Timeline for The first unstable homotopy group of $Sp(n)$
Current License: CC BY-SA 4.0
15 events
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| Nov 9, 2018 at 22:08 | comment | added | Jason DeVito - on hiatus | I just wanted to add that $\pi_{2k}(G)\otimes \mathbb{Q} = 0$ for any $k> 0$ and any Lie group $G$. To see this, one can reduce to connected and compact groups using the fact that a non-compact connected Lie group is topologically a product of $\mathbb{R}^n$ with a compact Lie group. Every compact Lie group is know to have the rational homotopy type of a product of spheres odd dimension (and we know exactly what spheres appear for each simple group), so we know how to compute the rational homotopy groups of any Lie group. | |
| May 16, 2018 at 15:43 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Mar 28, 2018 at 11:10 | vote | accept | Michael Albanese | ||
| Mar 27, 2018 at 12:10 | comment | added | Tyrone | The metastable groups $\pi_{4n+i}Sp(n)$, $i=2,\dots, 8$ are stated, with references, in Mimura's article "Homotopy Theory of Lie Groups", which appears in the Handbook of Algebraic Topology. I'm also aware that Kaoru Morisugi looked at the 2-primary homotopy of $Sp(n)$ above the metastable range, giving (pieces of) information up to $\pi_{4n+15}Sp(n)$ and $\pi_{8n+4}Sp(n)$. | |
| Mar 27, 2018 at 0:40 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
Added link to the nLab page 'orthogonal group' as referenced by the OP
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| Mar 26, 2018 at 16:01 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Mar 24, 2018 at 23:32 | comment | added | David Roberts♦ | Please feel free to extend the nLab table, Michael. I think it's important to have such data freely available, rather than locked away in the EDM, or scattered across multiple papers. | |
| Mar 24, 2018 at 21:16 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Mar 24, 2018 at 19:41 | answer | added | Matthias Wendt | timeline score: 24 | |
| Mar 24, 2018 at 19:39 | comment | added | mme | Follow the boundary map in the edge case. We have $\pi_{4n+3}\text{Sp}(n+1) \to \Bbb Z \to \pi_{4n+2} \text{Sp}(n) \to \pi_{4n+2} \text{Sp}(n+1)$. The other terms are in the computable range: this reduces to $\Bbb Z \to \Bbb Z \to G \to 0$. In particular $G$ is cyclic and computing its order is equivalent to computing the image of that boundary map. | |
| Mar 24, 2018 at 19:35 | answer | added | Will Sawin | timeline score: 27 | |
| Mar 24, 2018 at 19:27 | comment | added | Michael Albanese | @WillSawin: Yeah, I just edited the question to include a similar remark just before you posted your comment. Maybe I will ask about this as another question later. | |
| Mar 24, 2018 at 19:25 | comment | added | Will Sawin | If you tensor the $\pi_{n-1} (SO(n))$ with $\mathbb Q$, it is $4$-periodic, and there is no torsion for $4n$, at least in your data. So there may be some degree of pattern here. | |
| Mar 24, 2018 at 19:22 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Mar 24, 2018 at 19:04 | history | asked | Michael Albanese | CC BY-SA 3.0 |