I'm looking for basic examples that show the usefulness of spectral sequences even in the simplest case of spectral sequence of a filtered complex.

All I know are certain "extreme cases", where the spectral sequences collapses very early to yield the acyclicity of the given complex or some quasi-isomorphism to another easier complex (balancing tor, for example).

Is there an example of a useful filtration where one really computes something nontrivial also in the higher sheets?

The examples I have in mind come from topology. For example, the calculation of $H_{\ast}(\Omega{\mathbb S}^n;{\mathbb Z})$ is simply beautiful using the Serre spectral sequence, and one needs to pass to the $n$-th sheet until something happens. Another more difficult example would be the computation of the rational cohomology of $K({\mathbb Z},n)$ by induction on $n$ (depending on the parity of $n$, we get a polynomial algebra or an exterior algebra, if I remember correctly).

Are there similar, but purely algebraic examples which could show the usefulness of spectral sequences to those seeing them the first time?

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    $\begingroup$ Here's a powerful one with no acyclicity, to "localize" an isom assertion involving Ext's. Since Hom = global sections of sheaf-Hom, one gets spectral seq. for Ext whose terms are cohomology of sheaf Ext's. So to prove map of Ext's is isom (e.g., for analytic vs. algebraic with coherent sheaves, as in GAGA) when cohomologies match up nicely, by functoriality of formation of spectral seq. the isom problem reduces to case of sheaf-Ext's. That's awesome: a problem for global invariants reduces to sheaf version, which is "local". Used all the time (etale cohom, duality,...) to prove isoms. $\endgroup$
    – BCnrd
    May 3, 2010 at 0:38
  • $\begingroup$ This is an answer to a similar question. Use the Serre SS to compute the ring structure on various spaces, like your example for loops S^n and you can also do this for CP^inf and RP^inf $\endgroup$ May 3, 2010 at 20:24
  • $\begingroup$ PS, neither SS collapses, but you do not need to work with more than one sheet. I mention this example since it is the one I use to point out that not only can we compute more things, but the computation is in fact easier. $\endgroup$ May 3, 2010 at 20:25
  • $\begingroup$ One can prove the five-lemma with the spectral sequence of a double-complex ;). But I must admit that this spectral sequence collapses very soon as well. $\endgroup$ Aug 25, 2010 at 21:22
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    $\begingroup$ Look up the answers on mathoverflow.net/questions/22188/… they are probably close to identical to what you have here. $\endgroup$ Nov 6, 2010 at 9:45

6 Answers 6


This isn't exactly what you asked, but its a very simple example that (to me) demonstrates some of the necessity of the complexities of spectral sequences. Consider the ring $R=\mathbb{C}[x,y]$, and consider the module $M=(Rx+Ry)\oplus R/x$. Then the double dual spectral sequence converges to the original module: $$ Ext^{-i}_R(Ext^j_R(M,R),R) \Rightarrow M $$ The second page of this spectral sequence has

  • $R$ in degree $(0,0)$
  • $R/x$ in degree $(-1,1)$
  • $R/(Rx+Ry)$ in degree $(-1,2)$
  • A non-trivial knights-move map (differential on the second page) from $(0,0)$ to $(-1,2)$ which is the natural quotient map.

The spectral sequence collapses on the third page, with $(Rx+Ry)$ in degree $(0,0)$ and $R/x$ in degree $(-1,1)$. One shortcoming of this example is that you get the same module back, split apart into different components; rather than the associated graded of some interesting filtration. If memory serves, there was a way to tinker with this example to give it that property too, but it escapes me at the moment.


The best example I can think of is the Lyndon-Hochschild-Serre spectral sequence in group cohomology. See for instance Chapter VII Section 6 of Brown's Cohomology of Groups.
The spectral sequence, for a group extension $1\to H\to G\to Q \to 1$ (I'm using Brown's notation) and for a G-module M, is of the form $$E^2_{pq}=H_p(Q,H_q(H,M))\Rightarrow H_{p+q}(G,M).$$ Let's use this to calculate the third integral homology of the dihedral group D2n, for $n$ an odd integer. The group extension is $$1\to C_n\to D_{2n}\to C_2 \to 1,$$ and the corresponding Lyndon-Hochschild-Serre spectral sequence is $$E^2_{pq}=H_p(C_2,H_q(C_n,\mathbb{Z}))\Rightarrow H_{3}(D_{2n},\mathbb{Z}),$$ where p and q add up to 3. The integral homology of a cyclic group Cm is $\mathbb{Z}$ for $q=0$, vanishes for $q\in\mathbb{N}^\ast$ even, and is $\mathbb{Z}/m\mathbb{Z}$ for $q$ odd. Plug this information into the Lyndon-Hochschild-Serre spectral sequence, and you find $$H_{3}(D_{2n},\mathbb{Z})=H_0(C_2,\mathbb{Z}/n\mathbb{Z})\oplus H_3(C_2,\mathbb{Z})\simeq \mathbb{Z}/2n\mathbb{Z}.$$ This is easy and elegant calculation in my opinion, and occurs in practice in knot theory ($H_{3}(D_{2n},\mathbb{Z})$ turns out to be isomorphic to the relative bordism group of Fox n-coloured knots).

  • $\begingroup$ This is related to my answer and in some sense it's much better, because I just said "I can calculate an H^2 in my head using H-S" and you're saying "given a piece of paper, the same trick goes much much further". In both cases we have a mildly complex group which fortunately has simpler J-H factors. $\endgroup$ May 3, 2010 at 8:50
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    $\begingroup$ This calculation is wrong: The $\mathbb{Z}/n$ contribution to $\mathbb{Z}/(2n)=H_3(BD_{2n},\mathbb{Z})$ is not from $E^\infty_{0,3}=E^2_{0,3}=H_0(BC_2, \mathscr{H}_3(BC_n))$. You are ignoring the local coefficient system$\pi_1 BC_2 \curvearrowright H_*(BC_n)$. Thus $0=H_0(BC_2, \mathscr{H}_3(BC_n)) \neq H_0(BC_2, \mathbb{Z}/n)=\mathbb{Z}/n$. A second minor point is one that comes up in all spectral sequence calculations: $\mathbb{Z}/n \oplus \mathbb{Z}/2$ is not isomorphic to $\mathbb{Z}/2n$. This is just true up to extension. $\endgroup$ Nov 18, 2016 at 12:47

This is not the most profound answer but it's something that came up last week when I was reading a paper. The author wanted to prove that a certain obstruction to a problem was zero, and the obstruction lived in an $H^2$ that looked a bit scary: it was $H^2(W,V)$ with $W$ a local Weil group and $V$ a finite-dimensional vector space over a field of characteristic zero (I'll tell you all you need to know about this Weil group in a sec, in case you don't know what one is; it's a topological group coming up in number theory). But then I realised the obstruction was really in the image of $H^2(W/C,V)$ with $C$ a compact open subgroup of $W$ (a finite index subgroup of inertia).

But now I'm done because $W/C$ has a two-step filtration with a finite sub and a quotient isomorphic to $\mathbf{Z}$, and I know finite groups have no cohomology in char 0 in degrees 1 or more, and $\mathbf{Z}$ is the fundamental group of a 1-dimensional thing so it has no cohomology in degree 2 or more, and so by Hochschild-Serre, a calculation I can even do in my head in this example, there are no non-zero terms to build $H^2(W/C,V)$ from in $E_2$ and hence in $E_\infty$ so this group vanishes. I can do this calculation without even pulling out a piece of paper.

I'm sure if I were more "group-theoretic" I would have a much clearer picture about what was going on, but Hochschild-Serre just explains to me in a very concrete way how the cohomology of a group is built from the cohomology of its subs and quotients, and is definitely something I carry around in my "useful tools" bag.

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    $\begingroup$ I wish I could do these calculations in my head! $\endgroup$ May 26, 2015 at 23:49

Non-trivial spectral sequences occur when calculating the homotopy groups of tmf at 2 or 3: http://www.math.uiuc.edu/~rezk/512-spr2001-notes.pdf §16 ff., see also Tilman Bauer's article http://arxiv.org/abs/math/0311328


I can't recommend the following document by Tom Weston enough. It introduces spectral sequences rapidly and at a comfortable level of generality. It then applies the Hochschild-Serre sequence to group cohomology.



This is not an answer in algebra, but I wrote the following without noticing that you had asked for examples in algebra. So here it is.

I like the Mayer-Vietoris spectral sequence for giving easy examples. For simplicity, suppose that $K$ is a simplicial complex expressed as a union of $n$ subcomplexes $K_1,\ldots,K_n$. You can make a long exact sequence of chain complexes, with the degree zero bit being the chain complex $C_*(K)$, the degree one bit being the direct sum $\bigoplus_i C_*(K_i)$, the degree two bit begin the direct sum $\bigoplus_{i<j}C_*(K_i\cap K_j)$, the degree three bit being $\bigoplus_{i<j<k}C_*(K_i\cap K_j\cap K_k)$, etc.

For one of the two spectral sequences arising from this double chain complex, the $E^1$-page is all zero. For the other one, the $E^1$-page is the homologies of $K$ and all of the intersections of the pieces. If you only have two pieces, the final differential is $d^2$, which carries the same information as the connecting homomorphism in the standard Mayer-Vietoris long exact sequence.

How can the spectral sequence be helpful? I was once shown a `proof' that $H_m(K)=0$ for some $K$ made from $n$ pieces in this way, because $H_{m}(K_i)=0$ for all $i$ and $H_{m-1}$ of each intersection was 0. Knowing about the spectral sequence makes it easy to see that there is a mistake in this argument: what you really need is that $H_{m-k}=0$ for each of the $k$-fold intersections for each $k$.

For an example where a higher differential is non-zero, consider this spectral sequence for the reduced homology of the $(n-2)$-sphere, made as the boundary of an $(n-1)$-simplex. This triangulation of the $(n-2)$-sphere consists of $n$ top dimensional simplices $K_1,\ldots,K_n$ and each intersection of some proper subset of these is a simplex, while the intersection of all $n$ of them is empty. In the $E^1$-page of the non-trivial spectral sequence, every entry is zero except the zero column, which contains the reduced homology of $S^{n-2}$ and the $n$th column which contains the reduced homology of the empty set (i.e., $\mathbb{Z}$ in dimension $-1$ and 0 elsewhere). The differential $d^n$ must therefore be an isomorphism, showing that the reduced homology of $S^{n-2}$ is zero except in degree $n-2$ where it is $\mathbb{Z}$.


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