Timeline for Double coset formulas for Orthogonal groups [Solved]
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jun 24, 2015 at 18:27 | vote | accept | user43326 | ||
| Dec 10, 2013 at 16:38 | answer | added | user43326 | timeline score: 2 | |
| Dec 9, 2013 at 18:46 | comment | added | András Bátkai | You should post this as an answer. | |
| Dec 9, 2013 at 18:00 | history | edited | user43326 | CC BY-SA 3.0 |
added 229 characters in body
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| Dec 7, 2013 at 16:18 | history | edited | user43326 | CC BY-SA 3.0 |
Marked as solved, with the solution.
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| Dec 2, 2013 at 8:48 | comment | added | user76758 | @user43326: I know almost nothing about modern homotopy theory, but I've seen other settings where a global cohomological computation encounters complications because certain signs are merely "locally constant", and so I wondered if the disconnectedness of orthogonal groups might be contributing discrepancies governed by local signs and not a single overall global sign. But this is just pure speculation, so feel free to ignore it (or maybe Oscar R-W sees something about it). | |
| Dec 2, 2013 at 7:48 | comment | added | user43326 | @user76758: Maybe I misunderstood your comment. So basically what I am saying is that there are contradictions whatever the choice of signs are, which are only "locally constant". | |
| Nov 30, 2013 at 6:58 | history | edited | user43326 | CC BY-SA 3.0 |
Added a second argument showing $2id\neq 2Switch$.
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| Nov 28, 2013 at 11:16 | comment | added | user43326 | @OscarRandal-Williams I edited the question, added a paragraph explaining why $2\tau \neq 2⋅Id$. | |
| Nov 28, 2013 at 11:05 | history | edited | user43326 | CC BY-SA 3.0 |
Removed the markar [Solved], added the reason why $2(id-Switch)$ is non-null
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| Nov 28, 2013 at 10:34 | history | edited | user43326 | CC BY-SA 3.0 |
Marked as solved
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| Nov 27, 2013 at 17:09 | comment | added | Oscar Randal-Williams | The issue would be resolved if $Id - \tau = \tau - Id$ (so $2\tau = 2\cdot Id$), where $\tau$ is the switch map. I suspect that this is true, but I don't see an independent way to prove it. | |
| Nov 26, 2013 at 20:37 | comment | added | user43326 | @user76758 The connectedness of the group is irrelevant, we have the double coset formula for discrete groups as well. Besides, the classifying space is connected. | |
| Nov 26, 2013 at 20:00 | review | First posts | |||
| Nov 26, 2013 at 20:00 | |||||
| Nov 26, 2013 at 20:00 | comment | added | user76758 | How do you know that you have "globally constant" signs when working with a disconnected group? | |
| Nov 26, 2013 at 19:43 | history | asked | user43326 | CC BY-SA 3.0 |