Unimodality of h*-vectors

DefinitionLattice polytope spanned by integer points#✓fully proved

The lattice polytope in ℝ^d spanned by the finite set S ⊆ ℤ^d of integer points.

def latticePolytope {d : ℕ} (S : Finset (Fin d → ℤ)) : Set (Fin d → ℝ) :=
  convexHull ℝ ((fun v : Fin d → ℤ => fun i => (v i : ℝ)) '' (S : Set (Fin d → ℤ)))

Used by 1

Used byHibi's conjecture on symmetric \(h^*\)-vectors

DefinitionHibi's conjecture on symmetric \(h^*\)-vectors#✓fully proved

Hibi's conjecture on symmetric \(h^*\)-vectors. For every natural number \(d\), every finite set \(S \subseteq \mathbb{Z}^d\) whose convex hull \(P \subseteq \mathbb{R}^d\) has nonempty interior, and every polynomial \(h = \sum_i h^*_i z^i \in \mathbb{Z}[z]\) that is an \(h^*\)-polynomial of \(P\), that is,

\[\Big(\sum_{t \ge 0} \#(tP \cap \mathbb{Z}^d)\, z^t\Big) (1 - z)^{d+1} = h(z)\]

as formal power series, if \(h^*_i = h^*_{d-i}\) for all \(0 \le i \le d\), then \(h\) is unimodal.

This is Hibi's conjecture, Conjecture 2 of C. A. Athanasiadis's Oberwolfach report (Oberwolfach Reports 1 (2004), no. 39, p. 2078), stated there for \(d\)-dimensional integral polytopes; every such polytope is lattice-equivalent to a full-dimensional lattice polytope in \(\mathbb{R}^d\).

def hibiSymmetricUnimodal : Prop :=
  ∀ (d : ℕ) (S : Finset (Fin d → ℤ)), (interior (latticePolytope S)).Nonempty →
    ∀ h : Polynomial ℤ, IsHStarPolynomial (latticePolytope S) h →
      (∀ i ≤ d, h.coeff i = h.coeff (d - i)) → IsUnimodal h

Uses 3 · Used by 1

Statement usesLattice polytope spanned by integer points · The \(h^*\)-polynomial · Unimodal polynomials

Used byHibi's conjecture on symmetric \(h^*\)-vectors is false

TheoremHibi's conjecture on symmetric \(h^*\)-vectors is false#●ready to prove

Hibi's conjecture on symmetric \(h^*\)-vectors is false. There are a natural number \(d\), a finite set \(S \subseteq \mathbb{Z}^d\) whose convex hull \(P \subseteq \mathbb{R}^d\) has nonempty interior, and a polynomial \(h = \sum_i h^*_i z^i \in \mathbb{Z}[z]\) that is an \(h^*\)-polynomial of \(P\), that is,

\[\Big(\sum_{t \ge 0} \#(tP \cap \mathbb{Z}^d)\, z^t\Big) (1 - z)^{d+1} = h(z)\]

as formal power series, such that \(h^*_i = h^*_{d-i}\) for all \(0 \le i \le d\) but \(h\) is not unimodal.

Mustaţă and Payne (Example 1.1) give a reflexive \(6\)-dimensional lattice polytope with \(h^*\)-vector \((1, 6, 8, 6, 8, 6, 1)\), which is symmetric and not unimodal.

public theorem_wanted hibiSymmetricUnimodal_refuted : ¬ hibiSymmetricUnimodal

Sources

• C. A. Athanasiadis, On the unimodality of h*-vectors, in Ehrhart Quasipolynomials: Algebra, Combinatorics, and Geometry, Oberwolfach Reports 1 (2004), no. 39, p. 2078, Conjecture 2, DOI 10.4171/OWR/2004/39; resolution: M. Mustaţă and S. Payne, Ehrhart polynomials and stringy Betti numbers, Math. Ann. 333 (2005), 787-795, DOI 10.1007/s00208-005-0691-x, arXiv:math/0504486, Example 1.1: for f = (e1 + ... + e6)/3, N = ℤ⁶ + ℤ f and P the polytope with vertices e1, ..., e6, e1 - f, ..., e6 - f, P is reflexive with δ_P = (1, 6, 8, 6, 8, 6, 1), a counterexample to Hibi's conjecture for d = 6 (abstract: 'counterexamples to Hibi's conjecture on the unimodality of delta-vectors of reflexive polytopes'), https://arxiv.org/abs/math/0504486.

Uses 1

Statement usesHibi's conjecture on symmetric \(h^*\)-vectors

2 fully proved1 ready to prove

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