# Serre duality in Cech cohomology

Let $$X$$ be a smooth Riemann Surface (to be simple) and let $$E\longrightarrow X$$ be a coherent sheaf. The Serre Duality states that there exists a pairing $$H^1(X,E) \otimes H^0(X, E^\vee \otimes K_X) \longrightarrow H^1(X,K_X)\simeq \mathbb C$$ This is usually (at least where I read) done in terms of a resolution of $$E$$ or using the Dolbeault cohomology when $$E$$ is a vector bundle.

My question is whether this pairing can be computed in terms of Cech cohomology.

It seems that for a acyclic cover $$\{ U_i \}$$ $$(U_i\cap U_j , s_{ij}) \otimes (U_i, f_i \otimes \omega_i) \mapsto (U_i\cap U_j , f_i|_{U_i\cap U_j}(s_{ij})\omega_i|_{U_i\cap U_j})$$ must work but I'm not sure.

The first thing to check is that your $$1$$-cochain is well-defined. So does $$\sigma_{ij}=f_j|_{U_i\cap U_j}(s_{ij})\omega_j|_{U_i\cap U_j}$$ equal $$f_i|_{U_i\cap U_j}(s_{ij})\omega_i|_{U_i\cap U_j}$$? (Well, recall that $$\{f_i\otimes\omega_i\}$$ is a $$0$$-cocycle.) Then it's a matter of checking that $$\{\sigma_{ij}\}$$ define a $$1$$-cocycle: You compute the coboundary $$\sigma_{ij}-\sigma_{ik}+\sigma_{jk}$$ on $$U_i\cap U_j\cap U_k$$. If you use the fact that $$\{s_{ij}\}$$ is a $$1$$-cocycle, this all turns to $$0$$.

(It's probably best to assume $$i in these computations. But if not, we have, e.g., $$s_{ij}=-s_{ji}$$ for a $$1$$-cocycle. Then $$\sigma_{ji}=f_j|_{U_j\cap U_i}(s_{ji})\omega_j|_{U_j\cap U_i} = -f_i|_{U_i\cap U_j}(s_{ij})\omega_i|_{U_i\cap U_j} = -\sigma_{ij}$$, as is consistent.)

• Any idea to prove it is a nondegenerate pairing? – Alan Muniz Oct 7 '18 at 15:43
• This is a deeper, global result. The usual proof for compact complex manifolds uses Hodge theory. You might look at Forster's lovely book on Riemann Surfaces to see a proof based on residues and Riemann-Roch. – Ted Shifrin Oct 7 '18 at 16:08
• I'm aware of those proofs. The question is exacly how to do it in Cech cohomology. – Alan Muniz Oct 7 '18 at 16:13
• Maybe you can try to modify Bott-Tu’s Cech-flavored proof of Poincaré duality. I haven’t thought about it. – Ted Shifrin Oct 7 '18 at 16:43