Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31
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A rank-jumping kernel is not a vector subbundle

Statement refuted

The kernel of a smooth bundle map is always a smooth vector subbundle.

Facts & Assumptions

Given: The displayed claim.

[L1]

The kernel conclusion holds only under a constant-rank hypothesis (Constant-rank kernels and images of bundle maps over one base are subbundles).

Counterexample

technique · direct
1.1

On the trivial line bundle R×R→R, define the smooth bundle map Φ(x,v)=(x,xv). For x≠0, the fibre map is injective, so ker⁡Φx={0}. At x=0, the fibre map is zero, so ker⁡Φ0=R.

L1givenconstruct
2.1

The fibre dimensions of ker⁡Φ jump from 0 to 1, so the kernel is not locally trivial and therefore not a smooth vector subbundle. This is exactly why [L1] requires constant rank.

L1step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources