Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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  • 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 bundle map is equivalent to smooth local matrices

Statement

Let Φ:E→F be a fibrewise linear map over a smooth base map f:M→N. Choose local frames (e1,…,er) for E on U⊆M and (u1,…,us) for F on V⊆N with f(U)⊆V. Then Φ is smooth on E∣U if and only if there are smooth scalar functions aji:U→R such that

Φ(ei(p))=∑j=1saji(p)uj(f(p))

for every p∈U.

Facts & Assumptions

Given: A fibrewise linear map Φ:E→F over a smooth map f:M→N and local frames on U and V as above.

[L1]

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

[L2]

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

Proof

technique · direct
1.1L1given

By [L1], the chosen frames identify E∣U with U×Rr and F∣V with V×Rs. In these trivializations, fibrewise linearity forces Φ to have the form Φ(p,v)=(f(p),A(p)v) for a unique matrix A(p)=(aji(p)).

2.1L2step 1.1algebra∎

The local representative (p,v)↦(f(p),A(p)v) is smooth exactly when its matrix entries aji are smooth on U. Equivalently, the images Φ(ei)=∑jajiuj∘f have smooth local components, which is the criterion in [L2].

Depends on

Used by

Dependency tree · two levels

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Sources