3
$\begingroup$

This question is a follow-up on another MO query here.

Question. For $r\geq$ an integer, is it true that there exists homogeneous symmetric polynomial $P_r(x_1,\dots,x_n)$ with positive coefficients such that $$(-1)^r\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2} =\frac{P_r(x_1,\dots,x_n)}{\prod_{j=1}^nx_j}\prod_{i<j}^{1.n}\frac1{x_i+x_j} \,\, ?$$ I would be content with $0\leq r\leq n$. Is there an explicit formulation of $P_r$?

Remark. The factor $\prod_jx_j$, in the denominator of the RHS, is needed only when $r=0$.

For example, when $n=3, r=0$, the RHS is $\frac{x_1+x_2+x_3}{x_1x_2x_3(x_1+x_2)(x_1+x_3)(x_2+x_3)}$.

If $n=4, r=2$ then the RHS takes the form $$\frac{x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4}{(x_1+x_2)(x_1+x_3)(x_1+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4)}.$$

$\endgroup$
5
$\begingroup$

Denote by $\delta_n$ the partition $(n-1,n-2,\dots,1)$ of $\binom{n}2$, and the associated schur function is $$s_{\delta_n}(x_1,\dots,x_n)=\prod_{i<j}(x_i+x_j).$$ Consider now the ratio of determinants $$\begin{vmatrix} x_1^{2r-1} & x_2^{2r-1} &\cdots & x_n^{2r-1} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix} \cdot \begin{vmatrix} x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix}^{-1}$$ This evaluates to $\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2}$. Proof: Expand the first determinant by the first row, and use the fact that the second is just a Vandermonde that evaluates to $\prod_{i<j} (x_j^2-x_i^2)$.

Therefore your polynomial $P_r(x_1,x_2,\dots,x_n)$ is equal to $$(-1)^r \begin{vmatrix} x_1^{2r} & x_2^{2r} &\cdots & x_n^{2r} \\ x_1^{2n-3} & x_2^{2n-3} &\cdots & x_n^{2n-3} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^3 & x_2^3 & \cdots & x_n^3 \\ x_1 & x_2& \cdots &x_n \end{vmatrix} \cdot \begin{vmatrix} x_1^{2n-2} & x_2^{2n-2} &\cdots & x_n^{2n-2} \\ x_1^{2n-4} & x_2^{2n-4} &\cdots & x_n^{2n-4} \\ \vdots &\vdots & \ddots & \vdots\\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ 1 & 1& \cdots &1 \end{vmatrix}^{-1}\cdot s_{\delta_n}(x_1,x_2,\dots,x_n)$$ When $0\le r\le n$ we can use the definition of Schur polynomials as ratios of determinants to simplify this expression as $$s_{\lambda_{r,n}}(x_1,x_2,\dots,x_n)$$ Where the partition $\lambda_{r,n}$ is given by $(n-2,n-3,\dots,r+1,r,r,r,r-1,\dots,2,1)$. Therefore, not only is $P_{r}(x_1,x_2,\dots,x_n)$ symmetric with positive integer coefficients, but we also get a combinatorial interpretation of these coefficients as counting semistandard tableaux of shape $\lambda_{r,n}$.

$\endgroup$
  • $\begingroup$ I liked your approach and even more the interpretation. Wonderful! One point: when $r=0$, there is a factor $\prod_{i=1}^nx_i$ in the denominator. Can you account for that? $\endgroup$ – T. Amdeberhan Feb 28 '17 at 22:22
  • $\begingroup$ Yes, I already did, that's why I increased the exponents by 1, in the first determinant. Combinatorially this just means that when $r>0$ the partition admits a tableaux where every label $1,2,...,n$ appears, which is easy since when $r>0$ then $\lambda_{r,n}$ has a column of length $n$, but not when $r=0$ which is why the $\prod x_i$ remains in the denominator for that case. $\endgroup$ – Gjergji Zaimi Feb 28 '17 at 22:27
  • $\begingroup$ Okay, thanks. You might like to look into my formulation of an answer at the link; perhaps it is possible to write one determinant for all before splitting into a sum. The link is mathoverflow.net/questions/263265/… $\endgroup$ – T. Amdeberhan Feb 28 '17 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.