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The j-invariant of the Legendre normal form

Statement

For a full lattice $Λ_{τ}$ with invariants $g_{2}, g_{3}$ one has $g_{2} = \frac{4 π^{4}}{3} E_{4} (τ)$ and $g_{3} = \frac{8 π^{6}}{27} E_{6} (τ)$ , hence $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = j (τ) .$ Writing $λ = λ (τ)$ and putting $J_{L e g} (x) : = \frac{256 (x^{2} - x + 1)^{3}}{x^{2} (x - 1)^{2}}$ for $x \in C ∖ {0, 1}$ , the affine normalisation of the associated Legendre cubic gives $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = J_{L e g} (λ (τ)) = j (τ);$ in particular $J_{L e g} (1 - λ) = J_{L e g} (1 / λ) = J_{L e g} (λ)$ .

Facts & Assumptions

Given: $Λ_{τ} = Z + Z τ$ , its invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ , the cubic relation $℘^{' 2} = 4 ℘^{3} - g_{2} ℘ - g_{3} = 4 (℘ - e_{1}) (℘ - e_{2}) (℘ - e_{3})$ with distinct $e_{j}$ , and $λ (τ) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}}$ (Weierstrass cubic differential equation, Degree two of ℘ and its four branch points, The modular lambda function).

[F1]

$G_{k} = 2 ζ (k) E_{k}$ . The unnormalised Fourier coefficient computed in Eisenstein series are modular forms; their Fourier coefficients, Proof 1.2, is $(- 2 π i)^{k} / ((k - 1)! ζ (k))$ ; comparing its values $240$ for $k = 4$ and $- 504$ for $k = 6$ gives $ζ (4) = π^{4} / 90$ and $ζ (6) = π^{6} / 945$ (The level-one Eisenstein series E_k and the weight-two series E_2, The Lipschitz formula for the reciprocal-power sums).

[F2]

The lattice discriminant $D_{Λ} : = g_{2}^{3} - 27 g_{3}^{2}$ is nonzero because the roots $e_{j}$ are distinct. The normalised modular discriminant is $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728$ , with $j = E_{4}^{3} / Δ$ ; these are different normalisations, related in step 1.1 (Degree two of ℘ and its four branch points, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F3]

Writing the Weierstrass cubic $Y^{2} = 4 x^{3} - g_{2} x - g_{3}$ as $y^{2} = x^{3} + A x + B$ uses $Y = 2 y$ , $A = - g_{2} / 4$ , $B = - g_{3} / 4$ . Hence $1728 \cdot 4 A^{3} / (4 A^{3} + 27 B^{2}) = 1728 g_{2}^{3} / (g_{2}^{3} - 27 g_{3}^{2})$ . Choose $h$ with $h^{2} = d \neq 0$ (Every complex number has a square root, by an explicit Cartesian formula). Under $x = d X$ , $y = h^{3} \tilde{y}$ , the coefficients become $A / d^{2}, B / d^{3}$ , so both numerator and denominator acquire $d^{- 6}$ and this ratio is unchanged.

[F4]

The affine map $x \mapsto (x - e_{2}) / (e_{1} - e_{2})$ carries the branch triple $(e_{1}, e_{2}, e_{3})$ to $(1, 0, λ)$ ; cross-ratios and the labelling of branch points are preserved by affine maps (The modular lambda function, The cross-ratio is invariant under Möbius transformations, Degree two of ℘ and its four branch points).

Proof

1.1F1F2givenalgebra

By [F1], $g_{2} = 60 G_{4} = 120 ζ (4) E_{4} = 120 \cdot \frac{π^{4}}{90} E_{4} = \frac{4 π^{4}}{3} E_{4}$ and $g_{3} = 140 G_{6} = 280 ζ (6) E_{6} = 280 \cdot \frac{π^{6}}{945} E_{6} = \frac{8 π^{6}}{27} E_{6}$ ; hence $g_{2}^{3} = \frac{64 π^{12}}{27} E_{4}^{3}$ and $27 g_{3}^{2} = 27 \cdot \frac{64 π^{12}}{729} E_{6}^{2} = \frac{64 π^{12}}{27} E_{6}^{2}$ , so $g_{2}^{3} - 27 g_{3}^{2} = \frac{64 π^{12}}{27} (E_{4}^{3} - E_{6}^{2}) = \frac{64 π^{12}}{27} \cdot 1728 Δ$ by [F2]. Dividing, $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = \frac{E_{4}^{3}}{Δ} = j (τ)$ .

2.1F3F4step 1.1givenalgebra

Put $d = e_{1} - e_{2} \neq 0$ and choose $h$ with $h^{2} = d$ . In $Y^{2} = 4 (x - e_{1}) (x - e_{2}) (x - e_{3})$ , the substitutions $x = e_{2} + d u$ , $Y = 2 h^{3} v$ give $v^{2} = u (u - 1) (u - λ)$ by [F4]. After the original Weierstrass cubic is written with leading coefficient one, the translation by $e_{2}$ and subsequent centring cancel each other, while the dilation divides the centred coefficients $- g_{2} / 4, - g_{3} / 4$ by $d^{2}, d^{3}$ . Thus its invariant ratio is unchanged by [F3]. Completing the cube by $x = X + \frac{1 + λ}{3}$ gives $y^{2} = X^{3} + A X + B$ with $A = λ - \frac{(1 + λ)^{2}}{3} = - \frac{λ^{2} - λ + 1}{3}$ and $B = \frac{(1 + λ) (9 λ - 2 (1 + λ)^{2})}{27} = - \frac{(λ^{2} - λ - 2) (2 λ - 1)}{27}$ . Therefore $4 A^{3} = - \frac{4 (λ^{2} - λ + 1)^{3}}{27}$ and $27 B^{2} = \frac{(λ^{2} - λ - 2)^{2} (2 λ - 1)^{2}}{27}$ , and the algebraic identity $4 (λ^{2} - λ + 1)^{3} - (λ^{2} - λ - 2)^{2} (2 λ - 1)^{2} = 27 λ^{2} (λ - 1)^{2}$ , which holds for all $λ$ by expanding both sides, gives $4 A^{3} + 27 B^{2} = - λ^{2} (λ - 1)^{2}$ . Hence $J : = \frac{1728 \cdot 4 A^{3}}{4 A^{3} + 27 B^{2}} = \frac{1728 \cdot (- 4 (λ^{2} - λ + 1)^{3} / 27)}{- λ^{2} (λ - 1)^{2}} = \frac{256 (λ^{2} - λ + 1)^{3}}{λ^{2} (λ - 1)^{2}} = J_{L e g} (λ)$ .

3.1F3step 2.1givenalgebra∎

Invariance under the cross-ratio substitutions: $J_{L e g} (1 - λ) = J_{L e g} (λ)$ because $(1 - λ)^{2} - (1 - λ) + 1 = λ^{2} - λ + 1$ and $(1 - λ)^{2} ((1 - λ) - 1)^{2} = λ^{2} (λ - 1)^{2}$ ; and $J_{L e g} (1 / λ) = J_{L e g} (λ)$ because $λ^{- 2} - λ^{- 1} + 1 = (λ^{2} - λ + 1) λ^{- 2}$ and $(λ^{- 2}) (λ^{- 1} - 1)^{2} = λ^{2} (λ - 1)^{2} λ^{- 6}$ . Combining 1.1 and 2.1, $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = j (τ) = J_{L e g} (λ (τ))$ , which is the assertion.

Depends on

Used by

The modular lambda function: Y(2) biholomorphic to the twice-punctured plane, and the slit-plane quadrilateral Example

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Sources

C. T. McMullen, Advanced Complex Analysis, Math 213a course notes (Harvard, 2010) (standard reference, not scraped)
J. S. Milne, Modular Functions and Modular Forms (v1.31, 2017) (standard reference, not scraped)