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Can anyone give me a hint on how to start solving this problem?

Let $\Gamma \subset \mathbb{C}$ be a lattice. Show that the form $dz$ on $\mathbb{C}$ defines a holomorphic $1$-form on $\mathbb{C}/\Gamma$.

Any help appreciated!

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Where are you stuck? How did you define holomorphic 1-forms on $\mathbb{C}/\Gamma$ (a Riemann surface)? This is what you have to check. Do you know what the complex charts are? –  Gregor Bruns Nov 20 '12 at 14:47
    
@Gregor: Thanks for the help! The def. of hol. $1$-forms $\omega$ on a RS $X$: $\omega$ is hol. if, w.r.t. every chart $(U,z)$, $\omega$ may be written $\omega = f dz$ on $U \cap Y$, where $f \in \mathcal{O}(U \cap Y)$, $Y \subseteq X$, $Y$ open. Charts: Let $V \subset \mathbb{C}$ be open s.t. no two points are equiv. under $\Gamma$. Then $U:= \pi(V)$ open in $\mathbb{C}/\Gamma$, and $\pi| V \rightarrow U$ is a homeom., its inverse $\varphi :U \rightarrow V$ is a chart. But I'm not sure what I have to prove. Should I let $\varphi : U \rightarrow V$ be any chart, and find an appropriate $f$? –  Maethor Nov 20 '12 at 18:21

1 Answer 1

up vote 1 down vote accepted

Thank you for the information provided. Here are some little hints to get you started:

  • Your idea is the right one. Choose a chart, find the appropriate function, and prove it.
  • Your holomorphic $f$ will be in fact $1$.
  • Your $Y$ is in fact the whole torus $\mathbb{C}/\Gamma$.

The solution, if you write it down, should look almost trivial.

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Thanks again for the help! Any $f \in \mathcal{O}(\mathbb{C}/\Gamma)$ has to be a constant function. So a $1$-form $\omega$ on $\mathbb{C}/\Gamma$ is holomorphic if, w.r.t. every chart $(U,z)$, $\omega$ may be written $\omega = cdz$, where $c \in \mathbb{C}$. In my assignment $\omega = dz$. So I have to show that ... $f = c = 1$ will do the trick for any chart $(U,z)$? Is this correct? –  Maethor Nov 21 '12 at 10:13
    
Wait ... So would the solution to this problem be: Let the charts on $\mathbb{C}/\Gamma$ be defined as above, and let $(U,z)$ be any such chart, then $\omega = dz$ can be written $\omega = dz = 1 \cdot dz$ on $U$, where $1 \in \mathcal{O}(U)$, hence $\omega$ is a holomorphic $1$-form on $\mathbb{C}/\Gamma$? –  Maethor Nov 21 '12 at 10:37
    
Yes, that's what I thought the exercise was about :). Does this seem right to you? Your first comment does some kind of converse. –  Gregor Bruns Nov 21 '12 at 10:53
    
Yes, seems right to me. Thanks a lot for the help :) I might have said the converse in the first comment, I'm a bit new to these questions, and thus easily confused. Btw, I don't know if this is appropriate, but could you have a look at another of my questions? –  Maethor Nov 21 '12 at 12:01

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