A lattice parallelepiped with no unimodular triangulation

A lattice polytope has a unimodular triangulation if it can be cut into lattice simplices of the smallest possible volume. Haase, Paffenholz, Piechnik and Santos asked in their 2017 survey whether every lattice parallelepiped, and more generally every lattice zonotope, has one. Such polytopes have the integer decomposition property, and Codenotti and Santos proved that every 3-dimensional lattice parallelepiped can be covered by unimodular simplices. We show that the answer is no in every dimension $d\ge 4$. The 4-dimensional parallelepiped $P$ spanned by $e_1,e_2,e_3$ and $(1,2,3,5)$ has no unimodular triangulation, and its products with unit cubes give examples in higher dimensions. Its edges at the origin span a cone that Firla and Ziegler showed cannot be triangulated into unimodular cones generated by its Hilbert basis (its minimal generators). Yet the cone of $P$ at each vertex can be so triangulated with lattice points of $P$: the obstruction is not seen at any vertex. The proof is by hand: points of $P$ with coordinates in $\frac12\mathbb{Z}$ and $\frac14\mathbb{Z}$ force faces into any unimodular triangulation, until one such point lies in none. Checked certificates show that $P$ has a unimodular cover and is, up to lattice equivalence, the unique 4-dimensional example of determinant at most 5. The question in dimension 3 remains open. This work has not yet been reviewed by experts. The proofs, the checking programs and the text were produced by AI models (from OpenAI and Anthropic), working under the author's direction. Computer-assisted; not yet reviewed by experts. Produced by AI models under the author's direction.