Math 2.0

Convex and discrete geometry Definition

Ehrhart polynomial

Builds on Lattice polytopePolynomial function

Definition

Let P⊆ℝnP \subseteq \R^n be a lattice polytope of dimension dd. For each integer t≥1t \ge 1, let LP(t)=#(tP∩ℤn),tP={tx:x∈P},L_P(t) = \#\left(tP \cap \Z^n\right), \qquad tP = \set{tx : x \in P}, be the number of integer points in the dilate tPtP. By Ehrhart's theorem, there is a unique polynomial EP(t)E_P(t) with rational coefficients such that EP(t)=LP(t)E_P(t) = L_P(t) for all integers t≥1t \ge 1. This polynomial EPE_P is the Ehrhart polynomial of PP.

Remarks

The polynomial EPE_P has degree dd, and EP(0)=1E_P(0) = 1. If d=nd = n, the leading coefficient of EPE_P is the volume of PP. In general the leading coefficient is the volume of PP measured inside its affine hull, normalised by the lattice of integer points of that affine hull.

Example: for the unit square P=[0,1]2P = [0,1]^2, the dilate tP=[0,t]2tP = [0,t]^2 contains (t+1)2(t+1)^2 integer points, so EP(t)=t2+2t+1E_P(t) = t^2 + 2t + 1. For the triangle with vertices (0,0)(0,0), (1,0)(1,0), (0,1)(0,1), the dilate contains the points (a,b)∈ℤ2(a,b) \in \Z^2 with a,b≥0a, b \ge 0 and a+b≤ta + b \le t, so EP(t)=(t+1)(t+2)/2E_P(t) = (t+1)(t+2)/2. In both cases the leading coefficient is the area, 11 and 1/21/2.

Non-example: the segment P=[0,1/2]P = [0, 1/2] is a rational polytope but not a lattice polytope. Here LP(t)=⌊t/2⌋+1L_P(t) = \floor{t/2} + 1, which is not a polynomial in tt; it is only a quasi-polynomial of period 22. So the hypothesis that the vertices are integer points is needed.

The generating function ∑t≥0EP(t)zt\sum_{t \ge 0} E_P(t) z^t is a rational function of the form h∗(z)/(1−z)d+1h^*(z)/(1-z)^{d+1}, where h∗h^* is a polynomial with nonnegative integer coefficients.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Ehrhart_polynomial
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MSC52B20, 11P21, 05A15
DefinesEhrhart polynomial
UsesLattice polytope, Polynomial function

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