Symmetric submodels of $M[G]$ are exactly the classes $(HOD(M[x]))^{M[G]}$

Question

I'm trying to understand Grigorieff's proof (Theorem 3, page 478, Intermediate Submodels and Generic Extensions in Set Theory) that symmetric submodels of $M[G]$ are exactly $(HOD(M[x]))^{M[G]}$ for $x\in M[G]$. I'm struggling to follow the proof (this is not helped by the fact that the notation is quite dated and the proof refers to many previous theorems each of which refers to many other previous theorems). I was wondering if anyone was either able to provide a more straightforward proof (I'm not sure if this is possible but the paper was written in 1975 so hopefully the proof can be cleaned up) or direct me towards a source which does this? Thanks.

Answer 1

In an upcoming work, Jonathan Schilhan and I have managed to clarify and improve some of the related work and concepts from that paper and related to that proof.

But let me give you a nutshell explanation of Grigorieff's argument, from a modern perspective.

Prelude.

Nowadays, when I write about symmetric extensions, $\Bbb P$ is my forcing, $\scr G$ is a group of automorphisms, and $\scr F$ is a normal filter of subgroups of $\scr G$. But we can change $\Bbb P$, and as long as it is forcing equivalent, the result "shouldn't matter", so the best way to deal with that is to consider the Boolean completion (of the separative quotient) of $\Bbb P$.

Next, we can notice that every filter on any automorphism group of $\Bbb P$ extends to a natural filter on the entire automorphism group of $\Bbb P$ (and of course, its Boolean completion, should it not be a complete Boolean algebra). But now, a normal filter of subgroups of $\Bbb P$ is not necessarily normal anymore. So we need to weaken that definition and require that the "normaliser" is in the filter, rather than it is the entire group. This means that the symmetric system depends only on the Boolean completion of $\Bbb P$ and the filter. So, let's assume we only work with complete Boolean algebras. Great. Now when we talk about a symmetric system, supported over a fixed forcing, we really just vary all the normal filters of groups over its automorphism group.

One direction (with regards to Harry Styles)

The idea that a symmetric extension has the form $\mathrm{HOD}(M(x))^{M[G]}$, to use a slightly more modern notation, is quite straightforward. That is, indeed, the "easy" part of the proof.

The notation $\mathrm{OD}(M(x))^{M[G]}$ is the class of all those sets in $M[G]$ which are definable from an ordinal parameter as well as an element of $M(x)$; and $M(x)$ is the smallest model of $\sf ZF$ which extends $M$ and contains $x$ as an element. $\mathrm{HOD}(M(x))^{M[G]}$ is simply those elements of $\mathrm{OD}(M(x))^{M[G]}$ whose transitive closure is also a subset of the class.

First, let's allow parameters in general from $M[G]$. If $\scr G$ is the normaliser of $\scr F$, that is, from a modern perspective, this would be the automorphisms group we wish to focus on, then a name $\dot x$ is $\scr F$-symmetric if and only if $\exists K\in\scr F$ such that for all $\pi\in K$, $\dot x^G=\dot x^{\pi``G}$.

This is because $\dot x^{\pi``G}=(\pi^{-1}\dot x)^G$. So if $\pi\dot x=\dot x$, then $\pi^{-1}\dot x=\dot x$, and then the whole thing works out. So the above, literally, says that $\dot x$ is stable under a large group. Since if this happens that a name is hereditarily symmetric, this happens within some $V_\alpha^M$, we get that the sets which hereditarily satisfy this property, of $\mathrm{HOD}(M\cup F)^{M[G]}$ is going to be the symmetric extension, where $F=\{\{\pi``G\mid\pi\in K\}\mid K\in\scr F\}$.

Grigorieff's point is that by using "Magic Reflection Argument" we can replace the st $F$ by an element of the symmetric extension itself which captures, in a way, a lot of the content relevant here.

The other way around

This is where it gets even trickier. First, Grigorieff says: $M[G]$ is a generic extension of $\mathrm{HOD}(M(x))^{M[G]}$, and the forcing used for that is homogeneous. Next, if $T$ is that forcing, then the model is actually $M(T)$.

Now, we can recover the filter of groups from $T$, at the very least knowing the forcing that was used. And this, ultimately, brings us back to $M[G]$.

This too involves a lot of "Reflection Theorem Magic" arguments.

Conclusion (of Grigorieff's work)

It turns out, if so, that since a given forcing has only set-many filters of groups on its automorphism group, there are only set-many possible symmetric extensions it can define.

However, from the work of Grigorieff that wasn't fully explored (and I daresay, understood) until quite recently, until the works of Usuba on ground model definability in the $\sf ZF$ context, we get that if $M\subseteq N\subseteq M[G]$, then $N$ is a symmetric extension of $N$ if and only if $N=M(x)$ for some $x\in N$. And that $M[G]$ is a generic extension of $N$.

However, from the work done on the Bristol model, $N$ is not necessarily a symmetric extension given by the same forcing that produced $G$ over $M$.

Upcoming work, in a nutshell

By a very careful analysis of Grigorieff's argument, as well as reforumlating the language of symmetric systems we (and frankly, Jonathan deserves a large share of this credit) improve the bounds given by those "Reflection Theorem Magic" theorems and provide a much more robust and clear candidate for the set $x$. This allows for a finer and better understanding of the arguments.

I hope that we can have a preprint out by the summer.