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The Thom class of a disk bundle pairs with the base generator to one

Statement

Assume the Axiom of Choice as inherited from the Thom, normal-Thom and duality suppliers. Let $ξ \to S^{4}$ be an oriented rank-four real bundle with Euler number $ε = \pm 1$ with respect to the base generator $u \in H^{4} (S^{4}; Z)$ , let $W = D (ξ)$ , $M = S (ξ) = \partial W$ , $x = π^{*} u$ and let $U \in H^{4} (W, M; Z)$ be the oriented Thom generator in the normalization of Thom isomorphism for oriented vector bundles. With the total orientation base followed by fibre and the induced boundary orientation, the relative evaluation satisfies $⟨ U ⌣ x, [W, M] ⟩ = 1.$

Facts & Assumptions

Given: The oriented rank-four bundle $ξ$ over $S^{4}$ with Euler number $ε = \pm 1$ , its disk and sphere bundles $W = D (ξ)$ , $M = S (ξ)$ , the projection $π$ , the zero section $s$ , the base generator $u$ , the class $x = π^{*} u$ , and the normalized Thom generator $U$ .

[A1]

The Axiom of Choice is assumed (The Axiom of Choice).

[L1]

The Thom isomorphism gives $H^{4} (W, M; Z) = Z \cdot U$ and $H^{4} (W; Z) = Z \cdot x$ with the stated normalization (Thom isomorphism for oriented vector bundles).

[L2]

The Euler class is $e (ξ) = s^{*} j^{*} (U)$ ; since $e (ξ) = ε u$ and $s^{*} π^{*} = i d$ , the absolute class $j (U)$ satisfies $j (U) = ε x$ (Euler class by zero-section pullback of the Thom class).

[L3]

Assume AC. For a closed embedded oriented four-manifold $Z$ in a closed oriented eight-manifold $N$ , with its normal bundle $ν$ oriented in tangent-first order, the absolute image of the normal Thom class under tubular excision is the Poincare dual of $[Z]$ : the supplier's shuffle sign is $(- 1)^{4 \cdot 4} = + 1$ (The normal Thom class realizes the Poincare dual of a closed submanifold).

[L4]

Assume ${A C}_{ω}$ . For fillings $W, W^{'}$ glued along an orientation-preserving boundary identification into $N = W \cup_{M} (- W^{'})$ , there are excision isomorphisms and evaluation comparisons carrying $[N]$ to $[W, M]$ on the first side and to $- [W^{'}, M]$ on the second (Collared gluing has relative excision and evaluation maps).

[L5]

Relative cap and cup evaluation satisfy $⟨ a ⌣ y, [W, M] ⟩ = ⟨ y, a \cap [W, M] ⟩$ (Relative cap and cup evaluation identity).

Proof

technique · direct

1.1L4A1given

Double $W$ along its collar and write $N = W \cup_{M} (- W)$ ; by [L4] with $W^{'} = W$ there are the excision isomorphisms and the evaluation comparison for $N$ , and the zero section $Z = s (S^{4})$ is a closed oriented embedded four-sphere in $N$ with normal bundle $ξ$ .

2.1step 1.1L3L4

Let $E_{W} : H^{4} (W, M; Z) \to H^{4} (N, V; Z)$ be the inverse of the first excision isomorphism of [L4]; by [L3] the absolute image of $E_{W} (U)$ in $H^{4} (N; Z)$ is the Poincare dual of $Z$ , and the evaluation comparison of [L4] identifies $⟨ j_{V} E_{W} (U) ⌣ Y, [N] ⟩$ with $⟨ U ⌣ y, [W, M] ⟩$ whenever $Y$ restricts to $y$ on the first side.

3.1step 2.1L1L2

By [L2] the class $j (U) = ε x$ with $ε = \pm 1$ ; because [L1] says $j$ is an isomorphism of infinite cyclic groups up to the sign $ε$ , the class $x$ has the unique relative lift $ε U$ , whose restriction to the zero section is $u$ : indeed $s^{*} π^{*} u = u$ and $s^{*} j^{*} (ε U) = ε \cdot ε u = u$ .

4.1step 2.1step 3.1L3L5

Let $X$ be the absolute class on $N$ obtained by extending $x$ from the first side and zero from the second side via the inverse excision map, as in [L4]; its restriction to the zero section is $u$ , and the closed cap/evaluation identity of [L5] applied to $N$ and the Poincare-dual identification of step 2.1 give $⟨ j_{V} E_{W} (U) ⌣ X, [N] ⟩ = ⟨ s^{*} X, [S^{4}] ⟩ = ⟨ u, [S^{4}] ⟩ = 1$ .

5.1step 4.1L4L5∎

The evaluation comparison of step 2.1 identifies the left-hand side with $⟨ U ⌣ x, [W, M] ⟩$ , because $X$ restricts to $x$ on the first side; hence $⟨ U ⌣ x, [W, M] ⟩ = 1$ , with the orientation signs checked by the rank-four base/fibre block swap being positive and by the induced boundary orientation of $M$ .

Depends on

Used by

Dependency tree · two levels

56 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

John Milnor, On Manifolds Homeomorphic to the 7-Sphere, Annals of Mathematics 64 (1956), 399-405 (standard reference, not scraped)
Allen Hatcher, Algebraic Topology, Cambridge University Press 2002 (complete book) (standard reference, not scraped)