# q-Vandermonde identity

In mathematics, in the field of combinatorics, the q-Vandermonde identity is a q-analogue of the Chu–Vandermonde identity. Using standard notation for q-binomial coefficients, the identity states that

${\displaystyle {\binom {m+n}{k}}_{\!\!q}=\sum _{j}{\binom {m}{k-j}}_{\!\!q}{\binom {n}{j}}_{\!\!q}q^{j(m-k+j)}.}$

The nonzero contributions to this sum come from values of j such that the q-binomial coefficients on the right side are nonzero, that is, max(0, km) ≤ j ≤ min(n, k).

## Other conventions

As is typical for q-analogues, the q-Vandermonde identity can be rewritten in a number of ways. In the conventions common in applications to quantum groups, a different q-binomial coefficient is used. This q-binomial coefficient, which we denote here by ${\displaystyle B_{q}(n,k)}$, is defined by

${\displaystyle B_{q}(n,k)=q^{-k(n-k)}{\binom {n}{k}}_{\!\!q^{2}}.}$

In particular, it is the unique shift of the "usual" q-binomial coefficient by a power of q such that the result is symmetric in q and ${\displaystyle q^{-1}}$. Using this q-binomial coefficient, the q-Vandermonde identity can be written in the form

${\displaystyle B_{q}(m+n,k)=q^{nk}\sum _{j}q^{-(m+n)j}B_{q}(m,k-j)B_{q}(n,j).}$

## Proof

As with the (non-q) Chu–Vandermonde identity, there are several possible proofs of the q-Vandermonde identity. The following proof uses the q-binomial theorem.

One standard proof of the Chu–Vandermonde identity is to expand the product ${\displaystyle (1+x)^{m}(1+x)^{n}}$ in two different ways. Following Stanley,[1] we can tweak this proof to prove the q-Vandermonde identity, as well. First, observe that the product

${\displaystyle (1+x)(1+qx)\cdots \left(1+q^{m+n-1}x\right)}$

can be expanded by the q-binomial theorem as

${\displaystyle (1+x)(1+qx)\cdots \left(1+q^{m+n-1}x\right)=\sum _{k}q^{\frac {k(k-1)}{2}}{\binom {m+n}{k}}_{\!\!q}x^{k}.}$

Less obviously, we can write

${\displaystyle (1+x)(1+qx)\cdots \left(1+q^{m+n-1}x\right)=\left((1+x)\cdots (1+q^{m-1}x)\right)\left(\left(1+(q^{m}x)\right)\left(1+q(q^{m}x)\right)\cdots \left(1+q^{n-1}(q^{m}x)\right)\right)}$

and we may expand both subproducts separately using the q-binomial theorem. This yields

${\displaystyle (1+x)(1+qx)\cdots \left(1+q^{m+n-1}x\right)=\left(\sum _{i}q^{\frac {i(i-1)}{2}}{\binom {m}{i}}_{\!\!q}x^{i}\right)\cdot \left(\sum _{i}q^{mi+{\frac {i(i-1)}{2}}}{\binom {n}{i}}_{\!\!q}x^{i}\right).}$

Multiplying this latter product out and combining like terms gives

${\displaystyle \sum _{k}\sum _{j}\left(q^{j(m-k+j)+{\frac {k(k-1)}{2}}}{\binom {m}{k-j}}_{\!\!q}{\binom {n}{j}}_{\!\!q}\right)x^{k}.}$

Finally, equating powers of ${\displaystyle x}$ between the two expressions yields the desired result.

This argument may also be phrased in terms of expanding the product ${\displaystyle (A+B)^{m}(A+B)^{n}}$ in two different ways, where A and B are operators (for example, a pair of matrices) that "q-commute," that is, that satisfy BA = qAB.

## Notes

1. ^ Stanley (2011), Solution to exercise 1.100, p. 188.