자코비 세타함수

간단한 소개

• 세타함수의 정의 (spectral decomposition of heat kernel)
  \(\theta(\tau)=\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2/2}= \sum_{n=-\infty}^\infty \exp(\pi i n^2\tau)\), \(q=e^{2\pi i \tau}\)
  

\(\theta_{2}(\tau)= \sum_{n=-\infty}^\infty q^{(n+\frac{1}{2})^2/2}\)

\(\theta_{4}(\tau)= \sum_{n=-\infty}^\infty (-1)^n q^{n^2/2}\)

 

 

여러가지 공식들

\(\theta_2^4(q)+\theta_4^4(q)=\theta_3^4(q)\)

\(\theta_3^2(q^2)+\theta_2^2(q^2)=\theta_3^2(q)\)

\(\theta_3^2(q^2)-\theta_2^2(q^2)=\theta_3^2(q)\)

 

 

 

세타함수의 Modularity

\(\theta(-\frac{1}{\tau})=\sqrt{\frac{\tau}{i}} \theta({\tau})\)

\(\tau=iy, y>0\) 으로 쓰면,

\(\theta(\frac{i}{y})=\sqrt{y} \theta({iy)\)

 

(증명)

포아송의 덧셈 공식을 사용한다.

\(\sum_{n\in \mathbb Z}f(n)=\sum_{n\in \mathbb Z}\hat{f}(n)\)

 

\(f(x)=e^{\pi i x^2\tau\)

\(\hat{f}(\xi)=\sqrt{\frac{i}{\tau}}e^{-\pi i\frac{\xi^2}{\tau}\)

\(\theta(\tau)= \sum_{\in \mathbb Z} \exp(\pi i n^2\tau)=\sum_{n\in \mathbb Z}f(n)=\sum_{n\in \mathbb Z}\hat{f}(n)=\sqrt{\frac{i}{\tau}}\sum_{n\in \mathbb Z}e^{-\pi i n^2 \frac{1}{\tau}}=\sqrt{\frac{i}{\tau}}\theta(-\frac{1}{\tau})\) (증명끝)

 

 

Triple product 공식

\(\sum_{n=-\infty}^\infty z^{n}q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + zq^{2m-1}\right) \left( 1 + z^{-1}q^{2m-1}\right)\)

\(z=1\) 인 경우

\(\sum_{n=-\infty}^\infty q^{n^2}= \prod_{m=1}^\infty \left( 1 - q^{2m}\right) \left( 1 + q^{2m-1}\right)^2\)

 

triple product 의 증명

q-초기하급수(q-hypergeometric series) 

\(\prod_{n=1}^{\infty}(1+zq^n)=1+\sum_{n\geq 1}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)

\(\prod_{n=1}^{\infty}\frac{1}{1-zq^n}=1+\sum_{n\geq 1}\frac{q^n}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)

를 활용하자. 

 

 

세타함수와 singular modulus

\(k=k(\tau)=\frac{\theta_2^2(\tau)}{\theta_3^2(\tau)}\)

\(k'=\sqrt{1-k^2}=\frac{\theta_4^2(\tau)}{\theta_3^2(\tau)}\)

 

 

세타함수, AGM iteration, 타원적분

\(\frac{\theta_3^2(q)+\theta_4^2(q)}{2}=\theta_3(q^2)\)

\(\sqrt{\theta_3^2(q)\theta_4^2(q)}=\theta_4^2(q^2)\)

따라서 \(a_n=\theta_3^2(q^{2^n}),b_n=\theta_4^2(q^{2^n})\) 라 하면, \(a_n, b_n\)은 AGM iteration 을 만족하고 \(\lim_{n\to\infty}a_n=1\)이고, \(1=M(\theta_3^2(q),\theta_4^2(q))\)가 된다.

 

 

(정리)

주어진 \(0<k<1\) 에 대하여, \(k=k(q)=\frac{\theta_2^2(q)}{\theta_3^2(q)}\)를 만족시키는 \(q\)가 존재한다. 이 때,

\(M(1,k')=\theta_3^{-2}(q)\) 와 \(K(k) = \frac{\pi}{2}\theta_3^2(q)\)가 성립한다.\(K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \frac{\pi}{2}\theta_3^2(q)\)

 

(증명)

\(1=M(\theta_3^2(q),\theta_4^2(q))=\theta_3^{2}(q)M(1,\frac{\theta_4^2(q)}{\theta_3^2(q)})=\theta_3^{2}(q)M(1,k')\)

그러므로, \(M(1,k')=\theta_3^{-2}(q)\)이다.

한편, 란덴변환에 의해 \(K(k)=\frac{\pi}{2M(1,\sqrt{1-k^2})}\)가 성립(타원적분과 AGM의 관계 , 란덴변환과 AGM 참조)하므로,  \(K(k) = \frac{\pi}{2}\theta_3^2(q)\)도 증명된다. (증명끝)

 

 

 

 

 

 

 

 

관련된 다른 주제들

• 타원함수
• 이차형식
• 모듈라 형식(modular forms)
• 격자의 세타함수

 

표준적인 도서 및 추천도서

• Brief Introduction to Theta Functions
  
  • BELLMAN, RICHARD
• Tata Lectures on Theta I,II,III
  
  • David Mumford

 

위키링크

• http://en.wikipedia.org/wiki/Theta_functions

 

 

관련논문

• On a classical theta-function
  
  • Tomio Kubota, Nagoya Math. J. Volume 37 (1970), 183-189
• Shorter Notes: A Simple Proof of Jacobi's Triple Product Identity
  
  • George E. Andrews, Proceedings of the American Mathematical Society, Vol. 16, No. 2 (Apr., 1965), pp. 333-334

• Applications of Theta Functions to Arithmetic
  
  • G. D. Nichols, The American Mathematical Monthly, Vol. 45, No. 6 (Jun. - Jul., 1938), pp. 363-368
• A First Course in Modular Forms (Graduate Texts in Mathematics)
  
  • Fred Diamond and Jerry Shurman, 18-19p four_square_theorem_and_theta_funtion.pdf
• Karl Gustav Jacob Jacobi
  
  • Jacobi's Four Square Theorem. (Also available in postscript format [11 pages].) [CONSTRUCTION IN PROGRESS]
• http://www.jstor.org/action/doBasicSearch?Query=jacobi's+triple+product
• http://www.jstor.org/action/doBasicSearch?Query=