Let $A$ be a unital $C^\*$-algebras and $K(H)$ is $C^\*$-algebras of compact operators on separable Hilbert space $H$. Is it true that $(A \otimes K(H))^{\*\*}= A^{\*\*} \overline{\otimes}B(H)$?



second dual of minimal tensor products of $C^*$-algebras
Yes (but maybe there is a more direct argument ? ):


$A$ and $K(A) = A \otimes K(H)$ are Morita equivalent so they have equivalente categories of representations, moreover this equivalence is implemented as following: any representation of $K(A)$ is of the form $H \otimes R$ with $R$ a representation of $A$.


$A^{\*\*}$ can be constructed as the weak closure of $A$ into some "universal" representation of $A$, i.e. a representation $V$ such that any other representation is a direct sum of subrepresentations of $V$.


A universal representation of $A$ and of $K(A)$ corresponds to each other under the above equivalence (because being universal is a purely categorical property)


SO this mean that there is a representation $V$ of $A$ such that:


$A^{\*\*}$ is the weak closure (double commutant) of $A$ into $B(V)$.
$K(A)^{\*\*}$ is the weak closure of $A \otimes K(H)$ in $B(V \otimes H)$.


Because the commutant of $A \otimes K(H)$ is $A' \otimes \mathbb{C}$ (still because of the above equivalence of catégores: the commutant is the set of endomorphism of representations) and the bi-comutant of $A \otimes K(H)$ is indeed $A'' \otimes B(H)$ (for the spatial tensor product).


