Let $n\in\mathbb{N}$. Then, for $i\in\mathbb{N}$, the $i-$th power sum if defined to be the polynomial $p_i^{(n)}:=\sum_{j=1}^n x_j^i$ in $n$ indeterminates $x_1,x_2,\ldots,x_n$.

Then let $\lambda:=(\lambda_1,\ldots,\lambda_l)$ be a partition of $d$ (written with no trailing zeroes). We can define $p^{(n)}_\lambda:=\prod_{j=1}^l p^{(n)}_{\lambda_j}$. Note that every symmetric polynomial in $n$ variables of given degree $d$ over a field of characteristic $0$ can be expressed as a linear combination of the $p^{(n)}_\lambda$, where $\lambda$ runs through all partitions of $d$.

Now introduce a new indeterminate $x_{n+1}$, and let the symmetric group $S_{n+1}$ on $n+1$ letters act on such a $p^{(n)}_\lambda$. We define $P^{(n+1)}_\lambda:=\sum_{\sigma\in S_{n+1}} \sigma(p^{(n)}_\lambda)$. Therefore $P^{(n+1)}_\lambda$ is in fact a symmetric polynomial in $n+1$ variables. What can one say about the coefficients $c_\lambda$ such that $P^{(n+1)}_\lambda:=\sum_{\lambda\vdash d} c_\lambda p^{(n+1)}_\lambda$? Is there a special name for these and a combinatorial interpretation/ a method of calculating the $c_\lambda$ ?

  • $\begingroup$ would be nice to fix the notation. Since $\lambda$ is fixed, it cannot be used in the equation $P_\lambda^{(n+1)}:=\sum ...$ as a summation index. I guess it should look like this $P_\lambda^{(n+1)}:=\sum_{\mu\vdash d}c_{\lambda,\mu}p_{\mu}^{(n+1)}$. $\endgroup$ – JimT Apr 15 '18 at 4:25

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