Non forking extensions of types as extensions of filters Given a set of parameters $A$ a type in $S_n(A)$ may be thought of as a maximal filter on the monster model which can be constructed from $A$-definable subsets.
Given a type $q\in S_n(B)$ saying that $p\subset q$ can be naturally expressed in terms of filter extensions.
I am trying to find a condition on the nature of the filter extension to reflect non forking extension. Given that $q=p|B$, do I know something about the nature of the filter extension, or vice versa?
Thanks in advance
 A: Nice question! I think that the language of Boolean algebra is not in general sufficient to capture the full complexity of a theory. As you say, a type over $A$ is a nontrivial homomorphism from the boolean algebra of $A$-definable sets onto the two-element boolean algebra, and an extension of a type to the Boolean algebra of $B$ definable sets is just an extension of this homomorphism. The notion of nonforking requires more information which is not present in this boolean algebra.
For example, suppose that we are working with vector spaces, $A$ is a finite subspace of the big model, and that $B$ is a bigger finite subspace. From the perspective of abstract boolean algebra, you cannot tell the difference between the different types over $B$ - they are all indistinguishable atoms. Consequently you can have an abstract automorphism of Boolean algebras that takes a forking extension to a nonforking extension (this automorphism cannot come of course from an automorphism of the big model).
I don't know if you can get such example when working with models. In the $\omega$-stable case, at least, you can't, but I don't think this answer is very satisfactory.
So suppose that $A=M$ is an $\omega$-saturated model and that we are working in an $\omega$-stable theory. Then nonforking extensions are characterized by the fact do not let the Morley rank drop. The notion of a Morley rank for formulas works fine for Boolean algebras. The nonforking extension of an ultrafilter $p$ is then an ultrafilter $q$ such that the minimum over the Morley ranks of all elements of $q$ is the same as the minimal rank of an element in $p$.
