Double Complex

Double Complex

Definition 1. Let $\mathcal{A}$ be an addictive category. A double chain complex $A$ is given by $\{A_{p,q},d_{p,q}^{1},d_{p,q}^{2}\}_{p,q\in\mathbb{Z}}$ where $d_{p,q}^{1}:A_{p,q}\to A_{p-1,q}$ and $d_{p,q}^{2}:A_{p,q}\to A_{p,q-1}$ are morphisms of $\mathcal{A}$ s.t. the following conditions hold:

1. $d_{p-1,q}^{1}\circ d_{p,q}^{1}=0$;

2. $d_{p,q-1}^{2}\circ d_{p,q}^{2}=0$;

3. $d_{p,q-1}^{1}\circ d_{p,q}^{2}=d_{p-1,q}^{2}\circ d_{p,q}^{1}$.

for all $p,q\in \mathbb{Z}$.

The definition of double cochain complex is similar.

Definition 2. For a double complex, its total complex $\mathrm{Tot}(C)$ is given by $\mathrm{Tot}(C)_{n}=\oplus_{p+q=n}C_{p,q}$ with the differential morphism $d_{n}^{\mathrm{Tot}}=\sum_{p+q=n}(d_{p,q}^{1}+d_{p,q}^{2})$. We can verify $d_{n}^{\mathrm{Tot}}\circ d_{n+1}^{\mathrm{Tot}}=0$.

Notice that in abelian category we may not have $\oplus_{p+q=n}C_{p,q}$. Then we can use $\prod_{p+q=n}C_{p,q}$ to substitute it. We can define the homology on the partial chain complex $(A_{p,q},d_{p,q}^{1})$ and
$(A_{p,q},d_{p,q}^{2})$.

Definition 3. For two complex $C, D$, we can define the tensor product of $C$ and $D$ as $\mathrm{Tot}B$, where $B_{p,q}=C_{p}\otimes D_{q}$, $d_{p,q}^{1}=d_{p,q}\otimes id$, $d_{p,q}^{2}=(-1)^{p} id\otimes d_{p,q}$, $d_{n}^{\mathrm{Tot}}=\sum_{p+q=n}d_{p,q}^{1}+d_{p,q}^{2}$.

Definition 4. For two complex $C, D$, we can define the Hom of $C$ and $D$ as $\mathrm{Tot}B$, where $B_{p,q}=\mathrm{Hom}(C_{-p},D_{q})$, $d_{p,q}^{1}(f)(c)=(-1)^{p+q+1}fd_{p-1}(c)$, $d_{p,q}^{2}(f)(c)=d_{p}f(c)$.

Proposition 1. Morphisms $f:C\to C'$, $g:D\to D'$ induce chain complex maps

$$f\otimes f:C\otimes D\to C'\otimes D'$$

$$\mathrm{Hom}(f,g):\mathrm{Hom}(C',D)\to \mathrm{Hom}(C,D')$$

If $f\simeq f'$ and $g\simeq g'$, then $f\otimes g\simeq f'\otimes g'$, $\mathrm{Hom}(f,g)\simeq \mathrm{Hom}(f',g')$.

Proposition 2. If $C\simeq C'$, $D\simeq D'$, then $C\otimes D\simeq C' \otimes D'$, $\mathrm{Hom}(C,D)\simeq \mathrm{Hom}(C',D')$.