Lemma \[redux\]. The assertion concerning the core holds, since the factor $C_0$ is bounded with bounded inverse. The formula shows that $$(\st_H)_r[h,k]=(C_0^{\half}T_\op h ,C_0^{\half}T_\op k),
 \quad h,k \in \dom \st = \dom T,$$ and hence $H_{\rm r}=(C_0^{1/2}T_\op)^{\times}C_0^{1/2}T_\op
=(T_\op)^{\times}C_0 T_\op$ (cf.the discussion above).

Recall that if $\{h,h'\}\in T^* (I + i B) T$, then $\{h,\varphi\}\in T$ and $\{(I + i B)\varphi,h'\}\in T^*$. The last inclusion implies the condition $(I+iB)\varphi \in \dom T^{*}\subset \cdom T^{*}$, giving rise to $\varphi_2=-i(I+i B_{22})^{-1}B_{12}^*\varphi_1$. Thus, for instance, when $B=\diag (B_{11}, B_{22})$, it follows that $\varphi_{2}=0$, so that it is immediately clear that $\gamma \leq \arctan \|B_{11}\|$, independent of $B_{22}$. Note that the following assertions are equivalent:

1.  $B=\diag (B_{11}, B_{22})$;

2.  $B_{12}=0$;

3.  $C_0=I$;

4.  $\mul T$ is invariant under $B$,

in which case $C=B_{11}$. Hence, if $\mul T$ is invariant under $B$, i.e., if any of the assertions (i)–(iv) hold, then Theorem \[s-repr\] gives the following corollary, which coincides with [@HSSW17 Theorem 5.1]. In the case where $\mul T=\{0\}$ the corollary reduces to Theorem \[s-repr0o\].

\[old\] Let $T$ be a closed linear relation from a Hilbert space $\sH$ to a Hilbert space $\sK$, let $T_\op$ be the orthogonal operator part of $T$, and let $\mul 