Toeplitz extension

Example

Let $S$ be the unilateral shift on $\ell^2(\N)$, $S e_k = e_{k+1}$, and let $\mathcal{T}$ be the Toeplitz algebra, the C*-algebra generated by $S$. Let $\mathcal{K}$ be the ideal of compact operators on $\ell^2(\N)$. The Toeplitz extension is the short exact sequence $$0 \to \mathcal{K} \to \mathcal{T} \xrightarrow{\sigma} C(\mathbb{T}) \to 0.$$ Here $\sigma$ is the symbol map, determined by $\sigma(S) = z$. Check of the ideal: $1 - S S^*$ is the rank-one projection onto $\C e_0$, so $\mathcal{T}$ contains a nonzero compact operator. Since $\mathcal{T}$ acts irreducibly, it contains all of $\mathcal{K}$. The quotient $\mathcal{T}/\mathcal{K}$ is commutative, because $S^* S - S S^* = 1 - S S^*$ is compact, and it is generated by a unitary with spectrum $\mathbb{T}$, so it is $C(\mathbb{T})$.

The index map $\partial \colon K_1(C(\mathbb{T})) \to K_0(\mathcal{K})$ sends the generator $[z]$ to $-1$. Indeed $S$ is a lift of $z$ and a partial isometry, so $\partial[z] = [1 - S^* S] - [1 - S S^*] = 0 - [e_0 \text{ projection}]$, which is $-1$ under $K_0(\mathcal{K}) \cong \Z$. This is the Fredholm index of $S$: $\dim \ker S - \dim \ker S^* = 0 - 1$. Since $\partial$ is an isomorphism $\Z \to \Z$, the six-term exact sequence, with $K_1(\mathcal{K}) = 0$ and $K_0(C(\mathbb{T})) \cong \Z$, gives $$K_0(\mathcal{T}) \cong \Z, \qquad K_1(\mathcal{T}) = 0.$$ The class $[1]$ generates $K_0(\mathcal{T})$, and the inclusion $\mathcal{K} \to \mathcal{T}$ is zero on $K_0$.