Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A smooth distribution is exactly a locally framed constant-rank family of tangent spaces

Statement

Let pDpTpM be a rank-k family of tangent subspaces on a smooth manifold M. Then the following are equivalent:

  1. D is a smooth distribution.
  2. Every point of M has a neighborhood U and smooth vector fields X1,,Xk on U such that the vectors X1(q),,Xk(q) are linearly independent and span Dq for all qU.

Facts & Assumptions

Given: A rank-k family pDpTpM.

[A1]

In item 1, smoothness means that D is a rank-k smooth vector subbundle of TM.

Proof

technique · direct
1.1

Assume D is a smooth distribution. By the local description of a [given] subbundle, each point has a neighborhood U and a frame of TMU whose first k members already frame DU. Those first k sections are smooth vector fields, pointwise independent, and span the prescribed subspaces.

given
1.2

Conversely, assume such local vector fields exist near every point. On a [given] neighborhood U where X1,,Xk are pointwise independent, their span is a rank-k subbundle of TMU, because in a local trivialization of TM the columns formed by the Xi have rank k everywhere. Since that subbundle has fibres exactly Dq, the family is a smooth distribution on U.

givenalgebra
2.1

The two implications establish the equivalence. [given] ∎

given

Depends on

Used by

Dependency tree · two levels

13 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