Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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.

Finite trace-coordinate model on the stated loci

Statement

Assume full AC and let G=π1(EK) have a finite presentation. There is a finite list of words w1,…,wN∈G whose trace functions generate the coordinate ring of X(EK), so the map [ρ]↦(tr⁡ρ(w1),…,tr⁡ρ(wN)) is a closed embedding into CN, with defining ideal the kernel of the induced map C[z1,…,zN]→C[Hom⁡(G,SL2)]SL2. On the irreducible-character locus, equal trace coordinates determine conjugate representations; on the reducible locus, the quotient records the semisimplified diagonal character, not the possibly non-diagonalizable representative.

Facts & Assumptions

Given: Full AC and the finite Wirtinger presentation of the knot group.

[F1]

The character quotient coordinate ring is generated by traces of words (Trace-coordinate algebra for the SL2 character quotient).

[F2]

The character quotient is affine with finitely generated reduced coordinate algebra, the quotient map is onto, and generators give its actual closed coordinate model (Finite generation of invariants and the affine categorical quotient).

[F3]

Equal irreducible SL2 trace characters determine conjugacy proves by the actual rank-two matrix-algebra and trace-pairing argument that equal irreducible trace characters are conjugate in SL2(C), and a reducible representation has the trace character of its diagonal semisimplification. This is a pointwise proof, not an assertion of a representation over a function field.

Proof

1.1F1F2givenconstructalgebra

By [F2], choose a finite algebra generating list f1,…,fs for the character coordinate algebra. Each fj is a polynomial in finitely many word traces by [F1]. The union of these finitely many word lists is finite; denote it by w1,…,wN. Their traces generate an algebra containing all fj, hence the entire coordinate algebra. This establishes one fixed finite list for the quotient, rather than just a separate list for each function.

2.1F2step 1.1constructalgebra

Send zi to twi. Step 1.1 makes this polynomial-ring map onto the reduced invariant coordinate algebra. Its kernel is therefore radical. The actual affine coordinate-ring construction in [F2] identifies the quotient with the closed zero locus of that kernel in CN and its point map with the displayed trace-coordinate map. Thus this is the asserted closed embedding with exactly the stated defining ideal.

3.1F1F2F3step 1.1step 2.1constructalgebra∎

Equal listed coordinates give equal values of every element of the coordinate algebra by step 1.1. In particular they give equal traces on every group word. The irreducible claim follows from [F3], including preservation of irreducibility. A reducible representation is triangular in an invariant-line basis, and [F3] gives its diagonal semisimplification with identical traces. Since the listed traces are the quotient coordinates, both have exactly the same quotient point. This records the semisimplified character and does not assert conjugacy of a nonsemisimple triangular representation to that diagonal one. All full-AC hypotheses are retained from [F1] and [F2]; the pointwise conjugacy argument adds no choice axiom.

Depends on

Used by

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources