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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.

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  • $\begingroup$ 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). $\endgroup$
    – sjvega
    Dec 21, 2014 at 16:50

2 Answers 2

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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 $\sigma[S]=\mathbb T$.

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  • $\begingroup$ Thanks as usual @Martin Argerami, I had solved this a long time ago but it was nice to recall this problem. $\endgroup$
    – sjvega
    Jul 6, 2023 at 18:20
  • $\begingroup$ It's a very nice question, and it might trip many who are starting to look at Fredholm operators. $\endgroup$ Jul 6, 2023 at 19:19
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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?

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  • $\begingroup$ 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. $\endgroup$
    – sjvega
    Dec 22, 2014 at 18:42
  • $\begingroup$ @sjvega Then you know the $\lambda$ that will expose your error. :) $\endgroup$ Dec 22, 2014 at 19:58

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