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Nov 27, 2010 at 8:29 comment added Donu Arapura Sasha: this is very a good explanation.
Nov 27, 2010 at 7:53 comment added domenico fiorenza Hi Harry. I meant the cohomology sheaf. sorry for the inaccuracy. Sasha, that was a great comment! Absolutely obvious now that I see it, but it was it that made me see the light! if only books had more caveats like this! :) Just to see if I got it: in general it can only be said $\varphi:F^pH^{p+q}(\mathcal{F}^\bullet)\to F^{p+1}H^{p+q}(\mathcal{G}^\bullet)$, right? But then I would have vanishing of $\varphi$ on global cohomology in the exceptional case the spectral sequences abutting to hypercohomologies of $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$ would degenerate at $E_1$, right?
Nov 27, 2010 at 4:25 comment added Sasha I just want to explain why the Donu's comment is wrong. This is because the spectral sequence converges not to the global cohomology, but to its factors with respect to a certin filtration. So, from vanishing of the map on local cohomology you can deduce its vanishing on factors of global cohomology. In other words you can deduce the fact that the filtration is shifted.
Nov 26, 2010 at 22:37 comment added Harry Gindi What do you mean by local cohomology?
Nov 26, 2010 at 22:10 comment added Donu Arapura Perhaps not (in view of Torsten's example). I suppose you should require $\phi$ to vanish in the derived category of sheaves....
Nov 26, 2010 at 21:48 vote accept domenico fiorenza
Nov 26, 2010 at 21:46 answer added Torsten Ekedahl timeline score: 6
Nov 26, 2010 at 21:46 comment added Donu Arapura By local cohomology, I assume you mean the cohomology sheaves of the complex $\mathcal{F}^\bullet$. So then yes, your surmise is correct. You can use the spectral sequence as you said.
Nov 26, 2010 at 21:17 history asked domenico fiorenza CC BY-SA 2.5