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Given two forcing notions $\mathbb{P}_0,\mathbb{P}_1$, we know that the product $\mathbb{P_0}\times\mathbb{P_1}$ will produce the following diagram of models ordered by inclusion:

where $M$ is the ground model, $G_i$ is a $\mathbb{P_i}$-generic filter over $M$ and $G_0\times G_1$ is $\mathbb{P_0}\times\mathbb{P_1}$-generic over $M$. Note that $M[G_0]\cap M[G_1] = M$.

My question is whether, given three forcing notions $\mathbb{P_0},\mathbb{P_1},\mathbb{P_2}$, we can define a forcing $\mathbb{Q}$ that induces the following lattice:

where, again, $G_i$ is $\mathbb{P_i}$-generic over $M$. I also would like $M[G_i]\cap M[G_j] = M$ for $i\neq j$. Note that, since the diagram represents a lattice, once we have $G_i$ and $G_j$ for some $i\neq j$, we would be able to construct (modulo $M$) also $G_k$ with $k\neq i,j$.
Is such a construction known to be possible in a reasonably general (i.e. definable) way? Any reference?
Thanks

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Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. So conditions are binary sequences $(p,q,r)$ all of the same length, such that $r=p\oplus q$, ordered by extension.

We can present this in a more symmetric manner as: conditions are triples of finite binary sequences $(p,q,r)$ with all the same length, such that $p\oplus q\oplus r=0$. Each of the three is the bitwise sum of the other two.

This forcing will add a Cohen real on each coordinate. And so we will get $M[G_0,G_1,G_2]$ above $M[G_0]$, $M[G_1]$, and $M[G_2]$, and any two of them will be mutually generic, so we will have $M[G_i]\cap M[G_j]=M$ for $i\neq j$. But once you have two of them, you can define the third, and so you get the whole thing.

Of course, this picture isn't the complete lattice of grounds, since every Cohen real can be split in two, and so if you had meant to ask about that one would need a more precise question.

Meanwhile, you ask about all possible $\mathbb{P}_i$, but the answer in this generality is false.

Here is one quick way to make an example. Take a Suslin tree $T$ that is $3$-fold Suslin off the generic branch. This is a strong form of rigidity, which means that after forcing with the tree, what remains above any node not on the generic branch is still Suslin. And again.

If we have any three generics for this forcing, then they will be automatically mutually generic, since they meet every level of the tree and hence all maximal antichains. And so we can combine any two to form the model $M[G_0,G_1]$, which will not have $G_2$, and so it won't conform with your lattice.

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  • $\begingroup$ Yes, this is a nice example of what I had in mind. I wasn't talking about the complete lattice of grouds, I just wanted the models to form a sublattice of the latter, as in your example. My question is really whether this is possible in general (i.e., for any three forcings). $\endgroup$ Commented Oct 30, 2023 at 11:50
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    $\begingroup$ I expanded my answer to show how things can go wrong. I used a Suslin tree, but I suspect that it is almost never true, and there should be a ZFC counterexample. $\endgroup$ Commented Oct 30, 2023 at 11:52
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    $\begingroup$ I think you can get a ZFC counterexample by taking three forcings that exhibit automatic mutual genericity by being increasingly closed. For example, $\mathrm{Add}(\omega,1)$, $\mathrm{Add}(\omega_1,1)$, and $\mathrm{Add}((2^\omega)^+,1)$. Any three generics for these three forcings are mutually generic, and then your argument goes through. $\endgroup$ Commented Oct 31, 2023 at 16:34
  • $\begingroup$ Excellent, Miha. $\endgroup$ Commented Oct 31, 2023 at 16:57
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    $\begingroup$ Marcia Grozek has some work analyzing the degrees of constructibility of reals, arranging nearly any desired order, and much of her work uses Sacks forcing and iterated Sacks forcing. (I also have a paper with her where we use her methods to control the implicitly-constructible universe Imp.) Perhaps those methods would have something to say about your question. $\endgroup$ Commented Nov 9, 2023 at 16:40

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