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  "body": "\n## Definition\n\nLet $\\mathcal{A}$ be an [[additive category|B2VP96NJSK]], for example the category of [[modules|0NZ8JA25WK]] over a ring.\nA *double complex* in $\\mathcal{A}$ consists of objects $C_{p,q}$ for all integers $p, q$ and morphisms\n$$d^h_{p,q} \\colon C_{p,q} \\to C_{p-1,q}, \\qquad d^v_{p,q} \\colon C_{p,q} \\to C_{p,q-1},$$\ncalled the horizontal and vertical differentials, such that:\n\n1. $d^h \\circ d^h = 0$;\n2. $d^v \\circ d^v = 0$;\n3. $d^h \\circ d^v + d^v \\circ d^h = 0$.\n\nSuppose that the needed direct sums exist, for example for modules.\nThen the *total complex* is the [[chain complex|WQ82C0M3QB]] with\n$$\\mathrm{Tot}(C)_n = \\bigoplus_{p + q = n} C_{p,q}, \\qquad d = d^h + d^v.$$\n\n## Remarks\n\nThe 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.\nThe sums are finite, and exist in any additive category, when only finitely many $C_{p,q}$ with $p + q = n$ are nonzero for each $n$.\nSome 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.\nUsing products instead of direct sums gives a second total complex, which differs when infinitely many terms are nonzero on a diagonal.\n\nExample: 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 $C_{p,q}$.\nFor $x \\in P_p$ 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.\n\nNon-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 $2 \\ne 0$.\n",
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