Continue with the lattice generators \(\lambda_i\), zeta quasi-periods \(\eta_i\), and the canonical product \(\sigma\). Show that \[ \sigma(z+\lambda_i)=\exp(a_i+b_i z)\sigma(z) \] for suitable constants \(a_i,b_i\in\mathbb C\), and express \(a_i\) and \(b_i\) in terms of \(\lambda_i\) and \(\eta_i\).
Solution
The quasi-periodicity is \[ \boxed{ \sigma(z+\lambda_i) =-\exp\!\left(\eta_i\left(z+\frac{\lambda_i}{2}\right)\right)\sigma(z). } \] Consequently one may take \[ \boxed{b_i=\eta_i, \qquad a_i\equiv \frac{\eta_i\lambda_i}{2}+\pi i\pmod{2\pi i}.} \]
1. The quotient \(\sigma(z+\lambda_i)/\sigma(z)\) is an everywhere nonzero entire function after removing its apparent singularities.
The zeros of \(\sigma\) are exactly the lattice points, all simple. Translation by \(\lambda_i\in\Lambda\) permutes those zeros. Hence the numerator and denominator of \[ R_i(z)=\frac{\sigma(z+\lambda_i)}{\sigma(z)} \] have the same simple zeros. Every apparent zero or pole of the quotient is therefore removable, and the resulting entire function \(R_i\) has no zeros.
2. Its logarithmic derivative is the constant \(\eta_i\).
Using the identity \[ \zeta=\frac{\sigma'}{\sigma} \] from E-HM213A17-138 and the quasi-period relation of E-HM213A17-139 \[ \zeta(z+\lambda_i)=\zeta(z)+\eta_i, \] we obtain \[ \frac{R_i'(z)}{R_i(z)} =\frac{\sigma'(z+\lambda_i)}{\sigma(z+\lambda_i)} -\frac{\sigma'(z)}{\sigma(z)} =\eta_i. \] Thus \[ \frac{d}{dz}\bigl(e^{-\eta_i z}R_i(z)\bigr)=0, \] so for some \(C_i\ne0\), \[ R_i(z)=C_i e^{\eta_i z}. \] Equivalently, \[ \sigma(z+\lambda_i)=C_i e^{\eta_i z}\sigma(z). \]
3. The canonical product \(\sigma\) is odd.
Pair in the defining product the factors indexed by \(\lambda\) and \(-\lambda\). Their product is \[ \left(1-\frac{z^2}{\lambda^2}\right) \exp\!\left(\frac{z^2}{\lambda^2}\right), \] which is even in \(z\). Since the remaining initial factor is \(z\), the full canonical product satisfies \[ \sigma(-z)=-\sigma(z). \]
4. Oddness determines the constant \(C_i\).
2. Since \(\lambda_i/2\notin\Lambda\), the value \(\sigma(\lambda_i/2)\) is nonzero. Therefore\[ \begin{aligned} \sigma(\lambda_i/2) &=C_i e^{-\eta_i\lambda_i/2}\sigma(-\lambda_i/2)\\ &=-C_i e^{-\eta_i\lambda_i/2}\sigma(\lambda_i/2). \end{aligned} \]Cancelling gives\[ C_i=-e^{\eta_i\lambda_i/2}. \]Hence\[ \sigma(z+\lambda_i) =-e^{\eta_i(z+\lambda_i/2)}\sigma(z). \]Writing the minus sign as \(e^{\pi i}\) yields\[ b_i=\eta_i, \qquad a_i\equiv\frac{\eta_i\lambda_i}{2}+\pi i\pmod{2\pi i}, \]which is exactly the ambiguity inherent in the expression \(e^{a_i+b_i z}\).