# Spectral Sequence of a Double Complex

I knew my posting would be sparser this quarter, but this was a pretty large gap. I sort of lost motivation for posting, since my class moved on from spectral sequences to group cohomology, so I wasn’t constantly reminded about things to post about. Now the class is coming back around to spectral sequences in order to compute the group cohomology, so I’m sort of in the mood again.

We’ve mostly been looking at how to get a spectral sequence out of a filtered complex, and last time I promised I’d talk about convergence. Since my posting is so sparse, and I only wanted to give an idea about it rather than a proof of all the details of the convergence, I’m going to keep it brief in order to post about something more important.

The theorem is that if we have a filtration ${\{K^p\}}$ of a complex ${K}$ and the following three conditions are satisfied

1. ${K^p=0}$ when ${p<0}$
2. ${H_{p+q}(K^p/K^{p-1})=0}$ when ${q\leq -1}$
3. ${K=\cup K^p}$

then if ${r\geq \max\{p, q+1\}+1}$, then ${E^r_{p,q}\simeq E^\infty_{p,q}}$. Where we define ${E^\infty_{p,q}:= F_{p,q}/F_{p-1, q+1}}$ where ${F_{p,q}:=im (H_{p+q}(K^p)\rightarrow H_{p+q}(K))}$.

Basically what this says is that under some “mild" conditions that were naturally satisfied in our examples, the spectral sequence converges. And not only does it converge, but it converges to an associated graded object of the homology of the original complex.

The fact that it converges you can work out without much trouble if you've followed the previous posts. It basically amounts to noticing the first two conditions force all non-zero ${E^r}$ terms to be in the “first quadrant" when ${r>1}$. Since the differential maps get longer and longer as ${r}$ increases, paying attention to some specific ${E^*_{p,q}}$, the differential pointing to this term will have to come from the fourth quadrant, and eventually the differential will point to the second quadrant. But we said that these two quadrants were filled with 0 groups. Thus our sequence we are taking homology of is ${0\rightarrow E^r_{p,q}\rightarrow 0}$ for all ${r}$ after some point. So they stabilize. Figuring out exactly what these things are that they stabilize to is far more tedious and I won't do it.

The next topic that should be addressed is how to get a spectral sequence out of a double complex. This is actually much more useful it turns out, since there are two ways to do it that will both converge to the same thing which will give us lots of information.

First, let's recall that a double complex ${C_{p,q}}$ is just sets of complexes where the ${q}$ index the rows and the ${p}$ index the columns. So for a fixed ${q}$, we get ${\cdots \leftarrow C_{p-1, q}\leftarrow C_{p, q}\leftarrow \cdots }$ a chain complex with differential, say ${d^h}$ for horizontal. Likewise we get them going down in columns for any fixed ${p}$ with differential ${d^v}$ for vertical. The differentials are compatible in the sense that ${d^h\circ d^v+d^v\circ d^h=0}$. These also have to fit together into what we call the total complex ${(Tot C)_n=\bigoplus_{p+q=n} C_{p,q}}$ where the differential is ${d^h+d^v: C_{p,q}\rightarrow C_{p-1, q}\bigoplus C_{p,q-1}}$.

Now if we consider just the single complex ${Tot C}$, and put a filtration on it, we'll have a spectral sequence as before. One natural filtration is at the ${p}$th filtered part to chop everything from the ${p}$-th column to the right off and replace it with ${0}$. Notationally, this is ${(F_pC)_n=\bigoplus_{j+q=n \ \text{and} \ j\leq p} C_{j, q}}$.

The other natural one is to chop everything from the ${p}$-th row and up off and replace it with ${0}$. We'll notate this with a prime on the left ${F_pC}$. Both of these certainly satisfy ${\cdots \subset F_pC\subset F_{p+1}C\subset \cdots}$ since from ${p}$ to ${p+1}$ everything stays exactly the same except the ${p+1}$ column of ${C_{p,q}}$ gets stuck on. We also satisfy ${\cup F_pC=Tot C}$. We do need to assume that ${C_{p,q}=0}$ when ${p<0}$ or ${q<0}$ to get all the convergence conditions above, but this isn't too bad since most naturally arising double complexes (that we'll look at) are zero before some point and we can just shift indices.

The spectral sequence we get has ${E^1_{p,q}=H_{p+q}(F_pC/F_{p-1}C)}$. Note that for the first spectral sequence the ${E^2}$ term is computed by first taking vertical homology, and then horizontal homology: ${E^2_{p,q}\simeq H_p^h(H_q^v(C))}$ and for the other filtration it is ${E^2_{p,q}\simeq H_q^v(H^h_p(C))}$.

As usual, the formal aspect of this might seem overwhelming and confusing right now, so I'll hold off on working out what the maps actually are and let this absorb a little. It should become more clear what is going on when we do some examples.

## 2 thoughts on “Spectral Sequence of a Double Complex”

1. hilbertthm90 says:

Actually, that’s what I used to post this. I’ve always had weird things happen with inequalities in the past as well.