0
$\begingroup$

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 $S^2$.

$\endgroup$
  • 1
    $\begingroup$ Have you computed the transition functions? You'll need to know the maps $f_{\pm}$ explicitly. $\endgroup$ – Michael Albanese Dec 30 '14 at 2:43
  • $\begingroup$ You did not specify which homeomorphisms you chose. Wrong choice will not give you a complex structure. $\endgroup$ – Moishe Kohan Dec 30 '14 at 2:44
  • $\begingroup$ @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. $\endgroup$ – a student Dec 30 '14 at 2:45
  • $\begingroup$ 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. $\endgroup$ – Moishe Kohan Dec 30 '14 at 2:48
  • $\begingroup$ "Neither" means "none". In other words, you will not get identity transition maps in any of the three standard models. $\endgroup$ – Moishe Kohan Dec 30 '14 at 3:00
2
$\begingroup$

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 $z \mapsto \frac{1}{z}$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – a student Dec 30 '14 at 2:52
  • $\begingroup$ @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. $\endgroup$ – Steven Gubkin Dec 30 '14 at 2:59
  • $\begingroup$ I should have said "and the transition functions will be holomorphic on their domains, which are $\mathbb C - \{0\}$ in each case." $\endgroup$ – John Hughes Dec 30 '14 at 3:01
  • $\begingroup$ Hm... why is the conjugate needed? I used stereographic projection... maybe it's better if I post my solution in a new question... $\endgroup$ – a student Jan 1 '15 at 2:35
  • $\begingroup$ @StevenGubkin Thanks, I understand it now. $\endgroup$ – a student Jan 1 '15 at 5:55
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.