§23.5 Special Lattices

The Weierstrass functions take real values on the real axis iff the lattice is fixed under complex conjugation: $𝕃=\overline{𝕃}$; equivalently, when $g_2,g_3\in ℝ$. This happens in the cases treated in the following four subsections.

This occurs when both ${\omega }_1$ and ${\omega }_3/\mathrm{i}$ are real and positive. Then $\mathrm{\Delta }>0$ and the parallelogram with vertices at $0$, $2{\omega }_1$, $2{\omega }_1+2{\omega }_3$, $2{\omega }_3$ is a rectangle.

In this case the lattice roots $e_1$, $e_2$, and $e_3$ are real and distinct. When they are identified as in (23.3.9)

Also, $e_2$ and $g_3$ have opposite signs unless ${\omega }_3=\mathrm{i}{\omega }_1$, in which event both are zero.

As functions of $\mathrm{\Im }{\omega }_3$, $e_1$ and $e_2$ are decreasing and $e_3$ is increasing.

This occurs when ${\omega }_1$ is real and positive and ${\omega }_3=\mathrm{i}{\omega }_1$. The parallelogram $0$, $2{\omega }_1$, $2{\omega }_1+2{\omega }_3$, $2{\omega }_3$ is a square, and

Note also that in this case $\tau =\mathrm{i}$. In consequence,

This occurs when ${\omega }_1$ is real and positive, $\mathrm{\Im }{\omega }_3>0$, $\mathrm{\Re }{\omega }_3=\frac{1}{2}{\omega }_1$, and . The parallelogram $0$, $2{\omega }_1-2{\omega }_3$, $2{\omega }_1$, $2{\omega }_3$, is a rhombus: see Figure 23.5.1.

The lattice root $e_1$ is real, and $e_3=\overline{e_2}$, with $\mathrm{\Im }{e}_2>0$. $e_1$ and $g_3$ have the same sign unless $2{\omega }_3=(1+\mathrm{i}){\omega }_1$ when both are zero: the pseudo-lemniscatic case. As a function of $\mathrm{\Im }{e}_3$ the root $e_1$ is increasing. For the case ${\omega }_3={\mathrm{e}}^{\pi \mathrm{i}/3}{\omega }_1$ see §23.5(v).

This occurs when ${\omega }_1$ is real and positive and ${\omega }_3={\mathrm{e}}^{\pi \mathrm{i}/3}{\omega }_1$. The rhombus $0$, $2{\omega }_1-2{\omega }_3$, $2{\omega }_1$, $2{\omega }_3$ can be regarded as the union of two equilateral triangles: see Figure 23.5.2.

and the lattice roots and invariants are given by

Note also that in this case $\tau ={\mathrm{e}}^{\mathrm{i}\pi /3}$. In consequence,