On Bergeron's positivity problem for $q$-binomial coefficients

F. Bergeron recently asked the intriguing question whether $\binom{b+c}{b}_q -\binom{a+d}{d}_q$ has nonnegative coefficients as a polynomial in $q$, whenever $a,b,c,d$ are positive integers, $a$ is the smallest, and $ad=bc$. We conjecture that, in fact, this polynomial is also always unimodal, and combinatorially show our conjecture for $a\le 3$ and any $b,c\ge 4$. The main ingredient will be a novel (and rather technical) application of Zeilberger's KOH theorem.


Introduction
An interesting problem recently posed by F. Bergeron [3], which naturally arose in his studies of the q-Foulkes conjecture, is whether the coefficients of the symmetric polynomial b+c b q − a+d d q always form a nonnegative sequence, for any choice of positive integers a, b, c, d where a is the smallest and ad = bc. Here, m+n m q as usual denotes the q-binomial coefficient It is easily seen that m+n m q is a symmetric polynomial in q of degree mn. We remark here that a special case of Bergeron's question already appeared in Abdesselam-Chipalkatti [2], as a consequence of a more general conjecture. The same authors proved in [1] the case a = 2 of their conjecture. (We thank A. Abdesselam for pointing this out to us.) In this note, we conjecture that not only nonnegativity but also unimodality does hold in this context, and provide a combinatorial proof of our conjecture for a ≤ 3 and any b, c ≥ 4. Recall that a sequence of numbers is unimodal if it does not increase strictly after a strict decrease. We have: Notice that symmetry is clear, since both b+c b q and a+d d q are symmetric polynomials of the same degree, bc = ad. Also note that the case a = 1 of the conjecture is trivial, thanks to the unimodality of b+c b q (see e.g. [5,7,9,12,15]). In this note, first we apply a recent result of Pak-Panova [6], that we reproved combinatorially in [14], to provide a short proof of Conjecture 1.1 when a = 2, for any values of b and c. Then, we employ Zeilberger's KOH theorem [15] to settle the case a = 3.
Unimodality results for suitable differences of two q-binomial coefficients have also appeared, for instance, in the work of Reiner and Stanton ([8], Theorems 1 and 5), who employed interesting methods of representation theory in their proofs. See also their Conjecture 9 (or [11], Conjecture 7), which provides a broad family of possible further identities, and is still wide open to this day. It would be interesting to investigate whether our combinatorial approach via the KOH theorem, which appears to be new in this context, might also help with the Reiner-Stanton conjecture.
where we can assume that c ≥ b ≥ 3. Proving the result is tantamount to showing that Thus, it suffices to show that b+c b q is strictly increasing in all even degrees up to bc/2.
If b ≥ 5, the result follows from Lemma 2.1, since one can easily verify computationally that all nine exceptional q-binomial coefficients of the lemma satisfy the inequality b i−1 < b i when i ≤ bc/2 is even.
If b ≤ 4, the conclusion can be obtained in a few different ways. For a combinatorial proof, one can rely on the fact that in a symmetric chain decomposition of the poset L(b, c) for b = 3, 4 (we can assume c even when b = 3), at least one new chain is introduced in every even degree ≤ bc/2 (see [4,13]). When b = 4, notice that the result also immediately follows from [10], Lemma 2.1 (b).
Before proving Conjecture 1.1 for a = 3, we recall Zeilberger's KOH theorem [15]. This result rephrases, in algebraic terms, O'Hara's celebrated combinatorial proof of the unimodality of q-binomial coefficients [5], by decomposing these latter into suitable finite sums of unimodal, symmetric polynomials with nonnegative coefficients.
Fix positive integers m and n, and for any partition Theorem 2.4. Conjecture 1.1 is true when a = 3.
where b, c ≥ 4. We can assume that b ≡ 0 (mod 3). With some abuse of notation, define the first difference of a q-binomial coefficient as its truncation in the middle degree; i.e., we denote by Thus, showing the result is equivalent to proving that where inequalities between polynomials are defined degreewise (i.e., by α i q i ≤ β i q i we mean α i ≤ β i for all i).
By Lemma 2.3, we can decompose d+3 d q as d + 3 3 where the first term on the right side corresponds to the partition (3) of 3, the second to (2, 1), and the third to (1, 1, 1). By iterating Lemma 2.3 a total of b/3 times on the right side, standard computations give us that d+3 3 q equals: If we now consider the partition (3, 3, . . . , 3) of b, we see that its contribution to the KOH decomposition of b+c b q , by Lemma 2.3, is given by: which is precisely the first summand in (1). Thus, by (1), we want to show that (2) where the sum on the right is as in the KOH decomposition of b+c b q , and is indexed over all partitions λ ⊢ b, λ = (3, 3, . . . , 3).
We have: Note that both 2d − 2i + 1 and 3d − 6i + 1 are larger than bc/2 = 3d/2 for all of our indices i, since c ≥ 4. Thus, since the first difference is defined to be up to degree bc/2 = 3d/2, the last displayed formula becomes: It is the polynomial in (3) that we will bound with the first difference of suitable families of terms appearing in the KOH decomposition of b+c b q . We begin by dominating Hence, for now, we are not concerned with (q 2 + q 3 + · · · + q d ) + 0≤i≤ b−3 3 q 6i . Consider the following partitions of b: for any indices Since b ≥ 6 (because b ≡ 0 (mod 3)) and for c ≥ 4, all partitions λ i,j contain at least one part of each size 1, 2, and 3.
We want to show that Employing Lemma 2.3, the contribution of λ i,j to the KOH decomposition of b+c b q is For all i and j as above, we have Therefore, Thus, our goal is to show that (4) i,j (1 + q + · · · + q ci−4i )(q 6i+2j + · · · + q ci+2i+cj ) ≥ 1≤i≤ b−3
Also notice the following fact, which will be useful later on: When b > 6 is even, the coefficient of degree 2b − 6 on the left side of (4), say l 2b−6 , is strictly greater than the corresponding coefficient on the right side, say r 2b−6 . Indeed, a standard computation shows that r 2b−6 = 2. As for l 2b−6 , one can see that it is at least 4, by considering the contribution, to the degree 2b − 6 coefficient on the left side of (4), of the following four pairs of indices (i, j): This completes the proof that (5) l 2b−6 > r 2b−6 .
A quick thought tells us that we are done whenever the following inequality is true: Clearly, (6) is verified if we show that for each i = ⌊b/6⌋ + 1, . . . , (b − 3)/3, the left side begins in degree no larger, and ends in degree no smaller, than the right side. Since j ≤ ⌊b/2 − 2i(c − 1)/c⌋, standard computations give us that this is indeed the case for any c ≥ 5, if we assume b ≥ 18 when c = 5 (the theorem is immediate to check directly for b < 18 when c = 5). This shows (4).
Notice that every even power of q that appears in (7) for a given index i < (b − 3)/3 also appears for i + 1, with the only exception of q 2i . This, however, appears for i − 1 if i ≥ 2, so the total number of times it is present in (7) is again at least 2. We conclude that a coefficient equal to 1 in some degree n ≡ 0 (mod 6), if it exists, can only be contributed by i = (b − 3)/3, and such a degree n must be in the range: c(⌈b/2⌉ − 2) + 1 ≤ n ≤ c(⌈b/2⌉ − 1).
It follows that the theorem is proven whenever c(⌈b/2⌉−2)+1 > 2b−6. This is immediately verified to be the case for any c ≥ 6. When c = 5, the only exception is b = n = 6, but for b = 6 the theorem can easily be checked directly. Finally, when c = 4, the only exceptions are b even and n = 2b − 6. If b = 6 the theorem is again verified directly; if b > 6, we employ inequality (5). The proof of the theorem is complete.