"리만 세타 함수"의 두 판 사이의 차이

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12번째 줄: 12번째 줄:
 
\right]
 
\right]
 
(\Omega ,\mathbf{z})
 
(\Omega ,\mathbf{z})
=\sum_{{\mathbf{n}\in{\mathbb Z}^g}} e^{2 \pi i  \left(\frac{1}{2}\left(\mathbf{\nu _1}+ \mathbf{n} \right)\Omega \left(\mathbf{\nu _1}+ \mathbf{n} \right)+2 \left(\mathbf{\nu _1}+ \mathbf{n}\right)\left(\mathbf{\nu _2}+\mathbf{z}\right)\right)}
+
=\sum_{{\mathbf{n}\in{\mathbb Z}^g}} e^{2 \pi i  \left(\frac{1}{2}\left(\mathbf{\nu _1}+ \mathbf{n} \right)\Omega \left(\mathbf{\nu _1}+ \mathbf{n} \right)+ \left(\mathbf{\nu _1}+ \mathbf{n}\right)\left(\mathbf{\nu _2}+\mathbf{z}\right)\right)}
 
$$
 
$$
 
* characteristic이 $\mathbf{\nu _1}=\mathbf{\nu _2}=0\in \mathbb{C}^g$인 경우
 
* characteristic이 $\mathbf{\nu _1}=\mathbf{\nu _2}=0\in \mathbb{C}^g$인 경우

2013년 6월 16일 (일) 09:18 판

개요

  • $\mathcal{H}_g=\left\{\tau \in M_{g \times g}(\mathbb{C}) \ \big| \ \tau^{\mathrm{T}}=\tau, \textrm{Im}(\tau) \text{ positive definite} \right\}$
  • $\Omega\in \mathcal{H}_g$, $\mathbb{z}\in \mathbb{C}^g$
  • 리만세타함수 $\Theta: \mathcal{H}_g\times \mathbb{C}^g\to \mathbb{C}$ 를 다음과 같이 정의 ($\mathbf{\nu _1}, \mathbf{\nu _2}\in \mathbb{C}^g$ : characteristic)

$$ \Theta \left[ \begin{array}{c} \mathbf{\nu _1} \\ \mathbf{\nu _2} \\ \end{array} \right] (\Omega ,\mathbf{z}) =\sum_{{\mathbf{n}\in{\mathbb Z}^g}} e^{2 \pi i \left(\frac{1}{2}\left(\mathbf{\nu _1}+ \mathbf{n} \right)\Omega \left(\mathbf{\nu _1}+ \mathbf{n} \right)+ \left(\mathbf{\nu _1}+ \mathbf{n}\right)\left(\mathbf{\nu _2}+\mathbf{z}\right)\right)} $$

  • characteristic이 $\mathbf{\nu _1}=\mathbf{\nu _2}=0\in \mathbb{C}^g$인 경우

$$ \Theta \left[ \begin{array}{c} \mathbf{0} \\ \mathbf{0} \\ \end{array} \right] (\Omega ,\mathbf{z})=\sum_{{\mathbf{n}\in{\mathbb Z}^g}}e^{{2\pi i\left(\frac{1}{2}\mathbf{n}\cdot\boldsymbol{\Omega}\cdot\mathbf{n}+\mathbf{n}\cdot\mathbf{z}\right)}} $$


자코비 세타함수

$$ \begin{align*} \theta_{11}(z;\tau) &:= \Theta \left[ \begin{array}{c} 1/2 \\ 1/2 \\ \end{array} \right](\tau ,z) = \sum_{n \in \mathbb{Z}} q^{\frac{1}{2} \left( n+ \frac{1}{2} \right)^2} \, \E^{2 \pi i \left(n+\frac{1}{2} \right) \, \left( z+\frac{1}{2} \right) } \\ \theta_{10}(z;\tau) &:= \Theta \left[ \begin{array}{c} 1/2 \\ 0 \\ \end{array} \right](\tau ,z) = \sum_{n \in \mathbb{Z}} q^{\frac{1}{2} \left( n + \frac{1}{2} \right)^2} \, \E^{2 \pi i \left( n+\frac{1}{2} \right) z} \\ \theta_{00} (z;\tau) &:= \Theta \left[ \begin{array}{c} 0 \\ 0 \\ \end{array} \right](\tau ,z) = \sum_{n \in \mathbb{Z}} q^{\frac{1}{2} n^2} \, \E^{2 \pi i n z} \\ \theta_{01} (z;\tau) &:= \Theta \left[ \begin{array}{c} 0 \\ 1/2 \\ \end{array} \right](\tau ,z) = \sum_{n \in \mathbb{Z}} q^{\frac{1}{2} n^2} \, \E^{2 \pi i n \left( z+\frac{1}{2} \right) } \end{align*} $$


오일러의 오각수정리(pentagonal number theorem)

\(\prod_{n=1}^\infty (1-x^n)=\sum_{k=-\infty}^\infty(-1)^kx^{k(3k-1)/2}\)

\((1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots\)


의 양변에 \(q^{1/24}\)를 곱하여, 데데킨트 에타함수의 세타함수 표현을 얻는다

\(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})=\sum_{n=-\infty}^\infty(-1)^n q^{\frac{(6n+1)^2}{24}}\)



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