1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $i<j<k$ 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.)

$\endgroup$
  • $\begingroup$ Any idea to prove it is a nondegenerate pairing? $\endgroup$ – Alan Muniz Oct 7 '18 at 15:43
  • $\begingroup$ 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. $\endgroup$ – Ted Shifrin Oct 7 '18 at 16:08
  • $\begingroup$ I'm aware of those proofs. The question is exacly how to do it in Cech cohomology. $\endgroup$ – Alan Muniz Oct 7 '18 at 16:13
  • $\begingroup$ Maybe you can try to modify Bott-Tu’s Cech-flavored proof of Poincaré duality. I haven’t thought about it. $\endgroup$ – Ted Shifrin Oct 7 '18 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.