Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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.

The standard circle and its annular tubular neighbourhood

Example

For the unit circle S1={p∈R2:∥p∥=1}, the normal line at p is the radial line Rp. Thus the normal bundle is identified with S1×R,(p,t)↔(p,tp), and the normal addition map is E(p,t)=(1+t)p. For ∣t∣<1/2, the image is the annulus {x∈R2:1/2<∥x∥<3/2}, and the tubular retraction is radial normalization r(x)=x∥x∥.

Facts & Assumptions

Given: The unit circle S1⊆R2 with its Euclidean normal bundle.

Verification

technique · direct
1.1givenalgebra

Every vector orthogonal to TpS1 is a scalar multiple of p, so the displayed identification of the normal bundle is correct. Under that identification the normal addition map is E(p,t)=p+tp=(1+t)p.

2.1step 1.1

If ∣t∣<1/2, then 1/2<1+t<3/2, so E lands in the stated annulus. Conversely, every nonzero x in that annulus can be written uniquely as x=∥x∥x∥x∥=E ⁣(x∥x∥,∥x∥−1). Hence this is exactly the tubular neighbourhood promised by The Euclidean tubular neighbourhood theorem.

3.1step 2.1∎

The inverse formula in step 2.1 sends x to (x/∥x∥,∥x∥−1), so the associated retraction is r(x)=x/∥x∥, in agreement with A closed Euclidean submanifold has a smooth neighborhood retraction.

Depends on

Used by

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Dependency tree · two levels

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Sources