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The Cauchy Identity $$ \sum_{\nu}s_{\nu}(x)s_{\nu}(y) = \prod_{j,k=1}^{\infty}\frac{1}{1-x_{j}y_{k}} $$ expresses the sum over all integer partitions of the product of pairs of Schur polynomials as the double product in the right hand side above. Since Schur polynomials verify $$ s_{\nu}(x_{1},...,x_{n})=0 $$ if $l(\nu)>n$, specializing all but $n$ variables from each set in the Cauchy identity to zero we obtain $$ \sum_{l(\nu)\leq n}s_{\nu}(x_{1},...,x_{n})s_{\nu}(y_{1},...,y_{n}) = \prod_{j,k=1}^{n}\frac{1}{1-x_{j}y_{k}}. $$ Note that the sum is now restricted to partitions of length not greater than $n$.

My question is the following: is there a simple way to express the sum $$ \sum_{\nu_{1}\leq n}s_{\nu}(x)s_{\nu}(y), $$ where the sum is now restricted to partitions with first part less than or equal to $n$? (possibly specializing variables, as above).

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The sum (in any number of variables) is equal to the determinant $$\det(B_{j-i})_{1\le i,j\le n},$$ where $$B_i=\sum_{l=0}^\infty e_{l+i}(x)e_l(y),$$ and $e_l$ is the elementary symmetric function, with $e_l=0$ for $l<0$. This follows by applying the involution $\omega$ in $x$ and $y$ to Theorem 16 of my paper Symmetric functions and P-recursiveness, Journal of Combinatorial Theory, Series A 53 (1990), 257–285. The theorem follows fairly easily from the Cauchy-Binet formula and the Jacobi-Trudi formula.

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  • $\begingroup$ Thank you! I was actually wondering if there is also a simpler product expression for this determinant. I know there is one for the determinant in the aforementioned Theorem 16 (namely $\prod_{j,k=1}^{n}(1-x_{j}y_{k})^{-1}$, the one in the question); this is Baxter's identity. There is also such an expression for the determinant resulting from applying the involution $\omega$ in only one of the sets of variables $x$ or $y$ (namely $\prod_{j,k=1}^{n}(1+x_{j}y_{k})$. Do we have an analogue of Baxter's identity for the determinant appearing in your answer? $\endgroup$
    – D. Donnelly
    Commented May 20, 2017 at 11:15
  • $\begingroup$ I don't think there is a simple product expression. You can check it by just computing the sum for small values of $n$ and small numbers of variables and seeing if it factors. $\endgroup$
    – Ira Gessel
    Commented May 20, 2017 at 15:23
  • $\begingroup$ @IraGessel In another case, if we put condition $λ_1<N$ for $ λ=(λ_1,..,λ_l)$ we arrive to another expression at the $h=1$ (unlike the $-κ^2 \log [1−h^2]$ that divergent in that point), and principally we can make expansion around $h=1$ . ($N$ and $N_f$ both tend to ∞ with $κ=N_f/N$ fixed.) First terms appears in another topic mathoverflow.net/questions/406707/… $\endgroup$
    – Sergii Voloshyn
    Commented Oct 25, 2021 at 19:30

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