Remark 12.18.5. 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 $\epsilon (p, q, a, b) = (-1)^{pb}$.
Comments (0)
There are also: