Fuss-nonnesting orders, a rank-one obstruction, and a Boolean q-model

Definitions and mathematical statement

The aid identifies this extraction as two genuine remarks/questions following AIM Problem 1.3, not an independent Problem 5.3. The leading "Problem 5.3, in free probability" is an internal reference from surrounding material. The first question asks for a natural generalized nonnesting poset $NN_W^{(k)}$: in type $A$, it suggests $k$-divisible nonnesting set partitions under refinement, perhaps matching an order on co-filtered root-poset multichains in general. The second names Christos Athanasiadis and Stavros Garoufallidis as suggesting a $q$-version of Catalan combinatorics, but ends with an incomplete "See" and no reference or definition.

For the aid's concrete test let $k\geq1$. A set partition is $k$-divisible if each block size is a multiple of $k$; nonnesting means that its standard arcs joining consecutive elements of each block contain no nested pair $a<c<d<b$ with arcs $(a,b)$ and $(c,d)$. Order partitions by refinement. In rank one, consider these partitions of $[2k]$. They consist of the one-block maximum and exactly $k$ incomparable two-block partitions, so their poset is the $k$-fork $F_k$, with $k$ minimal elements and one maximum.

In contrast, the positive-root poset of $A_1$ is $\{\alpha\}$. Its geometric decreasing $k$-filter chains have the form \[ F_1=\cdots=F_j=\{\alpha\},\qquad F_{j+1}=\cdots=F_k=\varnothing\qquad(0\leq j\leq k). \] Root-addition constraints are vacuous here. Componentwise inclusion makes these $k+1$ choices a chain $C_{k+1}$. Therefore for every $k\geq2$, \[ C_{k+1}\not\cong F_k, \] even though both have $k+1$ elements. This refutes only the immediate componentwise-inclusion order-isomorphism proposal, not every possible order on geometric multichains or every natural generalized nonnesting construction. For $k=1$ the rank-one obstruction disappears.

For $W=A_1^r$, each of the $r$ independent positive roots has a threshold $j_\ell\in\{0,\ldots,k\}$, so the same filter order is $C_{k+1}^r$ with natural rank $\rho=\sum_\ell j_\ell$. The aid's $q$-Fuss-Catalan normalization gives \[ \sum_{F}q^{2\rho(F)}=(1+q^2+\cdots+q^{2k})^r =\left(\frac{[2k+2]_q}{[2]_q}\right)^r, \qquad [m]_q=1+q+\cdots+q^{m-1}. \] Thus twice the natural rank realizes this polynomial in the $A_1^r$ test family. It is not a general statistic theorem for all root systems, and it does not identify the incomplete source's exact intended $q$-construction.

Short English statement

The fragment asks for a generalized nonnesting order and mentions an undefined $q$-Catalan proposal. The aid shows a rank-one obstruction: componentwise filter chains form a chain, while divisible nonnesting refinement forms a fork. A product of rank-one systems realizes a particular $q$-Fuss-Catalan polynomial by twice its filter-chain rank.

Sources

• Local UnsolvedMath registry, permanent ID $20002704$, catalogue entry; aim-probability-notes.json, notes[145], extracted source item 5.3.
• American Institute of Mathematics, Braid groups, clusters and free probability, https://aimath.org/WWN/braidgroups/braidgroups.pdf. The aid locates the two substantive remarks after Problem 1.3 rather than at a standalone Problem 5.3.
• Original attribution: Christos Athanasiadis and Stavros Garoufallidis for a $q$-version of Catalan combinatorics. The citation following "See" is missing. No authoritative separate reference for the rank-one obstruction or product enumerator is supplied.
• Research attempt 1: partial result, problem partially solved at run, rigor label "proved," novelty confidence low. Registry classification partial progress, status partially solved, statement status exact, despite the contextual extraction and truncated reference.