Notes on prismatic cohomology

A double complex can itself be viewed as a complex in the category $\mathbf{Comp}(\mathcal{A})\text{.}$ 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.

We can use this to explain the signs in Lemma 12.1.2 as follows. For $A\in \mathbf{Ring}\text{,}$ let $K^{\bullet }$ be a commutative $A$-algebra object in $D(A)\text{.}$ The multiplication map can be interpreted as a map $\mathrm{Tot}(L_1^{\bullet }\otimes L_2^{\bullet })\to L_3^{\bullet }$ 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.

Let $K^0\to K^1\to \cdots$ be a sequence of morphisms in $D(A)$ for some $A\in \mathbf{Ring}\text{,}$ 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 $\mathrm{\infty }$-categories. See [10] in particular for a description of totalization in this framework that properly accounts for the Toda brackets.