Diffeomorphism of $\mathbb{C}P^1$ and $S^2$.

In an exercise I am asked to find a (smooth) submersion of $S^3$ onto the sphere $S^2$. So far I have a submersion of $S^3$ onto $\mathbb{C}P^1$.

Are $\mathbb{C}P^1$ and $S^2$ diffeomorphic? If so, how does one construct a diffeomorphism between the two and what is the geometric intuition behind the construction?

Thank you.

• $\mathbb{C}P^1$ can be canonically identified with the Alexandrov compactification of $\mathbb{C}$, which is the Riemann sphere, a natural identification of that with $S^2$ is given by stereographic projection. – Daniel Fischer Aug 2 '13 at 16:04
• @Daniel Thank you. I understand the concept; however, in order to account for the whole Riemann sphere one needs to consider two mappings, in the same way one gives a differential structure to $\mathbb{C}P^1$. My difficulty is in finding the explicit diffeomorpshim. – Weltschmerz Aug 2 '13 at 16:59

Here's a way of explicitly constructing a diffeomorphism.

Recall the stereographic charts on $S^2$, $(S^2\smallsetminus\{N\},\varphi_1)$ and $(S^2\smallsetminus\{S\},\varphi_2)$, i.e. $\varphi_1(x,y,z)=\frac{(x,y)}{1-z}$ and $\varphi_2(x,y,z)=\frac{(x,y)}{1+z}$. The transition map is $$\varphi_2\circ\varphi_1^{-1}(u,v)=\left(\frac{u}{u^2+v^2},\frac{v}{u^2+v^2}\right).$$

Recall also $\mathbb{CP}^1=\{[z:w]\mid z\neq 0\text{ or }w\neq 0\},$ with $[z:w]$ denoting the line through the origin and $(z,w)\in\mathbb{C}^2$. The typical charts are $(U_1,\psi_1)$ and $(U_2,\psi_2)$ where $U_1=\{[z:w]\mid z\neq 0\}$, $U_2=\{[z:w]\mid w\neq 0\}$, $$\psi_1[z:w]=wz^{-1}\in\mathbb{C}\cong\mathbb{R}^2\qquad\text{and}\qquad\psi_2[z:w]=\overline{zw^{-1}}\in\mathbb{C}\cong\mathbb{R}^2.$$ The transition map is $$\psi_2\circ\psi_1^{-1}(z)=\psi_2[1:z]=\overline{z}^{-1}$$ as a map from $\mathbb{C}$ to $\mathbb{C}$; viewing it instead as a map from $\mathbb{R}^2$ to $\mathbb{R}^2$ (i.e. taking $z=u+iv\cong(u,v)\in\mathbb{R}^2$), we have $$\psi_2\circ\psi_1^{-1}(u,v)=\left(\frac{u}{u^2+v^2},\frac{v}{u^2+v^2}\right).$$ Note that we chose $\psi_2[z:w]=\overline{zw^{-1}}$ instead of the more typical $\psi_2[z:w]=zw^{-1}$ so that $\varphi_2\circ\varphi_1^{-1}=\psi_2\circ\psi_1^{-1}.$

This suggests a natural correspondence between $S^2$ and $\mathbb{CP}^1$: define $\Phi:S^2\to\mathbb{CP}^1$ by $\Phi(\varphi_1^{-1}(u,v))=\psi_1^{-1}(u,v)$ and $\Phi(\varphi_2^{-1}(u,v))=\psi_2^{-1}(u,v)$, for any $(u,v)\in\mathbb{R}^2.$ This is a well defined map since if $\varphi_1^{-1}(u,v)=\varphi_2^{-1}(x,y)$ then $$(x,y)=\varphi_2\circ\varphi_1^{-1}(u,v)=\psi_2\circ\psi_1^{-1}(u,v),$$ so that $\psi_2^{-1}(x,y)=\psi_1^{-1}(u,v).$ In the same way, we have a well-defined inverse $\Phi^{-1}:\mathbb{CP}^1\to S^2$ defined by $\Phi^{-1}(\psi_1^{-1}(u,v))=\varphi_1^{-1}(u,v)$ and $\Phi^{-1}(\psi_2^{-1}(u,v))=\varphi_2^{-1}(u,v)$ for any $(u,v)\in\mathbb{R}^2.$ I'll leave it to you to check that $\Phi$ and $\Phi^{-1}$ are both smooth; of course this is easiest to do by checking that the local coordinate maps in terms of $\varphi_1,\varphi_2,\psi_1,\psi_2$ are smooth.

Beyond all of the details above, this just reflects the general fact that two manifolds are diffeomorphic when you can give them each a coordinate atlas with the same transition maps.

• It doesn't make sense that the charts for $\mathbb CP^1$ are $(U_1,\psi_1)$ and $(U_2,\psi_2)$ since the maps $\psi_i$ are not homeomorphic to an open subset of $\mathbb R^2$. The charts should be $(U_i,f \circ \psi_i)$ where $f: \mathbb C \to \mathbb R^2$ is canonical homeomorphism. – user46372819 Oct 9 '18 at 4:51
• This changes the look of the transition maps and we don't have such a natural correspondence between $S^2$ and $\mathbb CP^1$ as you wrote. – user46372819 Oct 9 '18 at 4:53
• @AlJebr The charts do make sense, both $\psi_i$ are homeomorphisms onto all $\mathbb{C}=\mathbb{R}^2.$ For instance $\psi_1^{-1}:\mathbb{C}\to U_1$ is given by $z\mapsto [1:z].$ – positrón0802 Jan 14 '19 at 16:12

The explicit map you are looking for is the following. Write $$S^2=\big\{(z,\alpha) \in \mathbb{C} \times \mathbb{R} \, \big| \, |z|^2+\alpha^2=1\big\}.$$ Then, employing the standard homogeneous coordinates on projective space, we have that the map \begin{align*} \varphi: S^2 & \to \mathbb{CP}^1 \\ (z,\alpha) & \mapsto [2z:1-\alpha] \end{align*} is a diffeomorphism.
• I'm not sure this is a valid map. This sends $(0+ 0i, 1)\mapsto [0:0]$, and $[0:0]$ is not a valid element of $\mathbb{CP}^{1}$. – Maximal Ideal Jun 3 '20 at 7:04
• @MaximalIdeal -- Very fair point. I suppose I should have written the map as $(z,\alpha) \mapsto [2z/(1-\alpha):1]$ and said that the map extends smoothly to one sending $(0,1)\mapsto[1:0]$. – Or Eisenberg Jul 10 '20 at 3:52