5
$\begingroup$

Everyone: This is my first post. Sorry if I break some protocol.

I do know some complex analysis and how to tell when a function from $\mathbb C \rightarrow\mathbb C$ . But I am confused when I hear of analytic or meromorphic functions from (sorry, don't know the notation) Riemann Sphere to itself. I think this has something to see with Algebraic Geometry and Varieties, of which I know very little. Would someone please expand on how one determines if/when a function from the Riemann Sphere to itself is meromorphic or analytic? I have seen some Differential Geometry in which we compose functions with chart maps to determine if a function (from a real manifold to another real manifold) is differentiable, or $C^k$. Is that what we do for complex functions, and, if so, are there some theorems to avoid doing the chart composition? Thanks for any help.

  • 2
    I'd recommend you to take a look at this http://books.google.ca/books/about/Algebraic_curves_and_Riemann_surfaces.html?id=qjg6GOQaHNEC&redir_esc=y2011-11-18
  • 2
    Thanks, ehsanmo, I guess it does come down to composing with coordinate patches so as to turn the map into a map between $\mathbb C$ to itself. I thought I remembered something about composing with $1/z$ ; maybe this $1/z$ is the chart map for the chart that contains $\infty$2011-11-18
  • 0
    @Alphonse: Yes, that's essentially right. More precisely, $1/z$ is the transition map between the two stereographic projection charts.2011-11-18
  • 1
    That is, if the "bottom" stereographic projection chart is $\phi_1$ and the "top" one is $\phi_2$, then $(\phi_1 \circ \phi_2^{-1})(z) = 1/z$. In particular, this means that $$(f\circ \phi_2^{-1})(z) = (f \circ \phi_1^{-1}) \circ (\phi_1 \circ \phi_2^{-1})(z) = (f \circ \phi_1^{-1})(1/z),$$ and we usually _identify_ $f \circ \phi_1^{-1}$ with $f$.2011-11-18
  • 0
    By the way: Often the Riemann sphere is denoted by $\hat{\mathbb{C}}$.2011-11-18
  • 0
    A [video](http://www.youtube.com/watch?v=JX3VmDgiFnY) a day keeps the doctor away :P2011-11-18
  • 0
    Great link, percusse. And it is heart-warming that a mathematical video has been watched 1 868 012 times.2011-11-18
  • 0
    @GeorgesElencwajg Thanks and I agree. But I think that a small 68000 part of that number is just due to me ;)2011-11-18

1 Answers 1