# Proof of generalized Cauchy formula

I would like to know if there is a proof for the identity used in the superconformal index of 4d ${\cal N}=2$ gauge theory. In the paper by Rastelli el al, it was discovered that Eq. (10) is equal to the right hand side in the previous equation. $$\exp\left[\sum_{n>0}\frac{1}{n}\frac{q^{n/2}}{1-q^n}\chi_{[1]}(a_1^n)\chi_{[1]}(a_2^n)\chi_{[1]}(a_3^n) \right]=\frac{(q)_\infty}{1-q}\prod^3_i \eta_2^{-1/2}(a_i) \sum_{R}\frac{\chi_R(a_1)\chi_R(a_2)\chi_R(a_3)}{\dim_q R}$$ wheret $$\eta_2(x)=\exp\left[-2\sum_{n>0} \frac{1}{n}\frac{q^n}{1-q}(\chi_{[1]}^2(x)-1)\right]$$ Note that $\chi_R$ is a character of the irreducible representation of ${\mathfrak sl}_2$ with highest weight $R$. If there are only two characters involved, it could be seen as the Cauchy formula in representation theory. I wonder if somebody know a proof for this identity.

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general identity involving Macdonald polynomials. One can show that acting on the left hand side with Macdonald operator (properly conjugated) on any of the $a_i$ gives the same result independent on the choice of $a_i$. see eg 5.11 there. then one can expand the left hand side in terms of the Macdonald polynomials which is given by a single sum over representations since the spectrum of Macdonald operator is not degenerate. Using this observation and some simple manipulations the identity can be established - the complete argument is detailed in section 6 of the above mentioned paper.