the definitions and and the identity $$(I+i B_{22})^{-1} =
(I+B_{22}^2)^{-\half}(I-iB_{22})(I+B_{22}^2)^{-\half},$$ observe that $$\begin{split}
&I+i B_{11} + B_{12}(I+i B_{22})^{-1}B_{12}^*  \\
&\hspace{1cm}  =C_0 +
 i[B_{11}-B_{12}(I+B_{22}^2)^{-\half}B_{22}(I+B_{22}^2)^{-\half}B_{12}^*]\\
&\hspace{1cm}  = C_0^{1/2}(I+i C) C_0^{1/2}.
\end{split}$$Therefore, it follows from , via the equivalence in , that $$\label{eka+}
\{(I + i B)\varphi, h'\} \in  T^* \quad \Leftrightarrow \quad
 \left\{ \begin{array}{l} \{C_0^{1/2}(I+i C) C_0^{1/2} \varphi_1,h'\} \in (T_{s})^{\times}, \\
  \varphi_2=-i(I+i B_{22})^{-1}B_{12}^*\varphi_1.
           \end{array}
           \right.$$ Combining and , one sees that $$\{h,h'\} \in (T_{\s})^{\times} C_0^{1/2}(I+i C) C_0^{1/2} T_{\s}.$$ Conversely, if this inclusion holds, then there exists $\varphi_{1} \in \cdom T^{*}$, such that $$\{h, \varphi_{1}\} \in T_{\s} \quad \mbox{and}
 \quad \{C_0^{1/2}(I+i C) C_0^{1/2} \varphi_1,h'\} \in (T_{\s})^{\times}.$$ Then define $\varphi_2=-i(I+i B_{22})^{-1}B_{12}^*\varphi_1$, so that $\varphi_{2} \in \mul T$. Furthermore, define $\varphi=\varphi_{1}+\varphi_{2}$. Hence $\{h, \varphi\} \in T$, and it follows from that $$\{h,h'\}\in T^* (I + i B) T.$$ Therefore one can rewrite $T^* (I + i B) T$ in the form .

Observe that $C_{0}^{\half}T_{\s}$ is a closed linear operator from the Hilbert space $\sH$ to the Hilbert space $\cdom T^{*}$ whose adjoint is given by $$\label{C0Tadjoint}
 ( C_0^{1/2}T_{\rm s})^{\times}=(T_{\rm s})^{\times} \, C_0^{1/2}.$$ Hence, by Theorem \[s-repr0o\] $(T_{\s})^{\times} C_0^{1/2}(I+i C) C_0^{1/2} T_{\s}$ is a maximal sectorial relation in $\sH$ and by the identity the same 