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I have a very particular situation involving a (non-exact) complex $K$ of coherent sheaves on a nonsingular projective variety $X$, and I need to compute the hypercohomology of the complex. The associated spectral sequence is highly degenerate. Of course, in degenerate cases, one hopes that there are techniques for getting at the hypercohomology rapidly. In the most degenerate case, wherein every sheaf in the complex is acyclic, then the hypercohomology of the complex is the cohomology of the spaces of global sections (with respect to the maps induced by the complex's differential, say, $d$):

$\mathbb{H}^q(X,K^\bullet)\cong H^q_d(\Gamma(X,K^\bullet))$.

The problem with my situation is that the sheaves are not quite acyclic. I will be more formal now about what I am dealing with. I have a non-exact complex $(K^\bullet,d)$ of three sheaves:

$K^1\stackrel{d}{\rightarrow}K^2\stackrel{d}{\rightarrow}K^3$.

Here, $K^1$ has no nonzero sheaf cohomology except for $H^1(K^1)$, whilst $K^2$ and $K^3$ have nonzero zero-th and first cohomology and all higher cohomology vanishes.

I would think that there must be a way of dealing with such a specialized, degenerate situation, but I haven't found it yet.

Please feel free to specialize this further, if it helps (e.g. torsion-free instead of only coherent ones, or even vector bundles). Also, feel free to make this more general: I am only restricting myself to three sheaves because that is precisely the problem I am faced with. In similar spirit, if giving $K^1$ a nonzero $H^0$ doesn't harm the chances of finding a reasonable solution, then by all means do so. (This would give the problem at hand a uniform description: hypercohomology of a complex of sheaves in which each sheaf has nonvanishing zero-th and first cohomology, and vanishing cohomology elsewhere.)

If there is a specific reference for where this situation is worked out (as I'm sure it must be and I probably just haven't looked hard enough), then please do pass it along.

Thanks!!

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You've indicated that you already know about the spectral sequence computing the hypercohomology, which is indeed quite degenerate in this case - in any case where the complexes of cohomology groups are concentrated in two degrees, there is a long exact sequence analogous to the Gysin sequence. Is there some specific "faster" computation you were hoping for? –  Tyler Lawson Apr 15 '10 at 18:34
    
Thank you, Tyler. Can you please write out this long exact sequence? I took a look at the Wikipedia article for Gysin sequence, but upon first glance I am unable to reconcile what I found there with what I'm considering. If you want to post this as an answer, I'd be glad to accept it---that is to say, I'm not looking for anything faster than what you propose. Thanks again. –  user5395 Apr 15 '10 at 19:53
    
This kind of degenerate spectral sequence setup appears as example 1.D on page 8 of McCleary's "A user's guide to spectral sequences" here: amazon.com/Spectral-Sequences-Cambridge-Advanced-Mathematics/dp/… I don't really have the time to TeX this up, but if someone else wants to they should feel free. –  Tyler Lawson Apr 15 '10 at 20:12
    
A reference for the LES of a 2-row spectral sequence is Weibel's "An intro to homological algebra", exercise 5.2.2. –  Tim Perutz Apr 15 '10 at 20:16
    
The spectral sequence linking the cohomology of the sheaves $K^i$ with hypercohomology is an $E_1$ spectral sequence; with the stated restrictions, it could have differentials at levels $E_1$ and $E_2$, I don't think you can extract a Gysin type exact sequence from it. –  Angelo Apr 15 '10 at 21:12

1 Answer 1

up vote 3 down vote accepted

Ok, so what will the spectral sequence give you? This is a very easy exercise, but since Altgr is not experienced, here is the solution. The term $E_2^{p,q}$ is the $p^{\rm th}$ cohomology group of the complex $\mathrm{H}^q K^{\bullet}$ (the complex of groups obtained by applying the $q^{\rm th}$ cohomology functor to the complex $K^\bullet$). Then one has equalities $$\mathbb H^i (K^\bullet) = 0 \quad{\rm for}\ i \neq 2,3,4$$ $$\mathbb H^4 (K^\bullet) = E_2^{3,1}$$ and an exact sequence $$ 0 \to E_2^{2,0} \to \mathbb H^2 (K^\bullet) \to E_2^{1,1} \to E_2^{3,0} \to \mathbb H^3 (K^\bullet) \to E_2^{2,1} \to 0 $$

The homomorphism $E_2^{1,1} \to E_2^{3,0}$ is the only differential at the $E_2$ level that can possibly be non-zero.

Without further information, there is nothing else one can say, other than this: working with hypercohomology without knowing spectral sequences is like driving nails into a wall without a hammer.

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Thank you, Angelo, I really appreciate your taking the time to write this up. Actually, seeing this worked example clears up a few errors in my thinking about spectral sequences. –  user5395 Apr 17 '10 at 11:56
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I am very glad this helped. When learning spectral sequences, working out a lot of examples is the key. On second thought, maybe it's not so different from anything else in mathematics. –  Angelo Apr 17 '10 at 12:21

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