Timeline for local systems, duals, cohomology
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2013 at 16:37 | comment | added | ChrisLazda | I don't think so. If $n=0$, then $U$ is just $\mathbb{P}^1$, so the cohomology is just that of the sphere, and so we have $h^0=1$, $h^1=0$, $h^2=1$. But if $n\geq 1$ then $U$ is homotopic to the wedge product of $n-1$ spheres, and so we will have $h^0=1$, $h^1=n-1$ and $h^2=0$. | |
| Jun 25, 2013 at 15:25 | comment | added | local | Thanks Chris. If I understand correctly, there will be vanishing only for $i=1$, as $H^0(U, \mathbb{C})$ has dimension $1+n$ ($n$ being the number of deleted points) and $H^2(U, \mathbb{C})$ has dimension $1$. Is that true? | |
| Jun 25, 2013 at 15:18 | comment | added | Alicia Garcia-Raboso | @ChrisLazda: d'oh! I was thinking of the vector bundle $V \otimes_\mathbb{C} \mathcal{O}_U$. | |
| Jun 25, 2013 at 15:14 | comment | added | ChrisLazda | Isn't V a locally constant sheaf of $\mathbb{C}$-modules, rather than a coherent module? To answer 2), as Fernando points out, triviality of the monodromy representation means that $V$ is globally constant, and hence $H^i(U,V)\cong H^i(U,\mathbb{C})\otimes_{\mathbb{C}}V$, and $H^i(U,\mathbb{C})$ is easy to calculate. | |
| Jun 25, 2013 at 15:04 | comment | added | Alicia Garcia-Raboso | For 2: $U$ is affine and $V$ is a coherent $\mathcal{O}_V$-module, so all the higher cohomology groups vanish. | |
| Jun 25, 2013 at 14:57 | review | First posts | |||
| Jun 26, 2013 at 9:07 | |||||
| Jun 25, 2013 at 14:47 | comment | added | local | No, Fernando, it's not an exercise. Thanks for explaining 1). What about 2)? Is it true that $H^i(U, V)=0$ for $i=1$ and $2$? | |
| Jun 25, 2013 at 14:45 | comment | added | Fernando Muro | Is this an exercise? I mean, 1) is pretty obvious, since you assume that the action of the fundamental group on $V$ is trivial, then you're only looking at the underlying vector space, and any f.d. vector space is isomorphic to the linear dual. | |
| Jun 25, 2013 at 14:37 | history | asked | local | CC BY-SA 3.0 |