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Quantization of unitaries preserving the positive subspace onto itself

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QAlgorithms.Papers.Independent.Intakethird100.P92.unitary_onto_quantization

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let (E,E+,θ)(E,E_+,\theta)(E,E+​,θ) be reflection positive, let q:E+→E^q:E_+\to\widehat Eq:E+​→E be its canonical quotient, and let UUU be a unitary operator with UE+=E+UE_+=E_+UE+​=E+​. Then:

  1. If θUθ=U\theta U\theta=UθUθ=U, the induced operator U^\widehat UU is unitary.
  2. If θUθ=U−1\theta U\theta=U^{-1}θUθ=U−1, the induced operator satisfies U^2=1\widehat U^2=1U2=1. There are closed orthogonal complementary subspaces F,G⊆EF,G\subseteq EF,G⊆E, invariant under θ,U,U−1\theta,U,U^{-1}θ,U,U−1, giving reflection-positive triples (F,F∩E+,θ∣F)(F,F\cap E_+,\theta|_F)(F,F∩E+​,θ∣F​) and (G,G∩E+,θ∣G)(G,G\cap E_+,\theta|_G)(G,G∩E+​,θ∣G​), such that G^=0\widehat G=0G=0 and (U∣F)2=1(U|_F)^2=1(U∣F​)2=1.
E=F⊕G,q(G∩E+)={0},U2∣F=1.E=F\oplus G,\qquad q(G\cap E_+)=\{0\},\qquad U^2|_F=1.E=F⊕G,q(G∩E+​)={0},U2∣F​=1.

The equation involving qqq expresses G^=0\widehat G=0G=0, because the reflected form on G∩E+G\cap E_+G∩E+​ is the restricted ambient form. This statement keeps both alternatives and the existence of the invariant decomposition.

The positive subspace also splits as E+=(E+∩F)⊕(E+∩G)E_+=(E_+\cap F)\oplus(E_+\cap G)E+​=(E+​∩F)⊕(E+​∩G), so the decomposition is a direct sum of reflection-positive triples.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P92
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P92

open scoped ComplexConjugate BigOperators NNReal ENNReal

open MeasureTheory

attribute [local instance] Classical.propDecidable

universe u v

variable (E : Type u) [NormedAddCommGroup E] [InnerProductSpace ℂ E] [CompleteSpace E]

variable {E}

/-- Lemma 2.4, PDF p. 9. Both alternatives and the invariant decomposition
are retained. The Hilbert sum is orthogonal and Eplus splits into its
intersections with F and G. The zero OS space of G is expressed by vanishing of the canonical
quotient on G intersect Eplus. No erroneous formula for G from the proof is used. -/
theorem unitary_onto_quantization [hHilbert : CompleteSpace E] (r : ReflectionData E) (U : E ≃ₗᵢ[ℂ] E)
    (hOnto : r.plus.map U.toLinearMap = r.plus) :
    ((∀ x, r.theta (U (r.theta x)) = U x) →
      ∃ V : OS r ≃ₗᵢ[ℂ] OS r, ∀ v : r.plus,
        ∃ hv : U (v : E) ∈ r.plus, V (osQ r v) = osQ r ⟨U v, hv⟩) ∧
    ((∀ x, r.theta (U (r.theta x)) = U.symm x) →
      ∃ A : OS r →L[ℂ] OS r,
        (∀ v : r.plus, ∃ hv : U (v : E) ∈ r.plus,
          A (osQ r v) = osQ r ⟨U v, hv⟩) ∧
        A.comp A = ContinuousLinearMap.id ℂ (OS r) ∧
        ∃ F G : Submodule ℂ E,
          IsClosed (F : Set E) ∧ IsClosed (G : Set E) ∧
          Disjoint F G ∧ F ⊔ G = ⊤ ∧
          Submodule.IsOrtho F G ∧
          r.plus = (r.plus ⊓ F) ⊔ (r.plus ⊓ G) ∧
          F.map r.theta.toLinearMap = F ∧ G.map r.theta.toLinearMap = G ∧
          F.map U.toLinearMap = F ∧ G.map U.toLinearMap = G ∧
          F.map U.symm.toLinearMap = F ∧ G.map U.symm.toLinearMap = G ∧
          (∀ v : r.plus, (v : E) ∈ G → osQ r v = 0) ∧
          (∀ v ∈ F, U (U v) = v)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P92

end
Source
Reflection positive one-parameter groups and dilations, arXiv:1312.6161v2, PDF p. 9, Lemma 2.4
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let EEE be a complete complex inner-product space, let P⊆EP\subseteq EP⊆E be a closed complex linear subspace, and let θ:E→E\theta:E\to Eθ:E→E be a complex linear isometric bijection such that θ2x=x\theta^2x=xθ2x=x for every x∈Ex\in Ex∈E and Re⁡⟨θx,x⟩≥0\operatorname{Re}\langle\theta x,x\rangle\geq0Re⟨θx,x⟩≥0 for every x∈Px\in Px∈P. Let HHH be the separated completion of PPP equipped with the reflected form (x,y)↦⟨θx,y⟩(x,y)\mapsto\langle\theta x,y\rangle(x,y)↦⟨θx,y⟩ and its seminorm x↦Re⁡⟨θx,x⟩x\mapsto\sqrt{\operatorname{Re}\langle\theta x,x\rangle}x↦Re⟨θx,x⟩​, and write q:P→Hq:P\to Hq:P→H for the canonical complex linear map into this completion, which identifies vectors at zero seminorm distance. For every complex linear isometric bijection U:E→EU:E\to EU:E→E satisfying U(P)=PU(P)=PU(P)=P, both of the following implications hold: (i) if θ(U(θx))=Ux\theta(U(\theta x))=Uxθ(U(θx))=Ux for every x∈Ex\in Ex∈E, then there exists a complex linear isometric bijection V:H→HV:H\to HV:H→H such that, for every v∈Pv\in Pv∈P, UvUvUv belongs to PPP and V(q(v))=q(Uv)V(q(v))=q(Uv)V(q(v))=q(Uv); (ii) if θ(U(θx))=U−1x\theta(U(\theta x))=U^{-1}xθ(U(θx))=U−1x for every x∈Ex\in Ex∈E, then there exists a continuous complex linear map A:H→HA:H\to HA:H→H such that, for every v∈Pv\in Pv∈P, UvUvUv belongs to PPP and A(q(v))=q(Uv)A(q(v))=q(Uv)A(q(v))=q(Uv), such that A∘A=id⁡HA\circ A=\operatorname{id}_HA∘A=idH​, and such that there exist closed complex linear subspaces F,G⊆EF,G\subseteq EF,G⊆E satisfying all of

F∩G={0},F+G=E,⟨f,g⟩=0for every f∈F, g∈G,F\cap G=\{0\},\qquad F+G=E,\qquad \langle f,g\rangle=0\quad\text{for every }f\in F,\ g\in G,F∩G={0},F+G=E,⟨f,g⟩=0for every f∈F, g∈G, P=(P∩F)+(P∩G),P=(P\cap F)+(P\cap G),P=(P∩F)+(P∩G), θ(F)=F,θ(G)=G,U(F)=F,U(G)=G,U−1(F)=F,U−1(G)=G,\theta(F)=F,\quad \theta(G)=G,\quad U(F)=F,\quad U(G)=G,\quad U^{-1}(F)=F,\quad U^{-1}(G)=G,θ(F)=F,θ(G)=G,U(F)=F,U(G)=G,U−1(F)=F,U−1(G)=G, q(v)=0for every v∈P∩G,U(Uv)=vfor every v∈F.q(v)=0\quad\text{for every }v\in P\cap G,\qquad U(Uv)=v\quad\text{for every }v\in F.q(v)=0for every v∈P∩G,U(Uv)=vfor every v∈F.

Here each occurrence of q(Uv)q(Uv)q(Uv) uses UvUvUv as an element of PPP, with its membership supplied by the asserted existence of a membership proof. No nonzero or positive-dimensional assumption is imposed on EEE, PPP, HHH, FFF, or GGG; zero spaces are included.

Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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