I've been reading "Boolean algebras with operators. Part I." (Jonsson, Tarski) where, given a subalgebra of a Boolean Algebra, they define its perfect extension. As far as I understand it can be extended to the case of Boolean Algebras with Operators (BAOs) by giving a notion of "extended (n-ary) operators" in the form
$m^{\sigma}=\underset{x\geq y\in C^{n}}{\sum}\underset{y\leq z\in B^{n}}{\prod}m(z)$
where m is an operator in the BAO B, which in its turn is a subalgebra of the extended BAO $B^{\sigma}$ and C is the set of closed elements of B (that is, meets of elements in B).
I do not fully understand that definition of extended operators, the reason to be of that form and if it could be defined in a different way.
I wonder, on the other hand, if it has to do with the definition of tensor product of BAOs, which has a similar form according to the answer given in: https://mathoverflow.net/questions/122800/products-of-boolean-algebras-and-probability-measures-thereon. Since the tensor is a coproduct, could be every BAO (as a factor of the tensor) seen as a subalgebra of the tensor seen as a new BAO (and perhaps treated as a kind of extension of its factors)?