Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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 $2g-2=2\cdot0-2=-2$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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