Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-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.

Worked cases of three-manifold surgery presentations

Example

Assume the Axiom of Choice. The empty integral surgery diagram gives the standard oriented S3. Surgery on the standard unknot with framing 0 gives S1×S2. For every integer p>0, surgery on the standard unknot with framing p gives the oriented lens space L(p,1) in the convention L(p,q)=EU(p/q).

Facts & Assumptions

Given: Full AC, the empty link, and the standard oriented unknot U(t)=(e2πit,0) in S3⊂C2.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function; full AC also supplies the countable-choice hypotheses in the chosen tubular data.

[A2]

The Axiom of Countable Choice (ACω): ACω says that every sequence of nonempty sets has a choice function.

[F1]

Integral framed-link surgery diagram: each component with integer n is filled along n[μ]+[λ], and the empty link has surgery result S3.

[F2]

Oriented slopes and knot Dehn filling: the integer n filling attaches a solid torus whose meridian maps to n[μU]+[λU].

[F3]

Lens spaces in the p/q filling orientation convention: for p>0, L(p,q) is the oriented filling of the unknot exterior with meridian slope p[μU]+q[λU].

[F4]

Meridian and preferred longitude: a preferred longitude is a degree-one parallel curve whose homology class maps to zero in the chosen exterior.

[F5]

Homology and preferred longitude of a knot exterior: under full AC the positive meridian and unique preferred longitude form the peripheral basis of the unknot exterior.

Verification

1.1givenA1A2choose

Applying [A1] to the range of any sequence (Xn) in [A2] and composing the resulting choice function with n↦Xn gives the countable-choice function required by [A2]. Thus the chosen tubular and slope interfaces used below are available.

1.2givenF1construct

For the empty link there are no tubular neighbourhoods to remove and no solid tori to attach. The definition [F1] therefore leaves the standard oriented S3 unchanged.

1.3givenF2F4F5construct

Use N(U)={∣z2∣≤1/2}; its exterior is the solid torus EU={∣z2∣≥1/2}≅D2×S1, with disk coordinate z1 and core coordinate arg⁡z2. The loop in the z1 direction has degree one and bounds a disk in EU, so [F4]–[F5] identify it with λU; the positive meridian μU runs once around the exterior core. For 0-surgery, [F2] sends the attached meridian to λU, the meridian of EU. The gluing-map convention in the slope definition permits the product gluing: the two disk factors glue to S2 and the core circles glue to S1, giving S1×S2 with the induced surgery orientation.

2.1givenF2F3algebra∎

For p>0, [F2] identifies the p-framed unknot filling with the filling of EU along p[μU]+[λU]. This is exactly L(p,1) by [F3]. The case p=1 is included.

Diagram

Each circle denotes the standard unknot and the adjacent integer is its framing. The first panel has no components. The second and third panels show the zero and positive-p surgeries. The panels illustrate the examples only; the coordinate and gluing arguments in the verification establish the identifications.

\begin{tikzpicture}[>=stealth]
  \node at (0,1) {empty};
  \node at (3,1) {$0$-framed};
  \node at (6,1) {$p$-framed, $p>0$};
  \node at (0,0) {$\varnothing$};
  \draw[thick] (3,0) circle (0.42);
  \node at (3.8,0) {$0$};
  \draw[thick] (6,0) circle (0.42);
  \node at (6.8,0) {$p$};
  \draw[->] (0.8,0) -- (2.1,0);
  \draw[->] (3.9,0) -- (5.1,0);
\end{tikzpicture}

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