베버(Weber) 모듈라 함수

개요

\(\mathfrak{f}(\tau)=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})\)

\(\mathfrak{f}_ 1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)

\(\mathfrak{f}_ 2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)

\(\gamma_2(\tau)=\frac{\mathfrak{f}(\tau)^{24}-16}{\mathfrak{f}(\tau)^8}=\sqrt[3]{j(\tau)}\)

\(\gamma_3(\tau)= \frac{(\mathfrak{f}(z)^{24} + 8) (\mathfrak{f}_ 1(z)^8 - \mathfrak{f}_ 2(z)^8)}{\mathfrak{f}(z)^8}=\sqrt{j(\tau)-1728}\)

여기서 \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\) 는 데데킨트 에타함수

\(\mathfrak{f}(\tau+1)=\zeta_{48}^{-1}\mathfrak{f}_ 1(\tau)\)
\(\mathfrak{f}_ 1(\tau+1)=\zeta_{48}^{-1}\mathfrak{f}(\tau)\)
\(\mathfrak{f}_ 2(\tau+1)=\zeta_{24}\mathfrak{f}_ 2(\tau)\)
\(\mathfrak{f}(-\frac{1}{\tau})=\mathfrak{f}(\tau)\)
\(\mathfrak{f}_ 1(-\frac{1}{\tau})=\mathfrak{f}_ 2(\tau)\)
\(\mathfrak{f}_ 2(-\frac{1}{\tau})=\mathfrak{f}_ 1(\tau)\)

타원 모듈라 j-함수 (j-invariant)
\(\mathfrak{f}(\tau)^{24}\), \(-\mathfrak{f}_ 1(\tau)^{24}\), \(-\mathfrak{f}_ 2(\tau)^{24}\)는 \((x-16)^3-j(\tau)x=0\) 의 근이다

\(\mathfrak{f}(\tau)=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})\)

\(\mathfrak{f}_ 1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)

\(\mathfrak{f}_ 2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)

여기서 \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\) 는 데데킨트 에타함수

q-초기하급수(q-hypergeometric series) 의 공식
\(\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n (n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)
\(z=q^{1/2}\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n\geq 0}\frac{q^{n (n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} (q^{1/2})^n=\sum_{n\geq 0}\frac{q^{n^2/2}}{(1-q)(1-q^2)\cdots(1-q^n)} \)
\(\prod_{n=1}^{\infty} (1+q^{2n-1})=\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})} \)
\(z=q\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n})=\sum_{n\geq 0}\frac{q^{n (n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}q^n=\sum_{n\geq 0}\frac{q^{n (n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)
위의 결과로부터 다음을 얻을 수 있다
\(\mathfrak{f}(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\)
\(\mathfrak{f}_ 2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n (n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)