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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 $1/4$th of the first Pontryagin class:
$H^\bullet \big( B Spin(3), \mathbb{Z} \big) \;\simeq\; \mathbb{Z} \big[ \tfrac{1}{4}p_1 \big]$Moreover, the integral cohomology ring of the classifying space of Spin(4) is freely generated from the first fractional Pontryagin class $\tfrac{1}{2}p_1$ and the combination $\tfrac{1}{2}\big( e + \tfrac{1}{2}p_1 \big)$, where $\chi$ is the Euler class:
$H^\bullet \big( B Spin(4), \mathbb{Z} \big) \;\simeq\; \mathbb{Z} \big[ \tfrac{1}{2}p_1 \,, \tfrac{1}{2}\big( \chi+ \tfrac{1}{2}p_1 \big) \big]$Finally, under the exceptional isomorphism (eq:Spin3SquareToO4) $\vartheta \;\colon\; Spin(3) \times Spin(3) \overset{\simeq}{\to} Spin(4)$ these classes are related by
$\begin{aligned} \vartheta^\ast \left( \tfrac{1}{2}p_1 \right) & = \phantom{-} \tfrac{1}{4}p_1 \otimes 1 + 1 \otimes \tfrac{1}{4} p_1 \\ \vartheta^\ast \Big( \tfrac{1}{2} \big( \chi + \tfrac{1}{2} p_1 \big) \Big) & = \phantom{-} \phantom{ 1 \otimes \tfrac{1}{4}p_1 + } 1 \otimes \tfrac{1}{4}p_1 \\ \text{hence} \phantom{AAAA} \vartheta^\ast\big( \chi \big) \phantom{ + \tfrac{1}{2} p_1 \big) \Big) } & = - \tfrac{1}{4}p_1 \otimes 1 + 1 \otimes \tfrac{1}{4} p_1 \end{aligned}$Therefore, under the canonical diagonal inclusion $\iota \colon Spin(3) \overset{\Delta}{\hookrightarrow} Spin(3) \times Spin(3) \simeq Spin(4)$ (eq:Spin3Diagonally) we have
$\begin{aligned} \iota^\ast \left( \tfrac{1}{2}p_1 \right) & = \tfrac{1}{2}p_1 \\ \iota^\ast \big( \chi \big) & = 0 \end{aligned}$Is $e=\chi$?
Thanks for catching. Fixed now.
Question:
Might the second generator in
$H^\bullet \big( B Spin(4), \mathbb{Z} \big) \;\simeq\; \mathbb{Z} \big[ \tfrac{1}{2}p_1 \,, \tfrac{1}{2}\big( \chi+ \tfrac{1}{2}p_1 \big) \big]$actually come from $B SO(4)$?
In other words, is the rational combination $\tfrac{1}{2}\big( \chi+ \tfrac{1}{2}p_1 \big)$ possibly already integral for $SO(4)$-bundles, not just for $Spin(4)$-bundles?
[edit: ah, no, this cannot be…]
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