Timeline for How feasible is it to prove Kazhdan's property (T) by a computer?
Current License: CC BY-SA 4.0
11 events
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| Nov 16, 2022 at 8:47 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Nov 11, 2014 at 9:40 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Nov 11, 2014 at 7:35 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Jun 13, 2014 at 8:08 | comment | added | Andreas Thom | $Aut(F_2)$ and $Aut(F_3)$ are known not to have Kazhdan's property $(T)$. If $Aut(F_n)$ for high $n$ would have Kazhdan's property $(T)$, this would explain why the product replacement algorithm works so well -- this was explained in a paper by Alex Lubotzky and Igor Pak. | |
| Jun 12, 2014 at 14:12 | comment | added | Vladimir | Why in particular $\mathrm{Aut}(F_4)$? is the answer known for $\mathrm{Aut}(F_2)$ and $\mathrm{Aut}(F_3)$? | |
| S May 24, 2014 at 4:48 | history | suggested | ThiKu | CC BY-SA 3.0 |
two Typos corrected
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| May 24, 2014 at 4:30 | review | Suggested edits | |||
| S May 24, 2014 at 4:48 | |||||
| May 22, 2014 at 2:44 | vote | accept | Narutaka OZAWA | ||
| May 22, 2014 at 2:43 | comment | added | Narutaka OZAWA | Fantastic. I accept it as an answer (so unfortunately I cannot accept Speyer's) and wish you good luck on Aut($F_4$)! BTW, I learned sometime ago that semidecidability of property (T) had been observed by Silberman. metric2011.wordpress.com/2011/03/02/… | |
| May 21, 2014 at 16:43 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| May 21, 2014 at 16:30 | history | answered | Andreas Thom | CC BY-SA 3.0 |