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.

  • $\begingroup$ 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 '14 at 16:50

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?

  • $\begingroup$ 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 '14 at 18:42
  • $\begingroup$ @sjvega Then you know the $\lambda$ that will expose your error. :) $\endgroup$ – DisintegratingByParts Dec 22 '14 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.