# Unimodular triangulation and Ehrhart polynomials

Let $P$ be a convex lattice polytope. Then it has a polynomial Ehrhart function.

I am interested in what can be said about the Ehrhart polynomial when $P$ has any of the properties

• is integrally closed
• has a unimodular triangulation
• has a unimodular pulling triangulation
• is compressed

Specifically, does any of these properties imply that the coefficients of the Ehrhart polynomial are non-negative?

I don't think you can generally conclude the nonnegativity of Ehrhart polynomials when expressed in the standard monomial basis; however, the more natural basis for Ehrhart polynomials is given by the binomial coefficients $\binom{n+d} d$, $\binom{n+d-1} d$, ..., $\binom n d$ (here $n$ is the variable of the polynomial and $d$ is the dimension of $P$). Stanley proved in 1980 that the coefficients of the Ehrhart polynomial of $P$ written in terms of this basis are nonnegative. This coefficient vector is known as the $h^*$-vector, $\delta$-vector, or Ehrhart $h$-vector of $P$. Here are some things that are known and connected to your question:

If $P$ has a unimodular triangulation $T$ then the $h^*$-vector of $P$ equals the $h$-vector of $T$ (a result contained in the aforementioned 1980 paper by Stanley). (I don't think a pulling triangulation buys anything extra, short of nice properties of the $h$-vector.)

For the case that $P$ is compressed, I believe the state of the art is described in a 2007 paper by Bruns and Roemer in JCTA.

For the case that $P$ is integrally closed, you may consult a recent preprint by Braun and Davis.

• Thanks! I suspected as much... The last paper only consider reflexive polytopes unfortunately. – Per Alexandersson Oct 8 '14 at 13:59
• De Loera, Haws and Koppe have some fascinating conjectures about coefficients of Erhart polynomials of matroid polytopes in the monomial basis. arxiv.org/abs/0710.4346 – David E Speyer Oct 8 '14 at 14:11

To add a partial answer, something I should have thought of before:

Every lattice polytope $P$ can be made into an integral polytope, by maginfying it with a sufficiently large factor. It is known that the dimension of $P$ suffices, to $dP$ is integrally closed. However, if $f(x)$ is the Ehrhart polynomial for $P$, then naturally $f(dx)$ is the Ehrhart polynomial for $dP$.

Notice that the sign of the coefficients in $f(x)$ and $f(dx)$ are the same. Hence, just signs of coefficients are not more special for Ehrhart polynomials of integrally closed polytopes.