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Math Help - What is the dimension of U(n), the group of all n x n unitary matrices?

  1. #1
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    What is the dimension of U(n), the group of all n x n unitary matrices?

    I am not sure how to find the dimension of U(n).

    Suggestions i have been given are: ((n^2) - n)) / 2

    or: 2(n^2)


    But i don't know which (if either) of these is correct, and how they are found.

    Thanks for any help
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  2. #2
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    Quote Originally Posted by Siknature View Post
    I am not sure how to find the dimension of U(n).

    Suggestions i have been given are: ((n^2) - n)) / 2

    or: 2(n^2)


    But i don't know which (if either) of these is correct, and how they are found.

    Thanks for any help
    The tangent space at 0 is the space of antisymmetric matrices, which has dimension \frac{n(n-1)}{2} (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of U(n).
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  3. #3
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    Quote Originally Posted by Laurent View Post
    The tangent space at 0 is the space of antisymmetric matrices, which has dimension \frac{n(n-1)}{2} (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of U(n).
    Thanks for your reply.

    How do you find the tangent space at a point?

    I'm interested in finding the tangent space at the identity.
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  4. #4
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    Quote Originally Posted by Siknature View Post
    Thanks for your reply.

    How do you find the tangent space at a point?

    I'm interested in finding the tangent space at the identity.
    Yes, in fact I mean "tangent space at the identity" in my post...

    To get the tangent space at A, differentiate at A the equation defining implicitly the manifold (here, A^TA=I) : it is the space of matrices H such that A^T H+H^T A=0. If A=I, this reduces to H=-H^T. (More formally, the tangent space at A is the kernel of the differential at A of a (submersive?) map \phi such that the manifold is a level set of \phi (at least locally at A))
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