# Spectral Sequence Arrows Revisited

I don’t want to do much today. In fact, we’ll probably not cover any new ground, but start to fill in what is going on a little better. Notice that in order to work with the spectral sequence last time, we didn’t actually have to figure out what the ${i, j,}$ or ${k}$ or ${d}$ maps were. We only needed that they existed. The main thing of today is to work out all these sequences. Some are exact. Some are chain complexes. Some are commutative diagrams. Most of them fit together in a really organized way, other parts are just related. It is useful to see how these fit together, since sometimes you can get by without the maps as long as you know it is part of a commutative exact diagram.

Recall that our current situation is that given a filtered chain complex ${\cdots K^{p-1}\subset K^p\subset \cdots \subset K}$ we get an exact couple ${(D_{p,q}, E_{p,q})}$ where ${D_{p,q}= H_{p+q}(K^p)}$ and ${E_{p,q}=H_{p+q}(K^p/K^{p-1})}$ by taking the long exact sequence in homology associated to the short exact sequence ${0\rightarrow K^{p-1}\rightarrow K^p\rightarrow K^p/K^{p-1}\rightarrow 0}$.

Two times ago we saw exactly what the ${i_{p,q}}$, ${j_{p,q}}$ and ${k_{p,q}}$ maps were.

Let’s develop them again, but being more careful as to how they all fit together. Well ${i_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})}$ and it comes from the induced maps on homology from the injection ${K^p\hookrightarrow K^{p+1}}$. So we get infinite strings of ${i_{p,q}}$ for fixed ${q}$ that relate as injections ${p}$ changes. i.e we’ll get as part of a large diagram columns that look like ${\cdots \rightarrow H_{p+q}(K^{p-2})\rightarrow H_{p+q}(K^{p-1})\rightarrow H_{p+q}(K^p)\rightarrow H_{p+q}(K^{p+1})\rightarrow\cdots}$.

The ${j_{p,q}: H_{p+q}(K^p)\rightarrow H_{p+q}(K^p/K^{p-1})}$ are induced on homology from the projection map ${K^p\rightarrow K^p/K^{p-1}}$.

The ${k_{p,q}: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-1})}$ comes from the snake lemma. It is the boundary map in the long exact sequence.

Note that these last two will fit into rows: ${\cdots \rightarrow H_{p+q}(K^p)\stackrel{j}{\rightarrow} H_{p+q}(K^p/K^{p-1})\stackrel{k}{\rightarrow} H_{p+q-1}(K^{p-1})\stackrel{j}{\rightarrow} H_{p+q-1}(K^{p-1}/K^{p-2})\rightarrow\cdots}$.

Thus we can form a large commutative diagram from these long exact sequences, and figure out how to extrapolate the exact couple from it. Below is the diagram relating what was just said:

(Sorry about the image. I’ll figure this out eventually. For now you can click on it and zoom to get the full size).

The labelled arrows are the ${i,j,k}$ maps of just a single exact couple. Note that there are infinitely many exact couples going on here, and they are all related. Note also that the exact couples don’t just go across or down.

Now let’s figure out the differentials and how to get the spectral sequence. Well, take the ${H_{p+q}(K^p/K^{p-1})}$ spot. Recall that in the exact couple, this is the ${E}$ term which is where our chain complex comes from. Then doing ${j\circ k}$ is the ${d}$ map. Note this is the really the ${d^1}$ map. So our page of ${E^1_{p,q}}$ comes from this row only.

Take homology with respect to this, and due to the nature of the exact couples, we’ll get another page of these things. Now we are on the ${E^2_{p,q}}$ page of groups. The ${d^2}$ map now is to go right, then up, then right. This is the composition ${k\circ i^{-1} \circ j: H_{p+q}(K^p/K^{p-1})\rightarrow H_{p+q-1}(K^{p-2}/K^{p-3})}$.

Take homology again. In general the ${d^r}$ map is going right, then up ${r-1}$ times then right one more time. This gives us exactly what we said last time, ${d^r_{p,q}: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}$.

Hopefully this helps clarify some of the inner workings of what is going on here. Next time we’ll talk about why the conditions from last time imply that the spectral sequence converges. We won’t prove it, but there are some nice concepts behind why it works that should be enlightening.

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### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.