Timeline for Cheap, non-constructive, free group generating rotations for Banach-Tarski
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2010 at 4:27 | answer | added | Igor Rivin | timeline score: -1 | |
| Dec 15, 2010 at 0:34 | answer | added | none | timeline score: 0 | |
| Dec 14, 2010 at 22:09 | answer | added | Ian Agol | timeline score: 6 | |
| Dec 14, 2010 at 18:50 | answer | added | John Wiltshire-Gordon | timeline score: 2 | |
| Dec 14, 2010 at 14:32 | answer | added | Bill Thurston | timeline score: 10 | |
| Dec 14, 2010 at 13:35 | history | edited | BS. |
edited tags
|
|
| Dec 14, 2010 at 9:36 | answer | added | BS. | timeline score: 18 | |
| Dec 14, 2010 at 8:22 | answer | added | Qiaochu Yuan | timeline score: 2 | |
| Dec 14, 2010 at 8:13 | comment | added | David Feldman | Does it help to fix two axes, perhaps for simplicity 90 degrees apart and then look only at rotations very close to the identity? I think working up to first order takes care of words that don't become trivial if you add the relation $ab=ba$. Could working up to second order suffice for these? | |
| Dec 14, 2010 at 8:00 | comment | added | Qiaochu Yuan | In particular, the set of pairs of rotations satisfying a given word is Zariski closed, hence measure zero unless it is all of SO(3) x SO(3), so it's enough to exhibit, for each word, a single pair of rotations which does not satisfy it. The discussion at sbseminar.wordpress.com/2007/09/17/… includes several attempts to rule out the second case without writing down a free subgroup, but I don't think they got anywhere. | |
| Dec 14, 2010 at 7:58 | comment | added | Qiaochu Yuan | Related: mathoverflow.net/questions/47585/… | |
| Dec 14, 2010 at 7:31 | history | asked | David Feldman | CC BY-SA 2.5 |