Concept

Q-Vandermonde identity

Summary
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 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). 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 , is defined by 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 . Using this q-binomial coefficient, the q-Vandermonde identity can be written in the form 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 in two different ways. Following Stanley, we can tweak this proof to prove the q-Vandermonde identity, as well. First, observe that the product can be expanded by the q-binomial theorem as Less obviously, we can write and we may expand both subproducts separately using the q-binomial theorem. This yields Multiplying this latter product out and combining like terms gives Finally, equating powers of between the two expressions yields the desired result. This argument may also be phrased in terms of expanding the product 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.
About this result
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website. Please make sure to verify the information with EPFL's official sources.