Riemann surface $y=(1-x^3)^{1/3}$

Let $X\subset \mathbb C^2$ be the Riemann surface given by （the multivalued function) $y=(1-x^3)^{1/3}$, and let $\phi:X \to \mathbb C$ be the induced map. Let $X'\subset P(\mathbb C^2)$ be the complex curve $x^3+y^3=z^3$ (in homogeneous coordinates). I want to define a complex manifold structure on $X'$ and a holomorphic $\phi':X'\to \mathbb C$ that extends $\phi$.

So far I know the following,

(i) $(x,y) \mapsto [x:y:1]$ gives a homeomorphism $\psi$ of $\mathbb C^2$ onto its image.

(ii) $\psi(X)\bigcup \{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}=X'$.

So I want to first define a manifold structure for $X$ and push it to $X'$ by $\psi$ and use another chart to cover $\{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}$. I think $x$ is a coordinate chart around $x_0$ if we fix a branch (fix a $y_0$). So we need three charts to cover $X$ because we have three branches. Is this correct?

Also, how do I construct a compatible chart to cover the three points at infinity?

If I got well, $X$ is the zero locus $x^3+y^3=1$ in $\mathbb{C}^2$ and your $\phi$ is the projection on one of the coordinate (I will assume is the $y$ just to be clear).

Now you take $X'$ that is ,as you can easily prove, the projective closure of $X$ in $\mathbb{P}^2(\mathbb{C})$.

As you correctly said , you need three charts because intuitively there are three branches. A standard way to proceed in the projective spaces is to use the standard charts $x_i \neq 0$ that are (in the projective plane) biholomorphic to $\mathbb{C}^2$.

If you restrict $X'$ to this charts , you will get three varieties: the first one (in the chart $z \neq0$) is your $X$ (and so you get your immersion) and the other two are very closely related to $X$ (so you can easily find complex structures on them in the same way you did with $X$).

• One of my concern is that when we consider a point$x_0,0$, could x still be a local coordinate even though $x_0$ is a branch point? – user136592 Mar 25 '18 at 1:16
• I would use the general fact that the partial derivatives of $x^3+y^3-1$ do not both vanish on $X$ and use the Implicit Function theorem – Tommaso Scognamiglio Mar 25 '18 at 7:03
• I can differentiate to get $3x^2 dx +3y^2dy=0$ which I can rearrange to get $\frac{dx}{y}=-\frac{y}{x^2}$. From this I can determine the local coordinate depending on whether $x=0$ or $y=0$, but how would you construct the other two varieties? – Jade Mar 26 '18 at 4:19
• Start writing your curve in the other two standard charts and look if you can make something similar – Tommaso Scognamiglio Mar 26 '18 at 6:36
• If you take another chart where $x \neq 0$, then let the affine coordinate in that chart be $(y',z')=(y/x,z/x)$ and homogenize the equation wrt to that, you get $1+y'^3=z'^3$. You can differentiate to get $3y'^2=3z'^2$ and when $y_0' \neq 0$, take $z'$ is a local coordinate with differential $w=dz'/y'$. When $y_0' =0$, take $y'$ to be the local coordinate and let $w=y'dy'/dz'^2$. Is this what you mean? Is this how you are supposed to approach this problem? – Jade Mar 27 '18 at 4:03