Deficiency subspaces and deficiency indices

Assume the Axiom of Choice. Let $T$ be a densely defined closed symmetric operator on a complex Hilbert space $H$ (Symmetric, self-adjoint and essentially self-adjoint operators, Densely defined, closed and closable operators, and cores). Its deficiency subspaces are $K_{+}:=\ker (T^{\ast }-i),K_{-}:=\ker (T^{\ast }+i),$ and its deficiency indices are the Hilbert dimensions $d_{\pm }(T):=\dim K_{\pm }$, that is, the cardinalities of orthonormal bases of $K_{\pm }$ (Orthonormal families, complete orthonormal systems and Hilbert bases).

The subspaces are the orthocomplements of the ranges. By The adjoint is well defined, closed, and reverses inclusions $\mathrm{ran}(T\pm i{)}^{\perp }=\ker (T^{\ast }\mp i)$, so $K_{+}=\mathrm{ran}(T+i{)}^{\perp }$ and $K_{-}=\mathrm{ran}(T-i{)}^{\perp }$. Each $K_{\pm }$ is a closed linear subspace by Orthogonal complements are closed, hence is itself a Hilbert space. For $x\in D(T)$, symmetry and conjugate symmetry make $r:=⟨Tx,x⟩=⟨x,Tx⟩$ real. With the inner product linear in its first variable Real and complex inner-product spaces and their induced length, $\mathrm{\parallel }(T\pm i)x{\mathrm{\parallel }}^2=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2+\mathrm{\parallel }x{\mathrm{\parallel }}^2\mp ir\pm ir=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2+\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{.}$ Thus $T\pm i$ is injective. If $(T\pm i)x_n\to y$, applying the identity to $x_n-x_m$ makes $(x_n)$ Cauchy. Completeness gives $x_n\to x$, and $T{x}_n=(T\pm i)x_n\mp i{x}_n\to y\mp ix$. Closedness of $T$ now gives $x\in D(T)$ and $(T\pm i)x=y$, proving both ranges closed. The closed-subspace decomposition theorem Orthogonal decomposition by a closed subspace therefore gives $H=\mathrm{ran}(T+i)\oplus K_{+}=\mathrm{ran}(T-i)\oplus K_{-}\mathrm{.}$ Also $K_{+}\cap K_{-}=\{0\}$, since membership forces $T^{\ast }u=iu=-iu$. The two deficiency spaces need not be orthogonal to each other in $H$.

Dimension convention and well-definedness. An orthonormal basis exists in each $K_{\pm }$ by Existence of a maximal orthonormal family, and maximality as completeness, using full AC. Its cardinality is independent of the basis, as follows. Let $(e_i{)}_{i\in I}$ and $(f_j{)}_{j\in J}$ be two orthonormal bases of the same Hilbert space. By The Bessel inequality for an arbitrary orthonormal family, for each $i$ and integer $n\ge 1$ at most $n$ indices satisfy $\mathrm{\mid }⟨e_i,f_j⟩{\mathrm{\mid }}^2\ge 1\mathrm{/}n$. Thus the support in $J$ of each row is countable. Each column has a nonzero entry: otherwise $f_j$ is orthogonal to the dense span of all $e_i$, hence to itself, contradicting norm one. Here orthogonality passes to the closure by Orthogonal complements are closed. If $I$ is infinite, full AC lets us enumerate the row supports and assign each $j$ to one row containing it; this gives an injection $J\to I\times \mathbb{N}$. Cardinal absorption Absorption: for cardinals $\kappa ,\lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda =\kappa$, and $\kappa \otimes \lambda =\kappa$ when $\lambda \mathrm{\ne }0$ and cardinal comparison Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$, the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$ give $\mathrm{\mid }J\mathrm{\mid }\le \mathrm{\mid }I\mathrm{\mid }$. If $I$ has finite size $n$, the residual $f_j-{\sum }_{i\in I}⟨f_j,e_i⟩e_i$ is orthogonal to the dense span of the $e_i$ and so vanishes. Taking its norm gives ${\sum }_{i\in I}\mathrm{\mid }⟨f_j,e_i⟩{\mathrm{\mid }}^2=1$. For every finite $F\subseteq J$, Bessel in the other direction yields $\mathrm{\mid }F\mathrm{\mid }={\sum }_{j\in F}{\sum }_{i\in I}\mathrm{\mid }⟨f_j,e_i⟩{\mathrm{\mid }}^2={\sum }_{i\in I}{\sum }_{j\in F}\mathrm{\mid }⟨e_i,f_j⟩{\mathrm{\mid }}^2\le n\mathrm{.}$ Hence $\mathrm{\mid }J\mathrm{\mid }\le n$. Exchanging the two bases proves equality of cardinalities in all cases. In the finite case each basis also spans algebraically by the same residual argument, so this is the ordinary linear dimension. For the zero space the basis is empty and the dimension is zero. AC is used for basis existence and the simultaneous choices in the infinite comparison.

Cayley sign and domain convention. Define $C_T:\mathrm{ran}(T+i)⟶\mathrm{ran}(T-i),C_T((T+i)x)=(T-i)x\mathrm{.}$ Injectivity of $T+i$ makes this well defined; the norm identity makes it an isometry onto the stated range. On its domain, $(I-C_T)(T+i)x=2ix$, whence $\mathrm{ran}(I-C_T)=D(T)$, since $D(T)$ is a complex linear subspace. If a unitary $U:H\to H$ extends $C_T$, it maps the orthogonal complement $K_{+}$ of the initial range onto the orthogonal complement $K_{-}$ of the final range, by preservation of inner products and surjectivity. Conversely, any unitary $W:K_{+}\to K_{-}$ gives the unitary extension $U=C_T\oplus W$ on the two displayed orthogonal decompositions. This describes the free part of a unitary extension and fixes the signs; it does not assert that such a $W$ always exists.