Difference between revisions of "Aufgaben:Problem 15"

Revision as of 09:31, 6 January 2015

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): $$\sigma(z)=z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\left(1-\frac{z}{\gamma}\right )e^{\frac{z}{\gamma}+\frac{z^{2}}{2\gamma^{2}}}$$

Exercise 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 \(\theta(z)=\sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{\pi i\omega n^2}\) 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.

Solution a)

From Serie 9, ex 1 we know that \(\wp\) is elliptic and then \(\wp(z)-\wp(a)\) is elliptic too (elliptic fct \(\pm\) const: still meromorphic and 2-per)

Claim: \(\wp(z)-\wp(a)\) has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\Gamma_\omega\)

Proof:

1. $$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta\left( z+\frac{1+\omega}{2}\right)+\frac{d^2}{dz^2}\log\theta\left( z+\frac{1+\omega}{2}\right)|_{z=a}=$$$$-\frac{d}{dz}\frac{\theta '\left( z+\frac{1+\omega}{2}\right)}{\theta\left( z+\frac{1+\omega}{2}\right)}+const=$$$$-\frac{\theta" \left( z+\frac{1+\omega}{2}\right)\theta\left( z+\frac{1+\omega}{2}\right)-\theta '\left( z+\frac{1+\omega}{2}\right)^2}{\theta\left( z+\frac{1+\omega}{2}\right)^2}+const$$
2. \(\theta\) holomorphic \(\theta',\theta"\) holomorphic too, and then the uinique poles of \(\wp(z)-\wp(a)\equiv\) the zeros of \(\theta(z+\frac{1+\omega}{2})\)
3. 'Proposition on lecture notes (p. 37 Fourier-Heat-ecc): \(\theta(\gamma+\frac{1+\omega}{2})=0\forall\gamma\in\Gamma_\omega\)
4. $$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}(\frac{1}{z}+c_0+c_1z+\cdots)+c_1+const=$$ $$-(-\frac{1}{z^2}+2c_1+\cdots)+const$$ \(\Rightarrow z=0\) poles of second order.
5. \(\wp(z)-\wp(a)\) elliptic on \(\Gamma_\omega\Rightarrow\) every \(\gamma\in\Gamma_\omega\) is a pole of order 2.
6. NB: thanks to ellipticity (double periodicity) of \(\wp(z)\) we can "move" the parallelogram such that there are no poles / zeros on the boundary for both \(\wp(z)-\wp(a)\) and \((\wp(z)-\wp(a))'\). Consider a new parallelogram \(P\) with \(z=0\) inside it.
7. From complex analysis: Prop 1.6: \(f\ elliptic\Rightarrow \int_{\partial P} fdz=0\) and argument principle: \(\int_{\partial P} \frac{f'}{f}dz=2\pi i[N_0-N_\infty]\), where \(N_0,N_\infty\) are the number of zeros respectively poles with multiplicity counted. If \(f\ elliptic\Rightarrow \frac{f'}{f}\) is elliptic (as a combination of elliptic functions) and then, setting \(f=\wp(z)-\wp(a)\), from prop 1.6 follows \(\int_{\partial P} \frac{f'}{f}=\frac{\wp'(z)-\wp'(a)}{\wp(z)-\wp(a)}=0\Rightarrow 0=2\pi \left[ N_0-N_\infty\right] \Rightarrow N_0=N_\infty\). From 4. we have \(N_\infty=2\Rightarrow N_0=2\)
8. \(\wp(z)-\wp(a)=0\iff\wp(z)=\wp(a)\overset{\wp\ is\ even}{=}\wp(-a)\Rightarrow\ z=\pm a\) are zeros of \(\wp(z)-\wp(a)\). Because of ellipticity of \(\wp(z)\), all the points in \(\left\{\pm a+\gamma\mid\gamma\in\Gamma_\omega\right\}\) are zeros of \(\wp (z)-\wp (a)\). Consider the fundamental parallelogram \(P\) which contains \(a\). We can consider the translation of \(-a\) by a vector \(\xi\in\Gamma_\omega\) s.t. \(\left(-a+\xi\right)\in P\). We then have two different zeros of \(\wp(z)-\wp(a)\) in \(P\) and from 6. we know that they are the only ones and they must be of order 1.
9. Ellipticity \(\Rightarrow z\) zero \(\iff z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\)

\(\square\)

Claim: \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is elliptic and has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\left\{\Gamma_\omega\right\}\)

Proof:

1. Periodic in two directions: Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then $$-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}$$ $$-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=$$ $$-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$.
2. Meromorphic: $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{(z+a)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}}](z-a)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{(z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}})^2(a\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{a}{\gamma})e^{\frac{a}{\gamma}+\frac{a^2}{2\gamma ^2}})^2}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}+\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{a^2}{\gamma ^2}}]}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\gamma^2(\gamma-z-a)(\gamma-z+a)e^{\frac{-2a}{\gamma}}}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}(\gamma-z)^2 (\gamma-a)^2}=$$ $$-\frac{(z+a)(z-a)}{z^2a^2}\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\gamma^2 e^{-\frac{2a}{\gamma}}\frac{(z+a-\gamma)(z-a-\gamma)}{(\gamma-z)^2(\gamma-a)^2}$$ \(\Rightarrow\) it's meromorphic and we observe that zeros and poles are as claimed. Meromorphic + 2-periodic = elliptical

\(\square\)

Claim: \(f,g\) elliptic with same poles, zeros then \(f=Ag\), with \(A\) const

Proof: We know that a function \(k(z)\) with zeros \(a_1 ,\cdots ,a_m\), poles \(b_1, \cdots , b_r\) can be written as \(k(z)=q(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}\) with \(q(z)\) analitic and \(q(a_i)\neq 0\ \forall a_i, q(b_i)\neq 0\ \forall b_i\).

Then we write $$f(z)=h(z)\frac{\prod_{a_i zero}(z-a_i)}{\prod_{b_i pole}(z-b_i)},\ \ g(z)=l(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}$$ with \(h,l\) analytic and with prop. of above.

If \(f,g\) elliptic, then \(\frac{f}{g}\) is elliptic too: \(\frac{f}{g}=\frac{h(z)}{l(z)}\) analitic as a quotient of analitic functions, then \(\frac{h(z)}{l(z)}\) has no singularities and then no poles. But an elliptic function without poles is constant since \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C})=q(P)\) compact, where \(P\) fundamental parallelogram, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\)

\(\square\)

Claim: Assuming that $$\wp(z)-\wp(a)=K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$ we claim that \(K=-1\)

Proof: $$\lim_{z\rightarrow 0} z^2(\wp(z)-\wp(a))=$$ $$\lim_{z\rightarrow 0}z^2\wp(z)-\overset{=0}{\overbrace{\lim_{z\rightarrow 0}z^2\wp(a)}}=$$ $$\lim_{z\rightarrow 0}z^2(-\frac{d}{dz}(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})+c_1))=$$ $$-\lim_{z\rightarrow 0}z^2(\frac{d}{dz}(\frac{1}{z}+c_0+zc_1+\cdots))=$$ $$-\lim_{z\rightarrow 0}z^2(-\frac{1}{z^2}+c_1+\cdots)=1$$ but \(\lim_{z\rightarrow 0}\wp(z)-\wp(a)=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\Rightarrow 1=\lim_{z\rightarrow 0}z^2 K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)

Now we prove that \(\sigma\) is odd and that \(\lim_{z\rightarrow 0}\sigma(z)/z=1\):

\(\sigma(-z)=-z\prod(1+\frac{z}{\gamma})e^{\frac{-z}{\gamma}+\frac{z^2}{2\gamma^2}}\overset{\gamma '=-\gamma}{=}-z\prod(1-\frac{z}{\gamma '})e^{\frac{z}{\gamma '}+\frac{z^2}{2\gamma '^2}}=-\sigma(z)\) and \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=\lim_{z\rightarrow 0}\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}}=\prod1=1\Rightarrow\) $$1=\lim_{z\rightarrow 0}z^2(K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2})=$$ $$\lim_{z\rightarrow 0}K\frac{z^2}{\sigma(z)^2}\frac{\sigma(z+a)\sigma(z-a)}{\sigma(a)^2}=$$ $$K\frac{\sigma(a)\sigma(-a)}{\sigma(a)^2}=$$ $$=K\frac{-\sigma(a)^2}{\sigma(a)^2}=-K\Rightarrow K=-1$$ \(\Rightarrow \wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)

\(\blacksquare\)

Note: In the lecture we saw that \(\vartheta\) functions are analytic and then \(\sigma\) analytic: when we want to show that \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is meromorphic, it's enough write that as a product / combination of analytic function it's analytic too except points for which the denominator is zero. This means that the zeros of the function are the ones of the numerator, while the poles the ones of the denominator.

Exercise b)

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

Solution b)

$$\wp '(a)=\lim_{z\rightarrow a} \frac{\wp(z)-\wp(a)}{z-a}\overset{a)}{=}$$ $$\lim_{z\rightarrow a}-\frac{\sigma(z+a)\sigma(z-a)}{(z-a)\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow a}\frac{\sigma(z-a)}{z-a}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}\overset{a)}{=}$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}$$

\(\blacksquare\)