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I read a number of posts here on MO, but haven't quite found an answer to the question of where the differentials in a spectral sequence come from.

I came across a differential $d^{0,1}$ on the $E_2$-page from a 2-dim to a 3-dim vector space (in an application of the cohomological Lyndon-Hochschild-Serre s.s.). I know that $E^{0,1}_{\infty}=\mathrm{ker} d^{0,1}$.

A priori, the kernel has dimension 0, 1 or 2. How can I determine the dimension?

In general, my question is, whether there is a way of tracking down differentials in a s.s. explicitly. Where do they come from in the first place? In the books I have read, differentials seem to be given... Still, I imagine, somewhere hidden in the proof must be a construction that proves they exist and are determined. After all, in my calculation, even if 0, the kernel of my differential does not get lost, but does get promoted to star in the second cohomology group, rather than the first. And this surely has to be prevented by a properly converging s.s.

(And just out of curiosity: How would such an explicit construction be possible in the Grothendieck s.s.?)

Thank you for any insights.

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  • $\begingroup$ In the Grothendieck spectral sequence the differentials come from the injective (or projective) resolutions involved (the construction can be found in XX.9 of Lang: Algebra): They give rise to a double complex. By filtering the double complex on obtains a filtered complex and a filtered complex always yields a spectral sequence with $(E_r,d_r)$ given explicitly in terms of the filtered complex (see Prop. XX.9.1 in Lang's book). $\endgroup$
    – Ralph
    Commented Apr 19, 2012 at 11:35
  • $\begingroup$ There is a difference between defining what a spectral sequence is and constructing one. In the definition of `spectral sequence' the differentials are given. When constructing a spectral sequence like the Hochschild-Serre spectral sequence one must construct the differentials. This is of course exactly what Hochschild and Serre did. $\endgroup$
    – Wilberd van der Kallen
    Commented Apr 20, 2012 at 7:36

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Without meaning to be snarky, I think there's some confusion here about what spectral sequences are for. In particular, the sense in which a SpSeq (or even a long exact sequence!) serves as a computational tool isn't that it tells you how to calculate things --- particularly, not how to calculate the things in it. Rather, it gives you a way to organize things you already know how to calculate; maybe it suggests a helpful summary of your calculations; and it will certainly remind you of what things you don't already know --- it will provide helpful questions to consider --- if you try to squeeze modules out of it.

As a small example, the usual snake-shapped diagram chase proving the isomorphism of, e.g. de-Rham and real Čech cohomology, is neatly summarized by noting that there's a related double complex which is very nearly exact in every way, and thus all the later pages are mostly zero, while the transgression is an isomorphism. We get a proof, because we already know what enough of the modules and differentials are.

If you really want to ask what the pages and differentials of some SpSeq are, there are algorithmic tools (i.e., programmes --- usually not suitable for human consumption!), provided you can properly specify what your particular spectral sequence is about. See e.g. (Ana Romero's thesis)1, building on homological perturbation lemmata (start at 2).

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This answer by Ralph gives an explicit description of the differential in the LHSSS you mention.

Another thought: presumably you have a group extension $1\to K\to G\to G/K\to 1$. If you are using field coefficients (as you seem to imply) then there are no extension problems and $$H^1(G)\cong E_{\infty}^{0,1}\oplus E^{1,0}_\infty \cong \ker d^{0,1}\oplus E^{1,0}_2.$$

Presumably you can calculate the first cohomologies of $G$ and $G/K$, in which case a simple dimension count should give you the dimension of the kernel of $d^{0,1}$.

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  • $\begingroup$ That has it somewhat backwards. I want to calculate the dimension of $H^1(G,M)$, with twisted coefficients from $K$ and $G/K$, which happen to be abelian. I thought the best way of getting at the cohomology of $G$ would be via the s.s. Indeed, the terms are easy to calculate, but the result is obscured by the differential. Maybe it would have been wiser to bite the bullet and calculate $H^1(G,M)$ directly via the bar resolution...? $\endgroup$
    – Earthliŋ
    Commented Apr 19, 2012 at 12:10
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    $\begingroup$ $H^1(G;M)$ is just the quotient of the derivations $Der(G,M)$ by the subgroup $Ider(G,M)$ of inner derivations. Also note that the seven-term-exact sequence yields the exact sequence $0 \to H^1(G/K;M) \to H^1(G;M) \to H^1(K;M)^{G/K} \to H^2(G/K;M^K)$ that could be of help. $\endgroup$
    – Ralph
    Commented Apr 19, 2012 at 12:32

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