Tuesday, 14 November 2006

ag.algebraic geometry - Complex Projective Space as a U(1) quotient

My original answer was unsalvegable so I've deleted it and am posting a new "answer". As with the first one, I don't rate this as particularly an answer but more just trying to understand what's going on.



I was initially having trouble understanding Scott's answer, but now I think I do and I think it gives the matrix representation wanted which isn't quite what David wrote.



We have SU(n+1) and inside this we have SU(n) and quotient out to get S2n+1. We also have a slightly larger subgroup which is S(U(n)timesU(1)), which contains SU(n), such that the quotient is mathbbCPn.



Now, S(U(n)timesU(1)) is U(n) via Amapsto(detA1,A) and the inclusion SU(n)toS(U(n)timesU(1)) goes over to the standard inclusion. Here, SU(n) is a normal subgroup and U(n) is the semi-direct product of SU(n) and U(1) with the map U(1)toU(n) given by lambdamapsto(lambda,1,dots,1) (diagonal matrix). When taken over to S(U(n)timesU(1)) this becomes lambdamapsto(lambda1,lambda,1,dots,1).



So then SU(n+1)/S(U(n)timesU(1))cong(SU(n+1)/SU(n))/U(1) where U(1)toSU(n+1) is the map lambdato(lambda1,lambda,1,dots,1).



This isn't the same as David's, I know, so it may not be what you want (since that answer's been accepted). Presumably only one satisfies the condition that you want and presumably it's David's since that answer's been accepted. Still, I was confused and I think I've straightened myself out now.

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