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Back again, check this out, let $X$ be a topological space and let $F^{\bullet}$ be a cochain complex of sheaves, I'm trying to compute the cohomology of the complex of global sections of the sheaves

$F^0(X) \rightarrow F^1(X) \rightarrow F^2(X) \rightarrow \cdots$.

I'm trying to prove specifically that the cohomology of this complex is 0. Now there's a possibility that the sheaves $F^q$ MIGHT be ACYCLIC but I have yet to prove that, when I brought up this possibility some of you guys pointed me to Hypercohomology (which I thank you for). I was reading up on Hypercohomology and I'm kind of lost and don't know how to even get started on how to compute the Hypercohomology groups. Here's basically what I want to know:

1 - Let's say the sheaves $F^q$ are in fact acyclic, what I got from what I was reading in this case was

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(H^0(X,F^{\bullet}))$,

but $H^0(X,F^q) \cong \Gamma(X,F^q)$, so

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,

which is what I'm looking for, so all I have to do is obtain $\mathbb{H}^n(X,F^{\bullet})$? And this is the cohomology of the complex $CF^\bullet(X) = tot(C^\bullet(F^\bullet)(X))$, where $CF^n(X)= \oplus_{p+q=n} C^p(F^q)(X)$ right?

And $F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution. This is all part of the definition of the Cartan-Eilenberg resolution right? So how can I go from

$F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution to

$CF^\bullet(X)$ is an acyclic resolution?

2 - How do I even go about this if I don't know if the sheaves $F^q$ are acyclic?

Any ideas?

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I'm not sure I completely understand your question, but:

There is a double complex with $\Gamma(C^pF^q)$ in the $(p,q)$ place.

By taking cohomology first vertically and then horizontally, or vice versa, you get two spectral sequences.

One spectral sequence has $E_1^{p,q}=H^q(X,F^p)$. The other has $E_2^{p,q}=H^p(X,H^q(F^\bullet))$. Both converge to the hypercohomology of $F^\bullet$.

Without more information about $F$, that's pretty much all you can say. Of course if you do have some information about the terms in one of these spectral sequences, you might be able to use it, together with the fact that both sequences have the same abutment, to deduce something about some of the terms in the other.

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Thank you, yeah, I feel like I don't have much to go on to solve this, does the topology of $X$ affect at all the computation of these groups, $X$ is a topological vector spaces, or the fact that the sheaves $F^q$ are sheaves of vector spaces? – Louis A Apr 5 '12 at 2:57
If "sheaves of vector spaces" means "sheaves of sections of vector bundles", and (certainly at least if everything is finite dimensional), you are done, since over a TVS every vector bundle is trivial so all of your $F^p$ are surely acyclic. – Steven Landsburg Apr 5 '12 at 3:52
You mean if the sections of the sheaves $F^q$ on a set $U$ are continuous sections $g:U \rightarrow E$ where $E$ is a vector bundle? the TVS might be infinite-dimensional, thank you – Louis A Apr 5 '12 at 17:51
All vector bundles on any contractible space are trivial. – Steven Landsburg Apr 5 '12 at 17:55

I am not entirely sure what you are asking, but I think it is essentially about how one can compute $\mathbb{H}^n(X,F^{\bullet})$.

As you point out, if the $F^i$ are acyclic, then indeed

$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,

This is of course not true if they are not acyclic. Now this is the point where you have to do what you have to do when you want to compute any derived functor: You have to replace your object by an injective (or acyclic) resolution.

So perhaps your question is: What's a resolution of a complex? ? (nested question marks!) But it seems that you know about the total complex of "mashing" together the resolutions of the individual sheaves.

So perhaps you just need a simple hint: For each $F^q$ take an arbitrary acyclic resolution $A^{q,\bullet}$ and then put them together into one (big) complex, all of whose entries are acyclic. Then use this to compute $\mathbb H$ as you did when you had an acyclic complex.

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Thank you, as I was saying, I wonder if the fact that $X$ is a topological vector space or the type of sheaf i.e. the $F^q$'s are sheaves of vector spaces help at all in the computation, I'm taking a plunge into this – Louis A Apr 5 '12 at 3:02

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