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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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A transverse Banach bundle section has a split zero submanifold

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let π:E→M be a smooth Banach vector bundle over a Banach manifold M (Smooth Banach vector bundle and section), assume that the specified smooth atlas of M is maximal among compatible smooth charts, and let s:M→E be a smooth section which is transverse to the zero section, meaning that at every zero p of s the vertical derivative Dvs(p):TpM→Ep is surjective with complemented kernel. Then the zero set s−1(0) is a split smooth submanifold of M and

Tp(s−1(0))=ker⁡Dvs(p)for every p∈s−1(0).

Facts & Assumptions

Given: AC, a smooth Banach vector bundle π:E→M, a maximal specified smooth atlas on M, and a smooth section s whose vertical derivative is onto with complemented kernel at every zero.

[L1]

The definition of the vertical derivative and its independence of the trivialization, including the transformation DσΨ(p)=g(p)DσΦ(p) at a zero (Smooth Banach vector bundle and section).

[L2]

Regular value theorem for Banach manifolds (Regular value theorem for Banach manifolds), applied under the assumed AC and its maximal-domain-atlas hypothesis: a Ck map, k≥1 (including k=∞), whose derivative at every point of a level set is onto with complemented kernel has that level set as a split Ck submanifold with tangent equal to the kernel.

[L3]

Split submanifolds and their local character (Split Banach submanifold); tangents of open subsets of a Banach space are identified with the model space (Tangent space and differential on a Banach manifold); tensor and operator calculus as used for the transformation in [L1] (Chain sum product and composition rules for Banach derivatives).

Proof

technique · direct
1.1L1L3

Let p be a zero of s and let Φ be a smooth local trivialization over an open neighbourhood U of p; then σ:=pr⁡2∘Φ∘s:U→F is smooth and σ−1(0)=s−1(0)∩U. For every q∈σ−1(0), [L1] identifies Dσ(q):TqU→F with Dvs(q) followed by the fibre isomorphism Φq:Eq→F. Thus Dσ(q) is surjective with complemented kernel for every point of this local zero set. The atlas on U formed by restrictions of charts of the maximal smooth atlas on M is itself maximal: every compatible smooth chart on U is also compatible with the atlas of M (charts not meeting its domain are automatically compatible), and hence already belongs to that atlas.

2.1step 1.1L2

Apply [L2] with k=∞ to the smooth map σ:U→F at the value 0. The domain carries the maximal smooth atlas verified in [step 1.1], and every point of σ−1(0) has derivative onto with complemented kernel. Therefore σ−1(0) is a split smooth submanifold of U and Tq(σ−1(0))=ker⁡Dσ(q) for every q∈σ−1(0).

3.1step 2.1L3

Since U is open in M, the set σ−1(0)=s−1(0)∩U is a split submanifold of M as well, with the same tangent spaces: splitness is local by [L3] and the local charts of U are charts of M.

4.1step 3.1L3

The zeros of s are covered by such neighbourhoods U as p ranges over s−1(0); by [step 3.1] each point of s−1(0) has a split chart in M, so s−1(0) is a split smooth submanifold of M.

4.2step 3.1L1L3

For the tangent description, fix a zero p and two trivializations Φ over U and Ψ over V with p∈U∩V, and let σΦ,σΨ be the corresponding local representatives; [L1] gives DσΨ(p)=g(p)DσΦ(p), where g(p) is the invertible fibre isomorphism of the cocycle. Hence the kernels of DσΨ(p) and DσΦ(p) coincide, and the kernel of Dvs(p) is intrinsically characterised as the set of ξ∈TpM with DσΦ(p)ξ=0 for one, equivalently every, trivialization around p.

5.1step 2.1step 4.2L1L3∎

Combining [step 2.1] with [step 4.2]: for a zero p, Tp(s−1(0))=ker⁡DσΦ(p)=ker⁡Dvs(p), the first equality because σ−1(0) is the zero set of the local representative and the tangent of a split submanifold is computed in its charts, the second by the trivialization-independence just proved.

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