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Euler and first Pontryagin classes of ξh,j

Statement

Assume the Axiom of Choice as inherited from the characteristic-class suppliers. Under the fixed quaternionic, base and upper-to-lower clutching orientations of Quaternionic clutching bundles ξh,j over S4, the bundle ξh,j has e(ξh,j)=(h+j)u,p1(ξh,j)=2(h−j)u∈H4(S4;Z), where u is the positive base generator.

Facts & Assumptions

Given: Integers h,j, the bundle ξh,j and the generator u∈H4(S4;Z) with ⟨u,[S4]⟩=1.

[A1]

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

[L1]

The clutching map satisfies gh,j(a)v=ahvaj, so for h,j≥0 it is the pointwise product of h copies of g1,0 and j copies of g0,1; for negative exponents the same holds with the corresponding inverse maps, and g−1,0=g1,0−1, g0,−1=g0,1−1 (Quaternionic clutching bundles ξh,j over S4).

[L2]

The Euler and first Pontryagin evaluations of the bundle clutched by a pointwise product are the sums of the evaluations of the factors, and inversion negates them (Degree-four characteristic evaluations add under the clutching product).

[L3]

The basic left and right bundles satisfy e=u and p1=+2u (left) and e=u, p1=−2u (right) (Pontryagin calibration of the basic quaternionic clutchings).

[L4]

Evaluation against the fundamental class is additive and, since H4(S4;Z)=Z⋅u with ⟨u,[S4]⟩=1, determines a degree-four class (Kronecker evaluation pairing).

Proof

technique · direct
1.1L1L2L3A1

For h,j≥0, [L1] writes gh,j as the pointwise product of h copies of g1,0 and j copies of g0,1; applying [L2] inductively with the basic values [L3] gives ⟨e(ξh,j),[S4]⟩=h⋅1+j⋅1=h+j and ⟨p1(ξh,j),[S4]⟩=h⋅2+j⋅(−2)=2(h−j).

2.1step 1.1L1L2

For arbitrary integers h,j, write the clutching as the same product with the inverse maps for the negative exponents; the inverse clause of [L2] negates both evaluations, so the displayed evaluations remain ⟨e(ξh,j),[S4]⟩=h+j and ⟨p1(ξh,j),[S4]⟩=2(h−j).

3.1step 2.1L4∎

By [L4] a degree-four integral class on S4 is determined by its evaluation on [S4], and ⟨(h+j)u,[S4]⟩=h+j, ⟨2(h−j)u,[S4]⟩=2(h−j); comparing with step 2.1 gives e(ξh,j)=(h+j)u and p1(ξh,j)=2(h−j)u, as asserted.

Depends on

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