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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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A nowhere-zero section forces the Euler class to vanish

Statement

Assume AC. Let EB be an R-oriented numerable real vector bundle of rank n1 over a base in the scope of the general Thom theorem. If E admits a nowhere-zero section, then e(E)=0in Hn(B;R). No converse is asserted.

Facts & Assumptions

Given: AC, an R-oriented numerable rank-n bundle EB with n1 over a base in the general Thom scope, and a nowhere-zero section σ of E.

[F1]

The Euler class is e(E)=sj(uE), and the Gysin sequence of an oriented bundle in the Thom scope is the exact and natural sequence Hkn(B;R)e(E)Hk(B;R)pHk(S(E);R)GHkn+1(B;R), where p:S(E)B is the sphere bundle projection (Euler class by zero-section pullback of the Thom class, Gysin long exact sequence of an oriented sphere bundle).

[F2]

Under AC every numerable real bundle carries a bundle metric (Numerable vector bundles admit bundle metrics).

[F3]

Pullback of cohomology is contravariantly functorial, so (pσ)=σp and the identity map induces the identity on cohomology (Singular cohomology is contravariantly functorial).

[F4]

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).

[A1]

AC is the Axiom of Choice in the form fixed by The Axiom of Choice.

Proof

1.1

The section can be normalised. Choose a bundle metric on the numerable bundle E by [F2] and put σ^(b)=σ(b)/σ(b); this is continuous because σ is nowhere zero, and it is a section of the sphere bundle p:S(E)B in the sense of [F4], that is, pσ^=idB. The normalisation is a specified function of the supplied section and metric, not a choice.

F2F4
2.1

The sphere projection is injective on cohomology. Since pσ^=idB, functoriality [F3] gives σ^p=(idB)=id on H(B;R); a map with a left inverse is injective, so p:Hk(B;R)Hk(S(E);R) is injective for every k.

F3step 1.1
3.1

The Euler class vanishes. Take k=n in the Gysin sequence of [F1]: H0(B;R)e(E)Hn(B;R)pHn(S(E);R). Exactness at Hn(B;R) says that the kernel of p is the image of e(E). Whether or not B is connected, the particular element e(E)=1e(E) lies in this image. By step 2.1 the kernel is zero, so e(E)=0. No assertion that the whole image is cyclic is needed.

F1step 2.1
4.1

Boundary cases and the missing converse. The hypothesis n1 is used in two places: the sphere bundle has nonempty fiber Sn1, and the unit normalisation of step 1.1 divides by the positive norm of a nonzero vector of a positive-dimensional fiber. For n=0 every section is the zero section, and the statement is excluded; with the standard unit orientation its Euler class is e(0B,1)=1, while an arbitrary supplied rank-zero orientation o gives e(0B,o)=o. If σ 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 e(E) 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.

F1F2A1step 1.1step 2.1

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