Exact couple

Exact couple of a filtered complex

Let R be a ring, which is fixed throughout the discussion. Note if R is 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:

From the filtration one can form the associated graded complex:

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

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

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

With the notation D p , q = H p + q ( F p C ) , E p , q 1 = H p + q ( gr ⁡ ( C ) p ) , the above reads:

which is precisely an exact couple and E 1 is a complex with the differential d = j ∘ k . The derived couple of this exact couple gives the second page and we iterate. In the end, one obtains the complexes E ∗ , ∗ r with the differential d:

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: Remembering d = j ∘ k , it is easy to see:

where they are viewed as subcomplexes of E 1 .

We will write the bar for F p C → F p C / F p − 1 C . Now, if [ x ¯ ] ∈ Z p , q r − 1 ⊂ E p , q 1 , then k ( [ x ¯ ] ) = i r − 1 ( [ y ] ) for some [ y ] ∈ D p − r , q + r − 1 = H p + q − 1 ( F p C ) . On the other hand, remembering k is a connecting homomorphism, k ( [ x ¯ ] ) = [ d ( x ) ] where x is a representative living in ( F p C ) p + q . Thus, we can write: d ( x ) − i r − 1 ( y ) = d ( x ′ ) for some x ′ ∈ F p − 1 C . Hence, [ x ¯ ] ∈ Z p r ⇔ x ∈ A p r modulo F p − 1 C , yielding Z p r ≃ ( A p r + F p − 1 C ) / F p − 1 C .

Next, we note that a class in ker ⁡ ( i r − 1 : H p + q ( F p C ) → H p + q ( F p + r − 1 C ) ) is represented by a cycle x such that x ∈ d ( F p + r − 1 C ) . Hence, since j is induced by ⋅ ¯ , B p r − 1 = j ( ker ⁡ i r − 1 ) ≃ ( d ( A p + r − 1 r − 1 ) + F p − 1 C ) / F p − 1 C .

We conclude: since A p r ∩ F p − 1 C = A p − 1 r − 1 ,

Proof: See the last section of May. ◻

Exact couple of a double complex

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 K p , q be a double complex. With the notation G p = ⨁ i ≥ p K i , ∗ , for each with fixed p, we have the exact sequence of cochain complexes:

Taking cohomology of it gives rise to an exact couple:

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

Example: Serre spectral sequence

The Serre spectral sequence arises from a fibration:

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).