Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Every smooth vector bundle admits a smooth bundle metric

Statement

Assume the Axiom of Countable Choice. Every smooth vector bundle admits a smooth bundle metric.

Facts & Assumptions

Given: The Axiom of Countable Choice and a smooth vector bundle E→M.

[L1]

Under Countable Choice, the base manifold admits smooth partitions of unity subordinate to indexed open covers (Smooth partitions of unity exist on manifolds, The Axiom of Countable Choice (ACω)).

[L2]

Local frames are equivalent to local trivializations (Local frames and local trivializations are equivalent data).

Proof

technique · direct
1.1L2givenconstruct

Index an open cover by all pairs (Uα,τα) supplied by the bundle's local trivializations, so each cover member comes with its trivialization and no separate family of frame choices is needed. Pull back the Euclidean inner product on Rr through τα to obtain a smooth local bundle metric hα on E∣Uα.

2.1L1step 1.1constructalgebra∎

Under the stated Countable Choice, [L1] gives a smooth partition of unity (ρα) indexed by these same trivialization pairs, with supp⁡ρα⊆Uα. Define ραhα on Uα and extend it by zero outside Uα; this is smooth because every point outside the closed support of ρα has a neighbourhood where the term vanishes. The sum h:=∑αραhα is locally finite and smooth. At each point p, some ρα(p)>0, all weights are nonnegative, and ∑αρα(p)=1, so hp is positive definite. Thus h is a smooth bundle metric.

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