# Why is it biholomorphic?

The proof of Theorem 13.5 in "Lectures on Riemann Surfaces" by Otto Forster begins by saying

Set $U_1:={\mathbb P}^1 \backslash \infty$ and $U_2:={\mathbb P}^1 \backslash 0$. Since $U_1 = {\mathbb C}$ and $U_2$ is biholomorphic to ${\mathbb C}$, it follows from (13.4) that $H^1(U_i, {\cal O})=0$.

In "1.5 Examples of Riemann Surfaces" in the book the maps $\phi_i:U_i \rightarrow {\mathbb C}, i=1,2$ are defined as follows:

$\phi_1$ is the identity map and $$\phi_2(z) := \left\{ \begin{array}{ll} 1/z & \mbox{for} \; z \in {\mathbb C}^*\\ 0 & \mbox{for} \; z = \infty \end{array} \right.$$

But, It seems to me that $\phi_2$ cannot be biholomorphic at $\infty$, because since $\phi_2'(z)=-1/z^2$,

$$\lim_{z\rightarrow \infty} \phi_2'(z) = 0.$$

Could someone point out where I made a mistake ?

• Yes, the limit is correct. But how does this produce any contradiction? – Zhen Lin Jan 13 '12 at 8:43
• I thought that $\phi_2'(\infty)\neq 0$ is a necessary condition for the existense of the holomorphic inverse mapping of$\phi_2$ at $\infty$. – Aki Jan 13 '12 at 9:49

The "paradox" you are asking about is extremely interesting and I can only congratulate you on the dynamic way you are studyng mathematics.

1) The first confusing point is that for a holomorphic function $\phi$ on an open subset $U$ of a manifold, in your case $\phi_2$ and $U_2$ , the naïve notion of derivative $\phi'(a)$ at a point $a\in U$ as a number does not work: you would get different numbers according to the chart you use.
The correct notion is that of a linear form on the tangent space $$d_a\phi:T_a (U) \to T_{\phi(a)} \mathbb R = \mathbb R$$
The recipe for computing $d_a\phi$ is to choose a chart $w$ in a neighbourhood of $a$, to consider the composed function $\phi_w=\phi \circ w^{-1}$ and to decree that we have $$d_a\phi (t\cdot \frac {\partial}{\partial w}) =t\cdot \phi_w'(w(a)) \quad (t\in \mathbb R)$$
If you do that in your situation with $U=U_2, a=\infty, \phi=\phi_2=w$, you will find completely tautologically that $d_\infty (\phi_2):T_\infty (\mathbb P^1)\to \mathbb R$ is given by $d_a\phi_2 (t\cdot \frac {\partial}{\partial w})= t$, since $(\phi_2)_w=\phi_2 \circ w^{-1}$ is the identity.

2) The second confusing point is that you are not allowed to calculate $d_\infty\phi_2$ by means of the chart $\phi_1=z$ since its domain does not contain infinity: $\infty\notin U_1=dom(\phi_1)=\mathbb C$.

3) In the language of divisors (introduced on page 127 of your book) the divisor of the global meromorphic differential form $dw\in \Gamma ( \mathbb P^1, \Omega_X ^1 \otimes_{\mathcal O_X} \mathcal M_X)$ is $div(w)=-2\cdot (0)$ and for $dz\in \Gamma ( \mathbb P^1, \Omega_X ^1 \otimes_{\mathcal O_X} \mathcal M_X)$ it is $div(z)=-2\cdot (\infty)$.
Both results confirm that the line bundle of holomorphic $1$-forms on $\mathbb P^1$, a Riemann surface of genus $g=0$, has degree $2g-2=2\cdot0-2=-2$.

• But $\phi_2$ is a chart for $U_2$, isn't it? The only condition we need is that $\phi_2$ is a homeomorphism and the transition map $\phi_1 \circ \phi_2^{-1}$ is biholomorphic, and these hold. – Zhen Lin Jan 13 '12 at 11:41
• Dear @Zhen Lin, you are absolutely right: I had mixed up two charts in my preceding answer. I have edited my post now and I am very, very grateful to you for your attentive reading. Thanks a lot! – Georges Elencwajg Jan 13 '12 at 18:11
• Thank you for your detailed explanation. Then, I made a mistake in using the chart $(U_1,z)$ to evaluate the function $\phi_2$. And, each chart $\psi:U \rightarrow V$ on any Riemann surface is biholomorphic since $\psi \circ \psi^{-1} = \mbox{id}_V$ is biholomorphic. Am I right ? – Aki Jan 15 '12 at 7:46