# The Unapologetic Mathematician

## Galois Connections

I want to mention a topic I thought I’d hit back when we talked about adjoint functors. We know that every poset is a category, with the elements as objects and a single arrow from $a$ to $b$ if $a\leq b$. Functors between such categories are monotone functions, preserving the order. Contravariant functors are so-called “antitone” functions, which reverse the order, but the same abstract nonsense as usual tells us this is just a monotone function to the “opposite” poset with the order reversed.

So let’s consider an adjoint pair $F\dashv G$ of such functors. This means there is a natural isomorphism between $\hom(F(a),b)$ and $\hom(a,G(b))$. But each of these hom-sets is either empty (if $a\not\leq b$) or a singleton (if $a\leq b$). So the adjunction between $F$ and $G$ means that $F(a)\leq b$ if and only if $a\leq G(b)$. The analogous condition for an antitone adjoint pair is that $b\leq F(a)$ if and only if $a\leq G(b)$.

There are some immediate consequences to having a Galois connection, which are connected to properties of adjoints. First off, we know that $a\leq G(F(a))$ and $F(G(b))\leq b$. This essentially expresses the unit and counit of the adjunction. For the antitone version, let’s show the analogous statement more directly: we know that $F(a)\leq F(a)$, so the adjoint condition says that $a\leq G(F(a))$. Similarly, $b\leq F(G(b))$. This second condition is backwards because we’re reversing the order on one of the posets.

Using the unit and the counit of an adjunction, we found a certain quasi-inverse relation between some natural transformations on functors. For our purposes, we observe that since $a\leq G(F(a))$ we have the special case $G(b)\leq G(F(G(b)))$. But $F(G(b))\leq b$, and $G$ preserves the order. Thus $G(F(G(b)))\leq G(b)$. So $G(b)=G(F(G(b)))$. Similarly, we find that $F(G(F(a)))=F(a)$, which holds for both monotone and antitone Galois connections.

Chasing special cases further, we find that $G(F(G(F(a))))=G(F(a))$, and that $F(G(F(G(b))))=F(G(b))$ for either kind of Galois connection. That is, $F\circ G$ and $G\circ F$ are idempotent functions. In general categories, the composition of two adjoint functors gives a monad, and this idempotence is just the analogue in our particular categories. In particular, these functions behave like closure operators, but for the fact that general posets don’t have joins or bottom elements to preserve in the third and fourth Kuratowski axioms.

And so elements left fixed by $G\circ F$ (or $F\circ G$) are called “closed” elements of the poset. The images of $F$ and $G$ consist of such closed elements

May 18, 2009 - Posted by | Algebra, Category theory, Lattices

1. “In particular, these functions behave like closure operators”

In fact, under the standard terminology, they are closure operators. What have been called closure operators here and in the UM post linked to above are usually called topological closure operators.

Comment by Todd Trimble | May 18, 2009 | Reply

2. Well, yes. I just wanted to keep readers from getting confused by the other two axioms.

Comment by John Armstrong | May 18, 2009 | Reply

3. […] Orthogonal Complementation is a Galois Connection We now know how to take orthogonal complements of subspaces in an inner product space. It turns out that this process (and itself again) forms an antitone Galois connection. […]

Pingback by Orthogonal Complementation is a Galois Connection « The Unapologetic Mathematician | May 19, 2009 | Reply

4. Typo: b <= F(G(b)) should be the other way around at the end of paragraph 3.

Comment by Mike Stay | May 26, 2009 | Reply

5. No, at the end of paragraph 3 I’m doing the antitone version, not the monotone one. It’s like working with contravariant functors.

As a particular case, consider the motivating case: the closure of a subgroup of the Galois group contains the original subgroup, and the closure of a subfield of an extension field contains the original subfield. Both directions are increasing, because of the antitonicity.

Comment by John Armstrong | May 26, 2009 | Reply

6. […] Doesn’t this start to look a bit like a Galois connection? […]

Pingback by Regular Outer Measures « The Unapologetic Mathematician | April 2, 2010 | Reply