![]() These may all be special cases of what Grothendieck described. ![]() (b) There are some cases where just a single map or similar gives you a filtered complex. (It is an acyclic resolution as VA says.) If you have a stratification that you wish was CW but isn't, then you just have a general filtered complex. A CW complex is stratified, but it has a special structure so that its chain complex doesn't need a spectral sequence. For instance it could be a stratified topological space. To discuss things more concretely, (I am told that VA's interesting answer is not a generalization of this paragraph.) For question (1), here are two general ways that filtered complexes show up: (a) you might get a filtered complex if you are interested in a chain complex that comes from a filtered object. Probably it is just more general than what I was going to say. VA gives a very general motivation for why filtered complexes show up so much. Two questions that remain are, (1) why do filtered complexes show up so much, and (2) is there anything that you could do with a filtered complex other than compute its spectral sequence? Qiaochu links to a really nice article by Timothy Chow that says a lot about the mechanics of how to go from a filtered complex to its spectral sequence. The spectral sequence gives you a way to deal with this situation. You can do the first step and the second step separately but they are not exactly independent of each other. Instead, they are made (via an appropriate filtration) from some other elementary, "acyclic" objects. $F(A)$ in case 2) is made of parts which are not themselves elementary. The reason for the spectral sequences in both cases is the same. Answer: Grothendieck's spectral sequence.ฤก and 2 account for the vast majority of applications of spectral sequences, and provide plenty of motivation - I am sure you will agree. Let us say you know the derived functors for $F$ and $G$ and would like to compute them for the composition $GF$. Let us say you have two functors $F:\mathcal A\to\mathcal B$ and $G:\mathcal B\to \mathcal C$. ![]() If $J^n$ are not acyclic, you get a spectral sequence instead, and that's the best you can do. derived functors $R^nF$), you can use this resolution to compute the cohomologies of $A$ (resp. Let's say you have a resolution $0\to A\to J^0\to J^1\to\dots$ (of a module, a sheaf, etc.) If $J^n$ are acyclic (meaning, have trivial higher cohomology, resp. ![]()
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