Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (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.

The diagonal in the two-sphere has self-intersection two

Example

Assume AC. Let S2⊆R3 carry its induced orientation and give S2×S2 the product orientation. Then the diagonal Δ⊆S2×S2 is a closed oriented embedded surface with 2dim⁡Δ=dim⁡(S2×S2) and Δ⋅Δ=⟨e(TS2),[S2]⟩=2. The value 2 is computed from the explicit tangent field X(p)=e3−z p, whose zeros are the two poles with local index +1 each; it previews the Euler characteristic χ(S2)=2 of the later Euler/index pair (not used here).

Facts & Assumptions

Given: AC, the unit sphere S2⊆R3 with its induced orientation, the product S2×S2 with the product orientation, its diagonal and the explicit field X(p)=e3−zp.

[F1]

The diagonal Δ⊆S2×S2 is a closed embedded surface of dimension 2 with 2⋅2=4=dim⁡(S2×S2) (The diagonal is an embedded submanifold, Products of smooth manifolds have a canonical product smooth structure).

[F2]

The normal bundle of the diagonal is canonically TM, orientation-preservingly when ΔM carries the orientation transported from M, with the tangent-first normal orientation (The normal bundle of the diagonal is canonically the tangent bundle, Product orientations, Canonical tangent and cotangent splittings for products).

[F3]

The self-intersection is Δ⋅Δ=⟨e(νΔ),[Δ]⟩=⟨e(TM),[M]⟩ (The self-intersection number is the Euler number of the normal bundle).

[F4]

S2 is the unit sphere, the regular level ∣p∣2=1, with tangent space TpS2=ker⁡(v↦2p⋅v)=p⊥ (Euclidean spheres and closed balls as subspaces of Rn, A regular level set is an embedded submanifold, The tangent space of a regular level set is the kernel).

[F5]

The local sign of the push-off equals the local zero index of the section, so the signed zero count is the self-intersection number (Normal push-off zeros are the self-intersection points).

Verification

technique · identify the normal bundle and compute the two zero signs in pole charts
1.1F1F2F3given

Δ is closed embedded of dimension 2 with 2⋅2=4=dim⁡(S2×S2) [F1], so [F3] gives Δ⋅Δ=⟨e(νΔ),[Δ]⟩. By [F2] the normal identification is the canonical one and is orientation-preserving with the orientation of Δ transported from S2, so Δ⋅Δ=⟨e(TS2),[S2]⟩.

2.1F3F4F5step 1.1algebra∎

In the projection charts of the two poles (x,y)↦(x,y,±1−x2−y2) the given field X(p)=e3−zp, tangent because (e3−zp)⋅p=z−z∣p∣2=0 by [F4], has exactly the two zeros ±e3, with local components (−zx,−zy) and derivatives −I2 at the north pole and I2 at the south pole. Both determinants are +1, so each zero has index +1 and the signed zero count of TS2 is 2; [F5] and [F3] identify that count with ⟨e(TS2),[S2]⟩. Hence Δ⋅Δ=2, which previews the Euler characteristic χ(S2)=2 of the later Euler/index pair (not used here).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

77 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