Denote by $ \binom{n}{k}_q = \prod_{i=0}^{k-1} \frac{ q^{n-i} - 1 }{ q^{k-i} - 1 } $, $ k = 0, 1, \ldots, n $, the $ q $-binomial (Gaussian) coefficients. These numbers are symmetric, in the sense that $ \binom{n}{k}_q = \binom{n}{n-k}_q $, unimodal, and hence maximized for $ k = \lfloor n/2 \rfloor $ (or $ k = \lceil n/2 \rceil $). I am interested in the sum $$ \sum^{n}_{\substack{ k=0 \\ k \, \equiv \, h \pmod \ell }} \binom{n}{k}_q $$ where $ h $ is a parameter over which the sum is to be maximized. Question: Is it true that, for any fixed $ q $, $ n $, $ \ell $, the choice $ h = \lfloor n/2 \rfloor $ (or, equivalently, $ h = \lceil n/2 \rceil $) maximizes the sum?

Special cases:

  • This is known to be true for binomial coefficients ($ q \to 1 $). It follows from the result of Katona (1972) stating that $$ \sum^{n}_{\substack{ k=0 \\ k \, \equiv \, \lfloor n/2 \rfloor \pmod \ell }} \binom{n}{k} $$ is the maximum cardinality of a family of subsets $ \mathcal{F} \subseteq 2^{\{1, \ldots, n\}} $ satisfying the condition that if $ A, B \in \mathcal{F} $ and $ A \subsetneq B $, then $ |B| - |A| \geqslant \ell $. The proof relies on the theory of partial orders. A simpler and direct proof for this case would also be appreciated.
  • The case $ \ell = 2 $. I saw the following identity in one of the answers to this question: $$ \sum_{i=0}^n(-1)^i\binom ni_q=\begin{cases}0,&n\, \text{odd}\\ \prod_{j=1}^{n/2}(1-q^{2j-1}),&n\, \text{even}\end{cases} $$ If this is true, it implies the statement I am interested in when $ \ell = 2 $. Does someone know how to prove this identity?

Background: The sum represents the number of subspaces of $ \mathbb{F}_q^n $ whose dimension is congruent to $ h $ modulo $ \ell $. The set of all subspaces of $ \mathbb{F}_q^n $ partially ordered by inclusion satisfies the so-called LYM condition. It follows from this fact and a result of Kleitman (1975) that $$ \max_h \sum^{n}_{\substack{ k=0 \\ k \, \equiv \, h \pmod \ell }} \binom{n}{k}_q $$ is the maximum cardinality of a family of subspaces satisfying the condition that if $ U \subsetneq V $, then $ \dim V - \dim U \geqslant \ell $. I was wondering if the maximizer can be found explicitly.

  • $\begingroup$ Quick experimentation turns up the counterexample $n=11$, $\ell=3$, $q=-1$ where the $h$ which maximises is $h = 1 \not\equiv 5 \pmod 3$. However, given the background I suspect that you want to add an assumption that $q$ is an integer $> 0$. $\endgroup$ Feb 24, 2022 at 16:12
  • $\begingroup$ @PeterTaylor In the context of $q$-analogues, the number $q$ means the size of a finite field, so it's a prime power. $\endgroup$ Feb 24, 2022 at 16:18
  • $\begingroup$ @LeechLattice, I think it would be more accurate to say "in some contexts in which $q$-analogues are used..." There are other contexts where $q$ is complex; as I understand the history, in the first applications of $q$-analogues it was likely to be a root of unity. $\endgroup$ Feb 25, 2022 at 0:10
  • $\begingroup$ Yes, I had in mind prime powers, although it should be true for all reals $ \geqslant 1 $ (by Nate's answer below). $\endgroup$
    – aleph
    Feb 25, 2022 at 9:32
  • 1
    $\begingroup$ The identity $\sum_{i=0}^n \binom{2m}{i}_q (-1)^i = (1-q)(1-q^3) \ldots (1-q^{2m-1})$ is correct: see for instance (3.3.8) in Andrews' textbook, The theory of partitions. Of course if $n$ is odd then the left-hand side is zero because $\binom{n}{i}_q = \binom{n}{n-i}_q$. $\endgroup$ Feb 26, 2022 at 10:51

1 Answer 1


Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.

$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^{2a+1}$

In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $q=3$ with $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.


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