## 12.25 Spectral sequences: double complexes

Let $A^{\bullet , \bullet }$ be a double complex, see Section 12.18. It is customary to denote $H^ p_ I(A^{\bullet , \bullet })$ the complex with terms $\mathop{\mathrm{Ker}}(d_1^{p, q})/\mathop{\mathrm{Im}}(d_1^{p - 1, q})$ (varying $q$) and differential induced by $d_2$. Then $H^ q_{II}(H^ p_ I(A^{\bullet , \bullet }))$ denotes its cohomology in degree $q$. It is also customary to denote $H^ q_{II}(A^{\bullet , \bullet })$ the complex with terms $\mathop{\mathrm{Ker}}(d_2^{p, q})/\mathop{\mathrm{Im}}(d_2^{p, q - 1})$ (varying $p$) and differential induced by $d_1$. Then $H^ p_ I(H^ q_{II}(A^{\bullet , \bullet }))$ denotes its cohomology in degree $p$. It will turn out that these cohomology groups show up as the terms in the spectral sequence for a filtration on the associated total complex or simple complex, see Definition 12.18.3.

There are two natural filtrations on the simple complex $sA^\bullet$ associated to the double complex $A^{\bullet , \bullet }$. Namely, we define

$F_ I^ p(sA^ n) = \bigoplus \nolimits _{i + j = n, \ i \geq p} A^{i, j} \quad \text{and} \quad F_{II}^ p(sA^ n) = \bigoplus \nolimits _{i + j = n, \ j \geq p} A^{i, j}.$

It is immediately verified that $(sA^\bullet , F_ I)$ and $(sA^\bullet , F_{II})$ are filtered complexes. By Section 12.24 we obtain two spectral sequences. It is customary to denote $({}'E_ r, {}'d_ r)_{r \geq 0}$ the spectral sequence associated to the filtration $F_ I$ and to denote $({}''E_ r, {}''d_ r)_{r \geq 0}$ the spectral sequence associated to the filtration $F_{II}$. Here is a description of these spectral sequences.

Lemma 12.25.1. Let $\mathcal{A}$ be an abelian category. Let $K^{\bullet , \bullet }$ be a double complex. The spectral sequences associated to $K^{\bullet , \bullet }$ have the following terms:

1. ${}'E_0^{p, q} = K^{p, q}$ with ${}'d_0^{p, q} = (-1)^ p d_2^{p, q} : K^{p, q} \to K^{p, q + 1}$,

2. ${}''E_0^{p, q} = K^{q, p}$ with ${}''d_0^{p, q} = d_1^{q, p} : K^{q, p} \to K^{q + 1, p}$,

3. ${}'E_1^{p, q} = H^ q(K^{p, \bullet })$ with ${}'d_1^{p, q} = H^ q(d_1^{p, \bullet })$,

4. ${}''E_1^{p, q} = H^ q(K^{\bullet , p})$ with ${}''d_1^{p, q} = (-1)^ q H^ q(d_2^{\bullet , p})$,

5. ${}'E_2^{p, q} = H^ p_ I(H^ q_{II}(K^{\bullet , \bullet }))$,

6. ${}''E_2^{p, q} = H^ p_{II}(H^ q_ I(K^{\bullet , \bullet }))$.

Proof. Omitted. $\square$

These spectral sequences define two filtrations on $H^ n(sK^\bullet )$. We will denote these $F_ I$ and $F_{II}$.

Definition 12.25.2. Let $\mathcal{A}$ be an abelian category. Let $K^{\bullet , \bullet }$ be a double complex. We say the spectral sequence $({}'E_ r, {}'d_ r)_{r \geq 0}$ weakly converges to $H^ n(sK^\bullet )$, abuts to $H^ n(sK^\bullet )$, or converges to $H^ n(sK^\bullet )$ if Definition 12.24.9 applies. Similarly we say the spectral sequence $({}''E_ r, {}''d_ r)_{r \geq 0}$ weakly converges to $H^ n(sK^\bullet )$, abuts to $H^ n(sK^\bullet )$, or converges to $H^ n(sK^\bullet )$ if Definition 12.24.9 applies.

As mentioned above there is no consistent terminology regarding these notions in the literature. In the situation of the definition, we have weak convergence of the first spectral sequence if for all $n$

$\text{gr}_{F_ I}(H^ n(sK^\bullet )) = \oplus _{p + q = n} {}'E_\infty ^{p, q}$

via the canonical comparison of Lemma 12.24.6. Similarly the second spectral sequence $({}''E_ r, {}''d_ r)_{r \geq 0}$ weakly converges if for all $n$

$\text{gr}_{F_{II}}(H^ n(sK^\bullet )) = \oplus _{p + q = n} {}''E_\infty ^{p, q}$

via the canonical comparison of Lemma 12.24.6.

Lemma 12.25.3. Let $\mathcal{A}$ be an abelian category. Let $K^{\bullet , \bullet }$ be a double complex. Assume that for every $n \in \mathbf{Z}$ there are only finitely many nonzero $K^{p, q}$ with $p + q = n$. Then

1. the two spectral sequences associated to $K^{\bullet , \bullet }$ are bounded,

2. the filtrations $F_ I$, $F_{II}$ on each $H^ n(K^\bullet )$ are finite,

3. the spectral sequences $({}'E_ r, {}'d_ r)_{r \geq 0}$ and $({}''E_ r, {}''d_ r)_{r \geq 0}$ converge to $H^*(sK^\bullet )$,

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 $H^ n(sK^\bullet )$ is in $\mathcal{C}$. Similarly for $({}''E_ r, {}''d_ r)_{r \geq 0}$.

Proof. Follows immediately from Lemma 12.24.11. $\square$

Here is our first application of spectral sequences.

Lemma 12.25.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex. Let $A^{\bullet , \bullet }$ be a double complex. Let $\alpha ^ p : K^ p \to A^{p, 0}$ be morphisms. Assume that

1. For every $n \in \mathbf{Z}$ there are only finitely many nonzero $A^{p, q}$ with $p + q = n$.

2. We have $A^{p, q} = 0$ if $q < 0$.

3. The morphisms $\alpha ^ p$ give rise to a morphism of complexes $\alpha : K^\bullet \to A^{\bullet , 0}$.

4. The complex $A^{p, \bullet }$ is exact in all degrees $q \not= 0$ and the morphism $K^ p \to A^{p, 0}$ induces an isomorphism $K^ p \to \mathop{\mathrm{Ker}}(d_2^{p, 0})$.

Then $\alpha$ induces a quasi-isomorphism

$K^\bullet \longrightarrow sA^\bullet$

of complexes. Moreover, there is a variant of this lemma involving the second variable $q$ instead of $p$.

Proof. The map is simply the map given by the morphisms $K^ n \to A^{n, 0} \to sA^ n$, which are easily seen to define a morphism of complexes. Consider the spectral sequence $({}'E_ r, {}'d_ r)_{r \geq 0}$ associated to the double complex $A^{\bullet , \bullet }$. By Lemma 12.25.3 this spectral sequence converges and the induced filtration on $H^ n(sA^\bullet )$ is finite for each $n$. By Lemma 12.25.1 and assumption (4) we have ${}'E_1^{p, q} = 0$ unless $q = 0$ and $'E_1^{p, 0} = K^ p$ with differential ${}'d_1^{p, 0}$ identified with $d_ K^ p$. Hence ${}'E_2^{p, 0} = H^ p(K^\bullet )$ and zero otherwise. This clearly implies $d_2^{p, q} = d_3^{p, q} = \ldots = 0$ for degree reasons. Hence we conclude that $H^ n(sA^\bullet ) = H^ n(K^\bullet )$. We omit the verification that this identification is given by the morphism of complexes $K^\bullet \to sA^\bullet$ introduced above. $\square$

Lemma 12.25.5. Let $\mathcal{A}$ be an abelian category. Let $M^\bullet$ be a complex of $\mathcal{A}$. Let

$a : M^\bullet [0] \longrightarrow \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right)$

be a homotopy equivalence in the category of complexes of complexes of $\mathcal{A}$. Then the map $\alpha : M^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ induced by $M^\bullet \to A^{0, \bullet }$ is a homotopy equivalence.

Proof. The statement makes sense as a complex of complexes is the same thing as a double complex. The assumption means there is a map

$b : \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right) \longrightarrow M^\bullet [0]$

such that $a \circ b$ and $b \circ a$ are homotopic to the identity in the category of complexes of complexes. This means that $b \circ a$ is the identity of $M^\bullet [0]$ (because there is only one term in degree $0$). Also, observe that $b$ is given by a map $b^0 : A^{0, \bullet } \to M^\bullet$ and zero in all other degrees. Thus $b$ induces a map $\beta : \text{Tot}(A^{\bullet , \bullet }) \to M^\bullet$ and $\beta \circ \alpha$ is the identity on $M^\bullet$. Finally, we have to show that the map $\alpha \circ \beta$ is homotopic to the identity. For this we choose maps of complexes $h^ n : A^{n, \bullet } \to A^{n - 1, \bullet }$ such that $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$ which exist by assumption. Here $d_1 : A^{n, \bullet } \to A^{n + 1, \bullet }$ are the differentials of the complex of complexes. We will also denote $d_2$ the differentials of the complexes $A^{n, \bullet }$ for all $n$. Let $h^{n, m} : A^{n, m} \to A^{n - 1, m}$ be the components of $h^ n$. Then we can consider

$h' : \text{Tot}(A^{\bullet , \bullet })^ k = \bigoplus \nolimits _{n + m = k} A^{n, m} \to \bigoplus \nolimits _{n + m = k - 1} A^{n, m} = \text{Tot}(A^{\bullet , \bullet })^{k - 1}$

given by $h^{n, m}$ on the summand $A^{n, m}$. Then we compute that the map

$d_{sA} \circ h' + h' \circ d_{sA}$

restricted to the summand $A^{n, m}$ is equal to

$d_1^{n - 1, m} \circ h^{n, m} + (-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m} + h^{n + 1, m} \circ d_1^{n, m} + h^{n, m + 1} \circ (-1)^ nd_2^{n, m}$

Since $h^ n$ is a map of complexes, the terms $(-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m}$ and $h^{n, m + 1} \circ (-1)^ nd_2^{n, m}$ cancel. The other two terms give $(\alpha \circ \beta )|_{A^{n, m}} - \text{id}_{A^{n, m}}$ because $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$. This finishes the proof. $\square$

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