# Relation between non-integral polytopes, integrally closed polytopes and polynomial Erhart quasi-polynomials

By lattice points, I will always mean points in $\mathbb{Z}^n$ and all polytopes here are convex rational polytopes.

If $P$ is an integral polytope, the counting function for the number of lattice points inside $nP$ is a polynomial $p(n)$. This is the Erhart polynomial, and if $P$ does not have integral vertices, $p(n)$ is in general just a quasi-polynomial.

A integrally closed polytope is a polytope, such that each lattice point in $nP$ is a sum of exactly $n$ lattice points in $P$.

To the question: Now, let $P$ be a non-integral polytope, but its Erhart quasi-polynomial is in fact a polynomial. Let $P'$ be the convex hull of the lattice points in $P'$.

Are there examples $P$ for which $P'$ is not integrally closed?

Motivation: Being integrally closed is a quite rare property for a polytope (in high dimensions). Having a polynomial Erhart function, when only a quasi-polynomial is expected is also rare. Therefore, the chance that these coincide is even rarer, so one would expect a lot of examples above, or these properties are related somehow.

• Are your polytopes all rational? Mar 9, 2014 at 0:09
• Yes, they are. For example I suspect that Gelfand-Tsetlin polytopes and BZ polytopes P (which in general are non-integral), have the property that P' ARE integrally closed. This is also a consequence of a conjecture of T. MacAllister if I am not mistaken... Mar 9, 2014 at 9:58

A combination of the polytopes introduced in Exercise 2.24 of Polytopes, Rings, and K-Theory by W. Bruns and J. Gubeladze (see also Section 3 of this paper) with a 3-dimensional version of the polygons in this paper by T. McAllister and K. Woods should do the trick. To be more concrete, I believe the convex hull of $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0, \frac 1 a - 1, 1)$, $(1, \frac 1 a - 1, 1)$, $(0,0,a)$, $(1,0,a)$, $(0,1,a)$, $(1,1,b)$, $(1,1,a+b)$ will give you an example with the right choices of $a$ and $b$ (the integer $b$ has to be chosen large enough so that the polytope is not normal---try something like $b=4$---, and $a$ has to be chosen large enough so that these 10 points are actually vertices).