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(detA−1,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(lambda−1,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(lambda−1,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.
No comments:
Post a Comment