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Math Help - Group theory: Can you help me understand the statement of the theorem?

  1. #1
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    Group theory: Can you help me understand the statement of the theorem?

    Hello,


    Please take a look at the theorem of equation 3.33 http://i.imgur.com/mrWey.jpg
    I don't even understand the statement. What does it mean that a function "belongs" to an representation. Also, is the statement correct, or should there be some coefficients in front of each f_i?


    I have uploaded this definition, which might be useful http://i.imgur.com/ic80p.jpg


    Thanks in advance, Tony Bruguier
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonybruguier View Post
    Hello,


    Please take a look at the theorem of equation 3.33 http://i.imgur.com/mrWey.jpg
    I don't even understand the statement. What does it mean that a function "belongs" to an representation. Also, is the statement correct, or should there be some coefficients in front of each f_i?


    I have uploaded this definition, which might be useful http://i.imgur.com/ic80p.jpg


    Thanks in advance, Tony Bruguier
    The best I can tell here is what it is saying. Given a finite group G one can form the group algebra \mathcal{A}(G) which is the set \mathbb{C}^G=\left\{f:G\to\mathbb{C}\right\} with pointwise scalar multplication/addition and the 'convolution' product \displaystyle a\ast b where \displaystyle \left(a\ast b\right)(x)=\sum_{g\in G}a\left(xg^{-1}\right)b(g). Then, if \widehat{G} denotes the set of all equivalency classes of irreps of G one can select a representative \rho^{(\alpha)}:G\to\mathcal{U}\left(V^{(\alpha)}\  right) (one can always put an inner product on a space for which any representation \psi:G\to\text{GL}(V) is unitary--so we can a priori assume that V^{(\alpha)} is a pre-Hilbert space and that the image of \rho^{(\alpha)} lives inside the unitary group \mathcal{U}\left(V^{(\alpha)})). If we then choose a basis \mathcal{B}^{(\alpha)} (assume it's unitary) then for V^{(\alpha)} we can create the mappings D^{(\alpha)}:G\to\text{Mat}_{d_\alpha}\left(\mathb  b{C}\right) where d_\alpha=\deg\rho^{(\alpha)} by g\mapsto \left[\rho^{(\alpha)}(g)\right]_{\mathcal{B}^{(\alpha)}} (it turns out that D^{(\alpha)} is itself an irrep in the class \alpha). We can then consider that d_\alpha^2 functions D^{(\alpha)}_{i,j} for i,j\in[d_\alpha] where D^{(\alpha)}_{i,j}(g) is the complex number sitting in the (i,j)^{\text{th}} entry of D^{(\alpha)}(g). Thus for every \alpha\in\widehat{G} we have created d_\alpha^2 elements of \mathcal{A}(G). I think then what it's stating is the fact that \left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\tex  t{ and }i,j\in[d_\alpha]\right\} forms an (orthogonal) basis for \mathcal{A}(G).



    That's how I interpret it, but a lot of the notation is foreign to me...so take that with a grain of salt.
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