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The kernel distribution of a constant-rank submersion is integrable

Statement

Let $F : M \to N$ be a smooth submersion. Then the kernel distribution $\ker d F \subseteq T M$ is integrable, and its maximal connected integral manifolds are the connected components of the level sets of $F$ .

Facts & Assumptions

Given: A smooth submersion $F : M \to N$ .

[A1]

Fix $p \in M$ and write $q : = F (p)$ .

Proof

technique · direct

1.1given

Because $F$ is a submersion, $q$ is a regular value and the level set [given] $F^{- 1} (q)$ is an embedded submanifold. Its tangent space at each point is the kernel of the differential of $F$ . Therefore each connected component of $F^{- 1} (q)$ is an integral manifold of $\ker d F$ .

1.2given

Repeating the same argument at every point of $M$ shows that each point [given] lies on such a connected component, so $\ker d F$ is integrable. Since an integral manifold of $\ker d F$ stays inside one level set of $F$ , the maximal connected integral manifolds are exactly the connected components of the fibres.

2.1given∎

Hence the kernel of a submersion is integrable with leaves equal to fibre [given] components.

Depends on

Used by

Explicit Godbillon-Vey rescaling calculation Example
The kernel of a submersion as an integrable distribution Example
The level-set foliation of a regular function Example
The Reeb foliation of the solid torus has the boundary as a leaf Proposition

Dependency tree · two levels

10 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

Will J. Merry, Differential Geometry (standard reference, not scraped)