Difference between revisions of "Aufgaben:Problem 15"

Revision as of 22:50, 27 December 2014

Problem

Recall, \(\vartheta\) is a theta function relative to a lattice \(\Gamma_{\omega}=\mathbb{Z}+\mathbb{Z}\omega\), \(\omega\in\mathbb{H}\), if \(\vartheta\) is entire and \(\forall\gamma\in\Gamma_{\omega}\ \exists a_{\gamma},b_{\gamma}\in\mathbb{C}\) such that \(\vartheta\left(z+\gamma\right)=e^{a_{\gamma}z+b_{\gamma}}\vartheta(z),\ \forall z\in\mathbb{C}\). Consider the following theta function relative to \(\Gamma_{\omega}\) (you may assume that the product converges and that \(\sigma\) satisfies the above definition): $$\sigma(z)=z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\left(1-\frac{z}{\gamma}\right )e^{\frac{z}{\gamma}+\frac{z^{2}}{2\gamma^{2}}}$$

a) Show that \(\forall a\notin\Gamma_{\omega}\), $$\wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^{2}\sigma(a)^{2}}$$ where \(\wp=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2}+c_1)\) is the Weiertrass \(\wp\)-function, and \(\theta(z)=\sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{\pi i\omega z^2}\) is the Riemann theta function and \(c_1\) is the coefficient in the expansion \(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})=\frac{1}{z}+c_0+c_1z+\cdots\)

Hint: show that both sides of the equality are elliptic funtions with the same poles and zeros.

b) Show that $$\wp'(a)=-\frac{\sigma(2a)}{\sigma(a)^4},\ \forall a\notin\Gamma_{\omega}$$

Solutions

Coming soon, ab 27-28 erste Skizze.

a) First "idea": [1], to develop... or pages 351-2 of [2] (to see if possible to use it)

From Freitag, Busam - Funktionentheorie [Springer] V.6.4 is exactly the same problem (the hint is related too): only to add some proofs.

It says: "Auf beiden Seiten steht bei festem \(a\) eine elliptische Funktion mit denselben Nullstellen \((\pm a)\) und Polstellen (to prove 1). Daher stimmen sie bis auf einen konstanten Faktor überein (to prove 2). Für die Normierung benutzt man die Beziehung \(\lim_{z\rightarrow 0}z^2(\wp(z)-\wp(a))=1\). dass auf der rechten Seite dasselbe herauskommt, folgt aus den Relationen \(\sigma(a)=-\sigma(-a)\) und \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=1\), welche unmittelbar aus der Definition folgen (better see the passages)."