The notion of a Galois connection has its root in Galois theory. By the fundamental theorem of Galois theory, there is a one-to-one correspondence between the intermediate fields between a field $L$ and its subfield $F$ (with appropriate conditions imposed on the extension $L/F$ ), and the subgroups of the Galois group $\operatorname{Gal}(L/F)$ such that the bijection is inclusion-reversing: $$\operatorname{Gal}(L/F)\supseteq H\supseteq\langle e \rangle\quad \mbox{ iff }\quad F\subseteq L^H\subseteq L,\mbox{ and}$$ $$F\subseteq K\subseteq L\quad\mbox{ iff }\quad \operatorname{Gal}(L/F)\supseteq \operatorname{Gal}(L/K)\supseteq\langle e \rangle.$$

If the language of Galois theory is distilled from the above paragraph, what remains reduces to a more basic and general concept in the theory of ordered-sets:

Definition. Let $(P, \le_P)$ and $(Q, \le_Q)$ be two posets. A Galois connection between $(P,\le_P)$ and $(Q,\le_Q)$ is a pair of functions $f:=(f^*,f_*)$ with $f^*\colon P\to Q$ and $f_*\colon Q\to P$ , such that, for all $p\in P$ and $q\in Q$ , we have $$f^*(p)\leq_Q q\quad \mbox{ iff }\quad p\leq_P f_*(q).$$ We denote a Galois connection between $P$ and $Q$ by $P\stackrel{f}{\multimap}Q$ , or simply $P\multimap Q$ .

If we define $\le_P^{\prime}$ on $P$ by $a\le_P^{\prime}b$ iff $b\le_P a$ , and define $\le_Q^{\prime}$ on $Q$ by $c\le_Q^{\prime}d$ iff $d\le_Q c$ , then $(P,\le_P^{\prime})$ and $(Q,\le_Q^{\prime})$ are posets, (the duals of $(P,\le_P)$ and $(Q,\le_Q)$ ). The existence of a Galois connection between $(P,\le_P)$ and $(Q,\le_Q)$ is the same as the existence of a Galois connection between $(Q,\le_Q^{\prime})$ and $(P,\le_P^{\prime})$ . In short, we say that there is a Galois connection between $P$ and $Q$ if there is a Galois connection between two posets $S$ and $T$ where $P$ and $Q$ are the underlying sets (of $S$ and $T$ respectively). With this, we may say without confusion that ``a Galois connection exists between $P$ and $Q$ iff a Galois connection exists between $Q$ and $P$ ''.

Remarks.

1. Since $f^*(p)\leq_Q f^*(p)$ for all $p\in P$ , then by definition, $p\leq_P f_*f^*(p)$ . Alternatively, we can write \begin{eqnarray}1_P\leq_P f_*f^*, \end{eqnarray}where $1_P$ stands for the identity map on $P$ . Similarly, if $1_Q$ is the identity map on $Q$ , then \begin{eqnarray}f^*f_*\leq_Q 1_Q. \end{eqnarray}
2. Suppose $a\leq_P b$ . Since $b\leq_P f_*f^*(b)$ by the remark above, $a\leq_P f_*f^*(b)$ and so by definition, $f^*(a)\leq_Q f^*(b)$ . This shows that $f^*$ is monotone. Likewise, $f_*$ is also monotone.
3. Now back to Inequality (1), $1_P\leq_P f_*f^*$ in the first remark. Applying the second remark, we obtain \begin{eqnarray}f^*\leq_Q f^*f_*f^*. \end{eqnarray}Next, according to Inequality (2), $f^*f_*(q)\leq_Q q$ for any $q\in Q$ , it is true, in particular, when $q=f^*(p)$ . Therefore, we also have \begin{eqnarray}f^*f_*f^*\leq_Q f^*.\end{eqnarray}Putting Inequalities (3) and (4) together we have \begin{eqnarray}f^*f_*f^*=f^*.\end{eqnarray}Similarly, \begin{eqnarray}f_*f^*f_*=f_*.\end{eqnarray}
4. If $(f,g)$ and $(f,h)$ are Galois connections between $(P,\le_P)$ and $(Q,\le_Q)$ , then $g=h$ . To see this, observe that $p\le_P g(q)$ iff $f(p)\le_Q q$ iff $p \le_P h(q)$ , for any $p\in P$ and $q\in Q$ . In particular, setting $p=g(q)$ , we get $g(q)\le_P h(q)$ since $g(q)\le_P g(q)$ . Similarly, $h(q)\le_P g(q)$ , and therefore $g=h$ . By a similarly argument, if $(g,f)$ and $(h,f)$ are Galois connections between $(P,\le_P)$ and $(Q,\le_Q)$ , then $g=h$ . Because of this uniqueness property, in a Galois connection $f=(f^*,f_*)$ , $f^*$ is called the upper adjoint of $f_*$ and $f_*$ the lower adjoint of $f^*$ .

Examples.

• The most famous example is already mentioned in the first paragraph above: let $L$ is a finite-dimensional Galois extension of a field $F$ , and $G:=\operatorname{Gal}(L/F)$ is the Galois group of $L$ over $F$ . If we define

  a.

  $P=\lbrace K\mid K\mbox{ is a field such that }F\subseteq K\subseteq L \rbrace,$ with $\leq_P=\subseteq$ ,

  b.

  $Q=\lbrace H\mid H\mbox{ is a subgroup of }G \rbrace,$ with $\leq_Q=\supseteq$ ,

  c.

  $f^*:P\to Q$ by $f^*(K)=\operatorname{Gal}(L/K)$ , and

  d.

  $f_*:Q\to P$ by $f_*(H)=L^H$ , the fixed field of $H$ in $L$ .

  Then, by the fundamental theorem of Galois theory, $f^*$ and $f_*$ are bijections, and $(f^*,f_*)$ is a Galois connection between $P$ and $Q$ .
• Let $X$ be a topological space. Define $P$ be the set of all open subsets of $X$ and $Q$ the set of all closed subsets of $X$ . Turn $P$ and $Q$ into posets with the usual set-theoretic inclusion. Next, define $f^*:P\to Q$ by $f^*(U)=\overline{U}$ , the closure of $U$ , and $f_*:Q\to P$ by $f_*(V)=\operatorname{int}(V)$ , the interior of $V$ . Then $(f^*,f_*)$ is a Galois connection between $P$ and $Q$ . Incidentally, those elements fixed by $f_*f^*$ are precisely the regular open sets of $X$ , and those fixed by $f^*f_*$ are the regular closed sets.

Remark. The pair of functions in a Galois connection are order preserving as shown above. One may also define a Galois connection as a pair of maps $f^*:P\to Q$ and $f_*:Q\to P$ such that $f^*(p)\le_Q q$ iff $f_*(q)\le_P p$ , so that the pair $f^*,f_*$ are order reversing. In any case, the two definitions are equivalent in that one may go from one definition to another, (simply exchange $Q$ with $Q^{\partial}$ , the dual of $Q$ ).

Bibliography

1

T.S. Blyth, Lattices and Ordered Algebraic Structures, Springer, New York (2005).

2

B. A. Davey, H. A. Priestley, Introduction to Lattices and Order, 2nd Edition, Cambridge (2003)