Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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  • 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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A vector bundle quotient by a subbundle is a smooth vector bundle

Statement

If S⊆E is a smooth rank-k subbundle of a smooth rank-r vector bundle E→M, then the fibrewise quotient E/S→M is a smooth rank-(r−k) vector bundle.

Facts & Assumptions

Given: A smooth vector bundle E→M and a smooth rank-k subbundle S⊆E.

[L1]

Locally, a subbundle is spanned by part of a frame of the ambient bundle (Vector subbundles).

Proof

technique · direct
1.1L1givenchoose

Around each point of M, choose a local frame (e1,…,er) of E such that (e1,…,ek) is a local frame of S. Then the quotient classes of ek+1,…,er form a basis of each quotient fibre Ep/Sp.

2.1step 1.1algebra∎

Using the basis from step 1.1, identify the quotient fibre over p with Rr−k by reading the coefficients of the classes of ek+1,…,er. If one changes to another adapted frame, the change-of-frame matrix has block upper-triangular form [AB0C], so the quotient coordinates transform by C. Hence the quotient charts are smoothly compatible and define a smooth rank-(r−k) vector bundle.

Depends on

Used by

Dependency tree · two levels

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Sources