Given a vector $\vec k=(k_1,k_2,\cdots)$ with $k_i$ are non-negative integers, the Newton polynomial $p_{\vec k}(x)$ is defined as \begin{equation} p_{\vec k}(x)=\prod_{j=1}^n p_j^{k_j}(x)~, \end{equation} where \begin{equation} p_j(x)=\sum_{i=1}^N x_i^j \end{equation} are power-sum symmetric polynomials. The Newton polynomials are homogeneous of degree $\ell=\sum_{j=1}^n j k_j$. They are related to the Schur polynomials $s_R(x)$ through the Frobenius formula \begin{equation} p_{\vec k}(x)=\sum_{R}\chi_R(C_{\vec k})s_R(x)~. \end{equation} where $\chi_R(C_{\vec k})$ is the character of the symmetric group $S_\ell$ labeled by the representation $R$ at the conjugacy class $C_{\vec k}$. By using orthogonality of the characters, we can invert the above formula as \begin{equation} s_R(x) =\sum_{\vec k}\frac{\chi_R(C_{\vec k})}{z_{\vec k}} p_{\vec k}(x)~. \end{equation} with $z_{\vec k}=\prod_j k_j! j^{k_j}$.

My question is whether there is a natural generalization of the Frobenius formula replacing the Schur polynomials $s_R(x)$ by the Macdonald polynomials $P_R(x;q,t)$. What is the generalization of power-sum symmetric polynomial $p_{\vec k}(x)$ supposed to be? I think that the characters $\chi_R(C_{\vec k})$ is lifted to rational functions of $q$ and $t$.

  • $\begingroup$ Have you tried first to see what happens in the case of Jack polynomials? $\endgroup$ Dec 3, 2013 at 22:14
  • $\begingroup$ I have not looked at it yet. $\endgroup$ Dec 5, 2013 at 11:06

1 Answer 1


You might want to start with Corollary 3.3.1 in this paper. There, Haglund and Wilson gives the power-sum expansion of the Macdonald $J$ polynomials (and from there you can easily get the power-sum expansion of the Macdonald-$P$ polynomials).


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