- Prove that
\[ \frac {(\Theta^ {\prime} (z | \tau)) ^ {2} - \Theta (z | \tau) \Theta^ {\prime \prime} (z | \tau)}{\Theta (z | \tau) ^ {2}} = \wp_ {\tau} (z - 1 / 2 - \tau / 2) + c _ {\tau}, \]
where \(c _ { \tau }\) can be expressed in terms of the first three derivatives of \(\Theta ( z | \tau )\) , with respect to \(z\) , at \(z = 1 / 2 + \tau / 2\) . Compare this formula with the result in Exercise 5 in the previous chapter.
Solution
Let \[ z_0=\frac12+\frac\tau2, \qquad H(z)=\frac{(\Theta'(z|\tau))^2-\Theta(z|\tau)\Theta''(z|\tau)}{\Theta(z|\tau)^2} =-\left(\frac{\Theta'}{\Theta}\right)'(z). \] The theta transformation laws are \[ \Theta(z+1|\tau)=\Theta(z|\tau), \qquad \Theta(z+\tau|\tau)=e^{-\pi i\tau-2\pi iz}\Theta(z|\tau). \] Hence \[ \frac{\Theta'}{\Theta}(z+1)=\frac{\Theta'}{\Theta}(z), \qquad \frac{\Theta'}{\Theta}(z+\tau)=\frac{\Theta'}{\Theta}(z)-2\pi i, \] so \(H\) is elliptic with periods \(1\) and \(\tau\).
By the Jacobi product formula, \(\Theta(z|\tau)\) has simple zeros exactly at \[ z_0+\Lambda, \qquad \Lambda=\mathbb Z+\mathbb Z\tau. \] Thus \(H\) has a double pole at every point of \(z_0+\Lambda\). If \(w=z-z_0\), write \[ \Theta(z|\tau)=aw+\frac b2w^2+\frac c6w^3+O(w^4), \] where \[ a=\Theta'(z_0|\tau)\ne0, \quad b=\Theta''(z_0|\tau), \quad c=\Theta'''(z_0|\tau). \] Then \[ \frac{\Theta'}{\Theta} =\frac1w+\frac b{2a} +\left(\frac c{3a}-\frac{b^2}{4a^2}\right)w+O(w^2), \] and therefore \[ H(z)=\frac1{w^2} +\frac{b^2}{4a^2}-\frac c{3a}+O(w). \tag{1} \]
The function \(\wp_\tau(z-z_0)\) is elliptic with the same periods and has at each point of \(z_0+\Lambda\) the same principal part \(w^{-2}\). Hence \[ H(z)-\wp_\tau(z-z_0) \] has removable singularities everywhere and extends to an entire elliptic function. It is bounded on a fundamental parallelogram, hence constant by Liouville’s theorem. Thus \[ H(z)=\wp_\tau(z-z_0)+c_\tau. \] Since \[ \wp_\tau(w)=\frac1{w^2}+O(w^2) \] has no constant term at \(w=0\), comparison with (1) gives \[ \boxed{ c_\tau =\frac14\left(\frac{\Theta''(z_0|\tau)}{\Theta'(z_0|\tau)}\right)^2 -\frac13\frac{\Theta'''(z_0|\tau)}{\Theta'(z_0|\tau)} }. \] This is the theta-function analogue of the sigma-function identity of E-SS9.EX-5.