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q-Vandermonde identity

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A q-analogue of the Chu–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

( m + n k ) q = ∑ j ( m k − j ) q ( n j ) q q j ( m − k + j ) . {\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, k − m) ≤ 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 B q ( n , k ) {\displaystyle B_{q}(n,k)} , is defined by

B q ( n , k ) = q − k ( n − k ) ( n k ) q 2 . {\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 q − 1 {\displaystyle q^{-1}} . Using this q-binomial coefficient, the q-Vandermonde identity can be written in the form

B q ( m + n , k ) = q n k ∑ j q − ( m + n ) j B q ( m , k − j ) B q ( n , j ) . {\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 ( 1 + x ) m ( 1 + x ) n {\displaystyle (1+x)^{m}(1+x)^{n}} in two different ways. Following Stanley, we can tweak this proof to prove the q-Vandermonde identity, as well. First, observe that the product

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

can be expanded by the q-binomial theorem as

( 1 + x ) ( 1 + q x ) ⋯ ( 1 + q m + n − 1 x ) = ∑ k q k ( k − 1 ) 2 ( m + n k ) q x k . {\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

( 1 + x ) ( 1 + q x ) ⋯ ( 1 + q m + n − 1 x ) = ( ( 1 + x ) ⋯ ( 1 + q m − 1 x ) ) ( ( 1 + ( q m x ) ) ( 1 + q ( q m x ) ) ⋯ ( 1 + q n − 1 ( q m x ) ) ) {\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

( 1 + x ) ( 1 + q x ) ⋯ ( 1 + q m + n − 1 x ) = ( ∑ i q i ( i − 1 ) 2 ( m i ) q x i ) ⋅ ( ∑ i q m i + i ( i − 1 ) 2 ( n i ) q x i ) . {\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

∑ k ∑ j ( q j ( m − k + j ) + k ( k − 1 ) 2 ( m k − j ) q ( n j ) q ) x k . {\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 x {\displaystyle x} between the two expressions yields the desired result.

This argument may also be phrased in terms of expanding the product ( A + B ) m ( A + B ) n {\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.

References

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