I am currently reading Cohomology of Number Fields by Neukirch, Schmidt and Wingberg, and I am curious about the motivation behind the construction in section 2.2:
A biregularly filtered cochain complex $F^\bullet A^\bullet$ is a filtration of $A^\bullet$ such that for each $n$, the filtration $F^\bullet A^n$ of $A^n$ is finite.
Given a biregular filtration $F^\bullet A^\bullet$, we can obtain an $E_1$ spectral sequence through the construction:
$$\begin{align} Z^{pq}_r&=\ker (F^pA^{p+q}\rightarrow A^{p+q+1}/F^{p+r}A^{p+q+1})\\ B^{pq}_r&=d(F^{p-r}A^{p+q-1})\cap F^pA^{p+q}\\ E^{pq}_r&=Z^{pq}_r/(B^{pq}_{r-1}+Z^{p+1,q-1}_{r-1}) \end{align}$$
And the spectral sequence has limit terms $E^n=H^n(A^\bullet)$.
I am wondering if there is any motivation for the constructions of $Z^{pq}_r, B^{pq}_r$, and $E^{pq}_r$, as they seem somewhat contrived and unnatural to me. Of course they are defined in such a way where the resulting spectral sequence has limit terms $H^n(A^\bullet)$, but is there any intuitive way to see why this definition is the one that works?
Another way to word my question is: how did someone who was developing this theory for the first time manage to come up with this construction?
My comment was perhaps a bit too vague/incomplete. I think there are two steps to trying to motivate/gain intuition for the spectral sequence of a filtered cochain complex. The first is to understand why it even exists and what it does, the second is to understand why it actually works. Maybe this is answer turned out a bit too long for its own good, but I hope it provides thorough guidance through those two steps.
First, let's quickly recap filtered objects. If you have a (decreasing) filtration $F^{\bullet}A$ on some algebraic object, it induces a sequence of filtration quotients $F^pA/F^{p+1}A$. If we assume the filtration to be finite, only finitely many of these are non-trivial and the object $A$ is "reconstructed" from those by finitely many extensions (of course, understanding extensions is oftentimes a very difficult problem itself). The credo I wanna follow is that filtered objects usually come up in such a way that the filtration quotients are fairly easy to understand, whereas the filtered objects themselves are fairly hard to understand. Thus, we usually wanna understand the filtration quotients as a "first approximation" to the filtered object itself.
Now, assume we have a biregularly filtered cochain complex $F^{\bullet}A^{\bullet}$. The inclusion of cochain complexes $F^pA^{\bullet}\rightarrow A^{\bullet}$ induces maps on cohomology $H^n(F^pA^{\bullet})\rightarrow H^n(A^{\bullet})$, whose images define a decreasing filtration $F^{\bullet}H^n(A^{\bullet})$ on $H^n(A^{\bullet})$. If $\mathcal{A}$ is some category of modules (which we assume for the sake of intuition, though it doesn't even pose a loss of generality), then $F^pH^n(A^{\bullet})$ are just those cohomology classes in $H^n(A^{\bullet})$, which can be represented by a cocycle in $F^pA^n$. Following the credo from the first paragraph, we want to understand the filtration quotients $F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})$.
Conversely, our credo suggests that while we do not understand the cochain complex $A^{\bullet}$, we perhaps do understand the filtration quotient cochain complexes $F^pA^{\bullet}/F^{p+1}A^{\bullet}$ and, in particular, their cohomology groups $H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})$. Note that there is a fundamental asymmetry here: the objects we do understand are obtained by taking filtration quotients of the cochain complex and then taking cohomology, whereas the objects we want to understand are obtained by taking cohomology and then taking filtration quotients of the induced filtration on cohomology. The fact that these objects will generally be very different is why the spectral sequence exists.
What the spectral sequence does is that it interpolates between the objects we understand, $H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})$ (you might recognize these as terms of the $E_1$-page), and the objects we want to understand, $F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})$ (you might recognize these as terms of the $E_{\infty}$-page). Now, how does it do that?
The elements of $F^pH^n(A^{\bullet})$ and hence of $F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})$ can be represented by cocycles lying in $F^pA^n$, as previously stated. A cocycle in $F^pA^n$ represents the trivial element in $F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})$ precisely when there is a cocycle in $F^{p+1}A^n$ (representing an element of $F^{p+1}H^n(A^{\bullet})$), which differs from the first cocycle by a coboundary (so they represent the same cohomology class in $H^n(A^{\bullet})$). Let me introduce some suggestive notation: $Z^{p,n}_{\infty}$ for the cocycles in $F^pA^n$ and $B^{p,n}_{\infty}$ for the coboundaries in $F^pA^n$. Then, we have just observed that $F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})=Z^{p,n}_{\infty}/(B^{p,n}_{\infty}+Z^{p+1,n}_{\infty})$. This might look somewhat intimidating, but it's starting to look like the formulas we want.
(An aside for the sake of clarity: The degree $p$ is called the filtration degree and the degree $n$ is called the total degree. There is also a complementary degree $q$ defined by $p+q=n$. Usually, the spectral sequence is indexed by $(p,q)$ rather than $(p,n)$, but the latter felt more natural for this motivation. Of course, the difference is merely a matter of convention and should not impede your understanding.)
Now, let's also look at $H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})$. These cohomology classes are represented by cocycles of $F^pA^{\bullet}/F^{p+1}A^{\bullet}$, which in turn are represented by cochains in $F^pA^{\bullet}$ whose differential lands in $F^{p+1}A^{\bullet}$. Such a cochain represents the trivial cohomology class in $H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})$ precisely when it differs from the differential of a cochain in $F^pA^{n-1}$ by a cochain of $F^{p+1}A^n$. Suggestive notation again: $Z^{p,n}_1$ for the cochains in $F^pA^n$ whose differential lands in $F^{p+1}A^{n+1}$, $Z^{p,n}_0$ for the cochains in $F^pA^n$ whose differential lands in $F^pA^{n+1}$ (which is just $F^pA^n$ itself!) and $B^{p,n}_0$ for those cochains $F^pA^n$ which are differentials of cochains $F^pA^{n-1}$, i.e. $d(F^pA^{n-1})$. Then, we have just observed that $H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})=Z^{p,n}_1/(B^{p,n}_0+Z^{p+1,n}_0)$. The notation might still not be entirely transparent as of now, but it's starting to take shape.
These two observations at hand and having previously handwaved about "interpolating" between them, let's try to make that precise. A cochain in $F^pA^n$ is a cocycle, i.e. an element of $Z^{p,n}_{\infty}$, precisely when its differential vanishes. Since the filtration $F^{\bullet}A^n$ is finite, this is equivalent to the differential landing in $F^{p+r}A^n$ for some large enough $r$ (depending on $n$). The cochains in $F^pA^n$ whose differential lands in $F^{p+r}A^n$ are called "$r$-almost cocycles (of bidegree $(p,n)$)" and denoted $Z^{p,n}_r$. Since the filtration is decreasing, we can think of elements in higher filtration degrees as being "smaller". The $r$-almost cocycles then "approximate" the usual cocycles in that "their differentials are sufficiently (meaning $r$ filtration steps) smaller". The $r$-almost cocycles form a decreasing filtration $F^pA^n=Z^{p,n}_0\supseteq Z^{p,n}_1\supseteq\dotsc\supseteq Z^{p,n}_r\supseteq\dotsc\supseteq Z^{p,n}_{\infty}$, which stabilizes after finitely many steps.
Similarly, a cochain $F^pA^n$ is a coboundary, i.e. an element of $B^{p,n}_{\infty}$, precisely when it is the differential of some cochain. Since the filtration $F^{\bullet}A^n$ is finite, this is equivalent to being the differential of some cochain in $F^{p-r}A^{n-1}$ for some large enough $r$ (depending on $n$). The cochains in $F^pA^n$ which are differentials of cochains in $F^{p-r}A^{n-1}$ are called "$r$-almost coboundaries (of bidegree $(p,n)$)" and denoted $B^{p,n}_r$. In the previous analogy, the $r$-almost coboundaries "approximate" the usual coboundaries in that they are "coboundaries witnessed by cochains that aren't too much (meaning $r$ filtration steps) larger". The $r$-almost coboundaries form an increasing filtration $B^{p,n}_0\subseteq B^{p,n}_1\subseteq\dotsc\subseteq B^{p,n}_r\subseteq\dotsc\subseteq B^{p,n}_{\infty}$, which stabilizies after finitely many steps. Of course, $B^{p,n}_{\infty}\subseteq Z^{p,n}_{\infty}$.
Now, as desired, we try and interpolate between $E^{p,n}_1=H^n(F^pA^{\bullet}/F^{p+1}A^{\bullet})=Z^{p,n}_1/(B^{p,n}_0+Z^{p+1,n}_0)$ and $E^{[p,n}_{\infty}=F^pH^n(A^{\bullet})/F^{p+1}H^n(A^{\bullet})=Z^{p,n}_{\infty}/(Z^{p+1,n}_{\infty}+B^{p,n}_{\infty})$. The notation laid out so far was hopefully clear enough to suggest $E^{p,n}_r=Z^{p,n}_r/(B^{p,n}_{r-1}+Z^{p+1,n}_{r-1})$ (granted, we knew this was the answer all along). Thus, the $(p,n)$-entry on the $r$-th page is the approximation to the $(p,n)$-entry on the $\infty$-page that comes from "ignoring everything that is too different in size (i.e. $r$ filtration steps smaller/larger)" by considering $r$-almost cocycles instead of cocycles and $r$-almost coboundaries instead of coboundaries. Note that we're not approximating $H^n(A^{\bullet})$ itself, but the filtration quotients of $H^n(A^{\bullet})$, which my previous comment wasn't clear about.
It might seem weird that we have to quotient out both and coboundary-like and cocycle-like terms at each step (meaning $B^{p,n}_{r-1}$ and $Z^{p+1,n}_{r-1}$ respectively). If you look at the two discussed examples ($r=1$ and $r=\infty$) as well as later computations, it will become clear that quotienting out by $B^{p,n}_{r-1}$ actually has the effect of accounting for coboundaries that become trivial in cohomology, whereas quotienting out by $Z^{p+1,n}_{r-1}$ has the effect of making sure that we keep ignoring everything in filtration degree $p+1$ (since $Z^{p,n}_r\cap F^{p+1}A^n=Z^{p+1,n}_{r-1}$), i.e. that we stay compatible with the filtration quotients.
Lastly, it remains to discuss the differentials. This is the computational part and I will not carry out the computations myself (they are tedious rather than difficult). There is no high road to understanding the computations except doing them yourself, over and over, and I suggest doing just that.
Since $Z^{p,n}_r$ are precisely those cochains whose filtration degree gets lowered $r$ steps by the differential of the cochain complex, it carries $Z^{p,n}_r$ into $Z^{p+r,n+1}_r$ with image $B^{p+r,n+1}_r$. You can check that this induces a differential $E^{p,n}_r\rightarrow E^{p+r,n+1}_r$. This differential vanishes on classes in $E^{p,n}_r$ represented by cochains in $Z^{p,n}_{r+1}$ since it then lands in filtration degree $p+1$ and the converse holds by a computation. The image of the differential into $E^{p,n}_r$ are those classes in $E^{p,n}_r$ represented by cochains in $B^{p,n}_r$, since the differential lowers the filtration degree by $r$. Thus, we can compute $$H^{p,n}(E^{\bullet,\bullet}_r)=\frac{Z^{p,n}_{r+1}+(B^{p,n}_{r-1}+Z^{p+1,n}_{r-1})}{B^{p,n}_r+(B^{p,n}_{r-1}+Z^{p+1,n}_{r-1})}=\frac{Z^{p,n}_{r+1}}{Z^{p,n}_{r+1}\cap(B^{p,n}_r+Z^{p+1,n}_{r-1})}=\frac{Z^{p,n}_{r+1}}{B^{p,n}_r+Z^{p+1,n}_r}=E^{p,n}_{r+1}.$$ This is the fundamental property of a spectral sequence: the cohomology of the $r$-th page computes the $r+1$-st page. Intuitively, classes on the $r$-th page are represented by $r$-almost cocycles and the vanishing of the $r$-th differential is an obstruction to being represented by $r+1$-almost cocycles. Thus, the representatives successively approximate the real cocycles "from above" in the sense that they get finer and finer at each level. On the other hand, the images of the $r$-th differential at the level of representatives are the $r$-almost coboundaries, so what we quotient by gets coarser and coarser at each level, approximating the real coboundaries "from below". When passing from the level of representatives to the actual classes, we additionally have to quotient out by everything in filtration degree $p+1$, as previously discussed.
The $E_1$ and $E_{\infty}$-pages have already been identified and the convergence of the spectral sequence is clear from the biregularity of the filtration, so we have successfully obtained the desired spectral sequence. I hope this makes the construction more accessible, but, frankly, the fact that it does work at the end of the day still feels like nothing short of a miracle.