
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. $\square$

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