An introduction to spectral sequences [Lecture notes]

12 Pages·2011·0.152 MB·English

by  Steven Dale Cutkosky

AN INTRODUCTION TO SPECTRAL SEQUENCES
STEVEN DALE CUTKOSKY
1. Introduction
In these notes we give an introduction to spectral sequences. Our exposition owes much
to the treatments in Eisenbud’s book [2] and in Rotman’s book [4].
2. Complexes and Differential Modules
Suppose that R is a commutative ring with identity. A complex of R-modules is a
sequence of R-module homomorphisms
(1) F : ··· → F d→i+1 F →di F → ···
i+1 i i−1
where d2 = 0, or a a sequence of R-module homomorphisms
(2) F(cid:48) : ··· → Fi−1 d→i−1 Fi →di Fi+1 → ···
where d2 = 0. The differential d has degree -1 in (1) and degree +1 in (2).
We will often write F∗ (or (F(cid:48))∗) to denote these complexes.
Suppose that M is an R-module, and
(3) ··· → F → F → F → M → 0
2 1 0
is a free resolution of M. Suppose that N is an R-module. Then
Hom (F ,N) → Hom (F ,N) → ···
R 0 R 1
is a complex of degree 1, and
··· → Hom (N,F ) → Hom (N,F )
R 1 R 0
is a complex of degree -1.
The homology of F at F in (1) is H (F) = ker d /im d . The homology of F(cid:48) at Fi
i i i i+1
in (2) is Hi(F(cid:48)) = ker d /im d .
i i−1
The complex (1) (with differential of degree -1) can be converted to a complex of degree
+1 by setting F−i = F . For instance, (3) becomes the exact complex
i
··· → F−2 → F−1 → F0 → M → 0
of degree 1.
A differential module (F,ϕ) is an R-module F with and endomorphism ϕ such that
ϕ2 = 0. The homology of F is H(F) = ker ϕ/imϕ. The differential module F is exact
of H(F) = 0. If (F,ϕ) and (G,ψ) are differential modules, then a homomorphism of
differentiable modules Λ : F → G is an R-module homomorphism which satisfies Λ◦ϕ =
ψ◦Λ.
Suppose that F is a complex (which we will assume has degree -1, although the con-
struction with “upper indices” is valid when the degree is 1). Then F can be viewed as
(cid:76)
a differential module F, by setting F = F . The map ϕ : F → F is defined by the
i∈Z i
∼ (cid:76)
differentials d : F → F . We have that H(F) = H (F).
i i i−1 i∈Z i
1
If
β
0 → F(cid:48) →α F → F(cid:48)(cid:48)→ 0
is a short exact sequence of differential modules, then there is a connecting homorphism
δ : H(F(cid:48)(cid:48)) → H(F(cid:48)), making
H(F(cid:48)) →α H(F)
(4) δ (cid:45) (cid:46) β
H(F(cid:48)(cid:48))
an exact triangle; that is ker α = im δ, ker δ = im β and ker β = im α. Th construction
of δ is as in the construction of the connecting homomorphism of a short exact sequence
of complexes. If (4) is a short exact sequence of complexes, then δ is the direct sum of the
connecting homorphisms of the homology modules of the complexes.
3. Spectral Sequences
Definition 3.1. A Spectral sequence is a sequence of modules rE for r ≥ 1, each with a
differential d : rE → rE satisfying d d = 0, such that r+1E ∼= ker d /im d = H(rE).
r r r r r
Suppose that rE is a spectral sequence. We will define submodules
0 = 1B ⊂ 2B ⊂ ··· ⊂ rB ⊂ ··· ⊂ rZ ⊂ ··· ⊂ 2Z ⊂ 1Z = 1E
suchthatiE = iZ/iB foreachi. Todothis,let1Z = 1E and1B = 0,sothat1E = 1Z/1B.
Having defined iB and iZ for i ≤ r, we define r+1Z as the kernel of the composite map
rZ → rZ/rB = rE →dr rE = rZ/rB
and let the image of this map be r+1B/rB. We have that
r+1Z/r+1B = H(rE) = r+1E
and
rB ⊂ r+1B ⊂ r+1Z ⊂ rZ
as required. Having defined all the iZ and iB, we set
∞Z = ∩Z , ∞B = ∪B .
r r
We define the limit of the spectral sequence to be
∞E = ∞Z/∞B.
We say that the spectral sequence collapses at rE if rE = ∞E, or equivalently, the
differentials d = 0 for i ≥ r.
i
4. Exact Couples
Definition 4.1. An exact couple is a triangle
α
A → A
(5) γ (cid:45) (cid:46) β
E
where A and E are R-modules and α,β,γ are R-module homomorphisms which are exact
in the sense that ker α = im γ, ker γ = im β and ker β = im α.
Let d : E → E be the composite map d = βγ. Since γβ = 0, we have that d2 = 0, so E
is a differential module, with homology H(E) = ker d/im d.
2
Proposition 4.2. Let (5) be an exact couple. Then there is a derived exact couple
α(cid:48)
αA → αA
(6) γ(cid:48) (cid:45) (cid:46) β(cid:48)
H(E)
where
(1) α(cid:48) is α restricted to αA,
(2) β(cid:48) is βα−1 : αA → H(E), taking αa to the homology class of βa for a ∈ A,
(3) γ(cid:48) is the map induced by γ on ker d (which automatically kills im d).
β(cid:48) is well defined in the Proposition 4.2, since ker α = im γ is taken to im d by β.
We construct the spectral sequence of the exact couple (5) by iterating (6), so that
1E = E with differential d = d = βγ
1
2E = H(E) with differential d = β(cid:48)γ(cid:48) from the derived couple
2
3E = H(H(E)) from the derived couple of the derived couple
.
.
.
Proposition 4.3. With the notation after the definition of a spectral sequence, we have
r+1Z = γ−1(im αr)
r+1B = β(ker αr).
Thus
∞E = ∞Z/∞B
= γ−1(∩im αr)/β(∪ker αr).
Proof. We prove the proposition by induction on r, the case r = 0 being immediate.
Assume that r ≥ 1, and that rZ = γ−1(im αr−1) and rB = β(ker αr−1). We have that
r+1Z = {x ∈ rZ | βα1−rγ(x) ∈ rB}.
Suppose that x ∈ γ−1(im αr). Then γ(x) = αr(y) for some y ∈ A, so that x ∈
γ−1(im αr−1) = rZ and βα1−rγ(x) = βα(y) = 0. Thus γ−1(im αr) ⊂ r+1Z.
Now suppose that x ∈ r+1Z. Then βα1−rγ(x) ∈ rB implies βα1−rγ(x) = β(y) for some
y ∈ ker αr−1, so α1−rγ(x)−y ∈ ker β = im α and α1−rγ(x) = y+α(z) for some z ∈ A.
Thus γ(x) = αr−1(y)+αr(z) = αr(z) ∈ im αr, so that x ∈ γ−1(im αr).
We have established that r+1Z = γ−1(im αr).
We have that
r+1B = {βα1−rγ(x) | x ∈ γ−1(im αr−1)}.
Suppose that x ∈ β(ker αr). Then x = β(y) for some y ∈ ker αr. Thus αr−1(y) = γ(z)
for some z ∈ E, so that z ∈ γ−1(im αr−1) = rZ. Now y = α1−rγ(z) implies x =
βα1−rγ(z) ∈ r+1B. Thus β(ker αr) ⊂ r+1B.
Suppose that z ∈ r+1B. Then z = βα1−rγ(x) for some x ∈ γ−1(im αr−1). There exists
y ∈ A such that γ(x) = αr−1(y). We have 0 = αγ(x) = αr(y) implies y ∈ ker αr. We have
z = βα1−rγ(x) = β(y) ∈ β(ker αr). Thus r+1B ⊂ β(ker αr). (cid:3)
We have that
d : rE = rZ/rB → rZ/rB = rE
r
is defined by
d (x+rB) = βα1−rγ(x)+rB
r
for x ∈ rZ.
3
Suppose that (F,d) is a differential module, and let α : F → F be a monomorphism.
Set F = F/αF. The differential d of F induces a differential on F, giving a short exact
sequence of differential modules
α β
0 → F → F → F → 0.
By the construction of (4), we have an induced exact couple
α
H(F) → H(F)
(7) γ (cid:45) (cid:46) β
H(F)
The boundary map γ is defined by
(8) γ(x) = [α−1d(z)]
where z is a lift to F of a representative of x in F. β is the map on homology induced by
the projection of F onto F.
The induced spectral sequence of this exact couple is called the spectral sequence of α
on F.
We have
1E = H(F) = [{z ∈ F | dz ∈ αF}/αF]/[{dz | z ∈ F}+αF/αF]
= {z ∈ F | dz ∈ αF}/αF +dF,
and
H(F) = {z ∈ F | dz = 0}/dF.
α : H(F) → H(F) is defined by
α(x+dF) = α(x)+dF
for x ∈ {z ∈ F | dz = 0}. β : H(F) → H(F) is defined by
β(x+dF) = x+(αF +dF)
for x ∈ {z ∈ F | dz = 0}. γ : H(F) → H(F) is defined by
γ(x) = α−1dx+dF
for x ∈ {z ∈ F | dz ∈ αF}.
d = βγ : 1E → 1E
1
is defined by
d (x+(αF +dF)) = α−1dx+(αF +dF)
1
for x ∈ {z ∈ F | dz ∈ αF}.
By Proposition 4.3, we have
r+1Z = γ−1(im αr)
(9) = {z+(αF +dF) | z ∈ F and dz ∈ αr+1F}
= {z ∈ F | dz ∈ αr+1F}+αF/αF +dF
and
r+1B = β(ker αr)
= {z+(αF +dF) | z ∈ F and αrz ∈ dF}
(10)
= {z+(αF +dF) | z ∈ α−rdF}
= α−rdF +αF/αF +dF.
4
Thus
r+1E = r+1Z/r+1B = {z ∈ F | dz ∈ αr+1F}+αF/α−rdF +αF
with differential d : r+1E → r+1E defined by
r+1
(11) d (x+(α−rdF +αF)) = (α−r−1dx+(α−rdF +αF))
r+1
for x ∈ {z ∈ F | dz ∈ αr+1F}.
5. Filtered Differential Modules
A filtered differential module is a differential module (G,d) together with a sequence of
submodules Gp for p ∈ Z satisfying
G ⊃ ··· ⊃ Gp ⊃ Gp+1 ⊃ ···
that are preserved by d; that is, dGp ⊂ Gp for all p. Generally, we have that Gp = G for
p ≤ 0.
Suppose that G is a filtered differential module. We will associate a spectral sequence
to G. Let F = (cid:76) Gp. The sum of the inclusion maps Gp+1 → Gp defines a map
p∈Z
α : F → F that is a monomorphism. Its cokernel is
(cid:77)
coker α = gr G = Gp/Gp+1.
p∈Z
From the exact sequence of differential modules
α β
0 → F → F → gr G → 0
we obtain the exact couple (7), with associated spectral sequence. Thus the spectral
sequence of α on F starts with
(cid:77) (cid:77)
1E = H(gr(G)) = H(Gp/Gp+1) = 1Ep.
p∈Z p∈Z
We have that
(12) 1Ep = H(Gp/Gp+1) = {z ∈ Gp | dz ∈ Gp+1}/Gp+1+dGp.
Further, H(F) ∼= (cid:76) H(Gp) and
p∈Z
H(Gp) = {z ∈ Gp | dz = 0}/dGp.
The maps α,β,γ are graded; we have
γ : H(Gp/Gp+1) → H(Gp+1)
is defined by
γ(x+(Gp+1+dGp)) = dx+dGp
for a representative x ∈ {z ∈ Gp | dz ∈ Gp+1},
β : H(Gp) → H(Gp/Gp+1)
is defined by
β(x+dGp) = x+(Gp+1+dGp)
for a representative x ∈ {z ∈ Gp | dz = 0}.
α : H(Gp+1) → H(Gp)
is defined by
α(x+dGp+1) = x+dGp
5
for a representative x ∈ {z ∈ Gp+1 | dz = 0}.
Since α−r(Gp) = Gp−r and αr+1(Gp) = Gp+r+1, and by (9) and (10), we have
r+1Zp = {z+(Gp+1+dGp) ∈ 1Ep | z ∈ Gp and dz ∈ Gp+r+1}
(13)
= {z ∈ Gp | dz ∈ Gp+r+1}+Gp+1/Gp+1+dGp}
and
r+1Bp = {z+(Gp+1+dGp) ∈ 1Ep | z ∈ Gp and z = dy for some y ∈ Gp−r}
= (Gp∩dGp−r)+Gp+1/Gp+1+dGp.
We thus have that
(14) rZp/rBp = {z ∈ Gp | dz ∈ Gp+r}/Gp∩dGp−r+1+Gp+1.
We have by (11) that d : rEp → rEp+r is given by
r
d (z+(Gp∩dGp−r+1+Gp+1)) = dz+(Gp+r ∩dGp+1+dGp+r+1)
r
for z ∈ Gp such that dz ∈ Gp+r.
ThemoduleH(G)isfilteredbythesubmodules(H(G))p = im H(Gp) → H(G). Writing
the associated graded module as gr H(G) = (cid:76) (H(G))p/(H(G))p+1, and writing Kp =
p∈Z
{z ∈ Gp | dz = 0}, we have exact sequences
0 → dG∩Gp/dGp → H(Gp) = Kp/dGp → (H(G))p → 0
so that
(HG)p/(HG)p+1 = Kp/(Kp+1+(dG∩Gp)) = Kp+Gp+1/(Gp∩dG)+Gp+1.
since Kp∩((Gp∩dG)+Gp+1) = Kp+1+(dG∩Gp).
We have
∞Zp/∞Bp = ∩ ({z ∈ Gp | dz ∈ Gp+r}+Gp+1)/∪ ((Gp∩dGp−r)+Gp+1).
r r
Writing the quotient on the right as Mp/Np, we have Mp ⊃ Kp + Gp+1 and Np ⊂
(Gp∩dG)+Gp+1. Taking the direct sum over all p, we see that gr H(G) is a quotient of
a submodule of ∞Z/∞B.
Definition 5.1. The spectral sequence of the filtered differential module G converges and
for any term rE of the spectral sequence we write rE ⇒ gr H(G) if gr H(G) = ∞Z/∞B;
that is, for each p we have
∩ ({z ∈ Gp | dz ∈ Gp+r}+Gp+1) = {z ∈ Gp | dz = 0}+Gp+1,
r
and
∪ ((Gp∩dGp−r)+Gp+1) = (Gp∩dG)+Gp+1.
r
The second of these conditions is relatively trivial; it will be satisfied as soon as G =
∪ Gp.
p
6
6. Double Complexes
Definition 6.1. A double complex C is a commutative diagram
. .
. .
. .
↑ ↑
··· → Ci,j+1 d→hor Ci+1,j+1 → ···
d ↑ ↑ d
vert vert
··· → Ci,j d→hor Ci+1,j → ···
↑ ↑
. .
. .
. .
(extendinginfinitelyinallfourdimensions)whereeachrowandeachcolumnisanordinary
complex.
Let C be a double complex. The total complex of C is the complex tot(C) defined
by tot(C)k = (cid:76) Ci,j, with differential d : tot(C)k → tot(C)k+1 defined by d(x) =
i+j=k
(−1)jd (x)+d (x) for x ∈ Ci,j.
hor vert
LetC beadoublecomplex. Therearetwoimportantfiltrationsontot(C). Thefirstone
is { Gp}, where Gp = (cid:76) Ci,j. The second is { Gp}, where Gp = (cid:76) Ci,j.
hor hor j≥p vert vert i≥p
Gp is a subcomplex of tot(C), with ( Gk)p = (cid:76) Ci,j. Also, Gp is a
hor hor i+j=k,j≥p vert
subcomplex of tot(C), with ( Gp)k = (cid:76) Ci,j.
vert i+j=k,i≥p
Let{Gp}beoneofthesetwofiltrationsoftot(C). LetF = (cid:76) Gp, andletα : F → F
p∈Z
be the map determined by the inclusions of Gp+1 into Gp. This data determines an exact
couple (7)
α
H(F) → H(F)
(15) γ (cid:45) (cid:46) β
H(F/α(F))
.
Since the Gp are subcomplexes of the complex tot(C), (15) determines long exact co-
homology sequences
(16) ··· → Hi(Gp+1) →α Hi(Gp) →β Hi(Gp/Gp+1) →γ Hi+1(Gp+1) → ···
WecanconsiderH(F)andH(F/α(F))asbigradedmodulesbysettingH(F)p,q = Hp+q(Gp)
and H(F/α(F))p,q = Hp+q(Gp/Gp+1). The maps α, β, γ in (15) are then of respective
bidegrees (−1,1), (0,0) and (1,0).
WethenhavethattherE arebigraded, andd = βα−(r−1)γ isbihomogeneousofdegree
r
r in the p grading and of degree −(r−1) in the q grading; that is, d is the direct sum of
r
maps
dp,q : rEp,q → rEp+r,q−r+1.
r
The differential d goes “r steps to the right and r−1 steps down”.
r
Let
rDp−r+1,q+r−1 = αr−1Hp+q(Gp) = {im Hp+q(Gp) → Hp+q(Gp−r+1)}.
Taking the successive derived couples of (15), we obtain long exact sequences
(17) ··· → rDp−r+2,q+r−2 →α rDp−r+1,q+r−1 βα−→(r−1) rEp,q →γ rDp+1,q → ···
7
Let us consider the case when we are considering the horizontal filtration, Gp = Gp.
hor
Then (counting the total degree p+q of an element of Cq,p in the first homology group,
and the relative degree q of an element of Cq,p in the second homology group) we have
1Ep,q = Hp+q(Gp/Gp+1) = {ker d : Cq,p → Cq+1,p}/{im d : Cq−1,p → Cq,p}
hor hor
= Hq (C∗,p) = Hq(C∗,p)
d
hor
is the homology of d .
hor
From our calculation (12), we have
1Ep = {z ∈ Gp | dz ∈ Gp+1}/Gp+1+dGp →d1 1Ep+1 = {z ∈ Gp+1 | dz ∈ Gp+2}/Gp+2+dGp+1
is defined by d ([x]) = [d(x)] for x ∈ {z ∈ Gp | dz ∈ Gp+1}. Suppose that x ∈ Cq,p,
1
with dx ∈ Gp+1, represents an element of 1Ep,q. Then d (x) = 0, so d ([x]) ∈ 1Ep+1,q is
hor 1
represented by the element d (x) ∈ Cq,p+1. We thus have that
vert
2Ep,q = {ker d : 1Ep,q → 1Ep+1,q}/{im d : 1Ep−1,q → 1Ep,q}
vert vert
= Hp Hq (C∗,∗).
vert hor
From (14), we have the map d : 2Ep → 2Ep+2,
2
{z ∈ Gp | dz ∈ Gp+2}/Gp∩dGp−1+Gp+1} →d2 {z ∈ Gp+2 | dz ∈ Gp+4}/Gp+2∩dGp+1+Gp+3,
is defined by d ([x]) = [d(x)] for x ∈ {z ∈ Gp | dz ∈ Gp+2}. Suppose that x ∈ Cq,p
2
represents an element of 2Ep,q. Then there exists z(cid:48) ∈ Gp+1 such that d(x+z(cid:48)) ∈ Gp+2.
We can then achieve this condition with a choice of z(cid:48) ∈ Gq−1,p+1. We necessarily then
have that d (x) = 0 and (−1)p+1d (z(cid:48)) = −d (x). so d (x) ∈ 1Ep+1,q is represented
hor hor vert 2
by the element d (z(cid:48)) ∈ Cq−1,p+2.
vert
The map d takes the homology class of z to the homology class of d (z(cid:48)). For this
2 vert
to be zero means that d (z(cid:48)) = (−1)p+2d (z(cid:48)(cid:48)) for some z(cid:48)(cid:48), and in this case d carries
vert hor 3
the homology class of z to d (z(cid:48)(cid:48)).
vert
Theorem 6.2. Associated with the double complex C are two spectral sequences r E and
hor
r E, corresponding, respectively, to the horizontal and vertical filtrations of tot(C). The
vert
1E terms are bigraded with the components given by
1 Ep,q = Hq(C∗,p), 1 Ep,q = Hq(Cp,∗).
hor vert
If Ci,j = 0 for all i < 0 or for all j > 0, then the horizontal spectral sequence converges,
so that
∞ E = gr H(tot(C)).
hor hor
Symmetrically, if Ci,j = 0 for all i > 0 or for all j < 0, then the vertical spectral
sequence converges.
Proof. We have
1 Ep,q = H( Gp/ Gp+1)p,q = Hp+q(gr (C)p) = Hq(C∗,p).
hor hor hor tot
The case for the vertical spectral sequence is similar.
Theproofofconvergenceusesthebigrading. Wewillprovecovergenceforthehorizontal
spectral sequence. The proof for the vertical spectral sequence is similar. Let {Gp} be the
horizontal filtration of tot(C). We must show that
∩ ({z ∈ Gp | dz ∈ Gp+r}+Gp+1) = {z ∈ Gp | dz = 0}+Gp+1,
r
and
∪ ((Gp∩dGp−r)+Gp+1) = (Gp∩dG)+Gp+1.
r
8
Since tot(C) = ∪ Gp, the second condition is trivially satisfied. The first condition
p
means that if z ∈ Gp and for each r there is an element y ∈ Gp+1 such that d(z−y ) ≡
r r
0 mod Gp+r, then there is an element y ∈ Gp+1 such that d(z − y) = 0. Breaking z
up into a sum of elements of the same total degree, we may assume that z ∈ (Gq)p =
(cid:76) Ci,j for some q. We can then assume (by possibly replacing y with its part
i+j=q,j≥p r
which is homogeneous of total degree q) that
(cid:77)
d(z−y ) ∈ (Gq+1)p+r = Cij.
r
i+j=q+1,j≥p+r
If Ci,j = 0 for j > 0, then (Gq+1)p+r = 0 for r > −p. If on the other hand, Ci,j = 0 for
i < 0, then (Gq+1)p+r = 0 if r > q+1−p. Thus in either case d(z−y ) = 0 for suitable
r
r, and we may take y = y for this value of r.
r
(cid:3)
7. Balanced Tor
SupposethatM andN areR-modules. WewillshowthatTorR(M,N)canbecomputed
i
from a free resolution of either M or N. Let
P : ··· → P →ϕi P → ··· → P
i i−1 0
and
Q : ··· → Q →ψi Q → ··· → Q
i i−1 0
be free resolutions of M and N respectively.
We will show that
∼ ∼
H(P ⊗ N) = H(tot(P ⊗ Q)) = H(M ⊗ Q)
R R R
as R-modules. Since TorR(M,N) computed from a free resolution of M is the first of
i
these, and TorR(M,N) computed from a free resolution of N is the last, this will suffice.
i
Let E be the vertical spectral sequence associated with the third quadrant double
vert
complex F = P ⊗ Q, which may be written with upper indices, using the convention
R
that Pi = P and Qj = Q .
−i −j
. .
. .
. .
↑ ↑
··· → Pi⊗Qj+1 ϕ→⊗1 Pi+1⊗Qj+1 → ···
P ⊗Q : 1⊗ψ ↑ ↑ 1⊗ψ
··· → Pi⊗Qj ϕ→⊗1 Pi+1⊗Qj → ···
↑ ↑
. .
. .
. .
We have
1 E = 1 E−i,−j = H−j(P−i⊗ Q∗) = H (P ⊗ Q ).
vert i,j vert R j i R ∗
··· → P ⊗Q 1⊗→ψj P ⊗Q → ··· → P ⊗Q → P ⊗N → 0
i j i j−1 i 0 i
is exact since P is free. Thus H (P ⊗ Q ) = 0 for j > 0. Tensoring the short exact
i j i R ∗
sequence
∼
0 → ϕ (Q ) → Q → H (Q ) = N → 0
1 1 0 0 ∗
9
with P , we obtain the short exact sequence
i
∼
0 → P ⊗ϕ (Q ) → P ⊗Q → P ⊗H (Q ) = P ⊗N → 0,
i 1 1 i 0 i 0 ∗ i
from which we compute
∼
H (P ⊗ Q ) = P ⊗ N.
0 i R ∗ i R
Thus the only nonzero 1 E terms are those with j = 0. The differential d is induced
vert i,j 1
by d = ϕ⊗1. Thus 1 E is the complex P ⊗ N. Thus 1 E is the complex P ⊗ N
hor vert R vert R
and
(cid:26)
H (P ⊗ N) for j = 0
2 E = 2 E−i,−j = i R
vert i,j vert 0 for j (cid:54)= 0.
We have
d−i,−j : 2 E−i,−j → 2 E−i+2,−j−1
2 vert vert
so d = 0. We further calculate
2
3 E−i,−j = (ker 2 E−i,−j →d2 2 E−i+2,−j−1)/(im 2 E−i−2,−j+1 →d2 2 E−i,−j
vert vert vert vert vert
so that 3 E−i,−j = 0 if j (cid:54)= 0. Since
vert
d−i,−j : 3 E−i,−j → 3 E−i+3,−j−2
3 vert vert
we have that d = 0. Continuing in this way, we calculate that d = 0 for k ≥ 2.
3 k
It follows that the spectral sequence degenerates at 2E; that is ∞ E = 2 E. Since all
vert vert
nonzero terms have j = 0,
(cid:77)
∞ E = ∞E ,
vert i,j k,0
i+j=k
and the filtration of H(tot(P ⊗ Q)) has only one nonzero piece. Thus we get
R
H(P ⊗ N) = gr (H(tot(P ⊗ Q))) = H(tot(P ⊗ Q)).
R vert R R
By symmetry (calculating the horizontal spectral sequence of P ⊗ Q) we get
R
H(tot(P ⊗ Q)) = H(M ⊗ Q).
R R
8. Exact sequence of terms of low degree
Proposition 8.1. If Ci,j is a third quadrant double complex (Ci,j = 0 if i > 0 or j > 0),
then writing E for E, we have
vert
a. H0(tot(C)) ∼= H0 H0 (C∗,∗).
hor vert
b. There is a 5-term exact sequence
H−2(tot(C)) → H−2H0 (C∗,∗) →d2 H0 H−1 (C∗,∗) → H−1(tot(C)) → H−1H0 (C∗,∗) → 0
hor vert hor vert hor vert
Proof. We first prove a. We have an exact sequence (17)
2D0,0 → 2D−1,1 → 2E0,0 → 2D1,0.
Since Gp = 0 for p > 0, we compute that 2D0,0 = 2D1,0 = 0.
2D−1,1 = im H0(G0) → H0(G−1) = C0,0/d(C0,−1)+d(C−1,0) = H0(tot(C)).
Since 2E0,0 = H0 H0 (C∗,∗), a follows.
hor vert
We now prove b. From (17), we have exact sequences
2D−3,1 → 2E−2,0 →γ 2D−1,0 → 2D−2,1 → 2E−1,0 → 2D1,0
2D0,−1 → 2D−1,0 β→α−1 2E0,−1 → 2D1,−1
10