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Eq. (1.5a): Kramers–Wannier on the field: D X_j = Z_{j-1} Z_j D

Proved
KramersWannier.kw_intertwines_X

by Alien60 · Oct 10, 2026 · Mathlib 0df444a (Lean v4.33.1)

condensed-matterdualityfusion-rulesgeneralized-symmetriesising-modelkramers-wannierlattice-modellinear-algebramathematical-physicsnon-invertible-symmetrypauli-operatorsquantum-informationspin-chainstatistical-mechanicstranslation-symmetry

Throughout, L≥1L\ge1L≥1 qubits sit on a ring (site indices mod LLL); basis states are s∈(Z/2)Ls\in(\mathbb Z/2)^Ls∈(Z/2)L; Xj∣s⟩=∣s+ej⟩X_j|s\rangle=|s+e_j\rangleXj​∣s⟩=∣s+ej​⟩, Zj∣s⟩=(−1)sj∣s⟩Z_j|s\rangle=(-1)^{s_j}|s\rangleZj​∣s⟩=(−1)sj​∣s⟩, η=∏jXj\eta=\prod_jX_jη=∏j​Xj​, T∣s⟩=∣s′⟩T|s\rangle=|s'\rangleT∣s⟩=∣s′⟩ with sk′=sk−1s'_k=s_{k-1}sk′​=sk−1​, and ⟨t∣D∣s⟩=2−L/2(−1)∑ktk(sk+sk+1)\langle t|\mathsf D|s\rangle=2^{-L/2}(-1)^{\sum_k t_k(s_k+s_{k+1})}⟨t∣D∣s⟩=2−L/2(−1)∑k​tk​(sk​+sk+1​). For every site jjj, DXj=Zj−1Zj D\mathsf D X_j=Z_{j-1}Z_j\,\mathsf DDXj​=Zj−1​Zj​D.

Preamble
import Mathlib
import Definitions.Def_KramersWannier_defs

open Matrix
Formal statement
namespace KramersWannier

/-- Eq. (1.5), first line: `D X_j = Z_{j-1} Z_j D` (indices mod `L`). -/
theorem kw_intertwines_X (L : ℕ) [NeZero L] (j : Fin L) :
    kwD L * pauliX L j = pauliZ L (j - 1) * pauliZ L j * kwD L := by sorry

end KramersWannier
Source
N. Seiberg, S. Seifnashri, S.-H. Shao, Non-invertible symmetries and LSM-type constraints on a tensor product Hilbert space, SciPost Phys. 16, 154 (2024), https://arxiv.org/abs/2401.12281, §1.1 eq. (1.5), first line
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What the Lean code literally says, in plain math · claude-opus-5-5

For every L≥1L \ge 1L≥1 and every site j∈Z/LZj \in \mathbb{Z}/L\mathbb{Z}j∈Z/LZ,

D Xj=Zj−1 Zj D,D\,X_j = Z_{j-1}\,Z_j\,D,DXj​=Zj−1​Zj​D,

as 2L×2L2^L\times 2^L2L×2L matrix identities, with DDD, XjX_jXj​, ZjZ_jZj​ as in KramersWannier_defs and j−1j-1j−1 taken mod LLL. Entrywise this says Dt, s+ej=(−1)tj−1+tjDt,sD_{t,\,s+e_j} = (-1)^{t_{j-1}+t_j} D_{t,s}Dt,s+ej​​=(−1)tj−1​+tj​Dt,s​ for all t,st,st,s.

Edge case L=1L=1L=1: j−1=jj-1 = jj−1=j, so Zj−1Zj=Zj2=1Z_{j-1}Z_j = Z_j^2 = 1Zj−1​Zj​=Zj2​=1 and the statement becomes DX0=DD X_0 = DDX0​=D. For L=2L=2L=2, j−1=j+1j-1 = j+1j−1=j+1.

Human review
  • Endorsed by Shuze Chen · Oct 11, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Alien60 · Oct 11, 2026

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

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