There is a $q$-binomial identity that I encountered in one paper I am reading (<https://arxiv.org/abs/1910.06193>) which probably admits a very simple proof that I do not see: for two nonnegative integers $a,b$, we have
$$
q^{ab}=\sum\_{k\ge 0}(-1)^kq^{\binom{k}2}\binom{a}{k}\_q\binom{b}{k}\_q(q;q)\_k,
 $$
where $(q;q)\_k=(1-q)(1-q^2)\cdots(1-q^k)$, $\binom{n}{k}\_q=\frac{(q;q)\_n}{(q;q)\_k(q;q)\_{n-k}}$ for $0\le k\le n$ and $\binom{n}{k}\_q=0$ for $k>n$.


I am sure that this can be proved by a version of the A=B method, but for connecting this to my work not the identity as such but rather a conceptual proof or including it into some context is desirable. Since it is not an identity "with positive coefficients", it is not obvious if I should ask for a combinatorial interpretation, or some sort of Euler characteristic result, so I will just leave it slightly open-ended: have you seen any version of this identity?



(Conceptual) proof and/or interpretation of a $q$-binomial identity
Note that it will be enough to check the identity when $q$ is a prime power, in which case we can choose a field $F$ with $|F|=q$, and vector spaces $A$ and $B$ of dimension $a$ and $b$ over $F$. In this context it is more natural to consider the function $\pi\_q(k)=\prod\_{i=1}^k(q^i-1)=(-1)^k(q;q)\_k$ rather than $(q;q)\_k$, because $\pi\_q(k)$ is a positive integer and is more closely related to counting problems. We then have
$$ {\binom{n}{k}\_q}=\frac{\pi\_q(n)}{\pi\_q(k)\pi\_q(n-k)} $$
and the stated identity becomes
$$ q^{ab} = \sum\_{k\geq 0} q^{\binom{k}{2}}\binom{a}{k}\_q\binom{b}{k}\_q\pi\_q(k). $$
The left hand side is $|\text{Hom}(A,B)|$. To give a homomorphism $\alpha\colon A\to B$ we have to choose a rank $k$, a kernel $U\leq A$ of dimension $a-k$, an image $V\leq B$ of dimension $k$, and an isomorphism $\alpha\_1\colon A/U\to V$. The number of choices for $U$ is $\binom{a}{a-k}\_q=\binom{a}{k}\_q$. The number of choices for $V$ is $\binom{b}{k}\_q$. The number of choices for $\alpha\_1$ is $|GL\_k(F)|=q^{\binom{k}{2}}\pi\_q(k)$.


