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I see in this paper the following definition:

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How ${\bf A''}$ is defined? Is there a relation between $\sigma({\bf A})$ and $\sigma_H({\bf A})$? Note that $\sigma_H({\bf A})$ is defined as:

Definition: $(\lambda_1,\lambda_2,\cdots,\lambda_n)\notin \sigma_H({\bf A})$ if there exist operators $U_1,\cdots,U_n,V_1,\cdots,V_n \in \mathcal{B}(\mathcal{H})$ such that $$\sum_{1\leq k \leq n}U_k(A_k-\lambda_k I)=I\;\hbox{and}\;\;\sum_{1\leq k \leq n}(A_k-\lambda_k I)V_k =I.$$

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The double commutant $A''$ is defined as $(A')'$, where $$ A'=\{T\in B(H):\ TA_j=A_jT,\ j=1,\ldots,n\} $$ and $$ A''=\{S\in B(H):\ ST=TS\ \forall T\in A'\}. $$ The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.

The Double Commutant Theorem says that, if $M\subset B(H)$ is a $*$-algebra, then $$ M''=\overline{M}^{\rm sot}. $$

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  • $\begingroup$ Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators. $\endgroup$ – Student Dec 26 '18 at 7:53
  • $\begingroup$ Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/… $\endgroup$ – Student Dec 26 '18 at 7:54
  • $\begingroup$ Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used. $\endgroup$ – Martin Argerami Dec 26 '18 at 13:10
  • $\begingroup$ Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer. $\endgroup$ – Student Dec 29 '18 at 11:11
  • $\begingroup$ What I said about the inclusion seems to be wrong. I have deleted that paragraph. $\endgroup$ – Martin Argerami Dec 29 '18 at 18:15

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