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Surjective group homomorphism

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Show the map (call it phi) from U_n to C*
    defined by phi(X) = det(X) for all matrices X in U_n,
    is a surjective homomorphism, where

    U_n is the subgroup of GL(n,C) consisting of unitary matrices
    C* = C\{0} = invertible/nonzero complex numbers
    det(.) is the determinant of .

    2. Relevant equations

    A matrix X in GL(n,C) is unitary if ((X_bar)^T).X = I
    where X_bar is the conjugate of X (taking the conjugate of each entry in X)

    3. The attempt at a solution

    Homomorphism is easy to verify:
    phi(XY) = det(XY) = det(X) det(Y) = phi(X) phi(Y)

    I'm having trouble showing it's surjective.

    For it to be surjective we need
    for all (a+bi) in C* there exists X in U_n such that phi(X) = det(X) = a+bi
    where a and b not both zero

    The problem is X needs to be unitary, that is, X-conjugate-transpose times X needs to be the identity matrix.
    It follows the the required matrix X can't be a diagonal matrix, otherwise it is not unitary if its determinant is (a+bi), in particular every entry on that diagonal where there's a complex number x+yi, that entry will become (x^2 + y^2) after taking ((X_bar)^T).X instead of the required "1" on the diagonal of an identity matrix.

    Similarly I couldn't get anywhere with triangular matrices, and any non-diagonal and non-triangular matrix seems to get too complicated, not to mention that triangular matrices are messy enough...
  2. jcsd
  3. Nov 11, 2007 #2


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    It's definitely not going to be surjective -- the determinant of a unitary matrix lies on the unit circle.
  4. Nov 12, 2007 #3
    Hmm.... that's true. Maybe there was a typo on the question.
    Unless there is some other interpretation for what C* is...
  5. Nov 12, 2007 #4


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    Maybe it was indeed meant that
    C* = { z \in C : ||z|| = 1 }
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