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Quantisation of the real branch: a candidate with real exponential has ∣u∣2>π2|u|^2 > \pi^2∣u∣2>π2

Proved
Diaz.real_quantisation

by carlok · Sep 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theory

Source. This is Carlo Perassi's mathematics, from his unpublished note on Diaz's modulus conjecture, section Polar coordinates and the discreteness of the period, statement Theorem (Quantisation of the real branch). Published on his mission with his permission. No novelty is claimed for it here; the argument is elementary, and it is possibly known — it has not been checked against the literature.

Statement. Let u∈Cu \in \mathbb{C}u∈C lie on neither axis (ℜu≠0\Re u \neq 0ℜu=0, ℑu≠0\Im u \neq 0ℑu=0) and suppose eue^{u}eu is real. Then ℑu∈πZ\Im u \in \pi\mathbb{Z}ℑu∈πZ, and consequently

uuˉ=(ℜu)2+(ℑu)2>π2.u\bar u = (\Re u)^2 + (\Im u)^2 > \pi^2 .uuˉ=(ℜu)2+(ℑu)2>π2.

What it says in the mission's terms. On the Diaz locus the two axis exclusions are supplied by Diaz.not_on_axes, so every candidate whose exponential is real satisfies them. The conclusion is that the real branch of the locus is quantised: its points are ℓ+inπ\ell + i n \piℓ+inπ with n∈Z∖{0}n \in \mathbb{Z}\setminus\{0\}n∈Z∖{0}, and their norms are bounded below by π2\pi^2π2 — they cannot accumulate at 000.

Why the bound is not formal. It is exactly the place where the mission's Laurent shadow model (Diaz.model_falsifies, Diaz.Exp0_*) parts company with the actual exponential. There ker⁡Exp0\ker \mathrm{Exp}_0kerExp0​ is a divisible line, the shadow locus is stable under Q×\mathbb{Q}^\timesQ×, and its norms q2ρq^2\rhoq2ρ accumulate at 000; here ker⁡exp⁡∩L=2πiZ\ker \exp \cap \mathcal{L} = 2\pi i \mathbb{Z}kerexp∩L=2πiZ is a rank-one lattice inside the divisible line 2πiQ2\pi i\mathbb{Q}2πiQ, and both conclusions fail in the model. The real branch is also, unlike the full locus, not stable under Q×\mathbb{Q}^\timesQ×: quququ has real exponential only when qn∈Zqn \in \mathbb{Z}qn∈Z.

Reading the Lean. (Complex.exp u).im = 0 is "eue^{u}eu is real"; Complex.normSq u is uuˉu \bar uuuˉ.

Proof. Complex.exp_im gives (eu)im=eℜusin⁡(ℑu)(e^{u})_{\mathrm{im}} = e^{\Re u}\sin(\Im u)(eu)im​=eℜusin(ℑu), and eℜu≠0e^{\Re u} \neq 0eℜu=0, so sin⁡(ℑu)=0\sin(\Im u) = 0sin(ℑu)=0 and ℑu=nπ\Im u = n\piℑu=nπ for some integer nnn. Since ℑu≠0\Im u \neq 0ℑu=0 we have n≠0n \neq 0n=0, hence ∣ℑu∣≥π|\Im u| \geq \pi∣ℑu∣≥π and (ℑu)2≥π2(\Im u)^2 \geq \pi^2(ℑu)2≥π2; adding (ℜu)2>0(\Re u)^2 > 0(ℜu)2>0 gives the strict inequality.

Preamble
import Mathlib

open ComplexConjugate
Formal statement
theorem Diaz.real_quantisation {u : ℂ} (hexp : (Complex.exp u).im = 0)
    (hre : u.re ≠ 0) (him : u.im ≠ 0) :
    (∃ n : ℤ, u.im = n * Real.pi) ∧ Real.pi ^ 2 < Complex.normSq u := by sorry

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