[Paper Review] Plane quartic twists of X(5,3)

J.A. Fernandez, Josep González+1 more|ArXiv.org|Jan 28, 2005

Algebraic Geometry and Number Theory3 citations

TL;DR

This paper presents a method to construct explicit plane quartic models over ℚ for twists X(5,3)ρ of the modular curve X(5,3), corresponding to odd surjective Galois representations ϱ: Gℚ → PGL₂(𝔽₃). Using a degree-four polynomial defining the splitting field of ϱ, the method computes a rational model of the twisted curve via Galois cohomology and automorphism group actions, yielding a genus-three curve with explicit rational points. The key contribution is a computational framework for determining quadratic ℚ-curves of degree 5 realizing a given projective mod 3 Galois representation.

ABSTRACT

Given an odd representation of the absolute Galois group of Q onto PGL(2,3) and a positive integer N, there exists a twisted modular curve defined over Q whose rational points classify the quadratic Q-curves of degree N realizing the representation. The paper gives a method to provide an explicit plane quartic model for this curve in the genus-three case N=5.

Motivation & Objective

• To provide an explicit method for constructing rational plane quartic models of twisted modular curves X(N,p)ρ for N=5 and p=3.
• To address the classification of quadratic ℚ-curves of degree 5 realizing a given odd surjective Galois representation ϱ: Gℚ → PGL₂(𝔽₃).
• To develop a computational framework using Galois cohomology and automorphism group actions to produce such models from the splitting field of ϱ.
• To demonstrate the feasibility of the method via an explicit example with a degree-four polynomial, yielding a concrete quartic equation.

Proposed method

• The method constructs the twist X(5,3)ρ using a 1-cocycle ξ = ϱ ∘ ε in H¹(Gℚ, W(5,3)), where W(5,3) ≅ PGL₂(𝔽₃) is the automorphism group of X(5,3) → X⁺(5).
• It uses a rational model of X(5,3) as a starting point, derived from the canonical model of X₀(5) and X(3), with a specific choice of involution corresponding to a matrix in PGL₂(𝔽₃) of order 2.
• The construction involves computing a Galois-equivariant isomorphism Ψ: X(5,3)ρ → X(5,3) via a matrix Θ in GL₄(ℚ) that encodes the twist, derived from the action of Galois on the automorphism group.
• The final model is obtained by transforming the original plane quartic model of X(5,3) using Θ, yielding a new quartic equation over ℚ with coefficients in ℤ.
• The method relies on the splitting field of a given degree-four polynomial f(X) ∈ ℤ[X], which must have the same splitting field as ϱ, to define the required Galois action.
• The construction is validated using computational algebra systems like Magma, and the resulting model is verified by computing rational points and analyzing the Jacobian rank.

Experimental results

Research questions

• RQ1 How can one explicitly construct a plane quartic model over ℚ for the twist X(5,3)ρ of the modular curve X(5,3) associated to a given odd surjective Galois representation ϱ: Gℚ → PGL₂(𝔽₃)?
• RQ2 What is the relationship between the splitting field of ϱ and the field of definition of the rational points on X(5,3)ρ?
• RQ3 Can the Chabauty-Coleman method be applied to determine the full set of rational points on X(5,3)ρ when the rank of its Jacobian is at least three?
• RQ4 What computational techniques can be used to produce and verify such models in practice, especially when standard methods like Chabauty-Coleman fail?

Key findings

• The paper provides an explicit plane quartic equation over ℚ for X(5,3)ρ associated to the Galois representation defined by the irreducible quartic polynomial f(X) = X⁴ - 3X² + 2X + 3.
• The model is given by the equation: -9XY(2X+Y)(9X+8Y) + 9(6X³ + 62X²Y + 66XY² + 15Y³)Z + 3(27X² - 104XY - 83Y²)Z² - 3(94X + 7Y)Z³ + 191Z⁴ = 0.
• Four rational points are found at infinity: [0:1:0], [1:0:0], [1:-2:0], and [8:-9:0], each corresponding to a ℚ-curve of degree 5 realizing the representation ϱ.
• The j-invariants of the corresponding ℚ-curves are computed explicitly, with j₁ = (−8√11)³(10 + 3√11), j₂ = (6(110 + 31√11))³(10 + 3√11), j₃ = (12(10 − √11))³(10 + 3√11), and j₄ = (2(−6878815950 + 2118474913√11)/53⁵)³(10 + 3√11).
• The rank of the Jacobian J(X(5,3)ρ) is shown to be at least three, which invalidates the use of the Chabauty-Coleman method for determining all rational points.