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Laver introduced the concept of termspace forcing. If $\mathbb P * \dot{\mathbb Q}$ is a two-step iteration, then we can order the $\mathbb P$-names for elements of $\dot{\mathbb Q}$ by putting $\dot q_1 \leq \dot q_0$ when $1 \Vdash \dot q_1 \leq \dot q_0$. This defines the termspace partial order $T(\mathbb P,\dot{\mathbb Q})$. The key fact is that the identity map is a projection from $\mathbb P \times T(\mathbb P,\dot{\mathbb Q})$ to $\mathbb P * \dot{\mathbb Q}$. It is easy to see that if $\Vdash$ "$\dot{\mathbb Q}$ is $\kappa$-closed," then $T(\mathbb P,\dot{\mathbb Q})$ is $\kappa$-closed.

Question: If $\Vdash$ "$\dot{\mathbb Q}$ is $(\kappa,\lambda)$-distributive," is $T(\mathbb P,\dot{\mathbb Q})$ $(\kappa,\lambda)$-distributive?

Recall that $(\kappa,\lambda)$-distributive means the forcing adds no functions from $\kappa$ to $\lambda$.

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    $\begingroup$ In general - the answer is negative. For example, take $\mathbb P$ to be the Prikry forcing for singularizing a measurable cardinal $\kappa$ and take $\mathbb Q$ to be shooting a club in $\kappa^{+}$, disjoint from $(S^{\kappa^{+}}_{\kappa})^V$. Using the $\kappa^{+}$-c.c. of $\mathbb{P}$, we can see that the termspace forcing also shoots a club disjoint from $S^{\kappa^{+}}_{\kappa}$, and thus collapses $\kappa^{+}$. I think that there is also a ZFC example, but I don't have one currently. $\endgroup$ Commented Jun 21, 2018 at 20:36
  • $\begingroup$ Very nice. Is the answer positive when $| \mathbb P| < \kappa$? $\endgroup$ Commented Jun 21, 2018 at 20:44
  • $\begingroup$ I don't know the history exactly, but my understanding is that term forcing was introduced independently by several different people, including Laver, Woodin and I think a few others. Please post comments with names. $\endgroup$ Commented Jun 21, 2018 at 22:29
  • $\begingroup$ Does anybody here know the origin of the name "termspace" for this particular forcing notion? I am not aware of any quote of Laver (or others) on this. My wild guess is that he had something similar to "spacetime" in mind. (e.g. If one interprets the word "term" as "time") However, even with this interpretation it is not immediately clear to me what such a forcing notion has to do with the concept of spacetime fabric in physics. $\endgroup$ Commented Jun 22, 2018 at 6:48
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    $\begingroup$ I think it is just choice of terminology and has no analogy with physics. I don’t know the real origin, but Foreman and Cummings both say it is Laver’s idea. $\endgroup$ Commented Jun 22, 2018 at 6:57

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Here is a ZFC counterexample.

Let $\mathbb{P}$ be the forcing to add a Cohen real $c$, and let $\omega_1=\bigsqcup_n S_n$ be a partition of $\omega_1$ into disjoint stationary sets $S_n$. Let $\dot{\mathbb{Q}}$ be the forcing to kill the stationarity of all $S_n$, for $n\in c$, the generic Cohen real added by the $\mathbb{P}$ forcing. It is not difficult to see that $\dot{\mathbb{Q}}$ is forced to be $(\omega,\infty)$-distributive, since we are really just shooting a club through the other $S_m$'s.

The term forcing for $\dot{\mathbb{Q}}$ over $\mathbb{P}$ will have to add a generic for $\mathbb{Q}$ over any Cohen real in the extension, and so it will have to kill the stationarity of all the $S_n$'s. Thus, it will have to collapse $\omega_1$. So it cannot be $\omega$-distributive.

A modification to the argument, using higher cardinals instead of $\omega_1$, will have the property that $\mathbb{P}$ is small relative to the distributivity, as you asked for in the comments. For example, let $\kappa$ be any uncountable regular cardinal and consider an $\omega$-partition of the cofinality $\omega$ ordinals up to $\kappa^+$. The $\mathbb{Q}$ forcing should kill parts of the partition, depending on the digits of $c$, and this will be fine, but the term forcing will have to kill them all, since it must add generics over any Cohen real in the extension. So $|\mathbb{P}|<\kappa$ and $\dot{\mathbb{Q}}$ is $(\kappa,\infty)$-distributive, but the term forcing collapses $\kappa^+$ and hence is not $\kappa$-distributive.

Here is a more extreme example, which makes the same point perhaps more clearly. Let $S\subset\omega_1$ be a stationary co-stationary set. Let $\mathbb{P}$ be the lottery sum $\{\text{yes}\}\oplus\{\text{no}\}$ of two trivial forcing notions, so that the generic filter selects either the point yes or the point no, and both of these are generic. Let $\dot{\mathbb{Q}}$ be the forcing that kills the stationarity of $S$, if the generic filter for $\mathbb{P}$ selected yes, and otherwise kills the stationarity of the complement of $S$. This forcing is $\omega$-distributive. But the term forcing has to add a generic for both the generics for $\mathbb{P}$, and so it will have to kill the stationarity both of $S$ and its complement. So it will collapse $\omega_1$.

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