# Sum identities with immanants

For $\chi$ being an irreducible character of the symmetric group $S_n$ and being $M$ a complex $n\times n$-matrix, I would like to show

$$\sum_{\sigma, \rho \in S_n} \overline{\chi(\sigma)} \chi(\rho) \prod_{j=1}^n M_{\sigma_j, \rho_j} = \frac{ n! }{\chi(e)} \sum_{\sigma \in S_n} \chi(\sigma) \prod_{j=1}^n M_{j, \sigma_j} ,$$ where the right-hand-side is proportional to the immanant of $M$, $$\text{imm}(M) = \sum_{\sigma \in S_n} \chi(\sigma) \prod_{j=1}^n M_{j, \sigma_j}$$ and $e$ is the identity element in $S_n$. This expression appears as the scalar product of many-body quantum states with exotic exchange symmetries.

For one-dimensional representations (i.e. the trivial constant or the alternating representation), we get the identities $$\sum_{\sigma, \rho \in S_n} \prod_{j=1}^n M_{\sigma_j, \rho_j} = n! ~ \text{perm}(M) \\ \sum_{\sigma, \rho \in S_n} \text{sgn}(\sigma) \text{sgn}(\rho) \prod_{j=1}^n M_{\sigma_j, \rho_j} = n! ~ \text{det}(M)$$ for the permanent and the determinant (related to bosons and fermions), respectively, which are shown using $\chi(\sigma \rho)=\chi(\sigma) \chi(\rho)$ and $\chi(e)=1$ for one-dimensional $\chi$.

• thanks, is there a simple way to see/understand the very last equality, i.e. why that sum over $/sigma$ resolves this way? – Malte May 15 '15 at 14:11
• That result is an adaptation of en.wikipedia.org/wiki/Schur_orthogonality_relations -- so that in particular, if $G$ is a finite group and $\chi, \xi$ are irreducible chars, then $$\sum_{\sigma \in G} \overline{\chi(\sigma)}\xi(\sigma\tau) = \begin{cases} |G|\chi(\tau)/\chi(e), & \text{if}\ \chi=\xi;\\ 0 & \text{otherwise}.\end{cases}$$ – Suvrit May 15 '15 at 14:39