Weierstrass zeta and sigma functions

Remarks

The product converges absolutely and locally uniformly, because each factor is $1 + O(\abs{z/\omega}^3)$ and $\sum_{\omega \ne 0}\abs{\omega}^{-3} < \infty$. So $\sigma$ is an entire function, odd, with simple zeros exactly at the points of $\Lambda$. Also $\zeta'(z) = -\wp(z)$, where $\wp$ is the Weierstrass elliptic function of $\Lambda$. Since $\wp$ is periodic, $\eta_j = \zeta(z + \omega_j) - \zeta(z)$ is a constant, the quasi-period, for $j = 1, 2$. Then $\sigma(z + \omega_j) = -\exp\big(\eta_j(z + \omega_j/2)\big)\,\sigma(z)$. If $\operatorname{Im}(\omega_2/\omega_1) > 0$, the Legendre relation $\eta_1\omega_2 - \eta_2\omega_1 = 2\pi i$ holds; it follows by integrating $\zeta$ around a fundamental parallelogram, where $\zeta$ has one simple pole of residue $1$. Example: replace $\Lambda$ by the one-dimensional group $\Z\pi$ in the same formulas. Since $\sum_{n \ne 0} 1/(n\pi)^2 = 1/3$, they give $\zeta(z) = \cot z + z/3$ and $\sigma(z) = e^{z^2/6}\sin z$. This is the limit of the lattice case when $\omega_1 = \pi$ is fixed and $\omega_2 \to i\infty$. Non-example: $\zeta$ is not an elliptic function. A nonconstant elliptic function cannot have only one simple pole in a period parallelogram, and indeed the Legendre relation shows that $\eta_1, \eta_2$ are not both $0$.