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.)

  • $\begingroup$ 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). $\endgroup$ Mar 10, 2020 at 18:29

1 Answer 1


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

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


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

two polytopes

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




Second set of polytopes 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}];
  • 2
    $\begingroup$ 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. $\endgroup$ Mar 10, 2020 at 21:13
  • $\begingroup$ @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 :-) $\endgroup$
    – skr
    Mar 10, 2020 at 22:15
  • $\begingroup$ See Theorem 9.9 of math.sfsu.edu/beck/papers/zonotopes.pdf. $\endgroup$ Mar 10, 2020 at 22:18

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