Consequences of Deficiency indices theorem (Von Neumann theory)

Question

Let $T: \operatorname{dom}(T) \rightarrow \scr H$ be a symmetric operator.

1. $T$ admits self-adjoint extensions $\iff$ $d_+ = d_-$, where $d_\pm = \dim \ker(T^\dagger \pm i \mathbb{I})$
2. If $d_+ = d_-$, $T$ admits as many self-adjoint extensions as the number of unitary operators $U_0 : \ker(T^\dagger -i \mathbb{I}) \rightarrow \ker(T^\dagger +i \mathbb{I}) $

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A consequence of this theorem is:

$ \operatorname{dom}(T^\dagger) = \overline {\operatorname{dom}(T)} \oplus \ker(T^\dagger -i \mathbb{I}) \oplus \ker(T^\dagger +i \mathbb{I})$

But I can't see the reason. The only thing I got is $\overline {\operatorname{dom}(T)}$, because $T$ is symmetric and hence $T \subset T^{\dagger \dagger} = \overline T \subset T^\dagger.$ But what about the other two? And why should they be in direct sum ($\oplus$)?

Answer 1

If $x\in\ker(T^\dagger-i I)$, then $T^\dagger x=ix$. If $x$ is also in $\operatorname{dom}(T)$, then $$ i\|x\|^2=\langle ix,x\rangle=\langle T^\dagger x,x\rangle=\langle x,Tx\rangle=\langle Tx,x\rangle. $$ As $T$ is symmetric, the rightmost term is real; it follow that $x=0$. So $$ \operatorname{dom}(T)\cap \ker(T^\dagger\pm i I)=\{0\}. $$ If $x\in \ker (T^\dagger-i I)\cap\ker(T^\dagger +iI)$, then $T^\dagger x=ix=-ix$, so $x=0$. Then $$ \overline {\operatorname{dom}(T)} \oplus \ker(T^\dagger -i \mathbb{I}) \oplus \ker(T^\dagger +i \mathbb{I}) $$ is a direct sum.

The direct sum is inside $\operatorname{dom}(T^\dagger)$ by definition. I don't immediately see the equality, but I assume it has to come from the theorem, as otherwise the extensions wouldn't be what they are.