T$ be invariant under the selfadjoint operator $B \in \mathbf{B}(\sK)$, so that $B=\diag (B_{11}, B_{22})$. Then the form $\st$ defined by $$\st[h,k]=((I+iB_{11})T_{\s}h, T_{\s}k), \quad h,k \in \dom \st=\dom T,$$ is closed and sectorial with vertex at the origin and semi-angle $\gamma \leq \arctan \|B_{11}\|$. Moreover, the maximal sectorial relation $H$ corresponding to the form $\st$ is given by $$H=(T_{\s})^{\times} (I+iB_{11}) T_{\s}=T^{*}(I+iB)T.$$

In the case that $\mul T$ is not invariant under $B$, one has $C_0\neq I$, and the formulas are different: for instance, the real part $(\st_H)_r$ in Theorem \[s-repr\] is of the form $$(\st_H)_r[h,k]=(C_0^{\half}T_\op h ,C_0^{\half}T_\op k),
 \quad h,k \in \dom \st=\dom T_\op = \dom T.$$

\[examp\] Assume that $B_{11} \neq 0$ and $$B_{11}=B_{12}(I+B_{22}^2)^{-\half}B_{22}(I+B_{22}^2)^{-\half}B_{12}^*,$$ so that $C=0$. In this case the maximal sectorial relation $H=T^*(I+iB)T$ in Theorem \[s-repr\] is selfadjoint, i.e., $H=H_r$ and the associated form $\st$ is nonnegative. On the other hand, with such a choice of $B$ the operator part of $T$ determines the maximal sectorial relation $(T_{\rm s})^*(I + iB)T_{\rm s}$ with semi-angle $\arctan \|B_{11}\|>0$, while $T^*(I + i B)T$ has semi-angle $\gamma=0$.

Maximal sectorial relations and their representations
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Let $H$ be a maximal sectorial relation in $\sH$ and let the closed sectorial form $\st_H$ 