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.
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A vector bundle quotient by a subbundle is a smooth vector bundle

Statement

If $S \subseteq E$ is a smooth rank- $k$ subbundle of a smooth rank- $r$ vector bundle $E \to M$ , then the fibrewise quotient $E / S \to M$ is a smooth rank- $(r - k)$ vector bundle.

Facts & Assumptions

Given: A smooth vector bundle $E \to M$ and a smooth rank- $k$ subbundle $S \subseteq E$ .

[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 $(e_{1}, \dots, e_{r})$ of $E$ such that $(e_{1}, \dots, e_{k})$ is a local frame of $S$ . Then the quotient classes of $e_{k + 1}, \dots, e_{r}$ form a basis of each quotient fibre $E_{p} / S_{p}$ .

2.1step 1.1algebra∎

Using the basis from step 1.1, identify the quotient fibre over $p$ with $R^{r - k}$ by reading the coefficients of the classes of $e_{k + 1}, \dots, e_{r}$ . If one changes to another adapted frame, the change-of-frame matrix has block upper-triangular form $[\begin{array}{cc} A & B \\ 0 & C \end{array}]$ , so the quotient coordinates transform by $C$ . Hence the quotient charts are smoothly compatible and define a smooth rank- $(r - k)$ vector bundle.

A vector bundle quotient by a subbundle is a smooth vector bundle

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources