Sum of $q$-binomial coefficients

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.

• 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 $> 0$. Feb 24, 2022 at 16:12
• @PeterTaylor In the context of $q$-analogues, the number $q$ means the size of a finite field, so it's a prime power. Feb 24, 2022 at 16:18
• @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. Feb 25, 2022 at 0:10
• Yes, I had in mind prime powers, although it should be true for all reals $\geqslant 1$ (by Nate's answer below). Feb 25, 2022 at 9:32
• 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 $\binom{n}{i}_q = \binom{n}{n-i}_q$. Feb 26, 2022 at 10:51

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.