PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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Smoothness of a bundle map is equivalent to smooth local matrices

Statement

Let $Φ : E \to F$ be a fibrewise linear map over a smooth base map $f : M \to N$ . Choose local frames $(e_{1}, \dots, e_{r})$ for $E$ on $U \subseteq M$ and $(u_{1}, \dots, u_{s})$ for $F$ on $V \subseteq N$ with $f (U) \subseteq V$ . Then $Φ$ is smooth on $E ∣_{U}$ if and only if there are smooth scalar functions $a_{j i} : U \to R$ such that

$Φ (e_{i} (p)) = \sum_{j = 1}^{s} a_{j i} (p) u_{j} (f (p))$

for every $p \in U$ .

Facts & Assumptions

Given: A fibrewise linear map $Φ : E \to F$ over a smooth map $f : M \to N$ 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 $U \times R^{r}$ and $F ∣_{V}$ with $V \times R^{s}$ . In these trivializations, fibrewise linearity forces $Φ$ to have the form $Φ (p, v) = (f (p), A (p) v)$ for a unique matrix $A (p) = (a_{j i} (p))$ .

2.1L2step 1.1algebra∎

The local representative $(p, v) \mapsto (f (p), A (p) v)$ is smooth exactly when its matrix entries $a_{j i}$ are smooth on $U$ . Equivalently, the images $Φ (e_{i}) = \sum_{j} a_{j i} u_{j} \circ f$ have smooth local components, which is the criterion in [L2].

Smoothness of a bundle map is equivalent to smooth local matrices

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources