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Say that a partial order $P$ is forcing-rigid in a model $V$ if whenever $G \subseteq P$ is generic over $V$, then in $V[G]$, $G$ is the only filter which is $P$-generic over $V$. This implies there are no nontrivial automorphisms of $P$.

If $P$ is forcing-rigid and $P$ forces "$\dot Q$ is forcing-rigid," then is $P * \dot Q$ forcing-rigid?

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    $\begingroup$ I think Jensen's partial order $P$ for adding a $\Delta^1_3$ real over $L$ might provide a consistent "no" answer to your question. $P$ is certainly forcing-rigid over $L$. I think (but I'm not sure) it is still forcing-rigid in $L[a]$ where $a$ is $P$-generic over $L$. If so, then $P \times P$ is an example where the first p.o. is rigid in $L$, the second is rigid in the extension, but their composition (the product) is certainly not rigid, because if $(a,b)$ is a generic pair then so is $(b,a)$. It's all in chapter 28 of Jech if you want to take a look. $\endgroup$
    – Will Brian
    Feb 9 '18 at 19:59
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    $\begingroup$ No. In particular assuming $\diamondsuit$ there is a rigid Souslin tree $T$ with the property that $1 \Vdash_T T_s=\{ t \in T: t \le_{T} s \}$ is rigid and Souslin, for every $s \not\in \dot{G}$. See, for example, citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – Not Mike
    Feb 9 '18 at 21:47
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    $\begingroup$ @NotMike I'd suggest giving a fuller version of your comment as an answer, since indeed it does answer Monroe's question. This is the Souslin-off-the-generic branch property that Gunter Fuchs and I had investigated in that paper, which is a strong form of rigidity. $\endgroup$ Feb 9 '18 at 23:36
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    $\begingroup$ Well, it is perfectly sensible to talk about unique generics without going to the Boolean algebra, especially for Souslin trees, and this sufficed for our main application to the automorphism tower problem, and that was the origin of the problem. There is a huge literature on the rigidity of Souslin trees, considered as trees. Meanwhile, in our paper we also made that same point you mentioned, if one allows non-uniform tree (different degrees), as it is trivial to make rigid trees that way. But I agree that considering the rigidity of the Boolean algebra is very natural. $\endgroup$ Feb 10 '18 at 15:02
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    $\begingroup$ Unique generics for a complete Boolean algebra is equivalent to saying that no two distinct lower cones are isomorphic. So your conjecture is equivalent to saying: if $\mathbb{B}$ is rigid, then it is totally rigid. $\endgroup$ Feb 10 '18 at 15:09
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No, rigidity is not preserved in iterations. In particular,

Proposition: If $T$ is a rigid Souslin-tree with the property that $$1 \Vdash_T T_{s} =\{ t \in T : t\le_T s \}\text{ is rigid and Souslin for every }s \not\in \dot{G}$$

(rigid and Souslin off-the-generic-branch in the terminology of FuchsHam2008.) Then,

  • $1\Vdash_T $" $\check{T}$ is rigid and totally-proper."

Proof: Assume $T$ satisfies the required property.

To see that $T$ remains proper, fix some countable elementary sub-model $M\prec H(\lambda)$ with $T \in M$ and let $t\in T\cap M$ and $\dot{s} \in M^{T}$ be such that $t \Vdash \dot{s} \in \check{T}$. Noting that $T$ is c.c.c., we can find some $\dot{s}_0 \in M^T$ such that $t \Vdash_T \dot{s}_0 \le_T \dot{s}$ and $\dot{s}_0 \not\in \dot{G}$. Now let $\dot{D} \in M^{T}$ be a $T$-name for a dense-open subset of $T$ and $(u,\dot{v})\le (t,\dot{s}_0)$ be any extension with $Lev_{T}(u) \ge \delta = M\cap \omega_1$ and $u \Vdash Lev_{T}(\dot{v}) \ge \check{\delta}$.

Then, we must have $ u \Vdash (\exists r \in \dot{D} \cap \check{M})(\dot{v} \le r)$ (since otherwise, a standard reflection argument yields $u \Vdash (\exists a \in T \cap M)(a \not\in \dot{G}$ and $T_a$ is not Souslin$)$). It follows that $u \Vdash_T \dot{v}$ is totally $(M[\dot{G}], \check{T})$-generic; and so $1 \Vdash_{T} \check{T}$ is totally-proper.

To see that $T$ remains rigid, note that if $t \Vdash_{T} \check{T}$ is not rigid, then for some $s \in T$ with $s \perp t$, we must have $t \Vdash_{T} \check{T}_s$ is not rigid, or $\check{T}_s$ is not c.c.c. $\square$.

Remark: To see that this provides a counter-example, note that the two-step iteration of $T$ with itself is isomorphic to the square $T^2$ which admits the non-trivial automorphism $(s,t)\rightarrow (t,s)$. (being totally-proper in the extension didn't really matter, I just thought it was worth pointing out.)

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    $\begingroup$ For totally proper, perhaps it is easier to observe that $T$ remains locally c.c.c. after forcing with $T$, because once you have a condition off the generic branch, then it is Suslin below that condition. (Locally c.c.c. = for every condition, there is a stronger condition below which the forcing is c.c.c.) $\endgroup$ Feb 10 '18 at 11:48
  • $\begingroup$ @JoelDavidHamkins you're right. Thanks for pointing that out. I had been screwing around with the idea of weakening the souslin off-the-generic hypothesis, and lost perspective. $\endgroup$
    – Not Mike
    Feb 10 '18 at 12:24

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