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in continuation of my previous post: Question Regarding Riemann-Hurwitz Formula Proof

I need another help in understanding how to compute Cohomology of Sheaves (and right derived functors): Given a compact Riemann Surface $X$ , and the sheaf $O_X(U):= \{f:U \to \mathbb{C} |f -holomorphic \} $ , I want to compute the cohomology groups $H^n (X, O_X) = R^n \Gamma (O_X) $ where $\Gamma$ is the global section functor: $ \Gamma(O_X) = O_X(X) $ .

As far as I know, this computation should give: $H^0 (X,O_X) = \mathbb{C} $ , $H^1(X,O_X) = \mathbb{C} ^ {2g} $ , $H^2(X,O_X) = \mathbb{C} $ and the other cohomology groups vanish.

I can't figure out how to get these results, and I can't find an appropriate example for such a calculation.

Can someone help me understand how to compute such a thing?

Thanks!

Here is what I did for the n=0 case: So we obviously know that given a functor $F$ , we have $ R^0 F(X) = F(X)$ which implies: $H^0 (X,O_X) = \Gamma(O_X) = \{f:X\to \mathbb{C} | f-analytic \}$ Why is is excatly $\mathbb{C}$ ? In order to continue the calculation for bigger n's, I need to know what is my exact sequence and what are the maps betweenthe elements of this sequence. I hope someone will be able to help me figure it out

Thanks a lot again !

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    $\begingroup$ You want $H^1(X,O_X)=\mathbb{C}^g$, where $g$ is the genus, This is often taken as the definition of genus. Also $H^i(X,O_X)=0$ for $i>1$ including $2$. Anyway, these results are not obvious without a bit of study. You should probably consult a book on Riemann surfaces, such as Forster's. $\endgroup$
    – Donu Arapura
    Commented Jun 12, 2012 at 20:23
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    $\begingroup$ I forgot to say $H^0(X,\mathbb{C})=\mathbb{C}$ is the one thing that can be explained easily. Since $X$ is compact, the modulus of any section $|f|$ attains a maximum. Now use the maximum principle to conclude that it is constant.t. $\endgroup$
    – Donu Arapura
    Commented Jun 12, 2012 at 20:35
  • $\begingroup$ Your question has already been answered, but let me add a comment. It seems that you are confusing the cohomology groups $H^i(X,\mathcal O_X)$ and $H^i(X,\mathbb C)$. The latter ones have dimensions like what you write. To learn the relationship between the two types of cohomology group, you need a bit of Hodge theory. $\endgroup$
    – Dan Petersen
    Commented Jun 13, 2012 at 12:42

1 Answer 1

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$H^0 (X;O_X)=C$ holds by the maximum principle for holomorphic functions since your $X$ is assumed to be compact.

$H^i (X;O_X)=0$ for $i >1$; because $O_X \to C^{\infty} \stackrel{\bar{\partial}}\to \Omega^{0,1} \to 0$ is a fine resolution.

$H^1 (X;O_X)$ has dimension $g$ (not $2g$). This is a hard result that depends on the Hodge theorem, or another version of some elliptic PDE theory. More specifically, any de Rham cohomology class $H^1 (X;C) \cong C^{2g}$ (the last isomorphism by topology) has a unique harmonic representative. Any harmonic $1$-form is uniquely the sum of a holomorphic and an antiholomorphic form. Since the conjugate of a holomorphic form is antiholomorphic, the space of holomorphic forms is $g$-dimensional. On the other hand, this is $H^0 (X;K_X)$, which by Serre duality is dual to $H^1 (X;O_X)$.

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  • $\begingroup$ Thanks a lot ! Can you explain me just what do you mean by a "fine resolution"? Thanks ! $\endgroup$
    – jason mfash
    Commented Jun 13, 2012 at 6:13
  • $\begingroup$ A resolution consisting of fine sheaves (for their definition, consult Wikipedia, or Bredon, or any other book on sheaf theory). $\endgroup$
    – Martin Brandenburg
    Commented Jun 13, 2012 at 7:09

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