the subspace $\sF_{0} \subset \sH \times \sH$ is defined by $$\label{F0}
\sF_{0} = \left\{ \left\{ A_{1\s}^{\half} h , A_{2\s}^{\half} h
\right\} \, : \, h \in \dom A_{1}^{\half} \cap \dom A_{2}^{\half}
\right\}.$$ The closure of $\sF_{0}$ in $\sH \times \sH$ will be denoted by $\sF$. Define the relation $\Psi$ from $\sH$ to $\sH \times \sH$ by $$\label{s-s-qu}
 \Psi=\left\{\, \left\{h, \left\{A_{1s}^{\half} h, A_{2s}^{\half}
 h\right\}\right\} :\, h \in \dom H_{1} \cap \dom H_{2} \,\right\} \subset
 \sH \times (\sH \times \sH).$$ It follows from this definition that $$\dom \Psi=\dom H_{1} \cap \dom H_{2}, \quad
  \mul \Psi=\{0\}, \quad
 \ran \Psi=\sE_0,$$ where the space $\sE_0 \subset \sH \times \sH$ is defined by $$\label{E0}
 \sE_0  =  \left\{\, \left\{A_{1\s}^{\half}f , A_{2\s}^{\half}f\right\}:\,
        f\in \dom H_{1} \cap \dom H_{2} \,\right\}.$$ Observe that $\sE_0 \subset \sF_0$. The closure of $\sE_{0}$ in $\sH \times \sH$ will be denoted by $\sE$. Hence, $$\label{s-s-ef}
          \sE \subset \sF.$$ Comparison of and shows $$\label{Psiclos}
\Psi \subset (\Phi^*)_\s,$$ and thus the operator $\Psi$ is closable and $\Psi^{**} \subset (\Phi^*)_\s$. It follows from $\cdom \Psi^*=(\mul \Psi^{**})^\perp$ and $\mul \Psi^*=(\dom \Psi)^\perp$, that $$\cdom \Psi^*=\sH , \quad \mul \Psi^* =(\dom H_{1} \cap \dom H_{2})^\perp.$$ Next, define the relation $K $ from $\sH \times \sH$ to $\sH$ by $$\begin{aligned}
\label{s-s-opK}
