CAR-algebra contains a unitary with full spectrum

Question

How can I see that the UHF algebra $M_{2^\infty}$ contains a unitary element with spectrum $\Bbb{T}$?

Answer 1

Using the natural inclusions $M_2(\mathbb C)\subset M_4(\mathbb C)\subset\cdots\subset M_{2^\infty}$ via $$\tag{1} A\longmapsto \begin{bmatrix} A&0\\0&A\end{bmatrix}, $$ construct $$ U_1=\begin{bmatrix} e^{2\pi i 1/2}&0\\0&1\end{bmatrix},\ \ U_2=\begin{bmatrix} e^{2\pi i 1/4} &0&0&0\\ 0 &e^{2\pi i 2/4} &0&0 \\ 0&0& e^{2\pi i 3/4}&0\\ 0&0&0& 1\end{bmatrix}, $$ and $U_k\in M_2^{k}(\mathbb C)$ has diagonal $\{e^{2\pi i r/2^k}\}_{r=1}^{2^k}$. Using the embedding $(1)$, one can check that \begin{align} \|U_{k+1}-U_k\|^2&=\max\{|e^{2\pi i 2r/2^{k+1}}-e^{2\pi i (2r+1)/2^{k+1}} |^2: \ r=1,\ldots,2^{k+1} \}\\ \ \\ &=|1-e^{2\pi i 1/2^{k+1}} |^2={(1-\cos \pi/2^{k})^2+\sin^2\pi/2^k}\\ \ \\ &=2(1-\cos\pi/2^k)=O(2^{-2k}). \end{align} Then $$ \|U_{k+\ell}-U_k\|=\|\sum_{j=1}^{\ell}(U_{k+j}-U_{k+j-1})\|\leq\sum_{j=1}^{\ell}\|U_{k+j}-U_{k+j-1}\|\leq c\,\sum_{j=1}^\ell 2^{-k-j}\leq c\,2^{-k+1}. $$ So the sequence $\{U_k\}$ converges to a unitary $U\in M_{2^\infty}$. For each $r,k$ we have $\lambda_{k,r}=e^{\pi i r/2^k}\in\sigma(U_m)$ as long as $m>k$. So $U_m-\lambda_{k,r}I$ is not invertible for all $m>k$, and its limit $U-\lambda_{k,r}I$ cannot be invertible (the set of invertible elements is open, so an invertible element cannot be a limit of non-invertible). Thus $\lambda_{k,r}\in\sigma(U)$. As the set $\{\lambda_{k,r}:\ k\in\mathbb N,\ r=1,\ldots,2^k\}$ is dense in $\mathbb T$, we conclude that $\sigma(U)=\mathbb T$.