# Confused about Hypercohomology terminology and meaning

check this: Given a sheaf complex $F^\bullet$, let's say I want to compute the hypercohomology of this complex, if we consider the bicomplex of sheaves

$C^\bullet(F^\bullet) = (C^p(F^q))\quad (p,q\in\mathbb{Z})$,

where $C^\bullet(F^q)$ is the Godement resolution of the sheaf $F^q$. The hypercohomology of $F^\bullet$ is the cohomology of the complex

$K^\bullet(X) = tot(C^\bullet(F^\bullet)(X))$.

If we use spectral sequences to compute the hypercohomology I have two spectral sequences, let's look at the first spectral sequence {$'E^{p,q}_r$}, this sequence converges to the final term $'E^{p,q}_\infty$ right?

This term is at the same time

$'E^{p,q}_\infty = Gr^p_C \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$

right? This is commonly expressed as

$'E^{p,q}_2 = H^p(X,H^q(F^\bullet)) \Rightarrow \mathbb{H}^{p + q}(K^\bullet(X))$,

and the second spectral sequence {$''E^{p,q}_r$} also converges to this. Ok my questions now are:

1 - Some authors simply say that these spectral sequences converge to the hypercohomology $\mathbb{H}^{p + q}(K^\bullet(X))$, why do they say that if the spectral sequences clearly converge to $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$? Or where am I wrong?

2 - Let's say I have succesfully computed ALL the terms in the two spectral sequences, what do I gain from obtaining $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$? What is that telling me?, like what if $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = 0$ for some $p$, and $q$? What can I get from knowing that $C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$ is $0$? How can I use that to compute $\mathbb{H}^{p + q}(K^\bullet(X))$, which is actually what I'm looking for? I know it's dumb and am missing something but I can't see it, can anybody please help me understand this, thanks.

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You seem to be misunderstanding something about the convergence of spectral sequence, but I'm not sure what. Two facts: a spectral sequence can converge to many things, and one usually uses the definition which allows a spectral sequence to converge to a sequence of filtered modules, rather than a sequence of graded modules. – Zhen Lin May 12 '12 at 23:41
Reading through the 1st chapter of MacCleary's book on spectral sequence might be helpful to see how extract information from convergence. – Mariano Suárez-Alvarez May 13 '12 at 2:12
Thank you both, I'm reading it right now, great exposition – Mario Carrasco May 15 '12 at 1:39

2. Successfully running such a spectral sequence provides you with the information you want "up to extension problems". The most basic example of an extension problem is if say you're after an $R$-module $M$, and I tell you that there's an exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$, and I tell you what $M'$ and $M''$ are. There is a beautiful theory of such extension problems: The set of isomorphism classes of such extensions $0 \rightarrow M' \rightarrow ? \rightarrow M'' \rightarrow 0$ form a group $\mbox{Ext}^1_R(M'',M')$, the extensions of $M''$ by $M'$. The identity element is given by the trivial extension $M=M' \oplus M''$. When this group vanishes, then you know that this is the only possibility! For example, this group vanishes when $R$ is a field: then $R$-modules are just vector spaces, and these are completely classified by their dimension (and you can easily check that it must be that $\dim(M')+\dim(M'')=\dim(M)$).
You can read the wikipedia page on the "Ext functor" to find out how to compute it. Perhaps it will help to know that this is the so-called derived functor of $\mbox{Hom}$.
I haven't read McCleary's book, although I have heard it's very good. But is it different from what I said, that in fact the $E_\infty$-page just gives you the associated graded for the filtration of the actual thing they say you converge to? I'd imagine this is what he means by "approximation"... – Aaron Mazel-Gee May 15 '12 at 1:55
@MarioCarrasco: As you say, even when you can resolve to the $E_\infty$-page, you're still not done. From here on out, approaches vary from one application to the next. Nevertheless, this is still the best known technique. I don't really know what else to say. Math is hard. If you want better machinery (and lord knows I do), then invent it yourself! I'll be rooting for you. Seriously. And also seriously, I seriously recommend that you try to get a handle on the Ext functor, because it perfectly codifies the issue at hand and in nice cases will tell you a complete, final answer. – Aaron Mazel-Gee May 16 '12 at 1:35
@MarioCarrasco: Sure thing. Ext is precisely the derived functor of Hom. It's not explicitly or solely related to hypercohomology (actually it is, but that's for another time), but it's the tool you're looking for. If there's a finite filtration $X=F^n \supset F^{n-1} \supset \cdots \supset F^0=0$ and you have access to the associated graded $gr(X)=\bigoplus F^k/F^{k-1}$, then you can inductively attempt to reconstruct the $F^k$ (and hence $X=F^n$) via the short exact sequences $0 \rightarrow F^{k-1} \rightarrow F^k \rightarrow F^k/F^{k-1} \rightarrow 0$... – Aaron Mazel-Gee May 16 '12 at 8:11
But this needs $F^{k-1}$ itself as input, so you begin at the base step $F^1 \cong F^1/F^0$. So at $k=2$, you have (loosely speaking) that $F^2 \in \mbox{Ext}^1(F^2/F^1,F^1)$ (really it's the isomorphism class of short exact sequence that is the element of Ext). At the very least, you should be able to compute this Ext group; if it vanishes, then $F^2 \cong F^1 \oplus F^2/F^1$ and you can breathe a sigh of relief -- no extension problems here! But in general, the most you can do is get control on the possibilities and then try to use situation-specific techniques to rule some of them out. – Aaron Mazel-Gee May 16 '12 at 8:18