Exact couple

Let R be a ring, which is fixed throughout the discussion. Note if R is ${\displaystyle \mathbb {Z} }$, then modules over R are the same thing as abelian groups.

Each filtered chain complex of modules determines an exact couple, which in turn determines a spectral sequence, as follows. Let C be a chain complex graded by integers and suppose it is given an increasing filtration: for each integer p, there is an inclusion of complexes:

${\displaystyle F_{p-1}C\subset F_{p}C.}$

From the filtration one can form the associated graded complex:

${\displaystyle \operatorname {gr} C=\bigoplus _{-\infty }^{\infty }F_{p}C/F_{p-1}C,}$

which is doubly-graded and which is the zero-th page of the spectral sequence:

${\displaystyle E_{p,q}^{0}=(\operatorname {gr} C)_{p,q}=(F_{p}C/F_{p-1}C)_{p+q}.}$

To get the first page, for each fixed p, we look at the short exact sequence of complexes:

${\displaystyle 0\to F_{p-1}C\to F_{p}C\to (\operatorname {gr} C)_{p}\to 0}$

from which we obtain a long exact sequence of homologies: (p is still fixed)

${\displaystyle \cdots \to H_{n}(F_{p-1}C){\overset {i}{\to }}H_{n}(F_{p}C){\overset {j}{\to }}H_{n}(\operatorname {gr} (C)_{p}){\overset {k}{\to }}H_{n-1}(F_{p-1}C)\to \cdots }$

With the notation ${\displaystyle D_{p,q}=H_{p+q}(F_{p}C),\,E_{p,q}^{1}=H_{p+q}(\operatorname {gr} (C)_{p})}$, the above reads:

${\displaystyle \cdots \to D_{p-1,q+1}{\overset {i}{\to }}D_{p,q}{\overset {j}{\to }}E_{p,q}^{1}{\overset {k}{\to }}D_{p-1,q}\to \cdots ,}$

which is precisely an exact couple and ${\displaystyle E^{1}}$ is a complex with the differential ${\displaystyle d=j\circ k}$. The derived couple of this exact couple gives the second page and we iterate. In the end, one obtains the complexes ${\displaystyle E_{*,*}^{r}}$ with the differential d:

${\displaystyle E_{p,q}^{r}{\overset {k}{\to }}D_{p-1,q}^{r}{\overset {{}^{r}j}{\to }}E_{p-r,q+r-1}^{r}.}$

The next lemma gives a more explicit formula for the spectral sequence; in particular, it shows the spectral sequence constructed above is the same one in more traditional direct construction, in which one uses the formula below as definition (cf. Spectral sequence#The spectral sequence of a filtered complex).

Sketch of proof:[1][2] Remembering ${\displaystyle d=j\circ k}$, it is easy to see:

${\displaystyle Z^{r}=k^{-1}(\operatorname {im} i^{r}),\,B^{r}=j(\operatorname {ker} i^{r}),}$

where they are viewed as subcomplexes of ${\displaystyle E^{1}}$.

We will write the bar for ${\displaystyle F_{p}C\to F_{p}C/F_{p-1}C}$. Now, if ${\displaystyle [{\overline {x}}]\in Z_{p,q}^{r-1}\subset E_{p,q}^{1}}$, then ${\displaystyle k([{\overline {x}}])=i^{r-1}([y])}$ for some ${\displaystyle [y]\in D_{p-r,q+r-1}=H_{p+q-1}(F_{p}C)}$. On the other hand, remembering k is a connecting homomorphism, ${\displaystyle k([{\overline {x}}])=[d(x)]}$ where x is a representative living in ${\displaystyle (F_{p}C)_{p+q}}$. Thus, we can write: ${\displaystyle d(x)-i^{r-1}(y)=d(x')}$ for some ${\displaystyle x'\in F_{p-1}C}$. Hence, ${\displaystyle [{\overline {x}}]\in Z_{p}^{r}\Leftrightarrow x\in A_{p}^{r}}$ modulo ${\displaystyle F_{p-1}C}$, yielding ${\displaystyle Z_{p}^{r}\simeq (A_{p}^{r}+F_{p-1}C)/F_{p-1}C}$.

Next, we note that a class in ${\displaystyle \operatorname {ker} (i^{r-1}:H_{p+q}(F_{p}C)\to H_{p+q}(F_{p+r-1}C))}$ is represented by a cycle x such that ${\displaystyle x\in d(F_{p+r-1}C)}$. Hence, since j is induced by ${\displaystyle {\overline {\cdot }}}$, ${\displaystyle B_{p}^{r-1}=j(\operatorname {ker} i^{r-1})\simeq (d(A_{p+r-1}^{r-1})+F_{p-1}C)/F_{p-1}C}$.

We conclude: since ${\displaystyle A_{p}^{r}\cap F_{p-1}C=A_{p-1}^{r-1}}$,

${\displaystyle E_{p,*}^{r}={Z_{p}^{r-1} \over B_{p}^{r-1}}\simeq {A_{p}^{r}+F_{p-1}C \over d(A_{p+r-1}^{r-1})+F_{p-1}C}\simeq {A_{p}^{r} \over d(A_{p+r-1}^{r-1})+A_{p-1}^{r-1}}.\qquad \square }$

Proof: See the last section of May. ${\displaystyle \square }$

A double complex determines two exact couples; whence, the two spectral sequences, as follows. (Some authors call the two spectral sequences horizontal and vertical.) Let ${\displaystyle K^{p,q}}$ be a double complex.[3] With the notation ${\displaystyle G^{p}=\bigoplus _{i\geq p}K^{i,*}}$, for each with fixed p, we have the exact sequence of cochain complexes:

${\displaystyle 0\to G^{p+1}\to G^{p}\to K^{p,*}\to 0.}$

Taking cohomology of it gives rise to an exact couple:

${\displaystyle \cdots \to D^{p,q}{\overset {j}{\to }}E_{1}^{p,q}{\overset {k}{\to }}\cdots }$

By symmetry, that is, by switching first and second indexes, one also obtains the other exact couple.

The Serre spectral sequence arises from a fibration:

${\displaystyle F\to E\to B.}$

For the sake of transparency, we only consider the case when the spaces are CW complexes, F is connected and B is simply connected; the general case involves more technicality (namely, local coefficient system).

• May, J. Peter, A primer on spectral sequences (PDF)
• Massey, William S. (1952), "Exact couples in algebraic topology. I, II", Annals of Mathematics, Second Series, 56: 363–396, doi:10.2307/1969805, MR 0052770.
• Weibel, Charles A. (1994), An introduction to homological algebra, Cambridge Studies in Advanced Mathematics, vol. 38, Cambridge: Cambridge University Press, doi:10.1017/CBO9781139644136, ISBN 0-521-43500-5, MR 1269324