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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07
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Boundary submanifolds of a boundaryless manifold have half-slice charts

Statement

If Sk is an embedded manifold with boundary in a boundaryless n-manifold, then interior points have ordinary slice charts and boundary points have charts with S={xk+1==xn=0, xk0}.

Facts & Assumptions

Given: A smooth embedding i:SkMn, where S is a manifold with boundary and M is boundaryless, and a point pS.

[L1]

The differential of a smooth embedding of manifolds with boundary is injective on the full tangent space (Immersions and embeddings for manifolds with boundary).

[L2]

A smooth half-space map admits a smooth Euclidean extension near each point (Smooth functions on relatively open half-space sets).

[L3]

A rank-k smooth map from a k-manifold has local coordinates in which it is u(u,0) (The constant-rank theorem for manifolds).

Proof

technique · direct
1.1

At an interior point, [L1] and [L3] give an ordinary slice chart. At a boundary point, choose boundary coordinates x on S and ordinary ambient coordinates. By [L2], extend the coordinate embedding to a smooth map F~ on an open subset of Rk. Its derivative at the boundary point equals the injective differential from [L1], so after shrinking an invertible k×k minor stays nonzero and F~ has constant rank k.

givenL1L2L3
2.1

Apply [L3] to F~. In the resulting source coordinates u=α(x) and target coordinates (u,w), it has the form u(u,0). The source change α need not preserve the face, but it can be absorbed into the target chart: postcompose that chart with the local diffeomorphism (u,w)(α1(u),w). In the new target coordinates, F~(x)=(x,0) in the original boundary coordinates.

L3step 1.1
3.1

Restricting x back to the source half-space now gives S={xk+1==xn=0, xk0} near the boundary point, while step 1.1 gives the ordinary slice at interior points.

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources