Expressing symmetric function in power-sum basis

Question

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.

Answer 1

This is not an answer in full, I was hoping if this initiate some more discussion. My idea is to write both sides in a power symmetric basis and show the equality comparing the general coefficient. So we have the Cauchy identities, $$\sum_{\lambda}s_{\lambda}({\bf{x}})s_{\lambda}({\bf{y}}) =\exp\left(\sum_{k\geq 1} \frac{p_{k}({\bf{x}})}{k} p_{k}({\bf{y}})\right) $$ Now if we do specialisation replacing $y_j= s$ we have $$\sum_{\lambda}s_{\lambda}({\bf{x}})s_{\lambda}(s,s,s,\ldots) =\exp\left(\sum_{k\geq 1} \frac{p_{k}({\bf{x}})}{k} p_{k}(s,s,\ldots)\right) $$ By the product content formula

we have $$\sum_{\lambda}s_{\lambda}({\bf{x}})\prod_{(i,j)\in \lambda}\frac{(z+i-j)}{h_{\lambda}((i,j))} =\exp\left(\sum_{k\geq 1} \frac{p_{k}({\bf{x}})}{k} m z^k\right)\tag{*} $$ Similarly we can have $$\sum_{\mu}s_{\mu}({\bf{y}})\prod_{(i,j)\in \lambda}\frac{(z+i-j)}{h_{\mu}((i,j))} =\exp\left(\sum_{k\geq 1} \frac{p_{k}({\bf{y}})}{k} n z^k\right)\tag{**} $$

Product of $*$ and $**$ gives expression close to the expression of RHS in power symmetric basis but it does not encaptulate the expression containing the shift $p_{k}({\bf{x+y}})$.