# 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.]

• if you have cl(x_1) contained in cl(x_2) in your C2, and cl(cl(x))=cl(x) in C3, then I would recognize it as a closure operator. Gerhard "Ask Me About System Design" Paseman, 2010.08.16 Aug 16 '10 at 7:16
• Also, Marcel Erne edited a book on Galois connections, and wrote a good introductory chapter for it. I would look there first for an answer to your question. Gerhard "Ask Me About System Design" Paseman, 2010.08.16 Aug 16 '10 at 7:19
• @Gerhard: thanks for pointing out some unfortunate typos in the definition of closure operator. (When I was typing my question, my interface with the site inexplicably became very slow, so I spent less time proofreading than usual.) Regarding Erne's book and its first chapter: I have skimmed over it and not found anything of direct relevance to my question. Please let me know if you have a specific passage in mind (the article is 138 pages). Aug 16 '10 at 9:02
• @Pete: the Krull topology example needs to be modified a bit, I think. The closure operator we get from the Galois correspondence doesn't send a subset of the Galois group to its closure; instead I think it sends a subset of the Galois group to the closure of the subgroup generated by it. I think this is still okay (it's a notion of closure operator internal to groups, say) but it seems worth saying. Nov 6 '13 at 23:40
• Clark, Your question is very good. However, the additivity of the closure operation is usually not needed. See our papers on relator spaces. Sincerely yours, Arpad Szaz Jul 19 '17 at 17:15

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.

A standard fact about adjunctions is 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 $\Psi(y\wedge y')=\Psi(y)\vee\Psi(y')$.

All of the Galois connections I know involving a power set arise from a relation $R : X \times Y \to 2$ as described for example here. As you observe, this relation can often be used to define a topological space. However, it can always be used to define a Chu space, which is in a precise sense a kind of generalized topological space.

Some definitions. The category $\text{Chu}$ of Chu spaces (in $\text{Set}$, over $2$) has as objects triplets $(X, Y, R)$ where $X, Y$ are sets and $R$ is a function $X \times Y \to 2$. A morphism $(X_1, Y_1, R_1) \to (X_2, Y_2, R_2)$ of Chu spaces is an adjunction: that is, it is a pair of functions $f_{\ast} : X_1 \to X_2$ and $f^{\ast} : Y_2 \to Y_1$ such that

$$R_2(f_{\ast}(x_1), y_2) = R_1(x_1, f^{\ast}(y_2)).$$

Many familiar categories turn out to be full subcategories of $\text{Chu}$.

Example. Suppose the map $Y \ni y \mapsto \{ x : xRy \forall y \in Y \} \in 2^X$ is injective. The full subcategory of Chu spaces with this property may be regarded as sets together with a distinguished collection of subsets, where $R$ is the incidence relation "the point $x \in X$ lies in the subset $S \in Y$," and where $f^{\ast}$ is consequently always the preimage along $f_{\ast}$, so the morphisms are precisely maps of sets such that the preimage of a distinguished subset is distinguished.

Subexample. In particular, topological spaces are such Chu spaces, and the notion of morphism agrees. Hence $\text{Top}$ is a full subcategory of Chu spaces.

This is the precise sense in which Chu spaces generalize topological spaces: being a topological space is a property of a Chu space, in the same way that being an abelian group is a property of a group.

Subexample. Similarly, the category $\text{Meas}$ of measurable spaces is a full subcategory of Chu spaces.

Example. Let $V$ be a vector space over $\mathbb{F}_2$. Then the dual pairing $\text{eval} : V \times V^{\ast} \to \mathbb{F}_2$ defines a Chu space. This inclusion is full.

Example. Similarly, let $B$ be a Boolean algebra with Stone space $S(B)$. Then the evaluation map $\text{eval} : B \times S(B) \to 2$ defines a Chu space. This inclusion is full.

Example. Let $P$ be a poset and let $2^P$ denote the poset of order-preserving functions $P \to 2$, where $2 = \{ 0, 1 \}$ and $0 < 1$. The evaluation map $\text{eval} : P \times 2^P \to 2$ defines a Chu space. This inclusion is full.

Example. $\text{Chu}$ is equipped with a natural involution switching $X$ and $Y$. Applying this involution to any of the above examples gives other examples. When applied to various subcategories of $\text{Chu}$ this involution reproduces many familiar dualities. For example, when applied to Boolean algebras, the involution reproduces Stone duality.

I haven't played much with Chu spaces, but my intuition is that one can think of $X$ as a set of "states" and $Y$ as a set of "observations" that one can make about states, with $R$ confirming whether or not a given observation is true of a given state. But then one can switch the roles of states and observations!

Some comments, unrelated to the other answer, on what happens in the case of the Zariski topology. Here the extra axiom is

$$V(I(S_1 \cup S_2)) = V(I(S_1)) \cup V(I(S_2))$$

(where, to fix notation, the $S_i$ are subsets of $k^n$, $I$ sends a subset of $k^n$ to the ideal of functions vanishing on it in $k[x_1, ..., x_n]$, and $V$ sends a subset of $k[x_1, ..., x_n]$ to its zero locus in $k^n$.) Now, we have $I(S_1 \cup S_2) = I(S_1) \cap I(S_2)$ for abstract reasons (the adjointness implies that $V$ and $I$ both send colimits to limits, or equivalently suprema to infima), but we don't have $V(T_1 \cap T_2) = V(T_1) \cup V(T_2)$ for arbitrary subsets $T_1, T_2$ of $k[x_1, ..., x_n]$ for rather silly reasons, e.g. $T_1$ and $T_2$ could be disjoint but generate the same ideal. Hence the condition mentioned in Nacho Lopez's answer doesn't apply here.

At best we can hope to have $V(T_1 \cap T_2) = V(T_1) \cup V(T_2)$ for $T_1, T_2$ ideals, which would still be enough to get the extra axiom, but note that this fails when $k$ has zero divisors, e.g. if $k = \mathbb{Q}[a, b]/(ab)$ then we could take $T_1 = (ax_1), T_2 = (bx_2)$, and then $T_1 \cap T_2 = (0)$ has zero locus $k^n$ but $V(T_1) \cup V(T_2)$ is much smaller, e.g. it doesn't contain $(1, 1)$. That strongly suggests it can't be true for abstract reasons: at the very least we need to use somewhere the fact that $k$ is an integral domain.

What is true for abstract reasons is that $T_1 \cap T_2 \subseteq T_1, T_2$, hence $V(T_1), V(T_2) \subseteq V(T_1 \cap T_2)$, hence $V(T_1) \cup V(T_2) \subseteq V(T_1 \cap T_2)$. In particular, we get the inclusion

$$V(I(S_1)) \cup V(I(S_2)) \subseteq V(I(S_1) \cap I(S_2)) = V(I(S_1 \cup S_2))$$

for abstract reasons. But to get the other inclusion we need to actually do something. For example:

Suppose $v \in V(I(S_1 \cup S_2))$, so $f(v) = 0$ for all $f \in I(S_1 \cup S_2)$. In particular $g(v) h(v) = 0$ for all $g \in I(S_1), h \in I(S_2)$ (since $IJ \subseteq I \cap J$ for ideals). If $g(v) = 0$ for all $g \in I(S_1)$ then $v \in V(I(S_1))$; otherwise, there is some $g$ such that $g(v) \neq 0$, hence from $g(v) h(v) = 0$ for all $h$ we deduce that $h(v) = 0$ for all $h$, and then $v \in V(I(S_2))$.

So, abstractly, here is what we did:

• In the domain of $V$ and the range of $I$ we restricted our attention to ideals.
• On ideals we introduced a new operation, namely ideal product, satisfying $I(S_1) I(S_2) \subseteq I(S_1 \cup S_2)$.
• We showed that $V(IJ) \subseteq V(I) \cup V(J)$. (For this step we needed to use the fact that $k$ is an integral domain.)

This is enough: once these conditions are met we then have

$$V(I(S_1 \cup S_2)) \subseteq V(I(S_1) I(S_2)) \subseteq V(I(S_1)) \cup V(I(S_2)).$$

To formalize this argument, we can think of ideal product as a monoidal operation on the poset of ideals making it a monoidal category. The poset of subsets of $k^n$ also naturally has a monoidal operation given by union. The condition on $I$ above is that $I$ is a lax monoidal functor with respect to these monoidal operations, and the condition on $V$ above is that $V$ is an oplax monoidal functor with respect to these monoidal operations. All told, we get the following:

Let $F : P \to Q$ and $G : Q \to P$ be an antitone Galois connection between posets $P, Q$. Suppose furthermore that $P$ has finite coproducts and that $Q$ is equipped with a monoidal operation $\otimes$ with respect to which $F$ is lax monoidal (where $P$ is equipped with the coproduct) and $G$ is oplax monoidal. Then the closure operator $GF$ on $P$ preserves binary coproducts.

But this is somehow unsatisfying.

It is possibly worth adding that the lattice of closed sets constructed from a polarity (a contravariant galois connection on powersets) is always complete. That would imply that the topology produced, when a topology is produced, is closed under arbitrary unions, and thus an Alexandroff topology. The specialization order can then be used to form the polarity.

• But the lattice of closed subsets of a topological space is always complete. Once you have arbitrary meets, you have arbitrary joins. It is most certainly not the case that the family of closed sets of a Galois connection has to be closed under arbitrary unions. The join in this complete lattice is given by the closure of the set-theoretic union. Oct 14 '19 at 9:20

Galois connections in particular are adjunctions $F\dashv G$ between posets, and like all adjunctions satisfy the triangle identities $FGF\backsimeq F$ and $GFG\backsimeq G$.

Lagois Connections by Melton, Schroder & Strecker take this as their definition.

One would suspect that they have similar properties to Galois connections - and they do - but they are not identical to them.

Like them, they too have closure & interior operators:

They note that any topological space $X$ gives a Lagois Connection whose closure operation is $X-\overline{A}$, and interior operation is $\overline{X-A}$.

I haven't checked, but I expect the reverse is achievable.

• If every Galois connection is a Lagois Connection and (you "expect") every Lagois Connection induces a topology, then are you suggesting that every Galois connection induces a topology?? This is certainly not true: every Moore closure operator on a set comes from a Galois connection and not every Moore closure operator is a Kuratowski closure operator. So could you clarify what you mean? Jan 16 '14 at 13:44
• The 'expect' wasn't the mathematicians 'expect', but more the ordinary language 'expect' - as in I expect the bus is coming in ten minutes. But when it doesn't you're not exactly surprised. It should be translated as 'its worth asking'. Jan 16 '14 at 15:24
• So are you saying that you are not actually answering my question but only raising a related question? I guess I don't understand what you mean you speculate in any sense about something that the question explains is not possible. Jan 16 '14 at 20:38
• Its easy enough to understand - I was overhasty in reading the question. I see now its adding little if anything to the question. Jan 17 '14 at 1:00