Timeline for Solving equations in SO(3) : an open problem by Jan Mycielski
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2016 at 3:12 | vote | accept | Alexander Belov | ||
| Feb 1, 2016 at 3:12 | vote | accept | Alexander Belov | ||
| Feb 1, 2016 at 3:12 | |||||
| Feb 1, 2016 at 3:10 | vote | accept | Alexander Belov | ||
| Feb 1, 2016 at 3:12 | |||||
| Jan 31, 2016 at 19:26 | comment | added | Vladimir Dotsenko | @AndreasThom - I seem to have misread it, sorry. I am glad though that me mentioning your paper somehow it brought this thread to an answer :) | |
| Jan 31, 2016 at 15:18 | comment | added | Fedor Petrov | I would replace "by conjugacy and continuity" to "by conjugacy and continuity and compactness", to resolve the possible answer $[0,\alpha)$. And a naive question: are there examples of similar questions on non-compact groups with not closed answer, like 'rotations by angle different from $\pi$'? | |
| Jan 31, 2016 at 14:39 | answer | added | Andreas Thom | timeline score: 14 | |
| Jan 31, 2016 at 14:24 | comment | added | Andreas Thom | @VladimirDotsenko: Why do you think there are more conjectures than actual results? | |
| Jan 31, 2016 at 14:23 | comment | added | Andreas Thom | Thanks for posting. I was not aware that this was asked by Mycielski. | |
| Jan 31, 2016 at 13:59 | answer | added | Ash Malyshev | timeline score: 4 | |
| Jan 28, 2016 at 9:48 | comment | added | Alexander Belov | @AntonMalyshev : It seems you are right! Thank you a lot, and also YCor and VladimirDotsenko! If you post your comment as an answer, I will accept it. | |
| Jan 27, 2016 at 20:20 | comment | added | Ash Malyshev | Corollary 3.3 in the paper Vladimir linked (arxiv.org/abs/1003.4093) appears to answer the question: $\alpha$ can be arbitrarily small. | |
| Jan 27, 2016 at 13:07 | comment | added | user35593 | If $X$ is a rotation with angle $\beta\in (0,\pi/2)$ then $X^{\lceil \pi/(2\beta) \rceil}$ is a rotation with angle at least $\pi/2$ so $\alpha$ is at least $\pi/2$. | |
| Jan 27, 2016 at 10:58 | comment | added | Vladimir Dotsenko | @YCor : yes you're right of course. In the MO question you link, there is a paper of Andreas Thom mentioned in one of the comments (arxiv.org/abs/1003.4093), discussion of Remark 3.6 in that paper is relevant for the case of SO(3), and shows that probably there are more conjectures than actual results for this question. | |
| Jan 27, 2016 at 10:12 | history | edited | YCor |
edited tags
|
|
| Jan 27, 2016 at 10:11 | comment | added | YCor | @VladimirDotsenko Borel's result does not help, since for every $\alpha>0$ the set of rotations of angle in $[0,\alpha]$ is Zariski dense. | |
| Jan 27, 2016 at 10:04 | comment | added | YCor | Related: mathoverflow.net/questions/45483/…. The key word is "word map". | |
| Jan 27, 2016 at 8:45 | comment | added | Vladimir Dotsenko | If I remember correctly, Borel proved for any connected semisimple algebraic group that the image of this map is Zariski dense. Would not that settle your question, more or less? | |
| Jan 27, 2016 at 6:42 | comment | added | Peter McNamara | @Watson, the interesting case is when the p's sum to zero and the q's sum to zero. | |
| Jan 27, 2016 at 6:01 | comment | added | Watson Ladd | Why doesn't $X$=identity, $Y$=rotation by $alpha/(q_1+q_2+\ldots+q_m)$ give any rotation we desire? | |
| Jan 27, 2016 at 4:58 | history | asked | Alexander Belov | CC BY-SA 3.0 |