Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31
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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.

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Assuming countable choice, normal and conormal bundles are smooth vector bundles

Statement

Assume ACω. If SM is an embedded submanifold, then the normal bundle ν(S)=TMS/TS and the conormal bundle NSTMS are smooth vector bundles over S.

Facts & Assumptions

Given: The axiom ACω and an embedded submanifold SM.

[L0]
[L1]

Around each point of S there is a slice chart in which S is given by y1==yc=0 (Embedded submanifolds and slice charts).

[L2]

A quotient by a smooth vector subbundle is a smooth vector bundle (A vector bundle quotient by a subbundle is a smooth vector bundle).

Proof

technique · direct
1.1

In a slice chart (x1,,xk,y1,,yc) with S={y=0}, the induced charts of [L0] make TMS a smooth vector bundle with local frame x1,,xk,y1,,yc, while TS is spanned by the xi. Hence TS is a smooth subbundle and the classes of yj give a local frame of the quotient TMS/TS. By [L2], the normal bundle is smooth.

L0L1L2given
2.1

In the same slice chart, the induced cotangent charts of [L0] give the local coframe dx1,,dxk,dy1,,dyc, and the covectors annihilating TS are exactly the span of dy1,,dyc. These local frames vary smoothly, so the conormal bundle is a smooth subbundle of TMS, hence a smooth vector bundle over S.

L0L1step 1.1algebra

Depends on

Used by

Dependency tree · two levels

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