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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Open questions in knot polynomials and surgery

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Unique Laurent decomposition for the Jones–Vassiliev basis

Statement

Let x be an invertible indeterminate and put p=x−x−1. Every Laurent polynomial P(x)∈Z[x,x−1] has a unique expression P(x)=a(p)+b(p)x with a,b∈Z[p], where Z[p] is the ring of finite integer-coefficient polynomials in the indeterminate p.

Facts & Assumptions

Given: The commutative polynomial ring R=Z[p], the monic polynomial G(X)=X2−pX−1∈R[X], and the Laurent polynomial ring Z[x,x−1].

[F1]

The polynomial ring R[X] consists of finite-support coefficient sequences with coefficientwise addition and finite convolution multiplication (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[F2]

If R is commutative, then R[X] with these operations is a commutative ring (Polynomial convolution makes R[x] a commutative ring containing R as its constant subring).

[F3]

The ideal generated by G is a two-sided ideal; the quotient ring consists of its additive cosets with multiplication induced from R[X] (The ideal generated by a subset and principal ideals, The quotient ring R/I with (r+I)(s+I)=rs+I).

[F4]

Division by the monic degree-two polynomial G gives each polynomial in R[X] a unique remainder of degree less than two (Division by a monic polynomial over a commutative ring).

Proof

1.1givenF1F2F3F4construct

Let Q=R[X]/(G). By [F1] and [F2], R[X] is a commutative polynomial ring, and [F3] makes (G) a two-sided ideal and Q a quotient ring. Since G is monic of degree two, [F4] gives every class in Q a representative a(p)+b(p)X of degree at most one in X. If two such representatives define the same class, their difference is a multiple of G and has degree less than two; uniqueness in [F4] forces the difference to be zero. Thus this representative is unique.

2.1givenF1F2F3step 1.1algebra∎

Define ϕ:R[X]→Z[x,x−1] on finite polynomials by ϕ(∑jcj(p)Xj)=∑jcj(x−x−1)xj. Substitution and finite convolution make ϕ a ring homomorphism, and ϕ(G)=x2−(x−x−1)x−1=0, so ϕ is constant on cosets of (G) and induces ϕˉ:Q→Z[x,x−1]. In Q, X(X−p)=1, so X is a unit with inverse X−p. Define ψ:Z[x,x−1]→Q by ψ(∑knkxk)=∑k≥0nkXk+∑k<0nk(X−p)−k; because X−p=X−1, finite Laurent multiplication is preserved. The composite ϕˉψ fixes x, while ψϕˉ fixes X and p; since Q is generated by X and p, the maps are inverse. The unique representatives from step 1.1 therefore give the asserted unique Laurent decomposition.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Jones–Vassiliev coefficient vectors

Definition

Assume full AC for the library Jones invariant. An oriented link is a finite disjoint union of oriented smooth circles embedded in S3, considered up to ambient isotopy; an oriented classical knot is the one-component case. For a nonempty oriented link L⊂S3, let VL(x)∈Z[x±1] be the library-normalized Jones polynomial, with VU(x)=1 for the unknot U; write VK for its value on a knot K. Here x=t1/2 in the library convention t=A−4, so an oriented skein triple satisfies x−2VL+(x)−x2VL−(x)=(x−x−1)VL0(x). Set p=x−x−1. By the unique Laurent decomposition, write VK(x)=∑q≥0(aq(K)+bq(K)x)pq, where aq(K),bq(K)∈Z and only finitely many are nonzero. The Jones–Vassiliev coefficient vector of order q is fq(K):=(aq(K),bq(K))T, and cq(K):=aq(K)+bq(K). This is the algebraic coefficient-vector definition used by Carmi and Cohen. The claim that the coordinates of fq are finite-type invariants of order at most q is a separate theorem and is not part of this definition.

RemarkRemark: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Jones polynomial unknot-detection status

Historical question

Assume full AC for the library Jones invariant and ACω for the crossing number. A classical knot is a smooth embedding S1↪S3 considered up to ambient isotopy; U is the unknot represented by a crossing-free circle, and “nontrivial” means it is not ambient isotopic to U. A regular diagram is a generic plane projection with only transverse double points, each marked with over/under information. Let c(K) be the least crossing count among regular diagrams representing K. Let VK(x) be the library-normalized Jones polynomial with VU(x)=1, where x=t1/2 and t=A−4. Set p=x−x−1 and write the unique expansion VK(x)=∑q≥0(aq(K)+bq(K)x)pq, where the integer coefficients have finite support; set fq(K)=(aq(K),bq(K))T. The historical Jones unknot-detection question is whether VK(x)=1 implies K=U.

Remark

Roberts's 1999 notes record the implication as a conjecture. Carmi and Cohen's arXiv:2606.22410v1, submitted on 2026-06-21, claims that every nontrivial classical knot has a nonzero coefficient vector fq(K) for some integer q with 2≤q≤3c(K), and claims VK(x)=1⇒K=U. This scaffold has read the stated result and the paper's proof route but has not independently validated the claimed proof; it asserts neither proof acceptance nor a definitive current open status.

The source's machine argument still requires an independent diagram-level audit of Proposition 5.12, particularly its exceptional-state collapse, smoothing proxy, and two-machine locality. Its numerical estimate Baux≤Bhost+12 is consistent with the stated overhead κmach=11: the argument uses the next layer r+1 after r>Bhost+11. This arithmetic check does not verify the geometric reductions. No Jones-detection theorem is registered by this scaffold.

4 · Examples, counterexamples and false statements

None yet.

Sources