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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Smooth sections form a module over smooth functions

Statement

If $E \to M$ is a smooth vector bundle, then the smooth sections of $E$ form a module over $C^{\infty} (M)$ under pointwise addition and scalar multiplication.

Facts & Assumptions

Given: A smooth vector bundle $E \to M$ .

[L1]

A section is smooth exactly when its local frame components are smooth (Smoothness of a section is equivalent to smooth local components).

[L2]

Sums and products of smooth scalar functions are smooth (Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives).

Proof

technique · direct

1.1L1given

Let $σ, τ$ be smooth sections and let $f \in C^{\infty} (M)$ . On a local frame, write $σ = \sum a_{i} s_{i}$ and $τ = \sum b_{i} s_{i}$ with smooth components $a_{i}, b_{i}$ . Then $σ + τ = \sum (a_{i} + b_{i}) s_{i}$ and $f σ = \sum (f a_{i}) s_{i}$ .

2.1L1L2step 1.1∎

By [L2], the component functions $a_{i} + b_{i}$ and $f a_{i}$ are smooth, so [L1] shows that $σ + τ$ and $f σ$ are again smooth sections. The module axioms hold fibrewise because each fibre is a vector space.

Smooth sections form a module over smooth functions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources