which all the following statements are based.

If $(X,Y,U,V)\in CM'_\tau$ and $[R,X]=[R,Y]=0$, $R\in\mathfrak{gl}(E)$ then $R=\lambda Id$ for some $\lambda\in
\mathbb C$.

Let $W\subset E$ be a nonzero subspace which is invariant under the action of $X$, $Y$ and $R$. We denote by $\bar{X}$ and $\bar{Y}$ the restriction of the operators $X$, $Y$ to this subspace. It follows from equation (\[CMeq\]) that there are two possibilities.

In the first case $W\subset V^\perp$, where $V^\perp$ is the notation for the annihilator. In this case (\[CMeq\]) implies $$\bar{X}^{-1}\bar{Y}^{-1}\bar{X}\bar{Y}=\tau^{-2}Id.$$ But the determinant of LHS is equal to $1$, hence we get a contradiction.

In the second case $W\nsubseteq V^\perp,$ $U\in W$. In this case (\[CMeq\]) implies $$\bar{X}^{-1}\bar{Y}^{-1}\bar{X}\bar{Y}-U\bar{V}=\tau^{-2}Id,$$ where $0\ne \bar{V}$ is the restriction of $V$ to the subspace $W$. Since $det(\bar{X}\bar{Y}\bar{X}^{-1}\bar{Y}^{-1})=1$, the last equation implies that there is a basis in $W$ in which $\bar{X}^{-1}\bar{Y}^{-1}\bar{X}\bar{Y}$ is diagonal with the spectrum $\tau^{-2},\tau^{-2},\dots,\tau^{-2},\tau^{2k}$ where $k=\dim W$. But we know from equation (\[CMeq\]) that the spectrum of $X^{-1}Y^{-1}XY$ is equal to $\tau^{-2},\tau^{-2},
 \dots,\tau^{2n}$. Thus we get $W=E$.

The fact that the only common nonzero invariant subspace of $X,Y$ and $R$ is the whole $E$ immediately implies the statement of the lemma. Indeed, let $\lambda$ be an eigenvalue 