I see in this paper the following definition:
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
$$\sum_{1\leq k \leq n}U_k(A_k-\lambda_k I)=I\;\hbox{and}\;\;\sum_{1\leq k \leq n}(A_k-\lambda_k I)V_k =I.$$
Best Answer
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 $$ M''=\overline{M}^{\rm sot}. $$