definition of convergence of a spectral sequence

Question

In Lecture Notes in Algebraic Topology by Davis & Kirk, on page 241, there is written:

What do $E^\infty$ and $\lim_{r\to\infty}E^r_{p,q}$ mean? If a spectral sequence is not first-quadrant and is not bounded (each diagonal having only finitely nonzero modules), such as the Leray-Serre-Atiyah-Hirzebruch spectral sequence, how is convergence defined? Shouldn't condition 1 be written so that we get an isomorphism $E^r_{p,q}\cong E^{r+1}_{p,q}$ instead of only a surjection?

Answer 1

I was told by my professor that $\lim_{n\to\infty}E_{p,n-p}^r$ means the direct limit of the direct system of modules $(E_{p,n-p}^r)_{r=r_0}^\infty$ and quotient projection homomorphisms $E_{p,n-p}^r\rightarrow E_{p,n-p}^{r+1}$, which exists by condition 1.

Furthermore, for a 1st & 4th quadrant spectral sequence (such as the Leray-Serre-Atiyah-Hirzebruch spectral sequence), condition 1 is always satisfied, since for fixed $p$ and $q$, the homomorphism $d^r:E_{p,q}^r \rightarrow E_{p-r,q+r-1}^r$ will be $0$ for large enough $r$, because if $r > p$, then $p-r<0$ and $E_{p-r,q+r-1}^r=0$.

Answer 2

$E^{\infty}$ is grid of modules (without any arrows between them). The only condition it needs to satisfy is that the elements along the diagonal $E_{p,n-p}$ (of of total sum /weight $n$) must be graded pieces of some filtration of an $A$-module.

The expression $E^{\infty}_{p,q}=\lim_{r \rightarrow \infty} E^{r}_{p,q}$ means that by condition 1, for every $(p,q)$ the module $E_{p,q}^{r}$ stabilizes (becomes identical) after finitely many $r$'s. The value of $E^{\infty}_{p,q}$ (module sitting at position $(p,q)$ in that grid) is that stable limiting module.

In spectral sequence language: $$E^{r+1}_{p,q} = \ker(d^{r}_{p,q}: E^{r}_{p,q} \rightarrow E^{r}_{p+r, q-r+1})/ \operatorname{im}(d^{r}_{p-r,q+r-1}: E^{r}_{p-r,q+r-1} \rightarrow E^{r}_{p,q}).$$ Now vanishing of $d^{r}_{p,q}$ depends only on $(p,q)$ so you can only claim surjection and not isomorphism. But that is not a problem because increasing $r_{0}$ (which you are allowed to do ) you will get the isomorphism, but that $r_{0}$ wouldn't have a clean description as the condition (1) above.

I am not sure what you mean by Leray spectral sequence not being bounded, however a spectral sequence needn't converge (and then it is usually useless).