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Are there any results stating that a given family of convex polytopes have Ehrhart polynomials with non-negative coefficients?

What methods are available for proving such a property for some family of polytopes?

Remember, the Ehrhart polynomial $p(k)$ for a convex polytope $P$ with integer vertices is given by the property that $p(k)$ counts the number of integer lattice points in the $k$-dilation of $P$, where $k$ is a positive integer.

Per Alexandersson
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4 Answers4

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I have converted a comment (slightly modified) to the following answer as requested by Per Alexandersson. If $Ω_P(k)$ is the order polynomial of a poset $P$, then $Ω_P(k+1)$ is the Ehrhart polynomial of the order polytope $\mathcal{O}(P)$. For any n≥1, the Ehrhart polynomial of the order polytope of the poset $P_n$ with one minimal element covered by $n$ other elements is $\sum_{i=1}^{k+1}i^n$. For $n=20$ the coefficient of $k$ is $−168011/330$, so Ehrhart polynomials of 0/1 polytopes (or even order polytopes) need not have nonnegative coefficients. Incidentally, it was easy to check using Stembridge's posets package for Maple that the Ehrhart polynomial of the order polytope of any poset with at most eight elements has nonnegative coefficients.

Richard Stanley
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  • Richard Stanley's example of a non-Ehrhart positive order polytope is discussed in more detail here: https://arxiv.org/abs/1806.08403 – Sam Hopkins Jun 25 '18 at 23:52
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There seems to be a lot of information (including the conjecture that this holds for the Birkhoff polytope) in Richard Stanley's slides. Of course, since Richard is a frequent participant, he can (and surely will) say more.

Igor Rivin
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  • Oh, I did not know that this was a conjecture for the Birkhoff polytope. That is very interesting! – Per Alexandersson Oct 29 '14 at 17:29
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    All the known explicit Ehrhart polynomials for the Birkhoff polytope are at http://www.math.binghamton.edu/dennis/Birkhoff/ – Brendan McKay Oct 29 '14 at 22:57
  • Probably order 10 is now within range of computers. – Brendan McKay Oct 29 '14 at 22:58
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    The Ehrhart polynomial of an order polytope can have negative coefficients. See EC1, 2nd ed., Exericse 3.164. Ehrhart polynomials of integer zonotopes do have nonnegative coefficients (e.g., Theorem 2.2 of http://math.mit.edu/~rstan/pubs/pubfiles/83.pdf). I am not sure if it's known whether or not the generalized permutohedra of Postnikov (http://math.mit.edu/~apost/papers/permutohedron_full.pdf) have this property. – Richard Stanley Mar 18 '15 at 00:49
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    @RichardStanley: I might be mistaken, but the first poset Figure 3.87 in EC1 2nd ed, gives the inequalities $z \leq x_1 \leq y$, $z\leq x_2$ and $z\leq x_3$ together with the condition that all values are between $0$ and $k$. Asking Mathematica for counting solutions gives the polynomial $\frac{1}{120} (k+1) (k+2) (k+3) \left(12 k^2+33 k+20\right)$. The code I used for counting lattice points:

    Table[Length@List@ToRules@Reduce[ And[z <= x1 <= y, z <= x2, z <= x3] && And[0<=z<=k,0<=y<=k,0<=x1<=k,0<=x2<=k,0<=x3<= k], {z, y, x1, x2, x3}, Integers],{k, 0, 5}]

     – Per Alexandersson Mar 19 '15 at 14:44
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    @PerAlexandersson: yow, you are right! I have made this mistake before. If $\Omega_P(k)$ is the order polynomial of a poset $P$, then $\Omega_P(k+1)$ (not $\Omega_P(k)$) is the Ehrhart polynomial of the order polytope. However, for any $n\geq 1$, the order polytope of the poset $P_n$ with one minimal element covered by $n$ other elements is $\sum_{i=1}^{k+1}i^n$. For $n=20$ the coefficient of $k$ is $-168011/330$, so Ehrhart polynomials of 0/1 polytopes need not have nonnegative coefficients. – Richard Stanley Mar 20 '15 at 18:17
  • @RichardStanley Ah, I suspected that was something about that. I wonder what smallest counter-example there is: There is no $5$-dimensional such order polytope, (checked by computer). I got the conjecture about 01-polytopes from an earlier version of a paper I found online, and the final version does not include this conjecture. If you would make your comment into an answer, I would accept that, since this was really the information I was looking for (but not the answer I was hoping for). – Per Alexandersson Mar 20 '15 at 20:07
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    The zeros of the Ehrhart polynomial $H_n(k)$ of the Birkhoff polytope of $n\times n$ doubly stochastic matrices look very interesting. For $9\times 9$ matrices see http://math.mit.edu/~rstan/zeros/magic9.pdf. Assuming that this behavior generalizes to all $n$, we would have the following. Let $c(n,i)$ be the coefficient of $k^{(n-1)^2-i}$ in $H_n(k)$. Then for fixed $k$, $c(n,i)/c(n,0)\sim n^{3i}/2^ii!$. See EC1, 2nd ed., Exercise~4.54. – Richard Stanley Mar 21 '15 at 00:11
  • Another class of polytopes whose Ehrhart polynomials have nonnegative coefficients are the lattice-face polytopes of Fu Liu. See http://arxiv.org/abs/0810.4655. In particular, if $\mathcal{P}$ is any rational polytope, then there exists an integral polytope $\mathcal{Q}$ combinatorially equivalent to $\mathcal{P}$ such that the Ehrhart polynomial of $\mathcal{Q}$ has nonnegative coefficients. – Richard Stanley Mar 26 '15 at 21:01
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Fu Liu recently put on the arXiv a nice survey about this question: https://arxiv.org/abs/1711.09962v1

Sam Hopkins
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Proposition 4 of Morelli's paper "Pick's Theorem and the Todd class of a toric variety" gives a sufficient condition: it describes a setting in which there is a positive formula for coefficient of $x^k$ as a sum over $k$-dimensional faces.

Hugh Thomas
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