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I am trying to really get comfortable with sheaf cohomology by actually sitting down and computing a whole heap of examples. But I am wondering what schemes can we actually reasonably compute the cohomology of?

Let's just restrict ourselves to projective schemes over a field to start with. Probably the simplest case is the cohomology of a hypersurface by taking the long exact sequence corresponding to the SES: $$ 0 \longrightarrow \mathcal{O}(d) \longrightarrow \mathcal{O} \longrightarrow i_{*}\mathcal{O}_{V}(d) \longrightarrow 0 $$ for the closed immersion $i : V \rightarrow \mathbb{P}^{n}$. Complements of hypersurfaces are similarly easy. For the case of curves, we know that the cohomology must vanish in degrees higher than $1$, so we via a Serre duality argument we really only need to calculate the global sections, right?

So what about arbitrary projective varieties? I know we can obtain the Hilbert polynomial by resolving the sheaf by line bundles. But can we actually calculate the explicit cohomology in any more general cases? What about complete intersections or smooth varieties? Or is any real generality hopeless?

I tagged this as a soft question too since there may not be a concrete answer.

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    $\begingroup$ The method you describe for projective varieties does indeed compute their cohomology after splitting the resolution in to short exact sequences as described in this answer. $\endgroup$ – KReiser Nov 29 '19 at 3:47
  • $\begingroup$ @KReiser thank you for the answer. I actually recently asked about this exact answer because I didn't quite understand the last comment about splitting up the resolution. Is there any source that does some examples of this procedure? And is it just then a matter of an induction argument up along the sequence? $\endgroup$ – Luke Nov 29 '19 at 3:57
  • $\begingroup$ Here's a post talking about how to split up a long exact sequence (like our resolution $0\to F_m \to \cdots \to F_1 \to \mathcal{O}_{\Bbb P^n} \to \mathcal{O}_X \to 0$) into a series of short exact sequences. Once you do that, it is an induction argument up the sequence as you say (modulo some minor fussing about in degrees $0$ and $n$ sometimes). I'll see if I can find or write a couple worked examples here for you. $\endgroup$ – KReiser Dec 2 '19 at 4:33
  • $\begingroup$ Thank you @KReiser I would appreciate that so much if you could! $\endgroup$ – Luke Dec 2 '19 at 8:02

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