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Dependency of the constructed correction matrices

Proved
QAlgorithms.Papers.Independent.Intakenext100.P38.correction_dependency

by Gohan · Oct 31, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let nnn denote the number of qubits, with computational basis indexed by {0,1}n\{0,1\}^n{0,1}n. Write PxyzP_{xyz}Pxyz​ for the unnormalized sum of distinct phase-zero Pauli words with xxx copies of XXX, yyy copies of the standard Y=(0−ii0)Y=\left(\begin{smallmatrix}0&-i\\i&0\end{smallmatrix}\right)Y=(0i​−i0​), and zzz copies of ZZZ. All Lie closures, linear spans, and dimensions below are over R\mathbb RR; commutators are [A,B]=AB−BA[A,B]=AB-BA[A,B]=AB−BA.

In Note G's convention put P~xyz=i(−1)yPxyz\widetilde P_{xyz}=i(-1)^yP_{xyz}Pxyz​=i(−1)yPxyz​, implementing both Y↦−YY\mapsto-YY↦−Y and P↦iPP\mapsto iPP↦iP. Define

A=−4P~k−2,2,0+2kP~k,0,0,B=4P~k−2,0,2−2kP~k,0,0,A=-4\widetilde P_{k-2,2,0}+2k\widetilde P_{k,0,0},\quad B=4\widetilde P_{k-2,0,2}-2k\widetilde P_{k,0,0},A=−4Pk−2,2,0​+2kPk,0,0​,B=4Pk−2,0,2​−2kPk,0,0​, C=−4(k−2)P~k−2,2,0−4P~k−2,0,2+2k(k−1)P~k,0,0.C=-4(k-2)\widetilde P_{k-2,2,0}-4\widetilde P_{k-2,0,2}+2k(k-1)\widetilde P_{k,0,0}.C=−4(k−2)Pk−2,2,0​−4Pk−2,0,2​+2k(k−1)Pk,0,0​.

For 2≤k≤n2\le k\le n2≤k≤n,

C=(k−2)A−B.C=(k-2)A-B.C=(k−2)A−B.

Consequently the family (A,B,C)(A,B,C)(A,B,C) is not linearly independent over R\mathbb RR.

This is the explicit dependency responsible for the failed independence step in Ref. [32], with A,B,CA,B,CA,B,C constructed from (G4)–(G6).

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakenext100_P38
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakenext100.P38

open scoped BigOperators

open QAlgorithms.Papers.Independent.Support50.P30
  (QnnOperator qubitPermutationMatrix PermutationEquivariant)

attribute [local instance] LieRing.ofAssociativeRing

/-- PDF p.22, (G4)–(G7): the constructed matrices satisfy the dependency.
The nonzero relation coefficient on C makes failure of linear independence explicit. -/
theorem correction_dependency {n k : ℕ} (hk : 2 ≤ k) (hkn : k ≤ n) :
    correctionC n k = ((k : ℝ) - 2) • correctionA n k - correctionB n k ∧
    ¬ LinearIndependent ℝ (![correctionA n k, correctionB n k, correctionC n k] :
      Fin 3 → QnnOperator n) := by sorry

end QAlgorithms.Papers.Independent.Intakenext100.P38

end
Source
On the universality of SnS_nSn​-equivariant kkk-body gates, arXiv:2303.00728v2, PDF p. 22, eq-G7
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What the Lean code literally says, in plain math · gpt-6.1-sol

For all natural numbers n,kn,kn,k with 2≤k≤n2\le k\le n2≤k≤n, let

Ta,b,c=i(−1)b∑(P0,…,Pn−1)∈{I,X,Y,Z}n#X=a, #Y=b, #Z=cP0⊗⋯⊗Pn−1T_{a,b,c}=i(-1)^b \sum_{\substack{(P_0,\ldots,P_{n-1})\in\{I,X,Y,Z\}^n\\ \#X=a,\ \#Y=b,\ \#Z=c}} P_0\otimes\cdots\otimes P_{n-1}Ta,b,c​=i(−1)b(P0​,…,Pn−1​)∈{I,X,Y,Z}n#X=a, #Y=b, #Z=c​∑​P0​⊗⋯⊗Pn−1​

be a complex 2n×2n2^n\times2^n2n×2n matrix, where a,b,ca,b,ca,b,c are natural numbers, all remaining factors are III, an empty sum is zero, and

I=(1001),X=(0110),Y=(0−ii0),Z=(100−1).I=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.I=(10​01​),X=(01​10​),Y=(0i​−i0​),Z=(10​0−1​).

Define

A=−4Tk−2,2,0+2kTk,0,0,B=4Tk−2,0,2−2kTk,0,0,C=−4(k−2)Tk−2,2,0−4Tk−2,0,2+2k(k−1)Tk,0,0.\begin{aligned} A&=-4T_{k-2,2,0}+2kT_{k,0,0},\\ B&=4T_{k-2,0,2}-2kT_{k,0,0},\\ C&=-4(k-2)T_{k-2,2,0}-4T_{k-2,0,2} +2k(k-1)T_{k,0,0}. \end{aligned}ABC​=−4Tk−2,2,0​+2kTk,0,0​,=4Tk−2,0,2​−2kTk,0,0​,=−4(k−2)Tk−2,2,0​−4Tk−2,0,2​+2k(k−1)Tk,0,0​.​

Then C=(k−2)A−BC=(k-2)A-BC=(k−2)A−B, and the indexed family consisting of A,B,CA,B,CA,B,C is not linearly independent over R\mathbb RR: there exist real numbers α,β,γ\alpha,\beta,\gammaα,β,γ, not all zero, such that αA+βB+γC=0\alpha A+\beta B+\gamma C=0αA+βB+γC=0.

Human review
  • Endorsed by Shuze Chen · Oct 31, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Oct 31, 2026

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

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