All simply connected $6$-manifolds which admit effective $T^4$-actions have been classified by Hae Soo Oh (See Theorem 1.1). The classification shows that only the following simply connected $6$-manifolds admit effective $T^4$-actions
$M(n,k):= \#(k-4) (S^2 \times S^4) \# (k-3) (S^3\times S^3) $ for $k\geq 4$, and
$N(n,k):= (S^2 \tilde{\times}S^4)\#(k-5) (S^2 \times S^4) \# (k-3) (S^3\times S^3) $ for $k>4$,
where $S^2 \tilde{\times}S^4$ is the unique non-spin $S^4$-bundle over $S^2$.
All of these manifolds admit free $S^1$-actions. And given an effective action of $T^4$ on a simply connected $6$-manifold $M$, most subgroups $S^1\subset T^4$ act almost freely on $M$. And there are plenty of examples of $T^4$-actions which do have a subgroup $S^1\subset T^4$ which acts freely. But I want to know if for every effective $T^4$-action on $M$, there always exists a subgroup $S^1\subset T^4$ which acts freely.
I'm not convinced that these free $S^1$ subtorus actions always exist, but I also can't produce any counter examples. If anyone has a proof or a counter example I'd love to hear it.