Visualize a projective curve $X^3+Y^3=Z^3$ in $P_2(C)$ as a torus Let $P_2(C)$ be the 2 dimensional complex projective space, I want to prove that the projective curve defined by $M=\{[X,Y,Z]\in P_2(C)|X^3+Y^3=Z^3\}$ is a torus. I know that there is a theorem saying that its genus is C(3-1,2)=1 but I'd like to visualize that by constructing the Riemann surface of the multi-function $y^3=1-x^3$. I can prove that on the Riemann sphere $S$, except 3 points $\{[1,1], [\omega, 1], [\omega^2,1]\}$ where $\omega^3=1$, all other points has 3 pre-images under the map $f: M \rightarrow S$ defined by $f([X,Y,Z])=[X,Z]$. Then I'm stuck: how to cut and glue the 3 spheres to become a torus, based on the 3 points?
 A: The language about gluing "3 spheres" is misleading. If you remove the three branch points $1$, $\omega$, $\omega^2$ from $S$, then the preimage $f^{-1}(S \setminus \{ 1, \omega, \omega^2 \})$ is not three spheres but rather a nontrivial 3-fold cover of a sphere. If we want to glue together 3 identical objects, we need to remove a branch cut $C$ from $S$, touching all three branch points, such that $f^{-1}(S \setminus C)$ becomes $3$ copies of $S \setminus C$. I'll take the branch cut to be the three rays connecting the branch points to the point $0$ of the Riemann sphere:

Then $S \setminus C$ (remembering the point at infinity) is a "hexagon" with angles $(360^{\circ}, 120^{\circ}, 360^{\circ}, 120^{\circ}, 360^{\circ}, 120^{\circ})$. The cover $f$ is triply branched at the $360^{\circ}$ angles, so its preimage is three  $(120^{\circ},120^{\circ},120^{\circ},120^{\circ},120^{\circ},120^{\circ})$ hexagons.  
To see how the three hexagons fit together to form a torus, imagine taking the hexagonal tiling of the plane shown below and quotienting by the translations that preserve the coloring:

