# Error in the calulation of the spectrum of the image of right shift operator in the Calkin algebra

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?

• After thinking this a long while, the defect in this reasoning must be about the closedness of the image of $\lambda - S$, although due to different sources this shouldn't matter (see mathoverflow.net/questions/54981/…). To see this, just notice that if $im(\lambda - S)$ is closed, this calculation would also imply that $\sigma(S)= \{|\lambda| < 1 \}$ (which contradicts the fact that the spectrum is compact). Dec 21, 2014 at 16:50

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 $$\sigma[S]=\mathbb T$$.
Note that $S^{\star}S-SS^{\star} \in \mathcal{K}$, and $S^{\star}S=I$. So where do you think the spectrum will be in the quotient algebra?
• I'm aware that the spectrum of $[S]$ is $\mathbb{S}^1$ (for example, because is the universal C*-algebra generated by a unitary). The problem is that I don't see the error in my calculation of it. I'm guessing that the closedness condition must be satisfied although in some sources says its superfluous. Dec 22, 2014 at 18:42
• @sjvega Then you know the $\lambda$ that will expose your error. :) Dec 22, 2014 at 19:58