Timeline for Is inner product preserved only by the stabiliser in a finite reflection group?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2014 at 17:24 | answer | added | Peter Michor | timeline score: 2 | |
| Sep 25, 2014 at 10:08 | comment | added | Peter Michor | The same. For reflection groups the closed chamber is a fundamental domain = the orbit space. | |
| Sep 25, 2014 at 7:56 | comment | added | Violetta | What if they are in the closure of the same chamber? | |
| Sep 25, 2014 at 6:03 | comment | added | Peter Michor | Now yes: Each chamber is a fundamental domain, and each orbit meets the chamber exactly once. Thus $y=z$ since they are in the same orbit, thus the statement is true. | |
| Sep 24, 2014 at 22:37 | history | reopened |
José Figueroa-O'Farrill Ricardo Andrade Derek Holt Stefan Kohl♦ S. Carnahan♦ |
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| Sep 24, 2014 at 21:49 | history | edited | Violetta | CC BY-SA 3.0 |
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| Sep 24, 2014 at 21:48 | comment | added | Violetta | Thank you for the responses. I am sorry I forgot to add that $x$ and $y$ must be in the same chamber. Otherwise this is indeed a bit trivial. | |
| Sep 24, 2014 at 18:32 | comment | added | Peter Michor | No: Take the dihedral group $D_3$ acting on $\mathbb R^2$. You have 6 chambers. Take $y$ in the interior of a chamber, and rotate it to $z$, then $z$ is not in the next chamber, but the second next. Now $x\ne 0$ is perpendicular to $z-y$, so it lies in the interior of a chamber (the one between or its negative), so the stabilizer of $x$ is trivial. | |
| Sep 24, 2014 at 16:32 | comment | added | Derek Holt | No, the site is for research level problems, it is definitely not for asking experts to give hints. On the other hand, I am surprised that nobody has given any indication why they have voted to close, and the question does not seem completely trivial to me, so I have voted to reopen. | |
| Sep 24, 2014 at 15:05 | comment | added | Ryan Budney | I didn't vote to close the question but when I read it I'm a little confused. Your condition $<x,y>=<x,z>$ seems too easy to satisfy to have any conclusion. Can't you just choose any $x$ orthogonal to both $y$ and $z$? Or does $<\cdot, \cdot>$ not represent inner product? | |
| Sep 24, 2014 at 14:44 | review | Reopen votes | |||
| Sep 24, 2014 at 22:37 | |||||
| Sep 24, 2014 at 12:51 | comment | added | Violetta | What is off-topic here? If this is trivial for someone, why not giving a hint where to look for the answer? My co-author and I are genuinely stuck with this, and I thought this is what this web site is for: ask experts in the area for a hint. | |
| Sep 24, 2014 at 12:27 | history | closed |
HJRW Dima Pasechnik Stefan Waldmann Stefan Kohl♦ Chris Godsil |
Not suitable for this site | |
| Sep 24, 2014 at 8:48 | review | Close votes | |||
| Sep 24, 2014 at 12:27 | |||||
| Sep 24, 2014 at 8:32 | history | edited | Violetta | CC BY-SA 3.0 |
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| Sep 24, 2014 at 8:27 | review | First posts | |||
| Sep 24, 2014 at 8:52 | |||||
| Sep 24, 2014 at 8:27 | history | asked | Violetta | CC BY-SA 3.0 |