Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Boundary-defining functions exist locally and detect inward vectors

Statement

Every boundary point has a boundary-defining function; for any such ρ, a vector v is inward exactly when dρp(v)>0.

Facts & Assumptions

Given: A boundary point p of a smooth manifold M, a boundary chart at p, and, for the sign assertion, a boundary-defining function ρ near p and a vector v∈TpM.

[L1]

A boundary-defining function is nonnegative, vanishes exactly on the boundary, and has nonzero differential there (Boundary-defining functions).

[L2]

Inward, outward, and boundary-tangent vectors have respectively positive, negative, and zero last coordinate in a boundary chart (Inward, outward, and boundary-tangent vectors).

Proof

technique · direct
1.1givenL1construct

In the chosen boundary chart take ρ=xn. It is nonnegative, vanishes precisely on the face, and has nonzero differential, so it is a boundary-defining function by [L1].

2.1givenL1L2step 1.1∎

For any boundary-defining function ρ, its restriction to the face is zero, so dρp vanishes on the tangent hyperplane. Its derivative in the positive normal coordinate is nonnegative because ρ≥0 on the half-space and ρ(p)=0; by [L1] it is nonzero, hence positive. Therefore dρp(v) has the sign of the last coordinate of v, and [L2] gives dρp(v)>0 exactly for inward v.

Depends on

Used by

Dependency tree · two levels

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Sources