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Let $C \to \mathbf P^1$ be a hyperelliptic curve of genus $g \ge 2$ obtained as a double cover of $\mathbf P^1$ branched at $r$ points. Let $\tilde U\subset C$ be its open subset obtained by removing all ramification points and $U\subset \mathbf P^1$ its pre-image under the double cover map. What are invertible functions on $\tilde U$? In other words, I need to compute $H^0(\tilde U, \mathbf G_m)$.

Surely it is possible to pull-back any invertible function from $\mathbf P^1$, so we have at least the group $H^0(U, \mathbf G_m) = \mathbf C^\times \times \mathbf Z^{r-1}$. Are there any other? If not, how to show this?

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Since you wrote $\mathbb{C}^*$, I'll assume you are in characteristic zero. Then $r=2g+2$ and the group you want (modulo constants) has rank $r-1$ but is bigger than the group coming from $\mathbb{P}^1$. If the curve is given by $y^2=f(x)=\prod (x-a_i), \deg f = 2g+1$, then $y$ is also on your group. I think that's it, the group modulo constants is generated by $y,x-a_1,x-a_2,\ldots$ subject to the relation $y^2=f(x)$. Yes, you can prove that there are no more functions using Clifford's theorem.

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  • $\begingroup$ Thank you very much! It is clear now what I have been missing. $\endgroup$
    – lime
    Jun 22, 2014 at 21:38

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