This is related to my previous question here, and indirectly motivated by Andreas Blass intriguing answer therein:

start from a ZF transitive model $M$, and consider the category $CBA(M)$ of complete boolean algebras of $M$ with the obvious boolean maps.

This category corresponds to the category of boolean-valued extensions of $M$, via the Scott-Solovay-Vopenka (SSV) construction. $CBA(M)$ contains the initial algebra $2$ (which corresponds to the trivial extension), and the forcing extensions of $M$ are the maps $B\rightarrow 2$ (ie the ultrafilters; notice that these maps in general do not exist in $M$).

Now the questions:

1) We can also consider maps $B_1\rightarrow B_2$ (the $B_2$ "points"of $B_1$). What do they represent when one goes to the boolean valued models? I would think they stand for "relative extensions", but can one make this precise?

2) Similarly, one can construct, given two such algebras $B_1$, $B_2$ where $B_1$ is a subalgebra of $B_2$, the Galois group $Gal(B_2/B_1)$, of boolean automorphisms of $B_2$ leaving fixed the ground algebra $B_1$. Can one use Gal to classify relative extensions?

3) (change of base model) Let $M_1$ and $M_2$ two transitive models (ie two objects of the ZF-multiverse, here to be thought of as the large category of transitive models of ZF), and a map of the first to the second, it looks like this induces a functor between $CBA(M_1 )$ and $CBA(M_2 )$ (something akin to the change of base of bundles in geometry). Is this (alleged) functor strong enough to capture the isomorphisms of the underlying multiverse?

More generally, what is the relation of the 2-valued multiverse to the extended multiverse of boolean valued models?


My recent paper with Dan Seabold,

has various sections describing the basic algebraic connections between the Boolean ultrapower and the corresponding Boolean forcing extensions and how they relate to subalgebras, ideals, quotients, products limits and so on, and this may be a part of the solution you seek.

Meanwhile, to my way of thinking, the most important thing to say about the Galois connection aspect of forcing extensions is the following.

Theorem. Suppose that $V\subset V[G]$ is a forcing extension arising by forcing with a complete Boolean algebra $\mathbb{B}$, where $G\subset\mathbb{B}$ is $V$-generic. Then every intermediate model $M$ of ZFC with $V\subset M\subset V[G]$, is itself a forcing extension of $V$, and specifically $M=V[G_0]$ where $G=G\cap\mathbb{B}_0$ for some complete subalgebra $\mathbb{B}_0\subset\mathbb{B}$. Furthermore, $V[G]$ is also a forcing extension of $M$ by the quotient forcing $\mathbb{B}/G_0$. Conversely, every complete subalgebra $\mathbb{B}_0\subset\mathbb{B}$ gives rise to such an intermediate forcing extension $V[G_0]$, where $G_0=G\cap\mathbb{B}_0$.

Thus, there is a connection between the intermediate models of ZFC in a forcing extension and the complete subalgebras of the original Boolean algebra. The theorem is definitely not obvious, but is proved in any of the standard accounts of forcing, such as Jech's book Set Theory.

So the theorem provides a correspondence between intermediate models of a given forcing extension $V\subset M\subset V[G]$ and complete subalgebras $\mathbb{B}_0\subset \mathbb{B}$ of the complete Boolean algebra giving rise to the original extension, and the two intermediate extensions $V\subset M$ and $M\subset V[G]$ arising as forcing extensions by $\mathbb{B}_0$ and by the quotient $\mathbb{B}/G_0$, respectively.

But the theorem does not provide a Galois correspondence in terms of automorphism groups and fixed points of automorphisms. In part, this is because transitive sets simply have no nontrivial automorphisms. There are no nontrivial automorphisms of $V[G]$, since it is a transitive class, and so one wouldn't ordinarily want to consider the group of automorphisms of $V[G]$. Two transitive classes are isomorphic if and only if they are identical, and so that approach also is not fruitful in the context of forcing extensions of $V$. But you ask about automorphisms of the $\mathbb{B}$-valued structure, and here of course there will be a rich automorphism theory.

  • $\begingroup$ Thank Joel, the ref does indeed seem pertinent ! ( I will have to digest it slowly). As for your last point, yes, "squeezing down" to ordinary 2-valued extension does indeed limit (kills) the "galois" perspective: 2-valued models (ie ordinary models), are "rigid". However, I was thinking of classification without contraction: one thing that intrigues me a lot is that boolean valued models DO have plenty of non trivial automorphisms (no rigidity). So I guess there was one semantic inaccuracy in 2) above. When I used "Can one use Gal to classify...?" I meant without collapse via ultrafilters $\endgroup$ – Mirco A. Mannucci Jul 6 '12 at 23:35

Note that the appropriate notion of morphisms in $CBA(M)$ are ($M$-)complete embeddings and generic ultrafilters correspond to ($M$-)complete homomorphisms of $B$ onto $2$ with the caveat that these do not exist in $M$ when $B$ is non-atomic. (Though, as Joel pointed out, we can consider Boolean ultrapowers instead of generic extensions and relax the completeness requirements.)

  1. Yes and no. As described in this answer, if $G$ is a generic filter over $B$ and $M \subseteq N \subseteq M[G]$ is an intermediate model that is generated by a subset of $M$ in $M[G]$ then $N = M[G_0]$ where $G_0 = G \cap B_0$ and $B_0$ is a complete subalgebra of $B$ in $M$. When $M$ is a model of ZFC, then the intermediate models $N$ that are generated by a subset of $M$ are precisely the intermediate models of ZFC, so this does not always classify all the intermediate models of ZF.

  2. No. There are nontrivial complete Boolean algebras that have no nontrivial automorphisms but plenty of complete subalgebras (see references from Jech and Shelah, Simple Complete Boolean Algebras, arXiv:math.LO/0406438). In general, complete automorphisms of $B$ control which elements of $M[G]$ are definable using parameters from $M$: the more automorphisms there are the fewer elements are definable.

  3. Change of base model is always difficult because complete Boolean algebras and complete homomorphisms may fail to be complete in a larger universe and may fail to exist entirely in a smaller universe. One can always complete a Boolean algebra in a larger universe, but once a homomorphism fails to be complete it cannot be "corrected" to a complete homomorphism. (For example, the complete homomorphism $e:B\to 2$ corresponding to a generic ultrafilter $G$ on an atomless Boolean algebra $B$ fails to be complete in any universe that contains $e$ since $e(\bigwedge_{x \in G} x) = e(0) = 0$ but $\bigwedge_{x \in G} e(x) = \bigwedge_{x \in G} 1 = 1$.)

A good starting point for intermediate submodels of forcing extensions is Grigorieff's classic paper Intermediate Submodels and Generic Extensions in Set Theory, Annals of Mathematics 101 (1975), 447–490.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.