# Dimension of $H^1(S,E)$

Let $S$ be a Riemann surface of genus $g$, $p\in S$ and $E$ be the holomorphic line bundle associated with the divisor $p$. This means that $E$ admits a section $\sigma$ with a simple zero at $p$ and non vanishing everywhere else. Does this imply that the $\bar{\partial}$ cohomology group $H_{\bar{\partial}}^1$ is trivial?

I think so, because then you can consider ${\sigma}^{-1}$, and this is a section with a simple pole at $p$. Do you agree with me?

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No, I'm afraid I don't agree with you.

The cohomology vector space $H^1(S,E)$ is dual to $H^0(S,K_S(-p))$ by Serre duality.
Since the bundle $K_S(-p)$ has degree $2g-3$, that bundle has indeed no section $\neq0$ for $g=0$ or for $g=1$.
But for larger values of $g$ there is no reason why it can't have non-zero sections.
On the contrary: for $g\geq 3$, Riemann-Roch guarantees that $\dim H^0(S,K_S(-p))\geq (1-g)+(2g-3)\gt 0$ and consequently that $H^1(S,E)\neq 0$.

NB
You use $H^1_{\bar \partial}(S,E)$ but by Dolbeault's theorem that cohomology vector space is isomorphic to the $H^1(S,E)$ defined by Čech cohomology or by derived functor cohomology.

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Sorry, but isn't the degree of $K_s(-p) = -1$ , no matter what the genus is? – Abramo Jun 10 '12 at 8:39
Dear Abramodj, if $E$ is a line bundle of degree $d$, then $E(-p)=E\otimes_ {\mathcal O} \mathcal O(-p)$ has degreee $d-1$. In the present case we apply this to $K_S$ which has degree $deg (K_S)=2g-2$ – Georges Elencwajg Jun 10 '12 at 8:58
So please explain what $K_s(p)$ is. I thought it was just the line bundle with one simple pole at $p$ (i.e. $\mathbb{C}\otimes O(p)$), and so in that case the degree would be just one. From what you say about Serre duality I guess $K_S(-p) = {T^*}^{1,0}M\otimes O(-p)$, right? – Abramo Jun 10 '12 at 9:04
Dear Abramodj: yes, that's right.The notation $K_S$ is standard in algebraic geometry for the bundle of algebraic (or holomorphic) differential forms. I suppose you learn Riemann surfaces in a setting closer to differential geometry and analysis. – Georges Elencwajg Jun 10 '12 at 9:11
Ok now I see, thanks. By the way yes, my course was closer to differential geometry. I will try with algebraic geometry next semester! – Abramo Jun 10 '12 at 11:37