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I'm trying to give a more or less easy proof of Serre duality on Riemann surfaces (if you have any hint, a part from Otto Forsters book, go ahead). I have some notes where it says that Cech cohomology on a compact RS with coeficients on a line bundle (for instance $O_D$) can be computed using any open covering of proper open sets, is this really true? That is, by Leray the question is: how can any line bundle be acyclic on a (proper) open set of X? I think it is true also for any non compact RS.

PD:We can use this to define the map residue without going through Mittag Leffer and so on (I think it clarifies things)

Edited:

Ok, here is a proof, do you see any mistakes?:

For any non compact Riemann surface $S$, $H^1(S,O_D)=0$.

Let $\{p_1,p_2,\dots\}\in S$ be a discrete set of infinitely many points (which exists since $S$ is non-compact) and consider the divisor $D'=\sum_{i=1}^{\infty}(-1)p_i$. Let $T$ be defined by the exact sequence of sheaves: $$0\rightarrow O_{D'}\rightarrow O\rightarrow T\rightarrow 0$$ Then $T$ is a skyscraper sheaf in the points $\{p_1,p_2,\dots\}$ (so $H^1(S,T)=0$ and $T(S)$ is infinite dimensional). From the long exact sequence of cohomology we obtain the exact sequence: $$O(S)\rightarrow T(S)\rightarrow H^1(S,O_{D'})$$ where since $\dim_{\mathbb{C}}H^1(S,O_{D'})<\infty$ we deduce that $O(S)$ is infinite dimensional. Thus we can consider a non constant $f\in O(X)$ and from the exact sequence: $$0\rightarrow O_D\rightarrow O_D\rightarrow R\rightarrow 0$$ (multiplication by $f$ in $O_D$) we deduce that $H^1(\cdot f):H^1(S,O_D)\rightarrow H^1(S,O_D)$ is surjective. If $H^1(S,O_D)\neq 0$ then $H^1(\cdot f)$ has an eigenvalue $\lambda\in\mathbb{C}$ but then $H^1(\cdot f-\lambda)$ can't be surjective.

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Ok thanks for the corrections. What I meant with that sentece is that if you have any other reference with a "sorter" proof it would be nice to post it. –  Miguel Dec 6 '12 at 17:31
    
Yes, What I meant was -D', otherwise it wouldn't make sense. Anyway, after solving this I can't see why the proof doesn't hold –  Miguel Dec 6 '12 at 17:43
    
I think your $S$ is compact and $D=S\setminus \{p_1, p_2,...\}$. –  user18119 Dec 20 '12 at 11:03
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1 Answer 1

Note that any open Riemann surface $D$ is a Stein space, so any coherent sheaf on $D$ is acyclic.

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