I am trying to prove the following identity \begin{equation} \prod_{i=1}^{m}(1-x_{i}z)^{-u}\prod_{j=1}^{n}(1-y_{i}z)^{-v} \prod_{i=1}^{m}\prod_{j=1}^{n}(1-(x_i +y_j)z)^{-w}\\ = \sum_{\lambda, \mu}c_{\lambda, \mu, \nu, \eta}(u,v,w)\frac{s_{\lambda}({\bf{x}})s_{\mu}({\bf{y}})}{h_{\lambda}h_{\mu}}z^{\mid \mu \mid+ \mid \lambda \mid} \tag{*}\label{star} \end{equation} where $$ c_{\lambda, \mu, \nu, \eta}(u,v,w)= \sum_{\nu\subset \lambda, \,\eta\subset\mu\\ \lvert \nu\rvert= \lvert \eta \rvert} c_{\lambda, \mu}^{\nu, \eta}d_{\nu, \eta}P_{\lambda / \nu , \mu / \eta}(u,v,w,m,n)$$ where $ c_{\lambda, \mu}^{\nu, \eta}\in \mathbb{Q}$ and $ d_{\nu, \eta}\in \mathbb{Q}[w]$ is a monic polynomial of degree $\lvert \nu \rvert$ and $$ P_{\lambda / \nu , \mu / \eta}= \prod_{(i,j)\in \lambda / \nu}(u+nw +j-i)\prod_{(i,j)\in \mu / \eta}(v+mw+j-i)$$, $h_{\lambda}$ denoting the hook length.

For the case $m=n=1$ we can prove this formula, but in general I can't see the proof. Initially I am trying the following approach, as the LHS of \eqref{star} has an explicit form and we know

$$ \ln\Big(\prod_{i\geq 1}\frac{1}{1-x_i z}\Big) = \sum_{n\geq 1}p_n ({\bf{x}})\frac{z^n}{n}$$ where ${\mathbf{x}}:= (x_1, x_2, \dotsc)$, where $p_{n}({\mathbf{x}})$ denotes the power symmetric basis. We express the LHS in the power symmetric basis and evaluating to $x_i = 0 $ for $i> m $ and $y_j =0$ for $j> n$.

Similarly we can write

$$ \ln\Bigl(\prod_{i\geq 1}\prod_{j\geq 1}\frac{1}{1-(x_i +y_j) z}\Bigr) = \sum_{j\geq 1} \sum_{n\geq 1}p_n ({\mathbf{x}}+y_j)\frac{z^n}{n} $$ where ${\mathbf{x}}+y_j = (x_1 + y_j , x_2 +y_j,\dotsc)$ in power symmetric basis.

Now for the RHS we can we write it as a power symmetric basis? Cauchy identity comes to mind but this expression is more general. Any idea will be really helpful.

`\lvert\nu\rvert`

, not $\mid\nu\mid$`\mid\nu\mid`

. (Note the difference between, for example, $2\lvert\nu\rvert$`2\lvert\nu\rvert`

and $2\mid\nu\mid$`2\mid\nu\mid`

.) I have edited accordingly. $\endgroup$1more comment