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Shifts of double complex and total complex

Sign rules...
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aisejohan committed Oct 3, 2019
1 parent 3520110 commit ff09225dbdd215c0ac0c0c3cb0b9ba76608df3fb
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@@ -5945,6 +5945,64 @@ \section{Spectral sequences: double complexes}
last two variables and then combine the first with the sum of the last two.
\end{remark}

\begin{remark}
\label{remark-shift-double-complex}
Let $\mathcal{A}$ be an additive category. Let $A^{\bullet, \bullet}$
be a double complex with differentials $d_1^{p, q}$ and $d_2^{p, q}$.
Denote $A^{\bullet, \bullet}[a, b]$ the double complex with
$$
(A^{\bullet, \bullet}[a, b])^{p, q} = A^{p + a, q + b}
$$
and differentials
$$
d_{A^{\bullet, \bullet}[a, b], 1}^{p, q} = (-1)^a d_1^{p + a, q + b}
\quad\text{and}\quad
d_{A^{\bullet, \bullet}[a, b], 2}^{p, q} = (-1)^b d_2^{p + a, q + b}
$$
In this situation there is a well defined isomorphism
$$
\gamma :
\text{Tot}(A^{\bullet, \bullet})[a + b]
\longrightarrow
\text{Tot}(A^{\bullet, \bullet}[a, b])
$$
which in degree $n$ is given by the map
$$
\xymatrix{
(\text{Tot}(A^{\bullet, \bullet})[a + b])^n =
\bigoplus_{p + q = n + a + b} A^{p, q}
\ar[d]^{\epsilon(p, q, a, b)\text{id}_{A^{p, q}}} \\
\text{Tot}(A^{\bullet, \bullet}[a, b])^n =
\bigoplus_{p' + q' = n} A^{p' + a, q' + b}
}
$$
for some sign $\epsilon(p, q, a, b)$. Of course the summand $A^{p, q}$
maps to the summand $A^{p' + a, q' + b}$ when $p = p' + a$ and $q = q' + b$.
To figure out the conditions on these signs observe that on the source we have
$$
d|_{A^{p, q}} = (-1)^{a + b}\left(d_1^{p, q} + (-1)^pd_2^{p, q}\right)
$$
whereas on the target we have
$$
d|_{A^{p' + a, q' + b}} =
(-1)^ad_1^{p' + a, q' + b} + (-1)^{p'}(-1)^bd_2^{p' + a, q' + b}
$$
Thus our constraints are that
$$
(-1)^a \epsilon(p, q, a, b) = \epsilon(p + 1, q, a, b)(-1)^{a + b}
\Leftrightarrow
\epsilon(p + 1, q, a, b) = (-1)^b \epsilon(p, q, a, b)
$$
and
$$
(-1)^{p' + b}\epsilon(p, q, a, b) =
\epsilon(p, q + 1, a, b) (-1)^{a + b + p}
\Leftrightarrow
\epsilon(p, q, a, b) = \epsilon(p, q + 1, a, b)
$$
Thus we choose $\epsilon(p, q, a, b) = (-1)^{pb}$.
\end{remark}

\begin{lemma}
\label{lemma-homotopy-complex-complexes}
Let $\mathcal{A}$ be an abelian category.

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