Ultrafilters preserved by $\mathbb{P}$ but not by products?

Question

Let $U\in V$ be an ultrafilter on $\omega$. We say $U$ is preserved under forcing with $\mathbb{P}$ if $\Vdash \forall x\subset \omega \ \exists Z\in U \ Z\subset x \vee Z\subset x^c$. In other words, $U$ generates an ultrafilter in $V[G]$. Familiar examples of such ultrafilters:

• Ramsey ultrafilters, P-points are preserved under Sacks forcing and Miller forcing (also their products)
• P-points/Ramsey ultrafilters are also preserved under iterations of Sacks forcing with countable support
• There are also combinatorial characterizations of ultrafilters preserved by Sacks forcing, related to Halpern-Lauchli theorem, see https://www.math.wisc.edu/~miller/res/ultra-s.pdf

My question is: is it known that an ultrafilter preserved by Sacks forcing necessarily needs to be preserved by side-by-side products of Sacks forcing (finite/countable support)? There are probably more ad-hoc examples of forcing $\mathbb{P}$ that preserves an ultrafilter $U$ but not the product $\mathbb{P}\times\mathbb{P}$ (is there any)? Maybe there is something easy that I overlook.

Answer 1

Regarding your final request, here is an example of a forcing notion $\mathbb{P}$ that preserves all ground-model ultrafilters on $\omega$, but $\mathbb{P}\times\mathbb{P}$ destroys all ground model ultrafilters.

Namely, assume CH and let $\mathbb{P}=T$ be a self-specializing Suslin tree, which is a Suslin tree with the property that forcing to add a branch $g$ through the tree makes $T$ a special $\omega_1$-tree off the generic branch; that is, if $b$ is any node not on the generic branch $g$, then the subtree $T_b$ of conditions extending $b$ is special in $V[g]$. In particular, forcing with $T$ once adds no reals, since it is a Suslin tree, and therefore preserves all ultrafilters on $\omega$. But forcing with the tree twice $T\times T$ amounts to forcing with a special Aronszajn tree in the second step and therefore collapses $\omega_1$ and consequently destroys all ground model ultrafilters on $\omega$.

Answer 2

Let $\mathbb M$ be Miller forcing. Then $\mathbb M$ adds an unbounded real, so $\mathbb M\times \mathbb M\times\mathbb M$ adds a Cohen real (Velickovic; also an unpublished result of Shelah). Hence this forcing destroys every ultrafilter from the ground model.

So for every ultrafilter $U$ in the ground model, one of the following is true:

• $\mathbb M$ destroys $U$.
• $\mathbb M$ preserves $U$, but $M\times M$ destroys $U$
• $P:=\mathbb M\times\mathbb M$ preserves $U$, but $P\times P$ destroys $U$.

The first alternative cannot happen for P-points, but I am not sure which of the other two may hold.