Can a Suslin tree squared collapse $\aleph_1$?

Question

A standard fact is that if $T$ is a Suslin tree, then $T\times T$ flipped upside down and viewed as a forcing is not ccc. But can it collapse cardinals? At the very least, the answer is not always - for instance, suppose there are Suslin trees $T_1$ and $T_2$ such that the tree theoretic product $T_1\otimes T_2$ is Suslin (as happens, for instance, under $\diamondsuit$). Then take $T$ to be a tree with two nodes $t,u$ such that the collection of nodes extending $t$ is isomorphic to $T_1$ and the nodes extending $u$ is isomorphic to $T_2$. Then $T\times T$ does not force that $\aleph_1$ is collapsed since the suborder of conditions below $(t,u)$ is ccc.

Answer 1

Answering with the hint to look in Jech and some help from a friend. The answer is yes; relevant Jech exercise is 15.20 in constructing a homogeneous Suslin tree. Suppose there are $\langle f_\alpha\mid\alpha<\omega_1\rangle$ such that $f_\alpha\colon\alpha\to2$, for $\alpha<\beta$ we have $f_\alpha=^*f_\beta$ and $T=\{g\mid g=^*f_\alpha\text{ for some }\alpha\}$ is a Suslin tree; such objects exist under $\diamondsuit$ or can be forced. Then $T\otimes T$ collapses $\aleph_1$: let $\langle t_\alpha,u_\alpha\mid \alpha<\omega_1^V\rangle$ be the generic branch through $T\otimes T$. Then mapping $n$ to the least $\alpha$ such that $t_\alpha,u_\alpha$ disagree in at least $n$ coordinates is a function from $\omega$ to $\omega_1^V$ with cofinal range.