Reducing Spaces: Domain Problem

Given a Hilbert space $\mathcal{H}$.

Consider normal operators:
$$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$

Denote for readability:
$$\mathcal{D}:=\mathcal{D}(N)=\mathcal{D}(N^*)$$

Regard a closed space:
$$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^\perp$$

Suppose invariance:
$$N(\mathcal{S}\cap\mathcal{D})\subseteq\mathcal{S}\quad N(\mathcal{S}^\perp\cap\mathcal{D})\subseteq\mathcal{S}^\perp$$


  Then it may still happen:
  $$\mathcal{D}\supsetneq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$


Does someone have a nonexample?

Reference

For a preexample: Preliminary
If $PT = TP$, then for any $x\in Z$
$$
Tx = TP(x) = PT(x) \in Z
$$
and so $T(Z) \subset Z$. Now since
$$
(I-P)T = T-TP = T-PT = T(I-P)
$$
it follows that $T(Z^{\perp}) \subset Z^{\perp}$.

Conversely, if $Z$ is reducing, then for any $x\in H$, write $x = Px + (I-P)x$, then
$$
Tx = P(Tx) + (I-P)Tx = T(Px) + T(I-P)x
$$
and so
$$
P(Tx) - T(Px) = T(I-P)x - (I-P)Tx \in Z\cap Z^{\perp} = \{0\}
$$
and so $TP = PT$.
If $PT = TP$, then for any $x\in Z$
$$
Tx = TP(x) = PT(x) \in Z
$$
and so $T(Z) \subset Z$. Now since
$$
(I-P)T = T-TP = T-PT = T(I-P)
$$
it follows that $T(Z^{\perp}) \subset Z^{\perp}$.

Conversely, if $Z$ is reducing, then for any $x\in H$, write $x = Px + (I-P)x$, then
$$
Tx = P(Tx) + (I-P)Tx = T(Px) + T(I-P)x
$$
and so
$$
P(Tx) - T(Px) = T(I-P)x - (I-P)Tx \in Z\cap Z^{\perp} = \{0\}
$$
and so $TP = PT$.