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Two small-alpha roots of the reduced near-four-dimensional gap equation

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QAlgorithms.Papers.Independent.Intakethird100.P31.reduced_small_alpha_saddles

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let N≥1N\ge1N≥1 be an integer, 0<ϵ<20<\epsilon<20<ϵ<2, and n>0n>0n>0. Consider the retained small-α\alphaα equation of (6.33),

Qϵ,n(α)=αn+α2ϵ−16(1n2−1)=0,Q_{\epsilon,n}(\alpha)=\frac\alpha n+\frac{\alpha^2}{\epsilon}-\frac16\left(\frac1{n^2}-1\right)=0,Qϵ,n​(α)=nα​+ϵα2​−61​(n21​−1)=0,

and put Δ=ϵ2+23ϵ(1−n2)\Delta=\epsilon^2+\frac23\epsilon(1-n^2)Δ=ϵ2+32​ϵ(1−n2). Its real roots are exactly

α±=−ϵ2n±Δ2n,Δ≥0.\alpha_\pm=-\frac\epsilon{2n}\pm\frac{\sqrt\Delta}{2n},\qquad \Delta\ge0.α±​=−2nϵ​±2nΔ​​,Δ≥0.

There are no real roots when Δ<0\Delta<0Δ<0. At Δ=0\Delta=0Δ=0 the two displayed branches coincide. Under the retained normalization d=Nα2/(8π2ϵ)d=N\alpha^2/(8\pi^2\epsilon)d=Nα2/(8π2ϵ), their amplitudes are

d±=N8π2[16(1n2−1)+ϵ∓Δ2n2].d_\pm=\frac{N}{8\pi^2}\left[\frac16\left(\frac1{n^2}-1\right)+\frac{\epsilon\mp\sqrt\Delta}{2n^2}\right].d±​=8π2N​[61​(n21​−1)+2n2ϵ∓Δ​​].

The exact threshold of this reduced equation is nc(ϵ)=1+3ϵ/2n_c(\epsilon)=\sqrt{1+3\epsilon/2}nc​(ϵ)=1+3ϵ/2​, with lim⁡ϵ↓0(nc(ϵ)−1)/ϵ=3/4\lim_{\epsilon\downarrow0}(n_c(\epsilon)-1)/\epsilon=3/4limϵ↓0​(nc​(ϵ)−1)/ϵ=3/4.

These are the roots and amplitudes of the specified asymptotic truncation, not exact roots or amplitudes of the full integral saddle equation.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P31
Formal statement
noncomputable section

open Filter Set MeasureTheory

open scoped Topology BigOperators

namespace QAlgorithms.Papers.Independent.Intakethird100.P31

/-- PDF p.37, (6.33)--(6.35): complete real roots and corresponding
amplitudes of the retained small-alpha equation. Zero discriminant is
coalescence, not two simple roots. Its exact reduced threshold has leading
shift `3*eps/4`. These roots are not asserted to solve (6.25) exactly. -/
theorem reduced_small_alpha_saddles (N : ℕ) (hN : 0 < N)
    (eps n : ℝ) (heps : 0 < eps) (hD : eps < 2) (hn : 0 < n) :
    (∀ alpha : ℝ, reducedGap eps n alpha = 0 ↔
      0 ≤ reducedDiscriminant eps n ∧
        (alpha = reducedRoot eps n true ∨ alpha = reducedRoot eps n false)) ∧
    (0 ≤ reducedDiscriminant eps n ↔ n ≤ Real.sqrt (1 + 3 * eps / 2)) ∧
    (0 ≤ reducedDiscriminant eps n →
      ((reducedRoot eps n true = reducedRoot eps n false ↔
          reducedDiscriminant eps n = 0) ∧
       (∀ plus : Bool,
         reducedAmplitude N eps (reducedRoot eps n plus) =
           (N : ℝ) / (8 * Real.pi ^ 2) *
             ((1 / n ^ 2 - 1) / 6 +
               (eps - (if plus then 1 else -1) *
                 Real.sqrt (reducedDiscriminant eps n)) / (2 * n ^ 2))))) ∧
    Tendsto (fun e : ℝ => (Real.sqrt (1 + 3 * e / 2) - 1) / e)
      (𝓝[>] (0 : ℝ)) (𝓝 (3 / 4 : ℝ)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P31

end
Source
Entanglement Entropy in the O(N) model, arXiv:0904.4477v1, PDF p. 37, §VI (6.32)–(6.35): two small-alpha saddles near four dimensions
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What the Lean code literally says, in plain math · gpt-6.1-sol

For every natural number N>0N>0N>0 and real numbers ε,n\varepsilon,nε,n satisfying 0<ε<20<\varepsilon<20<ε<2 and n>0n>0n>0, write

Δ=ε2+23ε(1−n2),rσ=−ε2n+σΔ2n(σ∈{1,−1}),\Delta=\varepsilon^2+\frac23\varepsilon(1-n^2), \qquad r_\sigma=-\frac{\varepsilon}{2n}+\frac{\sigma\sqrt{\Delta}}{2n} \quad(\sigma\in\{1,-1\}),Δ=ε2+32​ε(1−n2),rσ​=−2nε​+2nσΔ​​(σ∈{1,−1}),

where σ=1\sigma=1σ=1 corresponds to the Boolean value true and σ=−1\sigma=-1σ=−1 to false, and the real square root of a negative number is defined to be zero. The following four assertions hold simultaneously: for every real α\alphaα,

αn+α2ε−1/n2−16=0⟺Δ≥0  and  (α=r1  or  α=r−1);\frac{\alpha}{n}+\frac{\alpha^2}{\varepsilon} -\frac{1/n^2-1}{6}=0 \quad\Longleftrightarrow\quad \Delta\ge0\ \text{ and }\ (\alpha=r_1\ \text{ or }\ \alpha=r_{-1});nα​+εα2​−61/n2−1​=0⟺Δ≥0  and  (α=r1​  or  α=r−1​);

also,

Δ≥0⟺n≤1+3ε2;\Delta\ge0 \quad\Longleftrightarrow\quad n\le\sqrt{1+\frac{3\varepsilon}{2}};Δ≥0⟺n≤1+23ε​​;

if Δ≥0\Delta\ge0Δ≥0, then both

r1=r−1⟺Δ=0r_1=r_{-1}\quad\Longleftrightarrow\quad\Delta=0r1​=r−1​⟺Δ=0

and, for each σ∈{1,−1}\sigma\in\{1,-1\}σ∈{1,−1},

Nrσ28π2ε=N8π2(1/n2−16+ε−σΔ2n2);\frac{Nr_\sigma^2}{8\pi^2\varepsilon} = \frac{N}{8\pi^2} \left( \frac{1/n^2-1}{6} +\frac{\varepsilon-\sigma\sqrt{\Delta}}{2n^2} \right);8π2εNrσ2​​=8π2N​(61/n2−1​+2n2ε−σΔ​​);

and finally the right-hand limit is

lim⁡e→0+1+3e/2−1e=34.\lim_{e\to0^+}\frac{\sqrt{1+3e/2}-1}{e}=\frac34.e→0+lim​e1+3e/2​−1​=43​.
Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

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

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