G=\begin{pmatrix} 0& 0\\ 0& G_{bb}\end{pmatrix}: \,
 \begin{pmatrix} \ker H_{\rm r} \\ \cran H_{\rm r} \end{pmatrix} \to
 \begin{pmatrix} \ker H_{\rm r} \\ \cran H_{\rm r} \end{pmatrix},$$ such that $$\label{sss--}
 \st_H[h,k]=((I +iG) (H_{r})^{\half} h, (H_{r})^{\half} k),
 \quad
 h,k \in \dom \st_H =\dom H_{r}^{\half}.$$ Moreover, the corresponding maximal sectorial operator $H$ is given by $$H=(H_{\rm r})^{\half}(I+iG)(H_{\rm r})^{\half},$$ with $\mul H=\mul  H_{\rm r}$.

The inequality $$| (\st_{H})_{\rm i}[h,k] |^{2} \leq C \st_{\rm r}[h] \st_{\rm r}[k]
 =C\|H_{\rm r}^{\half} h\| \|H_{\rm r}^{\half} k\|, \quad h,k \in \dom,$$ shows the existence of a selfadjoint operator $G$ in $\sH \ominus \ker H$ such that $$\label{henilllaa}
 (\st_{H})_{\rm i}[h,k]=(G (H_{\rm r})^{\half} h, (H_{\rm r})^{\half} k),
 \quad h, k \in \dom (H_{\rm r})^{\half}.$$ Extend $G$ to all of $\sH$ in a trivial way, so that the same formula remains valid; see Corollary \[newnew\]. It follows from the decomposition $\st=\st_{\r} +i \st_{\I}$, cf. , and the identities and , that $$\st_{H}= (\st_{H})_{\rm r}+i (\st_{H})_{\rm i},$$ so that $$\st_{H}=[h,k]= ( (H_{r})_{\s}^{\half}h, (H_{r})_{\s}^{\half} k)
+i ( G(H_{r})_{\s}^{\half}h, (H_{r})_{\s}^{\half} k).$$ This last identity immediately gives . The rest follows from Corollary \[old\].

Now let $H$ be a maximal sectorial relation, let $H_{\r}$ be its real part, and let $(H_{r})_{\s}$ be its orthogonal operator part. Then 