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The zero locus of a transverse section represents the Euler dual

Statement

Assume AC. Let E→M be a smooth R-oriented rank-r real vector bundle over a closed R-oriented smooth n-manifold, with R=Z or F2. Let s be a smooth section transverse to the zero section. Its zero locus Z is a closed embedded submanifold of dimension n−r when nonempty, and νZ≅E∣Z. Orient its normal bundle by this isomorphism and orient Z in tangent-first order. Then e(E)∩[M]=(−1)r(n−r)(iZ)∗[Z]∈Hn−r(M;R). In particular (iZ)∗[Z]≠0 is equivalent to e(E)≠0, and either implies Z≠∅. When r=n, the Euler number ⟨e(E),[M]⟩ is the signed zero count; a nonzero count forces a zero. On a disconnected base the total count may cancel even when the zero-cycle class is nonzero. For r≠n, no evaluation of e(E) on the n-dimensional fundamental class is asserted. Over F2 orientations are canonical and the sign disappears. The zero section defines the Euler class by absolute pullback, and, when M≠∅, is transverse to itself exactly in rank zero; for M=∅ transversality is vacuous.

Facts & Assumptions

Given: AC and M,E,s,Z with the orientations of the statement.

[F1]

The vertical differential induces νZ≅E∣Z, and this isomorphism defines the tangent-first induced orientation (Normal bundle of the zero locus of a transverse section).

[F2]

The absolute image αˉ of the tangent-first normal Thom class satisfies αˉ∩[M]=(−1)rz(iZ)∗[Z] (The normal Thom class realizes the Poincare dual of a closed submanifold).

[F3]

The relative pullback of the Thom class through the punctured-bundle pair is the normal Thom class, and its absolute image is e(E) (Pullback of the Thom class along a transverse section computes the Euler class).

[F4]

Cap duality is an isomorphism; degree-top cap followed by zero-chain augmentation is Kronecker evaluation (Poincaré duality for oriented topological manifolds, Cap product with cohomology written first, Kronecker evaluation pairing).

[F5]

Smooth manifolds are admissible bases and their bundles are numerable under AC. The Euler class of a rank-zero bundle is its supplied orientation unit (Smooth manifolds have CW homotopy type, Euler class by zero-section pullback of the Thom class).

Proof

technique · identify the relative Thom pullback and then forget supports before applying normal Thom duality
1.1F1F3F5given

By [F1], Z has codimension r and normal bundle E∣Z with the prescribed orientation. By [F5] every bundle and base used here meets the Thom hypotheses. Let α∈Hr(M,M∖Z;R) be its normal Thom extension and αˉ its absolute image. By [F3], αˉ=e(E).

2.1F2F3F4step 1.1algebra

Apply [F2] with z=n−r to obtain e(E)∩[M]=(−1)r(n−r)(iZ)∗[Z]. If Z is empty its relative group is zero and [F3] gives e(E)=0, including r>n, when transversality forces the zero locus to be empty. Cap duality [F4] gives the stated equivalence of nonzero classes.

3.1F1F4F5step 2.1algebra∎

If r=n, each zero is nondegenerate, and its point orientation is the sign of Dvs:TxM→Ex in the supplied orientations. The cap formula of [F4] followed by augmentation therefore gives ⟨e(E),[M]⟩=∑x∈Zsgn⁡(Dvsx). If r=0, Z=M has dimension n; its induced orientation is the ambient orientation multiplied by the supplied rank-zero orientation unit o (over Z, o−1=o). Thus [Z]=o[M] and the formula reads e(0M,o)∩[M]=o[M], as [F5] requires. For the zero section the vertical derivative is zero at every base point, so it is surjective exactly when r=0 if the base is nonempty; there are no points to check when the base is empty. Empty manifolds and the canonical mod-two orientations satisfy the same formulas. AC is inherited from the stated suppliers.

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