Understanding $\varphi: S^2\setminus \{(0,1)\}\to \mathbb C\mathbb P^1$ given by $(z,\alpha) \mapsto \left[\frac{2z}{1-\alpha}: 1\right]$

Question

I am reading this MSE answer. Writing $$S^2=\big\{(z,\alpha) \in \mathbb{C} \times \mathbb{R}: |z|^2+\alpha^2=1\big\},$$ the author claims that \begin{align*} \varphi: S^2\setminus \{(0,1)\} & \to \mathbb{CP}^1 \\ (z,\alpha) & \mapsto \left[\frac{2z}{1-\alpha}: 1\right] \end{align*} extends smoothly to a map $\varphi: S^2\to \Bbb C\Bbb P^1$ sending $(0,1)$ to $[1:0]$. Moreover, $\varphi$ is a diffeomorphism. I'd like some help with understanding this statement.

1. Why is the map defined by $(z,\alpha) \mapsto \left[\frac{2z}{1-\alpha}: 1\right]$ on $S^2\setminus\{(0,1)\}$? The choice seems quite unmotivated.

2. Why does $\varphi$ extend smoothly to a map on $S^2$?

Lastly, I believe showing that $\varphi$ is a diffeomorphism would involve explicitly writing $\varphi^{-1}$ and composing with charts to obtain that $\varphi$ and $\varphi^{-1}$ are smooth maps. Thanks!

Answer 1

Consider the Stereographic projection: \begin{align} \phi:S^2\smallsetminus (0,0,1)&\longrightarrow \mathbb R^2\\ (x,y,\alpha)&\longmapsto\left(\frac{x}{1-\alpha},\frac{y}{1-\alpha}\right) \end{align} This can be identified with a map to $\mathbb C$, by composing the $\mathbb R$ vector space isomorphism $(x,y)\mapsto x+iy$. Hence, if we identify $S^2$ as a subset of $\mathbb C^2$, then the stereographic projection becomes something of the form: \begin{align} \phi:S^2\smallsetminus (0,1)&\longrightarrow \mathbb C\smallsetminus 0\\ (z,\alpha)&\longmapsto\frac{z}{1-\alpha} \end{align} so you can just compose this with the chart for most of $\mathbb {CP}^1$, and then argue that this extends to a smooth map on the whole of $S^2$.

To show it is a diffeomorphism, it suffices to show that this is bijective, and that the differential is a linear isomorphism everywhere, which it is. This fact follows from the inverse function theorem (as smoothness and continuity is local)

Answer 2

This is essentially contained in youler's answer to Diffeomorphism of $\mathbb{C}P^1$ and $S^2$.

$S^2$ has a smooth atlas consisting of the stereographic projections $$\varphi_1 : S^2 \setminus \{N\} \to \mathbb C = \mathbb R^2, \varphi_1(z,\alpha) = \frac{z}{1-\alpha}, $$ $$\varphi_2 : S^2 \setminus \{S\} \to \mathbb C, \varphi_2(z,\alpha) = \frac{z}{1+\alpha}. $$

Concerning the stereorgraphic projection see for example Two equal smooth structures on $S^n$.

$\mathbb{CP}^1$ has a smooth atlas consisting of the maps $$\psi_1 : \mathbb{CP}^1 \setminus \{[0:1]\} \to \mathbb C,\psi_2[z:w]=w/z ,$$ $$\psi_2 : \mathbb{CP}^1 \setminus \{[1:0]\} \to \mathbb C, \psi_2[z:w]=\overline{z/w}.$$ The inverses of the $\psi_i$ are $$\psi_1^{-1}(\omega) = [1:\omega], $$ $$\psi_2^{-1}(\zeta) = [\overline \zeta:1]. $$ This gives diffeomorphisms $$\delta_1 = \psi_1^{-1} \circ \varphi_1 :S^2 \setminus \{N\} \to \mathbb{CP}^1 \setminus \{[0:1]\} ,$$ $$\delta_2 = \psi_2^{-1} \circ \varphi_2 :S^2 \setminus \{S\} \to \mathbb{CP}^1 \setminus \{[1:0]\} .$$ youler showed that they fit together to a diffeomorphism $$\Delta : S^2 \to \mathbb{CP}^1 $$ such that $\Delta(N) = [0:1]$ and $\Delta(S) = [1:0]$.

Thus $\delta_1$ is the restriction of the diffeomorphism $\Delta$. In other words, $\delta_1$ extends to a diffeomorphism via $N \mapsto [0:1]$. We have

$$\delta_1(z,\alpha) = \left[1 : \frac{z}{1-\alpha}\right] = [1-\alpha : z] .$$

This is not exactly the map $\varphi$ in your question. However, the map $\tau : \mathbb{CP}^1 \to \mathbb{CP}^1, \tau[z:w] = [2w:z]$, is easily seen to be a diffeomorphism such that $\tau[1:0] = [0:1]$ and $\tau[0:1] = [1:0]$ (use the above charts $\psi_i$ to see that the maps $\tau$ and $\tau^{1}[w:z] = [z:w/2]$ are smooth).

This proves the claim in your question.