Determine the zeros of the Weierstrass \(\wp\)-function for each of the lattices \[ \mathbb Z\oplus\mathbb Zi \qquad\text{and}\qquad \mathbb Z\oplus\mathbb Z\omega, \] where \[ \omega=\frac{1+\sqrt{-3}}2. \]
Solution
Modulo the relevant lattice, the zeros are: \[ \boxed{ \begin{array}{c|c} \Lambda & \text{zeros of }\wp_\Lambda\\ \hline \mathbb Z\oplus\mathbb Zi & (1+i)/2\quad\text{with multiplicity }2,\\[2mm] \mathbb Z\oplus\mathbb Z\omega & \pm(1+\omega)/3\quad\text{each simple.} \end{array} } \] Thus in \(\mathbb C\) one adds all lattice translates of the displayed points.
1. If \(u\Lambda=\Lambda\), then\[ \wp_\Lambda(uz)=u^{-2}\wp_\Lambda(z). \]
From the defining series, \[ \wp_\Lambda(z) =\frac1{z^2} +\sum_{\lambda\in\Lambda\setminus\{0\}} \left( \frac1{(z-\lambda)^2}-\frac1{\lambda^2} \right). \] If multiplication by \(u\) permutes \(\Lambda\), write \(\lambda=u\mu\) in the sum. Then \[ \begin{aligned} \wp_\Lambda(uz) &=\frac1{u^2z^2} +\sum_{\mu\in\Lambda\setminus\{0\}} \left( \frac1{(uz-u\mu)^2}-\frac1{(u\mu)^2} \right)\\ &=u^{-2}\wp_\Lambda(z). \end{aligned} \]
2. For the square lattice\[ \Lambda_\square=\mathbb Z\oplus\mathbb Zi, \]the point\[ c=\frac{1+i}{2} \]is a zero of \(\wp_{\Lambda_\square}\).
Multiplication by \(i\) preserves \(\Lambda_\square\), so 1 gives \[ \wp(iz)=i^{-2}\wp(z)=-\wp(z). \] But \[ ic=\frac{-1+i}{2}=c-1, \] so periodicity gives \[ \wp(ic)=\wp(c). \] Combining the two identities, \[ \wp(c)=-\wp(c), \] and hence \[ \wp(c)=0. \]
3. The square-lattice zero in 2 has multiplicity two and is the only zero modulo \(\Lambda_\square\).
The point \(c\) is a nonzero point of order two on \[ \mathbb C/\Lambda_\square, \] because \[ 2c=1+i\in\Lambda_\square. \] Since \(\wp'\) is odd and \(\Lambda_\square\)-periodic, \[ \wp'(c)=\wp'(c-2c)=\wp'(-c)=-\wp'(c), \] so \[ \wp'(c)=0. \] Thus the zero at \(c\) has multiplicity at least two.
On the torus, \(\wp\) has one pole, of order two, so it has exactly two zeros counted with multiplicity. Hence the zero at \(c\) has multiplicity exactly two and exhausts all zeros modulo the lattice.
4. For the hexagonal lattice\[ \Lambda_{\mathrm{hex}}=\mathbb Z\oplus\mathbb Z\omega, \qquad \omega=e^{i\pi/3}, \]the two points\[ z_0=\frac{1+\omega}{3}, \qquad -z_0 \]are zeros.
Set \[ \rho=\omega^2=e^{2\pi i/3}=\omega-1. \] Multiplication by \(\rho\) preserves the lattice, since \[ \rho\cdot1=\omega-1\in\Lambda_{\mathrm{hex}}, \qquad \rho\omega=\omega^3=-1\in\Lambda_{\mathrm{hex}}. \] By 1, \[ \wp(\rho z)=\rho^{-2}\wp(z)=\rho\wp(z), \] because \(\rho^3=1\).
Now \[ (\rho-1)z_0 =\frac{(\omega-2)(1+\omega)}3 =-1\in\Lambda_{\mathrm{hex}}. \] Thus \[ \rho z_0\equiv z_0\pmod{\Lambda_{\mathrm{hex}}}. \] Periodicity and the scaling identity give \[ \wp(z_0) =\wp(\rho z_0) =\rho\wp(z_0). \] Since \(\rho\ne1\), \[ \wp(z_0)=0. \] The function \(\wp\) is even, so \[ \wp(-z_0)=0 \] as well.
5. These are exactly the two hexagonal-lattice zeros, and both are simple.
The two points are distinct modulo the lattice: if \[ z_0\equiv-z_0, \] then \[ 2z_0=\frac{2+2\omega}{3} \] would lie in \(\mathbb Z\oplus\mathbb Z\omega\), which it does not.
Thus \(\wp\) has at least two distinct zeros on the torus. As in 3, its unique double pole implies that the total zero multiplicity is exactly two. Therefore these are all the zeros, and each has multiplicity one.