This is a question from the book Elliptic Curves by Henry McKean and Victor Moll:
Consider the cubic $X_1: y^2=x^3-x$. It admits the involution $(x,y) \mapsto (\frac{-1}{x}, \frac{y}{x^2})$. It is not the involution $j: (x,y) \mapsto (x,-y)$ so it comes from addition of some half-period. What is that half period?
We know that every automorphism of $\mathbb{C}/\Lambda$ where $\Lambda=\{n_1\omega_1+n_2\omega_2 \mid n_1,n_2\in\mathbb{Z}\}$ for two fixed vectors $\omega_1, \omega_2 \in \mathbb{C}$ is either a translation $f(z)=z+c$ or a glide reflection $f(z)=-z+c$. This can be shown easily by considering $\mathbb{C}$ as the universal cover of $\mathbb{C}/\Lambda$.
I know that I can find the corresponding point of a lattice on an elliptic curve using the $\wp$ function but I can not solve this problem.
Any help is appreciated.
The elliptic curve with equation $\,y^2=x^3-x\,$ is Cremona 32a2. The involution given is $\,f_3(x,y) := \left(\frac{-1}{x}, \frac{y}{x^2}\right).\,$ Other involutions are given by $\,f_2(x,y) := \left(\frac{1-x}{1+x}, \frac{-2y}{(1+x)^2}\right)\,$ and $\,f_1(x,y) := \left(\frac{x+1}{x-1}, \frac{-2y}{(x-1)^2}\right)\,$ corresponding to translation by half periods $\,\frac{\omega_3}2,\; \frac{\omega_2}2,\;\frac{\omega_1}2\,$ respectively.
The real half period $\,\frac{\omega_1}2\,$ is the lemniscate constant corresponding to the point $\,(1,0).\,$ The imaginary half period is $\,\frac{\omega_2}2 = -i\,\frac{\omega_1}2\,$ corresponding to the point $\,(-1,0).\,$ The third half period $\,\frac{\omega_3}2 = \frac{\omega_1}2+\frac{\omega_2}2\,$ and corresponds to the point $\,(0,0).\,$ These correspond to the cubic roots $\,e_1=1,\;e_2=-1,\;e_3=0.\,$
My PARI/GP code to verify some of the above is