7

The homogeneous polynomial $F(x,y,z)=x^3+y^3+z^3$ clearly defines a smooth projective curve $X\subset\mathbb{P}^2$.

It is easy to see that $\pi:X\rightarrow\mathbb{P}^1$ defined by $$\pi([x:y:z])=[x:y] \ , $$ is a well defined holomorphic map of degree $3$. Now if we let $\epsilon_1, \epsilon_2, \epsilon_3$ be the three cube root of $-1$, one can show that the only ramification points of $\pi$ are $[\epsilon_1:1:0],[\epsilon_2:1:0],[\epsilon_3:1:0]$, and they are all triple points, hence by Riemann-Hurwitz's formula we find that $$g(X)=1 \ , $$ therefore $X$ is isomorphic to a torus $\mathbb{C}/\Lambda$.

My question is: how does this isomorphism "work"? I just can't see it..

Davide
  • 93

1 Answers1

6

The covering map $f\colon \mathbb{C} \to X$ is defined by $$ f(z) \;=\; \bigl[-\mathrm{cm}(z) : -\mathrm{sm}(z) : 1\bigr] $$ where $\mathrm{cm}(z)$ and $\mathrm{sm}(z)$ are the Dixonian elliptic functions (which are known to satisfy $\mathrm{cm}(z)^3 +\mathrm{sm}(z)^3 = 1$). See this answer by J. M. for an overview of the Dixonian elliptic functions, including pictures of them on the complex plane.

The lattice $\Lambda$ in this case is $\pi_3\mathbb{Z} + \pi_3e^{2\pi i/3}\mathbb{Z}$, which is a regular lattice of equilateral triangles on the complex plane. Here $\pi_3 \approx 5.29992$ is the value of $B(1/3,1/3)$, where $B$ is the Euler beta function. (One could of course eliminate the $\pi_3$'s by using the function $f(\pi_3 z)$ instead.)

Note that $\mathrm{cm}(z)$ and $\mathrm{sm}(z)$ have poles, and that the function $f$ can be extended holomorphically to these poles so as to make the covering map onto. In particular, $$ f\bigl(-\pi_3/3\bigr) \;=\; [1:-1:0]\qquad\text{and}\qquad f\bigl(e^{\pm i\pi/3}\pi_3/3\bigr) \;=\; \bigl[1:e^{\pm i\pi/3}:0\bigr]. $$

Community
  • 1
Jim Belk
  • 49,278