(Conceptual) proof and/or interpretation of a $q$-binomial identity

Question

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?

Answer 1

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 $|GL_k(F)|=q^{\binom{k}{2}}\pi_q(k)$.

Answer 2

Denoting $q^a=x$ (and fixing $b$), this is an expansion of the polynomial $x^b$ in the basis of $f_k(x):=(1-x)(1-x/q)\ldots (1-x/q^{k-1})$: $$ x^b=\sum_{k=0}^b (-1)^k q^{{k\choose 2}}{b\choose k}_qf_k(x), $$ or, if we use instead $g_k(x)=(x-1)(x-q)\ldots (x-q^{k-1})$, this reads as $$ x^b=\sum_{k=0}^b {b\choose k}_k g_k(x). $$ The dual representation (of $g_b$ in the basis $1,x,\ldots,x^{b}$) is called $q$-binomial theorem, so you may use the $q$-binomial transformation to conclude. Alternatively, if you do not want to use all this $q$-theory, you may put $y=0$ (and $n=b$) in the following two-variables identity: $$ (x-y)(x-qy)\ldots(x-q^{n-1}y)= \sum_{k=0}^n{n\choose k}_q g_{n-k}(x) (1-y)(1-qy)\ldots (1-q^{k-1}y).\quad(\clubsuit) $$ In order to prove $(\clubsuit)$, note that both sides are polynomials of degree at most $n$ in $x,y$, thus it suffices to check it for $x=q^i$, $y=q^{-j}$ where $i,j\geqslant 0$ and $i+j\leqslant n$. At these points, it is obvious (both sides are 0 when $i+j<n$, and there is unique non-zero term in RHS when $i+j=n$, which just equals to LHS.)