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Grothendieck's Galois Theory, as developed in SGA I, V.4, or very gently in Lenstra's notes, establishes an equivalence between profinite groups and Galois categories. We can put this into the following more general framework: Let $\mathsf{sets}$ denote the category of finite sets; for a topological group $\pi$ let $\pi\mathsf{-sets}$ denote the category of finite sets on which $\pi$ acts continuously. There are functors

$\mathsf{Cat}/\mathsf{sets} \leftrightarrow \mathsf{TopGrp}^{\mathrm{op}}$

given by $(F : C \to \mathsf{sets}) \mapsto \mathrm{Aut}(F)$ (which is topologized as a closed subgroup of $\prod_{X \in C} \mathrm{Aut}(F(X))$) and in the other direction by $\pi \mapsto \pi\mathsf{-sets}$. In order to avoid set-theoretic problems, $C$ and $\pi$ should be essentially small. There are canonical maps $\eta_C : C \to \mathrm{Aut}(F)\mathsf{-sets}$ and $\varepsilon_{\pi} : \pi \to \mathrm{Aut}(\pi\mathsf{-sets})$ which satisfy the triangular identities; hence we have an adjunction! And every adjunction restricts to an equivalence of categories between its fixed points. Now it's fairly easy to recognize profinite groups as those fixed points on the right hand side, but Grothendieck's nontrivial insight is the classification of the fixed points on the left hand side, which he calls Galois categories (in short: $C$ has finite limits and colimits, which $F$ preserves, there are mono-epi decompositions, monos split off, and $F$ is conservative).

Question 1. Does this point of view of Grothendieck's Galois theory have already appeared somewhere?

My real question is the following: Actually $\mathsf{Cat}/\mathsf{sets}$ is a $2$-category. A morphism $(C,F) \to (C',F')$ is a functor $P : C \to C'$ together with a chosen isomorphism $F \cong F' P$. A $2$-morphism between morphisms $P,Q : C \to C'$ is a natural transformation of the underyling functors, which is base-point preservering in the obvious sense.

Question 2. How can we endow $\mathsf{TopGrp}^{\mathrm{op}}$ with the structure of a $2$-category in such a way that the adjunction above becomes an $2$-adjunction?

- - Edit - - The comments + answers make me believe that it is just the "trivial" $2$-categorical enrichment with identities as $2$-morphisms. And since there is no reaction to Q1, I expect that the answer is "no, this is new"?

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    $\begingroup$ Have you looked at the 2-category described in math.ucr.edu/home/baez/twf_ascii/week176 ? $\endgroup$
    – S. Carnahan
    Jun 12, 2012 at 8:01
  • $\begingroup$ Only a idea: let $I$ the category $0\to 1$ and $i: I\to Set$ the natural inclusion. Let $A: \mathcal{A}\to Set,\ B: \mathcal{A}\to Set$ and let $F_0, F_1: (A, \mathcal{A})\to (B, \mathcal{B})$, a 2-cell betwenn $F_0$ and $F_1$ is identified by a functor $\phi: (i\times A, I\times \mathcal{A})\to ( B, \mathcal{B})$ in $Cat\downarrow Set$. Then "traslate" this by the adjunction on $TopGrp^{op}$ and see if it can work .. $\endgroup$ Jun 12, 2012 at 8:20
  • $\begingroup$ A 2-cell between $P, Q$ give the the identity when composed by $F'$, then by Godement law on 2-cell composition induce the identity maps on the automorphism groups. $\endgroup$ Jun 12, 2012 at 8:38
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    $\begingroup$ It seems to me that the 2-category of Galois categories is in fact equivalent to a 1-category. $\endgroup$
    – Angelo
    Jun 12, 2012 at 11:31
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    $\begingroup$ I would like to understand groups first before diving into topos theory. Sometimes generalizations are necessary, of course, but I don't see why my question should need this in order to make sense or have a meaningful answer. $\endgroup$ Jun 13, 2012 at 10:40

2 Answers 2

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EDIT: answer is expanded on OP's request.

The category $\pi Set_{fin}$ of finite, continuous $\pi$-sets is just the functor category $TopCat(\mathbf{B}\pi,finSet)$ where $\mathbf{B}$ gives the one-object topological groupoid associated with a topological group, and $finSet$ is the category of finite sets, viewed as a topologically discrete category. The groupoid $\mathbf{B}\pi$ is canonically pointed, so induces the functor $\pi Set_{fin} \to finSet$ i.e. an object of $Cat/finSet$. If you replace $TopGrp$ with the equivalent category $TopGpd_{1obj}$ of one-object topological groupoids, then this naturally comes with a notion of 2-arrow, and it is, as John Baez points out, conjugation by an element of the codomain. Then 2-arrows in $\pi Set_{fin}$ are natural transformations of functors to $finSet$.

Now really you are working with $TopGrp^{op}$, so you have two choices as to the direction of the 2-arrows, if you are taking 2-arrows as specified above. This should fall out of the definitions. Also, since $TopGpd_{1obj}$ is a (2,1)-category, you need to restrict attention to the (2,1)-category $Cat/finSet_{(2,1)}$ underlying $Cat/finSet$. Here $Cat/finSet_{(2,1)}$ is the isocomma category: objects are categories over $finSet$, arrows are triangles commuting up to a natural isomorphism and 2-arrows are natural isomorphisms in $Cat$ that are compatible with the 2-arrows in the triangles.

In more detail: given two objects $F\colon C\to finSet$ and $G\colon D\to finSet$, an arrow $F\to G$ is a pair consisting of a functor $f\colon C\to D$ and a natural isomorphism $c\colon G\circ f \Rightarrow F$. Given two arrows, $(f,c),(g,d)\colon F\to G$, a 2-arrow between them is a natural isomorphism $a\colon f\Rightarrow g$ such that the obvious 3-dimensional diagram commutes. Since all 2-arrows are invertible, from this commuting 3-d diagram we can write down an invertible endo-arrow $c+(G\circ a)+d^{-1}$ of $F$ (here $+$ is vertical composition of natural transformations--from right to left--$G$ really denotes the identity transformation on the functor $G$ and $\circ$ is horizontal composition).

I'm fairly confident that this is a 2-arrow in $TopGpd_{1obj}$. If one doesn't restrict to the (2,1)-category $Cat/Set_{(2,1)}$ then this doesn't work. (So really you should be thinking about $Gpd$-enriched categories, not so much general 2-categories.) EDIT: Actually, it is easy to see that this is a 2-arrow in $TopGpd_{1obj}$, because the homomorphism induced by $(f,c)$ is conjugation by $c$ (with whiskering): $\alpha \mapsto c+(f\circ \alpha)+c^{-1}$.

Thus we have a 2-functor $Cat/finSet_{(2,1)}\to TopGpd_{1obj}^{op}$.

Now in going the other way, I think we actually need to use not just $TopGpd_{1obj}$, but the isococomma category $\ast/TopGpd_{1obj}$, where a functor between topological groupoids respects the basepoint up to a 2-arrow (which is an automorphism of the canonical basepoint). I haven't checked but this looks like the description of the functor in the preceeding paragraphs works better. Then the functor $\ast/TopGpd_{1obj}^{op} \to Cat/finSet_{(2,1)}$ is just exponentiation with $finSet$, and we don't have to think too hard about what the 2-arrows etc do.

Then you need to worry about the adjunction. But if you already have a 1-adjunction, half the work is done.

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  • $\begingroup$ The terminology I'm using is the following (see also Lenstra/Grothendieck): $\pi\mathsf{-sets}$ is the category of finite sets on which $\pi$ acts continuously. In your answer you seem to talk about all sets on which $\pi$ acts as if it were a discrete group. Could you explain how every arrow $f \rightarrow g : F \to G$ yields a canonical element in $\mathrm{Aut}(F)$? $\endgroup$ Jun 13, 2012 at 10:37
  • $\begingroup$ For a topological group one usually uses continuous G-sets (discrete sets with a cts action). For a profinite group this is the Ind-completion of the finite G-sets (which are the coherent objects of the topos) and so basically looking at finite or arbitrary G-sets doesn't matter in this case. $\endgroup$ Jun 13, 2012 at 13:21
  • $\begingroup$ Ah, finite sets. That's perfectly fine. Consider $\mathbf{B}\pi$ as a topological groupoid, $finSet$ as a topological category --actually just need its core--and look at functors between topological groupoids. Everything else works fine replacing $Set$ by $finSet$. As far as the canonical element of $Aut(F)$ goes, I'm taking (and I assumed you did too) $Cat/finSet$ to be the isocomma (2-)category, where arrows are triangles commuting up to a specified invertible 2-arrow, and 2-arrows are natural isomorphisms in $Cat$ that cooperate with the 2-commuting triangles. Just writing down what a.. $\endgroup$
    – David Roberts
    Jun 13, 2012 at 23:03
  • $\begingroup$ ...2-arrow in $Cat/finSet$ is gives you an automorphism of its 0-source. I can write more detail in my answer if you like. $\endgroup$
    – David Roberts
    Jun 13, 2012 at 23:04
  • $\begingroup$ Thanks a lot for expanding your answer! So in the definition of a $2$-arrow $a : f \to g$, we have a natural transformation of functors such that for all $x \in X$ the composite of $G(a_x) : G(f(x)) \to G(g(x))$ with $d : G(g(x)) \cong F(x)$ equals $c : G(f(x)) \cong F(x)$, right? This is what I already mentioned in my question ("natural transformation of the underyling functors, which is base-point preservering in the obvious sense."). It follows that $G(a_x)$ is an isomorphism; when $G$ is conservative (which is the case for Galois categories) the same follows for $a_x$. So yes to $(2,1)$! $\endgroup$ Jun 14, 2012 at 8:03
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This was going to be a comment but got too long. First of all your framework is I think slightly off. You want to associate to a topological group G its classifying topos of continuous G-sets and its canonical point (the underlying set functor). The underlying set functor is pro-representable. Thus I think you really want to look at the category consisting of pro-representable functors and you want to assign the automorphism group of this functor which is a strict pro-discrete group, whose projective limit should be seen in the category of locales. There is a paper of Moerdjik, Prodiscrete groups and Galois toposes, giving the corresponding Galois theory.

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  • $\begingroup$ This might be interesting, but is not really answering my question. You just say that my question is the wrong one ... $\endgroup$ Jun 13, 2012 at 6:32
  • $\begingroup$ It was an extended comment not an answer. In any event you asked if this viewpoint was ever taken and I think Moerdijk's view is related. $\endgroup$ Jun 13, 2012 at 13:18

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