The Stacks project

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)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FLG. Beware of the difference between the letter 'O' and the digit '0'.