A nontrivial forcing notion $\newcommand\Q{\mathbb{Q}}\Q$ exhibits
*automatic mutual genericity*, if whenever $G,H\subseteq\Q$ are
distinct $V$-generic filters (existing, say, in some forcing
extension of $V$), then they are mutually generic.

In any forcing extension $V[G]$ by such forcing, $G$ must be the only $V$-generic filter; so this kind of forcing leads to unique generics. Thus, it is a rigidity concept. In particular, $\Q$ cannot be forcing equivalent below incompatible conditions, even in a forcing extension by a filter containing one of them, since then the isomorphic filter would be distinct but not mutually generic with it.

Meanwhile, it is consistent with ZFC that there are forcing notions $\Q$ with automatic mutual genericity.

For example, consider a Suslin tree $T$ that is Suslin-off-the-generic-branch, which means that after forcing to add a $V$-generic branch $g$ through $T$, then $T\upharpoonright p$ remains Suslin in $V[g]$ for any node $p$ not on $g$. If $g$ and $h$ are two $V$-generic branches through $T$, then consider a node $p$ on $h$ that is not on $g$. Since $T\upharpoonright p$ is Suslin in $V[g]$, every maximal antichain in $V[g]$ is refined by a level of the tree, and since $h$ goes through every level of the tree, it follows that $h$ is $V[g]$-generic, and so they are mutually generic. Gunter Fuchs and I discuss this property of Suslin trees in our paper Degrees of rigidity for Suslin trees, J. Symbolic Logic, vol. 74, iss. 2, pp. 423-454, 2009. We prove there that a $V$-generic Suslin tree is Suslin-off-the-generic-branch in $V[T]$, and also one can construct such trees from the $\Diamond$ principle.

Because all the examples of automatic mutual genericity of which I am aware have the character of these Suslin tree examples, which do not exist in every model of ZFC, I was wondering whether it might be possible that there are no forcing notions with automatic mutual genericity.

**Question.** Is it relatively consistent with ZFC that no
nontrivial forcing notion exhibits automatic mutual genericity?

Alternatively, I would be delighted if someone could exhibit in ZFC that there is a forcing notion with automatic mutual genericity.

This is question 10 of my recent paper, Upward closure and
amalgamation in the generic multiverse of a countable
model of set theory.
That paper is focused on various amalgamation and upward closure
results for countable models of set theory. For example, if $W$ is
any countable transitive model of set theory, then there are
$W$-generic Cohen reals $c$ and $d$ for which $W[c]$ and $W[d]$
have no common forcing extension. One can generalize the argument
(see the paper) to many other notions of forcing, including any
forcing notion $\Q$ that is *wide*, in the sense that $\Q$ is not
$|\Q|$-c.c. below any condition. I had wondered which other
forcing notions exhibit this non-amalgamation property, and by
means of the rigid Suslin tree examples was led to the concept of
automatic mutual genericity, which have amalgamation rather than
non-amalgamation.

notnecessarily by $\mathbb{P}$) such that $M[G]\models$"$\mathbb{P}$ does not exhibit automatic mutual genericity"? (To make this nontrivial, let's demand at least that $M[G]$ has the same cardinals as $M$.) $\endgroup$ – Noah Schweber Nov 4 '15 at 3:562more comments