observe that then the subspace $\mul T$ is closed. The adjoint $T^{*}$ of $T$ is the set of all $\{h,h'\} \in \sK \times \sH$ for which $$(h',f)=(h,f') \quad \mbox{for all} \quad \{f,f'\} \in T.$$ Hence, the definition of $T^{*}$ depends on the Hilbert spaces $\sH$ and $\sK$ in which $T$ is assumed to act. Let $\sK$ have the orthogonal decomposition $$\label{hilbert}
\sK=\cdom T^* \oplus \mul T,$$ and let $P$ be the orthogonal projection onto $\cdom T^*$. Observe that $PT \subset T$, since $\{0\} \times \mul T \subset T$. Therefore $T^{*} \subset (PT)^{*}=T^*P$, where the last equality holds since $P \in \mathbf{B}(\sK)$. Then one has $$\label{hilbert1}
 (PT)^*=T^* \operatorname{\, \widehat \oplus \,}\, (\mul T \times \{0\}).$$ The *orthogonal operator part* $T_{\rm s}$ of $T$ is defined as $T_{\rm s}=PT$. Hence $T_\op$ is an operator from the Hilbert space $\sH$ to the Hilbert space $\sK$ and $T_\op \subset T$. Note that $\ran T_\op \subset \cdom T^*=\sK \ominus \mul T$. Thus one may interpret $T_\op$ as an operator from the Hilbert space $\sH$ to the Hilbert space $\cdom T^*$ and one may also consider the adjoint $(T_\op)^\times$ of $T_{\s}$ with respect to these spaces. It is not difficult to see the connection between these 