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Consider $l^2(\mathbb{N})$ and the shift operator $S: l^2(\mathbb{N}) \to l^2(\mathbb{N}): e_n \mapsto e_{n+1}$. It is easy to see that $C^*(S)$ contains the compact operators $K(l^2(\mathbb{N}))$

In some lecture notes I'm reading (refer to https://math.dartmouth.edu/~dana/bookspapers/cstar.pdf, p66) it is claimed that

$$C^*(S)/K(l^2(\mathbb{N})) \cong C(\textbf{T})$$

This is very non-obvious to me! I can't even seem to define a map $C^*(S) \to C(\textbf{T})$ (and then apply isomorphism theorem).

Any help is appreciated!

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Since $S^*S-SS^*$ is compact, it is zero in the quotient. So, in the quotient, $S$ is normal (actually, a unitary). Thus the quotient is $C^*([S])$, where $[S]$ is the class of $S$. With a little care one can show that $\sigma([S])=\mathbb T$. Then $$ C^*([S])=C(\sigma([S]))=C(\mathbb T). $$

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  • $\begingroup$ Perfect! Thanks so much for all the help! Any intuition how I should have come up with this solution? $\endgroup$
    – user745578
    May 17, 2020 at 21:14
  • $\begingroup$ Also, what is the $\sigma$ notation? It is spectrum right? And how do you define $\mathbb{T} \subseteq \mathbb{C}$ then? $\endgroup$
    – user745578
    May 17, 2020 at 21:18
  • $\begingroup$ No. The argument is simple, but before knowing the result I would have never imagined. People do more than this, they actually characterize C$^*(S)$ by looking at $S$ as the Toeplitz operator $T_z$. $\endgroup$
    – Martin Argerami
    May 17, 2020 at 21:19
  • $\begingroup$ $\sigma(S)$ denotes the spectrum of $S$; it's the notation used in the book you cited. And $\mathbb T$ is standard notation for the unit circle (which you also used! but with a slightly less common boldface $T$). $\endgroup$
    – Martin Argerami
    May 17, 2020 at 21:20
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    $\begingroup$ Well, they use $\mathbb T$ precisely because the unit circle is the $1$-dimensional torus. $\endgroup$
    – Martin Argerami
    May 17, 2020 at 21:24

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