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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 6th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexPage created, but author did not leave any comments. <a href="https://ncatlab.org/nlab/revision/Spin%284%29/1">v1</a>, <a href="https://ncatlab.org/nlab/show/Spin%284%29">current</a>

   Page created, but author did not leave any comments.

   v1, current

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 6th 2019
   • (edited Mar 7th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexgave this its own little entry in order to have a place where to record the following fact: *** The [[integral cohomology|integral]] [[cohomology ring]] of the [[classifying space]] of [[Spin(3)]] is [[polynomial ring|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|integral]] [[cohomology ring]] of the [[classifying space]] of [[Spin(4)]] is [[polynomial ring|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 $$ \begin{aligned} \iota^\ast \left( \tfrac{1}{2}p_1 \right) & = \tfrac{1}{2}p_1 \\ \iota^\ast \big( \chi \big) & = 0 \end{aligned} $$ <a href="https://ncatlab.org/nlab/revision/Spin%284%29/1">v1</a>, <a href="https://ncatlab.org/nlab/show/Spin%284%29">current</a>

   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

   $$\begin{aligned} \iota^\ast \left( \tfrac{1}{2}p_1 \right) & = \tfrac{1}{2}p_1 \\ \iota^\ast \big( \chi \big) & = 0 \end{aligned}$$

   v1, current

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 6th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexIs $e=\chi$?

   Is $e=\chi$?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 7th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks for catching. Fixed now.

   Thanks for catching. Fixed now.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMar 22nd 2019
   • (edited Mar 22nd 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexQuestion: 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...]

   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…]

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeAug 29th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexadded picture of D3 and A3 Dynkin diagrams with their central node removed, illustrating the exceptional iso $Spin(4) \simeq SU(2)\times SU(2)$. <a href="https://ncatlab.org/nlab/revision/diff/Spin%284%29/12">diff</a>, <a href="https://ncatlab.org/nlab/revision/Spin%284%29/12">v12</a>, <a href="https://ncatlab.org/nlab/show/Spin%284%29">current</a>

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

   diff, v12, current

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJun 19th 2021
   • PermaLink
   Author: Urs
   Format: MarkdownItexadded pointer to: * [[Marcel Berger]], Section 8.9 of: *Geometry I*, Springer 1987 ([doi:10.1007/978-3-540-93815-6](https://doi.org/10.1007/978-3-540-93815-6)) <a href="https://ncatlab.org/nlab/revision/diff/Spin%284%29/13">diff</a>, <a href="https://ncatlab.org/nlab/revision/Spin%284%29/13">v13</a>, <a href="https://ncatlab.org/nlab/show/Spin%284%29">current</a>

   added pointer to:

   • Marcel Berger, Section 8.9 of: Geometry I, Springer 1987 (doi:10.1007/978-3-540-93815-6)

   diff, v13, current