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?
> 
> 
> 



On $V$-decisive and weakly homogeneous forcings
$\newcommand\B{\mathbb{B}}$


**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](http://jdh.hamkins.org/common-forcing-extension-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](https://mathoverflow.net/a/154561/1946) 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. 


