I want to start by saying that I only have very basic notions about spectral sequences.

Consider a short exact sequence of chain complexes $$0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0\ .$$ We have the respective long exact sequence in homology $$\cdots\longrightarrow H_n(A)\longrightarrow H_n(B)\longrightarrow H_n(C)\longrightarrow H_{n-1}(A)\ .$$ On the other hand, we could consider the bicomplex $$D_{p,q}:=\cases{A_p&if $q=1$,\\B_p&if $q=2$,\\C_p&if $q=3$,\\ 0&else.}$$ As differentials we take the differentials of the complexes in the vertical direction, and the maps of the SES in the horizontal direction (adding a sign if you want the two differentials to anticommute). As this bicomplex is bounded in $q$, the spectral sequence will collapse almost immediately (at page $2$, if I'm not wrong).

Question: What is the relation between the two constructions? Do they give the same thing? Can we recover one from the other?


1 Answer 1


To answer the question in the title : yes, long exact sequences can be derived from the degeneracy of some spectral sequences. In fact, there are several spectral sequences leading to the long exact sequence. I will explain in details two of them because the point of view is quite different.

So let's start with your construction. The complex $A\rightarrow B\rightarrow C$ is acyclic and so is its total complex. Thus, the spectral sequences associated to this complex converge to ... $0$. But lets have a look to what is going on. The fact that the spectral sequence converges to zero does give some informations. In fact there are two spectral sequences. One is completely uninteresting because $E_1=0$. The other one has $E_1$ page : $$\require{AMScd}\begin{CD} ... @.\\ H^1(A)@>i_*>> H^1(B)@>p_*>> H^1(C)\\ @.@.@.\\ H^0(A)@>i_*>> H^0(B)@>p_*>> H^0(C)\\ ... @. \end{CD}$$ First, the only differential at the $E_2$ page will link the homology at $H^{k+1}(A)$ to the homology at $H^k(C)$. In other words, there will be no non zero differential from and to object of the second column. This means that everything in the second column should vanish at $E_2$, so the sequences above are exact at $H^k(B)$.

Now the $E_2$-page looks like $$\begin{CD} ...\\ \ker i_*@. 0@. \operatorname{coker}p_*\\ @.@.@.\\ \ker i_*@. 0@. \operatorname{coker}p_*\\ ... \end{CD}$$ with a differential from $\ker i_*$ to $\operatorname{coker} p_*$ (one row below). Since this will be the last differential, then this must be an isomorphism. In other words, there are isomophism $u:\ker i_*\overset\sim\rightarrow H^k(C)/\operatorname{im}p_*$. Then the map $\partial:H^k(C)\rightarrow H^k(C)/\operatorname{im}p_*\overset{u^{-1}}\rightarrow\ker i_*\rightarrow H^{k+1}(A)$ is the boundary map, and from its definition, it is clear that it induces the long exact sequence in cohomology.

Here is another construction/point of view. Indeed, spectral sequences (and long exact sequences) are used to compute the cohomology of a complicated object from the cohomology of simpler ones. In this regard, I will suggest to look at the following constructions :

Consider the short exact sequence $$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ as a 2-steps filtration of the object $B$. Namely $F^0=B, F^1=A, F^2=0$. The graded pieces are $\operatorname{Gr}^0=C$ and $\operatorname{Gr}^1=A$. Now there exist a spectral sequence $$ E_0^{p,q}=\operatorname{Gr}^p B^{p+q}=F^pB^{p+q}/F^{p+1}B^{p+q}\Rightarrow H^{p+q}(B)$$ which indeed compute the cohomology of the object $B$ from the cohomology of the "simpler ones" $A$ and $B$. The $E_0$ page looks like $$ \begin{CD} @AAA@. @AAA\\ @.C^1 @.@. A^2\\ @AAA@. @AAA \\ @.C^0@. @. A^1\\ @AAA@. @AAA \end{CD} $$ where the vertical arrows are the differential of the complexes. Thus the $E_1$ page looks like $$\begin{CD} @. ...\\ H^1(C)@>>> H^2(A)\\ @.@.\\ H^0(C)@>>> H^1(A)\\ @. ... \end{CD}$$ where the differential are the boundary of the long exact sequence (this is clear from the construction of the spectral sequence associated to a filtered complex). So finally the $E_2$ page looks like $$\begin{CD} @. ...\\ \ker\partial^1@. \operatorname{coker}\partial^1\\ @.@.\\ \ker\partial^0@. \operatorname{coker}\partial^0\\ @. ... \end{CD}$$ and there are no other differentials. The spectral sequence collapses and from the construction, you know that the limit is $H(B)$, or more precisely the graded part of $H(B)$ where $F^0H(B)=H(B)$, $F^1H(B)=\operatorname{im}(H(A)\rightarrow H(B))$ and $F^2H(B)=0$.

Thus you get isomorphisms $\operatorname{im}(H^k(A)\rightarrow H^k(B))=\operatorname{coker}\partial^{k-1}$ and $H^k(B)/\operatorname{im}(H^k(A)\rightarrow H^k(B))=\ker\partial^k$. This last isomorphisms can be rewritten as $$ 0\rightarrow H^k(A)/\operatorname{Im}\partial^{k-1}\rightarrow H^k(B)\rightarrow \ker\partial^k\rightarrow 0$$ so putting all these short exact sequences (when $k$ varies), we get the long exact sequence.

Here is another construction with the same point of view as before. Consider the double complex $A\rightarrow B$ ($B$ in degree 0). Its total complex is then quasi-isomorphic to $C$. In this situation $C$ plays the role of the complicated object, and we want to compute its cohomology from some simpler parts, here $A$ and $B$. And indeed, there is a spectral sequence (in fact 2!) for the double complex $A\rightarrow B$. I don't write the details, but one of them degenerates at $E_2$ and (almost) the same argument as before gives the long exact sequence.

Of course, you can also use the double complex $B\rightarrow C$ whose total complex is quasi-isomorphic to $A$.

  • $\begingroup$ Dear Roland, it's been several times your answers have been very useful for me, thanks a lot ! Keep going ! $\endgroup$ Apr 10, 2018 at 23:01
  • $\begingroup$ @NicolasHemelsoet Oh, thanks, you're welcome. I'm glad I helped. $\endgroup$
    – Roland
    Apr 11, 2018 at 5:19

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