Unimodality conjecture for Toda eigenfunction numerators

Counterexample: the conjecture fails in every non-simply-laced irreducible Dynkin type.

The exact current statement is Conjecture 7.3 of A. Labelle, arXiv:2502.10655v3, revised July 28, 2026. It asserts unimodality of

$$P_\alpha(q)=(q)_\alpha^2\mathfrak J_\alpha(q)$$

for every nonnegative root-lattice element of an arbitrary split semisimple root system.

Write

$$d_i=\frac{(\alpha_i,\alpha_i)}2, \qquad D_a=\prod_i(q^{d_i};q^{d_i})_{a_i}, \qquad Q(a)=\frac12\left(\sum_i a_i\alpha_i,\sum_i a_i\alpha_i\right).$$

The source's defining recursion is

$$(1-q^{Q(a)})\mathfrak J_a = \sum_{\substack{0\le b\le a\b\ne a}} \frac{q^{Q(b)}}{D_{a-b}}\mathfrak J_b, \qquad \mathfrak J_0=1. \tag{1}$$

Substituting $P_b=D_b^2\mathfrak J_b$ gives the exact integral triangular recurrence

$$(1-q^{Q(a)})P_a = \sum_{\substack{0\le b\le a\b\ne a}} q^{Q(b)}P_b \prod_i \left[ \binom{a_i}{b_i}_{q^{d_i}} \prod_{j=b_i+1}^{a_i}(1-q^{d_i j}) \right]. \tag{2}$$

Now take type $B_2$, ordering the long root first. Then

$$(d_1,d_2)=(2,1), \qquad \bigl((\alpha_i,\alpha_j)\bigr) = \begin{pmatrix}4&-2\\-2&2\end{pmatrix}, \qquad Q(a,b)=2a^2-2ab+b^2.$$

Applying (2) gives the complete small-index table

$$\begin{array}{c|ccc} &b=0&b=1&b=2\\ \hline a=0&1&1&1\\ a=1&1&1+q+q^2&1+q+3q^2+q^3+q^4\\ a=2&1&1+q+q^2+q^3+q^4& 1+q+3q^2+2q^3+5q^4+2q^5+3q^6+q^7+q^8 . \end{array}$$

In particular, for $\alpha=2\alpha_1+2\alpha_2$,

$$\boxed{ P_\alpha(q) = 1+q+3q^2+2q^3+5q^4+2q^5+3q^6+q^7+q^8. }$$

Every coefficient is strictly positive, and the polynomial is palindromic, but

$$[q^2]P_\alpha=3> [q^3]P_\alpha=2< [q^4]P_\alpha=5.$$

Hence $P_\alpha$ is not unimodal.

The obstruction extends to every doubly-laced irreducible type. If a coefficient $a_i=0$, every $b\le a$ in (1) also has $b_i=0$; thus the recursion depends only on the full Dynkin subdiagram supporting $a$. Each $B_r,C_r$ for $r\ge2$, and $F_4$, contains a $B_2=C_2$ double-edge subdiagram. Put coefficient $2$ on its two vertices and $0$ elsewhere to obtain exactly the same nonunimodal polynomial.

For the remaining non-simply-laced type $G_2$, take

$$(d_1,d_2)=(1,3), \qquad Q(a,b)=a^2-3ab+3b^2, \qquad \alpha=3\alpha_1+3\alpha_2.$$

The exact coefficient sequence from (2) is

$$(1,2,5,11,16,24,39,46,60,84,88,99,119,110, 110,119,99,88,84,60,46,39,24,16,11,5,2,1).$$

It again has positive full support and is palindromic, but contains

$$119>110=110<119.$$

Thus Conjecture 7.3 is false in every non-simply-laced irreducible Dynkin type. The separate simply-laced and type-$A$ questions are not claimed here.