 if and only if there exists $$\sA_0(\bfq)(U)\in\operatorname{End}_{Z(\cH_0(\bfq))}(\cA_0(\bfq))$$ satisfying  0.  $\sA_0(\bfq)(U)$ is invertible ;  0.  $\sA_0(\bfq)(U)^0=\sA_0(\bfq)(U^0)=\sA_0(\bfq)(z_0)=z_0\operatorname{Id}$ ;  0.  $\sA_0(\bfq)(U)V_s(\bfq)=\textrm{ multiplication by $X$}$