# Serre Spectral Sequence

Today we’re going to back up and not worry about the tediousness of the last post. Let’s try to form some motivation and see how these things work even if we have to just assume some of the theory works for now. I think it will be helpful. So why care about spectral sequences? Well, one thing is that sometimes it is really hard to compute say ${H_n(C)}$, but it isn’t so bad to do ${H_n(C\otimes A)}$ for some ${A}$‘s. There is no good way to convert the information you get out of ${H_n(C\otimes A)}$ to information about ${H_n(C)}$, but this is exactly what the Bockstein spectral sequence does.

Today, I’ll talk about the Serre spectral sequence. This will help us calculate the homology of the loop space of an ${n}$-sphere. There is a nice proof that Tor is independent of which resolution you take (i.e. ${Tor(A, B)}$ can be calculated by resolving ${A}$ or ${B}$). There is a nice proof of the Snake Lemma using spectral sequences. There is a proof of the K\:{u}nneth formula. For the algebraic geometers, the Leray spectral sequence tells us information about sheaf cohomology. Anyway, there are tons of examples hitting many areas of math of uses of spectral sequences.

Right now we’ll not worry about the technical details of when and how a spectral sequence converges, but recall that the situation we ended on was that given a complex ${K}$, we could form a spectral sequence from a filtration ${\cdots \subset K^p\subset K^{p+1}\subset \cdots \subset K}$. We’ll just accept for now that if we have ${K^p=0}$ for ${p\max(p, q+1)}$ then ${E^r_{p,q}\simeq E^\infty_{p,q}}$. This ${E^\infty}$ term is the associated graded group of a filtration of ${H_*(K)}$.

This should be exciting. It means that if we can formulate a statement that reads, “There is a spectral sequence with ${E^2_{p,q}=}$ (fill in the blank), differential map ${d^r: E^r_{p,q}\rightarrow E^r_{p-r, q+r-1}}$, that converges strongly to ${H_{p+q}}$(fill in again).” Then we can just chase around differentials and hope that everything collapses fast (lots of times these things only have a few terms or even ${E^2=E^\infty}$). If this happens then we can just read off what the known things are, and we’ll have figured out new information. Note that being able to say such a statement is not bad. We only need the mere existence to begin work, and taking those conditions above for granted, we actually can formulate lots of statements such as these.

Let ${\Omega S^n}$ be the loop space of ${S^n}$, namely the space of loops based at the north pole. Let ${PS^n}$ be the path space, the space of paths starting at the north pole. Note here that our standard tools of algebraic topology are not very useful in trying to calculate ${H_q(\Omega S^n)}$. But we know that ${PS^n}$ is contractible using the obvious map of retracting along all the paths simultaneously. And we also know something very useful, that there is a spectral sequence with ${E^2_{pq}=H_p(S^n)\otimes_\mathbb{Z} H_q(\Omega S^n)}$, differentials ${d^r : E^r_{pq}\rightarrow E^r_{p-r, q+r-1}}$ converging strongly to ${H_{p+q}(PS^n)}$.

So recall that we get an entire page of groups. The ${E^2}$ groups (${r=2}$). Since ${H_p(S^n)}$ is ${0}$ for all ${p}$ except ${p=0, n}$, and is ${\mathbb{Z}}$ in those spots. We also know that ${H_0(\Omega S^n)=\mathbb{Z}}$. This allows us to fill in the whole ${q=0}$ row, and all ${p}$ columns are completely ${0}$ except in ${p=0}$ and ${p=n}$.

The differential map ${d^r}$ goes to the left ${r}$ and up ${r-1}$.

Examine the bottom row. Due to all the ${0}$‘s in columns not ${p=0}$ or ${n}$, all the differential maps that don’t land in the ${p=0}$ column must be 0. Thus there is only one possible non-zero differential coming out of there. Namely, the ${d^n_{n,0}}$ map. It lands in the ${E^2_{0, n-1}=H_{n-1}(\Omega S^n)}$ spot. Now the complex of ${d^n}$ maps is ${\cdots \rightarrow 0\rightarrow \mathbb{Z}\stackrel{d^n_{n,0}}{\rightarrow} H_{n-1}(\Omega S^n)\rightarrow 0\rightarrow \cdots }$. Recall that to get to the different ${r}$ values in ${E^r_{p,q}}$ you take homology with respect to the ${d^r}$ maps. Since ${E^\infty_{p,q}=H_{p+q}(PS^n)=0}$ for ${p+q\neq 0}$ we must have ${d^n_{n, 0}}$ an isomorphism in order for it to vanish. This gives us a ${\mathbb{Z}}$ in the 0, ${n-1}$ spot and hence another in the ${n}$, ${n-1}$ spot. We can keep repeating this argument giving a ${\mathbb{Z}}$ in the ${p=0, n}$ and ${q=}$ multiples of ${n-1}$ spots.

Since the differentials are all ${0}$ before ${r=n}$, nothing changes for ${E^2=E^3=\cdots E^n}$. Then at ${r=n}$ we get isomorphisms except at the ${E^2_{0,0}}$ position, so everything vanishes except that position which stays a ${\mathbb{Z}}$. Thus ${H_q(\Omega S^n)=\begin{cases} \mathbb{Z} \ if \ (n-1)|q \\ 0 \ else\end{cases}}$.

This is a pretty typical situation. Hopefully that gives a better overview of the process so that going back to the general case will have some sort of reference. If there are questions, feel free to ask. Maybe I’ll do another example before going back to theory.