A tableaux formula for $q$-rook numbers

Aritra Bhattacharya; Tirtharaj Basu

REVIEW 2 major objections 3 minor 14 references

A tableaux formula for $q$-rook numbers

T0 review · 2 major / 3 minor · reviewed 2026-08-06 · deepseek-v4-flash

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Pith's one-line read The paper proves that every Garsia–Remmel q-rook number can be written as a weighted sum over standard Young tableaux constrained by a Dyck path, and that the same weights are the coefficients of unicellular LLT functions in the…

desk verdict New tableaux formula for q-rook numbers with a workable proof, but Lemma 3.4 needs a small fix before publication. read the letter →

arxiv 2507.00766 v1 pith:37HSFDKO submitted 2025-07-01 math.CO math.RT

Tirtharaj Basu , Aritra Bhattacharya This is my paper · ORCID

classification math.COmath.RT MSC 05E0505A1005A30

keywords q-rooknumbersGarsia-RemmelrookstandardYoungtableauxDyckpathsunicellularLLTfunctionsq-Whittakerq-Stirlingchromaticsymmetric

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The paper proves that the Garsia–Remmel $q$-rook numbers $R_k(\lambda;q)$, the polynomials that count ways to place $k$ non-attacking rooks on a Ferrers board of shape $\lambda$ with a $q$-weighted inversion statistic, can be written as a weighted sum over standard Young tableaux constrained by a Dyck path. Given any Dyck path $\pi$ whose shape above it is $\lambda$, the sum runs over tableaux of shapes $\mu \vdash n$ with first row $n-k$, and each tableau's weight is a power of $q$ times a product of $q$-integers, provided entries in each column respect the order the path imposes on the numbers $1,\dots,n$. The proof shows the tableau sum obeys the same two-term recursion that defines the $q$-rook numbers, by tracking what happens when the largest entry is deleted from a valid tableau and the path shrinks by one step. If right, the formula gives every $q$-rook number a uniform, direct combinatorial description, and it connects rook theory to the Macdonald universe of symmetric functions: the same tableaux weights are the coefficients of unicellular LLT functions in the $q$-Whittaker basis, and the abelian case recovers the known rectangle decomposition of unicellular LLT functions with $q$-hit-number coefficients.

What carries the argument

The objects doing the work are the Dyck path's poset and the tableaux that respect it. A Dyck path $\pi$ of semilength $n$ determines a partition $\lambda(\pi)$ (the cells above the path) and a partial order on $[n]$ by $i <_\pi j$ exactly when the cell $(i,j)$ is above the path; $\mathrm{SYT}^\pi_\mu$ collects the standard Young tableaux of shape $\mu$ whose vertical order is compatible with $<_\pi$. The weight on a tableau is $$\mathrm{wt}(T;q) = $q^{{n(\mu') - \#\mathrm{Area}}$(\pi) + \gamma(T)} \prod_{b: \, \mathrm{coleg}(b)>0} [\mathrm{arm}^{<\pi}_{T(b)}(\mathrm{up}(b)) + 1]_q,$$ where $\gamma(T)$ counts pairs of boxes (one above the other) whose entries lie in $\mathrm{Area}(\pi)$, and each factor is a q-integer $[m]_q = 1 + q + \cdots + q^{m-1}$ measuring how many boxes to the right of the box above $b$ remain below $T(b)$ in the path's order. The proof machinery is the Garsia–Remmel recursion: instead of counting rooks directly, the paper shows the tableau sum obeys the same two-term recurrence, with weight-ratio lemmas computing what happens when the box with entry $n$ is added to a tableau, a ratio that telescopes into the q-integer $[\lambda_1 - n + \nu_1 + 1]_q$, exactly the coefficient appearing in the recursion.

What would settle it

Compute the right-hand side of (3.11) for every Dyck path of semilength $n \le 6$ and compare with $R_k(\lambda(\pi); q)$ obtained directly from the rook-placement definition (3.1): the formula is wrong if any two paths with the same shape $\lambda$ give different sums, or if any sum disagrees with the rook count. The targeted stress test is the induction step: find a path $\pi$ and a tableau $T \in \mathrm{SYT}^\pi_\mu$ such that deleting the box with entry $n$ leaves a tableau that violates the order of the reduced path $\pi'$; that would break the premise behind Lemma 3.4 even if the final formula happened to hold.

Watch

Extended reading notes

Core claim

On the paper's own terms, the central discovery is Theorem 3.1: for any Dyck path $\pi$ of semilength $n$ with $\lambda(\pi) = \lambda$, the Garsia–Remmel q-rook number satisfies $$R_k(\$\lambda$; q) = \sum_{\mu \vdash n,\, \mu_1 = n-k} $q^{{n(\mu') - \#\mathrm{Area}}$(\pi)} \sum_{T \in \mathrm{SYT}^\pi_\mu} $q^{{\gamma(T)}}$ \prod_{b \in \mu,\, \mathrm{coleg}(b)>0} [\mathrm{arm}^{<\pi}_{T(b)}(\mathrm{up}(b)) + 1]_q .$$ Here $\mathrm{SYT}^\pi_\mu$ is the set of standard Young tableaux of shape $\mu$ whose column order refines the poset $i <_\pi j$ defined by the cell $(i,j)$ lying above $\pi$, and $\gamma(T)$ counts pairs of boxes whose entries invert that path order. The proof shows the right-hand side satisfies the Garsia–Remmel recursion $R_k(\lambda;q) = q^{\lambda_1 - k} R_k(\widehat\lambda;q) + [\lambda_1 - k + 1]_q R_{k-1}(\widehat\lambda;q)$, where $\widehat\lambda$ is the partition obtained by removing the first row; the mechanism is that deleting the largest entry from a $\pi$-tableau of shape $\mu$ gives a $\pi'$-tableau for the path $\pi'$ obtained by removing the last NE step. A second claim, Corollary 5.2, identifies the same weights with the coefficients $c_{\pi,\mu}(q)$ in the q-Whittaker expansion of the unicellular LLT function $\chi_\pi(q)$, so $R_k(\lambda(\pi);q)$ is the sum of those coefficients over shapes $\mu$ with $\mu_1 = n-k$.

Load-bearing premise

The proof rests on one geometric premise: when the Dyck path is shrunk by deleting its final step, the order it puts on the remaining numbers is exactly what it was before, so removing the largest entry from any valid tableau always leaves a valid tableau for the smaller path.

Editorial extensions

If this is right

Every $q$-rook number acquires a uniform tableau-theoretic description: one computes a single weighted sum over path-constrained standard Young tableaux, with no recursion choices and no explicit rook placement.
$q$-rook numbers are partial sums of the q-Whittaker coefficients of unicellular LLT functions, so results about either side transfer directly: $R_k(\lambda(\pi);q) = \sum_{\mu_1 = n-k} q^{n(\mu') - \#\mathrm{Area}(\pi)} c_{\pi,\mu}(q)$.
For abelian paths (semilength at least $\lambda_1 + \lambda'_1$), the tableau sum collapses to a single shape $(N-k,k)$, giving $R_k(\lambda;q) = q^{|\lambda|-(N-k)k} c_{\pi,(N-k,k)}(q)$, and Section 6.3 re-derives the rectangle decomposition of abelian unicellular LLT functions with $q$-hit-number coefficients.
Special cases recover clean formulas: the staircase partition yields a tableau formula for the $q$-Stirling numbers of the second kind, and the $n$-th $q$-rook number of a partition inside the square $(n^n)$ is expressed as a sum of $e$-expansion coefficients of a unicellular LLT function.
The identification of the weights with $c_{\pi,\mu}(q)$ shows these q-Whittaker coefficients are polynomials with nonnegative integer coefficients, since every summand in the tableaux sum is manifestly a nonnegative polynomial.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

An implication the paper leaves implicit is a testable path-independence identity: two Dyck paths with the same shape $\lambda$ must give equal weighted tableau sums, since both equal $R_k(\lambda;q)$; verifying this directly for small semilengths could expose symmetries of the weights, such as invariance under path reversal.
Read as a generating function, the weight $q^{\gamma(T)}\prod[\mathrm{arm}+1]_q$ behaves like a single defect statistic measuring how far a tableau's column order departs from the path's order; if that reading is right, $q$-rook numbers count all standard tableaux of shapes with fixed first-row length graded by one statistic, which may simplify comparisons with linked-rook placements.
The recursion-matching strategy suggests a general template: any rook statistic whose weight changes, upon deleting the largest entry, by a ratio that telescopes into a q-integer will satisfy a Garsia–Remmel-style recursion; the $q$-hit numbers are the natural next candidate, since the paper reaches them only through previously known formulas.
The paper's closing finite-field remark records the identity $\sum_\mu [\mathbf{fW}_\mu(q)]\, \widetilde{\chi}_\pi(q) = 1$; an extension the authors leave open is whether the tableaux weights give a basis-free handle on these coefficients, so the rank distribution of matrices supported above a Dyck path could be read directly from tableaux, a checkable small-$n$ experiment.

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the letter

Dear [Colleague],

Read the Basu–Bhattacharya paper on q-rook numbers. The headline: the main formula is new and the proof strategy is sound, but the write-up has a small hole in Lemma 3.4 that needs patching before I'd call it fully rigorous.

What's actually new: Theorem 3.1 gives a sum over standard Young tableaux (constrained by a Dyck path poset) for the Garsia–Remmel q-rook numbers. The proof via the Garsia–Remmel recursion is a real induction, not a repackaging. The LLT connection in Corollary 5.2 is honestly labeled as a restatement of [GMR+25], and the paper discloses that [KLY25] has an independent proof. The fresh proof of the Guay-Paquet theorem (Prop 6.4) is a nice byproduct.

The soft spot: Lemma 3.4's equation (3.14) contains a false first equality. It claims N(j) = arm^{<π}_n((j, ν_{j+1}+1)) + 1. For π = N^nE^n, with ν = (n-1) and T the single row, N(1)=0 but arm+1 = 1. The correct identity is the second equality: N(j) = ν_j - ν_{j+1} - #{b in row j with (T(b),n)∈Area}. That's what the γ computation actually uses, so the main induction can be repaired. The bigger presentation issue is i=1: (3.13) has [N(0)]_q = 0, so it cannot give the i=1 ratio; Lemma 3.5 needs a separate direct computation for i=1, which it in fact states correctly. Also, in the telescoping sum over i>1, the paper should say explicitly that invalid extensions contribute zero; this is true because the bad set {(T(b),n)∈Area} is a suffix, so N(i-1)=0 for any invalid i>1, but it is currently implicit.

Net: the theorem appears correct, the gaps are patchable with a short addendum, and the authors are transparent about what is new vs. known. This is a solid contribution to the q-rook/LLT area. It deserves a serious referee. I'd take it to a reading group and would cite the tableaux formula once a corrected version is out.

Recommendation: send to peer review, with a request that the authors fix Lemma 3.4 and the i=1 handling.

Referee Report

2 major / 3 minor

Summary. The paper proposes a formula expressing the Garsia-Remmel q-rook numbers R_k(λ;q) as a weighted sum over standard Young tableaux satisfying a Dyck-path order condition, with weights involving the area statistic, a statistic γ(T), and products of q-integers of arm lengths. The proof is by induction, showing that the proposed right-hand side satisfies the classical Garsia-Remmel recursion. The paper then connects this formula to unicellular LLT functions by restating a theorem from an external preprint, and derives consequences for Abelian Dyck paths, a proof of a result of Guay-Paquet, and a connection to e-expansion coefficients.

Significance. If the main theorem is correct, it gives a new, explicit tableau formula for q-rook numbers and a new bridge between rook theory and LLT functions. The paper is clearly written, includes worked examples, and the overall strategy (prove the recursion) is natural. However, the proof of the central theorem contains a serious gap: a key lemma used in the induction step is false as stated, and the derived ratio formula is contradicted by examples. Because the main theorem is load-bearing for all subsequent results, the manuscript needs a substantial correction before its claims can be accepted.

major comments (2)

[§3.8, Lemma 3.4] Equation (3.14) is false: the identity N(j) = arm^{<π}_n((j, ν_{j+1}+1)) + 1 fails whenever the leftmost leg-0 box b0 = (j, ν_{j+1}+1) satisfies (T(b0), n) ∈ Area(π), because then T(b0) is maximal in <_π (Lemma 3.2) and is not counted by N(j). For instance, for π = N^nE^n and π' = N^{n-1}E^{n-1}, with ν = (n-1) and T = 1 2 ⋯ n-1, one has N(1) = 0 while arm^{<π}_n((1,1)) + 1 = 1. Since (3.14) is used to prove the ratio formula (3.13), the latter is also false for valid extensions; e.g., for the path of Figure 1 with ν = (3,2), T = 1 2 3 / 4 5, and i = 2, the actual weight ratio is q^2, whereas (3.13) gives q^2 · q^{-N(1)} [N(1)]_q = q. Consequently, the induction step in §3.8 does not establish Theorem 3.1.
[§3.8, Lemma 3.5] The second statement of (3.15) is not a consequence of (3.13) with i = 1, since (3.13) yields [N(0)]_q = 0 for i = 1. The claimed value q^{ν_1-(n-1-λ_1)} is correct and must be proved directly; as written, the derivation in Lemma 3.5 is invalid. This is a second, independent gap in the recursion proof.

minor comments (3)

[§2.7 / §3.6] The notation up(b) is used for boxes in the first row with coleg(b) > 0, where no upper box exists; the product in (3.10) should state explicitly that such factors are taken to be 1, or the product should be restricted to boxes that have an upper neighbor.
[§8.1] In Proposition 8.1, the displayed formula contains a typographical error: it should read [arm_{T(b)}(up(b)) + 1]_q rather than [arm_{T(b)}(up(b) + 1)]_q.
[§5] Proposition 5.1 is explicitly a restatement of [GMR+25, Theorem 4.1], an external preprint; Corollary 5.2 therefore depends on unpublished work. The authors should either prove the needed case or clearly flag this dependency in the introduction.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: Theorem 3.1 is proved from the classical recursion and independent external theorems; the Lemma 3.4 gap is a correctness issue, not circularity.

full rationale

Theorem 3.1 is not assumed: the right-hand side of (3.11) is shown to satisfy the Garsia-Remmel recursion (3.3) and the base case R_0 = q^{|λ|}, so the target identity is derived from the defining recursion. Lemma 3.5's telescoping is algebra from the definitions and does not presuppose Theorem 3.1. Proposition 5.1 and Corollary 5.2 use [GMR+25, Theorem 4.1], from a disjoint set of authors, as an external input; the present tableaux formula is not used to prove the c-coefficient identity. Section 7 invokes [AN21a, Theorem 1.2] only after checking multiplicativity, the modular law, and agreement on N^nE^n; that uniqueness theorem is independent of the authors' own work. There are no fitted parameters, no self-citations, and no load-bearing reference to the authors' prior results. I flag the reviewer's Lemma 3.4 concern at (3.14): the displayed equality N(j)=arm(...)+1 can overcount when (T(b),n)∈Area(π), and Lemma 3.5 requires the convention that invalid extensions contribute zero. This is a patchable proof gap, not a circular reduction of the conclusion to an input.

Assumptions & free parameters 0 free parameters · 5 assumptions · 0 invented entities

The central claim introduces no free parameters and no new entities. All objects (Dyck paths, SYT, q-rook numbers, LLT functions) are standard. The paper's contribution is a new identity among known objects.

assumptions (5)

standard math Garsia-Remmel recursions (3.3): R_k(λ;q)=q^{λ1−k}R_k(eλ;q)+[λ1−k+1]_q R_{k−1}(eλ;q) with R_0=q^{|λ|}.
The defining recursion for q-rook numbers from [GR86], used as the target recursion in the proof of Theorem 3.1.
domain assumption [GMR+25, Theorem 4.1] formula for the chromatic symmetric function expansion in terms of Hessenberg tableaux with weight fwt.
Used in Proposition 5.1 to identify the tableaux weights with q-Whittaker coefficients. This is an external preprint result.
domain assumption [AN21a, Theorem 1.2] uniqueness of multiplicative symmetric functions satisfying the modular law.
Used in Section 7 to conclude equality of the two sides of (7.2) from agreement on N^n E^n.
standard math Greene's theorem [Gre76] relating chain/antichain decompositions to partitions.
Used in Lemma 6.1 to bound possible shapes in the sum.
domain assumption LLT function properties from [CM18]: symmetry, ω-involution (4.4), and the plethystic relation (4.3) to chromatic quasisymmetric functions.
Background facts used in Sections 4-6.

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Cite this review

Pith. "Pith review of A tableaux formula for $q$-rook numbers." pith.science (2026). https://pith.science/paper/37HSFDKO

@misc{pith2026250700766,
  author       = {Pith},
  title        = {Pith review of: A tableaux formula for $q$-rook numbers},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/37HSFDKO}},
  note         = {Machine review of arXiv:2507.00766}
}

abstract

We provide a formula for the Garsia-Remmel $q$-rook numbers as a sum over standard Young tableaux. We connect our formula with the coefficients in $q$-Whittaker expansion of unicellular LLT functions.

Figures

Figures reproduced from arXiv: 2507.00766 by the authors.

**Figure 1.** Example of a Dyck path 2.5. Partitions. The set of all integer partitions is denoted by Par. We think of the Young diagram in the English convention, as in Macdonald’s book [Mac95], and follow Macdonald’s definition and convention throughout the paper concerning partitions. In particular, for a partition λ, its conjugate is denoted λ ′ , the weighted size n(λ ′ ) = X i≥1 λi 2 . The arm, leg, coarm, coleg of a bo… view at source ↗

**Figure 2.** C ∈ C3((6, 4, 4, 2, 1)) with inv(C) = 6 3.2. Recursions for q-rook numbers. For λ = (λ1, λ2, λ3, . . .) ∈ Par, let λe = (λ2, λ3, . . .) be the partition obtained by removing the first row. The q-rook numbers Rk(λ; q) for 0 ≤ k ≤ λ1 are determined by the recursions [GR86, Theorem 1.1] Rk(λ; q) = q λ1−kRk(λe; q) + [λ1 − k + 1]qRk−1(λe; q), (3.3) with initial conditions R0(λ; q) = q |λ| . 3.3. q-Stirling numbers. Let n… view at source ↗

**Figure 3.** Removal of last occurence of NE from π Let π ∈ Dn with λ(π) = λ and π ′ ∈ Dn−1 be the path obtained by removing the first row of π, i.e, π ′ is obtained from π by removing the last occurence of NE in π. For the path π from [PITH_FULL_IMAGE:figures/full_fig_p007_3.png] view at source ↗

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Works this paper leans on

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Pith tools

Reviewed August 6, 2026 · model on record in the stance chip above.

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