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Strong six exponentials configurations in the span of u^{±k}, u^{±1}, 1 exist exactly for k = 2 and k = 3

Proved
DiazModulus.power_hull_strong_six_exp_configuration_iff

by carlok · Oct 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theorytranscendence

Let uuu be transcendental and k≥1k \geq 1k≥1. There exist x1,x2x_1, x_2x1​,x2​, linearly independent over Q‾\overline{\mathbb{Q}}Q​, and y1,y2,y3y_1, y_2, y_3y1​,y2​,y3​, linearly independent over Q‾\overline{\mathbb{Q}}Q​, with all six products xiyjx_iy_jxi​yj​ in

Q‾ u−k+Q‾ u−1+Q‾+Q‾ u+Q‾ uk\overline{\mathbb{Q}}\,u^{-k} + \overline{\mathbb{Q}}\,u^{-1} + \overline{\mathbb{Q}} + \overline{\mathbb{Q}}\,u + \overline{\mathbb{Q}}\,u^{k}Q​u−k+Q​u−1+Q​+Q​u+Q​uk

if and only if k=2k = 2k=2 or k=3k = 3k=3.

Why it matters for Diaz's conjecture. At a candidate uuu (∣u∣|u|∣u∣ and eue^ueu algebraic) the space L~\widetilde{\mathcal{L}}L contains 111, uuu and uˉ=∣u∣2/u\bar u = |u|^2/uuˉ=∣u∣2/u, and if it contained uku^kuk it would contain uˉk=∣u∣2ku−k\bar u^k = |u|^{2k}u^{-k}uˉk=∣u∣2ku−k too. Roy's strong six exponentials theorem refutes uk∈L~u^k \in \widetilde{\mathcal{L}}uk∈L through numbers built from uuu alone exactly when a configuration as above exists. So it excludes u2u^2u2 (DiazModulus.candidate_multiplier_module) and u3u^3u3 (DiazModulus.candidate_cube_and_axis_multiple_not_mem_logAlgTilde), and by this route nothing further: not u4u^4u4, nor any higher power. Configurations that also use logarithms unrelated to uuu are not covered.

Proof. For k=2k = 2k=2 take x=(1,u)x = (1, u)x=(1,u), y=(u−1,1,u)y = (u^{-1}, 1, u)y=(u−1,1,u); for k=3k = 3k=3 take x=(u,u−1)x = (u, u^{-1})x=(u,u−1), y=(1,u−2,u2)y = (1, u^{-2}, u^2)y=(1,u−2,u2) (Diaz's configuration for his Corollaire 5(2), up to the factor ∣u∣2|u|^2∣u∣2). For k=1k = 1k=1 and k≥4k \geq 4k≥4, multiply by uku^kuk: since uuu is transcendental, the products become polynomials supported on E={0,k−1,k,k+1,2k}E = \{0, k-1, k, k+1, 2k\}E={0,k−1,k,k+1,2k}, and the rank-one relations become polynomial identities. The orders at 000 of the three independent polynomials of the first row are three distinct exponents ttt with ttt and t+δt + \deltat+δ both in EEE, for a fixed δ≠0\delta \neq 0δ=0. But for k=1k = 1k=1 and for k≥4k \geq 4k≥4 at most two such ttt exist for every δ≠0\delta \neq 0δ=0.

Novelty. Not found in the sources read (Diaz 1997, 2004, 2007; Waldschmidt's Variations, Further variations and The role of complex conjugation; Roy–Waldschmidt 1995, and 1997 §§0 and 7; Diophantine Approximation on Linear Algebraic Groups, §§11.5–11.6 and chapter 12). Diaz notes that his list of conjectures contains no general one for λn\lambda^nλn (2007, p. 376), and that inside the strong six exponentials theorem one cannot hope to go very far (p. 390); this node is a precise instance of the second remark.

Preamble
import Definitions.Def_DiazModulus
Formal statement
open Complex ComplexConjugate

namespace DiazModulus

/-- For transcendental `u` and `k ≥ 1`, a configuration of the strong six exponentials theorem (two
`Q̄`-independent `xᵢ`, three `Q̄`-independent `yⱼ`) with all six products in
`Q̄ u⁻ᵏ + Q̄ u⁻¹ + Q̄ + Q̄ u + Q̄ uᵏ` exists exactly when `k = 2` or `k = 3`. -/
theorem power_hull_strong_six_exp_configuration_iff {u : ℂ} (hu : u ∉ Qbar) {k : ℕ} (hk : 1 ≤ k) :
    (∃ (x : Fin 2 → ℂ) (y : Fin 3 → ℂ), LinearIndependent (↥Qbar) x ∧ LinearIndependent (↥Qbar) y ∧
      ∀ i j, x i * y j ∈ Submodule.span Qbar ({(u ^ k)⁻¹, u⁻¹, 1, u, u ^ k} : Set ℂ)) ↔
    (k = 2 ∨ k = 3) := by
  sorry

end DiazModulus
Source
Context: G. Diaz, Produits et quotients de combinaisons linéaires de logarithmes de nombres algébriques : conjectures et résultats partiels, J. Théor. Nombres Bordeaux 19 (2007), 373–391, p. 376, Corollaire 5 (p. 383), Théorème 7 and §2.4 (p. 390). Not found in the sources read. Formal proof: Diaz modulus mission, 1 October 2026 (C. Perassi).

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