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I've been experiencing minor qualms about my preprint "A Galois Connection in the Social Network" (accepted by Mathematics Magazine, pending revisions), and one of them involves the way I describe the Galois connection underlying Galois theory in terms of a binary relation between individual elements of the field $E$ and individual elements of the group Gal($E/K$), rather than a binary relation between subfields of $E$ containing $K$ and subgroups of Gal($E/K$). Is there a problem with doing things in an element-by-element way? (The current draft of the article is on the web at http://jamespropp.org/galois.pdf , although you probably don't need to read it to think about this half of my question.)

Another qualm I have is that I suspect that some version of the social network "application" that I present ($K(K(K(S)))=K(S)$, where $K(S) = \ ${$t:s \sim t$ for all $s \in S$}, for some symmetric binary relation $\sim$) occurs in the contest-problem literature or the recreational math literature, and I'd appreciate relevant citations if anyone knows of them.

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Just skimming through your article, I don't see a connection between individual homomorphisms and individual elements. Rather, the binary relation is between individual field homomorphisms and sets of elements (under the usual partial-ordering by the subset relation). The fixed sets are fields, as you say. –  Pace Nielsen Jul 29 '11 at 17:15
I'm not sure what you mean by "problem." –  Qiaochu Yuan Jul 29 '11 at 17:48
My problem is, I was inclined to let $A$ and $B$ respectively be the power sets of $E$ and Gal($E/K$), rather than the set of subfields of $E$ and the set of subgroups of Gal($E/K$) (as is standard), and I want to know if there are pitfalls here that I'm not noticing. It seems to me that for any subset $S$ of $E$, the elements of Gal($E/K$) that fix $S$ form a subgroup of Gal($E/K$), and conversely, for any subset $T$ of Gal($E/K$), the elements of $E$ that are fixed by $T$ form a subfield of $E$. So I don't see anything to worry about. But there may be technicalities I'm not seeing. –  James Propp Jul 29 '11 at 19:51
@James: there's nothing to worry about. The closed sets of the corresponding closure operators turn out to be subgroups and subfields. –  Qiaochu Yuan Jul 29 '11 at 23:28
@jim It is a detail : in galois.pdf the property is symetric (not reflexive as given). –  Jérôme JEAN-CHARLES Jan 24 '13 at 1:20
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up vote 6 down vote accepted

You are right: there is nothing to worry about. What you are describing is quite commonplace with Galois connections. In fact, "elementwise" binary relations are arguably the number one source of Galois connections.

Let $R \subseteq X \times Y$ be any binary relation. Then there is an induced Galois connection, a pair of contravariant (i.e., order-reversing) maps

$$P(X) \stackrel{-\backslash R}{\to} P(Y): S \mapsto \{y \in Y: \forall_{x \in X} \ \ x \in S \Rightarrow R(x, y)\}$$

$$P(Y) \stackrel{R/-}{\to} P(X): T \mapsto \{x \in X: \forall_{y \in Y} \ \ y \in T \Rightarrow R(x, y)\}$$

for which

$$S \subseteq R/T \qquad \text{iff} \qquad S \times T \subseteq R \qquad \text{iff} \qquad T \subseteq S\backslash R$$

One may compose these operations in either direction,

$$S \mapsto R/(S \backslash R), \qquad T \mapsto (R/T)\backslash R$$

to get closure operations satisfying the standard axioms, e.g., $S \subseteq S'$ implies $\bar{S} \subseteq \bar{S'}$ and

$$S \subseteq \bar{S}, \qquad \bar{\bar{S}} = \bar{S}$$

for all $S \in P(X)$. This may be proven just by taking advantage of the three-way equivalence above together with the contravariance.

Given a Galois extension $K/k$ with Galois group $G$, the Galois connection of Galois theory is induced by the subset $R \subseteq G \times K$ defined by

$$\{(\sigma, x) \in G \times K: \sigma(x) = x\}$$

and there is nothing wrong with considering the induced Galois connection on power sets rather than the posets of subgroups and intermediate subfields. The closed elements in this case are the subgroups and intermediate subfields! This statement is more or less the substance of the fundamental theorem of Galois theory. But my real point here is the utter generality of these sorts of Galois connections.

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How general is this construction? Can I take a functor $X \times Y \to C$ for categories $X, Y, C$ and turn this into a pair of adjoint functors between $X$ and $Y$? If not, what extra conditions do I need? –  Qiaochu Yuan Jul 29 '11 at 23:29
The relation of satisfiability between algebraic structures and identities (universally quantifies equations which are sentences) is an example involving a class and a set that gives rise to a Galois connection of equationally closed theories and classes of structures closed under certain operators. If the original poster takes care to call a class a class, a set a set, and not mix the two, then again there are no worries. Gerhard "Ask Me About System Design" Paseman, 2011.07.29 –  Gerhard Paseman Jul 30 '11 at 0:17
@Qiaochu: it's very general, as you surmise. One generalization considers $V$-enriched profunctors $R: X \times Y \to V$ for say a bicomplete symmetric monoidal closed category $V$. (The classical case discussed above takes $V = \\{0 \leq 1\\}$.) Then you can get a contravariant enriched adjunction between the functor categories $V^X$ and $V^Y$. Briefly, using the monoidal structure on $V$, there's an obvious way of combining two functors $S: X \to V$, $T: Y \to V$ to get a functor $S \otimes T: X \otimes Y \to V$. Then, using completeness and the closed structure of $V$, it is possible [cont] –  Todd Trimble Jul 30 '11 at 0:42
to write down analogous formulas for $R/T$ and $S \backslash R$ so that one has $V$-isomorphisms $\hom(S, R/T) \cong \hom(S \otimes T, R) \cong \hom(T, S\backslash R)$$ between enriched homs. These give you the adjunctions you need to make a complete analogy. –  Todd Trimble Jul 30 '11 at 0:45
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