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Cook isometry and intertwining

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and adopt all hypotheses of Cook domain criterion, writing its Hilbert space as $H$ and renaming its operators $A_{0}, A$ as $H_{0}, H$ : these are self-adjoint operators on $H$ with $D (H_{0}) \subseteq D (H)$ , $D \subseteq D (H_{0})$ dense and $e^{- i t H_{0}}$ -invariant, $\int_{half-line} ∥ (H - H_{0}) e^{- i t H_{0}} ψ ∥ d t < \infty$ for $ψ \in D$ , $t \mapsto (H - H_{0}) e^{- i t H_{0}} ψ$ norm-continuous for each $ψ \in D$ , and $Ω_{\pm}$ the extensions constructed there. Then:

(a) Intertwining. $Ω_{\pm} e^{- i s H_{0}} = e^{- i s H} Ω_{\pm}$ for all $s \in R$ , as bounded operators on $H$ .

(b) Domain and commutation. $Ω_{\pm} D (H_{0}) \subseteq D (H)$ and $H Ω_{\pm} ψ = Ω_{\pm} H_{0} ψ$ for every $ψ \in D (H_{0})$ .

(c) Range invariance. The closed subspace $ran Ω_{\pm}$ is invariant under $e^{- i s H}$ for every $s$ , and $e^{- i s H} ∣_{ran Ω_{\pm}} = Ω_{\pm} e^{- i s H_{0}} Ω_{\pm}^{- 1}$ in the sense of the isometric identification $Ω_{\pm} : H \to ran Ω_{\pm}$ .

(d) $Ω_{\pm}^{*} Ω_{\pm} = I$ , $∥ Ω_{\pm} η ∥ = ∥ η ∥$ for every $η \in H$ ; no surjectivity onto $H$ and no identification of $ran Ω_{\pm}$ with an absolutely continuous subspace is asserted.

Facts & Assumptions

Given: the data and conclusions of Cook domain criterion; $s \in R$ ; vectors $ψ \in D$ and $η \in D (H_{0})$ .

[F1]

$Ω_{\pm}$ are bounded linear isometries extending the full strong limits on all vectors; $Ω_{\pm}^{*} Ω_{\pm} = I$ and $∥ Ω_{\pm} ∥ \leq 1$ (Cook domain criterion). Their norm is exactly $1$ when the Hilbert space is nonzero, since the norm of every unit vector is preserved, and exactly $0$ on the zero Hilbert space.

[F2]

The explicit autonomous construction and adjoint-testing generator characterization in Cook domain criterion give strongly continuous groups on arbitrary Hilbert spaces under CC, with $η \in D (A)$ exactly when $(e^{- i t A} η - η) / t$ has a strong limit, then equal to $- i A η$ . These are the full generator domains in Infinitesimal generator of a unitary group.

Proof

1.1F1F2givenalgebra

[Correct time shift on the full strong limits] For any real $s, t$ , the two group laws give $F (t) e^{- i s H_{0}} = e^{- i s H} F (t + s)$ , because both sides equal $e^{i t H} e^{- i (t + s) H_{0}}$ . The criterion gives full strong limits on every vector, and a fixed time shift preserves either infinite-time direction. Taking $t \to + \infty$ or $t \to - \infty$ and using boundedness of the fixed group factors therefore gives $Ω_{\pm} e^{- i s H_{0}} = e^{- i s H} Ω_{\pm}$ on the whole Hilbert space. This proves (a).

2.1step 1.1F1F2algebra

[Domain invariance and commutation] Fix $η \in D (H_{0})$ and consider the difference quotient $\frac{1}{t} (e^{- i t H} Ω_{+} η - Ω_{+} η)$ . By step 1.1 this equals $Ω_{+} \frac{1}{t} (e^{- i t H_{0}} η - η)$ ; since $η \in D (H_{0})$ , the quotient converges to $Ω_{+} (- i H_{0} η)$ by [F2] and [F1]. Applying [F2] to self-adjoint $H$ shows $Ω_{+} η \in D (H)$ and $- i H Ω_{+} η = - i Ω_{+} H_{0} η$ , that is $H Ω_{+} η = Ω_{+} H_{0} η$ . The argument for $Ω_{-}$ is identical.

2.2step 1.1F1algebra

[Range invariance] Fix either sign. For $η = Ω_{\pm} ψ \in ran Ω_{\pm}$ and $s \in R$ , step 1.1 gives $e^{- i s H} η = Ω_{\pm} (e^{- i s H_{0}} ψ)$ , which lies in $ran Ω_{\pm}$ ; hence the range is invariant, and the displayed operator identity holds on the range because $Ω_{\pm}$ is injective with $Ω_{\pm}^{- 1} η = ψ$ on the range.

3.1step 1.1step 2.1step 2.2F1∎

[Conclusion] Claims (a), (b), (c) are steps 1.1, 2.1, 2.2, and (d) is part of [F1]. No surjectivity or absolute-continuity statement has been used.

Remarks

For a bounded isometry $Ω$ , convergence $F (t) ψ \to Ω ψ$ implies $Ω F (t) ψ - F (t) ψ \to Ω^{2} ψ - Ω ψ$ . Isometry alone does not make this limit zero; no such assertion is used in the proof.

Cook isometry and intertwining

Statement

Facts & Assumptions

Proof

Remarks

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