Let \[ \Lambda=\mathbb Z\lambda_1\oplus\mathbb Z\lambda_2 \] be a lattice in \(\mathbb C\), with Weierstrass function \(\wp\), and define the canonical product \[ \sigma(z) =z\prod_{\lambda\in\Lambda\setminus\{0\}} \left(1-\frac z\lambda\right) \exp\!\left(\frac z\lambda+\frac12\left(\frac z\lambda\right)^2\right). \]
Prove that there is a unique odd meromorphic function \(\zeta\) on \(\mathbb C\) satisfying \[ \zeta'(z)=-\wp(z), \] and relate \(\zeta\) to \(\sigma\).
Solution
The required function is the logarithmic derivative of the canonical product: \[ \boxed{ \zeta(z)=\frac{\sigma'(z)}{\sigma(z)} =\frac1z+ \sum_{\lambda\in\Lambda\setminus\{0\}} \left( \frac1{z-\lambda}+\frac1\lambda+\frac z{\lambda^2} \right). } \]
1. The displayed series defines a meromorphic function on \(\mathbb C\).
Fix a compact set \(K\subset\mathbb C\setminus\Lambda\). For \(z\in K\) and \(|\lambda|\) sufficiently large, \[ \frac1{z-\lambda}+\frac1\lambda+\frac z{\lambda^2} =O_K(|\lambda|^{-3}). \] Since a rank-two lattice satisfies \[ \sum_{\lambda\in\Lambda\setminus\{0\}}|\lambda|^{-3}<\infty, \] the series converges normally on \(K\). It therefore defines a holomorphic function off \(\Lambda\) and has a simple pole of residue \(1\) at every lattice point.
2. This function is exactly \(\sigma'/\sigma\) and satisfies \(\zeta'=-\wp\).
Logarithmically differentiating the canonical product off its zeros gives \[ \begin{aligned} \frac{\sigma'(z)}{\sigma(z)} &=\frac1z+ \sum_{\lambda\ne0} \left( -\frac{1/\lambda}{1-z/\lambda} +\frac1\lambda+\frac z{\lambda^2} \right)\\ &=\frac1z+ \sum_{\lambda\ne0} \left( \frac1{z-\lambda}+\frac1\lambda+\frac z{\lambda^2} \right). \end{aligned} \] The normal convergence established in 1, and the analogous estimate \[ -\frac1{(z-\lambda)^2}+\frac1{\lambda^2} =O_K(|\lambda|^{-3}), \] allow termwise differentiation. Hence \[ \begin{aligned} \zeta'(z) &=-\frac1{z^2} +\sum_{\lambda\ne0} \left( -\frac1{(z-\lambda)^2}+\frac1{\lambda^2} \right)\\ &=-\left[ \frac1{z^2} +\sum_{\lambda\ne0} \left( \frac1{(z-\lambda)^2}-\frac1{\lambda^2} \right) \right]\\ &=-\wp(z). \end{aligned} \] Both sides are meromorphic, so the identity holds on all of \(\mathbb C\).
3. The function \(\zeta\) is odd.
The series in 1 is normally convergent away from the lattice, so it may be reindexed by \(\lambda\mapsto-\lambda\). Thus \[ \begin{aligned} \zeta(-z) &=-\frac1z+ \sum_{\lambda\ne0} \left( \frac1{-z-\lambda}+\frac1\lambda-\frac z{\lambda^2} \right)\\ &=-\frac1z- \sum_{\mu\ne0} \left( \frac1{z-\mu}+\frac1\mu+\frac z{\mu^2} \right)\\ &=-\zeta(z). \end{aligned} \]
4. The odd primitive of \(-\wp\) is unique.
Let \(\zeta_1\) and \(\zeta_2\) be odd meromorphic functions with \[ \zeta_1'=\zeta_2'=-\wp. \] Then the meromorphic function \(h=\zeta_1-\zeta_2\) has \(h'=0\). A pole of \(h\) would give a pole of \(h'\), so \(h\) has no poles; hence \(h\) is entire and constant. Since both \(\zeta_1\) and \(\zeta_2\) are odd, \(h\) is odd, and the only odd constant is \(0\). Therefore \[ \zeta_1=\zeta_2. \] Thus the unique function requested in the problem is \[ \boxed{\zeta=(\log\sigma)'=\sigma'/\sigma.} \]