Timeline for Line bundles with meromorphic transition functions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 23, 2023 at 11:20 | comment | added | Jason Starr | I completely agree with you! The group $H^2(X,\mathcal{M}^\times)$ is often nonzero and captures part of the Brauer group of $X$. However, the short exact sequence from $\mathcal{O}^\times$ to $\mathcal{M}^\times$ to the "sheaf of Cartier divisors" shows that $H^1(X,\mathcal{M}^\times)$ is the quotient that I mentioned. | |
| May 22, 2023 at 21:32 | comment | added | KuSi | @Jason: it seems to be a common misbelief that the sheaf $\mathcal{M}^{\times}$ is acyclic whenever $X$ is projective. In "The sheaf of nonvanishing meromorphic functions in the projective algebraic case is not acyclic" by Chen, Kerr, and Lewis, it is proven that this statement is false. Even more: it is almost the converse as it holds if and only if $\mathrm{dim} X = 1$. | |
| May 22, 2023 at 18:29 | history | edited | KuSi | CC BY-SA 4.0 |
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| May 22, 2023 at 13:20 | comment | added | KuSi | In my situation, I'm given a group cocycle $e \in H^1(\pi_1(X),H^0(\tilde{X},\mathcal{M}^{\times}))$, and I know that there exists a divisor $D \subset X$ such that for any $\gamma \in \pi_1(X)$, the function $e_\gamma$ has poles only poles along $\pi^{\ast} D$ where $\pi \colon \tilde{X} \to X$ is the universal covering. Therefore, I believe it makes sense to have a well-defined divisor for $f$. | |
| May 22, 2023 at 10:03 | comment | added | R. van Dobben de Bruyn | What is your definition of the poles of $f$? By definition (if you use Čech cohomology), $f$ describes glueing data $f_{ij} \in \mathcal M^\times$ with a cocycle condition, but it is unclear to me how to (globally) extract a divisor from this. | |
| May 22, 2023 at 8:49 | history | edited | KuSi | CC BY-SA 4.0 |
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| May 22, 2023 at 8:49 | comment | added | KuSi | In my case, $X$ is projective, and hence, the element $f \in H^1(X,\mathcal{M}^{\times})$ is trivial. Even more: the section $s$ trivializes the object. But $s$ is a section of a meromorphic bundle, so it has zeroes and poles. So the question remains, what is the obstruction for $s$ to be a section of a line bundle, etc.? | |
| May 20, 2023 at 23:37 | comment | added | Jason Starr | I believe that cohomology group is the quotient of the Picard group of $X$ by the group of "Cartier divisor classes." So if $X$ is projective, for instance, then this cohomology group is trivial. However, it is nontrivial for the Hopf manifold. | |
| S May 20, 2023 at 22:11 | review | First questions | |||
| May 21, 2023 at 8:22 | |||||
| S May 20, 2023 at 22:11 | history | asked | KuSi | CC BY-SA 4.0 |