PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 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.
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Smoothness of a section is equivalent to smooth local components

Statement

Let $E \to M$ be a smooth vector bundle, let $U \subseteq M$ be open, and let $(s_{1}, \dots, s_{r})$ be a local frame on $U$ . A local section $σ$ on $U$ is smooth if and only if there are smooth functions $f_{1}, \dots, f_{r} : U \to R$ with

$σ (p) = \sum_{i = 1}^{r} f_{i} (p) s_{i} (p) (p \in U) .$

Facts & Assumptions

Given: A local frame $(s_{1}, \dots, s_{r})$ on $U$ and a local section $σ : U \to E ∣_{U}$ .

[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

1.1L1given

By [L1], the chosen frame gives a bundle chart $Φ : E ∣_{U} \to U \times R^{r}$ in which $s_{i} (p)$ corresponds to the $i$ -th standard basis vector. Therefore $Φ (σ (p)) = (p, f_{1} (p), \dots, f_{r} (p))$ exactly when $σ (p) = \sum_{i} f_{i} (p) s_{i} (p)$ .

2.1L2step 1.1algebra∎

In this chart, $σ$ is smooth exactly when the coordinate map $p \mapsto (f_{1} (p), \dots, f_{r} (p))$ is smooth. Equivalently, each component $f_{i}$ is smooth, and the criterion is local on $U$ by [L2].

Smoothness of a section is equivalent to smooth local components

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources