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Let's first recall some notations:

  • The $q$-Pochhammer symbol is defined as $$(x)_n = (x;q)_n := \prod_{0\leq l\leq n-1}(1-q^l x).$$

  • The $q$-binomial coefficient (also known as the Gaussian binomial coefficient) is defined as $$\binom{n}{k}_q := \frac{(q)_{n}}{(q)_{n-k}(q)_{k}}.$$

I found the following curious identity on $q$-binomial coefficients, and I would like to ask for some ideas on how to prove it.

$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} (q)_n$$

I am familiar with the $q$-binomial theorem and $q$-Vandermonde identity, which I think might be useful. If there are some other well-known identities on $q$-binomial coefficients that might be helpful in figuring this out, I would like to know them too. Thank you!

Some thoughts:

  • The classical version of this identity is simply $$\sum_{0\leq j\leq 2n}(-1)^{j}(2n+1-j)\binom{2n}{j} = \sum_{0\leq j\leq 2n}(-1)^{j}(j+1)\binom{2n}{j} = \delta_{n,0},$$ which can be proved simply by $$\sum_{0\leq j\leq 2n}(-1)^{j}\binom{2n}{j} + \sum_{0\leq j\leq 2n}(-1)^{j}j\binom{2n}{j} = (1-x)^{2n}+\frac{d}{dx}(1-x)^{2n} \bigg\vert_{x=1} = \delta_{n,0}.$$ There is a natural $q$-analogue of the above identity, obtained by replacing $(1-x)^{2n}$ by $(x)_{2n}$ and the derivative by the $q$-derivative, but unfortunately this is not the desired identity.

  • There seems to be other closely related identities: $$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n-1)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} \delta_{n,0},$$ $$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n+1)+\frac{j(j-1)}{2}}\binom{2n}{j}_q \overset{?}{=} q^{n}(q)_{2n}.$$ Proving these other identities (or a family of similar identities) might be helpful if we were to try to use induction on $n$.

Henry
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    Perhaps $q-$Zeilbergs gets you a nice recursion? https://risc.jku.at/sw/qzeil/ – Phicar Apr 25 '22 at 16:03
  • Can you provide some source for these identities? I had a general look and unfortunately there's very little on double-sum identities that I found lying around for the $q$-binomial. – Sarvesh Ravichandran Iyer Apr 28 '22 at 06:30
  • @Phicar Can qZeil handle double summations? I thought it was mostly for single summations. – Henry Apr 29 '22 at 00:00
  • @SarveshRavichandranIyer Unfortunately I am not aware of any useful reference either. – Henry Apr 29 '22 at 00:01
  • @Phicar Okay, I have turned the desired identity into a single-sum identity and was able to verify it using qZeil. This is nice, and thanks for the suggestion. Still, it would be nice to have a non-algorithmic proof of the identity. – Henry Apr 29 '22 at 02:21
  • @Henry That is nice. I will have a look today at it. The quadratic form on the exponent is unusual. Nice identity tho! – Phicar Apr 29 '22 at 05:01
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    It might be useful to post the same question on Mathoverflow. It is definitely worth it. – Wolfgang Apr 29 '22 at 06:31
  • I managed to obtain a partial explanation using combinatorics. I don't have a full answer to fall back upon ,but I eliminated the role of $k$ and figured out some further cancellations using the ideas. I don't want to post a partial answer (I'd rather it be complete and worthy of evaluation), but if someone's interested I'll put the details up in the next comment(s). – Sarvesh Ravichandran Iyer May 06 '22 at 03:25
  • So we begin with a combinatorial interpretation : Consider all $n$-tilings using $k$ green tiles and $n-k$ red tiles, such that every green tile's weight is $q^r$ where $r$ is the number of red tiles preceding this green tile. The weight of a tiling is the product of weights of all green tiles, and the sum over all such tilings of these weights equals $\binom{n}{k}_q$. We get that the LHS is a sum over all tilings of length $2n$ and $k \geq $ the number of green tiles, of sth. involving $k$ and the green tiles. Using $(-1)^j$ and some work, cancellations can be figured out that eliminate $k$. – Sarvesh Ravichandran Iyer May 06 '22 at 03:30

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