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gave this its own little entry in order to have a place where to record the following fact:
The integral cohomology ring of the classifying space of Spin(3) is freely generated from 1/4th of the first Pontryagin class:
H•(BSpin(3),ℤ)≃ℤ[14p1]Moreover, the integral cohomology ring of the classifying space of Spin(4) is freely generated from the first fractional Pontryagin class 12p1 and the combination 12(e+12p1), where χ is the Euler class:
H•(BSpin(4),ℤ)≃ℤ[12p1,12(χ+12p1)]Finally, under the exceptional isomorphism (eq:Spin3SquareToO4) ϑ:Spin(3)×Spin(3)≃→Spin(4) these classes are related by
ϑ*(12p1)=−14p1⊗1+1⊗14p1ϑ*(12(χ+12p1))=−1⊗14p1+1⊗14p1henceAAAAϑ*(χ)+12p1))=−14p1⊗1+1⊗14p1Therefore, under the canonical diagonal inclusion ι:Spin(3)Δ↪Spin(3)×Spin(3)≃Spin(4) (eq:Spin3Diagonally) we have
ι*(12p1)=12p1ι*(χ)=0Is e=χ?
Thanks for catching. Fixed now.
Question:
Might the second generator in
H•(BSpin(4),ℤ)≃ℤ[12p1,12(χ+12p1)]actually come from BSO(4)?
In other words, is the rational combination 12(χ+12p1) possibly already integral for SO(4)-bundles, not just for Spin(4)-bundles?
[edit: ah, no, this cannot be…]
added pointer to:
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