# Commutation Relation

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• Sep 27th 2010, 07:38 AM
Mauritzvdworm
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?
• Sep 27th 2010, 10:21 AM
Opalg
Quote:

Originally Posted by Mauritzvdworm
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 $\displaystyle M_{\text{id}}R_{\theta}(g)(z)=M_{\text{id}}(R_{\th eta}(g)(z))=M_{\text{id}}(g(e^{2\pi i \theta}z))=zg(e^{2\pi i \theta}z).$
• Sep 27th 2010, 11:41 AM
Mauritzvdworm
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 $\displaystyle z$?
• Sep 27th 2010, 12:06 PM
Opalg
Quote:

Originally Posted by Mauritzvdworm
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 $\displaystyle h(z) = R_\theta g(z)$.