TITLE: Sum identities with immanants
QUESTION [6 upvotes]: 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 $\chi$.

REPLY [3 votes]: I hope this is not homework.
\begin{eqnarray*}
  \text{lhs}&=&\sum_{\sigma,\rho} \overline{\chi(\sigma)}\chi(\rho)\left(\prod_{i=1}^nM_{\sigma(i),\rho(i)}\right)\\
  &=&\sum_{\sigma,\rho} \overline{\chi(\sigma)}\chi(\rho)\left(\prod_{i=1}^nM_{i,\rho\sigma^{-1}(i)}\right)\\
  &=&\sum_{\tau}\left(\sum_{\sigma}\overline{\chi(\sigma)}\chi(\tau\sigma) \right)\prod_{i=1}^n M_{i,\tau(i)}\\
  &=&\sum_{\tau}\frac{n!}{\chi(e)} \chi(\tau)\prod_{i=1}^nM_{i,\tau(i)}.
\end{eqnarray*}