"자코비 형식"의 두 판 사이의 차이

수학노트
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(section '관련논문' updated)
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==정의==
 
==정의==
 
* 2변수 함수
 
* 2변수 함수
* $(\tau,z)\in \mathcal{H}\times \mathbb{C}$에 대하여, 다음과 같이 정의된다 ($q=e^{\pi i \tau}$)
+
* <math>(\tau,z)\in \mathcal{H}\times \mathbb{C}</math>에 대하여, 다음과 같이 정의된다 (<math>q=e^{\pi i \tau}</math>)
 
\begin{align*}
 
\begin{align*}
 
%  \theta_1(z;\tau)
 
%  \theta_1(z;\tau)
119번째 줄: 119번째 줄:
  
 
==자코비 형식==
 
==자코비 형식==
* $q=e^{2\pi i \tau}$, $y=e^{2\pi i z}$
+
* <math>q=e^{2\pi i \tau}</math>, <math>y=e^{2\pi i z}</math>
* 다음을 만족시키는 함수 $\phi: \mathcal{H}\times \mathbb{C}\to \mathbb{C}$ 를 자코비 형식(k : weight, m : index)이라 한다
+
* 다음을 만족시키는 함수 <math>\phi: \mathcal{H}\times \mathbb{C}\to \mathbb{C}</math> 를 자코비 형식(k : weight, m : index)이라 한다
$$
+
:<math>
 
\phi\left(\frac{a\tau+b}{c\tau+d},\frac{z}{c\tau+d}\right) = (c\tau+d)^ke^{\frac{2\pi i mcz^2}{c\tau+d}}\phi(\tau,z)=(c\tau+d)^k y^{mc \frac{z}{c\tau+d}}\phi(\tau,z)
 
\phi\left(\frac{a\tau+b}{c\tau+d},\frac{z}{c\tau+d}\right) = (c\tau+d)^ke^{\frac{2\pi i mcz^2}{c\tau+d}}\phi(\tau,z)=(c\tau+d)^k y^{mc \frac{z}{c\tau+d}}\phi(\tau,z)
$$ 여기서 ${a\ b\choose c\ d}\in SL_2(\mathbb{Z})$
+
</math> 여기서 <math>{a\ b\choose c\ d}\in SL_2(\mathbb{Z})</math>
  
$\lambda, \mu\in \mathbb{Z}$에 대하여  
+
<math>\lambda, \mu\in \mathbb{Z}</math>에 대하여  
$$
+
:<math>
 
\phi(\tau,z+\lambda\tau+\mu) = e^{-2\pi i m(\lambda^2\tau+2\lambda z)}\phi(\tau,z)=q^{-m \lambda^2} y^{-2m\lambda}\phi(\tau,z)
 
\phi(\tau,z+\lambda\tau+\mu) = e^{-2\pi i m(\lambda^2\tau+2\lambda z)}\phi(\tau,z)=q^{-m \lambda^2} y^{-2m\lambda}\phi(\tau,z)
$$
+
</math>
 
* 푸리에 전개
 
* 푸리에 전개
$$
+
:<math>
 
\phi(\tau,z) = \sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)e^{2\pi i (n\tau+rz)}=\sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)q^nz^r
 
\phi(\tau,z) = \sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)e^{2\pi i (n\tau+rz)}=\sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)q^nz^r
$$
+
</math>
  
 
==메모==
 
==메모==
159번째 줄: 159번째 줄:
 
==관련논문==
 
==관련논문==
 
* Nathan C. Ryan, Nicolás Sirolli, Nils-Peter Skoruppa, Gonzalo Tornaría, Computing Jacobi Forms, arXiv:1602.07021 [math.NT], February 23 2016, http://arxiv.org/abs/1602.07021
 
* Nathan C. Ryan, Nicolás Sirolli, Nils-Peter Skoruppa, Gonzalo Tornaría, Computing Jacobi Forms, arXiv:1602.07021 [math.NT], February 23 2016, http://arxiv.org/abs/1602.07021
* Labrande, Hugo. “Computing Jacobi’s $\theta$ in Quasi-Linear Time.” arXiv:1511.04248 [math], November 13, 2015. http://arxiv.org/abs/1511.04248.
+
* Labrande, Hugo. “Computing Jacobi’s <math>\theta</math> in Quasi-Linear Time.” arXiv:1511.04248 [math], November 13, 2015. http://arxiv.org/abs/1511.04248.
 
* Bringmann, Kathrin, Larry Rolen, and Sander Zwegers. “On the Fourier Coefficients of Negative Index Meromorphic Jacobi Forms.” arXiv:1501.04476 [hep-Th], January 19, 2015. http://arxiv.org/abs/1501.04476.
 
* Bringmann, Kathrin, Larry Rolen, and Sander Zwegers. “On the Fourier Coefficients of Negative Index Meromorphic Jacobi Forms.” arXiv:1501.04476 [hep-Th], January 19, 2015. http://arxiv.org/abs/1501.04476.

2020년 11월 13일 (금) 02:52 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*}


=='"`UNIQ--h-1--QINU`"'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



=='"`UNIQ--h-2--QINU`"'자코비 형식==
* \(q=e^{2\pi i \tau}\), \(y=e^{2\pi i z}\)
* 다음을 만족시키는 함수 \(\phi: \mathcal{H}\times \mathbb{C}\to \mathbb{C}\) 를 자코비 형식(k : weight, m : index)이라 한다
\[
\phi\left(\frac{a\tau+b}{c\tau+d},\frac{z}{c\tau+d}\right) = (c\tau+d)^ke^{\frac{2\pi i mcz^2}{c\tau+d}}\phi(\tau,z)=(c\tau+d)^k y^{mc \frac{z}{c\tau+d}}\phi(\tau,z)
\] 여기서 \({a\ b\choose c\ d}\in SL_2(\mathbb{Z})\)

\(\lambda, \mu\in \mathbb{Z}\)에 대하여 \[ \phi(\tau,z+\lambda\tau+\mu) = e^{-2\pi i m(\lambda^2\tau+2\lambda z)}\phi(\tau,z)=q^{-m \lambda^2} y^{-2m\lambda}\phi(\tau,z) \]

  • 푸리에 전개

\[ \phi(\tau,z) = \sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)e^{2\pi i (n\tau+rz)}=\sum_{n\ge 0} \sum_{r^2\le 4mn} c(n,r)q^nz^r \]

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관련논문

  • Nathan C. Ryan, Nicolás Sirolli, Nils-Peter Skoruppa, Gonzalo Tornaría, Computing Jacobi Forms, arXiv:1602.07021 [math.NT], February 23 2016, http://arxiv.org/abs/1602.07021
  • Labrande, Hugo. “Computing Jacobi’s \(\theta\) in Quasi-Linear Time.” arXiv:1511.04248 [math], November 13, 2015. http://arxiv.org/abs/1511.04248.
  • Bringmann, Kathrin, Larry Rolen, and Sander Zwegers. “On the Fourier Coefficients of Negative Index Meromorphic Jacobi Forms.” arXiv:1501.04476 [hep-Th], January 19, 2015. http://arxiv.org/abs/1501.04476.