# Weak equivalence over forcing notions

We know there are several definitions about forcing equivalence which imply that two forcings notions can be equivalent or not. In general we like to know the similarity between generic extensions by different forcings. Among them, we are familiar with complete embedding, densely embedding and so on. As a matter of fact If $\mathbb P$ and $\mathbb Q$ are two forcing notions, let's call them equivalent if their Boolean completion are isomorphic as Boolean algebras, then if it is the case we have V[G]=V[H], for any $\mathbb P$- generic filter $G$ and $\mathbb{Q}$- generic filter $H$.

I am interested to understand similarity between satisfaction of generic extension by different forcing notions forcing notions, let's unpack this:

Definition: Fix $V\models ZFC$, Let $P$ and $Q$ be two forcing notions in $V$, we say $\mathbb P$ is weakly equivalent to $\mathbb Q$, if $V[G]\equiv V[H]$(i.e $V[G]$ is elementary equivalnet to $V[H]$), for all $G$ and $H$ which are $\mathbb {P}$- generic and $\mathbb Q$- genric filters over V, respectively.

Remark 1: Weakly equivalence probably is not definable in $V$.

Question: Is there any characterization of the above-mentioned notion based on Boolean algebras or order structure of forcings? Or at least sufficient conditions for being weakly equivalent and there is nontrivial example i.e two foricngs which are not equivalent but weakly equivalent.

Remark 2: By a theorem of Golshani and Mitchell, there is a model of $ZFC$, such that $Coll(\omega, \kappa)$ is weakly equivalent to $Coll(\omega,\lambda)$, for all $\kappa$ and $\lambda$ infinite. Thus consistently there are forcing notions which are weakly equivalent, but not equivalent.

Remark 3: Note that in Golshani-Mitchell model, trivial forcing is also weakly equivalent to collapse forcing, so it seems the situation is complicated.

• Interesting question! But note that your notion of weak equivalence is not an equivalence relation, since it isn't reflexive. Namely, some forcing notions $\mathbb{P}$ will not be equivalent to themselves, if different filters give rise to different theories in the extension. – Joel David Hamkins Mar 19 '17 at 14:36
• @JoelDavidHamkins yes, you are right. I don't need to be reflexive, but thank you for pointing out. – Rahman. M Mar 19 '17 at 15:01

Let me provide the alternative kind of example.

Theorem. If ZFC is consistent, then there is a model of ZFC in which any two weakly equivalent forcing notions are forcing equivalent.


To see this, suppose that $G\subset\P$ and $H\subset\Q$ are $M$-generic. The theory of $M[G]$ includes the assertion that the universe was obtained by forcing over $M$ with the poset $\P$, and this is expressible since $M$ is definable in $M[G]$ and $\P$ is definable in $M$. Thus, since $M[H]$ has the same theory, it follows that this statement is also true in $M[H]$, and so $M[H]=M[G']$ for some $M$-generic filter $G'\subset\P$. So every forcing extension by $\Q$ can be realized as a forcing extension by $\P$ and vice versa, and so the forcing notions are equivalent. QED

Let me remark further that your notion of equivalence applies only to forcing notions $\newcommand\P{\mathbb{P}}\P$ with the property that all conditions force the same sentences, since in particular every forcing extension by $\P$ itself must have the same theory. Let us call these the sententially homogeneous forcing notions. These are the forcing notions $\P$ with $\P\equiv\P$. Note that $\P\equiv\Q$ implies $\P\equiv\P$ and $\Q\equiv\Q$.

In the model of my proof, two forcing notions are weakly equivalent $\P\equiv\Q$ just in case they are each sententially homogeneous and $\P$ and $\Q$ are forcing equivalent. So it is relatively consistent with ZFC that weak equivalence is the restriction of a definable relation to the class of forcing notions to which it applies, the sententially homogeneous forcing notions. Meanwhile, I suspect that the question of whether a forcing notion is sententially homogeneous is not generally expressible in the language of set theory.

Aside. You said that two forcing notions are equivalent if and only if their Boolean algebras are isomorphic, but that is too strong, since a large lottery sum of a partial order with itself is forcing equivalent, in the sense that it gives rise to exactly the same forcing extensions, but the Boolean algebra is much larger (and has a bad chain condition and so on). The solution is to restrict the forcing to the cone below a condition. You want to say that two notions are equivalent, if they give rise to exactly the same forcing extensions, and this can be expressed in terms of their Boolean algebras by the assertion: below any respective conditions in their Boolean algebras, one can find still deeper isomorphic lower cones. See my related blog post, Common forcing extension via different forcing notions.

• In the first sentence of your proof, where you assume that $M$ remains definable in all its set-forcing extensions, did you intend that it's definable by the same formula in all those extensions (as would be the case if it satisfies $V=L$)? It seems you need that at the step where you find that $M[H]$ is a $\mathbb P$-forcing extension of $M$ (not just of some possibly different submodel). If the definitions of $M$ in $M[G]$ and $M[H]$ were different, then this wouldn't seem to follow. – Andreas Blass Mar 19 '17 at 21:11
• @Andreas: I think that the formula $\varphi(x,y)$ stating that $x$ is a member of the unique ground model with $y$ being a "large enough initial segment" works for all models, and $M$ would be definable in all its extensions with it, although with different parameters. However, since we only have two extensions, we can find a parameter that works for both of them. – Asaf Karagila Mar 19 '17 at 21:18
• Maybe I was too hasty with my previous comment. It seems that. if $M$ is definable in each of its forcing extensions, by possibly different formulas, then, for any two of its forcings $P$ and $Q$, there will also be a single formula that defines $M$ in both of the resulting extensions. Namely, let $\phi$ be the formula that defines $M$ in its Coll$(\omega,\kappa)$ extension, for a $\kappa$ so big that the collapse forcing "absorbs" both $P$ and $Q$ when applied after them in an iteration; then the formula "$\phi$ is forced by the collapsing" should work in both $P$ and $Q$ extensions. – Andreas Blass Mar 19 '17 at 21:20
• I had had in mind that it would be the same definition in all extensions, as with the V=L case, but indeed, it seems to me by your second comment that one can make the argument work also in the non-uniform case. – Joel David Hamkins Mar 19 '17 at 21:33