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Question about group action.

  1. Jul 9, 2010 #1

    quasar987

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    Can anyone explain to me why the action of SU(n) on U(n)/O(n) is transitive? Thanks.
     
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  3. Jul 9, 2010 #2

    Office_Shredder

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    It seems like that shouldn't be true. Given [tex]A \in U[/tex], the determinant of every element of [tex]A*O(n)[/tex] will be [tex] \pm det(A)[/tex]. Multiplying by an element of [tex]SU[/tex] shouldn't change the determinants, which means that you can't get matrices of every unit determinant by acting via the special unitary group.

    I don't know, maybe I'm just confused. Is this definitely fact?
     
  4. Jul 10, 2010 #3

    quasar987

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    Eeeck, did I write that? Terribly sorry, I am tired.

    I meant to ask:

    Can anyone explain to me why the action of SU(n) on the subgroup of U(n)/O(n) consisting of the classes UO(n) with det(U)=[itex]\pm 1[/itex] is transitive?

    This is a definite fact in so far as it is stated explicitely by Arnold in a famous article of his.
     
    Last edited: Jul 10, 2010
  5. Jul 10, 2010 #4

    Office_Shredder

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    If det(U)=1, then U is in SU(n), so obviously SU(n) can send O(n) (Identity*O(n)) to every class of the form U(n).

    If det(U)=--1, then U=V*(-I) where I is the identity, and V is in SU(n). -I is in O(n) so UO(n)=V*(-I)O(n=VO(n). det(V)=1 so we have reduced to the case above

    EDIT: Woops. Obviously doesn't work for even values of n

    Instead of just -1's, which are primitive second roots of unity, you need to have appropriate kth roots of unity on the diagonal when making V so that the determinant of your scalar matrix is -1

    But then the matrix isn't orthogonal. I'm gonna have to think about it


    FINAL EDIT (I hope): Ok here it is If det(U)=1, then obviously U*O(n) is contained in SU(n)*O(n). If det(U)=-1, let V be any matrix from O(n) with determinant -1 (e.g. reflection over an axis). Then [tex]U=U*V^{-1}*V[/tex] and [tex]U*V^{-1}*V*O(n)=U*V^{-1}*O(n)[/tex]. Obviously [tex]det(U*V^{-1})=1[/tex] so we're done
     
    Last edited: Jul 10, 2010
  6. Jul 13, 2010 #5

    quasar987

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    Oh! Yes it looks so simple when we know the solution isn't it.

    Check this version, it makes it look utterly trivial.

    Take UO(n), and U'O(n) any two classes in the subgroup of interest. Because O(n) contains a matrix J of determinant -1 (for instance J:=diag(-1 1 ... 1)), then we may assume WLOG that U and U' are in SU(n): UO(n)=(UJ)O(n). But SU(n), as with any group, acts transitively on itself, so we're done.

    But of course you did all the hard work; thanks! ;)
     
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