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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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    Ok now I see, thanks. By the way yes, my course was closer to differential geometry. I will try with algebraic geometry next semester!2012-06-10