Let \(g_2(\tau),g_3(\tau)\) be the Weierstrass invariants and let \(\lambda(\tau)\) be the cross-ratio of the three roots of \[ 4x^3-g_2(\tau)x-g_3(\tau)=0 \] together with \(\infty\), in the source’s ordering. Using \[ J(\tau)=\frac4{27}\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(1-\lambda)^2}, \] prove that \[ J(\tau)=\frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}. \]
Solution
Let \(e_1,e_2,e_3\) be the three roots of \[ 4x^3-g_2x-g_3=0 \] ordered so that \[ \lambda=\frac{e_3-e_2}{e_1-e_2}. \] We suppress the argument \(\tau\) until the end.
1. Express the roots in terms of \(\lambda\) and \(d=e_1-e_2\).
The coefficient of \(x^2\) in the cubic vanishes, so \[ e_1+e_2+e_3=0. \] Set \[ d=e_1-e_2\ne0. \] By definition of the cross-ratio, \[ e_3-e_2=\lambda d. \] Thus \[ e_1=e_2+d, \qquad e_3=e_2+\lambda d. \] Substituting into the zero-sum relation gives \[ 3e_2+(1+\lambda)d=0. \] Hence \[ \boxed{ e_1=\frac{2-\lambda}{3}d, \qquad e_2=-\frac{1+\lambda}{3}d, \qquad e_3=\frac{2\lambda-1}{3}d. } \]
2. The invariant \(g_2\) is\[ g_2=\frac43d^2(\lambda^2-\lambda+1). \]
Factor the cubic as \[ 4(x-e_1)(x-e_2)(x-e_3) =4x^3-g_2x-g_3. \] Comparing the coefficient of \(x\) gives \[ 4(e_1e_2+e_1e_3+e_2e_3)=-g_2. \] Since \(e_1+e_2+e_3=0\), \[ e_1e_2+e_1e_3+e_2e_3 =-\frac12(e_1^2+e_2^2+e_3^2). \] Using the formulas from 1, \[ \begin{aligned} e_1^2+e_2^2+e_3^2 &=\frac{d^2}{9} \left((2-\lambda)^2+(1+\lambda)^2+(2\lambda-1)^2\right)\\ &=\frac23d^2(\lambda^2-\lambda+1). \end{aligned} \] Therefore \[ g_2 =2(e_1^2+e_2^2+e_3^2) =\frac43d^2(\lambda^2-\lambda+1). \]
3. The discriminant satisfies\[ g_2^3-27g_3^2 =16d^6\lambda^2(1-\lambda)^2. \]
Divide the cubic by \(4\): \[ x^3-\frac{g_2}{4}x-\frac{g_3}{4} =(x-e_1)(x-e_2)(x-e_3). \] For a monic depressed cubic \(x^3+px+q\), its discriminant is \[ -4p^3-27q^2 =(e_1-e_2)^2(e_1-e_3)^2(e_2-e_3)^2. \] Here this gives \[ \frac{g_2^3-27g_3^2}{16} =(e_1-e_2)^2(e_1-e_3)^2(e_2-e_3)^2. \] Now \[ e_1-e_2=d, \qquad e_1-e_3=(1-\lambda)d, \qquad e_2-e_3=-\lambda d. \] Consequently \[ \frac{g_2^3-27g_3^2}{16} =d^6\lambda^2(1-\lambda)^2, \] which is the claimed identity.
4. Substitution into the cross-ratio formula yields the desired expression for \(J\).
Cubing the formula from 2 gives \[ g_2^3 =\frac{64}{27}d^6(\lambda^2-\lambda+1)^3. \] Together with 3, \[ \begin{aligned} \frac{g_2^3}{g_2^3-27g_3^2} &=\frac{64d^6(\lambda^2-\lambda+1)^3/27} {16d^6\lambda^2(1-\lambda)^2}\\ &=\frac4{27} \frac{(\lambda^2-\lambda+1)^3} {\lambda^2(1-\lambda)^2}\\ &=J. \end{aligned} \] Restoring the dependence on \(\tau\) gives \[ \boxed{ J(\tau) =\frac{g_2(\tau)^3} {g_2(\tau)^3-27g_3(\tau)^2}. } \]
Source Collections
Dependencies
T-CRVMODJThe modular function $J(\tau) = g_2^3/\Delta$ and the $\SL_2(\ZZ)$-action on $\HH$uses