# $\mathbb{CP}^1$ is compact?

$\mathbb{CP}^1$ is the set of all one dimensional subspaces of $\mathbb{C}^2$, if $(z,w)\in \mathbb{C}^2$ be non zero , then its span is a point in $\mathbb{CP}^1$.let $U_0=\{[z:w]:z\neq 0\}$ and $U_1=\{[z:w]:w\neq 0\}$, $(z,w)\in \mathbb{C}^2$,and $[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$ is a point in $\mathbb{CP}^1$, the map is $\phi_0:U_0\rightarrow\mathbb{C}$ defined by $$\phi_0([z:w])=w/z$$ the map $\phi:U_1\rightarrow\mathbb{C}$ defined by $$\phi_1([z:w])=z/w$$ Now,Could any one tell me why $\mathbb{CP}^1$ is the union of two closed sets $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$, where $D$ is closed unit disk in $\mathbb{C}$,and why $\mathbb{CP}^1$ is compact?

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What is the topology over $\mathbb{CP}^1$? – Seirios Jul 25 '12 at 9:31
Dear Patience, seeing that you have asked 11 questions (about Riemann surfaces) in two says, you might like to know that there is a limit of 50 questions you can ask per month: see here – wildildildlife Jul 25 '12 at 12:10

The maps you gave are the coordinate charts on $\mathbb{C}\mathbb{P}^1$ that makes it into a manifold. In particular, they are bijective.

If we take $\phi_0^{-1}(D)$, we get all point $[1:z]$, with $z\in D$. Similarly, $\phi_0^{-1}(D)$ is all points of the form $[z:1]$. Together, these sets cover $\mathbb{C}\mathbb{P}^1$. The inverse maps are continuous, because the maps $\phi_i$ are bijective, so $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$ are compact, as the image of a compact set under a continuous map is compact. It's not hard to show that if a space is a union of two compact sets, it is compact, so we are done.

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these are holomorphic bijection right? – Un Chien Andalou Jul 25 '12 at 9:40
Which page is this on in Miranda? There's something I want to check. – Potato Jul 25 '12 at 9:42
page-8&9.The Projective Line – Un Chien Andalou Jul 25 '12 at 9:43
I think your argument is ok! – Un Chien Andalou Jul 25 '12 at 9:45
Ok, nevermind, this is right. To answer your question: it doesn't make sense to speak of coordinate charts being holomophic bijections. You want the transition functions to be holomorphic. – Potato Jul 25 '12 at 9:47

The complex projective line is the union of the two open subsets $$D_i=\{[z_0:z_1]\text{ such that }z_i\neq0\},\quad i=0, 1.$$ Each of them is homeomorphic to $\Bbb C$, hence to the open unit disk.

Compacteness follows from the fact that it is homeomorphic to the sphere $S^2$. You can see this as a simple elaboration of the observation that $$\Bbb P^1(\Bbb C)=D_0\cup[0:1].$$

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The first has been answered, but i would like to add some details. So you do see that the $U_i = \phi_i^{-1}(\mathbb{C})$ for $i=0,1$ cover $\mathbb{C} P^1$ right? My claim is now that we can restrict to the closed disks in $\mathbb{C}$. The cases $[0:1]$ and $[1:0]$ are simple and the only points outside $U_{0,1} := U_0 \cap U_1$, so we restict to $[z:w] \in U_{0,1}$. Now both $\phi_i$ are defined at this point, and since $\phi_0([z:w]) = \frac{1}{\phi_1([z:w])}$ we can pick the one that has norm smaller or equal to one, so that the image lies in the unit disk. Since the point was arbitrary, every point has as $i$ such that its $\phi_i$ image lies in the unit disk, q.e.d.

Regarding the second one, recall that the continuous image of a compact space is compact. Moreover it is well known that the sphere $S \subset \mathbb{C}^2$ is compact, and that $\mathbb{C}P^1$ is a quotient of this sphere. Summarizing, there is a surjective quotient map $\pi: S \rightarrow \mathbb{C}P^1$, hence $\mathbb{C}P^1$ is the image of a compact space, hence compact.

The reason i add the second answer is that it is insightful and easily generalized to complex or real projective spaces of any dimension, even to Grassmanians (although that requires some adjustments) and i believe also you can replace $\mathbb{C}$ by any normed division algebra over $\mathbb{R}$.

Hope this helps. Joachim

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By the way, note that the spheres are not complex submanifolds of $\mathbb{C}^n$, we just regard them as topological spaces here. Since compactness is purely topological this suffices. – Joachim Jul 25 '12 at 10:01
thank you for the detail and informative answer. simply pleased and delighted – Un Chien Andalou Jul 25 '12 at 10:01
Happy to hear that =) – Joachim Jul 25 '12 at 10:03

Yes.

In fact, the complex projective line "is" the Riemann Sphere.

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The argument you have outlined appeals to the following lemma of general topology.

Lemma. Let $X$ be a topological space. If there are (quasi)compact subsets $Y_1, \ldots, Y_n$ such that $Y_1 \cup \cdots \cup Y_n = X$, then $X$ is itself (quasi)compact.

Proof. Let $\{ U_\alpha : \alpha \in I \}$ be an open cover of $X$. For each $i$, $\{ U_\alpha \cap Y_i \}$ is an open cover of $Y_i$, so has a finite subcover $\{ U_{\alpha_j} \cap Y_i : \alpha_j \in J_i \}$. Thus we have a finite subcover $\{ U_{\alpha_j} : 1 \le j \le n, \alpha_j \in J_i \}$ for $X$.

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