Suppose \[ (x,y)=(\wp(z),\wp'(z)) \] satisfies \[ y^2=4x^3+ax+b, \qquad a,b\in\mathbb R, \] and the cubic \(4x^3+ax+b\) has exactly one real root. What can you conclude about the shape of the lattice \(\Lambda\) defining \(\wp\)?
Solution
The lattice is rhombic rather than rectangular. More precisely, after choosing a primitive positive real period \(\alpha\), one can choose a second period of the form \[ \boxed{ \beta=\frac{\alpha}{2}+i\gamma, \qquad \gamma>0, } \] so that \[ \Lambda=\mathbb Z\alpha\oplus\mathbb Z\beta. \] Equivalently, \[ \Lambda=\mathbb Z\beta\oplus\mathbb Z\overline\beta, \] and these two generators have equal length; a fundamental parallelogram is therefore a rhombus symmetric about the real axis.
1. The lattice is invariant under complex conjugation.
The Weierstrass equation is \[ y^2=4x^3-g_2x-g_3. \] Comparing with \[ y^2=4x^3+ax+b, \qquad a,b\in\mathbb R, \] gives \[ g_2=-a\in\mathbb R, \qquad g_3=-b\in\mathbb R. \]
For the conjugate lattice \[ \overline\Lambda=\{\overline\lambda:\lambda\in\Lambda\}, \] absolute convergence of the Eisenstein sums gives \[ g_2(\overline\Lambda)=\overline{g_2(\Lambda)}=g_2(\Lambda), \qquad g_3(\overline\Lambda)=\overline{g_3(\Lambda)}=g_3(\Lambda). \] By E-HM213A17-135, the pair \((g_2,g_3)\) determines the lattice uniquely. Hence \[ \overline\Lambda=\Lambda. \]
2. Any conjugation-invariant lattice has either rectangular or rhombic shape.
Since \(\Lambda\) has rank two and is not contained in the imaginary axis, there is some \(\lambda\in\Lambda\) with nonzero real part. Then \[ \lambda+\overline\lambda=2\operatorname{Re}\lambda \] is a nonzero real lattice vector. Thus \[ \Lambda\cap\mathbb R=\mathbb Z\alpha \] for a unique primitive generator \(\alpha>0\).
Choose \(\beta\in\Lambda\) with \(\operatorname{Im}\beta>0\) whose class generates \[ \Lambda/\mathbb Z\alpha\cong\mathbb Z, \] and replace \(\beta\) by \(\beta+k\alpha\) so that \[ 0\le\operatorname{Re}\beta<\alpha. \] Because \(\overline\beta\in\Lambda\), write \[ \overline\beta=m\alpha+n\beta, \qquad m,n\in\mathbb Z. \] Comparing imaginary parts gives \[ n=-1. \] Hence \[ \beta+\overline\beta=m\alpha, \] or \[ 2\operatorname{Re}\beta=m\alpha. \] The normalization \(0\le\operatorname{Re}\beta<\alpha\) forces \[ m=0\quad\text{or}\quad m=1. \]
If \(m=0\), then \(\operatorname{Re}\beta=0\), so \(\alpha\) is real and \(\beta\) is purely imaginary: the lattice is rectangular.
If \(m=1\), then \[ \operatorname{Re}\beta=\frac\alpha2, \] and \[ \overline\beta=\alpha-\beta. \] Thus \[ \Lambda=\mathbb Z\beta\oplus\mathbb Z\overline\beta, \] with \(|\beta|=|\overline\beta|\), giving a rhombic fundamental parallelogram.
3. The rectangular case is impossible under the one-real-root hypothesis.
Suppose instead that \[ \Lambda=\mathbb Z\alpha\oplus\mathbb Z\beta, \qquad \alpha\in\mathbb R, \quad \beta\in i\mathbb R. \] Since \(\overline\Lambda=\Lambda\) by 1, conjugating the defining series gives \(\wp(\overline z)=\overline{\wp(z)}\). Because \(\alpha\) is real and \(\beta\) is purely imaginary, each of the three nonzero order-two points \[ \frac\alpha2, \qquad \frac\beta2, \qquad \frac{\alpha+\beta}{2} \] satisfies \(\overline\omega\equiv\pm\omega\pmod\Lambda\); for instance \(\overline{(\alpha+\beta)/2}=(\alpha+\beta)/2-\beta\). Since \(\wp\) is even and \(\Lambda\)-periodic, these points have real \(\wp\)-values \[ e_1,e_2,e_3. \] These are the three distinct roots of \[ 4x^3-g_2x-g_3 =4x^3+ax+b. \] Hence the cubic would have three real roots, contrary to the assumption that it has exactly one.
Therefore the only remaining case in 2 is the rhombic one, proving the asserted shape of \(\Lambda\).