Ziegler's generic restriction conjecture for logarithmic forms

About 3 years old · traced to

Let A′{\mathcal{A}}' be an essential and free arrangement in V=KℓV={\mathbb{K}}^\ell with ℓ≥4\ell\geq 4. Let XX be a generic subspace with dim⁡X>2\dim X>2, and define the restriction

A′∩X:={H∩X∣H∈A′∖AX′}.{\mathcal{A}}'\cap X:=\{H\cap X\mid H\in{\mathcal{A}}'\setminus{\mathcal{A}}'_X\}.

Here, XX is generic if it is (ℓ−codim⁡X)(\ell-\operatorname{codim}X)-generic with respect to A′{\mathcal{A}}', meaning that

codim⁡V(X∩Y)=codim⁡X+codim⁡Y\operatorname{codim}_V(X\cap Y)=\operatorname{codim}X+\operatorname{codim}Y

for every relevant intersection subspace YY of A′{\mathcal{A}}'. Write Ω1(A′)\Omega^1({\mathcal{A}}') for the module of logarithmic 11-forms and res⁡H\operatorname{res}_H for the restriction map. Ziegler's generic restriction conjecture. The map

res⁡H ⁣:Ω1(A′)⟶Ω1(A′∩X)\operatorname{res}_H\colon\Omega^1({\mathcal{A}}')\longrightarrow\Omega^1({\mathcal{A}}'\cap X)

is surjective, and the minimal numbers of generators of Ω1(A′)\Omega^1({\mathcal{A}}') and Ω1(A′∩X)\Omega^1({\mathcal{A}}'\cap X) are equal. This conjecture concerns how logarithmic forms behave under generic cuts of free arrangements; Ziegler stated that it would imply that such a generic cut is never free. Its resolution is not supplied in the source.

References

Primary source

Takuro Abe and Graham Denham, “On Ziegler's conjectures for logarithmic derivations of arrangements”, arXiv:2307.08173 (2026).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.