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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Von Neumann closure, standard vectors and modular analytic tools

1 · Prerequisites

2 · Summary

This page fixes the concrete representation conventions used by the later modular-equilibrium page. Hilbert-space inner products are linear in the first argument, commutants are taken in the bounded operators, and the Axiom of Choice is stated where the double-commutant, projection, and compactness arguments use it. A cyclic GNS vector is not assumed to be separating.

The A page supplies the local operator algebra needed for those claims. It proves the finite-tuple projection argument for strong closure and the cyclic-separating duality between an algebra and its commutant. It also proves bounded monotone strong convergence in a concrete von Neumann algebra, then constructs its predual as a quotient of trace-class operators. The polar calculus first proves the closed densely defined linear case from the graph projection and spectral theorem, then transfers it to antilinear operators.

The antilinear polar statement is applied to the graph closure of a closable operator. No identity between an operator and its closure is asserted for a merely closable operator. The Tomita operator on $M Ω$ is proved well-defined, densely defined, and closable from cyclic-separating hypotheses; its later modular data use this closure.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Concrete von Neumann algebra and faithful vector

Definition

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and $B (H)$ the space of bounded linear operators on $H$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). For a set $S \subseteq B (H)$ , write

$S^{'} : = {T \in B (H) : T s = s T for every s \in S}, S^{''} : = (S^{'})^{'} .$

A directed preorder is a nonempty set $I$ with a reflexive, transitive relation $\leq$ such that every pair $i, j \in I$ has a common upper bound in $I$ . A net in a set $X$ is a function from a directed preorder into $X$ ; its values are written $(x_{i})_{i \in I}$ .

The strong operator topology on $B (H)$ is the topology of pointwise norm convergence: $T_{α} \to T$ strongly exactly when $∥ (T_{α} - T) ξ ∥ \to 0$ for every $ξ \in H$ . The weak operator topology is the topology of pointwise matrix-coefficient convergence: $T_{α} \to T$ weakly exactly when $⟨ T_{α} ξ, η ⟩ \to ⟨ T ξ, η ⟩$ for every $ξ, η \in H$ . Use the linear-first inner-product convention.

A concrete von Neumann algebra on $H$ is a unital star subalgebra $M \subseteq B (H)$ satisfying $M = M^{''}$ . Under Choice, the double commutant theorem makes this equivalent to being closed in the weak operator topology.

An operator $a \in B (H)$ is positive when $a = a^{*}$ and $⟨ a ξ, ξ ⟩ \geq 0$ for every $ξ \in H$ , using the bounded adjoint convention of Hilbert-adjoint identities. Write $M_{+}$ for the positive operators in $M$ and $a \leq b$ when $b - a$ is positive.

A vector $Ω \in H$ is cyclic for $M$ when $\overline{M Ω} = H$ , and is separating for $M$ when $x Ω = 0$ for $x \in M$ implies $x = 0$ . For a unit vector $Ω$ , define its vector functional by $ω_{Ω} (x) = ⟨ x Ω, Ω ⟩$ . It is faithful when $ω_{Ω} (x^{*} x) = 0$ implies $x = 0$ ; this is equivalent to $Ω$ being separating because $ω_{Ω} (x^{*} x) = ∥ x Ω ∥^{2}$ .

For a bounded linear functional $φ : M \to C$ , positive means $φ (a) \in [0, + \infty)$ for every $a \in M_{+}$ ; in particular, each such value is real. A positive functional is normal when $φ (\sup_{i} x_{i}) = \sup_{i} φ (x_{i})$ for each bounded increasing net $(x_{i})_{i \in I}$ in $M_{+}$ whose supremum is taken in $M$ .

Definition references

Von Neumann algebras and commutants

The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators

Hilbert-adjoint identities

Strong and weak operator topologies

The Axiom of Choice

AC implies DC implies countable choice

Remarks

For a quasi-local state arising from a consistent family of finite local density matrices, assume Countable Choice together with the GNS supplier's C*-algebra completion and finite-matrix positive-square-root interfaces, exactly as stated in that supplier's Statement. Under these hypotheses it gives a cyclic representation. That construction does not assert that the cyclic vector is separating for the represented algebra, and it does not supply a separating vector for every state.

Cyclic representation of a quasi-local state

The definition $M = M^{''}$ is the chosen concrete representation interface. The canonical imported definition starts with weak-operator closure; the cited double-commutant theorem proves the equivalence under the stated Choice hypothesis.

Facts & Assumptions

Given: Choice, a complex Hilbert space $H$ , a unital star subalgebra $M \subseteq B (H)$ , and a unit vector $Ω \in H$ .

[A1]

The Axiom of Choice implies Countable Choice (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The canonical interface defines concrete von Neumann algebras by weak operator closure and defines commutants in $B (H)$ (Von Neumann algebras and commutants).

[F2]

Weak-operator convergence is pointwise convergence of matrix coefficients (Strong and weak operator topologies).

[F3]

Under Choice, a unital star subalgebra is weak-operator closed exactly when it equals its double commutant (The double commutant theorem for concrete von Neumann algebras).

[F4]

Bounded adjoints satisfy $⟨ A ξ, η ⟩ = ⟨ ξ, A^{*} η ⟩$ and $(A^{*})^{*} = A$ (Hilbert-adjoint identities).

[F5]

Under Countable Choice, the C*-algebra completion interface and the finite-matrix positive-square-root interface specified in its Statement, the quasi-local-state GNS supplier constructs a cyclic representation of the extended algebraic state of its consistent finite-density family (Cyclic representation of a quasi-local state).

Proof

technique · direct verification of the representation interface

Given: [A1], [F1]–[F5], and the objects in the Definition.

1.1F4givenalgebra

For $x \in M$ , [F4] gives $ω_{Ω} (x^{*} x) = ⟨ x^{*} x Ω, Ω ⟩ = ⟨ x Ω, x Ω ⟩ = ∥ x Ω ∥^{2} .$ Thus $ω_{Ω} (x^{*} x) = 0$ exactly when $x Ω = 0$ ; requiring this to force $x = 0$ is exactly separatingness.

1.2A1F1F2F3

By [F3], under the declared Choice hypothesis the weak-operator-closed definition of a concrete von Neumann algebra is equivalent to $M = M^{''}$ . Therefore the double-commutant convention in the Definition matches the canonical weak-operator-closure convention.

2.1A1F5∎

For the consistent finite-density family and extended state in [F5], assume its C*-algebra completion and finite-matrix positive-square-root interfaces. Together with Countable Choice from [A1], these are the hypotheses of the cited GNS construction, which supplies a cyclic representing vector. Its statement gives no separating-vector conclusion; separatingness remains the separate representation property defined above.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Conjugate Hilbert space and adjoint of an antilinear operator

Definition

Assume Countable Choice. All inner products are linear in their first argument. If $H$ is a complex Hilbert space, its conjugate Hilbert space $\overline{H}$ consists of symbols $\overline{x}$ for $x \in H$ , with $\overline{x} + \overline{y} = \overline{x + y}$ , $a \overline{x} = \overline{\overline{a} x}$ for $a \in C$ , and $⟨ \overline{x}, \overline{y} ⟩_{\overline{H}} : = \overline{⟨ x, y ⟩_{H}} .$ The canonical map $c_{H} : H \to \overline{H}$ , $c_{H} x = \overline{x}$ , is a conjugate-linear surjective isometry.

For complex Hilbert spaces $H_{1}, H_{2}$ , a linear operator $A$ is a linear map $A : D (A) \to H_{2}$ on a linear domain $D (A) \subseteq H_{1}$ . Its graph is $Γ (A) = {(x, A x) : x \in D (A)} \subseteq H_{1} \oplus H_{2}$ , with the direct-sum Hilbert norm. It is densely defined when $D (A)$ is dense in $H_{1}$ , and closed when $Γ (A)$ is closed.

For a densely defined linear operator $A : D (A) \subseteq H_{1} \to H_{2}$ , define $D (A^{*})$ to consist of those $y \in H_{2}$ for which $x \mapsto ⟨ A x, y ⟩_{H_{2}}$ is bounded on $D (A)$ in the $H_{1}$ norm. The bounded functional extends uniquely to $H_{1}$ ; its Riesz vector $A^{*} y \in H_{1}$ is specified by $⟨ A x, y ⟩_{H_{2}} = ⟨ x, A^{*} y ⟩_{H_{1}} (x \in D (A)) .$ This defines the linear adjoint $A^{*} : D (A^{*}) \subseteq H_{2} \to H_{1}$ .

An operator $S : D (S) \subseteq H \to H$ is antilinear when $S (a x + b y) = \overline{a} S x + \overline{b} S y$ . Its associated linear operator $\tilde{S} : D (\tilde{S}) \subseteq \overline{H} \to H$ is $D (\tilde{S}) = c_{H} (D (S))$ and $\tilde{S} (\overline{x}) = S x$ . For densely defined antilinear $S$ , define $S^{*} : = c_{H}^{- 1} {\tilde{S}}^{*},$ on the domain $D (S^{*}) = D ({\tilde{S}}^{*})$ . Thus $S^{*}$ is antilinear and $⟨ S x, y ⟩_{H} = ⟨ S^{*} y, x ⟩_{H} (x \in D (S), y \in D (S^{*})) .$

Definition references

Hilbert space

Real and complex inner-product spaces and their induced length

Riesz representation for Hilbert spaces

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Remarks

The direct-sum inner product is $⟨ (x, u), (y, v) ⟩ = ⟨ x, y ⟩_{H_{1}} + ⟨ u, v ⟩_{H_{2}}$ . It is complete because each summand is complete. The scalar and inner-product rules above make $\overline{H}$ an inner-product space, and $c_{H}$ preserves norms and is onto; consequently $\overline{H}$ is complete as well.

For fixed $y \in H_{2}$ , the map $x \mapsto ⟨ A x, y ⟩_{H_{2}}$ is linear. Riesz representation on $H_{1}$ therefore gives the unique adjoint vector whenever this map is bounded. Linearity and uniqueness show that $D (A^{*})$ is a linear subspace and $A^{*}$ is linear. For the antilinear convention, write ${\tilde{S}}^{*} y = \overline{z}$ . Then $⟨ S x, y ⟩_{H} = ⟨ c_{H} x, {\tilde{S}}^{*} y ⟩_{\overline{H}} = \overline{⟨ x, z ⟩_{H}} = ⟨ z, x ⟩_{H},$ so the displayed formula gives $S^{*} y = z$ . The definition makes no claim that $S$ is closable or that either adjoint domain is dense.

Facts & Assumptions

Given: Countable Choice, complex Hilbert spaces, a densely defined linear operator $A$ , and a densely defined antilinear operator $S$ .

[F1]

The inner product is linear in its first argument and conjugate-linear in its second; conjugating both scalar coefficients defines the complex conjugate vector space (Real and complex inner-product spaces and their induced length).

[F2]

Each bounded linear functional on a Hilbert space has a unique Riesz representing vector (Riesz representation for Hilbert spaces).

[F3]

Countable Choice is the explicit standing choice assumption (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: [F1]–[F3], the spaces and operators in the Definition.

1.1F1algebra

The scalar rules for $\overline{H}$ follow from conjugation in the underlying scalar multiplication, and $⟨ c_{H} x, c_{H} y ⟩_{\overline{H}} = \overline{⟨ x, y ⟩_{H}}$ . Thus $c_{H}$ preserves norms and is onto. A Cauchy sequence in $\overline{H}$ pulls back under $c_{H}$ to a Cauchy sequence in $H$ , whose limit maps forward; so $\overline{H}$ is complete.

1.2F1F2F3given

For fixed $y \in H_{2}$ , the map $x \mapsto ⟨ A x, y ⟩$ is linear. If it is bounded on the dense domain, continuity extends it uniquely to $H_{1}$ , and [F2] gives a unique $A^{*} y$ with the displayed adjoint identity. Uniqueness and linearity of the form in $y$ show that $D (A^{*})$ is a linear subspace and $A^{*}$ is linear there.

2.1F1F2step 1.1algebra∎

The domain $c_{H} (D (S))$ is a complex linear subspace because $D (S)$ is linear and $c_{H}$ is conjugate-linear. Additivity of $S$ makes $\tilde{S}$ additive, and for $a \in C$ , $\tilde{S} (a \overline{x}) = \tilde{S} (\overline{\overline{a} x}) = S (\overline{a} x) = a S x$ , so $\tilde{S}$ is linear. Since $c_{H}^{- 1}$ is conjugate-linear and ${\tilde{S}}^{*}$ is linear, $S^{*}$ is antilinear on its declared domain. If $y \in D ({\tilde{S}}^{*})$ and ${\tilde{S}}^{*} y = c_{H} z$ , then $⟨ S x, y ⟩_{H} = ⟨ c_{H} x, {\tilde{S}}^{*} y ⟩_{\overline{H}} = \overline{⟨ x, z ⟩_{H}} = ⟨ z, x ⟩_{H} = ⟨ c_{H}^{- 1} {\tilde{S}}^{*} y, x ⟩_{H} .$ This proves the stated antilinear-adjoint identity with its exact domain.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Trace-class duality and finite-rank density

Statement

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space, and let $S_{1} (H)$ be its trace-class operators with trace norm $∥ \cdot ∥_{1}$ . Write $Tr : = tr$ for the basis-independent trace defined earlier. Finite-rank operators are dense in $(S_{1} (H), ∥ \cdot ∥_{1})$ . For every bounded linear functional $Φ : S_{1} (H) \to C$ , there is a unique $X \in B (H)$ such that $Φ (T) = Tr (T X) (T \in S_{1} (H)),$ and the correspondence is isometric: $∥ Φ ∥ = ∥ X ∥$ .

Statement references

Trace class operator

Trace of a trace class operator

Trace is absolutely convergent and basis independent

Singular value decomposition for compact operators

Riesz representation for Hilbert spaces

Hilbert-adjoint identities

The Axiom of Choice

Facts & Assumptions

Given: [A1], a complex Hilbert space $H$ , its trace-class space, trace norm and trace.

[A1]

Assume the Axiom of Choice; it implies Countable Choice, the stated choice input for the imported trace-class tools (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Trace class operator defines $S_{1} (H)$ by summability of singular values. Singular value decomposition for compact operators indexes exactly the positive singular values by a finite initial segment or the positive integers and gives norm-convergent partial sums; the trace-norm characterization Nuclear series characterizes trace norm identifies their sum as the trace norm.

[F2]

Trace class is a Banach space and a two-sided ideal: $∥ A Y B ∥_{1} \leq ∥ A ∥ ∥ Y ∥_{1} ∥ B ∥$ (Trace class is a two sided Banach operator ideal).

[F3]

The basis-independent trace equals the scalar sum of any nuclear representation and obeys $∣ Tr (Y) ∣ \leq ∥ Y ∥_{1}$ (Trace of a trace class operator, Trace is absolutely convergent and basis independent).

[F4]

The inner product is linear in its first argument; every bounded linear functional on $H$ has a unique Riesz vector (Real and complex inner-product spaces and their induced length, Riesz representation for Hilbert spaces).

[F5]

Bounded operators have Hilbert adjoints satisfying $⟨ X ξ, η ⟩ = ⟨ ξ, X^{*} η ⟩$ (Hilbert-adjoint identities).

Proof

technique · direct

Given: [F1]–[F5], a bounded linear functional $Φ$ on $S_{1} (H)$ , and its trace pairing.

1.1A1F1construct

Let $J = {1, \dots, r}$ if $T$ has finite rank $r$ , and $J = N_{> 0}$ otherwise, as in [F1]. Write its singular expansion $T = \sum_{j \in J} s_{j} ⟨ \cdot, e_{j} ⟩ f_{j}$ and let $T_{N} = \sum_{j \in J, j \leq N} s_{j} ⟨ \cdot, e_{j} ⟩ f_{j}$ . The SVD gives $T - T_{N} = \sum_{j \in J, j > N} s_{j} ⟨ \cdot, e_{j} ⟩ f_{j}$ as a norm-convergent nuclear representation whose coefficient products sum to the tail $\sum_{j \in J, j > N} s_{j} \to 0$ . The nuclear characterization in [F1] therefore gives $∥ T - T_{N} ∥_{1} \leq \sum_{j \in J, j > N} s_{j} \to 0$ . In finite rank the tail is empty once $N \geq r$ . Thus finite-rank operators are trace-norm dense.

1.2F1F4construct

For $ξ, η \in H$ , define the rank-one operator $R_{ξ, η} z = ⟨ z, η ⟩ ξ$ . Its only possible nonzero singular value is $∥ ξ ∥ ∥ η ∥$ , so $∥ R_{ξ, η} ∥_{1} = ∥ ξ ∥ ∥ η ∥$ . The form $b (ξ, η) : = Φ (R_{ξ, η})$ is linear in $ξ$ , conjugate-linear in $η$ , and satisfies $∣ b (ξ, η) ∣ \leq ∥ Φ ∥ ∥ ξ ∥ ∥ η ∥$ . For fixed $ξ$ , the map $g_{ξ} (η) : = \overline{b (ξ, η)}$ is bounded and linear in $η$ . Riesz representation [F4] gives a unique vector $X ξ$ such that $g_{ξ} (η) = ⟨ η, X ξ ⟩$ ; conjugating gives $b (ξ, η) = ⟨ X ξ, η ⟩$ . Uniqueness makes $X$ linear, and the bound gives $∥ X ξ ∥ \leq ∥ Φ ∥ ∥ ξ ∥$ , so $X \in B (H)$ and $∥ X ∥ \leq ∥ Φ ∥$ .

2.1F3F5step 1.2algebra

The nuclear representation $R_{ξ, η} X = R_{ξ, X^{*} η}$ and [F3] give $Tr (R_{ξ, η} X) = ⟨ ξ, X^{*} η ⟩ = ⟨ X ξ, η ⟩$ . Every finite-rank operator is a finite sum of rank-one operators, so linearity of $Φ$ and of the trace gives $Φ (T) = Tr (T X)$ for every finite-rank $T$ .

3.1F2F3step 1.1step 2.1

For arbitrary $T \in S_{1} (H)$ , use the finite-rank approximants $T_{N}$ from step 1.1. Boundedness of $Φ$ gives $Φ (T_{N}) \to Φ (T)$ . Also $∣ Tr ((T - T_{N}) X) ∣ \leq ∥ (T - T_{N}) X ∥_{1} \leq ∥ T - T_{N} ∥_{1} ∥ X ∥ \to 0$ by [F2] and [F3]. Hence the same trace identity holds for every trace-class $T$ .

4.1F2F3step 1.2step 3.1

If $H = {0}$ , then $S_{1} (H) = B (H) = {0}$ , so the only bounded functional and representing operator are zero and the isometry assertion holds. Otherwise the trace identity and [F2]–[F3] imply $∣ Φ (T) ∣ \leq ∥ T ∥_{1} ∥ X ∥$ , so $∥ Φ ∥ \leq ∥ X ∥$ . For unit $ξ, η$ , step 2.1 gives $∣ ⟨ X ξ, η ⟩ ∣ = ∣ Φ (R_{ξ, η}) ∣ \leq ∥ Φ ∥$ , since $∥ R_{ξ, η} ∥_{1} = 1$ . Taking suprema over unit $ξ, η$ gives $∥ X ∥ \leq ∥ Φ ∥$ . If $Y$ also represents $Φ$ , applying the identity to each $R_{ξ, η}$ gives $⟨ (X - Y) ξ, η ⟩ = 0$ for all $ξ, η$ , hence $X = Y$ .

5.1step 1.1step 1.2step 2.1step 3.1step 4.1∎

Steps 1.1–4.1 prove finite-rank trace-norm density and establish a unique operator $X$ with the stated pairing and norm equality for every bounded functional $Φ$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Polar decomposition of a closed densely defined operator

Statement

Here a partial isometry, including when either Hilbert space is zero, is a bounded linear map $U$ that is isometric on $(\ker U)^{⊥}$ and vanishes on $\ker U$ . The initial space is $(\ker U)^{⊥}$ and the final space is $ran U$ . On nonzero Hilbert spaces this is the earlier definition.

Assume the Axiom of Choice. Let $H_{1}, H_{2}$ be complex Hilbert spaces and let $A : D (A) \subseteq H_{1} \to H_{2}$ be a linear operator with dense domain and closed graph. Let $A^{*}$ be the unbounded adjoint defined earlier. For a densely defined linear operator $R : D (R) \subseteq K \to K$ on a complex Hilbert space $K$ , self-adjoint means $D (R) = D (R^{*})$ and $R x = R^{*} x$ on this domain; positive means self-adjoint and $⟨ R x, x ⟩ \in [0, + \infty)$ for every $x \in D (R)$ . A unitary is a surjective linear isometry. Then $T : = A^{*} A$ , on $D (T) = {x \in D (A) : A x \in D (A^{*})}$ , is positive and self-adjoint. Its positive square root $P : = T^{1 / 2}$ has domain $D (P) = D (A)$ and satisfies $∥ P x ∥ = ∥ A x ∥$ for every $x \in D (A)$ . There is a unique partial isometry $U : H_{1} \to H_{2}$ such that $A = U P, \ker U = \ker A .$ Its initial space is $\overline{ran P} = (\ker A)^{⊥}$ and its final space is $\overline{ran A}$ . If $A$ is injective and has dense range, then $U$ is unitary.

Statement references

Conjugate Hilbert space and adjoint of an antilinear operator

The Axiom of Choice

Spectral theorem for unbounded self-adjoint operators (PVM form)

Isometry coisometry and partial isometry

Facts & Assumptions

Given: Choice, complex Hilbert spaces $H_{1}, H_{2}$ , and a densely defined linear $A : D (A) \subseteq H_{1} \to H_{2}$ whose graph is closed.

[F1]

The graph is a closed linear subspace of the Hilbert direct sum $H_{1} \oplus H_{2}$ . Every closed subspace has an orthogonal projection under Countable Choice (Conjugate Hilbert space and adjoint of an antilinear operator, Orthogonal decomposition by a closed subspace, The Hilbert orthogonal projection onto a closed subspace).

[F2]

For densely defined $A$ , $y \in D (A^{*})$ exactly when $x \mapsto ⟨ A x, y ⟩$ is bounded in the $H_{1}$ norm; then $⟨ A x, y ⟩ = ⟨ x, A^{*} y ⟩$ (Conjugate Hilbert space and adjoint of an antilinear operator).

[F3]

The unbounded spectral theorem for a positive self-adjoint $T$ supplies its spectral measure, the domain calculus for $T^{1 / 2}$ , and bounded spectral cutoffs (Spectral theorem for unbounded self-adjoint operators (PVM form)).

[F4]

For a positive spectral measure $μ$ , nonnegative measurable functions $g_{n} ↑ g$ satisfy $\int g_{n} d μ ↑ \int g d μ$ (Monotone convergence for the integral).

[F5]

For a bounded linear operator $R$ between Hilbert spaces, $\overline{ran R} = (\ker R^{*})^{⊥}$ (Kernel–range orthogonality for Hilbert adjoints). This is used for the bounded resolvent in step 2.1. The corresponding range-kernel identity for the unbounded self-adjoint operator $P$ is proved directly in step 5.1.

[F6]

A partial isometry is an isometry on its initial space and zero on its orthogonal complement; its final space is the closure of its range (Isometry coisometry and partial isometry).

[A1]

Choice implies Countable Choice, the hypothesis for the graph projection and Hilbert orthogonal decompositions (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: [A1], $H_{1}, H_{2}$ , and $A$ as in the Statement.

1.1A1F1F2construct

If $H_{1} = {0}$ , density forces $D (A) = {0}$ and $A = 0$ . Put $T = P = U = 0$ with their zero-space domains. Then $T$ is positive and self-adjoint, $P = T^{1 / 2}$ , $\ker U = \ker A = {0}$ , both asserted range closures are ${0}$ , and $U$ is a partial isometry by the zero-space convention in the Statement. If $A$ also has dense range, then $H_{2} = {0}$ and $U$ is unitary. Thus the conclusion holds; assume henceforth $H_{1} \neq {0}$ . Project $(y, 0)$ orthogonally onto the closed graph $Γ (A)$ to obtain $(x, A x)$ . Orthogonality against every $(z, A z)$ with $z \in D (A)$ gives $⟨ A z, A x ⟩ = ⟨ z, y - x ⟩$ . By [F2], $A x \in D (A^{*})$ and $A^{*} A x = y - x$ , so $(I + T) x = y$ . Thus $I + T$ is onto; it is injective because $⟨ (I + T) x, x ⟩ = ∥ x ∥^{2} + ∥ A x ∥^{2}$ , which vanishes only for $x = 0$ .

2.1A1F2F5step 1.1algebra

Let $R = (I + T)^{- 1}$ . If $x = R y$ , the preceding identity and Cauchy–Schwarz give $∥ R y ∥ \leq ∥ y ∥$ , so $R$ is bounded. For $x = R y$ and $w = R z$ , the graph-projection identity applied with test vectors $w$ and $x$ gives $⟨ y - x, w ⟩ = ⟨ A x, A w ⟩$ and $⟨ z - w, x ⟩ = ⟨ A w, A x ⟩$ . Conjugating the second equality and comparing shows $⟨ R y, z ⟩ = ⟨ y, R z ⟩$ ; hence $R$ is self-adjoint. It is injective and its range is exactly $D (T)$ , so [F5] gives $\overline{D (T)} = \overline{ran R} = H_{1}$ . The operator $T$ is symmetric, because for $x, y \in D (T)$ , $⟨ T x, y ⟩ = ⟨ A x, A y ⟩ = ⟨ x, T y ⟩$ , and positive, because $⟨ T x, x ⟩ = ∥ A x ∥^{2}$ . Finally let $y \in D (T^{*})$ and choose $x \in D (T)$ with $(I + T) x = (I + T^{*}) y$ , possible by step 1.1. Then $(I + T^{*}) (y - x) = 0$ . If $v$ lies in this kernel, then $⟨ (I + T) z, v ⟩ = 0$ for every $z \in D (T)$ ; surjectivity of $I + T$ forces $v = 0$ . Thus $y = x \in D (T)$ and $T^{*} y = T y$ . Therefore $T$ is self-adjoint.

3.1F2F3F4step 2.1algebra

Let $E$ be the spectral measure of the positive self-adjoint $T$ . It is supported on $[0, + \infty)$ : if $E ([- N, - 1 / N]) \neq 0$ for some positive integer $N$ , a nonzero vector in its range lies in $D (T)$ and has $⟨ T z, z ⟩ \leq - N^{- 1} ∥ z ∥^{2} < 0$ , contradicting positivity; these intervals cover $(- \infty, 0)$ . Set $P = T^{1 / 2}$ . For $x \in D (T)$ , the spectral calculus and the adjoint identity give $∥ P x ∥^{2} = ⟨ T x, x ⟩ = ∥ A x ∥^{2}$ . Now take any $x \in D (A)$ and, for each integer $n \geq 1$ , define $f_{n} (λ) = n / (n + λ)$ on $[0, + \infty)$ and $x_{n} = f_{n} (T) x = n (n + T)^{- 1} x$ . Since $f_{n}$ and $λ f_{n}$ are bounded, $x_{n} \in D (T)$ and $T x_{n} = n (x - x_{n})$ . For $z \in D (A)$ , write $d = z - x_{n}$ . The adjoint identity gives $⟨ A d, A x_{n} ⟩ = ⟨ d, T x_{n} ⟩ = - n ⟨ d, x_{n} - x ⟩$ . Expanding squared norms therefore yields $∥ A z ∥^{2} + n ∥ z - x ∥^{2} = ∥ A x_{n} ∥^{2} + n ∥ x_{n} - x ∥^{2} + ∥ A d ∥^{2} + n ∥ d ∥^{2} .$ Taking $z = x$ shows $∥ A x_{n} ∥^{2} \leq ∥ A x ∥^{2}$ . If $μ_{x} (B) = ⟨ E (B) x, x ⟩$ , then $∥ P x_{n} ∥^{2} = ∥ A x_{n} ∥^{2} = \int_{[0, \infty)} λ f_{n} (λ)^{2} d μ_{x} (λ)$ . The integrands increase pointwise to $λ$ , so [F4] gives $\int λ d μ_{x} \leq ∥ A x ∥^{2}$ . By the domain formula for $P = T^{1 / 2}$ , $x \in D (P)$ and $∥ P x ∥ \leq ∥ A x ∥$ .

4.1F1F3step 3.1

Conversely, if $x \in D (P)$ , let $x_{n} = E ([0, n]) x$ . The spectral theorem gives $x_{n} \in D (T)$ and $x_{n} \to x$ ; also $P x_{n} \to P x$ , since the tails of the finite measure $μ_{x}$ and of the integrable function $λ$ with respect to $μ_{x}$ tend to zero. Step 3.1 applied to $x_{n} - x_{m} \in D (T)$ shows that $A x_{n}$ is Cauchy and $∥ A x_{n} - A x_{m} ∥ = ∥ P (x_{n} - x_{m}) ∥$ . Closedness of $A$ then gives $x \in D (A)$ and $A x = \lim_{n} A x_{n}$ ; taking norms yields $∥ A x ∥ = ∥ P x ∥$ . Together with step 3.1 this proves $D (P) = D (A)$ and the asserted norm identity.

5.1F1F2F5F6step 4.1algebra

First, $\ker P$ is closed because $P$ is closed. For $y \in H_{1}$ , $y ⊥ ran P$ exactly when $⟨ P x, y ⟩ = 0$ for every $x \in D (P)$ . The adjoint definition then gives $y \in D (P^{*})$ and $P^{*} y = 0$ ; since $P = P^{*}$ , this is equivalent to $y \in \ker P$ . Conversely, if $y \in \ker P$ , self-adjointness gives $⟨ P x, y ⟩ = ⟨ x, P y ⟩ = 0$ for all $x \in D (P)$ . Hence $(ran P)^{⊥} = \ker P$ . For $C = \overline{ran P}$ , we have $C^{⊥} = \ker P$ ; the orthogonal decomposition theorem applied to $C$ and $\ker P$ therefore gives $C = (\ker P)^{⊥}$ . The norm identity from step 4.1 also gives $\ker P = \ker A$ . On $ran P$ , define $V (P x) = A x$ . If $P x = P y$ , then $∥ A (x - y) ∥ = ∥ P (x - y) ∥ = 0$ , so $V$ is well defined; the norm identity makes it an isometry. Extend it continuously to $C$ and set it to zero on $\ker P$ . For $x = c + k$ with $c \in C$ and $k \in \ker P$ , $∥ U x ∥ = ∥ c ∥ \leq ∥ x ∥$ ; thus the resulting linear map $U$ is bounded. It has kernel $\ker P$ , is isometric on $C = (\ker A)^{⊥}$ , and $A = U P$ . Its range is $\overline{ran A}$ because the extension of $V$ has range the closure of $V (ran P) = ran A$ . If $A$ is injective with dense range, its initial and final spaces are all of $H_{1}$ and $H_{2}$ , respectively, so $U$ is unitary. Any partial isometry with the stated factorization and kernel agrees with $U$ on $ran P$ and vanishes on $\ker P$ ; continuity gives uniqueness.

6.1step 1.1step 2.1step 3.1step 4.1step 5.1∎

Steps 1.1–5.1 prove positivity and self-adjointness of $A^{*} A$ , the domain and norm identity for its positive square root, the polar factorization, and the asserted initial/final spaces and uniqueness.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Strong closure and cyclic-separating duality

Statement

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and let $R \subseteq B (H)$ be a unital star subalgebra. Its strong-operator closure is $R^{''}$ . If $M \subseteq B (H)$ is a concrete von Neumann algebra and $Ω \in H$ , then $Ω$ is cyclic for $M$ exactly when it is separating for $M^{'}$ .

Statement references

Concrete von Neumann algebra and faithful vector

The double commutant theorem for concrete von Neumann algebras

Strong and weak operator topologies

Facts & Assumptions

Given: Choice, a complex Hilbert space $H$ , a unital star subalgebra $R \subseteq B (H)$ , and a concrete von Neumann algebra $M \subseteq B (H)$ .

[A1]

Choice implies Countable Choice, the hypothesis used for Hilbert orthogonal projections (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $S \subseteq B (H)$ , $S^{'}$ consists of the bounded operators commuting with every element of $S$ , and $S^{''} = (S^{'})^{'}$ . Each commutant is weak-operator closed (Von Neumann algebras and commutants).

[F2]

Strong convergence means pointwise norm convergence on $H$ ; weak convergence means convergence of every matrix coefficient. The weak operator topology is coarser than the strong operator topology (Strong and weak operator topologies).

[F3]

The finite Hilbert direct sum $H^{\oplus n}$ has coordinate inclusions and projections. A diagonal amplification $r^{\oplus n}$ of a bounded operator is bounded (Hilbert direct sums of unitary representations, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

A closed subspace $C$ of a Hilbert space has an orthogonal projection $P_{C}$ with range $C$ and kernel $C^{⊥}$ (Orthogonal decomposition by a closed subspace, The Hilbert orthogonal projection onto a closed subspace).

[F5]

The adjoint satisfies $⟨ A ξ, η ⟩ = ⟨ ξ, A^{*} η ⟩$ (Hilbert-adjoint identities).

[F6]

In the chosen representation, $Ω$ is cyclic for $M$ exactly when $\overline{M Ω} = H$ , and separating for $M$ exactly when $x Ω = 0$ for $x \in M$ implies $x = 0$ (Concrete von Neumann algebra and faithful vector).

[F7]

If $R$ is a unital star subalgebra and $W = W^{*} (R)$ its weak-operator closure, then $W^{''} = W$ under Choice (Von Neumann algebras and commutants, The double commutant theorem for concrete von Neumann algebras).

Proof

technique · direct finite-amplification argument

Given: [A1], [F1]–[F7], the Hilbert space and algebras in the Statement.

1.1A1F3F4F5givenconstruct

Fix $T \in R^{''}$ , an integer $n \geq 1$ , and a tuple $ξ = (ξ_{1}, \dots, ξ_{n}) \in H^{\oplus n}$ , and define $D_{r} = r^{\oplus n}$ for $r \in R$ . Let $C = \overline{{D_{r} ξ : r \in R}}$ . This is a closed linear subspace because $R$ is linear and $r \mapsto D_{r} ξ$ is linear. For each $r \in R$ , $D_{r} C \subseteq C$ : it holds on the orbit because $D_{r} D_{s} ξ = D_{r s} ξ$ , and extends to its closure because $D_{r}$ is bounded. Applying this to $r^{*} \in R$ gives $D_{r}^{*} C \subseteq C$ . If $η \in C^{⊥}$ and $ζ \in C$ , then [F5] gives $⟨ D_{r} η, ζ ⟩ = ⟨ η, D_{r}^{*} ζ ⟩ = 0$ , so $D_{r} C^{⊥} \subseteq C^{⊥}$ . The orthogonal projection $P_{C}$ from [F4] therefore commutes with every $D_{r}$ .

1.2F1F6given

Suppose $Ω$ is cyclic for $M$ , and let $y \in M^{'}$ satisfy $y Ω = 0$ . For every $x \in M$ , commutation gives $y x Ω = x y Ω = 0$ . Thus $y$ vanishes on the dense subspace $M Ω$ and, being bounded, is zero on $H$ . This proves that $Ω$ is separating for $M^{'}$ .

1.3F1F4F5F6algebra

Suppose $Ω$ is separating for $M^{'}$ . Let $C = \overline{M Ω}$ and let $P_{C}$ be its orthogonal projection from [F4]. For $a \in M$ , both $a$ and $a^{*}$ preserve $C$ , because $M$ is an algebra closed under adjoints. If $η \in C^{⊥}$ and $ξ \in C$ , then [F5] gives $⟨ a η, ξ ⟩ = ⟨ η, a^{*} ξ ⟩ = 0$ ; hence $a$ also preserves $C^{⊥}$ . Thus $a P_{C} = P_{C} a$ for all $a \in M$ , so $P_{C} \in M^{'}$ by [F1]. Since $I \in M$ , $Ω \in C$ and $(I - P_{C}) Ω = 0$ . Separatingness for $M^{'}$ forces $I - P_{C} = 0$ , whence $C = H$ and $Ω$ is cyclic for $M$ .

2.1F1F3step 1.1givenalgebra

Write $P_{C} = (P_{i j})_{i, j = 1}^{n}$ in coordinate blocks. The identity $P_{C} D_{r} = D_{r} P_{C}$ says $P_{i j} r = r P_{i j}$ for all $r \in R$ , so $P_{i j} \in R^{'}$ by [F1]. For $T \in R^{''}$ , it follows that $T$ commutes with every $P_{i j}$ ; thus $D_{T} P_{C} = P_{C} D_{T}$ . Since $I \in R$ , $ξ = D_{I} ξ \in C$ and $P_{C} ξ = ξ$ . Consequently $D_{T} ξ = P_{C} D_{T} ξ \in C$ .

3.1F2F3step 2.1

By the definition of $C$ , for every $ε > 0$ there is $r \in R$ with $∥ D_{T} ξ - D_{r} ξ ∥_{H^{\oplus n}} < ε$ . The direct-sum norm is the square root of the sum of the squared coordinate norms, so $∥ T ξ_{j} - r ξ_{j} ∥ < ε$ for each $j$ . These are exactly the finite strong-operator neighborhood tests. The empty tuple imposes no conditions. Therefore $T \in {\overline{R}}^{S O T}$ .

4.1A1F1F2F7step 3.1

Let $W = W^{*} (R) = {\overline{R}}^{W O T}$ . For each $x \in R^{'}$ , the set of operators commuting with $x$ is weak-operator closed and contains $R$ ; therefore it contains $W$ . This proves $R^{'} \subseteq W^{'}$ , while $R \subseteq W$ gives $W^{'} \subseteq R^{'}$ . Hence $R^{''} = W^{''} = W$ by [F7]. Strong convergence implies weak convergence, so ${\overline{R}}^{S O T} \subseteq W = R^{''}$ . Step 3.1 gives the reverse inclusion, proving the strong-closure identity.

5.1step 4.1step 1.2step 1.3∎

Steps 1.2 and 1.3 prove both directions of cyclic-separating duality, and step 4.1 proves the strong-closure identity.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Bounded increasing positive nets have strong suprema

Statement

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and $M \subseteq B (H)$ a concrete von Neumann algebra. Let $I$ be a nonempty directed set, and let $(x_{i})_{i \in I}$ be a net in $M$ such that each $x_{i}$ is positive, $x_{i} \leq x_{j}$ whenever $i \leq j$ , and $∥ x_{i} ∥ \leq C$ for some finite $C \geq 0$ . Here $A \leq B$ means that $A, B$ are self-adjoint and $⟨ A ξ, ξ ⟩ \leq ⟨ B ξ, ξ ⟩$ for every $ξ \in H$ . Then the net has a least upper bound $x = \sup_{i} x_{i}$ in $M_{+}$ , and $x_{i} \to x$ in the strong operator topology.

Statement references

Concrete von Neumann algebra and faithful vector

Strong and weak operator topologies

Facts & Assumptions

Given: [A1], a complex Hilbert space $H$ , a concrete von Neumann algebra $M \subseteq B (H)$ , a nonempty directed set $I$ , and the increasing positive net of the Statement.

[A1]

Assume Choice, the hypothesis for the weak-closure/double-commutant interface in Concrete von Neumann algebra and faithful vector and The Axiom of Choice.

[F1]

$M = M^{''}$ and commutants are weak-operator closed, so $M$ is weak-operator closed (Concrete von Neumann algebra and faithful vector, Von Neumann algebras and commutants).

[F2]

Weak-operator convergence is convergence of every matrix coefficient (Strong and weak operator topologies).

[F3]

A bounded sesquilinear form on $H$ is represented by a unique bounded operator, by applying Riesz representation to one argument (Riesz representation for Hilbert spaces).

[F4]

The complex inner product is linear in the first argument, and Cauchy–Schwarz bounds it by the product of norms (Real and complex inner-product spaces and their induced length, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

Proof

technique · direct

Given: [A1], [F1]–[F4] and the net in the Statement.

1.1A1F1F2F3F4givenconstruct

For $ξ \in H$ , the real numbers $q_{i} (ξ) : = ⟨ x_{i} ξ, ξ ⟩$ increase and satisfy $0 \leq q_{i} (ξ) \leq C ∥ ξ ∥^{2}$ . Define $q (ξ) : = \sup_{i \in I} q_{i} (ξ)$ . For each $i$ , the complex polarization identity gives $⟨ x_{i} ξ, η ⟩ = \frac{1}{4} \sum_{k = 0}^{3} i^{k} q_{i} (ξ + i^{k} η) .$ Each term on the right converges as $i$ increases, so the left-hand side converges for every $ξ, η$ . Let $b (ξ, η)$ be its limit. The forms $⟨ x_{i} ξ, η ⟩$ are sesquilinear and uniformly bounded by $C ∥ ξ ∥ ∥ η ∥$ , so their pointwise limit $b$ is sesquilinear with the same bound. For fixed $ξ$ , Riesz representation applied to $η \mapsto \overline{b (ξ, η)}$ gives a unique vector $x ξ$ with $b (ξ, η) = ⟨ x ξ, η ⟩$ . Sesquilinearity and uniqueness make $ξ \mapsto x ξ$ linear, and the bound gives $∥ x ξ ∥ \leq C ∥ ξ ∥$ ; thus $x \in B (H)$ and $∥ x ∥ \leq C$ . Since each $x_{i}$ is self-adjoint, $b (ξ, η) = \overline{b (η, ξ)}$ , hence $⟨ x ξ, η ⟩ = ⟨ ξ, x η ⟩$ and $x = x^{*}$ . Also $⟨ x ξ, ξ ⟩ = q (ξ) \geq 0$ , so $x$ is positive. By [F2], $x_{i} \to x$ weakly. Since $M$ is weakly closed by [F1], $x \in M$ .

2.1step 1.1given

For every $ξ$ , $q_{i} (ξ) \leq q (ξ) = ⟨ x ξ, ξ ⟩$ , so $x_{i} \leq x$ . If $y \in M_{+}$ is any other upper bound, then $q_{i} (ξ) \leq ⟨ y ξ, ξ ⟩$ for every $i, ξ$ ; taking the supremum over $i$ gives $⟨ x ξ, ξ ⟩ \leq ⟨ y ξ, ξ ⟩$ for every $ξ$ , hence $x \leq y$ . Thus $x = \sup_{i} x_{i}$ in $M_{+}$ .

3.1F4step 1.1step 2.1algebra∎

If $H = {0}$ , strong convergence is immediate. Otherwise fix $ξ \in H$ and put $D_{i} = x - x_{i}$ . Then $0 \leq D_{i} \leq x \leq C I$ and $⟨ D_{i} ξ, ξ ⟩ = q (ξ) - q_{i} (ξ) \to 0$ . Positivity gives the Cauchy–Schwarz inequality for the form $(α, β) \mapsto ⟨ D_{i} α, β ⟩$ : applying positivity to $ξ + z η$ for arbitrary $z \in C$ yields the nonnegative quantity $u + 2 Re (\overline{z} a) + ∣ z ∣^{2} v,$ where $u = ⟨ D_{i} ξ, ξ ⟩$ , $a = ⟨ D_{i} ξ, η ⟩$ , and $v = ⟨ D_{i} η, η ⟩$ . If $v > 0$ , taking $z = - a / v$ gives $∣ a ∣^{2} \leq u v .$ If $v = 0$ and $a \neq 0$ , taking $z = - t a$ for sufficiently large real $t > 0$ makes the quantity negative; hence $a = 0$ and the same inequality holds. Therefore $∣ ⟨ D_{i} ξ, η ⟩ ∣^{2} \leq ⟨ D_{i} ξ, ξ ⟩ ⟨ D_{i} η, η ⟩ \leq C ∥ η ∥^{2} ⟨ D_{i} ξ, ξ ⟩ .$ Taking the supremum over unit $η$ gives $∥ D_{i} ξ ∥^{2} \leq C ⟨ D_{i} ξ, ξ ⟩ \to 0$ . Therefore $∥ (x_{i} - x) ξ ∥ \to 0$ for every $ξ$ , which is strong-operator convergence.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Closed antilinear polar calculus and Tomita-operator closability

Statement

For this item, let $H$ be a complex Hilbert space and let $S : D (S) \subseteq H \to H$ be an antilinear operator. Set $Γ (S) : = {(x, S x) : x \in D (S)} \subseteq H \oplus H,$ where $H \oplus H$ has its direct-sum Hilbert norm. Use the conjugate space $\overline{H}$ , canonical conjugate-linear surjective isometry $c_{H} : H \to \overline{H}$ , and linearization $\tilde{S} : D (\tilde{S}) \subseteq \overline{H} \to H$ from the earlier conjugate-Hilbert-space and antilinear-adjoint definition. The map $c_{H} \oplus I_{H} : H \oplus H \to \overline{H} \oplus H$ given by $(x, y) \mapsto (c_{H} x, y)$ , with $I_{H}$ the identity on $H$ , is a real-linear isometric bijection and sends $Γ (S)$ onto $Γ (\tilde{S})$ . Call $S$ closed or closable when $\tilde{S}$ is closed or closable, respectively, in the complex-linear sense already defined for densely defined operators. For closable $S$ , $\overline{S}$ denotes the antilinear operator whose graph is the norm closure of $Γ (S)$ ; the proof verifies that this closure is a graph and equivalently that $\tilde{\overline{S}}$ is the graph closure of $\tilde{S}$ . An antiunitary on $H$ means a conjugate-linear, surjective isometry $H \to H$ .

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and let $S : D (S) \subseteq H \to H$ be a densely defined closable antilinear operator. Its graph closure $\overline{S}$ is closed and has the polar decomposition $\overline{S} = J ∣ \overline{S} ∣, ∣ \overline{S} ∣ = ({\overline{S}}^{*} \overline{S})^{1 / 2} .$ Here $J$ is antilinear, vanishes on $\ker \overline{S}$ , and is isometric on $(\ker \overline{S})^{⊥}$ ; if $\overline{S}$ is injective and has dense range, $J$ is antiunitary.

For a concrete von Neumann algebra $M \subseteq B (H)$ and a vector $Ω$ cyclic and separating for $M$ , define $D (S_{0}) = M Ω$ and $S_{0} (x Ω) = x^{*} Ω$ for $x \in M$ . Then $S_{0}$ is well defined, densely defined, antilinear, and closable.

Statement references

Concrete von Neumann algebra and faithful vector

Conjugate Hilbert space and adjoint of an antilinear operator

Unbounded linear operators: domain, graph and extension

Densely defined, closed and closable operators, and cores

Polar decomposition of a closed densely defined operator

Strong closure and cyclic-separating duality

The Axiom of Choice

Facts & Assumptions

Given: [A1], a complex Hilbert space $H$ , a densely defined closable antilinear $S$ , and, for the Tomita clause, a concrete von Neumann algebra $M$ with a cyclic and separating vector $Ω$ .

[A1]

Assume the Axiom of Choice, used by the local closed-operator polar lemma (The Axiom of Choice).

[F1]

The conjugate space $\overline{H}$ , canonical conjugate-linear isometry $c_{H}$ , linearization $\tilde{S}$ , and antilinear adjoint $S^{*}$ are defined by Conjugate Hilbert space and adjoint of an antilinear operator. In particular, $S^{*} = c_{H}^{- 1} {\tilde{S}}^{*}$ .

[F2]

A closed densely defined linear operator $A$ has polar decomposition $A = U B$ , where $B = (A^{*} A)^{1 / 2}$ is positive self-adjoint with $D (B) = D (A)$ , and $U$ has the stated initial and final spaces; when $A$ is injective with dense range, $U$ is unitary (Polar decomposition of a closed densely defined operator).

[F3]

For a concrete von Neumann algebra, a vector is cyclic for $M^{'}$ exactly when it is separating for $M^{''} = M$ (Strong closure and cyclic-separating duality, Concrete von Neumann algebra and faithful vector).

[F4]

A densely defined operator is closable exactly when its graph closure is the graph of a closed operator. Here this criterion is applied to the complex-linear operator $\tilde{S}$ : by the linear definition, closability means that $\tilde{S}$ has a closed linear extension. The closure of its graph is then contained in the graph of every such extension, so its vertical fiber is zero and the closed graph itself defines the smallest closed extension. Conversely, if that graph closure is a graph, it is a closed extension. The isometry $c_{H} \oplus I_{H}$ transfers this criterion to $Γ (S)$ ; its inverse sends a graph of a complex-linear operator to the graph of an antilinear operator. Thus it also defines $\overline{S}$ and identifies its graph with $\overline{Γ (S)}$ (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores).

[F5]

If $T$ is a closed densely defined antilinear operator and $A = \tilde{T}$ , then $T^{*} = c_{H}^{- 1} A^{*}$ and, with the operator-composition domains, $T^{*} T = c_{H}^{- 1} (A^{*} A) c_{H} .$ Conjugation by $c_{H}$ carries positive self-adjoint operators and their positive square roots to positive self-adjoint operators on $H$ ; this follows by transporting the spectral projections in the unbounded spectral calculus (Polar decomposition of a closed densely defined operator).

Proof

technique · direct

Given: [A1], [F1]–[F4], the operator $S$ , and the cyclic-separating data for $S_{0}$ .

1.1A1F1F2F5constructalgebra

Put $C = c_{H}$ and $T = \overline{S}$ . The definition of graph closure above makes $A : = \tilde{T}$ a closed densely defined complex-linear operator $\overline{H} \to H$ . By [F2], write $A = U B$ , where $B = (A^{*} A)^{1 / 2}$ is positive self-adjoint with $D (B) = D (A)$ and $U$ is a partial isometry with initial space $\overline{ran B} = (\ker A)^{⊥}$ , final space $\overline{ran A}$ , and kernel $\ker A$ . Set $J : = U C$ and $P : = C^{- 1} B C$ , with domain $D (P) = C^{- 1} D (B)$ . The composition $P$ is linear. If $E$ is the spectral measure of $B$ , then $F (Λ) : = C^{- 1} E (Λ) C$ is a spectral measure of orthogonal projections on $H$ , and its spectral integral is $P$ ; hence $P$ is positive self-adjoint. More explicitly, $D (T^{*} T) = {x : C x \in D (A), A C x \in D (A^{*})} = C^{- 1} D (A^{*} A)$ , and the adjoint convention [F1] gives $T^{*} T = C^{- 1} A^{*} A C$ there. Also $D (P^{2}) = C^{- 1} D (B^{2})$ and $P^{2} = C^{- 1} B^{2} C = C^{- 1} A^{*} A C = T^{*} T$ on that domain; uniqueness of the positive self-adjoint square root gives $P = (T^{*} T)^{1 / 2}$ . For $x \in D (T)$ , $J P x = U C C^{- 1} B C x = U B C x = A C x = T x$ , so $T = J P$ . The operator $J$ is antilinear, vanishes on $\ker T$ , and is isometric on $(\ker T)^{⊥}$ , since $C$ maps $\ker T$ onto $\ker A$ and preserves orthogonality. Its final space is $\overline{ran T}$ . If $T$ is injective and has dense range, then $A$ is injective with dense range, so $U$ is unitary by [F2] and $J = U C$ is antiunitary.

1.2givenF3algebra

For $x, y \in M$ , if $x Ω = y Ω$ , then $(x - y) Ω = 0$ ; separatingness of $Ω$ for $M$ forces $x = y$ . Thus the rule $S_{0} (x Ω) = x^{*} Ω$ is well defined. It is antilinear because $(a x + b y)^{*} = \overline{a} x^{*} + \overline{b} y^{*}$ , and its domain is dense because $Ω$ is cyclic for $M$ .

2.1givenF3F4step 1.2algebra

Let $u_{n} \in D ({\tilde{S}}_{0})$ satisfy $u_{n} \to 0$ in $\overline{H}$ and ${\tilde{S}}_{0} u_{n} \to η$ in $H$ . Write $u_{n} = C (x_{n} Ω)$ with $x_{n} \in M$ ; then $x_{n} Ω \to 0$ and $x_{n}^{*} Ω = S_{0} (x_{n} Ω) = {\tilde{S}}_{0} u_{n} \to η$ . For every $y^{'} \in M^{'}$ , the linear-first adjoint identity and commutation give $⟨ η, y^{'} Ω ⟩ = \lim_{n} ⟨ x_{n}^{*} Ω, y^{'} Ω ⟩ = \lim_{n} ⟨ y^{' *} Ω, x_{n} Ω ⟩ = 0.$ Since $Ω$ is separating for $M$ , [F3] makes it cyclic for $M^{'}$ , so $M^{'} Ω$ is dense and $η = 0$ . Thus the graph closure of the linear operator ${\tilde{S}}_{0}$ has zero vertical fiber: because $\overline{H} \oplus H$ is a metric space, every point of that closure is the limit of a sequence of graph points, and the argument just given excludes a nonzero limit over $0$ . This closed linear subspace is therefore the graph of a closed linear extension of ${\tilde{S}}_{0}$ . By [F4], ${\tilde{S}}_{0}$ is closable, and the local convention above proves that $S_{0}$ is closable with graph closure defined there.

3.1step 1.1step 1.2step 2.1∎

Step 1.1 proves the polar decomposition for the closure of every densely-defined closable antilinear operator, and steps 1.2–2.1 prove the Tomita operator clause with its stated cyclic-separating hypotheses. The Choice assumption is inherited through the spectral theorem used in [F2].

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Trace-class quotient predual and normal functionals

Statement

Assume the Axiom of Choice. Let $M \subseteq B (H)$ be a concrete von Neumann algebra on a complex Hilbert space $H$ , and define $M_{⊥} : = {T \in S_{1} (H) : Tr (T x) = 0 for every x \in M} .$ The ideal property makes $x \mapsto Tr (T x)$ a bounded linear functional on $M$ for each $T \in S_{1} (H)$ . Equip the coset space $Q : = S_{1} (H) / M_{⊥}$ with the quotient norm. The trace pairing induces an injective map $ι : Q \to M^{*}, ι ([T]) (x) : = Tr (T x),$ where $M^{*}$ is the space of bounded complex-linear functionals on the operator-normed space $M$ , with the operator norm. A member of $Q$ is called positive when its functional under $ι$ takes values in $[0, + \infty)$ on $M_{+}$ . Then the trace pairing $Ψ : M \to Q^{*}, (Ψ (x)) ([T]) = Tr (T x),$ is a well-defined isometric bijection, and $Q$ is a Banach space. Thus $Q$ is an isometric predual of $M$ , and its dual unit ball is weak-star compact. Every vector functional $x \mapsto ⟨ x ξ, ξ ⟩$ and every positive functional in $Q$ preserves suprema of bounded increasing positive nets in $M$ .

Statement references

Concrete von Neumann algebra and faithful vector

Trace class operator

Trace-class duality and finite-rank density

Trace class is a two sided Banach operator ideal

The dual space X^* of a normed space and its dual norm

The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))

The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)

A quotient of a Banach space by a closed subspace is Banach

The dual of a quotient is its annihilator

Banach–Alaoglu

The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter

Facts & Assumptions

Given: [A1], a concrete von Neumann algebra $M \subseteq B (H)$ , the trace-class space $S_{1} (H)$ , its trace pairing, and the quotient space in the Statement.

[A1]

Assume the Axiom of Choice, including the ultrafilter lemma needed for the Banach–Alaoglu compactness conclusion (The Axiom of Choice, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[F1]

Trace class is Banach, is an ideal under multiplication by bounded operators, and $∣ Tr (T) ∣ \leq ∥ T ∥_{1}$ (Trace class is a two sided Banach operator ideal, Trace is absolutely convergent and basis independent).

[F2]

Finite-rank operators are dense in trace norm, and every bounded linear functional on $S_{1} (H)$ is uniquely $Tr (\cdot X)$ for $X \in B (H)$ , isometrically (Trace-class duality and finite-rank density).

[F3]

A quotient of a Banach space by a closed subspace is Banach, and its dual is isometrically the annihilator of that subspace (A quotient of a Banach space by a closed subspace is Banach, The dual of a quotient is its annihilator, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)), The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), The dual space X^* of a normed space and its dual norm, Annihilator notation and the preannihilator).

[F4]

A bounded increasing positive net in $M$ has a supremum in $M$ and converges strongly to it (Bounded increasing positive nets have strong suprema).

[F5]

The dual unit ball of a normed space is weak-star compact under the ultrafilter lemma, and Choice supplies that lemma (Banach–Alaoglu, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, The Axiom of Choice).

[F6]

Since $M = M^{''}$ and commutants are weak-operator closed, $M$ is weak-operator closed. The Hilbert-space weak operator topology in Concrete von Neumann algebra and faithful vector is the topology of pointwise matrix-coefficient convergence, so its basic neighborhoods impose finitely many inequalities $∣ ⟨ (Y - X) ξ_{j}, η_{j} ⟩ ∣ < ε_{j}$ with $ε_{j} > 0$ . This uses that Definition's matrix-coefficient interface directly, rather than a representation theorem for $H^{*}$ (Von Neumann algebras and commutants, Strong and weak operator topologies).

Proof

technique · direct

Given: [F1]–[F5], the trace pairing, and the concrete algebra $M$ .

1.1F1F3givenconstruct

For every $x \in M$ , the functional $T \mapsto Tr (T x)$ on $S_{1} (H)$ is bounded by $∣ Tr (T x) ∣ \leq ∥ T ∥_{1} ∥ x ∥$ , using [F1]. Thus its common kernel over $x \in M$ is a closed linear subspace, namely $M_{⊥}$ . If $S \in M_{⊥}$ , then $ι ([T]) = ι ([T + S])$ ; taking the infimum over representatives in the trace-norm estimate shows $∥ ι ([T]) ∥_{M^{*}} \leq ∥ [T] ∥_{Q}$ . Its kernel is zero by the definition of $M_{⊥}$ . The quotient $Q = S_{1} (H) / M_{⊥}$ is Banach by [F1] and [F3]. The formula $Ψ (x) ([T]) = Tr (T x)$ is independent of the representative because every element of $M_{⊥}$ annihilates $x$ .

1.2F2F3

By [F3], $Q^{*}$ is isometrically identified with $M_{⊥}^{⊥} \subseteq S_{1} (H)^{*}$ . By [F2], $S_{1} (H)^{*}$ is isometrically $B (H)$ under the trace pairing. Under this identification, $M_{⊥}^{⊥}$ consists exactly of the bounded operators $X$ satisfying $Tr (T X) = 0$ for all $T \in M_{⊥}$ . It remains to identify this double annihilator with $M$ .

1.3F2F6givenalgebraconstruct

Every $x \in M$ belongs to $M_{⊥}^{⊥}$ by the definition of $M_{⊥}$ . Conversely, let $X \in B (H) ∖ M$ . By [F6], there is a basic weak-operator neighborhood of $X$ , defined by finitely many pairs $(ξ_{j}, η_{j})$ and tolerances, that is disjoint from $M$ . The coordinate map $θ (Y) = (⟨ Y ξ_{j}, η_{j} ⟩)_{j = 1}^{m}$ has $θ (X) \notin θ (M)$ ; $θ (M)$ is a linear subspace of $C^{m}$ and hence is closed. Finite-dimensional linear algebra supplies a linear functional $ℓ (z) = \sum_{j = 1}^{m} c_{j} z_{j}$ that vanishes on $θ (M)$ but not on $θ (X)$ . Set $T = \sum_{j = 1}^{m} c_{j} R_{ξ_{j}, η_{j}}$ , where $R_{ξ, η} z = ⟨ z, η ⟩ ξ$ . Then $T$ is finite-rank trace class and $Tr (T Y) = ℓ (θ (Y))$ for every $Y \in B (H)$ . Thus $T \in M_{⊥}$ but $Tr (T X) \neq 0$ , so $X \notin M_{⊥}^{⊥}$ . Therefore $M_{⊥}^{⊥} = M$ .

2.1A1F2F3F5step 1.2step 1.3

Steps 1.2 and 1.3 identify the canonical map $Ψ$ with the isometric duality identification of [F2]–[F3], so $Ψ$ is an isometric bijection $M \to Q^{*}$ . The Banach–Alaoglu theorem [F5] applied to $Q$ then makes the closed dual unit ball of $Q^{*}$ weak-star compact.

2.2F1F2F4F6step 1.1algebra

Let $(x_{i})$ be a bounded increasing positive net in $M$ with supremum $x$ . By [F4], $x_{i} \to x$ strongly; choose $C \geq 0$ with $∥ x_{i} ∥ \leq C$ for every $i$ . For every $ξ \in H$ , strong convergence gives $∥ x ξ ∥ = \lim_{i} ∥ x_{i} ξ ∥ \leq C ∥ ξ ∥$ , so $∥ x ∥ \leq C$ . For each $T \in S_{1} (H)$ , choose finite-rank $T_{n} \to T$ in trace norm by [F2]. Strong convergence gives $Tr (T_{n} x_{i}) \to Tr (T_{n} x)$ for each fixed $n$ , since $T_{n}$ is a finite sum of rank-one operators. Uniformly in $i$ , $∣ Tr ((T - T_{n}) x_{i}) ∣ \leq C ∥ T - T_{n} ∥_{1}, ∣ Tr ((T - T_{n}) x) ∣ \leq C ∥ T - T_{n} ∥_{1} .$ Given $ε > 0$ , first fix $n$ so that the sum of these two uniform errors is less than $ε / 2$ , then choose a net index after which the finite-rank trace difference is less than $ε / 2$ . This proves $Tr (T x_{i}) \to Tr (T x)$ . Hence every $ϕ = [T] \in Q$ satisfies $ι (ϕ) (x_{i}) \to ι (ϕ) (x)$ . If $ϕ$ is positive, then $ι (ϕ) (x_{i})$ increases and converges to $ι (ϕ) (x)$ , so $ι (ϕ) (x) = \sup_{i} ι (ϕ) (x_{i})$ . In particular, for $ξ \in H$ the vector functional is represented by $[R_{ξ, ξ}]$ , since $ι ([R_{ξ, ξ}]) (x) = Tr (R_{ξ, ξ} x) = ⟨ x ξ, ξ ⟩$ ; it is positive and therefore normal as well.

3.1step 1.1step 1.2step 1.3step 2.1step 2.2∎

Steps 1.1–1.3 and 2.1–2.2 prove the quotient predual, its weak-star compact dual ball, normality of positive functionals in the quotient, and normality of vector functionals. The positive-functional normality proof uses trace-norm approximation by finite-rank operators; it does not assume a positive trace-class lift for every positive functional on $M$ .

5 · Examples, experiments and counterexamples

None yet.

Von Neumann closure, standard vectors and modular analytic tools

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Concrete von Neumann algebra and faithful vector

Definition

Definition references

Remarks

Facts & Assumptions

Proof

Conjugate Hilbert space and adjoint of an antilinear operator

Definition

Definition references

Remarks

Facts & Assumptions

Proof

Trace-class duality and finite-rank density

Statement

Statement references

Facts & Assumptions

Proof

Polar decomposition of a closed densely defined operator

Statement

Statement references

Facts & Assumptions

Proof

Strong closure and cyclic-separating duality

Statement

Statement references

Facts & Assumptions

Proof

Bounded increasing positive nets have strong suprema

Statement

Statement references

Facts & Assumptions

Proof

Closed antilinear polar calculus and Tomita-operator closability

Statement

Statement references

Facts & Assumptions

Proof

Trace-class quotient predual and normal functionals

Statement

Statement references

Facts & Assumptions

Proof

5 · Examples, experiments and counterexamples

Sources