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Jul 8, 2012 at 11:46 answer added Daniel Sommerhoff timeline score: 0
Jul 8, 2012 at 11:10 vote accept Steven Gro
Jul 8, 2012 at 10:22 answer added Filippo Alberto Edoardo timeline score: 3
Jul 8, 2012 at 10:10 comment added Steven Gro Hey Dan, thanks for your answer. To specify my question: The text I am reading is: math.utah.edu/~bertram/courses/hilbert/ps/hilbert.ps On page 6 Bertram is proving the existence of the hilbert scheme and defines a grassmannian $G(P'(d_0),H^0 (\mathbb{P}^{m}_{A}, \mathcal{O}^{n}_{\mathbb{P}^{m}_{A}} (l + d_0)))$. I think that he is using $H^0 (\mathbb{P}^{m}_{A}, \mathcal{O}^{n}_{\mathbb{P}^{m}_{A}} (l + d_0)))$ as a sheaf, otherwise this notation wouldn't fit his definition of the grassmannian from the beginning of the document. Thanks in advance to all of you!
Jul 8, 2012 at 9:22 comment added Dan Petersen I think this question is too vague to get a good answer; there are several situations where one can think of a cohomology group as a sheaf in some sense. But take a look at Dimca's book "Sheaves in topology" and I bet your question will be answered.
Jul 8, 2012 at 8:57 history asked Steven Gro CC BY-SA 3.0