Timeline for Generator of $\pi_3(SU(4))$ in Mimura-Toda
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2014 at 1:14 | vote | accept | David Roberts♦ | ||
| Oct 3, 2014 at 1:14 | answer | added | David Roberts♦ | timeline score: 1 | |
| May 29, 2014 at 23:00 | comment | added | David Roberts♦ | @AllenKnutson yep, I got that. | |
| May 29, 2014 at 14:49 | comment | added | Allen Knutson | In both your maps $i$ you have $SU(2) \to G \to G/SU(2)$ where the target is $3$-connected. Hence both are giving isomorphisms $\pi_3(SU(2)) \to \pi_3(G)$. This is a slightly more specific version of Neil's answer. | |
| May 29, 2014 at 8:28 | comment | added | David Roberts♦ | @NeilStrickland - ah, that was much easier than the Lie theoretic argument. I'm happy to not think too much about ±1. | |
| May 29, 2014 at 8:26 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
typo
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| May 29, 2014 at 8:05 | comment | added | Neil Strickland | The standard inclusions $SU(n)\to SU(n+1)$ induce isomorphisms on $\pi_3$ for $n\geq 3$, as one can see from the long exact sequence of the fibration $SU(n)\to SU(n+1)\to S^{2n+1}$. Thus, if you take either generator of $\pi_3(SU(2))\simeq\mathbb{Z}$ and apply $i_*$ to it, you will get a generator for $\pi_3(SU(n))$. Does this answer your question, or are you concerned about the $\pm$-sign? | |
| May 29, 2014 at 6:50 | history | asked | David Roberts♦ | CC BY-SA 3.0 |