"자코비 형식"의 두 판 사이의 차이

2015년 11월 21일 (토) 03:54 판

정의

2변수 함수
$(\tau,z)\in \mathcal{H}\times \mathbb{C}$에 대하여, 다음과 같이 정의된다 ($q=e^{\pi i \tau}$)

\begin{align*} % \theta_1(z;\tau) \theta_{11}(z;\tau) & = \sum_{n \in \mathbb{Z}} q^{ \left( n+ \frac{1}{2} \right)^2} \, \E^{2 \pi i \left(n+\frac{1}{2} \right) \, \left( z+\frac{1}{2} \right) } = - i \, \theta_1(z; \tau) \\ & = -2 q^{1/4} \sin (\pi z)+2 q^{9/4} \sin (3 \pi z)-2 q^{25/4} \sin (5 \pi z)+2 q^{49/4} \sin (7 \pi z)-2 q^{81/4} \sin (9 \pi z) +\cdots, % \\ % & = % i \, q^{\frac{1}{8}} \, \E^{\pi \I z} \, % \left( % q, \E^{-2 \pi \I z}, \E^{2 \pi \I z} \, q % \right)_\infty \\[2mm] % % \theta_{2}(z;\tau) \theta_{10}(z;\tau) & = \sum_{n \in \mathbb{Z}} q^{\left( n + \frac{1}{2} \right)^2} \, \E^{2 \pi i \left( n+\frac{1}{2} \right) z} = \theta_2(z;\tau) \\ & = 2 q^{1/4} \cos (\pi z)+2 q^{9/4} \cos (3 \pi z)+2 q^{25/4} \cos (5 \pi z)+2 q^{49/4} \cos (7 \pi z)+2 q^{81/4} \cos (9 \pi z)+\cdots , \\[2mm] % % \theta_3 (z;\tau) \theta_{00} (z;\tau) & = \sum_{n \in \mathbb{Z}} q^{n^2} \, \E^{2 \pi i n z} = \theta_3 (z;\tau) \\ & =1+2 q \cos (2 \pi z)+2 q^4 \cos (4 \pi z)+2 q^9 \cos (6 \pi z)+2 q^{16} \cos (8 \pi z)+2 q^{25} \cos (10 \pi z)+\cdots, \\[2mm] % % \theta_0 (z;\tau) \theta_{01} (z;\tau) & = \sum_{n \in \mathbb{Z}} q^{ n^2} \, \E^{2 \pi i n \left( z+\frac{1}{2} \right) } = \theta_4(z;\tau) \\ & =1-2 q \cos (2 \pi z)+2 q^4 \cos (4 \pi z)-2 q^9 \cos (6 \pi z)+2 q^{16} \cos (8 \pi z)+2 q^{25} \cos (5 \pi (2 z+1))+\cdots \end{align*}

모듈라 성질

다음과 같은 모듈라 변환 성질을 갖는다

\begin{equation} \begin{pmatrix} \theta_{11}\left( z; \tau+1 \right) \\ \theta_{10}\left( z; \tau+1 \right) \\ \theta_{00}\left( z; \tau+1 \right) \\ \theta_{01}\left( z; \tau+1 \right) \end{pmatrix}

 =
 \sqrt{ \frac{1}{i} } \,
 \begin{pmatrix}
 i & 0 & 0 & 0 \\
 0 & i & 0 & 0 \\
 0 & 0 & 0 & e^{\frac{i \pi }{4}} \\
 0 & 0 & e^{\frac{i \pi }{4}} & 0 \\
  \end{pmatrix} \,
 \begin{pmatrix}
    \theta_{11}(z;\tau) \\
    \theta_{10}(z;\tau) \\
    \theta_{00}(z;\tau) \\
    \theta_{01}(z;\tau) 
  \end{pmatrix}=
 \begin{pmatrix}
 e^{\frac{i \pi }{4}} & 0 & 0 & 0 \\
 0 & e^{\frac{i \pi }{4}} & 0 & 0 \\
 0 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 \\
  \end{pmatrix} \,
 \begin{pmatrix}
    \theta_{11}(z;\tau) \\
    \theta_{10}(z;\tau) \\
    \theta_{00}(z;\tau) \\
    \theta_{01}(z;\tau) 
  \end{pmatrix}

\end{equation}

\begin{equation} \begin{pmatrix} \theta_{11}\left( \frac{z}{\tau}; -\frac{1}{\tau} \right) \\ \theta_{10}\left( \frac{z}{\tau}; -\frac{1}{\tau} \right) \\ \theta_{00}\left( \frac{z}{\tau}; -\frac{1}{\tau} \right) \\ \theta_{01}\left( \frac{z}{\tau}; -\frac{1}{\tau} \right) \end{pmatrix}

 =
 \sqrt{ \frac{\tau}{i} } \, \E^{ \pi i \frac{z^2}{\tau}} \,
 \begin{pmatrix}
    -i & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 1 & 0 \\
    0 & 1 & 0 & 0
  \end{pmatrix} \,
 \begin{pmatrix}
    \theta_{11}(z;\tau) \\
    \theta_{10}(z;\tau) \\
    \theta_{00}(z;\tau) \\
    \theta_{01}(z;\tau) 
  \end{pmatrix}

\end{equation}

weight k=1/2이고, index m=1/2인 벡터 자코비 형식의 예이다