Continue with the lattice \[ \Lambda=\mathbb Z\lambda_1\oplus\mathbb Z\lambda_2 \] and its Weierstrass zeta function \(\zeta\).
Show that there are constants \(\eta_1,\eta_2\in\mathbb C\) such that \[ \zeta(z+\lambda_i)=\zeta(z)+\eta_i \qquad(i=1,2). \]
Using the residue theorem, prove the Legendre relation \[ \det \begin{pmatrix} \eta_1&\eta_2\\ \lambda_1&\lambda_2 \end{pmatrix} =2\pi i. \]
Solution
For the sign in the displayed Legendre relation, take the ordered basis \((\lambda_1,\lambda_2)\) to be positively oriented in the real plane \(\mathbb C\cong\mathbb R^2\). Then \[ \boxed{\eta_1\lambda_2-\eta_2\lambda_1=2\pi i.} \] If the ordered basis is reversed, the left side and the orientation of its fundamental parallelogram both reverse, so the corresponding formula has the opposite sign.
1. Translation by a lattice period changes \(\zeta\) by a constant.
By E-HM213A17-138, \[ \zeta'(z)=-\wp(z). \] For any \(\lambda\in\Lambda\), periodicity of \(\wp\) gives \[ \frac{d}{dz}\bigl(\zeta(z+\lambda)-\zeta(z)\bigr) =-\wp(z+\lambda)+\wp(z)=0 \] away from the lattice poles. Hence the difference is constant there, and therefore as a meromorphic identity on \(\mathbb C\). Thus there is a constant \(\eta(\lambda)\) such that \[ \zeta(z+\lambda)=\zeta(z)+\eta(\lambda). \] Taking \(\lambda=\lambda_i\) defines \[ \eta_i=\eta(\lambda_i), \qquad i=1,2. \]
2. The zeta function has residue \(1\) at every lattice point.
By E-HM213A17-138, \[ \zeta(z)=\frac{\sigma'(z)}{\sigma(z)}. \] The canonical product \(\sigma\) has a simple zero at every point of \(\Lambda\). The logarithmic derivative of a holomorphic function at a simple zero has residue \(1\). Therefore \[ \operatorname{Res}_{z=\lambda}\zeta(z)=1 \qquad(\lambda\in\Lambda). \]
3. The residue theorem on a centered fundamental parallelogram gives\[ \int_{\partial P}\zeta(z)\,dz=2\pi i. \]
Let \[ P=\left\{s\lambda_1+t\lambda_2: -\frac12\le s,t\le\frac12\right\}. \] No lattice point lies on \(\partial P\), and \(0\) is the unique lattice point in its interior. With the positive orientation induced by the positively oriented ordered basis \((\lambda_1,\lambda_2)\), the residue theorem and 2 give \[ \int_{\partial P}\zeta(z)\,dz =2\pi i\operatorname{Res}_{0}\zeta =2\pi i. \]
4. Pairing opposite edges evaluates the same boundary integral as\[ \eta_1\lambda_2-\eta_2\lambda_1. \]
Set \[ a=-\frac{\lambda_1+\lambda_2}{2}. \] Traverse the boundary as \[ a\to a+\lambda_1 \to a+\lambda_1+\lambda_2 \to a+\lambda_2 \to a. \] For the bottom and top edges, translate the top edge by \(-\lambda_2\). Using 1, \[ \begin{aligned} I_{\mathrm{bottom}}+I_{\mathrm{top}} &=\int_a^{a+\lambda_1}\zeta(z)\,dz +\int_{a+\lambda_1}^{a}\zeta(z+\lambda_2)\,dz\\ &=\int_a^{a+\lambda_1}\zeta(z)\,dz -\int_a^{a+\lambda_1}(\zeta(z)+\eta_2)\,dz\\ &=-\eta_2\lambda_1. \end{aligned} \] Similarly, translate the right edge by \(-\lambda_1\) and pair it with the left edge: \[ \begin{aligned} I_{\mathrm{right}}+I_{\mathrm{left}} &=\int_a^{a+\lambda_2}\zeta(z+\lambda_1)\,dz +\int_{a+\lambda_2}^{a}\zeta(z)\,dz\\ &=\int_a^{a+\lambda_2}(\zeta(z)+\eta_1)\,dz -\int_a^{a+\lambda_2}\zeta(z)\,dz\\ &=\eta_1\lambda_2. \end{aligned} \] Adding the two pairs gives \[ \int_{\partial P}\zeta(z)\,dz =\eta_1\lambda_2-\eta_2\lambda_1 =\det \begin{pmatrix} \eta_1&\eta_2\\ \lambda_1&\lambda_2 \end{pmatrix}. \] By 3 this equals \(2\pi i\), which proves the stated Legendre relation with the positive-orientation convention.