# Is it consistent with ZFC that no nontrivial forcing notion has automatic mutual genericity?

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.

• Maybe a slightly easier sub-question: fix a model $M$ of ZFC with a forcing notion $\mathbb{P}\in M$. Is it obvious that there is a forcing extension $M[G]$ (not necessarily 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 $M$.) – Noah Schweber Nov 4 '15 at 3:56
• If you don't insist on cardinal-preservation, then you can make $\mathbb{P}$ countable, in which case it will not have automatic mutual genericity since it will be Cohen forcing if nontrivial. But with cardinal preservation, I don't know how to do this. – Joel David Hamkins Nov 4 '15 at 4:08
• Maybe Prikry's conjecture (any non-trivial c.c.c either adds a Cohen real or a Random real) implies no c.c.c. forcing has such a property. – Mohammad Golshani Nov 4 '15 at 4:32
• @JoelDavidHamkins Maybe one can use the ideas from $0^{\sharp}$ and some forcing principles'' to show the following: Assume $0^{\sharp}$ exists. Then no $P\in L$ exhibits automatic mutual genericity over $V$. – Mohammad Golshani Nov 7 '15 at 5:13
• At least this is true if $L \models P$ is $\kappa$-distributive of size $\kappa.$ As then by a result of Mack Stanley, over $V$, $P$ is equivalent to $Col(\omega, \kappa).$ – Mohammad Golshani Nov 7 '15 at 5:16

I'll sketch the proof here. First we showed that a poset $\mathbb{Q}$ satisfies automatic mutual genericity if and only if for every $p\perp q$ in $\mathbb{Q}$, forcing below $p$ does not add a new maximal antichain below $q$. This is the main part of the proof: one direction is trivial and the other involves collapsing $2^\mathbb{Q}$ to be countable and using a name for a new antichain $\dot{A}$ to construct $V$-generics $G,H$ below $p$ and $q$ which do not meet the dense-below-$(p,q)$ subset $$\{(p',q'):\exists q^* \textrm{ such that }q'\le q^*\textrm{ and }p'\Vdash q^*\in\dot{A} \}.$$
Now suppose $\mathbb{Q}$ has automatic mutual genericity, and let $\kappa$ be least so that $\mathbb{Q}$ adds a new subset of $\kappa$. For simplicity, assume that $\mathbb{Q}$ is a complete Boolean algebra. If $\mathbb{Q}$ had an antichain $A$ of size $\ge\kappa$, then in the generic extension by $\mathbb{Q}$ we could take $p\in A$ and use the new subset of $\kappa$ to define a new antichain below $-p$. So $\mathbb{Q}$ is $\kappa$-c.c., and therefore it is a $\kappa$-Suslin algebra.
Now the characterization of posets with automatic mutual genericity implies that $\mathbb{Q}$ must remain "Suslin off the generic branch," since otherwise this is witnessed by a new antichain.
• There is something wrong with this argument. In $L$ there is a ccc forcing which adds a real and exhibits automatic mutual genericity; this is Jensen's $\Delta^1_3$ real forcing, see corollary 28.6 in Jech's book. – Miha Habič Nov 6 '18 at 15:10