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Average ensemble error and existence of a good frozen vector

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QAlgorithms.Papers.Independent.Intakethird100.P14.ensemble_error_bound

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

prove2quantum

Let WWW be any binary-input classical–quantum channel with positive trace-one density outputs, let N=2nN=2^nN=2n, and let A⊆{1,…,N}A\subseteq\{1,\ldots,N\}A⊆{1,…,N} be an arbitrary information set, with K=∣A∣K=|A|K=∣A∣. Encode by GNG_NGN​ and use the prescribed collective successive cancellation decoder: square-root-density spectral tests at information positions, identity at frozen positions, and uniform unknown future bits. Write Pe(N,K,A,f)P_e(N,K,A,f)Pe​(N,K,A,f) for block error averaged uniformly over information messages with frozen vector fff, and Pe(N,K,A)P_e(N,K,A)Pe​(N,K,A) for its further uniform average over fff. Then

Pe(N,K,A)≤2∑i∈A12F(WN(i)),∃f:Pe(N,K,A,f)≤2∑i∈A12F(WN(i)).P_e(N,K,A)\le2\sqrt{\sum_{i\in A}\frac12\sqrt{F(W_N^{(i)})}},\qquad \exists f:\quad P_e(N,K,A,f)\le2\sqrt{\sum_{i\in A}\frac12\sqrt{F(W_N^{(i)})}}.Pe​(N,K,A)≤2i∈A∑​21​F(WN(i)​)​​,∃f:Pe​(N,K,A,f)≤2i∈A∑​21​F(WN(i)​)​​.

The fidelity FFF is squared fidelity. The result requires no polar selection rule and includes A=∅A=\varnothingA=∅, when every bit is frozen and block error is zero. It transfers split-channel reliability to an actual sequential decoder.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P14
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P14

open Filter MeasureTheory

open scoped ENNReal NNReal ComplexOrder BigOperators

open QAlgorithms.RealRegisters (Hilbert)

open QAlgorithms.Papers.Independent.Dependency50.P33 (IsDensity)

open QAlgorithms.Papers.Independent.Dependency50.P13 (compression)

open QAlgorithms.Papers.Independent.Renew46.P44 (wordOutput)

attribute [local instance] Classical.propDecidable Classical.decEq

/-- Proposition 4, PDF p.6; proof p.9. Both the frozen-vector average and
existence of a frozen vector are conclusions, for every information set.
The factor 2 is OUTSIDE sqrt and the factor 1/2 is INSIDE the sum. -/
theorem ensemble_error_bound {I : Type} (W : Bit → OutputCoefficients I) (hW : IsBinaryCqChannel W)
    (n : ℕ) (A : Finset (PolarWire n)) :
    ensembleError W n A ≤
      2 * Real.sqrt (∑ i ∈ A, (1 / 2 : ℝ) * Real.sqrt (squaredCqFidelity (splitChannel W n i))) ∧
    ∃ f : Frozen n A, frozenError W n A f ≤
      2 * Real.sqrt (∑ i ∈ A, (1 / 2 : ℝ) * Real.sqrt (squaredCqFidelity (splitChannel W n i))) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P14

end
Source
Polar codes for classical-quantum channels, arXiv:1109.2591v3, PDF p. 6, Proposition 4: ensemble error bound and existence of a frozen vector
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let III be any type, let HI=ℓ2(I;C)\mathcal H_I=\ell^2(I;\mathbb C)HI​=ℓ2(I;C), and let WWW assign a complex coefficient array WbW_bWb​ on I×II\times II×I to each bit b∈F2b\in\mathbb F_2b∈F2​. Assume each WbW_bWb​ is a density array in the following precise sense: it has an expansion Wb(i,j)=∑r=0∞ur(i)vr(j)‾W_b(i,j)=\sum_{r=0}^{\infty}u_r(i)\overline{v_r(j)}Wb​(i,j)=∑r=0∞​ur​(i)vr​(j)​ with ur,vr∈HIu_r,v_r\in\mathcal H_Iur​,vr​∈HI​ and ∑r∥ur∥∥vr∥<∞\sum_r\|u_r\|\|v_r\|<\infty∑r​∥ur​∥∥vr​∥<∞; every quadratic form restricted to a finite subset of III is real and nonnegative; and its real diagonal is summable with sum 111. Fix any n∈Nn\in\mathbb Nn∈N, put N=2nN=2^nN=2n, and let A⊆{0,…,N−1}A\subseteq\{0,\ldots,N-1\}A⊆{0,…,N−1}. Define Gn(r,c)=∏k=0n−1κ(bit⁡k(rev⁡nr),bit⁡k(c))G_n(r,c)=\prod_{k=0}^{n-1}\kappa(\operatorname{bit}_k(\operatorname{rev}_n r),\operatorname{bit}_k(c))Gn​(r,c)=∏k=0n−1​κ(bitk​(revn​r),bitk​(c)) over F2\mathbb F_2F2​, where κ(a,0)=1\kappa(a,0)=1κ(a,0)=1, κ(a,1)=a\kappa(a,1)=aκ(a,1)=a, and rev⁡n\operatorname{rev}_nrevn​ reverses the nnn binary digits of an index. For u∈F2Nu\in\mathbb F_2^Nu∈F2N​, set xc(u)=∑rurGn(r,c)x_c(u)=\sum_r u_rG_n(r,c)xc​(u)=∑r​ur​Gn​(r,c) and Vu(y,z)=∏c=0N−1Wxc(u)(yc,zc)V_u(y,z)=\prod_{c=0}^{N-1}W_{x_c(u)}(y_c,z_c)Vu​(y,z)=∏c=0N−1​Wxc​(u)​(yc​,zc​), for y,z∈INy,z\in I^Ny,z∈IN. For an index iii, prefix h∈F2ih\in\mathbb F_2^ih∈F2i​, and bit bbb, let Ti,h,b=2−N∑v∈F2NVfill⁡(h,b,v)T_{i,h,b}=2^{-N}\sum_{v\in\mathbb F_2^N}V_{\operatorname{fill}(h,b,v)}Ti,h,b​=2−N∑v∈F2N​​Vfill(h,b,v)​, where the filled word equals hjh_jhj​ for j<ij<ij<i, equals bbb at iii, and equals vjv_jvj​ for j>ij>ij>i. The split array Wn,iW_{n,i}Wn,i​, indexed by F2i×IN\mathbb F_2^i\times I^NF2i​×IN, is Wn,ib((h,y),(h′,z))=1h=h′2−iTi,h,b(y,z)W_{n,i}^b((h,y),(h',z))=\mathbf1_{h=h'}2^{-i}T_{i,h,b}(y,z)Wn,ib​((h,y),(h′,z))=1h=h′​2−iTi,h,b​(y,z). For any coefficient array XXX, let O(X)O(X)O(X) be a chosen bounded operator having coefficients XXX in the coordinate basis if one exists, and 000 otherwise; let R(T)R(T)R(T) be a chosen positive bounded operator satisfying R(T)2=TR(T)^2=TR(T)2=T if one exists, and 000 otherwise; and put tr⁡bT=∑j(Tej)j\operatorname{tr}_bT=\sum_j(Te_j)_jtrb​T=∑j​(Tej​)j​. For a binary array VVV, define F(V)=[Re⁡tr⁡bR(R(O(V0))O(V1)R(O(V0)))]2F(V)=[\operatorname{Re}\operatorname{tr}_bR(R(O(V_0))O(V_1)R(O(V_0)))]^2F(V)=[Retrb​R(R(O(V0​))O(V1​)R(O(V0​)))]2. For decoding, let Pi,hP_{i,h}Pi,h​ be the orthogonal projection onto the closed span of all eigenvectors with negative real eigenvalue of R(O(Ti,h,0))−R(O(Ti,h,1))R(O(T_{i,h,0}))-R(O(T_{i,h,1}))R(O(Ti,h,0​))−R(O(Ti,h,1​)), and set Πi,h,0=1−Pi,h\Pi_{i,h,0}=1-P_{i,h}Πi,h,0​=1−Pi,h​, Πi,h,1=Pi,h\Pi_{i,h,1}=P_{i,h}Πi,h,1​=Pi,h​. Given uuu, form KuK_uKu​ by processing i=0,…,N−1i=0,\ldots,N-1i=0,…,N−1, left multiplying by Πi,u<i,ui\Pi_{i,u_{<i},u_i}Πi,u<i​,ui​​ when i∈Ai\in Ai∈A and by the identity otherwise, and set s(u)=Re⁡tr⁡b(Ku∗KuO(Vu))s(u)=\operatorname{Re}\operatorname{tr}_b(K_u^*K_uO(V_u))s(u)=Retrb​(Ku∗​Ku​O(Vu​)). For frozen bits f∈F2Acf\in\mathbb F_2^{A^c}f∈F2Ac​, define ef=1−2−∣A∣∑m∈F2As(m∪f)e_f=1-2^{-|A|}\sum_{m\in\mathbb F_2^A}s(m\cup f)ef​=1−2−∣A∣∑m∈F2A​​s(m∪f), and define eˉ=2−∣Ac∣∑fef\bar e=2^{-|A^c|}\sum_f e_feˉ=2−∣Ac∣∑f​ef​. Then

eˉ≤2∑i∈A12F(Wn,i),∃f∈F2Ac,ef≤2∑i∈A12F(Wn,i).\bar e\le 2\sqrt{\sum_{i\in A}\frac12\sqrt{F(W_{n,i})}}, \qquad \exists f\in\mathbb F_2^{A^c},\quad e_f\le 2\sqrt{\sum_{i\in A}\frac12\sqrt{F(W_{n,i})}}.eˉ≤2i∈A∑​21​F(Wn,i​)​​,∃f∈F2Ac​,ef​≤2i∈A∑​21​F(Wn,i​)​​.

This includes n=0n=0n=0, when N=1N=1N=1, and A=∅A=\varnothingA=∅ or A={0,…,N−1}A=\{0,\ldots,N-1\}A={0,…,N−1}; an assignment on an empty set has one possible value. The basis traces use the total infinite-sum operation, which is 000 for a nonsummable family.

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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