# Combinatorial identity probably related to Vandermonde

I'm quite stuck on exercise 2.3 in Sepanski's Compact Lie Groups (that deals with checking that a specific Hermitian inner product on the space of homogeneous polynomials of degree $$n$$ is invariant wrt the action of $$SU(2)$$) because i can't prove an identity involving multiple sums and factorials. If my calculations are fine I should get that \begin{align} \sum_{i=0}^m \sum_{j=0}^m \binom{k}{i} \binom{n-k}{m-i} \binom{q}{j} \binom{n-q}{m-j} a^{2(i+j)}(-b^2)^{-i-j}=\begin{cases} 0 &\text{if } k \neq q\\ \binom{n}{m}^2 a^{2(2m-n)} b^{-2(2m-n)} &\text{if } k=q \end{cases} \end{align} where $$a^2+b^2=1$$. This seems somewhat related to Vandermonde's identity but I can't figure out a way to get it.

I managed to rewrite the equation as \begin{align} \sum_{i=0}^m \sum_{j=0}^m \binom{m}{i} \binom{n-m}{k-i} \binom{m}{j} \binom{n-m}{q-j} a^{2(i+j)}(-b^2)^{-i-j}=\begin{cases} 0 &\text{if } k \neq q\\ \binom{n}{q} \binom{n}{k}a^{2(2m-n)} b^{-2(2m-n)} &\text{if } k=q \end{cases} \end{align} but it doesn't appear to be so much more helpful.

I'm not so strong on combinatorics so any help would be appreciated!

• This is definitely not true -- just plug in something simple like $i = j = 1$, $a = b= 1/\sqrt{2}$.
– JBL
Commented Jul 28, 2022 at 12:54
• (Sorry, that should be $k = q = 1$, not $i = j = 1$.)
– JBL
Commented Jul 29, 2022 at 2:58

The double sum is separable into a product of single sums of the same form: \begin{align} &\sum_{i=0}^m \sum_{j=0}^m \binom{k}{i} \binom{n-k}{m-i} \binom{q}{j} \binom{n-q}{m-j} a^{2(i+j)}(-b^2)^{-i-j} \\ &=\left(\sum_{i=0}^m \binom{k}{i} \binom{n-k}{m-i} \left(-\frac{a^2}{b^2}\right)^{i}\right) \left(\sum_{j=0}^m \binom{q}{j} \binom{n-q}{m-j} \left(-\frac{a^2}{b^2}\right)^{j}\right) \\ & = S(n,m,a,b,k) S(n,m,a,b,q) \end{align}