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Theorem 1: Algebraically independent Weierstrass values

Proved
WeierstrassEllipticZeta.senthil_kumar_theorem_one

by tomasz · Sep 4, 2026 · Mathlib 0df444a (Lean v4.33.1)

algebraic-independenceelliptic-functionsnumber-theory

Let Ω\OmegaΩ be the lattice of an arbitrary period pair, with associated Weierstrass functions ℘,ζ\wp,\zeta℘,ζ and invariants g2,g3g_2,g_3g2​,g3​. Let ω≠0\omega\ne0ω=0 be an element of Ω\OmegaΩ, and let u1,u2∈Cu_1,u_2\in\mathbb Cu1​,u2​∈C satisfy both

u1,u2,ω are linearly independent over Q,u_1,u_2,\omega\text{ are linearly independent over }\mathbb Q,u1​,u2​,ω are linearly independent over Q,

and

(Zu1+Zu2)∩Ω={0}.(\mathbb Zu_1+\mathbb Zu_2)\cap\Omega=\{0\}.(Zu1​+Zu2​)∩Ω={0}.

Then at least two entries of

(g2,g3,ω,η(ω),u1,u2,℘(u1),ζ(u1),℘(u2),ζ(u2))\bigl(g_2,g_3,\omega,\eta(\omega),u_1,u_2,\wp(u_1),\zeta(u_1),\wp(u_2),\zeta(u_2)\bigr)(g2​,g3​,ω,η(ω),u1​,u2​,℘(u1​),ζ(u1​),℘(u2​),ζ(u2​))

are algebraically independent over Q\mathbb QQ, where η(ω)=ζ(ω1/2+ω)−ζ(ω1/2)\eta(\omega)=\zeta(\omega_1/2+\omega)-\zeta(\omega_1/2)η(ω)=ζ(ω1​/2+ω)−ζ(ω1​/2) and ζ\zetaζ is the fixed canonical lattice series. Formally, there exist distinct indices i,j∈{0,…,9}i,j\in\{0,\ldots,9\}i,j∈{0,…,9} for which no nonzero rational polynomial in two variables vanishes at the selected pair. The hypotheses already exclude lattice poles at u1,u2u_1,u_2u1​,u2​. No algebraicity hypothesis is added for the invariants, and no particular pair is prescribed. This formalizes the published Theorem 1. Its checked proof has no Open theorem dependencies.

Preamble
import Definitions.Def_WeierstrassEllipticZeta_Defs
Formal statement
namespace WeierstrassEllipticZeta

/-- Senthil Kumar (2026), Theorem 1. This is an open proof target. -/
theorem senthil_kumar_theorem_one (L : PeriodPair) (ω u₁ u₂ : ℂ)
    (hω_ne : ω ≠ 0)
    (hω_period : ω ∈ L.lattice)
    (h_linearIndependent : LinearIndependent ℚ ![u₁, u₂, ω])
    (h_intersection : Submodule.span ℤ {u₁, u₂} ⊓ L.lattice = ⊥) :
    HasAlgebraicallyIndependentPair (theoremOneValues L ω u₁ u₂) := by sorry

end WeierstrassEllipticZeta
Source
Senthil Kumar K, Algebraic independence of values of Weierstrass elliptic and zeta functions, Proceedings of the Edinburgh Mathematical Society (online 17 June 2026), §1, Theorem 1, https://doi.org/10.1017/S001309152610145X.
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What the Lean code literally says, in plain math · GPT-6 (Codex)

For every ordered pair L=(ω1,ω2)L=(\omega_1,\omega_2)L=(ω1​,ω2​) of complex numbers linearly independent over R\mathbb RR, put Λ={mω1+nω2:m,n∈Z}\Lambda=\{m\omega_1+n\omega_2:m,n\in\mathbb Z\}Λ={mω1​+nω2​:m,n∈Z}. For every ω,u1,u2∈C\omega,u_1,u_2\in\mathbb Cω,u1​,u2​∈C, assume that ω≠0\omega\ne0ω=0, that ω∈Λ\omega\in\Lambdaω∈Λ, that u1,u2,ωu_1,u_2,\omegau1​,u2​,ω are linearly independent over Q\mathbb QQ (that is, for all a,b,c∈Qa,b,c\in\mathbb Qa,b,c∈Q, au1+bu2+cω=0au_1+bu_2+c\omega=0au1​+bu2​+cω=0 implies a=b=c=0a=b=c=0a=b=c=0), and that (Zu1+Zu2)∩Λ={0}(\mathbb Zu_1+\mathbb Zu_2)\cap\Lambda=\{0\}(Zu1​+Zu2​)∩Λ={0}, where Zu1+Zu2={mu1+nu2:m,n∈Z}\mathbb Zu_1+\mathbb Zu_2=\{mu_1+nu_2:m,n\in\mathbb Z\}Zu1​+Zu2​={mu1​+nu2​:m,n∈Z}. Define g2=60∑λ∈Λ′1/λ4g_2=60\sum_{\lambda\in\Lambda}'1/\lambda^4g2​=60∑λ∈Λ′​1/λ4, g3=140∑λ∈Λ′1/λ6g_3=140\sum_{\lambda\in\Lambda}'1/\lambda^6g3​=140∑λ∈Λ′​1/λ6, and, for every z∈Cz\in\mathbb Cz∈C, define PL(z)=∑λ∈Λ′(1/(z−λ)2−1/λ2)P_L(z)=\sum_{\lambda\in\Lambda}'\bigl(1/(z-\lambda)^2-1/\lambda^2\bigr)PL​(z)=∑λ∈Λ′​(1/(z−λ)2−1/λ2) and ZL(z)=1/z+∑λ∈Λ′sz(λ)Z_L(z)=1/z+\sum_{\lambda\in\Lambda}'s_z(\lambda)ZL​(z)=1/z+∑λ∈Λ′​sz​(λ), with sz(0)=0s_z(0)=0sz​(0)=0 and sz(λ)=1/(z−λ)+1/λ+z/λ2s_z(\lambda)=1/(z-\lambda)+1/\lambda+z/\lambda^2sz​(λ)=1/(z−λ)+1/λ+z/λ2 for λ≠0\lambda\ne0λ=0. Each primed sum means the limit of sums over finite subsets of its index set, and is assigned the value 000 if that limit does not exist; every division is interpreted in C\mathbb CC with a/0=0a/0=0a/0=0 for every a∈Ca\in\mathbb Ca∈C, so these formulas define values for every zzz, including lattice points. The hypotheses force u1,u2≠0u_1,u_2\ne0u1​,u2​=0 and u1,u2∉Λu_1,u_2\notin\Lambdau1​,u2​∈/Λ; they also ensure that ω1/2\omega_1/2ω1​/2 and ω1/2+ω\omega_1/2+\omegaω1​/2+ω lie outside Λ\LambdaΛ. Form the ordered ten-tuple (v0,…,v9)=(g2,g3,ω,ZL(ω1/2+ω)−ZL(ω1/2),u1,u2,PL(u1),ZL(u1),PL(u2),ZL(u2))(v_0,\ldots,v_9)=\bigl(g_2,g_3,\omega,Z_L(\omega_1/2+\omega)-Z_L(\omega_1/2),u_1,u_2,P_L(u_1),Z_L(u_1),P_L(u_2),Z_L(u_2)\bigr)(v0​,…,v9​)=(g2​,g3​,ω,ZL​(ω1​/2+ω)−ZL​(ω1​/2),u1​,u2​,PL​(u1​),ZL​(u1​),PL​(u2​),ZL​(u2​)). Then there exist distinct indices i,j∈{0,1,…,9}i,j\in\{0,1,\ldots,9\}i,j∈{0,1,…,9} such that vi,vjv_i,v_jvi​,vj​ are algebraically independent over Q\mathbb QQ: for every polynomial F∈Q[X,Y]F\in\mathbb Q[X,Y]F∈Q[X,Y], F(vi,vj)=0F(v_i,v_j)=0F(vi​,vj​)=0 implies that FFF is the zero polynomial.

Human review
  • Endorsed by Shuze Chen · Sep 29, 2026

    Confirmed by the moderator at approval.

  • Endorsed by tomasz · Sep 29, 2026

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

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