Freeness of arrangements with rational K(pi,1) complements

Determine whether every complex hyperplane arrangement whose complement is a rational K(pi,1) space is free.

Background

A prior claim that OS-Koszul arrangements must be free was based on an incorrect assertion. Because the paper establishes that OS-Koszulness is equivalent to the rational K(pi,1) property for arrangement complements, the question of freeness can be reformulated topologically.

The authors explicitly reopen this issue after disproving the earlier conjectural relationship between OS-Koszulness and supersolvability.

References

If $\mathcal{C}(\A)$ is a rational $K(\pi,1)$ space, must $\A$ be a free arrangement?

— Koszul Orlik--Solomon Algebras from Non-supersolvable Arrangements  (2609.03836 - Le et al., 3 Sep 2026) in Section 6, Questions; second Question