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Quadratic pullback · residual density — certificate

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HodgeComputer.QuadraticCubicPullback.certificate

by benediamond · Nov 5, 2026 · Mathlib 0df444a (Lean v4.33.1)

algebraic-geometryhodge-computerquadratic-cubic-pullback

For q=1q=1q=1, and the fixed (F,A)(F,A)(F,A) of “Quadratic pullback · residual density”, prove the existence of a rational vector field vvv with components homogeneous of degree 111 and common denominator relatively prime to FFF, satisfying F2∣A−1⋅dF(v)+F⋅div⁡(v)F^{2}\mid A-1\cdot dF(v)+F\cdot\operatorname{div}(v)F2∣A−1⋅dF(v)+F⋅div(v). Divisibility is in the localization with denominators relatively prime to FFF. There is no denominator degree bound. More details.

Preamble
import Definitions.Def_HodgeComputerQuadraticCubicPullback

Formal statement
theorem HodgeComputer.QuadraticCubicPullback.certificate :
    HodgeComputer.HasCertificate
      (n := HodgeComputer.QuadraticCubicPullback.n)
      HodgeComputer.QuadraticCubicPullback.F
      HodgeComputer.QuadraticCubicPullback.A
      HodgeComputer.QuadraticCubicPullback.d
      HodgeComputer.QuadraticCubicPullback.q := by
  sorry
Source
https://github.com/hodge-computer/sextics/blob/83fbf5c9a0afe9136ce4f1ed324d085a2c899bd0/data/quadratic-cubic-pullback/instance.rec
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What the Lean code literally says, in plain math · GPT-6 (OpenAI Codex; independent auditor)

In the polynomial ring R=C[X0,X1,X2,Y0,Y1,Y2]R=\mathbb C[X_0,X_1,X_2,Y_0,Y_1,Y_2]R=C[X0​,X1​,X2​,Y0​,Y1​,Y2​] in six algebraically independent variables, corresponding in order to indices 0,1,2,3,4,50,1,2,3,4,50,1,2,3,4,5, fix the following polynomial:

F=(X02+57(X1X2+Y0Y1))3+(X12+511(X0X2+Y1Y2))3+(X22+513(X0X1+Y0Y2))3+(Y02+517(X0X1+Y1Y2))3+(Y12+519(X1X2+Y0Y2))3+(Y22+523(X0X2+Y0Y1))3.\begin{aligned} F={}&\left(X_0^2+\frac57(X_1X_2+Y_0Y_1)\right)^3\\ &+\left(X_1^2+\frac5{11}(X_0X_2+Y_1Y_2)\right)^3\\ &+\left(X_2^2+\frac5{13}(X_0X_1+Y_0Y_2)\right)^3\\ &+\left(Y_0^2+\frac5{17}(X_0X_1+Y_1Y_2)\right)^3\\ &+\left(Y_1^2+\frac5{19}(X_1X_2+Y_0Y_2)\right)^3\\ &+\left(Y_2^2+\frac5{23}(X_0X_2+Y_0Y_1)\right)^3. \end{aligned}F=​(X02​+75​(X1​X2​+Y0​Y1​))3+(X12​+115​(X0​X2​+Y1​Y2​))3+(X22​+135​(X0​X1​+Y0​Y2​))3+(Y02​+175​(X0​X1​+Y1​Y2​))3+(Y12​+195​(X1​X2​+Y0​Y2​))3+(Y22​+235​(X0​X2​+Y0​Y1​))3.​

Set A=X03Y03A=X_0^3Y_0^3A=X03​Y03​. With the fixed natural-number parameters n=5n=5n=5, d=6d=6d=6, and q=1q=1q=1, the theorem asserts that there exist polynomials D,Q,N0,N1,N2,N3,N4,N5∈RD,Q,N_0,N_1,N_2,N_3,N_4,N_5\in RD,Q,N0​,N1​,N2​,N3​,N4​,N5​∈R such that all of the following hold simultaneously: F≠0F\ne0F=0; every monomial with nonzero coefficient in FFF has total degree 666; every monomial with nonzero coefficient in AAA has total degree 666; D≠0D\ne0D=0; writing e=deg⁡De=\deg De=degD for the total degree of DDD, every monomial with nonzero coefficient in DDD has total degree eee; for each index i∈{0,1,2,3,4,5}i\in\{0,1,2,3,4,5\}i∈{0,1,2,3,4,5}, every monomial with nonzero coefficient in NiN_iNi​ has total degree e+1e+1e+1; every polynomial dividing both FFF and DDD is a nonzero constant; and the following exact identity holds in RRR:

AD2−D∑i=05∂F∂ziNi+F(D∑i=05∂Ni∂zi−∑i=05∂D∂ziNi)=F2Q.AD^2-D\sum_{i=0}^{5}\frac{\partial F}{\partial z_i}N_i+F\left(D\sum_{i=0}^{5}\frac{\partial N_i}{\partial z_i}-\sum_{i=0}^{5}\frac{\partial D}{\partial z_i}N_i\right)=F^2Q.AD2−Di=0∑5​∂zi​∂F​Ni​+F(Di=0∑5​∂zi​∂Ni​​−i=0∑5​∂zi​∂D​Ni​)=F2Q.

Here (z0,z1,z2,z3,z4,z5)=(X0,X1,X2,Y0,Y1,Y2)(z_0,z_1,z_2,z_3,z_4,z_5)=(X_0,X_1,X_2,Y_0,Y_1,Y_2)(z0​,z1​,z2​,z3​,z4​,z5​)=(X0​,X1​,X2​,Y0​,Y1​,Y2​), and the partial derivatives are formal polynomial derivatives. The homogeneity conditions concern total monomial degree and allow any NiN_iNi​ to be zero. The nonnegative integer eee is unrestricted, so a nonzero constant denominator is allowed. There is no separately imposed homogeneity or degree condition on QQQ, and Q=0Q=0Q=0 is allowed. All coefficients of the existentially quantified polynomials may be complex. The assertion is existence, with no uniqueness claim and no additional hypotheses.

Human review
  • Endorsed by Shuze Chen · Nov 6, 2026

    Confirmed by the moderator at approval.

  • Endorsed by benediamond · Nov 6, 2026

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

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