Spectral Sequences
Montek Gill
Contents
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
Chapter 1. Spectral sequences as algebraic objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1. Spectral sequences and maps of spectral sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2. Infinity terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3. Spectral sequences from towers of submodules of a differential module . . . . . . . . . . . . . . . 3
1.4. Spectral sequences from filtered differential graded modules . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5. Spectral sequences from exact couples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Chapter 2. Spectral sequences in topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1. Leray-Serre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2. Eilenberg-Moore . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3. Adams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
14
14
14
14
Introduction
In these notes, we define spectral sequences as algebraic objects, show how to construct them from
various types of algebraic data and then present some of the well-known spectral sequences which
arise from topological contexts. The first chapter describes the algebra and the second chapter
describes the topology.
The notes mostly follow McCleary’s text A User’s Guide to Spectral Sequences and the only thing
original about them so far is all the detail provided filling in definitions and proofs.
iii
Chapter 1
Spectral sequences as algebraic objects
Let A be an abelian category which is “concrete” in the sense that we can work with elements and
define things like quotients in terms of elements.
1.1. Spectral sequences and maps of spectral sequences
Definition 1.1 (spectral sequences and maps of spectral sequences). A spectral sequence
in A is a sequence (Er , dr , ϕr ), r ≥ 0, of triples of objects Er , morphisms dr : Er → Er such that
dr ◦ dr = 0 and isomorphisms ϕr : ker dr /im dr → Er+1 . Given two spectral sequences (Er , dr , ϕr )
and (Er0 , d0r , ϕ0r ), a morphism (Er , dr , ϕr ) → (Er0 , d0r , ϕ0r ) is a collection of morphisms fr : Er → Er0
such that the diagrams
Er
fr
fr
fr
ker dr /im dr
d0r
dr
Er
Er0
ϕ0r
ϕr
and
Er0
ker d0r /im d0r
fr+1
Er+1
0
Er+1
commute; that is, d0r ◦ fr = fr ◦ dr and fr+1 is the composition
ϕ−1
r
ϕ0
fr
0
Er+1 → ker dr /im dr → ker d0r /im d0r →r Er+1
.
Note the following:
• in general, the index may start from an arbitrary integer, which we consider to be a spectral
sequence by re-indexing as we do with sequences of numbers an for example
• while the objects Er are connected via the isomorphisms ϕr , the differentials dr are entirely
independent of one another; this means that one can construct spectral sequences quite
artificially: given an object E together with a self-differential d, take the homology of
(E, d) and then choose some self-differential of this object and continue in this fashion; the
point however I guess is that entire sequences of objects and self-differentials where one
object is the homology of the previous pair occur naturally
• for morphisms of spectral sequences, the previous point has the result that while f0 above
determines the remaining maps fr (in fact any fr determines the remaining fs ), the commutativity requirement (with the differentials) on f0 is not sufficient to guarantee the commutativity requirement on the remaining fr
• one might wonder what happens if an object in a spectral sequence is replaced with an
isomorphic object, via some given isomorphism, and the corresponding differential changed
according to that same isomorphism, that is, when one “conjugates a term”; more specifically, consider a spectral sequence (Er , dr , ϕr ), suppose given isomorphisms θr : Er0 → Er
and define a spectral sequence (Er0 , d0r , ϕ0r ) as follows: d0r = θr−1 ◦ dr ◦ θr , which is easily
checked to be a differential, and ϕ0r is the isomorphism
θr
ker d0r /im d0r →
ϕr
−1
θr+1
0
ker dr /im dr → Er+1 → Er+1
1
2
1. Spectral sequences as algebraic objects
where we note that θr induces the first map because d0r = θr−1 ◦ dr ◦ θr gives θr ◦ d0r =
dr ◦ θr ; then we can immediately check that the θr themselves provide an isomorphism
(Er0 , d0r , ϕ0r ) → (Er , dr , ϕr )
• most often, a spectral sequence (Er , dr , ϕr ) in A is of the following form:
– the category A is the category whose objects are Z-bigraded R-modules for some ring
R and whose morphisms are graded R-module homomorphisms
– the differentials dr on Er = ⊕p,q Erp,q are all of bidegree (−r, r − 1) (for a spectral
sequence of homological type) or all of bidegree (r, 1 − r) (for a spectral sequence of
cohomological type)
– the Z-bigrading on the homology of (⊕p,q Erp,q , dr ) is the natural one and the isomorphisms ϕr have bidegree (0, 0).
1.2. Infinity terms
Every spectral sequence has associated to it an “infinity term”, also known as the “limit term”.
To define this term, consider a spectral sequence (Er , dr , ϕr ), r ≥ 0, with isomorphisms ϕr :
ker dr /im dr → Er+1 . Let Br = im dr and Zr = ker dr for each r ≥ 0 and then define submodules
Br,0 ⊆ Zr,0 ⊆ E0 , which will satisfy
B0,0 ⊆ B1,0 ⊆ B2,0 ⊆ · · · ⊆ Bi,0 ⊆ · · · ⊆ Zi,0 ⊆ · · · ⊆ Z2,0 ⊆ Z1,0 ⊆ Z0,0 ⊆ E0
as follows:
• for r = 0, set B0,0 = B0 and Z0,0 = Z0
• for r = 1, we have
B1 ⊆ Z1
in E1 and we expand this to
B0 ⊆ B1,0 ⊆ Z1,0 ⊆ Z0
ϕ−1
0 (B1 )
in E0 by setting
• for r = 2, we have
= B1,0 /B0 and ϕ−1
0 (Z1 ) = Z1,0 /B0
B2 ⊆ Z2
in E2 and we expand this to
B1 ⊆ B2,1 ⊆ Z2,1 ⊆ Z1
−1
in E1 by setting ϕ−1
1 (B2 ) = B2,1 /B1 and ϕ1 (Z2 ) = Z2,1 /B1 and then expand this to
B0 ⊆ B1,0 ⊆ B2,0 ⊆ Z2,0 ⊆ Z1,0 ⊆ Z0
−1
in E0 by setting ϕ−1
0 (B2,1 ) = B2,0 /B0 and ϕ0 (Z2,1 ) = Z2,0 /B0
• now repeat the process.
We now define B∞ = ∪r Br,0 and Z∞ = ∩r Zr,0 and call Z∞ /B∞ the infinity term of the spectral
sequence. We say that a spectral sequence converges to its infinity term.
Note the following:
• via an induction argument, we have Zr,0 /Br,0 ∼
= Zr /Br ∼
= Er+1 .
For the infinity term to be a natural construction in some sense, isomorphic spectral sequences
should have isomorphic infinity terms (in particular, conjugating terms does not change the infinity
term, up to isomorphism). To see that this is indeed the case, consider two spectral sequences
(Er , dr , ϕr ) and (Er0 , d0r , ϕ0r ) and an isomorphism (Er , dr , ϕr ) → (Er0 , d0r , ϕ0r ); recall that the latter
means a collection of morphisms (here, isomorphisms) fr : Er → Er0 such that d0r ◦ fr = fr ◦ dr and
fr+1 is given as follows:
ϕ−1
r
fr
ϕ0
0
Er+1 → ker dr /im dr → ker d0r /im d0r →r Er+1
.
1.3. Spectral sequences from towers of submodules of a differential module
3
In the construction of the corresponding infinity terms, recall that we have the towers of submodules
B0,0 ⊆ B1,0 ⊆ B2,0 ⊆ · · · ⊆ Bi,0 ⊆ · · · ⊆ Zi,0 ⊆ · · · ⊆ Z2,0 ⊆ Z1,0 ⊆ Z0,0 ⊆ E0
and
0
0
0
0
0
0
0
0
B0,0
⊆ B1,0
⊆ B2,0
⊆ · · · ⊆ Bi,0
⊆ · · · ⊆ Zi,0
⊆ · · · ⊆ Z2,0
⊆ Z1,0
⊆ Z0,0
⊆ E00 .
0 and f (B ) = B 0 ; it will then be clear that f induces an isomorWe claim that f0 (Zr,0 ) = Zr,0
0
r,0
0
r,0
0
0
phism Z∞ /B∞ → Z∞ /B∞ . These equalities in fact can be seen to hold fairly immediately via an
induction argument from the two diagrams above which define morphisms of spectral sequences.
Thus isomorphic spectral sequences have isomorphic infinity terms.
Note the following:
• one can only construct the tower of submodules when one knows not only that each term Er
is the homology of the previous term but exactly how this is indeed the case; this information
is exactly what the isomorphisms ϕr provide and so it seems that they need to be part of
the data and that the infinity term shouldn’t necessarily be expected to be independent of
these isomorphisms (especially given that the differentials are entirely independent of one
another)
• there should be a result saying something like that a truncation of a spectral sequence is
isomorphic to the original spectral sequence so that the infinity term can be formed within
any term.
In the case that a spectral sequence (Er , dr , ϕr ) collapses at the N th term, that is, dr = 0 for r ≥ N ,
it is easy to identify the limit term. To see this, note that if dr = 0, then Br = 0, Zr = Er , and so
Br,r−1 = Br−1 and Zr,r−1 = Zr−1 . Carrying out the procedure defining the Br,0 and Zr,0 , we see
that Br,0 = Br−1,0 and Zr,0 = Zr−1,0 . Thus the tower of submodules becomes
B0,0 ⊆ B1,0 ⊆ B2,0 ⊆ · · · ⊆ BN −1,0 = BN,0 = · · · = B∞
⊆ Z∞ = · · · = ZN,0 = ZN −1,0 ⊆ · · · ⊆ Z2,0 ⊆ Z1,0 ⊆ Z0,0 ⊆ E0
and so the limit term is ZN −1,0 /BN −1,0 ∼
= EN .
1.3. Spectral sequences from towers of submodules of a differential module
In this section, we will reverse the construction of a tower of submodules of the first term of a
spectral sequence, which was carried out in defining the infinity term. Suppose given the following
data: a differential module (over some ring) (M, d), a tower of submodules
B0 ⊆ B1 ⊆ · · · ⊆ Z1 ⊆ Z0 ⊆ M
such that B0 = im d and Z0 = ker d and differentials dr+1 : Zr /Br → Zr /Br with kernel Zr+1 /Br
and image Br+1 /Br .
To define our spectral sequence, first set Er = Zr−1 /Br−1 for r ≥ 1 and E0 = M . Second, for
r ≥ 1, let the differential Er → Er be the given one, dr , and for r = 0, let d0 = d. Finally,
third, the isomorphism ϕ0 : Z0 /B0 → E1 is just the identity and for r ≥ 1, the isomorphism
ϕr : ker dr /im dr == (Zr /Br−1 )/(Br /Br−1 ) → Zr /Br is the canonical map.
Now, what is the infinity term of this spectral sequence? In defining the infinity term, we used
the symbols Z∗ and B∗ which have already been used here and so instead, here we use P∗ and Q∗
respectively. Now P0 = Z0 and Q0 = B0 by definition. Further, P1 = Z1 /B0 and Q1 = B1 /B0 and
so P1,0 = Z1 and Q1,0 = B1 . Next, P2 = Z2 /B1 and Q2 = B2 /B1 and one can check that then
P2,1 = Z2 /B0 and Q2 = B2 /B0 ; then also P2,0 = Z2 and Q2,0 = B2 . Repeating this, we see that
Pr,0 = Zr and Qr,0 = Br and the infinity term is exactly what it should be, namely (∩r Zr )/(∪r Br ).
4
1. Spectral sequences as algebraic objects
1.4. Spectral sequences from filtered differential graded modules
Definition 1.2 (filtered differential graded modules). A filtered differential graded R-module
(A, d, F ) consists of an R-module A together with three types of data, which are consistent with
one another:
• the module A has an Z-grading, that is, a choice of submodules An , indexed by Z, such
that A = ⊕n An
• the module A possesses a differential d : A → A, that is, d ◦ d = 0, and the differential
respects the grading, in fact, is of degree ±1, that is, d : An → An±1
• the module A has a filtration F , which may decreasing, in which case d is required to have
degree one, or increasing, in which case d is required to have degree negative one, and this
filtration is compatible with the differential in that d : F p A → F p A.
Associated to a filtered differential graded R-module A, we have the following constructions:
• we have a filtration on each An given by F p An = F p A ∩ An
• we have the associated graded module, which is given by ⊕p F p A/F p+1 A when F is decreasing
and by ⊕p F p A/F p−1 A when F is increasing
• the associated graded module is not only Z-graded but in fact Z-bigraded if we form the
associated graded module for each graded component of M , given by
M F p A ∩ Ap+q
M F p Ap+q
=
F p+1 Ap+q
F p+1 A ∩ Ap+q
p,q
p,q
when F is decreasing and by
M F p Ap+q
p,q
F p−1 Ap+q
=
M F p A ∩ Ap+q
F p−1 A ∩ Ap+q
p,q
when F is increasing; these direct summands aren’t directly submodules of the associated
graded module, but note that the kernels of the maps
⊆
F p A ∩ Ap+q → F p A → F p A/F p+1 A
and
⊆
F p A ∩ Ap+q → F p A → F p A/F p−1 A
are exactly F p+1 A ∩ Ap+q and F p−1 A ∩ Ap+q respectively giving canonical injections
F p Ap+q /F p+1 Ap+q → F p A/F p+1 A
and
F p Ap+q /F p−1 Ap+q → F p A/F p−1 A
where a class of an element is sent to the class of the same element (it can also be checked
that the doubly infinite direct sum of these submodules does indeed give the associated
graded module)
• note that for each fixed p, d gives a differential on F p A/F p+1 A when F is decreasing and on
F p A/F p−1 A when F is increasing, so that we can consider the homology H(F p A/F p+1 A) or
H(F p A/F p−1 A) and this homology is in fact Z-graded: in the case of a decreasing filtration
(the increasing case is similar), let
H n (F p A/F p+1 A) = {[z] ∈ H(F p A/F p+1 A) | z = [w] ∈ F p A/F p+1 A, w ∈ An }
which is easily checked to be a submodule and it is also easily checked that H(F p A/F p+1 A)
is indeed the direct sum of the H n (F p A/F p+1 A)
• the homology of A with respect to d is Z-graded by taking the homology of each graded
part: H(A) = ⊕n ker dn /im dn−1 when d has degree one and H(A) = ⊕n ker dn /im dn+1
when d has degree negative one (note that the direct summands aren’t submodules directly
but inject canonically); the homology is also filtered by taking the homology of each filtered
part: F p H(A) = im(H(F p A) → H(A)).
1.4. Spectral sequences from filtered differential graded modules
5
Given a filtered differential graded R-module A, where say the differential is of degree one and the
filtration is decreasing (for the case of an increasing filtration and a differential of degree one, we
simply re-index things to get the analogous result), we will construct a spectral sequence (Er , dr , ϕr ),
with r ≥ 1, in the category of Z-bigraded R-modules, such that Erp,q ∼
= H p+q (F p A/F p+1 A) and
each dr is of bidegree (r, 1 − r).
To begin, note that the filtration
· · · ⊆ F p+1 A ⊆ F p A ⊆ F p−1 A ⊆ · · · ⊆ A
gives rise to a filtration
· · · ⊆ F p+1 An ⊆ F p An ⊆ F p−1 An ⊆ · · · ⊆ An
on each An for n ∈ Z where F p An = F p A ∩ An . Note also that d(F p An ) ⊆ F p An+1 . Now, for fixed
p, q ∈ Z and r ≥ 0, set
Zrp,q = F p Ap+q ∩ d−1 (F p+r Ap+q+1 )
Brp,q = F p Ap+q ∩ d(F p−r Ap+q−1 ).
Note the following:
• we have the inclusions
B0p,q ⊆ B1p,q ⊆ · · · ⊆ Z1p,q ⊆ Z0p,q
• we have
d(Zrp,q ) = Brp+r,q−r+1
and
d(Zrp−r,q+r−1 ) = Brp,q ;
to see this, note that the latter implies the former and that if z ∈ Z p−r,q+r−1 , then z ∈
F p−r Ap+q−1 so that d(z) ∈ d(F p−r Ap+q−1 ) and also z ∈ d−1 (F p Ap+q ) so that d(z) ∈
F p Ap+q ; further, if z ∈ Brp,q , then z = d(z1 ) for some z1 ∈ F p−r Ap+q−1 and also, since
z ∈ F p Ap+q , z1 also lies in d−1 (F p Ap+q ) so that z1 ∈ Zrp−r,q+r−1 and z ∈ d(Zrp−r,q+r−1 ).
Now, again for fixed p, q ∈ Z, r ≥ 1, set
p,q
p+1,q−1
)
+ Br−1
Erp,q = Zrp,q /(Zr−1
p+1,q−1
p,q
where the inclusions Zr−1
⊆ Zrp,q and Br−1
⊆ Zrp,q follow immediately from the definitions.
The objects of our spectral sequence are ⊕p,q Erp,q , which we also denote by Er∗,∗ .
Next, we want to define the differential on Er∗,∗ . To do so, first note that d : Zrp,q → Zrp+r,q−r+1 and
let ηrp,q : Zrp,q → Erp,q be the canonical projection. Now we want to solve the following problem:
Zrp,q
d
ηrp,q
Erp,q
Zrp+r,q−r+1
ηrp+r,q−r+1
dp,q
r
Erp+r,q−r+1
This problem will be solved if the kernel of ηrp,q lies inside the kernel of ηrp+r,q−r+1 ◦ d; this is
p+1,q−1
p,q
p+1,q−1
p,q
p+r,q−r+1
indeed the case as we have d(Zr−1
+ Br−1
) = d(Zr−1
) + d(Br−1
) ⊆ Br−1
+0 ⊆
p+r+1,q−r
p+r,q−r+1
∗,∗
∗,∗
Zr−1
+ Br−1
. This gives us morphisms dr : Er → Er and we have dr ◦ dr = 0 since
d ◦ d = 0. Thus we have defined our differentials.
Note the following:
6
1. Spectral sequences as algebraic objects
• since the above square commutes and ηrp,q is surjective, we have
p+r,q−r+1
im dp,q
(d(Zrp,q ))
r = ηr
or equivalently, upon replacing p with p − r and q with q + r − 1, we have
im dp−r,q+r−1
= ηrp,q (d(Zrp−r,q+r−1 )).
r
p,q
Next, we want to define isomorphisms ϕp,q
: H p,q (Er∗,∗ , dr ) → Er+1
. To do this, consider the
r
following diagram:
Zrp+1,q−1 + Brp,q
p,q
Zr+1
Zrp,q
d
ηrp,q
η∗
ker dp,q
r
Erp,q
Zrp+r,q−r+1
ηrp+r,q−r+1
dp,q
r
Erp+r,q−r+1
H p,q (Er∗,∗ , dr )
0
p,q
p+1,q−1
)
+ Br−1
The only arrow here which needs explanation is η∗ : we know that ker dp,q
r = Z/(Zr−1
p,q
p,q
p+1,q−1
+ Br−1 ⊆ Z ⊆ Zr and what we want to show is that
for some submodule Z satisfying Zr−1
p,q
p,q
p+r+1,q−r
Zr+1 ⊆ Z. This is true because if z ∈ Zr+1 , then d(z) ∈ F p+r+1 Ap+q+1 and so d(z) ∈ Zr−1
⊆
p+r,q−r+1
p+r+1,q−r
= ker ηrp+r,q−r+1 . Note that η∗ is just a restriction of ηrp,q .
+ Br−1
Zr−1
p,q
p,q
Now, while we may or may not have that Zr+1
= Z, we can at least show that Zr+1
contains at
p,q
least one representative of each class in ker dr ; that is, that η∗ is surjective. To see this, note
p+r+1,q−r
p+r,q−r+1
+ Br−1
. Thus
that if z ∈ Z, then z ∈ Zrp,q and so z ∈ F p Ap+q ; further, d(z) ∈ Zr−1
p+r+1,q−r
d(z) = z1 +d(z2 ) where z1 ∈ Zr−1
and z2 ∈ F p+1 Ap+q . We have d(z−z2 ) ∈ F p+r+1 Ap+q+1 and
p+q
also z − z2 ∈ F p Ap+q since z ∈ Zrp,q and z2 ∈ F p+1 Ap+q ⊆ F p Ap+q ; this shows that z − z2 ∈ Zr+1
.
p+r,q−r+1
p+1
p+q
p+r
p+q+1
Also, note that z2 ∈ F
A
as before and further, d(z2 ) ∈ Br−1
⊆ F
A
; this
p+1,q−1
p,q
p+1,q−1
shows that z2 ∈ Zr−1
. Thus z = (z − z2 ) + z2 ∈ Zr+1 + Zr−1
. We will be done if we can
show that η∗ (z − z2 ) and the class of z in ker dp,q
are
equivalent;
this
amounts to showing that
r
p+1,q−1
p,q
p+1,q−1
z2 ∈ Zr−1
+ Br−1 , but this is true because, as we have already seen, z2 ∈ Zr−1
.
η∗
p,q
Next, we note that the composition Zr+1
→ ker dp,q
→ H p,q (Er∗,∗ , dr ) is surjective and want to
r
p,q
identify its kernel. This kernel is exactly Zr+1
∩ (ηrp,q )−1 (im dp−r,q+r−1
), and we claim that this is
r
p+1,q−1
p,q
p,q −1
p−r,q+r−1
exactly Zr
+ Br . We first identify (ηr ) (im dr
). Recall that Brp,q = d(Zrp−r,q+r−1 )
p−r,q+r−1
p,q
p−r,q+r−1
and that im dr
= ηr (d(Zr
)) and then note that ηrp,q (Brp,q ) = ηrp,q (d(Zrp−r,q+r−1 )) =
p,q
im dp−r,q+r−1
. From this we see that (ηrp,q )−1 (im drp−r,q+r−1 ) = Brp,q + ker ηrp,q = Brp,q + Br−1
+
r
p+1,q−1
p,q
p+1,q−1
p,q
p+1,q−1
p,q
Zr−1
= Br + Zr−1
. Now, we want to find the intersection (Br + Zr−1
) ∩ Zr+1 . To
do this, note that, as is easily checked, given submodules A, B, C of some module, if A ⊆ C, then
p,q
(A + B) ∩ C = A + B ∩ C. In our case, we note that Brp,q ⊆ Zr+1
and so need only to identify
1.4. Spectral sequences from filtered differential graded modules
7
p+1,q−1
p,q
Zr−1
∩ Zr+1
. This intersection is given by
F p+1 Ap+q ∩ d−1 (F p+r Ap+q+1 ) ∩ F p Ap+q ∩ d−1 (F p+r+1 Ap+q+1 )
which reduces to
F p+1 Ap+q ∩ d−1 (F p+r+1 Ap+q+1 ) = Zrp+1,q−1 .
Thus, the kernel of our original composite map is Brp,q + Zrp+1,q−1 , as desired. What this shows is
p,q
that our composite map descends to an isomorphism Er+1
→ H p,q (Er∗,∗ , dr ) and we let ϕp,q
r be the
∗,∗
∗,∗
∗,∗
∗,∗
inverse of this isomorphism. Our isomorphisms H (Er , dr ) → Er+1 are then ϕr = ⊕p,q ϕp,q
r .
We now have a spectral sequence (Er∗,∗ , dr∗,∗ , ϕr∗,∗ ) and finally, we wish to identify the first page; in
particular, we want to show that E1p,q ∼
= H p+q (F p A/F p+1 A). We have
E1p,q =
Z1p,q
Z0p+1,q−1 + B0p,q
F p Ap+q ∩ d−1 (F p+1 Ap+q+1 )
F p+1 Ap+q ∩ d−1 (F p+1 Ap+q+1 ) + F p Ap+q ∩ d(F p Ap+q−1 )
F p Ap+q ∩ d−1 (F p+1 Ap+q+1 )
=
.
F p+1 Ap+q + d(F p Ap+q−1 )
=
Recall that
H p+q (F p A/F p+1 A) = {[z] ∈ H(F p A/F p+1 A) | z = [w] ∈ F p A/F p+1 A, w ∈ Ap+q }.
Now, consider the map
F p Ap+q ∩ d−1 (F p+1 Ap+q+1 ) → H p+q (F p A/F p+1 A) : w 7→ [z] ∈ H(F p A/F p+1 A)
where z = [w] ∈ F p A/F p+1 A
which is easily checked to be well-defined, a module homomorphism and surjective. Further, w
lies in the kernel of this map if and only if [w] = [dy] for some y ∈ F p A, where the classes are
taken modulo F p+1 A; then w − dy ∈ F p+1 A and also since w ∈ Ap+q and d(w − dy) = dw, we
must have w − dy ∈ F p+1 Ap+q . Further, then dy = w − (w − dy) ∈ Ap+q so that y ∈ Ap+q−1
and dy ∈ d(F p Ap+q−1 ) and so finally w = (w − dy) + dy ∈ F p+1 Ap+q + d(F p Ap+q−1 ). Conversely,
if w ∈ F p+1 Ap+q + d(F p Ap+q−1 ), it is clear that w lies in the kernel and so our map induces an
isomorphism
∼
=
E1p,q → H p+q (F p A/F p+1 A).
This completes the construction of our spectral sequence and the identification of its first page.
Next, we want to identify its infinity term. This can be done under the assumption that the
filtration F is bounded. That is, we assume that for each n ∈ Z, there are integers s, t ∈ Z (which
may vary with n, in which case we write s = s(n) and t = t(n)), such that
0 = F s An ⊆ · · · ⊆ F t An = An .
p,q ∼
Under this assumption, we we will show that E∞
= F p H p+q (A, d)/F p+1 H p+q (A, d). First note
that
Zrp,q = F p Ap+q ∩ ker d
when
r ≥ s(p + q + 1) − p
and
Brp,q = F p Ap+q ∩ d(Ap+q−1 ) = F p Ap+q ∩ im d
when
For this reason, we define
p,q
Z∞
= F p Ap+q ∩ ker d
r ≥ p − t(p + q − 1).
8
1. Spectral sequences as algebraic objects
and
p,q
B∞
= F p Ap+q ∩ im d.
Note the following:
• we have
p,q
p,q
B0p,q ⊆ B1p,q ⊆ · · · ⊆ B∞
⊆ Z∞
⊆ · · · ⊆ Z1p,q ⊆ Z0p,q
p,q
p+1,q−1
p,q
• for sufficiently large r, Erp,q = Z∞
/(Z∞
+ B∞
).
Now, in defining the infinity term of a spectral sequence, we named submodules Z∗ and B∗ ; since
we have already used these symbols, we instead name these P∗ and Q∗ respectively. That is, for
p−r,q+r−1
fixed p, q and r ≥ 1, we let Qp,q
and Prp,q = ker dp,q
r = im dr
r and then define submodules
p,q
p,q
p,q
Qr,1 ⊆ Pr,1 ⊆ E1 , which will satisfy
p,q
p,q
p,q
p,q
p,q
p,q
p,q
p,q
Qp,q
1,1 ⊆ Q2,1 ⊆ Q3,1 ⊆ · · · ⊆ Qr,1 ⊆ · · · ⊆ Pr,1 ⊆ · · · ⊆ P3,1 ⊆ P2,1 ⊆ P1,1 ⊆ E1
as follows:
p,q
p,q
p,q
• for r = 1, set Qp,q
1,1 = Q1 and P1,1 = P1
• for r = 2, we have
p,q
Qp,q
2 ⊆ P2
in E2p,q and we expand this to
p,q
p,q
p,q
Qp,q
1 ⊆ Q2,1 ⊆ P2,1 ⊆ P1
p,q
p,q
p,q
p,q −1
p,q
p,q
p,q
−1
in E1p,q by setting (ϕp,q
1 ) (Q2 ) = Q2,1 /Q1 and (ϕ1 ) (P2 ) = P2,1 /Q1 ; to see exactly
p,q
what Qp,q
2,1 and P2,1 are, recall the diagram
Zrp+1,q−1 + Brp,q
p,q
Zr+1
Zrp,q
d
ηrp,q
η∗
ker dp,q
r
Erp,q
Zrp+r,q−r+1
ηrp+r,q−r+1
dp,q
r
Erp+r,q−r+1
H p,q (Er∗,∗ , dr )
0
from which we see that taking preimages under ϕ is essentially just taking images under
p,q
η; in particular, Qp,q
2,1 and P2,1 are formed by just taking the classes of elements modulo
Z0p+1,q−1 + B0p,q rather than Z1p+1,q−1 + B1p,q
p,q
• more generally, Qp,q
r,1 and Pr,1 are formed by just taking the classes of elements modulo
p+1,q−1
p,q
Z0p+1,q−1 + B0p,q rather than Zr−1
+ Br−1
.
Now, we have:
p+1,q−1
p+r,q−r+1
p,q
p+r,q−r+1
Prp,q = ker(dp,q
) = classes of elements in{z|z ∈ Zrp,q anddz ∈ Zr−1
+Br−1
}
r : Er → Er
1.4. Spectral sequences from filtered differential graded modules
9
p+1,q−1
p+r,q−r+1
and since Zr−1
= F p+r+1 Ap+q+1 ∩d−1 (F p+2r Ap+q+2 ) and Br−1
= F p+r Ap+q+1 ∩d(F p+1 Ap+q ),
we have that for sufficiently large r,
Prp,q = classes of elements in {z | z ∈ Zrp,q and z ∈ ker d}
= classes of elements in Zrp,q ∩ ker d
= classes of elements in F p Ap+q ∩ d−1 (F p+r Ap+q+1 ) ∩ ker d
p,q
= classes of elements in F p Ap+q ∩ ker d = Z∞
.
Further,
p−r,q+r−1
Qp,q
: Erp−r,q+r−1 → Erp,q ) = classes of elements in d(Zrp−r,q+r−1 )
r = im (dr
and since Zrp−r,q+r−1 = F p−r Ap+q−1 ∩ d−1 (F p Ap+q ), we have that for sufficiently large r,
p+q−1
Qp,q
∩ d−1 (F p Ap+q ))
r = classes of elements in d(A
p,q
= classes of elements in F p Ap+q ∩ im d = B∞
.
p,q
All told, we have shown that the Pr,1
and Qp,q
r,1 stabilize and
p,q
E∞
=
=
p,q
P∞
Qp,q
∞
p,q
Z∞
+ Z0p+1,q−1 + B0p,q
,
Z0p+1,q−1 + B0p,q
p,q
B∞
+ Z0p+1,q−1 + B0p,q
Z0p+1,q−1 + B0p,q
p,q
Z∞
+ Z0p+1,q−1 + B0p,q
∼
= p,q
B∞ + Z0p+1,q−1 + B0p,q
=
p,q
+ Z0p+1,q−1 + B0p,q
Z∞
p,q
B∞
+ Z0p+1,q−1
.
p,q
p,q
p,q
p,q
+
+ Z0p+1,q−1 + B0p,q )/(B∞
+ Z0p+1,q−1 + B0p,q → (Z∞
→ Z∞
Note that the composite map Z∞
p,q
p,q
p,q
p,q
∩ Z0p+1,q−1 =
+ Z∞
+ Z0p+1,q−1 ) = B∞
∩ (B∞
Z0p+1,q−1 ) is clearly surjective and has kernel Z∞
p,q
p+1,q−1
p,q
∩ Z0p+1,q−1 = F p Ap+q ∩ ker d ∩ F p+1 Ap+q ∩
where the last equality comes from Z∞
+ Z∞
B∞
p+1,q−1
. Thus
d−1 (F p+1 Ap+q+1 ) = F p+1 Ap+q ∩ ker d = Z∞
p,q ∼ p,q
p+1,q−1
E = Z /(Z
+ B p,q ).
∞
∞
∞
p,q
Er
Note that this is a limit version of the identity
H(A, d) be the natural projection. We claim that
∞
p,q
p+1,q−1
p,q
= Zr /(Zr−1
+ Br−1
). Now, let π
p,q
p
p+q
π(Z∞ ) = F H (A, d). Recall that
: ker d →
F p H p+q (A, d) = F p H(A, d) ∩ H p+q (A, d)
= im(H(F p A) → H(A)) ∩ im(H(Ap+q ) → H(A)).
p,q
Now, if z ∈ Z∞
= F p Ap+q ∩ ker d, then since z ∈ F p A, π(z) ∈ im(H(F p A) → H(A)) and since z ∈
p+q
A , π(z) ∈ im(H(Ap+q ) → H(A)). Conversely, if [z] ∈ im(H(F p A) → H(A)) ∩ im(H(Ap+q ) →
H(A)), then since H(Ap+q ) injects into H(A), z ∈ Ap+q . Further, since [z] = [z1 ] for some z1 ∈
F p A ∩ ker d, we have z = z1 + z2 for some z2 ∈ im d. Then since dz = dz1 , we have z1 ∈ Ap+q . Thus
p,q
p,q
[z] = π(z1 ) ∈ π(F p Ap+q ∩ker d) = π(Z∞
). Thus π restricts to a surjection π : Z∞
→ F p H p+q (A, d)
p+1,q−1
p,q
and next we want to show that π(Z∞
+ B∞
) ⊆ F p+1 H p+q (A, d). To see this, note that if
p+1
p+q
p
p+q
z1 ∈ F
A
∩ ker d and z2 ∈ F A
∩ im d, then π(z1 + z2 ) = π(z1 ) ∈ F p+1 H p+q (A, d). Thus
p,q
p+1,q−1
p,q
π induces a surjection d∞ : Z∞ /(Z∞
+ B∞
) → F p H p+q (A, d)/F p+1 H p+q (A, d). Finally,
p+1
p+q
suppose that d∞ [z] = 0. Then π(z) ∈ F
H (A, d). We have z ∈ Ap+q since H(Ap+q ) injects
into H(A) and also z = z1 + z2 for some z1 ∈ F p+1 A ∩ ker d and z2 ∈ im d. Since dz1 = dz2 ,
p,q
z1 ∈ Ap+q and so also z2 ∈ Ap+q . Since z ∈ Z∞
and so z ∈ F p A and also z1 ∈ F p+1 A, we have
10
1. Spectral sequences as algebraic objects
p+1,q−1
p,q
z2 ∈ F p A and so z2 ∈ F p Ap+q ∩ im d. Thus z ∈ Z∞
+ B∞
and so [z] = 0. Thus d∞ is an
p,q ∼ p p+q
isomorphism and so, from what we have already shown, E∞ = F H (A, d)/F p+1 H p+q (A, d).
1.5. Spectral sequences from exact couples
Definition 1.3 (exact couples). An exact couple is a datum (A, E, α, f, g) of objects A, E and
morphisms α, f, g in the fixed abelian category A as in the diagram
α
A
A
g
f
E
such that this diagram is exact.
Spectral sequences from exact couples. Given an exact couple (A, E, α, f, g), we have the following:
• g◦f : E → E and (g◦f )◦(g◦f ) = g◦(f ◦g)◦f = 0 and so we may set E 0 = ker(g◦f )/im(g◦f )
• we can also set A0 = im(α)
0
• we necessarily have α(A0 ) ⊆ im(α) = A0 and so we may set α0 = α|A
A0
• given e ∈ ker(g ◦ f ), we have g(f (e)) = 0 so that f (e) ∈ ker(g) = im(α); we also have
that if e ∈ im(g ◦ f ), say e = g(f (e0 )), then f (e) = (f ◦ g)(f (e0 )) = 0; thus we have a map
f 0 : E 0 → A0 : [e] 7→ f (e)
• given a ∈ A0 , say a is equal to both α(a0 ) and α(a00 ), then α(a0 − a00 ) = 0 and so a0 − a00 ∈
ker(α) = im(f ), say a0 − a00 = f (e); then we have g(a0 ) − g(a00 ) = g(a0 − a00 ) = g(f (e)) ∈
im(g ◦ f ) and so we have a well-defined map g 0 : A0 → E 0 : α(a) 7→ [g(a)].
We now claim that (A0 , E 0 , α0 , f 0 , g 0 ), depicted as in the diagram
α0
A0
f0
A0
g0
E0
is an exact couple. To see this, we have the following:
• let α(a) ∈ ker α0 , then α(α(a)) = 0 so that α(a) ∈ ker α = im f , say α(a) = f (e). Since
(g ◦ f )(e) = g(f (e)) = g(α(a)) = 0, e ∈ ker(g ◦ f ) and so α(a) = f 0 [e]; this shows that
ker α0 ⊆ im f 0 ; next, consider [e] ∈ E 0 and note that α0 (f 0 [e]) = α0 (f (e)) = α(f (e)) = 0 so
that im f 0 ⊆ ker α0
• let α(a) ∈ ker g 0 , then [g(a)] = 0 and so g(a) ∈ im(g ◦ f ), say g(a) = g(f (e)); we have
a − f (e) ∈ ker g = im α, say a − f (e) = α(a0 ), then α(a) = α(α(a0 )) ∈ im α0 ; this shows that
ker g 0 ⊆ im α0 ; next, let α(α(a)) ∈ im α0 , then g 0 (α(α(a))) = [g(α(a))] = 0; this shows that
im α0 ⊆ ker g 0
• now let [e] ∈ ker f 0 , then f (e) = 0 so that e ∈ ker f = im g, say e = g(a), then g 0 (α(a)) =
[g(a)] = [e]; this shows that ker f 0 ⊆ im g 0 ; now consider g 0 (α(a)) = [g(a)] and note that
f 0 [g(a)] = f (g(a)) = 0; this shows that im g 0 ⊆ ker f 0 .
We call this new exact couple the derived couple of (A, E, α, f, g).
We thus we have a sequence of derived couples: (A0 , E0 , α0 , f0 , g0 ), (A1 , E1 , α1 , f1 , g1 ), . . . and from
this get a sequence of modules Er and of differentials dr = gr ◦fr . We also have Er+1 = ker dr /im dr
so that we can simply take the isomorphisms ϕr to be identities. Thus each exact couple gives rise
to a spectral sequence.
1.5. Spectral sequences from exact couples
11
The case of bigraded modules. Now suppose that we are in the category of bigraded modules, so
that A = A∗,∗ and E = E ∗,∗ are bigraded (note that then A0 = im(α) is then bigraded by setting
(A0 )p,q = Ap,q ∩ im(α) = α(Ap+1,q−1 )) and suppose that α, g, f have bidegrees (−1, 1), (0, 0), (1, 0)
respectively:
α, (−1, 1)
A
A
f, (1, 0)
g, (0, 0)
E
The reason we choose these bidegrees is of course because this is a case which occurs often naturally
in practice. We present such an exact couple sometimes as:
α
f
α
g
Ap+2,q−1
E p+2,q−1
f
Ap+3,q−1
α
f
α
g
Ap+1,q
E p+1,q
f
α
g
Ap,q+1
E p,q+1
f
Ap+1,q+1
α
and sometimes as:
α
g
α
α
Ap+1,?
g
Ap+2,q
α
f
g
g
f
E p,∗
α
Ap,∗
Ap−1,∗
α
g
f
E p−1,∗
where in the latter case we say that the exact couple has been unrolled (note that an asterisk always
means that a direct sum has been taken over that index).
In this case we number the couples in the sequence of derived couples as (A1 , E1 , α1 , f1 , g1 ),
(A2 , E2 , α2 , f2 , g2 ), . . . and then want to show that dr then has bidegree (r, 1 − r), so that the
spectral sequence if of cohomological type. What we will actually show is that gr has bidegree
(r − 1, 1 − r) and fr has bidegree (1, 0), from which it follows that dr = gr ◦ fr has bidegree
(r − 1, 1 − r) + (1, 0) = (r, 1 − r). For r = 1, we have this by assumption. Assume now that gr
p,q
and fr do indeed have bidegrees (r − 1, 1 − r) and (1, 0) respectively. Given [e] ∈ Er+1
, we have
p+1,q
fr+1 [e] = fr (e) ∈ Ap+1,q
∩
im(α)
=
A
and
so
f
has
bidegree
(1,
0).
Also,
given
α(a)
∈ Ap,q
r
r+1
r+1
r+1 ,
p+1,q−1
p+1+r−1,q−1+1−r
p+r,q−r
so that a ∈ Ar
, we have gr+1 (α(a)) = [g(a)] and g(a) ∈ Er
= Er
and
so gr+1 has bidegree (r, −r); this completes the induction. Note also that αr always has bidegree
12
1. Spectral sequences as algebraic objects
(−1, 1).
The infinity term. We now want to isolate the infinity term of the spectral sequence arising from
an exact couple; we can give some information towards this in the case of bigraded modules. We
let
Zrp,∗ = f −1 (im αr−1 : Ap+r,∗ → Ap+1,∗ )
Brp,∗ = g(ker αr−1 : Ap,∗ → Ap−r+1,∗ )
and
both of which are submodules of E p,∗ . Note that
B1p,∗ ⊆ B2p,∗ ⊆ · · · ⊆ Z2p,∗ ⊆ Z1p,∗ .
Now we first claim that
Erp,∗ ∼
= Zrp,∗ /Brp,∗
for each r ≥ 1. For r = 1 this is clear because E1p,∗ = E p,∗ , Z1p,∗ = f −1 (Ap+1,∗ ) = E p,∗ and
B1p,∗ = g(0) = 0. For r = 2, we note that
E2p,∗ =
ker(g p+1,∗ ◦ f p,∗ )
(f p,∗ )−1 (ker g p+1,∗ )
f −1 (im α : Ap+2,∗ → Ap+1,∗ )
=
=
im(g p,∗ ◦ f p−1,∗ )
g p,∗ (im f p−1,∗ )
g(ker α : Ap,∗ → Ap−1,∗ )
so that the result holds for this value of r as well. More generally, this same argument shows that
p,∗
=
Er+1
ker(grp+1,∗ ◦ frp,∗ )
im(grp−r+1,∗ ◦ frp−r,∗ )
=
(frp,∗ )−1 (ker grp+1,∗ )
grp−r+1,∗ (im frp−r,∗ )
=
fr−1 (im αr : Ap+2,∗
→ Ap+1,∗
)
r
r
gr (ker αr : Ap−r+1,∗
→ Arp−r,∗ )
r
.
p+r+1,∗
p+3,∗
) =
) = αr αr−1 αr−2 (Ap+4,∗
Note that αr (Ap+2,∗
) = αr αr−1 (Ar−1
r
r−2 ) = · · · = αr · · · α1 (A1
p−r+1,∗
r−1
0
0
p,∗
r
p+r+1,∗
, it is clear that a = α (a ) for some a ∈ A so that if
α (A
). Also, given a ∈ Ar
r−1
r
p,∗
) ⊆ αr−1 (ker αr :
a ∈ ker αr , then a ∈ α (ker α : A → Ap−r,∗ ); thus ker(αr : Arp−r+1,∗ → Ap−r,∗
r
p,∗
p−r,∗
r−1
r
p,∗
p−r,∗
A →A
). Conversely, if a ∈ α (ker α : A → A
), then firstly a ∈ Arp−r+1,∗ because
= αr−1 (Ap,∗ ) and also we have αr (a) = α(a) = αr (a0 ) for some a0 ∈ ker(αr : Ap,∗ →
Ap−r+1,∗
r
p−r,∗
A
), which is zero; thus αr−1 (ker αr : Ap,∗ → Ap−r,∗ ) ⊆ ker(αr : Ap−r+1,∗
→ Ap−r,∗
). This
r
r
shows that
f −1 (im αr : Ap+r+1,∗ → Ap+1,∗ )
p,∗
= r r−1
Er+1
.
gr (α (ker αr : Ap,∗ → Ap−r,∗ ))
Now, define a map
fr−1 (im αr
:A
p+r+1,∗
p+1,∗
→A
)→
−1
(im αr : Ap+r+1,∗ → Ap+1,∗ )
fr−1
p−r,∗
2
gr−1 (ker αr−1
: Ap−r+2,∗
→ Ar−1
)
r−1
p,∗
by mapping e ∈ fr−1 (im αr : Ap+r+1,∗ → Ap+1,∗ ) to a representative e0 in Er−1
which is well0
0
00
0
00
defined because fr (e) = fr−1 (e ) and if e = [e ] = [e ], then e − e ∈ gr−1 (ker αr−1 : Ap−r+2,∗
→
r−1
p−r+1,∗
p−r+2,∗
p−r,∗
−1
2
r
p+r+1,∗
p+1,∗
Ar−1
) ⊆ gr−1 (ker αr−1 : Ar−1
→ Ar−1 ). Now, given e ∈ fr−1 (im α : A
→A
)
we have that (gr−1 ◦ fr−1 )(e) ∈ gr−1 (im αr : Ap+r+1,∗ → Ap+1,∗ ) ⊆ (gr−1 ◦ α)(αr−1 (Ap+r+1,∗ )) = 0
and from this we see that our map is surjective. Now suppose that e is in the kernel of this
p−r,∗
2
map; then e = [e0 ] where e0 ∈ gr−1 (ker αr−1
: Ap−r+2,∗
→ Ar−1
), say e0 = gr−1 (a), then e =
r−1
p−r+2,∗
p−r+1,∗
2
gr (α(a)); since a ∈ A
, α(a) ∈ A
and since α (a) = 0, α(α(a)) = 0 and so we have
p−r+1,∗
p−r,∗
e ∈ gr (ker αr : Ar
→ Ar
). Conversely, suppose that e ∈ gr (ker αr : Ap−r+1,∗
→ Arp−r,∗ ),
r
p−r+2,∗
then e = gr (a) = gr (αr−1 (a0 )) = [gr−1 (a0 )] where a ∈ Ap−r+1,∗
and so a0 ∈ Ar−1
and we see that
r
2
0
2
0
e will lie in the kernel of our map if α (a ) = 0; but this holds because α (a ) = α(a) = 0. Thus
the kernel of our map is precisely gr (ker αr : Ap−r+1,∗
→ Arp−r,∗ ) and so we get an isomorphism
r
p,∗
Er+1
=
fr−1 (im αr : Ap+r+1,∗
→ Ap+1,∗
)
r
r
gr (ker αr : Ap−r+1,∗
→ Arp−r,∗ )
r
→
−1
fr−1
(im αr : Ap+r+1,∗ → Ap+1,∗ )
2
gr−1 (ker αr−1
: Ap−r+2,∗
→ Ap−r,∗
r−1
r−1 )
.
1.5. Spectral sequences from exact couples
13
Repeating this procedure, we find that
−1 (im αr : Ap+r+1,∗ → Ap+1,∗ )
p,∗ ∼ f
Er+1
.
=
g(ker αr : Ap,∗ → Ap−r,∗ )
Next, we claim the limiting version of the identity just proven, that is:
E p,∗ ∼
= ∩r Z p,∗ / ∪r E p,∗ .
∞
r
r
This proof will be completed later. Note that the Z and B terms here are not those defined in the
definition of the infinity terms.
The case of filtered differential graded modules. The spectral sequence arising from a filtered differential graded module can be seen as a special case of the construction in this section. To see this,
we first construct an exact couple out of a filtered differential graded module . . . to be completed.
Chapter 2
Spectral sequences in topology
2.1. Leray-Serre
2.2. Eilenberg-Moore
2.3. Adams
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