$\varphi_n \to \varphi$ with $\varphi \in \sK$. Thus $$\{h_n,\varphi_n\} \to \{h,\varphi\},  \quad
  \{(I+iB)\varphi_n, h_n'\} \to \{(I+iB)\varphi, h'\}.$$ Since $T$ and $T^*$ are closed, one concludes that $ \{h,\varphi\} \in T$ and $\{(I+iB)\varphi, h'\} \in T^*$, which implies that $\{h,h'\} \in T^*(I+iB)T$. Hence $T^*(I+iB)T$ is closed.

Now let $H$ be the maximal sectorial relation corresponding to $\st$ in . Assume that $\{h,h'\} \in H$, then for all $k \in \dom \st=\dom T$ $$\st[h,k]=(h',k) \quad  \mbox{or} \quad ( (I+iB)Th,Tk)= (h',k),$$ which implies that $$\{(I+iB)Th, h'\} \in T^* \quad \mbox{or} \quad \{h,h'\} \in T^*(I+iB)T.$$ Consequently, it follows that $H \subset T^*(I+iB)T$. Since $T^*(I+iB)T$ is sectorial and $H$ is maximal sectorial, it follows that $H=T^*(I+iB)T$. In particular, one sees that the closed relation $T^*(I+iB)T$ is maximal sectorial.

With the closed linear operator $T$ from $\sH$ to $\sK$ and the selfadjoint operator $B \in \bB(\sK)$, consider the following matrix decomposition of $B$ $$\label{Bdec1+}
 B=\begin{pmatrix} B_{aa}& B_{ab}\\ B_{ba}^*& B_{bb}\end{pmatrix}: \,
 \begin{pmatrix} \ker T^* \\ \cran T \end{pmatrix} \to
 \begin{pmatrix} \ker T^* \\ \cran T \end{pmatrix}.$$ Then it is clear that $$\label{Bbb}
\st[h,k]=((I+iB)T h ,T k)=((I+iB_{bb})T h ,T k), \quad h,h \in \dom \st=\dom T,$$ which shows that only the compression of $B$ to $\cran T$ plays 