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.

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    $\begingroup$ I'd recommend you to take a look at this books.google.ca/books/about/… $\endgroup$ – Ehsan M. Kermani Nov 18 '11 at 1:57
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    $\begingroup$ 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 $\infty$ $\endgroup$ – Alphonse Nov 18 '11 at 2:26
  • $\begingroup$ @Alphonse: Yes, that's essentially right. More precisely, $1/z$ is the transition map between the two stereographic projection charts. $\endgroup$ – Jesse Madnick Nov 18 '11 at 2:28
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    $\begingroup$ 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 $f$. $\endgroup$ – Jesse Madnick Nov 18 '11 at 2:30
  • $\begingroup$ By the way: Often the Riemann sphere is denoted by $\hat{\mathbb{C}}$. $\endgroup$ – Dirk Nov 18 '11 at 9:54

Here is the most down-to-earth explanation I can think of: no evil algebraic geometry, no varieties!
Let me denote by $\mathbb P^1$ the Riemann sphere and write $\mathbb P^1=\mathbb C\cup \lbrace\infty \rbrace$.
Suppose you are given a map $f:\mathbb P^1 \to \mathbb P^1$ and you want to investigate it near $a\in\mathbb P^1$.

First case: $a\neq\infty , f(a) \neq \infty$
Then by restriction you get $f_0: U\to \mathbb C$ for some neighbourhood $U$ of $a$ and you can already handle that.

Second case: $a\neq\infty , f(a) =\infty$
The behavior of $f$ at $a$ is the same as that of $g=1/f$ and we have come back to first case.
Example: $f(z)=1/(z-2)^5$, with $a=2$ . Here $g(z)=(z-2)^5$. We say that $f$ takes value $\infty$ with multiplicity $5$ at $z=2$.

Third case: $a= \infty , f(a) =\infty$
Replace $f$ by $h(z)= \frac{1}{f(\frac {1}{z})}$ and you get a function $h:U\to \mathbb C$ whose behaviour at zero is by definition the behaviour of $f$ at $\infty$.
Example: if $f(z)=(z-2)^5$ , then $h(z)=\frac{z^5}{(1-2z)^5}$ and we say that $f$ has multiplicity $5$ at $\infty$, since $\frac{z^5}{(1-2z)^5}$ has a zero of order $5$ at zero.

Fourth case: $a= \infty , f(a) \neq \infty$
Replace $f(z)$ by $f(1/z)$. I'll skip the details.

Important remark
There are no genuine meromorphic maps $\mathbb P^1 \to \mathbb P^1$: every such meromorphic map is actually holomorphic. The same is true for meromorphic maps between compact Riemann surfaces. This is underappreciated and sometimes misunderstood fact. I actually wrote this answer principally to emphasize this subtle point!

Edit In order to address some comments, let me clarify the above remark.
Given a connected open subset $X \subset \mathbb C$ or even an arbitrary Riemann surface $X$ (compact or not, connected by definition) there is a bijective correspondence between :
1) The set of meromorphic functions $f$ on $X$.
2) The set of holomorphic maps $F: X\to \mathbb P^1$ that are not constantly equal to $\infty$
The correspondence associates to $f$ its extension $F$ obtained by decreeing that at a pole $p\in X$ of $f$ we define $F(p)=\infty$.
I have gone into this discussion because Alphonse wrote in his question "meromorphic functions from (sorry, don't know the notation) Riemann Sphere to itself". My point is that you should only talk about holomorphic maps from from the Riemann sphere to itself.
Finally note that the bijective correspondence above only holds in dimension 1: there is no way you can consider the meromorphic function $f(z,w)=z/w$ on $\mathbb C^2$ as a holomorphic map $F:\mathbb C^2 \to \mathbb P^1$

  • $\begingroup$ In your final important remark, do you really need that the domain is compact? $\endgroup$ – Willie Wong Nov 18 '11 at 9:55
  • $\begingroup$ What do you mean by "there are no genuine meromorphic maps"? "Meromorphic" just means "holomorphic as a function taking values in the Riemann sphere", or equivalently "holomorphic (as a function taking values in $\mathbb{C}$) with isolated poles". What would it even mean for a function $\mathbb{P}^1\to\mathbb{P}^1$ to be "meromorphic"? $\endgroup$ – Lasse Rempe-Gillen Nov 18 '11 at 9:56
  • $\begingroup$ @mathstribble: By meromorphic map $X\to Y$ between analytic spaces, I mean what complex analytic geometers call meromorphic maps.They are defined in terms of an analytic subspace of $X\times Y$. The precise definition is given for example in Ueno's Classification of algebraic varieties and compact complex spaces, (Springer LNM 439) page 13. Anyway, here I wanted to emphasize that we do not need the concept of meromorphic map, which answers Alphonse's question "I hear of analytic or meromorphic functions from (sorry, don't know the notation) Riemann Sphere to itself" $\endgroup$ – Georges Elencwajg Nov 18 '11 at 10:18
  • $\begingroup$ Dear @Willie, no the domain needn't be compact. I'll add a few words in an edit to take your remark into account.Thanks! $\endgroup$ – Georges Elencwajg Nov 18 '11 at 10:22

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