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Bernstein-Sato ideals for free hyperplane arrangements

Published 22 Sep 2026 in math.AG and math.AC | (2609.26336v1)

Abstract: Let f=(f1,…,fr)f=(f_1,\dots,f_r) be a complete factorization of a central hyperplane arrangement DD in X=C<sup>nX=\mathbb{C}<sup>n. For a monoid ideal K⊆N<sup>rK\subseteq \mathbb{N}<sup>r we study the Bernstein-Sato ideal B<sup>KfB<sup>K_f of ff along KK, that is, the C[s]\mathbb{C}[s]-annihilator of D<em>X[s]f<sup>s/∑</sup></em>m∈KD<em>X[s]f<sup>s+m\mathcal{D}<em>X[s]f<sup>s/\sum</sup></em>{m\in K}\mathcal{D}<em>X[s]f<sup>{s+m}. When DD is free we compute two families of these ideals with the help of AI. For the unit shift K=⟨ei⟩K=\langle e_i\rangle we prove that B<sup>−eifB<sup>{-e_i}_f is generated by an explicit product of linear forms indexed by the dense edges of DD contained in DiD_i. This determines all the Bernstein-Sato ideals B<sup>a,bf=Ann⁡</sup></em>C[s]DX[s]f<sup>s−a/DX[s]f<sup>s−bB<sup>{a,b}_f=\operatorname{Ann}</sup></em>{\mathbb{C}[s]}\mathcal{D}_X[s]f<sup>{s-a}/\mathcal{D}_X[s]f<sup>{s-b}, a≥ba\geq b, of a free arrangement, generalizing formulas of Maisonobe (2016) and Bath (2020). The main new ingredient identifies the multiplicities of the relative characteristic cycle of DX[s]f<sup>s/DX[s]f<sup>s+ei\mathcal{D}_X[s]f<sup>s/\mathcal{D}_X[s]f<sup>{s+e_i} along the conormal bundle of the origin with the coefficients of the Hilbert series of an Artinian complete intersection attached to a generic Ziegler restriction of DD; the total multiplicity computed in Wu (2022) then forces all the resulting coefficientwise upper bounds to be equalities. For the coordinate monoid ideal K=⟨e1,…,er⟩K=\langle e_1,\dots,e_r\rangle we show that B<sup>KfB<sup>K_f is generated by one Euler relation for each irreducible factor of the essential quotient of DD. Finally, we show that the zero locus of a Bernstein-Sato ideal along a monoid ideal need not be a finite union of translated linear subvarieties, even for a reduced free arrangement in C<sup>2\mathbb{C}<sup>2: for f=(x,y,x+y,x+2y)f=(x,y,x+y,x+2y) and K=⟨3e1,3e2⟩K=\langle 3e_1,3e_2\rangle we compute B<sup>KfB<sup>K_f exactly and find an irreducible quadric component. This disproves a conjecture due to Budur.

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