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Say that a forcing notion $\mathbb{P}$ is slow iff there is some $f:\mathbb{R}\rightarrow\mathbb{R}$ (in $V$) such that for every $\mathbb{P}$-name for a real, $\nu$, we have $\Vdash_\mathbb{P}\exists r\in V(\nu\oplus \check{r}\not\ge_Tf(r))$. For example:

  • No forcing which collapses the continuum to be countable is slow.

  • Cohen forcing is slow: the Cohen real itself $g$ never has $g\oplus r\ge_Tr'$, and every real added by Cohen forcing is Turing-equivalent to a Cohen real $\oplus$ a ground-model real.

  • I have a tentative proof that Hechler forcing is also slow. This is actually a bit tricky: certainly the Hechler real $h$ itself has limited computational power (namely $h\oplus r\not\ge_T\mathcal{O}^r$), but one still has to handle all the other reals added by Hechler forcing, and it's not quite as nice (unless I'm missing something) as Cohen.

  • On the other hand, there are non-slow forcings which are as forcing-theoretically nice as possible, i.e. c.c.c. - see Andeas Lietz' answer to my recent question.

I'm interested in which forcings are slow. This is way too broad a question, so here's a concrete instance: is slowness preserved by finite products? That is:

If $\mathbb{P}_1,\mathbb{P}_2$ are slow, must $\mathbb{P}_1\times\mathbb{P}_2$ be slow?

A natural guess is if $f_1$ and $f_2$ are not captured in the appropriate sense by $\mathbb{P}_1$ and $\mathbb{P}_2$ respectively, then something like $x\mapsto 2^{f_1(x)}3^{f_2(x)}$ would not be captured by $\mathbb{P}_1\times\mathbb{P}_2$. However, I don't actually see why this should be true. Certainly the square of a "tame" forcing need not be "tame" in general.

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  • $\begingroup$ It seems that any forcing that doesn't add reals is slow, but do you want $\mathbb{P}$ to add reals? Otherwise you can get a forcing that doesn't add reals, but its square collapses $\omega_1$ (a self-specializing Suslin tree or similar). $\endgroup$ Commented Dec 2 at 23:55
  • $\begingroup$ @MihaHabič Couldn't you just take the product of such a forcing with the Cohen forcing, if you did want it to add reals? The result is still slow, because Cohen forcing is slow, and its square still collapses $\omega_1$. $\endgroup$ Commented Dec 3 at 0:08
  • $\begingroup$ @MihaHabič Ah jeez I'd forgotten about such forcings (although obviously they're the prime example of the phenomenon described in my last sentence). Given paste bee's strategy, if you add this as an answer I'll accept it. $\endgroup$ Commented Dec 3 at 0:29

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This is not true, basically because products of tame forcings might not be tame, as you said.

First, notice that any forcing not adding a real is slow (since $\nu$ must be in the ground model, you can take $r=\nu$ and then the Turing jump function cannot be captured).

Fix a stationary costationary set $S\subseteq\omega_1$ and let $\mathbb{P}_1$ and $\mathbb{P}_2$ be the forcings to shoot clubs through $S$ and its complement, respectively. Both of these are slow since they don't add reals, but their product collapses $\omega_1$. This almost says that $\mathbb{P}_1\times\mathbb{P}_2$ is not slow according to your first point, but CH might not hold. We could force CH ahead of time explicitly, but luckily each of the club-shooting forcings force CH anyway (as explained in the answers to this MSE question). So the product $\mathbb{P}_1\times\mathbb{P}_2$ collapses the continuum to be countable, and is therefore not slow.

If it seems like a cheat to use forcings that don't add reals, we can fix that using paste bee's suggestion from the comments: if $\mathbb{P}$ doesn't add reals then its product with any slow forcing is slow. So just replace each $\mathbb{P}_i$ with its product with Cohen forcing to get the same result with forcings that add reals.

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    $\begingroup$ Hi Miha! This might be my personal minimum for $t_{accepted}-t_{answered}$. $\endgroup$ Commented Dec 3 at 19:31
  • $\begingroup$ I did see it pop up surprisingly fast... $\endgroup$ Commented Dec 3 at 19:34

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