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Eq. (2.65), Table 1: TD = DT and D† = DT

Proved
KramersWannier.kw_translation

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​). TD=DTT\mathsf D=\mathsf DTTD=DT and D†=DT\mathsf D^\dagger=\mathsf DTD†=DT.

Preamble
import Mathlib
import Definitions.Def_KramersWannier_defs

open Matrix
Formal statement
namespace KramersWannier

/-- Eq. (2.65) and Table 1: `TD = DT = D̄`, where `D̄ = D†` is the Hermitian conjugate. -/
theorem kw_translation (L : ℕ) [NeZero L] :
    transl L * kwD L = kwD L * transl L ∧ (kwD L)ᴴ = kwD L * transl 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, §2.4.1 eq. (2.65) and Table 1 (D̄ = D† = DT)
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What the Lean code literally says, in plain math · claude-opus-5-5

For every L≥1L \ge 1L≥1, with DDD and TTT as in KramersWannier_defs,

T D=D TandD†=D T,T\,D = D\,T \quad\text{and}\quad D^{\dagger} = D\,T,TD=DTandD†=DT,

where D†D^{\dagger}D† is the conjugate transpose, (D†)t,s=Ds,t‾(D^\dagger)_{t,s} = \overline{D_{s,t}}(D†)t,s​=Ds,t​​ (which equals Ds,tD_{s,t}Ds,t​ since all entries are real). The statement asserts exactly these two equalities; the equality TD=D†TD = D^\daggerTD=D† follows from them but is not stated separately. Edge case L=1L=1L=1: T=1T = 1T=1, so it reads D=DD = DD=D and D†=DD^\dagger = DD†=D.

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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