Today we begin a long set of posts on spectral sequences. I must first off say that I only learned what a spectral sequence was about 2 weeks ago, so I am far from knowledgable about them. This means if you have questions, I may not know the answers. I’m also going to go incredibly slowly. I want to carefully develop everything and give lots of motivating examples. Lastly, all of this is coming from the notes I’ve taken from John Palmieri’s class this quarter, so any great insights should be attributed to him and mistakes should be attributed to myself.

Like I’ve already said, I’m not familiar with the literature, so I’m not sure what the standard development is. For people who know some of this already, we’re going to define the spectral sequence associated to an exact couple. Then anytime we want a spectral sequence, we just need to figure out the exact couple it comes from. This is a different approach than say Weibel.

In general, this is going to work in any abelian category, but we’ll use the category of -modules just to keep it less scary for those that haven’t worked with abelian categories.

Our first definition is that of an exact couple. This is a diagram of -modules, and such that:

is exact in every spot. Then we immediately get a map by . This may look strange, since we’re not just going around the diagram. We sort of jump, but this is fine since at this stage we aren’t identifying or anything. Now this map has a nice property, by exactness that middle term is making the whole thing .

This means that is actually a chain complex. Thus we can take homology: .

Now take . Define the map by . Take . We will now think of as , but to be precise we pick an element whose preimage is , and take an equivalence class, so . Lastly define by . Now we have made lots of choices, but as usual in homological things it turns out by examination that everything is well-defined. It is also the case that with these new maps, the corresponding diagram is exact at every place. We call this the derived exact couple.

Since the derived exact couple is an exact couple, there is nothing stopping us from defining the derived exact couple of the derived exact couple. Then do it again, and again, … . Obviously, if we aren’t careful we’ll lose track of where we are, so choosing notation is essential. Suppose form an exact couple. Then let and . Next let and . This inductively gives us that and . In each of these we get three maps which tell us that each stage is an exact couple , also , and .

Since the triangle is exact, defining (which technically already happened in the definition of ) by we get a chain complex and can take homology. We say that is the spectral sequence associated to the exact couple. Note that .

This is probably incredibly useless for people wanting to know what a spectral sequence is and how it is used. But in general we are going to have a lot more information floating around than just this abstract exact couple. More specifically, we have to get the exact couple somehow, and the process of forming it will give us things to work with. The other thing is that we want to know that converges in some sense to something, and hopefully we have information about what this “” term is.

I don’t want to go through an entire example today, but we’ll set up the situation for the example (for those who have seen this, it will be the Serre Spectral Sequence).

Suppose we have a chain complex . Then given a filtration (not necessarily starting at the 0 complex or ending at the total complex, but the th grade of any should be a submodule of the th grade of and the submodules should respect the differentials). Then we will get a short exact sequence of chain complexes . This is just because it works trivially on each grade.

Well, our old standby says that a short exact sequence of chain complexes induces a long exact sequence in homology, so .

From here we define our exact couple. Let and .

Now use the maps given in the long exact sequence.

So which is really .

Our which maps .

Lastly the map which maps .

Lastly, the differential .

Don’t worry if this seems horribly confusing. We’ll look at a very concrete example next time in which we’ll see that often everyting stays stationary or goes to 0 leaving us with very little. I’ll just try to summarize a little of this filtered chain complex example. We are getting infinitely many exact couples all mingled together somehow. If we think of a -plane, then at each integer valued place we get a group . The differential maps move us to the “left” on place. So we have rows of chain complexes for each fixed . Remember that when we form the spectral sequence this is really only the part. We can think of this term as an entire “sheet” or “page” of groups related to each other. We get a page for each .

Some things that would be nice (and to get you thinking about why this might be useful) is if for each fixed pair the groups stabilize, i.e. . Then we would have a notion of . It would be nice to know how this limit term relates to the original chain complex, or filtration. It would be nice if the spectral sequence limit were somehow independent of the filtration, that way taking a limit of different filtrations would give us information we might not know. Like I said earlier, don’t worry if this post was confusing. It just needed to be done. After some examples, this will probably not seem so bad.

January 25, 2010 at 4:12 am

A typo: “Now take D’ = …” paragraph, I think you mean .

January 26, 2010 at 2:14 pm

Thanks. That is definitely what I wanted.

May 26, 2014 at 6:12 pm

Hi,

I would tend to disagree with the earlier commenter: I think you mean to think of as , where you think of as a function . The reason is that when you’re looking for inverse images for elements of , you’ll find them for $i$ but not necessarily for .

For example, if and is multiplication by 2, then . Now, consider : It has no preimage under , since for all . However, does have (two) preimages under ; namely, the elements and .

Anyway, to be precise: We want to define , so given , we have to define . Since , choose such that . Then we define .

To make this work, we have to show two things: first, that is well-defined, and second, that for all so that for all .

Note: I do think that where you put , you probably meant .

Thanks for this helpful post!