Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every vector subbundle has a smooth complement

Statement

Assume the Axiom of Countable Choice. Every smooth vector subbundle S⊆E has a smooth complement in E.

Facts & Assumptions

Given: The Axiom of Countable Choice and a smooth vector subbundle S⊆E.

[L1]

Under Countable Choice, the bundle E admits a smooth bundle metric (Every smooth vector bundle admits a smooth bundle metric, The Axiom of Countable Choice (ACω)).

[L2]

Orthogonal complements of subbundles are smooth subbundles (Orthogonal complements of subbundles are smooth subbundles).

Proof

technique · direct
1.1L1givenchoose

Choose a smooth bundle metric on E by [L1].

2.1L2step 1.1algebra∎

With that metric, [L2] gives a smooth subbundle S⊥⊆E, and fibrewise one has Ep=Sp⊕Sp⊥. Thus S⊥ is a smooth complement of S.

Depends on

Used by

Dependency tree · two levels

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Sources