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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeMar 6th 2019

    Page created, but author did not leave any comments.

    v1, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMar 6th 2019
    • (edited Mar 6th 2019)

    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 1/41/4th of the first Pontryagin class:

    H (BSpin(3),)[14p 1] 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 12p 1\tfrac{1}{2}p_1 and the combination 12(e+12p 1)\tfrac{1}{2}\big( e + \tfrac{1}{2}p_1 \big), where χ\chi is the Euler class:

    H (BSpin(4),)[12p 1,12(χ+12p 1)] 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) ϑ:Spin(3)×Spin(3)Spin(4)\vartheta \;\colon\; Spin(3) \times Spin(3) \overset{\simeq}{\to} Spin(4) these classes are related by

    ϑ *(12p 1) =14p 11+114p 1 ϑ *(12(χ+12p 1)) =114p 1+114p 1 henceAAAAϑ *(χ)+12p 1)) =14p 11+114p 1 \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 ι:Spin(3)ΔSpin(3)×Spin(3)Spin(4)\iota \colon Spin(3) \overset{\Delta}{\hookrightarrow} Spin(3) \times Spin(3) \simeq Spin(4) (eq:Spin3Diagonally) we have

    ι *(12p 1) =12p 1 ι *(χ) =0 \begin{aligned} \iota^\ast \left( \tfrac{1}{2}p_1 \right) & = \tfrac{1}{2}p_1 \\ \iota^\ast \big( \chi \big) & = 0 \end{aligned}

    v1, current

    • CommentRowNumber3.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 6th 2019

    Is e=χe=\chi?

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeMar 6th 2019

    Thanks for catching. Fixed now.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeMar 22nd 2019
    • (edited Mar 22nd 2019)

    Question:

    Might the second generator in

    H (BSpin(4),)[12p 1,12(χ+12p 1)] 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 BSO(4)B SO(4)?

    In other words, is the rational combination 12(χ+12p 1)\tfrac{1}{2}\big( \chi+ \tfrac{1}{2}p_1 \big) possibly already integral for SO(4)SO(4)-bundles, not just for Spin(4)Spin(4)-bundles?

    [edit: ah, no, this cannot be…]

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2019

    added picture of D3 and A3 Dynkin diagrams with their central node removed, illustrating the exceptional iso Spin(4)SU(2)×SU(2)Spin(4) \simeq SU(2)\times SU(2).

    diff, v12, current

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