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Suppose that $\Bbb P$ is a forcing in $V$, we say that $\Bbb P$ is $V$-decisive if whenever $\varphi(x_1,\ldots,x_n)$ is a statement in the language of forcing, and $u_1,\ldots,u_n\in V$ then $1_{\Bbb P}$ decides the truth of $\varphi(\check u_1,\ldots,\check u_n)$. (Side question, was this property of $\Bbb P$ given a name in the past?)

It's a classical theorem that weakly homogeneous forcing is $V$-decisive. But we can show a bit more. The question is whether or not there exists a $V$-decisive forcing whose Boolean completion is rigid, or at least not weakly homogeneous.

What we know:

  1. $\Bbb P$ is $V$-decisive if and only if whenever $G$ is $V$-generic, we have that $\bigcup\{H\in V[G]\mid H\subseteq\Bbb P\text{ is }V\text{-generic}\}=\Bbb P$. (Note that we don't require that $V[G]=V[H]$, just that $H$ is generic over $V$.)

  2. Equivalently, this means that if $p,q\in\mathcal B(\Bbb P)$ (the complete Boolean algebra that contains $\Bbb P$ as a dense subset) there are embeddings of $\mathcal B(\Bbb P)\restriction p$ into $\mathcal B(\Bbb P)\restriction q$. (Correction: earlier a density requirement was added, after Joel's answer it dawned on us that there is too much here.)

  3. Equivalently, for every $p\in\Bbb P$ there is some $q\leq p$ and a projection of $\mathcal B(\Bbb P)$ which maps $\mathcal B(\Bbb P)\restriction q$ onto $\mathcal B(\Bbb P)$, then $\Bbb P$ is $V$-decisive. Note that we do not require the projection to be injective, which would essentially mean that $\Bbb P$ is weakly homogeneous.

Question. Is there a $V$-decisive forcing whose Boolean completion is not weakly homogeneous? If not, is it at least consistent that there is one? In either case, can we find one whose Boolean completion is rigid?

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  • $\begingroup$ I think you mean a $V-$decisive which is not weakly homogeneous? $\endgroup$ Oct 19, 2014 at 10:42
  • $\begingroup$ In light of Joel's nice answer, I am wondering whether one should ask: is every $V$-decisive poset forcing equivalent to some weakly homogeneous poset? $\endgroup$
    – Ali Enayat
    Oct 19, 2014 at 12:45
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    $\begingroup$ @Asaf: I have not seen a name before; which is somewhat surprising given the naturality of the concept, your suggested name is well-chosen. $\endgroup$
    – Ali Enayat
    Oct 19, 2014 at 15:16
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    $\begingroup$ See this related question mathoverflow.net/questions/154554/… $\endgroup$ Oct 19, 2014 at 17:38
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    $\begingroup$ Agnostic is not the same as decisive, since he has no parameters. But my examples there work for your question if one used only ordinal parameters. So there can be Ord-decisive forcing notions that are not weakly homogeneous. $\endgroup$ Oct 19, 2014 at 18:38

1 Answer 1

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$\newcommand\B{\mathbb{B}}$

Update. The answer is no to all three questions.

Theorem. The following are equivalent for any complete Boolean algebra $\B$.

  1. $\B$ is $V$-decisive.
  2. For any two conditions $b,c\in\B$, forcing with $\B$ below $b$ adds a generic filter below $c$ with the same extension. That is, in any forcing extension $V[G]$ via $G\subset\B$, and any $c\in \B$, there is $V$-generic $H\subset\B$ with $c\in H$ and $V[G]=V[H]$.
  3. $\B$ is weakly homogeneous.

Proof. The implication $3\to 1$ is standard.

For $1\to 2$, you noticed part of this with your statement (1), but you get more than you state there; you can actually get $V[G]=V[H]$. The reason is that every condition $c$ forces that "there is a $\check V$-generic filter containing $\check c$ and giving rise to the whole extension", and so this is forced by all conditions.

So let us consider the implication $2\to 3$. For this, one can use the ideas in my blog post on Common forcing extensions via different forcing notions in order to get the desired automorphisms.

Specifically, assume statement $2$. Fix any two nonzero incompatible conditions $b,c\in\B$. By the argument of my blog post, we get an isomorphism of cones $\pi:\B\upharpoonright b'\cong\B\upharpoonright c'$ for some $b'\leq b, c'\leq c$. Let $d=\neg(b'\vee c')$, so that $b', c', d$ form a maximal antichain in $\B$. Since every element of $\B$ is the unique join of its parts below $b', c'$ and $d$, it follows that $\B$ is simply the product of the respective cones below these three elements. Thus, we may extend $\pi$ to an automorphism $\pi:\B\cong\B$ of the whole algebra, with $\pi(b')=c'$. Namely, $$\pi(x)=\pi(x\wedge b')\vee\pi^{-1}(x\wedge c')\vee (x\wedge d).$$ In particular, $\pi(b)$ is thereby made compatible with $c$, and so $\B$ is weakly homogeneous. QED

The proof shows that you don't really need arbitrary parameters from $V$, but rather it suffices to be able to refer to any particular element $b\in\B$, and then you also need somehow to refer to the ground model in order to state $\check V$-genericity. This can be done either by having a parameter for the collection of dense subsets of $\B$ in $V$, but in some cases like $V=L$ the ground model may be definable without parameters.

Meanwhile, my answer to Miha Habič's question shows that if you fall back to ordinal parameters, then it is consistent that a Boolean algebra is Ord-decisive but not weakly homogeneous.


Original answer. This answer is just about partial orders, which is not what was desired.

If one considers just the partial order, then the answer is yes, because in fact every forcing notion (meaning every partial order) is equivalent to a rigid forcing notion. Rigidity is simply not invariant under forcing equivalence.

Theorem. Every partial order is forcing equivalent to a rigid partial order. In particular, every forcing notion is forcing equivalent to a non-weakly homogeneous forcing notion.

Proof: This is a nice exercise in partial order combinatorics, which I encourage you simply to try to prove on your own. But here is one way to do it. Suppose we are given a partial order $\mathbb{P}$. We want to construct a new partial order $\mathbb{P}'$, into which $\mathbb{P}$ will densely embed, but where $\mathbb{P}'$ is rigid.

As a first step, let us associate to each node $p$ a distinct rigid and pairwise non-isomorphic partial order $A_p$, having a largest and smallest element. We construct the partial order $\mathbb{P}'$ first by replacing each node $p\in \mathbb{P}$ with a copy of $A_p$, but also adding a new stubby maximal node sticking up above the largest node of $A_p$, incomparable with everything else that was placed above $p$, and also two new stubby maximal nodes sticking up above the minimal element of $A_p$. It follows that any automorphism of $\mathbb{P}'$ must respect the copies of $A_p$, since their largest and smallest elements were marked by these stubby nodes. But since the various $A_p$'s were chosen to be non-isomorphic and rigid, it follows that $\mathbb{P}'$ can have no automorphism at all. But clearly $\mathbb{P}$ densely embeds into $\mathbb{P}'$, by associating each $p\in\mathbb{P}$ with the least element of the copy of $A_p$ inside $\mathbb{P}'$. QED

I take this argument to show that most forcing homogeneity arguments are not fundamentally about automorphisms of the partial order, or even automorphisms of the Boolean algebra, but rather automorphisms of cones in the Boolean algebra, in the way that you have already described.

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  • $\begingroup$ But we (myself and Yair, that is) asked about rigid completions, not rigid posets. Also, note that we didn't mention isomorphisms between cones, but rather embeddings. We already came up with an example (which was known, apparently) of a rigid partial order whose completion is the same as the collapse of a certain cardinal to $\omega$. So the question is fundamentally about the Boolean completions. I'm sorry if it wasn't clear enough from the question! (But thanks for the answer, it's always nice to hear that we didn't make mistakes in our own arguments before! :-)) $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 12:07
  • $\begingroup$ Regarding the embeddings on cones, note that any dense embedding of complete Boolean algebras must be an isomorphism. So your embeddings on the cones actually are isomorphisms of those cones. $\endgroup$ Oct 19, 2014 at 12:40
  • $\begingroup$ It seems that your answer was cut short (and the LaTeX still needs some work). In either case, the meaning was that the Boolean completion of the forcing is not weakly homogeneous, and in your example this is not the case, since the Boolean completion is just the lottery sum of $\kappa$ copies again. In either case, this is not quite what we were looking for, and I've edited the question to clarify (and remove the mistake, it shouldn't be dense embedding, just an embedding). Thank you for your help clarifying the question, and for the answer so far! Our deepest apologies for the extra work. :) $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 12:47
  • $\begingroup$ I've now posted a new argument, showing that the answer comes out the other way. $\endgroup$ Oct 19, 2014 at 19:42
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    $\begingroup$ Thanks Joel! I think that the observation that you can assume $V[H]=V[G]$ is the key point that we missed today. We actually went through that blog post and through Grigorieff's paper, and everything seems to would have fallen into place, if only we could have assumed $V[H]=V[G]$. And indeed this is the case here! $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 20:24

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