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By definition of complex manifold, a complex manifold is a manifold with holomorphic charts $U \to D^2 \subseteq \mathbb C$.

I want to define a complex structure on $S^2$.

Can you tell me if this is correct?

Let $D^+$ and $D^-$ denote $S^2-S$ and $S^2 -N$ respectively where $S,N$ are the north and south pole. Define charts $f_+$ and $f_-$ in the obvious way: map $D^+$ and $D^-$ homeomorphically to the open unit disk. Then $\{(D^+,f_+), (D^-, f_-)\}$ is a complex atlas (complex structure) for $S^2$.

a student
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    Have you computed the transition functions? You'll need to know the maps $f_{\pm}$ explicitly. – Michael Albanese Dec 30 '14 at 02:43
  • You did not specify which homeomorphisms you chose. Wrong choice will not give you a complex structure. – Moishe Kohan Dec 30 '14 at 02:44
  • @studiosus I am unsure about how to write down the maps. They look like identity maps... can I choose $F_\pm$ to be the identity maps? Then it would be easy to prove that the transition maps are holomorphic. – a student Dec 30 '14 at 02:45
  • You have to decide how do you describe the sphere. There are at least three standard ways, in neither one of them your charts will be the identity maps. – Moishe Kohan Dec 30 '14 at 02:48
  • "Neither" means "none". In other words, you will not get identity transition maps in any of the three standard models. – Moishe Kohan Dec 30 '14 at 03:00

2 Answers2

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I'm going to suggest that you choose the chart that sends $P = (x, y, t)$ in $S^2 - S$ to the intersection $z$ of the segment $SP$ with the $t = 0$ plane. Do the same for the $S^2 - N$, but throw in a conjugate. Those are your charts $f_{\pm}$. Now write out the transition function, which should end up being something like $z \mapsto \frac{1}{z}$.

John Hughes
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  • Thank you but if the transition maps are ${1\over z}$ then that's terrible, they are not holomorphic. The comment above by studiosus seems to say that the transition maps should be the identity maps. Am I misunderstanding something? – a student Dec 30 '14 at 02:52
  • @astudent The transition maps are $\frac{1}{z}$, and these are holomorphic where they are defined. You omit "0" from the domain of each one. – Steven Gubkin Dec 30 '14 at 02:59
  • I should have said "and the transition functions will be holomorphic on their domains, which are $\mathbb C - {0}$ in each case." – John Hughes Dec 30 '14 at 03:01
  • Hm... why is the conjugate needed? I used stereographic projection... maybe it's better if I post my solution in a new question... – a student Jan 01 '15 at 02:35
  • @StevenGubkin Thanks, I understand it now. – a student Jan 01 '15 at 05:55
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Look up the standard homeomorphism you get by removing the "point at infinity" in $S^2$ and the complex plane. Two charts using this homeomorphism are enough to give you the complex manifold charts.

user203864
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