s_{i-1}(\tilde{s}_i
s_i+\tilde{s}_i s_i)s_{i-1}\dots s_1= 2\sum_{i=1}^{n-1} s_{1i},$$ where $s_{1i}=s_1\dots s_{i-1}s_is_{i-1}\dots s_1$ is a permutation of $1$ and $i$.

The operator $\tilde{Z}/2$ acts on $\mathbb C[S_n]^{S_{n-1}}$ (by the left multiplication) and in the basis $e_i=(\sum_{w'\in
S_{n-1}}w')s_{1i}$ it has the matrix $J-Id$, $J_{ij}=1$, $1\le i,j
\le n.$ Hence $Spec(\tilde{Z}/2)=(n-1,-1,\dots,-1)$. On the other hand $Spec(\tilde{Z})=(k_1,k_2,\dots,k_2)$. Thus $k_1=2(n-1)$, $k_2=-2$ and we proved Proposition \[rk1\].

The map $\Phi$ on the subset $\mathcal{U}\subset Irrep^{n!}$ {#PhiU}
------------------------------------------------------------

It is possible to calculate $\Phi(V_{\mu,\nu})$ explicitly. Indeed let us fix a basis in $V_{\mu,\nu}^{A^{n-1}_\tau}$: $e_i=(\sum_{w'\in S_{n-1}}w')s_{1i}$, $i=1,\dots,n$.

For the matrices of the operators $\bar{X}_1$ and $\bar{Y}_1$ written in the basis $e_i$ the following equations hold $$\begin{gathered}
\bar{X}_1=diag(\nu_1,\dots,\nu_n)\\ \bar{Y}_{ii} =\mu_i
\prod_{j\ne i}\frac{(\tau^{-1}\nu_j-\tau\nu_i)}{(\nu_j-\nu_i)},
\quad i=1,\dots,n.\end{gathered}$$

The first equation is obvious. The second formula is a result of direct calculation using formulas (\[Y\]) for $Y_1$ and explicit formulas for $T_i$.

Indeed let make this calculation for $i=1$. The expansion of the product expression for $Y_1$ consists of the terms of the form $s_{i_1,j_1}\dots s_{i_r,j_r}c^{-1}F(X)P_1$, where $i_l<j_l$, $j_m<i_{m+1}$, $l=1,\dots,r$, $m=1,\dots,r-1$ and $F\in
{\mathbb C}[X^{\pm 1}]_{\delta(X)}$. We know that $Y_1e_1$ is a linear combination of $e_i$, $i=1,\dots,n$. The terms of the expansion of $Y_1 e_1$ which contribute to the coefficient before $e_1$ satisfy the equation 