2

When trying to prove this q-binomial identity I had soon the idea that here we have a q-Vandermonde identity in disguise. I could transform the identity to

\begin{align*} \color{blue}{(-1)^mq^{\binom{m+1}{2}+mn} \sum_{j=0}^m\binom{n}{j}_q\binom{-n-1}{m-j}_qq^{j(j-m-n)}=1}\tag{1} \end{align*} for non-negative integers $n\leq m$. The identity (1) looks similar to the following q-Vandermonde identity: \begin{align*} \sum_{j=0}^m&\binom{n}{j}_q\binom{-n-1}{m-j}_qq^{j(j-m-n-1)} =\binom{-1}{m}_q=(-1)^mq^{-\binom{m+1}{2}}\tag{2.1} \end{align*} from which \begin{align*} \color{blue}{(-1)^mq^{\binom{m+1}{2}}\sum_{j=0}^m\binom{n}{j}_q\binom{-n-1}{m-j}_qq^{j(j-m-n-1)}=1}\tag{2.2} \end{align*} follows.

In (2.1) we use besides the q-Vandermonde identity \begin{align*} \binom{n+m}{t}_q=\sum_{j=0}^t\binom{n}{j}_q\binom{m}{t-j}_qq^{j(m-t+j)} \end{align*} the identity \begin{align*} \color{blue}{\binom{-n}{k}_q} &=\frac{\left(1-q^{-n}\right)\left(1-q^{-n-1}\right)\cdots\left(1-q^{-n-k+1}\right)} {\left(1-q\right)\left(1-q^2\right)\cdots\left(1-q^k\right)}\\ &=\left(\prod_{j=n}^{n+k-1}q^{-j}\right)(-1)^k \frac{\left(1-q^{n}\right)\left(1-q^{n+1}\right)\cdots\left(1-q^{n+k-1}\right)} {\left(1-q\right)\left(1-q^2\right)\cdots\left(1-q^k\right)}\\ &=q^{-\sum_{j=n}^{n+k-1}j}(-1)^k\binom{n+k-1}{k}_q\\ &=q^{-\sum_{j=0}^{k-1}(j+n)}(-1)^k\binom{n+k-1}{k}_q\\ &\,\,\color{blue}{=q^{-nk-\binom{k}{2}}(-1)^k\binom{n+k-1}{k}_q} \end{align*}

Question: How can we algebraically derive identity (1)? It seems that (2.2) could be helpful.

Markus Scheuer
  • 112,413

1 Answers1

2

We want to prove equation $(1)$, which is $$ (-1)^mq^{\binom{m+1}{2}+mn}\sum_{j=0}^m \binom{n}{j}_q\binom{-n-1}{m-j}_q q^{j(j-m-n)}=1 $$ It is sufficient to prove this with $q$ replaced by $q^{-1}$, in which case it is sufficient to prove that $$ \sum_{j=0}^m \binom{n}{j}_{q^{-1}}\binom{-n-1}{m-j}_{q^{-1}} q^{j(m+n-j)-mn}=(-1)^m q^{\binom{m+1}{2}} $$ which is equivalent to $$ \sum_{j=0}^m \binom{n}{j}_{q^{-1}}\binom{-n-1}{m-j}_{q^{-1}}q^{-(j-m)(j-n) - \binom{m+1}{2}} = (-1)^m $$

Note that $$ \sum_{k=0}^{\infty} (-1)^kt^k = \sum_{k=0}^{\infty} (-t)^k = \frac 1{1+t} $$

Therefore, the coefficient of $t^m$ in $\frac 1{1+t}$ is $(-1)^m$. On the other hand, $$ \frac{1}{1+t} = \prod_{i=0}^{n-1} (1+q^{-i}(q^{-1}t)) \prod_{i=0}^{n} \frac 1{1+q^{-i}t} $$ The RHS is evaluable by the $q$-binomial theorem :$$ \prod_{i=0}^{n-1} (1+q^{-i}(q^{-1}t)) = \sum_{i=0}^n q^{-\binom{i+1}{2}}\binom{n}{i}_{q^{-1}}t^i = \sum_{i=0}^{\infty} q^{-\binom{i+1}{2}}\binom{n}{i}_{q^{-1}}t^i $$ and the associated identity the inverse of such a polynomial, which can be found here : $$ \prod_{i=0}^n \frac 1{1+q^{-i}t} = \sum_{k=0}^{\infty} \binom{n+k}{k}_{q^{-1}}(-1)^kt^k = \sum_{k=0}^{\infty} \binom{-n-1}{k}_{q^{-1}}q^{-nk-\binom{k+1}{2}}t^k $$ We used the same identity as the OP to convert the $q$-binomial coefficient into one with a negative top part. Hence, the associated coefficient of $t^m$ is formed by taking all $j=0$ to $m$ in the first term's expansion, and then letting $k=m-j$ in the second term's expansion. This yields $$ \sum_{j=0}^m \binom{-n-1}{m-j}_{q^{-1}}q^{-n(m-j)-\binom{m-j+1}{2}}q^{-\binom{j+1}{2}}\binom{n}{j}_{q^{-1}}\\ = \sum_{j=0}^m \binom{-n-1}{m-j}_{q^{-1}}\binom{n}{j}_{q^{-1}}q^{-(j-m)(j-n)-\binom{m+1}{2}} $$

as desired.

Sarvesh Ravichandran Iyer
  • 77,817
  • 1
    @SarveshRavichandranlyer: Many thanks for this great answer! (+1) I appreciate the representation of $\frac{1}{1+t}$ which cleverly enables us to use the q-binomial theorem. I will go through the details after work and accept it afterwards. KR – Markus Scheuer Apr 14 '23 at 08:35
  • 1
    @epi163sqrt Thank you. Perhaps the main step was the recognition that I needed to use $q^{-1}$ in place of $q$. I actually used a similar argument and equality without replacing $q$. The identity which that give told me that somewhere, $q \to q^{-1}$ was the right thing to do, and voila, things are good. – Sarvesh Ravichandran Iyer Apr 14 '23 at 11:35