Weierstrass 楕円関数
Weierstrass elliptic function $\wp$
$\Lambda=\{ m\ \omega_0+n\ \omega_1 : m,n \in \mathbb{Z} \}$
$\Lambda'=\Lambda - \{ 0 \}$
$$\wp (u)=\frac{1}{u^2}+\sum_{w\in \Lambda'} \left \{ \frac{1}{(u-w)^2}-\frac{1}{w^2} \right \}$$
$$\wp'(u)^2=4\wp(u)^3-g_2\ \wp(u)-g_3$$
$\omega_2:=-\omega_0-\omega_1$
$$e_k:=\wp\left (\frac{\omega_k}{2} \right )\ (k=0,1,2)$$
$e_0,e_1,e_2$は$4z^3-g_2z-g_3=0$の解となる.
Weierstrass function $\zeta$
$$\zeta(u)=\frac{1}{u}+\sum_{w\in \Lambda'} \left \{ \frac{1}{u-w}+\frac{1}{w}+\frac{u}{w^2} \right \}$$
$$\eta_k:=2\ \zeta \left (\frac{\omega_k}{2} \right )$$
$\omega_0+\omega_1+\omega_2=0$
$e_0+e_1+e_2=0$
$\eta_0+\eta_1+\eta_2=0$
$\eta_0\omega_1-\eta_1\omega_0=2\pi i$
$\eta_1\omega_2-\eta_2\omega_1=2\pi i$
$\eta_2\omega_0-\eta_0\omega_2=2\pi i$
Weierstrass function $\sigma$
$$\sigma(u)=u \prod_{w\in \Lambda'} \left (1-\frac{u}{w}\right )\exp \left [ \frac{u}{w}+\frac{u^2}{2w^2} \right ]$$
$$\sigma (u+\omega_k)=-\exp \left [\eta_k \left (u+\frac{\omega_k}{2} \right ) \right ]\sigma(u)$$
$$\sigma (u+\omega_0)=-\exp \left [\eta_0 \left (u+\frac{\omega_0}{2} \right ) \right ]\sigma(u)$$
について
$u=\omega_0 v$とおく.
$f(v)=\sigma(\omega_0 v)$とおく.
$g(v)=\log f(v)$とおく.
このとき
$$g(v+1)-g(v)=\eta_0\omega_0 \left (v+\frac{1}{2} \right )+\pi i$$
整数$a,b$について$v=a$から$v=b-1$まで和をとると
$$g(b)-g(a)=\eta_0\omega_0\sum_{v=a}^{b-1} \left (v+\frac{1}{2} \right )+\pi i (b-a)$$
$$=\eta_0\omega_0 \frac{b^2-a^2}{2}+\pi i (b-a)$$
$f$にもどすと
$$\frac{f(b)}{f(a)}=\pm \exp \left [\eta_0\omega_0\frac{b^2-a^2}{2} \right ]$$
ここで符号は$b-a$の偶奇に依存する.
$$\varphi(v)=f(v)\exp \left [-\eta_0 \omega_0 \frac{v^2}{2} \right ]$$
とおくと
$$\frac{\varphi(v+1)}{\varphi(v)}=-1$$
を満たす.
$$\varphi(v+\tau)=-\exp \left [-2\pi i \left (v+\frac{\tau}{2}\right ) \right ] \varphi(v)$$
$$\varphi \left (v+\frac{\tau}{2} \right )=-\exp \left [-2\pi i v \right ] \varphi \left (v-\frac{\tau}{2} \right )$$
$$\wp(u)-\wp \left (\frac{\omega_k}{2} \right )=-\frac{\sigma(u+\frac{\omega_k}{2})\sigma(u-\frac{\omega_k}{2})}{\sigma(\frac{\omega_k}{2})^2 \sigma(u)^2}$$
$$\sigma_k(u):=-\exp \left [\eta_k \frac{u}{2} \right]\frac{\sigma(u-\frac{\omega_k}{2})}{\sigma(\frac{\omega_k}{2})}$$
$$\wp(u)-e_k=\frac{\sigma_k(u)^2}{\sigma(u)^2}$$
$$\sqrt{\wp(u)-e_k}:=\frac{\sigma_k(u)}{\sigma(u)}$$
$$\wp'(u)=-2\frac{\sigma_0(u)\sigma_1(u)\sigma_2(u)}{\sigma(u)^3}$$
$$\frac{dw}{dz}=\frac{1}{\sqrt{(1-z^2)(1-\lambda z^2)}}$$
ここで
$w=u \sqrt{e_l-e_k}\ \ (l\ne k)$
$$z=\frac{\sigma(u)}{\sigma_k(u)}\sqrt{e_l-e_k}\ \ (l\ne k)$$
$$\lambda=\frac{e_j-e_k}{e_l-e_k}$$
$l,j,k$は全て異なる.
$$K=\int_0^1 \frac{dz}{\sqrt{(1-z^2)(1-k^2z^2)}}$$
$$K'=\int_0^1 \frac{dz}{\sqrt{(1-z^2)(1-k'^2z^2)}}$$
$k^2=\lambda$
$k^2+k'^2=1$
$$\sigma_k(u+\omega_l)=-\exp \left [\eta_k \frac{u+\omega_l}{2} \right]\frac{\sigma(u+\omega_l-\frac{\omega_k}{2})}{\sigma(\frac{\omega_k}{2})}=$$
$$-\exp \left [\eta_k \frac{u+\omega_l}{2} \right]\frac{-\exp \left [\eta_l \left (u-\frac{\omega_k}{2}+\frac{\omega_l}{2} \right ) \right ]\sigma(u-\frac{\omega_k}{2})}{\sigma(\frac{\omega_k}{2})}=$$
$$-\exp \left [\eta_k \frac{\omega_l}{2}+\eta_l \left (u-\frac{\omega_k}{2}+\frac{\omega_l}{2} \right ) \right] \sigma_k(u)=$$
$$-\exp \left [ \frac{\eta_k\omega_l-\eta_l\omega_k}{2} \right]\exp \left [\eta_l \left (u+\frac{\omega_l}{2} \right ) \right] \sigma_k(u)=\pm \exp \left [\eta_l \left (u+\frac{\omega_l}{2} \right ) \right] \sigma_k(u)$$
- Jacobiの楕円関数の基本周期は$4K,2iK'$である.
- Weierstrassの楕円関数の基本周期は$\omega_0,\omega_1$である.
- $$\sqrt{\wp(u)-e_k}=\frac{\sigma_k(u)}{\sigma(u)}$$の基本周期は$2\omega_0,\omega_1$である.
- Jacobiの楕円関数の基本周期の比は$\displaystyle{\frac{\omega_1}{2\omega_0}}$に等しい.
$$\tau=\frac{\omega_1}{\omega_0}=2\frac{\omega_1}{2\omega_0}=2\frac{2iK'}{4K}=i\frac{K'}{K}$$