Double complex

Remarks

The three conditions give $d^2 = (d^h)^2 + d^h d^v + d^v d^h + (d^v)^2 = 0$, so the total complex is a complex. The sums are finite, and exist in any additive category, when only finitely many $C_{p,q}$ with $p + q = n$ are nonzero for each $n$. Some texts ask instead that the squares commute, $d^h d^v = d^v d^h$, and put the sign into the total complex as $d = d^h + (-1)^p d^v$; the two conventions are equivalent. Using products instead of direct sums gives a second total complex, which differs when infinitely many terms are nonzero on a diagonal.

Example: for chain complexes $(P_\bullet, d^P)$ and $(Q_\bullet, d^Q)$ of modules over a commutative ring, set $C_{p,q} = P_p \otimes Q_q$, $d^h = d^P \otimes 1$ and $d^v = (-1)^p \, 1 \otimes d^Q$ on $C_{p,q}$. For $x \in P_p$ and $y \in Q_q$, $d^h d^v (x \otimes y) = (-1)^p\, d^P x \otimes d^Q y$ and $d^v d^h (x \otimes y) = (-1)^{p-1}\, d^P x \otimes d^Q y$, so condition 3 holds.

Non-example: the same data without the sign, $d^v = 1 \otimes d^Q$, fails condition 3 in general, since then $d^h d^v + d^v d^h = 2\, d^P \otimes d^Q$; over $\Z$ with $P = Q = (\Z \xrightarrow{1} \Z)$ this is $2 \ne 0$.