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Stable splitting of the tangent bundle of the Milnor disk bundle

Statement

Assume the Axiom of Choice as inherited from the connection and Pontryagin-class suppliers. Let ξ=ξh,j→S4 be the quaternionic clutching bundle, W=D(ξh,j) its disk bundle and π:W→S4 the projection. Then TW⊕ε1≅π∗(ξh,j⊕ε5),p1(TW)=2(h−j)π∗u∈H4(W;Z).

Facts & Assumptions

Given: The clutchings ξ=ξh,j, the disk bundle W=D(ξ) with projection π, and the generator u∈H4(S4;Z).

[A1]

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

[L1]

The vertical tangent bundle of a smooth vector bundle total space is the pullback of the bundle along the projection; restricting to the disk bundle and splitting the tangent sequence by a connection gives TW≅π∗(TS4⊕ξ) (Every smooth vector bundle admits a connection, The Milnor sphere and disk bundles Mh,j and Wh,j).

[L2]

The explicit map (v,t)↦v+tb at a base point of the unit sphere identifies TS4⊕ε1 with the trivial rank-five bundle ε5 (the unit sphere lies in R5 with outward normal b).

[L3]

Assume AC. Pontryagin classes are natural and stable on CW bases (Naturality, stability, and mod-two reduction of Pontryagin classes). On a CW-type base they are defined by transport along a homotopy equivalence (Pontryagin classes by complexification); homotopic pullbacks of numerable bundles are isomorphic (Homotopy invariance of vector-bundle pullback). Here the zero section s:S4→W and projection π are explicit homotopy inverses, using the fibre contraction. Thus p1(E)=π∗p1(s∗E) for a bundle E on W, and stability can be checked on S4.

[L4]

p1(ξh,j)=2(h−j)u under the calibrated clutching conventions (Euler and first Pontryagin classes of ξh,j).

Proof

technique · direct
1.1L1A1given

The tangent sequence 0→π∗ξ→TW→π∗TS4→0 is split by the horizontal lifts of a smooth connection on ξ, so TW≅π∗(TS4⊕ξ).

2.1step 1.1L2

The map (v,t)↦v+tb identifies TS4⊕ε1 with ε5, so adding a trivial line to both sides of step 1.1 gives TW⊕ε1≅π∗(TS4⊕ξ)⊕ε1≅π∗((TS4⊕ε1)⊕ξ)≅π∗(ξ⊕ε5), the first assertion.

3.1step 2.1L3L4∎

Pulling the stable splitting of step 2.1 back along the zero section gives s∗TW⊕ε1≅ξ⊕ε5 on the actual CW sphere. By [L3] and [L4], p1(s∗TW)=p1(ξ)=2(h−j)u. Transporting back through the explicit homotopy equivalence gives p1(TW)=π∗p1(s∗TW)=2(h−j)π∗u.

Depends on

Used by

Dependency tree · two levels

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Sources