Nice coordinate presentation of the covering map $S^2 \to \mathbb{R}P^2$

Question

I would like to write down a nice coordinate presentation of the double cover map $S^2 \to \mathbb{R}P^2$, where $S^2$ carries a single complex coordinate $z$ as the Riemann sphere $\mathbb{C}P^1 \cong \mathbb{C} \cup \infty$ and we use affine coordinates $x=X/Z,y=Y/Z$ for the affine patch of $\mathbb{R}P^2 =\{[X:Y:Z]\}$ where $Z\neq 0.$

Ideally this map should behave nicely with these structures. For example map the two hemispheres separately to the whole real plane, map the equator to the line at infinity 2 to 1, and maybe map the point at infinity of the Riemann sphere to that of the projective plane.

I know that a disc is not isomorphic to a plane in the complex category, so I must not expect the map to be holomorphic.

Here’s what I tried.

Since this is a double cover it should be degree two. The most obvious candidate is $z \mapsto z^2.$ This is after all the double cover map for $S^1 \to \mathbb{R}P^1$. Maybe it extends natural to higher spheres. But no, it’s not a covering map at 0 and $\infty$. As a holomorphic map we knew this would not work.

One obvious way to identify antipodal points on a sphere is to map $z$ to $z\cdot A(z),$ where $A$ is the antipodal map, as this product manifestly identifies antipodal points. In coordinates $A(z)=-\frac{1}{\bar{z}}$, so our map candidate is $z\mapsto -\frac{z}{\bar{z}}$. Or $re^{i\theta} \mapsto -e^{2i\theta}$. This one does map the equator to the line at infinity 2 to 1. But then the rest of the cylinder it collapses onto the equator. And not clear how to define it at the poles.

The map $\frac{z}{1-\lvert z\rvert^2}$ is a nice map of the unit disc to the plane. It does appear to map the equator to the line at infinity, as desired. Actually it appears that this one is also invariant under the antipodal map.

I thought I had a reason that this map didn’t work, but actually now it appears to meet all my criteria. Is $\frac{z}{1-\lvert z\rvert^2}$ actually the “nice” double cover map from $S^2$ to $\mathbb{R}P^2$? Is it obviously continuous and a local homeomorphism and 2 to 1 everywhere?

Answer 1

As a subset of $\mathbb{R}^3$ the double cover is $(x,y,z)\mapsto [x:y:z]$. An identification of $S^2$ with the Riemann sphere is

$$\left[z:w\right]\in \mathbb{CP}^1\mapsto \left(\frac{2zw^\ast}{|w|^2+|z|^2},\frac{-|w|^2+|z|^2}{|w|^2+|z|^2}\right)\in \mathbb{C}\times \mathbb{R}\cong \mathbb{R}^3\tag 1$$

where the image of this map is $S^2$.

On the affine patch $\mathbb{CP}^1\setminus\{(1:0)\}\cong \mathbb{C}$, it is

$$z\in \mathbb{C}\mapsto \left(\frac{2z}{1+|z|^2},\frac{-1+|z|^2}{1+|z|^2}\right)\in \mathbb{C}\times \mathbb{R}\cong \mathbb{R}^3\tag 2$$

So, composing the homeomorphism $\mathbb{CP}^1\xrightarrow{\sim} S^2$ with the double cover, we get

$$[z:w]\mapsto \left[2\Re(zw^\ast):2\Im(zw^\ast):-|w|^2+|z|^2\right]\tag 3$$

Or again, in the affine coordinates this is

$$z\mapsto \left(\frac{2\Re(z)}{|z|^2-1},\frac{2\Im(z)}{|z|^2-1}\right)\tag 4$$

Note. I obtained $(2)$ by computing the stereographic projection. The obvious completion of $(2)$ to all of $\mathbb{CP}^1$ is $(1)$.