extension problem of a spectral sequence

Question

From Hatcher's SSAT,

If the coefficient group $G$ is a field, then $H_n(X;G)$ is the direct sum $\oplus_p E^\infty_{p,n-p}$ of the terms along the $n^\text{th}$ diagonal of the $E^\infty$ page. For a nonfield $G$ such as $\mathbb{Z}$ one can only say this is true 'modulo extensions'.

Firstly, What precisely is meant by 'modulo extensions'? Is $H_n(X;G)$ given as a twisted direct sum of the $E^\infty_{p,n-p}$ terms? At the very least, is $H_n(X;G)$ equal to the direct sum as a set?

Secondly, how does one go about solving the extension problem?

Answer 1

$\DeclareMathOperator{\gr}{gr}$No, it's more complicated than that. What you have is that there is a filtration $F_p H_n$ on $H_n(X;G)$ such that the associated graded space $\gr H_n(X;G) = \bigoplus_p F_p H_n / F_{p-1} H_n$ is isomorphic to $\bigoplus_p E^\infty_{p,n-p}$. For vector spaces, since the only invariant is dimension, everything is projective etc, this implies what Hatcher said (because then $H_n \cong \gr H_n$). In general the story is more complicated.

To give you an example, consider what happens when $n=2$. Then $\gr H_2 \cong E^\infty_{0,2} \oplus E^\infty_{1,1} \oplus E^\infty_{2,0}$. This means that there exist exact sequences (you may need to reverse $p$ and $q$ in there, I'm not completely sure): $$0 \to X \to H_2 \to E^\infty_{0,2} \to 0\\ 0 \to E^\infty_{0,2} \to X \to E^\infty_{1,1} \to 0$$ where you don't even know $X$.

Given such an exact sequence, you say that "$X$ is an extension of $E^\infty_{1,1}$ by $E^\infty_{0,2}$", hence the name. In general this kind of problem is hard to solve; it's not at all the case that then $X$ will be a direct sum $E^\infty_{1,1} \oplus E^\infty_{0,2}$ (and similarly for $H_2$). They are classified by extension groups.

There's no hard and fast rule that you can follow that will give you the answer. And the complexity increases with $n$: there will be $n$ extension problems to solve (for a spectral sequence in the first quadrant like this one; when it's not bounded it's even more gruesome).

See also these two questions: question 1, question 2.

$\newcommand{\Z}{\mathbb{Z}}$Here's something that can happen: $E^\infty_{0,2} = \Z$, $E^\infty_{1,1} = \Z/2\Z$ and $E^\infty_{2,0} = \Z/2Z$. Then: $$0 \to \Z \to X \to \Z/2\Z \to 0 \\ 0 \to X \to H_2 \to \Z/2\Z \to 0$$ Then it's possible that $H_2$ is simply $\Z$! Because you can take $X = \Z$ in the first exact sequence (and the first map is multiplication by $2$), and then $H_2 = \Z$ too (and multiplication by $2$ again). Whereas the associated graded space is $\gr H_2 = \Z \oplus \Z/2\Z \oplus \Z/2\Z$...