If $S \in \mathcal{B}(\ell^2(\mathbb{N}))$ is the right shift operator $$ S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots),$$ and $\mathcal{C} := \mathcal{B}(\ell^2(\mathbb{N}))/\mathcal{K}$ is the Calkin algebra (here, $\mathcal{K}$ are the compact operators), what is the spectrum of $[S]$ (the class of $S$ in $\mathcal{C}$)?
If my analysis is correct, $\sigma([S])$ should be the values $\lambda \in \mathbb{C}$ for which the operator $\lambda - S$ is not a Fredholm operator, that is, the values such that either $\ker(\lambda -S)$ or $\ker(\bar{\lambda} - S^*)$ is infinite dimensional. If that is correct, then considering the fact that $$ 0 = (\lambda - S) (x_n) \iff x_n = 0 \quad \forall n$$ and also that $$ 0 = (\bar{\lambda} - S^*)(x_n) \iff \begin{cases} (x_n) \in \langle (1,\bar{\lambda},\bar{\lambda}^2,\ldots)\rangle \quad &\mbox{if } |\lambda| < 1 \\ x_n =0 \quad \forall n \quad &\mbox{if } |\lambda|\geq 1 \end{cases},$$ this implies that $\lambda -S$ is always Fredholm, independently of $\lambda$, which in turn, doesn't make any sense (as it would imply that $\sigma([S])=\emptyset$). I can't find my mistake anywhere, could you point it out to me?
Thanks in advance.
The definition of Fredholm operator $T$ is not that $\ker T$ and $\ker T^*$ are finite-dimensional. The second condition should be that $H/\operatorname{ran}T$ is finite-dimensional. The two conditions together imply that $\operatorname{ran}T$ is closed and hence $$H/\operatorname{ran}T\simeq(\operatorname{ran}T)^\perp=\ker T^*.$$
But you cannot conclude that $T$ is Fredholm if $\ker T$ and $\ker T^*$ are finite-dimensional, as your example shows. In the particular case of the shift, any $\lambda$ with $|\lambda|=1$ is an approximate eigenvalue. This tells you that $\lambda-S$ is not bounded below, and hence $\operatorname{ran}(\lambda-S)$ is not closed. Which precludes $\lambda-S$ from being Fredholm. This, together with the fact that $[S]$ is a unitary, shows that $\sigma[S]=\mathbb T$.