- Let \(\sigma ( z )\) be the canonical product
\[ \sigma (z) = z \prod_ {j = 1} ^ {\infty} E _ {2} (z / \tau_ {j}), \]
where \(\tau _ { j }\) is an enumeration of the periods \(\{ n + m \tau \}\) with \(( n , m ) \neq ( 0 , 0 )\) , and \(E _ { 2 } ( z ) = ( 1 - z ) e ^ { z + z ^ { 2 } / 2 }\)
Show that \(\sigma ( z )\) is an entire function of order 2 that has simple zeros at all the periods \(n + m \tau\) , and vanishes nowhere else.
Show that
\[ \frac {\sigma^ {\prime} (z)}{\sigma (z)} = \frac {1}{z} + \sum_ {(n, m) \neq (0, 0)} \left[ \frac {1}{z - n - m \tau} + \frac {1}{n + m \tau} + \frac {z}{(n + m \tau) ^ {2}} \right], \]
and that this series converges whenever z is not a lattice point.
- Let \(L ( z ) = - \sigma ^ { \prime } ( z ) / \sigma ( z )\) . Then
\[ L ^ {\prime} (z) = \frac {(\sigma^ {\prime} (z)) ^ {2} - \sigma (z) \sigma^ {\prime \prime} (z)}{(\sigma (z)) ^ {2}} = \wp (z). \]
Solution
Let \(\Lambda=\{n+m\tau:n,m\in\mathbb Z\}\), with \(\Im\tau>0\). The number of lattice points in \(|\omega|\le R\) is \(O(R^2)\). Hence \[ \sum_{\omega\in\Lambda\setminus\{0\}}|\omega|^{-3}<\infty, \] so the canonical product \[ \sigma(z)=z\prod_{\omega\in\Lambda\setminus\{0\}}E_2(z/\omega), \qquad E_2(w)=(1-w)e^{w+w^2/2}, \] converges locally uniformly and defines an entire function. Its only zeros are \(0\) and the nonzero lattice points, and each is simple, since each factor \(E_2(z/\omega)\) has a simple zero at \(z=\omega\) and all remaining factors are nonzero there.
The exponent of convergence of the lattice is \(2\): \(\sum_{\omega\ne0}|\omega|^{-s}\) converges for \(s>2\) and diverges for \(s\le2\). Thus any entire function with these zeros has order at least \(2\). On the other hand, the standard canonical-product estimate for a genus-two product with counting function \(N(R)=O(R^2)\) gives, for every \(\varepsilon>0\), \[ \log M_\sigma(R)=O(R^{2+\varepsilon}). \] Hence the order of \(\sigma\) is at most \(2\), and therefore exactly \(2\).
For part (b), logarithmic differentiation is legitimate locally uniformly away from \(\Lambda\). Since \[ \frac{E_2'(w)}{E_2(w)}=-\frac1{1-w}+1+w, \] we obtain \[ \frac{\sigma'(z)}{\sigma(z)} =\frac1z+\sum_{\omega\ne0} \left(\frac1{z-\omega}+\frac1\omega+\frac z{\omega^2}\right). \] Indeed, on a compact set \(|z|\le R\) avoiding \(\Lambda\), the summand for \(|\omega|>2R\) equals \[ \frac{z^2}{\omega^2(z-\omega)}, \] so it is \(O_R(|\omega|^{-3})\), and the series converges normally.
Finally let \(L=-\sigma'/\sigma\). Differentiating the normally convergent series gives \[ L'(z)=\frac1{z^2}+\sum_{\omega\ne0} \left(\frac1{(z-\omega)^2}-\frac1{\omega^2}\right). \] Because \(\Lambda=-\Lambda\), this is exactly the defining series for the Weierstrass function \(\wp(z)\). Hence \[ L'(z)=\wp(z) =\frac{(\sigma'(z))^2-\sigma(z)\sigma''(z)}{\sigma(z)^2}. \]