Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I am reading the book From holomorphic functions to complex manifolds by Klaus Fritzsche and Hans Grauert. I have a question about a fiber bundle. On page 186, the last line. How to show that $$ \Gamma(U, \mathcal{O}^*_{X}) \cong \mathcal{O}^*(U):=\{f\in \mathcal{O}(U) : f(x) \neq 0 \text{ for every } x\in U\}? $$ Thank you very much.

share|cite|improve this question
This follows almost immediately from the definitions. Note that (1) a section of a trivial bundle $U\to U\times F$ is the same as a map $U\to F$, and (2) there is a natural inclusion $\mathbb C^* \to \mathbb C$. – Aaron Oct 11 '11 at 3:40
@Aaron, thank you very much. – LJR Oct 11 '11 at 17:56
up vote 1 down vote accepted

To get this off the Unanswered list, I'll restate Aaron's explanation as an answer:

The notation is $\mathbb C^*=\mathbb C\setminus \{0\}$ and $\mathcal{O}_X^* = X\times \mathbb C^*$, a trivial fibre bundle. By definition of a section, an element of $\Gamma(U,\mathcal{O}_X^*)$ is a holomorphic map $s:U\to \mathcal{O}_X^*$ such that $\pi\circ s = \mathrm{id}_U$. The fibres being $\mathbb C^*$, what we have is a nonvanishing holomorphic function.

share|cite|improve this answer

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.