Nakano-positive determinants outside the Hodge-Riemann cone

Source abstract

Dinh and Nguyên asked whether the determinant of a Griffiths positive matrix with $(1,1)$-form entries belongs to the Hodge--Riemann cone. For every $n\geqslant4$ and $2\leqslant k\leqslant n-2$, we present an explicit Nakano positive $k\times k$ matrix of constant $(1,1)$-forms on $\C^n$ whose determinant has a singular Lefschetz map in bidegree $(1,n-k-1)$. This gives a negative answer throughout this range. The boundary cases $n=k\geqslant4$ and $n=k+1\geqslant5$ remain open. We next study the question under the simultaneous diagonalizability (SD) condition. Under this condition, we prove the Hodge--Riemann property in every bidegree $(p,q)$ with $p+q=n-k$ and $\min(p,q)\leqslant1$. Consequently, SD gives an affirmative answer when $1\leqslant k=n-2$ or $n-3$. For every $n\geqslant6$ and $2\leqslant k\leqslant n-4$, however, we present SD examples whose Lefschetz map in bidegree $(2,n-k-2)$ is singular, showing that SD alone does not imply the Hodge--Riemann property in all bidegrees. The positive result uses the theory of dually Lorentzian polynomials developed by Ross, Süß, and Wannerer, in particular their generalized Alexandrov--Fenchel inequality and its equality characterization. Exact Python verification programs accompany the constructions.