acts by a constant on $A^n_\tau$-irreducible components of the right $A^n_\tau$-module $V^{A^{n-1}_\tau}$. That is, there exists a basis in the module in which $Z$ is diagonal and of the form $diag(\lambda_1(\tau),\lambda_2(\tau),\dots,\lambda_2(\tau))$. Thus we only need to calculate $\lambda_1(\tau)$, $\lambda_2(\tau)$.

The module $V^{A^{n-1}_\tau}$ exists for all $\tau\ne 0$. As the operator $Z$ is invertible for all nonzero values of $\tau$, we have $\lambda_1(\tau)\ne 0$, $\lambda_2(\tau)\ne 0$.

The functions $\lambda_i(\tau)$ are single valued. Indeed for $n=2$ it is obvious. So let us suppose $n>2$, then the eigenvalues $\lambda_i(\tau)$ have the different multiplicities. Hence the Galois group of the extension of the field of rational functions by $\lambda_1(\tau)$, $\lambda_2(\tau)$ is trivial because it cannot exchange $\lambda_1$ and $\lambda_2$. Thus the functions $\lambda_i(\tau)$ are rational and we have $\lambda_i(\tau)=C_i
\tau^{k_i}$, $i=1,2$.

When $\tau=1$, the algebra $A^n_\tau$ becomes the group algebra of $S_n$, and $Z=1$. Thus we have $C_1=C_2=1$. The calculation of $k_1,k_2$ uses the quasiclassical limit reasoning.

If $\tau=e^h$ then we can write the expansion of $T_i$ in terms of $h$ $$T_i=s_i+h \tilde{s}_i+O(h^2),\quad i=1,\dots,n-1,$$ where $s_i=(i,i+1)$ is a usual transposition. Relation (\[T\^2\]) inside $H_{1,\tau}$ implies $$s_i\tilde{s}_i+\tilde{s}_i s_i=2s_i, \quad i=1,\dots,n-1.$$

Let us calculate the first nontrivial term $\tilde{Z}$ of the expansion of $Z=1+h\tilde{Z}+O(h^2)$: $$\tilde{Z}=\sum_{i=1}^{n-1} s_1\dots 