q-Vandermonde identity

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In mathematics, in the field of combinatorics, the q-Vandermonde identity is the q-analogue of the Chu-Vandermonde identity

\begin{bmatrix} m + n\\ k \end{bmatrix}_q = \sum_{j} \begin{bmatrix} m\\ k - j \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_q q^{j(m-k+j)}.

The proof follows from observing the q-binomial identity with q-commuting operators (namely BA = qAB).

[edit] Other conventions

In the conventions common in applications to quantum groups, where the q-binomial is symmetric under exchanging q and q − 1, the q-Vandermonde identity reads

\begin{bmatrix} m + n\\ k \end{bmatrix}_q = q^{m k} \sum_{j=0}^{n} q^{-(m+n)j} \begin{bmatrix} m\\ k - j \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_q.

[edit] Proof

Assume that A and B are operators that q-commute:

BA = qAB.\,

Then:

(A + B)^m(A + B)^n \,
= \left(\sum_{i=0}^m \begin{bmatrix} m\\ i \end{bmatrix}_{q}A^{i}B^{m-i}\right) \left(\sum_{j=0}^n \begin{bmatrix} n\\ j \end{bmatrix}_{q}A^{j}B^{n-j}\right)
= \sum_{i,j} \begin{bmatrix} m\\ i \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_{q}A^{i}B^{m-i}A^{j}B^{n-j}
= \sum_{i,j} \begin{bmatrix} m\\ i \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_{q}q^{j(m-i)}A^{i+j}B^{m+n-i-j}.

This makes use of the fact that

BA^2 = BAA = qABA = q^2AAB = q^2A^2B. \,

Now, consider the coefficient of A^{k}B^{m+n-k}\, in this expression. This gives

\sum_{j} \begin{bmatrix} m\\ k-j \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_{q}q^{j(m-k+j)}.

Now, from the q-binomial theory, we recognize that (A+B)^m(A+B)^n=(A+B)^{m+n}\, And thus, the coefficient of A^{k}B^{m+n-k}\, is

\begin{bmatrix} m+n\\ k \end{bmatrix}_q.

Combining the results gives

\begin{bmatrix} m + n\\ k \end{bmatrix}_q = \sum_{j} \begin{bmatrix} m\\ k - j \end{bmatrix}_q \begin{bmatrix} n\\ j \end{bmatrix}_q q^{j(m-k+j)}.

\mathrm{QED}\,