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Group Theory Intro stuff

  1. Oct 8, 2004 #1
    What does it mean to say that a n x n orthogonal matrix has n(n-1)/2 independent parameters? And why is this so? Can this be shown using the equation the summation with respect to i of the product aij(aik)= bjk

    where j,k=1,2,3.

    And bjk has the property bjk=1 when j=k
    bjk=0 when j doesn't equal k

    And with this being said, why does n x n unitary matrix have n^2-1 independent parameters. Can someone help clear some stuff up?
  2. jcsd
  3. Oct 8, 2004 #2

    matt grime

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    I can't explain the unitary bit, but for the orthogonal one:

    Any orthogonal matrix can be written as a product of basic (my terminology, not standard) rotation matrices (plus some reflection, but let's not worry about that here) with respect to the standard basis

    What are these? Well, what is a basic rotation: it fixes n-2 basis vectors and rotates the two remaining ones by some angle, theta. How many ways are there to pick 2 from n? n(n-1)/2

    why are there more for unitary ones? Well, each entry has a real and an imaginary part, but I'm not going to attempt a more detailed explanation cos i'll muck it up.

    That's a start anyway, but I'd need to know what your book defined 'independent parameter' as.
    Last edited: Oct 8, 2004
  4. Oct 9, 2004 #3
    See the thing is, my crummy book never defined independent parameter. What you said makes sense to me, but I think this can be shown using the orthogonality condition.
  5. Oct 10, 2004 #4


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    well lets try using the orthogonality condition. Consider the space of all square nxn matrices and map it into itself by the map taking A to A.Atranspose.

    I suppose an orthogonal; matrix is one whose inverse equals its transpose, right? So they would be the matrices which map to the identity by this map. Now the image of this map seems to equal all symmetric matrices, which do have dimension (1/2)(n)(n+1). So the domain space has dimension n^2 hence the fiber over one point would be expected to have dimension n^2 - (1/2)(n)n+1) = (1/2)(n-1)(n).

    This same approach should do the unitary case too.
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