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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.2012-05-12
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    Reading through the 1st chapter of MacCleary's book on spectral sequence might be helpful to see how extract information from convergence.2012-05-13
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    Thank you both, I'm reading it right now, great exposition2012-05-15

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  1. People generally say that a spectral sequence "converges" to something even though it generally only provides the associated graded object of the actual (filtered) thing you're after.

  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 $\mbox{Hom}$.

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    I'm reading McCleary's 'A user's guide to spectral sequences', so far from what I'm reading it seems that $'E^{p,q}_\infty$ is an approximation to the Hypercohomology, didn't know that, it explains the confusion I had about the term 'converges', other books don't explain that which is important to a reader new to spectral sequences, I think that anyone's brain immediately goes to the common meaning of converging, wish they stopped doing that in maths, it wouldn't kill them to spend 2 more lines just explaining the confusion, thanks bro2012-05-15
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    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"...2012-05-15
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    @MarioCarrasco: I'd strongly recommend taking a look at this blog post by Eric Peterson: "Spectral sequences on one blackboard" (!). For me at least, this perspective is by far the cleanest way to think about these things. http://chromotopy.org/?p=7212012-05-15
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    Thanks, I read the post, he says "these groups ‘limit’ to something meaningful with respect to the homology $H_* A$". I don't want to compute 'something meaningful with respect to the homology $H_* A$' of the complex I want to actually compute $H_* A$, what does "something meaningful with respect to X" even mean? That's so vague. I can't tell my supervisor "You know I didn't compute the Hypercohomology of the complex, but I computed something meaningful with respect to it", I want somebody to tell me how to go from this graded thing $E^{p,q}_\infty$ to the actual $H_* A$, sorry about the rant2012-05-15
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    @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.2012-05-16
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    Thank you, I got carried away with the rant there, a good ol' math rant, I've read a little bit about the Ext functor and am trying to read more about it now. It's basically defined as the derived functor of the Hom functor, not sure how it relates to Hypercohomology exactly? I'll continue reading though, thanks2012-05-16
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    @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 $0 \rightarrow F^{k-1} \rightarrow F^k \rightarrow F^k/F^{k-1} \rightarrow 0$...2012-05-16
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    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.2012-05-16
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    Thank you, thank you so much, again, at the risk of saying something really dumb, am not really much of an expert, I kind of get what you're saying but how does knowing that $F^1 \cong F^1/F^0$ and $F^2 \cong F^1 \oplus F^2/F^1$ help me compute $F^1$, and then $F^2$, and then the remaining $F^n$'s, I know it's a fact from exact sequences or something but I've been reading about other unrelated math stuff too and it's going over my head right now. Thank you so very much, you've been of great help2012-05-22
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    @MarioCarrasco: You seem completely bent on ignoring extension problems. While I would like to reassure you once again that they are a VERY REAL issue, let's pretend for a moment that we're working in a category where all short exact sequences split (e.g., vector spaces over a field). Then given our assumtion that the filtration is finite (i.e. that $F^0=0$), $F^1 \cong F^1/F^0$, and then the short exact sequences from 3 comments above give you $F^2 \cong F^2/F^1 \oplus F^1 \cong F^2/F^1 \oplus F^1/ F^0$, and then $F^3 \cong F^3/F^2 \oplus F^2 \cong \bigoplus_{i=1}^3 F^i/F^{i-1}$...2012-05-22
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    ... all the way up to to $X=F^n \cong \bigoplus_{i=1}^n F^i/F^{i-1}$. Let me reiterate that this is not usually the case, and there are at best *context-specific* methods for dealing with extension problems. For completeness, let me refer you to Boardman's "Conditionally convergent spectral sequences", which deals with this as well as a whole host of other issues surrounding computations with spectral sequences: http://hopf.math.purdue.edu/Boardman/ccspseq.pdf2012-05-22
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    Thanks! So what you're saying is that if: 1. The filtration is finite, 2. I have access to the associated graded vector space $H^* = \bigoplus F^k/F^{k-1}$ then if all the $\mbox{Ext}^1(F^{i}/F^{i-1},F^{i-1}) = 0$ then I'm done for and $H^* \cong \bigoplus_{i=1}^n F^i/F^{i-1}$? So can I NOT have 'ACCESS' to the associated graded vector space? That means that my complex doesn't have an associated graded vector space???? The $F^n$'s are actually (possibly infinite-dimensional) vector spaces2012-05-25
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    @MarioCarrasco: You're converging towards the right statement, but you're not there yet. Suppose that you successfully run the spectral sequence. What this gives you is a sequence (which is a priori doubly-infinite) of nested objects in $X$, say $X \leftarrow \cdots \supset F^k \supset F^{k-1} \supset \cdots$. Then: (1) In good cases, this is actually finite, i.e. $F^k=X$ for all sufficiently large $k$ and $F^k=0$ for all sufficiently small $k$. (2) In those cases, if additionally $\mbox{Ext}^1(F^i/F^{i-1},F^{i-1})=0$ for all $i$, then $X=\bigoplus F^i/F^{i-1}$.2012-05-26
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    [However, (2) usually fails, and (1) fails in a number of important situations as well. The only case I can think of where (2) is guaranteed to go through is when your objects are vector spaces (or nearly by definition, more generally when they are objects of some other semisimple additive category).]2012-05-26
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    Thank you again! Do you have any references for this? Books or papers I can look at? My objects are vector spaces2012-05-26
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    I learned about spectral sequences from Bott & Tu's "Differential Forms in Algebraic Topology", but that's by no means the only reference. (However, they also are only considering real vector spaces (in the context of de Rham cohomology) so they have no extension problems either.) Actually, most of what I've learned has been through conversation rather than from reading, so I'm not such a good person to ask. However, there should be an abundance of suggestions if you search previous math.SE and mathoverflow questions (or just ask google, of course).2012-05-27
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    Thank you bro, gonna look into that2012-06-01