Real-rootedness conjecture for preorder-polytope h-polynomials

An eight-element counterexample

Conjecture 5.3 of Athanasiadis and Chapoton, arXiv:2605.26916, is false, even for an ordinary poset with eight elements. The broader question for support-enumerators is raised by Wang et al. in arXiv:2608.16037, Problem 5.3.

Let $P$ have lower elements $a_1,a_2,a_3,d$ and upper elements $b_0,b_1,b_2,b_3$. Its only strict relations are

$$a_i<b_0,\qquad a_i<b_i,\qquad d<b_i \qquad (i=1,2,3).$$

These relations define a height-two poset, hence a preorder of the kind allowed in the conjecture.

For clarity, the polynomial at issue counts lattice points by the size of their support:

$$Q_P=\left\{\mathbf{x}\in\mathbb R_{\ge0}^{P}: \sum_{v\in I}x_v\le |I| \text{ for every order ideal }I\right\},$$ $$h_P(t)=\sum_{\mathbf{x}\in Q_P\cap\mathbb Z^{P}} t^{|\{v:x_v>0\}|}.$$

It is not the Ehrhart $h^\ast$-polynomial. We first compute $h_P(t)$ from these defining inequalities and then give an exact obstruction to real-rootedness.

1. Reducing the lattice-point count

Every lower coordinate is either $0$ or $1$, since its singleton is an order ideal. Fix the set $Z$ of lower elements whose coordinates are zero. Write

$$r=|Z\cap\{a_1,a_2,a_3\}|,\qquad \epsilon=\mathbf1_{\{d\in Z\}}.$$

By permuting the three branches, assume that $Z\cap\{a_1,a_2,a_3\}=\{a_1,\ldots,a_r\}$. There are $\binom3r$ choices of this type, each with lower support $4-r-\epsilon$.

Write $y_i=x_{b_i}$, and let $N(S)$ denote the lower elements below at least one member of an upper subset $S$. The ideal consisting of $S$ together with $N(S)$ gives

$$\sum_{b_i\in S}y_i\le |S|+|N(S)\cap Z|. \tag{1}$$

These inequalities are sufficient as well: any other ideal with upper part $S$ only adds lower elements, each with coordinate at most $1$.

Set

$$e_i=\max(y_i-1,0)\qquad(0\le i\le3).$$

The inequalities (1) are equivalent to

$$\sum_{b_i\in S}e_i\le |N(S)\cap Z| \tag{2}$$

for every upper subset $S$. Indeed, (2) implies (1) because $y_i-1\le e_i$. Conversely, apply (1) to the members of $S$ with $y_i\ge2$; their excesses sum to the left side of (2), and their lower neighborhood is contained in $N(S)$.

Put $(q)_+=\max(q,0)$. For this particular poset, (2) reduces exactly to the following three bounds:

$$\begin{aligned} 0\le e_0&\le r,\\ \sum_{i=1}^{3}\bigl(e_i-\mathbf1_{\{i\le r\}}\bigr)_+ &\le\epsilon,\\ e_0+e_1+e_2+e_3&\le r+\epsilon. \end{aligned} \tag{3}$$

To see this, a nonempty subset of $\{b_1,b_2,b_3\}$, indexed by $K$, has $|K\cap\{1,\ldots,r\}|+\epsilon$ zero lower neighbors. Taking all indices with positive surplus gives the second bound. A subset containing $b_0$ has $r$ zero lower neighbors if it is just $\{b_0\}$, and $r+\epsilon$ otherwise. Nonnegativity of the excesses then gives precisely the first and third bounds.

2. The resulting polynomial

Let $F_{r,\epsilon}(u)$ count the nonnegative integer vectors satisfying (3), with weight $u$ for each positive excess coordinate. The eight small counts are

$$\begin{array}{c|l} (r,\epsilon)&F_{r,\epsilon}(u)\\ \hline (0,0)&1\\ (0,1)&1+3u\\ (1,0)&1+2u\\ (1,1)&1+5u+5u^2\\ (2,0)&1+4u+3u^2\\ (2,1)&1+7u+13u^2+4u^3\\ (3,0)&1+6u+9u^2+4u^3\\ (3,1)&1+9u+24u^2+16u^3+u^4 \end{array} \tag{4}$$

Here every coordinate is at most $r+\epsilon\le4$, so the table involves only bounded four-coordinate counts. More explicitly, when $\epsilon=0$, choose $q$ of the first $r$ private coordinates to equal $1$. The coordinate $e_0$ then has $r-q$ possible positive values, giving

$$F_{r,0}(u)=\sum_{q=0}^{r}\binom rq u^q\bigl(1+(r-q)u\bigr).$$

When $\epsilon=1$, the private coordinates may instead have exactly one of these first $r$ coordinates equal to $2$, or exactly one of the other $3-r$ coordinates equal to $1$. Together with the choices in which neither happens, these exhaust the second bound in (3). For any such private choice of total $w$, the possible center values are $0\le e_0\le\min(r,r+1-w)$, giving the $F_{r,1}$ entries of (4).

For example, in the largest case $r=3,\epsilon=1$, the private coordinates $e_1,e_2,e_3$ lie in $\{0,1,2\}$, with at most one equal to $2$, while $e_0\le3$ and the total is at most $4$. Counting by support size gives

$$\begin{gathered} 1,\quad 3+3\cdot2=9,\quad 3\cdot5+3\cdot3=24,\\ 3\cdot4+4=16,\quad 1. \end{gathered}$$

The other rows follow from the same bounds with smaller $r$ or $\epsilon$.

For a fixed excess vector with $s$ positive coordinates, each of those $s$ upper coordinates is forced to be positive. At each of the other $4-s$ positions, the original coordinate $y_i$ can independently be $0$ or $1$. Its upper-support contribution is therefore $t^s(1+t)^{4-s}$. Consequently,

$$h_P(t)= \sum_{r=0}^{3}\sum_{\epsilon=0}^{1} \binom3r\,t^{4-r-\epsilon}(1+t)^4 F_{r,\epsilon}\!\left(\frac{t}{1+t}\right). \tag{5}$$

Expanding (4) in (5) gives

$$\begin{aligned} h_P(t)={}&1+17t+106t^2+303t^3+427t^4\\ &\quad+303t^5+106t^6+17t^7+t^8. \end{aligned} \tag{6}$$

In particular,

$$\begin{aligned} h_P(t)&=(1+t)^8\gamma\!\left(\frac{t}{(1+t)^2}\right),\\ \gamma(z)&=1+9z+24z^2+16z^3+z^4. \end{aligned} \tag{7}$$

3. An exact nonreal-root certificate

If the polynomial (6) had only real roots, all of them would be negative, since its coefficients are positive. Its palindromicity pairs each root with its reciprocal; moreover, $h_P(-1)=1$, so $-1$ is not a root. Thus

$$h_P(t)=\prod_{j=1}^{4}(t^2+c_jt+1), \qquad c_j>2.$$

Substituting in (7) would give

$$\gamma(z)=\prod_{j=1}^{4}\bigl(1+(c_j-2)z\bigr),$$

so $\gamma$ would also have only real roots.

However, every real-rooted polynomial $q$ satisfies

$$q'(s)^2-q(s)q''(s) =q(s)^2\sum_j\frac{1}{(s-\rho_j)^2}\ge0$$

at every real $s$ which is not a root, where the real roots $\rho_j$ are listed with multiplicity. For the polynomial $\gamma$ in (7), exact evaluation gives

$$\begin{aligned} \gamma\!\left(-\frac14\right)&=\frac1{256},\\ \gamma'\!\left(-\frac14\right)&=-\frac1{16},\\ \gamma''\!\left(-\frac14\right)&=\frac{99}{4}, \end{aligned}$$

and hence

$$\gamma'\!\left(-\frac14\right)^2- \gamma\!\left(-\frac14\right)\gamma''\!\left(-\frac14\right) =-\frac{95}{1024}<0.$$

This contradiction shows that $h_P(t)$ has nonreal roots. The explicitly defined eight-element poset $P$ therefore disproves Conjecture 5.3.