# Cauchy identity for Schur functions

PRELIMINARY. The Cauchy identity for Schur polynomials reads $$\sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\prod_{i,j=1}^n\frac 1{1-x_iy_j},$$ where $$s_\lambda$$ are the Schur polynomials and the sum on the left-hand side runs over all partitions (of length $$\leq n$$).

Denoting $$p_k(x_1,...,x_n)=x_1^k+...+ x_n ^k$$, we also have the identity $$\exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)y^k}{k}\right) = \prod_{i=1}^n \frac 1{1-x_iy},$$ which implies that we can rewrite Cauchy identity as $$\sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)p_k(y_1,...,y_n)}{k}\right).\qquad (\star)$$

THE QUESTION. Let now $$p_1,p_2,...$$ be an infinite set of independent variables, and let $$S_\lambda(p_1,p_2...)$$ be the Schur functions, namely the expression of Schur polynomials in the power-sum polynomials. Can the identity $$\sum_{\lambda}S_\lambda(p_1,p_2,...)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\dfrac{p_k(y_1^k+...+y_n^k)}{k}\right)\qquad (\star\star)$$ be deduced directly from $$(\star)$$ by an $$n\to\infty$$ limit argument?

Maybe I am missing something about rigorously inferring identities in the ring of symmetric functions (which is the inverse limit as $$n\to\infty$$ of the rings of symmetric polynomials in $$n$$ variables) from identities in the rings of symmetric polynomials in $$n$$ variables.

Any help is greatly appreciated!