
## 12.21 Spectral sequences: filtered complexes

Definition 12.21.1. Let $\mathcal{A}$ be an abelian category. A filtered complex $K^\bullet$ of $\mathcal{A}$ is a complex of $\text{Fil}(\mathcal{A})$ (see Definition 12.16.1).

We will denote the filtration on the objects by $F$. Thus $F^ pK^ n$ denotes the $p$th step in the filtration of the $n$th term of the complex. Note that each $F^ pK^\bullet$ is a complex of $\mathcal{A}$. Hence we could also have defined a filtered complex as a filtered object in the (abelian) category of complexes of $\mathcal{A}$. In particular $\text{gr} K^\bullet$ is a graded object of the category of complexes of $\mathcal{A}$.

To describe the spectral sequence associated to such an object we assume, for the moment, that $\mathcal{A}$ is an abelian category which has countable direct sums and countable direct sums are exact (this is not automatic, see Remark 12.15.3). Let us denote $d$ the differential of $K$. Forgetting the grading we can think of $\bigoplus K^ n$ as a filtered differential object of $\mathcal{A}$. Hence according to Section 12.20 we obtain a spectral sequence $(E_ r, d_ r)_{r \geq 0}$. In this section we work out the terms of this spectral sequence, and we endow the terms of this spectral sequence with additional structure coming from the grading of $K$.

First we point out that $E_0^ p = \text{gr}^ p K^\bullet$ is a complex and hence is graded. Thus $E_0$ is bigraded in a natural way. It is customary to use the bigrading

$E_0 = \bigoplus \nolimits _{p, q} E_0^{p, q}, \quad E_0^{p, q} = \text{gr}^ p K^{p + q}$

The idea is that $p + q$ should be thought of as the total degree of the (co)homology classes. Also, $p$ is called the filtration degree, and $q$ is called the complementary degree. The differential $d_0$ is compatible with this bigrading in the following way

$d_0 = \bigoplus d_0^{p, q}, \quad d_0^{p, q} : E_0^{p, q} \to E_0^{p, q + 1}.$

Namely, $d_0^ p$ is just the differential on the complex $\text{gr}^ p K^\bullet$ (which occurs as $\text{gr}^ pE_0$ just shifted a bit).

To go further we identify the objects $B_ r^ p$ and $Z_ r^ p$ introduced in Section 12.20 as graded objects and we work out the corresponding decompositions of the differentials. We do this in a completely straightforward manner, but again we warn the reader that our notation is not the same as notation found elsewhere. We define

$Z_ r^{p, q} = \frac{F^ pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1}) + F^{p + 1}K^{p + q}}{F^{p + 1}K^{p + q}}$

and

$B_ r^{p, q} = \frac{F^ pK^{p + q} \cap d(F^{p - r + 1}K^{p + q - 1}) + F^{p + 1}K^{p + q}}{F^{p + 1}K^{p + q}}$

and of course $E_ r^{p, q} = Z_ r^{p, q}/B_ r^{p, q}$. With these definitions it is completely clear that $Z_ r^ p = \bigoplus _ q Z_ r^{p, q}$, $B_ r^ p = \bigoplus _ q B_ r^{p, q}$, and $E_ r^ p = \bigoplus _ q E_ r^{p, q}$. Moreover, we have

$0 \subset \ldots \subset B_ r^{p, q} \subset \ldots \subset Z_ r^{p, q} \subset \ldots \subset E_0^{p, q}$

Also, the map $d_ r^ p$ decomposes as the direct sum of the maps

$d_ r^{p, q} : E_ r^{p, q} \longrightarrow E_ r^{p + r, q - r + 1}, \quad z + F^{p + 1}K^{p + q} \mapsto dz + F^{p + r + 1}K^{p + q + 1}$

where $z \in F^ pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1})$.

Lemma 12.21.2. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. There is a spectral sequence $(E_ r, d_ r)_{r \geq 0}$ in the category of bigraded objects of $\mathcal{A}$ associated to $(K^\bullet , F)$ such that $d_ r$ has bidegree $(r, - r + 1)$ and such that $E_ r$ has bigraded pieces $E_ r^{p, q}$ and maps $d_ r^{p, q} : E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$ as given above. Furthermore, we have $E_0^{p, q} = \text{gr}^ p(K^{p + q})$, $d_0^{p, q} = \text{gr}^ p(d^{p + q})$, and $E_1^{p, q} = H^{p + q}(\text{gr}^ p(K^\bullet ))$.

Proof. If $\mathcal{A}$ has countable direct sums and if countable direct sums are exact, then this follows from the discussion above. In general, we proceed as follows; we strongly suggest the reader skip this proof. Consider the bigraded object $A = (F^{p + 1}K^{p + 1 + q})$ of $\mathcal{A}$, i.e., we put $F^{p + 1}K^{p + 1 + q}$ in degree $(p, q)$ (the funny shift in numbering to get numbering correct later on). We endow it with a differential $d : A \to A[0, 1]$ by using $d$ on each component. Then $(A, d)$ is a differential bigraded object. Consider the map

$\alpha : A \to A[-1, 1]$

which is given in degree $(p, q)$ by the inclusion $F^{p + 1}K^{p + q} \to F^ pK^{p + q}$. This is an injective morphism of differential objects $\alpha : (A, d) \to (A, d)[-1, 1]$. Hence, we can apply Remark 12.19.6 with $S = [0, 1]$ and $T = [1, -1]$. The corresponding spectral sequence $(E_ r, d_ r)_{r \geq 0}$ of bigraded objects is the spectral sequence we are looking for. Let us unwind the definitions a bit. First of all we have $E_ r = (E_ r^{p, q})$. Then, since $T^ rS = [r, -r + 1$ we have $d_ r : E_ r \to E_ r[r, -r + 1]$ which means that $d_ r^{p, q} : E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$.

To see that the description of the graded pieces hold, we argue as above. Namely, first we have

$E_0 = \mathop{\mathrm{Coker}}(\alpha : A \to A[-1, 1])[0, -1] = \mathop{\mathrm{Coker}}(\alpha [0, -1] : A[0, -1] \to A[-1, 0])$

and by our choice of numbering above this gives

$E_0^{p, q} = \mathop{\mathrm{Coker}}(F^{p + 1}K^{p + q} \to F^ pK^{p + q}) = \text{gr}^ pK^{p + q}$

The first differential is given by $d_0^{p, q} = \text{gr}^ pd^{p + q} : E_0^{p, q} \to E_0^{p, q + 1}$. Next, the description of the boundaries $B_ r$ and the cocycles $Z_ r$ in Remark 12.19.6 translates into a straightforward manner into the formulae for $Z_ r^{p, q}$ and $B_ r^{p, q}$ given above. $\square$

Lemma 12.21.3. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume $\mathcal{A}$ has countable direct sums. Let $(E_ r, d_ r)_{r \geq 0}$ be the spectral sequence associated to $(K^\bullet , F)$.

1. The map

$d_1^{p, q} : E_1^{p, q} = H^{p + q}(\text{gr}^ p(K^\bullet )) \longrightarrow E_1^{p + 1, q} = H^{p + q + 1}(\text{gr}^{p + 1}(K^\bullet ))$

is equal to the boundary map in cohomology associated to the short exact sequence of complexes

$0 \to \text{gr}^{p + 1}(K^\bullet ) \to F^ pK^\bullet /F^{p + 2}K^\bullet \to \text{gr}^ p(K^\bullet ) \to 0.$
2. Assume that $d(F^ pK) \subset F^{p + 1}K$ for all $p \in \mathbf{Z}$. Then $d$ induces the zero differential on $\text{gr}^ p(K^\bullet )$ and hence $E_1^{p, q} = \text{gr}^ p(K^\bullet )^{p + q}$. Furthermore, in this case

$d_1^{p, q} : E_1^{p, q} = \text{gr}^ p(K^\bullet )^{p + q} \longrightarrow E_1^{p + 1, q} = \text{gr}^{p + 1}(K^\bullet )^{p + q + 1}$

is the morphism induced by $d$.

Proof. This is clear from the formula given for the differential $d_1^{p, q}$ just above Lemma 12.21.2. $\square$

Lemma 12.21.4. Let $\mathcal{A}$ be an abelian category. Let $\alpha : (K^\bullet , F) \to (L^\bullet , F)$ be a morphism of filtered complexes of $\mathcal{A}$. Let $(E_ r(K), d_ r)_{r \geq 0}$, resp. $(E_ r(L), d_ r)_{r \geq 0}$ be the spectral sequence associated to $(K^\bullet , F)$, resp. $(L^\bullet , F)$. The morphism $\alpha$ induces a canonical morphism of spectral sequences $\{ \alpha _ r : E_ r(K) \to E_ r(L)\} _{r \geq 0}$ compatible with the bigradings.

Proof. Obvious from the explicit representation of the terms of the spectral sequences. $\square$

Definition 12.21.5. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. The induced filtration on $H^ n(K^\bullet )$ is the filtration defined by $F^ pH^ n(K^\bullet ) = \mathop{\mathrm{Im}}(H^ n(F^ pK^\bullet ) \to H^ n(K^\bullet ))$.

Writing out what this means we see that

12.21.5.1
$$\label{homology-equation-filtration-cohomology} F^ pH^ n(K^\bullet , d) = \frac{\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n}{\mathop{\mathrm{Im}}(d) \cap K^ n}$$

and hence we see that

12.21.5.2
$$\label{homology-equation-graded-cohomology} \text{gr}^ p H^ n(K^\bullet ) = \frac{\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n}{\mathop{\mathrm{Ker}}(d) \cap F^{p + 1}K^ n + \mathop{\mathrm{Im}}(d) \cap F^ pK^ n}$$

(one intermediate step omitted).

Lemma 12.21.6. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. If $Z_\infty ^{p, q}$ and $B_\infty ^{p, q}$ exist (see proof), then

1. the limit $E_\infty$ exists and is a bigraded object having $E_\infty ^{p, q} = Z_\infty ^{p, q}/B_\infty ^{p, q}$ in bidegree $(p, q)$,

2. the $p$th graded part $\text{gr}^ pH^ n(K^\bullet )$ of the $n$th cohomology object of $K^\bullet$ is a subquotient of $E_\infty ^{p, n - p}$.

Proof. The objects $Z_\infty$, $B_\infty$, and the limit $E_\infty = Z_\infty /B_\infty$ of Definition 12.17.2 are bigraded objects of $\mathcal{A}$ by our construction of the spectral sequence in Lemma 12.21.2. Since $Z_ r = \bigoplus Z_ r^{p, q}$ and $B_ r = \bigoplus B_ r^{p, q}$, if we assume that

$Z_\infty ^{p, q} = \bigcap \nolimits _ r Z_ r^{p, q} = \bigcap \nolimits _ r \frac{F^ pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1}) + F^{p + 1}K^{p + q}}{F^{p + 1}K^{p + q}}$

and

$B_\infty ^{p, q} = \bigcup \nolimits _ r B_ r^{p, q} = \bigcup \nolimits _ r \frac{F^ pK^{p + q} \cap d(F^{p - r + 1}K^{p + q - 1}) + F^{p + 1}K^{p + q}}{F^{p + 1}K^{p + q}}$

exist, then $Z_\infty$ and $B_\infty$ exist with bidegree $(p, q)$ parts $Z_\infty ^{p, q}$ and $B_\infty ^{p, q}$ (follows from an elementary argument about unions and intersections of bigraded objects). Thus

$E_\infty ^{p, q} = \frac{\bigcap _ r (F^ pK^{p + q} \cap d^{-1}(F^{p + r}K^{p + q + 1}) + F^{p + 1}K^{p + q})}{\bigcup _ r (F^ pK^{p + q} \cap d(F^{p - r + 1}K^{p + q - 1}) + F^{p + 1}K^{p + q})}.$

where the top and the bottom exist. With $n = p + q$ we have

12.21.6.1
$$\label{homology-equation-on-top-bigraded} \mathop{\mathrm{Ker}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n} \subset \bigcap \nolimits _ r \left( F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n} \right)$$

and

12.21.6.2
$$\label{homology-equation-at-bottom-bigraded} \bigcup \nolimits _ r \left( F^ pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} \right) \subset \mathop{\mathrm{Im}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n}.$$

Thus a subquotient of $E_\infty ^{p, q}$ is

$\frac{\mathop{\mathrm{Ker}}(d) \cap F^ pK^{n} + F^{p + 1}K^ n}{\mathop{\mathrm{Im}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n}} = \frac{\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n}{\mathop{\mathrm{Im}}(d) \cap F^ pK^ n + \mathop{\mathrm{Ker}}(d) \cap F^{p + 1}K^ n}$

Comparing with (12.21.5.2) we conclude. $\square$

Definition 12.21.7. Let $\mathcal{A}$ be an abelian category. Let $(E_ r, d_ r)_{r \geq r_0}$ be a spectral sequence of bigraded objects of $\mathcal{A}$ with $d_ r$ of bidegree $(r, -r + 1)$. We say such a spectral sequence is

1. regular if for all $p, q \in \mathbf{Z}$ there is a $b = b(p, q)$ such that the maps $d_ r^{p, q} : E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$ are zero for $r \geq b$,

2. coregular if for all $p, q \in \mathbf{Z}$ there is a $b = b(p, q)$ such that the maps $d_ r^{p - r, q + r - 1} : E_ r^{p - r, q + r - 1} \to E_ r^{p, q}$ are zero for $r \geq b$,

3. bounded if for all $n$ there are only a finite number of nonzero $E_{r_0}^{p, n - p}$,

4. bounded below if for all $n$ there is a $b = b(n)$ such that $E_{r_0}^{p, n - p} = 0$ for $p \geq b$.

5. bounded above if for all $n$ there is a $b = b(n)$ such that $E_{r_0}^{p, n - p} = 0$ for $p \leq b$.

Bounded below means that if we look at $E_ r^{p, q}$ on the line $p + q = n$ (whose slope is $-1$) we obtain zeros as $(p, q)$ moves down and to the right. As mentioned above there is no consistent terminology regarding these notions in the literature.

Lemma 12.21.8. In the situation of Definition 12.21.7. Let $Z_ r^{p, q}, B_ r^{p, q} \subset E_{r_0}^{p, q}$ be the $(p, q)$-graded parts of $Z_ r, B_ r$ defined as in Section 12.17.

1. The spectral sequence is regular if and only if for all $p, q$ there exists an $r = r(p, q)$ such that $Z_ r^{p, q} = Z_{r + 1}^{p, q} = \ldots$

2. The spectral sequence is coregular if and only if for all $p, q$ there exists an $r = r(p, q)$ such that $B_ r^{p, q} = B_{r + 1}^{p, q} = \ldots$

3. The spectral sequence is bounded if and only if it is both bounded below and bounded above.

4. If the spectral sequence is bounded below, then it is regular.

5. If the spectral sequence is bounded above, then it is coregular.

Proof. Omitted. Hint: If $E_ r^{p, q} = 0$, then we have $E_{r'}^{p, q} = 0$ for all $r' \geq r$. $\square$

Definition 12.21.9. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. We say the spectral sequence associated to $(K^\bullet , F)$

1. weakly converges to $H^*(K^\bullet )$ if $\text{gr}^ pH^ n(K^\bullet ) = E_{\infty }^{p, n - p}$ via Lemma 12.21.6 for all $p, n \in \mathbf{Z}$,

2. abuts to $H^*(K^\bullet )$ if it weakly converges to $H^*(K^\bullet )$ and $\bigcap _ p F^ pH^ n(K^\bullet ) = 0$ and $\bigcup _ p F^ p H^ n(K^\bullet ) = H^ n(K^\bullet )$ for all $n$,

3. converges to $H^*(K^\bullet )$ if it is regular, abuts to $H^*(K^\bullet )$, and $H^ n(K^\bullet ) = \mathop{\mathrm{lim}}\nolimits _ p H^ n(K^\bullet )/F^ pH^ n(K^\bullet )$.

Weak convergence, abutment, or convergence is symbolized by the notation $E_ r^{p, q} \Rightarrow H^{p + q}(K^\bullet )$. As mentioned above there is no consistent terminology regarding these notions in the literature.

Lemma 12.21.10. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. The associated spectral sequence

1. weakly converges to $H^*(K^\bullet )$ if and only if for every $p, q \in \mathbf{Z}$ we have equality in equations (12.21.6.2) and (12.21.6.1),

2. abuts to $H^*(K)$ if and only if it weakly converges to $H^*(K^\bullet )$ and we have $\bigcap _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Im}}(d) \cap K^ n$ and $\bigcup _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Ker}}(d) \cap K^ n$.

Proof. Immediate from the discussions above. $\square$

Lemma 12.21.11. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume that the filtration on each $K^ n$ is finite (see Definition 12.16.1). Then

1. the spectral sequence associated to $(K^\bullet , F)$ is bounded,

2. the filtration on each $H^ n(K^\bullet )$ is finite,

3. the spectral sequence associated to $(K^\bullet , F)$ converges to $H^*(K^\bullet )$,

4. if $\mathcal{C} \subset \mathcal{A}$ is a weak Serre subcategory and for some $r$ we have $E_ r^{p, q} \in \mathcal{C}$ for all $p, q \in \mathbf{Z}$, then $H^ n(K^\bullet )$ is in $\mathcal{C}$.

Proof. Part (1) follows as $E_0^{p, n - p} = \text{gr}^ p K^ n$. Part (2) is clear from Equation (12.21.5.1). We will use Lemma 12.21.10 to prove that the spectral sequence weakly converges. Fix $p, n \in \mathbf{Z}$. Looking at the right hand side of (12.21.6.1) we see that we get $F^ pK^ n \cap \mathop{\mathrm{Ker}}(d) + F^{p + 1}K^ n$ because $F^{p + r}K^ n = 0$ for $r \gg 0$. Thus (12.21.6.1) is an equality. Look at the left hand side of (12.21.6.1). The expression is equal to the right hand side since $F^{p - r + 1}K^{n - 1} = K^{n - 1}$ for $r \gg 0$. Thus (12.21.6.1) is an equality. Since the filtration on $H^ n(K^\bullet )$ is finite by (2) we see that we have abutment. To prove we have convergence we have to show the spectral sequence is regular which follows as it is bounded (Lemma 12.21.8) and we have to show that $H^ n(K^\bullet ) = \mathop{\mathrm{lim}}\nolimits H^ n(K^\bullet )/F^ pH^ n(K^\bullet )$ which follows from the fact that the filtration on $H^*(K^\bullet )$ is finite proved in part (2).

Proof of (4). Assume that for some $r \geq 0$ we have $E_ r^{p, q} \in \mathcal{C}$ for some weak Serre subcategory $\mathcal{C}$ of $\mathcal{A}$. Then $E_{r + 1}^{p, q}$ is in $\mathcal{C}$ as well, see Lemma 12.9.3. By boundedness proved above (which implies that the spectral sequence is both regular and coregular, see Lemma 12.21.8) we can find an $r' \geq r$ such that $E_\infty ^{p, q} = E_{r'}^{p, q}$ for all $p, q$ with $p + q = n$. Thus $H^ n(K^\bullet )$ is an object of $\mathcal{A}$ which has a finite filtration whose graded pieces are in $\mathcal{C}$. This implies that $H^ n(K^\bullet )$ is in $\mathcal{C}$ by Lemma 12.9.3. $\square$

Lemma 12.21.12. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume that the filtration on each $K^ n$ is finite (see Definition 12.16.1) and that for some $r$ we have only a finite number of nonzero $E_ r^{p, q}$. Then only a finite number of $H^ n(K^\bullet )$ are nonzero and we have

$\sum (-1)^ n[H^ n(K^\bullet )] = \sum (-1)^{p + q} [E_ r^{p, q}]$

in $K_0(\mathcal{A}')$ where $\mathcal{A}'$ is the smallest weak Serre subcategory of $\mathcal{A}$ containing the objects $E_ r^{p, q}$.

Proof. Denote $E_ r^{even}$ and $E_ r^{odd}$ the even and odd part of $E_ r$ defined as the direct sum of the $(p, q)$ components with $p + q$ even and odd. The differential $d_ r$ defines maps $\varphi : E_ r^{even} \to E_ r^{odd}$ and $\psi : E_ r^{odd} \to E_ r^{even}$ whose compositions either way give zero. Then we see that

\begin{align*} [E_ r^{even}] - [E_ r^{odd}] & = [\mathop{\mathrm{Ker}}(\varphi )] + [\mathop{\mathrm{Im}}(\varphi )] - [\mathop{\mathrm{Ker}}(\psi )] - [\mathop{\mathrm{Im}}(\psi )] \\ & = [\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\psi )] - [\mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi ] \\ & = [E_{r + 1}^{even}] - [E_{r + 1}^{odd}] \end{align*}

Note that all the intervening objects are in the smallest Serre subcategory containing the objects $E_ r^{p, q}$. Continuing in this manner we see that we can increase $r$ at will. Since there are only a finite number of pairs $(p, q)$ for which $E_ r^{p, q}$ is nonzero, a property which is inherited by $E_{r + 1}, E_{r + 2}, \ldots$, we see that we may assume that $d_ r = 0$. At this stage we see that $H^ n(K^\bullet )$ has a finite filtration (Lemma 12.21.11) whose graded pieces are exactly the $E_ r^{p, n - p}$ and the result is clear. $\square$

The following lemma is more a kind of sanity check for our definitions. Surely, if we have a filtered complex such that for every $n$ we have

$H^ n(F^ pK^\bullet ) = 0\text{ for }p \gg 0 \quad \text{and}\quad H^ n(F^ pK^\bullet ) = H^ n(K^\bullet )\text{ for }p \ll 0,$

then the corresponding spectral sequence should converge?

Lemma 12.21.13. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume

1. for every $n$ there exist $p_0(n)$ such that $H^ n(F^ pK^\bullet ) = 0$ for $p \geq p_0(n)$,

2. for every $n$ there exist $p_1(n)$ such that $H^ n(F^ pK^\bullet ) \to H^ n(K^\bullet )$ is an isomorphism for $p \leq p_1(n)$.

Then

1. the spectral sequence associated to $(K^\bullet , F)$ is bounded,

2. the filtration on each $H^ n(K^\bullet )$ is finite,

3. the spectral sequence associated to $(K^\bullet , F)$ converges to $H^*(K^\bullet )$.

Proof. Fix $n$. Using the long exact cohomology sequence associated to the short exact sequence of complexes

$0 \to F^{p + 1}K^\bullet \to F^ pK^\bullet \to \text{gr}^ pK^\bullet \to 0$

we find that $E_1^{p, n - p} = 0$ for $p \geq \max (p_0(n), p_0(n + 1))$ and $p < \min (p_1(n), p_1(n + 1))$. Hence the spectral sequence is bounded (Definition 12.21.7). This proves (1).

It is clear from the assumptions and Definition 12.21.5 that the filtration on $H^ n(K^\bullet )$ is finite. This proves (2).

Next we prove that the spectral sequence weakly converges to $H^*(K^\bullet )$ using Lemma 12.21.10. Let us show that we have equality in (12.21.6.1). Namely, for $p + r > p_0(n + 1)$ the map

$d : F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) \to F^{p + r}K^{n + 1}$

ends up in the image of $d : F^{p + r}K^ n \to F^{p + r}K^{n + 1}$ because the complex $F^{p + r}K^\bullet$ is exact in degree $n + 1$. We conclude that $F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) = d(F^{p + r}K^ n) + \mathop{\mathrm{Ker}}(d) \cap F^ pK^ n$. Hence for such $r$ we have

$\mathop{\mathrm{Ker}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n} = F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n}$

which proves the desired equality. To show that we have equality in (12.21.6.2) we use that for $p - r + 1 < p_1(n - 1)$ we have

$d(F^{p - r + 1}K^{n - 1}) = \mathop{\mathrm{Im}}(d) \cap F^{p - r + 1}K^ n$

because the map $F^{p - r + 1}K^\bullet \to K^\bullet$ induces an isomorphism on cohomology in degree $n - 1$. This shows that we have

$F^ pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} = \mathop{\mathrm{Im}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n}$

for such $r$ which proves the desired equality.

To see that the spectral sequence abuts to $H^*(K^\bullet )$ using Lemma 12.21.10 we have to show that $\bigcap _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Im}}(d) \cap K^ n$ and $\bigcup _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Ker}}(d) \cap K^ n$. For $p \geq p_0(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Im}}(d) \cap K^ n$ and for $p \leq p_1(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Ker}}(d) \cap K^ n$. Combining weak convergence, abutment, and boundedness we see that (2) and (3) are true. $\square$

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