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Let $ s_{\lambda} $ be the Schur function associated to the partition $ \lambda $. Cauchy's identity (as in Macdonald) states that

$$ \sum_{\lambda} s_{\lambda}(X)s_{\lambda}(Y) = \prod_{i,j}(1-x_iy_j)^{-1} $$

where the sum is over all partitions. There is a generalization appearing in a paper by Ishikawa and Tagawa Schur Function Identities and Hook Length Posets (Theorem 2.1(i)) which states that if $ X=(x_1,...,x_m) $ and $ Y=(y_1,...,y_m) $,

$$ \sum_{\lambda} w^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) = \frac{1-|X||Y|}{(1-w|X||Y|)\prod_{i,j=1}^m(1-x_iy_j)} $$

where the sum is over partitions $ \lambda=(\lambda_1,...,\lambda_m) $ and $ |X|: = x_1x_2\cdots x_m $.

I am curious if there is a result along these lines which gives a closed form product expression for the generating function

$$ \sum_{\lambda} z_1^{\lambda_1}\cdots z_{m-1}^{\lambda_{m-1}}z_m^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) $$

where the sum is still over partitions $ \lambda=(\lambda_1,...,\lambda_m) $. In particular, at least for now I care about such a generating function in the form

$$ \sum_{\lambda} w^{\lambda_{m-1}}z^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) $$

It is my understanding that such a result can likely be discovered via the RSK correspondence, but this is not really my field. Thus before I delve into possibly slowly reinventing the wheel I thought I'd ask.

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    $\begingroup$ What is $w$? Even in the underlying paper, it is never introduced. Supposedly it is a weight between $0$ and $1$, with $w=1$ yielding Cauchy's identity. But $w=0$ yields nonsense. Is there simply a factor $w$ missing on the RHS? $\endgroup$
    – Wolfgang
    Commented Mar 25, 2019 at 8:11
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    $\begingroup$ @Wolfgang why does $w=0$ yield nonsense? It reduces to a generating function over partitions with $\lambda_m=0$. $\endgroup$ Commented Apr 1, 2019 at 22:12
  • $\begingroup$ @GjergjiZaimi Oh sure, I didn't realize that terms with $0^0=:1$ would survive on the LHS. And of course I guess it is tacitly assumed as usual that $\lambda_1\ge\cdots\ge\lambda_m$,. $\endgroup$
    – Wolfgang
    Commented Apr 2, 2019 at 7:39

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