If $\psi|_{\Lambda}=0$ then the divisor $D_{\psi}$ intersects properly the orbit $V(\Lambda)$. Moreover, the set $\{m_{\sigma}\}_{\sigma\in\Pi(\Lambda)}$ is a set of defining vectors of $\psi(\Lambda)$ and the restriction of $D_{\psi}$ to $V(\Lambda)$ is the divisor $D_{\psi(\Lambda)}$.