Timeline for Decomposition of $S^7=\operatorname{Spin}(7)/G_2$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2021 at 11:52 | answer | added | Andrei Smilga | timeline score: 1 | |
| Jun 28, 2021 at 10:20 | history | edited | YCor | CC BY-SA 4.0 |
added tags, formatting
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| May 19, 2014 at 4:38 | comment | added | Oliver Jones | @Andre: I'm making no such assumption. I'm not even approaching the problem from the bundle point of view. As I already mentioned, the result I'm using is in Helgason and it uses the exponential map, not bundles. Results of this type are generally not easy; see my earlier post on Stiefel manifolds: <mathoverflow.net/questions/139542>. However, in the case of $S^7$, I wonder if the octonionic structure doesn't help. | |
| May 19, 2014 at 4:19 | comment | added | André Henriques | You're also implicitly assuming that there is a preferred way of trivializing the bundle $Spin(7)\to G_2\to S^7$ in a neighborhood of a given point $p\in S^7$. My guess is that there are many ways of locally trivializing that bundle, and that they are all pretty ugly to write down. | |
| May 19, 2014 at 3:06 | answer | added | José Figueroa-O'Farrill | timeline score: 8 | |
| May 19, 2014 at 2:33 | comment | added | Oliver Jones | @Ryan: I'm being sloppy; all these equalities are potentially just holding in a neighborhood of some chosen point of the manifold. They can be global however. I just don't know for $S^7$. | |
| May 19, 2014 at 2:30 | comment | added | Ryan Budney | The statement $SO(3) = SO(2) \times S^2$ isn't a local statement. What kind of "local" are you talking about? If you mean in the sense of fibre bundles, yes there is a fibre bundle, but it is not trivial (as your notation presumes). | |
| May 19, 2014 at 2:29 | comment | added | Oliver Jones | @Ryan: You're saying that that isn't true locally in some neighborhood? | |
| May 19, 2014 at 2:29 | comment | added | Ryan Budney | But that's wrong, Oliver. $SO(3)$ is not the product of $SO(2)$ and $S^2$. The homotopy groups don't work out. It's a non-trivial fibre bundle, $SO(2) \to SO(3) \to S^2$ | |
| May 19, 2014 at 2:27 | comment | added | Oliver Jones | @Andre: A result in Helgason says that such a decomposition exists for reductive spaces; at least locally. In the case of $S^2$ we have $SO(3)=SO(2)\times S^2$. | |
| May 19, 2014 at 2:09 | comment | added | André Henriques | What makes you think that $Spin(7)=G_2\times S^7$? Think of the Hopf fibration $S^2=S^3/S^1$: would you then conclude that $S^3=S^2\times S^1$? | |
| May 19, 2014 at 1:57 | history | asked | Oliver Jones | CC BY-SA 3.0 |