Unitary deficiency parameterization at a reference scale

The graph of $A$ is the image of the graph of $S$ under the bounded invertible map $(f,g)\mapsto (f,E_{\ast }^{-1}g)$ on $H\oplus H$, so it is closed; real scaling preserves symmetry. If $g\in D(S^{\ast })$, then for every $f\in D(S)$, $⟨Af,g⟩=E_{\ast }^{-1}⟨Sf,g⟩=⟨f,E_{\ast }^{-1}{S}^{\ast }g⟩$, so $g\in D(A^{\ast })$ and $A^{\ast }g=E_{\ast }^{-1}{S}^{\ast }g$. Conversely, if $g\in D(A^{\ast })$, then $⟨Sf,g⟩=E_{\ast }⟨Af,g⟩=⟨f,E_{\ast }{A}^{\ast }g⟩$, so $g\in D(S^{\ast })$ and $S^{\ast }g=E_{\ast }{A}^{\ast }g$. Therefore $D(A^{\ast })=D(S^{\ast })$ and $A^{\ast }=E_{\ast }^{-1}{S}^{\ast }$. Since $A^{\ast }\mp iI=E_{\ast }^{-1}(S^{\ast }\mp i{E}_{\ast }I)$, the ordinary deficiency spaces of $A$ at $\pm i$ are exactly $N_{\pm }$.

Suppose $n_{+}=n_{-}$. Choose orthonormal bases $(e_j{)}_{j\in J_{+}}$ of $N_{+}$ and $(f_j{)}_{j\in J_{-}}$ of $N_{-}$. By [F6], choose a bijection $\pi :J_{+}\to J_{-}$. The map $V_0{e}_j=f_{\pi (j)}$ preserves norms on finite linear combinations by orthonormality, so it extends continuously to an isometry $V:N_{+}\to N_{-}$. Its range contains the finite span of the $f_j$, which is dense; the range of an isometry from the complete space $N_{+}$ is closed, so $V$ is onto and unitary.

For $f\in D(A)$, symmetry makes $⟨Af,f⟩$ real; expanding the two norms gives $\mathrm{\parallel }(A\pm iI)f{\mathrm{\parallel }}^2=\mathrm{\parallel }Af{\mathrm{\parallel }}^2+\mathrm{\parallel }f{\mathrm{\parallel }}^2$. Thus $A\pm iI$ are injective. If $(A\pm iI)f_k$ converges, the norm identity applied to differences makes $(f_k)$ Cauchy and then $(A{f}_k)$ Cauchy; closedness of $A$ shows the limit is in the range. Hence $M_{\pm }:=\mathrm{ran}(A\pm iI)$ are closed. By [F2] and [F3], $H=M_{+}\oplus N_{+}=M_{-}\oplus N_{-}$, with orthogonal sums.

Equip $D(A^{\ast })$ with the graph inner product $⟨f,g{⟩}_{A^{\ast }}=⟨f,g⟩+⟨A^{\ast }f,A^{\ast }g⟩$. It is complete by [F2, F4], and $D(A)$ is a closed subspace because $A$ is closed and $A^{\ast }f=Af$ there. If $z\in D(A^{\ast })$ is graph-orthogonal to $D(A)$, then for every $f\in D(A)$, $⟨Af,A^{\ast }z⟩=-⟨f,z⟩$. By the adjoint definition, this means $A^{\ast }z\in D(A^{\ast })$ and $A^{\ast 2}z=-z$. Therefore $z_{+}=(z-i{A}^{\ast }z)\mathrm{/}2$ and $z_{-}=(z+i{A}^{\ast }z)\mathrm{/}2$ satisfy $A^{\ast }{z}_{+}=i{z}_{+}$ and $A^{\ast }{z}_{-}=-i{z}_{-}$, and $z=z_{+}+z_{-}$. Conversely, $D(A)$ is graph-orthogonal to both kernels: for $u\in N_{+}$, $⟨f,u⟩+⟨Af,A^{\ast }u⟩=⟨f,u⟩+⟨Af,iu⟩=0$ by the adjoint identity and conjugate-linearity in the second argument, and the same computation with $A^{\ast }u=-iu$ holds for $u\in N_{-}$. If $u\in N_{+}$ and $v\in N_{-}$, then $⟨u,v⟩+⟨A^{\ast }u,A^{\ast }v⟩=⟨u,v⟩+⟨iu,-iv⟩=0$. Thus the two kernels are graph-orthogonal to each other. Applying [F3] in the graph Hilbert space proves the orthogonal decomposition $D(A^{\ast })=D(A)\oplus N_{+}\oplus N_{-}$.

Define $C_A:M_{+}\to M_{-}$ by $C_A((A+iI)f)=(A-iI)f$. Injectivity of $A+iI$ makes this well defined, the norm identity in step 2.1 makes it an isometry, and it is onto $M_{-}$. Define $W(m+u)=C_{A}m-Vu$ for $m\in M_{+}$ and $u\in N_{+}$. The decompositions from step 2.1 make $W$ well defined; it is unitary because its two orthogonal summands are mapped isometrically onto $M_{-}$ and $N_{-}$. For $f\in D(A)$, $(I-W)(A+iI)f=2if$, so $\mathrm{ran}(I-W)$ is dense. If $Wz=z$, then $z$ is orthogonal to $\mathrm{ran}(I-W)$ by unitarity, hence to the dense domain $D(A)$, so $z=0$. The Cayley correspondence [F5] therefore gives a self-adjoint operator $A_V$ with Cayley transform $W$.

For $m=(A+iI)f$ and $u\in N_{+}$, direct calculation gives $(I-W)(m+u)=2if+u+Vu$ and $i(I+W)(m+u)=2iAf+iu-iVu$. The Cayley inverse is $A_V=i(I+W)(I-W{)}^{-1}$ on $\mathrm{ran}(I-W)$, so $D(A_V)=\mathrm{ran}(I-W)=D(A)+\{u+Vu:u\in N_{+}\}$ and $A_V(2if+u+Vu)=2iAf+iu-iVu$. Equivalently, after renaming $2if$ as $f$, $A_V(f+u+Vu)=Af+iu-iVu$. The sum is direct: if $f=u+Vu\in D(A)$, then $(I-W)(A+iI)f=2if=(I-W)(2iu)$; injectivity of $I-W$ gives $(A+iI)f=2iu$. Its left side lies in $M_{+}$ and its right side in $N_{+}$, whose intersection is zero, so $u=f=0$. Multiplication by $E_{\ast }$ gives the displayed domain and action for $S_V=E_{\ast }{A}_V$. The two-way scalar-adjoint calculation in step 1.1 applies to $A_V$ as well, so $S_V^{\ast }=E_{\ast }{A}_V^{\ast }=S_V$. Thus $S_V$ is a self-adjoint extension of $S$.

Conversely, let $T$ be any self-adjoint extension of $S$ and put $R=E_{\ast }^{-1}T$. The two-way scalar-adjoint calculation in step 1.1 gives $R^{\ast }=E_{\ast }^{-1}{T}^{\ast }=R$, so $R$ is a self-adjoint extension of $A$. Its Cayley transform $C_R$ is unitary by [F5] and agrees with $C_A$ on $M_{+}$, since $R$ equals $A$ on $D(A)$. A unitary carrying $M_{+}$ onto $M_{-}$ carries their orthogonal complements $N_{+}$ onto $N_{-}$ by step 2.1 and [F2]; hence $C_R(N_{+})=N_{-}$. Put $V=-C_R{\mathrm{\mid }}_{N_{+}}$; this is unitary. The unitary $W$ constructed from this $V$ agrees with $C_R$ on both orthogonal summands of $H$, so uniqueness in [F5] gives $R=A_V$. This determines $V$ uniquely from $T$ and proves that every self-adjoint extension of $S$ is uniquely $S_V$.

If $n_{+}=n_{-}$, step 1.2 supplies a unitary and step 4.1 supplies a self-adjoint extension. If an extension exists, step 4.2 supplies a unitary $N_{+}\to N_{-}$, so its orthonormal bases have bijective index sets and $n_{+}=n_{-}$. This proves the existence equivalence and the complete parameterization.