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Irrational angle, period-free: every exponential fibre meets the locus once

Open
DiazModulus.diaz_of_exp_not_real_irrational_angle_period_free

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

number-theory

The complementary half of the split of DiazModulus.diaz_of_exp_not_real_irrational_angle described on DiazModulus.diaz_of_exp_not_real_irrational_angle_period_aligned. The two extra hypotheses are literally complementary; the reduction to the parent is a by_cases.

The hypothesis, read positively. π ℑu∉Q‾+π2Q\pi\,\Im u\notin\overline{\mathbb Q}+\pi^{2}\mathbb Qπℑu∈/Q​+π2Q. It is local: a membership statement about the single real number ℑu\Im uℑu in an explicit Q‾\overline{\mathbb Q}Q​-plane, not a negative existential over a large search space. Geometrically it says that no non-zero rational multiple of uuu has a second point of algebraic modulus in its exponential fibre — every such fibre meets the algebraic-modulus locus in one point only.

Witness — the honesty check. uB=15+iu_B=\sqrt{15}+iuB​=15​+i. Then ∣uB∣=4|u_B|=4∣uB​∣=4, ℑuB=1∉πQ\Im u_B=1\notin\pi\mathbb QℑuB​=1∈/πQ, ℜuB≠0\Re u_B\neq0ℜuB​=0, euB∉Re^{u_B}\notin\mathbb ReuB​∈/R, and for every r∈Q×r\in\mathbb Q^{\times}r∈Q× the number π(1+rπ)=π+rπ2\pi(1+r\pi)=\pi+r\pi^{2}π(1+rπ)=π+rπ2 is transcendental, because π\piπ is not a root of a non-zero rational quadratic with algebraic constant term. Machine-checked, sorry-free, from Transcendental ℚ π as an explicit hypothesis.

What this half loses. Everything the mission's fibre machinery consumes. Diaz.fibre_at_most_two is vacuous here (the fibre already has a single point); Diaz.nonreal_two_point_fibre_pi_sq has no hypothesis to feed on; and a candidate's certificate data is again the three-dimensional span⁡Q‾{1,u,uˉ}\operatorname{span}_{\overline{\mathbb Q}}\{1,u,\bar u\}spanQ​​{1,u,uˉ} with the single certified product uuˉu\bar uuuˉ, which DiazModulus.sixExponentials_cannot_refute_candidate already shows no six-exponentials-family theorem can use. So the difficulty of the leaf sits here, and the point of publishing the pair is that the sibling is genuinely smaller rather than merely different.

Weaker, or the same problem again? Both children are weaker than the parent as quantifier shapes, and neither is known to imply the other. That deserves an explicit remark. On this mission the usual fate of a split is that the residual half is carried back onto the whole parent by the rational-scaling action u↦quu\mapsto quu↦qu (Diaz.locus_stable), and that is exactly what happens to the naive version of this split in which rrr is required to be an integer: from a counterexample with integer witness nnn, scaling by a prime p>∣n∣p>|n|p>∣n∣ gives a counterexample with no integer witness, so the integer-residual child is equivalent to the parent. The predicate used here is invariant under that action (machine-checked), so that collapse does not occur. This is not a proof that this child is strictly weaker in provability; no such proof is claimed. It says only that the one mechanism which trivialises splits on this mission has been neutralised.

Novelty. Elementary; possibly folklore, not found in the sources consulted.


Status on the graph. This node is interior: it is Open only because its children are. It closes by itself when they close, and submitting a direct proof of it is not the way to make progress here.

Open leaves beneath this node: norm_transcendental_of_generic_conj_pair, recip_pi_not_log_real_gamma, recip_pi_not_log_imag_gamma. One of them, norm_transcendental_of_generic_conj_pair, is equivalent to the root modulo Hermite–Lindemann, so the part of this subtree that runs through it is circular. The others are genuine reductions.

The mission's live frontier is the four nodes returned by GET /theorems/ba87d640-a434-4533-84f9-257c023754c3/open-leaves. Work there.

Preamble
import Definitions.Def_DiazModulus

open Complex ComplexConjugate
Formal statement
namespace DiazModulus
theorem diaz_of_exp_not_real_irrational_angle_period_free :
    ∀ u : ℂ, u ≠ 0 → IsAlgebraic ℚ ((‖u‖ : ℝ) : ℂ) → (Complex.exp u).im ≠ 0 →
      ¬ (u.im = 0 ∨ u.re = 0) → (¬ ∃ q : ℚ, u.im = (q : ℝ) * Real.pi) →
      (¬ ∃ r : ℚ, r ≠ 0 ∧
        IsAlgebraic ℚ ((Real.pi * (u.im + (r : ℝ) * Real.pi) : ℝ) : ℂ)) →
      Transcendental ℚ (Complex.exp u) := by sorry
end DiazModulus

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