its irreducible components have dimension $2n$.

Local coordinates on $CM_\tau$  {#coor}
-------------------------------

It is easy to see that matrices $X,Y\in\mathfrak{gl}(n,\mathbb
C)$, $$\begin{gathered}
\label{repf}
X=diag(\lambda_1,\dots,\lambda_n), \\ \label{repm}Y_{ii}=q_i,\quad
i=1,\dots,n,\\ Y_{ij}=\frac{(\tau-\tau^{-1})q_i\lambda_j}
{(\tau\lambda_i-\tau^{-1}\lambda_j)},\quad 1\le i\ne j\le
n,\label{repl}\end{gathered}$$ satisfy the equation $$\label{eqXY-YX} rk(\tau
XY-\tau^{-1}YX)= 1,$$ for all $\lambda\in(\mathbb
C^*)^n\setminus D_\tau$, $q\in(\mathbb{C}^*)^n$ where $$D_\tau=
\{\lambda|\delta_\tau(\lambda)=\prod_{i\ne
j}(\tau\lambda_i-\tau^{-1}\lambda_j)=0\}.$$

There is a well known formula: if $M=(M_{ij})$, where $M_{ij}=(\lambda_i-\mu_j)^{-1},$ $ 1\le i,j\le n$, then $$\det(M)=\frac{\prod_{i<j}(\lambda_i-\lambda_j)(\mu_j-\mu_i)}
{\prod_{i,j}(\lambda_i-\mu_j)}.$$ To prove this formula one can proceed by the induction on $n$ using the Gaussian method of calculation of the determinant for the step of the induction.

Applying the last formula to the matrix $Y$ we see that $\det(Y)$ is nonzero if and only if $\lambda_i\ne\lambda_j$, $i\ne j$.

Let us denote by $\pi'_{12}$: $CM'_\tau\to GL(E)\times
GL(E)$ the projection on the first two coordinates. The previous reasoning shows that $(X,Y)\in \pi_{12}(CM'_\tau)$, for $\lambda\in(\mathbb C^*)^n\setminus (D_\tau\cup D)$, $q\in(\mathbb
C^*)^n$ where $$D= \{\lambda|\delta(\lambda)=\prod_{i< j}
(\lambda_i-\lambda_j)=0\}.$$

Now we can state

\[propcoor\] Let $(X,Y,U,V)\in CM'_\tau$ and $X$ be diagonalizable with the different eigenvalues $\lambda_i$, $i=1,\dots,n$ such that $\tau\lambda_i\ne \tau^{-1}\lambda_j$. Then the $GL(n,\mathbb C)$ orbit of $(X,Y,U,V)$ contains a representative satisfying equations $V=\lambda^t$ and (\[repf\]-\[repl\]) for some $q\in(\mathbb C^*)^n$. Such a representative is unique up to (simultaneous) permutation of 