Differential on total complex

Question

Given a bicomplex $C_{\ast,\ast}$, i.e. objects $C_{p,q}$ in an abelian category with horizontal $$d_{p,q}^h:C_{p,q}\to C_{p-1,q}$$ and vertical $$d_{p,q}^v:C_{p,q}\to C_{p,q-1}$$ differentials that anticommute. Then there is the notion of the total complex that consists of objects $Tot_n(C_{\ast,\ast})=\bigoplus_{p+q=n}C_{p,q}$ and a differential $d:Tot_n(C_{\ast,\ast})\to Tot_{n-1}(C_{\ast,\ast})$ that is given by $d=d^h+d^v$, according to many sources. I do see that $d\circ d=0$, using the formulas of an abelian category and anticommutativity, but I do not understand the differential maps themselves.

My question how the differentials are defined on an arbitrary element $(x_0,\dots,x_n)$ in $Tot_n(C_{\ast,\ast})$ as the $d_{p,q}$ are only defined on $C_{p,q}$. I can sort of muddle through but would be glad if you can give me $d(x_0,...,x_n)$ explicitly and show how $d\circ d=0$ element-wise. Thanks!

Answer 1

For the sake of simplicity, let's assume $C_{p,q} = 0$ whenever $p < 0$ or $q < 0$. Note, the result is still true without this assumption.

Let $x \in \operatorname{Tot}_n(C_{\bullet,\bullet}) = \bigoplus_{p+q=n}C_{p,q}$, then $x = x_{n,0} + x_{n-1,1} + \dots + x_{0,n}$ where $x_{p,q} \in C_{p,q}$. Applying the differential $d$ to $x$ gives $dx \in \operatorname{Tot}_{n-1}(C_{\bullet,\bullet})$ given by

\begin{align*} dx &= d(x_{n,0} + x_{n-1,1} + \dots + x_{0,n})\\ &= d_{n,0}x_{n,0} + d_{n-1,1}x_{n-1,1} + \dots + d_{0,n}x_{0,n}\\ &= d^h_{n,0}x_{n,0} + d^v_{n,0}x_{n,0} + d^h_{n-1,1}x_{n-1,1} + d^v_{n-1,1}x_{n-1,1} + \dots + d^h_{0,n}x_{0,n} + d^v_{0,n}x_{0,n}\\ &= d^h_{n,0}x_{n,0} + 0 + d^h_{n-1,1}x_{n-1,1} + d^v_{n-1,1}x_{n-1,1} + \dots + 0 + d^v_{0,n}x_{0,n}\\ &= (d^h_{n,0}x_{n,0} + d^v_{n-1,1}x_{n-1,1}) + \dots + (d^h_{1,n-1}x_{1,n-1} + d^v_{0,n}x_{0,n}).\\ \end{align*}

That is, $dx \in \operatorname{Tot}_{n-1}(C_{\bullet, \bullet})$ and for $r, s \geq 0$ with $r + s = n - 1$, $(dx)_{r,s} = d^h_{r+1,s}x_{r+1,s} + d^v_{r,s+1}x_{r,s+1}$.

For $y = y_{n-1,0} + y_{n-2,1} + \dots + y_{0,n-1} \in \operatorname{Tot}_{n-1}(C_{\bullet,\bullet})$ the same calculation shows that $dy \in \operatorname{Tot}_{n-2}(C_{\bullet,\bullet})$ with $(dy)_{t,u} = d^h_{t+1,u}y_{t+1,u} + d^v_{t,u+1}y_{t,u+1}$ for any $t, u \geq 0$ with $t + u = n-2$. Taking $y = dx$, we get

\begin{align*} (d(dx))_{t,u} &= d^h_{t+1,u}(dx)_{t+1,u} + d^v_{t,u+1}(dx)_{t,u+1}\\ &= d^h_{t+1,u}(d^h_{t+2,u}x_{t+2,u} + d^v_{t+1,u+1}x_{t+1,u+1}) + d^v_{t,u+1}(d^h_{t+1,u+1}x_{t+1,u+1} + d^v_{t,u+2}x_{t,u+2})\\ &= d^h_{t+1,u}d^h_{t+2,u}x_{t+2,u} + d^h_{t+1,u}d^v_{t+1,u+1}x_{t+1,u+1} + d^v_{t,u+1}d^h_{t+1,u+1}x_{t+1,u+1} + d^v_{t,u+1}d^v_{t,u+2}x_{t,u+2}\\ &= d^h_{t+1,u}d^h_{t+2,u}x_{t+2,u} + (d^h_{t+1,u}d^v_{t+1,u+1} + d^v_{t,u+1}d^h_{t+1,u+1})x_{t+1,u+1} + d^v_{t,u+1}d^v_{t,u+2}x_{t,u+2}\\ &= 0 \end{align*}

and therefore $d(dx) = 0$, i.e. $d^2 = 0$.