Timeline for Are there good product rules on the $k$-sphere?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2011 at 17:36 | comment | added | rpotrie | Hi, S7 is not a Lie group because the octonions (which are the ones which induce the $H$-space structure) lack of associativity. | |
| Feb 4, 2011 at 14:11 | vote | accept | Mircea | ||
| Feb 4, 2011 at 14:09 | vote | accept | Mircea | ||
| Feb 4, 2011 at 14:11 | |||||
| Feb 4, 2011 at 14:09 | comment | added | Mircea | dear rpotrie, thanks for the answer and the link. I didn't want to have necessarily Lie groups, but I guess some two-sided identity is quite natural. Why is $S^7$ not a Lie group? | |
| Jul 23, 2010 at 18:24 | comment | added | Greg Kuperberg | Right, the H-space version of the result is extremely convincing. The product law $P$ in the question is "good" if it is (a) continuous, and (b) has an element $1$ such that $a1 = 1a = a$. Or even such that $a \mapsto 1a$ and $a \mapsto a1$ are homotopic to the identity. That's not asking for a whole lot, but it already restricts you to the four examples, up to homotopy. | |
| Jul 23, 2010 at 12:55 | history | edited | rpotrie | CC BY-SA 2.5 |
added 183 characters in body
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| Jul 23, 2010 at 12:53 | comment | added | rpotrie | Yes, sorry, I will correct that. | |
| Jul 23, 2010 at 12:47 | comment | added | Gjergji Zaimi | $S^7$ is not a Lie group, did you mean H-space? en.wikipedia.org/wiki/H-space | |
| Jul 23, 2010 at 12:32 | history | answered | rpotrie | CC BY-SA 2.5 |