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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMar 22nd 2019

in order to have a good place to record the diagram:

$\array{ ( q_1, q_2 ) &\mapsto& (x \mapsto q_1 \cdot x \cdot \overline{q}_2) \\ Sp(1) \times Sp(1) &\overset{\simeq}{\longrightarrow}& Spin(4) \\ \big\downarrow && \big\downarrow \\ Sp(1)\cdot Sp(1) &\overset{\simeq}{\longrightarrow}& SO(4) }$
• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeApr 8th 2019
• (edited Apr 8th 2019)

just for completeness, I added this statement:

The integral cohomology ring of the classifying space $B SO(4)$ is

$H^\bullet \big( p_1, \chi, W_3 \big) / \big( 2 W_3 \big)$

where

Notice that the cup product of the Euler class with itself is the second Pontryagin class

$\chi \smile \chi \;=\; p_2 \,,$

which therefore, while present, does not appear as a separate generator.

I hope I got this right that $W_5$ does not appear.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeApr 8th 2019
• (edited Apr 8th 2019)

Yes, it seems to me that you only have $W_3$ out of the integral SW classes, based on one of those sources I sent you.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJun 5th 2019

copied over the homotopy groups of $SO(4)$ in low degree

$G$ $\pi_1$ $\pi_2$ $\pi_3$ $\pi_4$ $\pi_5$ $\pi_6$ $\pi_7$ $\pi_8$ $\pi_9$ $\pi_10$ $\pi_11$ $\pi_12$ $\pi_13$ $\pi_14$ $\pi_15$
$SO(4)$ $\mathbb{Z}_2$ 0 $\mathbb{Z}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 2}$ $\mathbb{Z}_{12}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 2}$ $\mathbb{Z}_{3}^{\oplus 2}$ $\mathbb{Z}_{15}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 2}$ $\mathbb{Z}_{2}^{\oplus 4}$ $\mathbb{Z}_2^{\oplus 2}\oplus\mathbb{Z}_{12}^{\oplus 2}$ $\mathbb{Z}_2^{\oplus 4}\oplus\mathbb{Z}_{84}^{\oplus 2}$ $\mathbb{Z}_2^{\oplus 4}$