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It is well known that spectral sequence is very important in algebraic geometry and complex geometry, but its definition seems very unnatural. For example, in Voisin's book Hodge theory and complex algebraic geometry, I, p.202, the author defines

$E_r^{p,q}:=\frac{{Z_r}^{p,q}}{{B_r}^{p,q}}$,

where $Z_r^{p,q}=\{x\in F^pA^{p+q}|dx\in F^{p+r}A^{p+q}\}$, and $B_r^{p,q}=Z_{r-1}^{p+1,q-1}+dZ_{r-1}^{p-r+1,q+r-2}$.

It can be checked that $B_r^{p,q}\subset Z_r^{p,q}$. Note that $d$ sends $Z_r^{p,q}$ to $Z_r^{p+r,q-r+1}$ and $B_r^{p,q}$ to $B_r^{p+r,q-r+1}$, so we have a differential $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ satisfying $d_r^2=0$.

Although I can verify all the facts stated above, but I don't know the motivation behind this definition, especially why we should define $B_r^{p,q}$ as the sum of $Z_{r-1}^{p+1,q-1}$ and $dZ_{r-1}^{p-r+1,q+r-2}$ and why the indices appear so bizarre. I believe there should be a natural motivation under which all these terms become natural and easy to remember, does anyone know it?

Remark: In Griffiths & Harris, the authors say that the spectral sequence is a generalization of a long exact cohomology sequence and promise they will come back to this topic after introducing some more definitions, but seems never come back.

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    $\begingroup$ You have a cochain complex $C$. Its differential $d$ is complicated, so the cohomology of $C$ is difficult to compute. You notice that you can filter $C$ so that lots of terms in the formula for $d(x)$ are in higher filtration than $x$, so in the associated graded cochain complex, the differential is much simpler. So you can calculate the cohomology of the associated graded cochain complex. But how does that relate to the cohomology of the original cochain complex $C$? That's what the SS is for: its input is the cohomology of the associated graded, and it converges to $H^*(C)$! $\endgroup$
    – user164898
    Commented Oct 6, 2022 at 14:56
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    $\begingroup$ Have you ever seen "You could have invented spectral sequences"? ams.org/notices/200601/fea-chow.pdf $\endgroup$ Commented Oct 6, 2022 at 15:24
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    $\begingroup$ Try to understand the case of a first quadrant double complex, that is a nonnegatively bigraded vector space with two anticommuting differentials $d_1$ and $d_2$ of bidegrees $(1,0)$ and $(0,1)$. Imagine that you want to compute the homology of $d_1+d_2$, and you know the homology of $d_1$ (all horizontal rows). What would you do? What to do to figure out which homology classes of $d_1$ extend to homology classes of $d_1+d_2$? Once you do some diagram chasing in this case, the general case will be obvious. $\endgroup$ Commented Oct 6, 2022 at 17:00
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    $\begingroup$ I strongly recommend the nice pictures by Ravi Vakil depicting what each page of the spectral sequence is doing: math216.wordpress.com/2022/03/25/… $\endgroup$ Commented Oct 7, 2022 at 12:53
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    $\begingroup$ @GunnarÞórMagnússon I love that paper, but at the same time whenever I see it I'm tempted to write a rebuttal titled "No, I couldn't have." $\endgroup$ Commented Oct 7, 2022 at 14:24

1 Answer 1

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I suspect a previous comment of mine led to this question, so let me say a few words here. The basic problem is this: Suppose $(A^\bullet, F)$ is a filtered complex, then one wants to relate the (co)homology of the associated graded $E_1=H^*(Gr_F A^\bullet)$ to the associated graded of the (co)homology $E_\infty=Gr_F H^*(A)$. It would wonderful if they were equal, and sometimes they are*, but they usually aren't. Instead, the spectral sequence gives a procedure that starts with $E_1$ and eventually gets to $E_\infty$. A bit more precisely, $E_1$ carries a differential. Taking cohomology gives $E_2$, which again carries a differential. Taking cohomology gives $E_3$ and so on. With appropriate conditions, this process will stabilize to $E_\infty$. This is obviously a very rough sketch, and is just meant to supply motivation and not computational skill.

  • The good news is that in the case you seem to be interested in (in another question), namely the Frölicher spectral sequence for a compact $\partial\overline{\partial}$-manifold, $E_1$ does equal $E_\infty$.
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  • $\begingroup$ Thanks for this nice explanation, Prof Arapura. Yes, I must admit that recently I'm thinking about issues on $\partial\bar\partial$-manifolds, and whether we can establish a VHS theory for manifolds with Frölicher spectral sequence degenerating at $E_1$, I find most of my problems boil down to that my understanding of spectral sequence is too shallow, with all these helps and more computations, I believe I can understand it better. $\endgroup$
    – Tom
    Commented Oct 7, 2022 at 13:04
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    $\begingroup$ This is indeed the approach that finally got me on board with spectral sequences. It is easy to show that $H^*(Gr_F A^\bullet)$ is typically bigger than the correct answer, and then when you try to correct it... you could have invented spectral sequences. I'd also like to take this opportunity to point out that Dickens' "A Christmas Carol" features an important spectral sequence. $\endgroup$ Commented Oct 9, 2022 at 21:26

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