Let $C_\mu$ be the size of the conjugacy class in $S_n$ of permutations whose cycletype is the partition $\mu\vdash n$. Let $\chi$ be the characters of the irreducible representations of $S_n$.

Let $\omega\vdash m$ and let $\theta\vdash(n+m)$. I am interested in the sum $$ \frac{1}{n!}\sum_{\mu\vdash n} C_\mu\chi_\lambda(\mu)\chi_\theta(\mu\cup\omega),$$ where $\mu\cup\omega$ is a partition of $n+m$ containing the parts of $\mu$ and of $\omega$. Numerics suggest that for most pairs $(\lambda,\theta)$ this sum is zero. In the simplest case of $\omega=0$, for example, the only non-vanishing pair is $\theta=\lambda$ (and the result is 1).

To illustrate, these are the values of the sum when $\omega=(3)$, with $n=4$ and $n+m=7$ ($\lambda$ is labelling the rows and $\theta$ the columns, both in lexicographic order) $$\left[ \begin {array}{ccccccccccccccc} 1&0&0&1&0&-1&0&0&1&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&-1&0&0&0&1&0&0&0 \\ 0&0&1&-1&0&0&0&0&0&0&-1&0&1&0&0 \\ 0&0&0&0&1&0&-1&0&0&0&0&0&0&1&0 \\ 0&0&0&0&0&0&0&0&1&-1&1&0&0&0&1\end {array} \right] $$

A possible solution would be to write $\chi_\theta(\mu\cup\omega)=\sum_{\rho\vdash n} a_\rho(\omega) \chi_\rho(\mu)$. Is there a known way to accomplish this decomposition? (I'm thinking something like the Murnaghan-Nakayama rule)

  • $\begingroup$ you mean m=3 and m+n=7, right? $\endgroup$
    – Wolfgang
    Feb 1, 2016 at 20:17

1 Answer 1


Since $\chi_\theta(\mu\cup\omega)=\langle s_\theta,p_\mu p_\omega\rangle$, your sum is given by $$ \frac{1}{n!}\sum_{\mu\vdash n} C_\mu\chi_\lambda(\mu)\langle s_\theta,p_\mu p_\omega\rangle = \left\langle s_\theta,p_\omega\cdot \frac{1}{n!}\sum_{\mu\vdash n} C_\mu \chi_\lambda(\mu)p_\mu\right\rangle $$ $$ \qquad\qquad = \langle s_\theta,p_\omega s_\lambda\rangle. $$ We can then expand $p_\omega s_\lambda$ in terms of Schur functions by Theorem 7.17.1 (the basis for the Murnaghan-Nakayama rule) of Enumerative Combinatorics, vol. 2. Note also that $\langle s_\theta,p_\omega s_\lambda\rangle = \langle s_{\theta/\lambda},p_\omega\rangle = \chi_{\theta/\lambda}(\omega)$, a value of the skew character $\chi_{\theta/\lambda}$.

  • $\begingroup$ Aha, I knew it could be done. Thanks! $\endgroup$
    – thedude
    Feb 4, 2016 at 17:29
  • $\begingroup$ Is there such a thing as a "skew zonal spherical function" $\omega_{\theta/\lambda}$ such that $Z_{\theta/\lambda}=\sum_{\rho}\omega_{\theta/\lambda}(\rho)(2^{\ell(\rho)}z_\rho)^{-1} p_\rho$, where $Z$ are zonal polynomials? $\endgroup$
    – thedude
    Feb 25, 2016 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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