# Exact Couples

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 ${A}$-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 ${A}$-modules, ${D}$ and ${E}$ such that:

is exact in every spot. Then we immediately get a map ${d: E\rightarrow E}$ by ${d=j\circ k}$. 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 ${i(D)\subset D}$ or anything. Now this map has a nice property, ${d\circ d=j\circ k\circ j\circ k=j\circ (k\circ j)\circ k=j\circ (0)\circ k=0}$ by exactness that middle term is ${0}$ making the whole thing ${0}$.

This means that ${\cdots \stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\stackrel{d}{\longrightarrow} E\rightarrow \cdots}$ is actually a chain complex. Thus we can take homology: $\displaystyle {H(E, d)=\frac{ker d}{im d}}$.

Now take ${D'= im i}$. Define the map ${i': D'\rightarrow D'}$ by ${i'=i\big|_{D'}}$. Take ${E'=H(E, d)}$. We will now think of ${j': D'\rightarrow E'}$ as ${j\circ i^{'-1}}$, but to be precise we pick an element whose preimage is ${x=i^{-1}(y)}$, and take an equivalence class, so ${j'(x)=[j(y)]}$. Lastly define ${k': E'\rightarrow D'}$ by ${k'([z])=k(z)}$. 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 ${(D, E)}$ form an exact couple. Then let ${D^1=D}$ and ${E^1=E}$. Next let ${D^2=D'}$ and ${E^2=E'}$. This inductively gives us that ${D^{r+1}=(D^r)'}$ and ${E^{r+1}=(E^r)'}$. In each of these we get three maps which tell us that each stage is an exact couple ${i^r: D^r\rightarrow D^r}$, also ${j^r: D^r\rightarrow E^r}$, and ${k^r: E^r\rightarrow D^r}$.

Since the triangle is exact, defining (which technically already happened in the definition of ${E^{r+1}}$) ${d^r: E^r\rightarrow E^r}$ by ${d^r=j^r\circ k^r}$ we get a chain complex and can take homology. We say that ${\{(E^r, d^r)\}_{\{r\geq 1\}}}$ is the spectral sequence associated to the exact couple. Note that ${E^{r+1}=H(E^r, d^r)}$.

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 ${\displaystyle \lim_{r\rightarrow\infty}(E^r, d^r)}$ converges in some sense to something, and hopefully we have information about what this “${E^\infty}$” 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 ${K}$. Then given a filtration ${\cdots \subset K^n\subset K^{n+1}\subset \cdots \subset K}$ (not necessarily starting at the 0 complex or ending at the total complex, but the ${j}$th grade of any ${K^n}$ should be a submodule of the ${j}$th grade of ${K}$ and the submodules should respect the differentials). Then we will get a short exact sequence of chain complexes ${0\rightarrow K^{P-1}\rightarrow K^P\rightarrow K^P/K^{P-1}\rightarrow 0}$. 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 ${\cdots \rightarrow H_n(K^{P-1})\stackrel{i}{\rightarrow} H_n(K^P)\stackrel{j}{\rightarrow} H_n(K^P/K^{P-1})\stackrel{k}{\rightarrow} H_{n-1}(K^{P-1})\rightarrow \cdots}$.

From here we define our exact couple. Let ${D_{p,q}=H_{p+q}(K^p)}$ and ${E_{p,q}=H_{p+q}(K^p/K^{p-1})}$.

Now use the maps given in the long exact sequence.

So ${i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})}$ which is really ${D_{p,q}\rightarrow D_{p+1, q-1}}$.

Our ${j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})}$ which maps ${D_{p,q}\rightarrow E_{p,q}}$.

Lastly the map ${k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})}$ which maps ${E_{p,q}\rightarrow D_{p-1, q}}$.

Lastly, the differential ${d_{p,q} = j_{p-1, q}\circ k_{p,q}: E_{p,q}\rightarrow E_{p-1, q}}$.

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 ${(p,q)}$-plane, then at each integer valued place we get a group ${E_{p,q}}$. The differential maps move us to the “left” on place. So we have rows of chain complexes for each fixed ${q}$. Remember that when we form the spectral sequence this is really only the ${E^1_{p,q}}$ part. We can think of this term as an entire “sheet” or “page” of groups related to each other. We get a page ${E^r_{p,q}}$ for each ${r\geq 1}$.

Some things that would be nice (and to get you thinking about why this might be useful) is if for each fixed pair ${(p,q)}$ the groups stabilize, i.e. ${E^N_{p,q}=E^{N+1}_{p,q}=\cdots}$. Then we would have a notion of ${E^\infty_{p,q}}$. 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.

## 3 thoughts on “Exact Couples”

1. soarerz says:

A typo: “Now take D’ = …” paragraph, I think you mean $j' = j \circ i'^{-1}$.

2. hilbertthm90 says:

Thanks. That is definitely what I wanted.

3. mathmandan says:

Hi,

I would tend to disagree with the earlier commenter: I think you mean to think of $j'$ as $j \circ i^{-1}$, where you think of $i$ as a function $D \to D'$. The reason is that when you’re looking for inverse images for elements of $D'$, you’ll find them for $i$ but not necessarily for $i'$.

For example, if $D = E = \mathbb{Z} / 4\mathbb{Z}$ and $i = j = k$ is multiplication by 2, then $D' = \{ 0, 2 \}$. Now, consider $2 \in D'$: It has no preimage under $i'$, since $i'(y) = 0$ for all $y \in D'$. However, $2$ does have (two) preimages under $i: D \to D$; namely, the elements $1$ and $3$.

Anyway, to be precise: We want to define $j': D' \to E'$, so given $x \in D'$, we have to define $j'(x) \in E' = Ker(d) / Im(d)$. Since $x \in D' = Im(i)$, choose $y \in D$ such that $i(y) = x$. Then we define $j'(x) = j(y) + Im(d)$.

To make this work, we have to show two things: first, that $j'$ is well-defined, and second, that $j(y) \in Ker(d)$ for all $y \in D$ so that $j'(x) \in Ker(d) / Im(d)$ for all $x \in D'$.

Note: I do think that where you put $x = i^{-1}(y)$, you probably meant $x = i(y)$.