# Fundamental period of the Weierstrass $\wp$ elliptic function?

Consider the Weierstrass $\wp$ elliptic function $\wp(z, g_2, g_3)$ with the invariants $g_2\in\mathbb{R}$ and $g_3\in\mathbb{R}$:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3$$

According to Wikipedia when $\Delta=g_ 2^3-27g_3^2 > 0$, we have three real roots $e_1 > e_2 > e_3$ and $\wp(z, g_2, g_3)\in\mathbb{R}$.

My question is: in that case (where all values are real, including $z$), how to compute the period $w\in\mathbb{R}$ (or half-period) of $\wp(z, g_2, g_3)$ as a function of $g_2, g_3, e_1, e_2, e_3$ ? (I am searching for an analytical formula in terms of standard functions or special functions available in this list, not in terms of an integral).

Additional note : If understand correctly, the same Wikipedia article seems to say that : $$\frac{w_{1}}{2}=\int_{e_1}^{\infty}\frac{dz}{\sqrt{4z^3-g_2 z-g_3}}$$ but I am not sure that their $w_1$ is the $w$ I am searching for. In any case I would like a formula in terms of special functions as explained above.

By a change of variables, you can put your elliptic curve into Legendre form $y^2=x(x-1)(x-\lambda)$; you should have $\lambda = (e_3-e_2)/(e_1-e_2)$. Then you can evaluate the integral numerically using Jacobi's form of the complete elliptic integral, with $k^2=\lambda$.
• So it is simply $K(\sqrt{\frac{e_3-e_2}{e_1-e_2}})$ ? Commented May 4, 2014 at 22:28
• @Vincent Yes, that should be right, perhaps up to a multiplicative constant which might appear when doing the change of variables to pass from the Weierstrass form of the elliptic integral to the Jacobi form. If you want to make sure, I recommend that you either check numerically for a given curve whose periods you already know, or carry out explicitly the appropriate change of variables to see what happens (the transformation is a Mobius transformation taking the points $\infty, e_1, e_2, e_3$ to $-1, 1, -k, k$. You can find it in dusty books, or work it out from scratch if you are patient.) Commented May 4, 2014 at 22:35
• But in that case $\lambda$ is negative as $e_1 > e_2 > e_3$ so I cannot compute the $K(k)$... Commented May 4, 2014 at 22:35
• @Vincent It shouldn't matter, since the Jacobi integral involves $k^2$ rather than $k$. Commented May 4, 2014 at 22:39
• @Vincent Assume the three roots are real and ordered as $e_1 > e_2 > e_3$, If I didn't make any mistake, then the half periods are given by $$\omega_1 = \frac{1}{\sqrt{e_1 - e_3}}K\left(\sqrt{\frac{e_2-e_3}{e_1-e_3}}\right) \quad\text{ and }\quad \omega_3 = \frac{i}{\sqrt{e_1-e_3}}K\left(\sqrt{\frac{e_1-e_2}{e_1-e_3}}\right)$$ Commented May 4, 2014 at 22:43