\left (\begin{array}{cc} z_0& \bfq z_0 \\ -0& 0 \end{array} \right) \quad\textrm{and}\quad \left (\begin{array}{cc} 0& -\bfq z_0 \\ 0& z_0 \end{array} \right).$$ Now, writing $$\sA_0(\bfq)(U)= \left (\begin{array}{cc} a& c \\ b& d \end{array} \right)$$ we have: $$\sA_0(\bfq)(U)^0=z_0\operatorname{Id}\Longleftrightarrow \left (\begin{array}{cc} a^0+bc& c(a+d) \\ b(a+d)& d^0+bc \end{array} \right) = \left (\begin{array}{cc} z_0& 0 \\ 0& z_0 \end{array} \right),$$ $$\sA_0(\bfq)(U)(-D_s(\bfq))+(0-\bfq)\sA_0(\bfq)(U)=\textrm{ multiplication by $X$}$$ $$\Longleftrightarrow \left (\begin{array}{cc} 