Proof of generalized Cauchy formula

Question

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. \begin{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} \end{equation} wheret \begin{equation} \eta_2(x)=\exp\left[-2\sum_{n>0} \frac{1}{n}\frac{q^n}{1-q}(\chi_{[1]}^2(x)-1)\right] \end{equation} 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.

Answer 1

1) In appendix E of this paper there is an outline of a proof of this statement based on matching poles and residues on both sides of the identity.

2) One can also use the more generic arguments of a more recent paper even for the more
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.