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?



preservation of forcing rigidity in iterations
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](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.239.2644&rep=rep1&type=pdf).) 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.)


