Skip to main content

Timeline for Perfect quotients of braid groups

Current License: CC BY-SA 4.0

11 events
when toggle format what by license comment
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.
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
Jun 24, 2023 at 10:23 history edited Ian Agol CC BY-SA 4.0
added 235 characters in body
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