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Out of curiosity I have started a tutorial in Riemann surfaces. But since I am mostly trained in (stochastic) analysis a lot of the prelimenaries from differential geometry and algebraic geometry are not always clear or known to me. (So you can probably expect more questions.) But I hope to pick things up along the way. For instance I have dificulties understanding the following sentences:

Since $P^1$ is compact, we can view $P^1$ as the one-point compactification of $U_0 \backsimeq \mathbb{C}$. With this interpretation in mind , $P^1$ is also called the Riemann sphere.

Here $P^1$ is the complex projective line corresponding to $\mathbb{C}^2$ and $U_0=P^1\setminus\{[0:1]\}$. Where $[0:1]\in P^1$ is the equivalence class of all two dimensional complex vectors that differ a nonzero complex constant from vector $(0,1)$. We also have $U_1=P^1\setminus\{[1:0]\}$. I understand that the map $\pi:\mathbb{C^2}\setminus\{0\}\rightarrow P^1$, defined by $\pi(z,w)=[z:w]$ induces the quotient topology on $P^1$ In particularly this makes $\pi$ a surjective continuous funtion. Since the unit sphere $S^3$ in $\mathbb{C}^2$ is compact, we also have that $P^1=\pi\left(S^3\right)$ is compact. I also see how we can assign a differential structure to $P^1$.

So far I think I am good, but I don't understand how this precisely gives the other statements in the cursive sentences. I take it that `$U_0 \backsimeq \mathbb{C}$' means that $U_0$ is homeomorphic (even diffeomorphic) with $\mathbb{C}$. So I understand it's enough to check that $P^1$ is indeed the one-point compactification of $U_0$. Am I correct? How does this work?

Kindly appreciated.

Aris
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  • I'm not sure exactly what you're asking. Are you defining the Riemann sphere to be the one-point compactification of $\mathbb{C}$? Are you comfortable with the concept of one-point compactification? Have you proved (or convinced yourself of) the statement $U_0 \simeq \mathbb{C}$? – Rhys Jul 01 '13 at 13:15

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You go astray when you start talking about the unit ball $\overline{B(0,1)}$. This does not surject onto the projective line, and the compactness of the closed ball has nothing to do with the compactness of the projective line.

Every point of the projective line can be represented by either $[a,1]$ for arbitrary complex $a$, or $[1,0]$. The last point is the "point at infinity" and the point referred to in the expression "one-point compactification". In the real case things are a little easier to visualize; here $\mathbb{R}$ is compactified by adding a single point at infinity, to form a circle. The complex case is very similar.

Mikhail Katz
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  • Maybe you want to consider the unit sphere $S^3\subset\mathbb C^2$? – Ted Shifrin Jun 30 '13 at 16:51
  • That would certainly be sufficient for capturing all the directions, but the unit ball $B\subset \mathbb{C}^2$ is certainly not suitable since its center does not even define a point of the projective line, and therefore its compactness does not explain the compactness of the projective line. – Mikhail Katz Jul 01 '13 at 11:05
  • I think that this problem solved by means of what Ted suggested. However my point concerning the one-point compactification remains... – Aris Jul 01 '13 at 13:02