Is it possible to have an $\omega_2$-length sequence of ($\omega_1$-)Suslin trees such that if one builds the product of finitely many trees in that sequence, one ends up with a Suslin tree again?
The existence of such a sequence of length $\omega$ follows from $\diamondsuit$, as was shown by Jensen. By Shelah and independently Todorcevic, already a Cohen real gives rise to a Suslin tree, so it could be possible that a adding $\aleph_2$ many Cohen reals produces such a sequence.
The answer is yes, and indeed, one can even have that any countable number of the Suslin trees join to a Suslin tree.
To see this, simply force with countable support to add $\omega_2$ many Suslin trees. The forcing to add one Suslin tree has conditions consisting of countable normal $\alpha$-tree, for $\alpha<\omega_1$, and this tree will become an initial segment of the desired generic tree. This forcing is countably closed and isomorphic to $\text{Add}(\omega_1,1)$, adding a Cohen subset of $\omega_1$. This forcing also adds a $\diamondsuit$-sequence, and there is a tight connection between the argument from diamond that there is a Suslin tree and the proof that this forcing adds a Suslin tree. Namely, given any name for a maximal antichain, one undertakes a bootstrapping argument to decide more and more of the antichain, until one has a condition $t$ that decides $A\cap t$ and such that $A\cap t$ is maximal in $t$. Then, one extends $t$ to $\bar t$ with one more level in a way that seals that antichain in any further extension of $\bar t$. So $\bar t$ forces that the antichain is bounded and hence countable.
Consider now adding $\omega_2$ many such Suslin trees, with countable support. An essentially similar sealing argument shows that these trees are Suslin, and furthermore, that they are mutually Suslin over each other. If you force with all but one of these trees, with countable support, then the remaining tree will still be Suslin. The reason is that if $\dot A$ is a name for antichain in $T$, the unforced tree, where $\dot A$ is a name in the product of the other trees, then it will be dense in the product of the other tree-forcing conditions that forces that $\dot A$ is bounded.
The argument is fundamentally similar to the methods used in my paper
The point is that to add a generic Suslin tree $T$ and then force with it, is the same as forcing with conditions $(t,b)$ where $t$ is a normal $(\alpha+1)$-tree and $b$ is an element on the top level (determining the generic path). This forcing is very nice, and can be used to show that a generic Suslin tree is Suslin off the generic branch.
