Let $\Lambda$ be the algebra of symmetric functions in infinitely many variables over $\mathbb{C}$.

The $n$-th power sum symmetric function $p_n$ is defined (formally) as \begin{equation} p_n=\sum_i x_i^n\ . \end{equation} The set consisting of symmetric functions $p_\mu=p_{\mu_1}\cdots p_{\mu_t}$, for all partitions $\mu=(\mu_1, \ldots, \mu_t)$, is a basis of $\Lambda$.

For any partition $\lambda$, let us denote by $J_\lambda^\alpha$ the Jack symmetric function associated with $\alpha$. This is uniquely determined by a triangular expansion with respect to the monomial symmetric functions and by the condition \begin{equation} \langle J_\lambda^\alpha, J_\mu^\alpha\rangle_\alpha = 0 \mbox{ for } \lambda\neq \mu\ , \end{equation} where $\langle \cdot , \cdot \rangle_{\alpha}$ is defined over the basis of power sums as \begin{equation} \langle p_\lambda, p_\mu\rangle_\alpha=\delta_{\lambda,\mu} z_\lambda \alpha^{\ell(\lambda)}\ , \end{equation} where $\delta_{\lambda,\mu}=\prod_a \delta_{\lambda_a,\mu_a}$, $z_\lambda=\prod_j j^{\, m_j}\, m_j!$ ($m_j=\# \{a\in \mathbb{N}\,\vert\, \lambda_a=j\}$) and $\ell(\lambda)$ is the legth of the partition $\lambda$.

Question: is it possible to determine explicitly an expression of $p_1^n$, with $n\geq 1$, in terms of Jack symmetric functions $J_\lambda^\alpha$?

  • $\begingroup$ you meant to say $p_{1^n}$, right? $\endgroup$
    – Suvrit
    Commented Jan 22, 2014 at 0:03
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    $\begingroup$ See Proposition 2.3 and Theorem 5.8 of math.mit.edu/~rstan/pubs/pubfiles/73.pdf. $\endgroup$ Commented Jan 22, 2014 at 3:13
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    $\begingroup$ Yes, Suvrit. Indeed, $(p_1)^n=p_{\lambda}$ with $\lambda=(1^n)$. $\endgroup$
    – user17778
    Commented Jan 22, 2014 at 9:59

1 Answer 1


To turn Richard's comment into an answer:

$$ J_{1^n} = p_{1^n} = \alpha^n n! \sum_{\lambda \vdash n} \frac{J_\lambda}{j_\lambda} $$ where $j_\lambda = \langle J_\lambda, J_\lambda \rangle$ is an explicit $\alpha$-deformation of two products of hooks-lengths in the diagram $\lambda$.


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