Translation (or shift) on chain complexes and the translated (or shifted) differential

Question

On p. 31 of Sheaves on Manifolds by Kashiwara & Schapira, they define the translation or shift of degree $[k]$ on the category of chain complexes in an abelian category as follows. Given a morphism of chain complexes $f: X \to Y$, they define $$X[k]^n = X^{n+k},$$ $$d^n_{X[k]}= (-1)^k d^{n+k}_X,$$ and $$f[k]^n = f^{n+k}.$$ I am wondering why there is a factor of $(-1)^k$ in front of the shifted differential? Is this because chain complexes are coalgebras for the shift endofunctor, and they want this shift to give the right notion of a double complex?

Answer 1

Great question! I thought about it at some point, and I found a more down to earth explanation for myself. Maybe it is wrong, so I will be happy to see other answers!

First of all, the issue boils down to $k = 1$, since the translation by $k\in \mathbb{N}$ should be $\underbrace{[1]\cdots [1]}_k$.

Recall the construction of the cone of a morphism of complexes $f^\bullet\colon X^\bullet\to Y^\bullet$. We define it by taking objects $C^n = Y^n \oplus X^{n+1}$ and differentials $d^n = \begin{pmatrix} d^n_Y & f^{n+1} \\ 0 & -d^{n+1}_X \end{pmatrix}$. Here we already put a minus sign in front of $d^{n+1}_X$, but I don't think there's some deep reason behind it: without this sign, we simply don't get the identity $d^{n+1}\circ d^n = 0$.

Then, the cone sits in the short sequence of complexes $$0 \to Y^\bullet \rightarrowtail C (f) \twoheadrightarrow X^\bullet [1] \to 0$$ You can check that the natural projection $C (f) \twoheadrightarrow X^\bullet [1]$ is a morphism of complexes (the projections $Y^n \oplus X^{n+1} \to X^{n+1}$ commute with the differentials) only with the convention that $d^n_{X^\bullet [1]} = -d^{n+1}_X$.

It is clear that this sign convention is supposed to simplify things when one works with cones and distinguished triangles. When Verdier defines translation in his thesis, he changes the sign of the differentials, but he doesn't give an explanation.

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Weibel in his book says "this sign convention is designed to simplify notation later on" (p. 10 of his book). But then, for cohomological complexes he defines $X^\bullet [1]^n$ to be $X^{n-1}$ (probably because his book is more focused on the homological numeration?), which is not very common.

Answer 2

To my understanding, the introduction of the minus sign in $d^n_{X[k]}=(-1)^kd^{n+k}_X$ is to enable the definition of distinguished triangles in $K(\mathcal{A})$ and $D(\mathcal{A})$ (these categories are, resp., the homotopy category of cochain complexes with terms in an abelian category $\mathcal{A}$ and the derived category of $\mathcal{A}$). The class of d.t.'s can be equivalently defined as the respective equivalence class of triangles spawned from two possible classes: either from the class of triangles $$ \label{cone}\tag{1} X^\bullet\xrightarrow{f}Y^\bullet\to C(f)^\bullet\xrightarrow{-p} X^\bullet[1] $$ arising from the cone of any cochain map $f$ (cf. 014E and comments after) or from the class of triangles $$ \label{tws}\tag{2} A^\bullet\to B^\bullet\to C^\bullet\xrightarrow{\delta}A^\bullet[1] $$ arising from term-wise split ses's of cochain complexes (by 05SS the induced triangle in $K(\mathcal{A})$ does not depend on the choice of splittings in each term; by 014L, the triangle isomorphism classes spawned by \eqref{cone} and \eqref{tws} coincide). Here's the deal: if we hadn't defined $d_{X^\bullet[1]}=-d_{X^{\bullet+1}}$, then the last maps in \eqref{cone} and \eqref{tws} wouldn't be cochain maps. That is, the introduction of the minus sign allows the components of $p$ and $\delta$ to commute with differentials.

Answer 3

View $A[1]$ as $K[1]\otimes A$ where $K$ is the ground field (or ring). An element of this chain complex is written as $(1\otimes a)$ where the $1$ in the first factor is the unit $K$ and has cohomological degree $-1$. Thus, by the usual sign rule for interchanges we have $d(1\otimes a)=-1\otimes da$. This accounts for the sign change in the differential in $A[1]$.