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Two forcing notions $\Bbb P$ and $\Bbb Q$ could be defined to be forcing equivalent if the associated complete Boolean algebras are isomorphic (so, the CBA's formed by considering the regular opens of the separative quotients of the forcing notions).

I believe that an equivalent definition can be given by the existence of a $\Bbb Q$-name $\dot{\cal G}$ and a $\Bbb P$-name $\dot{\cal H}$ such that

  • $\Bbb Q$ forces that $\dot{\cal G}$ names a $\Bbb P$-generic filter,
  • $\Bbb P$ forces that $\dot{\cal H}$ names a $\Bbb Q$-generic filter,
  • If $G\subset\Bbb P$ is $\Bbb P$-generic, then $G=\dot{\cal G}{}^{\cal H}$, where $\mathcal H=\dot{\mathcal H}{}^{G}$,
  • If $H\subset\Bbb Q$ is $\Bbb Q$-generic, then $H=\dot{\cal H}{}^{\cal G}$, where $\mathcal G=\dot{\mathcal G}{}^{H}$.

This is for instance Shelah's definition of forcing equivalent (Def. 5.2 in Proper and Improper Forcing). One can use the above names $\dot{\cal G}$ and $\dot{\cal H}$ to define an isomorphism between the complete Boolean algebras, by $\iota:\Bbb P\to\Bbb Q$ defined as $p\mapsto ||p\in\dot{\cal G}||$.


Let us call $\Bbb P$ and $\Bbb Q$ almost equivalent if there exist a $\Bbb Q$-name $\dot{\cal G}$ and a $\Bbb P$-name $\dot{\cal H}$ such that

  • $\Bbb Q$ forces that $\dot{\cal G}$ names a $\Bbb P$-generic filter,
  • $\Bbb P$ forces that $\dot{\cal H}$ names a $\Bbb Q$-generic filter,
  • If $\dot G$ is the canonical $\Bbb P$-name for the $\Bbb P$-generic filter, then $\Bbb P$ forces that $\mathbf V[\dot G]=\mathbf V[\dot{\mathcal H}]$
  • If $\dot H$ is the canonical $\Bbb Q$-name for the $\Bbb Q$-generic filter, then $\Bbb Q$ forces that $\mathbf V[\dot{\mathcal G}]=\mathbf V[\dot H]$

Here $\dot G$ and $\dot H$ are respectively the cannonical $\Bbb P$ and $\Bbb Q$-names for their generic filters. It is clear that forcing equivalent implies almost equivalent.

The reverse does not hold. For instance, we can consider the forcing notion $\Bbb P$ to be Cohen forcing, and $\Bbb Q=(\omega_1\times\Bbb P)\cup\{1\}$, where $1$ is the top element and $(\alpha',p')\leq (\alpha,p)$ iff $\alpha=\alpha'$ and $p'\leq p$ (so $\Bbb Q$ is the lottery sum of $\aleph_1$-many copies of Cohen forcing).

Since $\Bbb P$ is c.c.c. and $\Bbb Q$ is not, they cannot be forcing equivalent. On the other hand, the names $\dot{\cal G}=\{(p,(\alpha,p))\mid (\alpha,p)\in\Bbb Q\}$ and $\dot{\cal H}=\{((0,p),p)\mid p\in\Bbb P\}$ show that $\Bbb P$ and $\Bbb Q$ are almost equivalent.


My question is if exceptions such as above are the only exceptions.

More precisely, if $\Bbb P$ and $\Bbb Q$ are almost equivalent, does this imply that $\Bbb P$ and $\Bbb Q$ are forcing equivalent to lottery sums of forcings from $\{\Bbb P_i\mid i\in I\}$ and $\{\Bbb Q_j\mid j\in J\}$ respectively, where each $\Bbb P_i$ is forcing equivalent to some $\Bbb Q_j$ , and vice versa?

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    $\begingroup$ Relevant background: jdh.hamkins.org/… $\endgroup$ Commented yesterday
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    $\begingroup$ You should clarify the meaning of almost equivalent, since you are quantifying over $V$-generic filters $G$ and $H$, but of course, in nontrivial cases there aren't any such filters (that is, there are no such filters in $V$). So what do you mean? You should express the definition in a way that makes sense inside ZFC. Probably what you mean is that below every condition in $\mathbb{P}$ there is a cone that is forcing equivalent to a cone in $\mathbb{Q}$ and vice versa. $\endgroup$ Commented yesterday
  • $\begingroup$ @Joel My apologies, I was indeed a bit sloppy. I've rephrased it in terms of forcing, which I find a more natural way of looking at it. If I'm not mistaken, Theorem 1 from your blog post implies that for almost equivalent complete BAs (under my rephrased definition), every condition has a cone equivalent to a cone from the other forcing. $\endgroup$ Commented 19 hours ago

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I had previously posted an answer, which was not correct.

The correct answer is negative, as shown in the first part of the answer of Calliope Ryan-Smith, following a suggestion of Andreas Leitz there in the comments.

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  • $\begingroup$ How to "iterate this process into the limit"? Does it not contradict the first answer to this question of mine? $\endgroup$ Commented 12 hours ago
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    $\begingroup$ @newaccount Yes, you are right, and I was too quick here with my initial post. I have now edited simply to point at that other answer. $\endgroup$ Commented 7 hours ago

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