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Thread: 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 $\displaystyle R_{\theta}(g)(z)=g(e^{2\pi i \theta}z)$ with $\displaystyle g\in C(\mathbb{T})$ and $\displaystyle z\in\mathbb{T}$
    Further, let $\displaystyle M_f$ be the multiplication operator on $\displaystyle B(L^2(\mathbb{T}))$ with $\displaystyle f\in C(\mathbb{T})$

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

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

    Where $\displaystyle id$ denotes the identity function

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

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

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

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

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

    Where $\displaystyle id$ denotes the identity function

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

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

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

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

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

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

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

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

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