Timeline for algebraic de rham cohomology of a curve
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2017 at 10:08 | comment | added | Nicola | @user251222: the first cohomology of the projective line is trivial, so in this case the boundary map from H^1(U)\to H^0(Z) is given by the residues at each point of Z and you easily deduce that \dim H^1(U)=\dimH^0(Z) -1. | |
| Aug 5, 2016 at 18:01 | comment | added | 54321user | What if I try and compute the de Rham cohomology of the $\mathbb{P}^1$ by taking $Z$ to be a point, or even better, let $Z$ be several points? This argument that $H_{Z,dR}^1(X) =0$ cannot hold. | |
| Jun 21, 2013 at 10:00 | history | answered | Nicola | CC BY-SA 3.0 |