Alphabeta Math
TheoremStatement: 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.

Boundary of the integer-framed knot surgery trace

Statement

Assume ACω. Let K be a smooth oriented knot in the oriented 3-sphere with a chosen exterior EK and peripheral basis (μ,λ), where λ is preferred. For n∈Z, attach one smooth 4-dimensional 2-handle D2×D2 to the 4-ball B4 along S1×D2 in ∂B4=S3 by the standard orientation-compatible handle attachment, using a product framing whose attaching parallel on ∂EK has slope n[μ]+[λ]. Denote the resulting oriented trace by Xn(K). After rounding corners, its oriented boundary is the filling EK(n/1). This theorem covers integer framings only: a rational p/q filling has a trace only after specifying an integral continued-fraction handle chain or a separately defined rational-surgery cobordism; an arbitrary rational coefficient is not the framing of a single smooth 2-handle.

Facts & Assumptions

Given: The oriented 4-ball, a framed attaching knot K⊂S3, the specified product identification of its tubular neighborhood, and the slope convention for EK(n/1).

[F1]

Oriented slopes and knot Dehn filling: the filling replaces the knot neighborhood by D2×S1 whose meridian maps to n[μ]+[λ], and its orientation agrees with the exterior orientation.

[A1]

ACω is included for the chosen exterior and peripheral-system interface The Axiom of Countable Choice (ACω). The product framing is part of the handle-attachment data; after those data are supplied, the boundary decomposition makes no further choice.

Proof

1.1givenA1construct

The attaching region of the 2-handle is S1×D2⊂∂(D2×D2). It is glued to the tubular neighborhood N(K)⊂S3, so the portion of ∂B4 left in the boundary after attachment is exactly EK=S3∖int⁡N(K).

2.1step 1.1A1F1construct

The free face of the handle boundary is D2×S1, a solid torus. Its meridian disk D2×{z0} has boundary ∂D2×{z0}; the specified product framing sends this curve to the attaching parallel of slope n[μ]+[λ] on ∂EK. Thus the outgoing boundary is the union of EK and a solid torus with precisely the meridian image in [F1].

3.1givenA1F1step 2.1construct∎

The orientation-compatible handle attachment identifies the oriented attaching region with N(K) by a boundary-orientation-reversing map. The free face D2×S1 inherits its boundary orientation from the oriented handle, so its gluing to EK reverses the induced orientations on the common torus; this is exactly the oriented filling convention in [F1]. Rounding corners changes the boundary only inside a collar and preserves its orientation. Thus ∂Xn(K) is the stated oriented filling.

Diagram

The left decomposition is the original boundary of the 4-ball. The handle removes the neighborhood N(K) from that boundary and contributes its free face D2×S1; the meridian of this new solid torus is attached along nμ+λ. The diagram records the boundary pieces, while the proof above establishes the identification without using the picture.

EKN(K)S3=EK[N(K)TKEKD2£S1@Xn(K)@D27!n¹+¸removeN(K)insertfreeface

Depends on

Used by

Dependency tree · two levels

17 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