Let notation be as above. If $\Psi|_{\sigma}=0$, then $D_{\Psi}$ intersects $V(\sigma)$ properly and $\iota_{\sigma}^{\ast}D_{\Psi}=D_{\Psi(\sigma)}$. Moreover, $\{m_{{\overline{\tau}}}\}_{{\overline{\tau}}\in\Sigma(\sigma)}$ is a set of defining vectors of $\Psi(\sigma)$.