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



    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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