of $R$, then the corresponding eigenspace $W_\lambda$ is invariant under the action of $X$ and $Y$, hence it coincides with $E$.

The action of $GL(E)$ on $CM'_{\tau}$ is free.

$CM'_\tau$ is smooth.

Let us introduce the map $\Psi$: $GL(E)\times GL(E)\times E\times
E^* \to \mathfrak{gl}(E)$: $$\Psi(X,Y,U,V)=X^{-1}Y^{-1}XY-U\otimes
V.$$

It is enough to show that $d\Psi$ is epimorphic at a point $(X,Y,U,V)\in CM'_\tau$. Let $x,y\in\mathfrak{gl}(E)$, $u\in
E,v\in E^*$ and $X(t)=Xe^{xt}$, $Y(t)=Ye^{yt}$, $U(t)=U+tu$, $V(t)=V+tv$. Then $$\begin{gathered}
 d\Psi_{(X,Y,U,V)}(x,y,u,v)=
\frac{d}{dt}(\Psi(X(t),Y(t),U(t),V(t))|_{t=0}=\\
-xX^{-1}Y^{-1}XY+X^{-1}Y^{-1}XxY-X^{-1}yY^{-1}XY+
X^{-1}Y^{-1}XYy-U\otimes v-u\otimes V.\end{gathered}$$

If $d\Psi$ is not an epimorphism, then there exists $0\ne
R\in\mathfrak{gl}(E)$ such that $$tr(d\Psi_{(X,Y,U,V)}(x,y,u,v)R)=0$$ for all $x,y\in\mathfrak{gl}(E)$, $u\in E,v\in E^*$. Using the cyclic invariance of the trace, we can rewrite the last condition in the form: $$\begin{gathered}
 tr(x(YRX^{-1}Y^{-1}X-X^{-1}Y^{-1}XYR))+\\
tr(y(RX^{-1}Y^{-1}XY-Y^{-1}XYRX^{-1}))-
v(RU)-VR(u)=0.\end{gathered}$$ As the bilinear form $tr(xy)$ is nondegenerate, the last equation implies $$\begin{gathered}
YRX^{-1}Y^{-1}X-X^{-1}Y^{-1}XYR=0,\label{def1}\\
RX^{-1}Y^{-1}XY-Y^{-1}XYRX^{-1}=0,\label{def2}\\ RU=0,\quad
VR=0.\label{def3}\end{gathered}$$ These equations together with equation (\[CMeq\]) imply $[R,X]=[R,Y]=0$. Indeed, let us derive the first equation.

Multiplying on the right formula (\[CMeq\]) by $R$ we get $$\label{multR}
X^{-1}Y^{-1}XYR=\tau^{-2}R.$$ Hence $$\begin{gathered}
\tau^{-2}XRX^{-1}=Y^{-1}XYRX^{-1}=RX^{-1}Y^{-1}XY=\\R(\tau^{-1}U\otimes
V+\tau^{-2} Id)=\tau^{-2}Id,\end{gathered}$$ here the first equation uses (\[multR\]), second (\[def2\]), third (\[CMeq\]) and fourth (\[def3\]).

By the previous lemma $R=\lambda Id$ and finally from (\[def3\]) we get $R=0$.

\[sm2n\] $CM_\tau$ is smooth algebraic variety, and all 