When does a Galois connection induce a topology?

Let $(X,\leq)$ and $(Y,\leq)$ by partially ordered sets. Recall that a(n antitone) Galois connection between $X$ and $Y$ is a pair of order-reversing maps

$\Phi: X \rightarrow Y, \ \Psi: Y \rightarrow X$

such that for all $x \in X,\ y \in Y$, $x \leq \Psi(y) \iff \Phi(x) \geq y$.

In this situation, the composite $\Psi \circ \Phi$ (resp. $\Phi \circ \Psi$) is a closure operator $\operatorname{cl}$ on $X$ (resp. $Y$): that is,

(C1) For all $x \in X$, $x \leq \operatorname{cl}(x)$.
(C2) For all $x_1,x_2 \in X$, $x_1 \leq x_2 \implies \operatorname{cl}(x_1) \leq \operatorname{cl}(x_2)$.
(C3) For all $x \in X$, $\operatorname{cl}(\operatorname{cl}(x)) = \operatorname{cl}(x)$.

Let us consider the special case where $X$ is the power set of some set $\mathbb{X}$, partially ordered by inclusion. In order for the above closure operator to define a topology on $\mathbb{X}$ -- i.e., be a "Kuratowski closure operator" -- we need also

(C4) For all $x,y \in X$, $\operatorname{cl}(x \cup y) = \operatorname{cl}(x) \cup \operatorname{cl}(y)$.

It is easy to see that (C4) is not automatic for the closure operator associated to a Galois connection -- in fact, every closure operator is induced by some Galois connection, and there are lots of closure operators which do not satisfy (C4).

However, in practice, it seems quite often to be the case that at least one of the two closure operators induced by a Galois connection is a topological closure operator. Examples:

(1) The Krull topology on the automorphism group of an infinite Galois extension (coming from the usual Galois correspondence).

(2) For a field $K$, the Zariski topology on $K^n$ (coming from the Galois connection arising in the Nullstellensatz).

(3) For a field $K$, the Harrison topology on the real spectrum of $K$ (coming from the Galois connection induced by the "incidence relation" $x \in P$ for $x$ an element of $K$ and $P$ an ordering on $K$).

Question 1: Is there some natural abstract condition on the Galois connection one can impose in order to ensure that one of the two closure operators is a Kuratowski closure operator?



Question 2: Are there are other examples of topologies arising from Galois connections in an interesting way? [As above, every topology arises from some Galois connection, yet in a tautological way.]

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I'll try to give an approximation from Category Theory. First, a partially ordered set $(X,\leq)$ can be regarded as a category with objects the elements of $X$ and one arrow $x\to x'$ iff $x\leq x'$. From this point of view, an order-preserving function between partially ordered sets is just a functor between the respective categories. Your Galois connection $\Phi,\Psi$ is a pair of functors $\Phi:X\to Y^{op}$, $\Psi:Y^{op}\to X$ and $\Phi$ is a left adjoint of $\Psi$.

Now to closure operators. A closure operator on $T$ on $X$ is just an idempotent monad on the category $X$. The fact that any closure operator comes from a Galois connection is a particular case of the general fact that any monad on a category is induced by an adjunction (usually there are several adjunctions that induce the same monad).

The condition $\operatorname{cl}(x \cup y) = \operatorname{cl}(x) \cup \operatorname{cl}(y)$ is saying that your idempotent monad $\operatorname{cl}$ on $X$ preserves binary coproducts. What you write as union would be the supremum of two elements $(x\vee x)$ in your partially ordered set, which is just the coproduct in the corresponding category.

It is a starndard fact on adjunctions that the left adjoint (in your case $\Phi$) always preserves colimits, and in particular coproducts. Then, one condition that ensures that $\operatorname{cl}$ preserves joins is to require $Y^{op}$ to have and the right adjoint $\Phi:Y^{op}\to X$ to preserve binary coproducts. $Y^{op}$ has binary coproducts when when $(Y,\leq)$ has meets of two elements $(y\wedge y')$, and $\Psi$ preserves them when $\Psi(y\wedge y')=\Psi(y)\vee\Psi(y')$.

Hope this answers to some extent your Question 1.

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