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Large-ball reduced entropy and its exact constant-term integral

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QAlgorithms.Papers.Independent.Intakethird100.P58.d4_large_radius_entropy

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

prove2quantum

Fix the positive D4 scales and Nc≥1N_c\ge1Nc​≥1 as above. There exist ρmin⁡>0\rho_{\min}>0ρmin​>0 and regular cylinder profiles rρ(z)r_\rho(z)rρ​(z) for every ρ≥ρmin⁡\rho\ge\rho_{\min}ρ≥ρmin​, anchored by rρ(0)=ρr_\rho(0)=\rhorρ​(0)=ρ. Let ggg be the explicit regular coefficient (4.23), put q(z)=−(1+z)/(1+z+z2)=g′(z)q(z)=-(1+z)/(1+z+z^2)=g'(z)q(z)=−(1+z)/(1+z+z2)=g′(z), and define

H(z)=4g(z)+z(1−z3)q(z)22z3.H(z)=\frac{4g(z)+z(1-z^3)q(z)^2}{2z^3}.H(z)=2z34g(z)+z(1−z3)q(z)2​.

For every fixed regulator 0<a<10<a<10<a<1, HHH is integrable on [a,1][a,1][a,1] and the actual reduced entropy satisfies

S(ρ,a)=K∫a1[ρ2z3+H(z)]dz+O(ρ−1)(ρ→∞).S(\rho,a)=K\int_a^1\left[\frac{\rho^2}{z^3}+H(z)\right]dz+O(\rho^{-1})\qquad(\rho\to\infty).S(ρ,a)=K∫a1​[z3ρ2​+H(z)]dz+O(ρ−1)(ρ→∞).

Here K=π3Rcurv6U0R4/(2GN(10))K=\pi^3R_{\mathrm{curv}}^6U_0R_4/(2G_N^{(10)})K=π3Rcurv6​U0​R4​/(2GN(10)​) is the chosen reduced prefactor, Sρ2=Kρ2∫a1z−3dzS_{\rho^2}=K\rho^2\int_a^1z^{-3}dzSρ2​=Kρ2∫a1​z−3dz, and γ(a)=−K∫a1H(z)dz\gamma(a)=-K\int_a^1H(z)dzγ(a)=−K∫a1​H(z)dz. The error constant and threshold may depend on the fixed regulator and model, and precede ρ\rhoρ.

The result constructs a stationary cylinder family and its area asymptotics without assuming them. It gives the integral coefficient before the disputed evaluations (4.25)/(4.28), and does not exchange the fixed-cutoff and large-radius limits.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P58
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P58

open Set Filter MeasureTheory

open scoped Topology

/-- PDF pp.19--20, (4.24): construct the large-boundary-radius D4 cylinder
family and the constant-term integral of its actual reduced regulatedAction. First fix
0<a<1, then take rho to infinity. Neither a profile expansion nor the resulting
area estimate is assumed. This is the integral coefficient, not (4.25)/(4.28). -/
theorem d4_large_radius_entropy (P : D4Parameters) (N : ℕ) (hN : 0 < N) :
    ∃ (rhoMin : ℝ) (C : ℝ → ℝ → ℝ), 0 < rhoMin ∧
      (∀ rho : ℝ, rhoMin ≤ rho → IsCylinder .d4 rho (C rho)) ∧
      ∀ a : ℝ, 0 < a → a < 1 →
        IntervalIntegrable d4ConstantDensity volume a 1 ∧
        ∃ B M : ℝ, 0 < B ∧ rhoMin ≤ M ∧
          ∀ rho : ℝ, M ≤ rho →
            |d4Entropy P N (C rho) a 1 - d4Prefactor P N *
              (rho ^ 2 * (∫ z in a..(1 : ℝ), 1 / z ^ 3) +
                ∫ z in a..(1 : ℝ), d4ConstantDensity z)| ≤ B / rho := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P58

end
Source
Topological Entanglement Entropy and Holography, arXiv:0805.1891v2, PDF p. 20, (4.24)
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What the Lean code literally says, in plain math · gpt-6.1-sol

For every parameter pair consisting of real numbers κ>0\kappa>0κ>0 and L>0L>0L>0, representing the coupling and circle radius in PPP, and every natural number N>0N>0N>0, define

K=π3(πκ)2(4πκ9L2)L2(8π6κ2N2),Dr(z)=r(z)2z31+z(1−z3)(r′(z))2,K=\frac{\pi^3(\pi\kappa)^2\left(\frac{4\pi\kappa}{9L^2}\right)L} {2\left(\frac{8\pi^6\kappa^2}{N^2}\right)}, \qquad D_r(z)=\frac{r(z)^2}{z^3}\sqrt{1+z(1-z^3)(r'(z))^2},K=2(N28π6κ2​)π3(πκ)2(9L24πκ​)L​,Dr​(z)=z3r(z)2​1+z(1−z3)(r′(z))2​,

and define

G(z)=−arctan⁡((1+2z)/3)3−log⁡(1+z+z2)2+π63,H(z)=4G(z)+z(1−z3)(−1+z1+z+z2)22z3.G(z)=-\frac{\arctan((1+2z)/\sqrt3)}{\sqrt3} -\frac{\log(1+z+z^2)}2+\frac{\pi}{6\sqrt3}, \qquad H(z)=\frac{4G(z)+z(1-z^3)\left(-\frac{1+z}{1+z+z^2}\right)^2}{2z^3}.G(z)=−3​arctan((1+2z)/3​)​−2log(1+z+z2)​+63​π​,H(z)=2z34G(z)+z(1−z3)(−1+z+z21+z​)2​.

There exist a real number ρmin⁡>0\rho_{\min}>0ρmin​>0 and a function C:R→(R→R)C:\mathbb R\to(\mathbb R\to\mathbb R)C:R→(R→R) with the following properties. For every real ρ≥ρmin⁡\rho\ge\rho_{\min}ρ≥ρmin​, the function r=C(ρ)r=C(\rho)r=C(ρ) satisfies r(0)=ρr(0)=\rhor(0)=ρ, is continuous and strictly decreasing on [0,1][0,1][0,1], is twice continuously differentiable at every point of [0,1][0,1][0,1], has r(1)>0r(1)>0r(1)>0, and, for every 0<z<10<z<10<z<1, has r(z)>0r(z)>0r(z)>0 and satisfies

ddu(r(u)2u(1−u3)r′(u)u31+u(1−u3)(r′(u))2)∣u=z=2r(z)z31+z(1−z3)(r′(z))2.\left.\frac{d}{du} \left(\frac{r(u)^2u(1-u^3)r'(u)} {u^3\sqrt{1+u(1-u^3)(r'(u))^2}}\right)\right|_{u=z} = \frac{2r(z)}{z^3}\sqrt{1+z(1-z^3)(r'(z))^2}.dud​(u31+u(1−u3)(r′(u))2​r(u)2u(1−u3)r′(u)​)​u=z​=z32r(z)​1+z(1−z3)(r′(z))2​.

Moreover, DrD_rDr​ is interval integrable with respect to Lebesgue measure from sss to 111 for every real 0<s<10<s<10<s<1. For every real aaa with 0<a<10<a<10<a<1, the function HHH is interval integrable with respect to Lebesgue measure from aaa to 111, and there exist real constants B>0B>0B>0 and M≥ρmin⁡M\ge\rho_{\min}M≥ρmin​ such that, for every real ρ≥M\rho\ge Mρ≥M,

∣K∫a1DC(ρ)(z) dz−K(ρ2∫a11z3 dz+∫a1H(z) dz)∣≤Bρ.\left| K\int_a^1 D_{C(\rho)}(z)\,dz - K\left( \rho^2\int_a^1\frac{1}{z^3}\,dz+\int_a^1H(z)\,dz \right) \right| \le \frac{B}{\rho}.​K∫a1​DC(ρ)​(z)dz−K(ρ2∫a1​z31​dz+∫a1​H(z)dz)​≤ρB​.

The family CCC and the threshold ρmin⁡\rho_{\min}ρmin​ are chosen before aaa; the constants BBB and MMM may depend on aaa. No condition is imposed on C(ρ)C(\rho)C(ρ) for ρ<ρmin⁡\rho<\rho_{\min}ρ<ρmin​. All displayed quotients are real division, which assigns value zero when the denominator is zero.

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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