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

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Assuming countable choice, an ambient metric identifies the two normal bundles

Statement

Assume ${A C}_{ω}$ . Let $S \subseteq M$ be an embedded submanifold and let $g$ be a Riemannian metric on $M$ . If $q : T M ∣_{S} \to ν (S) = T M ∣_{S} / T S$ is the quotient map, then

$q ∣_{T S^{⊥}} : T S^{⊥} ⟶ ν (S)$

is a smooth vector-bundle isomorphism. Thus the fixed metric $g$ canonically identifies the quotient normal bundle with its $g$ -orthogonal realization.

Facts & Assumptions

Given: The axiom ${A C}_{ω}$ , an embedded submanifold $S \subseteq M$ , and an ambient Riemannian metric $g$ on $M$ .

[L0]

Under ${A C}_{ω}$ , $T M$ has a smooth manifold structure for which the induced tangent-bundle charts form a smooth atlas (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure).

[L4]

An induced tangent-bundle chart sends $v = \sum_{i} v^{i} \partial_{x^{i}} ∣_{p}$ to $(x (p), (v^{1}, \dots, v^{n}))$ (The induced tangent bundle chart).

[L5]

A smooth vector bundle is locally trivialized by fibrewise linear charts (Smooth vector bundles, rank, fibres, and trivial bundles).

[L6]

The inclusion $i : S ↪ M$ is smooth (The inclusion of an embedded submanifold is a smooth embedding).

[L7]

Pullback along a smooth map carries a smooth vector bundle to a smooth vector bundle (The pullback fibre product is a smooth vector bundle).

[L8]

In a slice chart, $S$ is a coordinate slice (Embedded submanifolds and slice charts).

[L9]

A smooth subbundle is locally spanned by part of a smooth ambient frame (Vector subbundles).

[F0]

A smooth bundle metric is a fibrewise inner product whose pairing of any two smooth local sections is smooth (Smooth bundle metrics).

[L1]

The orthogonal complement of a smooth subbundle is a smooth subbundle (Orthogonal complements of subbundles are smooth subbundles).

[L2]

The quotient map $T M ∣_{S} \to ν (S)$ is a smooth bundle map (The canonical map to a quotient bundle is a smooth bundle map).

[L3]

A fibrewise bijective smooth bundle map over the identity is a bundle isomorphism (A fibrewise bijective smooth bundle map over a diffeomorphism is a bundle isomorphism).

Proof

technique · direct

1.1F0L0L1L4L5L6L7L8L9given

By [L0], [L4], and [L5], the induced charts make $T M \to M$ a smooth vector bundle. By [L6] and [L7], its pullback along $i$ is the smooth vector bundle $T M ∣_{S} \to S$ . In a slice chart from [L8], its coordinate frame is $\partial_{x_{1}}, \dots, \partial_{x_{k}}, \partial_{y_{1}}, \dots, \partial_{y_{c}}$ and $T S$ is spanned by the first $k$ vectors, so [L9] makes $T S$ a smooth subbundle. In that pulled-back frame, the coefficients of $g ∣_{S}$ are the smooth coefficient functions of $g$ composed with the smooth inclusion $i$ ; hence [F0] makes $g ∣_{S}$ a smooth bundle metric on $T M ∣_{S}$ . Applying [L1], the orthogonal complements $T S_{p}^{⊥}$ form a smooth subbundle and fibrewise $T_{p} M = T_{p} S \oplus T_{p} S^{⊥}$ .

2.1L2L3step 1.1algebra∎

Restrict the quotient map of [L2] to $T S^{⊥}$ . On each fibre this is the usual linear isomorphism from a chosen complement onto the quotient by $T_{p} S$ . Hence the restricted map $T S^{⊥} \to ν (S)$ is fibrewise bijective, so [L3] shows that it is a smooth bundle isomorphism.

Assuming countable choice, an ambient metric identifies the two normal bundles

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources