テータ関数
楕円テータ関数にも定義に 2 つの流儀があり、注意が必要である。 フルヴィッツ・クーランの「楕円関数論」の定義では添え字が 1 から 4 ではなく、 0 から 3 である。
θ 1 ( x + y , τ ) θ 1 ( x − y , τ ) θ 4 2 ( 0 , τ ) = θ 1 2 ( x , τ ) θ 4 2 ( y , τ ) − θ 4 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 3 2 ( x , τ ) θ 2 2 ( y , τ ) − θ 2 2 ( x , τ ) θ 3 2 ( y , τ ) θ 2 ( x + y , τ ) θ 2 ( x − y , τ ) θ 4 2 ( 0 , τ ) = θ 2 2 ( x , τ ) θ 4 2 ( y , τ ) − θ 3 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 4 2 ( x , τ ) θ 2 2 ( y , τ ) − θ 1 2 ( x , τ ) θ 3 2 ( y , τ ) θ 3 ( x + y , τ ) θ 3 ( x − y , τ ) θ 4 2 ( 0 , τ ) = θ 3 2 ( x , τ ) θ 4 2 ( y , τ ) − θ 2 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 4 2 ( x , τ ) θ 3 2 ( y , τ ) − θ 1 2 ( x , τ ) θ 2 2 ( y , τ ) θ 4 ( x + y , τ ) θ 4 ( x − y , τ ) θ 4 2 ( 0 , τ ) = θ 4 2 ( x , τ ) θ 4 2 ( y , τ ) − θ 1 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 3 2 ( x , τ ) θ 3 2 ( y , τ ) − θ 2 2 ( x , τ ) θ 2 2 ( y , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{1}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\&\vartheta _{2}\left(x+y,\tau \right)\vartheta _{2}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\&\vartheta _{3}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)\\&\vartheta _{4}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)\\\end{aligned}}}
無限乗積表示
θ 1 ( v , τ ) = 2 e π i τ / 4 sin  π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e 2 m π i τ e 2 π i v ) ( 1 − e 2 m π i τ e − 2 π i v ) {\displaystyle \vartheta _{1}(v,\tau )=2e^{{\pi }i{\tau }/4}\sin {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}}
の対数微分により
θ 2 ( 0 , τ ) 2 = ∑ m = − ∞ ∞ e π i τ ( m + 1 / 2 ) 2 ∑ n = − ∞ ∞ e π i τ ( n + 1 / 2 ) 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( m + 1 / 2 ) 2 + ( n + 1 / 2 ) 2 ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( m + n + 1 ) 2 + ( m − n ) 2 ) / 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m + 1 ) 2 + ( 2 n ) 2 ) / 2 + ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m + 2 ) 2 + ( 2 n + 1 ) 2 ) / 2 ( ⌊ m + n 2 ⌋ ↦ m , ⌊ m − n 2 ⌋ ↦ n ) = 2 θ 3 ( 0 , 2 τ ) θ 2 ( 0 , 2 τ ) {\displaystyle {\begin{aligned}\vartheta _{2}\left(0,\tau \right)^{2}&=\sum _{m=-\infty }^{\infty }e^{\pi {i}\tau {(m+1/2)^{2}}}\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(n+1/2)^{2}}}=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {\left((m+1/2)^{2}+(n+1/2)^{2}\right)}}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((m+n+1)^{2}+(m-n)^{2}\right)/2}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m+1)^{2}+(2n)^{2}\right)/2}+\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m+2)^{2}+(2n+1)^{2}\right)/2}\qquad (\left\lfloor \textstyle {\frac {m+n}{2}}\right\rfloor \mapsto {m},\left\lfloor \textstyle {\frac {m-n}{2}}\right\rfloor \mapsto {n})\\&=2\vartheta _{3}\left(0,2\tau \right)\vartheta _{2}\left(0,2\tau \right)\end{aligned}}}
が得られる。
である。他の関数の零点も同様にして求められる。
θ 1 ( v ; τ ) = 0 ⇔ v = n + m τ θ 2 ( v ; τ ) = 0 ⇔ v = 2 n + 1 2 + m τ θ 3 ( v ; τ ) = 0 ⇔ v = 2 n + 1 2 + 2 m + 1 2 τ θ 4 ( v ; τ ) = 0 ⇔ v = n + 2 m + 1 2 τ {\displaystyle {\begin{aligned}&\vartheta _{1}(v;\tau )=0\;\Leftrightarrow \;v=n+m\tau \\&\vartheta _{2}(v;\tau )=0\;\Leftrightarrow \;v={\frac {2n+1}{2}}+m\tau \\&\vartheta _{3}(v;\tau )=0\;\Leftrightarrow \;v={\frac {2n+1}{2}}+{\frac {2m+1}{2}}\tau \\&\vartheta _{4}(v;\tau )=0\;\Leftrightarrow \;v=n+{\frac {2m+1}{2}}\tau \\\end{aligned}}}
{\displaystyle {\begin{aligned}\vartheta _{1}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=ie^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{1}\left(v,\tau \right),\\\vartheta _{2}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{4}\left(v,\tau \right),\\\vartheta _{3}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{3}\left(v,\tau \right),\\\vartheta _{4}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{2}\left(v,\tau \right).\end{aligned}}}
他に τ を変換するものとして
θ 3 ( v , τ + 1 ) = ∑ n = − ∞ ∞ e π i ( τ + 1 ) n 2 + 2 π i n v = ∑ n = − ∞ ∞ ( − 1 ) n e π i τ n 2 + 2 π i n v = θ 4 ( v , τ ) {\displaystyle \vartheta _{3}\left(v,\tau +1\right)=\sum _{n=-\infty }^{\infty }e^{\pi {i}(\tau +1)n^{2}+2\pi {i}nv}=\sum _{n=-\infty }^{\infty }(-1)^{n}e^{\pi {i}\tau {n^{2}}+2\pi {i}nv}=\vartheta _{4}(v,\tau )}