Timeline for Cheap, non-constructive, free group generating rotations for Banach-Tarski
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2010 at 2:08 | history | edited | John Wiltshire-Gordon | CC BY-SA 2.5 |
added 2 characters in body
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| Dec 14, 2010 at 23:59 | comment | added | John Wiltshire-Gordon | @David Feldman: I hope my latest edit clarifies "corresponding matrices." | |
| Dec 14, 2010 at 23:57 | history | edited | John Wiltshire-Gordon | CC BY-SA 2.5 |
added 479 characters in body
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| Dec 14, 2010 at 22:53 | comment | added | David Feldman | I can't figure out what you mean by "corresponding matrices." The matrices you write down aren't even invertible, so they don't live in a matrix group. | |
| Dec 14, 2010 at 21:11 | comment | added | John Wiltshire-Gordon | @David: Thanks for fixing my TeX. @Scott: I added some more of what I have in mind. | |
| Dec 14, 2010 at 21:09 | history | edited | John Wiltshire-Gordon | CC BY-SA 2.5 |
added 303 characters in body
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| Dec 14, 2010 at 20:04 | comment | added | S. Carnahan♦ | I'm afraid I don't understand. What does a pair of 3-by-3 matrices over a field with 9 elements do for you in terms of free groups? It seems more likely you want a quadratic unramified extension of the 3-adic field $\mathbb{Q}_3$, so you can identify the matrices with hyperbolic isometries. | |
| Dec 14, 2010 at 19:11 | comment | added | David E Speyer | TeX fixed. Plus some more characters. | |
| Dec 14, 2010 at 19:11 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 14, 2010 at 18:57 | comment | added | John Wiltshire-Gordon | Sorry, I answered what I assumed the question to be. Also, my matrices are not showing up properly. | |
| Dec 14, 2010 at 18:50 | history | answered | John Wiltshire-Gordon | CC BY-SA 2.5 |