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Shadow Theory

Observable Quotients and Exact Projected Dynamics: Closure, Memory, Minimal Dynamical Completion, and Effective Field Operators

Descent of observable functionals

Jeremy RodgersSection 4 of 14Original PDF ↗Publication DOI ↗

Descent of observable functionals

Definition 3.1 (Exact observability of a functional).

Let VV be a set and A:Hsrc→VA:\mathcal{H}_{\mathrm{src}}\to V. We say AA is exactly observable through KK if there exists Aˉ:Ran⁡K→V\bar A:\operatorname{Ran}K\to V with A=Aˉ∘K.(3.1) A=\bar A\circ K .\tag{3.1}

Proposition 3.2 (Functional descent).

AA is exactly observable through KK if and only if A(Ψ+n)=A(Ψ)for all Ψ∈Hsrc, n∈ker⁡K.(3.2) A(\Psi+n)=A(\Psi) \qquad\text{for all }\Psi\in\mathcal{H}_{\mathrm{src}},\ n\in\ker K .\tag{3.2} When (3.2) holds, Aˉ\bar A is unique.

Proof. If (3.1) holds and n∈ker⁡Kn\in\ker K, then A(Ψ+n)=Aˉ(KΨ+Kn)=Aˉ(KΨ)=A(Ψ)A(\Psi+n)=\bar A(K\Psi+Kn)=\bar A(K\Psi)=A(\Psi), which is (3.2).

Conversely, assume (3.2). Given ω∈Ran⁡K\omega\in\operatorname{Ran}K, choose any Ψ\Psi with KΨ=ωK\Psi=\omega and set Aˉ(ω):=A(Ψ)\bar A(\omega):=A(\Psi). If KΨ=KΨ′=ωK\Psi=K\Psi'=\omega then Ψ′=Ψ+n\Psi'=\Psi+n with n∈ker⁡Kn\in\ker K, so (3.2) gives A(Ψ′)=A(Ψ)A(\Psi')=A(\Psi); the definition is therefore independent of the chosen preimage, and (3.1) holds by construction. Uniqueness is forced because KK is onto Ran⁡K\operatorname{Ran}K: any Aˉ′\bar A' satisfying (3.1) agrees with Aˉ\bar A at every ω=KΨ\omega=K\Psi. ◻

Proposition 3.2 is the fiberwise descent statement of [22] specialized to the affine fibers Ψ+ker⁡K\Psi+\ker K; we have proved it here so that no external result is load-bearing.

Corollary 3.3 (Linear descent and the adjoint).

Let A∈Hsrc∗A\in\mathcal{H}_{\mathrm{src}}^{*} be a continuous linear functional. Then AA descends through KK, in the sense of Definition 3.1, if and only if A∣ker⁡K=0.(3.3) A|_{\ker K}=0 .\tag{3.3} Identifying Hsrc∗≅Hsrc\mathcal{H}_{\mathrm{src}}^{*}\cong\mathcal{H}_{\mathrm{src}} by the Riesz representation A=⟨a,⋅⟩A=\langle a , \cdot \rangle, condition (3.3) says a∈(ker⁡K)⊥a\in(\ker K)^{\perp}, and (ker⁡K)⊥=Ran⁡K∗‾.(3.4) (\ker K)^{\perp}=\overline{\operatorname{Ran}K^{*}} .\tag{3.4}

Proof. Since AA is linear, (3.2) reads A(n)=0A(n)=0 for all n∈ker⁡Kn\in\ker K, which is (3.3). For (3.4): Ψ∈ker⁡K\Psi\in\ker K iff KΨ=0K\Psi=0 iff ⟨KΨ,ω⟩=0\langle K\Psi , \omega \rangle=0 for all ω∈Hobs\omega\in\mathcal{H}_{\mathrm{obs}} iff ⟨Ψ,K∗ω⟩=0\langle \Psi , K^{*}\omega \rangle=0 for all ω\omega, i.e. iff Ψ⊥Ran⁡K∗\Psi\perp\operatorname{Ran}K^{*}. Hence ker⁡K=(Ran⁡K∗)⊥\ker K=(\operatorname{Ran}K^{*})^{\perp}, and taking orthogonal complements and using M⊥⊥=M‾M^{\perp\perp}=\overline{M} for a subspace MM gives (ker⁡K)⊥=Ran⁡K∗‾(\ker K)^{\perp}=\overline{\operatorname{Ran}K^{*}}. ◻

It is essential to notice what Definition 3.1 does and does not require. It requires a factorization, A=Aˉ∘KA=\bar A\circ K, and nothing more: Aˉ\bar A is a map on the set Ran⁡K\operatorname{Ran}K, and no continuity of Aˉ\bar A with respect to any particular norm on Ran⁡K\operatorname{Ran}K has been demanded. Corollary 3.3 therefore characterizes algebraic descent. For a continuous linear A=⟨a,⋅⟩A=\langle a , \cdot \rangle with a∈(ker⁡K)⊥a\in(\ker K)^{\perp}, the induced Aˉ\bar A is automatically continuous in the transported quotient norm ∥⋅∥♯\lVert\cdot\rVert_{\sharp} of Remark 2.4 (because Aˉ∘K~\bar A\circ\widetilde{K} is the continuous functional [Ψ]↦⟨a,Ψ⟩[\Psi]\mapsto\langle a , \Psi \rangle on the Hilbert space HK\mathcal{H}_{K}). But continuity in the inherited observable norm, the norm an experimenter actually has access to on Hobs\mathcal{H}_{\mathrm{obs}}, is a strictly stronger demand, and it is governed by a different, sharper condition.

Theorem 3.4 (Stable observability of a linear functional).

Let A=⟨a,⋅⟩∈Hsrc∗A=\langle a , \cdot \rangle\in\mathcal{H}_{\mathrm{src}}^{*} with a∈(ker⁡K)⊥a\in(\ker K)^{\perp}, so that AA descends algebraically, with unique Aˉ:Ran⁡K→C\bar A:\operatorname{Ran}K\to\mathbb{C} satisfying A=Aˉ∘KA=\bar A\circ K. Then Aˉ is bounded on (Ran⁡K,∥⋅∥Hobs)⟺a∈Ran⁡K∗,(3.5) \bar A\ \text{is bounded on }(\operatorname{Ran}K,\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}}) \qquad\Longleftrightarrow\qquad a\in\operatorname{Ran}K^{*} ,\tag{3.5} i.e. if and only if there exists an exact observable probe ω∈Hobs\omega\in\mathcal{H}_{\mathrm{obs}} with a=K∗ωa=K^{*}\omega. In that case Aˉ\bar A extends to the bounded functional ⟨ω,⋅⟩\langle \omega , \cdot \rangle on all of Hobs\mathcal{H}_{\mathrm{obs}}, and A(Ψ)=⟨ω,KΨ⟩HobsA(\Psi)=\langle \omega , K\Psi \rangle_{\mathcal{H}_{\mathrm{obs}}} for every Ψ\Psi: the source quantity is read off the observable state by a bounded measurement.

Proof. (⇐\Leftarrow) Suppose a=K∗ωa=K^{*}\omega with ω∈Hobs\omega\in\mathcal{H}_{\mathrm{obs}}. Then for every Ψ\Psi, A(Ψ)=⟨a,Ψ⟩=⟨K∗ω,Ψ⟩=⟨ω,KΨ⟩Hobs,A(\Psi)=\langle a , \Psi \rangle=\langle K^{*}\omega , \Psi \rangle=\langle \omega , K\Psi \rangle_{\mathcal{H}_{\mathrm{obs}}}, so Aˉ(y)=⟨ω,y⟩Hobs\bar A(y)=\langle \omega , y \rangle_{\mathcal{H}_{\mathrm{obs}}} for y∈Ran⁡Ky\in\operatorname{Ran}K, which satisfies ∣Aˉ(y)∣≤∥ω∥ ∥y∥Hobs\lvert\bar A(y)\rvert\le\lVert\omega\rVert\,\lVert y\rVert_{\mathcal{H}_{\mathrm{obs}}} and is therefore bounded in the inherited norm.

(⇒\Rightarrow) Suppose Aˉ\bar A is bounded on (Ran⁡K,∥⋅∥Hobs)(\operatorname{Ran}K,\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}}), say ∣Aˉ(y)∣≤M∥y∥Hobs\lvert\bar A(y)\rvert\le M\lVert y\rVert_{\mathcal{H}_{\mathrm{obs}}} for all y∈Ran⁡Ky\in\operatorname{Ran}K. Since Ran⁡K\operatorname{Ran}K is a subspace of Hobs\mathcal{H}_{\mathrm{obs}}, the Hahn–Banach theorem [1, Cor. 1.2] extends Aˉ\bar A to a bounded linear functional Λ\Lambda on all of Hobs\mathcal{H}_{\mathrm{obs}} with the same norm. By the Riesz representation theorem there is ω∈Hobs\omega\in\mathcal{H}_{\mathrm{obs}} with Λ=⟨ω,⋅⟩\Lambda=\langle \omega , \cdot \rangle. Then for every Ψ∈Hsrc\Psi\in\mathcal{H}_{\mathrm{src}}, ⟨a,Ψ⟩=A(Ψ)=Aˉ(KΨ)=Λ(KΨ)=⟨ω,KΨ⟩=⟨K∗ω,Ψ⟩,\langle a , \Psi \rangle=A(\Psi)=\bar A(K\Psi)=\Lambda(K\Psi)=\langle \omega , K\Psi \rangle =\langle K^{*}\omega , \Psi \rangle , so ⟨a−K∗ω,Ψ⟩=0\langle a-K^{*}\omega , \Psi \rangle=0 for all Ψ\Psi, i.e. a=K∗ω∈Ran⁡K∗a=K^{*}\omega\in\operatorname{Ran}K^{*}. ◻

Remark 10 (Algebraic descent versus stable readout: exact preimage and closure).

Equation (3.4) contains a closure and one must not delete it, and Theorem 3.4 says exactly what the closure costs. Combining the two:

  • a∈Ran⁡K∗‾=(ker⁡K)⊥a\in\overline{\operatorname{Ran}K^{*}}=(\ker K)^{\perp}   ⟺  \iff AA descends algebraically through KK (Corollary 3.3), with Aˉ\bar A continuous in the transported quotient norm;

  • a∈Ran⁡K∗a\in\operatorname{Ran}K^{*}   ⟺  \iff AA descends to a functional that is bounded in the inherited observable norm (Theorem 3.4).

The two conditions coincide precisely when Ran⁡K∗\operatorname{Ran}K^{*} is closed, which by the closed range theorem is equivalent to Ran⁡K\operatorname{Ran}K being closed [1, Thm. 2.19], the same dichotomy as in Theorem 2.3. So for a compact KK with infinite-dimensional range, for instance Example 4.5, Ran⁡K∗\operatorname{Ran}K^{*} is dense in (ker⁡K)⊥(\ker K)^{\perp} but proper, and there exist source functionals which descend perfectly well as maps of the observable state, yet possess no exact representing probe ω\omega with a=K∗ωa=K^{*}\omega. Such a functional is a limit of representable ones, but the approximating probes ωk\omega_{k} have ∥ωk∥→∞\lVert\omega_{k}\rVert\to\infty: reading the quantity off the observable state amplifies observational error without bound. This is (D2) reappearing on the dual side. Accordingly we say that the elements of (ker⁡K)⊥(\ker K)^{\perp} are the observable linear source directions, and reserve stably observable for those in Ran⁡K∗\operatorname{Ran}K^{*}; the phrase should never be read as asserting inherited-readout stability where only algebraic descent has been established.

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