If \(\lambda=(\lambda_1\ge\lambda_2\ge\dots\ge\lambda_k)\) is a partition of \(n\), let \(A^\lambda\) denote the irreducible representation of the symmetric group \(S_n\) such that the Frobenius image of \(\chi^{A^\lambda}=\chi^\lambda\) is the Schur function \(S_\lambda(x_1,\ldots,x_N)\) where \(N>n\).
Given a partition \(\lambda\) of \(n\), let \(l(\lambda)\) denote the number of parts of \(\lambda\) and \(\lambda'\) denote its conjugate partition. Let \(\chi^\lambda_\mu\) denote the value of the character of the irreducible representation \(A^\lambda\) of \(S_n\) at the conjugacy class indexed by the partition \(\mu\). Show that \[ \chi^{\lambda'}_\mu=(-1)^{n-l(\mu)}\chi^\lambda_\mu. \]
Solution
1. For a class function \(\chi\) of \(S_n\), its Frobenius characteristic is\[ \operatorname{ch}(\chi) =\sum_{\mu\vdash n}\frac{\chi_\mu}{z_\mu}p_\mu, \]where\[ p_\mu=p_{\mu_1}\cdots p_{\mu_{l(\mu)}} \]and \(z_\mu\) is the usual centralizer factor.In particular,\[ s_\lambda =\sum_{\mu\vdash n}\frac{\chi^\lambda_\mu}{z_\mu}p_\mu. \]
This is the defining power-sum expansion of the Frobenius characteristic map. By the convention in the problem, the irreducible character \(\chi^\lambda\) has Frobenius image \(s_\lambda\).
2. Let \(\omega\) be the standard involution of the ring of symmetric functions determined by\[ \omega(h_r)=e_r. \]Then\[ \omega(s_\lambda)=s_{\lambda'}. \]
The Jacobi–Trudi identity gives \[ s_\lambda=\det(h_{\lambda_i-i+j}), \] while the dual Jacobi–Trudi identity gives \[ s_{\lambda'}=\det(e_{\lambda_i-i+j}). \] Applying \(\omega\) entrywise to the first determinant therefore gives the second. Hence \(\omega(s_\lambda)=s_{\lambda'}\).
3. For every \(r\ge1\),\[ \omega(p_r)=(-1)^{r-1}p_r. \]
Let \[ H(t)=\sum_{m\ge0}h_mt^m, \qquad E(t)=\sum_{m\ge0}e_mt^m. \] The standard generating-function identities are \[ H(t)=\exp\left(\sum_{r\ge1}\frac{p_r}{r}t^r\right), \] and \[ E(t)=\exp\left(\sum_{r\ge1}(-1)^{r-1}\frac{p_r}{r}t^r\right). \] Since \(\omega(H(t))=E(t)\), comparison of the logarithms gives \[ \sum_{r\ge1}\frac{\omega(p_r)}r t^r = \sum_{r\ge1}(-1)^{r-1}\frac{p_r}r t^r. \] Comparing coefficients of \(t^r\) yields the formula.
4. If \(\mu=(\mu_1,\ldots,\mu_{l(\mu)})\vdash n\), then\[ \omega(p_\mu)=(-1)^{n-l(\mu)}p_\mu. \]
Using multiplicativity of \(\omega\) and 3, \[ \omega(p_\mu) =\prod_{j=1}^{l(\mu)}(-1)^{\mu_j-1}p_{\mu_j} =(-1)^{\sum_j(\mu_j-1)}p_\mu. \] Because \(\sum_j\mu_j=n\), \[ \sum_j(\mu_j-1)=n-l(\mu). \]
5. Applying \(\omega\) to the Frobenius expansion of \(s_\lambda\) gives\[ s_{\lambda'} =\sum_{\mu\vdash n} \frac{(-1)^{n-l(\mu)}\chi^\lambda_\mu}{z_\mu}p_\mu. \]
By 1, \[ s_\lambda =\sum_{\mu\vdash n}\frac{\chi^\lambda_\mu}{z_\mu}p_\mu. \] Apply \(\omega\) to both sides. By 2 the left side becomes \(s_{\lambda'}\), and by 4 each \(p_\mu\) is multiplied by \((-1)^{n-l(\mu)}\).
6. Therefore, for every partition \(\mu\vdash n\),\[ \boxed{\chi^{\lambda'}_\mu=(-1)^{n-l(\mu)}\chi^\lambda_\mu}. \]
Applying 1 to the conjugate partition \(\lambda'\) also gives \[ s_{\lambda'} =\sum_{\mu\vdash n}\frac{\chi^{\lambda'}_\mu}{z_\mu}p_\mu. \] The power sums \(p_\mu\) for \(\mu\vdash n\) form a basis of the degree-\(n\) symmetric functions over \(\mathbb Q\). Comparing the coefficient of \(p_\mu\) with 5 yields \[ \frac{\chi^{\lambda'}_\mu}{z_\mu} = \frac{(-1)^{n-l(\mu)}\chi^\lambda_\mu}{z_\mu}. \] Since \(z_\mu\ne0\), the desired identity follows.