nLab
double complex

Idea

A double complex is a complex in a category of complexes. Accordingly, a double chain complex is a chain complex in a category of chain complexes.

In the presence of enough products and/or coproducts, there is a total complex associated to a double complex, and the interest in double complexes is usually that in these total complexes.

Double chain complexes usually arise from the application of bifunctors (additive functors of two variables) of additive categories $C_1,C_2,C_3$

$$F:C_1\times C_2\to C_3$$F : C_1 \times C_2 \to C_3

to complexes in their two arguments. Combining this with the formation of total complexes then yields bifunctors from categories of complexes to categories of complexes.

$$\tilde{F}:\mathrm{Ch}(C{)}^{\mathrm{op}}\times \mathrm{Ch}(C)\to \mathrm{Ch}(C).$$\tilde F : Ch(C)^{op} \times Ch(C) \to Ch(C) \,.

The most important examples of this are induced by the hom-functor and the tensor product functor.

Under the Dold-Kan correspondence then $\tilde{F}$ can be understood as the internal hom between higher groupoids.

Definition

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$$\begin{array}{ccccc}& & ⋮& & ⋮\\ & & {\downarrow }^{d_X^v}& & {\downarrow }^{d_X^v}\\ \cdots & \to & X_{n,m}& \stackrel{d_X^h}{\to }& X_{n,m-1}& \to & \cdots \\ & & {\downarrow }^{d_X^v}& & {\downarrow }^{d_X^v}\\ \cdots & \to & X_{n,m}& \stackrel{d_X^h}{\to }& X_{n,m-1}& \to & \cdots \\ & & {\downarrow }^{d_X^v}& & {\downarrow }^{d_X^v}\\ & & ⋮& & ⋮\end{array}$$\array{ && \vdots && \vdots \\ & & \downarrow^{d_X^v} && \downarrow^{d_X^v} \\ \cdots &\to & X_{n,m} &\stackrel{d_X^h}{\to}& X_{n,m-1} & \to & \cdots \\ & & \downarrow^{d_X^v} && \downarrow^{d_X^v} \\ \cdots &\to & X_{n,m} &\stackrel{d_X^h}{\to}& X_{n,m-1} & \to & \cdots \\ & & \downarrow^{d_X^v} && \downarrow^{d_X^v} \\ && \vdots && \vdots }

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$${\mathrm{tot}}_{\oplus }^k=\underset{m+n=k}{⨁}{X}_{n,m}$$tot_{\oplus}^k = \bigoplus_{m+n=k} X_{n,m}

$$d_{{\mathrm{tot}}_{\oplus }}^k{\mid }_{X_{n,m}}=d_X^v\bigsqcup (-1{)}^{•}{d}_X^h$$d^k_{tot_\oplus}|_{X_{n,m}} = d^v_X \sqcup (-1)^\bullet d_X^h

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$${\mathrm{tot}}_{\prod }^k=\prod _{m+n=k}{X}_{n,m}$$tot_{\prod}^k = \prod_{m+n=k} X_{n,m}

$$d_{{\mathrm{tot}}_{\prod }}^k{\mid }_{X_{n,m}}=d_X^v+(-1{)}^{•}{d}_X^h$$d^k_{tot_\prod}|_{X_{n,m}} = d^v_X + (-1)^\bullet d_X^h

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