayed classes are an R-basis.

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