# Coordinate-symmetric convex polytopes with equal Erhart (quasi)-polynomials

Recall that given a nondegenerate polytope $$P \subset \mathbb{R}^n$$ which is the convex set of some vectors with integral coordinates, the Erhart polynomial $$p_P(t)$$ a polynomial such that $$p_P(t)$$ for natural numbers $$t$$ is the number of integral lattice points in the $$t$$-fold dilation $$tP$$ of $$P$$.

Do there exist two polytopes $$P, P'$$ which have equal Erhart polynomials, are preserved under Euclidean reflections along all coordinate hyperplanes $$\{x_i = 0\} \subset \mathbb{R}^n$$, and are not $$GL(n, \mathbb{Z})$$ equivalent, i.e. there is no integral matrix $$A$$ with $$det(A) = \pm 1$$ such that $$AP = P'$$?

In fact I would be happy if there were two polytopes which are convex hulls of rational vectors and are symmetric under reflection across all coordinate hyperplanes, which are not $$GL(n, \mathbb{Z})$$ equivalent but have the same Erhart quasi-polynomials (these are certain functions characterized in the same way as the Erhart polynomials using the same definition of their values on the positive natural numbers; they just cease to be polynomial functions in the rational case.)

• A keyword that might be helpful: polytopes preserved by reflection across all coordinate hyperplanes are sometimes called "unconditional" (I think this is a weird term, but it is what it is). Mar 10, 2020 at 18:29

How about the following two polygons, defined by their convex hulls:

{-6, -4}, {-6, 4}, {0, -8}, {0, 8}, {6, -4}, {6, 4}

and

{-6, -2}, {-6, 2}, {-2, -8}, {-2, 8}, {2, -8}, {2, 8}, {6,-2}, {6, 2}

Both of these have (if my calculations are correct), the Ehrhart polynomial $$1 + 6 t + 36 t^2$$.

Since the number of vertices are different, well, there is no matrix taking the first to the second.

There are more examples of this type, the two sets of points

{{-8,-2},{-8,2},{-4,-8},{-4,8},{4,-8},{4,8},{8,-2},{8,2}}}

and

{{-8,-2},{-8,2},{-6,-6},{-6,6},{-2,-8},{-2,8},{0,-8},{0,8},{2,-8},{2,8},{6,-6},{6,6},{8,-2},{8,2}}

both have Ehrhart polynomial $$1 + 8 t + 52 t^2$$.

I used some Mathematica code to randomly generate some points in a symmetric fashion, then compute convex hull. Finally, the Ehrhart polynomial is just a matter of counting lattice points in dilations, followed by an interpolation. Then one can just look at the data.

RandomPolytope[] := Module[{pts, sympts, grid, f, region},
pts = 1 + RandomInteger[6, {5, 2}];
grid = Join @@ Table[{x, y}, {x, -30, 30}, {y, -30, 30}];
sympts =
Union[Join @@
Join[{{#1, #2}, {-#1, #2}, {#1, -#2}, {-#1, -#2}} & @@@ pts]];
poly = Table[
region = BoundaryMeshRegion[ConvexHullMesh[k sympts]];
Length@Select[grid, RegionMember[region, #] &]
, {k, 1, 2}];
(Round@MeshCoordinates@region) ->
Expand[InterpolatingPolynomial[Prepend[poly, 1], k] /.
k -> k + 1]
];
data = Table[RandomPolytope[], {150}];

• These are zonotopes, I believe; if you write them as a Minkowski sum of line segments then it becomes easy to check their Ehrhart polynomials from Stanley's formula. Mar 10, 2020 at 21:13
• @SamHopkins would you mind giving a reference to which formula of Stanley you're referring to? I'm unfamiliar with all the standard enumerative combinatorics here but I'm happy to learn more about the subject :-)
– skr
Mar 10, 2020 at 22:15
• See Theorem 9.9 of math.sfsu.edu/beck/papers/zonotopes.pdf. Mar 10, 2020 at 22:18