Let $\chi_1$ be the map on the unit circle defined by $\chi_1(e^{it})=e^{it}$. Let $T_{\chi_1}$ be the corresponding Toeplitz operator. Consider the map $T_{\chi_1}^* T_{\chi_1}- T_{\chi_1} T_{\chi_1}^*$ where $T_{\chi_1}^*$ is the adjoint of $T_{\chi_1}$. The book I am reading says the map $T_{\chi_1}^* T_{\chi_1}- T_{\chi_1} T_{\chi_1}^*$ is a nonzero rank one. It is easy to see it is nonzero but I cannot see why it is rank one.
Toeplitz Operator question
1 Answers
If you move your operators to $\ell^2(\mathbb N)$, your $T_{\chi_1}$ goes to the unilateral shift $S$. So you are looking at $S^*S-SS^*$. Now, as $S$ is an isometry, $S^*S=I$. And $SS^*=I-E_{11}$, where $E_{11}$ is the rank one operator onto the first vector in the canonical basis. So $ S^*S-SS^*=I-(I-E_{11})=E_{11}, $ rank-one.
If you are not happy with this, you can do the computation directly. Let $f=\sum_{n=0}^\infty c_ne^{itn}$. Then it is easy to see that $T_{\chi_1}^*T_{\chi_1}f=f$. Also $ T_{\chi_1}T_{\chi_1}^*f=T_{\chi_1}P_He^{-it}f=T_{\chi_1}P_H\sum_{n=0}^\infty c_ne^{it(n-1)}=T_{\chi_1}P_H\sum_{n=-1}^\infty c_{n+1} e^{itn}=T_{\chi_1}\sum_{n=0}^\infty c_{n+1} e^{itn}=\sum_{n=0}^\infty c_{n+1} e^{it(n+1)}=\sum_{n=1}^\infty c_{n} e^{itn}. $ We conclude that $ (T_{\chi_1}^*T_{\chi_1}-T_{\chi_1}T_{\chi_1}^*)f=f-\sum_{n=1}^\infty c_{n} e^{itn}=c_0. $
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0Let $e_1,e_2,\ldots$ be the canonical basis of $\ell^2(\mathbb N)$. Then $V:e^{int}\mapsto e_n$ maps one orthonormal basis into another one, and thus induces a unitary operator $H^2\mapsto\ell^2(\mathbb N)$. – 2012-12-17