[proofplan] We prove the theorem by working inside the adjoint domain $\mathcal{D}(A^*)$ with the graph [inner product](/page/Inner%20Product). The first step is the von Neumann decomposition: every vector in $\mathcal{D}(A^*)$ splits uniquely into a vector from $\mathcal{D}(A)$, a vector from $N_+(A)$, and a vector from $N_-(A)$. The obstruction to self-adjointness is then entirely encoded in the boundary form on $N_+(A)\oplus N_-(A)$, whose self-orthogonal neutral subspaces are exactly graphs of unitary maps $N_+(A)\to N_-(A)$. This gives both the existence criterion $n_+(A)=n_-(A)$ and the stated parametrisation. [/proofplan]

[step:Equip the adjoint domain with the graph Hilbert structure] Since $A$ is densely defined, its adjoint $A^*$ is well-defined. Since $A$ is closed and symmetric, $A \subset A^*$ and $\mathcal{D}(A)$ is a closed subspace of $\mathcal{D}(A^*)$ for the graph norm of $A^*$. Define the graph inner product on $\mathcal{D}(A^*)$ by \begin{align*} (x,y)_{A^*}:=(x,y)_H+(A^*x,A^*y)_H. \end{align*} Because $A^*$ is closed, $\mathcal{D}(A^*)$ is a [Hilbert space](/page/Hilbert%20Space) with this inner product. For later use define the boundary form \begin{align*} \omega:\mathcal{D}(A^*)\times \mathcal{D}(A^*)\to \mathbb{C} \end{align*} by \begin{align*} \omega(x,y):=(A^*x,y)_H-(x,A^*y)_H. \end{align*} The form $\omega$ vanishes whenever one argument lies in $\mathcal{D}(A)$, because $A$ is symmetric and $A^*|_{\mathcal{D}(A)}=A$. [/step]

[step:Decompose $\mathcal{D}(A^*)$ into $\mathcal{D}(A)$ and the two deficiency spaces]We claim that \begin{align*} \mathcal{D}(A^*)=\mathcal{D}(A)\oplus N_+(A)\oplus N_-(A) \end{align*} as an orthogonal direct sum for the graph inner product. First, $N_+(A)$ and $N_-(A)$ are closed subspaces of $\mathcal{D}(A^*)$, since they are kernels of the continuous maps $A^*-iI:\mathcal{D}(A^*)\to H$ and $A^*+iI:\mathcal{D}(A^*)\to H$ with respect to the graph norm. If $x\in \mathcal{D}(A)$ and $u\in N_+(A)$, then $A^*u=iu$ and \begin{align*} (x,u)_{A^*}=(x,u)_H+(Ax,iu)_H. \end{align*} Since the Hilbert inner product is linear in the first variable, \begin{align*} (Ax,iu)_H=-i(Ax,u)_H. \end{align*} Because $u\in \mathcal{D}(A^*)$ and $x\in \mathcal{D}(A)$, \begin{align*} (Ax,u)_H=(x,A^*u)_H=(x,iu)_H=-i(x,u)_H. \end{align*} Therefore \begin{align*} (x,u)_{A^*}=(x,u)_H+(-i)(-i)(x,u)_H=0. \end{align*} The same computation with $A^*v=-iv$ gives $(x,v)_{A^*}=0$ for every $v\in N_-(A)$. If $u\in N_+(A)$ and $v\in N_-(A)$, then \begin{align*} (u,v)_{A^*}=(u,v)_H+(iu,-iv)_H. \end{align*} Since the second slot is conjugate-linear, \begin{align*} (iu,-iv)_H=i\,i\,(u,v)_H=-(u,v)_H. \end{align*} Hence $(u,v)_{A^*}=0$. It remains to see that the orthogonal complement of $\mathcal{D}(A)\oplus N_+(A)\oplus N_-(A)$ in $\mathcal{D}(A^*)$ is zero. Let $z\in \mathcal{D}(A^*)$ be graph-orthogonal to all three subspaces. Orthogonality to $N_+(A)$ and $N_-(A)$ means that $z$ is orthogonal to the kernels of $A^*-iI$ and $A^*+iI$ in the graph Hilbert space. Orthogonality to $\mathcal{D}(A)$ gives \begin{align*} (z,x)_H+(A^*z,Ax)_H=0 \end{align*} for every $x\in \mathcal{D}(A)$. Define $h:=A^*z\in H$. The last identity says \begin{align*} (Ax,h)_H=-(x,z)_H \end{align*} for every $x\in \mathcal{D}(A)$. Thus $h\in \mathcal{D}(A^*)$ and $A^*h=-z$. Since $h=A^*z$, we have $z\in\mathcal{D}((A^*)^2)$ and $(A^*)^2z=-z$. Define \begin{align*} u:=\frac{1}{2}(z-iA^*z) \end{align*} and \begin{align*} v:=\frac{1}{2}(z+iA^*z). \end{align*} Then \begin{align*} A^*u=\frac{1}{2}(A^*z-i(A^*)^2z)=\frac{1}{2}(A^*z+iz)=iu \end{align*} and \begin{align*} A^*v=\frac{1}{2}(A^*z+i(A^*)^2z)=\frac{1}{2}(A^*z-iz)=-iv. \end{align*} Hence $u\in N_+(A)$, $v\in N_-(A)$, and $z=u+v$. Since $z$ is graph-orthogonal to both deficiency spaces, \begin{align*} (z,z)_{A^*}=(z,u)_{A^*}+(z,v)_{A^*}=0. \end{align*} Therefore $z=0$.[/step]

[guided]The point of this step is to isolate the part of $\mathcal{D}(A^*)$ not already controlled by $A$. We use the graph inner product because closedness of $A^*$ makes $\mathcal{D}(A^*)$ into a Hilbert space, and Hilbert-space orthogonal decompositions are then available. Define \begin{align*} (x,y)_{A^*}:=(x,y)_H+(A^*x,A^*y)_H. \end{align*} The deficiency spaces are \begin{align*} N_+(A)=\ker(A^*-iI) \end{align*} and \begin{align*} N_-(A)=\ker(A^*+iI). \end{align*} They are closed in the graph norm because $A^*-iI$ and $A^*+iI$ are continuous maps from the graph Hilbert space $\mathcal{D}(A^*)$ into $H$. We next check the orthogonality relations. Let $x\in\mathcal{D}(A)$ and $u\in N_+(A)$. Since $A^*u=iu$, \begin{align*} (x,u)_{A^*}=(x,u)_H+(Ax,iu)_H. \end{align*} The inner product is linear in its first argument and conjugate-linear in its second argument, so \begin{align*} (Ax,iu)_H=-i(Ax,u)_H. \end{align*} Because $u\in\mathcal{D}(A^*)$, the defining property of the adjoint gives \begin{align*} (Ax,u)_H=(x,A^*u)_H=(x,iu)_H=-i(x,u)_H. \end{align*} Substitution yields \begin{align*} (x,u)_{A^*}=(x,u)_H+(-i)(-i)(x,u)_H=0. \end{align*} The computation for $v\in N_-(A)$ is identical with $A^*v=-iv$, and gives $(x,v)_{A^*}=0$. Now take $u\in N_+(A)$ and $v\in N_-(A)$. Then \begin{align*} (u,v)_{A^*}=(u,v)_H+(iu,-iv)_H. \end{align*} Since the second slot is conjugate-linear, \begin{align*} (iu,-iv)_H=i\,i\,(u,v)_H=-(u,v)_H. \end{align*} Thus $(u,v)_{A^*}=0$. We have proved pairwise graph-orthogonality of $\mathcal{D}(A)$, $N_+(A)$, and $N_-(A)$. It remains to prove that these three orthogonal pieces exhaust $\mathcal{D}(A^*)$. Let $z\in\mathcal{D}(A^*)$ be graph-orthogonal to their sum. In particular, $z$ is graph-orthogonal to $\mathcal{D}(A)$, so for every $x\in\mathcal{D}(A)$, \begin{align*} 0=(z,x)_{A^*}=(z,x)_H+(A^*z,Ax)_H. \end{align*} Set $h:=A^*z\in H$. Since the inner product is conjugate-linear in the second variable, the displayed identity is equivalent to \begin{align*} (Ax,h)_H=(x,-z)_H \end{align*} for all $x\in\mathcal{D}(A)$. This is exactly the defining condition for $h\in\mathcal{D}(A^*)$ with $A^*h=-z$. Since $h=A^*z$, this means $z\in\mathcal{D}((A^*)^2)$ and \begin{align*} (A^*)^2z=-z. \end{align*} Now define two vectors in $\mathcal{D}(A^*)$ by \begin{align*} u:=\frac{1}{2}(z-iA^*z) \end{align*} and \begin{align*} v:=\frac{1}{2}(z+iA^*z). \end{align*} The reason for these definitions is that they diagonalise the equation $(A^*)^2z=-z$ into the two eigenspaces of $A^*$ with eigenvalues $i$ and $-i$. Indeed, \begin{align*} A^*u=\frac{1}{2}(A^*z-i(A^*)^2z)=\frac{1}{2}(A^*z+iz)=iu \end{align*} and \begin{align*} A^*v=\frac{1}{2}(A^*z+i(A^*)^2z)=\frac{1}{2}(A^*z-iz)=-iv. \end{align*} Hence $u\in N_+(A)$ and $v\in N_-(A)$. Also $z=u+v$. Since $z$ was assumed graph-orthogonal to both $N_+(A)$ and $N_-(A)$, we get \begin{align*} (z,z)_{A^*}=(z,u)_{A^*}+(z,v)_{A^*}=0. \end{align*} The graph inner product is positive definite, so $z=0$. This proves that the three pairwise orthogonal subspaces exhaust $\mathcal{D}(A^*)$.[/guided]

[step:Compute the boundary form on the deficiency summand] By the decomposition just proved, each $y\in \mathcal{D}(A^*)$ has a unique representation \begin{align*} y=x+u+v \end{align*} with $x\in\mathcal{D}(A)$, $u\in N_+(A)$, and $v\in N_-(A)$. Since $\omega$ vanishes whenever one argument belongs to $\mathcal{D}(A)$, it suffices to compute $\omega$ on $N_+(A)\oplus N_-(A)$. For $u_1,u_2\in N_+(A)$, \begin{align*} \omega(u_1,u_2)=(iu_1,u_2)_H-(u_1,iu_2)_H=2i(u_1,u_2)_H. \end{align*} For $v_1,v_2\in N_-(A)$, \begin{align*} \omega(v_1,v_2)=(-iv_1,v_2)_H-(v_1,-iv_2)_H=-2i(v_1,v_2)_H. \end{align*} For $u\in N_+(A)$ and $v\in N_-(A)$, \begin{align*} \omega(u,v)=(iu,v)_H-(u,-iv)_H=0. \end{align*} Thus, for $u_1,u_2\in N_+(A)$ and $v_1,v_2\in N_-(A)$, \begin{align*} \omega(u_1+v_1,u_2+v_2)=2i\big((u_1,u_2)_H-(v_1,v_2)_H\big). \end{align*} [/step]

[step:Identify symmetric extensions with neutral boundary data]Let $B:\mathcal{D}(B)\subset H\to H$ be an operator satisfying $A\subset B\subset A^*$. Since $A$ is densely defined and $A\subset B$, the operator $B$ is densely defined. Set \begin{align*} K:=N_+(A)\oplus N_-(A) \end{align*} inside the graph Hilbert space $\mathcal{D}(A^*)$, and let \begin{align*} \Pi_K:\mathcal{D}(A^*)\to K \end{align*} denote the graph-[orthogonal projection](/theorems/437) from the von Neumann decomposition. Define the boundary subspace of $B$ by \begin{align*} L_B:=\Pi_K(\mathcal{D}(B))\subset K. \end{align*} Then \begin{align*} \mathcal{D}(B)=\mathcal{D}(A)\oplus L_B \end{align*} as an algebraic direct sum. Indeed, if $y\in\mathcal{D}(B)$ and $y=x+k$ is its von Neumann decomposition with $x\in\mathcal{D}(A)$ and $k\in K$, then $k\in L_B$. Conversely, if $k\in L_B$, choose $y_0=x_0+k\in\mathcal{D}(B)$ with $x_0\in\mathcal{D}(A)$. For every $x\in\mathcal{D}(A)$, the vector $x+k=(x-x_0)+y_0$ belongs to $\mathcal{D}(B)$ because $x-x_0\in\mathcal{D}(A)\subset\mathcal{D}(B)$. The sum is direct because $\mathcal{D}(A)\cap K=\{0\}$ in the graph-[orthogonal decomposition](/theorems/436). The operator $B$ is symmetric if and only if $\omega$ vanishes on $\mathcal{D}(B)\times\mathcal{D}(B)$. Since $\omega$ vanishes whenever one argument lies in $\mathcal{D}(A)$, this is equivalent to \begin{align*} \omega(\ell_1,\ell_2)=0 \end{align*} for all $\ell_1,\ell_2\in L_B$. Thus symmetric extensions of $A$ inside $A^*$ correspond to neutral boundary subspaces $L_B\subset K$ obtained as graph-orthogonal projections of their domains. For a subspace $L\subset K$, define its boundary annihilator by \begin{align*} L^{\omega}:=\{m\in K:\omega(m,\ell)=0\text{ for every }\ell\in L\}. \end{align*} We now compute the adjoint domain. Since $A\subset B$, the order reversal property of adjoints gives $B^*\subset A^*$, so every $y\in\mathcal{D}(B^*)$ belongs to $\mathcal{D}(A^*)$. Write $y=x+m$ with $x\in\mathcal{D}(A)$ and $m\in K$. Every $z\in\mathcal{D}(B)$ has the form $z=x_0+\ell$ with $x_0\in\mathcal{D}(A)$ and $\ell\in L_B$. All terms involving $\mathcal{D}(A)$ vanish in $\omega$, hence \begin{align*} (A^*y,z)_H-(y,Bz)_H=\omega(y,z)=\omega(m,\ell). \end{align*} Therefore $y\in\mathcal{D}(B^*)$ exactly when $m\in L_B^{\omega}$, and \begin{align*} \mathcal{D}(B^*)=\mathcal{D}(A)\oplus L_B^{\omega}. \end{align*} Consequently $B$ is self-adjoint if and only if $L_B=L_B^{\omega}$.[/step]

[guided]The subtle point is that the boundary part of an intermediate domain is not its literal intersection with $K$. A domain may contain $x+k$ without containing $k$ separately. The correct boundary object is therefore the projection of the domain onto the deficiency summand. Define \begin{align*} K:=N_+(A)\oplus N_-(A), \end{align*} and let \begin{align*} \Pi_K:\mathcal{D}(A^*)\to K \end{align*} be the graph-orthogonal projection furnished by the von Neumann decomposition. For an operator $B:\mathcal{D}(B)\subset H\to H$ with $A\subset B\subset A^*$, define \begin{align*} L_B:=\Pi_K(\mathcal{D}(B))\subset K. \end{align*} We prove the domain formula. Take $y\in\mathcal{D}(B)$. Its von Neumann decomposition is $y=x+k$ with $x\in\mathcal{D}(A)$ and $k\in K$, and by definition $k=\Pi_K y\in L_B$. This proves $\mathcal{D}(B)\subset\mathcal{D}(A)+L_B$. Conversely, take $k\in L_B$. By definition of $L_B$, there is some $y_0\in\mathcal{D}(B)$ with von Neumann decomposition $y_0=x_0+k$, where $x_0\in\mathcal{D}(A)$. For any $x\in\mathcal{D}(A)$, we have \begin{align*} x+k=(x-x_0)+y_0. \end{align*} Since $x-x_0\in\mathcal{D}(A)\subset\mathcal{D}(B)$ and $y_0\in\mathcal{D}(B)$, linearity of $\mathcal{D}(B)$ gives $x+k\in\mathcal{D}(B)$. Hence \begin{align*} \mathcal{D}(B)=\mathcal{D}(A)\oplus L_B, \end{align*} where the sum is direct because $\mathcal{D}(A)\cap K=\{0\}$. Next, $B$ is symmetric exactly when \begin{align*} (Bp,q)_H-(p,Bq)_H=0 \end{align*} for all $p,q\in\mathcal{D}(B)$. Since $B$ is the restriction of $A^*$ to $\mathcal{D}(B)$, this condition is the same as $\omega(p,q)=0$ for all $p,q\in\mathcal{D}(B)$. Writing $p=x_1+\ell_1$ and $q=x_2+\ell_2$ with $x_1,x_2\in\mathcal{D}(A)$ and $\ell_1,\ell_2\in L_B$, all terms containing $x_1$ or $x_2$ vanish because $\omega$ vanishes when either argument lies in $\mathcal{D}(A)$. Therefore $B$ is symmetric exactly when \begin{align*} \omega(\ell_1,\ell_2)=0 \end{align*} for all $\ell_1,\ell_2\in L_B$. Now define the boundary annihilator of a subspace $L\subset K$ by \begin{align*} L^{\omega}:=\{m\in K:\omega(m,\ell)=0\text{ for every }\ell\in L\}. \end{align*} We compute the adjoint domain. Since $A\subset B$, adjoints reverse inclusions, so $B^*\subset A^*$. Hence every $y\in\mathcal{D}(B^*)$ has a von Neumann decomposition $y=x+m$ with $x\in\mathcal{D}(A)$ and $m\in K$. Every $z\in\mathcal{D}(B)$ has the form $z=x_0+\ell$ with $x_0\in\mathcal{D}(A)$ and $\ell\in L_B$. The defining condition for $y\in\mathcal{D}(B^*)$ is equivalent, because $B\subset A^*$, to \begin{align*} (A^*y,z)_H-(y,Bz)_H=0 \end{align*} for every $z\in\mathcal{D}(B)$. The left-hand side is $\omega(y,z)$, and the $\mathcal{D}(A)$ components vanish from the boundary form. Thus the condition reduces exactly to \begin{align*} \omega(m,\ell)=0 \end{align*} for every $\ell\in L_B$, namely $m\in L_B^{\omega}$. Therefore \begin{align*} \mathcal{D}(B^*)=\mathcal{D}(A)\oplus L_B^{\omega}. \end{align*} It follows that $B$ is self-adjoint precisely when its boundary subspace satisfies $L_B=L_B^{\omega}$.[/guided]

[step:Classify self-orthogonal boundary subspaces as graphs of unitary maps]Let $L\subset N_+(A)\oplus N_-(A)$ be a subspace satisfying $L=L^{\omega}$. Define the coordinate projections \begin{align*} P_+:N_+(A)\oplus N_-(A)\to N_+(A) \end{align*} and \begin{align*} P_-:N_+(A)\oplus N_-(A)\to N_-(A) \end{align*} by $P_+(u+v)=u$ and $P_-(u+v)=v$. If $u+v\in L$, neutrality gives \begin{align*} 0=\omega(u+v,u+v)=2i\big(\|u\|_H^2-\|v\|_H^2\big). \end{align*} Hence $\|u\|_H=\|v\|_H$. In particular, if $P_+(u+v)=0$, then $u=0$, so $v=0$, and $u+v=0$. Therefore $P_+|_L$ is injective. The same argument shows that $P_-|_L$ is injective. The condition $L=L^{\omega}$ implies that $L$ is neutral, since $L\subset L^{\omega}$. Let $M:=P_+(L)\subset N_+(A)$. Since $P_+|_L$ is injective, there is a unique [linear map](/page/Linear%20Map) \begin{align*} V:M\to N_-(A) \end{align*} such that \begin{align*} L=\{u+Vu:u\in M\}. \end{align*} The equality $\|u\|_H=\|Vu\|_H$ for $u\in M$ shows that $V$ is an isometry on $M$. We first prove that $M$ is closed. We equip $K=N_+(A)\oplus N_-(A)$ with the Hilbert norm inherited from the graph inner product on $\mathcal{D}(A^*)$. On each deficiency space this graph norm is $\sqrt{2}$ times the $H$-norm, so the formula \begin{align*} \omega(u_1+v_1,u_2+v_2)=2i\big((u_1,u_2)_H-(v_1,v_2)_H\big) \end{align*} shows that $m\mapsto\omega(m,\ell)$ is continuous on $K$ for each fixed $\ell\in K$. Since $L=L^{\omega}$ and $L^{\omega}$ is the intersection of the closed kernels of these continuous linear functionals, the subspace $L$ is closed in $K$. Let $(u_k)_{k=1}^{\infty}$ be a sequence in $M$ with $u_k\to u$ in $N_+(A)$. Since $V$ is an isometry, $(Vu_k)_{k=1}^{\infty}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in the Hilbert space $N_-(A)$, hence converges to some $v\in N_-(A)$. Therefore $u_k+Vu_k\to u+v$ in $K$, and closedness of $L$ gives $u+v\in L$. Hence $u\in M$, so $M$ is closed. We next prove that $M=N_+(A)$. If $a\in M^\perp\subset N_+(A)$, then for every $u+v\in L$ we have \begin{align*} \omega(a,u+v)=2i(a,u)_H=0. \end{align*} Thus $a\in L^{\omega}=L$. Since $L$ is neutral, \begin{align*} 0=\omega(a,a)=2i\|a\|_H^2, \end{align*} so $a=0$. Because $M$ is closed and $M^\perp=\{0\}$, we get $M=N_+(A)$. We also prove that $P_-(L)=N_-(A)$. Let $M_-:=P_-(L)\subset N_-(A)$. The same closedness argument used for $M$ shows that $M_-$ is closed. If $b\in M_-^\perp\subset N_-(A)$, then for every $u+v\in L$ we have \begin{align*} \omega(b,u+v)=-2i(b,v)_H=0. \end{align*} Thus $b\in L^{\omega}=L$. Since $L$ is neutral, \begin{align*} 0=\omega(b,b)=-2i\|b\|_H^2, \end{align*} so $b=0$. Because $M_-$ is closed and $M_-^\perp=\{0\}$, we get $P_-(L)=N_-(A)$. Therefore $L$ determines a map \begin{align*} U:N_+(A)\to N_-(A) \end{align*} by requiring $u+Uu\in L$. The preceding injectivity and surjectivity show that $U$ is well-defined and onto, and the equality $\|u\|_H=\|Uu\|_H$ shows that $U$ is unitary. Hence \begin{align*} L=\{u+Uu:u\in N_+(A)\}. \end{align*} Conversely, if $U:N_+(A)\to N_-(A)$ is unitary and \begin{align*} L_U:=\{u+Uu:u\in N_+(A)\}, \end{align*} then for $u_1,u_2\in N_+(A)$, \begin{align*} \omega(u_1+Uu_1,u_2+Uu_2)=2i\big((u_1,u_2)_H-(Uu_1,Uu_2)_H\big)=0. \end{align*} Thus $L_U\subset L_U^{\omega}$. Conversely, let $a+b\in L_U^{\omega}$ with $a\in N_+(A)$ and $b\in N_-(A)$. Then for every $u\in N_+(A)$, \begin{align*} 0=\omega(a+b,u+Uu)=2i\big((a,u)_H-(b,Uu)_H\big). \end{align*} Since $U$ is unitary and the Hilbert inner product is linear in the first variable, $(b,Uu)_H=(U^{-1}b,u)_H$. Hence $(a-U^{-1}b,u)_H=0$ for every $u\in N_+(A)$, so $a=U^{-1}b$ and $b=Ua$. Therefore $a+b\in L_U$, proving $L_U^{\omega}=L_U$.[/step]

[guided]This is the linear algebra heart of the theorem. The boundary form on $N_+(A)\oplus N_-(A)$ is positive on $N_+(A)$ and negative on $N_-(A)$: \begin{align*} \omega(u+v,u+v)=2i\big(\|u\|_H^2-\|v\|_H^2\big). \end{align*} A neutral subspace is therefore a subspace on which the positive and negative contributions exactly balance. Let $L\subset N_+(A)\oplus N_-(A)$ be neutral. Define the projections \begin{align*} P_+:N_+(A)\oplus N_-(A)\to N_+(A) \end{align*} and \begin{align*} P_-:N_+(A)\oplus N_-(A)\to N_-(A) \end{align*} by $P_+(u+v)=u$ and $P_-(u+v)=v$. If $u+v\in L$, then neutrality applied to this vector gives \begin{align*} 0=\omega(u+v,u+v)=2i\big(\|u\|_H^2-\|v\|_H^2\big). \end{align*} Hence $\|u\|_H=\|v\|_H$. This equality has two consequences. First, no nonzero vector of $L$ can have $u=0$, because then $\|v\|_H=0$. Thus $P_+|_L$ is injective. Second, no nonzero vector of $L$ can have $v=0$, so $P_-|_L$ is injective. Assume now that $L=L^{\omega}$. Write $M:=P_+(L)$. Since $P_+|_L$ is injective, each $u\in M$ has a unique partner $Vu\in N_-(A)$ with $u+Vu\in L$, and this defines a linear map \begin{align*} V:M\to N_-(A). \end{align*} Neutrality gives $\|Vu\|_H=\|u\|_H$, so $V$ is an isometry on $M$. We also need $M$ to be closed before using orthogonal complements. The topology here is the Hilbert topology on $K$ inherited from the graph inner product on $\mathcal{D}(A^*)$; on $N_+(A)$ and $N_-(A)$ it is the same topology as the $H$-norm topology, up to the factor $\sqrt{2}$. The boundary formula shows that, for each fixed $\ell\in K$, the functional $m\mapsto\omega(m,\ell)$ is continuous on $K$. Since $L=L^{\omega}$, the subspace $L$ is an intersection of closed kernels of continuous linear functionals, hence is closed in $K$. Let $(u_k)_{k=1}^{\infty}$ be a sequence in $M$ converging to $u\in N_+(A)$. Because $V$ is an isometry, $(Vu_k)_{k=1}^{\infty}$ is Cauchy in $N_-(A)$ and converges to some $v\in N_-(A)$. The vectors $u_k+Vu_k$ converge to $u+v$ in $K$. Since each $u_k+Vu_k$ lies in the closed subspace $L$, the limit $u+v$ lies in $L$. Hence $u\in M$. Thus $M$ is closed. The same reasoning shows that $P_-(L)$ is closed. We now use the stronger condition $L=L^{\omega}$ to prove that the projections are onto. Suppose $a\in P_+(L)^\perp\subset N_+(A)$. For every $u+v\in L$, \begin{align*} \omega(a,u+v)=2i(a,u)_H=0. \end{align*} Thus $a\in L^{\omega}=L$. But every vector in $L$ is neutral against itself, so \begin{align*} 0=\omega(a,a)=2i\|a\|_H^2. \end{align*} Hence $a=0$. Since $P_+(L)$ is closed and has zero orthogonal complement, $P_+(L)=N_+(A)$. The argument for $P_-(L)$ is identical: if $b\in P_-(L)^\perp$, then $b\in L^{\omega}=L$, and neutrality gives $0=\omega(b,b)=-2i\|b\|_H^2$, so $b=0$. Therefore $P_-(L)=N_-(A)$. Now every $u\in N_+(A)$ has a unique partner $v\in N_-(A)$ with $u+v\in L$. Define \begin{align*} U:N_+(A)\to N_-(A) \end{align*} by $Uu=v$. Linearity follows from linearity of $L$. The norm equality above gives $\|Uu\|_H=\|u\|_H$, and surjectivity follows from $P_-(L)=N_-(A)$. Thus $U$ is unitary and \begin{align*} L=\{u+Uu:u\in N_+(A)\}. \end{align*} Conversely, start with a unitary map $U:N_+(A)\to N_-(A)$ and define \begin{align*} L_U:=\{u+Uu:u\in N_+(A)\}. \end{align*} For $u_1,u_2\in N_+(A)$, the boundary form computation gives \begin{align*} \omega(u_1+Uu_1,u_2+Uu_2)=2i\big((u_1,u_2)_H-(Uu_1,Uu_2)_H\big). \end{align*} Since $U$ is unitary, $(Uu_1,Uu_2)_H=(u_1,u_2)_H$, and the right-hand side is zero. Hence $L_U$ is neutral. Conversely, if $a+b\in L_U^{\omega}$ with $a\in N_+(A)$ and $b\in N_-(A)$, then for every $u\in N_+(A)$, \begin{align*} 0=\omega(a+b,u+Uu)=2i\big((a,u)_H-(b,Uu)_H\big). \end{align*} Using unitarity of $U$, this becomes $(a-U^{-1}b,u)_H=0$ for every $u\in N_+(A)$. Hence $a=U^{-1}b$, so $b=Ua$ and $a+b\in L_U$. Therefore $L_U^{\omega}=L_U$.[/guided]

[step:Construct the self-adjoint extension from a unitary deficiency map]Let $U:N_+(A)\to N_-(A)$ be unitary, and define \begin{align*} \mathcal{D}(A_U):=\{x+u+Uu:x\in\mathcal{D}(A),\ u\in N_+(A)\}. \end{align*} Define \begin{align*} A_U:\mathcal{D}(A_U)\to H \end{align*} by \begin{align*} A_U(x+u+Uu):=Ax+iu-iUu. \end{align*} The von Neumann decomposition makes the representation $x+u+Uu$ unique, so $A_U$ is well-defined. Since $A^*u=iu$ and $A^*Uu=-iUu$, this is exactly the restriction of $A^*$ to $\mathcal{D}(A_U)$. The subspace $L_U=\{u+Uu:u\in N_+(A)\}$ satisfies $L_U=L_U^{\omega}$ by the previous step. Hence $\omega$ vanishes on $\mathcal{D}(A_U)$, so $A_U$ is symmetric. If $y\in\mathcal{D}(A_U^*)$, then $y\in\mathcal{D}(A^*)$ and $\omega(y,z)=0$ for every $z\in\mathcal{D}(A_U)$. Writing $y=x+a+b$ with $x\in\mathcal{D}(A)$, $a\in N_+(A)$, and $b\in N_-(A)$, the terms involving $x$ vanish and the condition becomes \begin{align*} 0=\omega(a+b,u+Uu)=2i\big((a,u)_H-(b,Uu)_H\big) \end{align*} for every $u\in N_+(A)$. Since $U$ is unitary and the Hilbert inner product is linear in the first variable, $(b,Uu)_H=(U^{-1}b,u)_H$. Hence \begin{align*} (a-U^{-1}b,u)_H=0 \end{align*} for every $u\in N_+(A)$. Taking $u=a-U^{-1}b$ gives $a=U^{-1}b$, so $b=Ua$ and $a+b\in L_U$. Therefore $y\in\mathcal{D}(A_U)$, so $\mathcal{D}(A_U^*)\subset\mathcal{D}(A_U)$. Since $A_U$ is symmetric, $\mathcal{D}(A_U)\subset\mathcal{D}(A_U^*)$. Thus $\mathcal{D}(A_U^*)=\mathcal{D}(A_U)$, and $A_U$ is self-adjoint.[/step]

[guided]Start with a unitary map \begin{align*} U:N_+(A)\to N_-(A). \end{align*} The corresponding extension should keep the original domain $\mathcal{D}(A)$ and add the graph of $U$ in the two deficiency spaces. Thus we define \begin{align*} \mathcal{D}(A_U):=\{x+u+Uu:x\in\mathcal{D}(A),\ u\in N_+(A)\}. \end{align*} The von Neumann decomposition makes this representation unique, so the formula \begin{align*} A_U(x+u+Uu):=Ax+iu-iUu \end{align*} defines a single-valued linear operator $A_U:\mathcal{D}(A_U)\to H$. This formula is not arbitrary: since $A^*u=iu$ and $A^*Uu=-iUu$, it says exactly that $A_U$ is the restriction of $A^*$ to $\mathcal{D}(A_U)$. Let \begin{align*} L_U:=\{u+Uu:u\in N_+(A)\}. \end{align*} The previous classification proves $L_U=L_U^{\omega}$. Therefore the boundary form vanishes on $\mathcal{D}(A_U)=\mathcal{D}(A)\oplus L_U$, so $A_U$ is symmetric. It remains to show that the adjoint has no larger domain. Let $y\in\mathcal{D}(A_U^*)$. Since $A\subset A_U$, the adjoint inclusion gives $A_U^*\subset A^*$, hence $y\in\mathcal{D}(A^*)$. Write its von Neumann decomposition as \begin{align*} y=x+a+b \end{align*} with $x\in\mathcal{D}(A)$, $a\in N_+(A)$, and $b\in N_-(A)$. The condition $y\in\mathcal{D}(A_U^*)$ is equivalent to $\omega(y,z)=0$ for every $z\in\mathcal{D}(A_U)$. Taking $z=x_0+u+Uu$ and using the fact that all terms involving $\mathcal{D}(A)$ vanish, this becomes \begin{align*} 0=\omega(a+b,u+Uu)=2i\big((a,u)_H-(b,Uu)_H\big) \end{align*} for every $u\in N_+(A)$. Because $U$ is unitary and the Hilbert inner product is linear in the first variable, \begin{align*} (b,Uu)_H=(U^{-1}b,u)_H. \end{align*} Thus \begin{align*} (a-U^{-1}b,u)_H=0 \end{align*} for every $u\in N_+(A)$. Choosing $u=a-U^{-1}b$ gives $a=U^{-1}b$, hence $b=Ua$. Therefore $a+b=a+Ua\in L_U$, and so $y\in\mathcal{D}(A_U)$. We have proved $\mathcal{D}(A_U^*)\subset\mathcal{D}(A_U)$. The reverse inclusion follows from symmetry of $A_U$, so $\mathcal{D}(A_U^*)=\mathcal{D}(A_U)$ and $A_U$ is self-adjoint.[/guided]

[step:Extract the existence criterion and the parametrisation]If $A$ has a self-adjoint extension $B$, then its projected boundary subspace $L_B:=\Pi_K(\mathcal{D}(B))\subset N_+(A)\oplus N_-(A)$ satisfies $L_B=L_B^{\omega}$. By the classification of self-orthogonal boundary subspaces, $L_B$ is the graph of a unitary map $U:N_+(A)\to N_-(A)$. Therefore $N_+(A)$ and $N_-(A)$ are unitarily isomorphic, so \begin{align*} n_+(A)=n_-(A). \end{align*} Conversely, if $n_+(A)=n_-(A)$, then the Hilbert spaces $N_+(A)$ and $N_-(A)$ admit a unitary map $U:N_+(A)\to N_-(A)$. The construction above gives a self-adjoint extension $A_U$ of $A$. Finally, the same construction shows uniqueness of the parametrisation: different unitary maps have different graph subspaces $L_U$, hence different domains $\mathcal{D}(A_U)$. Therefore the self-adjoint extensions of $A$ are in one-to-one correspondence with unitary maps $U:N_+(A)\to N_-(A)$, with domain and action given by \begin{align*} \mathcal{D}(A_U)=\{x+u+Uu:x\in\mathcal{D}(A),\ u\in N_+(A)\} \end{align*} and \begin{align*} A_U(x+u+Uu)=Ax+iu-iUu. \end{align*} This proves both the existence criterion and the stated parametrisation.[/step]

[guided]We now read the criterion off from the boundary classification. Suppose first that $A$ has a self-adjoint extension $B$. The previous domain analysis assigns to $B$ the projected boundary subspace \begin{align*} L_B:=\Pi_K(\mathcal{D}(B))\subset N_+(A)\oplus N_-(A), \end{align*} and proves that $L_B=L_B^{\omega}$. The classification of such self-orthogonal boundary subspaces says that $L_B$ must be the graph of a unitary map \begin{align*} U:N_+(A)\to N_-(A). \end{align*} Thus the two deficiency spaces are unitarily isomorphic, and their Hilbert dimensions agree: \begin{align*} n_+(A)=n_-(A). \end{align*} Conversely, assume \begin{align*} n_+(A)=n_-(A). \end{align*} By equality of Hilbert-space dimensions, there exists a unitary map $U:N_+(A)\to N_-(A)$. The construction in the preceding step then defines an operator $A_U$ with \begin{align*} \mathcal{D}(A_U)=\{x+u+Uu:x\in\mathcal{D}(A),\ u\in N_+(A)\} \end{align*} and \begin{align*} A_U(x+u+Uu)=Ax+iu-iUu. \end{align*} That step proves $A_U$ is self-adjoint, so equality of the deficiency indices is sufficient for existence. Finally, the parametrisation is injective and surjective. It is surjective because every self-adjoint extension produces a self-orthogonal boundary subspace, and every such subspace is the graph of a unitary $U:N_+(A)\to N_-(A)$. It is injective because if two unitary maps $U_1$ and $U_2$ have the same extension domain, then their graph subspaces in $N_+(A)\oplus N_-(A)$ agree. For every $u\in N_+(A)$, the vector $u+U_1u$ then lies in the graph of $U_2$, forcing $U_1u=U_2u$ by uniqueness of the direct-sum decomposition. Hence $U_1=U_2$. This proves the stated one-to-one correspondence.[/guided]