Jacobi elliptic function $\mathrm{sn}$ is defined as


$$\operatorname{sn}(u,k)=x\Leftrightarrow u=\int\_0^x \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}.$$


and Weierstrass sigma function $\sigma$ is defined as


$$ \sigma(u)=u\prod\_{0\neq\ell\in{L\_z}} \left(1-\frac u \ell\right)e^{\frac u \ell + \frac{1}{2} \left(\frac u \ell \right)^2}.$$


where


$$L\_z=\{m+nz \mid m,n\in{\mathbb{Z}}\}$$


in "The lecture of multiple trigonometric function theory" (Japanese: 多重三角関数論講義), there was a description called "on the case that $k$ and $L$ are special, $\operatorname{sn}=\sigma$"


How should I decide $z$ and $k$ to get the proof of this proposition?



Jacobi and Weierstrass elliptic function
Weierstrass $\sigma$-function is entire function, it has no poles. Jacobi elliptic sine function (in general case) is an elliptic function, so it has poles. [For each path $\gamma$ connecting $0$ and $\infty$ (in general case and for $k=1$) the integral
$$u\_\gamma=\int\_\gamma \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$$
is well defined. So $\operatorname{sn}(u\_\gamma)=\infty$ for every $\gamma$. And it is true for all $k\in\mathbb C\setminus\{0\}$.] It means that the situation $\mathrm{sn}=\sigma$ is impossible. 


The are two possibilities in the case of singular curve $y^2=4x^3-g\_2x-g\_3$:
$$\sigma(u)=\frac{2\omega}{ \pi}e^{\frac{1}{6 }\left(\frac{\pi u}{ 2\omega}\right)^2}\sin\frac{\pi u}{ 2\omega}\qquad(g\_2^3=27g\_3^2\ne 0),$$
$$\sigma(u)=u\qquad(g\_2=g\_3= 0).$$


Jacobi elliptic sine is an elementary fuction only for $k=0,1$. If $k=0$ then $\operatorname{sn} x=\sin x$, and this function can't coinside with degenerate $\sigma$-functions above.


