First, a rather broad question: has there been any work on what, given a model $M$ of set theory, we can say about those models of set theory $N$ and posets $\mathbb{P}$ such that $\mathbb{P}\in N$ and $M=N[G]$ for some $G$ $\mathbb{P}$-generic over $N$?

Second, a more specific question. Let a poset $\mathbb{P}$ be $detectable$ if there is some sentence $\phi_\mathbb{P}$ in the language of set theory + a constant denoting $\mathbb{P}$ such that for all $M$ with $\mathbb{P}\in M$, we have $(M, \mathbb{P})\models\phi\iff M=N[G]$ for some model $N$ with $\mathbb{P}\in N$ and $G$ $\mathbb{P}$-generic over $N$. What posets are detectable? [Answered in the comments by Amit Kumar Gupta.]

Finally, an incredibly general question. Let $M$ be a model of $ZFC$, $\mathcal{C}$ a class of posets in $M$. Say $\mathcal{C}$ is $consistent$ if there is some elementary extension $N$ of $M$ such that, for all $\mathbb{P}\in \mathcal{C}$, there is some $N_\mathbb{P}\models ZFC$ with $\mathbb{P}\in N_\mathbb{P}$ and some $G$ $\mathbb{P}$-generic over $N_\mathbb{P}$ such that $N=N_\mathbb{P}[G]$. What are the consistent classes like? Can we say anything interesting about them?

I'm not sure if these questions are meaningful, or - even assuming they are - if they are interesting. Basically, what I'm interested in is the notion of inverse forcing - similar in an aesthetic sense, at least to me, to inverse Galois theory - and I haven't run into anything along these lines yet.

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    $\begingroup$ Reitz and Hamkins have explored a possible axiom called the Ground Axiom, which states that $V$ is not a forcing extension of any inner model $W$ by a nontrivial forcing $\mathbb{P} \in W$. In your second paragraph you seem to define a notion of detectability but it doesn't appear that you ask an actual question. A result of Laver's is that if $V=M[G]$ is a forcing extension of $M$ by a set forcing $\mathbb{P} \in M$ then $M$ is definable in $V$ from parameters in $M$. So in a sense this says that if $V$ is a forcing extension, then it can detect the ground model. $\endgroup$ Nov 8 '10 at 6:36
  • $\begingroup$ Amit, could you repost your comment as an answer... $\endgroup$ Nov 8 '10 at 14:19
  • $\begingroup$ @Amit Kumar Gupta: Yeah, I forgot to finish typing the second paragraph. That's fixed now, and I've noted that you answered that question. I hope you expand your comment into an answer! $\endgroup$ Nov 8 '10 at 16:05
  • $\begingroup$ In the definition of consistent, do you mean to say "forcing extension" rather than "elementary extension"? $\endgroup$ Oct 13 '11 at 0:18

Amit Gupta's comment gives the answer to your first question, which I will expand on slightly:

Every poset is detectable, and even more, the class of all posets is uniformly detectable. That is, there is a single formula $\phi(x)$ that takes as input a poset $P$, such that $\phi(P)$ is true in a model $M$ exactly when $M$ is a forcing extension of an inner model $N$ by $P$ (let us call $N$ a ground model of $M$ by $P$). It follows that the same is true for any set of posets (or definable class $C$ of posets) in $M$: we can detect when $M$ is a $C$-forcing extension.

Regarding your second question ("consistent classes"), there are many examples of consistent classes. For example, let $C$ be the class of posets that add a subset to a regular cardinal using the poset as defined in $L$ (that is, $C$ is the class of posets of the form $ Add(\kappa,1)^L $ ). If we start in $V$ and force with the (class) product of $C$, then in the resulting model $V[G]$, for every member $P$ of $C$ there is a ground model of $V[G]$ by $P$. This is because, for a fixed $\kappa$, we can use the commutativity of the product to rewrite the extension as class forcing (adding a subset to each cardinal except $\kappa$), followed by set forcing adding a subset to $\kappa$. This allows us to rewrite $V[G]=V[G^\prime][g]$, and $V[G]$ is an extension of $V[G^\prime]$ by $Add(\kappa,1)^L$ as required.

We can modify this argument in various ways to handle other classes. It will work for any class of posets from $L$ whose product preserves $ZFC$. We can also take our posets from another absolute inner model, such as $L[mu]$, instead of from $L$. Dropping the requirement that the posets be from a designated inner model (for example, taking $C$ to be the class of posets $Add(\kappa,1)$) may still possible but would require a more delicate argument. The issue here is that the definition of the poset $Add(\kappa,1)$ is not absolute, and so the poset defined in $V$ may not exist in every ground model. Nonetheless I suspect that the same model defined in the previous example will work.

There are also some classes that are clearly not consistent, for example the the class $C$ of partial orders that collapse $\kappa$ to $\omega$ (for every $\kappa$). Any model containing generics for all such posets will fail to satisfy $ZFC$, as every cardinal will be collapsed.

For the more general case, one avenue of attack might be to use the Maximality Principle (introduced by Joel Hamkins). This principle says that any statement that can be forced in such a way that further forcing cannot make it false, is already true, and existence of a ground model for a particular poset is just such a statement.


The comment by Amit Kumar Gupta above should really be expanded into an answer to the question. I think he mentions what you are really interested in.

Of course, there has been a long history of statements that say that the universe looks like a forcing extension of a certain kind. By these I mean the familiar forcing axioms like Martin's Axiom and the Proper Forcing Axiom, but also CPA and its variants by Ciesielski and Pawlikowski. But these axioms do not explicitly say that the universe is a forcing extension, it just behaves like one as far as certain combinatorics are concerned. And it is known that the forcing axioms (and a lot of other things) are consistent with the universe not being a (set-) forcing extension of anything.
Namely, it is possible to start with a model of the axiom that you are interested in and then use class forcing to code information into, say, cardinal arithmetic that prevents the universe from being a generic extension of another universe by set forcing (a forcing notion that is a set), without affecting the axiom in question.

An example more in the direction that you are looking at is Woodin's axiom (*) that explicitly says that the universe is a forcing extension of some kind.


Here is a nice observation that is not mentioned in the previous answers:

Even if we know that a model is a forcing extension, we may still not know either the forcing or the original model. Of course, this is clear if $V=M[G]$ where $G$ is generic over $M$ for some poset ${\mathbb P}$ that admits some ${\mathbb Q}$ as a factor, so $G=H*I$ for some $H$ generic over $M$ for ${\mathbb Q}$ and $V=(M[H])[I]$ for $I$ generic over $M[H]$ for the poset that is usually denoted by ${\mathbb Q}/G$. But in fact, $V$ could be generic over different, ${\subseteq}$-incomparable models for very different posets.

Here is an example (probably due to Woodin?): Assume a tiny amount of determinacy ($\Sigma^1_2$-determinacy is an overkill). It follows that, for any real $x$, ${\mathcal P}(\omega)\cap L[x]$ is countable, so there is a real $w$ such that $x\le_T w$ and $L[w]$ is a forcing extension of $L[x]$ by Cohen forcing. Also, there is a $w'$ with $x\le_T w'$ and $L[w']$ a forcing extension of $L[x]$ by random forcing. The amount of determinacy we are assuming allows us to invoke Martin's cone theorem to conclude that (from a mild large cardinal hypothesis):

There is a real $x$ such that for any real $y$, if $x\le_T y$, then $L[y]$ is both a forcing extension of some model $L[z]$ by Cohen forcing, and a random extension of some model $L[z']$ by random forcing.

(I know a few years ago a student of Kunen was looking at a similar problem, but I did not keep up with their results.)

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    $\begingroup$ Andres, if one performs product forcing $\mathbb{P}\times\mathbb{Q}$, then the extension can be viewed as arising from either factor (by your first observation), even when $\mathbb{P}$ and $\mathbb{Q}$ are very different. $\endgroup$ Nov 15 '10 at 0:32
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    $\begingroup$ It is also true that even if you know that a model is a forcing extension by a specific partial order $\mathbb{P}$, this doesn't determine the ground model. For example, adding two Cohen reals is isomorphic to adding one, so $V[c][d]$ is a forcing extension both of $V$ and of $V[d]$ (among continuum many other ground models) in each case by adding a Cohen real. $\endgroup$ Nov 15 '10 at 0:40
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    $\begingroup$ @Joel: Yes, I guess the point is one can easily modify the example so that there is no "common" ground model from which to trace the extensions. If we used ${\mathbb P}\times{\mathbb Q}$ we would have a ground model $M$ and a final extension $V=M[G_1][G_2]=M[G_2][G_1]$, and similarly with infinite, countable products. But I admit I do not know how far this idea can be pushed, or to what generality. I think that the forthcoming "set-theoretic geology" paper by Schindler and Fuchs may address some of these issues at the level of a Woodin cardinal. $\endgroup$ Nov 15 '10 at 0:45

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