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Galois connection (Definition)

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:

$\displaystyle \operatorname{Gal}(L/F)\supseteq H\supseteq\langle e \rangle$    iff $\displaystyle \quad F\subseteq L^H\subseteq L,$ and
$\displaystyle F\subseteq K\subseteq L$    iff $\displaystyle \quad \operatorname{Gal}(L/F)\supseteq \operatorname{Gal}(L/K)\supseteq\langle e \rangle.$

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

$\displaystyle f^*(p)\leq_Q q$    iff $\displaystyle \quad p\leq_P f_*(q).$
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_P^{\prime}d$ iff $ c\le_P d$, 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
    $\displaystyle 1_P\leq_P f_*f^*,$     (1)

    where $ 1_P$ stands for the identity map on $ P$. Similarly, if $ 1_Q$ is the identity map on $ Q$, then
    $\displaystyle f^*f_*\leq_Q 1_Q.$     (2)

  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
    $\displaystyle f^*\leq_Q f^*f_*f^*.$     (3)

    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
    $\displaystyle f^*f_*f^*\leq_Q f^*.$     (4)

    Putting Inequalities (3) and (4) together we have
    $\displaystyle f^*f_*f^*=f^*.$     (5)

    Similarly,
    $\displaystyle f_*f^*f_*=f_*.$     (6)

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$ is a field such that $ F\subseteq K\subseteq L \rbrace,$ with $ \leq_P=\subseteq$,
    b.
    $ Q=\lbrace H\mid H$ 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.



"Galois connection" is owned by CWoo. [ full author list (2) ]
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See Also: interior axioms

Other names:  Galois correspondence, Galois connexion
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Cross-references: regular closed, regular open sets, fixed, interior, closure, inclusion, closed subsets, open subsets, topological space, fixed field, Galois extension, finite-dimensional, inequality, monotone, identity map, iff, functions, posets, theory, language, Galois group, subgroups, extension, subfield, fields, one-to-one correspondence
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This is version 5 of Galois connection, born on 2005-03-16, modified 2006-12-26.
Object id is 6881, canonical name is GaloisConnection.
Accessed 3255 times total.

Classification:
AMS MSC06A15 (Order, lattices, ordered algebraic structures :: Ordered sets :: Galois correspondences, closure operators)

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What does this mean? by zabrocki on 2005-08-26 10:11:12
I'm having a problem with the following sentence in this article:
"The existence of a Galois connection between P and Q is the same as the existence of a Galois connection between Q and P so the definition is unambiguous."

Is this right? Does it use the same maps? f^* and f_*, just interchanged?
I can't seem to prove this.
Perhaps I am just misundertanding.

-Mike Zabrocki
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