[117], tag 0FNB. Other references may use different sign conventions.

We gather a few key facts about double complexes that will come up in our cohomology computations. Throughout, let \(\calA\) be a fixed abelian category.

Subsection13.1Double complexes and totalization

Definition13.1.1.

A double complex in \(\calA\) consists of a collection of objects \(K^{p,q}\) together with morphisms \(d_1^{p,q}\colon K^{p,q} \to K^{p+1,q}\) and \(d_2^{p,q}\colon K^{p,q} \to K^{p,q+1}\) such that the resulting diagram commutes and each row and column is itself a complex.

Remark13.1.3.

A double complex can itself be viewed as a complex in the category \(\Comp(\calA)\text{.}\) There are of course two different ways to do this, which for the moment are symmetric; we will have to break symmetry to discuss totalization. While this symmetry break will have some curious side effects (e.g., the graded commutativity of cohomology, as per Remark 13.2.3), most of the statements we make asymmetrically will have straightforward counterparts with the orientation reversed.

Definition13.1.4.Totalization.

Given a double complex \(K^{\bullet,\bullet}\text{,}\) the associated total complex (or for short the totalization) is the complex \(\Tot(K^{\bullet,\bullet})\) with

More precisely, this should be called the direct sum totalization as distinct from the direct product totalization, in which we take the product rather than the sum. The two coincide if the original complex is bounded above in both directions, or bounded below in both directions. However, we will later (in Section 18) have reason to consider the mixed situation, in which the complex is bounded above in one direction and bounded below in the other direction, and in this case we must pay attention to this distinction.

Subsection13.2Interchanging the rows and columns

Remark13.2.1.

Let \(K^{\bullet,\bullet}\) be a double complex in \(\calA\text{.}\) Let \(L^{\bullet,\bullet}\) be the transposed complex with \(L^{p,q} = K^{p,q}\) (and similarly for differentials). Then there are natural isomorphisms

for each \(n\text{,}\) but we have to choose these carefully to make this an isomorphism of complexes: the identification of \(K^{p,q} \subset \Tot(K^{\bullet,\bullet})^n\) with \(L^{q,p} \subset \Tot(L^{\bullet,\bullet})^n\) is given by multiplication by \((-1)^{pq}\text{.}\)

Example13.2.2.

Suppose that \(\calA\) is a symmetric monoidal category (e.g., \(\Mod_A\) using the tensor product) and let \(K^\bullet\) and \(L^\bullet\) be two bounded-below complexes. Then \(K^\bullet \otimes L^\bullet\) and \(L^\bullet \otimes K^\bullet\) are transposed complexes of each other, so we may use Remark 13.2.1 to identify their totalizations; in the case \(\calA = \Mod_A\text{,}\) both of these are quasi-isomorphic to \(K^\bullet \otimes^L_A L^\bullet\text{.}\)

Remark13.2.3.

We can use this to explain the signs in Lemma 12.1.2 as follows. For \(A \in \Ring\text{,}\) let \(K^\bullet\) be a commutative \(A\)-algebra object in \(D(A)\text{.}\) The multiplication map can be interpreted as a map \(\Tot(L_1^\bullet \otimes L_2^\bullet) \to L_3^\bullet\) for some complexes \(L_1, L_2, L_3\) which are quasi-isomorphic to \(K^\bullet\text{.}\) (Note that we cannot necessarily take the same representative and get a genuine map of complexes; that is, we did not assume that \(K^\bullet\) is a commutative ring object at the level of complexes.) Given classes \(a \in H^n(L_1^\bullet)\text{,}\)\(b \in H^m(L_2^\bullet)\text{,}\) we compute their product in \(H^{m+n}(L_3^\bullet)\) by choosing representatives of \(a\) and \(b\) in their respective complexes, taking the product, putting that into the totalization, and then applying the map to \(L_3^{m+n}\text{.}\) From this, it is clear that switching the order of the terms should introduce a sign of \((-1)^{mn}\) in conformance with Remark 13.2.1.

Subsection13.3The spectral sequence(s) of a double complex

Rather than giving an axiomatic treatment of spectral sequences, we give a narrow treatment centered around a bounded-below double complex, this being the case of most pressing interest for prismatic cohomology. Our goal is to present the key ideas without drowning the reader in the notation needed to make everything completely precise.

Proposition13.3.1.

Let \(K^{\bullet,\bullet}\) be a double complex concentrated in nonnegative degrees (in both directions). Then there exist objects \(E^{p,q}_i\) for \(i,p,q \geq 0\) with the following properties (where \(E^{p,q}_i = 0\) if \(p,q\) are not both nonnegative).

We have \(E^{p,q}_0 = K^{p,q}\) for all \(p,q.\)

For each \(i\text{,}\) there exist maps \(d_{(i)}^{p,q}\colon E_i^{p,q} \to E_i^{p+i,q+1-i}\) such that the maps in and out of \(E_i^{p,q}\) compose to zero and the cohomology of the resulting complex there is \(E_{i+1}^{p,q}\text{.}\) In particular, for any given \(p,q\text{,}\) the terms \(E_i^{p,q}\) for \(i \gg 0\) stabilize to an object we call \(E_\infty^{p,q}\text{.}\)

For \(i=0\text{,}\)\(d_{(i)}^{p,q}\) equals the differential \((-1)^p d_2^{p,q}\) of \(K^{\bullet,\bullet}\text{.}\)

For \(i=1\text{,}\)\(d_{(i)}^{p,q}\) is the map induced by \(d_1^{p,q}\text{.}\)

For \(n \geq 0\text{,}\) there is a filtration on \(H^n(\Tot(K^{\bullet,\bullet}))\) whose successive quotients are the objects \(E_\infty^{p,q}\) for \(p+q=n\text{.}\)

Moreover, the construction is natural in \(K^{\bullet,\bullet}\text{.}\)

Let \(K^{\bullet,\bullet} \to L^{\bullet,\bullet}\) be a morphism of double complexes. If the induced maps \(E_\infty^{p,q}(K) \to E_\infty^{p,q}(L)\) is an isomorphism, then the map \(\Tot(K^{\bullet,\bullet}) \to \Tot(L^{\bullet,\bullet})\) is a quasi-isomorphism.

By Proposition 13.3.1 (and in particular the naturality), the map \(H^n(\Tot(K^{\bullet,\bullet})) \to H^n(\Tot(L^{\bullet,\bullet}))\) has the property that it induces isomorphisms on the successive quotients of some filtration. By the five lemma, this implies that it is itself an isomorphism.

Corollary13.3.8.

Let \(K^{\bullet,\bullet}\) be a double complex in which the single complexes \(K^{\bullet,q}\) are acyclic for all \(q \gt 0\text{.}\) Then the morphism \(\Tot(K^{\bullet,\bullet}) \to K^{\bullet,0}\) is a quasi-isomorphism.

Apply Corollary 13.3.7 after filling \(K^{\bullet,0}\) out to a double complex by adding zeroes.

Subsection13.4Totalization in the derived category

Remark13.4.1.

Let \(K^0 \to K^1 \to \cdots\) be a sequence of morphisms in \(D(A)\) for some \(A \in \Ring\text{,}\) with every two consecutive arrows composing to zero; that is, it is a “complex consisting of objects of \(D(A)\)”.

In order to work with this sequence, one would like to choose representatives in \(K(A)\) so that the terms \(K^\bullet\) fit into a double complex. In practice, this is obstructed by the construction of Toda brackets. To illustrate this point, suppose that we have managed to represent each \(K^i\) as a complex and each morphism \(K^i \to K^{i+1}\) as a morphism of complexes (without localization). We then have a diagram in which \(\alpha\) represents some homotopy witnessing the vanishing of \(d^1 \circ d^0\) in \(K(A)\) and \(\beta\) represents some homotopy witnessing the vanishing of \(d^2 \circ d^1\) in \(K(A)\text{.}\) Then \(d^2 \circ \alpha\) and \(\beta \circ d^0\) are both homotopies that witness the vanishing of \(d^2 \circ d^1 \circ d^0\) in \(K(A)\text{,}\) but it may not be possible to choose \(\alpha\) and \(\beta\) to make them equal. In fact, these two homotopies together define a loop in the \(\pi_1\) of the space of maps between simplicial realizations of \(K^0\) and \(K^3\text{;}\) the Toda bracket is the isotopy class of this loop, whose nonvanishing provides an obstruction to choosing the morphisms so that the compositions \(d^1 \circ d^0\) and \(d^2 \circ d^1\) vanish on the nose. (One can similarly make higher Toda brackets by considering longer chunks of the sequence, conditionally on the vanishing of the lower-order brackets.)

This gives an example of why it is easier in the long run to work with \(D(A)\) in the framework of stable \(\infty\)-categories. See [10] in particular for a description of totalization in this framework that properly accounts for the Toda brackets.