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

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Isomorphic cocycles define isomorphic vector bundles

Statement

Suppose two smooth rank- $r$ cocycles on the same open cover satisfy

$g_{β α}^{'} (p) = h_{β} (p) g_{β α} (p) h_{α} (p)^{- 1}$

for smooth maps $h_{α} : U_{α} \to G L (r, R)$ . Then the two cocycles define isomorphic smooth vector bundles.

Facts & Assumptions

Given: Two smooth cocycles $(g_{β α})$ and $(g_{β α}^{'})$ on the same cover, together with a gauge family $(h_{α})$ satisfying the displayed relation.

[L1]

A smooth cocycle on a countable cover determines a smooth vector bundle by the quotient construction (Construction of a vector bundle from a smooth cocycle).

Proof

technique · direct

1.1L1givenconstruct

On the $α$ -th trivializing piece define $H_{α} (p, v) = (p, h_{α} (p) v)$ . If $(p, v)_{α} \sim (p, g_{β α} (p) v)_{β}$ , then the gauge relation gives $H_{β} (p, g_{β α} (p) v) = (p, g_{β α}^{'} (p) h_{α} (p) v) = H_{α} (p, v)$ , so the local maps descend to a well-defined bundle map $H : E \to E^{'}$ .

2.1L1step 1.1algebra∎

In the quotient charts of [L1], the descended map is $(p, v) \mapsto (p, h_{α} (p) v)$ , hence smooth and fibrewise linear. Replacing $h_{α}$ by $h_{α}^{- 1}$ gives the inverse construction, so $H$ is a smooth bundle isomorphism.

Depends on

Used by

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Sources

John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)