frame and dual frame of modal logic

Question

I am trying to understand frame and dual frame of modal logic from wiki: https://en.wikipedia.org/w/index.php?title=General_frame&oldid=756905480

When discussing dual frame, I quote

The Boolean algebra $⟨ A ,\, ∧ ,\, ∨ ,\, − ⟩$ has a Stone space, whose underlying set $F$ is the set of all ultrafilters of $A$.

(SE doesn't show scripted letter. better read on wikipedia.)

My problem here is, I can read the sentence in two ways:

1. the underlying set $F$ is the set of all ultrafilters of the atomic propositions closed in boolean algebra, so there is no modality at all. But if it's the case, it makes no sense to define transition relation by referring to modality.
2. the underlying set $F$ is the set of all ultrafilters of the modal logic itself. But then I don't understand how the ultrafilters would look like here. For example, what's the minimum of the ultrafilters?

Answer 1

The set $F$ is just the set of all ultrafilters on the Boolean algebra $(A,\wedge,\vee,-)$. This definition does not involve modality at all. However, we are also given an operation $\square$ on the set $A$, and we can use that operation in defining $R$, even though we did not use it in the definition of $F$.

I think your real confusion is in the definition of $A$. Here $(A,\wedge,\vee,-,\square)$ is just an arbitrary modal algebra. It does not need to consist of propositions in modal logic. You can get an example of a modal algebra by starting with a set of propositional variables and taking all the propositions that can be formed from them in modal logic, modulo an appropriate equivalence relation. If you do that, then $A$ does include propositions formed using $\square$. However, in defining $F$, we still only make use of the Boolean algebra structure on such propositions.