Math 256B. Spectral Sequences

For concreteness, we work over the category 𝔄𝔟 of abelian groups; however, everything will still work over an arbitrary abelian category.

Mostly this follows Lang’s Algebra, and also Vakil’s FOAG. (Note that Vakil interchanges the roles of p and q.)

First, we give the definition of spectral sequence that is likely the most familiar to mathematicians in algebraic geometry.

Definition. A spectral sequence is a sequence {E_{r}, d_{r}}_{r≥0} of bigraded objects

E_{r} = ⨁_{p,q∈ℕ} E_{r}^{p,q} ,

together with homomorphisms (called differentials) d_{r} = d_{r}^{p,q}: E_{r}^{p,q} → E_{r}^{p+r,q−r+1} (hence of bidegree (r, 1 − r) ) for all r, p, q, such that (i). d_{r}^{2} = 0 , and (ii). H(E_{r}) = E_{r+1} (i.e.,

E_{r+1}^{p,q} = ker(d_{r}^{p,q}: E_{r}^{p,q} → E_{r}^{p+r,q−r+1})/ im(d_{r}^{p−r,q+r−1} : E_{r}^{p−r,q−r+1} → E_{r}^{p,q})

for all r, p, q).

In the above, we also let E_{r}^{(p,q)} = 0 for all r ∈ ℕ and all (p, q) ∈ ℤ^{2} \ ℕ^{2} .

Remark. Let p, q ∈ ℕ and n = p + q. Then, for all r > n + 1 , we have q − r + 1 < 0 and p − r < 0 (since p, q ≤ n); hence d_{r}^{p,q} = d_{r}^{p−r,q+r−1} = 0 , and consequently

E_{r}^{p,q} = E_{r+1}^{p,q} = E_{r+2}^{p,q} = ….

We let E_{∞}^{p,q} denote this limiting value.

Definition. Let (K·, D) be a (co)complex of abelian groups. Then a filtration of (K·, D) is an ℕ-graded filtration K^{n} = F^{0}K^{n} ⊇ F^{1}K^{n} ⊇ … of K^{n} for all n ∈ ℕ such that D(F^{p}K^{n}) ⊆ F^{p}K^{n+1} for all n, p. We also assume that F^{p}K^{n} = 0 for all sufficiently large p, depending on n.

Definition. A filtered complex is a complex (K·, D) with a filtration.

Definition. Let (K·, D) be a filtered complex. Then, for all n ∈ ℕ, we define a filtration {F^{p}H^{n}(K·)}_{p∈ℕ} of H^{n}(K·) as follows. By definition of filtration, the inclusions F^{p}K^{n} → K^{n} for all n induce a map of F^{p}K· → K· of complexes, hence a map H^{n}(F^{p}K·) → H^{n}(K·) for all n. We define F^{p}H^{n}(K·) to be the image of this map. Since F^{p+1}K· → F^{p}K· is a map of complexes, and since F^{0}K· = K·, we have

H^{n}(K·) = F^{0}H^{n}(K·) ⊇ F^{1}H^{n}(K·) ⊇ …

for all n. Moreover, for all n there is a p such that F^{p}K^{n} = 0 , which gives F^{p}H^{n}(K·) = 0 (for the same p).

The main theorem of this handout is the following.

Theorem. Let (K·, D) be a filtered complex. Assume that F^{p}K^{n} = 0 for all n ∈ ℕ and all p > n. For all r, n, p ∈ ℕ, define:

X_{−1}^{n;p} = F^{p}K^{n} ,
X_{r}^{n;p} = F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) ,
Y_{r}^{n;p} = D(X_{r−1}^{n−1;p−(r−1)}) + X_{r−1}^{n;p+1} ,  and
E_{r}^{n;p} = X_{r}^{n;p}/Y_{r}^{n;p} .

Then:

(a). Y_{r}^{n;p} ⊆ X_{r}^{n;p} (and therefore E_{r}^{n;p} is well defined) for all r, n, p; (b). D induces well-defined maps

d_{r} = d_{r}^{n;p}: E_{r}^{n;p} → E_{r}^{n+1;p+r}
for all r, n, p;

(c). with the above differentials, and letting E_{r}^{p,q} = E_{r}^{n;p} and d_{r}^{p,q} = d_{r}^{n;p} for all

r, n, p, q with p + q = n, {E_{r}, d_{r}}_{r≥0} is a spectral sequence; and

(d). we have F^{n+1}H^{n}(K·) = 0 for all n, and

F^{p}H^{n}(K·)/F^{p+1}H^{n}(K·) ≅ E_{∞}^{n;p}
for all n ∈ ℕ and all p = 0, …, n.

Proof. (a). When r = 0 ,

\begin{aligned}X_{0}^{n;p} = F^{p}K^{n} ∩ D^{−1}(F^{p}K^{n+1}) & = F^{p}K^{n}  and \\ Y_{0}^{n;p} = D(F^{p+1}K^{n−1}) + F^{p+1}K^{n} & = F^{p+1}K^{n} , \end{aligned}  (1)

so clearly Y_{0}^{n;p} ⊆ X_{0}^{n;p}.

Now assume r > 0 . Since X_{r−1}^{n−1;p−r+1} = F^{p−r+1}K^{n−1} ∩ D^{−1}(F^{p}K^{n}) , we have

D(X_{r−1}^{n−1;p−r+1}) ⊆ D(D^{−1}(F^{p}K^{n})) ⊆ F^{p}K^{n} ;

combining this with D(X_{r−1}^{n−1;p−r+1}) ⊆ D^{−1}(F^{p+r}K^{n+1}) (since D ∘ D = 0 ) gives

D(X_{r−1}^{n−1;p−r+1}) ⊆ F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) = X_{r}^{n;p} .  (2)

Also,

X_{r−1}^{n;p+1} = F^{p+1}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) ⊆ F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) = X_{r}^{n;p} . (3)

Combining (2) and (3) then gives

Y_{r}^{n;p} = D(X_{r−1}^{n−1;p−(r−1)}) + X_{r−1}^{n;p+1} ⊆ X_{r}^{n;p} .

(b). This amounts to checking that D(X_{r}^{n;p}) ⊆ X_{r}^{n+1;p+r} and D(Y_{r}^{n;p}) ⊆ Y_{r}^{n+1;p+r}. For the first of these,

D(X_{r}^{n;p}) ⊆ D(D^{−1}(F^{p+r}K^{n+1})) ⊆ F^{p+r}K^{n+1}

because X_{r}^{n;p} ⊆ D^{−1}(F^{p+r}K^{n+1}) by definition, and

D(X_{r}^{n;p}) ⊆ D^{−1}(F^{p+2r}K^{n+2})

because D ∘ D = 0 , so

D(X_{r}^{n;p}) ⊆ F^{p+r}K^{n+1} ∩ D^{−1}(F^{p+2r}K^{n+2}) = X_{r}^{n+1;p+r} .

As for D(Y_{r}^{n;p}) ,

D(Y_{r}^{n;p}) = D(D(X_{r−1}^{n−1;p−(r−1)}) + X_{r−1}^{n;p+1})
= D(X_{r−1}^{n;p+1})
⊆ D(X_{r−1}^{n;p+1}) + X_{r−1}^{n+1;p+r+1}
= Y_{r}^{n+1;p+r} .

Note that this holds also for r = 0 , because the value of X_{r−1}^{n+1;p+r+1} did not play a role here. (c). This is a matter of checking that d_{r}^{2} = 0 and that H(E_{r}) = E_{r+1} .

The fact that d_{r}^{2} = 0 is immediate from the fact that D^{2} = 0 .

To check that H(E_{r}) = E_{r+1} , we follow Vakil 1.7.13.

Claim. ker d_{r}^{n;p} = \frac{X_{r−1}^{n;p+1} + X_{r+1}^{n;p}}{Y_{r}^{n;p}} .

Proof. It is easy to check that ker d_{r}^{n;p} = (X_{r}^{n;p} ∩ D^{−1}(Y_{r}^{n+1;p+r}))/Y_{r}^{n;p}, so it suffices to show that

X_{r}^{n;p} ∩ D^{−1}(Y_{r}^{n+1;p+r}) = X_{r−1}^{n;p+1} + X_{r+1}^{n;p} .  (4)

Indeed, we have

X_{r−1}^{n;p+1} = F^{p+1}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) ⊆ F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) = X_{r}^{n;p} (5)

and

X_{r}^{n;p} ∩ D^{−1}(X_{r−1}^{n+1;p+r+1})
= F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) ∩ D^{−1}(F^{p+r+1}K^{n+1} ∩ D^{−1}(F^{p+2r}K^{n+2}))
= F^{p}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) ∩ D^{−1}(F^{p+r+1}K^{n+1})
= F^{p}K^{n} ∩ D^{−1}(F^{p+r+1}K^{n+1})
= X_{r+1}^{n;p} .  (6)

Therefore, by definition of Y_{r}^{n+1;p+r}, general properties of homomorphisms, (5), and (6), we have

X_{r}^{n;p} ∩ D^{−1}(Y_{r}^{n+1;p+r}) = X_{r}^{n;p} ∩ D^{−1}(D(X_{r−1}^{n;p+1}) + X_{r−1}^{n+1;p+r+1})
= X_{r}^{n;p} ∩ (X_{r−1}^{n;p+1} + D^{−1}(X_{r−1}^{n+1;p+r+1}))
= X_{r−1}^{n;p+1} + (X_{r}^{n;p} ∩ D^{−1}(X_{r−1}^{n+1;p+r+1}))
= X_{r−1}^{n;p+1} + X_{r+1}^{n;p} .

This is (4), so the claim is proved. □ Now consider the image of d_{r}^{n;p−r}: E_{r}^{n;p−r} → E_{r}^{n;p}. Since X_{r−1}^{n−1;p−r+1} ⊆ X_{r}^{n−1;p−r},

\begin{aligned}im d_{r}^{n;p−r} & = \frac{D(X_{r}^{n−1;p−r}) + Y_{r}^{n;p}}{Y_{r}^{n;p}} \\ & = \frac{D(X_{r}^{n−1;p−r}) + D(X_{r−1}^{n−1;p−r+1}) + X_{r−1}^{n;p+1}}{Y_{r}^{n;p}} \end{aligned}

= \frac{D(X_{r}^{n−1;p−r}) + X_{r−1}^{n;p+1}}{Y_{r}^{n;p}} .

It will suffice to show that there is a well-defined isomorphism

E_{r+1}^{n;p} = \frac{X_{r+1}^{n;p}}{Y_{r+1}^{n;p}} = \frac{X_{r+1}^{n;p}}{D(X_{r}^{n−1;p−r}) + X_{r}^{n;p+1}} ⟶^{ϕ} \frac{X_{r+1}^{n;p} + X_{r−1}^{n;p+1}}{D(X_{r}^{n−1;p−r}) + X_{r−1}^{n;p+1}} ≅ \frac{ker d_{r}^{n;p}}{im d_{r}^{n;p−r}} .

We first claim that

X_{r+1}^{n;p} ∩ (D(X_{r}^{n−1;p−r}) + X_{r−1}^{n;p+1}) = D(X_{r}^{n−1;p−r}) + X_{r}^{n;p+1} .  (7)

Since D(X_{r}^{n−1;p−r}) ⊆ D(D^{−1}(F^{p}K^{n})) and D ∘ D = 0 , we have

D(X_{r}^{n−1;p−r}) ⊆ F^{p}K^{n} ∩ D^{−1}(0) ⊆ F^{p}K^{n} ∩ D^{−1}(F^{p+r+1}K^{n+1}) = X_{r+1}^{n;p} . Also

\begin{aligned}X_{r+1}^{n;p} ∩ X_{r−1}^{n;p+1} & = F^{p}K^{n} ∩ D^{−1}(F^{p+r+1}K^{n+1}) ∩ F^{p+1}K^{n} ∩ D^{−1}(F^{p+r}K^{n+1}) \\ & = F^{p+1}K^{n} ∩ D^{−1}(F^{p+r+1}K^{n+1}) \end{aligned}

= X_{r}^{n;p+1} .

Combining these two facts gives (7), because

X_{r+1}^{n;p} ∩ (D(X_{r}^{n−1;p−r}) + X_{r−1}^{n;p+1}) = D(X_{r}^{n−1;p−r}) + (X_{r+1}^{n;p} ∩ X_{r−1}^{n;p+1})
= D(X_{r}^{n−1;p−r}) + X_{r}^{n;p+1} .

Therefore ϕ is well-defined and injective. Surjectivity of ϕ is clear, so ϕ is an isomorphism and (c) is proved. (d). First of all, the fact that F^{n+1}H^{n}(K·) = 0 follows immediately from the assumption that F^{n+1}K^{n} = 0 .

By definition of F^{p}H^{n}(K·) ,

F^{p}H^{n}(K^{·}) = im (H^{n}(F^{p}K·) ⟶ \frac{ker D^{n}}{im D^{n−1}} ) = \frac{(F^{p}K^{n} ∩ ker D) + im D}{im D} .

First consider E_{∞}^{n;p}. Since d_{r}^{n;p} = 0 for all r > n + 1 (since E_{r}^{n;p+r} = 0 ), we have X_{r}^{n;p} = X_{r+1}^{n;p} and Y_{r}^{n;p} = Y_{r+1}^{n;p} for all r > n + 1 ; call these groups X_{∞}^{n;p} and Y_{∞}^{n;p}, respectively. We have

X_{∞}^{n;p} = F^{p}K^{n} ∩ ker D

and

Y_{r}^{n;p} = D(F^{p−r+1}K^{n−1} ∩ D^{−1}(F^{p}K^{n})) + X_{r−1}^{n;p+1}

for all n, r, and p, so

\begin{aligned}Y_{∞}^{n;p} & = D(K^{n−1} ∩ D^{−1}(F^{p}K^{n})) + X_{∞}^{n;p+1} \\ & = D(D^{−1}(F^{p}K^{n})) + X_{∞}^{n;p+1} \end{aligned}

= (F^{p}K^{n} ∩ im D) + (F^{p+1}K^{n} ∩ ker D) .

And, naturally, E_{∞}^{n;p} = X_{∞}^{n;p}/Y_{∞}^{n;p}.

We then claim that there is a well-defined isomorphism

E_{∞}^{n;p} = \frac{F^{p}K^{n} ∩ ker D}{(F^{p}K^{n} ∩ im D) + (F^{p+1}K^{n} ∩ ker D)}

⟶^{ϕ} \frac{(F^{p}K^{n} ∩ ker D) + im D}{(F^{p+1}K^{n} ∩ ker D) + im D} = \frac{F^{p}H^{n}(K)}{F^{p+1}H^{n}(K)} .

To see this, we first note that

F^{p}K^{n} ∩ ker D ∩ ((F^{p+1}K^{n} ∩ ker D) + im D) = F^{p}K^{n} ∩ ((F^{p+1}K^{n} ∩ ker D) + im D)
= (F^{p+1}K^{n} ∩ ker D) + (F^{p}K^{n} ∩ im D) .

Indeed, the first step holds because im D ⊆ ker D, and so ker D contains the quantity in parentheses. The second step is true because F^{p+1}K^{n}∩ker D is contained in F^{p}K^{n}.

Therefore ϕ is well-defined and injective. It is clearly surjective, so it is the desired isomorphism. □