PropositionStatement: 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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Orthogonal complements of subbundles are smooth subbundles

Statement

Let $S \subseteq E$ be a smooth vector subbundle of a smooth vector bundle $E \to M$ equipped with a smooth bundle metric. Then the orthogonal complements

$S_{p}^{⊥} : = {v \in E_{p} : ⟨ v, w ⟩_{p} = 0 for every w \in S_{p}}$

form a smooth vector subbundle $S^{⊥} \subseteq E$ .

Facts & Assumptions

Given: A smooth vector bundle $E \to M$ , a smooth subbundle $S \subseteq E$ , and a smooth bundle metric on $E$ .

[L1]

Locally, $S$ is spanned by part of a frame of $E$ (Vector subbundles).

[L2]

Gram-Schmidt orthonormalisation depends smoothly on a smooth frame (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

Proof

technique · direct

1.1L1L2givenchoose

Around each point, choose a local frame $(e_{1}, \dots, e_{r})$ of $E$ such that $(e_{1}, \dots, e_{k})$ spans $S$ . Apply smooth Gram-Schmidt from [L2] to obtain a local orthonormal frame $(u_{1}, \dots, u_{r})$ . Because the first $k$ input vectors already lie in $S$ , the first $k$ orthonormalized vectors still span $S$ .

2.1step 1.1algebra∎

For each fibre, the orthogonal complement of $S_{p}$ is then spanned by $u_{k + 1} (p), \dots, u_{r} (p)$ . These vectors vary smoothly, so they give a local frame of $S^{⊥}$ . Hence $S^{⊥}$ is a smooth vector subbundle of $E$ .

Depends on

Used by

Every vector subbundle has a smooth complement Corollary
Mean curvature vector Definition
Normal bundle of a formal immersion Definition
Tangential and normal projections along a Riemannian submanifold Definition
Formal immersion gives the tangent normal-bundle identity Lemma
Stiefel-Whitney classes of the tangent bundle of real projective space Lemma
Assuming countable choice, an ambient metric identifies the two normal bundles Proposition

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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