# Power sums and Jack symmetric functions

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 $$p_n=\sum_i x_i^n\ .$$ 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 $$\langle J_\lambda^\alpha, J_\mu^\alpha\rangle_\alpha = 0 \mbox{ for } \lambda\neq \mu\ ,$$ where $\langle \cdot , \cdot \rangle_{\alpha}$ is defined over the basis of power sums as $$\langle p_\lambda, p_\mu\rangle_\alpha=\delta_{\lambda,\mu} z_\lambda \alpha^{\ell(\lambda)}\ ,$$ 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 $J_\lambda^\alpha$?

• you meant to say $p_{1^n}$, right? – Suvrit Jan 22 '14 at 0:03
• See Proposition 2.3 and Theorem 5.8 of math.mit.edu/~rstan/pubs/pubfiles/73.pdf. – Richard Stanley Jan 22 '14 at 3:13
• Yes, Suvrit. Indeed, $(p_1)^n=p_{\lambda}$ with $\lambda=(1^n)$. – Francesco Jan 22 '14 at 9:59

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