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There is a (very) long essay by Grothendieck with the ominous title La Longue Marche à travers la théorie de Galois (The Long March through Galois Theory). As usual, Grothendieck knew what he was talking about: Galois Theory, far from being confined to its primary example, namely field extensions, is pervasive throughout mathematics, and still to be fully understood.

Ever since I first heard of forcing I was struck by its compelling analogy with field extensions: the ground model (read Q), the generic G (read a new element, say $\sqrt{2}$), the new M[G] (read Q[$\sqrt{2}$]), etc.

I quote Joel Hamkins 's words here, in his sparkling paper on the multiverse:

In effect, the forcing extension has adjoined the “ideal” object G to V , in much the same way that one might build a field extension such as Q[$\sqrt{2}$]

Of course, matters are a bit more complicated in set theory, you have to make sure the extension satisfies the axioms, that it does not "bother" the ordinals, and so on.

Yet, one cannot really think that the analogy stops here.

And it does not: in Galois theory, the main thing is the central theorem, establishing the Galois Connection between subfields of the extension and the subgroups of the Galois group, ie the group of the automorphisms of the extension leaving fixed the underlying field.

No Galois connection, no Galois Theory. But wait, first we need the group. So where is it?

A hint is in the great classic result by Jech-Sochor on showing the independence of AC: by considering a group of automorphisms of P, the ordered set of forcing conditions, one can obtain a new model which is (essentially) the set of fixed points of the induced automorphisms. This is even clearer when one looks at it from the point of view of boolean valued models: each automorphism of the boolean algebra induces an automorphism of the extended universe.

Now my question: is there some systematic work on classifying forcing extensions by their Galois group? Can one develop a full machinery which will apply to relative extensions?

NOTE: I think this is no idle brooding: someone for instance has asked here on MO about the 2-category of the multiverse. That is a tough question, and my sense is that before giving it a satisfactory answer some preliminary work needs to be done. Which work? Well, one needs to re-think the classical set constructions from a structural standpoint, leaving behind its gory technical details. Now, forcing is a huge part of the multiverse, and understanding the structural algebra underpinning it would be, I trust, a huge step toward an algebraic understanding of the multiverse.

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I hope that you do not mind, but I added a "multiverse" tag. –  Spice the Bird Jun 24 '12 at 5:14
    
Why care about the Jech-Sochor theorem in this aspect? You can do this directly with forcing. –  Asaf Karagila Jun 24 '12 at 5:33
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Why restricting to field extensions? Abstract Galois theory is available in a bunch of categories, namely Galois categories (perhaps even in a more general topos-theoretic context, but I can't tell any details). You can read all about this in Lenstra's notes Galois schemes for schemes. A original reference is SGA1. So perhaps you should look for a Galois category in the context of Forcing. In any case, I hope that there won't just be an "analogy" between Forcing and Galois Theory. Rather it would be nice if Forcing can be embedded into Grothendieck's general picture of Galois theory. –  Martin Brandenburg Jun 24 '12 at 7:39
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Spice, I do nit mind at all. Asaf, the Jech-Sochor is not the focus here, it is simply a step toward the galoisation of forcing. Martin, I most emphatically do not think of forcing as field extensions! All the things you quoted are the direct offspring of Grothendieck's intuition on the pervasiveness of galois. and yes, the goal would be that forcing were a chapter of the universal galois theory, as my title indicates –  Mirco Mannucci Jun 24 '12 at 10:28
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@Jon I'm talking about the homotopical structure where the weak equivalences are "local isomorphisms", i.e. the morphisms that become isomorphisms after sheafifying. Kashiwara and Schapira's textbook talks about Grothendieck topologies in these terms, but I don't think it's all that helpful. –  Zhen Lin Jan 14 '13 at 8:45
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3 Answers

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Yes and No... There are strong parallels between forcing and symmetric extensions and field extensions and this way of thinking has been fruitful. However, like in the case of general ring extensions and group extensions and similar problems, this analogy is not perfect and pushing the similarity too far may actually obscure what is really going on.

That said, symmetric extensions do indeed show a great deal of similarity with Galois theory. Some work, notably that of Serge Grigorieff [Intermediate Submodels and Generic Extensions in Set Theory, Annals of Mathematics 101 (1975), 447–490] shows that there is indeed a way to understand intermediate forcing extensions in a manner extremely similar to the way understand field extensions through Galois theory. Some have even pushed this analogy so far as to study some problems roughly analogous to the Inverse Galois Problem in this context, for example [Groszek and Laver, Finite groups of OD-conjugates, Period. Math. Hungar. 18 (1987), 87–97].

There are some very algebraic ways of understanding forcing and symmetric models in a more global sense. For example, forcing extensions correspond in a well-understood way to the category of complete Boolean algebras under complete embeddings. Moreover, the automorphism groups of these algebras plays a crucial role in our understanding of the inner model structure of forcing extensions. An even farther reaching approach comes from transposing the sheaf constructions from topos theory into the set-theoretic context [Blass and Scedrov, Freyd's models for the independence of the axiom of choice, Mem. Amer. Math. Soc. 79 (1989), no. 404]. One could argue that this suggests a stronger analogy with topology rather than algebra, but there are plenty of very deep analogies between Galois theory and topology.

The above is not a complete survey of these types of connections, it is just to demonstrate that connections do exist and that they have been looked at and useful for a long time. Because of the relative sparsity of the literature, one could argue that these aspects are underdeveloped but that is hasty judgement. The truth is that there appear to be disappointingly few practical aspects to this kind of approach, perhaps because they are not relevant to most current questions in set theory or perhaps for deeper reasons. For example, the inner model structure of the first (and largely regarded as the simplest) forcing extension, namely the simple Cohen extension, is incredibly rich and complex [Abraham and Shore, The degrees of constructibility of Cohen reals, Proc. London Math. Soc. 53 (1986), 193–208] and there does not appear to be a reasonable higher-level approach that may help us sort through this quagmire in a similar way that Galois theory can help us sort through the complex structure of $\overline{\mathbb{Q}}$.

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Two questions: 1) is the Abraham/Shore paper the most recent paper on this kind of result? 2) As far as trying to construct an analogy with Galois theory might go, a thought occurs: fix a poset P in a ground model M, and a generic G. Then there is the "usual" forcing extension M[G]. However, there are also the intermediate symmetric submodels of M[G], given by normal filters on the group of automorphisms of P. Each normal filter corresponds to a symmetric submodel. Can we go the other way? That is, is there a natural way to assign to a symmetric submodel a normal filter generating it? (cont'd) –  Noah S Jun 24 '12 at 4:45
    
A stronger question: is there some property of normal filters (call it "excellence" for now) such that every symmetric submodel is generated by a single excellent normal filter? This would seem to be the exact analog of the main theorem of Galois theory. Is there any work on this? –  Noah S Jun 24 '12 at 4:46
    
Okay, assuming the ground model satisfies global choice, I think there is a way of assigning normal filters to symmetric submodels in a 1-1 way, but it seems silly and not very interesting. I'm still looking for a natural way. –  Noah S Jun 24 '12 at 5:34
    
1) No, it is not. There are more results (by Adamowicz, Groszek, Shore, and myself, among others). Not all is published; I guess it's time for a new survey... 2) Yes, what you describe is essentially what is in Grigorieff's paper. –  François G. Dorais Jun 24 '12 at 6:07
    
3) It's not completely clear to me what you want with your excellent normal filters. Some of my results may be along those lines, but I'm not sure. (Feel free to contact me via email if the comment boxes are too small.) –  François G. Dorais Jun 24 '12 at 6:14
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It seems to me that the groups involved in producing models without AC (whether as groups of permutations of atoms or as groups of automorphisms of complete Boolean algebras) and the forcing constructions themselves are two rather separate things. The most frequent use of forcing is to produce models of ZFC (not just ZF), and then groups are not involved. On the other hand, the Fraenkel-Mostowski-Specker method of permutation models for the negation of AC involves only permutation groups (and normal filters of subgroups), not forcing. Groups and forcing come together in Cohen's method of symmetric models. In the original (and still most common) presentation, one has a group acting on the forcing notion. But even here, the groups and the forcing are less entangled than they seem. Vopenka and Hajek showed (in their book "Theory of Semisets") how to get Cohen-style models by (1) starting with a permutation model, (2) forcing over it, and (3) passing to the pure (or well-founded) part. In this presentation, the groups (and filters of subgroups) are only in step (1), and the forcing is only in step (2).

It seems to me that, if one wants to analyze forcing from an algebraic point of view, one should begin with the simplest case, where no symmetries are involved. The algebraic side of this is the study of complete Boolean algebras. Afterward, one can embellish the picture by adding group actions, either acting on the Boolean algebras or producing permutation models over which to force. The Grigorieff paper that Francois cited is an excellent place to start.

Though the OP asked about forcing, let me also mention that the non-forcing context of permutation models might be a better place to look for Galois-like phenomena. To begin, note that a normal filter $\mathcal F$ of subgroups of a group $G$ makes $G$ a topological group, in which $\mathcal F$ generates the neighborhood filter at the identity. The notion of symmetry with respect to $\mathcal F$ that is used in defining the permutation model is just continuity of the standard action of $G$ on sets.

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Andreas, yes, I mentioned the Jech-Sochor embedding (ie doing mostowski without using native urelementen) only because it brings forth an interesting point, namely the fact that automorphisms of P (or the associated complete boolean algebra) induce automorphisms of the boolean valued model. Now, I think this simple fact alone has some potential for a galois theory of forcing. Start from a model M, and consider the category CBA(M) of complete boolean algebras of M. Maps are just the obvious ones. In particular 2 is there, as the initial object (it corresponds of course to M itself, –  Mirco Mannucci Jul 4 '12 at 18:21
    
. Now, I can do Galois inside CBA(M), by studying Gal(B1/B2), the automorphisms of B1 leaving fixed B2, as B1 and B2 are two objects in the cat. What does Gal(B1/B2) tell us when we go to the corresponding boolean expansions of M? And we can also go back to standard forcing (ie models in the traditional sense) by considering equivariant ultrafilters... –  Mirco Mannucci Jul 4 '12 at 18:25
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Just a minor remark about the possibility of classifications:

Galois theory can begin by examining finite-degree algebraic extensions whose Galois group is finite and well understood. Later you can start considering larger and larger groups, but these are less well understood than before.

With forcing you often begin with "locally uncountable" groups (read: uncountability in the ground model). We add to the mix the fact that there may be new automorphisms in the generic extension (and often there are), this is not a good idea.

There is, however, something to be said for sure. I recently talked about this with a friend who (much like me) spends most of his time in an inner model without choice. We agreed that there is probably something Galois-like to say on the stabilizer groups of every $P$-name. Whether or not it is deep, we haven't got a chance to investigate further.

Lastly, I would add that the analogy of field theory and forcing can only go so far. The algebraic closure is unique (assuming enough choice, at least) whereas generic extensions of the same forcing need not be isomorphic.

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"generic extensions of the same forcing need not be isomorphic." They cannot be isomorphic, unless they are equal. (By Mostowski's collapsing theorem.) –  Andres Caicedo Jun 24 '12 at 5:53
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Andres points at something important that hasn't been stressed enough in this discussion, I think. A transitive model of set theory does not have any non-trivial automorphism. Hence the literal Galois group of a forcing extension is always trivial and in order to come up with an interesting theory it is necessary to look stabilizers of names and so on, as pointed out above. –  Stefan Geschke Jun 24 '12 at 9:58
    
"does not have any non-trivial automorphism" what in this case is an automorphism? Given that the OP mentioned a structural viewpoint, one could imagine that a self-equivalence of the category of sets in the extension fixing the sets in the base up to isomorphism might be a more appropriate automorphism than what is usually used in (material) set theory. –  David Roberts Jun 24 '12 at 23:39
    
@Downvoter: I enjoyed the downvote, thanks! I hope you at least bothered to read the answer... –  Asaf Karagila Jun 28 '12 at 17:05
    
@Asaf: I apologize for not replying (yet) to the many answers and comments on this question. My only excuse is: I was simply overwhelmed, and needed time to process the data. Rest assured, though, That I will reply to all, when digestion is complete. –  Mirco Mannucci Jun 30 '12 at 19:38
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