PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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An oriented transverse normal bundle orients an embedded submanifold

Statement

Assume ${A C}_{ω}$ . For an embedded submanifold, any two of the orientations of the ambient tangent bundle, tangent bundle, and transverse normal bundle determine the third.

Facts & Assumptions

Given: The axiom ${A C}_{ω}$ , an embedded submanifold $S \subseteq M$ , and orientations of any two among $T M ∣_{S}$ , $T S$ , and the normal bundle $ν S$ .

[L1]

Under ${A C}_{ω}$ , the normal bundle is a smooth vector bundle (Assuming countable choice, normal and conormal bundles are smooth vector bundles).

[L2]

Its fibre is the quotient $ν_{p} S = T_{p} M / T_{p} S$ (Normal and conormal bundles of an embedded submanifold).

[L3]

An orientation is a positive ray in the one-dimensional determinant line (Determinant-line orientations of finite-dimensional real vector spaces).

Proof

technique · direct

1.1givenL1L2construct

By [L1] and [L2], $0 \to T S \to T M ∣_{S} \to ν S \to 0$ is an exact sequence of smooth vector bundles. Local frames of $T S$ extended to frames of $T M ∣_{S}$ give the ordered smooth determinant-line isomorphism $\det (T M ∣_{S}) ≅ \det (T S) \otimes \det (ν S)$ .

2.1L3step 1.1∎

Under this isomorphism, [L3] turns any two selected positive rays into a unique third ray: tensor the tangent and normal rays to obtain the ambient ray, or choose the unique tangent or normal ray whose tensor product is the prescribed ambient ray. Smoothness is local in the adapted frames, so each resulting ray field is an orientation.

An oriented transverse normal bundle orients an embedded submanifold

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources