Non-rank-symmetry conjecture for hull configurations without blank sides

Complete proof for every polygonal hull configuration, with an exact rank-reciprocity formula.

Let $P$ have $k\ge3$ polygon vertices, $c_i\ge0$ configuration points in the relative interior of side $i$, and $n=k+\sum_i c_i$. Define

$$F_P(q)=\sum_{\pi\in\operatorname{NC}(P)}q^{n-\#\operatorname{blocks}(\pi)},\qquad [0]_q=0,\qquad[c]_q=1+q+\cdots+q^{c-1}\quad(c\ge1).$$

I prove the stronger identity

$$\boxed{q^{n-1}F_P(q^{-1})-F_P(q)=(-1)^kq^{k-1}(q-1)\prod_{i=1}^k[c_i]_q.}\tag{1}$$

First suppose $c_1=\cdots=c_k=1$, with cyclic boundary order $v_0,m_0,v_1,m_1,\ldots,v_{k-1},m_{k-1}$. Temporarily perturb the $m_i$ outward, producing the classical lattice $\operatorname{NC}(2k)$. Restoring collinearity forbids precisely the events

$$A_i=\{v_i,v_{i+1}\text{ share a block and }m_i\text{ is a singleton}\}.$$

For a proper subset $S\subsetneq\{0,\ldots,k-1\}$ of size $s$, its selected corner edges form a forest. Deleting the $s$ forced singleton midpoints and contracting the $s$ same-block corner edges gives a rank-shift-$s$ bijection

$$\bigcap_{i\in S}A_i\cong\operatorname{NC}(2k-2s).$$

For the full edge cycle there is exactly one partition: all polygon vertices form one block and every midpoint is singleton, giving rank $k-1$. Inclusion-exclusion therefore yields

$$F_{1,\ldots,1}(q)=\sum_{s=0}^{k-1}(-1)^s\binom{k}{s}q^sN_{2k-2s}(q)+(-1)^kq^{k-1},\tag{2}$$

where

$$N_m(q)=\sum_{j=0}^{m-1}\frac1m\binom mj\binom m{j+1}q^j.$$

Narayana reciprocity $q^{m-1}N_m(q^{-1})=N_m(q)$ makes every proper-forest summand reciprocal of degree $2k-1$. Thus

$$D_{1,\ldots,1}(q)=(-1)^kq^{k-1}(q-1).\tag{3}$$

Now let one side contain $c\ge2$ interior points and write its first boundary points as $a,x,b$, where $a$ is a polygon vertex. Set $Q=P\setminus\{x\}$ and $R=P\setminus\{x,b\}$. A block's points on a polygon side must form a contiguous run.

Partitions where $x$ shares the block of $b$ contribute $qF_Q(q)$. Partitions where $x$ is singleton contribute $F_Q(q)-qF_R(q)$: a singleton can be inserted precisely when $a,b$ belong to distinct blocks, and the excluded same-block partitions correspond to $\operatorname{NC}(R)$ with rank shift one.

For every remaining partition, let $y$ be the first clockwise same-block partner of $x$ in the order $b,\ldots,a$. Side contiguity implies either $y=a$ or $y$ lies off the original side. The chord $xy$ separates the nonempty boundary arcs

$$I_y=(b,\ldots,\operatorname{pred}(y)),\qquad O'_y=(y,\ldots,a),\qquad |I_y|+|O'_y|=n-1.$$

Adjacent-corner contraction gives the exact rank polynomial

$$H_y(q)=qF_{I_y}(q)F_{O'_y}(q).\tag{4}$$

Each arc configuration is either collinear, hence Boolean, or has a genuinely blank closing side. Its rank polynomial is therefore symmetric by the blank-side theorem. Consequently

$$q^{n-1}H_y(q^{-1})=H_y(q).$$

Summing over all $y$ gives

$$F_P(q)=(1+q)F_Q(q)-qF_R(q)+H_P(q),\qquad q^{n-1}H_P(q^{-1})=H_P(q).\tag{5}$$

For a $t$-point configuration $T$, write $D_T(q)=q^{t-1}F_T(q^{-1})-F_T(q)$. Since $|Q|=n-1$ and $|R|=n-2$, equation (5) implies

$$D_c(q)=(1+q)D_{c-1}(q)-qD_{c-2}(q).\tag{6}$$

A blank side gives $D_0(q)=0$. Since

$$[0]_q=0,\qquad[1]_q=1,\qquad[c]_q=(1+q)[c-1]_q-q[c-2]_q,$$

induction yields $D_c(q)=[c]_qD_1(q)$. Applying this independently to every side and using (3) proves (1).

Therefore the rank polynomial is symmetric if and only if at least one polygon side is blank. When every $c_i\ge1$, the right-hand side of (1) is nonzero; indeed, writing $F_P(q)=\sum_j a_jq^j$ gives

$$a_{n-k}-a_{k-1}=(-1)^{k+1}\neq0.$$

This proves the conjecture in its entirety.

References: M. Dougherty and G. Root, Noncrossing Partitions From Hull Configurations, Theorem 2.7, https://arxiv.org/abs/2604.14458; Noncrossing Partitions From Cones and Semicircles, Lemma 2.7(1), https://arxiv.org/abs/2604.14462.