I'm trying to solve the following problem.
Let $C$ be a smooth projective curve and let $\Omega^1_C$ be the canonical bundle. Show that there is a canonical isomorphism, $H^1(C, \Omega^1_C)\cong\mathbb{C}.$
If I'm not mistaken, it's fairly straightforward to show that $\text{dim }H^1(C, \Omega^1_C)=1$ via the Riemann-Roch theorem. I'm getting a bit stuck with showing the canonical isomorphism however.
My attempt
Since $C$ is projective, we can define a closed embedding $C\hookrightarrow\mathbb{P}^n$. Let $\mathcal{I}_C$ be the corresponding ideal sheaf and consider the conormal short exact sequence
$$0\to \mathcal{I}_C/\mathcal{I}_C^2\to \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C\to \Omega^1_C\to 0.$$
This induces a long exact sequence in cohomology
$$\cdots\to H^1(C,\mathcal{I}_C/\mathcal{I}_C^2)\to H^1(C, \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C)\to H^1(C,\Omega^1_C)\to H^2(C,\mathcal{I}_C/\mathcal{I}_C^2)\to \cdots $$
But $ H^2(C,\mathcal{I}_C/\mathcal{I}_C^2)$ vanishes (along with all other higher terms) for dimension reasons. Furthermore, I think I can show that $H^1(C, \Omega^1_{\mathbb{P}^n}\otimes \mathcal{O}_C)\cong \mathbb{C}$ via an Euler sequence argument.
If I could also show that $H^1(C,\mathcal{I}_C/\mathcal{I}_C^2)=0$ then I would be done, but it's this part I'm stuck on. I thought perhaps of using Serre duality but I didn't have much success.
Any help is much appreciated!
Best Answer
This is a direct consequence of Serre duality: $$H^1(C, \Omega^1_C) \cong H^0(C, \mathcal{O}_C)^{\vee} \cong \mathbb{C}$$ where the last isomorphism comes from the projectivity of $C$. The duality pairing also gives you that the isomorphism is canonical.