Clarifying the definition of essential self-adjointness If a Hilbert space operator $T$ has a unique self-adjoint extension, must the extension be the closure of $T$?
 A: No. Consider the operator $T:\mathbb C^2\to \mathbb C^2,\ T=\left(\matrix{0 & \ast\\ 1 & \ast}\right).$ That is $D(T)=\mathbb C e_1$ and $Te_1=e_2.$ The operator is closed and the unique self-adjoint extension is $\left(\matrix{0 & 1\\ 1 & 0}\right).$
A: If $T : \mathcal{D}(T)\subseteq X\rightarrow X$ is densely-defined on a Hilbert Space $X$, and has a selfadjoint extension $A$, then $T$ must be symmetric on its domain with $T \preceq A= A^{\star} \preceq T^{\star}$, where $A \preceq B$ means non-strict inclusion of graphs. If the closure $T^{c}$ of $T$ (which exists because $T$ is symmetric) is not equal to $T^{\star}$, then one has the non-trivial direct sum domain decomposition:
$$
    \mathcal{D}(T^{\star})=\mathcal{D}(T^{c})\oplus\mathcal{N}(T^{\star}+iI)\oplus\mathcal{N}(T^{\star}-iI).
$$
So the reason that $T^{\star} \ne T^{c}$ for a densely-defined symmetric operator $T$ is because $T^{\star}$ has complex eigenvalues, something which a selfadjoint operator cannot have.
Assuming that $T^{c}$ has a selfadjoint extension $A$, then the Cayley map $(A+iI)(A-iI)^{-1}$ is a unitary map of $\mathcal{N}(T^{\star}+iI)$ onto $\mathcal{N}(T^{\star}-iI)$ because, for $x \in \mathcal{D}(T^{\star})$,
$$
   (T^{\star}-iI)(A+iI)(A-iI)^{-1}x=(T^{\star}-iI)\left\{I+2i(A-iI)^{-1}\right\}x=(T+iI)x.
$$
The selfadjoint extensions of $T^{c}$ are in one-to-one correspondence with such unitary maps; it is a bit tedious, but one can start with a unitary $V :\mathcal{N}(T^{\star}+iI)\rightarrow\mathcal{N}(T^{\star}-iI)$ and produce a unique selfadjoint extension $A$ of $T$ with $(A+iI)(A-iI)^{-1}=V$ on $\mathcal{N}(T^{\star}+iI)$. If $T^{c}\ne T^{\star}$ and $T^{c}$ has a selfadjoint extension $A$, then $T^{c}$ always has more than one. Indeed, if there is one unitary map between $\mathcal{N}(T^{\star}+iI)$ and $\mathcal{N}(T^{\star}-iI)$, then there has to be more than one such unitary map, even if the spaces in question are one-dimensional. So a densely-defined symmetric linear operator $T$ on a Hilbert space has only one selfadjoint extension iff the closure of $T$ is $T^{\star}$.
