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!!
Best Answer
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.