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A nowhere-zero section forces the Euler class to vanish
Statement
Assume AC. Let be an -oriented numerable real vector bundle of rank over a base in the scope of the general Thom theorem. If admits a nowhere-zero section, then No converse is asserted.
Facts & Assumptions
Given: AC, an -oriented numerable rank- bundle with over a base in the general Thom scope, and a nowhere-zero section of .
The Euler class is , and the Gysin sequence of an oriented bundle in the Thom scope is the exact and natural sequence where is the sphere bundle projection (Euler class by zero-section pullback of the Thom class, Gysin long exact sequence of an oriented sphere bundle).
Under AC every numerable real bundle carries a bundle metric (Numerable vector bundles admit bundle metrics).
Pullback of cohomology is contravariantly functorial, so and the identity map induces the identity on cohomology (Singular cohomology is contravariantly functorial).
For a bundle with a supplied metric, the sphere bundle is the subspace of unit vectors and its projection is the restriction of the bundle projection (Disk, sphere, and Thom spaces of a metric vector bundle).
AC is the Axiom of Choice in the form fixed by The Axiom of Choice.
Proof
The section can be normalised. Choose a bundle metric on the numerable bundle by [F2] and put ; this is continuous because is nowhere zero, and it is a section of the sphere bundle in the sense of [F4], that is, . The normalisation is a specified function of the supplied section and metric, not a choice.
The sphere projection is injective on cohomology. Since , functoriality [F3] gives on ; a map with a left inverse is injective, so is injective for every .
The Euler class vanishes. Take in the Gysin sequence of [F1]: Exactness at says that the kernel of is the image of . Whether or not is connected, the particular element lies in this image. By step 2.1 the kernel is zero, so . No assertion that the whole image is cyclic is needed.
Boundary cases and the missing converse. The hypothesis is used in two places: the sphere bundle has nonempty fiber , and the unit normalisation of step 1.1 divides by the positive norm of a nonzero vector of a positive-dimensional fiber. For every section is the zero section, and the statement is excluded; with the standard unit orientation its Euler class is , while an arbitrary supplied rank-zero orientation gives . If is the zero section of a positive-rank bundle the normalisation is undefined, and indeed the conclusion can fail. The proposition asserts no converse: the vanishing of does not in general produce a nowhere-zero section, and no such section is constructed here. Over the empty base the unique class is zero. AC is used through the metric of [F2] and the Gysin sequence, as recorded.
Depends on
Used by
Dependency tree · two levels
24 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
- Allen Hatcher, Vector Bundles & K-Theory (standard reference, not scraped)
- Haynes Miller, MIT 18.906 Algebraic Topology II lecture notes (standard reference, not scraped)