Skip to main content
MathOverflow

Return to Answer

Bounty Ended with 50 reputation awarded by Marcel
sentence addd at end
Source Link
Richard Stanley
Richard Stanley
  • 50.8k
  • 14
  • 155
  • 279

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 $\chi_{\theta/\lambda}$.

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.

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 $\chi_{\theta/\lambda}$.

Source Link
Richard Stanley
Richard Stanley
  • 50.8k
  • 14
  • 155
  • 279

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.