Timeline for Perfect quotients of braid groups
Current License: CC BY-SA 4.0
11 events
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Jul 2, 2023 at 23:39 | comment | added | Ian Agol | @IanGershonTeixeira Good point. | |
Jun 30, 2023 at 2:01 | comment | added | Ian Gershon Teixeira | Also perhaps this is obvious but you don't need to reduce the problem further from $ Sp(2m,p) $ to $ PSp(2m,p) $ since $ Sp(2m,p) $ is already perfect. Indeed it is even quasisimple groupprops.subwiki.org/wiki/Symplectic_group_is_quasisimple . The exceptions are the same for both families: $ Sp(2,2)=PSp(2,2) \cong S_3, Sp(4,2)=PSp(4,2)\cong S_6 $ and $ Sp(2,3)\cong 2.A_4, PSp(2,3)\cong A_4 $ are not perfect. | |
Jun 28, 2023 at 8:31 | comment | added | Ian Gershon Teixeira | Lovely! This is exactly the sort of thing I was hoping for. I've accepted your answer. I've also edited my question with an update expanding on the specifics of your answer, to the extent that I understand them. Could you explain more about how the value of $ m $ is determined by $ n $? For example does every $ m $ show up for some $ B_n $? Do the values of $ m $ increase strictly with increasing $ n $? | |
Jun 28, 2023 at 5:58 | vote | accept | Ian Gershon Teixeira | ||
Jun 25, 2023 at 13:50 | history | edited | Ian Agol | CC BY-SA 4.0 |
Added an explicit reference answering the question.
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Jun 25, 2023 at 8:27 | comment | added | Ian Gershon Teixeira | Does this generalize the $ B_3 $ case in some simple minded way like every $ PSL(n-1,p) $ is a quotient of $ B_n $? Or is perhaps some even stronger result true like $ SL(n-1,\mathbb{Z}) $ is always a quotient of $ B_n $? | |
Jun 24, 2023 at 22:15 | history | undeleted | Ian Agol | ||
Jun 24, 2023 at 22:15 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 182 characters in body
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Jun 24, 2023 at 10:23 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 235 characters in body
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Jun 24, 2023 at 9:21 | history | deleted | Ian Agol | via Vote | |
Jun 24, 2023 at 9:20 | history | answered | Ian Agol | CC BY-SA 4.0 |