One of the magnificent theorems of $\sf ZFC$ is that there exists an Aronszajn tree on $\omega_1$. Namely, a tree of height $\omega_1$ in which every level is countable, but no branch is cofinal.

On the other hand, assuming the consistency of a weakly compact, $\sf ZFC$ *does not* prove that every tree on $\omega_2$—or any successor of an uncountable regular cardinal—which has levels of size less than $\aleph_2$ has a cofinal branch.

This raises a very natural question:

Is there some natural combinatorial property which holds or fails at $\omega_1$ if and only if $\omega_1$ is weakly compact in $L$?

(We can replace "weakly compact in $L$" by "There is a weakly compact cardinal in $L$" if that holds somehow; but usually we expect the cardinal with the failure to be that weakly compact.)

Both in a deleted comment and a deleted answer the Shelah–Harrington theorem was suggested that showed that under $\sf MA+\lnot CH$, $\omega_1$ is weakly compact if and only if every $\mathbf\Sigma^1_3$ is measurable if and only if ever $\mathbf\Delta^1_3$ set has the Baire property.

These are not quite combinatorial characteristics of $\omega_1$, but rather [somewhat-]topological characteristics of the continuum, which is not even $\omega_1$ here. Moreover, if anything is to be quoted from the said paper, it should be the theorem stating that under $\sf MA+\lnot CH$, $\omega_1$ is weakly compact in $L$ if it is inaccessible to reals.

A weakening of the original question could be, suppose $\kappa$ is weakly compact, and we force with $\operatorname{Col}(\omega,<\kappa)$. What combinatorial properties of $\kappa$ survive the collapse, and whose consistency at $\omega_1$ implies weak compactness in $L$?

not$\omega_1$. But it is an interesting fact, so thank you both for bringing this up. $\endgroup$ – Asaf Karagila Nov 23 '16 at 22:40