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The propagation asymmetry ω(k)−ω(kˉ)=2βcsin⁡k0\omega(k) - \omega(\bar k) = 2\beta c \sin k_0ω(k)−ω(kˉ)=2βcsink0​ on a qqq-dimensional lattice

Proved
PinnedAsymmetryQ.asymmetry

by ShapeZero · Sep 24, 2026 · Mathlib 0df444a (Lean v4.33.1)

latticemathematical-physicstrigonometry

Let q≥1q \ge 1q≥1, let K,c,βK, c, \betaK,c,β be real and k∈Rqk \in \mathbb{R}^qk∈Rq, with kˉ=(−k0,k1,…,kq−1)\bar k = (-k_0, k_1, \dots, k_{q-1})kˉ=(−k0​,k1​,…,kq−1​) and

ω(k)=βcsin⁡k0+(βcsin⁡k0)2+K+2c∑a(1−cos⁡ka).\omega(k) = \beta c \sin k_0 + \sqrt{\bigl(\beta c \sin k_0\bigr)^2 + K + 2c \sum_{a} (1 - \cos k_a)} .ω(k)=βcsink0​+(βcsink0​)2+K+2ca∑​(1−coska​)​.

Then

ω(k)−ω(kˉ)=2βcsin⁡k0.\omega(k) - \omega(\bar k) = 2\beta c \sin k_0 .ω(k)−ω(kˉ)=2βcsink0​.

Neither the stiffness KKK nor any transverse wavenumber appears in the asymmetry. The statement concerns the linear asymmetry on a uniform lattice, with the gauge term along one axis.

Preamble
import Mathlib
import Definitions.Def_PinnedAsymmetryQ_omega

open Real BigOperators
Formal statement
namespace PinnedAsymmetryQ
theorem asymmetry (q : ℕ) [NeZero q] (K c β : ℝ) (k : Fin q → ℝ) :
    omega q K c β k - omega q K c β (flip0 q k) = 2 * β * c * sin (k 0) := by sorry
end PinnedAsymmetryQ
Source
Shape Zero LLC, "Formal Proofs of the C1 Verification Package" (August 2026), §7, Theorem 7.1 (extended to q axes): https://github.com/ShapeZeroSZ/shape-zero/blob/main/01_source/proofs/ShapeZero_C1_Formal_Proofs.pdf ; corrected in "Errata — C1 Formal Proofs (Sections 3 and 6)", "Related: the pinned asymmetry (Section 7)": https://github.com/ShapeZeroSZ/shape-zero/blob/main/01_source/proofs/ERRATUM_Theorem_6.1.md
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What the Lean code literally says, in plain math · claude-opus-5-5

Theorem PinnedAsymmetryQ.asymmetry. Fix a natural number qqq with q≠0q \neq 0q=0 (the only hypothesis; it guarantees that the index set {0,1,…,q−1}\{0, 1, \dots, q-1\}{0,1,…,q−1} is nonempty and contains 000; the case q=1q = 1q=1, where 000 is the only index, is included). Fix arbitrary real numbers KKK, ccc, β\betaβ (no sign, nonzero or other conditions: KKK may be negative, and ccc or β\betaβ may be zero or negative). Fix an arbitrary real vector k=(k0,k1,…,kq−1)∈Rqk = (k_0, k_1, \dots, k_{q-1}) \in \mathbb{R}^qk=(k0​,k1​,…,kq−1​)∈Rq, indexed by a∈{0,…,q−1}a \in \{0, \dots, q-1\}a∈{0,…,q−1}, with no conditions on its entries (they are not reduced modulo 2π2\pi2π or restricted in any way).

The function ω\omegaω used in the statement is defined, for such q,K,c,β,kq, K, c, \beta, kq,K,c,β,k, by

ω(k)  =  βcsin⁡(k0)  +   (βcsin⁡k0)2+K+2c∑a=0q−1(1−cos⁡ka)  ,\omega(k) \;=\; \beta c \sin(k_0) \;+\; \sqrt{\,(\beta c \sin k_0)^2 + K + 2c \sum_{a=0}^{q-1} \bigl(1 - \cos k_a\bigr)\,}\,,ω(k)=βcsin(k0​)+(βcsink0​)2+K+2ca=0∑q−1​(1−coska​)​,

where ⋅\sqrt{\cdot}⋅​ is Mathlib's real square root: for x≥0x \ge 0x≥0 it is the usual non-negative square root, and for x<0x < 0x<0 it returns 000 (no error, no complex value). So whenever the radicand is negative, ω(k)=βcsin⁡k0\omega(k) = \beta c \sin k_0ω(k)=βcsink0​. The sum runs over all qqq indices, including a=0a = 0a=0.

The operation flip0\mathrm{flip}_0flip0​ takes the vector kkk and replaces only its 000-th entry by its negative, leaving all other entries unchanged:

flip0(k)a={−k0a=0,kaa≠0.\mathrm{flip}_0(k)_a = \begin{cases} -k_0 & a = 0,\\ k_a & a \neq 0.\end{cases}flip0​(k)a​={−k0​ka​​a=0,a=0.​

(When q=1q = 1q=1, flip0(k)=(−k0)\mathrm{flip}_0(k) = (-k_0)flip0​(k)=(−k0​).) Thus, written out,

ω(flip0k)=βcsin⁡(−k0)+ (βcsin⁡(−k0))2+K+2c(1−cos⁡(−k0))+2c∑a=1q−1(1−cos⁡ka)  ,\omega(\mathrm{flip}_0 k) = \beta c \sin(-k_0) + \sqrt{\,(\beta c \sin(-k_0))^2 + K + 2c\bigl(1 - \cos(-k_0)\bigr) + 2c \sum_{a=1}^{q-1} (1 - \cos k_a)\,}\,,ω(flip0​k)=βcsin(−k0​)+(βcsin(−k0​))2+K+2c(1−cos(−k0​))+2ca=1∑q−1​(1−coska​)​,

with the same convention for the square root of a negative number.

The theorem asserts that, for every such qqq, KKK, ccc, β\betaβ and kkk, the following exact equality of real numbers holds:

ω(k)  −  ω(flip0k)  =  2 β c sin⁡(k0).\omega(k) \;-\; \omega(\mathrm{flip}_0 k) \;=\; 2\,\beta\, c\, \sin(k_0).ω(k)−ω(flip0​k)=2βcsin(k0​).

Here sin⁡\sinsin and cos⁡\coscos are the real sine and cosine. There are no further hypotheses; the equality is claimed for all parameter values, including those where the radicands are negative (square roots then equal 000) and degenerate values such as c=0c = 0c=0, β=0\beta = 0β=0, or k0∈πZk_0 \in \pi\mathbb{Z}k0​∈πZ, where the right-hand side is 000.

Human review
  • Endorsed by Shuze Chen · Sep 24, 2026

    Confirmed by the moderator at approval.

  • Endorsed by ShapeZero · Sep 24, 2026

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

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