$\mathbb CP^1 \approx S^2$ proof check

Question

I wanted to give a whole proof of this fact as I was not able to find a detailed one myself. I have the feeling that such a proof has been asked quite frequently by several users and I hope this may help other students as me who are touching on this topic for the first time.

For $\mathbb CP^1$ we take the definition $(\mathbb C^2-\{0\})/(z\tilde{}\lambda z)$ for any $\lambda$ non zero complex number.

Define $$F: \mathbb R^4 \to S^2/\tilde{}_{antipodal}, \space (x_1,x_2,x_3,x_4) \mapsto [\frac{1}{\Vert(x_1,x_2,x_3)\Vert_2}(x_1,x_2,x_3)] $$

Observe that $$F(\lambda x)=[\frac{\lambda}{\vert \lambda \vert\ Vert(x_1,x_2,x_3)\Vert_2}(x_1,x_2,x_3)]=\{\stackrel{+}-\frac{1}{\Vert(x_1,x_2,x_3)\Vert_2}(x_1,x_2,x_3)\}=[\frac{1}{\Vert(x_1,x_2,x_3)\Vert_2}(x_1,x_2,x_3)]=F(x)$$

hence $F$ descend to $\tilde{F}: \mathbb R P^3 \to S^2/\tilde{}_{antipodal}$ and is continuous since $F$ is continuous. $\tilde{F}$ is obviously surjective and also injective since $\tilde{F}([x])=\tilde{F}([y]) \implies x \space\tilde{}_{antipodal}\space y$ and therefore $x=-y \implies [x]=[y]$. The domain is compact and the target is Hausdorff so we obtain $$\mathbb R P^3 \approx S^2 / \tilde{}_{antipodal}$$By some similiar argument one can show that $S^2 \approx S^2/\tilde{}_{antipodal}$ and hence $\mathbb R P^3 \approx S^2$ and so we conclude $$\mathbb C P^1 \approx S^2$$

Question: Is this approach of viewing $\mathbb C P^1$ as $\mathbb R P^3$ right? Is the argument right, are there any flaws?

There is definitely a problem with this: $\mathbb{RP}^2$ is not homeomorphic to $S^2$. — Simon Rose, Feb 02 '16 at 13:23
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable — Riccardo, Feb 02 '16 at 13:25
$\mathbb{RP}^3$ is three dimensional. $\mathbb{CP}^1$ is two-dimensional. — Cheerful Parsnip, Feb 02 '16 at 13:29
there is something wrong here as well. I mean $\mathbb{R}P^3$ has fundamental group $\mathbb{Z}/2\mathbb{Z}$, but $S^2$ is simply connected — Riccardo, Feb 02 '16 at 13:31
Your $F$ is not defined on (and does not extend continuously over) the fourth coordinate axis, i.e., at the poles $\pm(0,0,0,1)$ in $S^{3}$. — Andrew D. Hwang, Feb 02 '16 at 13:57
If you'll pardon self-promotion, there's a possibly-helpful diagram and sketch of a proof at Flag manifold to Complex Projective line. — Andrew D. Hwang, Feb 02 '16 at 14:02

score 2 · Accepted Answer · answered Feb 02 '16 at 13:27

2

Personally, the way I think of this is to look at the following decomposition of $\mathbb{C}^2$ into two sets, and see how they fit together into the quotient.

Let $U = \{(x, y) \in \mathbb{C}^2 \mid y \neq 0\}$ and let $P = \{(x, 0) \in \mathbb{C}^2\}$. Note that these are disjoint and their union is all of $\mathbb{C}^2$.

When we quotient $P$ by the action of $\mathbb{C}^\times$, we get a single point (since $P$ can be easily identified with $\mathbb{C}^\times$ in a way that preserves the action). When we quotient $U$ by the action of $\mathbb{C}^\times$, one can show that it is homeomorphic to $\mathbb{C} \cong \mathbb{R}^2$. So the resulting quotient can be seen as the union of the plane with one point.

Can you see where to go from there?

answered Feb 02 '16 at 13:27

Simon Rose

6,793

Are you heading towards the stereographic projection? – noctusraid Feb 02 '16 at 13:28
Basically, yes. – Simon Rose Feb 02 '16 at 13:29
Well we can extend the stereographic projection $p: S^2-{0} \to \mathbb R^2$ to $\tilde{p}: S^2 \to \mathbb R^2 \cup {\infty}$ homeomorphically. – noctusraid Feb 02 '16 at 13:37

score 0 · Answer 2 · answered Mar 20 '19 at 16:41

Considering the CW complex structure of $\Bbb C P^1$ , note that $\Bbb C P^1$ is obtained from $\Bbb C P^0$ by attaching a $2-cell$ but $\Bbb C P^0 \cong *$ , thus $$\Bbb C P^1 \cong D^2 / \partial D^2 \cong S^2 $$

$\mathbb CP^1 \approx S^2$ proof check

2 Answers2