Finding an automorphism of the Riemann sphere that sends the branch points of the Weierstrass elliptic function $\wp$ to $(0, \infty, -1, 1)$

Question

I’ve read on my lecture notes that we can find an automorphism of the Riemann sphere that sends the branch points of the Weierstrass elliptic function over the complex torus $X=\mathbb{C} / {\mathbb{Z}[i]}$ (that is, the image by $\wp$ of its four ramification points) to a given set of four points.

More concretely, I know that the ramification points of $\wp$ are $e_0=\pi(0)$, $e_1=\pi(i/2)$, $e_2=\pi(1/2)$, $e_3=\pi((1+i)/2)$ (where $\pi:X \rightarrow \mathbb{C}$ is the canonical projection), and according to the lecture notes there exists an automorphism $\phi: \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}} $ such that $\phi \circ (\wp(e_1),\wp(e_2),\wp(e_3),\wp(e_0))= (0,\infty,1,-1)$.

I’ve noticed that since $\wp(iz)=-\wp(z)$, we have $\wp(e_3)=0$ and $\wp(e_1)=-\wp(e_2)$. I also know that $\wp(e_0)=\infty$. So I’m looking for an automorphism $\phi: \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}} $ such that $\phi \circ (\wp(e_1),-\wp(e_1),0,\infty) = (0,\infty,1,-1)$.

I know automorphisms of the Riemann sphere are of the form $z \rightarrow \frac{az+b}{cz+d}$, and I’ve tried solving the system but I didn’t find a solution. Did I go wrong somewhere, or could there be an error in the problem statement? It would make more sense to me if the image of the four points by $\wp$ were $(1,-1,0,\infty)$ instead of $(0,\infty,1,-1)$, because then I guess the automorphism could be given by $a=1$, $b=0$, $c=\wp(e_1)$, $d=0$.

Answer 1

A Möbius transformation $T$ is uniquely determined by the images $(w_1, w_2, w_3)$ at three distinct points $(z_1, z_2, z_3)$. $T$ maps a fourth point $z_4$ to $w_4$ if and only if the cross ratios $$ (z_1, z_2, z_3,z_4) = (w_1, w_2, w_3, w_4) $$ are equal, and that is the case for these particular values: $$ (\wp(e_1), \wp(e_2), \wp(e_3), \wp(e_0)) = (\wp(e_1), -\wp(e_1), 0, \infty ) = -1 = (0, \infty, 1, -1) \, . $$

Concretely: $$ \phi(z) = \frac{z-\wp(e_1)}{z-\wp(e_2)} \cdot \frac{\wp(e_3)-\wp(e_2)}{\wp(e_3)-\wp(e_1)} $$ is the (unique) Möbius transformation which maps $\wp(e_3), \wp(e_1), \wp(e_2)$ to $1, 0, \infty$, respectively. Here we have $\wp(e_3) = 0$ and $\wp(e_1) = -\wp(e_2)$, so that it simplifies to $$ \phi(z) = -\frac{z-\wp(e_1)}{z+\wp(e_1)} \, . $$ Then $\phi(\wp(e_0)) = \phi(\infty) = -1$, so that $\phi$ has the desired properties.