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Math Help - Commutation Relation

  1. #1
    Member Mauritzvdworm's Avatar
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    Commutation Relation

    This has to be something stupid which I'm just not seeing. Let R_{\theta}(g)(z)=g(e^{2\pi i \theta}z) with g\in C(\mathbb{T}) and z\in\mathbb{T}
    Further, let M_f be the multiplication operator on B(L^2(\mathbb{T})) with f\in C(\mathbb{T})

    R_{\theta} can be extended to a unitary operator on B(L^2(\mathbb{T}))

    show that the following commutation relation holds
    R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}

    Where id denotes the identity function

    This is what I have:
    (R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))=  R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_  {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    but here the commutation relation does not hold. Am I using the definitions wrongly?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    This has to be something stupid which I'm just not seeing. Let R_{\theta}(g)(z)=g(e^{2\pi i \theta}z) with g\in C(\mathbb{T}) and z\in\mathbb{T}
    Further, let M_f be the multiplication operator on B(L^2(\mathbb{T})) with f\in C(\mathbb{T})

    R_{\theta} can be extended to a unitary operator on B(L^2(\mathbb{T}))

    show that the following commutation relation holds
    R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}

    Where id denotes the identity function

    This is what I have:
    (R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))=  R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_  {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)

    but here the commutation relation does not hold. Am I using the definitions wrongly?
    The calculation for (R_{\theta}M_{\text{id}})(g)(z) is correct. The other one is not. The operator M_{\text{id}} takes a function and multiplies its value at z by the independent variable z. So M_{\text{id}}R_{\theta}(g)(z)=M_{\text{id}}(R_{\th  eta}(g)(z))=M_{\text{id}}(g(e^{2\pi i \theta}z))=zg(e^{2\pi i \theta}z).
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  3. #3
    Member Mauritzvdworm's Avatar
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    So then (R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})
    and
    (M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)
    in which case the commutation relation holds

    The only part that is bothering me is the multiplication operator M_{id} or more generally M_{f},f\in C(\mathbb{T})

    The way I have it is that it is defined in the following way
    M_{f}(g)(z)=f(z)g(z)

    now in the second case we had
    M_{f}(g)(e^{2\pi i \theta}z)
    so why do we not use the rotated coordinate but just z?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Mauritzvdworm View Post
    So then (R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})
    and
    (M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)
    in which case the commutation relation holds

    The only part that is bothering me is the multiplication operator M_{id} or more generally M_{f},f\in C(\mathbb{T})

    The way I have it is that it is defined in the following way
    M_{f}(g)(z)=f(z)g(z)

    now in the second case we had
    M_{f}(g)(e^{2\pi i \theta}z)
    so why do we not use the rotated coordinate but just z?
    The definition of M_f is M_f g(z) = f(z)g(z). Now suppose you have a function h(z) defined by h(z) = g(e^{2\pi i \theta}z). It is still true that M_f h(z) = f(z)h(z) = f(z)g(e^{2\pi i \theta}z). In this case h = R_\theta g so that h(z) = R_\theta g(z).
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