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

Question

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!

Answer 1

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 $$ C^*([S])=C(\sigma([S]))=C(\mathbb T). $$