Understanding a proof of $\eta_1\omega_2-\eta_2\omega_1=2\pi i$

complex-analysisdefinite integralselliptic functions

The Weierstrass zeta function $\zeta$ of a lattice $L$ is defined by
$$\zeta (z,L)=\frac{1}{z}+\sum_{\omega\in L\setminus\{0\}}\left(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\right)$$
and the quasi-period $\eta$ is defined by
$$\eta (\omega,L)=\zeta (z+\omega,L)-\zeta (z,L)$$
and is independent of $z$. Furthermore,
$$\overline{P}=\{s\omega_1+t\omega_2|0\le s,t\le 1\}$$
is the fundamental parallelogram.

Now Legendre's relation states

For the basic periods and the associated basic quasi-periods of a lattice $L=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ it holds
$$\eta_1\omega_2-\eta_2\omega_1=2\pi i.$$

The proof is the following:

We shift the fundamental parallelogram so there are no lattice points on the border of $\overline{P_v}=\overline{P}+v$ with $v\in\mathbb{C}$ and apply the residue theorem:
$$\oint_{\delta \overline{P_v}}\zeta (z,L)\,\mathrm dz=2\pi i$$
because $\zeta$ has (modulo $L$) only one pole with residue $1$.
One can combine the values of the integrals along opposite sides of the parallelogram: The sides parallel to $\omega_1$ contribute $-\eta_2\omega_1$; the sides parallel to $\omega_2$ contribute $\eta_1\omega_2$. In total, the integral is $\eta_1 \omega_2-\eta_2\omega_1=2\pi i$.

Questions

I don't understand the part of the proof which is typeset in bold.

  1. What exactly does "has modulo $L$" mean?
  2. What exactly does "contribute" mean and why does it contribute $\eta_2\omega_1$?
  3. I understand that the residue is $1$ at the origin because of the Laurent expansion at the origin:
    $$\zeta(z,L)=\frac{1}{z}-\sum_{n=1}^\infty G_{2n+2}(L)z^{2n+1}$$
    where $G$ is the Eisenstein series but what about the residues at other poles?

I'm new to complex analysis and the theory of elliptic functions.

Best Answer

$$\begin{eqnarray}2i\pi=2i \pi Res(\zeta(s),z_0) & = & \int_{\partial (v+P)} \zeta(s)ds\\ &= & \int_0^1 \zeta(v+\omega_1 t)d(\omega_1 t)\\&+& \int_0^1 \zeta(v+\omega_1+\omega_2 t)d(\omega_2 t)\\&+& \int_0^1 \zeta(v+\omega_1+\omega_2-\omega_1 t)d(\omega_1 t)\\&+& \int_0^1 \zeta(v+\omega_2-\omega_2 t)d(\omega_2 t)\end{eqnarray}\\ = \omega_1 \int_0^1 (\zeta(v+\omega_1 t)-\zeta(v+\omega_2+\omega_1 t))dt+\omega_2 \int_0^1(\zeta(v+\omega_1+\omega_2 t)-\zeta(v+\omega_2t))dt\\ = \omega_2 \eta_1-\omega_1 \eta_2 $$

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