Introduction to Theta Functions II

Infinite product representations of theta functions

Let f(z) = \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz}). Each of these two products converge absolutely and uniformly in any bounded domain of values of z. Hence f(z) is analytic throughout the finite part of the z plane. The zeros of f(z) are simple zeros at the points z=\frac{2n+1}{2}\pi \tau i + m\pi i where m,n\in \mathbb{Z}. So, the zeros of f(z) coincide with the zeros of \vartheta_4(z). Therefore, the function \frac{\vartheta_4(z)}{f(z)} has no poles or zeros in the finite part of the plane. Also, it is easy to see that f(z+\pi) = f(z) and \begin{aligned} f(z+\pi \tau) &= \prod_{n=1}^\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\ &= f(z) \frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\ &= -q^{-1}e^{-2iz} f(z) \end{aligned} This is, \frac{\vartheta_4(z)}{f(z)} is a doubly periodic function with periods \pi,\pi\tau and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say G. We have \vartheta_4(z) = G \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\quad (1) Incrementing z by the half periods \frac{\pi}{2}, \frac{\pi \tau}{2} and \frac{\pi +\pi\tau}{2}, we get the product representations for the other theta functions: \begin{aligned} \vartheta_3(z) &= G \prod_{n=1}^{\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\quad (2) \\ \vartheta_1(z) &= 2G q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \quad (3) \\ \vartheta_2(z) &= 2G q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \quad (4) \end{aligned} Now, all that left is to find the constant G. From identity (3), it easy to see that \vartheta_1'(0) = 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 Now, using the identity \vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0), we get: \begin{aligned} 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 &= \left\{2G q^{\frac{1}{4}}\prod_{n=1}^{\infty}(1+q^{2n})^2 \right\} \times \left\{G\prod_{n=1}^{\infty}(1+q^{2n-1})^2 \right\} \\ &\quad \times \left\{G \prod_{n=1}^{\infty}(1-q^{2n-1})^2 \right\} \\ \implies G &= \prod_{n=1}^\infty (1-q^{2n}) \end{aligned} Note that the rearrangements are justified since all products converge absolutely. Finally, we have \begin{aligned} \vartheta_1(z) &= 2 q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n} \cos(2z)+ q^{4n}) \quad (5) \\ \vartheta_2(z) &= 2 q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n} \cos(2z) + q^{4n}) \quad (6) \\ \vartheta_3(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (7) \\ \vartheta_4(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (8) \end{aligned}

Derivatives of Ratios of Theta Functions

Consider the function \phi(z) = \frac{\vartheta_1'(z)\vartheta_4(z) - \vartheta_1(z)\vartheta_4'(z)}{\vartheta_2(z)\vartheta_3(z)}. Now, \phi(z) is doubly periodic with periods \pi and \frac{\pi\tau}{2}. Relative to these periods, the only poles of \phi(z) are at points congruent to \pi\over 2. It follows that \phi(z) is a constant. Letting z\to 0, we see that \phi(z)=\vartheta_4^2(0). It is therefore established that \frac{d}{dz}\left(\frac{\vartheta_1(z)}{\vartheta_4(z)} \right)=\vartheta_4^2(0) \frac{\vartheta_2(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (9) Two other identities if this kind can be derived using the same technique: \begin{aligned} \frac{d}{dz}\left(\frac{\vartheta_2(z)}{\vartheta_4(z)} \right) &=-\vartheta_3^2(0) \frac{\vartheta_1(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (10) \\ \frac{d}{dz}\left(\frac{\vartheta_3(z)}{\vartheta_4(z)} \right) &=-\vartheta_2^2(0) \frac{\vartheta_1(z)\vartheta_2(z)}{\vartheta_4^2(z)} \quad (11) \end{aligned} By writing \xi = \frac{\vartheta_1(z)}{\vartheta_4(z)} and making use of identities (13) and (14) from part 1, we obtain: \left(\frac{d\xi}{dz}\right)^2 = (\vartheta_2^2(0) - \xi^2 \vartheta_3^2(0))(\vartheta_3^2(0)-\xi^2 \vartheta_2^2(0)) Using the change of variables y = \frac{\vartheta_3(0)}{\vartheta_2(0)}\xi and u = \vartheta_3^2(0) z, the above differential equation can be written as: \left(\frac{dy}{du}\right)^2 = (1-y^2)(1-k^2 y^2) where k = \frac{\vartheta_2^2(0)}{\vartheta_3^2(0)}. This differential equation has the particular solution: y = \frac{\vartheta_3(0)}{\vartheta_2(0)} \frac{\vartheta_1(u \vartheta_3^{-2}(0))}{\vartheta_4(u \vartheta_3^{-2}(0))} We can write y as a function of u and k as y = \text{sn}(u,k) or simply \text{sn}(u). Clearly, \text{sn}(u,k) has periods 2\pi \vartheta^2_3(0) and \pi \tau \vartheta_3^2(0). It has two simple poles congruent to \frac{1}{2}\pi\tau \vartheta_3^2(0) and \pi \vartheta_3^2(0) + \frac{1}{2}\pi\tau \vartheta_3^2(0). Also, it is easy to see that \int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-k^2 y^2)}} = \text{sn}^{-1}(1) = \frac{\pi}{2}\vartheta_3^2(0) The integral on the left is denoted by complete elliptic integral of the first kind and is denoted by K(k).

References

  1. Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
  2. E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press

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