CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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A smooth section transverse to the zero section has a submanifold zero set

Statement

Let $π : E \to M$ be a smooth vector bundle of rank $r$ , and let $s : M \to E$ be a smooth section. If $s$ is transverse to the zero section, then the zero set $Z (s)$ is an embedded submanifold of codimension $r$ .

Facts & Assumptions

Given: A smooth vector bundle $π : E \to M$ of rank $r$ and a smooth section $s : M \to E$ that is transverse to the zero section.

[L1]

The zero section is a smooth embedding (The zero section is a smooth embedding).

[L2]

A section with surjective vertical differential at every zero has a submanifold zero set (A vector bundle section with surjective vertical differential at every zero has a submanifold zero set).

Proof

technique · direct

1.1L1givenalgebra

By [L1], the zero section is an embedded submanifold of $E$ . At a zero $p \in Z (s)$ , transversality of $s$ to the zero section means exactly that the induced quotient map $T_{p} M ⟶ T_{0_{p}} E / T_{0_{p}} (0_{M} (M)) ≅ E_{p}$ is surjective, which is the vertical differential of $s$ at $p$ .

2.1L2step 1.1∎

Therefore the vertical differential of $s$ is surjective at every zero, so [L2] implies that $Z (s)$ is an embedded submanifold of codimension $r$ .

Depends on

Used by

Dependency tree · two levels

8 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Transversality (standard reference, not scraped)