Nakano-positive determinants outside the Hodge-Riemann cone

Abstract

Dinh and Nguy\^en asked whether the determinant of a Griffiths positive matrix of $(1,1)$-forms belongs to the Hodge--Riemann cone. For every $n\geqslant4$ and $2\leqslant k\leqslant n-2$, we present counterexamples constructed by AI: Nakano positive $k\times k$ matrices of constant $(1,1)$-forms on $\mathbb{C}^n$ whose determinants have singular Lefschetz maps in bidegree $(1,n-k-1)$. Under the simultaneous diagonalizability (SD) condition, we prove the Hodge--Riemann property in every bidegree $(p,q)$ with $p+q=n-k$ and $\min(p,q)\leqslant1$. The proof uses the generalized Alexandrov--Fenchel inequality of Ross, S\"u\ss, and Wannerer. For every $n\geqslant6$ and $2\leqslant k\leqslant n-4$, we also present SD examples whose determinants have singular Lefschetz maps in bidegree $(2,n-k-2)$. Their underlying idea comes from Ross--Toma's construction. The remaining cases $n=k\geqslant4$ and $n=k+1\geqslant5$ are open for general Griffiths positive matrices. They reduce to the top-Chern case of Griffiths'positivity question, equivalently to Finski's double mixed discriminant problem. We also ask whether a $2\times2$ counterexample on $\mathbb{C}^4$ can be both Nakano positive and dual Nakano positive.