# Total complex of complexes

When we have a double complex of vector spaces $$V^{p,q}$$, we can produce a complex either taking direct sums or products along the anti-diagonals. Then, the differential in this new complex will be $$d = d_{\text{vert}} + (-1)^{\text{vertical degree}} d_{\text{horizontal}}.$$ If the complex was made of complexes, I would reckon that the differential of them should appear in the differential of the double complex. However, I wasn’t able to find a reference for such a definition. Can anyone help me or point me to the definition?

• Sorry, just to be clear, you're asking for a totalization of $V^{\bullet, \bullet}$ where now each $V^{p,q}$ is a complex and the differentials are chain maps? Then (up to some sign conventions) you've just told me you have a trigraded complex and you'd like to know how to totalize. But then (again, up to careful choice of sign to make everything work) you just do the same thing: $(\text{Tot } V)_k = \bigoplus_{p+q+j=k} V^{p,q}_j$, with differential $$d_{\text{vert}} \pm d_{\text{horiz}} \pm d_V,$$ aka, sum the three differentials. There's nothing special about bi- or tri-gradings. – Mike Miller Jun 12 at 15:34
• Yes, you’re right, but how do I find the sign convention? – Federico Jun 12 at 15:36
• Square the differential and make sure it comes out to zero. :) If you're not comparing to someone else's use of these totalizations, you don't need to make sure your sign convention matches up with theirs. It's probably best if you set things up so that if $V^{p,q}_\bullet$ is just a module concentrated in degree zero, this recovers the standard conventions you dictated above. – Mike Miller Jun 12 at 15:39
• Ok, thank you very much! – Federico Jun 12 at 15:40
• To figure out a way to assign signs(:)!) fix $j$ first, write down the totalization with respect to $(p,q)$ and then do it again with $j$ added. In each step you only have to deal with a bigrading. – Sándor Kovács Jun 12 at 16:43