Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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Smoothness of a section is equivalent to smooth local components

Statement

Let EM be a smooth vector bundle, let UM be open, and let (s1,,sr) be a local frame on U. A local section σ on U is smooth if and only if there are smooth functions f1,,fr:UR with

σ(p)=i=1rfi(p)si(p)(pU).

Facts & Assumptions

Given: A local frame (s1,,sr) on U and a local section σ:UEU.

[L1]

A local frame determines a local trivialization, and conversely (Local frames and local trivializations are equivalent data).

[L2]

Smoothness is local on the source (Smoothness is local on the source).

Proof

technique · direct
1.1

By [L1], the chosen frame gives a bundle chart Φ:EUU×Rr in which si(p) corresponds to the i-th standard basis vector. Therefore Φ(σ(p))=(p,f1(p),,fr(p)) exactly when σ(p)=ifi(p)si(p).

L1given
2.1

In this chart, σ is smooth exactly when the coordinate map p(f1(p),,fr(p)) is smooth. Equivalently, each component fi is smooth, and the criterion is local on U by [L2].

L2step 1.1algebra

Depends on

Used by

Dependency tree · two levels

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