Orlik–Solomon Koszulness conjecture

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Let L\mathcal{L} be a geometric lattice, and let OS(L)\mathrm{OS}(\mathcal{L}) be its Orlik–Solomon algebra. Orlik–Solomon Koszulness conjecture. The algebra OS(L)\mathrm{OS}(\mathcal{L}) is Koszul if and only if L\mathcal{L} is supersolvable.

This is described as a classical conjecture and is attributed to Yuzvinsky. The source does not state a resolution; the paper aims to shed new light on the problem.

References

Primary source

Basile Coron, “Matroid complexes and Orlik-Solomon algebras”, arXiv:2506.15048 (2025).

Progress summary

Refreshed
Claimed solved

A September 2026 preprint claims the conjecture is false in every rank three or higher, but the result has not been independently verified.

Shelton and Yuzvinsky posed the question in 1997: whether an Orlik–Solomon algebra is Koszul exactly when its geometric lattice is supersolvable. Earlier literature recorded the converse as open.

Known results

  • Hultman (2014) proved the equivalence for root ideal arrangements and noted affirmative results for hypersolvable, graphic, and other classes.
  • For cones over Dirichlet arrangements, Koszulness is equivalent to supersolvability.
  • Supersolvable geometric lattices give Orlik–Solomon algebras with quadratic Gröbner bases, hence are Koszul.

September 2026 counterexample claim

On September 3, 2026, Tuong Le, Chayim Lowen, and Jason McCullough submitted Koszul Orlik–Solomon Algebras from Non-supersolvable Arrangements. They claim three constructions of non-supersolvable arrangements with Koszul Orlik–Solomon algebras, including irreducible arrangements realizable over Q\mathbb{Q} in every rank ≥3\ge 3. If correct, this refutes the conjecture; the claim remains unverified.

Current status (as of September 2026): The conjecture has a claimed counterexample in every rank ≥3\ge 3, but the preprint has not been independently verified; no settled result is recorded.

Sources

Solutions 0

No solutions have been posted yet.